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Omni-MATH

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Kelvin the Frog and 10 of his relatives are at a party. Every pair of frogs is either friendly or unfriendly. When 3 pairwise friendly frogs meet up, they will gossip about one another and end up in a fight (but stay friendly anyway). When 3 pairwise unfriendly frogs meet up, they will also end up in a fight. In all other cases, common ground is found and there is no fight. If all $\binom{11}{3}$ triples of frogs meet up exactly once, what is the minimum possible number of fights?

Consider a graph $G$ with 11 vertices - one for each of the frogs at the party - where two vertices are connected by an edge if and only if they are friendly. Denote by $d(v)$ the number of edges emanating from $v$; i.e. the number of friends frog $v$ has. Note that $d(1)+d(2)+\ldots+d(11)=2e$, where $e$ is the number of edges in this graph. Focus on a single vertex $v$, and choose two other vertices $u, w$ such that $uv$ is an edge but $wv$ is not. There are then $d(v)$ choices for $u$ and $10-d(v)$ choices for $w$, so there are $d(v)(10-d(v))$ sets of three frogs that include $v$ and do not result in a fight. Each set, however, is counted twice - if $uw$ is an edge, then we count this set both when we focus on $v$ and when we focus on $w$, and otherwise we count it when we focus on $v$ and when we focus on $u$. As such, there are a total of $\frac{1}{2} \sum_{v} d(v)(10-d(v))$ sets of 3 frogs that do not result in a fight. Note that $\frac{d(v)+10-d(v)}{2}=5 \geq \sqrt{d(v)(10-d(v))} \Longrightarrow d(v)(10-d(v)) \leq 25$ by AM-GM. Thus there are a maximum of $\frac{1}{2} \sum_{v} d(v)(10-d(v)) \leq \frac{1}{2}(25 \cdot 11)=\frac{275}{2}$ sets of three frogs that do not result in a fight; since this number must be an integer, there are a maximum of 137 such sets. As there are a total of $\binom{11}{3}=165$ sets of 3 frogs, this results in a minimum $165-137=28$ number of fights. It remains to show that such an arrangement can be constructed. Set $d(1)=d(2)=\ldots=d(10)=5$ and $d(11)=4$. Arrange these in a circle, and connect each to the nearest two clockwise neighbors; this gives each vertex 4 edges. To get the final edge for the first ten vertices, connect 1 to 10,2 to 9,3 to 8,4 to 7, and 5 to 6. Thus 28 is constructable, and is thus the true minimum.

28

HMMT_2

Cyclic pentagon $ABCDE$ has side lengths $AB=BC=5, CD=DE=12$, and $AE=14$. Determine the radius of its circumcircle.

Let $C^{\prime}$ be the point on minor arc $BCD$ such that $BC^{\prime}=12$ and $C^{\prime}D=5$, and write $AC^{\prime}=BD=C^{\prime}E=x, AD=y$, and $BD=z$. Ptolemy applied to quadrilaterals $ABC^{\prime}D, BC^{\prime}DE$, and $ABDE$ gives $$\begin{aligned} & x^{2}=12y+5^{2} \\ & x^{2}=5z+12^{2} \\ & yz=14x+5 \cdot 12 \end{aligned}$$ Then $$\left(x^{2}-5^{2}\right)\left(x^{2}-12^{2}\right)=5 \cdot 12yz=5 \cdot 12 \cdot 14x+5^{2} \cdot 12^{2}$$ from which $x^{3}-169x-5 \cdot 12 \cdot 14=0$. Noting that $x>13$, the rational root theorem leads quickly to the root $x=15$. Then triangle $BCD$ has area $\sqrt{16 \cdot 1 \cdot 4 \cdot 11}=8\sqrt{11}$ and circumradius $R=\frac{5 \cdot 12 \cdot 15}{4 \cdot 8\sqrt{11}}=$ $\frac{225\sqrt{11}}{88}$.

\frac{225\sqrt{11}}{88}

HMMT_2

For a positive integer $n$, let $\theta(n)$ denote the number of integers $0 \leq x<2010$ such that $x^{2}-n$ is divisible by 2010. Determine the remainder when $\sum_{n=0}^{2009} n \cdot \theta(n)$ is divided by 2010.

Let us consider the $\operatorname{sum} \sum_{n=0}^{2009} n \cdot \theta(n)(\bmod 2010)$ in a another way. Consider the sum $0^{2}+1^{2}+2^{2}+\cdots+2007^{2}(\bmod 2010)$. For each $0 \leq n<2010$, in the latter sum, the term $n$ appears $\theta(n)$ times, so the sum is congruent to $\sum_{n=0}^{2009} n \cdot \theta(n)$. In other words, $$\sum_{n=0}^{2009} n \cdot \theta(n)=\sum_{n=0}^{2009} n^{2}=\frac{(2009)(2009+1)(2 \cdot 2009+1)}{6} \equiv(-1) \cdot \frac{2010}{6} \cdot(-1)=335 \quad(\bmod 2010)$$

335

HMMT_2

The set of points $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ in $\mathbf{R}^{4}$ such that $x_{1} \geq x_{2} \geq x_{3} \geq x_{4}$ is a cone (or hypercone, if you insist). Into how many regions is this cone sliced by the hyperplanes $x_{i}-x_{j}=1$ for $1 \leq i<j \leq n$ ?

$C(4)=14$.

14

HMMT_2

A point in three-space has distances $2,6,7,8,9$ from five of the vertices of a regular octahedron. What is its distance from the sixth vertex?

By a simple variant of the British Flag Theorem, if $A B C D$ is a square and $P$ any point in space, $A P^{2}+C P^{2}=B P^{2}+D P^{2}$. Four of the five given vertices must form a square $A B C D$, and by experimentation we find their distances to the given point $P$ must be $A P=2, B P=6, C P=9, D P=7$. Then $A, C$, and the other two vertices $E, F$ also form a square $A E C F$, so $85=A P^{2}+C P^{2}=E P^{2}+F P^{2}=8^{2}+F P^{2} \Rightarrow F P=\sqrt{21}$.

\sqrt{21}

HMMT_2

Find the largest integer that divides $m^{5}-5 m^{3}+4 m$ for all $m \geq 5$.

5!=120.

120

HMMT_2

How many lattice points are enclosed by the triangle with vertices $(0,99),(5,100)$, and $(2003,500) ?$ Don't count boundary points.

Using the determinant formula, we get that the area of the triangle is $$\left|\begin{array}{cc} 5 & 1 \\ 2003 & 401 \end{array}\right| / 2=1$$ There are 4 lattice points on the boundary of the triangle (the three vertices and $(1004,300)$ ), so it follows from Pick's Theorem that there are 0 in the interior.

0

HMMT_2

Find the sum of the reciprocals of all the (positive) divisors of 144.

As $d$ ranges over the divisors of 144 , so does $144 / d$, so the sum of $1 / d$ is $1 / 144$ times the sum of the divisors of 144 . Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is 403, so the answer is 403/144.

403/144

HMMT_2

What is the smallest positive integer $x$ for which $x^{2}+x+41$ is not a prime?

40.

40

HMMT_2

Tessa picks three real numbers $x, y, z$ and computes the values of the eight expressions of the form $\pm x \pm y \pm z$. She notices that the eight values are all distinct, so she writes the expressions down in increasing order. How many possible orders are there?

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

96

HMMT_2

Alice, Bob, and Charlie roll a 4, 5, and 6-sided die, respectively. What is the probability that a number comes up exactly twice out of the three rolls?

There are $4 \cdot 5 \cdot 6=120$ different ways that the dice can come up. The common number can be any of $1,2,3,4$, or 5: there are $3+4+5=12$ ways for it to be each of $1,2,3$, or 4, because we pick one of the three people's rolls to disagree, and there are 3,4, and 5 ways that roll can come up (for Alice, Bob, and Charlie respectively). Finally, there are 4 ways for Bob and Charlie to both roll a 5 and Alice to roll any number. Thus there are 52 different ways to satisfy the problem condition, and our answer is $\frac{52}{120}=\frac{13}{30}$.

\frac{13}{30}

HMMT_2

Does there exist an irrational number $\alpha>1$ such that \(\left\lfloor\alpha^{n}\right\rfloor \equiv 0 \quad(\bmod 2017)\) for all integers $n \geq 1$ ?

Yes. Let $\alpha>1$ and $0<\beta<1$ be the roots of $x^{2}-4035 x+2017$. Then note that \(\left\lfloor\alpha^{n}\right\rfloor=\alpha^{n}+\beta^{n}-1\). Let $x_{n}=\alpha^{n}+\beta^{n}$ for all nonnegative integers $n$. It's easy to verify that $x_{n}=4035 x_{n-1}-2017 x_{n-2} \equiv x_{n-1}$ $(\bmod 2017)$ so since $x_{1}=4035 \equiv 1(\bmod 2017)$ we have that $x_{n} \equiv 1(\bmod 2017)$ for all $n$. Thus $\alpha$ satisfies the problem.

Yes

HMMT_2

Suppose that $A, B, C, D$ are four points in the plane, and let $Q, R, S, T, U, V$ be the respective midpoints of $A B, A C, A D, B C, B D, C D$. If $Q R=2001, S U=2002, T V=$ 2003, find the distance between the midpoints of $Q U$ and $R V$.

This problem has far more information than necessary: $Q R$ and $U V$ are both parallel to $B C$, and $Q U$ and $R V$ are both parallel to $A D$. Hence, $Q U V R$ is a parallelogram, and the desired distance is simply the same as the side length $Q R$, namely 2001.

2001

HMMT_2

Let $\varphi(n)$ denote the number of positive integers less than or equal to $n$ which are relatively prime to $n$. Let $S$ be the set of positive integers $n$ such that $\frac{2 n}{\varphi(n)}$ is an integer. Compute the sum $\sum_{n \in S} \frac{1}{n}$.

Let $T_{n}$ be the set of prime factors of $n$. Then $\frac{2 n}{\phi(n)}=2 \prod_{p \in T} \frac{p}{p-1}$. We can check that this is an integer for the following possible sets: $\varnothing,\{2\},\{3\},\{2,3\},\{2,5\},\{2,3,7\}$. For each set $T$, the sum of the reciprocals of the positive integers having that set of prime factors is $\prod_{p \in T}\left(\sum_{m=1}^{\infty} \frac{1}{p^{m}}\right)=\prod_{p \in T} \frac{1}{p-1}$. Therefore the desired sum is $1+1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{12}=\frac{10}{3}$.

\frac{10}{3}

HMMT_2

Find the real solution(s) to the equation $(x+y)^{2}=(x+1)(y-1)$.

Set $p=x+1$ and $q=y-1$, then we get $(p+q)^{2}=pq$, which simplifies to $p^{2}+pq+q^{2}=0$. Then we have $\left(p+\frac{q}{2}\right)^{2}+\frac{3q^{2}}{4}$, and so $p=q=0$. Thus $(x, y)=(-1,1)$.

(-1,1)

HMMT_2

Let $(x, y)$ be a pair of real numbers satisfying $$56x+33y=\frac{-y}{x^{2}+y^{2}}, \quad \text { and } \quad 33x-56y=\frac{x}{x^{2}+y^{2}}$$ Determine the value of $|x|+|y|$.

Observe that $$\frac{1}{x+yi}=\frac{x-yi}{x^{2}+y^{2}}=33x-56y+(56x+33y)i=(33+56i)(x+yi)$$ So $$(x+yi)^{2}=\frac{1}{33+56i}=\frac{1}{(7+4i)^{2}}=\left(\frac{7-4i}{65}\right)^{2}$$ It follows that $(x, y)= \pm\left(\frac{7}{65},-\frac{4}{65}\right)$.

\frac{11}{65}

HMMT_2

The rational numbers $x$ and $y$, when written in lowest terms, have denominators 60 and 70 , respectively. What is the smallest possible denominator of $x+y$ ?

Write $x+y=a / 60+b / 70=(7 a+6 b) / 420$. Since $a$ is relatively prime to 60 and $b$ is relatively prime to 70 , it follows that none of the primes $2,3,7$ can divide $7 a+6 b$, so we won't be able to cancel any of these factors in the denominator. Thus, after reducing to lowest terms, the denominator will still be at least $2^{2} \cdot 3 \cdot 7=84$ (the product of the powers of 2,3 , and 7 dividing 420 ). On the other hand, 84 is achievable, by taking (e.g.) $1 / 60+3 / 70=25 / 420=5 / 84$. So 84 is the answer.

84

HMMT_2

Let $A B C$ be a triangle and $D, E$, and $F$ be the midpoints of sides $B C, C A$, and $A B$ respectively. What is the maximum number of circles which pass through at least 3 of these 6 points?

All $\binom{6}{3}=20$ triples of points can produce distinct circles aside from the case where the three points are collinear $(B D C, C E A, A F B)$.

17

HMMT_2

Let $a, b, c$ be nonzero real numbers such that $a+b+c=0$ and $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$. Find the value of $a^{2}+b^{2}+c^{2}$.

Let $\sigma_{1}=a+b+c, \sigma_{2}=ab+bc+ca$ and $\sigma_{3}=abc$ be the three elementary symmetric polynomials. Since $a^{3}+b^{3}+c^{3}$ is a symmetric polynomial, it can be written as a polynomial in $\sigma_{1}, \sigma_{2}$ and $\sigma_{3}$. Now, observe that $\sigma_{1}=0$, and so we only need to worry about the terms not containing $\sigma_{1}$. By considering the degrees of the terms, we see that the only possibility is $\sigma_{3}$. That is, $a^{3}+b^{3}+c^{3}=k\sigma_{3}$ for some constant $k$. By setting $a=b=1, c=-2$, we see that $k=3$. By similar reasoning, we find that $a^{5}+b^{5}+c^{5}=h\sigma_{2}\sigma_{3}$ for some constant $h$. By setting $a=b=1$ and $c=-2$, we get $h=-5$. So, we now know that $a+b+c=0$ implies $$a^{3}+b^{3}+c^{3}=3abc \quad \text { and } \quad a^{5}+b^{5}+c^{5}=-5abc(ab+bc+ca)$$ Then $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$ implies that $3abc=-5abc(ab+bc+ca)$. Given that $a, b, c$ are nonzero, we get $ab+bc+ca=-\frac{3}{5}$. Then, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)=\frac{6}{5}$.

\frac{6}{5}

HMMT_2

Mrs. Toad has a class of 2017 students, with unhappiness levels $1,2, \ldots, 2017$ respectively. Today in class, there is a group project and Mrs. Toad wants to split the class in exactly 15 groups. The unhappiness level of a group is the average unhappiness of its members, and the unhappiness of the class is the sum of the unhappiness of all 15 groups. What's the minimum unhappiness of the class Mrs. Toad can achieve by splitting the class into 15 groups?

One can show that the optimal configuration is $\{1\},\{2\}, \ldots,\{14\},\{15, \ldots, 2017\}$. This would give us an answer of $1+2+\cdots+14+\frac{15+2017}{2}=105+1016=1121$.

