Datasets:

furonghuang-lab

/

Easy2Hard-Bench

like 0

Follow

Furong Huang's Lab at UMD 14

AMC

AMC10

10B

N/A

There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$. What is the sum of the digits of $n$?

10

\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\] Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$). The answer is therefore $1 + 9 = \fbox{10}$.

AMC10 First Half

AMC10 B

AIME

AIME

I

N/A

For positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$.

695

First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work). Suppose by contradiction that there is a smaller valid $n$, where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 + 9 < 20.) Clearly $n > 200$ because the smallest three-digit number whose digits sum to 20 is $299$. Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition $N = n + 864$ regroups in the ones place. Then the units digit of $N$ is clearly less than 4. But as $1000 < 200 + 864 < N < 600 + 864 = 1464$, the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of $N$ is at most $5 + 9 + 4 < 20$, contradiction. Hence $\fbox{695}$ is the answer.

Intermediate AIME Problems

AIME

AMC

AMC12

12B

N/A

For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the $2008$th term of the sequence?

4015

Letting $S_n$ be the nth partial sum of the sequence: $\frac{S_n}{n} = n$ $S_n = n^2$ The only possible sequence with this result is the sequence of odd integers. $a_n = 2n - 1$ $a_{2008} = 2(2008) - 1 = 4015 \Rightarrow $ $\fbox{4015}$.

AMC12 Second Half

AMC12 B

AMC

AMC12

12A

N/A

A set $S$ is constructed as follows. To begin, $S = \{0,10\}$. Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$, all of whose coefficients $a_i$ are elements of $S$, then $x$ is put into $S$. When no more elements can be added to $S$, how many elements does $S$ have?

9

At first, $S=\{0,10\}$. \[\begin{tabular}{r c l c l} $10x+10$ & has root & $x=-1$ & so now & $S=\{-1,0,10\}$ \\ $-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10$ & has root & $x=1$ & so now & $S=\{-1,0,1,10\}$ \\ $x+10$ & has root & $x=-10$ & so now & $S=\{-10,-1,0,1,10\}$ \\ $x^3+x-10$ & has root & $x=2$ & so now & $S=\{-10,-1,0,1,2,10\}$ \\ $x+2$ & has root & $x=-2$ & so now & $S=\{-10,-2,-1,0,1,2,10\}$ \\ $2x-10$ & has root & $x=5$ & so now & $S=\{-10,-2,-1,0,1,2,5,10\}$ \\ $x+5$ & has root & $x=-5$ & so now & $S=\{-10,-5,-2,-1,0,1,2,5,10\}$ \end{tabular}\] At this point, no more elements can be added to $S$. To see this, let \begin{align} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align} with each $a_i$ in $S$. $x$ is a factor of $a_0$, and $a_0$ is in $S$, so $x$ has to be a factor of some element in $S$. There are no such integers left, so there can be no more additional elements. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \fbox{9}$

AMC12 Final Problems

AMC12 A

HMMT

HMMT-Nov

guts

Nov

Let $A B C$ be a triangle with $A B=A C=\frac{25}{14} B C$. Let $M$ denote the midpoint of $\overline{B C}$ and let $X$ and $Y$ denote the projections of $M$ onto $\overline{A B}$ and $\overline{A C}$, respectively. If the areas of triangle $A B C$ and quadrilateral $A X M Y$ are both positive integers, find the minimum possible sum of these areas.

1201

By similar triangles, one can show that $[A X M Y]=2 \cdot[A M X]=\left(\frac{24}{25}\right)^{2} \cdot 2[A B M]=$ $\left(\frac{24}{25}\right)^{2} \cdot[A B C]$. Thus the answer is $25^{2}+24^{2}=1201$. $\fbox{1201}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

Let $A_{1} A_{2} \ldots A_{6}$ be a regular hexagon with side length $11 \sqrt{3}$, and let $B_{1} B_{2} \ldots B_{6}$ be another regular hexagon completely inside $A_{1} A_{2} \ldots A_{6}$ such that for all $i \in\{1,2, \ldots, 5\}, A_{i} A_{i+1}$ is parallel to $B_{i} B_{i+1}$. Suppose that the distance between lines $A_{1} A_{2}$ and $B_{1} B_{2}$ is 7 , the distance between lines $A_{2} A_{3}$ and $B_{2} B_{3}$ is 3 , and the distance between lines $A_{3} A_{4}$ and $B_{3} B_{4}$ is 8 . Compute the side length of $B_{1} B_{2} \ldots B_{6}$.

3 \sqrt{3}

Solution: Let $X=A_{1} A_{2} \cap A_{3} A_{4}$, and let $O$ be the center of $B_{1} B_{2} \ldots B_{6}$. Let $p$ be the apothem of hexagon $B$. Since $O A_{2} X A_{3}$ is a convex quadrilateral, we have \[ \begin{aligned} {\left[A_{2} A_{3} X\right] } & =\left[A_{2} X O\right]+\left[A_{3} X O\right]-\left[A_{2} A_{3} O\right] \\ & =\frac{11 \sqrt{3}(7+p)}{2}+\frac{11 \sqrt{3}(8+p)}{2}-\frac{11 \sqrt{3}(3+p)}{2} \\ & =\frac{11 \sqrt{3}(12+p)}{2} \end{aligned} \] Since $\left[A_{2} A_{3} X\right]=(11 \sqrt{3})^{2} \frac{\sqrt{3}}{4}$, we get that \[ \frac{12+p}{2}=(11 \sqrt{3}) \frac{\sqrt{3}}{4}=\frac{33}{4} \Longrightarrow p=\frac{9}{2} \] Thus, the side length of hexagon $B$ is $p \cdot \frac{2}{\sqrt{3}}=3 \sqrt{3}$. $\fbox{3 \sqrt{3}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

geo

Feb

Let $A B C D$ be a square, and let $\ell$ be a line passing through the midpoint of segment $\overline{A B}$ that intersects segment $\overline{B C}$. Given that the distances from $A$ and $C$ to $\ell$ are 4 and 7 , respectively, compute the area of $A B C D$.

185

Solution: Consider the line $\ell^{\prime}$ through $B$ parallel to $\ell$, and drop perpendiculars from $A$ to $\ell^{\prime}$ and $C$ to $\ell^{\prime}$. Note that because $\ell$ passes through the midpoint of segment $A B$, the distance from $B$ to $\ell$ is 4 . Thus, the distances from $A$ to $\ell^{\prime}$ and from $C$ to $\ell^{\prime}$ are $4+4=8$ and $4+7=11$, respectively. Let $P$ be the foot from $A$ to $\ell^{\prime}$. Rotating the square $90^{\circ}$ from $B$ to $A$ sends the altitude from $C$ to $\ell^{\prime}$ to the segment along $\ell^{\prime}$ between $B$ and the foot from $A$ to $\ell^{\prime}$; hence $B P=11$. So the side length of the square is $\sqrt{A P^{2}+B P^{2}}=\sqrt{8^{2}+11^{2}}$, which means the area of the square is $8^{2}+11^{2}=185$. $\fbox{185}$.

HMMT Feb Easy

HMMT-Feb Geometry

AMC

AMC12

12A

N/A

Let $f(z)= \frac{z+a}{z+b}$ and $g(z)=f(f(z))$, where $a$ and $b$ are complex numbers. Suppose that $\left| a \right| = 1$ and $g(g(z))=z$ for all $z$ for which $g(g(z))$ is defined. What is the difference between the largest and smallest possible values of $\left| b \right|$?

\sqrt{3}-1

By algebraic manipulations, we obtain \[h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}\] where \[P=(a+1)^2+a(b+1)^2\] \[Q=a(b+1)(b^2+2a+1)\] \[R=(b+1)(b^2+2a+1)\] \[S=a(b+1)^2+(a+b^2)^2\] In order for $h(z)=z$, we must have $R=0$, $Q=0$, and $P=S$. $R=0$ implies $b=-1$ or $b^2+2a+1=0$. $Q=0$ implies $a=0$, $b=-1$, or $b^2+2a+1=0$. $P=S$ implies $b=\pm1$ or $b^2+2a+1=0$. Since $|a|=1\neq 0$, in order to satisfy all 3 conditions we must have either $b=\pm1$ or $b^2+2a+1=0$. In the first case $|b|=1$. For the latter case note that \[|b^2+1|=|-2a|=2\] \[2=|b^2+1|\leq |b^2|+1\] and hence, \[1\leq|b|^2\Rightarrow1\leq |b|\]. On the other hand, \[2=|b^2+1|\geq|b^2|-1\] so, \[|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}\]. Thus $1\leq |b|\leq \sqrt{3}$. Hence the maximum value for $|b|$ is $\sqrt{3}$ while the minimum is $1$ (which can be achieved for instance when $|a|=1,|b|=\sqrt{3}$ or $|a|=1,|b|=1$ respectively). Therefore the answer is $\fbox{\sqrt{3}-1}$.

AMC12 Final Problems

AMC12 A

HMMT

HMMT-Nov

guts

Nov

A small fish is holding 17 cards, labeled 1 through 17, which he shuffles into a random order. Then, he notices that although the cards are not currently sorted in ascending order, he can sort them into ascending order by removing one card and putting it back in a different position (at the beginning, between some two cards, or at the end). In how many possible orders could his cards currently be?

256

Instead of looking at moves which put the cards in order, we start with the cards in order and consider possible starting positions by backtracking one move: each of 17 cards can be moved to 16 new places. But moving card $k$ between card $k+1$ and card $k+2$ is equivalent to moving card $k+1$ between card $k-1$ and card $k$. We note that these are the only possible pairs of moves which produce the same result, so we have double counted 16 moves. Thus, we have a total of $17 \times 16-16=256$ possible initial positions. $\fbox{256}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the contest?

28

Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\fbox{28}$

AMC8 First Half

AMC8

HMMT

HMMT-Feb

guts

Feb

Let $p$ denote the proportion of teams, out of all participating teams, who submitted a negative response to problem 5 of the Team round (e.g. "there are no such integers"). Estimate $P=\lfloor 10000 \mathrm{p}\rfloor$. An estimate of $E$ earns $\max (0,\lfloor 20-|P-E| / 20\rfloor)$ points. If you have forgotten, problem 5 of the Team round was the following: "Determine, with proof, whether there exist positive integers $x$ and $y$ such that $x+y, x^{2}+y^{2}$, and $x^{3}+y^{3}$ are all perfect squares."

5568

Solution: Of the 88 teams competing in this year's Team round, 49 of them answered negatively, 9 (correctly) provided a construction, 16 answered ambiguously or did not provide a construction, and the remaining 14 teams did not submit to problem 5 . Thus $p=\frac{49}{88} \approx 0.5568$. $\fbox{5568}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC12

12B

N/A

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

\frac{\sqrt{5}-1}{2}

Let $a$ be the short side of the rectangle, and $b$ be the long side of the rectangle. The diagonal, therefore, is $\sqrt{a^2 + b^2}$. We can get the equation $\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}$. Cross-multiplying, we get $a\sqrt{a^2 + b^2} = b^2$. Squaring both sides of the equation, we get $a^2 (a^2 + b^2) = b^4$, which simplifies to $a^4 + a^2b^2 - b^4 = 0$. Solving for a quadratic in $a^2$, using the quadratic formula we get $a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}$ which gives us $\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}$. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is $\fbox{\frac{\sqrt{5}-1}{2}}$. Solution by: vedadehhc

AMC12 First Half

AMC12 B

HMMT

HMMT-Nov

guts

Nov

An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the right or one unit up, but it will never move in the same direction more than twice in the row. In how many different ways can it get to the point $(5,5)$ ?

84

We can change the ant's sequence of moves to a sequence $a_{1}, a_{2}, \ldots, a_{10}$, with $a_{i}=0$ if the $i$-th step is up, and $a_{i}=1$ if the $i$-th step is right. We define a subsequence of moves $a_{i}, a_{i+1}, \ldots, a_{j}$, $(i \leq j)$ as an up run if all terms of the subsequence are equal to 0 , and $a_{i-1}$ and $a_{j+1}$ either do not exist or are not equal to 0 , and define a right run similarly. In a sequence of moves, up runs and right runs alternate, so the number of up rights can differ from the number of right runs by at most one. Now let $f(n)$ denote the number of sequences $a_{1}, a_{2}, \ldots, a_{n}$ where $a_{i} \in\{1,2\}$ for $1 \leq i \leq n$, and $a_{1}+a_{2}+\cdots+a_{n}=5$. (In essence, we are splitting the possible 5 up moves into up runs, and we are doing the same with the right moves). We can easily compute that $f(3)=3, f(4)=4, f(5)=1$, and $f(n)=0$ otherwise. For each possible pair of numbers of up runs and right runs, we have two choices of which type of run is first. Our answer is then $2\left(f(3)^{2}+f(3) f(4)+f(4)^{2}+f(4) f(5)+f(5)^{2}\right)=2(9+12+16+4+1)=84$. $\fbox{84}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

gen

Nov

Katie has a fair 2019-sided die with sides labeled $1,2, \ldots, 2019$. After each roll, she replaces her $n$-sided die with an $(n+1)$-sided die having the $n$ sides of her previous die and an additional side with the number she just rolled. What is the probability that Katie's $2019^{\text {th }}$ roll is a $2019 ?$

\frac{1}{2019}

Since Katie's original die is fair, the problem is perfectly symmetric. So on the 2019th roll, each number is equally probable as any other. Therefore, the probability of rolling a 2019 is just $\frac{1}{2019}$. $\fbox{\frac{1}{2019}}$.

HMMT Nov Easy

HMMT-Nov General

AMC

AMC10

10A

N/A

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements. Brian: "Mike and I are different species." Chris: "LeRoy is a frog." LeRoy: "Chris is a frog." Mike: "Of the four of us, at least two are toads." How many of these amphibians are frogs?

3

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog. As Mike is a frog, his statement is false, hence there is at most one toad. As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and $\fbox{3}$ frogs.

