{"year": "2015", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}$ be a sequence of real numbers, and let $m$ be a fixed positive integer less than $n$. We say an index $k$ with $1 \\leq k \\leq n$ is good if there exists some $\\ell$ with $1 \\leq \\ell \\leq m$ such that $$ a_{k}+a_{k+1}+\\cdots+a_{k+\\ell-1} \\geq 0 $$ where the indices are taken modulo $n$. Let $T$ be the set of all good indices. Prove that $$ \\sum_{k \\in T} a_{k} \\geq 0 $$", "solution": " First we prove the result if the indices are not taken modulo $n$. Call a number $\\ell$-good if $\\ell$ is the smallest number such that $a_{k}+a_{k+1}+\\cdots+a_{k+\\ell-1} \\geq 0$, and $\\ell \\leq m$. Then if $a_{k}$ is $\\ell$-good, the numbers $a_{k+1}, \\ldots, a_{k+\\ell-1}$ are good as well. Then by greedy from left to right, we can group all the good numbers into blocks with nonnegative sums. Repeatedly take the first good number, if $\\ell$-good, group it with the next $\\ell$ numbers. An example for $m=3$ : $$ \\langle 4\\rangle \\quad \\begin{array}{ccccccccc} -1 & -2 & 6\\rangle & -9 & -7 & \\langle 3\\rangle & \\langle-2 & 4\\rangle & \\langle-1 . \\end{array} $$ We can now return to the original problem. Let $N$ be a large integer; applying the algorithm to $N$ copies of the sequence, we deduce that $$ N \\sum_{k \\in T} a_{k}+c_{N} \\geq 0 $$ where $c_{N}$ represents some \"error\" from left-over terms. As $\\left|c_{N}\\right| \\leq \\sum\\left|a_{i}\\right|$, by taking $N$ large enough we deduce the problem.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $A B C$ be a scalene triangle. Let $K_{a}, L_{a}$, and $M_{a}$ be the respective intersections with $B C$ of the internal angle bisector, external angle bisector, and the median from $A$. The circumcircle of $A K_{a} L_{a}$ intersects $A M_{a}$ a second time at a point $X_{a}$ different from $A$. Define $X_{b}$ and $X_{c}$ analogously. Prove that the circumcenter of $X_{a} X_{b} X_{c}$ lies on the Euler line of $A B C$.", "solution": " The main content of the problem: Claim - $\\angle H X_{a} G=90^{\\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. First proof by Brokard. To do this, it suffices to show that $M$ has the same power with respect to the circle with diameter $\\overline{A H}$ and the circle with diameter $\\overline{K L}$. In fact I claim both circles are orthogonal to the circle with diameter $\\overline{B C}$ ! The former follows from Brokard's theorem, noting that $A$ is on the polar of $H$, and the latter follows from the harmonic bundle. ![](https://cdn.mathpix.com/cropped/2024_11_19_8fb7b517def5801eef36g-04.jpg?height=635&width=1109&top_left_y=1410&top_left_x=482) Then $\\overline{A M}$ is the radical axis, so $X$ lies on both circles.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $A B C$ be a scalene triangle. Let $K_{a}, L_{a}$, and $M_{a}$ be the respective intersections with $B C$ of the internal angle bisector, external angle bisector, and the median from $A$. The circumcircle of $A K_{a} L_{a}$ intersects $A M_{a}$ a second time at a point $X_{a}$ different from $A$. Define $X_{b}$ and $X_{c}$ analogously. Prove that the circumcenter of $X_{a} X_{b} X_{c}$ lies on the Euler line of $A B C$.", "solution": " The main content of the problem: Claim - $\\angle H X_{a} G=90^{\\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. Second proof by orthocenter reflection, Bendit Chan. As before, we know $M X \\cdot M A=$ $M K \\cdot M L=M B \\cdot M C$, but $X$ lies inside segment $A M$. Construct parallelogram $A B A^{\\prime} C$. Then $M X \\cdot M A^{\\prime}=M B \\cdot M C$, so $X B A^{\\prime} C$ is concyclic. However, it is well-known the circumcircle of $\\triangle B A^{\\prime} C$ (which is the reflection of $(A B C)$ across $\\overline{B C}$ ) passes through $H$ and in fact has diameter $\\overline{A^{\\prime} H}$. So this gives $\\angle H X A^{\\prime}=90^{\\circ}$ as needed.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $A B C$ be a scalene triangle. Let $K_{a}, L_{a}$, and $M_{a}$ be the respective intersections with $B C$ of the internal angle bisector, external angle bisector, and the median from $A$. The circumcircle of $A K_{a} L_{a}$ intersects $A M_{a}$ a second time at a point $X_{a}$ different from $A$. Define $X_{b}$ and $X_{c}$ analogously. Prove that the circumcenter of $X_{a} X_{b} X_{c}$ lies on the Euler line of $A B C$.", "solution": " The main content of the problem: Claim - $\\angle H X_{a} G=90^{\\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. Third proof by barycentric coordinates. Alternatively we may just compute $X=\\left(a^{2}\\right.