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[ "Note that if $\\tan \\theta$ is positive, then $\\theta$ is in the first or third quadrant, so $0^{\\circ} < \\theta < 90^{\\circ} \\pmod{180^{\\circ}}$\nFurthermore, the only way $\\tan{\\left(2^{n}\\theta\\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \\[2^0\\theta \\equiv 2^3\\theta \\equiv 2^6\\theta \\equiv ... \\pmod{180^{\\circ}}.\\] (This is because if it isn't the same value, the terminal angle will gradually shift from the first quadrant into different quadrants, making the condition for positive tan untrue. This must also be true in order for $\\theta$ to be unique.)\nThis is the case if $2^3\\theta \\equiv 2^0\\theta \\pmod{180^{\\circ}}$ , so $7\\theta \\equiv 0^{\\circ} \\pmod{180^{\\circ}}$ . Therefore, recalling that $0^{\\circ}<\\theta<90^{\\circ},$ the possible $\\theta$ are: \\[\\frac{180}{7}^{\\circ}, \\frac{360}{7}^{\\circ}, \\frac{540}{7}^{\\circ}.\\]\n$\\frac{180}{7}^{\\circ}$ does not work since $\\tan\\left(2 \\cdot \\frac{180}{7}^{\\circ}\\right)$ is positive.\n$\\frac{360}{7}^{\\circ}$ does not work because $\\tan\\left(4 \\cdot \\frac{360}{7}^{\\circ}\\right)$ is positive.\nThus, $\\theta = \\frac{540}{7}^{\\circ}$ , and a quick check verifies that it does work. Our desired answer is $540 + 7 = \\boxed{547}$", "As in the previous solution, we note that $\\tan \\theta$ is positive when $\\theta$ is in the first or third quadrant. In order for $\\tan\\left(2^n\\theta\\right)$ to be positive for all $n$ divisible by $3$ , we must have $\\theta$ $2^3\\theta$ $2^6\\theta$ , etc to lie in the first or third quadrants. We already know that $\\theta\\in(0,90)$ . We can keep track of the range of $2^n\\theta$ for each $n$ by considering the portion in the desired quadrants, which gives \\[n=1 \\implies (90,180)\\] \\[n=2\\implies (270,360)\\] \\[n=3 \\implies (180,270)\\] \\[n=4 \\implies (90,180)\\] \\[n=5\\implies(270,360)\\] \\[n=6 \\implies (180,270)\\] \\[\\cdots\\] at which point we realize a pattern emerging. Specifically, the intervals repeat every $3$ after $n=1$ . We can use these repeating intervals to determine the desired value of $\\theta$ since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.\nInitially, the lower bound is $0$ (at $n=0$ ), then increases to $\\frac{90}{2}=45$ at $n=1$ . This then becomes $45+\\frac{45}{2}$ at $n=2$ $45+\\frac{45}{2}$ at $n=3$ $45+\\frac{45}{2}+\\frac{45}{2^3}$ at $n=4$ $45+\\frac{45}{2}+\\frac{45}{2^3}+\\frac{45}{2^4}$ at $n=5$ , etc. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as $n$ approaches infinity, the lower bound converges to \\[\\sum_{k=0}^{\\infty}\\left(45+\\frac{45}{2}\\right)\\cdot \\left(\\frac{1}{8}\\right)^k=\\frac{45+\\frac{45}{2}}{1-\\frac{1}{8}}=\\frac{\\frac{135}{2}}{\\frac{7}{8}}=\\frac{540}{7}\\implies p+q=540+7=\\boxed{547}\\] -ktong", "Since $\\tan\\left(\\theta\\right) > 0$ $0 < \\theta < 90$ . Since $\\tan\\left(2\\theta\\right) < 0$ $\\theta$ has to be in the second half of the interval (0, 90) ie (45, 90). Since $\\tan\\left(4\\theta\\right) < 0$ $\\theta$ has to be in the second half of that interval ie (67.5, 90). And since $\\tan\\left(8\\theta\\right) > 0$ $\\theta$ has to be in the first half of (67.5, 90). Inductively, the pattern repeats: $\\theta$ is in the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be $0.11011011011..._2 = \\frac{6}{7}_{10}$ . So we want the number which is 6/7 of the way through the interval (0, 90) so $\\theta = \\frac{6}{7}\\cdot 90 = \\frac{540}{7}$ and $p+q = 540 + 7 = \\boxed{547}$", "With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the \"first round of $3$ powers of two\", which are $\\frac{540}{8} = 67.5$ and $\\frac{630}{8} = 78.75$ . Going on to the \"second round of $3$ powers of two, we set the new lower and upper bounds as $\\frac{360 \\times 12.5}{64} = 70.3125$ and $\\frac{360 \\times 13.75}{64} = 77.34375$ using some guess and check and bashing. Now, it is obvious that the bounds for the \"zeroth round of $3$ powers of two\" are $0$ and $90$ , and notice that $90 - 78.75 = 11.25$ and $78.75 - 77.34375 = 1.40625$ and $\\frac{11.25}{1.40625} = 8$ .\nThis is obviously a geometric series, so setting $11.25$ as $u$ , we obtain $90 - (u + \\frac{u}{8} + \\frac{u}{64} + ...)$ $90 - \\frac{u}{1-\\frac{1}{8}}$ $\\frac{u}{\\frac{7}{8}}$ $\\frac{45}{4} \\times \\frac{8}{7}$ which simplifies to $\\frac{90}{7}$ . We can now finally subtract $\\frac{90}{7}$ from $\\frac{630}{7}$ and then we get $\\frac{540}{7}$ as the unique angle, so $\\boxed{547}$ is our answer.\n-fidgetboss_4000", "Since $7$ is the only number n such that f(x) = $2^{\\lfloor x\\rfloor}$ $\\text{(mod 7)}$ has a period of 3, we find that $\\theta$ is a multiple of ${\\frac{180}{7}}^\\circ$ . Note that the tangents of ${\\frac{180}{7}}^\\circ$ ${\\frac{360}{7}}^\\circ$ ${\\frac{540}{7}}^\\circ$ are positive while those of ${\\frac{720}{7}}^\\circ$ ${\\frac{900}{7}}^\\circ$ , and ${\\frac{1080}{7}}^\\circ$ are negative. With a bit of trial and error, we find $\\theta$ is ${\\frac{540}{7}}^\\circ$ and $540+7$ is $\\boxed{547}$" ]

[ "We can use the fact that $\\log_{a^b} c = \\frac{1}{b} \\log_a c.$ This can be proved by using change of base formula to base $a.$\nSo, the original equation $\\log_2{(\\log_{2^4}{n})} = \\log_{2^2}{(\\log_{2^2}{n})}$ becomes \\[\\log_2\\left({\\frac{1}{4}\\log_{2}{n}}\\right) = \\frac{1}{2}\\log_2\\left({\\frac{1}{2}\\log_2{n}}\\right).\\] Using log property of addition, we expand both sides and then simplify: \\begin{align*} \\log_2{\\frac{1}{4}}+\\log_2{(\\log_{2}{n}}) &= \\frac{1}{2}\\left[\\log_2{\\frac{1}{2}} +\\log_{2}{(\\log_2{n})}\\right] \\\\ \\log_2{\\frac{1}{4}}+\\log_2{(\\log_{2}{n}}) &= \\frac{1}{2}\\left[-1 +\\log_{2}{(\\log_2{n})}\\right] \\\\ -2+\\log_2{(\\log_{2}{n}}) &= -\\frac{1}{2}+ \\frac{1}{2}(\\log_{2}{(\\log_2{n})}). \\end{align*} Subtracting $\\frac{1}{2}(\\log_{2}{(\\log_2{n})})$ from both sides and adding $2$ to both sides gives us \\[\\frac{1}{2}(\\log_{2}{(\\log_2{n})}) = \\frac{3}{2}.\\] Multiplying by $2,$ exponentiating, and simplifying gives us \\begin{align*} \\log_{2}{(\\log_2{n})} &= 3 \\\\ \\log_2{n}&=8 \\\\ n&=256. \\end{align*} Adding the digits together, we have $2+5+6=\\boxed{13}.$", "We will apply the following logarithmic identity: \\[\\log_{p^k}{q^k}=\\log_{p}{q},\\] which can be proven by the Change of Base Formula \\[\\log_{p^k}{q^k}=\\frac{\\log_{p}{q^k}}{\\log_{p}{p^k}}=\\frac{k\\log_{p}{q}}{k}=\\log_{p}{q}.\\] Note that $\\log_{16}{n}\\neq0,$ so we rewrite the original equation as follows: \\begin{align*} \\log_4{(\\log_{16}{n})^2}&=\\log_4{(\\log_4{n})} \\\\ (\\log_{16}{n})^2&=\\log_4{n} \\\\ (\\log_{16}{n})^2&=\\log_{16}{n^2} \\\\ (\\log_{16}{n})^2&=2\\log_{16}{n} \\\\ \\log_{16}{n}&=2, \\end{align*} from which $n=16^2=256.$ The sum of its digits is $2+5+6=\\boxed{13}.$", "Using the change of base formula on the RHS of the initial equation yields \\[\\log_2{(\\log_{16}{n})} = \\frac{\\log_2{(\\log_4{n})}}{\\log_2{4}}.\\] This means we can multiply each side by $2$ for \\[\\log_2{(\\log_{16}{n})^2} = \\log_2{(\\log_4{n})}.\\] Canceling out the logs gives \\[(\\log_{16}{n})^2=\\log_4{n}.\\] We use change of base on the RHS to see that \\begin{align*} (\\log_{16}{n})^2&=\\frac{ \\log_{16}{n}}{\\log_{16}{4}} \\\\ (\\log_{16}{n})^2&=2 \\log_{16}{n}. \\end{align*} Substituting in $m = \\log_{16}{n}$ gives $m^2=2m,$ so $m$ is either $0$ or $2.$ Since $m=0$ yields no solution for $n$ (since this would lead to use taking the log of $0$ ), we get $2 = \\log_{16}{n},$ or $n=16^2=256,$ for the digit-sum of $2 + 5 + 6 = \\boxed{13}.$", "Suppose $\\log_2(\\log_{16}n)=k\\implies\\log_{16}n=2^k\\implies n=16^{2^k}.$ Similarly, we have $\\log_4(\\log_4 n)=k\\implies \\log_4 n=4^k\\implies n=4^{4^k}.$ Thus, we have \\[16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}\\] and \\[4^{4^k}=4^{2^{2k}},\\] so $k+1=2k\\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$ , and the requested answer is $2+5+6=\\boxed{13}.$", "We know that, as the answer is an integer, $n$ must be some power of $16.$ Testing $16$ yields \\begin{align*} \\log_2{(\\log_{16}{16})} &= \\log_4{(\\log_4{16})} \\\\ \\log_2{1} &= \\log_4{2} \\\\ 0 &= \\frac{1}{2}, \\end{align*} which does not work. We then try $256,$ giving us \\begin{align*} \\log_2{(\\log_{16}{256})} &= \\log_4{(\\log_4{256})} \\\\ \\log_2{2} &= \\log_4{4} \\\\ 1 &= 1, \\end{align*} which holds true. Thus, $n = 256,$ so the answer is $2 + 5 + 6 = \\boxed{13}.$" ]

[ "Since these form a geometric series, $\\frac{\\log_2{x}}{\\log_4{x}}$ is the common ratio. Rewriting this, we get $\\frac{\\log_x{4}}{\\log_x{2}} = \\log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\\frac{\\log_4{x}}{\\log_8{2x}} = 2 \\implies \\log_4{x} = 2\\log_8{2} + 2\\log_8{x} \\implies \\frac{1}{2}\\log_2{x} = \\frac{2}{3} + \\frac{2}{3}\\log_2{x}$\n$\\implies -\\frac{1}{6}\\log_2{x} = \\frac{2}{3} \\implies \\log_2{x} = -4 \\implies x = \\frac{1}{16}$ . Therefore, $1 + 16 = \\boxed{017}$", "If we set $x=2^y$ , we can obtain three terms of a geometric sequence through logarithm properties. The three terms are \\[\\frac{y+1}{3}, \\frac{y}{2}, y.\\] In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: \\[\\frac{y^2+y}{3} = \\frac{y^2}{4},\\] which can be solved to reveal $y = -4$ . Therefore, $x = 2^{-4} = \\frac{1}{16}$ , so our answer is $\\boxed{017}$", "Let $r$ be the common ratio. We have \\[r = \\frac{\\log_4{(x)}}{\\log_8{(2x)}} = \\frac{\\log_2{(x)}}{\\log_4{(x)}}\\] Hence we obtain \\[(\\log_4{(x)})(\\log_4{(x)}) = (\\log_8{(2x)})(\\log_2{(x)})\\] Ideally we change everything to base $64$ and we can get: \\[(\\log_{64}{(x^3)})(\\log_{64}{(x^3)}) = (\\log_{64}{(x^6)})(\\log_{64}{(4x^2)})\\] Now divide to get: \\[\\frac{\\log_{64}{(x^3)}}{\\log_{64}{(4x^2)}} = \\frac{\\log_{64}{(x^6)}}{\\log_{64}{(x^3)}}\\] By change-of-base we obtain: \\[\\log_{(4x^2)}{(x^3)} = \\log_{(x^3)}{(x^6)} = 2\\] Hence $(4x^2)^2 = x^3 \\rightarrow 16x^4 = x^3 \\rightarrow x = \\frac{1}{16}$ and we have $1+16 = \\boxed{017}$ as desired.", "Let $r$ be the common ratio, and let $a$ be the starting term ( $a=\\log_{8}{(2x)}$ ). We then have: \\[\\log_{8}{(2x)}=a, \\log_{4}{(x)}=ar, \\log_{2}{(x)}=ar^2\\] Rearranging these equations gives: \\[8^a=2x, 4^{ar}=x, 2^{ar^2}=x\\] Deal with the last two equations first: Setting them equal gives: \\[4^{ar}=2^{ar^2} \\implies 2^{2ar}=2^{ar^2} \\implies 2ar=ar^2 \\implies r=2\\] Using this value of $r$ , substitute into the first and second equations (or the first and third, it doesn't really matter) to get: \\[8^a=2x, 4^{2a}=x\\] Changing these to a common base gives: \\[2^{3a}=2x, 2^{4a}=x\\] Dividing the first equation by 2 on both sides yields: \\[2^{3a-1}=x\\] Setting these equations equal to each other and removing the exponent again gives: \\[3a-1=4a \\implies a=-1\\] Substituting this back into the first equation gives: \\[8^{-1}=2x \\implies 2x=\\frac{1}{8} \\implies x=\\frac{1}{16}\\] Therefore, $m+n=1+16=\\boxed{017}$", "We can relate the logarithms as follows:\n\\[\\frac{\\log_4{x}}{\\log_8{(2x)}}=\\frac{\\log_2{x}}{\\log_4{x}}\\] \\[\\log_8{(2x)}\\log_2{x}=\\log_4{x}\\log_4{x}\\]\nNow we can convert all logarithm bases to $2$ using the identity $\\log_a{b}=\\log_{a^c}{b^c}$\n\\[\\log_2{\\sqrt[3]{2x}}\\log_2{x}=\\log_2{\\sqrt{x}}\\log_2{\\sqrt{x}}\\]\nWe can solve for $x$ as follows:\n\\[\\frac{1}{3}\\log_2{(2x)}\\log_2{x}=\\frac{1}{4}\\log_2{x}\\log_2{x}\\] \\[\\frac{1}{3}\\log_2{(2x)}=\\frac{1}{4}\\log_2{x}\\] \\[\\frac{1}{3}\\log_2{2}+\\frac{1}{3}\\log_2{x}=\\frac{1}{4}\\log_2{x}\\] We get $x=\\frac{1}{16}$ . Verifying that the common ratio is positive, we find the answer of $\\boxed{017}$", "If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\\frac{1+\\log_2{x}}{3}$ and $\\frac{1}{2}\\log_2{x}$ , respectively. Therefore: \\[\\frac{1}{2}\\log_2{x}=\\sqrt{\\left(\\frac{1+\\log_2{x}}{3}\\right)\\left(\\log_2{x}\\right)}\\] Let $n=\\log_2{x}$ . We can rewrite the expression as: \\[\\frac{n}{2}=\\sqrt{\\frac{n(n+1)}{3}}\\] \\[\\frac{n^2}{4}=\\frac{n(n+1)}{3}\\] \\[4n(n+1)=3n^2\\] \\[4n^2+4n=3n^2\\] \\[n^2+4n=0\\] \\[n(n+4)=0\\] \\[n=0 \\text{ and } -4\\] Zero does not work in this case, so we consider $n=-4$ $\\log_2{x}=-4 \\rightarrow x=\\frac{1}{16}$ . Therefore, $1+16=\\boxed{017}$", "Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let $y$ be the exponent of $\\log_8 (2x)$ , then we have $8^y=2x;\\:4^{2y}=x;\\:2^{4y}=x.$ Wee can then divide the first equation by two to have the right side be $x$ . Also, $2^{4y}=\\left(2^{4}\\right)^y=16^y$ . Setting this equal to $\\frac{8^y}{2}$ , we can divide the two equations to get $2^y=\\frac12$ . Therefore, $y=-1$ . After that, we can raise $16$ to the $-1$ th power to get $x=16^{-1}\\Rightarrow x=\\frac1{16}$ . We then get our sum of $1+16=\\boxed{017}$" ]

[ "\\begin{align*} \\cos 2023 x + i \\sin 2023 x &= (\\cos x + i \\sin x)^{2023}\\\\ &= \\cos^{2023} x + \\binom{2023}{1} \\cos^{2022} x (i\\sin x) + \\binom{2023}{2} \\cos^{2021} x (i \\sin x)^{2} +\\binom{2023}{3} \\cos^{2023} x (i \\sin x)^{3}\\\\ &+ \\dots + \\binom{2023}{2022} \\cos x (i \\sin x)^{2022} + (i \\sin x)^{2023}\\\\ &= \\cos^{2023} x + i \\binom{2023}{1} \\cos^{2022} x \\sin x - \\binom{2023}{2} \\cos^{2021} x \\sin^{2} x - i\\binom{2023}{3} \\cos^{2020} x \\sin^{3} x + \\dots\\\\ &- \\binom{2023}{2022} \\cos x \\sin^{2022} x - i \\sin^{2023} x\\\\ \\end{align*}\nBy equating real and imaginary parts:\n\\[\\cos 2023 x = \\cos^{2023} x - \\binom{2023}{2} \\cos^{2021} x \\sin^{2} x + \\dots - \\binom{2023}{2022} \\cos x \\sin^{2022} x\\]\n\\[\\sin 2023 x = \\binom{2023}{1} \\cos^{2022} x \\sin x - \\binom{2023}{3} \\cos^{2020} x \\sin^{3} x + \\dots - \\sin^{2023} x\\]\n\\begin{align*} \\tan2023x &= \\frac{ \\sin2023x }{ \\cos2023x } = \\frac{ \\binom{2023}{1} \\cos^{2022} x \\sin x - \\binom{2023}{3} \\cos^{2020} x \\sin^{3} x + \\dots - \\sin^{2023} x }{ \\cos^{2023} x - \\binom{2023}{2} \\cos^{2021} x \\sin^{2} x + \\dots - \\binom{2023}{2022} \\cos x \\sin^{2022} x }\\\\ &= \\frac{ \\binom{2023}{1} \\frac{\\cos^{2022} x \\sin x}{\\cos^{2023} x} - \\binom{2023}{3} \\frac{\\cos^{2020} x \\sin^{3} x}{\\cos^{2023} x} + \\dots - \\frac{\\sin^{2023} x}{\\cos^{2023} x} }{ \\frac{\\cos^{2023} x}{\\cos^{2023} x} - \\binom{2023}{2} \\frac{\\cos^{2021} x \\sin^{2} x}{\\cos^{2023} x} + \\dots - \\binom{2023}{2022} \\frac{\\cos x \\sin^{2022} x}{\\cos^{2023} x} }\\\\ &= \\frac{ \\binom{2023}{1} \\tan x - \\binom{2023}{3} \\tan^{3}x + \\dots - \\tan^{2023}x }{ 1 - \\binom{2023}{2} \\tan^{2}x + \\dots - \\binom{2023}{2022} \\tan^{2022} x }\\\\ \\end{align*}\n\\[a_{2023} = \\boxed{1}\\]", "Note that $\\tan{kx} = \\frac{\\binom{k}{1}\\tan{x} - \\binom{k}{3}\\tan^{3}{x} + \\cdots \\pm \\binom{k}{k}\\tan^{k}{x}}{\\binom{k}{0}\\tan^{0}{x} - \\binom{k}{2}\\tan^{2}{x} + \\cdots + \\binom{k}{k-1}\\tan^{k-1}{x}}$ , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of $\\tan{2x}, \\tan{3x},$ and $\\tan{4x}$ , and can notice the pattern from that. The expression given essentially matches the formula of $\\tan{kx}$ exactly. $a_{2023}$ is evidently equivalent to $\\pm\\binom{2023}{2023}$ , or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of $\\binom{k}{k}\\tan^{k}{x}$ is $\\boxed{1}$", "For odd $n$ , we have \\begin{align*} \\tan nx & = \\frac{\\sin nx}{\\cos nx} \\\\ & = \\frac{\\frac{1}{2i} \\left( e^{i n x} - e^{-i n x} \\right)} {\\frac{1}{2} \\left( e^{i n x} + e^{-i n x} \\right)} \\\\ & = - i \\frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\\\ & = - i \\frac{\\left( \\cos x + i \\sin x \\right)^n - \\left( \\cos x - i \\sin x \\right)^n} {\\left( \\cos x + i \\sin x \\right)^n + \\left( \\cos x - i \\sin x \\right)^n} \\\\ & = \\frac{ - 2 i \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m + 1} \\left( \\cos x \\right)^{n - 2m - 1} \\left( i \\sin x \\right)^{2m + 1}} {2 \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m} \\left( \\cos x \\right)^{n - 2m} \\left( i \\sin x \\right)^{2m}} \\\\ & = \\frac{ \\frac{1}{\\left( \\cos x \\right)^n} \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m + 1} \\left( \\cos x \\right)^{n - 2m - 1} \\left( i \\sin x \\right)^{2m + 1}} {i \\frac{1}{\\left( \\cos x \\right)^n} \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m} \\left( \\cos x \\right)^{n - 2m} \\left( i \\sin x \\right)^{2m}} \\\\ & = \\frac{ \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m + 1} \\left( i \\tan x \\right)^{2m + 1}} {i \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m} \\left( i \\tan x \\right)^{2m}} \\\\ & = \\frac{ \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m + 1} \\left( \\tan x \\right)^{2m + 1} i^{2m + 1}} {\\sum_{m=0}^{(n-1)/2} \\binom{n}{2m} \\left( \\tan x \\right)^{2m} i^{2m + 1}} \\\\ & = \\frac{ \\sum_{m=0}^{(n-1)/2} \\binom{n}{2m + 1} \\left( \\tan x \\right)^{2m + 1} \\left( -1 \\right)^m} {\\sum_{m=0}^{(n-1)/2} \\binom{n}{2m} \\left( \\tan x \\right)^{2m} \\left( -1 \\right)^m} . \\end{align*}\nThus, for $n = 2023$ , we have \\begin{align*} a_{2023} & = \\binom{2023}{2023} \\left( -1 \\right)^{(2023-1)/2} \\\\ & = \\left( -1 \\right)^{1011} \\\\ & = \\boxed{1}" ]

[ "If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.\nThere are 6 possible colors for the center triangle.\nThus, in total we have $6\\cdot(20 + 30 + 6) = \\boxed{336}$ total possibilities.", "We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.\nCase 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.\nCase 2: 120 degree rotation. Note that the three \"outer\" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.\nCase 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.\nCase 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.\nNow, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\\boxed{336}$ distinguishable ways to color the triangle.", "There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \\[6 \\binom{5 + 3}{3} = 6\\binom{8}{3} = \\boxed{336}.\\]" ]

[ "Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \\[\\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\\] Solving gives: \\[\\frac{48 + x}{6} = 10\\]\n\\[48 + x = 60\\]\n\\[x = 12\\]\nSo, the answer is $\\boxed{12}$", "Use average deviation:\nThe average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.\nSo we got\n\\[-2+1-3+2+0=-2\\]\nThe last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so $\\boxed{12}$" ]

[ "The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \\boxed{11}$" ]

[ "For simplicity, suppose $\\Delta = a$ $\\diamondsuit = b$ and $\\bullet = c$ . Then, \\[3a+b=9c\\] \\[a=b+c\\] and we want to know what $2b$ is in terms of $c$ . Substituting the second equation into the first, we have \\[4b=6c\\Rightarrow 2b=3c\\]\nThus, we need $3$ $\\bullet$ 's $\\rightarrow \\boxed{3}$" ]

[ "There are $2$ ways to place the remaining $\\text{As}$ $2$ ways to place the remaining $\\text{Bs}$ , and $1$ way to place the remaining $\\text{Cs}$ for a total of $(2)(2)(1) = \\boxed{4}$" ]