1121

HMMT_2

You have 128 teams in a single elimination tournament. The Engineers and the Crimson are two of these teams. Each of the 128 teams in the tournament is equally strong, so during each match, each team has an equal probability of winning. Now, the 128 teams are randomly put into the bracket. What is the probability that the Engineers play the Crimson sometime during the tournament?

There are $\binom{128}{2}=127 \cdot 64$ pairs of teams. In each tournament, 127 of these pairs play. By symmetry, the answer is $\frac{127}{127 \cdot 64}=\frac{1}{64}$.

\frac{1}{64}

HMMT_2

For the sequence of numbers $n_{1}, n_{2}, n_{3}, \ldots$, the relation $n_{i}=2 n_{i-1}+a$ holds for all $i>1$. If $n_{2}=5$ and $n_{8}=257$, what is $n_{5}$ ?

33.

33

HMMT_2

In some squares of a $2012\times 2012$ grid there are some beetles, such that no square contain more than one beetle. At one moment, all the beetles fly off the grid and then land on the grid again, also satisfying the condition that there is at most one beetle standing in each square. The vector from the centre of the square from which a beetle $B$ flies to the centre of the square on which it lands is called the [i]translation vector[/i] of beetle $B$. For all possible starting and ending configurations, find the maximum length of the sum of the [i]translation vectors[/i] of all beetles.

In a \(2012 \times 2012\) grid, we place beetles such that no square contains more than one beetle. When the beetles fly off and land again, each beetle has a translation vector from its initial to its final position. We aim to find the maximum length of the sum of these translation vectors for all possible starting and ending configurations. The answer is \(\frac{2012^3}{4}\), which is achieved by moving \(\frac{2012^2}{2}\) beetles that start on the left half of the grid each \(\frac{2012}{2}\) units to the right. We now prove that this is maximal. Suppose the beetles start at positions \(X_i\) and end at positions \(Y_i\), and let \(O\) denote the center of the board. By the triangle inequality, \[ \left|\sum \overrightarrow{X_i Y_i}\right| \leq \left|\sum \overrightarrow{O X_i}\right| + \left|\sum \overrightarrow{O Y_i}\right|. \] We will prove that \(\left| \sum \overrightarrow{O X_i}\right| \leq \frac{2012^3}{8}\). This will be done by applying the triangle inequality smartly. Assume \(O = (0,0)\) and scale the board so that the gridlines are of the form \(x = 2i\) and \(y = 2i\) for integers \(i\). The closest square centers to \(O\) are \((\pm 1, \pm 1)\). For \(1 \leq n \leq 1006\), consider the set \(S\) of points \((x,y)\) such that \(\max\{|x|, |y|\} = 2n-1\), which forms a square-shaped "frame". Define \(f(A) = \sum_{P \in A} \overrightarrow{OP}\), where \(A \subseteq S\). We claim that \[ |f(A)| \leq 6n^2 - 6n + 2. \] To prove this, we observe the following "smoothing"-type facts: - If \(A\) contains two opposite points of the form \((a,b)\) and \((-a,-b)\), we can delete both without changing anything. - If \(A\) does not contain \((a,b)\) nor \((-a,-b)\), then one of them must form a non-obtuse angle with \(f(A)\), so adding that one to \(A\) will increase \(|f(A)|\). - If \(A\) contains some \((a,b)\) which forms an obtuse angle with \(f(A)\), then removing it from \(A\) will increase \(|f(A)|\). Hence, if \(|f(A)|\) is maximal, we must have \(|A| = |S|/2 = 4n-2\), and the range of the arguments of the vectors formed by elements of \(A\) is at most \(180^\circ\). It cannot be exactly \(180^\circ\) by the first property. Thus, \(A\) must be formed from a contiguous run of points along the frame. The rest of the problem is essentially algebraic. We only consider \(A\) which satisfy the above requirements, meaning that some entire "side" of the frame must be contained in \(A\). Without loss of generality, assume the right side (\(x = 2n-1\)). Suppose the rightmost point on the top side has \(x\)-coordinate \(2(n-k)-1 > 0\), so the rightmost point on the bottom side has \(x\)-coordinate \(2(k-n)+1 < 0\) (where \(k \geq 0\)). In this case, the \(x\)-component of \(f(A)\) equals \[ 2 \sum_{i=n-k}^{n-1} (2i-1) + 2n(2n-1) = 2k(2n-k-2) + 2n(2n-1). \] The \(y\)-component of \(f(A)\) is \[ -2(2n-1)(n-1-k). \] Therefore, if \(m = n-1-k\), we have \[ |f(A)|^2 = (2m(2n-1))^2 + (2(n-1-m)(n-1+m) + 2n(2n-1))^2 = 36n^4 - 72n^3 + 60n^2 - 24n + 4 + 4m^2(m^2 - 2n^2 + 2n - 1). \] Because \(k \geq 0\), \(m \leq n-1\), so \[ m^2 - 2n^2 + 2n - 1 \leq (n-1)^2 - 2n^2 + 2n - 1 = -n^2 \leq 0, \] thus, \[ 36n^4 - 72n^3 + 60n^2 - 24n + 4 + 4m^2(m^2 - 2n^2 + 2n - 1) \leq 36n^4 - 72n^3 + 60n^2 - 24n + 4 = (6n^2 - 6n + 2)^2, \] which is the desired bound. To finish, by summing our bound over \(1 \leq n \leq 1006\), we have \[ \left|\sum \overrightarrow{OX_i}\right| \leq \sum_{n=1}^{1006} (6n^2 - 6n + 2) = \frac{2012^3}{4}. \] Remembering that we scaled by a factor of \(2\), this implies that we actually have \(\left| \sum \overrightarrow{OX_i}\right| \leq \frac{2012^3}{8}\), which is the desired result. The answer is \(\boxed{\frac{2012^3}{4}}\).

\frac{2012^3}{4}

china_team_selection_test

Let $z=1-2 i$. Find $\frac{1}{z}+\frac{2}{z^{2}}+\frac{3}{z^{3}}+\cdots$.

Let $x=\frac{1}{z}+\frac{2}{z^{2}}+\frac{3}{z^{3}}+\cdots$, so $z \cdot x=\left(1+\frac{2}{z}+\frac{3}{z^{2}}+\frac{4}{z^{3}}+\cdots\right)$. Then $z \cdot x-x=$ $1+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\cdots=\frac{1}{1-1 / z}=\frac{z}{z-1}$. Solving for $x$ in terms of $z$, we obtain $x=\frac{z}{(z-1)^{2}}$. Plugging in the original value of $z$ produces $x=(2 i-1) / 4$.

(2i-1)/4

HMMT_2

Let $P(x)$ be a polynomial with degree 2008 and leading coefficient 1 such that $P(0)=2007, P(1)=2006, P(2)=2005, \ldots, P(2007)=0$. Determine the value of $P(2008)$. You may use factorials in your answer.

Consider the polynomial $Q(x)=P(x)+x-2007$. The given conditions tell us that $Q(x)=0$ for $x=0,1,2, \ldots, 2007$, so these are the roots of $Q(x)$. On the other hand, we know that $Q(x)$ is also a polynomial with degree 2008 and leading coefficient 1 . It follows that $Q(x)=x(x-1)(x-2)(x-3) \cdots(x-2007)$. Thus $$P(x)=x(x-1)(x-2)(x-3) \cdots(x-2007)-x+2007$$ Setting $x=2008$ gives the answer.

2008!-1

HMMT_2

(a) Can 1000 queens be placed on a $2017 \times 2017$ chessboard such that every square is attacked by some queen? A square is attacked by a queen if it lies in the same row, column, or diagonal as the queen. (b) A $2017 \times 2017$ grid of squares originally contains a 0 in each square. At any step, Kelvin the Frog chooses two adjacent squares (two squares are adjacent if they share a side) and increments the numbers in both of them by 1. Can Kelvin make every square contain a different power of 2? (c) A tournament consists of single games between every pair of players, where each game has a winner and loser with no ties. A set of people is dominated if there exists a player who beats all of them. Does there exist a tournament in which every set of 2017 people is dominated? (d) Every cell of a $19 \times 19$ grid is colored either red, yellow, green, or blue. Does there necessarily exist a rectangle whose sides are parallel to the grid, all of whose vertices are the same color? (e) Does there exist a $c \in \mathbb{R}^{+}$such that $\max (|A \cdot A|,|A+A|) \geq c|A| \log ^{2}|A|$ for all finite sets $A \subset \mathbb{Z}$? (f) Can the set $\{1,2, \ldots, 1093\}$ be partitioned into 7 subsets such that each subset is sum-free (i.e. no subset contains $a, b, c$ with $a+b=c)$?

Answer: NNYYYY

NNYYYY

HMMT_2

Let $P$ be a polyhedron where every face is a regular polygon, and every edge has length 1. Each vertex of $P$ is incident to two regular hexagons and one square. Choose a vertex $V$ of the polyhedron. Find the volume of the set of all points contained in $P$ that are closer to $V$ than to any other vertex.

Observe that $P$ is a truncated octahedron, formed by cutting off the corners from a regular octahedron with edge length 3. So, to compute the value of $P$, we can find the volume of the octahedron, and then subtract off the volume of truncated corners. Given a square pyramid where each triangular face an equilateral triangle, and whose side length is $s$, the height of the pyramid is $\frac{\sqrt{2}}{2}s$, and thus the volume is $\frac{1}{3} \cdot s^{2} \cdot \frac{\sqrt{2}}{2}s=\frac{\sqrt{2}}{6}s^{3}$. The side length of the octahedron is 3, and noting that the octahedron is made up of two square pyramids, its volume must be is $2 \cdot \frac{\sqrt{2}(3)^{3}}{6}=9\sqrt{2}$. The six "corners" that we remove are all square pyramids, each with volume $\frac{\sqrt{2}}{6}$, and so the resulting polyhedron $P$ has volume $9\sqrt{2}-6 \cdot \frac{\sqrt{2}}{6}=8\sqrt{2}$. Finally, to find the volume of all points closer to one particular vertex than any other vertex, note that due to symmetry, every point in $P$ (except for a set with zero volume), is closest to one of the 24 vertices. Due to symmetry, it doesn't matter which $V$ is picked, so we can just divide the volume of $P$ by 24 to obtain the answer $\frac{\sqrt{2}}{3}$.

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

HMMT_2

The sequences of real numbers $\left\{a_{i}\right\}_{i=1}^{\infty}$ and $\left\{b_{i}\right\}_{i=1}^{\infty}$ satisfy $a_{n+1}=\left(a_{n-1}-1\right)\left(b_{n}+1\right)$ and $b_{n+1}=a_{n} b_{n-1}-1$ for $n \geq 2$, with $a_{1}=a_{2}=2015$ and $b_{1}=b_{2}=2013$. Evaluate, with proof, the infinite sum $\sum_{n=1}^{\infty} b_{n}\left(\frac{1}{a_{n+1}}-\frac{1}{a_{n+3}}\right)$.

First note that $a_{n}$ and $b_{n}$ are weakly increasing and tend to infinity. In particular, $a_{n}, b_{n} \notin\{0,-1,1\}$ for all $n$. For $n \geq 1$, we have $a_{n+3}=\left(a_{n+1}-1\right)\left(b_{n+2}+1\right)=\left(a_{n+1}-1\right)\left(a_{n+1} b_{n}\right)$, so $\frac{b_{n}}{a_{n+3}}=\frac{1}{a_{n+1}\left(a_{n+1}-1\right)}=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}}$. Therefore, $\sum_{n=1}^{\infty} \frac{b_{n}}{a_{n+1}}-\frac{b_{n}}{a_{n+3}} =\sum_{n=1}^{\infty} \frac{b_{n}}{a_{n+1}}-\left(\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}}\right) =\sum_{n=1}^{\infty} \frac{b_{n}+1}{a_{n+1}}-\frac{1}{a_{n+1}-1}$. Furthermore, $b_{n}+1=\frac{a_{n+1}}{a_{n-1}-1}$ for $n \geq 2$. So the sum over $n \geq 2$ is $\sum_{n=2}^{\infty}\left(\frac{1}{a_{n-1}-1}-\frac{1}{a_{n+1}-1}\right) =\lim _{N \rightarrow \infty} \sum_{n=2}^{N}\left(\frac{1}{a_{n-1}-1}-\frac{1}{a_{n+1}-1}\right) =\frac{1}{a_{1}-1}+\frac{1}{a_{2}-1}-\lim _{N \rightarrow \infty}\left(\frac{1}{a_{N}-1}+\frac{1}{a_{N+1}-1}\right) =\frac{1}{a_{1}-1}+\frac{1}{a_{2}-1}$. Hence the final answer is $\left(\frac{b_{1}+1}{a_{2}}-\frac{1}{a_{2}-1}\right)+\left(\frac{1}{a_{1}-1}+\frac{1}{a_{2}-1}\right)$. Cancelling the common terms and putting in our starting values, this equals $\frac{2014}{2015}+\frac{1}{2014}=1-\frac{1}{2015}+\frac{1}{2014}=1+\frac{1}{2014 \cdot 2015}$

1+\frac{1}{2014 \cdot 2015}

HMMT_2

Let $S$ be the set of $3^{4}$ points in four-dimensional space where each coordinate is in $\{-1,0,1\}$. Let $N$ be the number of sequences of points $P_{1}, P_{2}, \ldots, P_{2020}$ in $S$ such that $P_{i} P_{i+1}=2$ for all $1 \leq i \leq 2020$ and $P_{1}=(0,0,0,0)$. (Here $P_{2021}=P_{1}$.) Find the largest integer $n$ such that $2^{n}$ divides $N$.

From $(0,0,0,0)$ we have to go to $( \pm 1, \pm 1, \pm 1, \pm 1)$, and from $(1,1,1,1)$ (or any of the other similar points), we have to go to $(0,0,0,0)$ or $(-1,1,1,1)$ and its cyclic shifts. If $a_{i}$ is the number of ways to go from $(1,1,1,1)$ to point of the form $( \pm 1, \pm 1, \pm 1, \pm 1)$ in $i$ steps, then we need to find $\nu_{2}\left(16 a_{2018}\right)$. To find a recurrence relation for $a_{i}$, note that to get to some point in $( \pm 1, \pm 1, \pm 1, \pm 1)$, we must either come from a previous point of the form $( \pm 1, \pm 1, \pm 1, \pm 1)$ or the point $(0,0,0,0)$. In order to go to one point of the form $( \pm 1, \pm 1, \pm 1, \pm 1)$ through $(0,0,0,0)$ from the point $( \pm 1, \pm 1, \pm 1, \pm 1)$, we have one way of going to the origin and 16 ways to pick which point we go to after the origin. Additionally, if the previous point we visit is another point of the form $( \pm 1, \pm 1, \pm 1, \pm 1)$ then we have 4 possible directions to go in. Therefore the recurrence relation for $a_{i}$ is $a_{i}=4 a_{i-1}+16 a_{i-2}$. Solving the linear recurrence yields $a_{i}=\frac{1}{\sqrt{5}}(2+2 \sqrt{5})^{i}-\frac{1}{\sqrt{5}}(2-2 \sqrt{5})^{i}=4^{i} F_{i+1}$ so it suffices to find $\nu_{2}\left(F_{2019}\right)$. We have $F_{n} \equiv 0,1,1,2,3,1(\bmod 4)$ for $n \equiv 0,1,2,3,4,5(\bmod 6)$, so $\nu_{2}\left(F_{2019}\right)=1$, and the answer is $4+2 \cdot 2018+1=4041$.