AMC10 Second Half

AMC10 A

HMMT

HMMT-Nov

guts

Nov

The points $A=\left(4, \frac{1}{4}\right)$ and $B=\left(-5,-\frac{1}{5}\right)$ lie on the hyperbola $x y=1$. The circle with diameter $A B$ intersects this hyperbola again at points $X$ and $Y$. Compute $X Y$.

\sqrt{\frac{401}{5}}

\section*{Solution:} Let $A=(a, 1 / a), B=(b, 1 / b)$, and $X=(x, 1 / x)$. Since $X$ lies on the circle with diameter $\overline{A B}$, we have $\angle A X B=90^{\circ}$. Thus, $\overline{A X}$ and $\overline{B X}$ are perpendicular, and so the product of their slopes must be -1 . We deduce: \[ \frac{a-x}{\frac{1}{a}-\frac{1}{x}} \frac{b-x}{\frac{1}{b}-\frac{1}{x}}=-1 \Longrightarrow(a x)(b x)=-1 \] so $x= \pm \sqrt{-a b}$. Plugging in $a=4$ and $b=-5$ gives $X=(\sqrt{20}, 1 / \sqrt{20})$ and $Y=(-\sqrt{20},-1 / \sqrt{20})$, giving the answer. $\fbox{\sqrt{\frac{401}{5}}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

gen

Nov

Triangle $A B C$ has $A B=5, B C=7$, and $C A=8$. New lines not containing but parallel to $A B$, $B C$, and $C A$ are drawn tangent to the incircle of $A B C$. What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines?

\frac{31}{5} \sqrt{3}

From the law of cosines we compute $\measuredangle A=\cos ^{-1}\left(\frac{5^{2}+8^{2}-7^{2}}{2(5)(8)}\right)=60^{\circ}$. Using brackets to denote the area of a region, we find that \[ [A B C]=\frac{1}{2} A B \cdot A C \cdot \sin 60^{\circ}=10 \sqrt{3} \] The radius of the incircle can be computed by the formula \[ r=\frac{2[A B C]}{A B+B C+C A}=\frac{20 \sqrt{3}}{20}=\sqrt{3} \] Now the height from $A$ to $B C$ is $\frac{2[A B C]}{B C}=\frac{20 \sqrt{3}}{7}$. Then the height from $A$ to $D E$ is $\frac{20 \sqrt{3}}{7}-2 r=\frac{6 \sqrt{3}}{7}$. Then $[A D E]=\left(\frac{6 \sqrt{3} / 7}{20 \sqrt{3} / 7}\right)^{2}[A B C]=\frac{9}{100}[A B C]$. Here, we use the fact that $\triangle A B C$ and $\triangle A D E$ are similar. Similarly, we compute that the height from $B$ to $C A$ is $\frac{2[A B C]}{C A}=\frac{20 \sqrt{3}}{8}=\frac{5 \sqrt{3}}{2}$. Then the height from $B$ to $H J$ is $\frac{5 \sqrt{3}}{2}-2 r=\frac{\sqrt{3}}{2}$. Then $[B H J]=\left(\frac{\sqrt{3} / 2}{5 \sqrt{3} / 2}\right)^{2}[A B C]=\frac{1}{25}[A B C]$. Finally, we compute that the height from $C$ to $A B$ is $\frac{2[A B C]}{5}=\frac{20 \sqrt{3}}{5}=4 \sqrt{3}$. Then the height from $C$ to $F G$ is $4 \sqrt{3}-2 r=2 \sqrt{3}$. Then $[C F G]=\left(\frac{2 \sqrt{3}}{4 \sqrt{3}}\right)^{2}[A B C]=\frac{1}{4}[A B C]$. Finally we can compute the area of hexagon $D E F G H J$. We have \[ \begin{gathered} {[D E F G H J]=[A B C]-[A D E]-[B H J]-[C F G]=[A B C]\left(1-\frac{9}{100}-\frac{1}{25}-\frac{1}{4}\right)=[A B C]\left(\frac{31}{50}\right)=} \\ 10 \sqrt{3}\left(\frac{31}{50}\right)=\frac{31}{5} \sqrt{3} \end{gathered} \] $\fbox{\frac{31}{5} \sqrt{3}}$.

HMMT Nov Easy

HMMT-Nov General

HMMT

HMMT-Feb

geo

Feb

For $i=0,1, \ldots, 5$ let $l_{i}$ be the ray on the Cartesian plane starting at the origin, an angle $\theta=i \frac{\pi}{3}$ counterclockwise from the positive $x$-axis. For each $i$, point $P_{i}$ is chosen uniformly at random from the intersection of $l_{i}$ with the unit disk. Consider the convex hull of the points $P_{i}$, which will (with probability 1) be a convex polygon with $n$ vertices for some $n$. What is the expected value of $n$ ?

2+4 \ln (2)

A vertex $P_{i}$ is part of the convex hull if and only if it is not contained in the triangle formed by the origin and the two adjacent vertices. Let the probability that a given vertex is contained in the aforementioned triangle be $p$. By linearity of expectation, our answer is simply $6(1-p)$. Say $\left|P_{0}\right|=a,\left|P_{2}\right|=b$. Stewart's Theorem and the Law of Cosines give that $p$ is equal to the probability that $\left|P_{1}\right|<\sqrt{a b-a b \frac{a^{2}+b^{2}+a b}{(a+b)^{2}}}=\frac{a b}{a+b}$; alternatively this is easy to derive using coordinate methods. The corresponding double integral evaluates to $p=\frac{2}{3}(1-\ln (2))$, thus telling us our answer. $\fbox{2+4 \ln (2)}$.

HMMT Feb Hard

HMMT-Feb Geometry

HMMT

HMMT-Nov

guts

Nov

There are five people in a room. They each simultaneously pick two of the other people in the room independently and uniformly at random and point at them. Compute the probability that there exists a group of three people such that each of them is pointing at the other two in the group.

\frac{5}{108}

Solution: The desired probability is the number of ways to pick the two isolated people times the probability that the remaining three point at each other. So, \[ P=\left(\begin{array}{l} 5 \\ 2 \end{array}\right) \cdot\left(\frac{\left(\begin{array}{l} 2 \\ 2 \end{array}\right)}{\left(\begin{array}{l} 4 \\ 2 \end{array}\right)}\right)^{3}=10 \cdot\left(\frac{1}{6}\right)^{3}=\frac{5}{108} \] is the desired probability. $\fbox{\frac{5}{108}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10B

Nov

Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

27

Let Alice have the number A, Bob B. When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$. Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$, $256$, and $289$. So, $A$ could equal $25$, $56$, or $89$, so $A+B$ is $27$, $58$, or $91$, of only $\fbox{27}$ is an answer choice.

AMC10 First Half

AMC10 B

AMC

AMC10

10B

N/A

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?

2

Substituting each $x$ and $y$ with $1$, we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$. Thus, $\frac{x+3y}{3x-y}=\fbox{2}$

AMC10 First Half

AMC10 B

HMMT

HMMT-Feb

alg

Feb

Let $\alpha, \beta$, and $\gamma$ be three real numbers. Suppose that \[ \begin{aligned} \cos \alpha+\cos \beta+\cos \gamma & =1 \\ \sin \alpha+\sin \beta+\sin \gamma & =1 \end{aligned} \] Find the smallest possible value of $\cos \alpha$.

\frac{-1-\sqrt{7}}{4}

Let $a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta$, and $c=\cos \gamma+i \sin \gamma$. We then have \[ a+b+c=1+i \] where $a, b, c$ are complex numbers on the unit circle. Now, to minimize $\cos \alpha=\operatorname{Re}[a]$, consider a triangle with vertices $a, 1+i$, and the origin. We want $a$ as far away from $1+i$ as possible while maintaining a nonnegative imaginary part. This is achieved when $b$ and $c$ have the same argument, so $|b+c|=|1+i-a|=2$. Now $a, 0$, and $1+i$ form a $1-2-\sqrt{2}$ triangle. The value of $\cos \alpha$ is now the cosine of the angle between the 1 and $\sqrt{2}$ sides plus the $\frac{\pi}{4}$ angle from $1+i$. Call the first angle $\delta$. Then \[ \begin{aligned} \cos \delta & =\frac{1^{2}+(\sqrt{2})^{2}-2^{2}}{2 \cdot 1 \cdot \sqrt{2}} \\ & =\frac{-1}{2 \sqrt{2}} \end{aligned} \] and \[ \begin{aligned} \cos \alpha & =\cos \left(\frac{\pi}{4}+\delta\right) \\ & =\cos \frac{\pi}{4} \cos \delta-\sin \frac{\pi}{4} \sin \delta \\ & =\frac{\sqrt{2}}{2} \cdot \frac{-1}{2 \sqrt{2}}-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{7}}{2 \sqrt{2}} \\ & =\frac{-1-\sqrt{7}}{4} \end{aligned} \] $\fbox{\frac{-1-\sqrt{7}}{4}}$.

HMMT Feb Hard

HMMT-Feb Algebra

AMC

AMC8

8

N/A

The product of the two $99$-digit numbers $303,030,303,...,030,303$ and $505,050,505,...,050,505$ has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?

8

We can first make a small example to find out $A$ and $B$. So, $303\times505=153015$ The ones digit plus thousands digit is $5+3=8$. Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\fbox{8}$ This is a direct multiplication way.

AMC8 Second Half

AMC8

AMC

AMC10

10B

N/A

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$

150

[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,2), B=(0,0), C=(2,0), D=(2+sqrt(3),1), E=(2,2); draw(A--B--C--D--E--cycle); draw(E--C,gray); draw(rightanglemark(B,A,E)); draw(rightanglemark(A,B,C)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,E); label("$E$",E,NE); [/asy] $AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: \[\angle E = \angle AEC + \angle CED = 90 + 60 = \fbox{150}\]

AMC10 First Half

AMC10 B

AMC

AMC10

10B

N/A

A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?

100

\begin{align} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \fbox{100}. \end{align} Note: Revenue is a gain.

AMC10 First Half

AMC10 B

AMC

AMC12

12A

N/A

Let $S$ be a subset of $\{1,2,3,\dots,30\}$ with the property that no pair of distinct elements in $S$ has a sum divisible by $5$. What is the largest possible size of $S$?

13

Of the integers from $1$ to $30$, there are six each of $0,1,2,3,4\ (\text{mod}\ 5)$. We can create several rules to follow for the elements in subset $S$. No element can be $1\ (\text{mod}\ 5)$ if there is an element that is $4\ (\text{mod}\ 5)$. No element can be $2\ (\text{mod}\ 5)$ if there is an element that is $3\ (\text{mod}\ 5)$. Thus we can pick 6 elements from either $1\ (\text{mod}\ 5)$ or $4\ (\text{mod}\ 5)$ and 6 elements from either $2\ (\text{mod}\ 5)$ or $3\ (\text{mod}\ 5)$ for a total of $6+6=12$ elements. Considering $0\ (\text{mod}\ 5)$, there can be one element that is so because it will only be divisible by $5$ if paired with another element that is $0\ (\text{mod}\ 5)$. The final answer is $\fbox{13}$. == Solution 2 == ( Similar to solution 1 but more detailed) [Errors to correct below: We don't totally discard 0 (mod 5), since we can still have one of those. The $k$ and $x$ values are largest values, not quantities. So there are $6$ numbers in each category. We take 2 complete categories plus one from the $0$ category for a total of $13.$] Since we are taking the sum of distinct numbers mod(5), the only values that can be the remainder is $0$,$1$,$2$,$3$,$4$. We dismiss the case of remainder $0$ because this would make the sum a multiple of 5, which is not allowed. In addition, since we can't have 0 or 5 mod(5), we make some restraints. There are four different cases for the "allowed" mods in the sequence, or the remainders that the sum of distinct elements are allowed to be in the sequence. They are: 1) 1mod(5), and 3mod(5) 2) 1mod(5), and 2mod(5) 3) 2mod(5), and 4mod(5), 4) 2mod(5), and 1mod(5) Now, we can calculate the number of elements in each of these cases. For the first case, we rewrite 1mod(5) in the form 5k+1. Because 5k+1 < 30, which is the largest element in the sequence possible, we obtain k=5. In addition, for 3mod(5), we have 3x+5<30, x=5. This gives us 5+5=10 solutions. The reason why we can add these elements is because all of these elements, no matter which two you add, can never be a multiple of 5. The only possible remainders for the sum are 1+1, 1+3, or 3+3. In a similar manner, we calculate the element using the same logic. We find that 2mod(5) has 5 elements, and 1mod(5) has 8 elements, meaning that there are a total of 13 elements, which is the answer.

AMC12 Second Half

AMC12 A

HMMT

HMMT-Nov

guts

Nov

All positive integers whose binary representations (excluding leading zeroes) have at least as many 1 's as 0 's are put in increasing order. Compute the number of digits in the binary representation of the 200th number.

9

Solution: We do a rough estimation. There are 255 positive integers with at most 8 digits and a majority of them, but not more than 200, satisfy the property. Meanwhile, there are 511 positive integers with at most 9 digits, and a majority of them satisfy this property. Thus, the answer must be greater than 8 and at most 9 . $\fbox{9}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

Pascal has a triangle. In the $n$th row, there are $n+1$ numbers $a_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n}$ where $a_{n, 0}=a_{n, n}=1$. For all $1 \leq k \leq n-1, a_{n, k}=a_{n-1, k}-a_{n-1, k-1}$. Let $N$ be the value of the sum \[ \sum_{k=0}^{2018} \frac{\left|a_{2018, k}\right|}{\left(\begin{array}{c} 2018 \\ k \end{array}\right)} \] Estimate $N$. An estimate of $E>0$ earns $\left\lfloor 20 \cdot 2^{-|N-E| / 70}\right\rfloor$ points.

780.9280674537

A good estimate for this question is to use the fact that \[ \sum_{k=0}^{2018}\left|a_{2018, k}\right|=\frac{2^{2018}+2}{3} \] the answer to Guts 17. This suggests that each $\left|a_{2018, k}\right|$ is roughly $\frac{1}{3}$ of its corresponding entry $\left(\begin{array}{c}2018 \\ k\end{array}\right)$ in the usual Pascal's triangle, as the sum of the terms in the 2018th row of Pascal's triangle is $2^{2018}$. This then gives an estimate of $\frac{2018}{3}$, which earns 6 points. Code for computing answer in Python 3: $\fbox{780.9280674537}$.

