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Furong Huang's Lab at UMD 14

AMC

AMC12

12A

N/A

A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$?

240

The first box has volume $2\times3\times5=30\text{ cm}^3$, and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$. The second has a volume that is $6$ times greater, so it holds $6\times40=\fbox{240}$ grams.

AMC12 First Half

AMC12 A

HMMT

HMMT-Nov

team

Nov

Consider five-dimensional Cartesian space \[ \mathbb{R}^{5}=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \mid x_{i} \in \mathbb{R}\right\} \] and consider the hyperplanes with the following equations: \begin{itemize} $x_{i}=x_{j}$ for every $1 \leq i<j \leq 5$; $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=-1$; $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0$; $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1$. \end{itemize} Into how many regions do these hyperplanes divide $\mathbb{R}^{5}$ ?

480

(Joint with Junyao Peng) Note that given a set of plane equations $P_{i}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=0$, for $i=1,2, \ldots, n$, each region that the planes separate the space into correspond to a $n$-tuple of -1 and 1 , representing the sign of $P_{1}, P_{2}, \ldots P_{n}$ for all points in that region. Therefore, the first set of planes separate the space into $5 !=120$ regions, with each region representing an ordering of the five coordinates by numerical size. Moreover, the next three planes are parallel to each other and perpendicular to all planes in the first set, so these three planes separate each region into 4. Therefore, a total of $4 \cdot 120=480$ regions is created. $\fbox{480}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC10

10A

N/A

For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$?

3

We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$: \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$. Solving the equation gives us $x = \fbox{3}.$

AMC10 First Half

AMC10 A

HMMT

HMMT-Nov

guts

Nov

Rebecca has twenty-four resistors, each with resistance $1 \mathrm{ohm}$. Every minute, she chooses any two resistors with resistance of $a$ and $b$ ohms respectively, and combine them into one by one of the following methods: \begin{itemize} Connect them in series, which produces a resistor with resistance of $a+b$ ohms; Connect them in parallel, which produces a resistor with resistance of $\frac{a b}{a+b}$ ohms; Short-circuit one of the two resistors, which produces a resistor with resistance of either $a$ or $b$ ohms. \end{itemize} Suppose that after twenty-three minutes, Rebecca has a single resistor with resistance $R$ ohms. How many possible values are there for $R$ ? If the correct answer is $C$ and your answer is $A$, you get $\max \left(\left\lfloor 30\left(1-\left|\log _{\log _{2} C} \frac{A}{C}\right|\right)\right\rfloor, 0\right)$ points.

1015080877

This is the same problem as in OEIS A153588. It is helpful to see (or guess) that neither the numerator or the denominator of the final resistance exceed the $(n+1)$-th Fibonacci number, which in this case is $F_{25}=75025$, using concepts on the line of continued fractions. So $75025^{2} \approx 5.6 \times 10^{9}$ is an upper bound for the total number, which is already close to the final answer. Multiplying by some constant factor to remove non-reduced fractions (such as $\frac{3}{4}$ to deal with parity) will improve this result. $\fbox{1015080877}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

alg

Feb

Let $p(x)$ and $q(x)$ be two cubic polynomials such that $p(0)=-24, q(0)=30$, and \[ p(q(x))=q(p(x)) \] for all real numbers $x$. Find the ordered pair $(p(3), q(6))$.

(3,-24)

Note that the polynomials $f(x)=a x^{3}$ and $g(x)=-a x^{3}$ commute under composition. Let $h(x)=x+b$ be a linear polynomial, and note that its inverse $h^{-1}(x)=x-b$ is also a linear polynomial. The composite polynomials $h^{-1} f h$ and $h^{-1} g h$ commute, since function composition is associative, and these polynomials are also cubic. We solve for the $a$ and $b$ such that $\left(h^{-1} f h\right)(0)=-24$ and $\left(h^{-1} g h\right)(0)=30$. We must have: \[ a b^{3}-b=-24,-a b^{3}-b=30 \Rightarrow a=1, b=-3 \] These values of $a$ and $b$ yield the polynomials $p(x)=(x-3)^{3}+3$ and $q(x)=-(x-3)^{3}+3$. The polynomials take on the values $p(3)=3$ and $q(6)=-24$. Remark: The pair of polynomials found in the solution is not unique. There is, in fact, an entire family of commuting cubic polynomials with $p(0)=-24$ and $q(0)=30$. They are of the form \[ p(x)=t x(x-3)(x-6)-24, q(x)=-t x(x-3)(x-6)+30 \] where $t$ is any real number. However, the values of $p(3)$ and $q(6)$ are the same for all polynomials in this family. In fact, if we give the initial conditions $p(0)=k_{1}$ and $q(0)=k_{2}$, then we get a general solution of \[ \begin{gathered} p(x)=t\left(x^{3}-\frac{3}{2}\left(k_{1}+k_{2}\right) x^{2}+\frac{1}{2}\left(k_{1}+k_{2}\right)^{2} x\right)+\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{1} \\ q(x)=-t\left(x^{3}-\frac{3}{2}\left(k_{1}+k_{2}\right) x^{2}+\frac{1}{2}\left(k_{1}+k_{2}\right)^{2} x\right)-\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{2} \end{gathered} \] $\fbox{(3,-24)}$.

HMMT Feb Hard

HMMT-Feb Algebra

AIME

AIME

I

N/A

Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$.

53

Let $d$ be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer, $d$ must be a factor of the total difference, which is $374-319=55$. Also note that the number of pages Anh reads is $dn+\frac{d(d-1)}{2}$. Similarly, the number of minutes she reads for is $dt+\frac{d(d-1)}{2}$. When $d$ is odd (which it must be), both of these numbers are multiples of $d$. Therefore, $d$ must be a factor of $55$, $319$, and $374$. The only such numbers are $1$ and $11$. We know that Anh reads for at least $2$ days. Therefore, $d=11$. Using this, we find that she reads $55$ "additional" pages and $55$ "additional" minutes. Therefore, $n=\frac{374-55}{11}=29$, while $t=\frac{319-55}{11}=24$. The answer is therefore $29+24=\fbox{53}$.

Easy AIME Problems

AIME

AMC

AMC10

10B

N/A

Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?

\frac {15}{32}\sqrt{3}

[asy] /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); [/asy] The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$). Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD}$ into a $1:1$ ratio (This can be shown by similar triangles.). Thus, it must also divide transversal $\overline {AC}$ and transversal $\overline {CO}$ into a $1:1$ ratio. By symmetry, the same applies for $\overline {CE}$ and $\overline {EA}$ as well as $\overline {EM}$ and $\overline {AN}.$ In $\triangle ACE,$ we see that $\frac{[MNO]}{[ACE]} = \frac{1}{4}$ and $\frac{[MPN]}{[ACE]} = \frac{1}{8}.$ Our desired area becomes \[\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \fbox{\frac {15}{32}\sqrt{3}}.\]

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Feb

comb

Feb

A $4 \times 4$ window is made out of 16 square windowpanes. How many ways are there to stain each of the windowpanes, red, pink, or magenta, such that each windowpane is the same color as exactly two of its neighbors? Two different windowpanes are neighbors if they share a side.

24

For the purpose of explaining this solution, let's label the squares as \[ \begin{array}{llll} 11 & 12 & 13 & 14 \\ 21 & 22 & 23 & 24 \\ 31 & 32 & 33 & 34 \\ 41 & 42 & 43 & 44 \end{array} \] Note that since the corner squares $11,14,41,44$ each only have two neighbors, each corner square is the same color as both of its neighbors (for example, 11,12, and 21 are the same color, 31,41, and 42 are the same color, etc.). This corner square constraint heavily limits the possible colorings. We will now use casework. Case 1: Suppose two corner squares on the same side (without loss of generality, let them be 11 and 14) have the same color (without loss of generality, red). Then $21,11,12,13,14,24$ are all red, and 12 has two red neighbors (11 and 13) so its third neighbor (22) is a color different from red (without loss of generality, magenta). But 22 has two red neighbors (12 and 21), so its other two neighbors (23 and 32)must be magenta. Applying the same logic symmetrically, we find that all four interior squares $(22,23,32,33)$ have the same color. Furthermore, 21 has one magenta neighbor 22 , so 31 must be red. Symmetrically, 34 is red, and by the corner square constraint we have that all the exterior squares are\\ the same color. Thus in general, this case is equivalent to a window taking the following form (with distinct colors $A$ and $B$ ) The number of choices of $A$ and $B$ is $3 \cdot 2=6$. Case 2: No two corner squares on the same side have the same color. Then from the corner square constraint 12 has neighbor 11 of the same color and neighbor 13 of a different color, so its neighbor 22 must be the same color as 12 . Therefore, this case is equivalent to coloring each quadrant entirely in one color such two quadrants sharing a side have different colors. (A quadrant refers to the four squares on one vertical half and one horizontal half, e.g. 13,14, 23, 24). If only two colors are used, the window will take the form (with distinct colors $A$ and $B$ ): \[ \begin{array}{llll} A & A & B & B \\ A & A & B & B \\ B & B & A & A \\ B & B & A & A \end{array} \] Again there are $3 \cdot 2=6$ ways to chose $A$ and $B$. If all three colors are used, the window will take the form (with distinct colors $A, B$ and $C$ ) \[ \begin{array}{llll} A & A & B & B \\ A & A & B & B \\ C & C & A & A \\ C & C & A & A \end{array} \] or $A A B B$ $A A B B$ $B B C C$ $B B C$ There are $3 \cdot 2 \cdot 1=6$ ways to select colors for each of these forms. Therefore, there are 6 colorings in Case 1 and $6+6+6$ in Case 2, for a total of 24 colorings. $\fbox{24}$.

HMMT Feb Easy

HMMT-Feb Combinatorics

HMMT

HMMT-Nov

team

Nov

For how many positive integers $n \leq 100$ is it true that $10 n$ has exactly three times as many positive divisors as $n$ has?

28

Let $n=2^{a} 5^{b} c$, where $2,5 \nmid c$. Then, the ratio of the number of divisors of $10 n$ to the number of divisors of $n$ is $\frac{a+2}{a+1} \frac{b+2}{b+1}=3$. Solving for $b$, we find that $b=\frac{1-a}{2 a+1}$. This forces $(a, b)=(0,1),(1,0)$. Therefore, the answers are of the form $2 k$ and $5 k$ whenever $\operatorname{gcd}(k, 10)=1$. There are 50 positive numbers of the form $2 k$ and 20 positive numbers of the form $5 k$ less than or equal to 100 . Of those 70 numbers, only $\frac{1}{2} \cdot \frac{4}{5}$ have $k$ relatively prime to 10 , so the answer is $70 \cdot \frac{1}{2} \cdot \frac{4}{5}=28$. $\fbox{28}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC10

10B

N/A

A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

24

The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \fbox{24}$

AMC10 First Half

AMC10 B

HMMT

HMMT-Nov

gen

Nov

Find the maximum value of $x+y$, given that $x^{2}+y^{2}-3 y-1=0$.

\frac{\sqrt{26}+3}{2}

We can rewrite $x^{2}+y^{2}-3 y-1=0$ as $x^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{4}$. We then see that the set of solutions to $x^{2}-y^{2}-3 y-1=0$ is the circle of radius $\frac{\sqrt{13}}{2}$ and center $\left(0, \frac{3}{2}\right)$. This can be written as $x=\frac{\sqrt{13}}{2} \cos (\theta)$ and $y=\frac{\sqrt{13}}{2} \sin (\theta)+\frac{3}{2}$. Thus, $x+y=\frac{3}{2}+\frac{\sqrt{13}}{2}(\cos (\theta)+\sin (\theta))=\frac{3}{2}+\frac{\sqrt{13}}{2} \sqrt{2} \sin \left(\theta+45^{\circ}\right)$, which is maximized for $\theta=45^{\circ}$ and gives $\frac{\sqrt{26}+3}{2}$. (We could also solve this geometrically by noting that if $x+y$ attains a maximum value of $s$ then the line $x+y=s$ is tangent to the circle.) $\fbox{\frac{\sqrt{26}+3}{2}}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

guts

Nov

Find the smallest positive integer $n$ such that the divisors of $n$ can be partitioned into three sets with equal sums.

