{"year": "2010", "tier": "T4", "problem_label": "1", "problem_type": "Algebra", "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-alg-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "2", "problem_type": "Algebra", "exam": "HMMT", "problem": "The rank of a rational number $q$ is the unique $k$ for which $q=\\frac{1}{a_{1}}+\\cdots+\\frac{1}{a_{k}}$, where each $a_{i}$ is the smallest positive integer such that $q \\geq \\frac{1}{a_{1}}+\\cdots+\\frac{1}{a_{i}}$. Let $q$ be the largest rational number less than $\\frac{1}{4}$ with rank 3 , and suppose the expression for $q$ is $\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\frac{1}{a_{3}}$. Find the ordered triple $\\left(a_{1}, a_{2}, a_{3}\\right)$.", "solution": "$(5,21,421)$ Suppose that $A$ and $B$ were rational numbers of rank 3 less than $\\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\frac{1}{a_{3}}$ and $B=\\frac{1}{b_{1}}+\\frac{1}{b_{2}}+\\frac{1}{b_{3}}$ are the expressions for $A$ and $B$ as stated in the problem. If $b_{1}<a_{1}$ then $A<\\frac{1}{a_{1}-1} \\leq \\frac{1}{b_{1}}<B$. In other words, of all the rationals less than $\\frac{1}{4}$ with rank 3 , those that have $a_{1}=5$ are greater than those that have $a_{1}=6,7,8, \\ldots$ Therefore we can \"build\" $q$ greedily, adding the largest unit fraction that keeps $q$ less than $\\frac{1}{4}$ :\n$\\frac{1}{5}$ is the largest unit fraction less than $\\frac{1}{4}$, hence $a_{1}=5$;\n$\\frac{1}{27}$ is the largest unit fraction less than $\\frac{1}{4}-\\frac{1}{5}$, hence $a_{2}=21$;\n$\\frac{1}{421}$ is the largest unit fraction less than $\\frac{1}{4}-\\frac{1}{5}-\\frac{1}{21}$, hence $a_{3}=421$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "3", "problem_type": "Algebra", "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-alg-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "4", "problem_type": "Algebra", "exam": "HMMT", "problem": "Suppose that there exist nonzero complex numbers $a, b, c$, and $d$ such that $k$ is a root of both the equations $a x^{3}+b x^{2}+c x+d=0$ and $b x^{3}+c x^{2}+d x+a=0$. Find all possible values of $k$ (including complex values).", "solution": "$1,-1, i,-i$ Let $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4}-a=0$, which means that $k$ is either $1,-1, i$, or $-i$. If $a=b=c=d=1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a=b=c=1$ and $d=-3$, then 1 is a root of both polynomials. So $k$ can be $1,-1, i$, and $-i$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "5", "problem_type": "Algebra", "exam": "HMMT", "problem": "Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$.", "solution": "-90 We have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\\frac{\\sqrt{2 b-1} \\pm 1}{2}$. Then\n\n$$\n\\begin{aligned}\nx^{20}+y^{20} & =\\left(\\frac{\\sqrt{2 b-1}+1}{2}\\right)^{20}+\\left(\\frac{\\sqrt{2 b-1}-1}{2}\\right)^{20} \\\\\n& =\\frac{1}{2^{20}}\\left[(\\sqrt{2 b-1}+1)^{20}+(\\sqrt{2 b-1}-1)^{20}\\right] \\\\\n& =\\frac{2}{2^{20}}\\left[(\\sqrt{2 b-1})^{20}+\\binom{20}{2}(\\sqrt{2 b-1})^{18}+\\binom{20}{4}(\\sqrt{2 b-1})^{16}+\\ldots\\right] \\\\\n& =\\frac{2}{2^{20}}\\left[(2 b-1)^{10}+\\binom{20}{2}(2 