HMMT Nov Guts

HMMT-Nov Guts

AIME

AIME

II

N/A

An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

180

By counting the number of green cubes $2$ different ways, we have $12a=20b$, or $a=\dfrac{5}{3} b$. Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$. We check a few small values of $a,b$ and solve for $c$, checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$, $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$, and $(a,b)=(15,9)$ gives $c=4$, with a volume of $540$. Any higher $(a,b)$ will $ab>180$, so therefore, the minimum volume is $\fbox{180}$.

Easy AIME Problems

AIME

AIME

AIME

II

N/A

A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

13

Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$, $1$, $2$, $3$, $4$, $5$, $6$, and $7$, in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$'s. In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$s. The number of ways for this to happen is $2 \cdot 6 = 12$ (the first card she draws has to be a $1$ (2 choices), while the second card can be any card but the remaining card with a $1$ (6 choices)). Each of these assignments is equally likely, so desired probability is $\frac{12}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \fbox{13}$.

Intermediate AIME Problems

AIME

HMMT

HMMT-Feb

comb

Feb

There are 800 marbles in a bag. Each marble is colored with one of 100 colors, and there are eight marbles of each color. Anna draws one marble at a time from the bag, without replacement, until she gets eight marbles of the same color, and then she immediately stops. Suppose Anna has not stopped after drawing 699 marbles. Compute the probability that she stops immediately after drawing the 700th marble.

99 / 101

Solution: In order to not stop after 699 marbles, the last 101 marbles must consist of 2 marbles of one color, and one marble from each other color. Since each of these marbles is equally likely to be the next to be drawn, and we stop after drawing the next marble as long as it's not one of the two of the same color, the desired probability is simply $\frac{99}{101}$. $\fbox{99 / 101}$.

HMMT Feb Easy

HMMT-Feb Combinatorics

HMMT

HMMT-Feb

geo

Feb

Points $X, Y$, and $Z$ lie on a circle with center $O$ such that $X Y=12$. Points $A$ and $B$ lie on segment $X Y$ such that $O A=A Z=Z B=B O=5$. Compute $A B$.

2 \sqrt{13}

Solution: Let the midpoint of $X Y$ be $M$. Because $O A Z B$ is a rhombus, $O Z \perp A B$, so $M$ is the midpoint of $A B$ as well. Since $O M=\frac{1}{2} O X, \triangle O M X$ is a $30-60-90$ triangle, and since $X M=6$, $O M=2 \sqrt{3}$. Since $O A=5$, the Pythagorean theorem gives $A M=\sqrt{13}$, so $A B=2 \sqrt{13}$. $\fbox{2 \sqrt{13}}$.

HMMT Feb Easy

HMMT-Feb Geometry

AMC

AMC12

12A

N/A

Square $ABCD$ has area $36,$ and $\overline{AB}$ is parallel to the x-axis. Vertices $A,$ $B$, and $C$ are on the graphs of $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x,$ respectively. What is $a?$

\sqrt [6]{3}

Notice that all of the graphs $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x$ have a domain of $(0, \infty]$. Also notice that $y = \log_{a}x$ is the furthest to the right, as adding coefficients in front of the $\log$ part only makes the graph steeper. Since $A$ is on the graph of $y = \log_{a}x$ and $B$ is on the graph of $y = 2\log_{a}x$, $\,$ $A$ must be to the right of $B$. We are told that $\overline{AB}$ is parallel to the x-axis. Let $A$ be the point $(x, y)$. Then the points $B$ and $C$ are $(x-6, y)$ and $(x-6, y+6)$ respectively. Substituting these coordinates into the equations given yields $y = \log_{a}x,$ $y = 2\log_{a}x-6,$ and $y+6 = 3\log_{a}x-6$. Rearranging a bit, we get the following equations: $1) a^y = x$ $2) a^y = (x-6)^2$ $3) a^{y+6} = (x-6)^3$ Using equations 1 and 2, we get $x=(x-6)^2$. Solving yields $x=9, x=-4$. However, $-4$ is extraneous so $x=9$ (also because we know $A$ has to have a positive $x$-coordinate). Using the second and third equations, we get $\frac{a^{y+6}}{a^y} = \frac{(x-6)^3}{(x-6)^2}$ $\Rightarrow$ $a^6 = x-6$. Plugging in $9$ for $x$ yields $a^6 = 3$, or $a= \sqrt[6]{3} \Rightarrow \fbox{\sqrt [6]{3}}$.

AMC12 Final Problems

AMC12 A

HMMT

HMMT-Nov

team

Nov

A function $f(x, y)$ is linear in $x$ and in $y . f(x, y)=\frac{1}{x y}$ for $x, y \in\{3,4\}$. What is $f(5,5)$ ?

\frac{1}{36}

The main fact that we will use in solving this problem is that $f(x+2, y)-f(x+1, y)=$ $f(x+1, y)-f(x, y)$ whenever $f$ is linear in $x$ and $y$. Suppose that $f(x, y)=a x y+b y+c x+d=$ $x(a y+c)+(b y+d)$ for some constants $a, b, c$, and $d$. Then it is easy to see that \[ \begin{aligned} f(x+2, y)-f(x+1, y) & =(x+2)(a y+c)+(b y+d)-(x+1)(a y+c)-(b y+d)=a y+c \\ f(x+1, y)-f(x, y) & =(x+1)(a y+c)+(b y+d)-x(a y+c)-(b y+d)=a y+c \end{aligned} \] which implies that $f(x+2, y)-f(x+1, y)=f(x+1, y)-f(x, y)$. In particular, $f(5, y)-f(4, y)=$ $f(4, y)-f(3, y)$, so $f(5, y)=2 f(4, y)-f(3, y)$. Similarly, $f(x, 5)=2 f(x, 4)-f(x, 3)$. Now we see that: \[ \begin{aligned} f(5,5) & =2 f(5,4)-f(5,3) \\ & =2[2 f(4,4)-f(3,4)]-[2 f(4,3)-f(3,3)] \\ & =4 f(4,4)-2 f(3,4)-2 f(4,3)+f(3,3) \\ & =\frac{4}{16}-\frac{4}{12}+\frac{1}{9} \\ & =\frac{1}{4}-\frac{1}{3}+\frac{1}{9} \\ & =\frac{1}{9}-\frac{1}{12} \\ & =\frac{1}{36} \end{aligned} \] so the answer is $\frac{1}{36}$. $\fbox{\frac{1}{36}}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC12

12B

N/A

A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?

2024

Let $x=e^{i\pi/6}$, a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is: \[1+2x+3x^2+\cdots+(k+1)x^k\] We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series. \begin{align} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=x+2x^2+3x^3+\cdots+2015x^{2015} \\ (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\ S &= \frac{1-x^{2015}}{(1-x)^2}-\frac{2015x^{2015}}{1-x} \end{align} We want to find $|S|$. First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$. Therefore \[S =\frac{1-\frac{1}{x}}{(1-x)^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.\] Hence, since $|x|=1$, we have $|S| = \frac{2016}{|1-x|}.$ Now we just have to find $|1-x|$. This can just be computed directly: \[1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i\] \[|1-x|^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2\] \[|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}\] Therefore $|S|=2016\cdot\frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008 \sqrt{6}$. Thus the answer is $1008+2+1008+6=\fbox{2024}$.

AMC12 Final Problems

AMC12 B

AMC

AMC8

8

N/A

The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner $\text{A}$, added to the number on Spinner $\text{B}$. What is the probability that $N$ is a perfect square number? [asy] //diagram by pog give me 1 billion dollars for this size(6cm); usepackage("mathptmx"); filldraw(arc((0,0), r=4, angle1=0, angle2=90)--(0,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((0,0), r=4, angle1=90, angle2=180)--(0,0)--cycle,lightgray); filldraw(arc((0,0), r=4, angle1=180, angle2=270)--(0,0)--cycle,mediumgray); filldraw(arc((0,0), r=4, angle1=270, angle2=360)--(0,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$5$", (-1.5,1.7)); label("$6$", (1.5,1.7)); label("$7$", (1.5,-1.7)); label("$8$", (-1.5,-1.7)); label("Spinner A", (0, -5.5)); filldraw(arc((12,0), r=4, angle1=0, angle2=90)--(12,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((12,0), r=4, angle1=90, angle2=180)--(12,0)--cycle,lightgray); filldraw(arc((12,0), r=4, angle1=180, angle2=270)--(12,0)--cycle,mediumgray); filldraw(arc((12,0), r=4, angle1=270, angle2=360)--(12,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$1$", (10.5,1.7)); label("$2$", (13.5,1.7)); label("$3$", (13.5,-1.7)); label("$4$", (10.5,-1.7)); label("Spinner B", (12, -5.5)); [/asy]

\dfrac{1}{8}

First, we calculate that there are a total of $4\cdot4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=\fbox{\dfrac{1}{8}}$ probability that the number formed by the two spinners is a perfect square.

AMC8 First Half

AMC8

AMC

AMC8

8

N/A

When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by $6$?

1

We have six cases: each different case, every one where a different number cannot be seen. The rolls that omit numbers one through five are all something times six: an example would be where the number you cannot see is one, so the product should be 2 x 3 x 4 x 5 x 6, and so product should be divisible by six. The roll that omits six on the other hand is 1 x 2 x 3 x 4 x 5, which has 2 x 3, also equal to six. We can see that all of them have a factor of 6 and therefore are divisible by six, so the solution should be E,1. $\fbox{1}$.

AMC8 First Half

AMC8

AMC

AMC10

10A

N/A

A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$?

\frac{1}{8}

The rectangle has a width of $4$ and a height of $1$. The area of this rectangle is $4\cdot1=4$. The line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$. The area which $x<y$ is the right isosceles triangle with side length $1$ that has vertices at $(0,0)$, $(1,1)$, and $(0,1)$. The area of this triangle is $\frac{1}{2}\cdot1^{2}=\frac{1}{2}$ Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \fbox{\frac{1}{8}}$

AMC10 Second Half

AMC10 A

AMC

AMC10

10B

N/A

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$

385

There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle: [asy]size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy] But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles. Thus the answer is $1330-945=385$ which is $\fbox{385}$.

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Feb

guts

Feb

Suppose $\triangle A B C$ has lengths $A B=5, B C=8$, and $C A=7$, and let $\omega$ be the circumcircle of $\triangle A B C$. Let $X$ be the second intersection of the external angle bisector of $\angle B$ with $\omega$, and let $Y$ be the foot of the perpendicular from $X$ to $B C$. Find the length of $Y C$.

\frac{13}{2}

Extend ray $\overrightarrow{A B}$ to a point $D$, Since $B X$ is an angle bisector, we have $\angle X B C=\angle X B D=180^{\circ}-$ $\angle X B A=\angle X C A$, so $X C=X A$ by the inscribed angle theorem. Now, construct a point $E$ on $B C$ so that $C E=A B$. Since $\angle B A X \cong \angle B C X$, we have $\triangle B A X \cong \triangle E C X$ by SAS congruence. Thus, $X B=X E$, so $Y$ bisects segment $B E$. Since $B E=B C-E C=8-5=3$, we have $Y C=E C+Y E=$ $5+\frac{1}{2} \cdot 3=\frac{13}{2}$. (Archimedes Broken Chord Thoerem). $\fbox{\frac{13}{2}}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

guts

Nov

Lily and Sarah are playing a game. They each choose a real number at random between -1 and 1 . They then add the squares of their numbers together. If the result is greater than or equal to 1, Lily wins, and if the result is less than 1, Sarah wins. What is the probability that Sarah wins?

\frac{\pi}{4}

If we let $x$ denote Lily's choice of number and $y$ denote Sarah's, then all possible outcomes are represented by the square with vertices $(-1,-1),(-1,1),(1,-1)$, and $(1,1)$. Sarah wins if $x^{2}+y^{2} \leq 1$, which is the area inside the unit circle. Since this has an area of $\pi$ and the entire square has an area of 4 , the probability that Sarah wins is $\frac{\pi}{4}$. $\fbox{\frac{\pi}{4}}$.

HMMT Nov Guts

HMMT-Nov Guts

AIME

AIME

I

N/A

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$ [asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

145

Lemma: If $x,y$ satisfy $px+qy=1$, then the minimal value of $\sqrt{x^2+y^2}$ is $\frac{1}{\sqrt{p^2+q^2}}$. Proof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}$. In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\frac{1}{\sqrt{p^2+q^2}}$. --- Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3}),$ and let $(a,0),(0,b)$ be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is $\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)$. This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$, i.e. $a,b$ must satisfy \[\frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,\] which can be simplified to \[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\] By the lemma, the minimal value of $\sqrt{a^2+b^2}$ is \[\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},\] so the minimal area of the equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\] and hence the answer is $75+3+67=\fbox{145}$.

Very Hard AIME Problems

AIME

AMC

AMC10

10A

N/A

What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\]where $x$ is a real number?

2018

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $\fbox{2018}$.

AMC10 Second Half

AMC10 A

AMC

AMC8

8

N/A

A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon? [asy] defaultpen(linewidth(0.7)); draw(polygon(3)); pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30); draw(D--E--F--cycle); [/asy]

1:1

The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is $\fbox{1:1}$

AMC8 First Half

AMC8

HMMT

HMMT-Feb

guts

Feb

Consider the graph in 3 -space of \[ 0=x y z(x+y)(y+z)(z+x)(x-y)(y-z)(z-x) \] This graph divides 3 -space into $N$ connected regions. What is $N$ ?

48

Note that reflecting for each choice of sign for $x, y, z$, we get new regions. Therefore, we can restrict to the case where $x, y, z>0$. In this case, the sign of the expression only depends on $(x-y)(y-z)(z-x)$. It is easy to see that for this expression, every one of the $3 !=6$ orderings for $\{x, y, z\}$ contributes a region. Therefore, our answer is $2^{3} \cdot 3 !=48$. $\fbox{48}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

guts

Nov

Given an angle $\theta$, consider the polynomial \[ P(x)=\sin (\theta) x^{2}+(\cos (\theta)+\tan (\theta)) x+1 \] Given that $P$ only has one real root, find all possible values of $\sin (\theta)$.