[ "Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\\circ$ from each other) yields a cube , so the volume is $\\frac12 \\cdot 12^3 = 864$ , so our answer is $\\boxed{864}$" ]

[ "In bag $A$ , there are $26$ jellybeans, and $50\\%$ are yellow. That means there are $26\\times 50\\% = 26\\times 0.50 = 13$ yellow jelly beans in this bag.\nIn bag $B$ , there are $28$ jellybeans, and $25\\%$ are yellow. That means there are $28\\times 25\\% = 28\\times 0.25 = 7$ yellow jelly beans in this bag.\nIn bag $C$ , there are $30$ jellybeans, and $20\\%$ are yellow. That means there are $30\\times 20\\% = 30\\times 0.20 = 6$ yellow jelly beans in this bag.\nIn all three bags, there are $13 + 7 + 6 = 26$ yellow jelly beans in total , and $26 + 28 + 30 = 84$ jelly beans of all types in total.\nThus, $\\frac{26}{84} = \\frac{26}{84}\\cdot 100\\% = \\frac{13 }{ 42} \\cdot 100\\% = 30.9\\%$ of all jellybeans are yellow. Thus, the correct answer is $\\boxed{31}$" ]

[ "\"Evenly spaced\" just means the bins form an arithmetic sequence.\nSuppose the middle bin in the sequence is $x$ . There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\\cdot 2^{-3x} = 6\\cdot \\frac{1}{8^x}$ , so the probability $x$ is the middle bin is $6\\cdot\\frac{x-1}{8^x}$ . Then, we want the sum \\begin{align*} 6\\sum_{x=2}^{\\infty}\\frac{x-1}{8^x} &= \\frac{6}{8}\\left[\\frac{1}{8} + \\frac{2}{8^2} + \\frac{3}{8^3}\\cdots\\right]\\\\ &= \\frac34\\left[\\left(\\frac{1}{8} + \\frac{1}{8^2} + \\frac{1}{8^3}+\\cdots \\right) + \\left(\\frac{1}{8^2} + \\frac{1}{8^3} + \\frac{1}{8^4} + \\cdots \\right) + \\cdots\\right]\\\\ &= \\frac34\\left[\\frac17\\cdot \\left(1 + \\frac{1}{8} + \\frac{1}{8^2} + \\frac{1}{8^3} + \\cdots \\right)\\right]\\\\ &= \\frac34\\cdot \\frac{8}{49}\\\\ &= \\frac{6}{49} \\end{align*} The answer is $6+49=\\boxed{55}.$", "As in solution 1, note that \"evenly spaced\" means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$ . Further note that each $(a, d)$ pair uniquely determines a set of $3$ bins.\nWe have $a\\geq1$ because the leftmost bin in the sequence can be any bin, and $d\\geq1$ , because the bins must be distinct.\nThis gives us the following sum for the probability: \\begin{align*} 6 \\sum_{a=1}^{\\infty} \\sum_{d=1}^{\\infty} 2^{-3a-3d} &= 6 \\sum_{a=1}^{\\infty} \\sum_{d=1}^{\\infty} 2^{-3a} \\cdot 2^{-3d} \\\\ &= 6 \\left( \\sum_{a=1}^{\\infty} 2^{-3a} \\right) \\left( \\sum_{d=1}^{\\infty} 2^{-3d} \\right) \\\\ &= 6 \\left( \\sum_{a=1}^{\\infty} 8^{-a} \\right) \\left( \\sum_{d=1}^{\\infty} 8^{-d} \\right) \\\\ &= 6 \\left( \\frac{1}{7} \\right) \\left( \\frac{1}{7} \\right) \\\\ &= \\frac{6}{49} .\\end{align*} Therefore the answer is $6 + 49 = \\boxed{55}$", "This is a slightly messier variant of solution 2. If the first ball is in bin $i$ and the second ball is in bin $j>i$ , then the third ball is in bin $2j-i$ . Thus the probability is \\begin{align*} 6\\sum_{i=1}^{\\infty}\\sum_{j=i+1}^\\infty2^{-i}2^{-j}2^{-2j+i}&=6\\sum_{i=1}^{\\infty}\\sum_{j=i+1}^\\infty2^{-3j}\\\\ &=6\\sum_{i=1}^{\\infty}\\left(\\frac{2^{-3(i+1)}}{1-\\tfrac{1}{8}}\\right)\\\\ &=6\\sum_{i=1}^\\infty\\frac{8}{7}\\cdot2^{-3}\\cdot2^{-3i}\\\\ &=\\frac{6}{7}\\sum_{i=1}^\\infty2^{-3i}\\\\ &=\\frac{6}{7}\\cdot\\frac{2^{-3}}{1-\\tfrac18}\\\\ &=\\frac{6}{49}. \\end{align*} Therefore the answer is $6 + 49 = \\boxed{55}$", "Based on the value of $n,$ we construct the following table: \\[\\begin{array}{c|c|c|c} & & & \\\\ [-1.5ex] \\textbf{Exactly }\\boldsymbol{n}\\textbf{ Spaces Apart} & \\textbf{Bin \\#s} & \\textbf{Expression} & \\textbf{Prob. of One Such Perm.} \\\\ [1ex] \\hline\\hline & & & \\\\ [-1.5ex] n=1 & 1,2,3 & 2^{-1}\\cdot2^{-2}\\cdot2^{-3} & 2^{-6} \\\\ [1ex] & 2,3,4 & 2^{-2}\\cdot2^{-3}\\cdot2^{-4} & 2^{-9} \\\\ [1ex] & 3,4,5 & 2^{-3}\\cdot2^{-4}\\cdot2^{-5} & 2^{-12} \\\\ [1ex] & 4,5,6 & 2^{-4}\\cdot2^{-5}\\cdot2^{-6} & 2^{-15} \\\\ [1ex] & \\cdots & \\cdots & \\cdots \\\\ [1ex] \\hline & & & \\\\ [-1.5ex] n=2 & 1,3,5 & 2^{-1}\\cdot2^{-3}\\cdot2^{-5} & 2^{-9} \\\\ [1ex] & 2,4,6 & 2^{-2}\\cdot2^{-4}\\cdot2^{-6} & 2^{-12} \\\\ [1ex] & 3,5,7 & 2^{-3}\\cdot2^{-5}\\cdot2^{-7} & 2^{-15} \\\\ [1ex] & 4,6,8 & 2^{-4}\\cdot2^{-6}\\cdot2^{-8} & 2^{-18} \\\\ [1ex] & \\cdots & \\cdots & \\cdots \\\\ [1ex] \\hline & & & \\\\ [-1.5ex] n=3 & 1,4,7 & 2^{-1}\\cdot2^{-4}\\cdot2^{-7} & 2^{-12} \\\\ [1ex] & 2,5,8 & 2^{-2}\\cdot2^{-5}\\cdot2^{-8} & 2^{-15} \\\\ [1ex] & 3,6,9 & 2^{-3}\\cdot2^{-6}\\cdot2^{-9} & 2^{-18} \\\\ [1ex] & 4,7,10 & 2^{-4}\\cdot2^{-7}\\cdot2^{-10} & 2^{-21} \\\\ [1ex] & \\cdots & \\cdots & \\cdots \\\\ [1ex] \\hline & & & \\\\ [-1.5ex] \\cdots & \\cdots & \\cdots & \\cdots \\\\ [1ex] \\end{array}\\] Since three balls have $3!=6$ permutations, the requested probability is \\begin{align*} 6\\left(\\sum_{k=0}^{\\infty}2^{-6-3k}+\\sum_{k=0}^{\\infty}2^{-9-3k}+\\sum_{k=0}^{\\infty}2^{-12-3k}+\\cdots\\right)&=6\\left(2^{-6}\\sum_{k=0}^{\\infty}2^{-3k}+2^{-9}\\sum_{k=0}^{\\infty}2^{-3k}+2^{-12}\\sum_{k=0}^{\\infty}2^{-3k}+\\cdots\\right) \\\\ &=\\left(6\\sum_{k=0}^{\\infty}2^{-3k}\\right)\\cdot\\left(2^{-6}+2^{-9}+2^{-12}+\\cdots\\right) \\\\ &=\\left(6\\sum_{k=0}^{\\infty}2^{-3k}\\right)\\cdot\\left(\\sum_{k=0}^{\\infty}2^{-6-3k}\\right) \\\\ &=\\left(6\\sum_{k=0}^{\\infty}2^{-3k}\\right)\\cdot\\left(2^{-6}\\sum_{k=0}^{\\infty}2^{-3k}\\right) \\\\ &=\\frac{6}{1-2^{-3}}\\cdot\\frac{2^{-6}}{1-2^{-3}} \\\\ &=\\frac{6}{49} \\end{align*} by infinite geometric series, from which the answer is $6+49=\\boxed{55}.$" ]

[ "Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$\nIf Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$\nAt this moment, if the last card contained a $10$ $9$ , or a $6$ , Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$ , she would know that the other two are $2$ and $5$ .) This eliminates three more possibilities.\nThus before Stacy took her look, we are left with four possible cases: $(2,4,7)$ $(1,4,8)$ $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\\boxed{4}$" ]

[ "1984 AIME-6.png\nThe line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle.\nDraw the midpoint of $\\overline{AC}$ (the centers of the other two circles), and call it $M$ . If we draw the feet of the perpendiculars from $A,C$ to the line (call $E,F$ ), we see that $\\triangle AEM\\cong \\triangle CFM$ by HA congruency ; hence $M$ lies on the line. The coordinates of $M$ are $\\left(\\frac{19+14}{2},\\frac{84+92}{2}\\right) = \\left(\\frac{33}{2},88\\right)$\nThus, the slope of the line is $\\frac{88 - 76}{\\frac{33}{2} - 17} = -24$ , and the answer is $\\boxed{024}$", "First of all, we can translate everything downwards by $76$ and to the left by $14$ . Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem:\nTwo circles, each of radius $3$ , are drawn with centers at $(0, 16)$ , and $(5, 8)$ . A line passing through $(3,0)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?\nNote that this is equivalent to finding a line such that the distance from $(0,16)$ to the line is the same as the distance from $(5,8)$ to the line. Let the line be $y - ax - b = 0$ . Then, we have that: \\[\\frac{|-5a + 8 - b|}{\\sqrt{a^2+1}}= \\frac{|16 - b|}{\\sqrt{a^2+1}} \\Longleftrightarrow |-5a+8-b| = |16-b|\\] We can split this into two cases.\nCase 1: $16-b = -5a + 8 - b \\Longleftrightarrow a = -\\frac{8}{5}$\nIn this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible.\nCase 2: $b-16 = -5a + 8 - b \\Longleftrightarrow 2b + 5a = 24$\nBut we also know that it passes through the point $(3,0)$ , so $-3a-b = 0 \\Longleftrightarrow b = -3a$ . Plugging this in, we see that $2b + 5a = 24 \\Longleftrightarrow a = -24$ $\\boxed{24}$", "Consider the region of the plane between $x=16$ and $x=17$ . The parts of the circles centered at $(14,92)$ and $(19,84)$ in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$ . We then find that the slope of the line is $-24$ and our answer is $\\boxed{024}$", "We can redefine the coordinate system so that the center of the center circle is the origin, for easier calculations, as the slope of the line and the congruence of the circles do not depend on it. $O_1=(-3, 16)$ $O_2=(0,0)$ , and $O_3=(2,8)$ . A line bisects a circle iff it passes through the center. Therefore, we can ignore the bottom circle because it contributes an equal area with any line. A line passing through the centroid of any plane system with two perpendicular lines of reflectional symmetry bisects it. We have defined two points of the line, which are $(0,0)$ and $(-\\frac{1}{2},12)$ . We use the slope formula to calculate the slope, which is $-24$ , leading to an answer of $\\boxed{024}$ $QED \\blacksquare$", "Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\\ell$ is the desired line. Draw lines $\\ell_1$ and $\\ell_2$ both parallel to $\\ell$ such that $\\ell_1$ passes through $(14,92)$ and $\\ell_2$ passes through $(19,84)$ . Clearly, $\\ell$ must be the \"average\" of $\\ell_1$ and $\\ell_2$ . Suppose $\\ell:=y=mx+b, \\ell_1:=y=mx+c, \\ell_2:=y=mx+d$ . Then $b=76-17m, c=92-14m, d=84-19m$ . So we have that \\[76-17m=\\frac{92-14m+84-19m}{2},\\] which yields $m=-24$ for an answer of $\\boxed{024}$" ]

[ "Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\\overline{AO}$ , where $O$ is the center of the large semicircle.\nThese two squares have a total area of $2 \\cdot 5^2$ , but have two quarter circle \"bites\" of radius $5$ that must be removed. Thus, the bottom part of the figure has area\n$2\\cdot 25 - 2 \\cdot \\frac{1}{4}\\pi \\cdot 5^2$\n$50 - \\frac{25\\pi}{2}$\nThis is the area of the part of the figure underneath $\\overline{BD}$ . The part of the figure over $\\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\\frac{1}{2}\\pi\\cdot 5^2 = \\frac{25\\pi}{2}$\nAdding the two areas gives a total area of $\\boxed{50}$" ]

[ "Denote the number of bananas the first monkey took from the pile as $b_1$ , the second $b_2$ , and the third $b_3$ ; the total is $b_1 + b_2 + b_3$ . Thus, the first monkey got $\\frac{3}{4}b_1 + \\frac{3}{8}b_2 + \\frac{11}{24}b_3$ , the second monkey got $\\frac{1}{8}b_1 + \\frac{1}{4}b_2 + \\frac{11}{24}b_3$ , and the third monkey got $\\frac{1}{8}b_1 + \\frac{3}{8}b_2 + \\frac{1}{12}b_3$\nTaking into account the ratio aspect, say that the third monkey took $x$ bananas in total. Then,\n$x = \\frac{1}{4}b_1 + \\frac{1}{8}b_2 + \\frac{11}{72}b_3 = \\frac{1}{16}b_1 + \\frac{1}{8}b_2 + \\frac{11}{48}b_3 = \\frac{1}{8}b_1 + \\frac{3}{8}b_2 + \\frac{1}{12}b_3$\nSolve this to find that $\\frac{b_1}{11} = \\frac{b_2}{13} = \\frac{b_3}{27}$ . All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that $b_1$ is divisible by $8$ $b_2$ is divisible by $8$ , and $b_3$ is divisible by $72$ (however, since the denominator contains a $27$ , the factors of $3$ cancel, and it only really needs to be divisible by $8$ ). Thus, the minimal value is when each fraction is equal to $8$ , and the solution is $8(11 + 13 + 27) = \\boxed{408}$", "Let the first monkey take $8x$ bananas, the second monkey take $8y$ , and the third monkey take $24z$ . I chose these numbers to make it so, when each monkey splits his bananas, they will get an integer amount of each variable. So, when the first monkey distributes his bananas, he gets $6x$ , and the other monkeys get $1x$ . So, we can make expressions for how many bananas each monkey gets. Also, the variables have to be integers too.\nMonkey 1: $6x+3y+11z$\nMonkey 2: $1x+2y+11z$\nMonkey 3: $1x+3y+2z$\nSo, they are in a ratio $3:2:1$ . But, we can turn it into an equation by multiplying the amount of bananas each monkey has by 2, 3, 6. Now, the ratio is $6:6:6$ , so, $12x+6y+22z = 3x+6y+33z = 6x+18y+12z$ . Subtracting $3x+6y+12z$ from all, we get $9x+10z = 21z = 3x+12y$ . Let's split this into 3 equations.\n\\[9x+10z = 21z\\] \\[21z = 3x+12y\\] \\[9x+10z = 3x+12y\\]\nLet's look at the first equation. Rearranging, it gets us $9x = 11z$\nWe can rearrange the third equation, then divide by 2, then subtract the second equation. \\[21z-6y = 12y-5z\\] \\[26z = 18y\\] \\[13z = 9y\\]\nIt is clear $z$ is a multiple of 9, so let $z = 9$ . Then we get the $x = 11$ , and $y = 13$ . Testing, we confirm this will get the first monkey 204 bananas, the second 136, and the third, 68. Adding them up, we get that there were $\\boxed{408}$ bananas originally in the pile.", "In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get\n$27x=11z$ $27y=13z$\nWhile solving, make sure to leave a list of numbers that must divide $x$ $y$ , and $z$ . For example, just by looking at the basic fractions you receive from writing the starting equations, 24 divides $z$ , 8 divides $y$ and $x$ . In the expression above, it's also clear that 27 divides $z$ , 13 divides $y$ , and 11 divides $z$ . You might be wondering why I wrote the expression in terms of z. That's because $z$ has the largest divisor. The LCM of all it's divisors shows that $z$ must be divisible by 216. The total amount of bananas can be found to equal to $17z/9$ . This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest possible solution to be $\\boxed{408}$", "Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \\[\\frac{3}{4}A+\\frac{3}{8}B+\\frac{11}{24}C=\\frac{1}{2}\\] \\[\\frac{1}{8}A+\\frac{1}{4}B+\\frac{11}{24}C=\\frac{1}{3}\\] \\[\\frac{1}{8}A+\\frac{3}{8}B+\\frac{2}{24}C=\\frac{1}{6}.\\] Solve this your favorite way to get that \\[(A,B,C)=\\left( \\frac{11}{51}, \\frac{13}{51}, \\frac{9}{17} \\right).\\] We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \\cdot 51 = \\boxed{408}.$" ]

[ "Let $x=\\angle{BOC}$\nWe see that the shaded region is the inner ring plus a sector $x^\\circ$ of the outer ring. The area of this in terms of $x$ is $\\left( 4 \\pi - \\pi \\right)+\\frac{x}{360} \\left( 9 \\pi - 4 \\pi \\right)$ . This simplifies to $3 \\pi + \\frac{x}{360}(5 \\pi)$\nAlso, the unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\\circ$ of the outer ring. The area of this is $\\pi + \\frac{360-x}{360}(5 \\pi)$\nWe are told these are equal, therefore $\\pi + \\frac{x}{360}(5 \\pi) = 3 \\pi + \\frac{360-x}{360}(5 \\pi)$ . Solving for $x$ reveals $x=\\boxed{108}$", "Notice for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of $\\angle{BOC}$ to $360-\\angle{BOC}$ . With that, all we need to do is solve for the shaded region.\nThe inner most circle has radius $1$ , and the second circle has radius 2. Therefore, the first shaded area has $4 \\pi - \\pi = 3 \\pi$ area. The circle has total area $9 \\pi$ , so the other shaded region must have $1.5 \\pi$ area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is $9 \\pi - 4 \\pi = 5 \\pi$ , so the non-shaded part of the outer ring is $5 \\pi - 1.5 \\pi = 3.5 \\pi$\nNow as said before, the ratio of these two areas is the ratio of $\\angle{BOC}$ and $360 - \\angle{BOC}$ . So, $\\frac{3.5}{1.5} = \\frac{7}{3}$ . We have $7x:3x$ where $7x+3x = 360$ $x = 36$ , so our answer is $3x = 108, \\boxed{108}$", "The AMC 8s allow a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is $180$ degrees), and we let the angle of desire be $x$ . We can estimate that $180-x$ is just about $30$ degrees short of $x$ itself, so $x-30=180-x$ , solving gives $x=105$ , therefore the closest answer is $\\boxed{108}$", "Suppose the desired angle is some fraction $x$ of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius $1$ is completely unshaded, so it contributes $1$ to the unshaded area. (Everything will be a multiple of $\\pi$ , so we omit it.) The inner annulus has area $2^2 - 1^2 = 3$ , which it contributes to the shaded area. The outer annulus has a total area of $3^2 - 2^2 = 5$ ; the fraction $x$ is shaded, so the shaded portion of the outer annulus contributes $5x$ to the shaded area, while the other $1 - x$ fraction is unshaded, so the unshaded portion contributes $5(1-x)$ to the unshaded area. We now equate and solve. \\[1 + 5(1-x) = 3 + 5x\\] Upon solving, we find that $x = \\frac{3}{10}$ , so the degree measure is $360 \\cdot \\frac{3}{10} = \\boxed{108}$" ]

[ "Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ $60^\\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$\nNote that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$ $4$ $5$ Pythagorean triple. Now note that $\\angle ABC + \\angle ACB = 120^\\circ$ and $\\angle XCA + \\angle XBA = 90^\\circ,$ so $\\angle XCB+\\angle XBC = 30^\\circ$ and $\\angle BXC = 150^\\circ.$ Applying the law of cosines on triangle $BXC$ yields\n\\[BC^2 = BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}\\]\nand thus the area of $ABC$ equals \\[\\frac{\\sqrt{3}}{4}\\cdot BC^2 = 25\\frac{\\sqrt{3}}{4}+9.\\]\nso our final answer is $3+4+25+9 = \\boxed{041}.$", "Let's call the circle center $X$ . It has a distance of 3, 4, 5 to an equilateral triangle $LMN$ . Consider $X$ ’s pedal triangle $ABC$ . Since $X$ ’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$ . We’ll take the one inside $ABC$ , i.e., the Fermat point, because it leads to larger $\\triangle LMN$ . Now we construct the three equilateral triangles $ABD$ $ACE$ , and $BCF$ , the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$ . Since $AEMCX$ is concyclic with $XM$ =4 as diameter, we have $AC=4\\sin(60)$ . Similarly, $AB=3\\sin(60)$ , and $BC=5\\sin(60)$ . So $\\triangle ABC$ is a 3-4-5 right triangle with $\\angle BAC=90$ . With some more angle chasing we get \\[\\angle MXC+\\angle LXB = \\angle MAC + \\angle LAB = 180 – \\angle BAC = 90\\] \\[\\angle LXM = 360 – (\\angle MXC + \\angle LXB + \\angle BXC) = 360 –(90+120)=150\\] By Law of Cosines, we have \\[LM^2 = 3^2+4^2-2*3*4\\cos(150)=25+12\\sqrt 3\\] And the area follows. \\[[LMN] = \\frac{25}{4}\\sqrt{3} + 9; \\boxed{041}.\\] By Mathdummy", " We have $x=3$ $y=4$ , and $z=5$ . Because $AD=m$ is the median of $\\triangle AOB$ , by Stewart's Theorem we have \\[m^2=\\frac 12 (x^2+y^2)-\\frac 14 \\cdot z^2\\quad \\Rightarrow \\quad m = \\frac 52.\\] Because $CD$ is the altitude of equilateral triangle $OBC$ , we have $CD=\\frac{\\sqrt{3}}2\\cdot z$ . Then in $\\triangle ADC$ , we have $\\angle ADC=\\varphi+90^\\circ$ , so $\\cos(\\angle ADC)=-\\sin\\varphi$ , and the Law of Cosines gives \\[s^2=m^2+\\frac 34\\cdot z^2 + mz\\sqrt{3}\\sin\\varphi = 25 \\left(1+\\frac {\\sqrt{3}}{2}\\cdot\\sin\\varphi \\right)\\] To calculate $\\sin\\phi$ we apply the Law of Cosines to $\\triangle ADO$ to get \\[mz\\cos\\varphi = m^2+\\frac 14\\cdot z^2 - x^2 \\quad \\Rightarrow \\quad \\cos\\varphi = \\frac 7{25}\\quad \\Rightarrow \\quad \\sin\\varphi = \\frac {24}{25}.\\] Finally, we get $s^2=25+12\\sqrt{3}$ and and thus the area of $\\triangle ABC$ equals \\[\\frac{\\sqrt{3}}{4}\\cdot s^2 = 25\\frac{\\sqrt{3}}{4}+9.\\] so our final answer is $3+4+25+9 = \\boxed{041}.$" ]

[ "If circle $Q$ has diameter $4$ , then so do congruent circles $P$ and $R$ . Draw a diameter through $P$ parallel to $AD$ . The diameter will be congruent to $AD$ , and thus $AD = 4$ , which is the height of the rectangle.\nDraw a horizontal line $PQR$ that extends to the sides of the rectangle. This line is $2$ diameters long, so it has length $4\\cdot 2 = 8$ . It is parallel and congruent to $AB$ , so the width of the rectangle is $8$\nThus, the area of the rectangle is $lw = 4\\cdot 8 = 32$ , and the answer is $\\boxed{32}$", "The area of circle $P$ and circle $R$ almost fill the rectangle. Circle $P$ has radius $\\frac{4}{2} = 2$ , and so does circle $R$ . Thus, the sum of their areas is $\\pi \\cdot 2^2 + \\pi \\cdot 2^2 = 8\\pi$ . Since $\\pi \\approx 3.14$ , the area of the circles is just over $24$ . The area of the rectangle is greater than the area of the circles. $64$ would be too high, as the two circles appear to take up much more than half the area of the rectangle. But $24$ is too low, as it is less than the area of the two circles. Thus, the only reasonable answer is $\\boxed{32}$ , which is $32$" ]

[ "Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed face. $2(51)+32+16+8+4+2=164 \\implies \\boxed{164}$" ]