b-1)^{9}+\\binom{20}{4}(2 b-1)^{8}+\\ldots\\right] \\\\\n& =20\n\\end{aligned}\n$$\n\nWe want to find the sum of distinct roots of the above polynomial in $b$; we first prove that the original polynomial is square-free. The conditions $x+y=1$ and $x^{20}+y^{20}=20$ imply that $x^{20}+(1-x)^{20}-20=0$; let $p(x)=x^{20}+(1-x)^{20}-20 . p$ is square-free if and only if $G C D\\left(p, p^{\\prime}\\right)=c$ for some constant $c$ :\n\n$$\n\\begin{aligned}\n\\operatorname{GCD}\\left(p, p^{\\prime}\\right) & =\\operatorname{GCD}\\left(x^{20}+(1-x)^{20}-20,20\\left(x^{19}-(1-x)^{19}\\right)\\right) \\\\\n& =\\operatorname{GCD}\\left(x^{20}-x(1-x)^{19}+(1-x)^{19}-20,20\\left(x^{19}-(1-x)^{19}\\right)\\right) \\\\\n& =G C D\\left((1-x)^{19}-20, x^{19}-(1-x)^{19}\\right) \\\\\n& =G C D\\left((1-x)^{19}-20, x^{19}-20\\right)\n\\end{aligned}\n$$\n\nThe roots of $x^{19}-20$ are $\\sqrt[19]{20^{k}} \\exp \\left(\\frac{2 \\pi i k}{19}\\right)$ for some $k=0,1, \\ldots, 18$; the roots of $(1-x)^{19}-20$ are $1-\\sqrt[19]{20^{k}} \\exp \\left(\\frac{2 \\pi i k}{19}\\right)$ for some $k=0,1, \\ldots, 18$. If $x^{19}-20$ and $(1-x)^{19}-20$ share a common root, then there exist integers $m, n$ such that $\\sqrt[19]{20^{m}} \\exp \\left(\\frac{2 \\pi i m}{19}\\right)=1-\\sqrt[19]{20^{n}} \\exp \\left(\\frac{2 \\pi i n}{19}\\right)$; since the imaginary parts of both sides must be the same, we have $m=n$ and $\\sqrt[19]{20^{m}} \\exp \\left(\\frac{2 \\pi i m}{19}\\right)=\\frac{1}{2} \\Longrightarrow 20^{m}=\\frac{1}{2^{19}}$, a contradiction. Thus we have proved that the polynomial in $x$ has no double roots. Since for each $b$ there exists a unique pair ( $x, y$ ) (up to permutations) that satisfies $x^{2}+y^{2}=b$ and $(x+y)^{2}=1$, the polynomial in $b$ has no double roots.\nLet the coefficient of $b^{n}$ in the above equation be $\\left[b^{n}\\right]$. By Vieta's Formulas, the sum of all possible values of $b=x^{2}+y^{2}$ is equal to $-\\frac{\\left[b^{9}\\right]}{\\left[b^{0}\\right]} .\\left[b^{10}\\right]=\\frac{2}{2^{20}}\\left(2^{10}\\right)$ and $\\left[b^{9}\\right]=\\frac{2}{2^{20}}\\left(-\\binom{10}{1} 2^{9}+\\binom{20}{2} 2^{9}\\right)$; thus $-\\frac{\\left[b^{9}\\right]}{\\left[b^{10}\\right]}=-\\frac{\\binom{10}{1} 2^{9}-\\binom{20}{2_{2}}^{9}}{2^{10}}=-90$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n5. [5]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "6", "problem_type": "Algebra", "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. $p(x)$ 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-alg-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "7", "problem_type": "Algebra", "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-alg-solutions.jsonl", "problem_match": "\n7. [5]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "8", "problem_type": "Algebra", "exam": "HMMT", "problem": "How many polynomials of degree exactly 5 with real coefficients send the set $\\{1,2,3,4,5,6\\}$ to a permutation of itself?", "solution": "714 For every permutation $\\sigma$ of $\\{1,2,3,4,5,6\\}$, Lagrange Interpolation ${ }^{1}$ gives a polynomial of degree at most 5 with $p(x)=\\sigma(x)$ for every $x=1,2,3,4,5,6$. Additionally, this polynomial is unique: assume that there exist two polynomials $p, q$ of degree $\\leq 5$ such that they map $\\{1,2,3,4,5,6\\}$ to the same permutation. Then $p-q$ is a nonzero polynomial of degree $\\leq 5$ with 6 distinct roots, a contradiction. Thus an upper bound for the answer is $6!