0, \frac{\sqrt{5}-1}{2}

Solution: Note that if $\sin (\theta)=0$, then the polynomial has 1 root. Now assume this is not the case then the polynomial is a quadratic in $x$. Factor the polynomial as $(\tan (\theta) x+1)(x+\sec (\theta))$. Then the condition is equivalent to $\sec (\theta)=\frac{1}{\tan (\theta)}$, which is equivalent to $\sin (\theta)=\cos ^{2}(\theta)=1-\sin ^{2}(\theta)$. Solving now gives $\sin (\theta)=\frac{\sqrt{5}-1}{2}$ as the only solution. $\fbox{0, \frac{\sqrt{5}-1}{2}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

The skeletal structure of circumcircumcircumcoronene, a hydrocarbon with the chemical formula $\mathrm{C}_{150} \mathrm{H}_{30}$, is shown below. Each line segment between two atoms is at least a single bond. However, since each carbon $(\mathrm{C})$ requires exactly four bonds connected to it and each hydrogen $(\mathrm{H})$ requires exactly one bond, some of the line segments are actually double bonds. How many arrangements of single/double bonds are there such that the above requirements are satisfied? If the correct answer is $C$ and your answer is $A$, you get $\left.\max \left(\left\lvert\, 30\left(1-\left|\log _{\log _{2} C} \frac{A}{C}\right|\right)\right.\right\rfloor, 0\right)$ points.

267227532

The problem is equivalent to the one in OEIS A008793, a.k.a. "number of ways to tile hexagon of edge $\mathrm{n}$ with diamonds of side 1." Notice that there is a bjiection between such a tiling and the number of ways to stack some unit cubes alongside a corner of an $n \times n \times n$ box (see the Art of Problem Solving logo as an example, also known as 3 -dimensional Young diagrams), where this problem $n=5$. It is known that there are $\left(\begin{array}{c}2 n \\ n\end{array}\right)=252$ ways to stack one layer (since each way correspond a way to walk from a corner of a 5 by 5 grid to the opposite one), so $\frac{252^{5}}{5 !} \approx 8 \times 10^{9}$ gives a somewhat loose upper bound (generate five layers and sort them by size, and hope that it will be a valid stack in general). This result can be improved by dividing out a reasonable constant factor after considering the probability that sorting by size indeed gives a valid stack (for example, it would be fair to guess that there is about a $\frac{1}{4 !}$ chance that the first row of each layer will be in the right order, given that each row has a small role in determining the final size of the layer; dividing 24 from the previous result gives a very close guess). In general, a guess anywhere between $10^{8}$ and $10^{9}$ is a fair guess. $\fbox{267227532}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

geo

Feb

Let $A B C$ be a triangle, and let $B C D E, C A F G, A B H I$ be squares that do not overlap the triangle with centers $X, Y, Z$ respectively. Given that $A X=6, B Y=7$, and $C Z=8$, find the area of triangle $X Y Z$.

\frac{21 \sqrt{15}}{4}

By the degenerate case of Von Aubel's Theorem we have that $Y Z=A X=6$ and $Z X=B Y=7$ and $X Y=C Z=8$ so it suffices to find the area of a $6-7-8$ triangle which is given by $\frac{21 \sqrt{15}}{4}$. To prove that $A X=Y Z$, note that by $\mathrm{LoC}$ we get \[ Y X^{2}=\frac{b^{2}}{2}+\frac{c^{2}}{2}+b c \sin \angle A \] and \[ \begin{aligned} A X^{2} & =b^{2}+\frac{a^{2}}{2}-a b(\cos \angle C-\sin \angle C) \\ & =c^{2}+\frac{a^{2}}{2}-a c(\cos \angle B-\sin \angle B) \\ & =\frac{b^{2}+c^{2}+a(b \sin \angle C+c \sin \angle B)}{2} \\ & =\frac{b^{2}}{2}+\frac{c^{2}}{2}+a h \end{aligned} \] where $h$ is the length of the $A$-altitude of triangle $A B C$. In these calculations we used the well-known fact that $b \cos \angle C+c \cos \angle B=a$ which can be easily seen by drawing in the $A$-altitude. Then since $b c \sin \angle A$ and $a h$ both equal twice the area of triangle $A B C$, we are done. $\fbox{\frac{21 \sqrt{15}}{4}}$.

HMMT Feb Hard

HMMT-Feb Geometry

AMC

AMC10

10B

N/A

Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$, then the average value (arithmetic mean) of the integers remaining is $32$. If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$. If the greatest integer is then returned to the set, the average value of the integers rises to $40$. The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$. What is the average value of all the integers in the set $S$?

36.8

Let the lowest value be $L$ and the highest $G$, and let the sum be $Z$ and the amount of numbers $n$. We have $\frac{Z-G}{n-1}=32$, $\frac{Z-L-G}{n-2}=35$, $\frac{Z-L}{n-1}=40$, and $G=L+72$. Clearing denominators gives $Z-G=32n-32$, $Z-L-G=35n-70$, and $Z-L=40n-40$. We use $G=L+72$ to turn the first equation into $Z-L=32n+40$. Since $Z-L=40(n-1)$ we substitute it into the equation which gives $n=10$. Turning the second into $Z-2L=35n+2$ using $G=L+72$ we see $L=8$ and $Z=368$ so the average is $\frac{Z}{n}=\fbox{36.8}$

AMC10 Second Half

AMC10 B

AMC

AMC8

8

N/A

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price? [asy] import graph; size(16.38cm); real lsf=2; pathpen=linewidth(0.7); pointpen=black; pen fp = fontsize(10); pointfontpen=fp; real xmin=-1.33,xmax=11.05,ymin=-9.01,ymax=-0.44; pen ycycyc=rgb(0.55,0.55,0.55); pair A=(1,-6), B=(1,-2), D=(1,-5.8), E=(1,-5.6), F=(1,-5.4), G=(1,-5.2), H=(1,-5), J=(1,-4.8), K=(1,-4.6), L=(1,-4.4), M=(1,-4.2), N=(1,-4), P=(1,-3.8), Q=(1,-3.6), R=(1,-3.4), S=(1,-3.2), T=(1,-3), U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,-6), P_1=(14,-5), Q_1=(14,-4), R_1=(14,-3), S_1=(14,-2), C_1=(1.4,-6), D_1=(1.8,-6), G_1=(2.4,-6), H_1=(2.8,-6), I_1=(3.4,-6), J_1=(3.8,-6), K_1=(4.4,-6), L_1=(4.8,-6), M_1=(5.4,-6), N_1=(5.8,-6), T_1=(6.4,-6), U_1=(6.8,-6), V_1=(7.4,-6), W_1=(7.8,-6), Z_1=(8.4,-6), A_2=(8.8,-6), B_2=(9.4,-6), C_2=(9.8,-6), D_2=(10.4,-6), E_2=(10.8,-6), L_2=(2.4,-3.2), M_2=(2.8,-3.2), N_2=(3.4,-4), O_2=(3.8,-4), P_2=(4.4,-3.6), Q_2=(4.8,-3.6), R_2=(5.4,-3.6), S_2=(5.8,-3.6), T_2=(6.4,-3.4), U_2=(6.8,-3.4), V_2=(7.4,-3.8), W_2=(7.8,-3.8), Z_2=(8.4,-2.8), A_3=(8.8,-2.8), B_3=(9.4,-3.2), C_3=(9.8,-3.2), D_3=(10.4,-3.8), E_3=(10.8,-3.8); filldraw(C_1--E_1--F_1--D_1--cycle,ycycyc); filldraw(G_1--L_2--M_2--H_1--cycle,ycycyc); filldraw(I_1--N_2--O_2--J_1--cycle,ycycyc); filldraw(K_1--P_2--Q_2--L_1--cycle,ycycyc); filldraw(M_1--R_2--S_2--N_1--cycle,ycycyc); filldraw(T_1--T_2--U_2--U_1--cycle,ycycyc); filldraw(V_1--V_2--W_2--W_1--cycle,ycycyc); filldraw(Z_1--Z_2--A_3--A_2--cycle,ycycyc); filldraw(B_2--B_3--C_3--C_2--cycle,ycycyc); filldraw(D_2--D_3--E_3--E_2--cycle,ycycyc); D(B--A,linewidth(0.4)); D(H--(8,-5),linewidth(0.4)); D(N--(8,-4),linewidth(0.4)); D(T--(8,-3),linewidth(0.4)); D(B--(8,-2),linewidth(0.4)); D(B--S_1); D(T--R_1); D(N--Q_1); D(H--P_1); D(A--O_1); D(C_1--E_1); D(E_1--F_1); D(F_1--D_1); D(D_1--C_1); D(G_1--L_2); D(L_2--M_2); D(M_2--H_1); D(H_1--G_1); D(I_1--N_2); D(N_2--O_2); D(O_2--J_1); D(J_1--I_1); D(K_1--P_2); D(P_2--Q_2); D(Q_2--L_1); D(L_1--K_1); D(M_1--R_2); D(R_2--S_2); D(S_2--N_1); D(N_1--M_1); D(T_1--T_2); D(T_2--U_2); D(U_2--U_1); D(U_1--T_1); D(V_1--V_2); D(V_2--W_2); D(W_2--W_1); D(W_1--V_1); D(Z_1--Z_2); D(Z_2--A_3); D(A_3--A_2); D(A_2--Z_1); D(B_2--B_3); D(B_3--C_3); D(C_3--C_2); D(C_2--B_2); D(D_2--D_3); D(D_3--E_3); D(E_3--E_2); D(E_2--D_2); label("0",(0.88,-5.91),SE*lsf,fp); label(" 5",(0.3,-4.84),SE*lsf,fp); label(" 10",(0.2,-3.84),SE*lsf,fp); label(" 15",(0.2,-2.85),SE*lsf,fp); label(" 20",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95),SE*lsf,fp); label("$3$",(3.53,-5.94),SE*lsf,fp); label("$4$",(4.55,-5.94),SE*lsf,fp); label("$5$",(5.49,-5.95),SE*lsf,fp); label("$6$",(6.53,-5.95),SE*lsf,fp); label("$7$",(7.55,-5.95),SE*lsf,fp); label("$8$",(8.52,-5.95),SE*lsf,fp); label("$9$",(9.57,-5.97),SE*lsf,fp); label("$10$",(10.56,-5.94),SE*lsf,fp); label("Month",(7.14,-6.43),SE*lsf,fp); D(A,linewidth(1pt)); D(B,linewidth(1pt)); D(D,linewidth(1pt)); D(E,linewidth(1pt)); D(F,linewidth(1pt)); D(G,linewidth(1pt)); D(H,linewidth(1pt)); D(J,linewidth(1pt)); D(K,linewidth(1pt)); D(L,linewidth(1pt)); D(M,linewidth(1pt)); D(N,linewidth(1pt)); D(P,linewidth(1pt)); D(Q,linewidth(1pt)); D(R,linewidth(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth(1pt)); D(F_1,linewidth(1pt)); D(O_1,linewidth(1pt)); D(P_1,linewidth(1pt)); D(Q_1,linewidth(1pt)); D(R_1,linewidth(1pt)); D(S_1,linewidth(1pt)); D(C_1,linewidth(1pt)); D(D_1,linewidth(1pt)); D(G_1,linewidth(1pt)); D(H_1,linewidth(1pt)); D(I_1,linewidth(1pt)); D(J_1,linewidth(1pt)); D(K_1,linewidth(1pt)); D(L_1,linewidth(1pt)); D(M_1,linewidth(1pt)); D(N_1,linewidth(1pt)); D(T_1,linewidth(1pt)); D(U_1,linewidth(1pt)); D(V_1,linewidth(1pt)); D(W_1,linewidth(1pt)); D(Z_1,linewidth(1pt)); D(A_2,linewidth(1pt)); D(B_2,linewidth(1pt)); D(C_2,linewidth(1pt)); D(D_2,linewidth(1pt)); D(E_2,linewidth(1pt)); D(L_2,linewidth(1pt)); D(M_2,linewidth(1pt)); D(N_2,linewidth(1pt)); D(O_2,linewidth(1pt)); D(P_2,linewidth(1pt)); D(Q_2,linewidth(1pt)); D(R_2,linewidth(1pt)); D(S_2,linewidth(1pt)); D(T_2,linewidth(1pt)); D(U_2,linewidth(1pt)); D(V_2,linewidth(1pt)); D(W_2,linewidth(1pt)); D(Z_2,linewidth(1pt)); D(A_3,linewidth(1pt)); D(B_3,linewidth(1pt)); D(C_3,linewidth(1pt)); D(D_3,linewidth(1pt)); D(E_3,linewidth(1pt)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]

70

The highest price was in Month 1, which was $17. The lowest price was in Month 3, which was $10. 17 is $\frac{17}{10}\cdot100=170\%$ of 10, and is $170-100=70\%$ more than 10. Therefore, the answer is $\fbox{70}$

AMC8 First Half

AMC8

HMMT

HMMT-Nov

gen

Nov

Two circles $\omega$ and $\gamma$ have radii 3 and 4 respectively, and their centers are 10 units apart. Let $x$ be the shortest possible distance between a point on $\omega$ and a point on $\gamma$, and let $y$ be the longest possible distance between a point on $\omega$ and a point on $\gamma$. Find the product $x y$.

51

Let $\ell$ be the line connecting the centers of $\omega$ and $\gamma$. Let $A$ and $B$ be the intersections of $\ell$ with $\omega$, and let $C$ and $D$ be the intersections of $\ell$ with $\gamma$, so that $A, B, C$, and $D$ are collinear, in that order. The shortest distance between a point on $\omega$ and a point on $\gamma$ is $B C=3$. The longest distance is $A D=3+10+4=17$. The product is 51 . $\fbox{51}$.

HMMT Nov Easy

HMMT-Nov General

AMC

AMC8

8

N/A

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? [asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]

8

You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns: +−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\fbox{8}$.