120

Solution: I claim the answer is 120 . First, note that $120=2^{3} \cdot 3 \cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5)=15 \cdot 4 \cdot 6=360$. Thus, we need to split the divisors into groups summing to 120 . But then we can just take $\{120\},\{20,40,60\},\{1,2,3,4,5,6,8,10,12,15,24,30\}$. Thus, 120 works. Now we need to show 120 is the lowest. Let $s(n)$ be the sum of divisors. Since $n$ will be in one of the piles, we need $s(n) \geq 3 n$. First, we claim that $n$ must have at least 3 distinct prime divisors. Surely, if it had 2 distinct prime divisors, say $p$ and $q$, so that $n=p^{a} q^{b}$, then the sum of divisors is \[ \left(1+p+p^{2}+\ldots+p^{a}\right)\left(1+q+q^{2}+\ldots+q^{b}\right)=p^{a} q^{b}\left(1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}\right)\left(1+\frac{1}{q}+\ldots+\frac{1}{q^{b}}\right) \] However, the expression $1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}$ is maximized when $p$ is minimized, and further, as $a$ is finite must be at most $\frac{1}{1-\frac{1}{p}}=\frac{p}{p-1}$. Thus, the sum of divisors is less than \[ p^{a} q^{b} \frac{p}{p-1} \frac{q}{q-1} \leq n \cdot 2 \cdot \frac{3}{2}=3 n \] Thus, $n$ can't have 2 distinct prime divisors and must have at least 3 distinct prime divisors. As we already discovered 120 works, we need not worry about 4 distinct prime divisors, as the value of $n$ would be at least $2 \cdot 3 \cdot 5 \cdot 7=210$. We now work through the numbers with 3 distinct divisors. If 2 is not one of them, then the only number that works is $105=3 \cdot 5 \cdot 7$, which has a sum of divisors that is not large enough. Therefore, 2 must be a prime divisor of $n$. Additionally, if 3 is not a divisor, then our options are $2 \cdot 5 \cdot 7$ and $2 \cdot 5 \cdot 11$, which also do not work. Therefore, 3 must also be a prime divisor. Then, if 5 is not a prime divisor, then if $n$ is $2 \cdot 3 \cdot p$, it has a sum of divisors of $(1+2)(1+3)(1+p)=n \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{p+1}{p}$, which is only at least $3 n$ if $p$ is exactly 2 , which is not feasible. Additionally, if we use $2^{2}$, then the sum of divisors is $(1+2+4)(1+3)(1+p)=n \cdot \frac{7}{4} \cdot \frac{4}{3} \cdot \frac{p}{p+1}$, so $\frac{p+1}{p}>\frac{9}{7} \Longrightarrow p<4.5$, which also can't happen. Further, we can't have $3^{2}$ be a divisor of $n$ as $2 \cdot 3^{2} \cdot 5$ is the only value less than 120 with this, and that also does not work. Lastly, we just need to check $2^{3} \cdot 3 \cdot p$, which has a sum of divisors of $(1+2+4+8)(1+3)(1+p)=n \cdot \frac{15}{8} \cdot \frac{4}{3} \cdot \frac{p+1}{p}=n \cdot \frac{5}{2} \cdot \frac{p}{p+1}$, so $p=5$ and that works. This means that $n=120$ is the smallest value for which $s(n) \geq 3 n$, and thus is our answer. $\fbox{120}$.

HMMT Nov Guts

HMMT-Nov Guts

AIME

AIME

I

N/A

Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$. Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$, where $x$ is in $\mathcal{S}$. In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$.

170

It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \not\equiv s-16 \pmod{5},$ one term must have all the factors of $5$ and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and thus the corresponding element in $\mathcal{T}$ is $\frac{2484^2 - 256}{1000} = 6170 \rightarrow \fbox{170}$

Hard AIME Problems

AIME

AMC

AMC12

12A

N/A

For each positive integer $n$, let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$, with no more than three $A$s in a row and no more than three $B$s in a row. What is the remainder when $S(2015)$ is divided by $12$?

8

We can start off by finding patterns in $S(n)$. When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$. Rearranging the expression we realize that the terms aside from $2^{2015}$ are congruent to $0$ mod $12$(Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod $12$). Using patterns we can see that $2^{2015}$ is congruent to $8$ mod $12$. Therefore $\fbox{8}$ is our answer.

AMC12 Final Problems

AMC12 A

HMMT

HMMT-Feb

guts

Feb

Tessa picks three real numbers $x, y, z$ and computes the values of the eight expressions of the form $\pm x \pm y \pm z$. She notices that the eight values are all distinct, so she writes the expressions down in increasing order. For example, if $x=2, y=3, z=4$, then the order she writes them down is \[ -x-y-z,+x-y-z,-x+y-z,-x-y+z,+x+y-z,+x-y+z,-x+y+z,+x+y+z \] How many possible orders are there?

96

Solution: There are $2^{3}=8$ ways to choose the sign for each of $x, y$, and $z$. Furthermore, we can order $|x|,|y|$, and $|z|$ in $3 !=6$ different ways. Now assume without loss of generality that $0<x<y<z$. Then there are only two possible orders depending on the sign of $x+y-z$ : \[ \begin{aligned} & -x-y-z,+x-y-z,-x+y-z,-x-y+z, x+y-z, x-y+z,-x+y+z, x+y+z \\ & -x-y-z,+x-y-z,-x+y-z, x+y-z,-x-y+z, x-y+z,-x+y+z, x+y+z \end{aligned} \] Thus, the answer is $8 \cdot 6 \cdot 2=96$. $\fbox{96}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10B

N/A

The entries in a $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

42

Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right. Case 1: Center 4 \[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\] 3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$ Case 2: Center 5 \[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\] Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$ Case 3: Center 6 By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$ \[12+18+12=\fbox{42}\]

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Nov

guts

Nov

Isabella writes the expression $\sqrt{d}$ for each positive integer $d$ not exceeding 8 ! on the board. Seeing that these expressions might not be worth points on HMMT, Vidur simplifies each expression to the form $a \sqrt{b}$, where $a$ and $b$ are integers such that $b$ is not divisible by the square of a prime number. (For example, $\sqrt{20}, \sqrt{16}$, and $\sqrt{6}$ simplify to $2 \sqrt{5}, 4 \sqrt{1}$, and $1 \sqrt{6}$, respectively.) Compute the sum of $a+b$ across all expressions that Vidur writes. Submit a positive real number $A$. If the correct answer is $C$ and your answer is $A$, you get $\max \left(0,\left\lceil 20\left(1-|\log (A / C)|^{1 / 5}\right)\right\rceil\right)$ points.

534810086

Solution: Let $\sqrt{n}$ simplifies to $a_{n} \sqrt{b_{n}}$, and replace 8 ! by $x$. First, notice that $\sum_{n \leq x} a_{n}$ is small $\left(O\left(x^{3 / 2}\right)\right.$ in particular) because each term cannot exceed $\sqrt{x}$. On the other hand, $\sum_{n \leq x} b_{n}$ will be large; we have $b_{n}=n$ when $n$ is squarefree, and squarefree numbers occurs $\frac{6}{\pi^{2}}$ over the time. Thus, it suffices to consider $\sum_{n \leq x} b_{n}$. We first explain how to derive the formula heuristically. Then, we will provide a rigorous proof that \[ B(x):=\sum_{n \leq x} b_{n}=\frac{\pi^{2}}{30} x^{2}+O\left(x^{3 / 2}\right) \] For heuristic explanation, we first rewrite the sum as \[ B(x)=\sum_{\substack{a^{2} b \leq x \\ b \text { squarefree }}} b=\sum_{a \leq x} \sum_{\substack{b \leq x / a^{2} \\ b \text { squarefree }}} b \] We estimate the inner sum as follows: first, recall that the density of squarefree numbers is $\frac{6}{\pi^{2}}$. The sum of first $k$ positive integers is approximately $k^{2} / 2$, so the sum of squarefree numbers from $1,2, \ldots, k$ should roughly be about $\frac{6}{\pi^{2}} \cdot \frac{k^{2}}{2}=\frac{3}{\pi^{2}} k^{2}$. Knowing this, we estimate \[ \begin{aligned} B(x) & \approx \sum_{a \leq x} \frac{3}{\pi^{2}}\left(\frac{x}{a^{2}}\right)^{2} \\ & =x^{2} \sum_{a \leq x} \frac{3}{\pi^{2}} \frac{1}{a^{4}} \\ & \approx \frac{3}{\pi^{2}} x^{2} \sum_{a=1}^{\infty} \frac{1}{a^{4}} \\ & =\frac{3}{\pi^{2}} x^{2} \cdot \frac{\pi^{4}}{90}=\frac{\pi^{2}}{30} x^{2} \end{aligned} \] The estimate $\frac{\pi^{2}}{30} \cdot(8 !)^{2}=534834652$ is good enough for 18 points. We now give a rigorous proof, which is essentially the above proof, but the errors are properly treated. To do that, we need several lemmas and some standard techniques in analytic number theory. Lemma 1. The number of squarefree integers not exceeding $x$ is $\frac{6}{\pi^{2}} x+O(\sqrt{x})$. Proof. This is a standard result in analytic number theory, but we give the full proof for completeness. \[ \mu(n)= \begin{cases}(-1)^{r} & n \text { is the product of } r \geq 0 \text { distinct primes } \\ 0 & \text { otherwise. }\end{cases} \] Then, by Inclusion-Exclusion, we have \[ \begin{aligned} \#\{\text { squarefree } \leq x\} & =\sum_{k \leq \sqrt{x}} \mu(k)\left\lfloor\frac{x}{k^{2}}\right\rfloor \\ & =\sum_{k \leq \sqrt{x}} \mu(k) \frac{x}{k^{2}}+O(\sqrt{x}) \\ & =x\left(\sum_{k \leq \sqrt{x}} \frac{\mu(k)}{k^{2}}\right)+O(\sqrt{x}) \end{aligned} \] The inner summation is \[ \begin{aligned} \sum_{k=1}^{\infty} \frac{\mu(k)}{k^{2}}+O\left(\sum_{k \geq \sqrt{x}} \frac{1}{k^{2}}\right) & =\prod_{p \text { prime }}\left(1-\frac{1}{p^{2}}\right)+O\left(\frac{1}{\sqrt{x}}\right) \\ & =\frac{1}{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots}+O\left(\frac{1}{\sqrt{x}}\right) \\ & =\frac{6}{\pi^{2}}+O\left(\frac{1}{\sqrt{x}}\right) \end{aligned} \] so putting it together, we get that \[ \#\{\text { squarefree } \leq x\}=x\left(\frac{6}{\pi^{2}}+O\left(\frac{1}{\sqrt{x}}\right)\right)+O(\sqrt{x})=\frac{6}{\pi^{2}} x+O(\sqrt{x}) \] Lemma 2. We have \[ \sum_{\substack{n \text { squarefree } \\ n \leq x}} n=\frac{3}{\pi^{2}} x^{2}+O\left(x^{3 / 2}\right) \] Proof. We apply Abel's summation formula on the sequence $a_{n}=\mathbf{1}_{\text {squarefree }}(n)$ and weight $\phi(n)=n$. Define the partial summation $A(x)=\sum_{n \leq x} a_{n}$. Applying Abel's summation, we get that \[ \begin{aligned} \sum_{\substack{n \text { squarefree } \\ n \leq x}} n & =\sum_{n \leq x} a_{n} \phi(n) \\ & =A(x) \phi(x)-\int_{1}^{x} A(t) \phi^{\prime}(t) d t \\ & =\left(\frac{6}{\pi^{2}} x+O(\sqrt{x})\right) x-\int_{1}^{x}\left(\frac{6}{\pi^{2}} t+O(\sqrt{t})\right) d t \\ & =\left(\frac{6}{\pi^{2} x}+O\left(x^{3 / 2}\right)\right)-\left(\frac{6}{\pi^{2}} \cdot \frac{x^{2}}{2}+O\left(x^{3 / 2}\right)\right) \\ & =\frac{3}{\pi^{2}} x^{2}+O\left(x^{3 / 2}\right) \end{aligned} \] Main Proof. Once we have Lemma 2., it is easy to get the desired estimate. We have \[ \begin{aligned} B(x) & =\sum_{\substack{a^{2} b \leq x \\ b \text { squarefree }}} b \\ & =\sum_{a \leq x} \sum_{\substack{b \leq x / a^{2} \\ b \text { squarefree }}} b \\ & =\sum_{a \leq x} \frac{3}{\pi^{2}} \frac{x^{2}}{a^{4}}+O\left(\frac{x^{3 / 2}}{a^{3}}\right) \\ & =\frac{3}{\pi^{2}} x^{2}\left(\sum_{a \leq x} \frac{1}{a^{4}}\right)+O\left(x^{3 / 2} \sum_{a \leq x} \frac{1}{a^{3}}\right) \end{aligned} \] Since $\sum_{a=1}^{\infty} \frac{1}{a^{3}}$ converges, we get that the big- $O$ term is indeed $O\left(x^{3 / 2}\right)$. Now, we only need to deal with the main term. Note the estimate \[ \sum_{a \leq x} \frac{1}{a^{4}}=\sum_{a=1}^{\infty} \frac{1}{a^{4}}-\sum_{a \geq x} \frac{1}{a^{4}}=\frac{\pi^{4}}{90}+O\left(\frac{1}{x^{3}}\right) \] Hence, we have \[ B(x)=\frac{3}{\pi^{2}} x^{2} \cdot \frac{\pi^{4}}{90}+O\left(x^{3 / 2}\right)=\frac{\pi^{2}}{30} x^{2}+O\left(x^{3 / 2}\right) \] as desired. Here is the code that gives the exact answer: $\fbox{534810086}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10A

