{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": "Algebra", "exam": "HMMT", "problem": "Positive real numbers $x, y$ satisfy the equations $x^{2}+y^{2}=1$ and $x^{4}+y^{4}=\\frac{17}{18}$. Find $x y$.", "solution": "$\\sqrt[\\frac{1}{6}]{ }$ We have $2 x^{2} y^{2}=\\left(x^{2}+y^{2}\\right)^{2}-\\left(x^{4}+y^{4}\\right)=\\frac{1}{18}$, so $x y=\\frac{1}{6}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $f(n)$ be the number of times you have to hit the $\\sqrt{ }$ key on a calculator to get a number less than 2 starting from $n$. For instance, $f(2)=1, f(5)=2$. For how many $1<m<2008$ is $f(m)$ odd?", "solution": "242 This is $\\left[2^{1}, 2^{2}\\right) \\cup\\left[2^{4}, 2^{8}\\right) \\cup\\left[2^{16}, 2^{32}\\right) \\ldots$, and $2^{8}<2008<2^{16}$ so we have exactly the first two intervals.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": "Algebra", "exam": "HMMT", "problem": "Determine all real numbers $a$ such that the inequality $\\left|x^{2}+2 a x+3 a\\right| \\leq 2$ has exactly one solution in $x$.", "solution": "1,2 Let $f(x)=x^{2}+2 a x+3 a$. Note that $f(-3 / 2)=9 / 4$, so the graph of $f$ is a parabola that goes through $(-3 / 2,9 / 4)$. Then, the condition that $\\left|x^{2}+2 a x+3 a\\right| \\leq 2$ has exactly one solution means that the parabola has exactly one point in the strip $-1 \\leq y \\leq 1$, which is possible if and only if the parabola is tangent to $y=1$. That is, $x^{2}+2 a x+3 a=2$ has exactly one solution. Then, the discriminant $\\Delta=4 a^{2}-4(3 a-2)=4 a^{2}-12 a+8$ must be zero. Solving the equation yields $a=1,2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": "Algebra", "exam": "HMMT", "problem": "The function $f$ satisfies\n\n$$\nf(x)+f(2 x+y)+5 x y=f(3 x-y)+2 x^{2}+1\n$$\n\nfor all real numbers $x, y$. Determine the value of $f(10)$.", "solution": "$\\quad-49$ Setting $x=10$ and $y=5$ gives $f(10)+f(25)+250=f(25)+200+1$, from which we get $f(10)=-49$.\n\nRemark: By setting $y=\\frac{x}{2}$, we see that the function is $f(x)=-\\frac{1}{2} x^{2}+1$, and it can be checked that this function indeed satisfies the given equation.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $f(x)=x^{3}+x+1$. Suppose $g$ is a cubic polynomial such that $g(0)=-1$, and the roots of $g$ are the squares of the roots of $f$. Find $g(9)$.", "solution": "899 Let $a, b, c$ be the zeros of $f$. Then $f(x)=(x-a)(x-b)(x-c)$. Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$, so $g(x)=k\\left(x-a^{2}\\right)\\left(x-b^{2}\\right)\\left(x-c^{2}\\right)$ for some constant $k$. Since $a b c=-f(0)=-1$, we have $k=k a^{2} b^{2} c^{2}=-g(0)=1$. Thus,\n\n$$\ng\\left(x^{2}\\right)=\\left(x^{2}-a^{2}\\right)\\left(x^{2}-b^{2}\\right)\\left(x^{2}-c^{2}\\right)=(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)=-f(x) f(-x)\n$$\n\nSetting $x=3$ gives $g(9)=-f(3) f(-3)=-(31)(-29)=899$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n5. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "6", "problem_type": "Algebra", "exam": "HMMT", "problem": "A root of unity is a complex number that is a solution to $z^{n}=1$ for some positive integer $n$. Determine the number of roots of unity that are also roots of $z^{2}+a z+b=0$ for some integers $a$ and $b$.", "solution": "8 The only real roots of unity are 1 and -1 . If $\\zeta$ is a complex root of unity that is also a root of the equation $z^{2}+a z+b$, then its conjugate $\\bar{\\zeta}$ must also be a root. In this case, $|a|=|\\zeta+\\bar{\\zeta}| \\leq|\\zeta|+|\\bar{\\zeta}|=$ 2 and $b=\\zeta \\bar{\\zeta}=1$. So we only need to check the quadratics $z^{2}+2 z+1, z^{2}+z+1, z^{2}+1, z^{2}-z+1, z^{2}-2 z+1$. We find 8 roots of unity: $\\pm 1, \\pm i, \\frac{1}{2}( \\pm 1 \\pm \\sqrt{3} i)$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "7", "problem_type": "Algebra", "exam": "HMMT", "problem": "Compute $\\sum_{n=1}^{\\infty} \\sum_{k=1}^{n-1} \\frac{k}{2^{n+k}}$.", "solution": "$\\frac{4}{9}$ We change the order of summation:\n\n$$\n\\sum_{n=1}^{\\infty} \\sum_{k=1}^{n-1} \\frac{k}{2^{n+k}}=\\sum_{k=1}^{\\infty} \\frac{k}{2^{k}} \\sum_{n=k+1}^{\\infty} \\frac{1}{2^{n}}=\\sum_{k=1}^{\\infty} \\frac{k}{4^{k}}=\\frac{4}{9} .\n$$\n\n(The last two steps involve the summation of an infinite geometric series, and what is sometimes called an infinite arithmetico-geometric series. These summations are quite standard, and thus we omit the details here.)", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n7. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "8", "problem_type": "Algebra", "exam": "HMMT", "problem": "Compute $\\arctan \\left(\\tan 65^{\\circ}-2 \\tan 40^{\\circ}\\right)$. (Express your answer in degrees as an angle between $0^{\\circ}$ and $180^{\\circ}$.)", "solution": "$25^{\\circ}$ First Solution: We have\n\n$$\n\\tan 65^{\\circ}-2 \\tan 40^{\\circ}=\\cot 25^{\\circ}-2 \\cot 50^{\\circ}=\\cot 25^{\\circ}-\\frac{\\cot ^{2} 25^{\\circ}-1}{\\cot 25^{\\circ}}=\\frac{1}{\\cot 25^{\\circ}}=\\tan 25^{\\circ} .\n$$\n\nTherefore, the answer is $25^{\\circ}$.\nSecond Solution: We have\n$\\tan 65^{\\circ}-2 \\tan 40^{\\circ}=\\frac{1+\\tan 20^{\\circ}}{1-\\tan 20^{\\circ}}-\\frac{4 \\tan 20^{\\circ}}{1-\\tan ^{2} 20^{\\circ}}=\\frac{\\left(1-\\tan 20^{\\circ}\\right)^{2}}{\\left(1-\\tan 20^{\\circ}\\right)\\left(1+\\tan 20^{\\circ}\\right)}=\\tan \\left(45^{\\circ}-20^{\\circ}\\right)=\\tan 25^{\\circ}$.\nAgain, the answer is $25^{\\circ}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "9", "problem_type": "Algebra", "exam": "HMMT", "problem": "Let $S$ be the set of points $(a, b)$ with $0 \\leq a, b \\leq 1$ such that the equation\n\n$$\nx^{4}+a x^{3}-b x^{2}+a x+1=0\n$$\n\nhas at least one real root. Determine the area of the graph of $S$.", "solution": "$\\frac{1}{4}$ After dividing the equation by $x^{2}$, we can rearrange it as\n\n$$\n\\left(x+\\frac{1}{x}\\right)^{2}+a\\left(x+\\frac{1}{x}\\right)-b-2=0\n$$\n\nLet $y=x+\\frac{1}{x}$. We can check that the range of $x+\\frac{1}{x}$ as $x$ varies over the nonzero reals is $(-\\infty,-2] \\cup[2, \\infty)$. Thus, the following equation needs to have a real root:\n\n$$\ny^{2}+a y-b-2=0 .