{"year": "2007", "tier": "T4", "problem_label": "1", "problem_type": "Algebra", "exam": "HMMT", "problem": "Compute\n\n$$\n\\left\\lfloor\\frac{2007!+2004!}{2006!+2005!}\\right\\rfloor\n$$\n\n(Note that $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.)", "solution": "2006. We have\n\n$$\n\\left\\lfloor\\frac{2007!+2004!}{2006!+2005!}\\right\\rfloor=\\left\\lfloor\\frac{\\left(2007 \\cdot 2006+\\frac{1}{2005}\\right) \\cdot 2005!}{(2006+1) \\cdot 2005!}\\right\\rfloor=\\left\\lfloor\\frac{2007 \\cdot 2006+\\frac{1}{2005}}{2007}\\right\\rfloor=\\left\\lfloor 2006+\\frac{1}{2005 \\cdot 2007}\\right\\rfloor\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "2", "problem_type": "Algebra", "exam": "HMMT", "problem": "Two reals $x$ and $y$ are such that $x-y=4$ and $x^{3}-y^{3}=28$. Compute $x y$.", "solution": "$-\\mathbf{3}$. We have $28=x^{3}-y^{3}=(x-y)\\left(x^{2}+x y+y^{2}\\right)=(x-y)\\left((x-y)^{2}+3 x y\\right)=4 \\cdot(16+3 x y)$, from which $x y=-3$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "3", "problem_type": "Algebra", "exam": "HMMT", "problem": "Three real numbers $x, y$, and $z$ are such that $(x+4) / 2=(y+9) /(z-3)=(x+5) /(z-5)$. Determine the value of $x / y$.", "solution": "1/2. Because the first and third fractions are equal, adding their numerators and denominators produces another fraction equal to the others: $((x+4)+(x+5)) /(2+(z-5))=(2 x+9) /(z-3)$. Then $y+9=2 x+9$, etc.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "4", "problem_type": "Algebra", "exam": "HMMT", "problem": "Compute\n\n$$\n\\frac{2^{3}-1}{2^{3}+1} \\cdot \\frac{3^{3}-1}{3^{3}+1} \\cdot \\frac{4^{3}-1}{4^{3}+1} \\cdot \\frac{5^{3}-1}{5^{3}+1} \\cdot \\frac{6^{3}-1}{6^{3}+1}\n$$", "solution": "$\\frac{\\mathbf{4 3} \\text {. }}{63}$ Use the factorizations $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$ and $n^{3}+1=(n+1)\\left(n^{2}-n+1\\right)$ to write\n\n$$\n\\frac{1 \\cdot 7}{3 \\cdot 3} \\cdot \\frac{2 \\cdot 13}{4 \\cdot 7} \\cdot \\frac{3 \\cdot 21}{5 \\cdot 13} \\cdot \\frac{4 \\cdot 31}{6 \\cdot 21} \\cdot \\frac{5 \\cdot 43}{7 \\cdot 31}=\\frac{1 \\cdot 2 \\cdot 43}{3 \\cdot 6 \\cdot 7}=\\frac{43}{63}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "5", "problem_type": "Algebra", "exam": "HMMT", "problem": "A convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$; determine its area.", "solution": "$\\mathbf{4 \\sqrt { 1 7 }}$. By symmetry, the quadrilateral is a rectangle having $x=y$ and $x=-y$ as axes of symmetry. Let $(a, b)$ with $a>b>0$ be one of the vertices. Then the desired area is\n\n$$\n(\\sqrt{2}(a-b)) \\cdot(\\sqrt{2}(a+b))=2\\left(a^{2}-b^{2}\\right)=2 \\sqrt{a^{4}-2 a^{2} b^{2}+b^{4}}=2 \\sqrt{100-2 \\cdot 4^{2}}=4 \\sqrt{17}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n5. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "6", "problem_type": "Algebra", "exam": "HMMT", "problem": "Consider the polynomial $P(x)=x^{3}+x^{2}-x+2$. Determine all real numbers $r$ for which there exists a complex number $z$ not in the reals such that $P(z)=r$.", "solution": "$\\mathbf{r}>\\mathbf{3}, \\mathbf{r}<\\frac{\\mathbf{4 9}}{\\mathbf{2 7}}$. Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x)=r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x)-r$ has a repeated zero. Thus, we write\n\n$$\nP(x)-r=x^{3}+x^{2}-x+(2-r)=(x-p)^{2}(x-q)=x^{3}-(2 p+q) x^{2}+p(p+2 q) x-p^{2} q\n$$\n\nThen $q=-2 p-1$ and $1=p(p+2 q)=p(-3 p-2)$ so that $p=1 / 3$ or $p=-1$. It follows that the graph of $P(x)$ is horizontal at $x=1 / 3$ (a maximum) and $x=-1$ (a minimum), so the desired values $r$ are $r>P(-1)=3$ and $r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "7", "problem_type": "Algebra", "exam": "HMMT", "problem": "An infinite sequence of positive real numbers is defined by $a_{0}=1$ and $a_{n+2}=6 a_{n}-a_{n+1}$ for $n=0,1,2, \\cdots$ Find the possible value(s) of $a_{2007}$.", "solution": "$\\mathbf{2}^{\\mathbf{2 0 0 7}}$. The characteristic equation of the linear homogeneous equation is $m^{2}+m-6=$ $(m+3)(m-2)=0$ with solutions $m=-3$ and $m=2$. Hence the general solution is given by $a_{n}=A(2)^{n}+B(-3)^{n}$ where $A$ and $B$ are constants to be determined. Then we have $a_{n}>0$ for $n \\geq 0$, so necessarily $B=0$, and $a_{0}=1 \\Rightarrow A=1$. Therefore, the unique solution to the recurrence is $a_{n}=2^{n}$ for all n .", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n7. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "8", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $A:=\\mathbb{Q} \\backslash\\{0,1\\}$ denote the set of all rationals other than 0 and 1. A function $f: A \\rightarrow \\mathbb{R}$ has the property that for all $x \\in A$,\n\n$$\nf(x)+f\\left(1-\\frac{1}{x}\\right)=\\log |x|\n$$\n\nCompute the value of $f(2007)$.", "solution": "$\\log (\\mathbf{2 0 0 7} / \\mathbf{2 0 0 6})$. Let $g: A \\rightarrow A$ be defined by $g(x):=1-1 / x$; the key property is that\n\n$$\ng(g(g(x)))=1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{x}}}=x\n$$\n\nThe given equation rewrites as $f(x)+f(g(x))=\\log |x|$. Substituting $x=g(y)$ and $x=g(g(z))$ gives the further equations $f(g(y))+f(g(g(y)))=\\log |g(x)|$ and $f(g(g(z)))+f(z)=\\log |g(g(x))|$. Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives\n\n$$\nf(x)=\\frac{1}{2} \\cdot(\\log |x|-\\log |g(x)|+\\log |g(g(x))|)\n$$\n\nFor $x=2007$, we have $g(x)=\\frac{2006}{2007}$ and $g(g(x))=\\frac{-1}{2006}$, so that\n\n$$\nf(2007)=\\frac{\\log |2007|-\\log \\left|\\frac{2006}{2007}\\right|+\\log \\left|\\frac{-1}{2006}\\right|}{2}=\\log (2007 / 2006)\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "9", "problem_type": "Algebra", "exam": "HMMT", "problem": "The complex numbers $\\alpha_{1}, \\alpha_{2}, \\alpha_{3}$, and $\\alpha_{4}$ are the four distinct roots of the equation $x^{4}+2 x^{3}+2=0$. Determine the unordered set\n\n$$\n\\left\\{\\alpha_{1} \\alpha_{2}+\\alpha_{3} \\alpha_{4}, \\alpha_{1} \\alpha_{3}+\\alpha_{2} \\alpha_{4}, \\alpha_{1} \\alpha_{4}+\\alpha_{2} \\alpha_{3}\\right\\} .\n$$", "solution": "$\\{\\mathbf{1} \\pm \\sqrt{\\mathbf{5}}, \\mathbf{- 2}\\}$. Employing the elementary symmetric polynomials $\\left(s_{1}=\\alpha_{1}+\\alpha_{2}+\\alpha_{3}+\\alpha_{4}=\\right.