N/A

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

\frac{3\sqrt{3}}{\pi}

Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle. In a circle with a radius $r$, the side of an inscribed equilateral triangle is $r\sqrt{3}$. So $s=r\sqrt{3}$. The perimeter of the triangle is $3s=3r\sqrt{3}$ The area of the circle is $\pi r^{2}$ So: $\pi r^{2} = 3r\sqrt{3}$ $\pi r=3\sqrt{3}$ $r=\frac{3\sqrt{3}}{\pi} \Rightarrow\fbox{\frac{3\sqrt{3}}{\pi}}$

AMC10 Second Half

AMC10 A

AMC

AMC12

12A

N/A

$ABCD$ is a square of side length $\sqrt{3} + 1$. Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$. The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$, sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$, where $a$, $b$, and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. What is $a + b + c$?

19

We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$. It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3} + 1$. The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$. The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{PA'}$) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{PC'}$). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$. Call the point of intersection $E$. From the previous paragraph, it is clear that $PE = \sqrt{3}$ and $BE = 1$. This means $PB = 2$, and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$, and so the angle of the segment is $60^{\circ}$. Imagining the process in reverse, it is clear that the situation is the same with point $D$. The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1$ and $\frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3$. The other two segments both have area $\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3}$. The total area is therefore \[(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})\] \[= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}\] \[= \frac{10 \pi}{3} + 2\] \[= \frac{1}{3} (10 \pi + 6)\] Since $a = 10$, $b = 6$, and $c = 3$, the answer is $a + b + c = 10 + 6 + 3 = \fbox{19}$.

AMC12 Final Problems

AMC12 A

HMMT

HMMT-Feb

geo

Feb

Let $A B C$ be an obtuse triangle with circumcenter $O$ such that $\angle A B C=15^{\circ}$ and $\angle B A C>90^{\circ}$. Suppose that $A O$ meets $B C$ at $D$, and that $O D^{2}+O C \cdot D C=O C^{2}$. Find $\angle C$.

35

Let the radius of the circumcircle of $\triangle A B C$ be $r$. \[ \begin{gathered} O D^{2}+O C \cdot C D=O C^{2} \\ O C \cdot C D=O C^{2}-O D^{2} \\ O C \cdot C D=(O C+O D)(O C-O D) \\ O C \cdot C D=(r+O D)(r-O D) \end{gathered} \] By the power of the point at $\mathrm{D}$, \[ \begin{gathered} O C \cdot C D=B D \cdot D C \\ r=B D \end{gathered} \] Then, $\triangle O B D$ and $\triangle O A B$ and $\triangle A O C$ are isosceles triangles. Let $\angle D O B=\alpha . \angle B A O=90-\frac{\alpha}{2}$. In $\triangle A B D, 15+90-\frac{\alpha}{2}=\alpha$. This means that $\alpha=70$. Furthermore, $\angle A C B$ intercepts minor arc $A B$, thus $\angle A C B=\frac{\angle A O B}{2}=\frac{70}{2}=35$ $\fbox{35}$.

HMMT Feb Hard

HMMT-Feb Geometry

AMC

AMC8

8

N/A

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?

36

The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$, the sidelength is $24/4=6$, and the area is $6^2=\fbox{36}$.

AMC8 First Half

AMC8

AMC

AMC10

10B

N/A

A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?

\frac{2\sqrt{3}}{3}

This tetrahedron has the 4 vertices in these positions: on a corner (lets call this $A$) of the cube, and the other three corners ($B$, $C$, and $D$) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height $x$) by finding the volume of the tetrahedron in two ways. $\frac{1 \times 1}{2}$ is the area of the isosceles base of the tetrahedron. Multiply by the height, $1$, and divide by $3$, we have the volume of the tetrahedron as $\frac{1}{6}$. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: $[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}$. So we have: $\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}$, and so $x=\frac{\sqrt{3}}{3}$. Since we know what the height is, we can find the height of the remaining structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, $\sqrt{3}$, so we just subtract $\frac{\sqrt{3}}{3}$ from $\sqrt{3}$ to get $\frac{2\sqrt{3}}{3}$, or $\fbox{\frac{2\sqrt{3}}{3}}.$

AMC10 Final Problems

AMC10 B

AMC

AMC8

8

N/A

Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

37

If at least half the guesses are too low, then Norb's age must be greater than $36.$ If two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$ Lastly, Norb's age is a prime number so the answer must be $\fbox{37}$

AMC8 Second Half

AMC8

HMMT

HMMT-Nov

guts

Nov

Over all real numbers $x$ and $y$, find the minimum possible value of \[ (x y)^{2}+(x+7)^{2}+(2 y+7)^{2} \]

45

Solution 1: Rewrite the given expression as $\left(x^{2}+4\right)\left(1+y^{2}\right)+14(x+2 y)+94$. By Cauchy-Schwartz, this is at least $(x+2 y)^{2}+14(x+2 y)+94=(x+2 y+7)^{2}+45$. The minimum is 45 , attained when $x y=2, x+2 y=-7$. Solution 2: Let $z=2 y, s=x+z, p=x z$. We seek to minimize \[ \begin{aligned} \left(\frac{x z}{2}\right)^{2}+(x+7)^{2}+(z+7)^{2} & =\frac{p^{2}}{4}+\left(x^{2}+z^{2}\right)+14(x+z)+98 \\ & =\frac{p^{2}}{4}+s^{2}-2 p+14 s+98 \\ & =\left(\frac{p}{2}-2\right)^{2}+(s+7)^{2}+45 \\ & \geq 45 \end{aligned} \] Equality holds when $s=-7, p=4$. Since $s^{2} \geq 4 p$, this system has a real solution for $x$ and $z$. $\fbox{45}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

Consider a square, inside which is inscribed a circle, inside which is inscribed a square, inside which is inscribed a circle, and so on, with the outermost square having side length 1 . Find the difference between the sum of the areas of the squares and the sum of the areas of the circles.

2-\frac{\pi}{2}

The ratio of the area of each square and the circle immediately inside it is $\frac{4}{\pi}$. The total sum of the areas of the squares is $1+\frac{1}{2}+\frac{1}{4}+\ldots=2$. Difference in area is then $2-2 \cdot \frac{4}{\pi}$. $\fbox{2-\frac{\pi}{2}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

thm

Nov

On a blackboard a stranger writes the values of $s_{7}(n)^{2}$ for $n=0,1, \ldots, 7^{20}-1$, where $s_{7}(n)$ denotes the sum of digits of $n$ in base 7. Compute the average value of all the numbers on the board.

3680

Solution 1: We solve for 0 to $b^{n}-1$ and $s_{b}(n)^{2}$ (i.e. base $b$ ). Let $n=d_{1} \ldots d_{n}$ in base $b$, where there may be leading zeros. Then $s_{b}(n)=d_{1}+\cdots+d_{n}$, regardless of the leading zeros. \[ \mathbb{E}\left[s_{d}(n)^{2}\right]=\mathbb{E}\left[\left(d_{1}+\cdots+d_{n}\right)^{2}\right]=\sum_{1 \leq i \leq n} \mathbb{E}\left[d_{i}^{2}\right]+2 \sum_{1 \leq i<j \leq n} \mathbb{E}\left[d_{i} d_{j}\right] \] and now notice that we can treat choosing $n$ uniformly as choosing the $d_{i}$ uniformly independently from $\{0, \ldots, b-1\}$. Thus this simplifies to \[ \mathbb{E}\left[s_{d}(n)^{2}\right]=n \mathbb{E}\left[d_{1}^{2}\right]+n(n-1) \mathbb{E}\left[d_{1}\right]^{2} \] Now \[ \begin{gathered} \mathbb{E}\left[d_{1}^{2}\right]=\frac{0^{2}+\cdots+(b-1)^{2}}{b}=\frac{(b-1)(2 b-1)}{6} \\ \mathbb{E}\left[d_{1}\right]=\frac{0+\cdots+(b-1)}{b}=\frac{b-1}{2} \end{gathered} \] so the answer is \[ n \cdot \frac{(b-1)(2 b-1)}{6}+n(n-1) \cdot\left(\frac{b-1}{2}\right)^{2} \] Plugging in $b=7, n=20$ yields the result. Solution 2: There are two theorems we will cite regarding variance and expected value. The first is that, for any variable $X$, \[ \operatorname{Var}(X)=E\left[X^{2}\right]-E[X]^{2} \] The second is that, for two independent variables $X$ and $Y$, \[ \operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y) \] Let $X$ be the sum of all of the digits. We want to find $E\left[X^{2}\right]$. The expected of a single digit is $\frac{1}{7}(0+1+2+3+4+5+6)=3$. Thus, the expected value of the sum of the digits is $E[X]=20 \times 3=60$, so $E[X]^{2}=3600$. The variance of a single digit is $\frac{1}{7}\left[(0-3)^{2}+(1-3)^{2}+\ldots+(6-3)^{2}\right]=\frac{9+4+1+0+1+4+9}{7}=4$. Since the digits are independent, their variances add by the second theorem above. Therefore, the variance of the sum of all of the digits is $\operatorname{Var}(X)=20 \times 4=80$. Finally, using the first theorem, we have $E\left[X^{2}\right]=E[X]^{2}+\operatorname{Var}(X)=3680$. $\fbox{3680}$.