$ : $\\left.2 S_{A}: 2 S_{A}\\right)$. Let $F=\\left(0: S_{C}: S_{B}\\right)$ be the foot from $H$. Then we check that $X H F M$ is cyclic, which is power of a point from $A$.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $P$ be the set of all primes, and let $M$ be a non-empty subset of $P$. Suppose that for any non-empty subset $\\left\\{p_{1}, p_{2}, \\ldots, p_{k}\\right\\}$ of $M$, all prime factors of $p_{1} p_{2} \\ldots p_{k}+1$ are also in $M$. Prove that $M=P$.", "solution": " Obviously $|M|=\\infty$. Assume for contradiction $p \\notin M$. We say a prime $q \\in M$ is sparse if there are only finitely many elements of $M$ which are $q(\\bmod p)$ (in particular there are finitely many sparse primes). Now let $C$ be the product of all sparse primes (note $p \\nmid C$ ). First, set $a_{0}=1$. For $k \\geq 0$, consider then the prime factorization of the number $$ C a_{k}+1 $$ No prime in its factorization is sparse, so consider the number $a_{k+1}$ obtained by replacing each prime in its factorization with some arbitrary representative of that prime's residue class. In this way we select a number $a_{k+1}$ such that - $a_{k+1} \\equiv C a_{k}+1(\\bmod p)$, and - $a_{k+1}$ is a product of distinct primes in $M$. In particular, $a_{k} \\equiv C^{k}+C^{k-1}+\\cdots+1(\\bmod p)$ But since $C \\not \\equiv 0(\\bmod p)$, we can find a $k$ such that $a_{k} \\equiv 0(\\bmod p)($ namely, $k=p-1$ if $C \\equiv 1$ and $k=p-2$ else) which is clearly impossible since $a_{k}$ is a product of primes in $M$ !", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $x, y, z$ be real numbers (not necessarily positive) such that $x^{4}+y^{4}+z^{4}+x y z=4$. Prove that $x \\leq 2$ and $$ \\sqrt{2-x} \\geq \\frac{y+z}{2} $$", "solution": " We prove that the condition $x^{4}+y^{4}+z^{4}+x y z=4$ implies $$ \\sqrt{2-x} \\geq \\frac{y+z}{2} $$ We first prove the easy part. Claim - We have $x \\leq 2$. $$ \\begin{aligned} 5=x^{4}+y^{4}+\\left(z^{4}+1\\right)+x y z & =\\frac{3 x^{4}}{4}+\\left(\\frac{x^{4}}{4}+y^{4}\\right)+\\left(z^{4}+1\\right)+x y z \\\\ & \\geq \\frac{3 x^{4}}{4}+x^{2} y^{2}+2 z^{2}+x y z . \\end{aligned} $$ We evidently have that $x^{2} y^{2}+2 z^{2}+x y z \\geq 0$ because the quadratic form $a^{2}+a b+2 b^{2}$ is positive definite, so $x^{4} \\leq \\frac{20}{3} \\Longrightarrow x \\leq 2$. Now, the desired statement is implied by its square, so it suffices to show that $$ 2-x \\geq\\left(\\frac{y+z}{2}\\right)^{2} $$ We are going to proceed by contradiction (it seems that many solutions do this) and assume that $$ 2-x<\\left(\\frac{y+z}{2}\\right)^{2} \\Longleftrightarrow 4 x+y^{2}+2 y z+z^{2}>8 $$ By AM-GM, $$ \\begin{aligned} x^{4}+3 & \\geq 4 x \\\\ \\frac{y^{4}+1}{2} & \\geq y^{2} \\\\ \\frac{z^{4}+1}{2} & \\geq z^{2} \\end{aligned} $$ which yields that $$ x^{4}+\\frac{y^{4}+z^{4}}{2}+2 y z+4>8 $$ If we replace $x^{4}=4-\\left(y^{4}+z^{4}+x y z\\right)$ now, this gives $$ -\\frac{y^{4}+z^{4}}{2}+(2-x) y z>0 \\Longrightarrow(2-x) y z>\\frac{y^{4}+z^{4}}{2} $$ Since $2-x$ and the right-hand side are positive, we have $y z \\geq 0$. Now $$ \\frac{y^{4}+z^{4}}{2 y z}<2-x<\\left(\\frac{y+z}{2}\\right)^{2} \\Longrightarrow 2 y^{4}+2 z^{4}2015$ we're done.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}} {"year": "2015", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "USA_TSTST", "problem": "Let $\\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\\varphi(n)=m$ has at least 2015 solutions in $n$.", "solution": " 【 Second solution with smallest primes, by Yang Liu. Let $2=p_{1}1$. Then Alice can complete the phase without losing if and only if $x$ is even; if so she begins a $Y$-phase with $(X, Y)=(0, x / 2)$. Now note that unless $X=0$, Bob now has a winning move WrongX. Conversely he may only play Sleep if $X=0$. We have an analogous claim for $Y$-phases. Thus if $n$ is not a power of 2 , we see that Alice eventually loses. Now suppose $n=2^{k}$; then Alice reaches $(X, Y)=\\left(0,2^{k-1}\\right),\\left(2^{k-2}, 0\\right), \\ldots$ until either reaching $(1,0)$ or $(0,1)$. At this point she can play ClaimX or ClaimY, respectively; the game is now in state Cl. Bob cannot play either FakeX or FakeY, so he must play Sleep, and then Alice wins by playing Win. Thus Alice has a winning strategy when $n=2^{k}$.", "metadata": {"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl"}}