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HarvardMIT/md/en-282-2025-feb-algnt-solutions.md

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+
+# HMMT February 2025  
+
+February 15, 2025  
+
+# Algebra and Number Theory Round  
+
+1. Compute the sum of the positive divisors (including 1) of 9! that have units digit 1.  
+
+Proposed by: Jackson Dryg  
+
+Answer: 103  
+
+Solution: The prime factorization of 9! is \(2^{7} \cdot 3^{4} \cdot 5 \cdot 7\) . Every divisor of 9! has prime factorization \(2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d}\) , where \(0 \leq a \leq 7\) , \(0 \leq b \leq 4\) , \(0 \leq c \leq 1\) , and \(0 \leq d \leq 1\) . If the divisor has units digit 1, it cannot be divisible by 2 or 5, so \(a = c = 0\) .  
+
+Now take cases on the value of \(d\) :  
+
+- If \(d = 0\) , then the divisor is \(3^{b}\) for some \(0 \leq b \leq 4\) . The possible divisors are 1, 3, 9, 27, and 81, of which 1 and 81 work.  
+
+- If \(d = 1\) , then the divisor is \(3^{b} \cdot 7\) for some \(0 \leq b \leq 4\) . The possible divisors are then 7, 3 \(\cdot 7\) , 9 \(\cdot 7\) , 27 \(\cdot 7\) , and 81 \(\cdot 7\) . Of these, only \(3 \cdot 7 = 21\) works.  
+
+The answer is \(1 + 21 + 81 = \boxed{103}\) .  
+
+2. Mark writes the expression \(\sqrt{abcd}\) on the board, where \(abcd\) is a four-digit number and \(a \neq 0\) . Derek, a toddler, decides to move the \(a\) , changing Mark's expression to \(a\sqrt{bc}\) . Surprisingly, these two expressions are equal. Compute the only possible four-digit number \(abcd\) .  
+
+Proposed by: Pitchayut Saengrungkongka  
+
+Answer: 3375  
+
+Solution: Let \(x = bcd\) . Then, we rewrite the given condition \(\sqrt{abcd} = a\sqrt{bc}\) as  
+
+\[1000a + x = a^{2}x,\]  
+
+which simplifies as  
+
+\[(a^{2} - 1)x = 1000a.\]  
+
+In particular, \(a^{2} - 1\) divides \(1000a\) . Since \(\gcd (a^{2} - 1, a) = 1\) , it follows that \(a^{2} - 1 \mid 1000\) . The only \(a \in \{1, 2, \ldots , 9\}\) that satisfies this is \(a = 3\) . Then \(8x = 3000\) , so \(x = 375\) . Thus \(abcd = \boxed{3375}\) .  
+
+3. Given that \(x\) , \(y\) , and \(z\) are positive real numbers such that  
+
+\[x^{\log_{2}(yz)} = 2^{8} \cdot 3^{4}, \quad y^{\log_{2}(zx)} = 2^{9} \cdot 3^{6}, \quad \text{and} \quad z^{\log_{2}(xy)} = 2^{5} \cdot 3^{10},\]  
+
+compute the smallest possible value of \(xyz\) .  
+
+Proposed by: Derek Liu  
+
+Answer: 1 576  
+
+Solution: Let \(k = \log_{2}3\) for brevity. Taking the base- 2 log of each equation gives  
+
+\[(\log_{2}x)(\log_{2}y + \log_{2}z) = 8 + 4k,\] \[(\log_{2}y)(\log_{2}z + \log_{2}x) = 9 + 6k,\] \[(\log_{2}z)(\log_{2}x + \log_{2}y) = 5 + 10k.\]
+
+
+
+Adding the first two equations and subtracting the third yields \(2 \log_{2} x \log_{2} y = 12\) , so \(\log_{2} x \log_{2} y = 6\) . Similarly, we get  
+
+\[\log_{2} x \log_{2} y = 6,\] \[\log_{2} y \log_{2} z = 3 + 6 k,\] \[\log_{2} z \log_{2} x = 2 + 4 k.\]  
+
+Multiplying the first two equations and dividing by the third yields \((\log_{2} y)^{2} = 9\) , so \(\log_{2} y = \pm 3\) . Then, the first and last equations tell us \(\log_{2} x = \pm 2\) and \(\log_{2} z = \pm (1 + 2 k)\) , with all signs matching. Thus  
+
+\[\log_{2} x + \log_{2} y + \log_{2} z = \pm (3 + 2 + (1 + 2 k)) = \pm (6 + 2 k),\]  
+
+so  
+
+\[x y z = 2^{\pm (6 + 2 k)} = 2^{6} \cdot 3^{2} \quad \text{or} \quad 2^{-6} \cdot 3^{-2}.\]  
+
+Clearly, the smallest solution is \(2^{- 6} \cdot 3^{- 2} = \left[\frac{1}{576}\right]\) .  
+
+4. Let \(\lfloor z\rfloor\) denote the greatest integer less than or equal to \(z\) . Compute  
+
+\[\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor .\]  
+
+Proposed by: Linus Yifeng Tang  
+
+Answer: \(- 984\)  
+
+Solution: The key idea is to pair up the terms \(\left\lfloor \frac{2025}{x} \right\rfloor\) and \(\left\lfloor \frac{2025}{x} \right\rfloor\) . There are 1000 such pairs and one lone term, \(\left\lfloor \frac{2025}{1000.5} \right\rfloor = 2\) . Thus,  
+
+\[\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor = 2 + \sum_{x \in \{0.5,1.5, \ldots , 999.5\}} \left(\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{-x} \right\rfloor \right).\]  
+
+We note that  
+
+  
+
+Therefore,  
+
+  
+
+As \(x\) ranges in the set \(\{0.5, 1.5, 2.5, \ldots , 999.5\}\) , \(2 x\) ranges in the set \(\{1, 3, 5, \ldots , 1999\}\) . This set includes all 15 odd divisors of 4050 except for 2025. Thus, there are 14 values of \(x\) for which \(\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{- x} \right\rfloor\) evaluates to 0, and the remaining \(1000 - 14 = 986\) values of \(x\) make it evaluate to \(- 1\) . Therefore,  
+
+\[\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor = 2 + \sum_{x \in \{0.5,1.5, \ldots , 999.5\}} \left(\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{-x} \right\rfloor \right) = 2 + 986 \cdot (-1) = \boxed{- 984}.\]
+
+
+
+5. Let \(\mathcal{S}\) be the set of all nonconstant monic polynomials \(P\) with integer coefficients satisfying \(P\left(\sqrt{3} + \sqrt{2}\right) = P\left(\sqrt{3} - \sqrt{2}\right)\) . If \(Q\) is an element of \(\mathcal{S}\) with minimal degree, compute the only possible value of \(Q(10) - Q(0)\) .  
+
+Proposed by: David Dong  
+
+Answer: 890  
+
+Solution: First, note that the polynomial \(x^{4} - 10x^{2} + 1\) has both \(\sqrt{3} +\sqrt{2}\) and \(\sqrt{3} - \sqrt{2}\) as roots. It suffices to check whether a polynomial of degree at most 3 belongs in \(\mathcal{S}\) . Suppose \(f(x) = a x^{3} + b x^{2} + c x + d\in \mathcal{S}\) . We compute  
+
+\[(\sqrt{3} +\sqrt{2})^{3} - (\sqrt{3} -\sqrt{2})^{3} = 22\sqrt{2\] \[(\sqrt{3} +\sqrt{2})^{2} - (\sqrt{3} -\sqrt{2})^{2} = 4\sqrt{6\] \[(\sqrt{3} +\sqrt{2})^{1} - (\sqrt{3} -\sqrt{2})^{1} = 2\sqrt{2,\]  
+
+so we get that  
+
+\[f(\sqrt{3} +\sqrt{2}) - f(\sqrt{3} -\sqrt{2}) = (22\sqrt{2})a + (4\sqrt{6})b + (2\sqrt{2})c.\]  
+
+By resolving linear dependencies, it's clear that \(b = 0\) and \(c = - 11a\) . It follows that if \(f\) is not the zero polynomial, it must be cubic. It is then clear that \(f(x) = x^{3} - 11x + d\) has minimal degree in \(\mathcal{S}\) , and thus \(Q(10) - Q(0) = f(10) - f(0) = \boxed{890}\) .  
+
+6. Let \(r\) be the remainder when \(2017^{2025!} - 1\) is divided by 2025!. Compute \(\frac{r}{2025!}\) . (Note that 2017 is prime.)  
+
+Proposed by: Srinivas Arun  
+
+Answer: \(\frac{1311}{2017}\)  
+
+Solution: Let \(N = 2017^{2025!}\) . Let \(p\) be a prime dividing 2025! other than 2017. Let \(p^{k}\) be the largest power of \(p\) dividing 2025!. Clearly, \(\phi (p^{k}) = (p - 1)p^{k - 1}\) divides 2025! and \(\gcd (2017, p^{k}) = 1\) , so by Euler's Totient Theorem,  
+
+\[N\equiv 1\pmod {p^{k}}.\]  
+
+Repeating for all such primes \(p\) , we obtain  
+
+\[N\equiv 1\pmod {2025! / 2017}.\]  
+
+Therefore, \(\frac{2025!}{2017} |N - 1\) , so \(r = \frac{2025!}{2017} s\) for some \(0\leq s< 2017\) . Also, since \(N\equiv 0\) (mod 2017), we have \(r = \frac{2025!}{2017} s\equiv - 1\) (mod 2017).  
+
+By Wilson's,  
+
+\[\frac{2025!}{2017} = 2016!(2018)(2019)\ldots (2025)\equiv -8!\equiv 20\pmod {2017}.\]  
+
+Therefore, \(s\) is negative the inverse of 20 (mod 2017), which is 1311. Our answer is  
+
+\[\frac{r}{2025!} = \frac{(2025! / 2017)(1311)}{2025!} = \frac{1311}{2017}.\]  
+
+7. There exists a unique triple \((a,b,c)\) of positive real numbers that satisfies the equations  
+
+\[2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1)\quad \mathrm{and}\quad ab + bc + ca = 1.\]  
+
+Compute \(a + b + c\) .
+
+
+
+Proposed by: David Wei  
+
+Answer: \(\frac{9\sqrt{23}}{23} = \frac{9}{\sqrt{23}}\)  
+
+Solution 1: The crux of this problem is to apply the trigonometric substitutions \(a = \cot \alpha\) , \(b = \cot \beta\) , and \(c = \cot \gamma\) , with \(0 < \alpha , \beta , \gamma < \pi / 2\) . Then, the given equations translate to  
+
+\[\frac{2}{\sin^{2}\alpha} = \frac{3}{\sin^{2}\beta} = \frac{4}{\sin^{2}\gamma}\quad \mathrm{and}\quad \cot \alpha \cot \beta +\cot \beta \cot \gamma +\cot \gamma \cot \alpha = 1.\]  
+
+From the second equation, we get  
+
+\[\cot \gamma = \frac{1 - \cot\alpha\cot\beta}{\cot\alpha + \cot\beta} = -\cot (\alpha +\beta).\]  
+
+Since \(\alpha\) , \(\beta\) , and \(\gamma\) all between 0 and \(\pi / 2\) , we discover that  
+
+\[\alpha +\beta +\gamma = \pi .\]  
+
+Let \(\triangle ABC\) be the (acute) triangle with side lengths \(BC = \sqrt{2}\) , \(CA = \sqrt{3}\) , and \(AB = \sqrt{4}\) . By Law of Sines, setting \(\alpha = \angle A\) , \(\beta = \angle B\) , and \(\gamma = \angle C\) will satisfy both equations. Thus, Law of Cosines gives  
+
+\[\cos \alpha = \frac{3 + 4 - 2}{2\cdot\sqrt{3}\cdot\sqrt{4}} = \frac{5}{\sqrt{48}}\Rightarrow a = \cot \alpha = \frac{5}{\sqrt{23}}\]  
+
+Similar calculations give \(b = \frac{3}{\sqrt{23}}\) and \(c = \frac{1}{\sqrt{23}}\) , so the answer is \(a + b + c = \left[\frac{9}{\sqrt{23}}\right]\) .  
+
+Solution 2: Let \(2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1) = x\) . Then, since \(ab + bc + ca = 1\) , we have the following system of equations:  
+
+\[(a + b)(c + a) = a^{2} + ab + bc + ca = a^{2} + 1 = x / 2\] \[(b + c)(a + b) = b^{2} + ab + bc + ca = b^{2} + 1 = x / 3\] \[(c + a)(b + c) = c^{2} + ab + bc + ca = c^{2} + 1 = x / 4.\]  
+
+Taking advantage of symmetry, we discover that  
+
+\[a + b = \sqrt{\frac{2x}{3}},\quad b + c = \sqrt{\frac{x}{6}},\quad \mathrm{and}\quad c + a = \sqrt{\frac{3x}{8}}.\]  
+
+To solve for \(x\) , notice that  
+
+\[2 = 2(ab + bc + ca)\] \[\quad = (a + b)^{2} + (b + c)^{2} + (c + a)^{2} - 2(a^{2} + b^{2} + c^{2})\] \[\quad = \frac{2x}{3} +\frac{x}{6} +\frac{3x}{8} -2\left(\frac{x}{2} -1 + \frac{x}{3} -1 + \frac{x}{4} -1\right)\] \[\quad = -\frac{23x}{24} +6,\]  
+
+so \(x = \frac{96}{23}\) . Therefore,  
+
+\[a + b + c = \frac{1}{2}\left(\sqrt{\frac{2x}{3}} +\sqrt{\frac{x}{6}} +\sqrt{\frac{3x}{8}}\right)\] \[\qquad = \frac{1}{2}\left(\frac{8 + 4 + 6}{\sqrt{23}}\right) = \left[\frac{9}{\sqrt{23}}\right].\]
+
+
+
+8. Define \(\operatorname {sgn}(x)\) to be 1 when \(x\) is positive, \(-1\) when \(x\) is negative, and 0 when \(x\) is 0. Compute  
+
+\[\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}}.\]  
+
+(The arguments to sin are in radians.)  
+
+Proposed by: Karthik Venkata Vedula  
+
+Answer: \(\boxed {1 - \frac{2}{\pi}}\)  
+
+Solution: Note that each of following is equivalent to the next.  
+
+\(\cdot \operatorname {sgn}(\sin (2^{n})) = +1\)  
+
+\(\cdot 0< 2^{n}\bmod 2\pi < \pi\)  
+
+\(\cdot 0< \frac{2^{n}}{\pi}\bmod 2< 1\)  
+
+The \(n\) th digit after the decimal point in the binary representation of \(\frac{1}{\pi}\) is 0.  