AMC8 Second Half

AMC8

HMMT

HMMT-Nov

guts

Nov

Find the sum of all positive integers $n \leq 2015$ that can be expressed in the form $\left\lceil\frac{x}{2}\right\rceil+y+x y$, where $x$ and $y$ are positive integers.

2029906

Lemma: $n$ is expressible as $\left\lceil\frac{x}{2}\right\rceil+y+x y$ iff $2 n+1$ is not a Fermat Prime. Proof: Suppose $n$ is expressible. If $x=2 k$, then $2 n+1=(2 k+1)(2 y+1)$, and if $x=2 k-1$, then $n=k(2 y+1)$. Thus, if $2 n+1$ isn't prime, we can factor $2 n+1$ as the product of two odd integers $2 x+1,2 y+1$ both greater than 1, resulting in positive integer values for $x$ and $y$. Also, if $n$ has an odd factor greater than 1 , then we factor out its largest odd factor as $2 y+1$, giving a positive integer value for $x$ and $y$. Thus $n$ is expressible iff $2 n+1$ is not prime or $n$ is not a power of 2 . That leaves only the $n$ such that $2 n+1$ is a prime one more than a power of two. These are well-known, and are called the Fermat primes. It's a well-known fact that the only Fermat primes $\leq 2015$ are $3,5,17,257$, which correspond to $n=1,2,8,128$. Thus the sum of all expressible numbers is $\frac{2015 \cdot 2016}{2}-(1+2+8+128)=2029906$. $\fbox{2029906}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10A

N/A

The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

7

Let the three numbers be equal to $a$, $b$, and $c$. We can now write three equations: $a+b=12$ $b+c=17$ $a+c=19$ Adding these equations together, we get that $2(a+b+c)=48$ and $a+b+c=24$ Substituting the original equations into this one, we find $c+12=24$ $a+17=24$ $b+19=24$ Therefore, our numbers are 12, 7, and 5. The middle number is $\fbox{7}$

AMC10 First Half

AMC10 A

HMMT

HMMT-Feb

guts

Feb

Danny has a set of 15 pool balls, numbered $1,2, \ldots, 15$. In how many ways can he put the balls in 8 indistinguishable bins such that the sum of the numbers of the balls in each bin is 14,15 , or $16 ?$

122

Clearly, the balls numbered $15,14, \ldots, 9,8$ must be placed in separate bins, so we number the bins $15,14, \ldots, 9,8$. Note that bins 15 and 14 may contain only one ball while all other bins must contain at least two balls. We have two cases to examine. Case 1: Only one bin contains exactly one ball. Let $a_{i}$ denote the number of ways to place the balls numbered $1,2, \ldots, i-1$ into the bins numbered $15,14, \ldots, 15-i+1$. We can place either $i-1$ or $i-2$ into the bin numbered $15-i+1$. If we place $i-1$ in there, then there are $a_{i-1}$ ways to finish packing the rest. If we place $i-2$ in this bin, then $i-1$ must be placed in the bin numbered $15-i+2$, so there are $a_{i-2}$ ways to place the rest of the balls. Therefore $a_{i}=a_{i-1}+a_{i-2}$. Since $a_{1}=2$ and $a_{2}=3$, the sequence $\left\{a_{i}\right\}$ is the Fibonacci sequence, and $a_{7}=34$. Case 2: Both bins 14 and 15 contain only one ball. A pair of balls from 1-7 must be put together to one of the bins numbered 8 through 13. This pair has sum at most 8 , so we can count for all the cases. \begin{center} \begin{tabular}{c|c} Balls & Number of packings \\ \hline 1,2 & 16 \\ \hline 1,3 & 10 \\ \hline 1,4 & 12 \\ \hline 1,5 & 12 \\ \hline 1,6 & 10 \\ \hline 1,7 & 8 \\ \hline 2,3 & 6 \\ \hline 2,4 & 4 \\ \hline 2,5 & 4 \\ \hline 2,6 & 3 \\ \hline 3,4 & 2 \\ \hline 3,5 & 1 \\ \hline \end{tabular} \end{center} Therefore, there are 88 possibilities in this case, and the total number of possibilities is 122 . $\fbox{122}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Feb

team

Feb

Nine distinct positive integers summing to 74 are put into a $3 \times 3$ grid. Simultaneously, the number in each cell is replaced with the sum of the numbers in its adjacent cells. (Two cells are adjacent if they share an edge.) After this, exactly four of the numbers in the grid are 23. Determine, with proof, all possible numbers that could have been originally in the center of the grid.

18

Solution: Suppose the initial grid is of the format shown below: \[ \left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \] After the transformation, we end with \[ \left[\begin{array}{ccc} a_{n} & b_{n} & c_{n} \\ d_{n} & e_{n} & f_{n} \\ g_{n} & h_{n} & i_{n} \end{array}\right]=\left[\begin{array}{ccc} b+d & a+c+e & b+f \\ a+e+g & b+d+f+h & c+e+i \\ d+h & g+e+i & f+h \end{array}\right] \] Since $d \neq f, a_{n}=b+d \neq b+f=c_{n}$. By symmetry, no two corners on the same side of the grid may both be 23 after the transformation. Since $c \neq g, b_{n}=a+c+e \neq a+e+g=d_{n}$. By symmetry, no two central-edge squares sharing a corner may both be 23 after the transformation. Assume for the sake of contradiction that $e_{n}=23$. Because $a_{n}, c_{n}, g_{n}, i_{n}<e_{n}$, none of $a_{n}, c_{n}, g_{n}, i_{n}$ can be equal to 23 . Thus, 3 of $b_{n}, d_{n}, f_{n}, h_{n}$ must be 23 . WLOG assume $b_{n}=d_{n}=f_{n}=23$. Thus is a contradiction however, as $b_{n} \neq d_{n}$. Thus, $e_{n} \neq 23$. This leaves the case with two corners diametrically opposite and two central edge squares diametrically opposite being 23. WLOG assume $a_{n}=b_{n}=h_{n}=i_{n}=23$. Thus, $92=4 \cdot 23=a_{n}+b_{n}+h_{n}+i_{n}=(b+d)+(a+c+e)+(e+g+i)+(f+h)=(a+b+c+d+$ $e+f+g+h+i)+e$. Since $a+b+c+d+e+f+g+h+i=74$, this means that $e=92-74=18$. One possible example of 18 working is $\left[\begin{array}{lll}4 & 16 & 2 \\ 6 & 18 & 7 \\ 1 & 17 & 3\end{array}\right]$. Thus the only possible value for the center is 18. $\fbox{18}$.

HMMT Feb Team

HMMT-Feb Team

HMMT

HMMT-Feb

gen

Feb

Let $H$ be a regular hexagon of side length $x$. Call a hexagon in the same plane a "distortion" of $H$ if and only if it can be obtained from $H$ by translating each vertex of $H$ by a distance strictly less than 1. Determine the smallest value of $x$ for which every distortion of $H$ is necessarily convex.

4

Let $H=A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be the hexagon, and for all $1 \leq i \leq 6$, let points $A_{i}^{\prime}$ be considered such that $A_{i} A_{i}^{\prime}<1$. Let $H^{\prime}=A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime} A_{5}^{\prime} A_{6}^{\prime}$, and consider all indices modulo 6 . For any point $P$ in the plane, let $D(P)$ denote the unit disk $\{Q \mid P Q<1\}$ centered at $P$; it follows that $A_{i}^{\prime} \in D\left(A_{i}\right)$. Let $X$ and $X^{\prime}$ be points on line $A_{1} A_{6}$, and let $Y$ and $Y^{\prime}$ be points on line $A_{3} A_{4}$ such that $A_{1} X=$ $A_{1} X^{\prime}=A_{3} Y=A_{3} Y^{\prime}=1$ and $X$ and $X^{\prime}$ lie on opposite sides of $A_{1}$ and $Y$ and $Y^{\prime}$ lie on opposite sides of $A_{3}$. If $X^{\prime}$ and $Y^{\prime}$ lie on segments $A_{1} A_{6}$ and $A_{3} A_{4}$, respectively, then segment $A_{1}^{\prime} A_{3}^{\prime}$ lies between the lines $X Y$ and $X^{\prime} Y^{\prime}$. Note that $\frac{x}{2}$ is the distance from $A_{2}$ to $A_{1} A_{3}$. If $\frac{x}{2} \geq 2$, then $C\left(A_{2}\right)$ cannot intersect line $X Y$, since the distance from $X Y$ to $A_{1} A_{3}$ is 1 and the distance from $X Y$ to $A_{2}$ is at least 1. Therefore, $A_{1}^{\prime} A_{3}^{\prime}$ separates $A_{2}^{\prime}$ from the other 3 vertices of the hexagon. By analogous reasoning applied to the other vertices, we may conclude that $H^{\prime}$ is convex. If $\frac{x}{2}<2$, then $C\left(A_{2}\right)$ intersects $X Y$, so by choosing $A_{1}^{\prime}=X$ and $A_{3}^{\prime}=Y$, we see that we may choose $A_{2}^{\prime}$ on the opposite side of $X Y$, in which case $H^{\prime}$ will be concave. Hence the answer is 4 , as desired. $\fbox{4}$.

HMMT Feb Guts

HMMT-Feb General

AMC

AMC12

12B

N/A

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

\frac{\sqrt5}{10}

Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle{NPM}=90^{\circ}$ due to Thales's Theorem. [asy] /* Made by QIDb602; edited by MRENTHUSIASM */ size(180); pair A, B, C, D, M, N, O, P; O = origin; A = (-1/2,-1/2); B = (-1/2,1/2); C = (1/2,1/2); D = (1/2,-1/2); M = midpoint(C--D); N = midpoint(A--B); path p; p = Circle(O,1/2); P = intersectionpoints(A--M,p)[0]; fill(N--A--M--cycle,yellow); dot("$\omega$",O,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); [/asy] Using the Pythagorean theorem, $AM=\frac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $\triangle APN\sim\triangle ANM:$ \begin{align} \frac{AP}{AN}&=\frac{AN}{AM} \\ \frac{AP}{1/2}&=\frac{1/2}{\sqrt5/2} \\ AP&=\fbox{\frac{\sqrt5}{10}}. \end{align}

AMC12 First Half

AMC12 B

HMMT

HMMT-Nov

guts

Nov

Points $A, B, C$, and $D$ lie on a line in that order such that $\frac{A B}{B C}=\frac{D A}{C D}$. If $A C=3$ and $B D=4$, find $A D$.

6

Let $B C=x$, then the equation becomes $\frac{3-x}{x}=\frac{7-x}{4-x}$. This simplifies to a quadratic equation with solutions $x=1$ and $x=6$. Since $x<3$, we have $x=1$ and $A D=7-x=6$. $\fbox{6}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

Let $T$ be the set of numbers of the form $2^{a} 3^{b}$ where $a$ and $b$ are integers satisfying $0 \leq a, b \leq 5$. How many subsets $S$ of $T$ have the property that if $n$ is in $S$ then all positive integer divisors of $n$ are in $S$ ?

924

Consider the correspondence $(a, b) \leftrightarrow 2^{a} 3^{b}$ for non-negative integers $a$ and $b$. So we can view $T$ as the square of lattice points $(a, b)$ where $0 \leq a, b \leq 5$, and subsets of $T$ as subsets of this square. Notice then that the integer corresponding to $\left(a_{1}, b_{1}\right)$ is a divisor of the integer corresponding to $\left(a_{2}, b_{2}\right)$ if and only if $0 \leq a_{1} \leq a_{1}$ and $0 \leq b_{1} \leq b_{2}$. This means that subsets $S \subset T$ with the desired property,\\ correspond to subsets of the square where if a point is in the set, then so are all points to the left and south of it. Consider any such subset $S$. For each $0 \leq x \leq 5$, let $S_{x}$ be the maximum $y$ value of any point $(x, y) \in S$, or -1 if there is no such point. We claim the values $S_{x}$ uniquely characterize $S$. This is because each $S_{x}$ characterizes the points of the form $(x, y)$ in $S$. In particular, $(x, z)$ will be in $S$ if and only if $z \leq S_{x}$. If $(x, z) \in S$ with $z>S_{x}$, then $S_{x}$ is not the maximum value, and if $(x, z) \notin S$ with $z \leq S_{x}$, then $S$ fails to satisfy the desired property. We now claim that $S_{x} \geq S_{y}$ for $x<y$, so the sequence $S_{0}, \ldots, S_{1}$ is decreasing. This is because if $\left(y, S_{y}\right)$ is in the set $S$, then so must be $\left(x, S_{y}\right)$. Conversely, it is easy to see that if $S_{0}, \ldots, S_{1}$ is decreasing, then $S$ is a set satisfying the desired property. We now claim that decreasing sequences $S_{0}, \ldots, S_{5}$ are in bijective correspondence with walks going only right and down from $(-1,5)$ to $(5,-1)$. The sequence $S_{0}, \ldots, S_{5}$ simply corresponds to the walk $(-1,5) \rightarrow\left(-1, S_{0}\right) \rightarrow\left(0, S_{0}\right) \rightarrow\left(0, S_{1}\right) \rightarrow\left(1, S_{1}\right) \rightarrow \cdots \rightarrow\left(4, S_{5}\right) \rightarrow\left(5, S_{5}\right) \rightarrow(5,-1)$. Geometrically, we are tracing out the outline of the set $S$. The number of such walks is simply $\left(\begin{array}{c}12 \\ 6\end{array}\right)$, since we can view it as choosing the 6 of 12 steps at which to move right. Thus the number of subsets $S$ of $T$ with the desired property is $\left(\begin{array}{c}12 \\ 6\end{array}\right)=924$. $\fbox{924}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Feb

guts

Feb

A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots x_{8}$ is square-free if $i_{1}, i_{2}, \ldots i_{k}$ are distinct. (A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product? \[ \prod_{1 \leq i<j \leq 8}\left(1+x_{i} x_{j}\right) \]

764

Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+$ $x_{i} x_{j}$ ). Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$. We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both 1 . Then $a_{2}=2, a_{3}=4, a_{4}=10$, $a_{5}=26, a_{6}=76, a_{7}=232$, and finally, $a_{8}=764$. $\fbox{764}$.