HMMT Nov Hard

HMMT-Nov Theme

AMC

AMC10

10B

N/A

How many integer values of $x$ satisfy $|x|<3\pi$?

19

Since $3\pi\approx9.42$, we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\fbox{19}$.

AMC10 First Half

AMC10 B

HMMT

HMMT-Nov

gen

Nov

Let $n$ be the 200th smallest positive real solution to the equation $x-\frac{\pi}{2}=\tan x$. Find the greatest integer that does not exceed $\frac{n}{2}$.

314

Drawing the graphs of the functions $y=x-\frac{\pi}{2}$ and $y=\tan x$, we may observe that the graphs intersect exactly once in each of the intervals $\left(\frac{(2 k-1) \pi}{2}, \frac{(2 k+1) \pi}{2}\right)$ for each $k=1,2, \cdots$. Hence, the 200th intersection has $x$ in the range $\left(\frac{399 \pi}{2}, \frac{401 \pi}{2}\right)$. At this intersection, $y=x-\frac{\pi}{2}$ is large, and thus, the intersection will be slightly less than $\frac{401 \pi}{2}$. We have that $\left\lfloor\frac{401 \pi}{4}\right\rfloor=\left\lfloor 100 \pi+\frac{\pi}{4}\right\rfloor=\left\lfloor 314.16+\frac{\pi}{4}\right\rfloor=314$. $\fbox{314}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Feb

guts

Feb

Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$.

\frac{72+32 \sqrt{2}}{7}

Let $u=x^{2}+2 y^{2}$. By AM-GM, $u \geq \sqrt{8} x y$, so $x y \leq \frac{u}{\sqrt{8}}$. If we let $x y=k u$ where $k \leq \frac{1}{\sqrt{8}}$, then we have \[ \begin{gathered} u(1-k)=8 \\ u(1+k)=x^{2}+x y+2 y^{2} \end{gathered} \] that is, $u(1+k)=8 \cdot \frac{1+k}{1-k}$. It is not hard to see that the maximum value of this expression occurs at $k=\frac{1}{\sqrt{8}}$, so the maximum value is $8 \cdot \frac{1+\frac{1}{\sqrt{8}}}{1-\frac{1}{\sqrt{8}}}=\frac{72+32 \sqrt{2}}{7}$. $\fbox{\frac{72+32 \sqrt{2}}{7}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

Find the smallest possible area of an ellipse passing through $(2,0),(0,3),(0,7)$, and $(6,0)$.

\frac{56 \pi \sqrt{3}}{9}

Let $\Gamma$ be an ellipse passing through $A=(2,0), B=(0,3), C=(0,7), D=(6,0)$, and let $P=(0,0)$ be the intersection of $A D$ and $B C$. Area of $\Gamma$ As is inchanged under an affine transformation, so we just have to minimize this quantity over situations where $\Gamma$ is a circle and $\frac{P A}{P D}=\frac{1}{3}$ and $\frac{P B}{B C}=\frac{3}{7}$. In fact, we may assume that $P A=\sqrt{7}, P B=3, P C=7, P D=3 \sqrt{7}$. If $\angle P=\theta$, then we can compute lengths to get \[ r=\frac{\text { Area of } \Gamma}{\text { Area of } A B C D}=\pi \frac{32-20 \sqrt{7} \cos \theta+21 \cos ^{2} \theta}{9 \sqrt{7} \cdot \sin ^{3} \theta} \] Let $x=\cos \theta$. Then if we treat $r$ as a function of $x$, \[ 0=\frac{r^{\prime}}{r}=\frac{3 x}{1-x^{2}}+\frac{42 x-20 \sqrt{7}}{32-20 x \sqrt{7}+21 x^{2}} \] which means that $21 x^{3}-40 x \sqrt{7}+138 x-20 \sqrt{7}=0$. Letting $y=x \sqrt{7}$ gives \[ 0=3 y^{3}-40 y^{2}+138 y-140=(y-2)\left(3 y^{2}-34 y+70\right) \] The other quadratic has roots that are greater than $\sqrt{7}$, which means that the minimum ratio is attained when $\cos \theta=x=\frac{y}{\sqrt{7}}=\frac{2}{\sqrt{7}}$. Plugging that back in gives that the optimum $\frac{\text { Area of } \Gamma}{\text { Area of } A B C D}$ is\\ $\frac{28 \pi \sqrt{3}}{81}$, so putting this back into the original configuration gives Area of $\Gamma \geq \frac{56 \pi \sqrt{3}}{9}$. If you want to check on Geogebra, this minimum occurs when the center of $\Gamma$ is $\left(\frac{8}{3}, \frac{7}{3}\right)$. $\fbox{\frac{56 \pi \sqrt{3}}{9}}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC8

8

N/A

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$? [asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]

\frac{12}{5}

Because $\overline{BD} + \overline{CD} = 5,$ we can see that when we draw a line from point $B$ to imaginary point $D$ that line applies to both triangles. Let us say that $x$ is that line. Perimeter of $\triangle{ABD}$ would be $\overline{AD} + 4 + x$, while the perimeter of $\triangle{ACD}$ would be $\overline{AD} + 3 + (5 - x)$. Notice that we can find $x$ from these two equations by setting them equal and then canceling $\overline{AD}$. We find that $x = 2$, and because the height of the triangles is the same, the ratio of the areas is $2:3$, so that means that the area of $\triangle ABD = \frac{2 \cdot 6}{5} = \fbox{\frac{12}{5}}$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

geo

Feb

Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Suppose $P Q R S$ is a square such that $P$ and $R$ lie on line $B C, Q$ lies on line $C A$, and $S$ lies on line $A B$. Compute the side length of this square.

42 \sqrt{2}

\section*{Solution:} Let $A^{\prime}$ be the reflection of $A$ across $B C$. Since $Q$ and $S$ are symmetric across $B C$, we get that $Q \in B A^{\prime}$, $S \in C A^{\prime}$. Now, let $X$ and $M$ be the midpoints of $A A^{\prime}$ and $P R$. Standard altitude computation gives $B X=5, C X=9, A X=12$. Moreover, from similar triangles, $C X: C Y=A A^{\prime}: P R=B X: B M$, so $B M: C M=5: 9$, so we easily get that $B M=35 / 2$. Now, $P M=\frac{12}{9} \cdot B Y=42$, so the side length is $42 \sqrt{2}$. $\fbox{42 \sqrt{2}}$.

HMMT Feb Easy

HMMT-Feb Geometry

HMMT

HMMT-Feb

calc

Feb

Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k !}$. Calculate $\sum_{n=2}^{\infty} f(n)$.

3-e

\[ \begin{aligned} \sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k !} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \cdot \frac{1}{k(k-1)} \\ & =\sum_{k=2}^{\infty} \frac{1}{(k-1) !} \cdot \frac{1}{k^{2}(k-1)} \end{aligned} \] \[ \begin{aligned} & =\sum_{k=2}^{\infty} \frac{1}{(k-1) !}\left(\frac{1}{k-1}-\frac{1}{k^{2}}-\frac{1}{k}\right) \\ & =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1) !}-\frac{1}{k \cdot k !}-\frac{1}{k !}\right) \\ & =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1) !}-\frac{1}{k \cdot k !}\right)-\sum_{k=2}^{\infty} \frac{1}{k !} \\ & =\frac{1}{1 \cdot 1 !}-\left(e-\frac{1}{0 !}-\frac{1}{1 !}\right) \\ & =3-e \end{aligned} \] $\fbox{3-e}$.

HMMT Feb Hard

HMMT-Feb Calculus

AMC

AMC10

10A

N/A

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?

996

The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$, which means they would be able to cancel with the factors in $n!$. Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$, so we subtract $1$ to get $n=\fbox{996}$.

AMC10 First Half

AMC10 A

HMMT

HMMT-Feb

comb

Feb

A tourist is learning an incorrect way to sort a permutation $\left(p_{1}, \ldots, p_{n}\right)$ of the integers $(1, \ldots, n)$. We define a fix on two adjacent elements $p_{i}$ and $p_{i+1}$, to be an operation which swaps the two elements if $p_{i}>p_{i+1}$, and does nothing otherwise. The tourist performs $n-1$ rounds of fixes, numbered $a=1,2, \ldots, n-1$. In round $a$ of fixes, the tourist fixes $p_{a}$ and $p_{a+1}$, then $p_{a+1}$ and $p_{a+2}$, and so on, up to $p_{n-1}$ and $p_{n}$. In this process, there are $(n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2}$ total fixes performed. How many permutations of $(1, \ldots, 2018)$ can the tourist start with to obtain $(1, \ldots, 2018)$ after performing these steps?

1009 ! \cdot 1010 !

Note that the given algorithm is very similar to the well-known Bubble Sort algorithm for sorting an array. The exception is that in the $i$-th round through the array, the first $i-1$ pairs are not checked. We claim a necessary and sufficient condition for the array to be sorted after the tourist's process is: for all $i$, after $i$ rounds, the numbers $1, \cdots, i$ are in the correct position. Firstly, this is necessary because these indices of the array are not touched in future rounds - so if a number was incorrect, then it would stay incorrect. On the other hand, suppose this condition holds. Then, we can "add" the additional fixes during each round (of the first $i-1$ pairs during the $i$-th round) to make the process identical to bubble sort. The tourist's final result won't change because by our assumption these swaps won't do anything. However, this process is now identical to bubble sort, so the resulting array will be sorted. Thus, our condition is sufficient. Now, there are two positions the 1 can be in $\left(p_{1}, p_{2}\right)$. There are three positions the 2 can be in $\left(p_{1}, \cdots, p_{4}\right.$ except for the position of 1$)$. Similarly, for $1 \leq i \leq 1009$ there are $2 i-(i-1)=i+1$ positions $i$ can be in, and after that the remaining 1009 numbers can be arranged arbitrarily. Thus, the answer is $1010 ! \cdot 1009$ !. $\fbox{1009 ! \cdot 1010 !}$.

HMMT Feb Hard

HMMT-Feb Combinatorics

AIME

AIME

II

N/A

Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

167

The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\frac{1}{17}$, and the probability that he is using the biased die is $\frac{16}{17}$. The probability of rolling a third six is \[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\] Therefore, our desired $p+q$ is $65+102= \fbox{167}$

Intermediate AIME Problems

AIME

AMC

AMC10

10A

N/A

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

15

To find how many small bottles we need, we can simply divide $500$ by $35$. This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$, we have to round up to $15$ bottles, or $\fbox{15}$

AMC10 First Half

AMC10 A

AMC

AMC10

10B

N/A

Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$, and that $u_3=9$ and $u_6=128$. What is $u_5$?

53

If we plug in $n=4$, we get $128=2u_5+u_4.$ By plugging in $n=3$, we get $u_5=2u_4+9.$ This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$, therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\fbox{53}}$.

AMC10 Second Half

AMC10 B

AMC

AMC10

10B

N/A

Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$

3

Manipulate the powers of $5$ in order to get a clean expression. \[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\] \[25^{-2} = (5^2)^{-2} = 5^{-4}\] \[5^{-4} = 5^{-\frac{12}{x}}\] If two numbers are equal, and their bases are equal, then their exponents are equal as well. Set the two exponents equal to each other. \begin{align}-4&=\frac{-12}{x}\\ -4x&=-12\\ x&=\fbox{3}\end{align}

AMC10 First Half

AMC10 B

HMMT

HMMT-Nov

guts

Nov

Let $x, y, z$ be real numbers satisfying \[ \begin{aligned} 2 x+y+4 x y+6 x z & =-6 \\ y+2 z+2 x y+6 y z & =4 \\ x-z+2 x z-4 y z & =-3 \end{aligned} \] Find $x^{2}+y^{2}+z^{2}$.

