{"year":2021,"label":"1","problem":"Determine all integers $s \\geq 4$ for which there exist positive integers $a, b, c, d$ such that $s=a+b+c+d$ and $s$ divides $a b c+a b d+a c d+b c d$.","solution":"  The answer is $s$ composite.  Composite construction. Write $s=(w+x)(y+z)$, where $w, x, y, z$ are positive integers. Let $a=w y, b=w z, c=x y, d=x z$. Then  $$ a b c+a b d+a c d+b c d=w x y z(w+x)(y+z) $$  so this works.   $$ (a+b)(a+c)(a+d)=(a b c+a b d+a c d+b c d)+a^{2}(a+b+c+d) \\equiv 0 \\quad(\\bmod s) $$  Hence $s$ divides a product of positive integers less than $s$, so $s$ is composite. Suppose that $s$ is prime. Then the polynomial $(x-a)(x-b)(x-c)(x-d) \\in \\mathbb{F}_{s}[x]$ is even, so the roots come in two opposite pairs in $\\mathbb{F}_{s}$. Thus the sum of each pair is at least $s$, so the sum of all four is at least $2 s>s$, contradiction."}
{"year":2021,"label":"2","problem":"Points $A, V_{1}, V_{2}, B, U_{2}, U_{1}$ lie fixed on a circle $\\Gamma$, in that order, and such that $B U_{2}>A U_{1}>B V_{2}>A V_{1}$. Let $X$ be a variable point on the $\\operatorname{arc} V_{1} V_{2}$ of $\\Gamma$ not containing $A$ or $B$. Line $X A$ meets line $U_{1} V_{1}$ at $C$, while line $X B$ meets line $U_{2} V_{2}$ at $D$. Prove there exists a fixed point $K$, independent of $X$, such that the power of $K$ to the circumcircle of $\\triangle X C D$ is constant.","solution":"  For brevity, we let $\\ell_{i}$ denote line $U_{i} V_{i}$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\\Gamma$ such that $\\overline{A E} \\| \\ell_{1}$ and $\\overline{B F} \\| \\ell_{2}$. The problem conditions imply that $E$ lies between $U_{1}$ and $A$ while $F$ lies between $U_{2}$ and $B$. Then we let  $$ K=\\overline{A F} \\cap \\overline{B E} $$  This point exists because $A E F B$ are the vertices of a convex quadrilateral. Remark (How to identify the fixed point). If we drop the condition that $X$ lies on the arc, then the choice above is motivated by choosing $X \\in\\{E, F\\}$. Essentially, when one chooses $X \\rightarrow E$, the point $C$ approaches an infinity point. So in this degenerate case, the only points whose power is finite to $(X C D)$ are bounded are those on line $B E$. The same logic shows that $K$ must lie on line $A F$. Therefore, if the problem is going to work, the fixed point must be exactly $\\overline{A F} \\cap \\overline{B E}$. \u3010 First approach by Vincent Huang. We need the following claim: Claim - Suppose distinct lines $A C$ and $B D$ meet at $X$. Then for any point $K$  $$ \\operatorname{pow}(K, X A B)+\\operatorname{pow}(K, X C D)=\\operatorname{pow}(K, X A D)+\\operatorname{pow}(K, X B C) $$   Construct the points $P=\\ell_{1} \\cap \\overline{B E}$ and $Q=\\ell_{2} \\cap \\overline{A F}$, which do not depend on $X$. Claim - Quadrilaterals $B P C X$ and $A Q D X$ are cyclic. ![](https:\/\/cdn.mathpix.com\/cropped\/2024_11_19_9733e686be42a6dd888ag-5.jpg?height=803&width=792&top_left_y=244&top_left_x=638)  Now, for the particular $K$ we choose, we have  $$ \\begin{aligned} \\operatorname{pow}(K, X C D) & =\\operatorname{pow}(K, X A D)+\\operatorname{pow}(K, X B C)-\\operatorname{pow}(K, X A B) \\\\ & =K A \\cdot K Q+K B \\cdot K P-\\operatorname{pow}(K, \\Gamma) . \\end{aligned} $$"}