HMMT Feb Guts

HMMT-Feb Guts

HMMT

HMMT-Nov

gen

Nov

$A B C D$ is a parallelogram satisfying $A B=7, B C=2$, and $\angle D A B=120^{\circ}$. Parallelogram $E C F A$ is contained in $A B C D$ and is similar to it. Find the ratio of the area of $E C F A$ to the area of $A B C D$.

\frac{39}{67}

First, note that $B D$ is the long diagonal of $A B C D$, and $A C$ is the long diagonal of $E C F A$. Because the ratio of the areas of similar figures is equal to the square of the ratio of their side lengths, we know that the ratio of the area of $E C F A$ to the area of $A B C D$ is equal to the ratio $\frac{A C^{2}}{B D^{2}}$. Using law of cosines on triangle $A B D$, we have $B D^{2}=A D^{2}+A B^{2}-2(A D)(A B) \cos \left(120^{\circ}\right)=2^{2}+$ $7^{2}-2(2)(7)\left(-\frac{1}{2}\right)=67$ Using law of cosines on triangle $A B C$, we have $A C^{2}=A B^{2}+B C^{2}-2(A B)(B C) \cos \left(60^{\circ}\right)=7^{2}+2^{2}-$ $2(7)(2)\left(\frac{1}{2}\right)=39$. Finally, $\frac{A C^{2}}{B D^{2}}=\frac{39}{67}$. $\fbox{\frac{39}{67}}$.

HMMT Nov Hard

HMMT-Nov General

HMMT

HMMT-Nov

gen

Nov

Call a positive integer $n$ quixotic if the value of \[ \operatorname{lcm}(1,2,3, \ldots, n) \cdot\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right) \] is divisible by 45 . Compute the tenth smallest quixotic integer.

573

Solution: Let $L=\operatorname{lcm}(1,2,3, \ldots, n)$, and let $E=L\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)$ denote the expression. In order for $n$ to be quixotic, we need $E \equiv 0(\bmod 5)$ and $E \equiv 0(\bmod 9)$. We consider these two conditions separately. Claim: $E \equiv 0(\bmod 5)$ if and only if $n \in\left[4 \cdot 5^{k}, 5^{k+1}\right)$ for some nonnegative integer $k$. Proof. Let $k=\left\lfloor\log _{5} n\right\rfloor$, which is equal to $\nu_{5}(L)$. In order for $E$ to be divisible by 5 , all terms in $\frac{L}{1}, \frac{L}{2}, \ldots, \frac{L}{n}$ that aren't multiples of 5 must sum to a multiple of 5 . The potential terms that are not going to be multiples of 5 are $L / 5^{k}, L /\left(2 \cdot 5^{k}\right), L /\left(3 \cdot 5^{k}\right)$, and $L /\left(4 \cdot 5^{k}\right)$, depending on the value of $n$. \begin{itemize} \item If $n \in\left[5^{k}, 2 \cdot 5^{k}\right)$, then only $L / 5^{k}$ appears. Thus, the sum is $L / 5^{k}$, which is not a multiple of 5 . \item If $n \in\left[2 \cdot 5^{k}, 3 \cdot 5^{k}\right)$, then only $L / 5^{k}$ and $L /\left(2 \cdot 5^{k}\right)$ appear. The sum is $3 L /\left(2 \cdot 5^{k}\right)$, which is not a multiple of 5 . \item If $n \in\left[3 \cdot 5^{k}, 4 \cdot 5^{k}\right)$, then only $L / 5^{k}, L /\left(2 \cdot 5^{k}\right)$, and $L /\left(3 \cdot 5^{k}\right)$ appear. The sum is $11 L /\left(6 \cdot 5^{k}\right)$, which is not a multiple of 5 . \item If $n \in\left[4 \cdot 5^{k}, 5^{k+1}\right)$, then $L / 5^{k}, L /\left(2 \cdot 5^{k}\right), L /\left(3 \cdot 5^{k}\right)$, and $L /\left(4 \cdot 5^{k}\right)$ all appear. The sum is $25 L /\left(12 \cdot 5^{k}\right)$, which is a multiple of 5 . Thus, this case works. \end{itemize} Only the last case works, implying the claim. Claim: $E \equiv 0(\bmod 9)$ if and only if $n \in\left[7 \cdot 3^{k-1}, 8 \cdot 3^{k-1}\right)$ for some positive integer $k$. Proof. This is a repeat of the previous proof, so we will only sketch it. Let $k=\left\lfloor\log _{3} n\right\rfloor$, which is equal to $\nu_{3}(L)$. This time, the terms we need to consider are those that are not multiples of 9 , which are \[ \frac{L}{3^{k-1}}, \frac{L}{2 \cdot 3^{k-1}}, \cdots, \frac{L}{8 \cdot 3^{k-1}} \] Similar to the above, we need to check that the sum of the first $j$ terms is divisible by 9 if and only if $j=7$. There are 8 cases, but we could reduce workload by showing first that it is divisible by 3 if and only if $j \in\{6,7,8\}$ (there are only $L / 3^{k}$ and $L /\left(2 \cdot 3^{k}\right)$ to consider), then eliminate 6 and 8 by using $(\bmod 9)$. Doing a little bit of arithmetic, we'll get the first 10 quixotic numbers: $21,22,23,567,568,569,570$, $571,572,573$. $\fbox{573}$.

HMMT Nov Hard

HMMT-Nov General

AMC

AMC12

12B

N/A

A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?

20\sqrt{2}

Let the side lengths of the rectangular box be $x, y$ and $z$. From the information we get \[4(x+y+z) = 48 \Rightarrow x+y+z = 12\] \[2(xy+yz+xz) = 94\] The sum of all the lengths of the box's interior diagonals is \[4 \sqrt{x^2+y^2+z^2}\] Squaring the first expression, we get: \[144 =(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+xz)\] \[144 = x^2+y^2+z^2 + 94\] Hence \[x^2+y^2+z^2 = 50\] \[4 \sqrt{x^2+y^2+z^2} = \fbox{20\sqrt{2}}\]

AMC12 Second Half

AMC12 B

HMMT

HMMT-Nov

guts

Nov

Find the number of positive integers less than 1000000 which are less than or equal to the sum of their proper divisors. If your answer is $X$ and the actual value is $Y$, your score will be $\max \left(0,20-80\left|1-\frac{X}{Y}\right|\right)$ rounded to the nearest integer.

247548

$\mathrm{N}=1000000$ $\mathrm{s}=[0] * \mathrm{~N}$ ans $=0$ for $i$ in range $(1, N)$ : if $i<=s[i]$ : ans $+=1$ for $j$ in range( $i+i, N, i)$ : $s[j]+=i$ print(ans) $\fbox{247548}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC12

12B

N/A

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

\; \frac {17}{28}

This solution is pretty similar to the above. So there are 4 spots remaining, and it would be hard to count all the combinations where Auntie Em could park, but you can count all the combinations where Auntie Em can't park. Since the 12 cars are indistinguishable, we can use distributions here. There must be $a_1 \geq 0$ cars, then an empty spot, then $d_2 \geq 1$ cars, then an empty spot, then $d_3 \geq 1$ cars, then an empty spot, then $d_4 \geq 1$ cars, then an empty spot, then $a_5 \geq 0$ cars. \[a_1 + d_2 + d_3 + d_4 + a_5 = 12\] To remove all the restrictions, let $a_n = d_n - 1$. \[a_1 + a_2 + a_3 + a_4 + a_5 = 9\] We can now use stars and bars on these values (9 "stars", and 5 - 1 = 4 dividers), to get $\binom{13}{4}$ possibilities where she can't park. There are $\binom{16}{4}$ possibilities in total. \[\frac{\binom{13}{4}}{\binom{16}{4}} = \frac{11}{28}\] Subtracting that from 1 to get the probability she can park, the correct answer is $\fbox{\; \frac {17}{28}}$.

AMC12 Final Problems

AMC12 B

AMC

AMC10

10A

N/A

Pablo buys popsicles for his friends. The store sells single popsicles for \[1$ each, $3$-popsicle boxes for \]2$ each, and $5$-popsicle boxes for \[3$. What is the greatest number of popsicles that Pablo can buy with \]8$?

13

\[3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$, we have \]2$ left. We cannot buy a third \[3$ box, so we opt for the \]2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ popsicles, so the answer is $\fbox{13}$.

AMC10 First Half

AMC10 A

AMC

AMC10

10B

N/A

Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$. What is the area of rhombus $BFDE$? [asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]

8

Using the property that opposite angles are equal in a rhombus, $\angle DAB = \angle DCB = 60 ^\circ$ and $\angle ADC = \angle ABC = 120 ^\circ$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$. The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD = s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3}$. The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}$. Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $\left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3}$. Let $x$ be the area of rhombus $BFDE$. Then $\frac{x}{24} = \frac{1}{3}$, so $x = \fbox{8}$.

AMC10 Second Half

AMC10 B

HMMT

HMMT-Feb

comb

Feb

Starting with the number 0 , Casey performs an infinite sequence of moves as follows: he chooses a number from $\{1,2\}$ at random (each with probability $\frac{1}{2}$ ) and adds it to the current number. Let $p_{m}$ be the probability that Casey ever reaches the number $m$. Find $p_{20}-p_{15}$.

\frac{11}{\frac{11}{2^{20}}}

We note that the only way $n$ does not appear in the sequence is if $n-1$ and then $n+1$ appears. Hence, we have $p_{0}=1$, and $p_{n}=1-\frac{1}{2} p_{n-1}$ for $n>0$. This gives $p_{n}-\frac{2}{3}=-\frac{1}{2}\left(p_{n-1}-\frac{2}{3}\right)$, so that \[ p_{n}=\frac{2}{3}+\frac{1}{3} \cdot\left(-\frac{1}{2}\right)^{n} \] so $p_{20}-p_{15}$ is just \[ \frac{1-(-2)^{5}}{3 \cdot 2^{20}}=\frac{11}{2^{20}} \] $\fbox{\frac{11}{\frac{11}{2^{20}}}}$.

HMMT Feb Easy

HMMT-Feb Combinatorics

HMMT

HMMT-Nov

team

Nov

Call an ordered pair $(a, b)$ of positive integers fantastic if and only if $a, b \leq 10^{4}$ and \[ \operatorname{gcd}(a \cdot n !-1, a \cdot(n+1) !+b)>1 \] for infinitely many positive integers $n$. Find the sum of $a+b$ across all fantastic pairs $(a, b)$.

5183

Solution: We first prove the following lemma, which will be useful later. Lemma: Let $p$ be a prime and $1 \leq n \leq p-1$ be an integer. Then, $n !(p-1-n) ! \equiv(-1)^{n-1}(\bmod p)$. Proof. Write \[ \begin{aligned} n !(p-n-1) ! & =(1 \cdot 2 \cdots n)((p-n-1) \cdots 2 \cdot 1) \\ & \equiv(-1)^{p-n-1}(1 \cdot 2 \cdots n)((n+1) \cdots(p-2)(p-1)) \quad(\bmod p) \\ & =(-1)^{n}(p-1) ! \\ & \equiv(-1)^{n-1} \quad(\bmod p) \end{aligned} \] (where we have used Wilson's theorem). This implies the result. Now, we begin the solution. Suppose that a prime $p$ divides both $a \cdot n !-1$ and $a \cdot(n+1) !+b$. Then, since \[ -b \equiv a \cdot(n+1) ! \equiv(n+1) \cdot(a \cdot n !) \equiv(n+1) \quad(\bmod p) \] we get that $p \mid n+b+1$. Since we must have $n<p$ (or else $p \mid n$ !), we get that, for large enough $n$, $n=p-b-1$. However, by the lemma, \[ a(-1)^{b-1} \equiv a \cdot b !(p-1-b) !=a \cdot b ! n ! \equiv b ! \quad(\bmod p) \] This must hold for infinitely many $p$, so $a=(-1)^{b-1} b$ !. This forces all fantastic pairs to be in form $((2 k-1) !, 2 k-1)$. Now, we prove that these pairs all work. Take $n=p-2 k$ for all large primes $p$. Then, we have \[ \begin{aligned} a \cdot n ! & \equiv(2 k-1) !(p-2 k) ! \\ & \equiv(-1)^{2 k} \equiv 1 \quad(\bmod p) \\ a \cdot(n+1) ! & \equiv(n+1) \cdot(a \cdot n !) \\ & \equiv(p-2 k+1) \cdot 1 \equiv-(2 k-1) \quad(\bmod p) \end{aligned} \] so $p$ divides the gcd. The answer is $(1+1)+(6+3)+(120+5)+(5040+7)=5183$. $\fbox{5183}$.

HMMT Nov Team

HMMT-Nov Team

AMC

AMC12

12B

N/A

The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$-intercept $(0,-p)$, where $p\ne 0$. What is $b$?

4

A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$. Because the $y$-intercept is $(0,-p)$ and $p\ne 0$, it follows that $a=-2/p$. Thus\[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\fbox{4}$.

AMC12 Second Half

AMC12 B

AMC

AMC10

10B

N/A

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? [asy] defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); [/asy]

25\pi

Since the area of the square is $40$, the length of a side is $\sqrt{40}=2\sqrt{10}$. The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\sqrt{10}$. Using the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$. So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = \fbox{25\pi}$.

AMC10 First Half

AMC10 B

AMC

AMC8

8

N/A

Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?

\frac{1}{2}

When Elisa started, she finished a lap in $\frac{25}{10}=2.5$ minutes. Now, she finishes a lap is $\frac{24}{12}= 2$ minutes. The difference is $2.5-2=\fbox{\frac{1}{2}}$.

AMC8 First Half

AMC8

HMMT

HMMT-Nov

guts

Nov

Pascal has a triangle. In the $n$th row, there are $n+1$ numbers $a_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n}$ where $a_{n, 0}=a_{n, n}=1$. For all $1 \leq k \leq n-1, a_{n, k}=a_{n-1, k}-a_{n-1, k-1}$. What is the sum of all numbers in the 2018th row?