+
+Similarly, \(\operatorname {sgn}(\sin (2^{n})) = - 1\) if and only if the \(n\) - th digit after the decimal point in the binary representation of \(\frac{1}{\pi}\) is 1. In particular, if \(a_{n}\) is the \(n\) - th digit, then \(\operatorname {sgn}(\sin (2^{n})) = 1 - 2a_{n}\) . Thus, the desired sum is  
+
+\[\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}} = \sum_{n = 1}^{\infty}\frac{1 - 2a_{n}}{2^{n}} = \left(\sum_{n = 1}^{\infty}\frac{1}{2^{n}}\right) - 2\left(\sum_{n = 1}^{\infty}\frac{a_{n}}{2^{n}}\right) = \left[\frac{1 - \frac{2}{\pi}}{2}\right].\]  
+
+9. Let \(f\) be the unique polynomial of degree at most 2026 such that for all \(n\in \{1,2,3,\ldots ,2027\}\)  
+
+  
+
+Suppose that \(\frac{a}{b}\) is the coefficient of \(x^{2025}\) in \(f\) , where \(a\) and \(b\) are integers such that \(\gcd (a,b) = 1\) . Compute the unique integer \(r\) between 0 and 2026 (inclusive) such that \(a - rb\) is divisible by 2027. (Note that 2027 is prime.)  
+
+Proposed by: Pitchayut Saengrungkonga  
+
+Answer: \(\boxed {1037}\)  
+
+Solution 1: Let \(p = 2027\) . We work in \(\mathbb{F}_{p}\) for the entire solution. Recall the well- known fact that  
+
+  
+
+assuming \(0^{0} = 1\) . In particular, for any polynomial \(g(x) = b_{0} + b_{1}x + \dots +b_{n}x^{n}\) , we have  
+
+\[-\sum_{x\in \mathbb{F}_{p}}g(x) = b_{p - 1} + b_{2(p - 1)} + \dots +b_{\lfloor n / (p - 1)\rfloor (p - 1)}.\]  
+
+We apply this fact on \(g(x) = xf(x)\) . As \(\deg x f(x)\leq p\) , the right hand side is simply the coefficient of \(x^{2025}\) , which is what we want. Hence, the answer is  
+
+\[-\sum_{x\in \mathbb{F}_{p}}x f(x) = -(1^{2} + 2^{2} + \dots +45^{2}) = -\frac{45\cdot 46\cdot 91}{6}\equiv \boxed {1037}\pmod {2027}.\]
+
+
+
+Solution 2: Again, let \(p = 2027\) and work in \(\mathbb{F}_{p}\) . By the Lagrange Interpolation formula, we get that  
+
+\[f(x) = \sum_{i\in \mathbb{F}_{p}}f(i)\prod_{j\neq i}{\frac{x - j}{i - j}}.\]  
+
+We now simplify the polynomial in the product sign on the right- hand side. First, recall the identity  
+
+\[\prod_{j\in \mathbb{F}_{p}}(x - j) = x^{p} - x = (x - i)^{p} - (x - i).\]  
+
+The denominator \(\prod_{j\neq i}(i - j)\) becomes \((p - 1)! = - 1\) by Wilson's. Thus, we get that  
+
+\[\prod_{j\neq i}{\frac{x - j}{i - j}} = -{\frac{(x - i)^{p} - (x - i)}{x - i}} = -(x - i)^{p - 1} + 1.\]  
+
+The coefficient of \(x^{p - 2}\) in the above expression is \(- i\) . Therefore, the first equation gives that the coefficient of \(x^{p - 2}\) in \(f(x)\) is  
+
+\[\sum_{i\in \mathbb{F}_{p}} - if(i) = -(1^{2} + 2^{2} + \dots +45^{2}) = -\frac{45\cdot 46\cdot 91}{6}\equiv \boxed {1037}\pmod {2027}.\]  
+
+10. Let \(a\) , \(b\) , and \(c\) be pairwise distinct complex numbers such that  
+
+\[a^{2} = b + 6,\quad b^{2} = c + 6,\quad \mathrm{and}\quad c^{2} = a + 6.\]  
+
+Compute the two possible values of \(a + b + c\)  
+
+Proposed by: Vasawat Rawangwong  
+
+Answer: \(\boxed {\frac{- 1 + \sqrt{17}}{2}},\frac{- 1 - \sqrt{17}}{2}\)  
+
+Solution 1: Notice that any of \(a\) , \(b\) , or \(c\) being 3 or \(- 2\) implies \(a = b = c\) , which is invalid. Thus,  
+
+\[(a^{2} - 9)(b^{2} - 9)(c^{2} - 9) = (b - 3)(c - 3)(a - 3)\implies (a + 3)(b + 3)(c + 3) = 1,\] \[(a^{2} - 4)(b^{2} - 4)(c^{2} - 4) = (b + 2)(c + 2)(a + 2)\implies (a - 2)(b - 2)(c - 2) = 1.\]  
+
+Therefore, 2 and \(- 3\) are roots of the polynomial \((x - a)(x - b)(x - c) + 1\) , and so there exists some \(t\) such that  
+
+\[(x - t)(x - 2)(x + 3) = (x - a)(x - b)(x - c) + 1.\]  
+
+Comparing coefficients gives \(a + b + c = t - 1\) and \(a b + b c + c a = - (t + 6)\) . We can then solve for \(t\) by noting \(a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = a + b + c + 18\) , so  
+
+\[a b + b c + c a = \frac{1}{2} ((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})) = \frac{1}{2} ((a + b + c)^{2} - (a + b + c + 18)).\]  
+
+Hence,  
+
+\[- (t + 6) = \frac{1}{2} ((t - 1)^{2} - (t + 17))\Longrightarrow t^{2} - t - 4 = 0\Longrightarrow t = \frac{1\pm\sqrt{17}}{2}.\]  
+
+Therefore \(a + b + c = \boxed {\frac{- 1\pm\sqrt{17}}{2}}\) are the two possible values of \(a + b + c\)  
+
+Solution 2: Let \(s = a + b + c\) . Subtracting two adjacent equations gives \(a^{2} - b^{2} = b - c\) , or \((a - b)(a + b) = (b - c)\) . Multiplying this and its cyclic variants gives  
+
+\[(a + b)(b + c)(c + a) = 1.\]
+
+
+
+Now, we recall the identity  
+
+\[(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3(a + b)(b + c)(c + a)\] \[\qquad \Rightarrow \qquad s^{3} = a^{3} + b^{3} + c^{3} + 3.\]  
+
+To simplify \(a^{3} + b^{3} + c^{3}\) , we add \(a\) times the first equation, \(b\) times the second, and \(c\) times the third to obtain  
+
+\[a^{3} + b^{3} + c^{3} = a(b + 6) + b(c + 6) + c(a + 6)\] \[\qquad = (a b + b c + c a) + 6 s\] \[\qquad = \frac{1}{2}\Big((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})\Big) + 6 s\] \[\qquad = \frac{1}{2} s^{2} - \frac{1}{2}\Big((b + 6) + (c + 6) + (a + 6)\Big) + 6 s\] \[\qquad = \frac{1}{2} s^{2} + \frac{11}{2} s - 9.\]  
+
+Therefore,  
+
+\[s^{3} = \frac{1}{2} s^{2} + \frac{11}{2} s - 6 \Rightarrow \left(s - \frac{3}{2}\right)\left(s^{2} + s - 4\right) = 0.\]  
+
+At this point, the only reasonable guess is that \(s = \frac{3}{2}\) is an extra solution, and the remaining two roots \(s = \frac{- 1\pm\sqrt{17}}{2}\) are the possible answers. We now justify this guess. Assume for sake of contradiction that \(s = \frac{3}{2}\) . Then,  
+
+\[a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = \frac{39}{2\] \[ab + bc + ca = \frac{1}{2}\left(\frac{9}{4} -\frac{39}{2}\right) = -\frac{69}{8}.\]  
+
+Then, observe  
+
+\[a b c = (a + b + c)(a b + b c + c a) - (a + b)(b + c)(c + a)\] \[= -\frac{207}{16} -1 = -\frac{223}{16}.\]  
+
+On the other hand,  
+
+\[(a + 6)(b + 6)(c + 6) = 216 + 36(a + b + c) + 6(ab + bc + ca) + abc\] \[\qquad = 216 + 36\cdot \frac{3}{2} -6\cdot \frac{69}{8} -\frac{223}{16},\]  
+
+which is a rational number of denominator 16. But \((a + 6)(b + 6)(c + 6) = b^{2}c^{2}a^{2} = \left(-\frac{223}{16}\right)^{2}\) has denominator \(16^{2} = 256\) , a contradiction. Thus \(s = \frac{3}{2}\) is impossible. (It arises from \(a = b = c = \frac{1}{2}\) which satisfies \((a + b)(b + c)(c + a) = 1\) but not the given conditions.)  
+
+Solution 3: Subtracting any two adjacent equations gives \(a^{2} - b^{2} = b - c\) , which is equivalent to both \((a - b)(a + b) = (b - c)\) and \((a - b)(a + b + 1) = (a - c)\) . Multiplying each of these with its respective cyclic variants and canceling the \((a - b)(b - c)(c - a)\) factor (which is given to be nonzero), we get  
+
+\[(a + b)(b + c)(c + a) = 1\quad \mathrm{and}\quad (a + b + 1)(b + c + 1)(c + a + 1) = -1.\]  
+
+Expanding the latter equation and using the given equations gives the following result.  
+
+\[(a + b)(b + c)(c + a) + (a^{2} + b^{2} + c^{2}) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1\] \[1 + (b + 6 + c + 6 + a + 6) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1\] \[3(a + b + c) + 3(ab + bc + ca) = -21\]  
+
+\[a + b + c + ab + bc + ca = -7.\]
+
+
+
+Let \(s = a + b + c\) . We can then solve for \(s\) by considering the following:  
+
+\[s^{2} = (a^{2} + b^{2} + c^{2}) + 2(ab + bc + ca)\] \[\quad = (b + 6 + c + 6 + a + 6) + 2(-7 - a - b - c)\] \[\quad = -s + 4,\]  
+
+so \(s = \frac{- 1\pm\sqrt{17}}{2}\) .  
+
+Solution 4: Let \(s = a + b + c\) and consider the polynomial  
+
+\[x + (x^{2} - 6) + ((x^{2} - 6)^{2} - 6) - s = x^{4} - 11x^{2} + x + 24 - s.\]  
+
+This polynomial has roots \(a\) , \(b\) , and \(c\) . By Vieta's, the sum of all four roots is 0, so its fourth root must be \(- s\) . Using Vieta's again, we have \(ab + bc + ca - sa - sb - sc = - 11\) . We can now solve for \(s\) .  
+
+\[a b + b c + c a - (a + b + c)^{2} = -11\] \[a^{2} + b^{2} + c^{2} + a b + b c + c a = 11\] \[\frac{1}{2} ((a + b + c)^{2} + (a^{2} + b^{2} + c^{2})) = 11\] \[(a + b + c)^{2} + (b + 6 + c + 6 + a + 6) = 22\] \[s^{2} + s - 4 = 0\Longrightarrow s = \frac{-1\pm\sqrt{17}}{2}.\]  
+
+Remark. Another way to finish using this approach is to substitute \(- s\) directly into \(x^{4} - 11x^{2} + x + 24 - s = 0\) to get \((s - 3)(s + 2)(x^{2} + x - 4) = 0\) , then discard the solutions \(s = 3\) and \(s = - 2\) , which arise from the invalid values \(a = b = c = 3\) and \(a = b = c = - 2\) . (In the invalid cases, \(s \neq a + b + c\) because \(a = b = c\) is only a single root to the polynomial.)
+
+

HarvardMIT/md/en-282-2025-feb-comb-solutions.md

@@ -0,0 +1,196 @@
+
+# HMMT February 2025  
+
+February 15, 2025  
+
+# Combinatorics Round  
+
+1. Compute the number of ways to arrange the numbers 1, 2, 3, 4, 5, 6, and 7 around a circle such that the product of every pair of adjacent numbers on the circle is at most 20. (Rotations and reflections count as different arrangements.)  
+
+Proposed by: Karthik Venkata Vedula  
+
+Answer: 56  
+
+Solution: Fix the position of the number 7. Note that the only numbers that can be next to 7 are 1 and 2, so they must occupy the two slots adjacent to 7.  
+
+Now, the number 6 can only be adjacent to 1, 2, and 3. As 6 can no longer be adjacent to both 1 and 2, we conclude that 6 must be adjacent to 3 as well as either 1 or 2.  
+
+Finally, the numbers 4 and 5 may be placed arbitrarily in the remaining two slots.  
+
+There are 7 choices for the number 7's position, 2 ways to place 1 and 2 next to it, 2 ways to place 6 and 3, and 2 ways to place 4 and 5, giving us a total count of \(7\cdot 2\cdot 2\cdot 2 = \boxed{56}\) .  
+
+2. Kelvin the frog is on the bottom-left lily pad of a \(3\times 3\) grid of lily pads, and his home is at the topright lily pad. He can only jump between two lily pads which are horizontally or vertically adjacent. Compute the number of ways to remove 4 of the lily pads so that the bottom-left and top-right lily pads both remain, but Kelvin cannot get home.  
+
+Proposed by: Srinivas Arun  
+
+Answer: 29  
+
+Solution: We instead count the arrangements for which Kelvin can get home. Note that at minimum, Kelvin must use 5 lilypads to get home, leaving 4 lily pads that are not on the path. This means that if we were to remove 4 lilypads and Kelvin can still get home, the non- removed lilypads form a shortest path from the bottom- left to the top- right. As there are \(\binom{4}{2} = 6\) of these shortest paths, our answer is \(\binom{7}{4} - 6 = \boxed{29}\) .  
+
+3. Ben has 16 balls labeled 1, 2, 3, ..., 16, as well as 4 indistinguishable boxes. Two balls are neighbors if their labels differ by 1. Compute the number of ways for him to put 4 balls in each box such that each ball is in the same box as at least one of its neighbors. (The order in which the balls are placed does not matter.)  
+
+Proposed by: Benjamin Shimabukuro  
+
+Answer: 105  
+
+Solution: Each box must either contain a single group of four consecutive balls (e.g. 5, 6, 7, 8) or two groups of two consecutive balls (e.g. 5, 6, 9, 10). Since all groups have even lengths, this means that 1 and 2 are in the same group, 3 and 4 are in the same group, and so on. We can think of each of these 8 pairs of balls as an individual unit, so the answer is equal to the number of ways to put 8 objects in 4 indistinguishable boxes, where each box has 2 objects without any additional restrictions. The number of ways to do this is \(\frac{8!}{2^{4}\cdot 4!} = \boxed{105}\) .  