29

Solution: We multiply the first, second, and third equations by $\frac{1}{2},-\frac{1}{2}$, and -1 , respectively, then add the three resulting equations. This gives $x y+x z+y z=-2$. Doing the same with the coefficients $-1,2$, and 3 gives $x+y+z=5$, from which $(x+y+z)^{2}=25$. So $x^{2}+y^{2}+z^{2}=25-2 \cdot-2=29$. $\fbox{29}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$? [asy] defaultpen(linewidth(1)); for ( int x = 0; x &lt; 5; ++x ) { draw((0,x)--(4,x)); draw((x,0)--(x,4)); } fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,3)--(1,3)--(1,4)--(0,4)--cycle); fill((2,3)--(4,3)--(4,4)--(2,4)--cycle); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle); label("$A$", (0, 4), NW); label("$B$", (4, 4), NE); label("$C$", (4, 0), SE); label("$D$", (0, 0), SW); [/asy]

4

Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\fbox{4}$ squares need to be colored to match its corresponding square.

AMC8 First Half

AMC8

HMMT

HMMT-Nov

team

Nov

How many ways are there to color the vertices of a triangle red, green, blue, or yellow such that no two vertices have the same color? Rotations and reflections are considered distinct.

24

There are 4 ways to color the first vertex, then 3 ways to color the second vertex to be distinct from the first, and finally 2 ways to color the third vertex to be distinct from the earlier two vertices. Multiplying gives 24 ways. $\fbox{24}$.

HMMT Nov Team

HMMT-Nov Team

HMMT

HMMT-Feb

guts

Feb

Let $A B C D$ be a rectangle such that $A B=20$ and $A D=24$. Point $P$ lies inside $A B C D$ such that triangles $P A C$ and $P B D$ have areas 20 and 24, respectively. Compute all possible areas of triangle $P A B$.

98,118,122,142

Solution: There are four possible locations of $P$ as shown in the diagram. Let $O$ be the center. Then, $[P A O]=10$ and $[P B O]=12$. Thus, $[P A B]=[A O B] \pm[P A O] \pm[P B O]=120 \pm 10 \pm 12$, giving the four values $98,118,122$, and 142 . $\fbox{98,118,122,142}$.

HMMT Feb Guts

HMMT-Feb Guts

AIME

AIME

I

N/A

There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

17

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 $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}$ $\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$. Therefore, $1 + 16 = \fbox{17}$. ~ JHawk0224

Easy AIME Problems

AIME

AMC

AMC12

12B

N/A

Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?

\frac 76

Joe has 2 ounces of cream, as stated in the problem. JoAnn had 14 ounces of liquid, and drank $\frac{1}{7}$ of it. Therefore, she drank $\frac{1}{7}$ of her cream, leaving her $2*\frac{6}{7}$. $\frac{2}{2*\frac{6}{7}}=\frac{7}{6} \Rightarrow \fbox{\frac 76}$

AMC12 Second Half

AMC12 B

AMC

AMC12

12A

N/A

The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$?

16

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers. The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$ As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$, for some nonnegative integer $k$. $a^2-8a=k^2$ $a(a-8)=k^2$ $((a-4)+4)((a-4)-4)=k^2$ $(a-4)^2-4^2=k^2$ $(a-4)^2=k^2+4^2$ From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$. In the case $k=0$, the equation simplifies to $|a-4|=4$. From this equation, we have $a=0,8$. For both $a=0$ and $a=8$, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.") If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$. Hence, $a=-1,9$. Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$, so these two values also satisfy the original constraints. There are a total of four possible values for $a$: $-1,0,8,$ and $9$. Hence, the sum of all of the possible values of $a$ is $\fbox{16}$.

AMC12 Second Half

AMC12 A

AMC

AMC12

12B

N/A

Let $x=-2016$. What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$?

4032

By: dragonfly First of all, lets plug in all of the $x$'s into the equation. $\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$ Then we simplify to get $\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$ which simplifies into $\bigg|$ $2016$ $\bigg|$ $+2016$ and finally we get $\fbox{4032}$

AMC12 First Half

AMC12 B

HMMT

HMMT-Feb

guts

Feb

There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, one random frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find the expected number of time-steps before all frogs are in the pool for the first time.

2^{2018}-1

Consider the general case of $n$ frogs. Let $E_{i}$ be the expected time for all frogs to enter the pool when $i$ frogs are on the shore and $n-i$ frogs are in the pool. We have $E_{0}=0, E_{n}=1+E_{n-1}$, and \[ E_{i}=\frac{i}{n} E_{i-1}+\frac{n-i}{n} E_{i+1}+1 \] for $0<i<n$. Define $f_{i}$ so that \[ E_{i}=\frac{f_{i}}{(n-1)(n-2) \cdots(i)}+E_{i-1} \] Then by plugging this equation into the first equation, we can show that \[ f_{i}=n(n-1) \cdots(i+1)+(n-i) f_{i+1} \] Furthermore, we know that $f_{n}=1$. Therefore \[ \begin{aligned} f_{1} & =\sum_{i=1}^{n} \frac{n !}{i !} \frac{(n-1) !}{(n-i) !} \\ & =(n-1) ! \sum_{i=1}^{n}\left(\begin{array}{c} n \\ i \end{array}\right) \\ & =(n-1) !\left(2^{n}-1\right) \end{aligned} \] Therefore \[ E_{1}=\frac{(n-1) !\left(2^{n}-1\right)}{(n-1) !}+E_{0}=2^{n}-1 \] Plugging in $n=2018$ yields $E_{1}=2^{2018}-1$. $\fbox{2^{2018}-1}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

alg

Feb

Find the sum of all positive integers $n$ for which \[ \frac{15 \cdot n !^{2}+1}{2 n-3} \] is an integer.

90

Solution: It is clear that $n=1$ and $n=2$ work so assume that $n>2$. If $2 n-3$ is composite then its smallest prime factor is at most $\frac{2 n-3}{2}<n$ so will be coprime to $15 \cdot n !^{2}+1$. Therefore assume that $2 n-3=p$ is prime. We can rewrite the numerator as \[ (-1)^{n} \cdot 15 \cdot\left(1 \cdot 2 \cdots \frac{p+3}{2}\right) \cdot\left(\frac{p-3}{2} \cdot \frac{p-1}{2} \cdots(p-1)\right)+1 \quad(\bmod p) \] By Wilson's Theorem, $(p-1) ! \equiv-1(\bmod p)$, so the expression simplifies to \[ (-1)^{n+1} \cdot 15 \cdot \frac{p-3}{2} \cdot \frac{p-1}{2} \cdot \frac{p+1}{2} \cdot \frac{p+3}{2}+1 \equiv(-1)^{n+1} \cdot \frac{135}{16}+1 \quad(\bmod p) \] If $p \equiv 3(\bmod 4)$, then we have \[ \frac{135+16}{16} \equiv \frac{151}{16} \equiv 0 \quad(\bmod p) \] If $p \equiv 1(\bmod 4)$, then we have \[ \frac{135-16}{16} \equiv \frac{119}{16} \equiv 0 \quad(\bmod p) \] So $p$ must be a prime divisor of 151 or 119 , which means that $p \in\{7,17,151\}$. All of these numbers work aside from $7($ because $7 \equiv 3(\bmod 4))$ and the corresponding values of $n$ are 10 and 77 . The sum of the solutions is then $1+2+10+77=90$. $\fbox{90}$.

HMMT Feb Hard

HMMT-Feb Algebra

AMC

AMC12

12B

N/A

The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$. What is the remainder when $S$ is divided by $2017?$

1024

Note that $2014\equiv -3 \mod2017$. We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{ if n is odd}\] and \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{ if n is even}\] Obviously, $62$ falls in the second category, so our desired value is \[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \fbox{1024}\]

AMC12 Final Problems

AMC12 B

AMC

AMC12

12A

N/A

Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a,b)$ lies above the parabola $y=ax^2-bx$?

\frac{19}{81}

If $(a,b)$ lies above the parabola, then $b$ must be greater than $y(a)$. We thus get the inequality $b>a^3-ba$. Solving this for $b$ gives us $b>\frac{a^3}{a+1}$. Now note that $\frac{a^3}{a+1}$ constantly increases when $a$ is positive. Then since this expression is greater than $9$ when $a=4$, we can deduce that $a$ must be less than $4$ in order for the inequality to hold, since otherwise $b$ would be greater than $9$ and not a single-digit integer. The only possibilities for $a$ are thus $1$, $2$, and $3$. For $a=1$, we get $b>\frac{1}{2}$ for our inequality, and thus $b$ can be any integer from $1$ to $9$. For $a=2$, we get $b>\frac{8}{3}$ for our inequality, and thus $b$ can be any integer from $3$ to $9$. For $a=3$, we get $b>\frac{27}{4}$ for our inequality, and thus $b$ can be any integer from $7$ to $9$. Finally, if we total up all the possibilities we see there are $19$ points that satisfy the condition, out of $9 \times 9 = 81$ total points. The probability of picking a point that lies above the parabola is thus $\frac{19}{81} \rightarrow \fbox{\frac{19}{81}}$

AMC12 Second Half

AMC12 A

AMC

AMC12

12A

Nov

Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$

74

By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = \fbox{74}.$

AMC12 First Half

AMC12 A

HMMT

HMMT-Feb

guts

Feb

Let $A, E, H, L, T$, and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$. Compute the probability that $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A$.

\frac{55}{81}

Solution: There are $3^{3}-2^{3}=19$ ways to choose $L, A$, and $V$ such that $L \cdot A \cdot V \cdot A=0$, since at least one of $\{L, A, V\}$ must be 0 , and $3^{3}-1=26$ ways to choose $T, H$, and $E$ such that $\lfloor T \cdot H \cdot E\rfloor=0$, since at least one of $\{T, H, E\}$ must not be 1 , for a total of $19 \cdot 26=494$ ways. There is only one way to make $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A=1$, namely setting every variable equal to 1 , so there are 495 total ways that work out of a possible $3^{6}=729$, for a probability of $\frac{55}{81}$. $\fbox{\frac{55}{81}}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10A

N/A

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

23

From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\fbox{23}$

AMC10 First Half

AMC10 A

HMMT

HMMT-Nov

guts

Nov

Let $P(x)$ be a quadratic polynomial with real coefficients. Suppose that $P(1)=20, P(-1)=22$, and $P(P(0))=400$. Compute the largest possible value of $P(10)$.

2486

Solution: Let $P(x)=a x^{2}+b x+c$. The given equations give us: \[ \begin{aligned} & a+b+c=20 \\ & a-b+c=22 \end{aligned} \] Hence $b=-1, a+c=21$, and so the final equation gives us $a c^{2}=400$. Substituting $a=21-c$ and solving the cubic in $c$, we get $c=-4,5,20$. Of these, the smallest value $c=-4$ (and hence $\left.P(x)=25 x^{2}-x-4\right)$ ends up giving the largest value of $P(10)$. $\fbox{2486}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC12

12A

N/A

For a certain complex number $c$, the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\]has exactly 4 distinct roots. What is $|c|$?

\sqrt{10}

The polynomial can be factored further broken down into $P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$ by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$. We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$, or $(x - [1+i])$ and $(x - [2-2i])$. When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \pm i$ and $|c| = \fbox{\sqrt{10}}$.