4041

HMMT_2

Let $S=\{(x, y) \mid x>0, y>0, x+y<200$, and $x, y \in \mathbb{Z}\}$. Find the number of parabolas $\mathcal{P}$ with vertex $V$ that satisfy the following conditions: - $\mathcal{P}$ goes through both $(100,100)$ and at least one point in $S$, - $V$ has integer coordinates, and - $\mathcal{P}$ is tangent to the line $x+y=0$ at $V$.

We perform the linear transformation $(x, y) \rightarrow(x-y, x+y)$, which has the reverse transformation $(a, b) \rightarrow\left(\frac{a+b}{2}, \frac{b-a}{2}\right)$. Then the equivalent problem has a parabola has a vertical axis of symmetry, goes through $A=(0,200)$, a point $B=(u, v)$ in $S^{\prime}=\{(x, y) \mid x+y>0, x>y, y<200, x, y \in \mathbb{Z}, \text { and } x \equiv y \bmod 2\}$ and a new vertex $W=(w, 0)$ on $y=0$ with $w$ even. Then $\left(1-\frac{u}{w}\right)^{2}=\frac{v}{200}$. The only way the RHS can be the square of a rational number is if $\frac{u}{w}=\frac{v^{\prime}}{10}$ where $v=2\left(10-v^{\prime}\right)^{2}$. Since $v$ is even, we can find conditions so that $u, w$ are both even: $v^{\prime} \in\{1,3,7,9\} \Longrightarrow\left(2 v^{\prime}\right)|u, 20| w$, $v^{\prime} \in\{2,4,6,8\} \Longrightarrow v^{\prime}|u, 10| w$, $v^{\prime}=5 \Longrightarrow 2|u, 4| w$. It follows that any parabola that goes through $v^{\prime} \in\{3,7,9\}$ has a point with $v^{\prime}=1$, and any parabola that goes through $v^{\prime} \in\{4,6,8\}$ has a point with $v^{\prime}=2$. We then count the following parabolas: - The number of parabolas going through $(2 k, 162)$, where $k$ is a nonzero integer with $|2 k|<162$. - The number of parabolas going through $(2 k, 128)$ not already counted, where $k$ is a nonzero integer with $|2 k|<128$. (Note that this passes through $(k, 162)$.) - The number of parabolas going through $(2 k, 50)$ not already counted, where $k$ is a nonzero integer with $|2 k|<50$. (Note that this passes through $\left(\frac{2 k}{5}, 162\right)$, and any overlap must have been counted in the first case.) The number of solutions is then $2\left(80+\frac{1}{2} \cdot 64+\frac{4}{5} \cdot 25\right)=264$.

264

HMMT_2

Let $\triangle A B C$ be a triangle with $A B=7, B C=1$, and $C A=4 \sqrt{3}$. The angle trisectors of $C$ intersect $\overline{A B}$ at $D$ and $E$, and lines $\overline{A C}$ and $\overline{B C}$ intersect the circumcircle of $\triangle C D E$ again at $X$ and $Y$, respectively. Find the length of $X Y$.

Let $O$ be the cirumcenter of $\triangle C D E$. Observe that $\triangle A B C \sim \triangle X Y C$. Moreover, $\triangle A B C$ is a right triangle because $1^{2}+(4 \sqrt{3})^{2}=7^{2}$, so the length $X Y$ is just equal to $2 r$, where $r$ is the radius of the circumcircle of $\triangle C D E$. Since $D$ and $E$ are on the angle trisectors of angle $C$, we see that $\triangle O D E, \triangle X D O$, and $\triangle Y E O$ are equilateral. The length of the altitude from $C$ to $A B$ is $\frac{4 \sqrt{3}}{7}$. The distance from $C$ to $X Y$ is $\frac{X Y}{A B} \cdot \frac{4 \sqrt{3}}{7}=\frac{2 r}{7} \cdot \frac{4 \sqrt{3}}{7}$, while the distance between lines $X Y$ and $A B$ is $\frac{r \sqrt{3}}{2}$. Hence we have $\frac{4 \sqrt{3}}{7}=\frac{2 r}{7} \cdot \frac{4 \sqrt{3}}{7}+\frac{r \sqrt{3}}{2}$. Solving for $r$ gives that $r=\frac{56}{65}$, so $X Y=\frac{112}{65}$.

\frac{112}{65}

HMMT_2

In a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, the host opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door. What is the probability that his new door is hiding a prize?

If Bob initially chooses a door with a prize, then he will not find a prize by switching. With probability $5 / 7$ his original door does not hide the prize. After the host opens the three doors, the remaining three doors have equal probability of hiding the prize. Therefore, the probability that Bob finds the prize is $\frac{5}{7} \times \frac{1}{3}=\frac{5}{21}$.

\frac{5}{21}

HMMT_2

A collection $\mathcal{S}$ of 10000 points is formed by picking each point uniformly at random inside a circle of radius 1. Let $N$ be the expected number of points of $\mathcal{S}$ which are vertices of the convex hull of the $\mathcal{S}$. (The convex hull is the smallest convex polygon containing every point of $\mathcal{S}$.) Estimate $N$.

Here is C++ code by Benjamin Qi to estimate the answer via simulation. It is known that the expected number of vertices of the convex hull of $n$ points chosen uniformly at random inside a circle is $O\left(n^{1 / 3}\right)$. See "On the Expected Complexity of Random Convex Hulls" by Har-Peled.

72.8

HMMT_2

How many solutions in nonnegative integers $(a, b, c)$ are there to the equation $2^{a}+2^{b}=c!\quad ?$

We can check that $2^{a}+2^{b}$ is never divisible by 7 , so we must have $c<7$. The binary representation of $2^{a}+2^{b}$ has at most two 1 's. Writing 0 !, 1 !, 2 !, \ldots, 6$ ! in binary, we can check that the only possibilities are $c=2,3,4$, giving solutions $(0,0,2),(1,2,3),(2,1,3)$, $(3,4,4),(4,3,4)$.

5

HMMT_2

A palindrome is a positive integer that reads the same backwards as forwards, such as 82328. What is the smallest 5 -digit palindrome that is a multiple of 99 ?

Write the number as $X Y Z Y X$. This is the same as $10000 X+1000 Y+100 Z+10 Y+X=$ $99(101 X+10 Y+Z)+20 Y+2 X+Z$. We thus want $20 Y+2 X+Z$ to be a multiple of 99 , with $X$ as small as possible. This expression cannot be larger than $20 \cdot 9+2 \cdot 9+9=207$, and it is greater than 0 (since $X \neq 0$ ), so for this to be a multiple of 99 , it must equal 99 or 198. Consider these two cases. To get 198, we must have $Y=9$, which then leaves $2 X+Z=18$. The smallest possible $X$ is 5 , and then $Z$ becomes 8 and we have the number 59895 . To get 99 , we must have $Y=4$. Then, $2 X+Z=19$, and, as above, we find the minimal $X$ is 5 and then $Z=9$. This gives us the number 54945 . This is smaller than the other number, so it is the smallest number satisfying the conditions of the problem.

54945

HMMT_2

We call a positive integer $t$ good if there is a sequence $a_{0}, a_{1}, \ldots$ of positive integers satisfying $a_{0}=15, a_{1}=t$, and $a_{n-1} a_{n+1}=\left(a_{n}-1\right)\left(a_{n}+1\right)$ for all positive integers $n$. Find the sum of all good numbers.

By the condition of the problem statement, we have $a_{n}^{2}-a_{n-1} a_{n+1}=1=a_{n-1}^{2}-a_{n-2} a_{n}$. This is equivalent to $\frac{a_{n-2}+a_{n}}{a_{n-1}}=\frac{a_{n-1}+a_{n+1}}{a_{n}}$. Let $k=\frac{a_{0}+a_{2}}{a_{1}}$. Then we have $\frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_{n-2}+a_{n}}{a_{n-1}}=\frac{a_{n-3}+a_{n-1}}{a_{n-2}}=\cdots=\frac{a_{0}+a_{2}}{a_{1}}=k$. Therefore we have $a_{n+1}=k a_{n}-a_{n-1}$ for all $n \geq 1$. We know that $k$ is a positive rational number because $a_{0}, a_{1}$, and $a_{2}$ are all positive integers. We claim that $k$ must be an integer. Suppose that $k=\frac{p}{q}$ with $\operatorname{gcd}(p, q)=1$. Since $k a_{n}=a_{n-1}+a_{n+1}$ is always an integer for $n \geq 1$, we must have $q \mid a_{n}$ for all $n \geq 1$. This contradicts $a_{2}^{2}-a_{1} a_{3}=1$. Conversely, if $k$ is an integer, inductively all $a_{i}$ are integers. Now we compute $a_{2}=\frac{t^{2}-1}{15}$, so $k=\frac{t^{2}+224}{15 t}$ is an integer. Therefore $15 k-t=\frac{224}{t}$ is an integer. Combining with the condition that $a_{2}$ is an integer limits the possible values of $t$ to $1,4,14,16,56$, 224. The values $t<15$ all lead to $a_{n}=0$ for some $n$ whereas $t>15$ leads to a good sequence. The sum of the solutions is $16+56+224=296$.

296

HMMT_2

Anastasia is taking a walk in the plane, starting from $(1,0)$. Each second, if she is at $(x, y)$, she moves to one of the points $(x-1, y),(x+1, y),(x, y-1)$, and $(x, y+1)$, each with $\frac{1}{4}$ probability. She stops as soon as she hits a point of the form $(k, k)$. What is the probability that $k$ is divisible by 3 when she stops?

The key idea is to consider $(a+b, a-b)$, where $(a, b)$ is where Anastasia walks on. Then, the first and second coordinates are independent random walks starting at 1, and we want to find the probability that the first is divisible by 3 when the second reaches 0 for the first time. Let $C_{n}$ be the $n$th Catalan number. The probability that the second random walk first reaches 0 after $2 n-1$ steps is $\frac{C_{n-1}}{2^{2 n-1}}$, and the probability that the first is divisible by 3 after $2 n-1$ steps is $\frac{1}{2^{2 n-1}} \sum_{i \equiv n \bmod 3}\binom{2 n-1}{i}$ (by letting $i$ be the number of -1 steps). We then need to compute $\sum_{n=1}^{\infty}\left(\frac{C_{n-1}}{4^{2 n-1}} \sum_{i \equiv n \bmod 3}\binom{2 n-1}{i}\right)$. By a standard root of unity filter, $\sum_{i \equiv n \bmod 3}\binom{2 n-1}{i}=\frac{4^{n}+2}{6}$. Letting $P(x)=\frac{2}{1+\sqrt{1-4 x}}=\sum_{n=0}^{\infty} C_{n} x^{n}$ be the generating function for the Catalan numbers, we find that the answer is $\frac{1}{6} P\left(\frac{1}{4}\right)+\frac{1}{12} P\left(\frac{1}{16}\right)=\frac{1}{3}+\frac{1}{12} \cdot \frac{2}{1+\sqrt{\frac{3}{4}}}=\frac{3-\sqrt{3}}{3}$.

\frac{3-\sqrt{3}}{3}

HMMT_2

Let $A B C$ be a triangle and $\omega$ be its circumcircle. The point $M$ is the midpoint of arc $B C$ not containing $A$ on $\omega$ and $D$ is chosen so that $D M$ is tangent to $\omega$ and is on the same side of $A M$ as $C$. It is given that $A M=A C$ and $\angle D M C=38^{\circ}$. Find the measure of angle $\angle A C B$.

By inscribed angles, we know that $\angle B A C=38^{\circ} \cdot 2=76^{\circ}$ which means that $\angle C=104^{\circ}-\angle B$. Since $A M=A C$, we have $\angle A C M=\angle A M C=90^{\circ}-\frac{\angle M A C}{2}=71^{\circ}$. Once again by inscribed angles, this means that $\angle B=71^{\circ}$ which gives $\angle C=33^{\circ}$.

33^{\circ}

HMMT_2

Square $A B C D$ has side length 1. A dilation is performed about point $A$, creating square $A B^{\prime} C^{\prime} D^{\prime}$. If $B C^{\prime}=29$, determine the area of triangle $B D C^{\prime}$.

$29^{2}-2 \cdot \frac{1}{2}(29)\left(\frac{29}{2}\right)-\frac{1}{2}=420$.

420

HMMT_2

How many ways are there to label the faces of a regular octahedron with the integers 18, using each exactly once, so that any two faces that share an edge have numbers that are relatively prime? Physically realizable rotations are considered indistinguishable, but physically unrealizable reflections are considered different.

Well, instead of labeling the faces of a regular octahedron, we may label the vertices of a cube. Then, as no two even numbers may be adjacent, the even numbers better form a regular tetrahedron, which can be done in 2 ways (because rotations are indistinguishable but reflections are different). Then 3 must be opposite 6, and the remaining numbers $-1,5,7-$ may be filled in at will, in $3!=6$ ways. The answer is thus $2 \times 6=12$.

12

HMMT_2

Cyclic pentagon $ABCDE$ has a right angle $\angle ABC=90^{\circ}$ and side lengths $AB=15$ and $BC=20$. Supposing that $AB=DE=EA$, find $CD$.

By Pythagoras, $AC=25$. Since $\overline{AC}$ is a diameter, angles $\angle ADC$ and $\angle AEC$ are also right, so that $CE=20$ and $AD^{2}+CD^{2}=AC^{2}$ as well. Beginning with Ptolemy's theorem, $$\begin{aligned} & (AE \cdot CD+AC \cdot DE)^{2}=AD^{2} \cdot EC^{2}=\left(AC^{2}-CD^{2}\right)EC^{2} \\ & \quad \Longrightarrow CD^{2}\left(AE^{2}+EC^{2}\right)+2 \cdot CD \cdot AE^{2} \cdot AC+AC^{2}\left(DE^{2}-EC^{2}\right)=0 \\ & \quad \Longrightarrow CD^{2}+18 CD-175=0 \end{aligned}$$ It follows that $CD^{2}+18CD-175=0$, from which $CD=7$.

7

HMMT_2

A vertex-induced subgraph is a subset of the vertices of a graph together with any edges whose endpoints are both in this subset. An undirected graph contains 10 nodes and $m$ edges, with no loops or multiple edges. What is the minimum possible value of $m$ such that this graph must contain a nonempty vertex-induced subgraph where all vertices have degree at least 5?