+
+4. Sophie is at \((0,0)\) on a coordinate grid and would like to get to \((3,3)\) . If Sophie is at \((x,y)\) , in a single step she can move to one of \((x + 1,y)\) , \((x,y + 1)\) , \((x - 1,y + 1)\) , or \((x + 1,y - 1)\) . She cannot revisit any points along her path, and neither her \(x\) - coordinate nor her \(y\) - coordinate can ever be less than 0 or greater than 3. Compute the number of ways for Sophie to reach \((3,3)\) .
+
+
+
+Proposed by: Derek Liu  
+
+Answer: 2304  
+
+Solution: Let a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Kelvin's sum of coordinates by 1, while all other moves do not change the sum, so Kelvin must make 6 of them, one to increase this sum from \(i\) to \(i + 1\) for each \(i \in [0, 5]\) .  
+
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)
+  
+
+We claim there exists a unique path corresponding to each set of 6 lateral moves chosen in this way. Indeed, the diagonal moves allow Kelvin to get from any point on \(x + y = i\) to any second point on \(x + y = i\) in exactly one way.  
+
+Observe that when \(i \leq 2\) , the number of lateral moves that increase the sum of coordinates from \(i\) to \(i + 1\) is \(2i\) , as is the number that increase it from \(5 - i\) to \(6 - i\) . Thus the answer is \(2 \cdot 4 \cdot 6 \cdot 6 \cdot 4 \cdot 2 = \lfloor 2304 \rfloor\) .  
+
+5. In an \(11 \times 11\) grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maximum number of doors he can go through in such a path.  
+
+Proposed by: Derek Liu  
+
+Answer: 200  
+
+Solution:  
+
+![](data:image/jpeg;base64,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)
+  
+
+This is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has \(4 \cdot 9 = 36\) cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then 20 of these 36 cells are white. An Eulerian circuit uses an even number of doors from each cell, so at least one door from each of these cells goes unused. No door connects two white cells, so at least 20 doors are unused, leaving at most \(2 \cdot 10 \cdot 11 - 20 = \lfloor 200 \rfloor\) doors crossed.
+
+
+
+To see this is achievable, we first delete the bottom- right cell and its 2 doors, as well as the top- left cell and its 2 doors. This leaves 8 odd- degree cells along each side of the grid; we can delete 4 doors along each to cover all remaining odd- degree cells, for 20 total doors deleted. The resulting graph is connected and has no odd- degree cells, so it must have an Eulerian circuit. This circuit is our desired path.  
+
+6. Compute the number of ways to pick two rectangles in a \(5 \times 5\) grid of squares such that the edges of the rectangles lie on the lines of the grid and the rectangles do not overlap at their interiors, edges, or vertices. The order in which the rectangles are chosen does not matter.  
+
+Proposed by: Jacob Paltrowitz  
+
+Answer: 6300  
+
+Solution: A rectangle can be specified by two intervals, one specifying its horizontal extent ( \(x\) - coordinates of left and right sides) and one specifying its vertical extent ( \(y\) - coordinates of bottom and top sides). For the rectangles to not overlap, we need either the horizontal intervals or the vertical intervals to be disjoint (possibly both).  
+
+First, we will count the number of ways for the horizontal intervals to be disjoint. Let these intervals be \([a,b]\) and \([c,d]\) . As the order of the rectangles does not matter, we can assume without loss of generality that \(a< c\) , so \(a< b< c< d\) . Then there are \(\binom{6}{4}\) choices for \(a\) , \(b\) , \(c\) , and \(d\) . There are no restrictions on the vertical intervals, so the number of ways to choose them is \(\binom{6}{2}^{2}\) . Thus, the total number of pairs of rectangles for which the horizontal intervals are disjoint is \(\binom{6}{4}\binom{6}{2}^{2}\) .  
+
+By symmetry, the total number of pairs of rectangles for which the vertical intervals are disjoint is the same. It remains to count the number of ways for both the horizontal and vertical intervals to be disjoint. Again, let the horizontal intervals be \([a,b]\) and \([c,d]\) , and let the vertical intervals be \([e,f]\) and \([g,h]\) . We can assume \(a< b< c< d\) without loss of generality, so there are \(\binom{6}{4}\) ways to choose the horizontal intervals. However, the cases \(e< f< g< h\) and \(g< h< e< f\) are now distinct, so there are \(2\binom{6}{4}\) ways to choose the vertical intervals. Therefore, there are \(2\binom{6}{4}^{2}\) pairs of rectangles for which both the horizontal and vertical intervals are disjoint.  
+
+Through inclusion- exclusion, we get the final answer of  
+
+\[2\cdot \binom{6}{4}\binom{6}{2}^{2} - 2\cdot \binom{6}{4}^{2} = \boxed{6300}.\]  
+
+7. Compute the number of ways to arrange 3 copies of each of the 26 lowercase letters of the English alphabet such that for any two distinct letters \(x_{1}\) and \(x_{2}\) , the number of \(x_{2}\) 's between the first and second occurrences of \(x_{1}\) equals the number of \(x_{2}\) 's between the second and third occurrences of \(x_{1}\) .  
+
+Proposed by: Derek Liu  
+
+Answer: 22526!  
+
+Solution: First, we prove such a string can be divided into blocks where each block consists of the same substring written three times. We prove the following lemma.  
+
+Lemma 1. For any letter \(x_{1}\) , the strings between the first and second occurrences of \(x_{1}\) and between the second and third occurrences of \(x_{1}\) are the same.  
+
+Proof. Call these two strings \(s_{1}\) and \(s_{2}\) . We know they must be permutations of each other, and if a letter appears twice in \(s_{1}\) , it would also have to appear twice in \(s_{2}\) , for four appearances in all, which is impossible. Thus, no letter appears twice in \(s_{1}\) (and likewise in \(s_{2}\) ).  
+
+Assume for sake of contradiction that for some letters \(x_{2}\) and \(x_{3}\) in these strings, \(x_{2}\) appears before \(x_{3}\) in \(s_{1}\) , but after \(x_{3}\) in \(s_{2}\) . Then, between these two appearances of \(x_{2}\) (which are consecutive, because
+
+
+
+no other \(x_{2}\) 's appear in either \(s_{1}\) or \(s_{2}\) ), there must be two \(x_{3}\) 's. This implies there must also be two \(x_{3}\) 's between the other pair of consecutive \(x_{2}\) 's, contradiction.  
+
+We conclude any two letters in \(s_{1}\) and \(s_{2}\) appear in the same order in both strings, so \(s_{1} = s_{2}\) . \(\square\)  
+
+Let the first letter of our 78- character string be \(x_{1}\) , and suppose the next appearance of \(x_{1}\) is the \((k + 1)\) - th letter. Let \(x_{2}\) , \(x_{3}\) , ..., \(x_{k}\) be the letters in between. Then, \(x_{2}x_{3}\ldots x_{k}\) is the string between the first and second \(x_{1}\) 's, so it must also be the string between the second and third \(x_{1}\) 's. Thus, after the second \(x_{1}\) , we must have \(x_{2}x_{3}\ldots x_{k}x_{1}\) .  
+
+Now, between the first and second \(x_{k}\) 's is the string \(x_{1}x_{2}x_{3}\ldots x_{k - 1}\) , so this must also be between the second and third \(x_{k}\) 's. Thus, after the second \(x_{k}\) , we must have \(x_{1}x_{2}x_{3}\ldots x_{k}\) .  
+
+Thus, the first \(3k\) letters are simply \(x_{1}x_{2}x_{3}\ldots x_{k}\) repeated three times. We can remove this block of \(3k\) letters and repeat to show that the whole string can be divided into such blocks.  
+
+To count the number of such strings, we first note that there are \(2^{25}\) ways to divide the strings into such blocks. This is because there are 25 possible places which can divide two blocks (after the 3rd, 6th, 9th, etc. letters), and we can choose any subset of these to divide blocks.  
+
+The string is then uniquely determined by the first one- third of each block, which must consist of every letter exactly once (as the whole string is just three copies of these thirds spliced together). These thirds can consist of any ordering of the 26 letters, so there are 26! strings with any given partition of blocks.  
+
+We conclude the total number of strings is \(\boxed{2^{25}\cdot 26!}\) .  
+
+Remark. Another way to see the characterization of strings is as follows. First, solve the problem for the case of two letters \(a\) and \(b\) : it is straightforward to show that the only strings are \(ababab\) , \(bababa\) , \(aaabbb\) , \(bbbaaa\) . Going back to the original problem, the six appearances of any pair of letters must be in the order of one of the four strings above; say two letters are interleaved if they alternate appearances. Construct a graph on letters with an edge between two letters if and only if they are interleaved. Note that if \(a\) and \(b\) are interleaved, and \(b\) and \(c\) are interleaved, then \(a\) and \(c\) must be interleaved. Thus, the graph is transitive, implying that it is a union of disjoint cliques. We can now finish as before.  
+
+8. Albert writes 2025 numbers \(a_{1}\) , ..., \(a_{2025}\) in a circle on a blackboard. Initially, each of the numbers is uniformly and independently sampled at random from the interval \([0,1]\) . Then, each second, he simultaneously replaces \(a_{i}\) with \(\max (a_{i - 1}, a_{i}, a_{i + 1})\) for all \(i = 1, 2, \ldots , 2025\) (where \(a_{0} = a_{2025}\) and \(a_{2026} = a_{1}\) ). Compute the expected value of the number of distinct values remaining after 100 seconds.  
+
+Proposed by: Isaac Zhu  
+
+Answer: \(\boxed{2025}\)  
+
+Solution: We can assume that the initial numbers are all distinct, since this occurs with probability 1. For clarity, we denote the value of \(a_{i}\) after \(t\) seconds as \(a_{i,t}\) . The index \(i\) is taken mod 2025.  
+
+In general, after \(k< 1012\) seconds, we claim the expected number of distinct values remaining is \(\frac{2025}{k + 1}\) . To show this, we first prove that for any remaining value, its appearances are consecutive. Indeed, note that for all \(i\) and \(k\) ,  
+
+\[a_{i,k} = \max (a_{i - 1,k - 1}, a_{i,k - 1}, a_{i + 1,k - 1}) = \max (a_{i - 2,k - 2}, \ldots , a_{i + 2,k - 2}) = \dots = \max (a_{i - k,0}, \ldots , a_{i + k,0}).\]  
+
+Given an initial number \(a_{c,0}\) , let \(j_{1}\) and \(j_{2}\) be the smallest positive integers such that \(a_{c - j_{1},0} > a_{c,0}\) and \(a_{c + j_{2},0} > a_{c,0}\) . As the initial numbers are distinct, we conclude \(a_{i,k} = a_{c,0}\) if and only if \(\{i - k, i - k + 1, \ldots , i + k\}\) contains \(c\) but neither \(c - j_{1}\) nor \(c + j_{2}\) (mod 2025). The indices \(i\) that satisfy this are clearly consecutive.
+
+
+
+Now consider the indicator variables  
+
+\[\mathbf{1}_{i} = \left\{ \begin{array}{ll}1 & \mathrm{if~}a_{i,k}\neq a_{i + 1,k}\\ 0 & \mathrm{otherwise} \end{array} \right.\]  
+
+for \(1\leq i\leq 2025\) . Note that the number of distinct values on the board after \(k\) seconds is simply \(\textstyle \sum_{i = 1}^{2025}\mathbf{1}_{i}\) . By linearity of expectation, it suffices to compute \(\mathbb{E}[\mathbf{1}_{i}]\) for each \(i\) . Recall  
+
+\[a_{i,k} = \max (a_{i - k,0},\ldots ,a_{i + k,0})\]  
+
+for each \(1\leq i\leq 2025\) . Hence, the condition \(a_{i,k}\neq a_{i + 1,k}\) can be written as  
+
+\[\max (a_{i - k,0},\ldots ,a_{i + k,0})\neq \max (a_{i + 1 - k,0},\ldots ,a_{i + 1 + k,0}).\]  
+
+This is the case if and only if \(\max (a_{i - k,0},\ldots ,a_{i + k + 1,0})\) is either \(a_{i - k,0}\) or \(a_{i + k + 1,0}\) (as \(a_{i - k,0}\neq a_{i + k + 1,0}\) and \(2k + 2< 2025\) ). Since the \(2k + 2\) values \(a_{i - k,0}\) , ..., \(a_{i + k + 1,0}\) were sampled independently from the same distribution, each of \(a_{i - k,0}\) and \(a_{i + k + 1,0}\) has a \(\frac{1}{2k + 2}\) probability of being their maximum. Hence,  
+
+\[\mathbb{E}[\mathbf{1}_{i}] = \operatorname *{Pr}[a_{i,k}\neq a_{i + 1,k}] = \frac{2}{2k + 2} = \frac{1}{k + 1}.\]  
+
+Thus, the expected number of distinct values is  
+
+\[\sum_{i = 1}^{2025}\mathbb{E}[\mathbf{1}_{i}] = \frac{2025}{k + 1}.\]  
+
+Substituting \(k = 100\) yields the answer \(\left[\frac{2025}{101}\right]\)  
+
+9. Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon.  
+
+Proposed by: Albert Wang  
+
+Answer: \(\frac{4}{9}\)  
+
+Solution:  
+
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+  
+
+First, we compute the probability that the line through two random points in a triangle \(A B C\) passes through segments \(\overline{{A B}}\) and \(\overline{{A C}}\) . We can take an affine transform of the two random points and the triangle such that \(A B C\) becomes equilateral. Since the distribution of the two points is still uniform and independent, the probability of the line intersecting any two given sides is \(\frac{1}{3}\) by symmetry.  
+
+Next, we compute the probability that the line through two random points in a rectangle \(A B C D\) passes through opposite edges \(\overline{{A B}}\) and \(\overline{{C D}}\) .
+
+
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+  
+
+If the line passes through \(\overline{{A B}}\) and \(\overline{{B C}}\) , the points must both lie in triangle \(A B C\) , whose area is half that of \(A B C D\) . Given this, the probability the line passes through those two sides is \(\frac{1}{3}\) , as computed before. Thus the probability the line passes through \(\overline{{A B}}\) and \(\overline{{B C}}\) is \(\left(\frac{1}{2}\right)^{2}\cdot \frac{1}{3} = \frac{1}{12}\) . The same goes for the other pairs of adjacent edges. By symmetry, the line is equally likely to pass through either pair of opposite edges, each with probability \(\frac{1}{2}\left(1 - 4\cdot \frac{1}{12}\right) = \frac{1}{3}\) .  
+
+![](data:image/jpeg;base64,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+  
+
+We now return to the original problem. If the line passes through a pair of opposite edges, then both points must be in the rectangle formed by these edges, which has area \(\frac{2}{3}\) that of the hexagon. Given this, the probability the line passes through those two edges is \(\frac{1}{3}\) as computed before. Thus, the probability that the line passes through the given pair of opposite edges is \(\left(\frac{2}{3}\right)^{2}\cdot \frac{1}{3} = \frac{4}{27}\) . Hence, the probability the line passes through any of the three pairs of opposite edges is \(3\cdot \frac{4}{27} = \left[\frac{4}{9}\right]\) .  