AMC12 Second Half

AMC12 A

HMMT

HMMT-Nov

guts

Nov

At Easter-Egg Academy, each student has two eyes, each of which can be eggshell, cream, or cornsilk. It is known that $30 \%$ of the students have at least one eggshell eye, $40 \%$ of the students have at least one cream eye, and $50 \%$ of the students have at least one cornsilk eye. What percentage of the students at Easter-Egg Academy have two eyes of the same color?

80 \%

For the purposes of this solution, we abbreviate "eggshell" by "egg", and "cornsilk" by "corn". We know that there are only six combinations of eye color possible: egg-cream, egg-corn, egg-egg, creamcorn, cream-cream, corn-corn. If we let the proportions for each of these be represented by $a, b, c, d$, $e$, and $f$ respectively, we have the following four equalities: \[ \begin{aligned} a+b+c & =.3 \\ a+d+e & =.4 \\ b+d+f & =.5 \\ a+b+c+d+e+f & =1 \end{aligned} \] where the first three equalities come from the given conditions. Adding the first three equations and subtracting the fourth, we obtain that \[ a+b+d=.2 \] which is the proportion of people with different colored eyes. The proportion of people with the same eye color is thus $1-.2=.8$. $\fbox{80 \%}$.

HMMT Nov Guts

HMMT-Nov Guts

AIME

AIME

II

N/A

Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon into 7 triangles by drawing segments $\overline{JB}$, $\overline{JC}$, $\overline{JD}$, $\overline{JE}$, $\overline{JF}$, and $\overline{JG}$. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles. [asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]

184

We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$. Recall that the centroid is $\frac{1}{3}$ way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point $J$. Furthermore, we can take advantage of the reflective symmetry across the line parallel to $BC$ going through $J$ by dealing with less coordinates and ommiting the $\frac{1}{2}$ in the shoelace formula. By doing some basic algebra, we find that the coordinates of the centroids of $\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE$ are $\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),$ and $\left(18, -\frac{23}{2}\right)$, respectively. We'll have to throw in the projection of the centroid of $\bigtriangleup JAB$ to the line of reflection to apply shoelace, and that point is $\left( \frac{8}{3}, -\frac{23}{2}\right)$ Finally, applying Shoelace, we get: $\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|$ $=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|$ $=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|$ $=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\fbox{184}$ Solution by ktong note: for a slightly simpler calculation, notice that the heptagon can be divided into two trapezoids of equal area and a small triangle.

Intermediate AIME Problems

AIME

AMC

AMC10

10B

N/A

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?

62

Let the sides of the rectangular parking lot be $a$ and $b$. Then $a^2 + b^2 = 625$ and $ab = 168$. Add the two equations together, then factor. \begin{align} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \fbox{62}$

AMC10 Second Half

AMC10 B

AMC

AMC10

10B

N/A

Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$, together with its interior. For real $r\geq0$, let $S(r)$ be the set of points in $3$-dimensional space that lie within a distance $r$ of some point in $B$. The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$, where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\frac{bc}{ad}?$

19

Split $S(r)$ into 4 regions: 1. The rectangular prism itself 2. The extensions of the faces of $B$ 3. The quarter cylinders at each edge of $B$ 4. The one-eighth spheres at each corner of $B$ Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$, so $d=12$. Region 2: This volume is equal to the surface area of $B$ times $r$ (these "extensions" are just more boxes). The volume is then $\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r$ to get $c=38$. Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is $4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2$. Therefore, $b=8\pi$. Region 4: There is an eighth-sphere of radius $r$ at each corner of $B$. Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius $r$. The volume of this sphere is then $\frac{4}{3}\pi \cdot r^3$, so $a=\frac{4\pi}{3}$. Using these values, $\dfrac{bc}{ad}=\frac{(8\pi)(38)}{(4\pi/3)(12)} = \fbox{19}$. To see a diagram of $S(r)$, view TheBeautyofMath's explanation video (Video Solution 1). ~DrJoyo

AMC10 Second Half

AMC10 B

AMC

AMC12

12B

N/A

A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars?

9

The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \fbox{9}$

AMC12 First Half

AMC12 B

AMC

AMC10

10A

N/A

Six distinct positive integers are randomly chosen between $1$ and $2006$, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$?

1

For two numbers to have a difference that is a multiple of $5$, the numbers must be congruent $\bmod{5}$ (their remainders after division by $5$ must be the same). $0, 1, 2, 3, 4$ are the possible values of numbers in $\bmod{5}$. Since there are only $5$ possible values in $\bmod{5}$ and we are picking $6$ numbers, by the Pigeonhole Principle, two of the numbers must be congruent $\bmod{5}$. Therefore the probability that some pair of the $6$ integers has a difference that is a multiple of $5$ is $\fbox{1}$.

AMC10 Second Half

AMC10 A

HMMT

HMMT-Nov

guts

Nov

Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points?

\frac{1}{2}

Label the vertices of the cube $A, B, C, D, E, F, G, H$, such that $A B C D$ is the top face of the cube, $E$ is directly below $A, F$ is directly below $B, G$ is directly below $C$, and $H$ is directly below $D$. We can obtain a volume of $\frac{1}{2}$ by taking the vertices $A, B, C, F$, and $H$. To compute the volume of $A C B F H$, we will instead compute the volume of the parts of the cube that are not part of $A B C F H$. This is just the three tetrahedrons $C F G H, A E F H$, and $A C D H$, which each have volume $\frac{1}{6}$ (by using the $\frac{1}{3} b h$ formula for the area of a pyramid). Therefore, the volume not contained in $A C B F H$ is $3 \cdot \frac{1}{6}=\frac{1}{2}$, so the volume contained in $A C B F H$ is $1-\frac{1}{2}=\frac{1}{2}$. $\fbox{\frac{1}{2}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC12

12B

N/A

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

-2

Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as \begin{align} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ -12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\ \left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)&=0 \\ (z+\bar{z}+2)^2+(z\bar{z}-6)^2&=0. \end{align} As the two quantities in the parentheses are real, both quantities must equal $0$ so \[z+\frac6z=z+\bar{z}=\fbox{-2}.\]

AMC12 Second Half

AMC12 B

AMC

AMC12

12A

N/A

Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?

\frac{223}{286}

Suppose $p$ is the answer. We calculate $1-p$. Assume that the circumradius of the 12-gon is $1$, and the 6 different lengths are $a_1$, $a_2$, $\cdots$, $a_6$, in increasing order. Then $a_k = 2\sin ( \frac{k\pi}{12} )$. So $a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5$, $a_2=1$, $a_3=\sqrt{2}\approx 1.4$, $a_4=\sqrt{3}\approx 1.7$, $a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3$, $a_6 = 2$. Now, Consider the following inequalities: $a_3>2a_1 > a_2$ $a_4> a_1 + a_2>a_3$ $a_4<a_1 + a_3=a_5$ $a_1 + a_4 > a_6$ $2a_2 = 2 = a_6$. Thus any two segments with at least one them longer than $a_2$ have a sum greater than $a_6$. Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means $(a_x, a_y, a_z)$: 1-1-3, 1-1-4, 1-1-5, 1-1-6, 1-2-4, 1-2-5, 1-2-6, 1-3-5, 1-3-6, 2-2-6 Note that there are $12$ segments of each length of $a_1$, $a_2$, $\cdots$, $a_5$, respectively, and $6$ segments of length $a_6$. There are $66$ segments in total. In the above list there are $3$ triples of the type a-a-b without 6, $2$ triples of a-a-6 where a is not 6, $3$ triples of a-b-c without 6, and $2$ triples of a-b-6 where a, b are not 6. So, \[1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)\] \[= \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286}\] $\fbox{\frac{223}{286}}$.

AMC12 Final Problems

AMC12 A

AMC

AMC10

10B

N/A

In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?

12+12\sqrt{2}

The center of the circle lies on the intersection between the perpendicular bisectors of chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let $AC=a$ and $BC=b$. Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$. Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$. But by Pythagorean Theorem on $\triangle ABC$, we know $a^2+b^2=144$, because $AB=12$. Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$. So our equation simplifies further to $a^2+ab=144$. However $a^2+b^2=144$, so $a^2+ab=a^2+b^2$, which means $ab=b^2$, or $a=b$. Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, and thus the perimeter is $\fbox{12+12\sqrt{2}}$. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); /* dots and labels */ dot((3.46,0.96),dotstyle); label("$A$", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label("$C$", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label("$B$", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label("$Z$", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label("$W$", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label("$X$", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label("$Y$", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label("$O$", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

AMC10 Second Half

AMC10 B

HMMT

HMMT-Feb

comb

Feb

Kelvin the Frog likes numbers whose digits strictly decrease, but numbers that violate this condition in at most one place are good enough. In other words, if $d_{i}$ denotes the $i$ th digit, then $d_{i} \leq d_{i+1}$ for at most one value of $i$. For example, Kelvin likes the numbers 43210, 132, and 3, but not the numbers 1337 and 123. How many 5-digit numbers does Kelvin like?

14034

Suppose first that no digit violates the constraint; i.e. the digits are in strictly decreasing order. There are $\left(\begin{array}{c}10 \\ 5\end{array}\right)$ ways to choose the digits of the number, and each set of digits can be arranged in exactly one way, so there are $\left(\begin{array}{c}10 \\ 5\end{array}\right)$ such numbers. We now perform casework on which digit violates the constraint. If it is the final digit, the first four digits must be arranged in decreasing order, which there are $\left(\begin{array}{c}10 \\ 4\end{array}\right)$ ways to do. The final digit can then be any digit, but we have overcounted the ones in which the number is in fully decreasing order (this can happen, for example, if the first 4 digits we chose were $5,4,3$, and 2 - a last digit of 1 was already counted in the first case). Therefore, there are $\left(\begin{array}{c}10 \\ 4\end{array}\right)\left(\begin{array}{c}10 \\ 1\end{array}\right)-252$ new numbers in this case. If the offending digit is second from the right, the first 3 digits must be decreasing, as must the last 2 digits. There are $\left(\begin{array}{c}10 \\ 3\end{array}\right)\left(\begin{array}{c}10 \\ 2\end{array}\right)$ ways to do this. As before, we overcount the case where the second digit from the right is not actually an offender, so we again overcount the case where all 5 digits decrease. Hence there are $\left(\begin{array}{c}10 \\ 3\end{array}\right)\left(\begin{array}{c}10 \\ 2\end{array}\right)-252$ new numbers in this case. The case where the third digit is the offender is identical to the previous case, so there are another $\left(\begin{array}{c}10 \\ 3\end{array}\right)\left(\begin{array}{c}10 \\ 2\end{array}\right)-252$ numbers to account for. The final case is when the second digit is the offending digit, in which case there are $\left(\begin{array}{c}10 \\ 4\end{array}\right)$ ways to choose the final 4 digits, but only 9 ways to choose the opening digit (as 0 cannot be a leading digit). Accounting for the usual overcounting, our final answer is \[ 252+\left[\left(\begin{array}{c} 10 \\ 4 \end{array}\right)\left(\begin{array}{c} 10 \\ 1 \end{array}\right)-252\right]+2\left[\left(\begin{array}{c} 10 \\ 3 \end{array}\right)\left(\begin{array}{c} 10 \\ 2 \end{array}\right)-252\right]+\left[\left(\begin{array}{c} 10 \\ 4 \end{array}\right) \cdot 9-252\right] \] which is easily calculated as 14034 . $\fbox{14034}$.

HMMT Feb Easy

HMMT-Feb Combinatorics

AMC

AMC10

10B

N/A

Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $$$1$ more than a pink pill, and Al's pills cost a total of $\textdollar 546$ for the two weeks. How much does one green pill cost?

20

Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\fbox{20}$.