Suppose that we want to find the vertex-induced subgraph of maximum size where each vertex has degree at least 5. To do so, we start with the entire graph and repeatedly remove any vertex with degree less than 5. If there are vertices left after this process terminates, then the subgraph induced by these vertices must have all degrees at least 5. Conversely, if there is a vertex-induced subgraph where all degrees are at least 5, then none of these vertices can be removed during the removing process. Thus, there are vertices remaining after this process if and only if such a vertex-induced subgraph exists. If the process ends with an empty graph, the largest possible number of edges are removed when the first 5 removed vertices all have 4 edges at the time of removal, and the last 5 vertices are all connected to each other, resulting in $5 \times 4+4+3+2+1+0=30$ removed edges. The answer is $30+1=31$.

31

HMMT_2

There exist several solutions to the equation $1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x}$ where $x$ is expressed in degrees and $0^{\circ}<x<180^{\circ}$. Find the sum of all such solutions.

We first apply sum-to-product and product-to-sum: $\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x}$. $2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)$. Factoring out $\sin (2 x)=0$, $\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x)$. Factoring out $\cos (1.5 x)=0$ (which gives us $60^{\circ}$ as a solution), $\sin (2.5 x)=2 \cos (2 x) \sin (1.5 x)$. Convert into complex numbers, we get $\left(x^{3.5}-x^{-3.5}\right)-\left(x^{0.5}-x^{-0.5}\right)=\left(x^{2.5}-x^{-2.5}\right)$. $x^{7}-x^{6}-x^{4}+x^{3}+x-1=0$. $(x-1)\left(x^{6}-x^{3}+1\right)=0$. We recognize the latter expression as $\frac{x^{9}+1}{x^{3}+1}$, giving us $x=0^{\circ}, 20^{\circ}, 100^{\circ}, 140^{\circ}, 220^{\circ}, 260^{\circ}, 340^{\circ}$. The sum of the solutions is $20^{\circ}+60^{\circ}+100^{\circ}+140^{\circ}=320^{\circ}$.

320^{\circ}

HMMT_2

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

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 $2^{3} \cdot 3!=48$.

48

HMMT_2

A snake of length $k$ is an animal which occupies an ordered $k$-tuple $\left(s_{1}, \ldots, s_{k}\right)$ of cells in a $n \times n$ grid of square unit cells. These cells must be pairwise distinct, and $s_{i}$ and $s_{i+1}$ must share a side for $i=1, \ldots, k-1$. If the snake is currently occupying $\left(s_{1}, \ldots, s_{k}\right)$ and $s$ is an unoccupied cell sharing a side with $s_{1}$, the snake can move to occupy $\left(s, s_{1}, \ldots, s_{k-1}\right)$ instead. Initially, a snake of length 4 is in the grid $\{1,2, \ldots, 30\}^{2}$ occupying the positions $(1,1),(1,2),(1,3),(1,4)$ with $(1,1)$ as its head. The snake repeatedly makes a move uniformly at random among moves it can legally make. Estimate $N$, the expected number of moves the snake makes before it has no legal moves remaining.

Let $n=30$. The snake can get stuck in only 8 positions, while the total number of positions is about $n^{2} \times 4 \times 3 \times 3=36 n^{2}$. We can estimate the answer as $\frac{36 n^{2}}{8}=4050$, which is good enough for 13 points. Let's try to compute the answer as precisely as possible. For each head position $(a, b)$ and tail orientation $c \in[0,36)$, let $x=36(n a+b)+c$ be an integer denoting the current state of the snake. Let $E_{x}$ by the expected number of moves the snake makes if it starts at state $x$. If from state $x$ the snake can transition to any of states $y_{1}, y_{2}, \ldots, y_{k}$, then add an equation of the form $E_{x}-\frac{1}{k} \sum_{i=1}^{k} E_{y_{i}}=1$. Otherwise, if there are no transitions out of state $x$ then set $E_{x}=0$. It suffices to solve a system of $36 n^{2}$ linear equations for $E_{0}, E_{1}, \ldots, E_{36 n^{2}-1}$. Then the answer will equal $E_{i}$, where $i$ corresponds to the state described in the problem statement. Naively, using Gaussian elimination would require about $\left(36 n^{2}\right)^{3} \approx 3.4 \cdot 10^{13}$ operations, which is too slow. Also, it will require too much memory to store $\left(36 n^{2}\right)^{2}$ real numbers at once. We can use the observation that initially, the maximum difference between any two indices within the same equation is at most $\approx 72 n$, so Gaussian elimination only needs to perform approximately $\left(36 n^{2}\right) \cdot(72 n)^{2} \approx 1.5 \cdot 10^{11}$ operations. Furthermore, we'll only need to store $\approx\left(36 n^{2}\right) \cdot(72 n)$ real numbers at a time. Benjamin Qi's solution ends up finishing in less than two minutes for $n=30$ (C++ code).

4571.8706930

HMMT_2

Let $ABCD$ be a convex quadrilateral with $AC=7$ and $BD=17$. Let $M, P, N, Q$ be the midpoints of sides $AB, BC, CD, DA$ respectively. Compute $MN^{2}+PQ^{2}$.

$MPNQ$ is a parallelogram whose side lengths are 3.5 and 8.5 so the sum of squares of its diagonals is $\frac{7^{2}+17^{2}}{2}=169$.

169

HMMT_2

Find all integers $m$ such that $m^{2}+6 m+28$ is a perfect square.

We must have $m^{2}+6 m+28=n^{2}$, where $n$ is an integer. Rewrite this as $(m+3)^{2}+19=$ $n^{2} \Rightarrow n^{2}-(m+3)^{2}=19 \Rightarrow(n-m-3)(n+m+3)=19$. Let $a=n-m-3$ and $b=n+m+3$, so we want $a b=19$. This leaves only 4 cases: - $a=1, b=19$. Solve the system $n-m-3=1$ and $n+m+3=19$ to get $n=10$ and $m=6$, giving one possible solution. - $a=19, b=1$. Solve the system, as above, to get $n=10$ and $m=-12$. - $a=-1, b=-19$. We get $n=-10$ and $m=-12$. - $a=-19, b=-1$. We get $n=-10$ and $m=6$. Thus the only $m$ are 6 and -12 .

6,-12

HMMT_2

At a nursery, 2006 babies sit in a circle. Suddenly each baby pokes the baby immediately to either its left or its right, with equal probability. What is the expected number of unpoked babies?

The probability that any given baby goes unpoked is $1 / 4$. So the answer is $2006 / 4=1003 / 2$.

\frac{1003}{2}

HMMT_2

Point $A$ lies at $(0,4)$ and point $B$ lies at $(3,8)$. Find the $x$-coordinate of the point $X$ on the $x$-axis maximizing $\angle AXB$.

Let $X$ be a point on the $x$-axis and let $\theta=\angle AXB$. We can easily see that the circle with diameter $AB$ does not meet the $x$-axis, so $\theta \leq \pi$. Thus, maximizing $\theta$ is equivalent to maximizing $\sin \theta$. By the Law of Sines, this in turn is equivalent to minimizing the circumradius of triangle $ABX$. This will occur when the circumcircle of $ABX$ is the smaller of the two circles through $A$ and $B$ tangent to the $x$-axis. So let $X$ now be this point of tangency. Extend line $AB$ to meet the $x$-axis at $C=(-3,0)$; by Power of a Point $CX^{2}=CA \cdot CB=50$ so $CX=5 \sqrt{2}$. Clearly $X$ has larger $x$-coordinate than $C$, so the $x$-coordinate of $X$ is $5 \sqrt{2}-3$.

5 \sqrt{2}-3

HMMT_2

Given right triangle $ABC$, with $AB=4, BC=3$, and $CA=5$. Circle $\omega$ passes through $A$ and is tangent to $BC$ at $C$. What is the radius of $\omega$?

Let $O$ be the center of $\omega$, and let $M$ be the midpoint of $AC$. Since $OA=OC$, $OM \perp AC$. Also, $\angle OCM=\angle BAC$, and so triangles $ABC$ and $CMO$ are similar. Then, $CO/CM=AC/AB$, from which we obtain that the radius of $\omega$ is $CO=\frac{25}{8}$.

\frac{25}{8}

HMMT_2

Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $n$ to $5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the number of beans Alice has is greater than 2008 and has last two digits 42. What is the minimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

Consider the number of beans Alice has in base 5. Note that $2008=31013_{5}, 42=132_{5}$, and $100=400_{5}$. Now, suppose Alice has $d_{k} \cdots d_{2} d_{1}$ beans when she wins; the conditions for winning mean that these digits must satisfy $d_{2} d_{1}=32, d_{k} \cdots d_{3} \geq 310$, and $d_{k} \cdots d_{3}=4i+1$ for some $i$. To gain these $d_{k} \cdots d_{2} d_{1}$ beans, Alice must spend at least $5\left(d_{1}+d_{2}+\cdots+d_{k}\right)+k-1$ cents (5 cents to get each bean in the "units digit" and $k-1$ cents to promote all the beans). We now must have $k \geq 5$ because $d_{k} \cdots d_{2} d_{1}>2008$. If $k=5$, then $d_{k} \geq 3$ since $d_{k} \cdots d_{3} \geq 3100$; otherwise, we have $d_{k} \geq 1$. Therefore, if $k=5$, we have $5\left(d_{1}+d_{2}+\cdots+d_{k}\right)+k-1 \geq 44>36$; if $k>5$, we have $5\left(d_{1}+d_{2}+\cdots+d_{k}\right)+k-1 \geq 30+k-1 \geq 35$. But we can attain 36 cents by taking $d_{k} \cdots d_{3}=1000$, so this is indeed the minimum.

35

HMMT_2

Determine all \(\alpha \in \mathbb{R}\) such that for every continuous function \(f:[0,1] \rightarrow \mathbb{R}\), differentiable on \((0,1)\), with \(f(0)=0\) and \(f(1)=1\), there exists some \(\xi \in(0,1)\) such that \(f(\xi)+\alpha=f^{\prime}(\xi)\).

First consider the function \(h(x)=\frac{e^{x}-1}{e-1}\), which has the property that \(h^{\prime}(x)=\frac{e^{x}}{e-1}\). Note that \(h \in V\) and that \(h^{\prime}(x)-h(x)=1 /(e-1)\) is constant. As such, \(\alpha=1 /(e-1)\) is the only possible value that could possibly satisfy the condition from the problem. For \(f \in V\) arbitrary, let \(g(x)=f(x) e^{-x}+h(-x)\), with \(g(0)=0\) and also \(g(1)=e^{-1}+\frac{e^{-1}-1}{e-1}=0\). We compute that \(g^{\prime}(x)=f^{\prime}(x) e^{-x}-f(x) e^{-x}-h^{\prime}(-x)\). Now apply Rolle's Theorem to \(g\) on the interval \([0,1]\); it yields some \(\xi \in(0,1)\) with the property that \(g^{\prime}(\xi)=0 \Longrightarrow f^{\prime}(\xi) e^{-\xi}-f(\xi) e^{-\xi}-\frac{e^{-\xi}}{e-1}=0 \Longrightarrow f^{\prime}(\xi)=f(\xi)+\frac{1}{e-1}\) showing that \(\alpha=1 /(e-1)\) indeed satisfies the condition from the problem.

\(\alpha = \frac{1}{e-1}\)

imc

In how many ways can one fill a \(4 \times 4\) grid with a 0 or 1 in each square such that the sum of the entries in each row, column, and long diagonal is even?

First we name the elements of the square as follows: \(a_{11}, a_{12}, a_{13}, a_{14}, a_{21}, a_{22}, a_{23}, a_{24}, a_{31}, a_{32}, a_{33}, a_{34}, a_{41}, a_{42}, a_{43}, a_{44}\). We claim that for any given values of \(a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{32}\), and \(a_{33}\) (the + signs in the diagram below), there is a unique way to assign values to the rest of the entries such that all necessary sums are even. Taking additions mod 2, we have \(a_{14}=a_{11}+a_{12}+a_{13}\), \(a_{24}=a_{21}+a_{22}+a_{23}\), \(a_{44}=a_{11}+a_{22}+a_{33}\), \(a_{42}=a_{12}+a_{22}+a_{32}\), \(a_{43}=a_{13}+a_{23}+a_{33}\). Since the 4th column, the 4th row, and the 1st column must have entries that sum to 0, we have \(a_{34}=a_{14}+a_{24}+a_{44}=a_{12}+a_{13}+a_{21}+a_{23}+a_{33}\), \(a_{41}=a_{42}+a_{43}+a_{44}=a_{11}+a_{12}+a_{13}+a_{23}+a_{32}\), \(a_{31}=a_{11}+a_{21}+a_{41}=a_{12}+a_{13}+a_{21}+a_{23}+a_{32}\). It is easy to check that the sum of entries in every row, column, and the main diagonal is even. Since there are \(2^{8}=256\) ways to assign the values to the initial 8 entries, there are exactly 256 ways to fill the board.

256

HMMT_2

Determine the number of four-digit integers $n$ such that $n$ and $2n$ are both palindromes.

Let $n=\underline{a} \underline{b} \underline{b} \underline{a}$. If $a, b \leq 4$ then there are no carries in the multiplication $n \times 2$, and $2n=(2a)(2b)(2b)(2a)$ is a palindrome. We shall show conversely that if $n$ and $2n$ are palindromes, then necessarily $a, b \leq 4$. Hence the answer to the problem is $4 \times 5=\mathbf{20}$ (because $a$ cannot be zero). If $a \geq 5$ then $2n$ is a five-digit number whose most significant digit is 1, but because $2n$ is even, its least significant digit is even, contradicting the assumption that $2n$ is a palindrome. Therefore $a \leq 4$. Consequently $2n$ is a four-digit number, and its tens and hundreds digits must be equal. Because $a \leq 4$, there is no carry out of the ones place in the multiplication $n \times 2$, and therefore the tens digit of $2n$ is the ones digit of $2b$. In particular, the tens digit of $2n$ is even. But if $b \geq 5$, the carry out of the tens place makes the hundreds digit of $2n$ odd, which is impossible. Hence $b \leq 4$ as well.

20

HMMT_2

Let $A B C D$ be a cyclic quadrilateral, and let segments $A C$ and $B D$ intersect at $E$. Let $W$ and $Y$ be the feet of the altitudes from $E$ to sides $D A$ and $B C$, respectively, and let $X$ and $Z$ be the midpoints of sides $A B$ and $C D$, respectively. Given that the area of $A E D$ is 9, the area of $B E C$ is 25, and $\angle E B C-\angle E C B=30^{\circ}$, then compute the area of $W X Y Z$.