\n$$\n\nIts discriminant, $a^{2}+4(b+2)$, is always positive since $a, b \\geq 0$. Then, the maximum absolute value of the two roots is\n\n$$\n\\frac{a+\\sqrt{a^{2}+4(b+2)}}{2} .\n$$\n\nWe need this value to be at least 2 . This is equivalent to\n\n$$\n\\sqrt{a^{2}+4(b+2)} \\geq 4-a .\n$$\n\nWe can square both sides and simplify to obtain\n\n$$\n2 a \\geq 2-b\n$$\n\nThis equation defines the region inside $[0,1] \\times[0,1]$ that is occupied by $S$, from which we deduce that the desired area is $1 / 4$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "10", "problem_type": "Algebra", "exam": "HMMT", "problem": "Evaluate the infinite sum\n\n$$\n\\sum_{n=0}^{\\infty}\\binom{2 n}{n} \\frac{1}{5^{n}}\n$$", "solution": "$\\sqrt{5}$ First Solution: Note that\n\n$$\n\\begin{aligned}\n\\binom{2 n}{n} & =\\frac{(2 n)!}{n!\\cdot n!}=\\frac{(2 n)(2 n-2)(2 n-4) \\cdots(2)}{n!} \\cdot \\frac{(2 n-1)(2 n-3)(2 n-5) \\cdots(1)}{n!} \\\\\n& =2^{n} \\cdot \\frac{(-2)^{n}}{n!}\\left(-\\frac{1}{2}\\right)\\left(-\\frac{1}{2}-1\\right)\\left(-\\frac{1}{2}-2\\right) \\cdots\\left(-\\frac{1}{2}-n+1\\right) \\\\\n& =(-4)^{n}\\binom{-\\frac{1}{2}}{n}\n\\end{aligned}\n$$\n\nThen, by the binomial theorem, for any real $x$ with $|x|<\\frac{1}{4}$, we have\n\n$$\n(1-4 x)^{-1 / 2}=\\sum_{n=0}^{\\infty}\\binom{-\\frac{1}{2}}{n}(-4 x)^{n}=\\sum_{n=0}^{\\infty}\\binom{2 n}{n} x^{n} .\n$$\n\nTherefore,\n\n$$\n\\sum_{n=0}^{\\infty}\\binom{2 n}{n}\\left(\\frac{1}{5}\\right)^{n}=\\frac{1}{\\sqrt{1-\\frac{4}{5}}}=\\sqrt{5}\n$$\n\nSecond Solution: Consider the generating function\n\n$$\nf(x)=\\sum_{n=0}^{\\infty}\\binom{2 n}{n} x^{n}\n$$\n\nIt has formal integral given by\n\n$$\ng(x)=I(f(x))=\\sum_{n=0}^{\\infty} \\frac{1}{n+1}\\binom{2 n}{n} x^{n+1}=\\sum_{n=0}^{\\infty} C_{n} x^{n+1}=x \\sum_{n=0}^{\\infty} C_{n} x^{n}\n$$\n\nwhere $C_{n}=\\frac{1}{n+1}\\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\\sum_{n=0}^{\\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that\n\n$$\n1+x h(x)^{2}=1+x \\sum_{i, j \\geq 0} C_{i} C_{j} x^{i+j}=1+x \\sum_{k \\geq 0}\\left(\\sum_{i=0}^{k} C_{i} C_{k-i}\\right) x^{k}=1+\\sum_{k \\geq 0} C_{k+1} x^{k+1}=h(x)\n$$\n\nwhere we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain\n\n$$\nh(x)=\\frac{1 / x \\pm \\sqrt{1 / x^{2}-4 / x}}{2}=\\frac{1 \\pm \\sqrt{1-4 x}}{2 x}\n$$\n\nNote that we must choose the - sign in the $\\pm$, since the + would lead to a leading term of $\\frac{1}{x}$ for $h$ (by expanding $\\sqrt{1-4 x}$ into a power series). Therefore, we see that\n\n$$\nf(x)=D(g(x))=D(x h(x))=D\\left(\\frac{1-\\sqrt{1-4 x}}{2}\\right)=\\frac{1}{\\sqrt{1-4 x}}\n$$\n\nand our answer is hence $f(1 / 5)=\\sqrt{5}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-alg-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}