| # $13^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament <br> Saturday 20 February 2010 <br> <br> General Test, Part 1 |
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| <br> <br> General Test, Part 1} |
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| 1. [3] Suppose that $x$ and $y$ are positive reals such that |
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| x-y^{2}=3, \quad x^{2}+y^{4}=13 |
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| Find $x$. |
| Answer: $\frac{3+\sqrt{17}}{2}$ Squaring both sides of $x-y^{2}=3$ gives $x^{2}+y^{4}-2 x y^{2}=9$. Subtract this equation from twice the second given to get $x^{2}+2 x y^{2}+y^{4}=17 \Longrightarrow x+y^{2}= \pm 17$. Combining this equation with the first given, we see that $x=\frac{3 \pm \sqrt{17}}{2}$. Since $x$ is a positive real, $x$ must be $\frac{3+\sqrt{17}}{2}$. |
| 2. [3] Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2 x$ is in $S$ then $2 x$ is also in $T$ ? |
| Answer: 180 We partition the elements of $S$ into the following subsets: $\{1,2,4,8\},\{3,6\},\{5,10\}$, $\{7\},\{9\}$. Consider the first subset, $\{1,2,4,8\}$. Say 2 is an element of $T$. Because $2 \cdot 2=4$ is in $S, 4$ must also be in $T$. Furthermore, since $4 \cdot 2=8$ is in $S, 8$ must also be in $T$. So if $T$ contains 2 , it must also contain 4 and 8 . Similarly, if $T$ contains 1 , it must also contain 2,4 , and 8 . So $T$ can contain the following subsets of the subset $\{1,2,4,8\}$ : the empty set, $\{8\},\{4,8\},\{2,4,8\}$, or $\{1,2,4,8\}$. This gives 5 possibilities for the first subset. In general, we see that if $T$ contains an element $q$ of one of these subsets, it must also contain the elements in that subset that are larger than $q$, because we created the subsets for this to be true. So there are 3 possibilities for $\{3,6\}, 3$ for $\{5,10\}, 2$ for $\{7\}$, and 2 for $\{9\}$. This gives a total of $5 \cdot 3 \cdot 3 \cdot 2 \cdot 2=180$ possible subsets $T$. |
| 3. [4] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle. |
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| Answer: $\sqrt{5}$ |
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| Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since |
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| $$ |
| \begin{aligned} |
| {[A B C D] } & =[A B C]+[A C D] \\ |
| & =\left[A B^{\prime} C\right]+[A C D] \\ |
| & =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\ |
| & =\left[A C D E B^{\prime}\right]+[A E C] \\ |
| & =\frac{7}{10}[A B C D]+[A E C], |
| \end{aligned} |
| $$ |
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| That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$. |
| Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C, \triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. The $C E=3 x, D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$. |
| 4. [4] Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$ ? |
| Answer: 1092 Suppose that $S_{N}=0$ for some $N \geq 0$. Then $a_{N+1}=1$ because $N+1 \geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \geq N$ : |
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| | $k$ | $a_{k}$ | $S_{k}$ | |
| | :--- | ---: | :--- | |
| | $N$ | | 0 | |
| | $N+1$ | 1 | $N+1$ | |
| | $N+2$ | 1 | $2 N+3$ | |
| | $N+3$ | -1 | $N$ | |
| | $N+4$ | 1 | $2 N+4$ | |
| | $N+5$ | -1 | $N-1$ | |
| | $N+6$ | 1 | $2 N+5$ | |
| | $N+7$ | -1 | $N-2$ | |
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| We see inductively that, for every $i \geq 1$, |
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| S_{N+2 i}=2 N+2+i |
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| S_{N+1+2 i}=N+1-i |
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| thus $S_{3 N+3}=0$ is the next $k$ for which $S_{k}=0$. The values of $k$ for which $S_{k}=0$ satisfy the recurrence relation $p_{n+1}=3 p_{n}+3$, and we compute that the first terms of the sequence are $0,3,12,39,120,363,1092$; hence 1092 is our answer. |
| 5. [4] Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only blocks that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya's stack (the order in which they are removed matters). |