[ "If the lower cells contain $A, B$ and $C$ , then the second row will contain $A + B$ and $B + C$ , and the top cell will contain $A + 2B + C$ . To obtain the\nsmallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top\nnumber will be $7$ . For the largest sum, place $9$ in the center cell and $7$ and $8$ in\nthe outer ones. This top number will be $33$ . The difference is $33 - 7 = \\boxed{26}$" ]

[ "We will try to use symmetry as much as possible.\nPick the first vertex $A$ , its choice clearly does not influence anything.\nPick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.\nIn the first case, there are $2$ faces that contain the edge $AB$ . In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$ , the entire triangle $ABC$ will be on a face. On the other hand, if $C$ is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good $C$ is $2/6 = 1/3$\nIn the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$ . Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$\nIn the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$\nSumming up all cases, the resulting probability is: \\[\\frac 37\\cdot\\frac 13 + \\frac 37\\cdot \\frac 23 + \\frac 17\\cdot 1 = \\boxed{47}\\]", "There are $\\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.\nThere are $\\binom{4}{3}=4$ ways to choose three points from the vertices of a single face. Since there are six faces, $4 \\times 6 = 24$\nThus, the probability of what we don't want is $\\frac{24}{56} = \\frac{3}{7}$ . Using complementary probability,\n\\[1- \\frac 37 = \\boxed{47}\\]" ]

[ "\nSince two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is\n\\[2d + d = 3d\\]\nand the distance between each of the chords is just $2d$ . Let the radius of the circle be $r$ . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:\nBy the Pythagorean theorem, we can create the following system of equations:\n\\[19^2 + d^2 = r^2\\]\n\\[17^2 + (2d + d)^2 = r^2\\]\nSolving, we find $d = 3$ , so $2d = \\boxed{6}$", " If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives \\[19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.\\] This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$ . Similarly, another round of Stewart's Theorem applied to $\\triangle OEF$ with cevian $\\overleftrightarrow{OQ}$ gives \\[17\\cdot 34\\cdot 17 + \\tfrac{3}{2}d\\cdot 34\\cdot\\tfrac{3}{2}d=17r^{2}+17r^{2}.\\] This simplifies to $9826+\\tfrac{153}{2}d^{2}=34r^{2}$ . Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \\begin{align*} 361+\\tfrac{1}{4}d^{2} &= r^{2} \\\\ 289+\\tfrac{9}{4}d^{2} &= r^{2} \\\\ \\end{align*} By transitive, $361+\\tfrac{1}{4}d^{2}=289+\\tfrac{9}{4}d^{2}$ . Therefore $(\\tfrac{9}{4}-\\tfrac{1}{4})d^{2}=361-289\\rightarrow 2d^{2}=72\\rightarrow d^{2}=36\\rightarrow d=\\boxed{6}.$" ]

[ "\nSince two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is\n\\[2d + d = 3d\\]\nand the distance between each of the chords is just $2d$ . Let the radius of the circle be $r$ . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:\nBy the Pythagorean theorem, we can create the following system of equations:\n\\[19^2 + d^2 = r^2\\]\n\\[17^2 + (2d + d)^2 = r^2\\]\nSolving, we find $d = 3$ , so $2d = \\boxed{6}$", " If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives \\[19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.\\] This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$ . Similarly, another round of Stewart's Theorem applied to $\\triangle OEF$ with cevian $\\overleftrightarrow{OQ}$ gives \\[17\\cdot 34\\cdot 17 + \\tfrac{3}{2}d\\cdot 34\\cdot\\tfrac{3}{2}d=17r^{2}+17r^{2}.\\] This simplifies to $9826+\\tfrac{153}{2}d^{2}=34r^{2}$ . Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \\begin{align*} 361+\\tfrac{1}{4}d^{2} &= r^{2} \\\\ 289+\\tfrac{9}{4}d^{2} &= r^{2} \\\\ \\end{align*} By transitive, $361+\\tfrac{1}{4}d^{2}=289+\\tfrac{9}{4}d^{2}$ . Therefore $(\\tfrac{9}{4}-\\tfrac{1}{4})d^{2}=361-289\\rightarrow 2d^{2}=72\\rightarrow d^{2}=36\\rightarrow d=\\boxed{6}.$" ]

[ "Plants shared by two beds have been counted\ntwice, so the total is $500 + 450 + 350 - 50 - 100 = \\boxed{1150}$", "Bed A has $350$ plants it doesn't\nshare with B or C. Bed B has $400$ plants it doesn't\nshare with A or C. And C has $250$ it doesn't share\nwith A or B. The total is $350 + 400 + 250 + 50 + 100 = \\boxed{1150}$ plants." ]

[ "Each cup is filled with $\\frac{3}{4} \\cdot \\frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\\frac{3}{20} \\cdot 100 = \\boxed{15}$", "The pitcher is $\\frac{3}{4}$ full, i.e. $75\\%$ full. Therefore each cup receives $\\frac{75}{5}=\\boxed{15}$ percent of the total capacity.", "Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that each cup received $15\\%$ of the total capacity of the pitcher, yielding $\\boxed{15}$ as the answer." ]

[ "Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework. \nLet the three friends be $a$ $b$ $c$ repectively.\n$a + b + c = 3$\nCase $1:a=0$\n$b + c = 3$\n$b = 0,1,2,3$\n$c = 3,2,1,0$\n$\\boxed{4}$ solutions." ]

[ "For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\\binom{3+3-1}{3-1} = \\binom{5}{2} = \\boxed{10}$", "Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negative integral solutions. \nLet the three friends be $a, b, c$ respectively.\n$a + b + c = 3$ The total being 3 and 2 plus signs, which implies $\\binom{3+2}{2} = \\binom{5}{2} = \\boxed{10}$", "For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:\nfor n = number of items, and s = slots:\nNow we can plug in our values,\nnumber of items = 6, and slots = 3:\n$\\binom{6-1}{3-1} = \\binom{5}{2} = \\boxed{10}$" ]

[ "If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \\times 2 = 72$ dollars after the first step.\nThen Jan doubles his money, and Toy has $72 \\times 2 = 144$ dollars after the second step.\nThen Toy doubles whatever Amy and Jan have. Since Toy ended up with $36$ , he spent $144 - 36 = 108$ to double their money. Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total. And, just before this step, Toy had $144$ dollars. Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\\boxed{252}$" ]

[ "Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\\boxed{37}$ dots.", "The dots in the next hexagon have four bands. From innermost to outermost:\nTogether, the answer is $1+6+12+18=\\boxed{37}.$", "The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon has $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus, the number of dots added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has $1+6+12+18=\\boxed{37}$ dots.", "From the full diagram below, the answer is $\\boxed{37} draw(shift(R)*scale(7.25)*hex,black+linewidth(1.25)); [/asy] ~MRENTHUSIASM" ]

[ "We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \\cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\\boxed{150}$ . ~avamarora", "Using the diagram we find that the larger side of the small rectangle is $2$ times the length of the smaller side. Therefore, the longer side is $5 \\cdot 2 = 10$ . So the area of the identical rectangles is $5 \\cdot 10 = 50$ . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \\cdot 3 = \\boxed{150}$ . ~~fath2012", "We see that if the short sides are 5, the long side has to be $5\\cdot2=10$ because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle $ABCD$ )\nis $10+5=15$ because long side + short side of the small rectangle is $15$ . The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\\cdot10$ which is $\\boxed{150}$ .\n~~mathboy282", "We see that the $2$ rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the $2$ rectangles put together is a square. So, we can infer that the length of the rectangles is $10$ . Adding that to the width of the third rectangle which is $5$ , we get that the length of the rectangle is $15$ . Multiplying $10$ and $15$ gives us $15\\cdot10$ which is $\\boxed{150}$ .\n~~awesomepony566", "There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer $\\boxed{150}$ .\n~elenafan" ]

[ " The $24$ -sided polygon is made out of $24$ shapes like $\\triangle ABC$ . Then $\\angle BAC=360^\\circ/24=15^\\circ$ , and $\\angle EAC = 45^\\circ$ , so $\\angle{EAB} = 30^{\\circ}$ . Then $EB=AE\\tan 30^\\circ = \\sqrt{3}$ ; therefore $BC=EC-EB=3-\\sqrt{3}$ . Thus \\[[ABC] = \\frac{BC}{EC}\\cdot [ACE] = \\frac{3-\\sqrt{3}}{3}\\cdot \\frac 92\\] and the required area is $24\\cdot[ABC] =108-36\\sqrt{3}$ . Finally $108+36+3=\\boxed{147}$ . \n~lopkiloinm", "As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).\nDenote by $O$ the center of this dodecagon.\nHence, $\\angle AOB = \\frac{360^\\circ}{12} = 30^\\circ$\nBecause the length of a side of a square is 6, $AO = 3 \\sqrt{2}$\nHence, $AB = 2 AO \\sin \\frac{\\angle AOB}{2} = 3 \\left( \\sqrt{3} - 1 \\right)$\nWe notice that $\\angle MAB = \\angle MBA = 30^\\circ$ .\nHence, $AM = \\frac{AB}{2\\cos \\angle MAB} = 3 - \\sqrt{3}$\nTherefore, the area of the region that three squares cover is \\begin{align*} & {\\rm Area} \\ ABCDEFGHIJKL - 12 {\\rm Area} \\ \\triangle MAB \\\\ & = 12 {\\rm Area} \\ \\triangle OAB - 12 {\\rm Area} \\ \\triangle MAB \\\\ & = 12 \\cdot \\frac{1}{2} OA \\cdot OB \\sin \\angle AOB - 12 \\cdot \\frac{1}{2} MA \\cdot MB \\sin \\angle AMB \\\\ & = 6 OA^2 \\sin \\angle AOB - 6 MA^2 \\sin \\angle AMB \\\\ & = 108 - 36 \\sqrt{3} . \\end{align*}\nTherefore, the answer is $\\boxed{147}$" ]

[ "First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\\sqrt{2}.$ Thus the area of the dodecagon is $3 \\cdot (3\\sqrt{2})^2 = 3 \\cdot 18 = 54.$ But, the problem asked for the area of the combined figure which was made from the rotated squares. This area is the area of the dodecagon, which was found, subtracting the $12$ isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this $x.$ Then, we can use the Law of Cosines on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that $x = 3\\sqrt{3} -3.$ Realize that the $12$ isosceles are congruent with an angle measure of $120^{\\circ},$ this means that we can create $4$ congruent equilateral triangles with side length of $3\\sqrt3 - 3.$ The area of the equilateral triangle is $\\frac{\\sqrt{3}}{4} \\cdot (3\\sqrt{3} -3)^2 = \\frac{\\sqrt{3}}{4} \\cdot (36 - 18\\sqrt{3}) = \\frac{36\\sqrt{3} - 54}{4}.$ Thus, the area of all the twelve small equilateral traingles are $4 \\cdot \\frac{36\\sqrt{3} - 54}{4} = 36\\sqrt{3} - 54$ . Thus, the requested area is $54 - (36\\sqrt{3} - 54) = 108 - 36\\sqrt{3}.$ Thus, $a+b+c = 108 + 36 + 3 = 147,$ so the answer is $\\boxed{147}.$", "To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;\n\nThe area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated $30^{\\circ}$ , each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:\n\nSince the side lengths of the squares must be the same, and they are both 6, we have a system of equations; \\[y+x+y\\sqrt{3} = 6\\] \\[\\frac{x\\sqrt{3}}{2} + 2y + \\frac{x}{2} = 6\\]\nWe solve this to get $x = 6-2\\sqrt{3}$ and $y = 3-\\sqrt{3}$\nThe area of each triangle is $\\frac{x}{2} \\cdot \\frac{x\\sqrt{3}}{2} \\cdot \\frac{1}{2} = 6\\sqrt{3} - 9$ by plugging in $x$\nThe rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.\nUnfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.\n\nThe area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE. Now we only need to find the area of $\\bigtriangleup BCF$\n$\\angle GBD \\cong \\angle ECA \\cong 30^{\\circ}$ and by symmetry $\\bigtriangleup BCF$ is isosceles, so it is a 30-30-120 triangle. If we draw a perpendicular, we split it into two 30-60-90 triangles;\n\nBy symmetry, the distance from A to the edge of the square is equal to the distance from D to the edge of the square is equal to $y$ . AC = BD = $x$ , and the side length of the square is 6, so we use PIE to obtain \\[x+x-BC = 6-y-y \\implies BC = 12 - 6\\sqrt{3}\\] To find the height of $\\bigtriangleup BFC$ , we see that $HC = \\frac{BC}{2} = 6-3\\sqrt{3}$ . Then by 30-60-90 triangles, $HF = \\frac{HC}{\\sqrt{3}} = 2\\sqrt{3} - 3$ . Finally, the area of $\\bigtriangleup BFC = \\frac{BC \\cdot HF}{2} = 21\\sqrt{3}-36$\nPutting it all together, the area of the entire diagram is the area of the square plus four of these triangle-triangle intersections. The area of these intersections by PIE is $2 \\cdot [ACE] - [BFC] = 12\\sqrt{3}-18-(21\\sqrt{3}-36) = 18-9\\sqrt{3}$ . \nTherefore the total area is $36 + 4 \\cdot(18-9\\sqrt{3}) = 36 + 72 - 36 \\sqrt{3} = 108 - 36\\sqrt{3}$\nThus $a + b + c = 108+36+3 = 147 = \\boxed{147}$", "As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).\nDenote by $O$ the center of this dodecagon.\nHence, $\\angle AOB = \\frac{360^\\circ}{12} = 30^\\circ$\nBecause the length of a side of a square is 6, $AO = 3 \\sqrt{2}$\nHence, $AB = 2 AO \\sin \\frac{\\angle AOB}{2} = 3 \\left( \\sqrt{3} - 1 \\right)$\nWe notice that $\\angle MAB = \\angle MBA = 30^\\circ$ .\nHence, $AM = \\frac{AB}{2\\cos \\angle MAB} = 3 - \\sqrt{3}$\nTherefore, the area of the region that three squares cover is \\begin{align*} & \\ [ABCDEFGHIJKL] - 12[MAB] \\\\ & = 12 [OAB] - 12 [MAB] \\\\ & = 12 \\cdot \\frac{1}{2} OA \\cdot OB \\sin \\angle AOB - 12 \\cdot \\frac{1}{2} MA \\cdot MB \\sin \\angle AMB \\\\ & = 6 OA^2 \\sin \\angle AOB - 6 MA^2 \\sin \\angle AMB \\\\ & = 108 - 36\\sqrt{3} \\end{align*}\nTherefore, the answer is $\\boxed{147}$", "Let $O$ be the center of the polygon, $A$ be the bottom right corner of the first square, $C$ be the next vertex to the left of $A$ , and $M$ be the midpoint between $A$ and $B$ , where $B$ is the bottom left corner of the first square. Note that because there are three $90^{\\circ}$ squares separated by $\\frac{90^{\\circ}}{3} = 30^{\\circ}$ , each side of the 24-sided polygon is equal in length, meaning to calculate the area of the whole polygon, we find the area of $\\bigtriangleup OAC$ and multiply by 24.\nTo find $[\\bigtriangleup OAC]$ , we already know the height $\\overline{OM}$ is the sidelength of the square over $2$ , or $\\frac{6}{2}=3$ , so we just need the length of the base $\\overline{AC}$ . Notice that $\\bigtriangleup OCM$ is a $30-60-90$ triangle since $\\angle COM = \\frac{360^{\\circ}}{12} = 30^{\\circ}$ , so $\\overline{CM} = \\frac{\\overline{OM}}{\\sqrt{3}} = \\frac{3}{\\sqrt{3}} = \\sqrt{3}$ . Then $\\overline{AC} = \\overline{AM} - \\overline{CM} = \\frac{6}{2} - \\sqrt{3} = 3 - \\sqrt{3}$ , so \\begin{align*} & [\\bigtriangleup AOC] = \\frac{1}{2} \\cdot \\overline{OM} \\cdot \\overline{AC} \\\\ & = \\frac{1}{2} (3)(3 - \\sqrt{3}) \\\\ & = \\frac{9 - 3\\sqrt{3}}{2} \\end{align*}\nThen the whole area of the polygon is $\\frac{9 - 3\\sqrt{3}}{2} \\cdot 24 = 108 - 36\\sqrt{3}$ . The desired solution is then $108 + 36 + 3 = 147$ , so the answer is $\\boxed{147}$" ]

[ "Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.\nWe also know that all three working together take $x$ hours. \nNow the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \\[x=\\frac1{\\frac1{x+6}+\\frac1{x+1}+\\frac1{2x}}\\] \\[\\frac{x}{x+6}+\\frac{x}{x+1}+\\frac{x}{2x}=1\\] \\[\\frac{x}{x+6}+\\frac{x}{x+1}=\\frac12\\] \\[2x(x+1)+2x(x+6)=(x+1)(x+6)\\] \\[2x^2+2x+2x^2+12x=x^2+7x+6\\] \\[3x^2+7x-6=0\\] \\[(3x-2)(x+3)=0\\] \\[x\\in\\{\\frac23, -3\\}\\] Obviously, the number of hours is positive, so the answer is $\\boxed{23}$" ]

[ "We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\\circ}$\nLet the number of sides in our polygons be $3\\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$\nThe integral angle of a regular $k$ -gon is $180 \\frac{k-2}k$ . Therefore we are looking for integer solutions to:\n\\[360 = 180\\left( \\frac{a-2}a + \\frac{b-2}b + \\frac{c-2}c \\right)\\]\nWhich can be simplified to:\n\\[2 = \\left( \\frac{a-2}a + \\frac{b-2}b + \\frac{c-2}c \\right)\\]\nFurthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes\n\\[2 = \\left( 2\\cdot\\frac{a-2}a + \\frac{b-2}b \\right)\\]\nMultiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$\nUsing the standard technique for Diophantine equations , we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$\nRemembering that $a,b\\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ $(2,4)$ $(4,2)$ , and $(8,1)$\nThese correspond to the following pairs $(a,b)$ $(5,10)$ $(6,6)$ $(8,4)$ , and $(12,3)$\nThe perimeters of the resulting polygon for these four cases are $14$ $12$ $14$ , and $\\boxed{21}$", "We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\\boxed{21}$ sides." ]

[ "Let $d$ be the common difference of the arithmetic sequence , so $a = b-d$ and $c = b+d$\nSince increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence \\[\\frac{b}{b-d+1} = \\frac{b+d}{b}\\] \\[\\frac{b}{b-d} = \\frac{b+d+2}{b}\\] Cross-multiply in both equations to get a system of equations \\[b^2 = b^2 - d^2 + b + d\\] \\[b^2 = b^2 - d^2 + 2b - 2d\\] Rearranging terms results in \\[d^2 = b+d\\] \\[d^2 = 2b-2d\\] Substitute and solve for $d$ \\[b+d = 2b-2d\\] \\[d = \\frac{b}{3}\\] Finally, substitute $d$ and solve for $b$ . Since $b \\ne 0$ , dividing by $b$ is allowed. \\[(\\frac{b}{3})^2 = b + \\frac{b}{3}\\] \\[\\frac{b^2}{9} = \\frac{4b}{3}\\] \\[\\frac{b}{9} = \\frac{4}{3}\\] \\[b = 12\\] The answer is $\\boxed{12}$" ]

[ "Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$ . WLOG assume $a$ is the largest. Then, $x+y>a$ , meaning the solution is $\\boxed{12}$ , as shown in the graph below. ", "WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $\\boxed{12}$", "This problem is going to require some geometric probability, so let's dive right in.\nTake three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true\n\\[x + y > z\\] \\[y + z > x\\] \\[x + z > y\\]\nNow, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.\nNow, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.\n\\[y > -x + z\\] \\[y > x - z\\] \\[y < x + z\\]\nNow, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ $y \\in [0, 1]$\nI recommend drawing the following figure to get an idea of what is going on.\nThe first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.\nNote: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$\nUsing the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to\nFind the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.\nExcept, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!\nTo find the area of the triangles, our plan of attack will be\nFind the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$\nThe area of the triangles is\n\\[\\frac{z^2}{2} + (1-z)^2 = \\frac{1}{2}\\left(-3x^2 + 4x\\right)\\]\nThe integral from $0$ to $1$ is\n\\[\\frac{1}{2}\\int_0^1\\left(-3x^2 + 4x\\right)dx = \\frac{1}{2}\\]\nThe total area of all the possible unit squares is quite obviously\n\\[\\int_0^11dx = 1\\]\nThus, the area not enclosed by the triangles is $1 - \\dfrac{1}{2} = \\dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is\n\\[\\frac{\\frac{1}{2}}{1} = \\boxed{12}\\]", "Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\\boxed{12}$ probability that the second cut is on the opposite side." ]

[ "There is a total of $P(1000,6)$ possible ordered $6$ -tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$\nThere are $C(1000,6)$ possible sets $\\{a_1,a_2,a_3,b_1,b_2,b_3\\}.$ We have five valid cases for the increasing order of these six elements:\nNote that the $a$ 's are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$ 's are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$\nSo, there is a total of $C(1000,6)\\cdot5\\cdot6^2$ valid ordered $6$ -tuples. The requested probability is \\[p=\\frac{C(1000,6)\\cdot5\\cdot6^2}{P(1000,6)}=\\frac{C(1000,6)\\cdot5\\cdot6^2}{C(1000,6)\\cdot6!}=\\frac14,\\] from which the answer is $1+4=\\boxed{005}.$", "Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$ . Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick.\nThe total number of arrangements is ${6\\choose3} = 20$ ; therefore, $p = \\frac{3 + 2}{20} = \\frac{1}{4}$ , and the answer is $1 + 4 = \\boxed{005}$", "As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the $n$ th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \\[\\frac{\\binom{2n}{n}}{n+1}\\] Thus, there are $\\frac{\\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are $\\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\\frac{\\binom{6}{3}/4}{\\binom{6}{3}}=\\frac14\\Longrightarrow \\boxed{005}$" ]

[ "Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \\frac{t \\pi}{30}$ $b(t) = \\frac{t \\pi}{42}$ $c(t) = \\frac{t \\pi}{70}$\nIn order for the planets and the central star to be collinear, $a(t)$ $b(t)$ , and $c(t)$ must differ by a multiple of $\\pi$ . Note that $a(t) - b(t) = \\frac{t \\pi}{105}$ and $b(t) - c(t) = \\frac{t \\pi}{105}$ , so $a(t) - c(t) = \\frac{ 2 t \\pi}{105}$ . These are simultaneously multiples of $\\pi$ exactly when $t$ is a multiple of $105$ , so the planets and the star will next be collinear in $\\boxed{105}$ years." ]

[ "The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\\boxed{160}$" ]

[ "Let the smallest number be $x.$ It follows that the largest number is $4x.$\nSince $x,15,$ and $4x$ are equally spaced on a number line, we have \\begin{align*} 4x-15 &= 15-x \\\\ 5x &= 30 \\\\ x &= \\boxed{6} ~MRENTHUSIASM", "Let the common difference of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\\implies d=9$ . Finally, we find that the smallest number is $15-9=\\boxed{6}$", "Let the smallest number be $x$ . Since the integers are equally spaced, and there are three of them, the middle number ( $15$ ) is the arithmetic mean of the other two numbers ( $x$ and $4x$ ). Thus, we set up the equation $(4x + x)/3 = 15$ , and, solving for $x$ , get $x = 6$ . Since $6$ is the smallest number out of the list $6, 15, 24$ $24$ because it equals $4x$ ), the answer is $\\boxed{6}$", "Let the smallest number be $x$ . Because $x$ and $4x$ are equally spaced from $15$ $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$ , we get that the mean is also $2.5x$ . We get that $2.5x=15$ , and solving gets $x=\\boxed{6}$" ]

[ "We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$ , and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$ , and as $p < q$ , we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\\boxed{2}$" ]

[ "There are $\\frac{6!}{3!\\cdot2!\\cdot1!}=60$ total orderings.\nSuppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:\n$R\\ W R\\ W R$\n$R\\ R\\ W R\\ W$\n$W R\\ R\\ W R$\n$R\\ W R\\ R\\ W$\n$W R\\ W R\\ R$\nFor the first case, there are $6$ possible places we can put the blue bead. For the other 4 cases, there is only one place we can put the blue bead, which is between the pair of red beads. The desired probability is $\\frac{6+4(1)}{60}=\\frac{10}{60}=\\boxed{16}$" ]

[ "First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.\n\\[4.8x-4.4x=500 \\Rightarrow x=1250\\]\nBecause $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line.\nNow we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\\boxed{2,500}$ seconds.", "Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively. The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.\nIf they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively. The last two runners have a difference of $250$ meters, which is not a multiple of $500$\nIf they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively. The distance separating each pair of runners is a multiple of $500$ , so the answer is $\\boxed{2,500}$ seconds.", "Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$ . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$ , so $t=\\boxed{2,500}$" ]