Reflect $E$ across $D A$ to $E_{W}$, and across $B C$ to $E_{Y}$. As $A B C D$ is cyclic, $\triangle A E D$ and $\triangle B E C$ are similar. Thus $E_{W} A E D$ and $E B E_{Y} C$ are similar too. Now since $W$ is the midpoint of $E_{W} E, X$ is the midpoint of $A B, Y$ is the midpoint of $E E_{Y}$, and $Z$ is the midpoint of $D C$, we have that $W X Y Z$ is similar to $E_{W} A E D$ and $E B E_{Y} C$. From the given conditions, we have $E W: E Y=3: 5$ and $\angle W E Y=150^{\circ}$. Suppose $E W=3 x$ and $E Y=5 x$. Then by the law of cosines, we have $W Y=\sqrt{34+15 \sqrt{3}} x$. Thus, $E_{W} E: W Y=6: \sqrt{34+15 \sqrt{3}}$. So by the similarity ratio, $[W X Y Z]=\left[E_{W} A E D\right]\left(\frac{\sqrt{34+15 \sqrt{3}}}{6}\right)^{2}=2 \cdot 9 \cdot\left(\frac{34+15 \sqrt{3}}{36}\right)=17+\frac{15}{2} \sqrt{3}$.

17+\frac{15}{2} \sqrt{3}

HMMT_2

Consider an equilateral triangular grid $G$ with 20 points on a side, where each row consists of points spaced 1 unit apart. More specifically, there is a single point in the first row, two points in the second row, ..., and 20 points in the last row, for a total of 210 points. Let $S$ be a closed non-selfintersecting polygon which has 210 vertices, using each point in $G$ exactly once. Find the sum of all possible values of the area of $S$.

Imagine deforming the triangle lattice such that now it looks like a lattice of 45-45-90 right triangles with legs of length 1. Note that by doing this, the area has multiplied by $\frac{2}{\sqrt{3}}$, so we need to readjust out answer on the isosceles triangle lattice by a factor of $\frac{\sqrt{3}}{2}$ at the end. By Pick's Theorem, the area in the new lattice is given by $I+\frac{P}{2}-1=0+105-1=104$. Therefore, the answer is $104 \cdot \frac{\sqrt{3}}{2}=52 \sqrt{3}$.

52 \sqrt{3}

HMMT_2

A Vandal and a Moderator are editing a Wikipedia article. The article originally is error-free. Each day, the Vandal introduces one new error into the Wikipedia article. At the end of the day, the moderator checks the article and has a $2 / 3$ chance of catching each individual error still in the article. After 3 days, what is the probability that the article is error-free?

Consider the error that was introduced on day 1. The probability that the Moderator misses this error on all three checks is $1 / 3^{3}$, so the probability that this error gets removed is $1-\frac{1}{3^{3}}$. Similarly, the probability that the moderator misses the other two errors are $1-\frac{1}{3^{2}}$ and $1-\frac{1}{3}$. So the probability that the article is error-free is $$\left(1-\frac{1}{3^{3}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{3}\right)=\frac{416}{729}$$

\frac{416}{729}

HMMT_2

Begining at a vertex, an ant crawls between the vertices of a regular octahedron. After reaching a vertex, it randomly picks a neighboring vertex (sharing an edge) and walks to that vertex along the adjoining edge (with all possibilities equally likely.) What is the probability that after walking along 2006 edges, the ant returns to the vertex where it began?

For each nonnegative integer $n$, let $a_{n}, b_{n}$, and $c_{n}$ denote the respective probabilities that the ant is where it began, at a neighbor of where it began, or is opposite where it began after moving along $n$ edges. We seek $a_{2006}$. We have $a_{0}=1$ and $b_{0}=c_{0}=0$. We also have the recursive system $$\begin{aligned} a_{n} & =\frac{b_{n-1}}{4} \\ b_{n} & =a_{n-1}+\frac{b_{n-1}}{2}+c_{n-1} \\ c_{n} & =\frac{b_{n-1}}{4} \end{aligned}$$ for integers $n \geq 1$. Substituting into the second equation we have $b_{n}=\frac{b_{n-1}}{2}+\frac{b_{n-2}}{2}$ for $n \geq 2$. Solving the characteristic equation $x^{2}-\frac{x}{2}-\frac{1}{2}=0$ for $x=1, \frac{-1}{2}$, we write $b_{n}=a \cdot 1^{n}+b(-1 / 2)^{n}$. Using $b_{0}=0, b_{1}=1$, we compute $$b_{n}=\frac{2}{3} \cdot\left(1-(-1 / 2)^{n}\right)$$ From which we find $a_{2006}=\frac{b_{2005}}{4}=\frac{1}{6}\left(1+\frac{1}{2^{2005}}\right)=\frac{2^{2005}+1}{3 \cdot 2^{2006}}$.

\frac{2^{2005}+1}{3 \cdot 2^{2006}}

HMMT_2

For odd primes $p$, let $f(p)$ denote the smallest positive integer $a$ for which there does not exist an integer $n$ satisfying $p \mid n^{2}-a$. Estimate $N$, the sum of $f(p)^{2}$ over the first $10^{5}$ odd primes $p$. An estimate of $E>0$ will receive $\left\lfloor 22 \min (N / E, E / N)^{3}\right\rfloor$ points.

Note that the smallest quadratic nonresidue $a$ is always a prime, because if $a=b c$ with $b, c>1$ then one of $b$ and $c$ is also a quadratic nonresidue. We apply the following heuristic: if $p_{1}$, $p_{2}, \ldots$ are the primes in increasing order, then given a "uniform random prime" $q$, the values of $\left(\frac{p_{1}}{q}\right),\left(\frac{p_{2}}{q}\right), \ldots$ are independent and are 1 with probability $\frac{1}{2}$ and -1 with probability $\frac{1}{2}$. Of course, there is no such thing as a uniform random prime. More rigorously, for any $n$, the joint distributions of $\left(\frac{p_{1}}{q}\right), \ldots,\left(\frac{p_{n}}{q}\right)$ where $q$ is a uniform random prime less than $N$ converges in distribution to $n$ independent coin flips between 1 and -1 as $N \rightarrow \infty$. For ease of explanation, we won't adopt this more formal view, but it is possible to make the following argument rigorous by looking at primes $q<N$ and sending $N \rightarrow \infty$. Given any $n$, the residue of $q \bmod n$ is uniform over the $\varphi(n)$ residues $\bmod n$ that are relatively prime to $n$. By quadratic reciprocity, conditioned on either $q \equiv 1(\bmod 4)$ or $q \equiv 3(\bmod 4)$, exactly half of the nonzero residues $\bmod p_{n}$ satisfy $\left(\frac{p_{n}}{q}\right)=1$ and exactly half satisfy $\left(\frac{p_{n}}{q}\right)=-1$ for odd $p_{n}$ (the case of $p_{n}=2$ is slightly different and one must look mod 8, but the result is the same). The residue of $q \bmod 8, p_{2}, p_{3}, \ldots, p_{n}$ are independent as these are pairwise relatively prime, yielding our heuristic. Thus, we may model our problem of finding the smallest quadratic nonresidue with the following process: independent fair coins are flipped for each prime, and we take the smallest prime that flipped heads. We can estimate the expected value of $f(p)^{2}$ as $\sum_{n=1}^{\infty} \frac{p_{n}^{2}}{2^{n}}$. Looking at the first few terms gives us $\frac{2^{2}}{2}+\frac{3^{2}}{4}+\frac{5^{2}}{8}+\frac{7^{2}}{16}+\frac{11^{2}}{32}+\frac{13^{2}}{64}+\frac{17^{2}}{128}+\frac{19^{2}}{256}+\frac{23^{2}}{512}+\frac{29^{2}}{1024} \approx 22$. The terms after this decay rapidly, so a good approximation is $E=22 \cdot 10^{5}$, good enough for 20 points. The more inaccurate $E=20 \cdot 10^{5}$ earns 15 points. This Python code computes the exact answer: ``` def smallest_nqr(p): for a in range(1,p): if pow(a, (p-1)//2,p)==p-1: return a ```import sympyprint(sum([smallest_nqr(p)**2 for p in sympy.ntheory.primerange(3,sympy.prime(10**5+2))])) Remark. In 1961, Erdős showed that as $N \rightarrow \infty$, the average value of $f(p)$ over odd primes $p<N$ will converge to $\sum_{n=1}^{\infty} \frac{p_{n}}{2^{n}} \approx 3.675$.

2266067

HMMT_2

In the alphametic $W E \times E Y E=S C E N E$, each different letter stands for a different digit, and no word begins with a 0. The $W$ in this problem has the same value as the $W$ in problem 31. Find $S$.

Problems 31-33 go together. See below.

5

HMMT_2

Two sides of a regular $n$-gon are extended to meet at a $28^{\circ}$ angle. What is the smallest possible value for $n$?

We note that if we inscribe the $n$-gon in a circle, then according to the inscribed angle theorem, the angle between two sides is $\frac{1}{2}$ times some $x-y$, where $x$ and $y$ are integer multiples of the arc measure of one side of the $n$-gon. Thus, the angle is equal to $\frac{1}{2}$ times an integer multiple of $\frac{360}{n}$, so $\frac{1}{2} \cdot k \cdot \frac{360}{n}=28$ for some integer $k$. Simplifying gives $7 n=45 k$, and since all $k$ are clearly attainable, the smallest possible value of $n$ is 45.

45

HMMT_2

Let $A B C$ be a triangle with $A B=2, C A=3, B C=4$. Let $D$ be the point diametrically opposite $A$ on the circumcircle of $A B C$, and let $E$ lie on line $A D$ such that $D$ is the midpoint of $\overline{A E}$. Line $l$ passes through $E$ perpendicular to $\overline{A E}$, and $F$ and $G$ are the intersections of the extensions of $\overline{A B}$ and $\overline{A C}$ with $l$. Compute $F G$.

Using Heron's formula we arrive at $[A B C]=\frac{3 \sqrt{15}}{4}$. Now invoking the relation $[A B C]=\frac{a b c}{4 R}$ where $R$ is the circumradius of $A B C$, we compute $R^{2}=\left(\frac{2 \cdot 3}{[A B C]^{2}}\right)=$ $\frac{64}{15}$. Now observe that $\angle A B D$ is right, so that $B D E F$ is a cyclic quadrilateral. Hence $A B \cdot A F=A D \cdot A E=2 R \cdot 4 R=\frac{512}{15}$. Similarly, $A C \cdot A G=\frac{512}{15}$. It follows that $B C G F$ is a cyclic quadrilateral, so that triangles $A B C$ and $A G F$ are similar. Then $F G=B C \cdot \frac{A F}{A C}=4 \cdot \frac{512}{2 \cdot 15 \cdot 3}=\frac{1024}{45}$

\frac{1024}{45}

HMMT_2

Compute the positive integer less than 1000 which has exactly 29 positive proper divisors.

Recall that the number $N=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}$ (where the $p_{i}$ are distinct primes) has exactly $(e_{1}+1)(e_{2}+1) \cdots(e_{k}+1)$ positive integer divisors including itself. We seek $N<1000$ such that this expression is 30. Since $30=2 \cdot 3 \cdot 5$, we take $e_{1}=1, e_{2}=2, e_{3}=4$. Then we see that $N=5^{1} 3^{2} 4^{2}=720$ is satisfactory.

720

HMMT_2

Let $\triangle A B C$ be a triangle inscribed in a unit circle with center $O$. Let $I$ be the incenter of $\triangle A B C$, and let $D$ be the intersection of $B C$ and the angle bisector of $\angle B A C$. Suppose that the circumcircle of $\triangle A D O$ intersects $B C$ again at a point $E$ such that $E$ lies on $I O$. If $\cos A=\frac{12}{13}$, find the area of $\triangle A B C$.

Consider the following lemma: Lemma. $A D \perp E O$. Proof. By the Shooting Lemma, the reflection of the midpoint $M$ of arc $B C$ not containing $A$ over $B C$ lies on $(A D O)$. Hence $\measuredangle A D E+\measuredangle D E O=\measuredangle M D C+\measuredangle D M^{\prime} O=\measuredangle M D C+\measuredangle M^{\prime} M D=90^{\circ}$. This is enough to imply $A D \perp E O$. Thus $I$ is the foot from $O$ onto $A D$. Now $A I^{2}+I O^{2}=A O^{2}$. By Euler's formula, $\left(\frac{r}{\sin \frac{A}{2}}\right)^{2}+R^{2}-2 R r=R^{2}$. Hence $r=2 R \sin ^{2} \frac{A}{2}$. Then $s=a+\frac{r}{\tan \frac{A}{2}}=a+R \sin A=3 R \sin A$ and $[A B C]=r s=\left(2 R \sin ^{2} \frac{A}{2}\right)(3 R \sin A)$. Since $R=1$, we get $[A B C]=3(1-\cos A) \sin A$. Plugging in $\sin A=\frac{5}{13}$ and $\cos A=\frac{12}{13}$, we get $[A B C]=3 \cdot \frac{1}{13} \cdot \frac{5}{13}=\frac{15}{169}$.

\frac{15}{169}

HMMT_2

Find all positive integers $n>1$ for which $\frac{n^{2}+7 n+136}{n-1}$ is the square of a positive integer.

Write $\frac{n^{2}+7 n+136}{n-1}=n+\frac{8 n+136}{n-1}=n+8+\frac{144}{n-1}=9+(n-1)+\frac{144}{(n-1)}$. We seek to find $p$ and $q$ such that $p q=144$ and $p+q+9=k^{2}$. The possibilities are seen to be $1+144+9=154,2+72+9=83,3+48+9=60,4+36+9=49,6+24+9=39$, $8+18+9=35,9+16+9=34$, and $12+12+9=33$. Of these, $\{p, q\}=\{4,36\}$ is the only solution to both equations. Hence $n-1=4,36$ and $n=5,37$.

5, 37

HMMT_2

There are 100 houses in a row on a street. A painter comes and paints every house red. Then, another painter comes and paints every third house (starting with house number 3) blue. Another painter comes and paints every fifth house red (even if it is already red), then another painter paints every seventh house blue, and so forth, alternating between red and blue, until 50 painters have been by. After this is finished, how many houses will be red?

House $n$ ends up red if and only if the largest odd divisor of $n$ is of the form $4 k+1$. We have 25 values of $n=4 k+1 ; 13$ values of $n=2(4 k+1)$ (given by $k=0,1,2, \ldots, 12$ ); 7 values of $n=4(4 k+1)(k=0,1, \ldots, 6) ; 3$ values of $n=8(4 k+1)(k=0,1,2) ; 2$ of the form $n=16(4 k+1$ ) (for $k=0,1)$; 1 of the form $n=32(4 k+1)$; and 1 of the form $n=64(4 k+1)$. Thus we have a total of $25+13+7+3+2+1+1=52$ red houses.

52

HMMT_2

Points $A, C$, and $B$ lie on a line in that order such that $A C=4$ and $B C=2$. Circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have $\overline{B C}, \overline{A C}$, and $\overline{A B}$ as diameters. Circle $\Gamma$ is externally tangent to $\omega_{1}$ and $\omega_{2}$ at $D$ and $E$ respectively, and is internally tangent to $\omega_{3}$. Compute the circumradius of triangle $C D E$.