+
+10. The circumference of a circle is divided into 45 arcs, each of length 1. Initially, there are 15 snakes, each of length 1, occupying every third arc. Every second, each snake independently moves either one arc left or one arc right, each with probability \(\frac{1}{2}\) . If two snakes ever touch, they merge to form a single snake occupying the arcs of both of the previous snakes, and the merged snake moves as one snake. Compute the expected number of seconds until there is only one snake left.  
+
+Proposed by: Srinivas Arun  
+
+Answer: \(\frac{448}{3}\)  
+
+Solution: We solve the problem generally for \(n\) snakes and \(3n\) arcs. Without loss of generality, fix the two snakes \(A\) and \(B\) that will eventually form the ends of the last snake. Note that \(A\) and \(B\) must be consecutive in the initial configuration; assume \(B\) lies immediately clockwise from \(A\) .  
+
+Let \(d\) be the arclength between \(A\) and \(B\) (measuring clockwise from \(A\) to \(B\) ). As long as \(d \neq 0\) and \(d \neq 2n\) , we know that \(A\) and \(B\) lie in different snakes and thus move independently. Therefore, we can consider \(d\) to be on a random walk starting at 2, where
+
+

HarvardMIT/md/en-284-tournaments-2025-hmic-solutions.md

@@ -0,0 +1,183 @@
+
+# HMIC 2025 April 20-27, 2025  
+
+1. [5] Let \(A B C D\) be a convex quadrilateral. Define parabolas \(\mathcal{P}_{A}\) \(\mathcal{P}_{B}\) \(\mathcal{P}_{C}\) , and \(\mathcal{P}_{D}\) to have directrices \(B D\) \(C A\) \(D B\) , and \(A C\) , and foci \(A\) \(B\) \(C\) , and \(D\) , respectively. Prove that no two of these parabolas intersect more than once.  
+
+(A parabola with directrix \(\ell\) and focus \(P\) consists of all points \(X\) for which \(P X\) equals the distance from \(P\) to \(\ell\) .)  
+
+Proposed by: Albert Wang  
+
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+  
+
+Solution 1: Let \(d(P, X Y)\) be the distance from \(P\) to line \(X Y\) . We will first prove \(\mathcal{P}_{A}\) and \(\mathcal{P}_{B}\) intersect at most once.  
+
+Claim 1. Let \(\ell_{A B}\) be the perpendicular bisector of \(A B\) . Then, \(\mathcal{P}_{A}\) is tangent to \(\ell_{A B}\) .  
+
+Proof. Consider any point \(X\) on \(\mathcal{P}_{A}\) . Then,  
+
+\[X A = d(X,B D)\leq X B,\]  
+
+with equality only holding at the unique point \(X\) for which \(X B\perp B D\) . Thus, \(\mathcal{P}_{A}\) lies entirely on one side of \(\ell_{A B}\) , touching it once at this point \(X\) . \(\square\)  
+
+It follows that \(\mathcal{P}_{A}\) and \(\mathcal{P}_{B}\) are on different sides of \(\ell_{A B}\) and hence can intersect at most once (possibly at a common tangency point to \(\ell_{A B}\) ).  
+
+Since \(A B C D\) is convex, \(A\) and \(C\) lie on opposite sides of line \(B D\) , the common directrix of \(\mathcal{P}_{A}\) and \(\mathcal{P}_{C}\) . Thus, \(\mathcal{P}_{A}\) and \(\mathcal{P}_{C}\) lie on opposite sides of line \(B D\) and cannot intersect at all.  
+
+The remaining pairs of parabolas are handled similarly.  
+
+Solution 2: Let \(d(P, X Y)\) be the distance from \(P\) to line \(X Y\) .  
+
+Claim 2. \(\mathcal{P}_{A}\) and \(\mathcal{P}_{B}\) intersect at most once.  
+
+Proof. Let \(P\) be a common point of both parabolas. Then, \(d(P, B D) = P A\) and \(d(P, A C) = P B\) , which combined imply  
+
+\[P A = d(P,B D)\leq P B = d(P,A C)\leq P A.\]  
+
+Thus, the inequalities above are equalities, i.e., \(P A = P B\) , \(P A\perp A C\) , and \(P B\perp B D\) . Such \(P\) , if it exists, is unique. \(\square\)
+
+
+
+The remaining pairs of parabolas are handled similarly. As in solution 1, \(\mathcal{P}_{A}\) and \(\mathcal{P}_{C}\) cannot intersect, nor can \(\mathcal{P}_{B}\) and \(\mathcal{P}_{D}\) .  
+
+2. [7] Find all polynomials \(P\) with real coefficients for which there exists a polynomial \(Q\) with real coefficients such that for all real \(t\) ,  
+
+\[\cos (P(t)) = Q(\cos t).\]  
+
+Proposed by: Karthik Venkata Vedula  
+
+Answer: All constant functions and \(P(x) = a x + b\pi\) for all nonzero integers \(a\) and integers \(b\)  
+
+Solution 1: It is well- known that these polynomials work by taking \(Q\) to be a Chebyshev polynomial (if \(P\) is linear) or a constant (if \(P\) is constant).  
+
+Suppose that \(\deg P \geq 2\) . Now consider the density of the roots of \(\cos (P(t))\) , i.e.  
+
+\[\lim_{n \to \infty} \frac{\text{number of roots in the interval} [-n, n]}{n}.\]  
+
+Since \(\cos (P(t)) = Q(\cos t)\) , the density is finite, because for each interval of length \(2\pi\) , there can only be a finite number of roots (i.e. twice the degree of \(Q\) ). However, we claim that \(\cos (P(t))\) has an infinite density of roots. In particular, consider the solutions to \(P(x) = \pm (2k - 1)\pi /2\) over positive integers \(k\) . Asymptotically, for large \(k\) , such \(x\) will always exist and be \(\Theta (k^{1 / \deg P})\) . As \(k \to \infty\) , such \(x\) become infinitely dense, contradicting the finite density of roots of \(Q(\cos t)\) . Therefore, \(\deg P \leq 1\) .  
+
+If \(\deg P = 1\) , let \(P(t) = at + b\) . Observe that \(Q(\cos t)\) is periodic with period \(2\pi\) , so \(\cos (P(t))\) must be as well. This is only the case when \(a\) is an integer. Furthermore,  
+
+\[Q(\cos t) = \cos (at + b) = \cos (at)\cos (b) - \sin (at)\sin (b)\]  
+
+must be an even function. Note that \(\cos (at)\cos (b)\) is even and \(\sin (at)\sin (b)\) is odd, so \(\sin (at)\sin (b) = 0\) for all \(t\) , and hence \(b\) is an integer multiple of \(\pi\) .  
+
+Thus, the only polynomials that work are the ones claimed above.  
+
+Solution 2: Taking the derivative of both sides,  
+
+\[\sin (P(t))P^{\prime}(t) = (- \sin t)Q^{\prime}(\cos t).\]  
+
+The right- hand side is bounded in \(t\) , so the left- hand side must also be bounded. If \(\deg P \geq 2\) , then as \(t\) approaches \(\infty\) , \(P^{\prime}(t)\) approaches \(\pm \infty\) and \(\sin (P(t))\) does not approach 0, contradiction. Thus, \(\deg P \leq 1\) , and we finish as before.  
+
+3. [8] Let \(ABCD\) be a parallelogram, and let \(O\) be a point inside \(ABCD\) . Suppose the circumcircles of triangles \(OAB\) and \(OCD\) intersect at \(P \neq O\) , and the circumcircles of triangles \(OBC\) and \(OAD\) intersect at \(Q \neq O\) . Prove \(\angle POQ\) equals one of the angles of quadrilateral \(ABCD\) .  
+
+Proposed by: Derek Liu
+
+
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+  
+
+Solution: In what follows, all angles are directed.  
+
+Claim 1. The points \(P\) and \(Q\) are symmetric over the center of \(ABCD\) .  
+
+Proof. Note that  
+
+\[\angle A P B = \angle A O B = \angle O A D + \angle C B O = \angle O Q D + \angle C Q O = \angle C Q D.\]  
+
+Similar equalities hold for each pair of opposite sides, so \(P\) and \(Q\) are symmetric across the parallelogram's center. \(\square\)  
+
+Consequently, \(AP\) and \(CQ\) are parallel, so  
+
+\[\angle P O Q = \angle P O B + \angle B O Q = \angle P A B + \angle B C Q = \angle C B A,\]  
+
+as desired. (Once we undirect the angles, \(\angle P O Q\) is either \(\angle B\) or \(\pi - \angle B = \angle A\) .)  
+
+4. [9] Determine whether there exist infinitely many pairs of distinct positive integers \(m\) and \(n\) such that \(2^{m} + n\) divides \(2^{n} + m\) .  
+
+Proposed by: Carlos Rodriguez, Jordan Lefkowitz  
+
+Answer: Yes  
+
+Solution: Let \(k\) be a positive integer, and set \(m = 2^{k}\) and \(n = p - 2^{2^{k}}\) for prime \(p\) to be chosen later. We want \(2^{m} + n = p\) to divide \(2^{p - 2^{2^{k}}} + 2^{k}\) , which is equivalent to having  
+
+\[0\equiv 2^{p - 2^{2^{k}}} + 2^{k}\equiv 2^{1 - 2^{2^{k}}} + 2^{k}\equiv 2^{1-2^{2^{k}}}\left(2^{2^{2^{k}} + k - 1} + 1\right)\pmod {p}.\]  
+
+Let \(r = 2^{2^{k}} + k - 1\) . Since \(r \neq 3\) , by Zsigmondy, we can pick a prime \(p\) that divides \(2^{r} + 1\) but not \(2^{s} + 1\) for any nonnegative integer \(s < r\) . Let \(d = \operatorname{ord}_{p}(2)\) . Then, \(2^{|d - r|} \equiv -1 \pmod {p}\) , so by definition of \(p\) , we have \(|d - r| \geq r\) . Hence, \(d \geq 2r\) . As \(d \mid p - 1\) , we conclude  
+
+\[p > 2r = 2\left(2^{2^{k}} + k - 1\right) > 2^{2^{k}} + 2^{k},\]  
+
+so \(n = p - 2^{2^{k}} > m\) . Since \(p \mid 2^{r} + 1\) by definition, \((m, n)\) is a pair of distinct positive integers with \(2^{m} + n \mid 2^{n} + m\) .  
+
+As \(k\) was arbitrary (and \(m = 2^{k}\) ), there exist infinitely many such pairs.
+
+
+
+5. [13] Compute the smallest positive integer \(k > 45\) for which there exists a sequence \(a_{1}\) , \(a_{2}\) , \(a_{3}\) , ..., \(a_{k - 1}\) of positive integers satisfying the following conditions:  
+
+\(a_{i} = i\) for all integers \(1\leq i\leq 45\) \(a_{k - i} = i\) for all integers \(1\leq i\leq 45\) , and for any odd integer \(1\leq n\leq k - 45\) , the sequence \(a_{n}\) , \(a_{n + 1}\) , ..., \(a_{n + 44}\) is a permutation of \(\{1,2,\ldots ,45\}\) .  
+
+Proposed by: Derek Liu  
+
+Answer: 1059  
+
+Solution: First, we show 1059 is optimal. Assume for sake of contradiction that \(k< 1059\)  
+
+The given condition ensures that \(\{a_{1},a_{2}\} = \{a_{46},a_{47}\}\) , \(\{a_{3},a_{4}\} = \{a_{48},a_{49}\}\) , and so on. In particular, if \(a_{i} = j\) , the next appearance of \(j\) must either be \(a_{i + 44}\) , \(a_{i + 45}\) , or \(a_{i + 46}\) ; working modulo 45, these indices all differ from \(i\) by at most 1. Furthermore, \(a_{i} = a_{i + 44}\) is only possible if \(i\) is even, and \(a_{i} = a_{i + 46}\) is only possible if \(i\) is odd.  
+
+Also, \(a_{1}\neq a_{45}\) by definition. Thus, if the sequence contains the same number at least 25 times, the 25th appearance has index at least \(2 + 24\cdot 44 = 1058\) , implying \(k\geq 1059\) . Hence, we can assume every number appears at most 24 times in the sequence.  
+
+Observe that there exists a unique integer \(1\leq j\leq 45\) such that  
+
+\[(k - j) - j = k - 2j\equiv 22\pmod {45}.\]  
+
+Let \(k_{1} = j\) , \(k_{2}\) , \(k_{3}\) , ..., \(k_{\ell} = k - j\) be the indices of where \(j\) appears in the sequence; as assumed above, \(\ell \leq 24\) . For any \(i\) , we proved \(k_{i} - k_{i - 1}\) is either 44, 45, or 46, and thus either 0 or \(\pm 1\) modulo 45. Since \(k - j\) and \(j\) differ by \(22\equiv - 23\) modulo 45, either \(k_{i} - k_{i - 1} = 46\) for at least 22 different \(i\) , or \(k_{i} - k_{i - 1} = 44\) for at least 23 different \(i\) . We split into cases based on which.  
+
+If \(k_{i} - k_{i - 1} = 46\) at least 22 times, we claim \(\ell = 23\) . Indeed, if \(\ell = 24\) , then  
+
+\[(k - j)\geq j + 22\cdot 46 + 44 = j + 1056.\]  
+
+Since \((k - j) - j\equiv 22\) (mod 45), we actually have \((k - j)\geq j + 1057\) , so \(k\geq 2j + 1057\geq 1059\) , contradiction.  
+
+Thus, \(\ell = 23\) , so \(k_{i} - k_{i - 1} = 46\) for all \(i\) , and \(k = 2j + 22\cdot 46 = 2j + 1012\) , which means we can assume \(j\leq 23\) (otherwise \(k > 1059\) ).  
+
+Since \(a_{j} = a_{j + 46}\) , we must also have \(a_{j + 1} = a_{j + 45}\) be the second appearance of \(j + 1\) , which means \(j\) must be odd. Then, as \(j + 45\) is even, \(a_{j + 45} = j + 1\) must either be equal to either \(a_{j + 89}\) or \(a_{j + 90}\) . Now, \(a_{k - j - 1} = j + 1\) as well, and  
+
+\[(k - j - 1) - (j + 90)\equiv 21\pmod {45},\]  
+
+so if \(a_{j + 90}\) is the 3rd appearance of \(j + 1\) , then \(a_{k - j - 1}\) must be at least the 24th. Thus, it must be exactly the 24th, with each appearance of \(j + 1\) being exactly 46 terms after the previous. Thus,  
+
+\[k - j - 1\geq (j + 90) + 21\cdot 46 = j + 1056,\]  
+
+and \(k\geq 2j + 1057\geq 1059\) , as desired. If \(a_{j + 89}\) is the 3rd appearance of \(j + 1\) , then since \((k - j - 1) - (j + 89)\equiv 22\) (mod 45), it follows that \(a_{k - j - 1}\) must be at least the 25th appearance, contradiction.  