[ "This solution refers to the Diagram section.\nWe let $\\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third, as shown below: Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \\sqrt{560}$ . Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$ . Then we have $x^2 = 576-560 \\implies x = 4$\nWe have $AO_A = BD$ because of the rectangle, so $\\sqrt{11^2-r^2} = \\sqrt{13^2-r^2}-4$ .\nSquaring, we have $121-r^2 = 169-r^2 + 16 - 8 \\cdot \\sqrt{169-r^2}$ .\nSubtracting, we get $8 \\cdot \\sqrt{169-r^2} = 64 \\implies \\sqrt{169-r^2} = 8 \\implies 169-r^2 = 64 \\implies r^2 = 105$ .\nWe also notice that since we had $\\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$ $AO_A = 4$\nWe take a cross-section that contains $A$ and $C$ , which contains these two spheres but not the third, as shown below: We have $CO_C = \\sqrt{19^2-r^2} = \\sqrt{361 - 105} = \\sqrt{256} = 16$ . Since $AO_A = 4$ , we have $EO_C = 16-4 = 12$ . Using Pythagorean theorem, $O_AE = \\sqrt{30^2 - 12^2} = \\sqrt{900-144} = \\sqrt{756}$ . Therefore, $O_AE^2 = AC^2 = \\boxed{756}$", "Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.\nAccording to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$\nThe desired value is $(11+19)^2-(16-4)^2=\\boxed{756}.$", "Denote by $r$ the radius of three congruent circles formed by the cutting plane.\nDenote by $O_A$ $O_B$ $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$ $B$ $C$ , respectively.\nBecause three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ $O_A O_C = 11 + 19 = 30$\nWe have $O_A A^2 = 11^2 - r^2$ $O_B B^2 = 13^2 - r^2$ $O_C C^2 = 19^2 - r^2$\nBecause $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\\angle O_A A B = \\angle O_B BA = 90^\\circ$\nHence, \\begin{align*} O_B B - O_A A & = \\sqrt{O_A O_B^2 - AB^2} \\\\ & = 4 . \\hspace{1cm} (1) \\end{align*}\nRecall that \\begin{align*} O_B B^2 - O_A A^2 & = \\left( 13^2 - r^2 \\right) - \\left( 11^2 - r^2 \\right) \\\\ & = 48 . \\hspace{1cm} (2) \\end{align*}\nHence, taking $\\frac{(2)}{(1)}$ , we get \\[ O_B B + O_A A = 12 . \\hspace{1cm} (3) \\]\nSolving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$\nThus, $r^2 = 11^2 - O_A A^2 = 105$\nThus, $O_C C = \\sqrt{19^2 - r^2} = 16$\nBecause $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\\angle O_A A C = \\angle O_C CA = 90^\\circ$\nTherefore, \\begin{align*} AC^2 & = O_A O_C^2 - \\left( O_C C - O_A A \\right)^2 \\\\ & = \\boxed{756}" ]

[ "\\[t=2h-3d\\] \\[3d^3+t^3=2h^3\\]\nFirst, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$ . That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$ . Then $t=8-3=5$ . Everything appears to be relatively prime already. The answer is thus $1+16+25=\\boxed{42}.$ ~lopkiloinm" ]

[ "$AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\\triangle ABC \\sim \\triangle BED$ . The hypotenuse of right triangle BED is $\\sqrt{1^2+2^2}=\\sqrt{5}$\n\\[\\frac{AC}{BC}=\\frac{BD}{ED} \\Rightarrow \\frac{AC}{BC} = \\frac21 \\Rightarrow AC=2BC\\]\n\\[\\frac{AC}{AB}=\\frac{BD}{BE} \\Rightarrow \\frac{AC}{1}=\\frac{2}{\\sqrt{5}} \\Rightarrow AC=\\frac{2}{\\sqrt{5}}\\]\nSince $AC=2BC$ $BC=\\frac{1}{\\sqrt{5}}$ $\\triangle ABC$ is a right triangle so the area is just $\\frac12 \\cdot AC \\cdot BC = \\frac12 \\cdot \\frac{2}{\\sqrt{5}} \\cdot \\frac{1}{\\sqrt{5}} = \\boxed{15}$", "Let $\\text{E}$ be the origin. Then, $\\text{D}=(1, 0)$ $\\text{A}=(0, 2)$ $\\text{B}=(1, 2)$ $\\text{F}=(2, 1)$\n${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\\frac{1}{2}x+2$\nSubtracting the second equation from the first gives us $\\frac{5}{2}x-2=0$ . \nThus, $x=\\frac{4}{5}$ .\nPlugging this into the first equation gives us $y=\\frac{8}{5}$\nSince $\\text{C} (0.8, 1.6)$ $G$ is $(0.8, 2)$\n${AB}=1$ and ${CG}=0.4$\nThus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot 0.4=0.2=\\frac{1}{5}$ . The answer is $\\boxed{15}$", "Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$\nTriangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \\frac{3}{2} = 2: 3$\nBased on similarity the length of the height of $GC$ is thus $\\frac{2}{5}\\cdot1 = \\frac{2}{5}$\nThus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{5}=\\frac{1}{5}$ . \nThe answer is $\\boxed{15}$", "Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.\nBecause of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\\sqrt{5}$\nTriangle $ALK$ is clearly a similar triangle to triangle $ABC$ . Segment $AB$ is the hypotenuse of triangle $ABC$ . So, we can write down:\n\\[AK/AB = LK/BC\\] , which is equal to: \\[\\frac{\\sqrt{5}}{1} = \\frac{1}{BC}\\] Solving this equation yields:\n\\[BC = \\frac{1}{\\sqrt{5}}\\]\nBy Pythagorean theorem, we can now find segment $AC$ \\[(\\frac{1}{\\sqrt{5}})^2 + AC^2 = 1\\] Solving this yields:\n\\[AC^2 = \\frac{4}{5}\\] , so $AC = \\frac{2}{\\sqrt{5}}$\nSo then we can use \\[A = \\frac{1}{2} * b * a.\\] So \\[A = \\frac{1}{2} * \\frac{1}{\\sqrt{5}} * \\frac{2}{\\sqrt{5}}\\]\n\\[= \\boxed{15}\\]" ]

[ " Graph the three points on the coordinate grid. Noting that the opposite sides of a parallelogram are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$ . The sum of the x-coordinates and y-coordinates is $9$ , so the answer is $\\boxed{9}$" ]

[ "We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.\nCase $1$ : \nSecond Row: a b c\nThird Row: b c a\nCase $2$ :\nSecond Row: a c b\nThird Row: c b a\nCase $3$ : \nSecond Row: a b c\nThird Row: c a b\nCase $4$ : \nSecond Row: a c b\nThird Row: b a c\nFor each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has $2 \\cdot 2 \\cdot 2 = 8$ possibilities. Since there are four cases, when pair $a$ has someone in the leftmost seat of the second row, there are $32$ ways to rearrange it. However, someone from either pair $a$ $b$ , or $c$ could be sitting in the leftmost seat of the second row. So, we have to multiply it by $3$ to get our answer of $32 \\cdot 3 = 96$ . So, the correct answer is $\\boxed{96}$", "Lets call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$ . We can split our problem into two cases:\nThere is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.\nStarting off with the first case, we see that there are $3!=6$ ways to arrange the A,B,C. Then, we have to choose which sibling sits. There are $2$ choices for each set of siblings meaning we have $2^3=8$ ways to arrange that. So, there are $48$ ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are $2$ ways to place the A sibling and each placement yields only $1$ possibility. So, our first case has $48\\cdot2=96$ possibilities.\nIn our second case, there are $3$ ways to choose which set of siblings will be in the same row and $4$ ways to choose the person in between. So, there $3*4 = 12$ ways to arrange the first row. In the second row, however, we see that it is impossible to make everything work out. So, there are $0$ possibilities for this case.\nThus, there are $96+0 = \\boxed{96}$ possibilities for this trip.", "Call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$\nThere are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it ( $\\circ$ denotes an empty seat):\n\\[A_1 \\circ \\circ\\] \\[\\circ \\ \\circ \\ \\circ\\]\nThen there are 4 choices for the second seat ( $B_1$ $B_2$ $C_1$ , or $C_2$ ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$\n\\[A_1 B_1 \\circ \\\\\\] \\[\\circ \\ \\circ \\ \\circ\\]\nThe last seat in the first row cannot be $A_2$ because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be $C_1$ or $C_2$ . Let's say WLOG that it is $C_1$ . There are two ways to create a second row:\n\\[A_1 B_1 C_1 \\\\\\] \\[B_2 C_2 A_2\\]\nand\n\\[A_1 B_1 C_1 \\\\\\] \\[C_2 A_2 B_2\\]\nTherefore, there are $6 \\cdot 4 \\cdot 2 \\cdot 2= \\boxed{96}$ possible seating arrangements.", "WLOG, define the three pairs of siblings to be: $A$ $B$ , and $C$ . Now, notice that you can only form a correct grouping either like this:\n\\[A B C\\]\n\\[B C A\\]\nor this:\n\\[C B A\\]\n\\[A C B\\]\nHowever, we need to consider the different orders. There are $3!$ ways to order each pair (eg. the same letters) and $2^3$ ways to order the people each of the three pairs. Now, we can just multiply everything together, yielding:\n\\[2\\cdot3!\\cdot2^3\\]\nWhich is $\\boxed{96}$", "Let the families be $A$ $B$ $C$ . In any given possible arrangement, there are $3! = 6$ ways to arrange the families and $2 \\cdot 2 \\cdot 2 = 8$ ways to arrange the siblings. This means the answer has to be divisble by $6 \\cdot 8 = 48$ . The only answer choice that satisfies this is $\\boxed{96}$", "If a pair of siblings are in the same row, the other $2$ pairs of siblings cannot fit in the remaining $4$ seats to meet the requirements. Therefore, the siblings must be in different rows.\nLet the first pair of siblings be $a_1$ $a_2$ , the second pair of siblings be $b_1$ $b_2$ , and the third pair of siblings be $c_1$ $c_2$ . For each row there must be $1$ child from each family. WLOG, let $a_1$ $b_1$ $c_1$ be in the first row. There are $3!$ arrangements. $a_2$ $b_2$ $c_2$ are in the second row, and they cannot be in the same column as their sibling. Now the problem becomes a $3$ element Derangement problem.\nAn $n$ element derangement problem is to find the number of permutations of $n$ elements where each element has a specified location, and no element is in it's specified location. For example, there are $3$ elements $1$ $2$ $3$ $1$ cannot be in the first location, $2$ cannot be in the second location, and $3$ cannot be in the third location. The derangements are: $(2,3,1)$ $(3,1,2)$ $D_3=2$\nEach pair of siblings can swap their position. There are $2^3$ swaps. The total arrangements are: $3! \\cdot D_3 \\cdot 2^3=6 \\cdot 2 \\cdot 8 =\\boxed{96}$" ]

[ "The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .\nThe only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with\na $\\boxed{6}$" ]

[ "The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\\boxed{2}$", "Note how the first and second row already contain a $2$ . Since the third row, last column already has a $4$ , the only possible place a $2$ could be in is the bottom right square. Thus our answer is $\\boxed{2}$" ]

[ "If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \\times 4=240$ days which is about $\\boxed{8}$" ]

[ "The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\\times3(7)$ , or $18\\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle. Thus, the area of the waste material is $378/2=\\boxed{189}$" ]

[ "Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $18$ and $21$ , and the amount of bracing Genevieve needs is the sum of these lengths, which is $39, \\boxed{39}$" ]

[ "The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ , so the area is $\\frac{(6)(7)}{2}=21, \\boxed{21}$", "Drawing in the diagonals of the kite will form four right triangles on the \"inside\" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the \"outside\" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or $\\frac{(6)(7)}{2}=21, \\boxed{21}$", "Pick's Theorem states: \\[\\frac{\\text{number of boundary points}}{2}+\\text{number of interior points}-1\\] as the area of a figure on a grid. Counting, we see there are $4$ boundary points and $20$ interior points. Therefore, we have \\[\\frac{4}{2}+20-1\\implies 20+1\\implies 21.\\] Hence, the answer is $\\boxed{21}$" ]

[ "Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \\begin{align*}T-N&=2(T-3N)\\\\ T-N&=2T-6N\\\\ T&=5N\\\\ T/N&=\\boxed{5} Note that actual values were not found." ]

[ "Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \\begin{align*}T-N&=2(T-3N)\\\\ T-N&=2T-6N\\\\ T&=5N\\\\ T/N&=\\boxed{5} Note that actual values were not found." ]

[ "The total amount paid is $105 + 125 + 175 = 405$ . To get how much each should have paid, we do $405/3 = 135$\nThus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that $t-d = 30 - 10 = \\boxed{20}$", "The difference in the money that Tom paid and Dorothy paid is $20$ . In order for them both to have paid the same amount, Tom must pay $20$ more than Dorothy. The answer is $\\boxed{20}$", "The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had $200$ dollars to spend. This means that after the vacation we want the amount of money, they currently have. After the trip, Tom would've $95$ dollars, Dorothy would've $75$ dollars, and Sammy had $25$ dollars. This gives us a total of $95+75+25=195$ dollars.\nWe want to equally split this money, as that is what happens after splitting the cost equally. This means that we want Dorothy, Tom, and Sammy to each have $65$ dollars. We know that Tom gave Sammy $t$ dollars meaning that we want to split this money first. As Tom gives money to no one else, we want him to reach $65$ dollars in this trade, meaning that as Tom has $95$ dollars and Sammy has $25$ dollars, we can do a trade of $30$ so $t=30$ . After this trade, we get that Tom has $65$ dollars, Sammy has $55$ dollars, and Dorothy has $75$ dollars.\nNext trade is where Dorothy gives $d$ dollars to Sammy. Dorothy has $75$ dollars and Sammy has $55$ dollars. As both of these don't have $65$ dollars and this is the last trade, we need to make sure both have $65$ dollars at the end. This is possible if $d=10$\nWe want to find $t-d=30-10=\\boxed{20}$" ]

[ "Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \\[105 + t = 125 + d.\\] Rearranging, we have \\[t-d = \\boxed{20}.\\]", "Add up the amounts that Tom, Dorothy, and Sammy paid to get $ $405$ , and divide by 3 to get $ $135$ , the amount that each should have paid.\nTom, having paid $ $105$ , owes Sammy $ $30$ , and Dorothy, having paid $ $125$ , owes Sammy $ $10$\nThus, $t - d = 30 - 10 = 20$ , which is $\\boxed{20}$" ]

[ "Tony works $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the beginning of his working period, he would have earned $12 * 50 = 600$ dollars. If he was $13$ years old at the beginning of his working period, he would have earned $13 * 50 = 650$ dollars. Because he earned $630$ dollars, we know that he was $13$ for some period of time, but not the whole time, because then the money earned would be greater than or equal to $650$ . This is why he was $12$ when he began, but turned $13$ sometime during the six month period and earned $630$ dollars in total. So the answer is $13$ .The answer is $\\boxed{13}$ . We could find out for how long he was $12$ and $13$ $12 \\cdot x + 13 \\cdot (50-x) = 630$ . Then $x$ is $20$ and we know that he was $12$ for $20$ days, and $13$ for $30$ days. Thus, the answer is $13$", "Let $x$ equal Tony's age at the end of the period. We know that his age changed during the time period (since $630$ does not evenly divide $50$ ). Thus, his age at the beginning of the time period is $x - 1$\nLet $d$ be the number of days Tony worked while his age was $x$ . We know that his earnings every day equal his age (since $2 \\cdot 0.50 = 1$ ). Thus, \\[x \\cdot d + (x - 1)(50 - d) = 630\\] \\[x\\cdot d + 50x - x\\cdot d - 50 + d = 630\\] \\[50x + d = 680\\] \\[x = \\dfrac{680 - d}{50}\\] Since $0 < d <50$ $d = 30$ . Then we know that $50x = 650$ and $x = \\boxed{13}$", "Since Tony worked for $50$ days, he has worked for $100$ hours. Let $k$ be his hourly wage. Then, we know that $100k = 630$\nDividing both sides by $50$ , we get that $2k$ (Daily wage, which is equal to his age) $\\approx$ $12.6$ . Since we now know that his age was either 12 or 13 during the 6 month span, and we are asked to find his age at the end of this time, the answer is $\\boxed{13}$" ]

[ "There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\\cdot10+11\\cdot20=\\boxed{430}$" ]

[ "First, calculate how many of each type of problem she got right:\nArithmetic: $70\\% \\cdot 10 = 0.70 \\cdot 10 = 7$\nAlgebra: $40\\% \\cdot 30 = 0.40 \\cdot 30 = 12$\nGeometry: $60\\% \\cdot 35 = 0.60 \\cdot 35 = 21$\nAltogether, Tori answered $7 + 12 + 21 = 40$ questions correct.\nTo get a $60\\%$ on her test overall, she needed to get $60\\% \\cdot 75 = 0.60 \\cdot 75 = 45$ questions right.\nTherefore, she needed to answer $45 - 40 = 5$ more questions to pass, so the correct answer is $\\boxed{5}$" ]

[ "First, let's consider a section $\\mathcal{P}$ of the solids, along the axis.\nBy some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\\mathcal{P}$ we took crosses one of the equator of the sphere.\nHere I drew two graphs, the first one is the case when $T$ is internally tangent to $S$\n\nand the second one is when $T$ is externally tangent to $S$\n\nFor both graphs, point $O$ is the center of sphere $S$ , and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\\mathcal{P}$ . Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\\mathcal{P}$ $EF\\bot CD$ $HG\\bot CD$\nAnd then, we can start our calculation.\nIn both cases, we know $\\Delta OEF\\sim \\Delta OGH\\Longrightarrow \\frac{EF}{OE} =\\frac{GH}{OG}$\nHence, in the case of internal tangent, $\\frac{E_iF_i}{OE_i} =\\frac{G_iH_i}{OG_i}\\Longrightarrow \\frac{6}{11-3} =\\frac{r_i}{11}\\Longrightarrow r_i=\\frac{33}{4}$\nIn the case of external tangent, $\\frac{E_oF_o}{OE_o} =\\frac{G_oH_o}{OG_o}\\Longrightarrow \\frac{6}{11+3} =\\frac{r_o}{11}\\Longrightarrow r_o=\\frac{33}{7}$\nThereby, $r_i-r_o=\\frac{33}{4}-\\frac{33}{7}=\\frac{99}{28}$ . And there goes the answer, $99+28=\\boxed{127}$" ]

[ "Extend $\\overline {AB}$ and $\\overline {DC}$ to meet at $E$ . Then\n\\begin{align*} \\angle BED &= 180^{\\circ} - \\angle EDB - \\angle DBE\\\\ &= 180^{\\circ} - 134^{\\circ} -23^{\\circ} = 23^{\\circ}. \\end{align*}\nThus $\\triangle BDE$ is isosceles with $DE = BD$ . Because $\\overline {AD} \\parallel \\overline {BC}$ , it follows that the triangles $BCE$ and $ADE$ are similar. Therefore \\[\\frac 95 = \\frac {BC}{AD} = \\frac {CD + DE}{DE} = \\frac {CD}{BD} + 1 = CD + 1,\\] so $CD = \\boxed{45}.$" ]

[ "Angle chasing* reveals that $\\triangle BPC\\sim\\triangle BDA$ , therefore \\[2=\\frac{BD}{BP}=\\frac{AB}{BC}=\\frac{AB}{43},\\] or $AB=86$\nAdditional angle chasing shows that $\\triangle ABO\\sim\\triangle CDO$ , therefore \\[2=\\frac{AB}{CD}=\\frac{BO}{OD}=\\frac{BP+11}{BP-11},\\] or $BP=33$ and $BD=66$\nSince $\\triangle ADB$ is right, the Pythagorean theorem implies that \\[AD=\\sqrt{86^2-66^2}=4\\sqrt{190}.\\] The answer is $4+190=\\boxed{194}$", "Since $\\triangle BCD$ is isosceles with base $\\overline{BD},$ it follows that median $\\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$\nSince $\\angle AOD=\\angle COP$ by vertical angles, we conclude that $\\triangle AOD\\sim\\triangle COP$ by AA, from which $\\frac{AD}{CP}=\\frac{OD}{OP},$ or \\[AD=CP\\cdot\\frac{OD}{OP}=h\\cdot\\frac{x}{11}.\\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \\begin{align*} 2[AOD]&=2[BOC] \\\\ OD\\cdot AD&=OB\\cdot CP \\\\ x\\left(\\frac{hx}{11}\\right)&=(x+22)h \\\\ x^2&=11(x+22). \\end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$\nApplying the Pythagorean Theorem to right $\\triangle CPB,$ we have \\[h=\\sqrt{43^2-33^2}=\\sqrt{(43+33)(43-33)}=\\sqrt{760}=2\\sqrt{190}.\\] Finally, we get \\[AD=h\\cdot\\frac{x}{11}=4\\sqrt{190},\\] so the answer is $4+190=\\boxed{194}.$", "Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$\n(1) $\\triangle CPO \\sim \\triangle ADO,$ so we get $\\frac{AD}{x} = \\frac{y}{11},$ or $AD = \\frac{xy}{11}.$\n(2) Applying Pythagorean Theorem on $\\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$\n(3) $\\triangle BPC \\sim \\triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$\nThus, $\\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \\sqrt{43^2 - 33^2} = 2 \\sqrt{190},$ so $AD = 4 \\sqrt{190}$ and the answer is $4+190=\\boxed{194}.$", "Observe that $\\triangle BPC$ is congruent to $\\triangle DPC$ ; both are similar to $\\triangle BDA$ . Let's extend $\\overline{AD}$ and $\\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\\angle BDE$ is $90$ degrees, and that $\\angle DBE \\cong \\angle PBC \\cong \\angle DBA \\implies \\angle DBE \\cong \\angle DBA$ . Thus, by ASA, we know that $\\triangle ABD \\cong \\triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .\nLet $M$ be the midpoint of $\\overline{DE}$ . Note that $\\triangle CME$ is congruent to $\\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\\overline{BE}.$\nTherefore, $\\overline{AC}$ and $\\overline{BD}$ are both medians of $\\triangle ABE$ . This means that $O$ is the centroid of $\\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\\frac{BO}{2} = DO = \\frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \\frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\\frac{DB}{2}-\\frac{DB}{3} = \\frac{DB}{6} = 11 \\implies DB = 66$\nWe are almost finished. Each side length of $\\triangle ABD$ is twice as long as the corresponding side length $\\triangle CBP$ or $\\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \\cdot 43 = 86$ . Now, by Pythagorean theorem on $\\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \\implies 86^{2}-66^{2} = AD^{2} \\implies AD = \\sqrt{3040} \\implies AD = 4 \\sqrt{190}$\nThe answer is $4+190 = \\boxed{194}$", "Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\\perp BD$ . Since $AD\\perp BD$ as well, $AD\\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$\nSince $P$ is the midpoint of $BD$ , the height of $\\triangle APB$ on side $AB$ is half that of $\\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$\nAs a basic property of a trapezoid, $\\triangle AOB \\sim \\triangle COD$ , so $\\frac{OB}{OD}=\\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$\nSince $\\triangle AOD \\sim \\triangle COP$ $\\frac{AD}{PC}=\\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\\sqrt{43^2-33^2}=\\sqrt{760}$\nSo, $AD=2\\sqrt{760}=4\\sqrt{190}$ . The correct answer is $\\boxed{194}$", "Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\\frac{b}{a}.$ Since $AB \\parallel CD$ the slope of $CD$ is the same. Note that as $\\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \\frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \\frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\\left(\\frac{2}{3}a, 0\\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \\frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\\triangle DPC$ $\\triangle DBC$ is isosceles), we have $b = \\sqrt{43^2 - 33^2} = 2\\sqrt{190},$ so $2b=4\\sqrt{190}.$\nFinally, the answer is $4+190=\\boxed{194}.$" ]