Let the center of $\omega_{i}$ be $O_{i}$ for $i=1,2,3$ and let $O$ denote the center of $\Gamma$. Then $O, D$, and $O_{1}$ are collinear, as are $O, E$, and $O_{2}$. Denote by $F$ the point of tangency between $\Gamma$ and $\omega_{3}$; then $F, O$, and $O_{3}$ are collinear. Writing $r$ for the radius of $\Gamma$ we have $O O_{1}=r+2, O O_{2}=r+1, O O_{3}=3-r$. Now since $O_{1} O_{3}=1$ and $O_{3} O_{2}=2$, we apply Stewart's theorem: $$\begin{aligned} O O_{1}^{2} \cdot O_{2} O_{3}+O O_{2}^{2} \cdot O_{1} O_{3} & =O O_{3}^{2} \cdot O_{1} O_{2}+O_{1} O_{3} \cdot O_{3} O_{2} \cdot O_{1} O_{2} \\ 2(r+2)^{2}+(r+1)^{2} & =3(3-r)^{2}+1 \cdot 2 \cdot 3 \end{aligned}$$ We find $r=\frac{6}{7}$. Now the key observation is that the circumcircle of triangle $C D E$ is the incircle of triangle $O O_{1} O_{2}$. We easily compute the sides of $O O_{1} O_{2}$ to be $\frac{13}{7}, \frac{20}{7}$, and 3. By Heron's formula, the area of $O O_{1} O_{2}$ is $\frac{18}{7}$, but the semiperimeter is $\frac{27}{7}$, so the desired radius is $\frac{2}{3}$.

\frac{2}{3}

HMMT_2

There are \(n\) girls \(G_{1}, \ldots, G_{n}\) and \(n\) boys \(B_{1}, \ldots, B_{n}\). A pair \((G_{i}, B_{j})\) is called suitable if and only if girl \(G_{i}\) is willing to marry boy \(B_{j}\). Given that there is exactly one way to pair each girl with a distinct boy that she is willing to marry, what is the maximal possible number of suitable pairs?

We represent the problem as a graph with vertices \(G_{1}, \ldots, G_{n}, B_{1}, \ldots, B_{n}\) such that there is an edge between vertices \(G_{i}\) and \(B_{j}\) if and only if \((G_{i}, B_{j})\) is suitable, so we want to maximize the number of edges while having a unique matching. We claim the answer is \(\frac{n(n+1)}{2}\). First, note that this can be achieved by having an edge between \(G_{i}\) and \(B_{j}\) for all pairs \(j \leq i\), because the only possible matching in this case is pairing \(G_{i}\) with \(B_{i}\) for all \(i\). To prove that this is maximal, we first assume without loss of generality that our unique matching consists of pairing \(G_{i}\) with \(B_{i}\) for all \(i\), which takes \(n\) edges. Now, note that for any \(i, j\), at most one of the two edges \(G_{i} B_{j}\) and \(G_{j} B_{i}\) can be added, because if both were added, we could pair \(G_{i}\) with \(B_{j}\) and \(G_{j}\) with \(B_{i}\) instead to get another valid matching. Therefore, we may add at most \(\binom{n}{2} \cdot 1=\frac{n(n-1)}{2}\) edges, so the maximal number of edges is \(n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}\) as desired.

\frac{n(n+1)}{2}

HMMT_2

Find the maximum possible number of diagonals of equal length in a convex hexagon.

First, we will prove that 7 is possible. Consider the following hexagon \(A B C D E F\) whose vertices are located at \(A(0,0), B\left(\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right), C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), D(0,1), E\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), F\left(-\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right)\). One can easily verify that all diagonals but \(B E\) and \(C F\) have length 1. Now suppose that there are at least 8 diagonals in a certain convex hexagon \(A B C D E F\) whose lengths are equal. There must be a diagonal such that, with this diagonal taken out, the other 8 have equal length. There are two cases. Case I. The diagonal is one of \(A C, B D, C E, D F, E A, F B\). WLOG, assume it is \(A C\). We have \(E C= E B=F B=F C\). Thus, \(B\) and \(C\) are both on the perpendicular bisector of \(E F\). Since \(A B C D E F\) is convex, both \(B\) and \(C\) must be on the same side of line \(E F\), but this is impossible as one of \(B\) or \(C\), must be contained in triangle \(C E F\). Contradiction. Case II: The diagonal is one of \(A D, B E, C F\). WLOG, assume it is \(A D\). Again, we have \(E C=E B= F B=F C\). By the above reasoning, this is a contradiction. Thus, 7 is the maximum number of possible diagonals.

7

HMMT_2

Let $Y$ be as in problem 14. Find the maximum $Z$ such that three circles of radius $\sqrt{Z}$ can simultaneously fit inside an equilateral triangle of area $Y$ without overlapping each other.

We first find that, in problem 15, each of the circles of radius $\sqrt{Z}$ is the incircle of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle formed by cutting the equilateral one in half. The equilateral triangle itself has sidelength $\frac{2 \sqrt{Y}}{\sqrt[4]{3}}$, so the said inradius is $$\sqrt{Z}=\frac{1+\sqrt{3}-2}{2} \cdot \frac{1}{2} \cdot \frac{2 \sqrt{Y}}{\sqrt[4]{3}}$$ so that $$Z=\frac{(-1+\sqrt{3})^{2}}{4 \sqrt{3}} Y=\frac{4-2 \sqrt{3}}{4 \sqrt{3}} Y=\frac{2 \sqrt{3}-3}{6} Y$$ Now we guess that $X=2$ and see that, miraculously, everything works: in the problem 14, say a crimson flower is placed first. Then there are 2 possibilities for $C_{-} C_{-{ }_{-1}}, 4$ for $C_{-} C_{--}, 2$ for $C_{-\ldots} C_{-}$, and 2 for $C_{\ldots_{-}} C$, giving a total of 10. Of course, the first flower can be of any of the three hues, so $Y=3 \cdot 10=30$. We compute $Z$ and check $X$ in a straightforward manner. If $X>2$, then $Y>30$, and $Z>10 \sqrt{3}-15$, with the result that $X \leq 2$, a contradiction. Assuming $X<2$ results in a similar contradiction.

10 \sqrt{3}-15

HMMT_2

Spencer is making burritos, each of which consists of one wrap and one filling. He has enough filling for up to four beef burritos and three chicken burritos. However, he only has five wraps for the burritos; in how many orders can he make exactly five burritos?

Spencer's burrito-making can include either 3, 2, or 1 chicken burrito; consequently, he has $\binom{5}{3}+\binom{5}{2}+\binom{5}{1}=25$ orders in which he can make burritos.

25

HMMT_2

Let $a \geq b \geq c$ be real numbers such that $$\begin{aligned} a^{2} b c+a b^{2} c+a b c^{2}+8 & =a+b+c \\ a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3 a b c & =-4 \\ a^{2} b^{2} c+a b^{2} c^{2}+a^{2} b c^{2} & =2+a b+b c+c a \end{aligned}$$ If $a+b+c>0$, then compute the integer nearest to $a^{5}$.

We factor the first and third givens, obtaining the system $$\begin{aligned} a^{2} b c+a b^{2} c+a b c^{2}-a-b-c=(a b c-1)(a+b+c) & =-8 \\ a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3 a b c=(a b+b c+c a)(a+b+c) & =-4 \\ a^{2} b^{2} c+a b^{2} c^{2}+a^{2} b c^{2}-a b-b c-c a=(a b c-1)(a b+b c+c a) & =2 \end{aligned}$$ Writing $X=a+b+c, Y=a b+b c+c a, Z=a b c-1$, we have $X Z=-8, X Y=-4, Y Z=$ 2. Multiplying the three yields $(X Y Z)^{2}=64$ from which $X Y Z= \pm 8$. Since we are given $X>0$, multiplying the last equation by $X$ we have $2 X=X Y Z= \pm 8$. Evidently $X Y Z=8$ from which $X=4, Y=-1, Z=-2$. We conclude that $a, b, c$ are the roots of the polynomial $P(t)=t^{3}-4 t^{2}-t+1$. Thus, $P(a)=a^{3}-4 a^{2}-a+1=0$, and also $P(b)=P(c)=0$. Now since $P(1 / 2)=-\frac{3}{8}, P(0)=1$ and $P(-2 / 3)=-\frac{11}{27}$, we conclude that $-2 / 3<c<0<b<1 / 2<a$. It follows that $\left|b^{5}+c^{5}\right|<\frac{1}{2}$. Thus, we compute $a^{5}+b^{5}+c^{5}$. Defining $S_{n}=a^{n}+b^{n}+c^{n}$, we have $S_{n+3}=4 S_{n+2}+S_{n+1}-S_{n}$ for $n \geq 0$. Evidently $S_{0}=3, S_{1}=4, S_{2}=(a+b+c)^{2}-2(a b+b c+c a)=18$. Then $S_{3}=4 \cdot 18+4-3=73$, $S_{4}=4 \cdot 73+18-4=306$, and $S_{5}=4 \cdot 306+73-18=1279$. Since $\left|b^{5}+c^{5}\right|<\frac{1}{2}$, we conclude that $\left|S_{5}-a^{5}\right|<\frac{1}{2}$, or that 1279 is the integer nearest to $a^{5}$.

1279

HMMT_2

A sequence is defined by $A_{0}=0, A_{1}=1, A_{2}=2$, and, for integers $n \geq 3$, $$A_{n}=\frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\frac{1}{n^{4}-n^{2}}$$ Compute $\lim _{N \rightarrow \infty} A_{N}$.

If we sum the given equation for $n=3,4,5, \ldots, N$, we obtain $$\sum_{n=3}^{N} A_{n}=\sum_{n=3}^{N} \frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\frac{1}{n^{4}-n^{2}}$$ This reduces dramatically to $$\begin{equation*} A_{N}+\frac{2 A_{N-1}}{3}+\frac{A_{N-2}}{3}=A_{2}+\frac{2 A_{1}}{3}+\frac{A_{0}}{3}+\sum_{n=3}^{N} \frac{1}{n^{4}-n^{2}} \tag{*} \end{equation*}$$ Let $\lim _{N \rightarrow \infty} A_{N}=L$. Under this limit, the left hand side of $\left(^{*}\right)$ is simply $2 L$. We compute the sum on the right with the help of partial fractions $$\begin{aligned} \lim _{N \rightarrow \infty} \sum_{n=3}^{N} \frac{1}{n^{4}-n^{2}} & =\sum_{n=3}^{\infty} \frac{1}{n^{2}-1}-\frac{1}{n^{2}} \\ & =\left(\sum_{n=3}^{\infty} \frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right)+\frac{1}{1^{2}}+\frac{1}{2^{2}}-\sum_{n=1}^{\infty} \frac{1}{n^{2}} \\ & =\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{5}{4}-\frac{\pi^{2}}{6} \\ & =\frac{5}{3}-\frac{\pi^{2}}{6} \end{aligned}$$ With this we easily find $L=\frac{1}{2} \cdot\left(2+\frac{2}{3} \cdot 1+\frac{1}{3} \cdot 0+\frac{5}{3}-\frac{\pi^{2}}{6}\right)=\frac{13}{6}-\frac{\pi^{2}}{12}$, and we are done.

\frac{13}{6}-\frac{\pi^{2}}{12}

HMMT_2

In a game, there are three indistinguishable boxes; one box contains two red balls, one contains two blue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will draw two balls of the same color or two of different colors. Then, he picks a box, draws a ball at random, looks at the color, and replaces the ball in the same box. Finally, he repeats this; however, the boxes are not shuffled between draws, so he can determine whether he wants to draw again from the same box. Raj wins if he predicts correctly; if he plays optimally, what is the probability that he will win?

If Bob predicts that he will draw two balls of the same color, then there are two possible plays: he draws from the same box, or he draws from a different box. If he draws from the same box, then in the $\frac{2}{3}$ chance that he originally picked box 1, he will always win, and in the $\frac{1}{3}$ chance that he picked box 2, he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3}+\frac{1}{3} \cdot \frac{1}{2}=\frac{5}{6}$. If he draws from a different box, then if he originally picked box 1, he will win with probability $\frac{1}{4}$ and if he originally picked box 2, he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3} \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{3}$. If Bob predicts that he will draw two balls of different colors, then we can consider the same two possible plays. Using similar calculations, if he draws from the same box, then he will win with probability $\frac{1}{6}$, and if he draws from a different box, then he will win with probability $\frac{2}{3}$. Looking at all cases, Bob's best play is to predict that he will draw two balls of the same color and then draw the second ball from the same box, with a winning probability of $\frac{5}{6}$.

\frac{5}{6}

HMMT_2

Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of real numbers defined by $a_{0}=21, a_{1}=35$, and $a_{n+2}=4 a_{n+1}-4 a_{n}+n^{2}$ for $n \geq 2$. Compute the remainder obtained when $a_{2006}$ is divided by 100.

No pattern is evident in the first few terms, so we look for a formula for $a_{n}$. If we write $a_{n}=A n^{2}+B n+C+b_{n}$ and put $b_{n+2}=4 b_{n+1}-4 b_{n}$. Rewriting the original recurrence, we find $$\begin{aligned} A n^{2}+(4 A+B) n+(4 A+2 B+C)+b_{n+2} & \\ =4\left(A n^{2}+(2 A+B) n+(A+B+C)\right. & \left.+b_{n+1}\right)-4\left(A n^{2}+B n+C+b_{n}\right)+n^{2} \\ & =n^{2}+8 A n+(4 A+4 B)+4 b_{n+1}-4 b_{n} \end{aligned}$$ Solving, $A=1, B=4, C=8$. With this information, we can solve for $b_{0}=1$ and $b_{1}=6$. Since the characteristic equation of the recurrence of the $b_{i}$ is $x^{2}-4 x+4=$ $(x-2)^{2}=0$, we have $b_{n}=(D n+E) \cdot 2^{n}$ for some constants $D$ and $E$. Using the known values $b_{0}$ and $b_{1}$, we compute $D=2$ and $E=1$, and finally $$a_{n}=n^{2}+4 n+8+(2 n+1) \cdot 2^{n}$$ Now, taking modulo 100, we have $a_{2006} \equiv 6^{2}+4 \cdot 6+8+13 \cdot 2^{2006}(\bmod 100)$. Evidently $2^{2006} \equiv 0(\bmod 4)$, but by Euler's theorem $2^{\phi(25)} \equiv 2^{20} \equiv 1(\bmod 25)$, and so $2^{2006} \equiv$ $2^{6} \equiv 14(\bmod 25)$. Now the Chinese remainder theorem yields $2^{2006} \equiv 64(\bmod 100)$, and we compute $a_{2006} \equiv 36+24+8+13 \cdot 64 \equiv 0(\bmod 100)$.