+
+The other case, where \(k_{i} - k_{i - 1} = 44\) at least 23 times (i.e., for all \(i\) ) plays out similarly. Since \(k_{2} = k_{1} + 44\) , we immediately have \(j = k_{1}\) is even. Furthermore, \(a_{j - 1} = a_{j + 45} = j - 1\) is the second appearance of \(j - 1\) , and since \(j + 45\) is odd, the third appearance is either \(a_{j + 90}\) or \(a_{j + 91}\) . However,  
+
+\[(k - j + 1) - (j + 90)\equiv -22\pmod {45},\]
+
+
+
+so regardless of whether \(a_{j + 90}\) or \(a_{j + 91}\) is the 3rd appearance of \(j - 1\) , we know \(a_{k - j + 1}\) must be at least the 25th, contradiction.  
+
+Thus \(k \geq 1059\) , and it suffices to provide a construction.  
+
+Given an integer \(m\) , let \(\alpha\) represent the permutation on \(\{1,2,\ldots ,m\}\) given by swapping 1 and 2, 3 and 4, etc. and let \(\beta\) represent the permutation given by swapping 2 and 3, 4 and 5, etc.  
+
+Claim 1. Both of the products \(\dots \beta \alpha \beta \alpha\) and \(\dots \alpha \beta \alpha \beta\) , with exactly \(m\) terms in the products, equal the reverse permutation.  
+
+Proof. If we track any number \(i\) , observe that if \(i\) is odd,  
+
+\[\alpha (i) = i + 1\] \[\beta \alpha (i) = i + 2\] \[\vdots\] \[\dots \beta \alpha (i) = m\] \[\underset {m - i}{\underbrace{\dots \beta \alpha (i) = m}}\] \[\underset {m - i + 1}{\underbrace{\dots \beta \alpha (i) = m - 1}}\] \[\underset {m - i + 2}{\underbrace{\dots \beta \alpha (i) = m - i + 1}}.\]  
+
+A similar chain occurs if \(i\) is even, so \(\dots \beta \alpha\) is indeed the reverse permutation. The same argument shows that the product \(\dots \alpha \beta\) is also the reverse permutation. \(\square\)  
+
+Our construction now goes as follows: for any odd \(n\) , let \(a_{n + 46} = a_{n}\) and \(a_{n + 45} = a_{n + i}\) unless \(n \equiv 0\) or 23 mod 45. Then, we can note that \(a_{46}\) through \(a_{68}\) are the permutation \(\alpha\) on \(\{1,2,\ldots ,23\}\) , and \(a_{91}\) through \(a_{113}\) are \(\alpha \beta\) , etc. By our claim, \(a_{1 + 23\cdot 45} = a_{1036}\) through \(a_{1058}\) are 1 through 23 reversed. Similarly, as there are only 22 numbers from 24 through 45, we know \(a_{24 + 22\cdot 45} = a_{1014}\) through \(a_{1035}\) are 24 through 45 reversed. It follows that this sequence satisfies the above conditions for \(k = \left\lfloor 1059\right\rfloor\) .  
+
+The diagram below shows what the construction would look like when 45 is replaced with 9. The sequence can be read row- by- row. The vertical line separates 1 through 5 from 6 through 9.  
+
+\[1 2 3 4 5 6 7 8 9\] \[2 1 4 3 5 6 8 7 9\] \[2 4 1 5 3 8 6 9 7\] \[4 2 5 1 3 8 9 6 7\] \[4 5 2 3 1 9 8 7 6\] \[5 4 3 2 1\]  
+
+Remark. The exact same proof and construction work for all odd \(n \geq 5\) , yielding an answer of \(\frac{n^2 + 2n + 3}{2}\) . Notably, for \(n = 3\) the answer is 6 instead of 9.
+
+

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+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Compute the sum of the positive divisors (including 1) of 9! that have units digit 1.", "solution": "Answer: 103", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nProposed by: Jackson Dryg  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Compute the sum of the positive divisors (including 1) of 9! that have units digit 1.", "solution": "The prime factorization of 9! is \\(2^{7} \\cdot 3^{4} \\cdot 5 \\cdot 7\\) . Every divisor of 9! has prime factorization \\(2^{a} \\cdot 3^{b} \\cdot 5^{c} \\cdot 7^{d}\\) , where \\(0 \\leq a \\leq 7\\) , \\(0 \\leq b \\leq 4\\) , \\(0 \\leq c \\leq 1\\) , and \\(0 \\leq d \\leq 1\\) . If the divisor has units digit 1, it cannot be divisible by 2 or 5, so \\(a = c = 0\\) .  \n\nNow take cases on the value of \\(d\\) :  \n\n- If \\(d = 0\\) , then the divisor is \\(3^{b}\\) for some \\(0 \\leq b \\leq 4\\) . The possible divisors are 1, 3, 9, 27, and 81, of which 1 and 81 work.  \n\n- If \\(d = 1\\) , then the divisor is \\(3^{b} \\cdot 7\\) for some \\(0 \\leq b \\leq 4\\) . The possible divisors are then 7, 3 \\(\\cdot 7\\) , 9 \\(\\cdot 7\\) , 27 \\(\\cdot 7\\) , and 81 \\(\\cdot 7\\) . Of these, only \\(3 \\cdot 7 = 21\\) works.  \n\nThe answer is \\(1 + 21 + 81 = \\boxed{103}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n1. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Mark writes the expression \\(\\sqrt{abcd}\\) on the board, where \\(abcd\\) is a four-digit number and \\(a \\neq 0\\) . Derek, a toddler, decides to move the \\(a\\) , changing Mark's expression to \\(a\\sqrt{bc}\\) . Surprisingly, these two expressions are equal. Compute the only possible four-digit number \\(abcd\\) .", "solution": "Answer: 3375", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nProposed by: Pitchayut Saengrungkongka  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Mark writes the expression \\(\\sqrt{abcd}\\) on the board, where \\(abcd\\) is a four-digit number and \\(a \\neq 0\\) . Derek, a toddler, decides to move the \\(a\\) , changing Mark's expression to \\(a\\sqrt{bc}\\) . Surprisingly, these two expressions are equal. Compute the only possible four-digit number \\(abcd\\) .", "solution": "Let \\(x = bcd\\) . Then, we rewrite the given condition \\(\\sqrt{abcd} = a\\sqrt{bc}\\) as  \n\n\\[1000a + x = a^{2}x,\\]  \n\nwhich simplifies as  \n\n\\[(a^{2} - 1)x = 1000a.\\]  \n\nIn particular, \\(a^{2} - 1\\) divides \\(1000a\\) . Since \\(\\gcd (a^{2} - 1, a) = 1\\) , it follows that \\(a^{2} - 1 \\mid 1000\\) . The only \\(a \\in \\{1, 2, \\ldots , 9\\}\\) that satisfies this is \\(a = 3\\) . Then \\(8x = 3000\\) , so \\(x = 375\\) . Thus \\(abcd = \\boxed{3375}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Given that \\(x\\) , \\(y\\) , and \\(z\\) are positive real numbers such that  \n\n\\[x^{\\log_{2}(yz)} = 2^{8} \\cdot 3^{4}, \\quad y^{\\log_{2}(zx)} = 2^{9} \\cdot 3^{6}, \\quad \\text{and} \\quad z^{\\log_{2}(xy)} = 2^{5} \\cdot 3^{10},\\]  \n\ncompute the smallest possible value of \\(xyz\\) .", "solution": "Answer: 1 576", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nProposed by: Derek Liu  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Given that \\(x\\) , \\(y\\) , and \\(z\\) are positive real numbers such that  \n\n\\[x^{\\log_{2}(yz)} = 2^{8} \\cdot 3^{4}, \\quad y^{\\log_{2}(zx)} = 2^{9} \\cdot 3^{6}, \\quad \\text{and} \\quad z^{\\log_{2}(xy)} = 2^{5} \\cdot 3^{10},\\]  \n\ncompute the smallest possible value of \\(xyz\\) .", "solution": "Let \\(k = \\log_{2}3\\) for brevity. Taking the base- 2 log of each equation gives  \n\n\\[(\\log_{2}x)(\\log_{2}y + \\log_{2}z) = 8 + 4k,\\] \\[(\\log_{2}y)(\\log_{2}z + \\log_{2}x) = 9 + 6k,\\] \\[(\\log_{2}z)(\\log_{2}x + \\log_{2}y) = 5 + 10k.\\]\n\n\n\nAdding the first two equations and subtracting the third yields \\(2 \\log_{2} x \\log_{2} y = 12\\) , so \\(\\log_{2} x \\log_{2} y = 6\\) . Similarly, we get  \n\n\\[\\log_{2} x \\log_{2} y = 6,\\] \\[\\log_{2} y \\log_{2} z = 3 + 6 k,\\] \\[\\log_{2} z \\log_{2} x = 2 + 4 k.\\]  \n\nMultiplying the first two equations and dividing by the third yields \\((\\log_{2} y)^{2} = 9\\) , so \\(\\log_{2} y = \\pm 3\\) . Then, the first and last equations tell us \\(\\log_{2} x = \\pm 2\\) and \\(\\log_{2} z = \\pm (1 + 2 k)\\) , with all signs matching. Thus  \n\n\\[\\log_{2} x + \\log_{2} y + \\log_{2} z = \\pm (3 + 2 + (1 + 2 k)) = \\pm (6 + 2 k),\\]  \n\nso  \n\n\\[x y z = 2^{\\pm (6 + 2 k)} = 2^{6} \\cdot 3^{2} \\quad \\text{or} \\quad 2^{-6} \\cdot 3^{-2}.\\]  \n\nClearly, the smallest solution is \\(2^{- 6} \\cdot 3^{- 2} = \\left[\\frac{1}{576}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\lfloor z\\rfloor\\) denote the greatest integer less than or equal to \\(z\\) . Compute  \n\n\\[\\sum_{j = -1000}^{1000} \\left\\lfloor \\frac{2025}{j + 0.5} \\right\\rfloor .\\]", "solution": "Answer: \\(- 984\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nProposed by: Linus Yifeng Tang  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\lfloor z\\rfloor\\) denote the greatest integer less than or equal to \\(z\\) . Compute  \n\n\\[\\sum_{j = -1000}^{1000} \\left\\lfloor \\frac{2025}{j + 0.5} \\right\\rfloor .\\]", "solution": "The key idea is to pair up the terms \\(\\left\\lfloor \\frac{2025}{x} \\right\\rfloor\\) and \\(\\left\\lfloor \\frac{2025}{x} \\right\\rfloor\\) . There are 1000 such pairs and one lone term, \\(\\left\\lfloor \\frac{2025}{1000.5} \\right\\rfloor = 2\\) . Thus,  \n\n\\[\\sum_{j = -1000}^{1000} \\left\\lfloor \\frac{2025}{j + 0.5} \\right\\rfloor = 2 + \\sum_{x \\in \\{0.5,1.5, \\ldots , 999.5\\}} \\left(\\left\\lfloor \\frac{2025}{x} \\right\\rfloor + \\left\\lfloor \\frac{2025}{-x} \\right\\rfloor \\right).\\]  \n\nWe note that  \n\n  \n\nTherefore,  \n\n  \n\nAs \\(x\\) ranges in the set \\(\\{0.5, 1.5, 2.5, \\ldots , 999.5\\}\\) , \\(2 x\\) ranges in the set \\(\\{1, 3, 5, \\ldots , 1999\\}\\) . This set includes all 15 odd divisors of 4050 except for 2025. Thus, there are 14 values of \\(x\\) for which \\(\\left\\lfloor \\frac{2025}{x} \\right\\rfloor + \\left\\lfloor \\frac{2025}{- x} \\right\\rfloor\\) evaluates to 0, and the remaining \\(1000 - 14 = 986\\) values of \\(x\\) make it evaluate to \\(- 1\\) . Therefore,  \n\n\\[\\sum_{j = -1000}^{1000} \\left\\lfloor \\frac{2025}{j + 0.5} \\right\\rfloor = 2 + \\sum_{x \\in \\{0.5,1.5, \\ldots , 999.5\\}} \\left(\\left\\lfloor \\frac{2025}{x} \\right\\rfloor + \\left\\lfloor \\frac{2025}{-x} \\right\\rfloor \\right) = 2 + 986 \\cdot (-1) = \\boxed{- 984}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathcal{S}\\) be the set of all nonconstant monic polynomials \\(P\\) with integer coefficients satisfying \\(P\\left(\\sqrt{3} + \\sqrt{2}\\right) = P\\left(\\sqrt{3} - \\sqrt{2}\\right)\\) . If \\(Q\\) is an element of \\(\\mathcal{S}\\) with minimal degree, compute the only possible value of \\(Q(10) - Q(0)\\) .", "solution": "Answer: 890", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nProposed by: David Dong  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let \\(\\mathcal{S}\\) be the set of all nonconstant monic polynomials \\(P\\) with integer coefficients satisfying \\(P\\left(\\sqrt{3} + \\sqrt{2}\\right) = P\\left(\\sqrt{3} - \\sqrt{2}\\right)\\) . If \\(Q\\) is an element of \\(\\mathcal{S}\\) with minimal degree, compute the only possible value of \\(Q(10) - Q(0)\\) .", "solution": "First, note that the polynomial \\(x^{4} - 10x^{2} + 1\\) has both \\(\\sqrt{3} +\\sqrt{2}\\) and \\(\\sqrt{3} - \\sqrt{2}\\) as roots. It suffices to check whether a polynomial of degree at most 3 belongs in \\(\\mathcal{S}\\) . Suppose \\(f(x) = a x^{3} + b x^{2} + c x + d\\in \\mathcal{S}\\) . We compute  \n\n\\[(\\sqrt{3} +\\sqrt{2})^{3} - (\\sqrt{3} -\\sqrt{2})^{3} = 22\\sqrt{2\\] \\[(\\sqrt{3} +\\sqrt{2})^{2} - (\\sqrt{3} -\\sqrt{2})^{2} = 4\\sqrt{6\\] \\[(\\sqrt{3} +\\sqrt{2})^{1} - (\\sqrt{3} -\\sqrt{2})^{1} = 2\\sqrt{2,\\]  \n\nso we get that  \n\n\\[f(\\sqrt{3} +\\sqrt{2}) - f(\\sqrt{3} -\\sqrt{2}) = (22\\sqrt{2})a + (4\\sqrt{6})b + (2\\sqrt{2})c.\\]  \n\nBy resolving linear dependencies, it's clear that \\(b = 0\\) and \\(c = - 11a\\) . It follows that if \\(f\\) is not the zero polynomial, it must be cubic. It is then clear that \\(f(x) = x^{3} - 11x + d\\) has minimal degree in \\(\\mathcal{S}\\) , and thus \\(Q(10) - Q(0) = f(10) - f(0) = \\boxed{890}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n5. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(r\\) be the remainder when \\(2017^{2025!