[ "Angle chasing* reveals that $\\triangle BPC\\sim\\triangle BDA$ , therefore \\[2=\\frac{BD}{BP}=\\frac{AB}{BC}=\\frac{AB}{43},\\] or $AB=86$\nAdditional angle chasing shows that $\\triangle ABO\\sim\\triangle CDO$ , therefore \\[2=\\frac{AB}{CD}=\\frac{BO}{OD}=\\frac{BP+11}{BP-11},\\] or $BP=33$ and $BD=66$\nSince $\\triangle ADB$ is right, the Pythagorean theorem implies that \\[AD=\\sqrt{86^2-66^2}=4\\sqrt{190}.\\] The answer is $4+190=\\boxed{194}$", "Since $\\triangle BCD$ is isosceles with base $\\overline{BD},$ it follows that median $\\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$\nSince $\\angle AOD=\\angle COP$ by vertical angles, we conclude that $\\triangle AOD\\sim\\triangle COP$ by AA, from which $\\frac{AD}{CP}=\\frac{OD}{OP},$ or \\[AD=CP\\cdot\\frac{OD}{OP}=h\\cdot\\frac{x}{11}.\\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \\begin{align*} 2[AOD]&=2[BOC] \\\\ OD\\cdot AD&=OB\\cdot CP \\\\ x\\left(\\frac{hx}{11}\\right)&=(x+22)h \\\\ x^2&=11(x+22). \\end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$\nApplying the Pythagorean Theorem to right $\\triangle CPB,$ we have \\[h=\\sqrt{43^2-33^2}=\\sqrt{(43+33)(43-33)}=\\sqrt{760}=2\\sqrt{190}.\\] Finally, we get \\[AD=h\\cdot\\frac{x}{11}=4\\sqrt{190},\\] so the answer is $4+190=\\boxed{194}.$", "Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$\n(1) $\\triangle CPO \\sim \\triangle ADO,$ so we get $\\frac{AD}{x} = \\frac{y}{11},$ or $AD = \\frac{xy}{11}.$\n(2) Applying Pythagorean Theorem on $\\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$\n(3) $\\triangle BPC \\sim \\triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$\nThus, $\\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \\sqrt{43^2 - 33^2} = 2 \\sqrt{190},$ so $AD = 4 \\sqrt{190}$ and the answer is $4+190=\\boxed{194}.$", "Observe that $\\triangle BPC$ is congruent to $\\triangle DPC$ ; both are similar to $\\triangle BDA$ . Let's extend $\\overline{AD}$ and $\\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\\angle BDE$ is $90$ degrees, and that $\\angle DBE \\cong \\angle PBC \\cong \\angle DBA \\implies \\angle DBE \\cong \\angle DBA$ . Thus, by ASA, we know that $\\triangle ABD \\cong \\triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .\nLet $M$ be the midpoint of $\\overline{DE}$ . Note that $\\triangle CME$ is congruent to $\\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\\overline{BE}.$\nTherefore, $\\overline{AC}$ and $\\overline{BD}$ are both medians of $\\triangle ABE$ . This means that $O$ is the centroid of $\\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\\frac{BO}{2} = DO = \\frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \\frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\\frac{DB}{2}-\\frac{DB}{3} = \\frac{DB}{6} = 11 \\implies DB = 66$\nWe are almost finished. Each side length of $\\triangle ABD$ is twice as long as the corresponding side length $\\triangle CBP$ or $\\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \\cdot 43 = 86$ . Now, by Pythagorean theorem on $\\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \\implies 86^{2}-66^{2} = AD^{2} \\implies AD = \\sqrt{3040} \\implies AD = 4 \\sqrt{190}$\nThe answer is $4+190 = \\boxed{194}$", "Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\\perp BD$ . Since $AD\\perp BD$ as well, $AD\\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$\nSince $P$ is the midpoint of $BD$ , the height of $\\triangle APB$ on side $AB$ is half that of $\\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$\nAs a basic property of a trapezoid, $\\triangle AOB \\sim \\triangle COD$ , so $\\frac{OB}{OD}=\\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$\nSince $\\triangle AOD \\sim \\triangle COP$ $\\frac{AD}{PC}=\\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\\sqrt{43^2-33^2}=\\sqrt{760}$\nSo, $AD=2\\sqrt{760}=4\\sqrt{190}$ . The correct answer is $\\boxed{194}$", "Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\\frac{b}{a}.$ Since $AB \\parallel CD$ the slope of $CD$ is the same. Note that as $\\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \\frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \\frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\\left(\\frac{2}{3}a, 0\\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \\frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\\triangle DPC$ $\\triangle DBC$ is isosceles), we have $b = \\sqrt{43^2 - 33^2} = 2\\sqrt{190},$ so $2b=4\\sqrt{190}.$\nFinally, the answer is $4+190=\\boxed{194}.$" ]

[ "\nIn the diagram, $\\overline{DE} \\perp \\overline{AB}, \\overline{FC} \\perp \\overline{AB}$ . \nDenote $\\overline{AE} = x$ and $\\overline{DE} = h$ . In right triangle $AED$ , we have from the Pythagorean theorem: $x^2+h^2=100$ . Note that since $EF = DC$ , we have $BF = 33-DC-x = 12-x$ . Using the Pythagorean theorem in right triangle $BFC$ , we have $(12-x)^2 + h^2 = 196$\nWe isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$\nSetting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \\implies 24x = 48 \\implies x = 2$ . Now, we can determine that $h^2 = 100-4 \\implies h = 4\\sqrt{6}$\n\nThe two diagonals are $\\overline{AC}$ and $\\overline{BD}$ . Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$ , we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$ . Since $\\sqrt{96+529}<\\sqrt{96+961}$ $25$ is the shorter length*, so the answer is $\\boxed{25}$", "\nThe area of $\\Delta AED$ is by Heron's, $4\\sqrt{9(4)(3)(2)}=24\\sqrt{6}$ . This makes the length of the altitude from $D$ onto $\\overline{AE}$ equal to $4\\sqrt{6}$ . One may now proceed as in Solution $1$ to obtain an answer of $\\boxed{25}$" ]

[ "Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$\nThen $XD=xy-70, XC=y(92-x)-50,$ thus \\[\\frac{xy-70}{y(92-x)-50} = \\frac{XD}{XC} = \\frac{ED}{EC}=\\frac{AP}{PB} = \\frac{x}{92-x},\\] which we can rearrange, expand and cancel to get $120x=70\\cdot 92,$ hence $AP=x=\\frac{161}{3}$ . This gives us a final answer of $161+3=\\boxed{164}$", "From $(1)$ above, $x = \\frac{70r}{h}$ and $92-x = \\frac{50r}{h}$ . Adding these equations yields $92 = \\frac{120r}{h}$ . Thus, $x = \\frac{70r}{h} = \\frac{7}{12}\\cdot\\frac{120r}{h} = \\frac{7}{12}\\cdot92 = \\frac{161}{3}$ , and $m+n = \\boxed{164}$", "The area of the trapezoid is $\\frac{(19+92)h}{2}$ , where $h$ is the height of the trapezoid.\nDraw lines $CP$ and $BP$ . We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$ $CPD$ , and $PBA$\n$[BPC] = \\frac{1}{2} \\cdot 50 \\cdot r$ (where $r$ is the radius of the tangent circle.)\n$[CPD] = \\frac{1}{2} \\cdot 19 \\cdot h$\n$[PBA] = \\frac{1}{2} \\cdot 70 \\cdot r$\n$[BPC] + [CPD] + [PBA] = 60r + \\frac{19h}{2} = [ABCD] = \\frac{(19+92)h}{2}$\n$60r = 46h$\n$r = \\frac{23h}{30}$\nFrom Solution 1 above, $\\frac{h}{70} = \\frac{r}{x}$\nSubstituting $r = \\frac{23h}{30}$ , we find $x = \\frac{161}{3}$ , hence the answer is $\\boxed{164}$" ]

[ "Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).\nWe create a table to keep track of what numbers each child says for each round.\n$\\begin{tabular}{||c c c c||} \\hline Round & Tadd & Todd & Tucker \\\\ [0.5ex] \\hline\\hline 1 & 1 & 2-3 & 4-6 \\\\ \\hline 2 & 7-10 & 11-15 & 16-21 \\\\ \\hline 3 & 22-28 & 29-36 & 37-45 \\\\ \\hline 4 & 46-55 & 56-66 & 67-78 \\\\ [1ex] \\hline \\end{tabular}$\nTadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + \\dots + (3n - 2) = \\frac{n(3n-1)}{2}$ , by the sum of arithmetic series formula.\nWe find that $\\dfrac{37(110)}{2}=2035$ , so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$ , the nth triangular number is said, or $\\dfrac{n(n+1)}{2}$ . So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\\dfrac{109(110)}{2}=5995$ . Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$\nThus, the answer is $\\boxed{5979}$", "Firstly, as in Solution 1, we list how many numbers Tadd says, Todd says, and Tucker says in each round.\nTadd: $1, 4, 7, 10, 13 \\cdots$\nTodd: $2, 5, 8, 11, 14 \\cdots$\nTucker: $3, 6, 9, 12, 15 \\cdots$\nWe can find a general formula for the number of numbers each of the kids say after the $n$ th round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get $\\sum_{i=1}^n 3n-2=-2n+3\\sum_{i=1}^n n=-2n+\\frac{3n(n+1)}{2}=\\frac{3n^2-n}{2}$\nNow, to find the number of rotations Tadd and his siblings go through before Tadd says his $2019$ th number, we know the inequality $\\frac{3n^2-n}{2}<2019$ must be satisfied, and testing numbers gives the maximum integer value of $n$ as $36$\nThe next main insight, in order to simplify the computation process, is to notice that the $2019$ th number Tadd says is simply the number of numbers Todd and Tucker say plus the $2019$ Tadd says, which will be the answer since Tadd goes first.\nCarrying out the calculation thus becomes quite simple:\n\\[\\left(\\sum_{i=1}^{36} 3n+\\sum_{i=1}^{36} 3n-1\\right)+2019=\\left(\\sum_{i=1}^{36} 6n-1\\right)+2019=(5+11+17...+215)+2019=\\frac{36(220)}{2}+2019\\]\nAt this point, we can note that the last digit of the answer is $9$ , which gives $\\boxed{5979}$ . (Completing the calculation will confirm the answer, if you have time.)", "Think of a turn as adding a block or an interval of numbers to each person. Here are the first couple \"blocks\" that Tadd has (a block will be denoted in interval notation): [ $1$ ], [ $7$ $10$ ], [ $22$ $28$ ], [ $46$ $55$ ]. From simple inspection, we can notice two things.\n$1$ ) The start points of each interval is increasing arithmetically by $9$\n$2$ ) The length of each \"block\" is increasing by $3$\nSince each block length is increasing by 3, then we want $1$ $4$ $7$ +.. $3n-2$ $2019$ or $n(3n-1)/2 = 2019$ , where $n$ is the amount of blocks we have. In a short amount of time, you will see that $n$ is not an integer from that equation, but thats perfectly okay since we can just find an $n$ that gets it close to 2019, then we can work from there. From inspection, you will see that a good n value that does that exact job is $37$\nWe can then deduce that the start point of the $37$ th block will be $1+(6+15+24..+321) = 5887$ if we refer to fact $1$ . We can easily find the length of 37th block by $36*3+1 = 109$ if we refer to fact $2$ . Remember that the length increases by $3$ for each block. So we then can deduce that our endpoint for the 37th block is $5887+109-1 = 5995$ . The endpoint of the $37$ th block will be the $2035$ th number said because we can plug $37$ into $n(3n-1)/2$ , which tells us how many numbers Tadd has said where $n$ is the number of blocks Tadd has. From there we know that the $2019$ th number is in $37$ th block, so that means we can take our endpoint and subtract $2035-2019 = 16$ from the endpoint of the $37$ th block, which is $5995$ , so we get $5995-16 = 5979$ or $\\boxed{5979}$ .\n~~triggod" ]

[ "Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\\cdot60 + 45 = 165$ seconds. We then set up a ratio: \\[\\frac{3}{5}=\\frac{x}{165}\\] \\[3(165)=5x\\] \\[x=3(33)=99\\approx\\boxed{100}.\\]", "$2$ minutes and $45$ seconds is equal to $120+45=165\\text{ seconds}$\nSince $6$ cars pass at around $10$ seconds, there are about $\\left \\lfloor{\\dfrac{165}{10}}\\right \\rfloor =16$ groups of $6$ cars. There are about $16\\cdot6=96\\text{ cars}$ , so the closest answer choice is $\\boxed{100}$" ]

[ "Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem \\begin{align*} a^2+b^2 &=5^2 , \\\\ b^2+c^2&=6^2, \\\\ c^2+a^2 &=7^2 , \\end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$ ; similarly, $b^2 = 6$ and $c^2 = 30$ . Since $OA$ $OB$ , and $OC$ are mutually perpendicular, the tetrahedron's volume is \\[abc/6\\] because we can consider the tetrahedron to be a right triangular pyramid. \\[abc/6 = \\sqrt{a^2b^2c^2}/6 = \\frac{\\sqrt{19 \\cdot 6 \\cdot 30}}{6} = \\sqrt{95},\\] which is answer choice $\\boxed{95}$" ]

[ "The product of two lengths with a common point brings to mind the Power of a Point Theorem\nSince $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle. Since $\\angle APB = \\widehat {AB} = 2 \\angle ACB$ , we can see that point $C$ will also lie on the circle, since the measure of arc $\\widehat {AB}$ is twice the measure of inscribed angle $\\angle ACB$ , which is true for all inscribed angles.\n\nSince $PDB$ is a line, we have $PD + DB = PB$ , which gives $3 = DB + 2$ , or $DB = 1$\nWe now extend radius $PB = 3$ to diameter $EB = 6$ . Since $EDB$ is a line, we have $ED + DB = EB$ , which gives $ED + 1 = 6$ , or $ED = 5$\nFinally, we apply the power of a point theorem to point $D$ . This states that $AD \\cdot DC = DB \\cdot DE$ . Since $DB = 1$ and $DE = 5$ , the desired product is $5$ , which is $\\boxed{5}$", "Construct the angle bisector of $\\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\\angle EPD = \\angle BCD = x,$ so $EPCB$ is cyclic. Then, $\\frac{PD}{ED} = \\frac{CD}{BD}$ or $\\frac{2}{2x} = \\frac{CD}{1}.$ Thus, $AD \\cdot DC = 5x \\cdot DC = 5,$ or $\\boxed{5}.$" ]

[ "Let $D$ be the foot of the altitude to $\\overline BC$ . Then $BD = \\sqrt {13^2 - 12^2} = 5$ and $DC = \\sqrt {15^2 - 12^2} = 9$ . Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ . The sum of the two possible values is $14 + 4 = \\boxed{18}$" ]

[ "Let $\\angle BAE = \\angle ACD = x$\n\\begin{align*}\\angle BCD &= \\angle AEC = 60^\\circ\\\\ \\angle EAC + \\angle FCA + \\angle ECF + \\angle AEC &= \\angle EAC + x + 60^\\circ + 60^\\circ = 180^\\circ\\\\ \\angle EAC &= 60^\\circ - x\\\\ \\angle BAC &= \\angle EAC + \\angle BAE = 60^\\circ - x + x = 60^\\circ\\end{align*}\nSince $\\frac{AC}{AB} = \\frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\\angle BCA = \\boxed{90}$" ]

[ "Let $\\angle BAE = \\angle ACD = x$\n\\begin{align*}\\angle BCD &= \\angle AEC = 60^\\circ\\\\ \\angle EAC + \\angle FCA + \\angle ECF + \\angle AEC &= \\angle EAC + x + 60^\\circ + 60^\\circ = 180^\\circ\\\\ \\angle EAC &= 60^\\circ - x\\\\ \\angle BAC &= \\angle EAC + \\angle BAE = 60^\\circ - x + x = 60^\\circ\\end{align*}\nSince $\\frac{AC}{AB} = \\frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\\angle BCA = \\boxed{90}$" ]

[ "Let $I$ be the incenter of $\\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\\angle ABC$ and $\\angle ACB$ respectively. Then, $\\angle BID = \\angle CBI = \\angle DBI,$ so $\\triangle BDI$ is isosceles , and similarly $\\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $\\triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $\\triangle ADE$ and $\\triangle ABC$ is $\\frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = \\frac{43}{63} \\times 20 = \\frac{860}{63}$ . Thus, $m + n = \\boxed{923}$", "The semiperimeter of $ABC$ is $s = \\frac{20 + 21 + 22}{2} = \\frac{63}{2}$ . By Heron's formula , the area of the whole triangle is $A = \\sqrt{s(s-a)(s-b)(s-c)} = \\frac{21\\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $r = \\frac{A}{s} = \\frac{\\sqrt{1311}}6$ . Since $\\triangle ADE \\sim \\triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $A=\\frac{1}{2}bh$ , we find $h_{ABC} = \\frac{21\\sqrt{1311}}{40}$ . Thus, we have\nSolving for $DE$ gives $DE=\\frac{860}{63},$ so the answer is $m+n=\\boxed{923}$", "Let $P$ be the incenter ; then it is be the intersection of all three angle bisectors . Draw the bisector $AP$ to where it intersects $BC$ , and name the intersection $F$\nUsing the angle bisector theorem , we know the ratio $BF:CF$ is $21:22$ , thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$ , giving $F$ a weight of $43$ . In the same manner, using another bisector, we find that $A$ has a weight of $20$ . So, now we know $P$ has a weight of $63$ , and the ratio of $FP:PA$ is $20:43$ . Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$ . So, $DE$ is $43/63$ the size of $BC$ . Multiplying this ratio by the length of $BC$ , we find $DE$ is $860/63 = m/n$ . Therefore, $m+n=\\boxed{923}$", "More directly than Solution 2, we have \\[DE=BC\\left(\\frac{h_a-r}{h_a}\\right)=20\\left(1-\\frac{r}{\\frac{[ABC]}{\\frac{BC}{2}}}\\right)=20\\left(1-\\frac{10r}{sr}\\right)=20\\left(1-\\frac{10}{\\frac{63}{2}}\\right)=\\frac{860}{63}\\implies \\boxed{923}.\\]", "Diagram borrowed from Solution 3.\nLet the angle bisector of $\\angle{A}$ intersects $BC$ at $F$\nApplying the Angle Bisector Theorem on $\\angle{A}$ we have \\[\\frac{AB}{BF}=\\frac{AC}{CF}\\] \\[BF=BC\\cdot(\\frac{AB}{AB+AC})\\] \\[BF=\\frac{420}{43}\\] Since $BP$ is the angle bisector of $\\angle{B}$ , we can once again apply the Angle Bisector Theorem on $\\angle{B}$ which gives \\[\\frac{BA}{AP}=\\frac{BF}{FP}\\] \\[\\frac{AP}{PF}=\\frac{AB}{BF}=\\frac{41}{20}\\] Since $\\bigtriangleup ADE\\sim\\bigtriangleup ABC$ we have \\[\\frac{DE}{BC}=\\frac{AP}{AF}\\] \\[DE=BC\\cdot(\\frac{AP}{(\\frac{61}{41})\\cdot AP})\\] Solving gets $DE=\\frac{860}{63}$ . Thus $m+n=860+63=\\boxed{923}$", "Label $P$ the point the angle bisector of $A$ intersects ${BC}$ . First we find ${BP}$ and ${PC}$ . By the Angle Bisector Theorem, $\\frac{BP}{PC} = \\frac{21}{22}$ and solving for each using the fact that ${BC} = 20$ , we see that ${BP} = \\frac{420}{43}$ and $PC = \\frac{440}{43}$\n\\[{AP} = 21*22 - \\frac{440}{43}\\cdot\\frac{420}{43}\\] \\[{AP} = 21*22 - \\frac{440\\cdot420}{43^2}\\]\nNow we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \\frac{43}{63} \\cdot[21\\cdot22 - \\frac{420*440}{43^2}] = \\frac{43^2\\cdot22 - 20\\cdot440}{43\\cdot3}$\nNow because $\\triangle{APE} ~ \\triangle{ABC}$ , we can find ${DE}$ by $\\frac{AO}{AP} \\cdot 20$ . Dividing and simplifying, we see that $\\frac{1}{21}\\cdot\\frac{43}{3}\\cdot20 = \\frac{860}{63}$ . So the answer is $\\boxed{923}$" ]

[ "Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$ $BC$ at $R$ , and $AC$ at $S$ . Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$ (also $I$ ) of circle $C$ . Let $x=QB$ $y=RC$ and $z=AS$ . Note that $BR=x$ $SC=y$ and $AQ=z$ . Hence $x+z=27$ $x+y=25$ , and $z+y=26$ . Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$\nBy Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\\sqrt{39(14)(13)(12)}$ . On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$ . Then $39r=\\sqrt{39(14)(13)(12)}$ so that $r^2=56$\nSince the radius of circle $O$ is perpendicular to $BC$ at $R$ , we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=\\boxed{15}$", "We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label $A$ with a mass of $25$ $B$ with $26$ , and $C$ with $27$ . We also label where the angle bisectors intersect the opposite side $A'$ $B'$ , and $C'$ correspondingly. It follows then that point $B'$ has mass $52$ . Which means that $\\overline{BB'}$ is split into a $2:1$ ratio. We can then use Stewart's to find $\\overline{BB'}$ . So we have $25^2\\frac{27}{2} + 27^2\\frac{25}{2} = \\frac{25 \\cdot 26 \\cdot 27}{4} + 26\\overline{BB'}^2$ . Solving we get $\\overline{BB'} = \\frac{45}{2}$ . Plugging it in we get $\\overline{BI} = 15$ . Therefore the answer is $\\boxed{15}$", "We can use POP(Power of a point) to solve this problem. First, notice that the area of $\\triangle ABC$ is $\\sqrt{39(39 - 27)(39 - 26)(39 - 25)} = 78\\sqrt{14}$ . Therefore, using the formula that $sr = A$ , where $s$ is the semi-perimeter and $r$ is the length of the inradius, we find that $r = 2\\sqrt{14}$\nDraw radii to the three tangents, and let the tangent hitting $BC$ be $T_1$ , the tangent hitting $AB$ be $T_2$ , and the tangent hitting $AC$ be $T_3$ . Let $BI = x$ . By the pythagorean theorem, we know that $BT_1 = \\sqrt{x^2 - 56}$ . By POP, we also know that $BT_2$ is also $\\sqrt{x^2 - 56}$ . Because we know that $BC = 25$ , we find that $CT_1 = 25 - \\sqrt{x^2 - 56}$ . We can rinse and repeat and find that $AT_2 = 26 - (25 - \\sqrt{x^2 - 56}) = 1 + \\sqrt{x^2 - 56}$ . We can find $AT_2$ by essentially coming in from the other way. Since $AB = 27$ , we also know that $AT_3 = 27 - \\sqrt{x^2 - 56}$ . By POP, we know that $AT_2 = AT_3$ , so $1 + \\sqrt{x^2 - 56} = 27 - \\sqrt{x^2 - 56}$\nLet $\\sqrt{x^2 - 56} = A$ , for simplicity. We can change the equation into $1 + A = 27 - A$ , which we find $A$ to be $13$ . Therefore, $\\sqrt{x^2 - 56} = 13$ , which further implies that $x^2 - 56 = 169$ . After simplifying, we find $x^2 = 225$ , so $x = \\boxed{15}$" ]

[ "First, consider the triangle in a coordinate system with vertices at $(0,0)$ $(9,0)$ , and $(a,b)$ . Applying the distance formula , we see that $\\frac{ \\sqrt{a^2 + b^2} }{ \\sqrt{ (a-9)^2 + b^2 } } = \\frac{40}{41}$\nWe want to maximize $b$ , the height, with $9$ being the base.\nSimplifying gives $-a^2 -\\frac{3200}{9}a +1600 = b^2$\nTo maximize $b$ , we want to maximize $b^2$ . So if we can write: $b^2=-(a+n)^2+m$ , then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality , because if ${x^2 \\ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \\ge 0}$ ).\n$b^2=-a^2 -\\frac{3200}{9}a +1600=-\\left(a +\\frac{1600}{9}\\right)^2 +1600+\\left(\\frac{1600}{9}\\right)^2$\n$\\Rightarrow b\\le\\sqrt{1600+\\left(\\frac{1600}{9}\\right)^2}=40\\sqrt{1+\\frac{1600}{81}}=\\frac{40}{9}\\sqrt{1681}=\\frac{40\\cdot 41}{9}$\nThen the area is $9\\cdot\\frac{1}{2} \\cdot \\frac{40\\cdot 41}{9} = \\boxed{820}$", "Let $A, B$ be the endpoints of the side with length $9$ . Let $\\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$ ; let this intersect $AB$ at $P$ and $Q$ , where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\\implies \\dfrac{\\frac{41}{9}}{BQ+9}=\\dfrac{\\frac{40}{9}}{BQ}\\implies BQ=360$ . Finally, this means that the radius of $\\Gamma$ is $\\dfrac{360+\\frac{40}{9}}{2}=180+\\dfrac{20}{9}$\nSince the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\\Gamma$ , which just makes the altitude have length $180+\\dfrac{20}{9}$ . Thus, the area of the triangle is $\\dfrac{9\\cdot \\left(180+\\frac{20}{9}\\right)}{2}=\\boxed{820}$", "We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$ . Heron's gives\n$\\sqrt{\\left(\\frac{81x+9}{2} \\right) \\left(\\frac{81x-9}{2} \\right) \\left(\\frac{x+9}{2} \\right) \\left(\\frac{-x+9}{2} \\right)}$\nThis can be simplified to\n$\\frac{9}{4} \\cdot \\sqrt{(81x^2-1)(81-x^2)}$\nWe can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.\nWe have that $-324x^3+13124x=0$ , so $x=\\frac{\\sqrt{3281}}{9}$\nPlugging this into the expression, we have that the area is $\\boxed{820}$", "We can start how we did above in solution 4 to get $\\frac{9}{4} * \\sqrt{(81x^2-1)(81-x^2)}$ .\nThen, we can notice the inside is a quadratic in terms of $x^2$ , which is $-81(x^2)^2+6562x^2-81$ . This is maximized when $x^2 = \\frac{3281}{81}$ .If we plug it into the equation, we get $\\frac{9}{4} *\\frac{9}{4}*\\frac{3280}{9} = \\boxed{820}$" ]