AMC10 First Half

AMC10 B

AIME

AIME

II

N/A

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n.$

28

Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$. Notice that there are $10!$ possible pairings. Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad. Note that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people. Thus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$), check the left shoe it is paired with (say $L_i$), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$, find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$. We can imagine each right shoe "sending" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times. Effectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$; thus, we need to count pairings where every cycle has length at least $5$. This is only possible if there is a single cycle of length $10$ or two cycles of length $5$. The first case yields $9!$ working pairings. The second case yields $\frac{{10\choose 5}}{2}\cdot{4!}^2=\frac{10!}{2 \cdot {5!}^2} \cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$, the probability is $\frac{1}{10}+\frac{1}{50} = \frac{3}{25}$, for an answer of $\fbox{28}$.

Very Hard AIME Problems

AIME

HMMT

HMMT-Nov

team

Nov

Joey wrote a system of equations on a blackboard, where each of the equations was of the form $a+b=c$ or $a \cdot b=c$ for some variables or integers $a, b, c$. Then Sean came to the board and erased all of the plus signs and multiplication signs, so that the board reads: \[ \begin{array}{ll} x & z=15 \\ x & y=12 \\ x & x=36 \end{array} \] If $x, y, z$ are integer solutions to the original system, find the sum of all possible values of $100 x+10 y+z$.

2037

Solution: The bottom line gives $x=-6, x=6$ or $x=18$. If $x=-6, y$ can be -2 or 18 and $z$ must be 21 , so the possible values for $100 x+10 y+z$ are -599 and -399 . If $x=6, y$ can be 2 or 6 and $z$ must be 9 , so the possible values are 629 and 669 . If $x=18, y$ must be -6 and $z$ must be -3 , so the only possible value is 1737 . The total sum is 2037 . $\fbox{2037}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC8

8

N/A

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ , $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?

48

We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\fbox{48}$.

AMC8 First Half

AMC8

AMC

AMC8

8

N/A

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?

58

We compute $2^8+1=257$. We're all familiar with what $6^3$ is, namely $216$, which is too small. The smallest cube greater than it is $7^3=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$, which therefore will clearly be the largest cube less than $2^{18}+1$. So, the required number of cubes is $64-7+1= \fbox{58}$.

AMC8 Second Half

AMC8

AMC

AMC12

12B

N/A

Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?

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 $\frac 12 \cdot 4 \cdot 3 = \fbox{6}$.

AMC12 First Half

AMC12 B

AMC

AMC10

10A

N/A

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?

4

Since there are $n^2$ little faces on each face of the big wooden cube, there are $6n^2$ little faces painted red. Since each unit cube has $6$ faces, there are $6n^3$ little faces total. Since one-fourth of the little faces are painted red, $\frac{6n^2}{6n^3}=\frac{1}{4}$ $\frac{1}{n}=\frac{1}{4}$ $n=\fbox{4}$

AMC10 Second Half

AMC10 A

AMC

AMC10

10B

N/A

Joe and JoAnn each bought $12$ ounces of coffee in a $16$ ounce cup. Joe drank $2$ ounces of his coffee and then added $2$ ounces of cream. JoAnn added $2$ ounces of cream, stirred the coffee well, and then drank $2$ ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?

\frac{7}{6}

After drinking and adding cream, Joe's cup has $2$ ounces of cream. After adding cream to her cup, JoAnn's cup had $14$ ounces of liquid. By stirring and then drinking $2$ ounces out of the $14$ ounces of liquid, she drank $\frac{2}{14}=\frac{1}{7}$th of the cream. So there are $2\cdot\frac{6}{7}=\frac{12}{7}$ ounces of cream left. So the desired ratio is: $2 \div \frac{12}{7}= \fbox{\frac{7}{6}}$.

AMC10 Second Half

AMC10 B

AMC

AMC12

12B

N/A

The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?

250,500

Let the roots be $r,s,r + s$, and let $t = rs$. Then $(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$ and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$. Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$. Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer. Since $r > 0$ and $t > 0$, it follows that $0 < t = r(1002-r) < 501^2 = 251001$, with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \fbox{250,500}$ possible values of $n = -1002t$.

AMC12 Final Problems

AMC12 B

AMC

AMC10

10A

N/A

A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?

70

Let the total amount of flowers be $x$. Thus, the number of pink flowers is $0.6x$, and the number of red flowers is $0.4x$. The number of pink carnations is $\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\frac{3}{4}(0.4x) = 0.3x$. Summing these, the total number of carnations is $0.4x+0.3x=0.7x$. Dividing, we see that $\frac{0.7x}{x} = 0.7 = \fbox{70}$

AMC10 First Half

AMC10 A

AMC

AMC8

8

N/A

When three positive integers $a$, $b$, and $c$ are multiplied together, their product is $100$. Suppose $a < b < c$. In how many ways can the numbers be chosen?

4

The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$ If $c=10,$ then $(a,b,c)=(2,5,10).$ If $c=20,$ then $(a,b,c)=(1,5,20).$ If $c=25,$ then $(a,b,c)=(1,4,25).$ If $c=50,$ then $(a,b,c)=(1,2,50).$ Together, the numbers $a,b,$ and $c$ can be chosen in $\fbox{4}$ ways.

AMC8 First Half

AMC8

HMMT

HMMT-Feb

comb

Feb

In each cell of a $4 \times 4$ grid, one of the two diagonals is drawn uniformly at random. Compute the probability that the resulting 32 triangular regions can be colored red and blue so that any two regions sharing an edge have different colors.

\frac{1}{512}

Solution: Give each cell coordinates from $(1,1)$ to $(4,4)$. Claim. The grid has a desired coloring if and only if every vertex not on the boundary meets an even number of edges and diagonals. Proof. If this were not the case, the odd number of regions around the vertex would have to alternate between the two colors, which is clearly impossible. In the event that every vertex has an even number of incident edges, it is not hard to show that the grid is always colorable. We claim the diagonals drawn in the cells of form $(1, a)$ and $(a, 1)$ for $1 \leq a \leq 4$ uniquely determine the rest (for a valid coloring to exist). Indeed, given the diagonals for any three cells around a vertex, we can uniquely determine the fourth one using the parity in the claim above. If $(1,1),(1,2),(2,1)$ are fixed, so is $(2,2)$; likewise so are $(2,3)$ and $(2,4)$, etc. until the whole grid is fixed. The solid lines force the dotted lines as described above. Thus, once the seven cells along the top row and leftmost column are determined, the remaining nine have a $\frac{1}{2^{9}}=\frac{1}{512}$ chance of being selected in a way that admits a coloring. $\fbox{\frac{1}{512}}$.

HMMT Feb Hard

HMMT-Feb Combinatorics

AMC

AMC8

8

N/A

Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?

\text{Dan}

The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$ The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count. Assuming the six people start with $1$, Arn counts $7$ so he leaves first. Then, Cyd counts $14$ as there are $7$ numbers to be counted from this point. Then, Fon, Bob, and Eve, count, $17,$ $21,$ and $27$, respectively, so the last one standing is Dan. Hence, the answer would be $\fbox{\text{Dan}}$.

AMC8 First Half

AMC8

AMC

AMC12

12B

N/A

For some real numbers $a$ and $b$, the equation \[8x^3 + 4ax^2 + 2bx + a = 0\] has three distinct positive roots. If the sum of the base-$2$ logarithms of the roots is $5$, what is the value of $a$?

-256

Let the three roots be $x_1,x_2,x_3$. \[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\] By Vieta’s formulas, \[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\] gives us that $a = -8x_1x_2x_3 = -256 $. $\fbox{-256}$.

AMC12 Second Half

AMC12 B

HMMT

HMMT-Nov

guts

Nov

Regular hexagon $A B C D E F$ has side length 2. A laser beam is fired inside the hexagon from point $A$ and hits $\overline{B C}$ at point $G$. The laser then reflects off $\overline{B C}$ and hits the midpoint of $\overline{D E}$. Find $B G$.

\frac{2}{5}

Look at the diagram below, in which points $J, K, M, T$ and $X$ have been defined. $M$ is the midpoint of $\overline{D E}, B C J K$ is a rhombus with $J$ lying on the extension of $\overline{C D}, T$ is the intersection of lines $\overline{C D}$ and $\overline{G M}$ when extended, and $X$ is on $\overline{J T}$ such that $\overline{X M} \| \overline{J K}$. It can be shown that $m \angle M D X=m \angle M X D=60^{\circ}$, so $\triangle D M X$ is equilateral, which yields $X M=1$. The diagram indicates that $J X=5$. One can show by angle-angle similarity that $\triangle T X M \sim \triangle T J K$, which yields $T X=5$. One can also show by angle-angle similarity that $\triangle T J K \sim \triangle T C G$, which yields the proportion $\frac{T J}{J K}=\frac{T C}{C G}$. We know everything except $C G$, which we can solve for. This yields $C G=\frac{8}{5}$, so $B G=\frac{2}{5}$. $\fbox{\frac{2}{5}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC12

12A

N/A

Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x$, $y$, and $z$ are each chosen from the set $\{0,1,2\}$. How many equilateral triangles all have their vertices in $S$?

80

For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left. First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacent vertices into a triangle. This triangle will have a side length of $\sqrt{2}$; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and $9 \cdot 8 = 72$ equilateral triangles. [asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((2,0,0)--(0,2,0)--(0,0,2)--cycle,green); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle,red); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE); [/asy] NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\fbox{80}$. [asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE); label("$A$",(0,2,0), NW); label("$B$",(2,0,2), NW); [/asy]

AMC12 Final Problems

AMC12 A

AMC

AMC8

8

N/A

A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$

4

A number is divisible by $15$ precisely if it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$, and the number is of the form $5\square 5\square 5$. If the unknown digit is $x$, we deduce $5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$. We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$. It can be $0$, $3$, $6$, or $9$, so there are $\fbox{4}$ options: $50505$, $53535$, $56565$, and $59595$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

guts

Feb

Suppose $a$ and $b$ are positive integers such that $a^{b}=2^{2023}$. Compute the smallest possible value of $b^{a}$.

1

Solution: By taking $a=2^{2023}$ and $b=1$, we get $b^{a}=1$, which is clearly the minimum. $\fbox{1}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

guts

Nov

Ten points are equally spaced on a circle. A graph is a set of segments (possibly empty) drawn between pairs of points, so that every two points are joined by either zero or one segments. Two graphs are considered the same if we can obtain one from the other by rearranging the points. Let $N$ denote the number of graphs with the property that for any two points, there exists a path from one to the other among the segments of the graph. Estimate the value of $N$. If your answer is a positive integer $A$, your score on this problem will be the larger of 0 and $\lfloor 20-5|\ln (A / N)|\rfloor$. Otherwise, your score will be zero.

11716571

The question asks for the number of isomorphism classes of connected graphs on 10 vertices. This is enumerated in \href{http://oeis.org/A001349}{http://oeis.org/A001349} the answer is 11716571. In fact, of the $2^{45} \approx 3.51 \cdot 10^{13} \approx 3 \cdot 10^{13}$ graphs on 10 labelled vertices, virtually all (about $3.45 \cdot 10^{13}$ ) are connected. You might guess this by noticing that an "average" graph has 22.5 edges, which is fairly dense (and virtually all graphs with many edges are connected). Moreover, a "typical" isomorphism class contains $10 ! \approx 3 \cdot 10^{6}$ elements, one for each permutation of the vertices. So estimating the quotient $\frac{3 \cdot 10^{13}}{3 \cdot 10^{6}}=10^{7}$ gives a very close estimate. $\fbox{11716571}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

Triangles $ABC$, $ADE$, and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. If $AB = 4$, what is the perimeter of figure $ABCDEFG$? [asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy]

15

The large triangle $ABC$ has sides of length $4$. The medium triangle has sides of length $2$. The small triangle has sides of length $1$. There are $3$ segment sizes, and all segments depicted are one of these lengths. Starting at $A$ and going clockwise, the perimeter is: $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 + 2 + 2 + 1 + 1 + 1$ $15$, thus the answer is $\fbox{15}$

AMC8 Second Half

AMC8

AMC

AMC8

8

N/A

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$?