{"year":2021,"label":"2","problem":"Points $A, V_{1}, V_{2}, B, U_{2}, U_{1}$ lie fixed on a circle $\\Gamma$, in that order, and such that $B U_{2}>A U_{1}>B V_{2}>A V_{1}$. Let $X$ be a variable point on the $\\operatorname{arc} V_{1} V_{2}$ of $\\Gamma$ not containing $A$ or $B$. Line $X A$ meets line $U_{1} V_{1}$ at $C$, while line $X B$ meets line $U_{2} V_{2}$ at $D$. Prove there exists a fixed point $K$, independent of $X$, such that the power of $K$ to the circumcircle of $\\triangle X C D$ is constant.","solution":"  For brevity, we let $\\ell_{i}$ denote line $U_{i} V_{i}$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\\Gamma$ such that $\\overline{A E} \\| \\ell_{1}$ and $\\overline{B F} \\| \\ell_{2}$. The problem conditions imply that $E$ lies between $U_{1}$ and $A$ while $F$ lies between $U_{2}$ and $B$. Then we let  $$ K=\\overline{A F} \\cap \\overline{B E} $$  This point exists because $A E F B$ are the vertices of a convex quadrilateral. Remark (How to identify the fixed point). If we drop the condition that $X$ lies on the arc, then the choice above is motivated by choosing $X \\in\\{E, F\\}$. Essentially, when one chooses $X \\rightarrow E$, the point $C$ approaches an infinity point. So in this degenerate case, the only points whose power is finite to $(X C D)$ are bounded are those on line $B E$. The same logic shows that $K$ must lie on line $A F$. Therefore, if the problem is going to work, the fixed point must be exactly $\\overline{A F} \\cap \\overline{B E}$. \u3010 Second approach by authors. Let $Y$ be the second intersection of $(X C D)$ with $\\Gamma$. Let $S=\\overline{E Y} \\cap \\ell_{1}$ and $T=\\overline{F Y} \\cap \\ell_{2}$.  Claim - Points $S$ and $T$ lies on ( $X C D)$ as well.  Now let $X^{\\prime}$ be any other choice of $X$, and define $C^{\\prime}$ and $D^{\\prime}$ in the obvious way. We are going to show that $K$ lies on the radical axis of $(X C D)$ and $\\left(X^{\\prime} C^{\\prime} D^{\\prime}\\right)$. ![](https:\/\/cdn.mathpix.com\/cropped\/2024_11_19_9733e686be42a6dd888ag-5.jpg?height=798&width=798&top_left_y=1828&top_left_x=629)  The main idea is as follows: Claim - The point $L=\\overline{E Y} \\cap \\overline{A X^{\\prime}}$ lies on the radical axis. By symmetry, so does the point $M=\\overline{F Y} \\cap \\overline{B X^{\\prime}}$ (not pictured).   $$ \\operatorname{pow}\\left(L, X^{\\prime} C^{\\prime} D^{\\prime}\\right)=L C^{\\prime} \\cdot L X^{\\prime}=L S \\cdot L Y=\\operatorname{pow}(L, X C D) $$  To conclude, note that by Pascal theorem on  $$ E Y F A X^{\\prime} B $$  it follows $K, L, M$ are collinear, as needed. Remark. All the conditions about $U_{1}, V_{1}, U_{2}, V_{2}$ at the beginning are there to eliminate configuration issues, making the problem less obnoxious to the contestant.  In particular, without the various assumptions, there exist configurations in which the point $K$ is at infinity. In these cases, the center of $X C D$ moves along a fixed line."}
{"year":2021,"label":"3","problem":"Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ that satisfy the inequality  $$ f(y)-\\left(\\frac{z-y}{z-x} f(x)+\\frac{y-x}{z-x} f(z)\\right) \\leq f\\left(\\frac{x+z}{2}\\right)-\\frac{f(x)+f(z)}{2} $$  for all real numbers $x<y<z$.","solution":"  Answer: all functions of the form $f(y)=a y^{2}+b y+c$, where $a, b, c$ are constants with $a \\leq 0$.  If $I=(x, z)$ is an interval, we say that a real number $\\alpha$ is a supergradient of $f$ at $y \\in I$ if we always have  $$ f(t) \\leq f(y)+\\alpha(t-y) $$  for every $t \\in I$. (This inequality may be familiar as the so-called \"tangent line trick\". A cartoon of this situation is drawn below for intuition.) We will also say $\\alpha$ is a supergradient of $f$ at $y$, without reference to the interval, if $\\alpha$ is a supergradient of some open interval containing $y$. ![](https:\/\/cdn.mathpix.com\/cropped\/2024_11_19_9733e686be42a6dd888ag-7.jpg?height=375&width=984&top_left_y=1320&top_left_x=546)  Claim - The problem condition is equivalent to asserting that $\\frac{f(z)-f(x)}{z-x}$ is a supergradient of $f$ at $\\frac{x+z}{2}$ for the interval $(x, z)$, for every $x<z$.  