} - 1\\) is divided by 2025!. Compute \\(\\frac{r}{2025!}\\) . (Note that 2017 is prime.)", "solution": "Answer: \\(\\frac{1311}{2017}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nProposed by: Srinivas Arun  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let \\(r\\) be the remainder when \\(2017^{2025!} - 1\\) is divided by 2025!. Compute \\(\\frac{r}{2025!}\\) . (Note that 2017 is prime.)", "solution": "Let \\(N = 2017^{2025!}\\) . Let \\(p\\) be a prime dividing 2025! other than 2017. Let \\(p^{k}\\) be the largest power of \\(p\\) dividing 2025!. Clearly, \\(\\phi (p^{k}) = (p - 1)p^{k - 1}\\) divides 2025! and \\(\\gcd (2017, p^{k}) = 1\\) , so by Euler's Totient Theorem,  \n\n\\[N\\equiv 1\\pmod {p^{k}}.\\]  \n\nRepeating for all such primes \\(p\\) , we obtain  \n\n\\[N\\equiv 1\\pmod {2025! / 2017}.\\]  \n\nTherefore, \\(\\frac{2025!}{2017} |N - 1\\) , so \\(r = \\frac{2025!}{2017} s\\) for some \\(0\\leq s< 2017\\) . Also, since \\(N\\equiv 0\\) (mod 2017), we have \\(r = \\frac{2025!}{2017} s\\equiv - 1\\) (mod 2017).  \n\nBy Wilson's,  \n\n\\[\\frac{2025!}{2017} = 2016!(2018)(2019)\\ldots (2025)\\equiv -8!\\equiv 20\\pmod {2017}.\\]  \n\nTherefore, \\(s\\) is negative the inverse of 20 (mod 2017), which is 1311. Our answer is  \n\n\\[\\frac{r}{2025!} = \\frac{(2025! / 2017)(1311)}{2025!} = \\frac{1311}{2017}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n6. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique triple \\((a,b,c)\\) of positive real numbers that satisfies the equations  \n\n\\[2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1)\\quad \\mathrm{and}\\quad ab + bc + ca = 1.\\]  \n\nCompute \\(a + b + c\\) .", "solution": "Answer: \\(\\frac{9\\sqrt{23}}{23} = \\frac{9}{\\sqrt{23}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nProposed by: David Wei  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique triple \\((a,b,c)\\) of positive real numbers that satisfies the equations  \n\n\\[2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1)\\quad \\mathrm{and}\\quad ab + bc + ca = 1.\\]  \n\nCompute \\(a + b + c\\) .", "solution": "The crux of this problem is to apply the trigonometric substitutions \\(a = \\cot \\alpha\\) , \\(b = \\cot \\beta\\) , and \\(c = \\cot \\gamma\\) , with \\(0 < \\alpha , \\beta , \\gamma < \\pi / 2\\) . Then, the given equations translate to  \n\n\\[\\frac{2}{\\sin^{2}\\alpha} = \\frac{3}{\\sin^{2}\\beta} = \\frac{4}{\\sin^{2}\\gamma}\\quad \\mathrm{and}\\quad \\cot \\alpha \\cot \\beta +\\cot \\beta \\cot \\gamma +\\cot \\gamma \\cot \\alpha = 1.\\]  \n\nFrom the second equation, we get  \n\n\\[\\cot \\gamma = \\frac{1 - \\cot\\alpha\\cot\\beta}{\\cot\\alpha + \\cot\\beta} = -\\cot (\\alpha +\\beta).\\]  \n\nSince \\(\\alpha\\) , \\(\\beta\\) , and \\(\\gamma\\) all between 0 and \\(\\pi / 2\\) , we discover that  \n\n\\[\\alpha +\\beta +\\gamma = \\pi .\\]  \n\nLet \\(\\triangle ABC\\) be the (acute) triangle with side lengths \\(BC = \\sqrt{2}\\) , \\(CA = \\sqrt{3}\\) , and \\(AB = \\sqrt{4}\\) . By Law of Sines, setting \\(\\alpha = \\angle A\\) , \\(\\beta = \\angle B\\) , and \\(\\gamma = \\angle C\\) will satisfy both equations. Thus, Law of Cosines gives  \n\n\\[\\cos \\alpha = \\frac{3 + 4 - 2}{2\\cdot\\sqrt{3}\\cdot\\sqrt{4}} = \\frac{5}{\\sqrt{48}}\\Rightarrow a = \\cot \\alpha = \\frac{5}{\\sqrt{23}}\\]  \n\nSimilar calculations give \\(b = \\frac{3}{\\sqrt{23}}\\) and \\(c = \\frac{1}{\\sqrt{23}}\\) , so the answer is \\(a + b + c = \\left[\\frac{9}{\\sqrt{23}}\\right]\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "There exists a unique triple \\((a,b,c)\\) of positive real numbers that satisfies the equations  \n\n\\[2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1)\\quad \\mathrm{and}\\quad ab + bc + ca = 1.\\]  \n\nCompute \\(a + b + c\\) .", "solution": "Let \\(2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1) = x\\) . Then, since \\(ab + bc + ca = 1\\) , we have the following system of equations:  \n\n\\[(a + b)(c + a) = a^{2} + ab + bc + ca = a^{2} + 1 = x / 2\\] \\[(b + c)(a + b) = b^{2} + ab + bc + ca = b^{2} + 1 = x / 3\\] \\[(c + a)(b + c) = c^{2} + ab + bc + ca = c^{2} + 1 = x / 4.\\]  \n\nTaking advantage of symmetry, we discover that  \n\n\\[a + b = \\sqrt{\\frac{2x}{3}},\\quad b + c = \\sqrt{\\frac{x}{6}},\\quad \\mathrm{and}\\quad c + a = \\sqrt{\\frac{3x}{8}}.\\]  \n\nTo solve for \\(x\\) , notice that  \n\n\\[2 = 2(ab + bc + ca)\\] \\[\\quad = (a + b)^{2} + (b + c)^{2} + (c + a)^{2} - 2(a^{2} + b^{2} + c^{2})\\] \\[\\quad = \\frac{2x}{3} +\\frac{x}{6} +\\frac{3x}{8} -2\\left(\\frac{x}{2} -1 + \\frac{x}{3} -1 + \\frac{x}{4} -1\\right)\\] \\[\\quad = -\\frac{23x}{24} +6,\\]  \n\nso \\(x = \\frac{96}{23}\\) . Therefore,  \n\n\\[a + b + c = \\frac{1}{2}\\left(\\sqrt{\\frac{2x}{3}} +\\sqrt{\\frac{x}{6}} +\\sqrt{\\frac{3x}{8}}\\right)\\] \\[\\qquad = \\frac{1}{2}\\left(\\frac{8 + 4 + 6}{\\sqrt{23}}\\right) = \\left[\\frac{9}{\\sqrt{23}}\\right].\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n7. ", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Define \\(\\operatorname {sgn}(x)\\) to be 1 when \\(x\\) is positive, \\(-1\\) when \\(x\\) is negative, and 0 when \\(x\\) is 0. Compute  \n\n\\[\\sum_{n = 1}^{\\infty}\\frac{\\operatorname{sgn}(\\sin(2^{n}))}{2^{n}}.\\]  \n\n(The arguments to sin are in radians.)", "solution": "Answer: \\(\\boxed {1 - \\frac{2}{\\pi}}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nProposed by: Karthik Venkata Vedula  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Define \\(\\operatorname {sgn}(x)\\) to be 1 when \\(x\\) is positive, \\(-1\\) when \\(x\\) is negative, and 0 when \\(x\\) is 0. Compute  \n\n\\[\\sum_{n = 1}^{\\infty}\\frac{\\operatorname{sgn}(\\sin(2^{n}))}{2^{n}}.\\]  \n\n(The arguments to sin are in radians.)", "solution": "Note that each of following is equivalent to the next.  \n\n\\(\\cdot \\operatorname {sgn}(\\sin (2^{n})) = +1\\)  \n\n\\(\\cdot 0< 2^{n}\\bmod 2\\pi < \\pi\\)  \n\n\\(\\cdot 0< \\frac{2^{n}}{\\pi}\\bmod 2< 1\\)  \n\nThe \\(n\\) th digit after the decimal point in the binary representation of \\(\\frac{1}{\\pi}\\) is 0.  \n\nSimilarly, \\(\\operatorname {sgn}(\\sin (2^{n})) = - 1\\) if and only if the \\(n\\) - th digit after the decimal point in the binary representation of \\(\\frac{1}{\\pi}\\) is 1. In particular, if \\(a_{n}\\) is the \\(n\\) - th digit, then \\(\\operatorname {sgn}(\\sin (2^{n})) = 1 - 2a_{n}\\) . Thus, the desired sum is  \n\n\\[\\sum_{n = 1}^{\\infty}\\frac{\\operatorname{sgn}(\\sin(2^{n}))}{2^{n}} = \\sum_{n = 1}^{\\infty}\\frac{1 - 2a_{n}}{2^{n}} = \\left(\\sum_{n = 1}^{\\infty}\\frac{1}{2^{n}}\\right) - 2\\left(\\sum_{n = 1}^{\\infty}\\frac{a_{n}}{2^{n}}\\right) = \\left[\\frac{1 - \\frac{2}{\\pi}}{2}\\right].\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n8. ", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be the unique polynomial of degree at most 2026 such that for all \\(n\\in \\{1,2,3,\\ldots ,2027\\}\\)  \n\n  \n\nSuppose that \\(\\frac{a}{b}\\) is the coefficient of \\(x^{2025}\\) in \\(f\\) , where \\(a\\) and \\(b\\) are integers such that \\(\\gcd (a,b) = 1\\) . Compute the unique integer \\(r\\) between 0 and 2026 (inclusive) such that \\(a - rb\\) is divisible by 2027. (Note that 2027 is prime.)", "solution": "Answer: \\(\\boxed {1037}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nProposed by: Pitchayut Saengrungkonga  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be the unique polynomial of degree at most 2026 such that for all \\(n\\in \\{1,2,3,\\ldots ,2027\\}\\)  \n\n  \n\nSuppose that \\(\\frac{a}{b}\\) is the coefficient of \\(x^{2025}\\) in \\(f\\) , where \\(a\\) and \\(b\\) are integers such that \\(\\gcd (a,b) = 1\\) . Compute the unique integer \\(r\\) between 0 and 2026 (inclusive) such that \\(a - rb\\) is divisible by 2027. (Note that 2027 is prime.)", "solution": "Let \\(p = 2027\\) . We work in \\(\\mathbb{F}_{p}\\) for the entire solution. Recall the well- known fact that  \n\n  \n\nassuming \\(0^{0} = 1\\) . In particular, for any polynomial \\(g(x) = b_{0} + b_{1}x + \\dots +b_{n}x^{n}\\) , we have  \n\n\\[-\\sum_{x\\in \\mathbb{F}_{p}}g(x) = b_{p - 1} + b_{2(p - 1)} + \\dots +b_{\\lfloor n / (p - 1)\\rfloor (p - 1)}.\\]  \n\nWe apply this fact on \\(g(x) = xf(x)\\) . As \\(\\deg x f(x)\\leq p\\) , the right hand side is simply the coefficient of \\(x^{2025}\\) , which is what we want. Hence, the answer is  \n\n\\[-\\sum_{x\\in \\mathbb{F}_{p}}x f(x) = -(1^{2} + 2^{2} + \\dots +45^{2}) = -\\frac{45\\cdot 46\\cdot 91}{6}\\equiv \\boxed {1037}\\pmod {2027}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let \\(f\\) be the unique polynomial of degree at most 2026 such that for all \\(n\\in \\{1,2,3,\\ldots ,2027\\}\\)  \n\n  \n\nSuppose that \\(\\frac{a}{b}\\) is the coefficient of \\(x^{2025}\\) in \\(f\\) , where \\(a\\) and \\(b\\) are integers such that \\(\\gcd (a,b) = 1\\) . Compute the unique integer \\(r\\) between 0 and 2026 (inclusive) such that \\(a - rb\\) is divisible by 2027. (Note that 2027 is prime.)", "solution": "Again, let \\(p = 2027\\) and work in \\(\\mathbb{F}_{p}\\) . By the Lagrange Interpolation formula, we get that  \n\n\\[f(x) = \\sum_{i\\in \\mathbb{F}_{p}}f(i)\\prod_{j\\neq i}{\\frac{x - j}{i - j}}.\\]  \n\nWe now simplify the polynomial in the product sign on the right- hand side. First, recall the identity  \n\n\\[\\prod_{j\\in \\mathbb{F}_{p}}(x - j) = x^{p} - x = (x - i)^{p} - (x - i).\\]  \n\nThe denominator \\(\\prod_{j\\neq i}(i - j)\\) becomes \\((p - 1)! = - 1\\) by Wilson's. Thus, we get that  \n\n\\[\\prod_{j\\neq i}{\\frac{x - j}{i - j}} = -{\\frac{(x - i)^{p} - (x - i)}{x - i}} = -(x - i)^{p - 1} + 1.\\]  \n\nThe coefficient of \\(x^{p - 2}\\) in the above expression is \\(- i\\) . Therefore, the first equation gives that the coefficient of \\(x^{p - 2}\\) in \\(f(x)\\) is  \n\n\\[\\sum_{i\\in \\mathbb{F}_{p}} - if(i) = -(1^{2} + 2^{2} + \\dots +45^{2}) = -\\frac{45\\cdot 46\\cdot 91}{6}\\equiv \\boxed {1037}\\pmod {2027}.\\]", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n9. ", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct complex numbers such that  \n\n\\[a^{2} = b + 6,\\quad b^{2} = c + 6,\\quad \\mathrm{and}\\quad c^{2} = a + 6.\\]  \n\nCompute the two possible values of \\(a + b + c\\)", "solution": "Answer: \\(\\boxed {\\frac{- 1 + \\sqrt{17}}{2}},\\frac{- 1 - \\sqrt{17}}{2}\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nProposed by: Vasawat Rawangwong  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct complex numbers such that  \n\n\\[a^{2} = b + 6,\\quad b^{2} = c + 6,\\quad \\mathrm{and}\\quad c^{2} = a + 6.\\]  \n\nCompute the two possible values of \\(a + b + c\\)", "solution": "Notice that any of \\(a\\) , \\(b\\) , or \\(c\\) being 3 or \\(- 2\\) implies \\(a = b = c\\) , which is invalid. Thus,  \n\n\\[(a^{2} - 9)(b^{2} - 9)(c^{2} - 9) = (b - 3)(c - 3)(a - 3)\\implies (a + 3)(b + 3)(c + 3) = 1,\\] \\[(a^{2} - 4)(b^{2} - 4)(c^{2} - 4) = (b + 2)(c + 2)(a + 2)\\implies (a - 2)(b - 2)(c - 2) = 1.\\]  \n\nTherefore, 2 and \\(- 3\\) are roots of the polynomial \\((x - a)(x - b)(x - c) + 1\\) , and so there exists some \\(t\\) such that  \n\n\\[(x - t)(x - 2)(x + 3) = (x - a)(x - b)(x - c) + 1.\\]  \n\nComparing coefficients gives \\(a + b + c = t - 1\\) and \\(a b + b c + c a = - (t + 6)\\) . We can then solve for \\(t\\) by noting \\(a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = a + b + c + 18\\) , so  \n\n\\[a b + b c + c a = \\frac{1}{2} ((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})) = \\frac{1}{2} ((a + b + c)^{2} - (a + b + c + 18)).