2

In general, the sum of the numbers on the $n$th row will be \[ \sum_{k=0}^{n} a_{n, k}=a_{n, 0}+\sum_{k=1}^{n-1}\left(a_{n-1, k}-a_{n-1, k-1}\right)+a_{n, n}=a_{n, 0}+\left(a_{n-1, n-1}-a_{n-1,0}\right)+a_{n, n}=2 \] $\fbox{2}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

Danielle picks a positive integer $1 \leq n \leq 2016$ uniformly at random. What is the probability that $\operatorname{gcd}(n, 2015)=1$ ?

\frac{1441}{2016}

We split the interval $[1,2016]$ into [1,2015] and 2016. The number of integers in $[1,2015]$ that are relatively prime to 2015 is $\phi(2015)=\frac{4}{5} \cdot \frac{12}{13} \cdot \frac{30}{31} \cdot 2015=1440$. Also, 2016 is relatively prime to 2015 , so there are a total of 1441 numbers in $[1,2016]$ that are relatively prime to 2015 . Then the probability of picking a number relatively prime to 2015 is $\frac{1441}{2016}$. $\fbox{\frac{1441}{2016}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

guts

Nov

A polynomial $P$ with integer coefficients is called tricky if it has 4 as a root. A polynomial is called teeny if it has degree at most 1 and integer coefficients between -7 and 7 , inclusive. How many nonzero tricky teeny polynomials are there?

2

If a degree 0 polynomial has 4 as a root, then it must be the constant zero polynomial. Thus, we will only consider polynomials of degree 1 . If $P$ has degree 1, integer coefficients, and 4 as a root, then it must be of the form $P(x)=a(x-4)=$ $a x-4 a$ for some nonzero integer $a$. Since all integer coefficients are between -7 and 7 , inclusive, we require $-7 \leq 4 a \leq 7$, which gives us $a=-1,1$. Note that for both values, the coefficient of $x$ is also between -7 and 7 , so there are 2 tricky teeny polynomials. $\fbox{2}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10B

N/A

A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair $(a,b)$?

2

Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$. With this information we can make the equation: \begin{eqnarray} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0 \end{eqnarray} Applying Simon's Favorite Factoring Trick, we get \begin{eqnarray}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray} Since $b > a$, then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$, or $(a-4) = 2$ and $(b-4) = 4$. This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\fbox{2}$

AMC10 Final Problems

AMC10 B

HMMT

HMMT-Nov

guts

Nov

In a $k \times k$ chessboard, a set $S$ of 25 cells that are in a $5 \times 5$ square is chosen uniformly at random. The probability that there are more black squares than white squares in $S$ is $48 \%$. Find $k$.

9

Solution: We know that there must be fewer black squares than white squares, and $k$ must be odd. Additionally, we know that there are $k-4$ ways to pick the left column of the $5 \times 5$ square so that the right column can fit within the $k \times k$ grid, and $k-4$ ways to pick the top row by similar logic. Therefore, there are $(k-4)^{2}$ of these $5 \times 5$ squares on this chessboard, and because there will be more black squares than white squares whenever there exists a black square in the top left corner, there are $\frac{(k-4)^{2}-1}{2}$ of them have more black squares than white squares, corresponding to the number of black squares in the upper $(k-4) \times(k-4)$ grid. Thus, we have \[ \frac{\frac{(k-4)^{2}-1}{2}}{(k-4)^{2}}=0.48 \Longrightarrow k=9 \] $\fbox{9}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC12

12A

N/A

Circle $C_1$ has its center $O$ lying on circle $C_2$. The two circles meet at $X$ and $Y$. Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ=13$, $OZ=11$, and $YZ=7$. What is the radius of circle $C_1$?

\sqrt{30}

Let $r$ denote the radius of circle $C_1$. Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$. Let $t$ be the measure of angle $YOX$. Since $YO=OX=r$, the law of cosines on triangle $YOX$ gives us $\cos t =-79/121$. Again since $ZYOX$ is cyclic, the measure of angle $YZX=180-t$. We apply the law of cosines to triangle $ZYX$ so that $XY^2=7^2+13^2-2(7)(13)\cos(180-t)$. Since $\cos(180-t)=-\cos t=79/121$ we obtain $XY^2=12000/121$. But$XY^2=400r^2/121$ so that $r=\fbox{\sqrt{30}}$.

AMC12 Second Half

AMC12 A

AMC

AMC8

8

N/A

A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$-by-$3$ grid. Shown below is a sample configuration with three $\triangle$s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

84

Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end. We take casework: Case 1: 3 lines: In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$ Case 2: 2 lines: In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$ Finally, we add and multiply: $2(36+6)=2(42)=\fbox{84}$.

AMC8 Second Half

AMC8

HMMT

HMMT-Feb

team2

Feb

Let $a, b$, and $c$ be complex numbers such that $|a|=|b|=|c|=1$. If \[ \frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=1 \] as well, determine the product of all possible values of $|a+b+c|$.

2

Let $s=a+b+c$. Then \[ \begin{aligned} s^{3} & =a^{3}+b^{3}+c^{3}+3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)+6 a b c \\ & =a b c\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}+3\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)+6\right) \\ & =a b c\left(1+\left(3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9\right)+6\right) \\ & =a b c\left(3 s\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\right) \\ & =a b c(3 s \bar{s}-2) \quad(\text { because } \bar{s}=\bar{a}+\bar{b}+\bar{c}=1 / a+1 / b+1 / c) \\ & =a b c\left(3|s|^{2}-2\right) \end{aligned} \] Taking absolute values, we find $|s|^{3}=\left.|3| s\right|^{2}-2 \mid$. It follows that $|s|$ must be a positive real root of $x^{3}-3 x^{2}+2=0$ or $x^{3}+3 x^{2}-2=0$. However, since the negative real roots of $x^{3}-3 x^{2}+2=0$ are exactly the additive inverses of the positive real roots of $x^{3}-3 x^{2}+2=0$, and all three roots of $x^{3}-3 x^{2}+2=0$ are real $\left(x^{3}-3 x^{2}+2=0\right.$ may be factored as $(x-1)\left(x^{2}-2 x-2\right)=0$, and the discriminant of $x^{2}-2 x-2$ is positive), the product of all possible values of $|s|$ is $(-2) \cdot(-1)^{n}$, where $n$ denotes the number of negative real roots of $x^{3}-3 x^{2}+2=0$. By Descartes's Rule of Signs, we see that $n$ is odd, so the answer is 2 , as desired. \section*{Warm Up Your Proof Skills! [40]} The problems in this section require complete proofs. $\fbox{2}$.

HMMT Feb Team

HMMT-Feb Team

AMC

AMC8

8

N/A

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a depth of $37$ cm. A rock with volume $1000\text{cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?

0.25

The water level will rise $1$cm for every $100 \cdot 40 = 4000\text{cm}^2$. Since $1000$ is $\frac{1}{4}$ of $4000$, the water will rise $\frac{1}{4}\cdot1 = \fbox{0.25}$

AMC8 Second Half

AMC8

HMMT

HMMT-Nov

gen

Nov

Mario has a deck of seven pairs of matching number cards and two pairs of matching Jokers, for a total of 18 cards. He shuffles the deck, then draws the cards from the top one by one until he holds a pair of matching Jokers. The expected number of complete pairs that Mario holds at the end (including the Jokers) is $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100 m+n$.

1003

Solution: Considering ordering the nine pairs by the time they are first complete. Since the pairs are treated equally by the drawing process, this ordering is a uniform ordering. Therefore the problem becomes the following: consider ordering $7 \mathrm{~N}$ 's and $2 \mathrm{~J}$ 's randomly. What is the expected position of the first J? We may solve this by linearity of expectation. Every $\mathrm{N}$ has exactly a $1 / 3$ chance of being in front of the $2 \mathrm{~J}$ 's, so the expected number of $\mathrm{N}$ 's before the first $\mathrm{J}$ is $7 / 3$. Thus the expected position of the first $\mathrm{J}$ is $7 / 3+1=10 / 3$. $\fbox{1003}$.

HMMT Nov Hard

HMMT-Nov General

AMC

AMC8

8

N/A

At Central Middle School the $108$ students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists this items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes. They learn that a big concert is scheduled for the same night and attendance will be down $25\%$. How many recipes of cookies should they make for their smaller party?

11

If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\% - 25\% = 75\%$ of these cookies, or $75\% \cdot 216 = 0.75\cdot 216 = 162$ cookies. $162$ cookies requires $\frac{162}{15} = 10.8$ batches. However, since half-batches are forbidden, we must round up to get $\left\lceil \frac{162}{15} \right\rceil = 11$ batches, and the correct answer is $\fbox{11}$.

AMC8 Second Half

AMC8

AMC

AMC8

8

N/A

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?

1:2

Using the formula for the volume of a cylinder, we get Alex, $108\pi$, and Felicia, $216\pi$. We can quickly notice that $\pi$ cancels out on both sides and that Alex's volume is $1/2$ of Felicia's leaving $1/2 = \fbox{1:2}$ as the answer.

AMC8 First Half

AMC8

AMC

AMC10

10B

N/A

A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?

729

Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday. There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$, or $\fbox{729}$.

AMC10 Second Half

AMC10 B

HMMT

HMMT-Nov

guts

Nov

Two unit squares $S_{1}$ and $S_{2}$ have horizontal and vertical sides. Let $x$ be the minimum distance between a point in $S_{1}$ and a point in $S_{2}$, and let $y$ be the maximum distance between a point in $S_{1}$ and a point in $S_{2}$. Given that $x=5$, the difference between the maximum and minimum possible values for $y$ can be written as $a+b \sqrt{c}$, where $a, b$, and $c$ are integers and $c$ is positive and square-free. Find $100 a+10 b+c$.

472

Solution: Consider what must happen in order for the minimum distance to be exactly 5 . Let one square, say $S_{1}$ have vertices of $(0,0),(0,1),(1,0)$, and $(1,1)$. Further, assume WLOG that the center of $S_{2}$ is above the line $y=\frac{1}{2}$ and to the right of the line $x=\frac{1}{2}$, determined by the center of $S_{1}$. There are three cases to consider: \begin{itemize} \item the right side of $S_{1}$ and the left side of $S_{2}$ are 5 units apart, and the bottom left vertex of $S_{2}$ lies under the line $y=1$; \item the top side of $S_{1}$ and the bottom side of $S_{2}$ are 5 units apart, and the bottom left vertex of $S_{2}$ lies to the left of the line $x=1$; \item the bottom left coordinate of $S_{2}$ is $(a, b)$ with $a, b \geq 1$, and $5=\sqrt{(a-1)^{2}+(b-1)^{2}}$. \end{itemize} We see that the first two cases are symmetric, so consider the case where the left edge of $S_{2}$ lies on the line $x=6$. When this is true, the maximum distance will be achieved between $(0,0)$ and the upper right vertex of $S_{2}$. The upper right vertex can achieve the points $(7, c)$ where $1 \leq c \leq 2$, and so $y \in[\sqrt{50}, \sqrt{53}]$. The other case we have to consider is when the bottom left vertex of $S_{2},(a, b)$, is above $y=1$ and to the right of $x=1$, in which case the maximum distance is achieved from $(0,0)$ and the upper right vertex of $S_{2}$. This distance is $\sqrt{(a+1)^{2}+(b+1)^{2}}$, which, by the triangle inequality, is at most $\sqrt{(a-1)^{2}+(b-1)^{2}}+\sqrt{2^{2}+2^{2}}=5+2 \sqrt{2}$. Since equality holds when $a=b=5 / \sqrt{2}+1$, the largest possible maximum here is $5+2 \sqrt{2}$, and the difference between the largest and smallest possible values of $y$ is $5+2 \sqrt{2}-\sqrt{50}=5-3 \sqrt{2}$. $\fbox{472}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC10

10A

N/A

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square? [asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3*sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("$2$",midpoint(O0--X),NW); label("$1$",midpoint(O2--Y),SE); [/asy]

22 + 12\sqrt {2}

We draw the diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$. We know that the side length $s$ in terms of the diagonal $d$ is $s=\frac{d}{\sqrt{2}}$, so our side length is $\frac{2\sqrt{2}+6}{\sqrt{2}}$. However, we are trying to look for the area, so squaring $\frac{2\sqrt{2}+6}{\sqrt{2}}$ yields $\frac{44+24\sqrt{2}}{2}=\fbox{22 + 12\sqrt {2}}$

AMC10 Second Half

AMC10 A

AMC

AMC8

8

N/A

Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?

6

Let the smallest number be $x.$ It follows that the largest number is $4x.$ Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \fbox{6}. \end{align}

AMC8 First Half

AMC8

HMMT

HMMT-Nov

gen

Nov

A polyhedron has faces that are all either triangles or squares. No two square faces share an edge, and no two triangular faces share an edge. What is the ratio of the number of triangular faces to the number of square faces?

\frac{4}{3}

Let $s$ be the number of square faces and $t$ be the number of triangular faces. Every edge is adjacent to exactly one square face and one triangular face. Therefore, the number of edges is equal to $4 s$, and it is also equal to $3 t$. Thus $4 s=3 t$ and $\frac{t}{s}=\frac{4}{3}$ $\fbox{\frac{4}{3}}$.

HMMT Nov Easy

HMMT-Nov General

HMMT

HMMT-Feb

guts

Feb

Let $A B C$ be a triangle with incircle tangent to the perpendicular bisector of $B C$. If $B C=A E=$ 20 , where $E$ is the point where the $A$-excircle touches $B C$, then compute the area of $\triangle A B C$.

