{"year": "2010", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Suppose that $x$ and $y$ are positive reals such that\n\n$$\nx-y^{2}=3, \\quad x^{2}+y^{4}=13\n$$\n\nFind $x$.", "solution": "$\\frac{3+\\sqrt{17}}{2}$ 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 $\\frac{3+\\sqrt{17}}{2}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "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 $T$ ?", "solution": "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 $T$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "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.\n![](https://cdn.mathpix.com/cropped/2025_01_24_9ba978224411db7b77a2g-1.jpg?height=201&width=432&top_left_y=1515&top_left_x=890)", "solution": "$\\sqrt{5}$\n![](https://cdn.mathpix.com/cropped/2025_01_24_9ba978224411db7b77a2g-1.jpg?height=346&width=421&top_left_y=1876&top_left_x=901)\n\nGiven 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\n\n$$\n\\begin{aligned}\n{[A B C D] } & =[A B C]+[A C D] \\\\\n& =\\left[A B^{\\prime} C\\right]+[A C D] \\\\\n& =\\left[A B^{\\prime} E\\right]+2[A E C]+[E D C] \\\\\n& =\\left[A C D E B^{\\prime}\\right]+[A E C] \\\\\n& =\\frac{7}{10}[A B C D]+[A E C],\n\\end{aligned}\n$$\n\nThat is, $[A E C]=\\frac{3}{10}[A B C D]=\\frac{3}{10}$.\nSince $\\triangle E C D$ 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 $\\frac{A D}{D C}=\\frac{A E+E D}{D C}=\\frac{3+2}{\\sqrt{5}}=\\sqrt{5}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "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 $S_{k}=0$ ?", "solution": "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 $k \\geq N$ :\n\n| $k$ | $a_{k}$ | $S_{k}$ |\n| :--- | ---: | :--- |\n| $N$ |  | 0 |\n| $N+1$ | 1 | $N+1$ |\n| $N+2$ | 1 | $2 N+3$ |\n| $N+3$ | -1 | $N$ |\n| $N+4$ | 1 | $2 N+4$ |\n| $N+5$ | -1 | $N-1$ |\n| $N+6$ | 1 | $2 N+5$ |\n| $N+7$ | -1 | $N-2$ |\n\nWe see inductively that, for every $i \\geq 1$,\n\n$$\nS_{N+2 i}=2 N+2+i\n$$\n\nand\n\n$$\nS_{N+1+2 i}=N+1-i\n$$\n\nthus $S_{3 N+3}=0$ 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.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "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).\n![](https://cdn.mathpix.com/cropped/2025_01_24_9ba978224411db7b77a2g-2.jpg?height=318&width=315&top_left_y=2194&top_left_x=943)", "solution": "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 $1 \\cdot 4 \\cdot 7 \\cdot 10 \\cdot 13-1 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4=3384$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n5. [4]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "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?", "solution": "20503\n\nLet 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.\nThe 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 $x^{\\prime}+y^{\\prime}+z=201$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\n## Answer: "}}
{"year": "2010", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "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 $p$ ?", "solution": "1005 Let $p(x)$ be a polynomial with the maximum number of minus signs.\n$p(x)$ 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.\nLet $p(x)=\\frac{x^{2011}+1}{x+1}=x^{2010}-x^{2009}+x^{2008}-\\ldots-x+1 .-1$ 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.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "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 $n$.", "solution": "2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.\nNote that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.\nBase 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.\nWe 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:\n\n$$\n\\begin{aligned}\n& \\left(x_{1}-a_{1}\\right)^{2}+x_{2}^{2}+\\ldots+x_{k}^{2}=r_{1}^{2} \\\\\n& \\left(x_{1}-a_{2}\\right)^{2}+x_{2}^{2}+\\ldots+x_{k}^{2}=r_{2}^{2}\n\\end{aligned}\n$$\n\nIf $a_{1}=a_{2}$, 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.\nAssume 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.\nWe 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).", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "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 $S$.", "solution": "$\\frac{\\sqrt{6}+\\sqrt{2}+8}{4}$\n![](https://cdn.mathpix.com/cropped/2025_01_24_9ba978224411db7b77a2g-4.jpg?height=679&width=673&top_left_y=1520&top_left_x=764)\n\nBy 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\nmust 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 $\\omega_{3}$.\nNow, 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\n\n$$\n\\begin{aligned}\nM N & =P R \\cdot \\cos \\left(180^{\\circ}-\\angle E P R\\right) \\\\\n& =\\cos \\left(180^{\\circ}-E P Q-R P Q\\right) \\\\\n& =-\\cos \\left(\\left(270^{\\circ}-60^{\\circ}\\right) / 2+60^{\\circ}\\right) \\\\\n& =-\\cos \\left(165^{\\circ}\\right) \\\\\n& =\\cos \\left(15^{\\circ}\\right) \\\\\n& =\\frac{\\sqrt{6}+\\sqrt{2}}{4} .\n\\end{aligned}\n$$\n\nThus $A D=A M+M N+N D=1+\\frac{\\sqrt{6}+\\sqrt{2}}{4}+1=\\frac{\\sqrt{6}+\\sqrt{2}+8}{4}$ as claimed.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let $a, b, c, x, y$, and $z$ be complex numbers such that\n\n$$\na=\\frac{b+c}{x-2}, \\quad b=\\frac{c+a}{y-2}, \\quad c=\\frac{a+b}{z-2} .\n$$\n\nIf $x y+y z+z x=67$ and $x+y+z=2010$, find the value of $x y z$.", "solution": "-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\n\n$$\n\\begin{aligned}\n\\frac{1}{x-1}+\\frac{1}{y-1}+\\frac{1}{z-1} & =1 \\\\\n(y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) & =(x-1)(y-1)(z-1) \\\\\nx y+y z+z x-2(x+y+z)+3 & =x y z-(x y+y z+z x)+(x+y+z)-1 \\\\\nx y z-2(x y+y z+z x)+3(x+y+z)-4 & =0 \\\\\nx y z-2(67)+3(2010)-4 & =0 \\\\\nx y z & =-5892\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}