\\]  \n\nHence,  \n\n\\[- (t + 6) = \\frac{1}{2} ((t - 1)^{2} - (t + 17))\\Longrightarrow t^{2} - t - 4 = 0\\Longrightarrow t = \\frac{1\\pm\\sqrt{17}}{2}.\\]  \n\nTherefore \\(a + b + c = \\boxed {\\frac{- 1\\pm\\sqrt{17}}{2}}\\) are the two possible values of \\(a + b + c\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct complex numbers such that  \n\n\\[a^{2} = b + 6,\\quad b^{2} = c + 6,\\quad \\mathrm{and}\\quad c^{2} = a + 6.\\]  \n\nCompute the two possible values of \\(a + b + c\\)", "solution": "Let \\(s = a + b + c\\) . Subtracting two adjacent equations gives \\(a^{2} - b^{2} = b - c\\) , or \\((a - b)(a + b) = (b - c)\\) . Multiplying this and its cyclic variants gives  \n\n\\[(a + b)(b + c)(c + a) = 1.\\]\n\n\n\nNow, we recall the identity  \n\n\\[(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3(a + b)(b + c)(c + a)\\] \\[\\qquad \\Rightarrow \\qquad s^{3} = a^{3} + b^{3} + c^{3} + 3.\\]  \n\nTo simplify \\(a^{3} + b^{3} + c^{3}\\) , we add \\(a\\) times the first equation, \\(b\\) times the second, and \\(c\\) times the third to obtain  \n\n\\[a^{3} + b^{3} + c^{3} = a(b + 6) + b(c + 6) + c(a + 6)\\] \\[\\qquad = (a b + b c + c a) + 6 s\\] \\[\\qquad = \\frac{1}{2}\\Big((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})\\Big) + 6 s\\] \\[\\qquad = \\frac{1}{2} s^{2} - \\frac{1}{2}\\Big((b + 6) + (c + 6) + (a + 6)\\Big) + 6 s\\] \\[\\qquad = \\frac{1}{2} s^{2} + \\frac{11}{2} s - 9.\\]  \n\nTherefore,  \n\n\\[s^{3} = \\frac{1}{2} s^{2} + \\frac{11}{2} s - 6 \\Rightarrow \\left(s - \\frac{3}{2}\\right)\\left(s^{2} + s - 4\\right) = 0.\\]  \n\nAt this point, the only reasonable guess is that \\(s = \\frac{3}{2}\\) is an extra solution, and the remaining two roots \\(s = \\frac{- 1\\pm\\sqrt{17}}{2}\\) are the possible answers. We now justify this guess. Assume for sake of contradiction that \\(s = \\frac{3}{2}\\) . Then,  \n\n\\[a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = \\frac{39}{2\\] \\[ab + bc + ca = \\frac{1}{2}\\left(\\frac{9}{4} -\\frac{39}{2}\\right) = -\\frac{69}{8}.\\]  \n\nThen, observe  \n\n\\[a b c = (a + b + c)(a b + b c + c a) - (a + b)(b + c)(c + a)\\] \\[= -\\frac{207}{16} -1 = -\\frac{223}{16}.\\]  \n\nOn the other hand,  \n\n\\[(a + 6)(b + 6)(c + 6) = 216 + 36(a + b + c) + 6(ab + bc + ca) + abc\\] \\[\\qquad = 216 + 36\\cdot \\frac{3}{2} -6\\cdot \\frac{69}{8} -\\frac{223}{16},\\]  \n\nwhich is a rational number of denominator 16. But \\((a + 6)(b + 6)(c + 6) = b^{2}c^{2}a^{2} = \\left(-\\frac{223}{16}\\right)^{2}\\) has denominator \\(16^{2} = 256\\) , a contradiction. Thus \\(s = \\frac{3}{2}\\) is impossible. (It arises from \\(a = b = c = \\frac{1}{2}\\) which satisfies \\((a + b)(b + c)(c + a) = 1\\) but not the given conditions.)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct complex numbers such that  \n\n\\[a^{2} = b + 6,\\quad b^{2} = c + 6,\\quad \\mathrm{and}\\quad c^{2} = a + 6.\\]  \n\nCompute the two possible values of \\(a + b + c\\)", "solution": "Subtracting any two adjacent equations gives \\(a^{2} - b^{2} = b - c\\) , which is equivalent to both \\((a - b)(a + b) = (b - c)\\) and \\((a - b)(a + b + 1) = (a - c)\\) . Multiplying each of these with its respective cyclic variants and canceling the \\((a - b)(b - c)(c - a)\\) factor (which is given to be nonzero), we get  \n\n\\[(a + b)(b + c)(c + a) = 1\\quad \\mathrm{and}\\quad (a + b + 1)(b + c + 1)(c + a + 1) = -1.\\]  \n\nExpanding the latter equation and using the given equations gives the following result.  \n\n\\[(a + b)(b + c)(c + a) + (a^{2} + b^{2} + c^{2}) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1\\] \\[1 + (b + 6 + c + 6 + a + 6) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1\\] \\[3(a + b + c) + 3(ab + bc + ca) = -21\\]  \n\n\\[a + b + c + ab + bc + ca = -7.\\]\n\n\n\nLet \\(s = a + b + c\\) . We can then solve for \\(s\\) by considering the following:  \n\n\\[s^{2} = (a^{2} + b^{2} + c^{2}) + 2(ab + bc + ca)\\] \\[\\quad = (b + 6 + c + 6 + a + 6) + 2(-7 - a - b - c)\\] \\[\\quad = -s + 4,\\]  \n\nso \\(s = \\frac{- 1\\pm\\sqrt{17}}{2}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 3: "}}
+{"year": "2025", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let \\(a\\) , \\(b\\) , and \\(c\\) be pairwise distinct complex numbers such that  \n\n\\[a^{2} = b + 6,\\quad b^{2} = c + 6,\\quad \\mathrm{and}\\quad c^{2} = a + 6.\\]  \n\nCompute the two possible values of \\(a + b + c\\)", "solution": "Let \\(s = a + b + c\\) and consider the polynomial  \n\n\\[x + (x^{2} - 6) + ((x^{2} - 6)^{2} - 6) - s = x^{4} - 11x^{2} + x + 24 - s.\\]  \n\nThis polynomial has roots \\(a\\) , \\(b\\) , and \\(c\\) . By Vieta's, the sum of all four roots is 0, so its fourth root must be \\(- s\\) . Using Vieta's again, we have \\(ab + bc + ca - sa - sb - sc = - 11\\) . We can now solve for \\(s\\) .  \n\n\\[a b + b c + c a - (a + b + c)^{2} = -11\\] \\[a^{2} + b^{2} + c^{2} + a b + b c + c a = 11\\] \\[\\frac{1}{2} ((a + b + c)^{2} + (a^{2} + b^{2} + c^{2})) = 11\\] \\[(a + b + c)^{2} + (b + 6 + c + 6 + a + 6) = 22\\] \\[s^{2} + s - 4 = 0\\Longrightarrow s = \\frac{-1\\pm\\sqrt{17}}{2}.\\]  \n\nRemark. Another way to finish using this approach is to substitute \\(- s\\) directly into \\(x^{4} - 11x^{2} + x + 24 - s = 0\\) to get \\((s - 3)(s + 2)(x^{2} + x - 4) = 0\\) , then discard the solutions \\(s = 3\\) and \\(s = - 2\\) , which arise from the invalid values \\(a = b = c = 3\\) and \\(a = b = c = - 2\\) . (In the invalid cases, \\(s \\neq a + b + c\\) because \\(a = b = c\\) is only a single root to the polynomial.)", "metadata": {"resource_path": "HarvardMIT/segmented/en-282-2025-feb-algnt-solutions.jsonl", "problem_match": "\n10. ", "solution_match": "\nSolution 4: "}}

HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl

@@ -0,0 +1,12 @@
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once.  \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": 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"metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nProposed by: Albert Wang  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once.  \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "Let \\(d(P, X Y)\\) be the distance from \\(P\\) to line \\(X Y\\) . We will first prove \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) intersect at most once.  \n\nClaim 1. Let \\(\\ell_{A B}\\) be the perpendicular bisector of \\(A B\\) . Then, \\(\\mathcal{P}_{A}\\) is tangent to \\(\\ell_{A B}\\) .  \n\nProof. Consider any point \\(X\\) on \\(\\mathcal{P}_{A}\\) . Then,  \n\n\\[X A = d(X,B D)\\leq X B,\\]  \n\nwith equality only holding at the unique point \\(X\\) for which \\(X B\\perp B D\\) . Thus, \\(\\mathcal{P}_{A}\\) lies entirely on one side of \\(\\ell_{A B}\\) , touching it once at this point \\(X\\) . \\(\\square\\)  \n\nIt follows that \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) are on different sides of \\(\\ell_{A B}\\) and hence can intersect at most once (possibly at a common tangency point to \\(\\ell_{A B}\\) ).  \n\nSince \\(A B C D\\) is convex, \\(A\\) and \\(C\\) lie on opposite sides of line \\(B D\\) , the common directrix of \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) . Thus, \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) lie on opposite sides of line \\(B D\\) and cannot intersect at all.  \n\nThe remaining pairs of parabolas are handled similarly.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let \\(A B C D\\) be a convex quadrilateral. Define parabolas \\(\\mathcal{P}_{A}\\) \\(\\mathcal{P}_{B}\\) \\(\\mathcal{P}_{C}\\) , and \\(\\mathcal{P}_{D}\\) to have directrices \\(B D\\) \\(C A\\) \\(D B\\) , and \\(A C\\) , and foci \\(A\\) \\(B\\) \\(C\\) , and \\(D\\) , respectively. Prove that no two of these parabolas intersect more than once.  \n\n(A parabola with directrix \\(\\ell\\) and focus \\(P\\) consists of all points \\(X\\) for which \\(P X\\) equals the distance from \\(P\\) to \\(\\ell\\) .)", "solution": "Let \\(d(P, X Y)\\) be the distance from \\(P\\) to line \\(X Y\\) .  \n\nClaim 2. \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{B}\\) intersect at most once.  \n\nProof. Let \\(P\\) be a common point of both parabolas. Then, \\(d(P, B D) = P A\\) and \\(d(P, A C) = P B\\) , which combined imply  \n\n\\[P A = d(P,B D)\\leq P B = d(P,A C)\\leq P A.\\]  \n\nThus, the inequalities above are equalities, i.e., \\(P A = P B\\) , \\(P A\\perp A C\\) , and \\(P B\\perp B D\\) . Such \\(P\\) , if it exists, is unique. \\(\\square\\)\n\n\n\nThe remaining pairs of parabolas are handled similarly. As in solution 1, \\(\\mathcal{P}_{A}\\) and \\(\\mathcal{P}_{C}\\) cannot intersect, nor can \\(\\mathcal{P}_{B}\\) and \\(\\mathcal{P}_{D}\\) .", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n1. [5]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) ,  \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "Answer: All constant functions and \\(P(x) = a x + b\\pi\\) for all nonzero integers \\(a\\) and integers \\(b\\)", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nProposed by: Karthik Venkata Vedula  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) ,  \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "It is well- known that these polynomials work by taking \\(Q\\) to be a Chebyshev polynomial (if \\(P\\) is linear) or a constant (if \\(P\\) is constant).  \n\nSuppose that \\(\\deg P \\geq 2\\) . Now consider the density of the roots of \\(\\cos (P(t))\\) , i.e.  \n\n\\[\\lim_{n \\to \\infty} \\frac{\\text{number of roots in the interval} [-n, n]}{n}.\\]  \n\nSince \\(\\cos (P(t)) = Q(\\cos t)\\) , the density is finite, because for each interval of length \\(2\\pi\\) , there can only be a finite number of roots (i.e. twice the degree of \\(Q\\) ). However, we claim that \\(\\cos (P(t))\\) has an infinite density of roots. In particular, consider the solutions to \\(P(x) = \\pm (2k - 1)\\pi /2\\) over positive integers \\(k\\) . Asymptotically, for large \\(k\\) , such \\(x\\) will always exist and be \\(\\Theta (k^{1 / \\deg P})\\) . As \\(k \\to \\infty\\) , such \\(x\\) become infinitely dense, contradicting the finite density of roots of \\(Q(\\cos t)\\) . Therefore, \\(\\deg P \\leq 1\\) .  \n\nIf \\(\\deg P = 1\\) , let \\(P(t) = at + b\\) . Observe that \\(Q(\\cos t)\\) is periodic with period \\(2\\pi\\) , so \\(\\cos (P(t))\\) must be as well. This is only the case when \\(a\\) is an integer. Furthermore,  \n\n\\[Q(\\cos t) = \\cos (at + b) = \\cos (at)\\cos (b) - \\sin (at)\\sin (b)\\]  \n\nmust be an even function. Note that \\(\\cos (at)\\cos (b)\\) is even and \\(\\sin (at)\\sin (b)\\) is odd, so \\(\\sin (at)\\sin (b) = 0\\) for all \\(t\\) , and hence \\(b\\) is an integer multiple of \\(\\pi\\) .  \n\nThus, the only polynomials that work are the ones claimed above.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nSolution 1: "}}