$ $-2, s_{2}=\\alpha_{1} \\alpha_{2}+\\alpha_{1} \\alpha_{3}+\\alpha_{1} \\alpha_{4}+\\alpha_{2} \\alpha_{3}+\\alpha_{2} \\alpha_{4}+\\alpha_{3} \\alpha_{4}=0, s_{3}=\\alpha_{1} \\alpha_{2} \\alpha_{3}+\\alpha_{2} \\alpha_{3} \\alpha_{4}+\\alpha_{3} \\alpha_{4} \\alpha_{1}+\\alpha_{4} \\alpha_{1} \\alpha_{2}=0$, and $s_{4}=\\alpha_{1} \\alpha_{2} \\alpha_{3} \\alpha_{4}=2$ ) we consider the polynomial\n\n$$\nP(x)=\\left(x-\\left(\\alpha_{1} \\alpha_{2}+\\alpha_{3} \\alpha_{4}\\right)\\right)\\left(x-\\left(\\alpha_{1} \\alpha_{3}+\\alpha_{2} \\alpha_{4}\\right)\\right)\\left(x-\\left(\\alpha_{1} \\alpha_{4}+\\alpha_{2} \\alpha_{3}\\right)\\right)\n$$\n\nBecause $P$ is symmetric with respect to $\\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4}$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials. We compute\n\n$$\n\\begin{aligned}\nP(x) & =x^{3}-s_{2} x^{2}+\\left(s_{3} s_{1}-4 s_{4}\\right) x+\\left(-s_{3}^{2}-s_{4} s_{1}^{2}+s_{4} s_{2}\\right) \\\\\n& =x^{3}-8 x-8 \\\\\n& =(x+2)\\left(x^{2}-2 x-4\\right)\n\\end{aligned}\n$$\n\nThe roots of $P(x)$ are -2 and $1 \\pm \\sqrt{5}$, so the answer is $\\{1 \\pm \\sqrt{5},-2\\}$.\nRemarks. It is easy to find the coefficients of $x^{2}$ and $x$ by expansion, and the constant term can be computed without the complete expansion and decomposition of $\\left(\\alpha_{1} \\alpha_{2}+\\alpha_{3} \\alpha_{4}\\right)\\left(\\alpha_{1} \\alpha_{3}+\\alpha_{2} \\alpha_{4}\\right)\\left(\\alpha_{1} \\alpha_{4}+\\right.$ $\\alpha_{2} \\alpha_{3}$ ) by noting that the only nonzero 6 th degree expressions in $s_{1}, s_{2}, s_{3}$, and $s_{4}$ are $s_{1}^{6}$ and $s_{4} s_{1}^{2}$. The general polynomial $P$ constructed here is called the cubic resolvent and arises in Galois theory.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "10", "problem_type": "Algebra", "exam": "HMMT", "problem": "The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \\ldots, r_{2007}$. A polynomial $P$ of degree 2007 has the property that $P\\left(r_{j}+\\frac{1}{r_{j}}\\right)=0$ for $j=1, \\ldots, 2007$. Determine the value of $P(1) / P(-1)$.", "solution": "| $\\mathbf{2 8 9}$. |\n| :---: |\n| For some constant $k$, we have |\n\n$$\nP(z)=k \\prod_{j=1}^{2007}\\left(z-\\left(r_{j}+\\frac{1}{r_{j}}\\right)\\right)\n$$\n\nNow writing $\\omega^{3}=1$ with $\\omega \\neq 1$, we have $\\omega^{2}+\\omega=-1$. Then\n\n$$\n\\begin{gathered}\nP(1) / P(-1)=\\frac{k \\prod_{j=1}^{2007}\\left(1-\\left(r_{j}+\\frac{1}{r_{j}}\\right)\\right)}{k \\prod_{j=1}^{2007}\\left(-1-\\left(r_{j}+\\frac{1}{r_{j}}\\right)\\right)}=\\prod_{j=1}^{2007} \\frac{r_{j}^{2}-r_{j}+1}{r_{j}^{2}+r_{j}+1}=\\prod_{j=1}^{2007} \\frac{\\left(-\\omega-r_{j}\\right)\\left(-\\omega^{2}-r_{j}\\right)}{\\left(\\omega-r_{j}\\right)\\left(\\omega^{2}-r_{j}\\right)} \\\\\n=\\frac{f(-\\omega) f\\left(-\\omega^{2}\\right)}{f(\\omega) f\\left(\\omega^{2}\\right)}=\\frac{\\left(-\\omega^{2007}+17 \\omega^{2006}+1\\right)\\left(-\\left(\\omega^{2}\\right)^{2007}+17\\left(\\omega^{2}\\right)^{2006}+1\\right)}{\\left(\\omega^{2007}+17 \\omega^{2006}+1\\right)\\left(\\left(\\omega^{2}\\right)^{2007}+17\\left(\\omega^{2}\\right)^{2006}+1\\right)}=\\frac{\\left(17 \\omega^{2}\\right)(17 \\omega)}{\\left(2+17 \\omega^{2}\\right)(2+17 \\omega)} \\\\\n=\\frac{289}{4+34\\left(\\omega+\\omega^{2}\\right)+289}=\\frac{289}{259} .\n\\end{gathered}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-alg-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}