11

We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$, because of the above statement. Now we need to test the remaining answer choices. Case 1: $x = 6$ Mode: $6$ Median: $6$ Mean: $\frac{37}{7}$ Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated. Case 2: $x = 11$ Mode: $6$ Median: $6$ Mean: $6$ We are done with this problem, because we have found when $x = 11$, the condition is satisfied. Therefore, the answer is $\fbox{11}$.

AMC8 First Half

AMC8

AMC

AMC10

10B

N/A

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?

12

The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$. The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$. Clearly $L=60$ and the answer is $720/60=\fbox{12}$.

AMC10 First Half

AMC10 B

HMMT

HMMT-Feb

guts

Feb

Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in[0,1]$ uniformly at random, and (a) If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ with her marker. (b) If $x>\frac{1}{2}$ she colors the intervals $[x, 1]$ and $\left[0, x-\frac{1}{2}\right]$ with her marker. What is the expected value of the number of steps Natalie will need to color the entire interval black?

5

The first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half. Solution 1 (non-calculus): We assume the interval has $2 n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expected value of the number of turns we need if there are $k$ white points left. So we must calculate $f(n)$. We observe that \[ \begin{gathered} f(k)=1+\frac{(n-k+1) \cdot 0+(n-k+1) \cdot f(k)+2 \sum_{i=1}^{k-1} f(i)}{2 n} \\ f(k) \frac{n+k-1}{2 n}=1+\frac{\sum_{i=1}^{k-1} f(i)}{n} \\ f(k+1) \frac{n+k}{2 n}=1+\frac{\sum_{i=1}^{k} f(i)}{n} \\ f(k+1)=f(k) \frac{n+k+1}{n+k} \\ f(k)=f(1) \frac{n+k}{n+1} \end{gathered} \] And note that $f(1)=2$ so $f(n)=\frac{4 n}{n+1}$ and $\lim _{n \rightarrow \infty} f(n)=4$. Therefore adding the first turn, the expected value is 5 . Solution 2 (calculus): We let $f(x)$ be the expected value with length $x$ uncolored. Like above, $\lim _{x \rightarrow 0} f(x)=2$. Similarly we have the recursion \[ \begin{gathered} f(x)=1+\left(\frac{1}{2}-x\right) f(x)+2 \int_{0}^{x} f(y) d y \\ f^{\prime}(x)=0+\frac{1}{2} f^{\prime}(x)-f(x)-x f^{\prime}(x)+2 f(x) \\ \frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+\frac{1}{2}} \end{gathered} \] And solving yields $f(x)=c\left(\frac{1}{2}+x\right)$ and since $\lim _{x \rightarrow 0} f(x)=2, c=4$. So $f(x)=2+4 x$ and $f\left(\frac{1}{2}\right)=4$. Therefore adding the first turn, our expected value is 5 . $\fbox{5}$.

HMMT Feb Guts

HMMT-Feb Guts

AIME

AIME

II

N/A

A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

187

Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes). Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring. Case 2: Two 5's are rolled. Case 3: No 5's are rolled. To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$, let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities. To find $a_{n+1}$ given $a_n$ (where $n \ge 1$), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ($5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$. Computing $a_2$, $a_3$, $a_4$ gives $a_2 = 7$, $a_3 = 32$, and $a_4 = 157$. Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is \[\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \fbox{187}\] -scrabbler94

Easy AIME Problems

AIME

HMMT

HMMT-Feb

guts

Feb

Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3 x^{2}+10 x y+$ $3 y^{2}$ is the power of some prime.

29

Solution: We can factor as $(3 x+y)(x+3 y)$. If $x \geq y$, we need $\frac{3 x+y}{x+3 y} \in\{1,2\}$ to be an integer. So we get the case where $x=y$, in which we need both to be a power of 2 , or the case $x=5 y$, in which case we need $y$ to be a power of 2 . This gives us $11+9+9=29$ solutions, where we account for $y=5 x$ as well. $\fbox{29}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

guts

Feb

Let $A B C$ be a triangle with area 1. Let points $D$ and $E$ lie on $A B$ and $A C$, respectively, such that $D E$ is parallel to $B C$ and $D E / B C=1 / 3$. If $F$ is the reflection of $A$ across $D E$, find the area of triangle $F B C$.

\frac{1}{3}

Let $A F$ intersect $B C$ at $H$. Since $D E / B C=1 / 3$ and $F$ and $A$ are equidistant from $D E$, we have $A F=\frac{2}{3} A H$ and $F H=A H-A F=\frac{1}{3} A H$. Furthermore, since $A F$ is perpendicular to $D E$, we have $A H$ and $F H$ are the altitudes of triangles $A B C$ and $F B C$ respectively. Therefore the area of triangle $F B C$ is $\frac{1}{2} \cdot F H \cdot B C=\frac{1}{2} \cdot \frac{1}{3} \cdot A H \cdot B C=\frac{1}{3}$. $\fbox{\frac{1}{3}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

geo

Feb

Let $A B C D$ be a regular tetrahedron with side length 1 . Let $X$ be the point in triangle $B C D$ such that $[X B C]=2[X B D]=4[X C D]$, where $[\varpi]$ denotes the area of figure $\varpi$. Let $Y$ lie on segment $A X$ such that $2 A Y=Y X$. Let $M$ be the midpoint of $B D$. Let $Z$ be a point on segment $A M$ such that the lines $Y Z$ and $B C$ intersect at some point. Find $\frac{A Z}{Z M}$.

\frac{4}{7}

We apply three-dimensional barycentric coordinates with reference tetrahedron $A B C D$. The given conditions imply that \[ \begin{aligned} X & =(0: 1: 2: 4) \\ Y & =(14: 1: 2: 4) \\ M & =(0: 1: 0: 1) \\ Z & =(t: 1: 0: 1) \end{aligned} \] for some real number $t$. Normalizing, we obtain $Y=\left(\frac{14}{21}, \frac{1}{21}, \frac{2}{21}, \frac{4}{21}\right)$ and $Z=\left(\frac{t}{t+2}, \frac{1}{t+2}, 0, \frac{1}{t+2}\right)$. If $Y Z$ intersects line $B C$ then there exist parameters $\alpha+\beta=1$ such that $\alpha Y+\beta Z$ has zero $A$ and $D$ coordinates, meaning \[ \begin{aligned} \frac{14}{21} \alpha+\frac{t}{t+2} \beta & =0 \\ \frac{4}{21} \alpha+\frac{1}{t+2} \beta & =0 \\ \alpha+\beta & =1 \end{aligned} \] Adding twice the second equation to the first gives $\frac{22}{21} \alpha+\beta=0$, so $\alpha=-22, \beta=21$, and thus $t=\frac{7}{2}$. It follows that $Z=(7: 2: 0: 2)$, and $\frac{A Z}{Z M}=\frac{2+2}{7}=\frac{4}{7}$. $\fbox{\frac{4}{7}}$.

HMMT Feb Hard

HMMT-Feb Geometry

AMC

AMC12

12B

Nov

A bug starts at a vertex of a grid made of equilateral triangles of side length $1$. At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit away from the starting position?

\frac{13}{108}

Let $S(n)$ be the number of paths of $n$ moves such that the bug never will have been more than $1$ unit away from the starting position. Clearly, by symmetry, there are two possible states here, the bug being on the center and the bug being on one of the vertices of the unit hexagon around the center. Let $C(n)$ be the number of paths with the aforementioned restriction that end on the center. Let $V(n)$ be the number of paths with the aforementioned restriction that end on a vertex of the surrounding unit hexagon. We have $S(n) = 6C(n-1) + 3V(n-1),$ since from the center, there are $6$ possible points to land to and from a vertex there are $3$ possible points to land to (the two adjacent vertices and the center). We also have $C(n) = V(n-1)$, since to get to the center the bug must have come from a vertex, and $V(n) = 2V(n-1) + 6C(n-1),$ since from a vertex there are two vertices to move to, and from the center there are $6$ vertices to move to. We can construct a recursion table using the base cases $V(1) = 6$ and $C(1) = 0$ and our recursive rules for $C(n)$ and $V(n)$ as follows: \[\begin{array}{c|c|c} n & V(n) & C(n) \\ \hline & & \\ [-2ex] 1 & 6 & 0 \\ 2 & 12 & 6 \\ 3 & 60 & 12 \\ 4 & 192 & 60 \\ \end{array}\] Then, $S(5) = 6C(4) + 3V(4) = 6 \cdot 60 + 3 \cdot 192 = 936,$ and the desired probability is thus $\frac{936}{6^5} = \fbox{\frac{13}{108}}.$

AMC12 Second Half

AMC12 B

AMC

AMC10

10A

N/A

What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]

80

We write \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.\] Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\fbox{80}$.

AMC10 Second Half

AMC10 A

AMC

AMC12

12A

N/A

A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?

64

So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$ Observe that since there are $280$ pieces of fruit, \[b+r+g+c=280.\] Since there are twice as many raspberries as blueberries, \[2b=r.\] The fact that there are three times as many grapes as cherries implies, \[3c=g.\] Because there are four times as many cherries as raspberries, we deduce the following: \[4r=c.\] Note that we are looking for $c.$ So, we try to rewrite all of the other variables in terms of $c.$ The third equation gives us the value of $g$ in terms of $c$ already. We divide the fourth equation by $4$ to get that $r=\frac{c}{4}.$ Finally, substituting this value of $r$ into the first equation provides us with the equation $b=\frac{c}{8}$ and substituting yields: \[\frac{c}{4}+\frac{c}{8}+3c+c=280\] Multiply this equation by $8$ to get: \[2c+c+24c+8c=8\cdot 280,\] \[35c=8\cdot 280,\] \[c=64.\] \[\fbox{64}\]

AMC12 First Half

AMC12 A

HMMT

HMMT-Nov

guts

Nov

Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ be a polynomial whose roots are all negative integers. If $a+b+c+d=2009$, find $d$.

528

Call the roots $-x_{1},-x_{2},-x_{3}$, and $-x_{4}$. Then $f(x)$ must factor as $\left(x+x_{1}\right)\left(x+x_{2}\right)(x+$ $\left.x_{3}\right)\left(x+x_{4}\right)$. If we evaluate $f$ at 1 , we get $\left(1+x_{1}\right)\left(1+x_{2}\right)\left(1+x_{3}\right)\left(1+x_{4}\right)=a+b+c+d+1=$ $2009+1=2010$. $2010=2 \cdot 3 \cdot 5 \cdot 67$. $d$ is the product of the four roots, so $d=(-1) \cdot(-2) \cdot(-4) \cdot(-66)$. $\fbox{528}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids? [asy] size((70)); draw((0,0)--(7.5,13)--(15,0)--(0,0)); draw((1.88,3.25)--(9.45,3.25)); draw((11.2,0)--(7.5,6.5)); draw((9.4,9.7)--(5.6,3.25)); [/asy]

5

The area outside the small triangle but inside the large triangle is $16-1=15$. This is equally distributed between the three trapezoids. Each trapezoid has an area of $15/3 = \fbox{5}$.

AMC8 First Half

AMC8

AMC

AMC8

8

N/A

What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?

10

Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$, the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$. One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is $\fbox{10}$.

AMC8 Second Half

AMC8