100 \sqrt{2}

Solution: Let the incircle and $B C$ touch at $D$, the incircle and perpendicular bisector touch at $X, Y$ be the point opposite $D$ on the incircle, and $M$ be the midpoint of $B C$. Recall that $A, Y$, and $E$ are collinear by homothety at $A$. Additionally, we have $M D=M X=M E$ so $\angle D X Y=\angle D X E=90^{\circ}$. Therefore $E, X$, and $Y$ are collinear. Since $M X \perp B C$, we have $\angle A E B=45^{\circ}$. The area of $A B C$ is \[ \frac{1}{2} B C \cdot A E \cdot \sin \angle A E B=100 \sqrt{2} \] $\fbox{100 \sqrt{2}}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10B

N/A

Point $P$ is inside equilateral $\triangle ABC$. Points $Q$, $R$, and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, respectively. Given that $PQ=1$, $PR=2$, and $PS=3$, what is $AB$?

4\sqrt{3}

Drawing $\overline{PA}$, $\overline{PB}$, and $\overline{PC}$, $\triangle ABC$ is split into three smaller triangles. The altitudes of these triangles are given in the problem as $PQ$, $PR$, and $PS$. Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get: \[\frac{s}{2} + \frac{2s}{2} + \frac{3s}{2} = \frac{s^2\sqrt{3}}{4}\] where $s$ is the length of a side of the equilateral triangle \[s = \fbox{4\sqrt{3}}\] Note - This is called Viviani's Theorem.

AMC10 Second Half

AMC10 B

HMMT

HMMT-Feb

guts

Feb

$\triangle P N R$ has side lengths $P N=20, N R=18$, and $P R=19$. Consider a point $A$ on $P N . \triangle N R A$ is rotated about $R$ to $\triangle N^{\prime} R A^{\prime}$ so that $R, N^{\prime}$, and $P$ lie on the same line and $A A^{\prime}$ is perpendicular to $P R$. Find $\frac{P A}{A N}$.

\frac{19}{18}

Denote the intersection of $P R$ and $A A^{\prime}$ be $D$. Note $R A^{\prime}=R A$, so $D$, being the altitude of an isosceles triangle, is the midpoint of $A A^{\prime}$. Thus, \[ \angle A R D=\angle A^{\prime} R D=\angle N R A \] so $R A$ is the angle bisector of $P N R$ through $R$. By the angle bisector theorem, we have $\frac{P A}{A N}=\frac{P R}{R N}=\frac{19}{18}$ $\fbox{\frac{19}{18}}$.

HMMT Feb Guts

HMMT-Feb Guts

AMC

AMC10

10A

N/A

For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$

4

Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$. Proof of claim: We examine the positive integers mod $9$. Here are the cases. Case 1. $n \equiv 1 \pmod 9$. Now, we examine $S(n)$ modulo $9$. Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$. (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$.) Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$, the units digit decreases by $9$ while the tens digit increases by $1$. This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$, so the "mod 9" of the sum increases by $1$. This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$. Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$. This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$, $S(S(n)) \equiv 1 \pmod 9$. Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$. Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$. Case 2. $n \equiv 2 \pmod 9$. In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$. Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$, so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$. It is trivial to see that for $n \equiv k \pmod 9$, for $0 \leq k \leq 8$, we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$. Only when $k=0, 3, 6$ is $3k$ a multiple of $9$, which means that $n$ must be a multiple of $3$. Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$. Case 1. $n$ is a multiple of $9$. An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$. The valid sums of the digits of $n$ are $18$ and $27$ in this case. Case 1.1. The sum of the digits is $18$. This means that $S(n)=18, S(S(n))=9$, so $n=2007-18-9=1980$. Clearly this number satisfies our constraints. Case 1.2. The sum of the digits is $27$. This means that $S(n)=27, S(S(n))=9$, ,so $n=2007-27-9=1971$. Since the sum of the digits of $1971$ is not $27$, this does not work. This means that there is $1$ integer in this case. Case 2. $n$ is a multiple of $3$, not $9$. . Case 2.1. Integers of the form $\overline{20ab}$. Then $S(n)=3$ or $S(n)=6$; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$. Case 2.2. Integers of the form $\overline{19ab}$. Then $S(n)=12, 15, 21, 24$ and we consider each case separately. Case 2.2.1. Integers with $S(n)=12$. That means $n=2007-12-3=1992$ which clearly does not work. Case 2.2.2. Integers with $S(n)=15$. That means $n=2007-15-6=1986$ which also does not work Case 2.2.3. Integers with $S(n)=21$. That means $n=2007-21-3=1983$ which is valid. Case 2.2.4. Integers with $S(n)=24$. That means $n=2007-24-6=1977$ which is also valid. We have considered every case, so there are $\fbox{4}$ integers that satisfy the given condition.

AMC10 Final Problems

AMC10 A

AMC

AMC8

8

N/A

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

338

The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$. Therefore the numbers are $333$, $334$, $335$, $336$, $337$, $338$, so the answer is $\fbox{338}$

AMC8 Second Half

AMC8

HMMT

HMMT-Nov

guts

Nov

A polygon $\mathcal{P}$ is drawn on the 2D coordinate plane. Each side of $\mathcal{P}$ is either parallel to the $x$ axis or the $y$ axis (the vertices of $\mathcal{P}$ do not have to be lattice points). Given that the interior of $\mathcal{P}$ includes the interior of the circle $x^{2}+y^{2}=2022$, find the minimum possible perimeter of $\mathcal{P}$.

8 \sqrt{2022}

Solution: The minimum possible perimeter is achieved by an axis-aligned square with all four sides tangent to the circle, which has area $8 \sqrt{2022}$. To see why this is true, notice that there must be at least $2 \sqrt{2022}$ length of total perimeter facing left, $2 \sqrt{2022}$ length facing up, $2 \sqrt{2022}$ facing right, and $2 \sqrt{2022}$ facing down in order for the polygon to be closed and have a shadow of length at least $2 \sqrt{2022}$ in both the $x$ and $y$ directions. $\fbox{8 \sqrt{2022}}$.

HMMT Nov Guts

HMMT-Nov Guts

AMC

AMC8

8

N/A

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$, $11$, $7$, $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?

12

Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\] Solving gives: \[\frac{48 + x}{6} = 10\] \[48 + x = 60\] \[x = 12\] So, the answer is $\fbox{12}$

AMC8 First Half

AMC8

AIME

AIME

I

N/A

Let $S$ be the set of positive integers $k$ such that the two parabolas\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]intersect in four distinct points, and these four points lie on a circle with radius at most $21$. Find the sum of the least element of $S$ and the greatest element of $S$.

285

Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately: The two parabolas intersect at four distinct points. By a quick sketch, we have two subconditions: The point $(-k,20)$ is on or below the parabola $y=x^2-k.$ We need $20\leq(-k)^2-k,$ from which $k\geq5.$ Moreover, the point $(-k,20)$ is on the parabola $y=x^2-k$ when $k=5.$ We will prove that the two parabolas intersect at four distinct points at this value of $k:$ Substituting $y=x^2-5$ into $x=2(y-20)^2-5,$ we get $x=2\left(\left(x^2-5\right)-20\right)^2-5.$ Expanding and rearranging give \[2x^4-100x^2-x+1245=0. \hspace{20mm}(\bigstar)\] By either the graphs of the parabolas or the Rational Root Theorem, we conclude that $x=-5$ is a root of $(\bigstar).$ So, we factor its left side: \[(x+5)\left(2x^3-10x^2-50x+249\right)=0.\] By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that $2x^3-10x^2-50x+249=0$ has two positive roots and one negative root such that $x\neq-5.$ So, $(\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points. For Subcondition A, we deduce that $k\geq5.$ Remark for Subcondition A Recall that if $1\leq k\leq 4,$ then the point $(-k,20)$ is above the parabola $y=x^2-k.$ It follows that for $-k\leq x\leq0:$ The maximum value of $y$ for the parabola $y=x^2-k$ occurs at $x=-k,$ from which $y=k^2-k\leq12.$ The minimum value of $y$ for the parabola $x=2(y-20)^2-k$ occurs at $x=0,$ from which $y=20-\sqrt{\frac k2}>18.$ Clearly, the parabola $x=2(y-20)^2-k$ and the left half of the parabola $y=x^2-k$ do not intersect. Therefore, the two parabolas do not intersect at four distinct points. The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$ The lower half of the parabola $x=2(y-20)^2-k$ is $y=20-\sqrt{\frac{x+k}{2}}.$ We need $-k\leq20-\sqrt{\frac k2},$ which holds for all values of $k.$ For Subcondition B, we deduce that $k$ can be any positive integer. For Condition 1, we obtain $\boldsymbol{k\geq5}$ by taking the intersection of Subconditions A and B. The four points of intersection lie on a circle with radius at most $21.$ For equations of circles, the coefficients of $x^2$ and $y^2$ must be the same. So, we add the equation $y=x^2-k$ to half the equation $x=2(y-20)^2-k:$ \[y+\frac12x=x^2+(y-20)^2-\frac32k.\] We expand, rearrange, and complete the squares: \begin{align} y+\frac12x&=x^2+y^2-40y+400-\frac32k \\ \frac32k-400&=\left(x^2-\frac12x\right)+\left(y^2-41y\right) \\ \frac32k-400+\frac{1}{16}+\frac{1681}{4}&=\left(x-\frac14\right)^2+\left(y-\frac{41}{2}\right)^2. \end{align} We need $\frac32k-400+\frac{1}{16}+\frac{1681}{4}\leq21^2,$ from which $k\leq\left\lfloor\frac{6731}{24}\right\rfloor=280.$ For Condition 2, we obtain $\boldsymbol{k\leq280.}$ Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\fbox{285}.$ ~MRENTHUSIASM

Very Hard AIME Problems

AIME

HMMT

HMMT-Nov

guts

Nov

Let $X Y Z$ be an equilateral triangle, and let $K, L, M$ be points on sides $X Y, Y Z, Z X$, respectively, such that $X K / K Y=B, Y L / L Z=1 / C$, and $Z M / M X=1$. Determine the ratio of the area of triangle $K L M$ to the area of triangle $X Y Z$.

\frac{1}{5}

First, we note that \[ [K L M]=[X Y Z]-[X K M]-[Y L K]-[Z M L] \] Then, note that \[ \begin{gathered} {[X K M]=\frac{X K}{X Y} \cdot \frac{X M}{X Z} \cdot[X Y Z]=\frac{B}{B+1} \cdot \frac{1}{2} \cdot[X Y Z]} \\ {[Y L K]=\frac{Y L}{Y Z} \cdot \frac{Y K}{Y X} \cdot[X Y Z]=\frac{1}{C+1} \cdot \frac{1}{B+1} \cdot[X Y Z]} \\ {[Z M L]=\frac{Z M}{Z X} \cdot \frac{Z L}{Z Y} \cdot[X Y Z]=\frac{1}{2} \cdot \frac{1}{C+1} \cdot[X Y Z]} \end{gathered} \] Consequently, \[ \begin{aligned} A & =\frac{[K L M]}{[X Y Z]} \\ & =1-\frac{B}{B+1} \cdot \frac{1}{2}-\frac{1}{C+1} \cdot \frac{1}{B+1}-\frac{C}{C+1} \cdot \frac{1}{2} \\ & =\frac{B+C}{(B+1)(C+1)(2)} \end{aligned} \] If we solve our system of equations for $A, B, C$, we get that $A=\frac{1}{5}$. $\fbox{\frac{1}{5}}$.

HMMT Nov Guts

HMMT-Nov Guts

HMMT

HMMT-Nov

thm

Nov

Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime on every positive multiple of 10 , and a nickel on every multiple of 5 that is not a multiple of 10 . Crisp then starts at 0 . Every second, he has a $\frac{2}{3}$ chance of jumping from his current location $x$ to $x+3$, and a $\frac{1}{3}$ chance of jumping from his current location $x$ to $x+7$. When Crisp jumps on either a dime or a nickel, he stops jumping. What is the probability that Crisp stops on a dime?

\frac{20}{31}

Let "a 3" mean a move in which Crisp moves from $x$ to $x+3$, and "a 7 " mean a move in which Crisp moves from $x$ to $x+7$. Note that Crisp stops precisely the first time his number of 3 's and number of 7's differs by a multiple of 5 , and that he'll stop on a dime if they differ by 0 , and stop on a nickel if they differ by 5 . This fact will be used without justification. We split into two cases: (a) Crisp begins with a 3. Rather than consider the integer Crisp is on, we'll count the difference, $n$, between his number of 3's and his number of 7's. Each 3 increases $n$ by 1, and each 7 decreases $n$ by 1 . Currently, $n$ is 1 . The probability of stopping on a dime, then, is the probability $n$ reaches 0 before $n$ reaches 5 , where $n$ starts at 1 . Let $a_{i}$ be the probability $n$ reaches 0 first, given a current position of $i$, for $i=1,2,3,4$. We desire $a_{1}$. We have the system of linear equations \[ \begin{aligned} a_{1} & =\frac{2}{3} a_{2}+\frac{1}{3} \cdot 1 \\ a_{2} & =\frac{2}{3} a_{3}+\frac{1}{3} a_{1} \\ a_{3} & =\frac{2}{3} a_{4}+\frac{1}{3} a_{2} \\ a_{4} & =\frac{2}{3} \cdot 0+\frac{1}{3} a_{3} \end{aligned} \] From which we determine that $a_{1}=\frac{15}{31}$.\\ (b) Crisp begins with a 7. Now, let $m$ be the difference between his number of 7's and his number of 3 's. Let $b_{i}$ denote his probability of stopping on a dime, given his current position of $m=i$. We desire $b_{1}$. We have the system of linear equations \[ \begin{aligned} b_{1} & =\frac{1}{3} b_{2}+\frac{2}{3} \cdot 1 \\ b_{2} & =\frac{1}{3} b_{3}+\frac{2}{3} b_{1} \\ b_{3} & =\frac{1}{3} b_{4}+\frac{2}{3} b_{2} \\ b_{4} & =\frac{1}{3} \cdot 0+\frac{2}{3} b_{3} \end{aligned} \] From which we determine that $b_{1}=\frac{30}{31}$. We conclude that the answer is $\frac{2}{3} a_{1}+\frac{1}{3} b_{1}=\frac{2}{3} \cdot \frac{15}{31}+\frac{1}{3} \cdot \frac{30}{31}=\frac{20}{31}$. $\fbox{\frac{20}{31}}$.

HMMT Nov Hard

HMMT-Nov Theme