At this point, we may readily verify that all claimed quadratic functions $f(x)=$ $a x^{2}+b x+c$ work - these functions are concave, so the graphs lie below the tangent line at any point. Given $x<z$, the tangent at $\\frac{x+z}{2}$ has slope given by the derivative $f^{\\prime}(x)=2 a x+b$, that is  $$ f^{\\prime}\\left(\\frac{x+z}{2}\\right)=2 a \\cdot \\frac{x+z}{2}+b=\\frac{f(z)-f(x)}{z-x} $$  as claimed. (Of course, it is also easy to verify the condition directly by elementary means, since it is just a polynomial inequality.)  Now suppose $f$ satisfies the required condition; we prove that it has the above form. Claim - The function $f$ is concave.   $$ \\begin{aligned} \\frac{z-y}{z-x} f(x)+\\frac{y-x}{z-x} f(z) & \\leq \\frac{z-y}{z-x}(f(y)+\\alpha(x-y))+\\frac{y-x}{z-x}(f(y)+\\alpha(z-y)) \\\\ & =f(y) \\end{aligned} $$  That is, $f$ is a concave function. Continuity follows from the fact that any concave function on $\\mathbb{R}$ is automatically continuous.  Lemma (see e.g. https:\/\/math.stackexchange.com\/a\/615161 for picture) Any concave function $f$ on $\\mathbb{R}$ is continuous.   $$ f(p)+\\frac{f(b)-f(p)}{b-p} \\varepsilon \\leq f(p+\\varepsilon) \\leq f(p)+\\frac{f(p)-f(a)}{p-a} \\varepsilon $$   Claim - The function $f$ cannot have more than one supergradient at any given point.   $$ g(t)=\\frac{f(y)-f(y-t)}{t}-\\frac{f(y+t)-f(y)}{t} . $$  We contend that $g(\\varepsilon) \\leq \\frac{3}{5} g(3 \\varepsilon)$ for any $\\varepsilon>0$. Indeed by the problem condition,  $$ \\begin{aligned} & f(y) \\leq f(y-\\varepsilon)+\\frac{f(y+\\varepsilon)-f(y-3 \\varepsilon)}{4} \\\\ & f(y) \\leq f(y+\\varepsilon)-\\frac{f(y+3 \\varepsilon)-f(y-\\varepsilon)}{4} . \\end{aligned} $$  ![](https:\/\/cdn.mathpix.com\/cropped\/2024_11_19_9733e686be42a6dd888ag-8.jpg?height=223&width=609&top_left_y=1699&top_left_x=1029)  Summing gives the desired conclusion. Now suppose that $f$ has two supergradients $\\alpha<\\alpha^{\\prime}$ at point $y$. For small enough $\\varepsilon$, we should have we have $f(y-\\varepsilon) \\leq f(y)-\\alpha^{\\prime} \\varepsilon$ and $f(y+\\varepsilon) \\leq f(y)+\\alpha \\varepsilon$, hence  $$ g(\\varepsilon)=\\frac{f(y)-f(y-\\varepsilon)}{\\varepsilon}-\\frac{f(y+\\varepsilon)-f(y)}{\\varepsilon} \\geq \\alpha^{\\prime}-\\alpha . $$  This is impossible since $g(\\varepsilon)$ may be arbitrarily small.  Claim - The function $f$ is quadratic on the rational numbers.  $$ f(x+3 d)-3 f(x+2 d)+3 f(x+d)-f(x)=0 . $$  If we fix $d=1 \/ n$, it follows inductively that $f$ agrees with a quadratic function $\\widetilde{f}_{n}$ on the set $\\frac{1}{n} \\mathbb{Z}$. On the other hand, for any $m \\neq n$, we apparently have $\\widetilde{f}_{n}=\\widetilde{f}_{m n}=\\widetilde{f}_{m}$, so the quadratic functions on each \"layer\" are all equal.  Since $f$ is continuous, it follows $f$ is quadratic, as needed. Remark (Alternate finish using differentiability due to Michael Ren). In the proof of the main claim (about uniqueness of supergradients), we can actually notice the two terms $\\frac{f(y)-f(y-t)}{t}$ and $\\frac{f(y+t)-f(y)}{t}$ in the definition of $g(t)$ are both monotonic (by concavity). Since we supplied a proof that $\\lim _{t \\rightarrow 0} g(t)=0$, we find $f$ is differentiable.  Now, if the derivative at some point exists, it must coincide with all the supergradients; (informally, this is why \"tangent line trick\" always has the slope as the derivative, and formally, we use the mean value theorem). In other words, we must have  $$ f(x+y)-f(x-y)=2 f^{\\prime}(x) \\cdot y $$  holds for all real numbers $x$ and $y$. By choosing $y=1$ we obtain that $f^{\\prime}(x)=f(x+1)-f(x-1)$ which means $f^{\\prime}$ is also continuous.  Finally differentiating both sides with respect to $y$ gives  $$ f^{\\prime}(x+y)-f^{\\prime}(x-y)=2 f^{\\prime}(x) $$  which means $f^{\\prime}$ obeys Jensen's functional equation. Since $f^{\\prime}$ is continuous, this means $f^{\\prime}$ is linear. Thus $f$ is quadratic, as needed."}