0

HMMT_2

Wesyu is a farmer, and she's building a cao (a relative of the cow) pasture. She starts with a triangle $A_{0} A_{1} A_{2}$ where angle $A_{0}$ is $90^{\circ}$, angle $A_{1}$ is $60^{\circ}$, and $A_{0} A_{1}$ is 1. She then extends the pasture. First, she extends $A_{2} A_{0}$ to $A_{3}$ such that $A_{3} A_{0}=\frac{1}{2} A_{2} A_{0}$ and the new pasture is triangle $A_{1} A_{2} A_{3}$. Next, she extends $A_{3} A_{1}$ to $A_{4}$ such that $A_{4} A_{1}=\frac{1}{6} A_{3} A_{1}$. She continues, each time extending $A_{n} A_{n-2}$ to $A_{n+1}$ such that $A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$. What is the smallest $K$ such that her pasture never exceeds an area of $K$?

First, note that for any $i$, after performing the operation on triangle $A_{i} A_{i+1} A_{i+2}$, the resulting pasture is triangle $A_{i+1} A_{i+2} A_{i+3}$. Let $K_{i}$ be the area of triangle $A_{i} A_{i+1} A_{i+2}$. From $A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$ and $A_{n} A_{n+1}=A_{n} A_{n-2}+A_{n-2} A_{n+1}$, we have $A_{n} A_{n+1}=\left(1+\frac{1}{2^{n}-2}\right) A_{n} A_{n-2}$. We also know that the area of a triangle is half the product of its base and height, so if we let the base of triangle $A_{n-2} A_{n-1} A_{n}$ be $A_{n} A_{n-2}$, its area is $K_{n-2}=\frac{1}{2} h A_{n} A_{n-2}$. The area of triangle $A_{n-1} A_{n} A_{n+1}$ is $K_{n-1}=\frac{1}{2} h A_{n} A_{n+1}$. The $h$ 's are equal because the distance from $A_{n-1}$ to the base does not change. We now have $\frac{K_{n-1}}{K_{n-2}}=\frac{A_{n} A_{n+1}}{A_{n} A_{n-2}}=1+\frac{1}{2^{n}-2}=\frac{2^{n}-1}{2^{n}-2}$. Therefore, $\frac{K_{1}}{K_{0}}=\frac{3}{2}, \frac{K_{2}}{K_{0}}=\frac{K_{2}}{K_{1}} \frac{K_{1}}{K_{0}}=\frac{7}{6} \cdot \frac{3}{2}=\frac{7}{4}$, $\frac{K_{3}}{K_{0}}=\frac{K_{3}}{K_{2}} \frac{K_{2}}{K_{0}}=\frac{15}{14} \cdot \frac{7}{4}=\frac{15}{8}$. We see the pattern $\frac{K_{n}}{K_{0}}=\frac{2^{n+1}-1}{2^{n}}$, which can be easily proven by induction. As $n$ approaches infinity, $\frac{K_{n}}{K_{0}}$ grows arbitrarily close to 2, so the smallest $K$ such that the pasture never exceeds an area of $K$ is $2 K_{0}=\sqrt{3}$.

\sqrt{3}

HMMT_2

A function $f: A \rightarrow A$ is called idempotent if $f(f(x))=f(x)$ for all $x \in A$. Let $I_{n}$ be the number of idempotent functions from $\{1,2, \ldots, n\}$ to itself. Compute $\sum_{n=1}^{\infty} \frac{I_{n}}{n!}$.

Let $A_{k, n}$ denote the number of idempotent functions on a set of size $n$ with $k$ fixed points. We have the formula $A_{k, n}=\binom{n}{k} k^{n-k}$ for $1 \leq k \leq n$ because there are $\binom{n}{k}$ ways to choose the fixed points and all $n-k$ remaining elements must map to fixed points, which can happen in $k^{n-k}$ ways. Hence $\sum_{n=1}^{\infty} \frac{I_{n}}{n!} =\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{A_{k, n}}{n!} =\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{k^{n-k}}{k!(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=k}^{\infty} \frac{k^{n-k}}{(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=0}^{\infty} \frac{k^{n}}{n!} =\sum_{k=1}^{\infty} \frac{1}{k!} e^{k} =e^{e}-1

e^{e}-1

HMMT_2

Determine all triplets of real numbers $(x, y, z)$ satisfying the system of equations $x^{2} y+y^{2} z =1040$, $x^{2} z+z^{2} y =260$, $(x-y)(y-z)(z-x) =-540$.

Call the three equations $(1),(2),(3) \cdot(1) /(2)$ gives $y=4 z .(3)+(1)-(2)$ gives $\left(y^{2}-z^{2}\right) x=15 z^{2} x=240$ so $z^{2} x=16$. Therefore $z(x+2 z)^{2}=x^{2} z+z^{2} y+4 z^{2} x=\frac{81}{5}$, $z(x-2 z)^{2}=x^{2} z+z^{2} y-4 z^{2} x=\frac{49}{5}$ so $\left|\frac{x+2 z}{x-2 z}\right|=\frac{9}{7}$. Thus either $x=16 z$ or $x=\frac{z}{4}$. If $x=16 z$, then (1) becomes $1024 z^{3}+16 z^{3}=1040$, so $(x, y, z)=(16,4,1)$. If $x=\frac{z}{4}$, then $(1)$ becomes $\frac{1}{4} z^{3}+16 z^{3}=1040$, so $(x, y, z)=(1,16,4)$.

(16,4,1),(1,16,4)

HMMT_2

Find, as a function of $\, n, \,$ the sum of the digits of \[9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),\] where each factor has twice as many digits as the previous one.

The answer is $9 \cdot 2^n$ . Let us denote the quantity $\prod_{k=0}^n \bigl( 10^{2^k}-1 \bigr)$ as $P_n$ . We wish to find the sum of the digits of $P_n$ . We first note that \[P_{n-1} < \prod_{k=0}^{n-1} 10^{2^k} = 10^{2^n-1},\] so $P_{n-1}$ is a number of at most $2^n$ digits. We also note that the units digit is not equal to zero. We may thus represent $P_{n-1}$ as \[\sum_{k=0}^{2^n-1} 10^k d_k ,\] where the $d_k$ are digits and $d_0 \neq 0$ . Then \begin{align*} P_n &= \bigl( 10^{2^n}-1 \bigr) P_{n-1} = \sum_{k=0}^{2^n-1} - 10^k d_k + \sum_{k=0}^{2^n-1} 10^{2^n+k} d_k \\ &= (10-d_0) + \sum_{k=1}^{2^n-1} 10^k(9-d_k) + 10^{2^n}(d_0-1) + \sum_{k=1}^{2^n-1} 10^{2^n+k} d_k . \end{align*} Thus the digits of $P_n$ are \[10-d_0, 9-d_1, 9-d_2, \dotsc, 9-d_{2^n-1}, d_0-1, d_1, d_2, \dotsc, d_{2^n-1} ,\] and the sum of these is evidently $9 \cdot 2^n$ , as desired. $\blacksquare$ Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

\[ 9 \cdot 2^n \]

usamo

Let $ABC$ be a right triangle with $\angle A=90^{\circ}$. Let $D$ be the midpoint of $AB$ and let $E$ be a point on segment $AC$ such that $AD=AE$. Let $BE$ meet $CD$ at $F$. If $\angle BFC=135^{\circ}$, determine $BC/AB$.

Let $\alpha=\angle ADC$ and $\beta=\angle ABE$. By exterior angle theorem, $\alpha=\angle BFD+\beta=$ $45^{\circ}+\beta$. Also, note that $\tan \beta=AE/AB=AD/AB=1/2$. Thus, $$1=\tan 45^{\circ}=\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}=\frac{\tan \alpha-\frac{1}{2}}{1+\frac{1}{2} \tan \alpha}$$ Solving for $\tan \alpha$ gives $\tan \alpha=3$. Therefore, $AC=3AD=\frac{3}{2}AB$. Using Pythagorean Theorem, we find that $BC=\frac{\sqrt{13}}{2}AB$. So the answer is $\frac{\sqrt{13}}{2}$.

\frac{\sqrt{13}}{2}

HMMT_2

Find the smallest positive integer $n$ such that $\frac{5^{n+1}+2^{n+1}}{5^{n}+2^{n}}>4.99$.

Writing $5^{n+1}=5 \cdot 5^{n}$ and $2^{n+1}=2 \cdot 2^{n}$ and cross-multiplying yields $0.01 \cdot 5^{n}>2.99 \cdot 2^{n}$, and re-arranging yields $(2.5)^{n}>299$. A straightforward calculation shows that the smallest $n$ for which this is true is $n=7$.

7

HMMT_2

Let $X$ be as in problem 13. Let $Y$ be the number of ways to order $X$ crimson flowers, $X$ scarlet flowers, and $X$ vermillion flowers in a row so that no two flowers of the same hue are adjacent. (Flowers of the same hue are mutually indistinguishable.) Find $Y$.

Problems 13-15 go together. See below.

30

HMMT_2

Compute $$\sum_{k=1}^{\infty} \frac{3 k+1}{2 k^{3}+k^{2}} \cdot(-1)^{k+1}$$

Via partial fraction decomposition we write the sum as $$\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{2}{1+2 k}+\frac{1}{k^{2}}\right)(-1)^{k+1}$$ Now recall that $$\begin{aligned} \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} & =S_{1} \\ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln (2) & =S_{2} \\ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2 k-1}=\frac{\pi}{4} & =S_{3} \end{aligned}$$ Manipulating (1), we deduce $$\begin{aligned} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}} & =\left(\sum_{k=1}^{\infty} \frac{1}{k^{2}}\right)-2 \cdot \sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}} \\ =\frac{\pi^{2}}{6}-2 / 4 \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{12} & =S_{4} \end{aligned}$$ It is then easily seen that the answer is equal to $S_{2}+2 \cdot S_{3}-2+S_{4}$.

\frac{\pi^{2}}{12}+\frac{\pi}{2}-2+\ln 2

HMMT_2

Let $ABCD$ be a regular tetrahedron, and let $O$ be the centroid of triangle $BCD$. Consider the point $P$ on $AO$ such that $P$ minimizes $PA+2(PB+PC+PD)$. Find $\sin \angle PBO$.

We translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA+6PB$. Construct the line $\ell$ through $A$ but outside the triangle $ABO$ so that $\sin \angle(AO, \ell)=\frac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\ell$, then $PQ=\frac{1}{6}AP$. Then, since $PA+6PB=6(PQ+PB)$, it is equivalent to minimize $PQ+PB$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\ell$ (so that $PQ+PB$ is simply the distance from $B$ to $\ell$). Then, $\angle AQB=90^{\circ}$, and since $\angle AOB=90^{\circ}$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\angle PBO=\angle OPA=\angle(AO, \ell)$, and the sine of this angle is therefore $\frac{1}{6}$.

\frac{1}{6}

HMMT_2

When a single number is added to each member of the sequence 20, 50, 100, the sequence becomes expressible as $x, a x, a^{2} x$. Find $a$.

$\frac{5}{3}$.

\frac{5}{3}

HMMT_2

In bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention, each hand is assigned a number of "points" based on the formula $$4 \times(\# \mathrm{~A} \text { 's })+3 \times(\# \mathrm{~K} \text { 's })+2 \times(\# \mathrm{Q} \text { 's })+1 \times(\# \mathrm{~J} \text { 's })$$ Given that a particular hand has exactly 4 cards that are A, K, Q, or J, find the probability that its point value is 13 or higher.

Obviously, we can ignore the cards lower than J. Simply enumerate the ways to get at least 13 points: AAAA (1), AAAK (16), AAAQ (16), AAAJ (16), AAKK (36), AAKQ (96), AKKK (16). The numbers in parentheses represent the number of ways to choose the suits, given the choices for the values. We see that there are a total of $1+16+16+16+36+96+16=197$ ways to get at least 13. There are a total of $\binom{16}{4}=1820$ possible ways to choose 4 cards from the 16 total A's, K's, Q's, and J's. Hence the answer is $197 / 1820$.

\frac{197}{1820}

HMMT_2

An isosceles trapezoid $A B C D$ with bases $A B$ and $C D$ has $A B=13, C D=17$, and height 3. Let $E$ be the intersection of $A C$ and $B D$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $A B E$ and $C D E$. Compute the sum of the radii of $\Omega$ and $\omega$.

Let $\Omega$ have center $O$ and radius $R$ and let $\omega$ have center $P$ and radius $M$. Let $Q$ be the intersection of $A B$ and $O E$. Note that $O E$ is the perpendicular bisector of $A B$ because the trapezoid is isosceles. Also, we see $O E$ is the circumradius of $\Omega$. On the other hand, we know by similarity of $\triangle A E B$ and $\triangle C E D$ that $Q E=\frac{13}{13+17} \cdot 3=\frac{13}{30} \cdot 3$. And, because $B Q=13 / 2$ and is perpendicular to $O Q$, we can apply the Pythagorean theorem to $\triangle O Q B$ to see $O Q=\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Since $O E=O Q+Q E, R=\frac{13}{30} \cdot 3+\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Solving this equation for $R$ yields $R=\frac{13}{30} \cdot 39$. Since by similarity $M=\frac{17}{13} R$, we know $R+M=\frac{30}{13} R$, so $R+M=39$.

39

HMMT_2

Suppose $A B C$ is a triangle with incircle $\omega$, and $\omega$ is tangent to $\overline{B C}$ and $\overline{C A}$ at $D$ and $E$ respectively. The bisectors of $\angle A$ and $\angle B$ intersect line $D E$ at $F$ and $G$ respectively, such that $B F=1$ and $F G=G A=6$. Compute the radius of $\omega$.

Let $\alpha, \beta, \gamma$ denote the measures of $\frac{1}{2} \angle A, \frac{1}{2} \angle B, \frac{1}{2} \angle C$, respectively. We have $m \angle C E F=90^{\circ}-\gamma, m \angle F E A=90^{\circ}+\gamma, m \angle A F G=m \angle A F E=180^{\circ}-\alpha-\left(90^{\circ}+\gamma\right)=$ $\beta=m \angle A B G$, so $A B F G$ is cyclic. Now $A G=G F$ implies that $\overline{B G}$ bisects $\angle A B F$. Since $\overline{B G}$ by definition bisects $\angle A B C$, we see that $F$ must lie on $\overline{B C}$. Hence, $F=D$. If $I$ denotes the incenter of triangle $A B C$, then $\overline{I D}$ is perpendicular to $\overline{B C}$, but since $A, I, F$ are collinear, we have that $\overline{A D} \perp \overline{B C}$. Hence, $A B C$ is isoceles with $A B=A C$. Furthermore, $B C=2 B F=2$. Moreover, since $A B F G$ is cyclic, $\angle B G A$ is a right angle. Construct $F^{\prime}$ on minor $\operatorname{arc} G F$ such that $B F^{\prime}=6$ and $F^{\prime} G=1$, and let $A B=x$. By the Pythagorean theorem, $A F^{\prime}=B G=\sqrt{x^{2}-36}$, so that Ptolemy applied to $A B F^{\prime} G$ yields $x^{2}-36=x+36$. We have $(x-9)(x+8)=0$. Since $x$ is a length we find $x=9$. Now we have $A B=A C=9$. Pythagoras applied to triangle $A B D$ now yields $A D=\sqrt{9^{2}-1^{2}}=4 \sqrt{5}$, which enables us to compute $[A B C]=\frac{1}{2} \cdot 2 \cdot 4 \sqrt{5}=4 \sqrt{5}$. Since the area of a triangle is also equal to its semiperimeter times its inradius, we have $4 \sqrt{5}=10 r$ or $r=\frac{2 \sqrt{5}}{5}$.