+{"year": "2025", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Find all polynomials \\(P\\) with real coefficients for which there exists a polynomial \\(Q\\) with real coefficients such that for all real \\(t\\) ,  \n\n\\[\\cos (P(t)) = Q(\\cos t).\\]", "solution": "Taking the derivative of both sides,  \n\n\\[\\sin (P(t))P^{\\prime}(t) = (- \\sin t)Q^{\\prime}(\\cos t).\\]  \n\nThe right- hand side is bounded in \\(t\\) , so the left- hand side must also be bounded. If \\(\\deg P \\geq 2\\) , then as \\(t\\) approaches \\(\\infty\\) , \\(P^{\\prime}(t)\\) approaches \\(\\pm \\infty\\) and \\(\\sin (P(t))\\) does not approach 0, contradiction. Thus, \\(\\deg P \\leq 1\\) , and we finish as before.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n2. [7]", "solution_match": "\nSolution 2: "}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(O\\) be a point inside \\(ABCD\\) . Suppose the circumcircles of triangles \\(OAB\\) and \\(OCD\\) intersect at \\(P \\neq O\\) , and the circumcircles of triangles \\(OBC\\) and \\(OAD\\) intersect at \\(Q \\neq O\\) . Prove \\(\\angle POQ\\) equals one of the angles of quadrilateral \\(ABCD\\) .", "solution": 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"metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n3. [8]", "solution_match": "\nProposed by: Derek Liu\n\n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let \\(ABCD\\) be a parallelogram, and let \\(O\\) be a point inside \\(ABCD\\) . Suppose the circumcircles of triangles \\(OAB\\) and \\(OCD\\) intersect at \\(P \\neq O\\) , and the circumcircles of triangles \\(OBC\\) and \\(OAD\\) intersect at \\(Q \\neq O\\) . Prove \\(\\angle POQ\\) equals one of the angles of quadrilateral \\(ABCD\\) .", "solution": "In what follows, all angles are directed.  \n\nClaim 1. The points \\(P\\) and \\(Q\\) are symmetric over the center of \\(ABCD\\) .  \n\nProof. Note that  \n\n\\[\\angle A P B = \\angle A O B = \\angle O A D + \\angle C B O = \\angle O Q D + \\angle C Q O = \\angle C Q D.\\]  \n\nSimilar equalities hold for each pair of opposite sides, so \\(P\\) and \\(Q\\) are symmetric across the parallelogram's center. \\(\\square\\)  \n\nConsequently, \\(AP\\) and \\(CQ\\) are parallel, so  \n\n\\[\\angle P O Q = \\angle P O B + \\angle B O Q = \\angle P A B + \\angle B C Q = \\angle C B A,\\]  \n\nas desired. (Once we undirect the angles, \\(\\angle P O Q\\) is either \\(\\angle B\\) or \\(\\pi - \\angle B = \\angle A\\) .)", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n3. [8]", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Determine whether there exist infinitely many pairs of distinct positive integers \\(m\\) and \\(n\\) such that \\(2^{m} + n\\) divides \\(2^{n} + m\\) .", "solution": "Answer: Yes", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n4. [9]", "solution_match": "\nProposed by: Carlos Rodriguez, Jordan Lefkowitz  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Determine whether there exist infinitely many pairs of distinct positive integers \\(m\\) and \\(n\\) such that \\(2^{m} + n\\) divides \\(2^{n} + m\\) .", "solution": "Let \\(k\\) be a positive integer, and set \\(m = 2^{k}\\) and \\(n = p - 2^{2^{k}}\\) for prime \\(p\\) to be chosen later. We want \\(2^{m} + n = p\\) to divide \\(2^{p - 2^{2^{k}}} + 2^{k}\\) , which is equivalent to having  \n\n\\[0\\equiv 2^{p - 2^{2^{k}}} + 2^{k}\\equiv 2^{1 - 2^{2^{k}}} + 2^{k}\\equiv 2^{1-2^{2^{k}}}\\left(2^{2^{2^{k}} + k - 1} + 1\\right)\\pmod {p}.\\]  \n\nLet \\(r = 2^{2^{k}} + k - 1\\) . Since \\(r \\neq 3\\) , by Zsigmondy, we can pick a prime \\(p\\) that divides \\(2^{r} + 1\\) but not \\(2^{s} + 1\\) for any nonnegative integer \\(s < r\\) . Let \\(d = \\operatorname{ord}_{p}(2)\\) . Then, \\(2^{|d - r|} \\equiv -1 \\pmod {p}\\) , so by definition of \\(p\\) , we have \\(|d - r| \\geq r\\) . Hence, \\(d \\geq 2r\\) . As \\(d \\mid p - 1\\) , we conclude  \n\n\\[p > 2r = 2\\left(2^{2^{k}} + k - 1\\right) > 2^{2^{k}} + 2^{k},\\]  \n\nso \\(n = p - 2^{2^{k}} > m\\) . Since \\(p \\mid 2^{r} + 1\\) by definition, \\((m, n)\\) is a pair of distinct positive integers with \\(2^{m} + n \\mid 2^{n} + m\\) .  \n\nAs \\(k\\) was arbitrary (and \\(m = 2^{k}\\) ), there exist infinitely many such pairs.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n4. [9]", "solution_match": "\nSolution: "}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the smallest positive integer \\(k > 45\\) for which there exists a sequence \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{k - 1}\\) of positive integers satisfying the following conditions:  \n\n\\(a_{i} = i\\) for all integers \\(1\\leq i\\leq 45\\) \\(a_{k - i} = i\\) for all integers \\(1\\leq i\\leq 45\\) , and for any odd integer \\(1\\leq n\\leq k - 45\\) , the sequence \\(a_{n}\\) , \\(a_{n + 1}\\) , ..., \\(a_{n + 44}\\) is a permutation of \\(\\{1,2,\\ldots ,45\\}\\) .", "solution": "Answer: 1059", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n5. [13]", "solution_match": "\nProposed by: Derek Liu  \n\n"}}
+{"year": "2025", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Compute the smallest positive integer \\(k > 45\\) for which there exists a sequence \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{k - 1}\\) of positive integers satisfying the following conditions:  \n\n\\(a_{i} = i\\) for all integers \\(1\\leq i\\leq 45\\) \\(a_{k - i} = i\\) for all integers \\(1\\leq i\\leq 45\\) , and for any odd integer \\(1\\leq n\\leq k - 45\\) , the sequence \\(a_{n}\\) , \\(a_{n + 1}\\) , ..., \\(a_{n + 44}\\) is a permutation of \\(\\{1,2,\\ldots ,45\\}\\) .", "solution": "First, we show 1059 is optimal. Assume for sake of contradiction that \\(k< 1059\\)  \n\nThe given condition ensures that \\(\\{a_{1},a_{2}\\} = \\{a_{46},a_{47}\\}\\) , \\(\\{a_{3},a_{4}\\} = \\{a_{48},a_{49}\\}\\) , and so on. In particular, if \\(a_{i} = j\\) , the next appearance of \\(j\\) must either be \\(a_{i + 44}\\) , \\(a_{i + 45}\\) , or \\(a_{i + 46}\\) ; working modulo 45, these indices all differ from \\(i\\) by at most 1. Furthermore, \\(a_{i} = a_{i + 44}\\) is only possible if \\(i\\) is even, and \\(a_{i} = a_{i + 46}\\) is only possible if \\(i\\) is odd.  \n\nAlso, \\(a_{1}\\neq a_{45}\\) by definition. Thus, if the sequence contains the same number at least 25 times, the 25th appearance has index at least \\(2 + 24\\cdot 44 = 1058\\) , implying \\(k\\geq 1059\\) . Hence, we can assume every number appears at most 24 times in the sequence.  \n\nObserve that there exists a unique integer \\(1\\leq j\\leq 45\\) such that  \n\n\\[(k - j) - j = k - 2j\\equiv 22\\pmod {45}.\\]  \n\nLet \\(k_{1} = j\\) , \\(k_{2}\\) , \\(k_{3}\\) , ..., \\(k_{\\ell} = k - j\\) be the indices of where \\(j\\) appears in the sequence; as assumed above, \\(\\ell \\leq 24\\) . For any \\(i\\) , we proved \\(k_{i} - k_{i - 1}\\) is either 44, 45, or 46, and thus either 0 or \\(\\pm 1\\) modulo 45. Since \\(k - j\\) and \\(j\\) differ by \\(22\\equiv - 23\\) modulo 45, either \\(k_{i} - k_{i - 1} = 46\\) for at least 22 different \\(i\\) , or \\(k_{i} - k_{i - 1} = 44\\) for at least 23 different \\(i\\) . We split into cases based on which.  \n\nIf \\(k_{i} - k_{i - 1} = 46\\) at least 22 times, we claim \\(\\ell = 23\\) . Indeed, if \\(\\ell = 24\\) , then  \n\n\\[(k - j)\\geq j + 22\\cdot 46 + 44 = j + 1056.\\]  \n\nSince \\((k - j) - j\\equiv 22\\) (mod 45), we actually have \\((k - j)\\geq j + 1057\\) , so \\(k\\geq 2j + 1057\\geq 1059\\) , contradiction.  \n\nThus, \\(\\ell = 23\\) , so \\(k_{i} - k_{i - 1} = 46\\) for all \\(i\\) , and \\(k = 2j + 22\\cdot 46 = 2j + 1012\\) , which means we can assume \\(j\\leq 23\\) (otherwise \\(k > 1059\\) ).  \n\nSince \\(a_{j} = a_{j + 46}\\) , we must also have \\(a_{j + 1} = a_{j + 45}\\) be the second appearance of \\(j + 1\\) , which means \\(j\\) must be odd. Then, as \\(j + 45\\) is even, \\(a_{j + 45} = j + 1\\) must either be equal to either \\(a_{j + 89}\\) or \\(a_{j + 90}\\) . Now, \\(a_{k - j - 1} = j + 1\\) as well, and  \n\n\\[(k - j - 1) - (j + 90)\\equiv 21\\pmod {45},\\]  \n\nso if \\(a_{j + 90}\\) is the 3rd appearance of \\(j + 1\\) , then \\(a_{k - j - 1}\\) must be at least the 24th. Thus, it must be exactly the 24th, with each appearance of \\(j + 1\\) being exactly 46 terms after the previous. Thus,  \n\n\\[k - j - 1\\geq (j + 90) + 21\\cdot 46 = j + 1056,\\]  \n\nand \\(k\\geq 2j + 1057\\geq 1059\\) , as desired. If \\(a_{j + 89}\\) is the 3rd appearance of \\(j + 1\\) , then since \\((k - j - 1) - (j + 89)\\equiv 22\\) (mod 45), it follows that \\(a_{k - j - 1}\\) must be at least the 25th appearance, contradiction.  \n\nThe other case, where \\(k_{i} - k_{i - 1} = 44\\) at least 23 times (i.e., for all \\(i\\) ) plays out similarly. Since \\(k_{2} = k_{1} + 44\\) , we immediately have \\(j = k_{1}\\) is even. Furthermore, \\(a_{j - 1} = a_{j + 45} = j - 1\\) is the second appearance of \\(j - 1\\) , and since \\(j + 45\\) is odd, the third appearance is either \\(a_{j + 90}\\) or \\(a_{j + 91}\\) . However,  \n\n\\[(k - j + 1) - (j + 90)\\equiv -22\\pmod {45},\\]\n\n\n\nso regardless of whether \\(a_{j + 90}\\) or \\(a_{j + 91}\\) is the 3rd appearance of \\(j - 1\\) , we know \\(a_{k - j + 1}\\) must be at least the 25th, contradiction.  \n\nThus \\(k \\geq 1059\\) , and it suffices to provide a construction.  \n\nGiven an integer \\(m\\) , let \\(\\alpha\\) represent the permutation on \\(\\{1,2,\\ldots ,m\\}\\) given by swapping 1 and 2, 3 and 4, etc. and let \\(\\beta\\) represent the permutation given by swapping 2 and 3, 4 and 5, etc.  \n\nClaim 1. Both of the products \\(\\dots \\beta \\alpha \\beta \\alpha\\) and \\(\\dots \\alpha \\beta \\alpha \\beta\\) , with exactly \\(m\\) terms in the products, equal the reverse permutation.  \n\nProof. If we track any number \\(i\\) , observe that if \\(i\\) is odd,  \n\n\\[\\alpha (i) = i + 1\\] \\[\\beta \\alpha (i) = i + 2\\] \\[\\vdots\\] \\[\\dots \\beta \\alpha (i) = m\\] \\[\\underset {m - i}{\\underbrace{\\dots \\beta \\alpha (i) = m}}\\] \\[\\underset {m - i + 1}{\\underbrace{\\dots \\beta \\alpha (i) = m - 1}}\\] \\[\\underset {m - i + 2}{\\underbrace{\\dots \\beta \\alpha (i) = m - i + 1}}.\\]  \n\nA similar chain occurs if \\(i\\) is even, so \\(\\dots \\beta \\alpha\\) is indeed the reverse permutation. The same argument shows that the product \\(\\dots \\alpha \\beta\\) is also the reverse permutation. \\(\\square\\)  \n\nOur construction now goes as follows: for any odd \\(n\\) , let \\(a_{n + 46} = a_{n}\\) and \\(a_{n + 45} = a_{n + i}\\) unless \\(n \\equiv 0\\) or 23 mod 45. Then, we can note that \\(a_{46}\\) through \\(a_{68}\\) are the permutation \\(\\alpha\\) on \\(\\{1,2,\\ldots ,23\\}\\) , and \\(a_{91}\\) through \\(a_{113}\\) are \\(\\alpha \\beta\\) , etc. By our claim, \\(a_{1 + 23\\cdot 45} = a_{1036}\\) through \\(a_{1058}\\) are 1 through 23 reversed. Similarly, as there are only 22 numbers from 24 through 45, we know \\(a_{24 + 22\\cdot 45} = a_{1014}\\) through \\(a_{1035}\\) are 24 through 45 reversed. It follows that this sequence satisfies the above conditions for \\(k = \\left\\lfloor 1059\\right\\rfloor\\) .  \n\nThe diagram below shows what the construction would look like when 45 is replaced with 9. The sequence can be read row- by- row. The vertical line separates 1 through 5 from 6 through 9.  \n\n\\[1 2 3 4 5 6 7 8 9\\] \\[2 1 4 3 5 6 8 7 9\\] \\[2 4 1 5 3 8 6 9 7\\] \\[4 2 5 1 3 8 9 6 7\\] \\[4 5 2 3 1 9 8 7 6\\] \\[5 4 3 2 1\\]  \n\nRemark. The exact same proof and construction work for all odd \\(n \\geq 5\\) , yielding an answer of \\(\\frac{n^2 + 2n + 3}{2}\\) . Notably, for \\(n = 3\\) the answer is 6 instead of 9.", "metadata": {"resource_path": "HarvardMIT/segmented/en-284-tournaments-2025-hmic-solutions.jsonl", "problem_match": "\n5. [13]", "solution_match": "\nSolution: "}}