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| Answer: 3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3 . She removes 5 blocks, for a total of $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a block on the bottom layer does not expose any new blocks. There are $1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=256$ of these (the last factor of 4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answer is $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13-1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=3384$. |
| 6. [5] John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay? |
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| ## Answer: 20503 |
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| Let the number of 2,5 , and 10 dollar notes John can use be $x, y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2 x+5 y+10 z=2010$. Consider this equation $\bmod 2$. Because $2 x, 10 z$, and 2010 are even, $5 y$ must also be even, so $y$ must be even. Now consider the equation $\bmod 5$. Because $5 y, 10 z$, and 2010 are divisible by $5,2 x$ must also be divisible by 5 , so $x$ must be divisible by 5 . So both $2 x$ and $5 y$ are divisible by 10 . So the equation is equivalent to $10 x^{\prime}+10 y^{\prime}+10 z=2010$, or $x^{\prime}+y^{\prime}+z=201$, with $x^{\prime}, y^{\prime}$, and $z$ nonnegative integers. There is a well-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table (explained in further detail below), so there are $\binom{203}{2}=20503$ ways. |
| The bijection between solutions of $x^{\prime}+y^{\prime}+z=201$ and arrangements of 203 balls in a row is as follows. Given a solution of the equation, we put $x^{\prime}$ white balls in a row, then a black ball, then $y^{\prime}$ white balls, then a black ball, then $z$ white balls. This is like having 203 balls in a row on a table and picking two of them to be black. To go from an arrangement of balls to a solution of the equation, we just read off $x^{\prime}, y^{\prime}$, and $z$ from the number of white balls in a row. There are $\binom{203}{2}$ ways to choose 2 of 203 balls to be black, so there are $\binom{203}{2}$ solutions to $x^{\prime}+y^{\prime}+z=201$. |
| 7. [6] Suppose that a polynomial of the form $p(x)=x^{2010} \pm x^{2009} \pm \cdots \pm x \pm 1$ has no real roots. What is the maximum possible number of coefficients of -1 in $p$ ? |
| Answer: 1005 Let $p(x)$ be a polynomial with the maximum number of minus signs. |
| $p(x)$ cannot have more than 1005 minus signs, otherwise $p(1)<0$ and $p(2) \geq 2^{2010}-2^{2009}-\ldots-2-1=$ 1 , which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than 1. |
| Let $p(x)=\frac{x^{2011}+1}{x+1}=x^{2010}-x^{2009}+x^{2008}-\ldots-x+1 .-1$ is the only real root of $x^{2011}+1=0$ but $p(-1)=2011$; therefore $p$ has no real roots. Since $p$ has 1005 minus signs, it is the desired polynomial. |
| 8. [6] A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$. |
| Answer: 2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension. |
| Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles. |
| Base case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identical, and in the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus, in dimension 2 , the largest finite number of intersection points is 2 , because the number of pairwise intersection points is 0,1 , or 2 for distinct circles. |
| We now prove that the intersection of two $k$-dimensional spheres is either the empty set, a $(k-1)$ dimensional sphere, a $k$-dimensional sphere (which only occurs if the original spheres are identical and coincident). Consider two spheres in $k$-dimensional space, and impose a coordinate system such that the centers of the two spheres lie on one coordinate axis. Then the equations for the two spheres become identical in all but one coordinate: |
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| $$ |
| \begin{aligned} |
| & \left(x_{1}-a_{1}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{1}^{2} \\ |
| & \left(x_{1}-a_{2}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{2}^{2} |
| \end{aligned} |
| $$ |