[ "Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$\nThus,\n\\[\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1\\]\nNow let's apply the angle bisector theorem.\n\\[\\frac {AL}{LB}=\\frac {AC}{BC}=\\frac {450}{300}=\\frac {3}{2}\\]\n\\[\\frac {AM}{LP}=\\frac {AL}{LB}+1=\\frac {5}{2}\\]\n\\[\\frac {180}{LP}=\\frac {5}{2}\\]\n\\[LP=\\boxed{072}\\]", "Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \\[\\frac{BL}{CB}=\\frac{AL}{CA}\\implies\\frac{BL}{300}=\\frac{AL}{450}\\implies 3BL=2AL\\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$ . Since $K$ is the midpoint of $A$ and $C$ , the weight of $A$ is equal to the weight of $C$ , which equals $2$ .\nAlso, since the weight of $L$ is $5$ and $C$ is $2$ , we can weight $P$ as $7$\nBy the definition of mass points, \\[\\frac{LP}{CP}=\\frac{2}{5}\\implies LP=\\frac{2}{5}CP\\] By vertical angles, angle $MKA =$ angle $PKC$ . \nAlso, it is given that $AK=CK$ and $PK=MK$\nBy the SAS congruence, $\\triangle MKA$ $\\triangle PKC$ . So, $MA$ $CP$ $180$ .\nSince $LP=\\frac{2}{5}CP$ $LP = \\frac{2}{5}(180) = \\boxed{072}$", "Using the diagram from solution $1$ , we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$\nApplying the angle bisector theorem to $\\triangle CKB$ , we get that $\\frac{KP}{PB} = \\frac{225}{300} = \\frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$\nNow, apply law of cosines on $\\triangle CKP$ and $\\triangle CPB.$\nIf we let $\\angle KCP = \\angle PCB = \\alpha$ , then the law of cosines gives the following system of equations:\n\\[9x^2 = 225^2 + 180^2 - 2\\cdot 225 \\cdot 180 \\cdot \\cos \\alpha\\] \\[16x^2 = 180^2 + 300^2 - 2 \\cdot 180 \\cdot 300 \\cdot \\cos \\alpha.\\]\nBashing those out, we get that $x = 15 \\sqrt{13}$ and $\\cos \\alpha = \\frac{7}{10}.$\nSince $\\cos \\alpha = \\frac{7}{10}$ , we can use the double angle formula to calculate that $\\cos \\left(2 \\alpha \\right) = -\\frac{1}{50}.$\nNow, apply Law of Cosines on $\\triangle ABC$ to find $AB$\nWe get: \\[AB^2 = 450^2 + 300^2 - 2 \\cdot 450 \\cdot 300 \\cdot \\left(- \\frac{1}{50} \\right).\\]\nBashing gives $AB = 30 \\sqrt{331}.$\nFrom the angle bisector theorem on $\\triangle ABC$ , we know that $\\frac{AL}{BL} = \\frac{450}{300} = \\frac{3}{2}.$ So, $AL = 18 \\sqrt{331}$ and $BL = 12 \\sqrt{331}.$\nNow, we apply Law of Cosines on $\\triangle ALC$ and $\\triangle BLC$ in order to solve for the length of $LC$\nWe get the following system:\n\\[(18 \\sqrt{331})^2 = 450^2 + LC^2 - 2 \\cdot 450 \\cdot LC \\cdot \\frac{7}{10}\\] \\[(12 \\sqrt{331})^2 = LC^2 + 300^2 - 2 \\cdot 300 \\cdot LC \\cdot \\frac{7}{10}\\]\nThe first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$\nThe only value that satisfies both equations is $LC = 252$ , and since $LP = LC - PC$ , we have \\[LC = 252 - 180 = \\boxed{072}.\\]" ]

[ "Let $E$ $F$ and $G$ be the points of tangency of the incircle with $BC$ $AC$ and $AB$ , respectively. Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ . Let the length of the median be $3m$ . Then by two applications of the Power of a Point Theorem $DE^2 = 2m \\cdot m = AF^2$ , so $DE = AF$ . Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$ . Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$\nNow, by Stewart's Theorem in triangle $\\triangle ABC$ with cevian $\\overline{AD}$ , we have\n\\[(3m)^2\\cdot 20 + 20\\cdot10\\cdot10 = 10^2\\cdot10 + (30 - 2c)^2\\cdot 10.\\]\nOur earlier result from Power of a Point was that $2m^2 = (10 - c)^2$ , so we combine these two results to solve for $c$ and we get\n\\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \\quad \\Longrightarrow \\quad c^2 - 12c + 20 = 0.\\]\nThus $c = 2$ or $= 10$ . We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$ , so our triangle has area $\\sqrt{28 \\cdot 18 \\cdot 8 \\cdot 2} = 24\\sqrt{14}$ and so the answer is $24 + 14 = \\boxed{038}$", "WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$ . We use power of a point to get that $AG = DE = \\sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\\sqrt{2} m$\nSince now we have $AC = 10$ $BC = 20, AB = 10+2\\sqrt{2} m$ in triangle $\\triangle ABC$ and cevian $AD = 3m$ . Now, we can apply Stewart's Theorem\n\\[2000 + 180 m^2 = 10(10+2\\sqrt{2}m)^{2} + 1000\\] \\[1000 + 180 m^2 = 1000 + 400\\sqrt{2}m + 80 m^{2}\\] \\[100 m^2 = 400\\sqrt{2}m\\]\n$m = 4\\sqrt{2}$ or $m = 0$ if $m = 0$ , we get a degenerate triangle, so $m = 4\\sqrt{2}$ , and thus $AB = 26$ . You can now use Heron's Formula to finish. The answer is $24 \\sqrt{14}$ , or $\\boxed{038}$", "Let $E, F$ , and $G$ be the point of tangency (as stated in Solution 1). We can now let $AD$ be $3m$ . By using Power of a Point Theorem on A to the incircle, you get that $AG^2 = 2m^2$ . We can use it again on point D to the incircle to get the equation $(10 - CE)^2 = 2m^2$ . Setting the two equations equal to each other gives $(10 - CE)^2 = AG^2$ , and it can be further simplified to be $10 - CE = AG$\nLet lengths $AC$ and $AB$ be called $b$ and $c$ , respectively. We can write $AG$ as $\\frac{b + c - 20}{2}$ and $CE$ as $\\frac{b + 20 - c}{2}$ . Plugging these into the equation, you get:\n\\[10 - \\frac{b + 20 - c}{2} = \\frac{b + c - 20}{2}\\] \\[b + c - 20 + b + 20 - c = 20 \\rightarrow b = 10\\]\nAdditionally, by Median of a triangle formula, you get that $3m = \\frac{\\sqrt{2c^2 - 200}}{2}$\nRefer back to the fact that $AG^2 = 2m^2$ . We can now plug in our variables.\n\\[\\left(\\frac{b + c - 20}{2}\\right)^2 = 2m^2 \\rightarrow (c - 10)^2 = 8m^2\\] \\[c^2 - 20c + 100 = 8 \\cdot \\frac{2c^2 - 200}{36}\\] \\[9c^2 - 180c + 900 = 4c^2 - 400\\] \\[5c^2 - 180c + 1300 = 0\\] \\[c^2 - 36c + 260 = 0\\]\nSolving, you get that $c = 26$ or $10$ , but the latter will result in a degenerate triangle, so $c = 26$ .\nFinally, you can use Heron's Formula to get that the area is $24\\sqrt{14}$ , giving an answer of $\\boxed{038}$" ]

[ "Let $\\angle CAB=A$ $\\angle ABC=B$ $\\angle BCA=C$ for convenience.\nIt's well-known that $\\angle BOC=2A$ $\\angle BIC=90+\\frac{A}{2}$ , and $\\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$ , it follows that $\\angle BOC=\\angle BIC=\\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic. Observe that $BC=1$ is fixed, hence the circumcircle of cyclic pentagon $BCOIH$ is also fixed. Similarly, as $OB=OC$ (both are radii), it follows that $O$ and also $[BCO]$ is fixed. Since $[BCOIH]=[BCO]+[BOIH]$ is maximal, it suffices to maximize $[BOIH]$\nVerify that $\\angle IBC=\\frac{B}{2}$ $\\angle HBC=90-C$ by angle chasing; it follows that $\\angle IBH=\\angle HBC-\\angle IBC=90-C-\\frac{B}{2}=\\frac{A}{2}-\\frac{C}{2}=30-\\frac{C}{2}$ since $A+B+C=180\\implies\\frac{A}{2}+\\frac{B}{2}+\\frac{C}{2}=90$ by Triangle Angle Sum. Similarly, $\\angle OBC=(180-120)/2=30$ (isosceles base angles are equal), hence \\[\\angle IBO=\\angle IBC-\\angle OBC=\\frac{B}{2}-30=60-\\frac{A}{2}-\\frac{C}{2}=30-\\frac{C}{2}\\] Since $\\angle IBH=\\angle IBO$ $IH=IO$ by Inscribed Angles.\nThere are two ways to proceed.\nLetting $O'$ and $R$ be the circumcenter and circumradius, respectively, of cyclic pentagon $BCOIH$ , the most straightforward is to write $[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]$ , whence \\[[BOIH]=\\frac{1}{2}R^2(\\sin(60-C)+\\sin(60-C)+\\sin(2C-60)-\\sin(60))\\] and, using the fact that $R$ is fixed, maximize $2\\sin(60-C)+\\sin(2C-60)$ with Jensen's Inequality.\nA more elegant way is shown below.\nLemma: $[BOIH]$ is maximized only if $HB=HI$\nProof by contradiction: Suppose $[BOIH]$ is maximized when $HB\\neq HI$ . Let $H'$ be the midpoint of minor arc $BI$ be and $I'$ the midpoint of minor arc $H'O$ . Then $[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]$ since the altitude from $H'$ to $BI$ is greater than that from $H$ to $BI$ ; similarly $[BH'I'O]>[BOIH']>[BOIH]$ . Taking $H'$ $I'$ to be the new orthocenter, incenter, respectively, this contradicts the maximality of $[BOIH]$ , so our claim follows. $\\blacksquare$\nWith our lemma( $HB=HI$ ) and $IH=IO$ from above, along with the fact that inscribed angles that intersect the same length chords are equal, \\[\\angle ABC=2\\angle IBC=2(\\angle OBC+\\angle OBI)=2(30+\\frac{1}{3}\\angle OCB)=80\\implies\\boxed{80}\\]" ]

[ "Let $D$ be the midpoint of $\\overline{BC}$ . Then by SAS Congruence, $\\triangle ABD \\cong \\triangle ACD$ , so $\\angle ADB = \\angle ADC = 90^o$\nNow let $BD=y$ $AB=x$ , and $\\angle IBD = \\dfrac{\\angle ABD}{2} = \\theta$\nThen $\\mathrm{cos}{(\\theta)} = \\dfrac{y}{8}$\nand $\\mathrm{cos}{(2\\theta)} = \\dfrac{y}{x} = 2\\mathrm{cos^2}{(\\theta)} - 1 = \\dfrac{y^2-32}{32}$\nCross-multiplying yields $32y = x(y^2-32)$\nSince $x,y>0$ $y^2-32$ must be positive, so $y > 5.5$\nAdditionally, since $\\triangle IBD$ has hypotenuse $\\overline{IB}$ of length $8$ $BD=y < 8$\nTherefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$ $6.5$ $7$ , and $7.5$\nHowever, only one of these values, $y=6$ , yields an integral value for $AB=x$ , so we conclude that $y=6$ and $x=\\dfrac{32(6)}{(6)^2-32}=48$\nThus the perimeter of $\\triangle ABC$ must be $2(x+y) = \\boxed{108}$", "Angle bisectors motivate trig bash. \nDefine angle $IBC = x$ . Foot of perpendicular from $I$ to $BC$ is point $P$ $\\overline{BC} = 2\\overline{BP} = 2(8\\cos(x)) = N$ , where $N$ is an integer. Thus, $\\cos(x) = \\frac{N}{16}$ . Via double angle, we calculate $\\overline{AB}$ to be $\\frac{8\\cos(x)}{2\\cos(x)^2 - 1} = \\frac{64N}{N^2 - 128}$ . This is to be an integer. We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\\overline{AB} = 48, \\overline{BC} = 12$ .\nOur answer is $2 * 48 + 12 = \\boxed{108}$ .\n- whatRthose", "Let $M$ be midpoint $BC, BM = x, AB = y, \\angle IBM = \\alpha.$\n$BI$ is the bisector of $\\angle ABM$ in $\\triangle ABM.$ $BI = \\frac {2 xy \\cos \\alpha}{x+y} = 8, \\cos \\alpha = \\frac {x}{8} \\implies \\frac {x^2 y}{x+y} = 32.$ \\[y = \\frac {32 x} {x^2 - 32}.\\] $BC = 2x$ is integer, $5.5^2 < 32 \\implies x \\ge 6.$ $BM < BI \\implies x =\\{ 6, 6.5, 7, 7.5 \\}.$\nIf $x > 6$ then $y$ is not integer. \\[x = 6 \\implies y = 48 \\implies 2(x+y) = \\boxed{108}.\\] [email protected], vvsss" ]

[ "Let $PC = x$ . Since $\\angle APB = \\angle BPC = \\angle CPA$ , each of them is equal to $120^\\circ$ . By the Law of Cosines applied to triangles $\\triangle APB$ $\\triangle BPC$ and $\\triangle CPA$ at their respective angles $P$ , remembering that $\\cos 120^\\circ = -\\frac12$ , we have\n\\[AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x\\]\nThen by the Pythagorean Theorem $AB^2 + BC^2 = CA^2$ , so\n\\[x^2 + 10x + 100 = x^2 + 6x + 36 + 196\\]\nand\n\\[4x = 132 \\Longrightarrow x = \\boxed{033}.\\]" ]

[ "By the Law of Cosine $\\cos A = \\frac{AC^2 + AB^2 - BC^2}{ 2 \\cdot AC \\cdot AB} = \\frac{20^2 + 11^2 - 24^2}{2\\cdot20\\cdot11} = -\\frac18$\nAs $ABEC$ is a cyclic quadrilateral, $\\angle CEA = \\angle CBA$ . As $BDEF$ is a cyclic quadrilateral, $\\angle CBA = \\angle FEA$\n$\\because \\quad \\angle CEA = \\angle FEA \\quad \\text{and} \\quad \\angle CAE = \\angle FAE$\n$\\therefore \\quad \\triangle AFE \\cong \\triangle ACE$ by $ASA$\nHence, $AF = AC = 20$\nBy the Law of Cosine $CF = \\sqrt{20^2 + 20^2 - 2 \\cdot 20 \\cdot 20 (-\\frac18)} = \\sqrt{900} = \\boxed{30}$", "Construct the $E$ -antipode, $E^{\\prime}\\in(ABC)$ . Notice $\\triangle CE^{\\prime}A\\stackrel{+}{\\sim}\\triangle CBF$ by spiral similarity at $C$ , thus $CF=\\dfrac{CB\\cdot CA}{CE^{\\prime}}=\\frac{480}{CE^{\\prime}}$ . Let $CE^{\\prime}=x$ ; by symmetry $BE^{\\prime}=x$ as well and $\\cos\\angle BE^{\\prime}C=\\cos\\angle A=\\tfrac{11^{2}+20^{2}-24^{2}}{2\\cdot 11\\cdot 20}=-\\tfrac{1}{8}$ from Law of Cosines in $\\triangle ABC$ , so by Law of Cosines in $\\triangle BE^{\\prime}C$ we have \\[x^{2}+x^{2}+\\left(2x^{2}\\right)\\left(-\\dfrac{1}{8}\\right)=24^{2}\\] from which $x=16$ . Now, $CF=\\dfrac{480}{16}=\\boxed{30}$", "Applying Stewart's theorem on $\\triangle ABC$ with cevian $\\overline{CF}$ using the directed lengths $AF = AC = 20$ and $FB = 11-20 = -9$ , we obtain \\begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\\\ 11CF^{2} - 1980 &= 11520 - 3600\\end{align*} so $CF=\\sqrt{\\frac{11520 - 3600 + 1980}{11}}=\\sqrt{\\frac{9900}{11}}=\\sqrt{900}=\\boxed{30}$", "Note that $\\angle CAF = \\angle CAB$ so we may plug into Law of Cosines to find the angle's cosine: \\[AB^2+AC^2-2\\cdot AB \\cdot AC \\cdot \\cos(\\angle CAB) = BC^2 \\to \\cos(\\angle CAB) = -\\frac{1}{8}.\\]\nSo, we observe that we can use Law of Cosines again to find $CF$ \\[CF^2 = AF^2+AC^2-2 \\cdot AF \\cdot AC \\cdot \\cos(\\angle CAF) = 900 \\to CF=\\boxed{30}\\] both ways.", "This solution is based on this figure: 2021 AMC 12B (Nov) Problem 24, sol.png\nDenote by $O$ the circumcenter of $\\triangle BED$ .\nDenote by $R$ the circumradius of $\\triangle BED$\nIn $\\triangle BCF$ , following from the law of cosines, we have \\begin{align*} CF^2 & = BC^2 + BF^2 - 2 BC \\cdot BF \\cos \\angle CBF \\\\ & = BC^2 + BF^2 + 2 BC \\cdot BF \\cos \\angle ABC . \\hspace{1cm} (1) \\end{align*} For $BF$ , we have \\begin{align*} BF & = 2 R \\cos \\angle FBO \\\\ & = 2 R \\cos \\left( 180^\\circ - \\angle ABC - \\angle CBO \\right) \\\\ & = 2 R \\cos \\left( 180^\\circ - \\angle ABC - \\frac{180^\\circ - \\angle BOD}{2} \\right) \\\\ & = 2 R \\cos \\left( 180^\\circ - \\angle ABC - \\frac{180^\\circ - 2 \\angle BED}{2} \\right) \\\\ & = 2 R \\cos \\left( 180^\\circ - \\angle ABC - \\frac{180^\\circ - 2 \\angle BCA}{2} \\right) \\\\ & = 2 R \\cos \\left( 90^\\circ - \\angle ABC + \\angle BCA \\right) \\\\ & = 2 R \\sin \\left( \\angle ABC - \\angle BCA \\right) \\\\ & = \\frac{BD}{\\sin \\angle BED} \\sin \\left( \\angle ABC - \\angle BCA \\right) \\\\ & = \\frac{BD}{\\sin \\angle BCA} \\sin \\left( \\angle ABC - \\angle BCA \\right) \\\\ & = BD \\left( \\sin \\angle ABC \\cot \\angle BCA - \\cos \\angle ABC \\right) . \\hspace{1cm} (2) \\end{align*} The fourth equality follows from the property that $B$ $D$ $E$ are concyclic.\nThe fifth and the ninth equalities follow from the property that $A$ $B$ $C$ $E$ are concyclic.\nBecause $AD$ bisects $\\angle BAC$ , following from the angle bisector theorem, we have \\[ \\frac{BD}{CD} = \\frac{AB}{AC} . \\] Hence, $BD = \\frac{24 \\cdot 11}{31}$\nIn $\\triangle ABC$ , following from the law of cosines, we have \\begin{align*} \\cos \\angle ABC & = \\frac{AB^2 + BC^2 - AC^2}{2 AB \\cdot BC} \\\\ & = \\frac{9}{16} \\end{align*} and \\begin{align*} \\cos \\angle BCA & = \\frac{AC^2 + BC^2 - AB^2}{2 AC \\cdot BC} \\\\ & = \\frac{57}{64} . \\end{align*} Hence, $\\sin \\angle ABC = \\frac{5 \\sqrt{7}}{16}$ and $\\sin \\angle BCA = \\frac{11 \\sqrt{7}}{64}$ .\nHence, $\\cot \\angle BCA = \\frac{57}{11 \\sqrt{7}}$\nNow, we are ready to compute $BF$ whose expression is given in Equation (2).\nWe get $BF = 9$\nNow, we can compute $CF$ whose expression is given in Equation (1).\nWe have $CF = 30$\nTherefore, the answer is $\\boxed{30}$" ]

[ "If $\\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so\n\\[\\alpha\\omega - \\beta\\omega^2 = 25\\alpha - 625\\beta = 0\\]\nand $\\alpha = 25\\beta$ . If $\\omega = \\frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \\frac{12 + 17 + 25}{2} = 27$\n\\[[ABC] = \\sqrt{27 \\cdot 15 \\cdot 10 \\cdot 2} = 90\\]\nso\n\\[45 = \\alpha\\omega - \\beta\\omega^2 = \\frac{625}{2} \\beta - \\beta\\frac{625}{4} = \\beta\\frac{625}{4}\\]\nand\n\\[\\beta = \\frac{180}{625} = \\frac{36}{125}\\]\nso the answer is $m + n = 36 + 125 = \\boxed{161}$", "\nSimilar triangles can also solve the problem.\nFirst, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$\nAfter finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \\frac{36}{5}$ . \nSolving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$ . We then know that $CE = \\frac{77}{5}$\nNow consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$\nLet $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$ . Since $PF+FQ=PQ=\\omega$ . We can solve for $PF$ and $FQ$ in terms of $\\omega$ . We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$\nLet's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$ . We can set up the proportion:\n$\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$ . Solving for $AF$ $AF = \\frac{3}{4} PF = \\frac{3}{4} \\cdot \\frac{48}{125} \\omega = \\frac{36}{125} \\omega$\nWe can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \\frac{36}{5} - \\frac{36}{125} \\omega$\nTherefore, $[PQRS] = PQ \\cdot PS = \\omega (\\frac{36}{5} - \\frac{36}{125} \\omega) = \\frac{36}{5}\\omega - \\frac{36}{125} \\omega^2$\nThis means that $\\beta = \\frac{36}{125} \\Rightarrow (m,n) = (36,125) \\Rightarrow m+n = \\boxed{161}$", "Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\\frac{36}{5}$ . Now, through the use of similarity between $\\triangle APQ$ and $\\triangle ABC$ , you find $\\frac{w}{25}=\\frac{h}{36/5}$ . Thus, $h=\\frac{36w}{125}$ . To find the height of the rectangle, subtract $h$ from $\\frac{36}{5}$ to get $\\left(\\frac{36}{5}-\\frac{36w}{125}\\right)$ , and multiply this by the other given side $w$ to get $\\frac{36w}{5}-\\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\\boxed{161}$", "Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\\frac{36}{5}$\nNow, we see that $\\sin{B}=\\frac{PS}{12-L}\\implies PS=\\sin{B}(12-L)$ We easily find that $\\sin{B}=\\frac{3}{5}$\nHence, $PS=\\frac{3}{5}(12-L)$\nNow, we see that $[PQRS]=\\frac{3}{5}(12-L)(w)$\nNow, it is obvious that we want to find $L$ in terms of $W$\nLooking at the diagram, we see that because $PQRS$ is a rectangle, $\\triangle{APQ}\\sim{\\triangle{ABC}}$\nHence.. we can now set up similar triangles.\nWe have that $\\frac{AP}{AB}=\\frac{PQ}{BC}\\implies \\frac{L}{12}=\\frac{W}{25}\\implies 25L=12W\\implies L=\\frac{12W}{25}$\nPlugging back in..\n$[PQRS]=\\frac{3w}{5}(12-(\\frac{12W}{25}))\\implies \\frac{3w}{5}(\\frac{300-12W}{25})\\implies \\frac{900W-36W^2}{125}$\nSimplifying, we get $\\frac{36W}{5}-\\frac{36W^2}{125}$\nHence, $125+36=\\boxed{161}$", "Proceed as in solution 1. When $\\omega$ is equal to zero, $\\alpha - \\beta\\omega=\\alpha$ is equal to the altitude. This means that $25\\beta$ is equal to $\\frac{36}{5}$ , so $\\beta = \\frac{36}{125}$ , yielding $\\boxed{161}$" ]