\frac{2 \sqrt{5}}{5}

HMMT_2

Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During each move of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another to turn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips both cards face-down and keeps repeating this process. Initially, Rahul doesn't know which cards are which. Assuming that he has perfect memory, find the smallest number of moves after which he can guarantee that the game has ended.

Label the 10 cards $a_{1}, a_{2}, \ldots, a_{5}, b_{1}, b_{2}, \ldots, b_{5}$ such that $a_{i}$ and $b_{i}$ match for $1 \leq i \leq 5$. First, we'll show that Rahul cannot always end the game in less than 4 moves, in particular, when he turns up his fifth card (during the third move), it is possible that the card he flips over is not one which he has yet encountered; consequently, he will not guarantee being able to match it, so he cannot guarantee that the game can end in three moves. However, Rahul can always end the game in 4 moves. To do this, he can turn over 6 distinct cards in his first 3 moves. If we consider the 5 sets of cards $\left\{a_{1}, b_{1}\right\},\left\{a_{2}, b_{2}\right\},\left\{a_{3}, b_{3}\right\},\left\{a_{4}, b_{4}\right\},\left\{a_{5}, b_{5}\right\}$, then by the pigeonhole principle, at least 2 of the 6 revealed cards must be from the same set. Rahul can then turn over those 2 cards on the fourth move, ending the game.

4

HMMT_2

Let $S=\{2,3,4, \ldots\}$ denote the set of integers that are greater than or equal to 2. Does there exist a function $f: S \rightarrow S$ such that $f(a) f(b)=f\left(a^{2} b^{2}\right)$ for all $a, b \in S$ with $a \neq b$?

We prove that there is no such function. For arbitrary elements $a$ and $b$ of $S$, choose an integer $c$ that is greater than both of them. Since $bc>a$ and $c>b$, we have $$ f\left(a^{4} b^{4} c^{4}\right)=f\left(a^{2}\right) f\left(b^{2} c^{2}\right)=f\left(a^{2}\right) f(b) f(c) $$ Furthermore, since $ac>b$ and $c>a$, we have $$ f\left(a^{4} b^{4} c^{4}\right)=f\left(b^{2}\right) f\left(a^{2} c^{2}\right)=f\left(b^{2}\right) f(a) f(c) $$ Comparing these two equations, we find that for all elements $a$ and $b$ of $S$, $$ f\left(a^{2}\right) f(b)=f\left(b^{2}\right) f(a) \quad \Longrightarrow \quad \frac{f\left(a^{2}\right)}{f(a)}=\frac{f\left(b^{2}\right)}{f(b)} $$ It follows that there exists a positive rational number $k$ such that $$ f\left(a^{2}\right)=k f(a), \quad \text { for all } a \in S $$ Substituting this into the functional equation yields $$ f(ab)=\frac{f(a) f(b)}{k}, \quad \text { for all } a, b \in S \text { with } a \neq b $$ Now combine the functional equation with equations (1) and (2) to obtain $$ f(a) f\left(a^{2}\right)=f\left(a^{6}\right)=\frac{f(a) f\left(a^{5}\right)}{k}=\frac{f(a) f(a) f\left(a^{4}\right)}{k^{2}}=\frac{f(a) f(a) f\left(a^{2}\right)}{k}, \quad \text { for all } a \in S $$ It follows that $f(a)=k$ for all $a \in S$. Substituting $a=2$ and $b=3$ into the functional equation yields $k=1$, however $1 \notin S$ and hence we have no solutions.

There is no such function \( f: S \rightarrow S \).

apmoapmo_sol

Triangle $ABC$ obeys $AB=2AC$ and $\angle BAC=120^{\circ}$. Points $P$ and $Q$ lie on segment $BC$ such that $$\begin{aligned} AB^{2}+BC \cdot CP & =BC^{2} \\ 3AC^{2}+2BC \cdot CQ & =BC^{2} \end{aligned}$$ Find $\angle PAQ$ in degrees.

We have $AB^{2}=BC(BC-CP)=BC \cdot BP$, so triangle $ABC$ is similar to triangle $PBA$. Also, $AB^{2}=BC(BC-2CQ)+AC^{2}=(BC-CQ)^{2}-CQ^{2}+AC^{2}$, which rewrites as $AB^{2}+CQ^{2}=$ $BQ^{2}+AC^{2}$. We deduce that $Q$ is the foot of the altitude from $A$. Thus, $\angle PAQ=90^{\circ}-\angle QPA=90^{\circ}-$ $\angle ABP-\angle BAP$. Using the similar triangles, $\angle PAQ=90^{\circ}-\angle ABC-\angle BCA=\angle BAC-90^{\circ}=40^{\circ}$.

40^{\circ}

HMMT_2

Find the number of positive divisors $d$ of $15!=15 \cdot 14 \cdots 2 \cdot 1$ such that $\operatorname{gcd}(d, 60)=5$.

Since $\operatorname{gcd}(d, 60)=5$, we know that $d=5^{i} d^{\prime}$ for some integer $i>0$ and some integer $d^{\prime}$ which is relatively prime to 60. Consequently, $d^{\prime}$ is a divisor of $(15!) / 5$; eliminating common factors with 60 gives that $d^{\prime}$ is a factor of $\left(7^{2}\right)(11)(13)$, which has $(2+1)(1+1)(1+1)=12$ factors. Finally, $i$ can be 1,2 , or 3, so there are a total of $3 \cdot 12=36$ possibilities.

36

HMMT_2

How many ways can you color the squares of a $2 \times 2008$ grid in 3 colors such that no two squares of the same color share an edge?

Denote the colors $A, B, C$. The left-most column can be colored in 6 ways. For each subsequent column, if the $k$th column is colored with $AB$, then the $(k+1)$th column can only be colored with one of $BA, BC, CA$. That is, if we have colored the first $k$ columns, then there are 3 ways to color the $(k+1)$th column. It follows that the number of ways of coloring the board is $6 \times 3^{2007}$.

2 \cdot 3^{2008}

HMMT_2

Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove: (a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !; (b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ ! .

(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} X(n-k)$ !, so $2 p_{k} \nmid(n-k)$ !. (b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)$ !. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$. If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$, unless $b<3$ or $b=a$. Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq 2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$ !. Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\{1,2, \ldots, n-k\}$. Hence $n=a^{2}$ divides into $(n-k)!$.

(a) $2 p_{k} \nmid (n-k)!$ (b) $n \mid (n-k)!$

apmoapmo_sol

I have 8 unit cubes of different colors, which I want to glue together into a $2 \times 2 \times 2$ cube. How many distinct $2 \times 2 \times 2$ cubes can I make? Rotations of the same cube are not considered distinct, but reflections are.

Our goal is to first pin down the cube, so it can't rotate. Without loss of generality, suppose one of the unit cubes is purple, and let the purple cube be in the top left front position. Now, look at the three positions that share a face with the purple cube. There are $\binom{7}{3}$ ways to pick the three cubes that fill those positions and two ways to position them that are rotationally distinct. Now, we've taken care of any possible rotations, so there are simply 4 ! ways to position the final four cubes. Thus, our answer is $\binom{7}{3} \cdot 2 \cdot 4!=1680$ ways.

1680

HMMT_2

Let $f(r)=\sum_{j=2}^{2008} \frac{1}{j^{r}}=\frac{1}{2^{r}}+\frac{1}{3^{r}}+\cdots+\frac{1}{2008^{r}}$. Find $\sum_{k=2}^{\infty} f(k)$.

We change the order of summation: $$\sum_{k=2}^{\infty} \sum_{j=2}^{2008} \frac{1}{j^{k}}=\sum_{j=2}^{2008} \sum_{k=2}^{\infty} \frac{1}{j^{k}}=\sum_{j=2}^{2008} \frac{1}{j^{2}\left(1-\frac{1}{j}\right)}=\sum_{j=2}^{2008} \frac{1}{j(j-1)}=\sum_{j=2}^{2008}\left(\frac{1}{j-1}-\frac{1}{j}\right)=1-\frac{1}{2008}=\frac{2007}{2008}$$

\frac{2007}{2008}

HMMT_2

Given a point $p$ and a line segment $l$, let $d(p, l)$ be the distance between them. Let $A, B$, and $C$ be points in the plane such that $A B=6, B C=8, A C=10$. What is the area of the region in the $(x, y)$-plane formed by the ordered pairs $(x, y)$ such that there exists a point $P$ inside triangle $A B C$ with $d(P, A B)+x=d(P, B C)+y=d(P, A C)$?

Place $A B C$ in the coordinate plane so that $A=(0,6), B=(0,0), C=(8,0)$. Consider a point $P=(a, b)$ inside triangle $A B C$. Clearly, $d(P, A B)=a, d(P, B C)=b$. Now, we see that the area of triangle $A B C$ is $\frac{6 \cdot 8}{2}=24$, but may also be computed by summing the areas of triangles $P A B, P B C, P C A$. The area of triangle $P A B$ is $\frac{6 \cdot a}{2}=3 a$, and similarly the area of triangle $P B C$ is $4 b$. Thus, it follows easily that $d(P, C A)=\frac{24-3 a-4 b}{5}$. Now, we have $$(x, y)=\left(\frac{24}{5}-\frac{8}{5} a-\frac{4}{b} b, \frac{24}{5}-\frac{3}{5} a-\frac{9}{5} b\right)$$ The desired region is the set of $(x, y)$ obtained by those $(a, b)$ subject to the constraints $a \geq 0, b \geq$ $0,6 a+8 b \leq 48$. Consequently, our region is the triangle whose vertices are obtained by evaluating $(x, y)$ at the vertices $(a, b)$ of the triangle. To see this, let $f(a, b)$ output the corresponding $(x, y)$ according to the above. Then, we can write every point $P$ in $A B C$ as $P=m(0,0)+n(0,6)+p(8,0)$ for some $m+n+p=1$. Then, $f(P)=m f(0,0)+n f(0,6)+p f(8,0)=m\left(\frac{24}{5}, \frac{24}{5}\right)+n(-8,0)+p(0,-6)$, so $f(P)$ ranges over the triangle with those three vertices. Therefore, we need the area of the triangle with vertices $\left(\frac{24}{5}, \frac{24}{5}\right),(0,-6),(-8,0)$, which is easily computed (for example, using determinants) to be $\frac{288}{5}$.

\frac{288}{5}

HMMT_2

Let $\mathcal{H}$ be the unit hypercube of dimension 4 with a vertex at $(x, y, z, w)$ for each choice of $x, y, z, w \in \{0,1\}$. A bug starts at the vertex $(0,0,0,0)$. In how many ways can the bug move to $(1,1,1,1)$ by taking exactly 4 steps along the edges of $\mathcal{H}$?

You may think of this as sequentially adding 1 to each coordinate of $(0,0,0,0)$. There are 4 ways to choose the first coordinate, 3 ways to choose the second, and 2 ways to choose the third. The product is 24.

24

HMMT_2

Let $A B C D$ be a parallelogram with $A B=8, A D=11$, and $\angle B A D=60^{\circ}$. Let $X$ be on segment $C D$ with $C X / X D=1 / 3$ and $Y$ be on segment $A D$ with $A Y / Y D=1 / 2$. Let $Z$ be on segment $A B$ such that $A X, B Y$, and $D Z$ are concurrent. Determine the area of triangle $X Y Z$.

Let $A X$ and $B D$ meet at $P$. We have $D P / P B=D X / A B=3 / 4$. Now, applying Ceva's Theorem in triangle $A B D$, we see that $$\frac{A Z}{Z B}=\frac{D P}{P B} \cdot \frac{A Y}{Y D}=\frac{3}{4} \cdot \frac{1}{2}=\frac{3}{8}$$ Now, $$\frac{[A Y Z]}{[A B C D]}=\frac{[A Y Z]}{2[A B D]}=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{11}=\frac{1}{22}$$ and similarly $$\frac{[D Y X]}{[A B C D]}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4}$$ Also, $$\frac{[X C B Z]}{[A B C D]}=\frac{1}{2}\left(\frac{1}{4}+\frac{8}{11}\right)=\frac{43}{88}$$ The area of $X Y Z$ is the rest of the fraction of the area of $A B C D$ not covered by the three above polygons, which by a straightforward calculation $19 / 88$ the area of $A B C D$, so our answer is $$8 \cdot 11 \cdot \sin 60^{\circ} \cdot \frac{19}{88}=\frac{19 \sqrt{3}}{2}$$

\frac{19 \sqrt{3}}{2}

HMMT_2

For how many unordered sets $\{a, b, c, d\}$ of positive integers, none of which exceed 168, do there exist integers $w, x, y, z$ such that $(-1)^{w} a+(-1)^{x} b+(-1)^{y} c+(-1)^{z} d=168$? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 e^{\left.-3 \frac{|C-A|}{C}\right\rfloor}\right.$.

As an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by 24 later) and add to 168 with a unique choice of signs; then, it suffices to count $e+f+g+h=168$ with each $e, f, g, h$ in $[-168,168]$ and then divide by 24 (we drop the condition that none of them can be zero because it shouldn't affect the answer that much). One way to do this is generating functions. We want the coefficient of $t^{168}$ in the generating function $\left(t^{-168}+t^{-167}+\ldots+t^{167}+t^{168}\right)^{4}=\left(t^{169}-t^{-168}\right)^{4} /(t-1)^{4}$. Clearing the negative powers, it suffices to find the coefficient of $t^{840}$ in $\left(t^{337}-1\right)^{4} /(t-1)^{4}=\left(1-4 t^{337}+6 t^{674}-\ldots\right) \frac{1}{(t-1)^{4}}$. To do this we expand the bottom as a power series in $t$: $\frac{1}{(t-1)^{4}}=\sum_{n \geq 0}\binom{n+3}{3} t^{n}$. It remains to calculate $\binom{840+3}{3}-4 \cdot\binom{840-337+3}{3}+6 \cdot\binom{840-674+3}{3}$. This is almost exactly equal to $\frac{1}{6}\left(843^{3}-4 \cdot 506^{3}+6 \cdot 169^{3}\right) \approx 1.83 \times 10^{7}$. Dividing by 24, we arrive at an estimation 762500. Even if we use a bad approximation $\frac{1}{6 \cdot 24}\left(850^{3}-4\right.$. $500^{3}+6 \cdot 150^{3}$) we get approximately 933000, which is fairly close to the answer.

761474

HMMT_2