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| If $a_{1}=a_{2}$, the spheres are concentric, and so they are either nonintersecting or coincident, intersecting in a $k$-dimensional sphere. If $a_{1} \neq a_{2}$, then subtracting the equations and solving for $x_{1}$ yields $x_{1}=\frac{r_{1}^{2}-a_{1}^{2}-r_{2}^{2}+a_{2}^{2}}{2\left(a_{2}-a_{2}\right)}$. Plugging this in to either equation above yields a single equation equation that describes a $(k-1)$-dimensional sphere. |
| Assume we are in dimension $d$, and suppose for induction that for all $k$ less than $d$, any two distinct $k$-dimensional spheres intersecting in a finite number of points intersect in at most two points. Suppose we have a collection of $d$-dimensional spheres $s_{1}, s_{2}, \ldots, s_{m}$. Without loss of generality, suppose the $s_{i}$ are distinct. Let $t_{i}$ be the intersection of $s_{i}$ and $s_{i+1}$ for $1 \leq i<m$. If any $t_{i}$ are the empty set, then the intersection of the $t_{i}$ is empty. None of the $t_{i}$ is a $d$-dimensional sphere because the $s_{i}$ are distinct. Thus each of $t_{1}, t_{2}, \ldots, t_{m-1}$ is a $(d-1)$-dimensional sphere, and the intersection of all of them is the same as the intersection of the $d$-dimensional spheres. We can then apply the inductive hypothesis to find that $t_{1}, \ldots, t_{m-1}$ intersect in at most two points. Thus, by induction, a set of spheres in any dimension which intersect at only finitely many points intersect at at most two points. |
| We now exhibit a set of $2^{2009}$ 2010-dimensional spheres, and prove that their intersection contains exactly two points. Take the spheres with radii $\sqrt{2013}$ and centers $(0, \pm 1, \pm 1, \ldots, \pm 1)$, where the sign of each coordinate is independent from the sign of every other coordinate. Because of our choice of radius, all these spheres pass through the points $( \pm 2,0,0, \ldots 0)$. Then the intersection is the set of points $\left(x_{1}, x_{2}, \ldots, x_{2010}\right)$ which satisfy the equations $x_{1}^{2}+\left(x_{2} \pm 1\right)^{2}+\cdots+\left(x_{2010} \pm 1\right)^{2}=2013$. The only solutions to these equations are the points $( \pm 2,0,0, \ldots, 0)$ (since $\left(x_{i}+1\right)^{2}$ must be the same as $\left(x_{i}-1\right)^{2}$ for all $i>1$, because we may hold all but one of the $\pm$ choices constant, and change the remaining one). |
| 9. [7] Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$. |
| Answer: $\frac{\sqrt{6}+\sqrt{2}+8}{4}$ |
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| By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry ( since $A B=A D$ ) it |
| must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$. |
| Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have |
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| $$ |
| \begin{aligned} |
| M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\ |
| & =\cos \left(180^{\circ}-E P Q-R P Q\right) \\ |
| & =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\ |
| & =-\cos \left(165^{\circ}\right) \\ |
| & =\cos \left(15^{\circ}\right) \\ |
| & =\frac{\sqrt{6}+\sqrt{2}}{4} . |
| \end{aligned} |
| $$ |
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| Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed. |
| 10. [8] Let $a, b, c, x, y$, and $z$ be complex numbers such that |
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| a=\frac{b+c}{x-2}, \quad b=\frac{c+a}{y-2}, \quad c=\frac{a+b}{z-2} . |
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| If $x y+y z+z x=67$ and $x+y+z=2010$, find the value of $x y z$. |
| Answer: -5892 Manipulate the equations to get a common denominator: $a=\frac{b+c}{x-2} \Longrightarrow x-2=$ $\frac{b+c}{a} \Longrightarrow x-1=\frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1}=\frac{a}{a+b+c}$; similarly, $\frac{1}{y-1}=\frac{b}{a+b+c}$ and $\frac{1}{z-1}=\frac{c}{a+b+c}$. Thus |
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| $$ |
| \begin{aligned} |
| \frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1} & =1 \\ |
| (y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) & =(x-1)(y-1)(z-1) \\ |
| x y+y z+z x-2(x+y+z)+3 & =x y z-(x y+y z+z x)+(x+y+z)-1 \\ |
| x y z-2(x y+y z+z x)+3(x+y+z)-4 & =0 \\ |
| x y z-2(67)+3(2010)-4 & =0 \\ |
| x y z & =-5892 |
| \end{aligned} |
| $$ |
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