[ "For each $k$ between $2$ and $2450$ , the area of the trapezoid with $\\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\\triangle{ABC}$ . Let $d_k$ be the length of segment $\\overline{P_kQ_k}$ . The area of the trapezoid with bases $\\overline{P_{k-1}Q_{k-1}}$ and $P_kQ_k$ is $\\left(\\frac{d_k}{5\\sqrt{3}}\\right)^2 - \\left(\\frac{d_{k-1}}{5\\sqrt{3}}\\right)^2 = \\frac{d_k^2-d_{k-1}^2}{75}$ times the area of $\\triangle{ABC}$ . (This logic also applies to the topmost triangle if we notice that $d_0 = 0$ .) However, we also know that the area of each shape is $\\frac{1}{2450}$ times the area of $\\triangle{ABC}$ . We then have $\\frac{d_k^2-d_{k-1}^2}{75} = \\frac{1}{2450}$ . Simplifying, $d_k^2-d_{k-1}^2 = \\frac{3}{98}$ . However, we know that $d_0^2 = 0$ , so $d_1^2 = \\frac{3}{98}$ , and in general, $d_k^2 = \\frac{3k}{98}$ and $d_k = \\frac{\\sqrt{\\frac{3k}{2}}}{7}$ . The smallest $k$ that gives a rational $d_k$ is $6$ , so $d_k$ is rational if and only if $k = 6n^2$ for some integer $n$ .The largest $n$ such that $6n^2$ is less than $2450$ is $20$ , so $k$ has $\\boxed{020}$ possible values.", "We have that there are $2449$ trapezoids and $1$ triangle of equal area, with that one triangle being $AP_1Q_1$ . Notice, if we \"stack\" the trapezoids on top of $\\bigtriangleup AP_1Q_1$ the way they already are, we'd create a similar triangle, all of which are similar to $\\bigtriangleup ABC$ , and since the trapezoids and $\\bigtriangleup AP_1Q_1$ have equal area, each of these similar triangles $AP_kQ_k$ have area $\\frac{k}{2450}\\left[ ABC\\right]$ , and so $\\frac{\\left[ AP_kQ_k\\right]}{\\left[ABC\\right]}=\\frac{k}{2450}$ . We want the ratio of the side lengths $P_kQ_k:BC$ . Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or \\[\\frac{P_kQ_k}{BC}=\\sqrt{\\frac{k}{2450}}\\] \\[\\implies P_kQ_k=BC\\cdot \\sqrt{\\frac{k}{2450}}=5\\sqrt{3}\\cdot\\sqrt{\\frac{k}{2450}}=\\frac{1}{7}\\cdot \\sqrt{\\frac{3k}{2}}=\\frac{3}{7}\\sqrt{\\frac{k}{6}}\\] \\[\\implies k=6n^2<2450\\] \\[\\implies 0<n\\leq 20\\] so there are $\\boxed{020}$ solutions.", "Let $T_1$ stand for $AP_1Q_1$ , and $T_k = AP_kQ_k$ . All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$ . The next trapezoid will also have an area of $x$ , as given. Therefore, $T_k$ has an area of $kx$ . The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, $P_k Q_k=P_1 Q_1\\cdot \\sqrt{k}$ , and the same if $Q$ is substituted for $P$ throughout. We want the side $P_k Q_k$ to be rational. Setting up proportions: \\[5\\sqrt{3} : \\sqrt{2450}=35\\sqrt{2}\\] \\[\\sqrt{6} : 14\\] which shows that $P_1 Q_1=\\frac{\\sqrt{6}}{14}$ . In order for $\\sqrt{k} P_1 Q_1$ to be rational, $\\sqrt{k}$ must be some rational multiple of $\\sqrt{6}$ . This is achieved at $\\sqrt{k}=\\sqrt{6}, 2\\sqrt{6}, \\ldots, 20\\sqrt{6}$ . We end there as $21\\sqrt{6}=\\sqrt{2646}$ . There are 20 numbers from 1 to 20, so there are $\\boxed{020}$ solutions." ]

[ "Let the points of intersection of $\\ell_A, \\ell_B,\\ell_C$ with $\\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$ . Furthermore, let the desired triangle be $\\triangle XYZ$ , with $X$ closest to side $BC$ $Y$ closest to side $AC$ , and $Z$ closest to side $AB$ . Hence, the desired perimeter is $XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115$ since $HG=55$ $EF=15$ , and $ID=45$\nNote that $\\triangle AHG\\sim \\triangle BID\\sim \\triangle EFC\\sim \\triangle ABC$ , so using similar triangle ratios, we find that $BI=HA=30$ $BD=HG=55$ $FC=\\frac{45}{2}$ , and $EC=\\frac{55}{2}$\nWe also notice that $\\triangle EFC\\sim \\triangle YFG\\sim \\triangle EXD$ and $\\triangle BID\\sim \\triangle HIZ$ . Using similar triangles, we get that \\[FY+YG=\\frac{GF}{FC}\\cdot \\left(EF+EC\\right)=\\frac{225}{45}\\cdot \\left(15+\\frac{55}{2}\\right)=\\frac{425}{2}\\] \\[DX+XE=\\frac{DE}{EC}\\cdot \\left(EF+FC\\right)=\\frac{275}{55}\\cdot \\left(15+\\frac{45}{2}\\right)=\\frac{375}{2}\\] \\[HZ+ZI=\\frac{IH}{BI}\\cdot \\left(ID+BD\\right)=2\\cdot \\left(45+55\\right)=200\\] Hence, the desired perimeter is $200+\\frac{425+375}{2}+115=600+115=\\boxed{715}$ -ktong", "Let the diagram be set up like that in Solution 1.\nBy similar triangles we have \\[\\frac{AH}{AB}=\\frac{GH}{BC}\\Longrightarrow AH=30\\] \\[\\frac{IB}{AB}=\\frac{DI}{AC}\\Longrightarrow IB=30\\] Thus \\[HI=AB-AH-IB=60\\]\nSince $\\bigtriangleup IHZ\\sim\\bigtriangleup ABC$ and $\\frac{HI}{AB}=\\frac{1}{2}$ , the altitude of $\\bigtriangleup IHZ$ from $Z$ is half the altitude of $\\bigtriangleup ABC$ from $C$ , say $\\frac{h}{2}$ . Also since $\\frac{EF}{AB}=\\frac{1}{8}$ , the distance from $\\ell_C$ to $AB$ is $\\frac{7}{8}h$ . Therefore the altitude of $\\bigtriangleup XYZ$ from $Z$ is \\[\\frac{1}{2}h+\\frac{7}{8}h=\\frac{11}{8}h\\]\nBy triangle scaling, the perimeter of $\\bigtriangleup XYZ$ is $\\frac{11}{8}$ of that of $\\bigtriangleup ABC$ , or \\[\\frac{11}{8}(220+180+120)=\\boxed{715}\\]", "Notation shown on diagram. By similar triangles we have \\[k_1 = \\frac{EF}{BC} = \\frac{AE}{AB} = \\frac {AF}{AC} = \\frac {1}{4},\\] \\[k_2 = \\frac{F''E''}{AC} = \\frac {BF''}{AB} = \\frac{1}{4},\\] \\[k_3 = \\frac{E'F'}{AB} = \\frac{E'C }{AC} = \\frac{1}{8}.\\] So, \\[\\frac{ZE}{BC} = \\frac{F''E}{AB} = \\frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,\\] \\[\\frac{FY}{BC} = \\frac{FE'}{AC} = \\frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.\\] \\[k = \\frac{ZY}{BC} = \\frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)\\] \\[k = 2 - k_1 - k_2 - k_3 = 2 - \\frac{1}{4} - \\frac{1}{4} - \\frac{1}{8} = \\frac{11}{8}.\\] \\[\\frac{ZY+YX +XZ}{BC +AB + AC} = k \\implies ZY + YX + XZ =\\frac{11}{8} (220 + 120 + 180) = \\boxed{715}.\\] [email protected], vvsss" ]

[ "\nNotice that \\[\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.\\] By the Law of Cosines, \\[\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.\\] Then, \\[DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.\\] Let $X=\\overline{AB}\\cap\\overline{CF}$ $a=XB$ , and $b=XD$ . Then, \\[XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.\\] However, since $\\triangle XFD\\sim\\triangle XAC$ $XF=\\tfrac{4+a}3$ , but since $\\triangle XFE\\sim\\triangle XBC$ \\[\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,\\] and the requested sum is $5+21+2+4=\\boxed{032}$", "Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$\nBecause of exterior angles,\n$\\angle ACB = \\angle CBE - \\angle CAD$\nBut $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$ . But $\\angle CFE - \\angle CFD = \\angle DFE$ , so $\\angle ACB = \\angle DFE$ . Using Law of Cosines on $\\triangle ABC$ , we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$ . Since $\\angle ACB = \\angle DFE$ $\\cos(\\angle DFE) = \\frac{3}{4}$ . We are given that $DF = 2$ and $FE = 7$ , so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$\nLet $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$ . Using Power of a Point with respect to $G$ within $\\omega_1$ , we find that $AG \\cdot GD = CG \\cdot GF$ . We can also apply Power of a Point with respect to $G$ within $\\omega_2$ to find that $CG \\cdot GF = BG \\cdot GE$ . Therefore, $AG \\cdot GD = BG \\cdot GE$\n$AG \\cdot GD = BG \\cdot GE$\n$(AB + BG) \\cdot GD = BG \\cdot (GD + DE)$\n$AB \\cdot GD + BG \\cdot GD = BG \\cdot GD + BG \\cdot DE$\n$AB \\cdot GD = BG \\cdot DE$\n$4 \\cdot GD = BG \\cdot 4\\sqrt{2}$\n$GD = BG \\cdot \\sqrt{2}$\nNote that $\\triangle GAC$ is similar to $\\triangle GFD$ $GF = \\frac{BG + 4}{3}$ . Also note that $\\triangle GBC$ is similar to $\\triangle GFE$ , which gives us $GF = \\frac{7 \\cdot BG}{5}$ . Solving this system of linear equations, we get $BG = \\frac{5}{4}$ . Now, we can solve for $BE$ , which is equal to $BG(\\sqrt{2} + 1) + 4\\sqrt{2}$ . This simplifies to $\\frac{5 + 21\\sqrt{2}}{4}$ , which means our answer is $\\boxed{032}$", "Construct $FC$ and let $FC\\cap AE=K$ . Let $FK=x$ . Using $\\triangle FKE\\sim \\triangle BKC$ \\[BK=\\frac{5}{7}x\\] Using $\\triangle FDK\\sim ACK$ , it can be found that \\[3x=AK=4+\\frac{5}{7}x\\to x=\\frac{7}{4}\\] This also means that $BK=\\frac{21}{4}-4=\\frac{5}{4}$ . It suffices to find $KE$ . It is easy to see the following: \\[180-\\angle ABC=\\angle KBC=\\angle KFE\\] Using reverse Law of Cosines on $\\triangle ABC$ $\\cos{\\angle ABC}=\\frac{1}{8}\\to \\cos{180-\\angle ABC}=\\frac{-1}{8}$ . Using Law of Cosines on $\\triangle EFK$ gives $KE=\\frac{21\\sqrt 2}{4}$ , so $BE=\\frac{5+21\\sqrt 2}{4}\\to \\boxed{032}$ .\n-franchester", "Let $P=AE\\cap CF$ . Let $CP=5x$ and $BP=5y$ ; from $\\triangle{CBP}\\sim\\triangle{EFP}$ we have $EP=7x$ and $FP=7y$ . From $\\triangle{CAP}\\sim\\triangle{DFP}$ we have $\\frac{6}{4+5y}=\\frac{2}{7y}$ giving $y=\\frac{1}{4}$ . So $BP=\\frac{5}{4}$ and $FP=\\frac{7}{4}$ . These similar triangles also gives us $DP=\\frac{5}{3}x$ so $DE=\\frac{16}{3}x$ . Now, Stewart's Theorem on $\\triangle{FEP}$ and cevian $FD$ tells us that \\[\\frac{560}{9}x^3+28x=\\frac{49}{3}x+\\frac{245}{3}x,\\] so $x=\\frac{3\\sqrt{2}}{4}$ . Then $BE=\\frac{5}{4}+7x=\\frac{5+21\\sqrt{2}}{4}$ so the answer is $\\boxed{032}$ as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)", "Connect $CF$ meeting $AE$ at $J$ . We can observe that $\\triangle{ACJ}\\sim \\triangle{FJD}$ Getting that $\\frac{AJ}{FJ}=\\frac{AC}{FD}=3$ . We can also observe that $\\triangle{CBJ}\\sim \\triangle{EFJ}$ , getting that $\\frac{CB}{EF}=\\frac{BJ}{FJ}=\\frac{CJ}{EJ}=\\frac{5}{7}$\nAssume that $BJ=5x;FJ=7x$ , since $\\frac{AJ}{FJ}=3$ , we can get that $\\frac{AJ}{FJ}=\\frac{AB+BJ}{FJ}=\\frac{4+5x}{7x}=3$ , getting that $x=\\frac{1}{4}; BJ=\\frac{5}{4}; FJ=\\frac{7}{4}$\nUsing Power of Point, we can get that $BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ$ Assume that $DJ=5k,CJ=15k$ , getting that $JE=21k, DE=16k$\nNow applying Law of Cosine on two triangles, $\\triangle{ACJ};\\triangle{FJE}$ separately, we can get two equations\n$(1): (15k)^2+(\\frac{21}{4})^2-2*15k *\\frac{21}{4} * cos\\angle{CJA}=36$\n$(2):(21k)^2+(\\frac{7}{4})^2-2*\\frac{7}{4} * 21k*cos\\angle{FJE}=49$\nSince $\\angle{CJA}=\\angle{FJE}$ , we can use $15(2)-7(1)$ to eliminate the $cos$ term\nThen we can get that $5040k^2=630$ , getting $k=\\frac{\\sqrt{2}}{4}$\n$BE=21k=\\frac{21\\sqrt{2}}{4}; BJ=\\frac{5}{4}$ , so the desired answer is $\\frac{21\\sqrt{2}+5}{4}$ , which leads to the answer $\\boxed{032}$", "Nice problem!\nFirst, let $AE$ and $CF$ intersect at $X$ . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that \\[\\angle DFE = \\angle XFE - \\angle XFD = \\angle CBE - \\angle CAB = 180 - \\angle ABC - \\angle CAB = \\angle BAC\\] By the so-called \"Reverse Law of Cosines\" on $\\triangle ABC$ we have \\[\\cos(\\angle BAC) = \\frac{4^2 - 5^2 - 6^2}{-2 \\cdot 5 \\cdot 6} = \\frac{3}{4}\\] Applying on $\\triangle DFE$ gives \\[DE^2 = 2^2 + 7^2 - 2 \\cdot 2 \\cdot 7 \\cos(\\angle DFE)\\] \\[= 2^2 + 7^2 - 2 \\cdot 2 \\cdot 7 \\cdot \\frac{3}{4}\\] \\[=32\\] So $DE = 4 \\sqrt{2}$ , now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to $BX$ and $XD$ , which are crucial lengths in the problem. Suppose $BX = r, XD = s$ for simplicity. We have:\n$\\bullet~~~~\\triangle AXC \\sim \\triangle FXD$ $\\bullet~~~~\\triangle BXC \\sim \\triangle FXE$\nSo \\[\\frac{AX}{FX} = \\frac{XC}{XD} = \\frac{AC}{FD} \\implies \\frac{4 + r}{FX} = \\frac{XC}{s} = 3\\] \\[\\frac{BX}{FX} = \\frac{XC}{XE} = \\frac{BC}{FE} \\implies \\frac{r}{FX} = \\frac{XC}{s + 4 \\sqrt{2}} = \\frac{5}{7}\\] \\[\\implies \\frac{4 + r}{r} = \\frac{s + 4 \\sqrt{2}}{s} = \\frac{21}{5}\\] \\[\\implies r = \\frac{5}{4}, s = \\frac{5 \\sqrt{2}}{4}\\] So $BE = r + s + 4 \\sqrt{2} = \\frac{5 + 21 \\sqrt{2}}{4}$ . The requested sum is $5 + 21 + 2 + 4 = \\boxed{032}$" ]

[ " -Diagram by Brendanb4321\nNote that from the tangency condition that the supplement of $\\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\\angle AKB$ and $\\angle AKC$ , respectively, so from tangent-chord, \\[\\angle AKC=\\angle AKB=180^{\\circ}-\\angle BAC\\] Also note that $\\angle ABK=\\angle KAC$ $^{(*)}$ , so $\\triangle AKB\\sim \\triangle CKA$ . Using similarity ratios, we can easily find \\[AK^2=BK*KC\\] However, since $AB=7$ and $CA=9$ , we can use similarity ratios to get \\[BK=\\frac{7}{9}AK, CK=\\frac{9}{7}AK\\]\nGiving us \\[AK^2+\\frac{49}{81}AK^2+\\frac{22}{27}AK^2=49\\] \\[\\implies \\frac{196}{81}AK^2=49\\] \\[AK=\\frac{9}{2}\\] so our answer is $9+2=\\boxed{011}$", "Consider an inversion with center $A$ and radius $r=AK$ . Then, we have $AB\\cdot AB^*=AK^2$ , or $AB^*=\\frac{AK^2}{7}$ . Similarly, $AC^*=\\frac{AK^2}{9}$ . Notice that $AB^*KC^*$ is a parallelogram, since $\\omega_1$ and $\\omega_2$ are tangent to $AC$ and $AB$ , respectively. Thus, $AC^*=B^*K$ . Now, we get that \\[\\cos(\\angle AB^*K)=\\cos(180-\\angle BAC)=-\\frac{11}{21}\\] so by Law of Cosines on $\\triangle AB^*K$ we have \\[(AK)^2=(AB^*)2+(B^*K)^2-2\\cdot AB^*\\cdot B^*K \\cdot \\cos(\\angle AB^*K)\\] \\[\\Rightarrow (AK)^2=\\frac{AK^4}{49}+\\frac{AK^4}{81}-2\\cdot \\frac{AK^2}{7}\\frac{AK^2}{9}\\frac{-11}{21}\\] \\[\\Rightarrow 1=\\frac{AK^2}{49}+\\frac{AK^2}{81}+\\frac{22AK^2}{63\\cdot21}\\] \\[\\Rightarrow AK=\\frac{9}{2}\\] Then, our answer is $9+2=\\boxed{11}$ . \n-brianzjk", "Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\\angle BAC =x.$ This implies \\[\\angle O_{1}AB = \\angle O_{1}BA = \\angle O_{2}AC = \\angle O_{2}CA = 90^{\\circ}-x\\] by the angle by by tangent. Then we also know that \\[\\angle AO_{1}B = \\angle AO_{2}C = 2x\\] Now we first find $\\cos x.$ We use law of cosines on $\\bigtriangleup ABC$ to obtain \\[64 = 81 + 48 - 2 \\cdot 9 \\cdot 7 \\cdot \\cos{x}\\] \\[\\implies \\cos{x} =\\frac{11}{21}\\] \\[\\implies \\sin{x} =\\frac{8\\sqrt{5}}{21}\\] Then applying law of sines on $\\bigtriangleup AO_{1}B$ we obtain \\[\\frac{7}{\\sin{2x}} =\\frac{O_{1}B}{\\sin{90^{\\circ}-x}}\\] \\[\\implies\\frac{7}{2\\sin{x}\\cos{x}} =\\frac{O_{1}B}{\\cos{x}}\\] \\[\\implies O_{1}B = O_{1}A=\\frac{147}{16\\sqrt{5}}\\] Using similar logic we obtain $O_{2}A =\\frac{189}{16\\sqrt{5}}.$\nNow we know that $\\angle O_{1}AO_{2}=180^{\\circ}-x.$ Thus using law of cosines on $\\bigtriangleup O_{1}AO_{2}$ yields \\[O_{1}O_{2} =\\sqrt{\\left(\\frac{147}{16\\sqrt{5}}\\right)^2+\\left(\\frac{189}{16\\sqrt{5}}\\right)^2-2\\:\\cdot \\left(\\frac{147}{16\\sqrt{5}}\\right)\\cdot \\frac{189}{16\\sqrt{5}}\\cdot -\\frac{11}{21}}\\] While this does look daunting we can write the above expression as \\[\\sqrt{\\left(\\frac{189+147}{16\\sqrt{5}}\\right)^2 - 2\\cdot \\left(\\frac{147}{16\\sqrt{5}}\\right)\\cdot \\frac{189}{16\\sqrt{5}}\\cdot \\frac{10}{21}} =\\sqrt{\\left(\\frac{168}{8\\sqrt{5}}\\right)^2 - \\left(\\frac{7 \\cdot 189 \\cdot 5}{8 \\sqrt{5} \\cdot 8\\sqrt{5}}\\right)}\\] Then factoring yields \\[\\sqrt{\\frac{21^2(8^2-15)}{(8\\sqrt{5})^2}} =\\frac{147}{8\\sqrt{5}}\\] The area \\[[O_{1}AO_{2}] =\\frac{1}{2} \\cdot\\frac{147}{16\\sqrt{5}} \\cdot\\frac{189}{16\\sqrt{5}} \\cdot \\sin(180^{\\circ}-x) =\\frac{1}{2} \\cdot\\frac{147}{16\\sqrt{5}} \\cdot\\frac{189}{16\\sqrt{5}} \\cdot\\frac{8\\sqrt{5}}{21}\\] Now $AK$ is twice the length of the altitude of $\\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have \\[\\frac{1}{2} \\cdot h \\cdot\\frac{147}{8\\sqrt{5}} =\\frac{1}{2} \\cdot\\frac{147}{16\\sqrt{5}} \\cdot\\frac{189}{16\\sqrt{5}} \\cdot\\frac{8\\sqrt{5}}{21}\\] \\[\\implies h =\\frac{9}{4}\\] Thus our desired length is $\\frac{9}{2} \\implies m+n = \\boxed{11}.$", "By the definition of $K$ , it is the spiral center mapping $BA\\to AC$ , which means that it is the midpoint of the $A$ -symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$ , we have $\\triangle ABM'\\sim\\triangle AMC$ . By Stewart's Theorem, it then follows that \\[AK = \\frac{AM'}{2} = \\frac{AC\\cdot AB}{2AM} = \\frac{7\\cdot 9}{2\\sqrt{\\frac{9^2\\cdot 4 + 7^2\\cdot 4 - 4^2\\cdot 8}{8}}} = \\frac{7\\cdot 9}{2\\sqrt{49}} = \\frac{9}{2}\\implies m + n = \\boxed{11}.\\]", "We start by assigning coordinates to point $A$ , labeling it $(0,0)$ and point $B$ at $(7,0)$ , and letting point $C$ be above the $x$ -axis. Through an application of the Pythagorean Theorem and dropping an altitude to side $AB$ , it is easy to see that $C$ has coordinates $(33/7, 24\\sqrt{5}/7)$\nLet $O1$ be the center of circle $\\omega_1$ and $O2$ be the center of circle $\\omega_2$ . Since circle $\\omega_1$ contains both points $A$ and $B$ $O1$ must lie on the perpendicular bisector of line $AB$ , and similarly $O2$ must lie on the perpendicular bisector of line $AC$ . Through some calculations, we find that the perpendicular bisector of $AB$ has equation $x = 3.5$ , and the perpendicular bisector of $AC$ has equation $y = {-11\\sqrt{5}/40 \\cdot x} + 189\\sqrt{5}/80$\nSince circle $\\omega_1$ is tangent to line $AC$ at $A$ , its radius must be perpendicular to $AC$ at $A$ . \nTherefore, the radius has equation $y = {{-11\\cdot\\sqrt{5}/40} \\cdot x}$ . Substituting the $x$ -coordinate of $O1$ into this, we find the y-coordinate of $O1 = {{-11 \\cdot \\sqrt{5}/40} \\cdot 7/2} = {-77 \\cdot \\sqrt{5}/80}$\nSimilarly, since circle $\\omega_2$ is tangent to line $AB$ at $A$ , its radius must be perpendicular to $AB$ at $A$ . Therefore, the radius has equation $x = 0$ and combining with the previous result for $O2$ we get that the coordinates of $O2$ are $(0, 189\\sqrt{5}/80)$\nWe now find the slope of $O1O2$ , the line joining the centers of circles $\\omega_1$ and $\\omega_2$ , which turns out to be ${({266 \\cdot \\sqrt{5} / 80}) \\cdot -2/7} = {-19 \\cdot \\sqrt{5}/20}$ . Since the $y$ -intercept of that line is at $O2(0,189\\sqrt{5}/80)$ , the equation is $y = {{-19 \\cdot \\sqrt{5}/20} \\cdot x} + {189 \\cdot \\sqrt{5}/80}$ . Since circles $\\omega_1$ and $\\omega_2$ intersect at points $A$ and $K$ , line $AK$ is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, $AK$ has slope ${4 \\cdot \\sqrt{5}/19}$ . Since point $A$ is $(0,0)$ , this line has a $y$ -intercept of $0$ , so it has equation $y$ ${{4 \\cdot \\sqrt{5}/19} \\cdot x}$\nWe set ${{4 \\cdot \\sqrt{5}/19} \\cdot x} = {{-19 \\cdot \\sqrt{5}/20} \\cdot x} + {189 \\cdot \\sqrt{5}/80}$ in order to find the intersection $I$ of the radical axis $AK$ and $O1O2$ . Through some moderate bashing, we find that the intersection point is $I(57/28, {3 \\cdot \\sqrt{5}/7})$ . We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting $A$ over $I$ yields $K$ and $AK$ $2AI$ = (This is the most tedious part of the bash) ${2 \\cdot \\sqrt{(57/28)^2 + ({3 \\cdot \\sqrt{5}/7)^2)}}} = {2 \\cdot \\sqrt{3969/784}} = {2 \\cdot 63/28} = {2 \\cdot 9/4} = 9/2$ . Therefore the answer is $9 + 2 = \\boxed{011}.$" ]