=720$ polynomials.\nHowever, not every polynomial obtained by Lagrange interpolation is of degree 5 (for example, $p(x)=$ $x$ ). We can count the number of invalid polynomials using finite differences ${ }^{2}$ A polynomial has degree less than 5 if and only if the sequence of 5 th finite differences is 0 . The 5 th finite difference of $p(1), p(2), p(3), p(4), p(5), p(6)$ is $p(1)-5 p(2)+10 p(3)-10 p(4)+5 p(5)-p(6)$; thus we want to solve $p(1)-5 p(2)+10 p(3)-10 p(4)+5 p(5)-p(6)=0$ with $\\{p(1), p(2), p(3), p(4), p(5), p(6)\\}=\\{1,2,3,4,5,6\\}$.\nTaking the above equation modulo 5 , we get $p(1)=p(6)(\\bmod 5) \\Longrightarrow\\{p(1), p(6)\\}=\\{1,6\\}$. Note that $1-5 p(2)+10 p(3)-10 p(4)+5 p(5)-6=0$ if and only if $6-5 p(5)+10 p(4)-10 p(3)+5 p(2)-1=0$, so we may assume that $p(1)=1$ and double our result later. Then we have $\\{p(2), p(3), p(4), p(5)\\}=\\{2,3,4,5\\}$ and\n\n$$\n-p(2)+2 p(3)-2 p(4)+p(5)=1\n$$\n\nThe above equation taken modulo 2 implies that $p(2), p(5)$ are of opposite parity, so $p(3), p(4)$ are of opposite parity. We do casework on $\\{p(2), p(5)\\}$ :\n(a) $p(2)=2, p(5)=3 ; 2 p(3)-2 p(4)=0$ is a contradiction\n(b) $p(2)=2, p(5)=5 ; 2 p(3)-2 p(4)=-2 \\Longrightarrow p(3)-p(4)=-1 \\Longrightarrow p(3)=3, p(4)=4$\n(c) $p(2)=3, p(5)=2 ; 2 p(3)-2 p(4)=-2 \\Longrightarrow p(3)-p(4)=-1 \\Longrightarrow p(3)=4, p(4)=5$\n(d) $p(2)=3, p(5)=4 ; 2 p(3)-2 p(4)=0$ is a contradiction\n(e) $p(2)=4, p(5)=3 ; 2 p(3)-2 p(4)=2 \\Longrightarrow p(3)-p(4)=1$ but $\\{p(3), p(4)\\}=\\{2,5\\}$, contradiction\n(f) $p(2)=4, p(5)=5 ; 2 p(3)-2 p(4)=0$ is a contradiction\n(g) $p(2)=5, p(5)=2 ; 2 p(3)-2 p(4)=4 \\Longrightarrow p(3)-p(4)=2$, contradiction\n(h) $p(2)=5, p(5)=4 ; 2 p(3)-2 p(4)=2 \\Longrightarrow p(3)-p(4)=1 \\Longrightarrow p(3)=3, p(4)=2$\n\nHence there are a total of $720-2(3)=714$ polynomials.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "9", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $f(x)=c x(x-1)$, where $c$ is a positive real number. We use $f^{n}(x)$ to denote the polynomial obtained by composing $f$ with itself $n$ times. For every positive integer $n$, all the roots of $f^{n}(x)$ are real. What is the smallest possible value of $c$ ?", "solution": "2 We first prove that all roots of $f^{n}(x)$ are greater than or equal to $-\\frac{c}{4}$ and less than or equal to $1+\\frac{c}{4}$. Suppose that $r$ is a root of $f^{n}(x)$. If $r=-\\frac{c}{4}, f^{-1}(r)=\\left\\{\\frac{1}{2}\\right\\}$ and $-\\frac{c}{4}<\\frac{1}{2}<1+\\frac{c}{4}$ since $c$ is positive. Suppose $r \\neq-\\frac{c}{4}$; by the quadratic formula, there exist two complex numbers $r_{1}, r_{2}$ such that $r_{1}+r_{2}=1$ and $f\\left(r_{1}\\right)=f\\left(r_{2}\\right)=r$. Thus all the roots of $f^{n}(x)$ (except $\\frac{1}{2}$ ) come in pairs that sum to 1 . No root $r$ of $f^{n}(x)$ can be less than $-\\frac{c}{4}$, otherwise $f^{n+1}(x)$ has an imaginary root, $f^{-1}(r)$. Also, no root $r$ of $f^{n}(x)$ can be greater than $1+\\frac{c}{4}$, otherwise its \"conjugate\" root will be less than $-\\frac{c}{4}$.