[ "Let $O$ be the incenter of $\\triangle{ABC}$ . Because $\\overline{MO} \\parallel \\overline{BC}$ and $\\overline{BO}$ is the angle bisector of $\\angle{ABC}$ , we have\n\\[\\angle{MBO} = \\angle{CBO} = \\angle{MOB} = \\frac{1}{2}\\angle{MBC}\\]\nIt then follows due to alternate interior angles and base angles of isosceles triangles that $MO = MB$ . Similarly, $NO = NC$ . The perimeter of $\\triangle{AMN}$ then becomes \\begin{align*} AM + MN + NA &= AM + MO + NO + NA \\\\ &= AM + MB + NC + NA \\\\ &= AB + AC \\\\ &= 30 \\rightarrow \\boxed{30}", "Let $O$ be the incenter. $AO$ is the angle bisector of $\\angle MAN$ . Let the angle bisector of $\\angle BAC$ meets $BC$ at $P$ and the angle bisector of $\\angle ABC$ meets $AC$ at $Q$ . By applying both angle bisector theorem and Menelaus' theorem,\n$\\frac{AO}{OP} \\times \\frac{BP}{BC} \\times \\frac{CQ}{QA} = 1$\n$\\frac{AO}{OP} \\times \\frac{12}{30} \\times \\frac{24}{12} = 1$\n$\\frac{AO}{OP}=\\frac{5}{4}$\n$\\frac{AO}{AP}=\\frac{5}{9}$\nPerimeter of $\\triangle AMN = \\frac{12+24+18}{9} \\times 5 = 30 \\rightarrow \\boxed{30}$", "Like in other solutions, let $O$ be the incenter of $\\triangle ABC$ . Let $AO$ intersect $BC$ at $D$ . By the angle bisector theorem, $\\frac{BD}{DC} = \\frac{AB}{AC} = \\frac{12}{18} = \\frac{2}{3}$ . Since $BD+DC = 24$ , we have $\\frac{BD}{24-BD} = \\frac{2}{3}$ , so $3BD = 48 - 2BD$ , so $BD = \\frac{48}{5} = 9.6$ . By the angle bisector theorem on $\\triangle ABD$ , we have $\\frac{DO}{OA} = \\frac{BD}{BA} = \\frac{4}{5} = 0.8$ , so $\\frac{DA}{OA} = 1 + \\frac{DO}{OA} = \\frac{9}{5} = 1.8$ , so $\\frac{AO}{AD} = \\frac{5}{9}$ . Because $\\triangle AMN \\sim \\triangle ABC$ , the perimeter of $\\triangle AMN$ must be $\\frac{5}{9} (12 + 18 + 24) = 30$ , so our answer is $\\boxed{30}$", "We know that the ratio of the perimeter of $\\triangle AMN$ and $\\triangle ABC$ is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from $A$ to $BC$ is $\\frac{9\\sqrt{15}}{4}$ from Herons and then $A=\\frac{bh}{2}$ and also that the height from $A$ to $MN$ is simply the height from $A$ to $BC$ minus the inradius. We know the area and the semiperimeter so $r=\\frac{A}{s}$ which gives us $r=\\sqrt{15}$ . Now we know that the altitude from $A$ to $MN$ is $\\frac{5\\sqrt{15}}{4}$ so the ratios of the heights from $A$ for $\\triangle AMN$ and $\\triangle ABC$ is $\\frac{5}{9}$ . Thus the perimeter of $\\triangle AMN$ is $\\frac{5}{9} \\times 54 = 30$ so our answer is $\\boxed{30}$" ]

[ "The base of the triangle is $AB = \\sqrt{3^2 + 3^2} = 3\\sqrt 2$ . Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\\frac 4{\\sqrt 2} = 2\\sqrt 2$ . Therefore its area is $\\frac 12 \\cdot 3\\sqrt 2 \\cdot 2\\sqrt 2 = \\boxed{6}$" ]

[ "Because the line $x + y = 7$ is parallel to $\\overline {AB}$ , the area of $\\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$ . In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\\frac 12 \\cdot 4 \\cdot 3 = \\boxed{6}$" ]

[ "We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$\nLet $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\\frac{24 \\cdot 7} {2} =\\frac{25 \\cdot (CN)} {2}.$\nFrom the third equation, we get $CN=\\frac{168} {25}.$\nBy the Pythagorean Theorem in $\\Delta BCN,$ we have\n$BN=\\sqrt{\\left(\\frac{24 \\cdot 25} {25}\\right)^2-\\left(\\frac{24 \\cdot 7} {25}\\right)^2}=\\frac{24} {25}\\sqrt{25^2-7^2}=\\frac{576} {25}.$\nThus, $MN=\\frac{576} {25}-\\frac{25} {2}=\\frac{527} {50}.$\nIn $\\Delta ADM$ , we use the Pythagorean Theorem to get $DM=\\sqrt{15^2-\\left(\\frac{25} {2}\\right)^2}=\\frac{5} {2} \\sqrt{11}.$\nThus, $[CDM]=\\frac{527 \\cdot 5\\sqrt{11}} {50 \\cdot 2 \\cdot 2}= \\frac{527\\sqrt{11}} {40}.$\nHence, the answer is $527+11+40=\\boxed{578}.$", "Let $E$ be the intersection of lines $BC$ and $DM$ . Since triangles $\\Delta CME$ and $\\Delta CMD$ share a side and height, the area of $\\Delta CDM$ is equal to $\\frac{DM}{EM}$ times the area of $\\Delta CME$ .\nBy AA similarity, $\\Delta EMB$ is similar to $\\Delta ACB$ $\\frac{EM}{AC}=\\frac{MB}{CB}$ . Solving yields $EM=\\frac{175}{48}$ . Using the same method but for $EB$ yields $EB=\\frac{625}{48}$ . As in previous solutions, by the Pythagorean Theorem $DM=\\frac{5\\sqrt{11}}{2}$ . So, $\\frac{DM}{EM}=\\frac{24\\sqrt{11}}{35}$ .\nNow, since we know both $CB$ and $EB$ , we can find that $CE=\\frac{527}{48}$ . The height of $\\Delta CME$ is the length from point $M$ to $CB$ . Since $M$ is the midpoint of $AB$ , the height is just $\\frac{1}{2}\\cdot7=\\frac{7}{2}$ . Using this, we can find that the area of $\\Delta CMD$ is $\\frac{1}{2}\\cdot\\frac{7}{2}\\cdot\\frac{527}{48}\\cdot\\frac{24\\sqrt{11}}{35}=\\frac{527\\sqrt{11}}{40}$ , giving our answer of $\\boxed{578}$" ]

[ "Observe that $\\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ since $m\\angle MEI = m\\angle MAI = 45^\\circ$ ), so $MI=2,MC=\\frac{3}{\\sqrt{2}}$ . With $\\angle{MCI}=45^\\circ$ , we can use Law of Cosines to determine that $CI=\\frac{3\\pm\\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\\boxed{12}$", "We first claim that $\\triangle{EMI}$ is isosceles and right.\nProof: Construct $\\overline{MF}\\perp\\overline{AB}$ and $\\overline{MG}\\perp\\overline{AC}$ . Since $\\overline{AM}$ bisects $\\angle{BAC}$ , one can deduce that $MF=MG$ . Then by AAS it is clear that $MI=ME$ and therefore $\\triangle{EMI}$ is isosceles. Since quadrilateral $AIME$ is cyclic, one can deduce that $\\angle{EMI}=90^\\circ$ . Q.E.D.\nSince the area of $\\triangle{EMI}$ is 2, we can find that $MI=ME=2$ $EI=2\\sqrt{2}$\nSince $M$ is the mid-point of $\\overline{BC}$ , it is clear that $AM=\\frac{3\\sqrt{2}}{2}$\nNow let $AE=a$ and $AI=b$ . By Ptolemy's Theorem, in cyclic quadrilateral $AIME$ , we have $2a+2b=6$ . By Pythagorean Theorem, we have $a^2+b^2=8$ . One can solve the simultaneous system and find $b=\\frac{3+\\sqrt{7}}{2}$ . Then by deducting the length of $\\overline{AI}$ from 3 we get $CI=\\frac{3-\\sqrt{7}}{2}$ , giving the answer of $\\boxed{12}$ . (Surefire2019)", "Like above, notice that $\\triangle{EMI}$ is isosceles and right, which means that $\\dfrac{ME \\cdot MI}{2} = 2$ , so $MI^2=4$ and $MI = 2$ . Then construct $\\overline{MF}\\perp\\overline{AB}$ and $\\overline{MG}\\perp\\overline{AC}$ as well as $\\overline{MI}$ . It's clear that $MG^2+GI^2 = MI^2$ by Pythagorean, so knowing that $MG = \\dfrac{AB}{2} = \\dfrac{3}{2}$ allows one to solve to get $GI = \\dfrac{\\sqrt{7}}{2}$ . By just looking at the diagram, $CI=AC-MF-GI=\\dfrac{3-\\sqrt{7}}{2}$ . The answer is thus $3+7+2=\\boxed{12}$", "Let $A$ lie on $(0,0)$ $E$ on $(0,y)$ $I$ on $(x,0)$ , and $M$ on $\\left(\\frac{3}{2},\\frac{3}{2}\\right)$ . Since ${AIME}$ is cyclic, $\\angle EMI$ (which is opposite of another right angle) must be a right angle; therefore, $\\overrightarrow{ME} \\cdot \\overrightarrow{MI} = \\left<\\frac{-3}{2}, y-\\frac{3}{2}\\right> \\cdot \\left<x-\\frac{3}{2}, -\\frac{3}{2}\\right> = 0$ . Compute the dot product to arrive at the relation $y=3-x$ . We can set up another equation involving the area of $\\triangle EMI$ using the Shoelace Theorem . This is \\[2=\\frac{1}{2}\\left[\\frac{3}{2}\\left(y-\\frac{3}{2}\\right)-xy+\\frac{3}{2}\\left(x+\\frac{3}{2}\\right)\\right].\\] Multiplying, substituting $3-x$ for $y$ , and simplifying, we get $x^2 -3x + \\frac{1}{2}=0$ . Thus, $(x,y)=\\left(\\frac{3 \\pm \\sqrt{7}}{2},\\frac{3 \\mp \\sqrt{7}}{2}\\right)$ . But $AI>AE$ , meaning $x=AI=\\frac{3 + \\sqrt{7}}{2}$ and $CI = 3-\\frac{3 + \\sqrt{7}}{2}=\\frac{3 - \\sqrt{7}}{2}$ , and the final answer is $3+7+2=\\boxed{12}$", "From $AIME$ cyclic we get $\\angle{MEI} = \\angle{MAI} = 45^\\circ$ and $\\angle{MIE} = \\angle{MAE} = 45^\\circ$ , so $\\triangle{EMI}$ is an isosceles right triangle.\nFrom $[EMI]=2$ we get $EM=MI=2$\nNotice $\\triangle{AEM} \\cong \\triangle{CIM}$ , because $\\angle{AEM}=180-\\angle{AIM}=\\angle{CIM}$ $EM=IM$ , and $\\angle{EAM}=\\angle{ICM}=45^\\circ$\nLet $CI=AE=x$ , so $AI=3-x$\nBy Pythagoras on $\\triangle{EAI}$ we have $x^2+(3-x)^2=EI^2=8$ , and solve this to get $x=CI=\\dfrac{3-\\sqrt{7}}{2}$ for a final answer of $3+7+2=\\boxed{12}$", "Let $CI=a$ $BE=b$ . Because opposite angles in a cyclic quadrilateral are supplementary, we have $\\angle EMI=90^{\\circ}$ . By the law of cosines, we have $MI^2=a^2+\\frac{9}{4}-\\frac{3}{2}a$ , and $ME^2=b^2+\\frac{9}{4}-\\frac{3}{2}b$ . Notice that $EI=2MO$ , where $O$ is the origin of the circle mentioned in the problem. Thus $\\frac{2MO*MO}{2}=2\\implies MO=\\sqrt{2}, EI=2\\sqrt{2}$ . By the Pythagorean Theorem, we have $ME^2+MI^2=EI^2\\implies a^2+\\frac{9}{4}-\\frac{3}{2}a+b^2+\\frac{9}{4}-\\frac{3}{2}b=(2\\sqrt{2})^2=8$ . By the Pythagorean Theorem, we have $AE^2+AI^2=EI^2\\implies (3-a)^2+(3-b)^2=(2\\sqrt{2})^2=8\\implies 18-6a-6b+a^2+b^2=8$ . Thus we have $18-6a-6b+a^2+b^2=a^2+\\frac{9}{4}-\\frac{3}{2}a+b^2+\\frac{9}{4}-\\frac{3}{2}b\\implies 18-6a-6b=\\frac{9}{2}-\\frac{3}{2}a-\\frac{3}{2}b\\implies \\frac{27}{2}$ $=\\frac{9}{2}a+\\frac{9}{2}b\\implies a+b=3\\implies 3-b=a$ . We know that \\begin{align*} (3-a)^2+(3-b)^2&=8 \\\\ (3-a)^2+a^2&=8 \\\\ 2a^2-6a+9&=8 \\\\ 2a^2-6a+1&=0 \\\\ a&=\\frac{6\\pm \\sqrt{36-8}}{2}=\\frac{3\\pm\\sqrt{7}}{2}. \\end{align*} We take the smaller solution because we have $AI>AE\\implies 3-AI<3-AE\\implies CI<CE$ , and we want $CI$ , not $CE$ , thus $CI=\\frac{3-\\sqrt{7}}{2}$ . Thus our final answer is $3+7+2=\\boxed{12}$" ]

[ "Since triangle $ABD$ is congruent to triangle $ECD$ and $\\overline{CE} =11$ $\\overline{AB}=11$ . Since $\\overline{AB}=\\overline{BC}$ $\\overline{BC}=11$ . Because point $D$ is the midpoint of $\\overline{BC}$ $\\overline{BD}=\\frac{\\overline{BC}}{2}=\\frac{11}{2}=\\boxed{5.5}$" ]

[ "Construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$\nAlso notice that $\\overline{OA}$ and $\\overline{OC}$ are the angle bisectors of angle $\\angle BAC$ and $\\angle BCA$ respectively. We then deduce $\\angle AOC=120^\\circ$\nConsider another point $M$ on Circle $X$ opposite to point $O$\nAs $AOCM$ is an inscribed quadrilateral of Circle $X$ $\\angle AMC=180^\\circ-120^\\circ=60^\\circ$\nAfterward, deduce that $\\angle AXC=2·\\angle AMC=120^\\circ$\nBy the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$\n$2r^2(1-\\cos(120^\\circ))=6^2$\n$r^2=12$\nThe area is therefore $\\pi r^2 = \\boxed{12}$", "We have $\\angle AOC = 120^\\circ$\nDenote by $R$ the circumradius of $\\triangle AOC$ .\nIn $\\triangle AOC$ , the law of sines implies \\[ 2 R = \\frac{AC}{\\sin \\angle AOC} = 4 \\sqrt{3} . \\]\nHence, the area of the circumcircle of $\\triangle AOC$ is \\[ \\pi R^2 = 12 \\pi . \\]\nTherefore, the answer is $\\boxed{12}$", "As in the previous solution, construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$ . Let $Y$ be the intersection of $\\overline{OX}$ and $\\overline{AB}$\nNote that since $\\overline{OA}$ is the angle bisector of $\\angle BAC$ that $\\angle OAC=30^\\circ$ . Also by symmetry, $\\overline{OX}$ $\\perp$ $\\overline{AB}$ and $AY = 3$ . Thus $\\tan(30^\\circ) = \\frac{OY}{3}$ so $OY = \\sqrt{3}$\nLet $r$ be the radius of circle $X$ , and note that $AX = OX = r$ . So $\\triangle AYX$ is a right triangle with legs of length $3$ and $r - \\sqrt{3}$ and hypotenuse $r$ . By Pythagoras, $3^2 + (r - \\sqrt{3})^2 = r^2$ . So $r = 2\\sqrt{3}$\nThus the area is $\\pi r^2 = \\boxed{12}$", "The semiperimeter is $\\frac{6+6+6}{2}=9$ units.\nThe area of the triangle is $9\\sqrt{3}$ units squared. By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius $r=\\sqrt{3}$ . As $\\angle{AOC}=120^\\circ$ , we can form an altitude from point $O$ to side $AC$ at point $M$ , forming two 30-60-90 triangles. As $CM=MA=3$ , we can solve for $OC=2\\sqrt{3}$ . Now, the area of the circle is just $\\pi*(2*\\sqrt{3})^2 = 12\\pi$ . Select $\\boxed{12}$" ]

[ "Call the diameter of the circle $d$ . If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$ , then $\\bigtriangleup OAE=\\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\\sqrt{d^2-12}$ . We know this since $OC=OA=OB$ and $OC$ is the hypotenuse of a $30-60-90$ right triangle, with the longer leg being $\\frac{6}{2}=3$ so $OC=2\\sqrt{3}$ . Applying Ptolemy's Theorem on cyclic quadrilateral $OCEA$ , we get $2({\\sqrt{d^2-12}})\\cdot{2\\sqrt{3}}=6d$ . Squaring and solving we get $d^2=48 \\Longrightarrow (2r)^2=48$ so $r^2=12$ . Therefore, the area of the circle is $\\boxed{12}$" ]

[ "We solve for $\\angle A$ as follows: \\[4\\angle A+4\\angle A+\\angle A=180\\implies 9\\angle A=180\\implies \\angle A=20.\\] That means that minor arc $\\widehat{BC}$ has measure $40^\\circ$ . We can fit a maximum of $\\frac{360}{40}=\\boxed{9}$ of these arcs in the circle." ]

[ " Because the triangle is inscribed, the sum of the minor arcs equals $360^\\circ$ . Thus, \\[x+75+2x+25+3x-22=360\\] \\[6x+78=360\\] Solving this yields $x = 47$ , so the inscribed angles are $122^\\circ$ $99^\\circ$ , and $119^\\circ$ . Noting that an angle of $\\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\\triangle ABC$ are $59.5^\\circ$ $49.5^\\circ$ , and $\\boxed{61}$" ]

[ " Let $\\angle ACP=\\alpha$ $\\angle PCQ=\\beta$ , and $\\angle QCB=\\gamma$ . Note that since $\\triangle ACQ\\sim\\triangle TBQ$ we have $\\tfrac{AC}{CQ}=\\tfrac56$ , so by the Ratio Lemma \\[\\dfrac{AP}{PQ}=\\dfrac{AC}{CQ}\\cdot\\dfrac{\\sin\\alpha}{\\sin\\beta}\\quad\\implies\\quad \\dfrac{\\sin\\alpha}{\\sin\\beta}=\\dfrac{24}{15}.\\] Similarly, we can deduce $\\tfrac{PC}{CB}=\\tfrac47$ and hence $\\tfrac{\\sin\\beta}{\\sin\\gamma}=\\tfrac{21}{24}$\nNow Law of Sines on $\\triangle ACS$ $\\triangle SCT$ , and $\\triangle TCB$ yields \\[\\dfrac{AS}{\\sin\\alpha}=\\dfrac{ST}{\\sin\\beta}=\\dfrac{TB}{\\sin\\gamma}.\\] Hence \\[\\dfrac{ST^2}{\\sin^2\\beta}=\\dfrac{TB\\cdot AS}{\\sin\\alpha\\sin\\gamma},\\] so \\[TS^2=TB\\cdot AS\\left(\\dfrac{\\sin\\beta}{\\sin\\alpha}\\dfrac{\\sin\\beta}{\\sin\\gamma}\\right)=\\dfrac{15\\cdot 21}{24^2}\\cdot 5\\cdot 7=\\dfrac{35^2}{8^2}.\\] Hence $ST=\\tfrac{35}8$ and the requested answer is $35+8=\\boxed{43}$", "Projecting through $C$ we have \\[\\frac{3}{4}\\times \\frac{13}{6}=(A,Q;P,B)\\stackrel{C}{=}(A,T;S,B)=\\frac{ST}{7}\\times \\frac{13}{5}\\] which easily gives $ST=\\frac{35}{8}\\Longrightarrow 35+8=\\boxed{043}$", "By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \\[5\\cdot 7+13\\cdot ST=AT\\cdot BS.\\] Therefore, in order to find $ST$ , it suffices to find $AT\\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.\nAs $\\triangle APS\\sim \\triangle CPB$ , we find \\[\\frac{4}{PC}=\\frac{7}{BC}.\\] Therefore, $\\frac{BC}{PC}=\\frac{7}{4}$\nAs $\\triangle BQT\\sim\\triangle CQA$ , we find \\[\\frac{6}{CQ}=\\frac{5}{AC}.\\] Therefore, $\\frac{AC}{CQ}=\\frac{5}{6}$\nAs $\\triangle ATQ\\sim\\triangle CBQ$ , we find \\[\\frac{AT}{BC}=\\frac{7}{CQ}.\\] Therefore, $AT=\\frac{7\\cdot BC}{CQ}$\nAs $\\triangle BPS\\sim \\triangle CPA$ , we find \\[\\frac{9}{PC}=\\frac{BS}{AC}.\\] Therefore, $BS=\\frac{9\\cdot AC}{PC}$ . Thus we find \\[AT\\cdot BS=\\left(\\frac{7\\cdot BC}{CQ}\\right)\\left(\\frac{9\\cdot AC}{PC}\\right).\\] But now we can substitute in our previously found values for $\\frac{BC}{PC}$ and $\\frac{AC}{CQ}$ , finding \\[AT\\cdot BS=63\\cdot \\frac{7}{4}\\cdot \\frac{5}{6}=\\frac{21\\cdot 35}{8}.\\] Substituting this into our original expression from Ptolemy's Theorem, we find \\begin{align*}35+13ST&=\\frac{21\\cdot 35}{8}\\\\13ST&=\\frac{13\\cdot 35}{8}\\\\ST&=\\frac{35}{8}.\\end{align*} Thus the answer is $\\boxed{43}$", "Extend $\\overline{AB}$ past $B$ to point $X$ so that $CPTX$ is cyclic. Then, by Power of a Point on $CPTX$ $(CQ)(QT) = (PQ)(QX)$ . By Power of a Point on $CATB$ $(CQ)(QT) = (AQ)(QB) = 42$ . Thus, $(PQ)(QX) = 42$ , so $BX = 8$\nBy the Inscribed Angle Theorem on $CPTX$ $\\angle SCT = \\angle BXT$ . By the Inscribed Angle Theorem on $ASTC$ $\\angle SCT = \\angle SAT$ , so $\\angle BXT = \\angle SAT$ . Since $ASTB$ is cyclic, $\\angle AST = \\angle TBX$ . Thus, $\\triangle AST \\sim \\triangle XBT$ , so $AS/XB = ST/BT$ . Solving for $ST$ yields $ST = \\frac{35}{8}$ , for a final answer of $35+8 = \\boxed{043}$", "By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \\[AS\\cdot BT+AB\\cdot ST=AT\\cdot BS.\\] Projecting through $C$ we have \\[\\frac{AQ \\cdot PB}{PQ \\cdot AB} = (A,Q; P,B)\\stackrel{C}{=}(A,T; S,B)=\\frac{AT \\cdot BS}{ST \\cdot AB}.\\] Therefore \\[AT \\cdot BS = \\frac {AQ \\cdot PB}{PQ} \\times ST \\implies\\] \\[\\left(\\frac {AQ \\cdot PB}{PQ} - AB\\right)\\times ST = AS \\cdot BT \\implies\\] \\[ST = \\frac {AS \\cdot BT \\cdot PQ}{AQ \\cdot PB – AB \\cdot PQ}\\] \\[ST = \\frac {7\\cdot 5 \\cdot 3}{7\\cdot 9 – 13 \\cdot 3 } = \\frac {35}{8} \\implies 35 + 8 = \\boxed{43}.\\]", "Connect $AT$ and $\\angle{SCT}=\\angle{SAT}, \\angle{ACS}=\\angle{ATS}, \\frac{ST}{\\sin \\angle{SAT}}=\\frac{AS}{\\sin \\angle{ATS}}$\nSo we need to get the ratio of $\\frac{\\sin \\angle{ACS}}{\\sin \\angle{SCT}}$\nBy clear observation $\\triangle{CAQ}\\sim \\triangle{BTQ}$ , we have $\\frac{CQ}{AC}=\\frac{6}{5}$ , LOS tells $\\frac{AC}{\\sin \\angle{CPA}}=\\frac{4}{\\sin \\angle{ACS}}; \\frac{CQ}{\\sin \\angle{CPQ}}=\\frac{3}{\\sin \\angle{PCQ}}$ so we get $\\frac{\\sin \\angle{PCQ}}{\\sin \\angle{ACS}}=\\frac{5}{8}$ , the desired answer is $7\\cdot \\frac{\\sin \\angle{SAT}}{\\sin \\angle{ATS}}=\\frac{35}{8}$ leads to $\\boxed{043}$" ]