\nDefine $g(x)=\\frac{1}{2}\\left(1+\\sqrt{1+\\frac{4 x}{c}}\\right)$, the larger inverse of $f(x)$. Note that $g^{n}(x)$ is the largest element of $f^{-n}(x)$ (which is a set). $g^{n}(0)$ should be less than or equal to $1+\\frac{c}{4}$ for all $n$. Let $x_{0}$ be the nonzero real number such that $g\\left(x_{0}\\right)=x_{0}$; then $c x_{0}\\left(x_{0}-1\\right)=x_{0} \\Longrightarrow x_{o}=1+\\frac{1}{c}$. $x_{0}<g(x)<x$ if $x>x_{0}$ and $x<g(x)<x_{0}$ if $x<x_{0}$; it can be proved that $g^{n}$ converges to $x_{0}$. Hence we have the requirement that $x_{0}=1+\\frac{1}{c} \\leq 1+\\frac{c}{4} \\Longrightarrow c \\geq 2$.\nWe verify that $c=2$ is possible. All the roots of $f^{-} n(x)$ will be real if $g(0) \\leq 1+\\frac{c}{4}=\\frac{3}{2}$. We know that $0<\\frac{3}{2} \\Longrightarrow g(0)<\\frac{3}{2}$, so $g^{2}(0)<\\frac{3}{2}$ and $g^{n}(0)<g^{n+1}(0)<\\frac{3}{2}$ for all $n$. Therefore all the roots of $f^{n}(x)$ are real.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}}
{"year": "2010", "tier": "T4", "problem_label": "10", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $p(x)$ and $q(x)$ be two cubic polynomials such that $p(0)=-24, q(0)=30$, and\n$$\np(q(x))=q(p(x))\n$$\nfor all real numbers $x$. Find the ordered pair $(p(3), q(6))$.", "solution": "$(3,-24)$ Note that the polynomials $f(x)=a x^{3}$ and $g(x)=-a x^{3}$ commute under composition. Let $h(x)=x+b$ be a linear polynomial, and note that its inverse $h^{-1}(x)=x-b$ is also a linear polynomial. The composite polynomials $h^{-1} f h$ and $h^{-1} g h$ commute, since function composition is associative, and these polynomials are also cubic.\nWe solve for the $a$ and $b$ such that $\\left(h^{-1} f h\\right)(0)=-24$ and $\\left(h^{-1} g h\\right)(0)=30$. We must have:\n$$\na b^{3}-b=-24,-a b^{3}-b=30 \\Rightarrow a=1, b=-3\n$$\n\nThese values of $a$ and $b$ yield the polynomials $p(x)=(x-3)^{3}+3$ and $q(x)=-(x-3)^{3}+3$. The polynomials take on the values $p(3)=3$ and $q(6)=-24$.\nRemark: The pair of polynomials found in the solution is not unique. There is, in fact, an entire family of commuting cubic polynomials with $p(0)=-24$ and $q(0)=30$. They are of the form\n\n$$\np(x)=t x(x-3)(x-6)-24, q(x)=-t x(x-3)(x-6)+30\n$$\n\nwhere $t$ is any real number. However, the values of $p(3)$ and $q(6)$ are the same for all polynomials in this family. In fact, if we give the initial conditions $p(0)=k_{1}$ and $q(0)=k_{2}$, then we get a general solution of\n\n$$\n\\begin{gathered}\np(x)=t\\left(x^{3}-\\frac{3}{2}\\left(k_{1}+k_{2}\\right) x^{2}+\\frac{1}{2}\\left(k_{1}+k_{2}\\right)^{2} x\\right)+\\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{1} \\\\\nq(x)=-t\\left(x^{3}-\\frac{3}{2}\\left(k_{1}+k_{2}\\right) x^{2}+\\frac{1}{2}\\left(k_{1}+k_{2}\\right)^{2} x\\right)-\\frac{k_{2}-k_{1}}{k_{2}+k_{1}} x+k_{2}\n\\end{gathered}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-alg-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}