problem: Given a triangle $A B C$, let $P$ and $Q$ be points on segments $\overline{A B}$ and $\overline{A C}$, respectively, such that $A P=A Q$. Let $S$ and $R$ be distinct points on segment $\overline{B C}$ such that $S$ lies between $B$ and $R, \angle B P S=\angle P R S$, and $\angle C Q R=\angle Q S R$. Prove that $P, Q$, $R, S$ are concyclic.

solution: Assume for contradiction that $(P R S)$ and $(Q R S)$ are distinct. Then $\overline{R S}$ is the radical axis of these two circles. However, $\overline{A P}$ is tangent to $(P R S)$ and $\overline{A Q}$ is tangent to $(Q R S)$, so point $A$ has equal power to both circles, which is impossible since $A$ does not lie on line $B C$.


problem: Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_{1}, a_{2}, \ldots$, $a_{n}$ with  $$ \max \left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq n \cdot \min \left(a_{1}, a_{2}, \ldots, a_{n}\right), $$  there exist three that are the side lengths of an acute triangle.

solution: The answer is all $n \geq 13$. Define $\left(F_{n}\right)$ as the sequence of Fibonacci numbers, by $F_{1}=F_{2}=1$ and $F_{n+1}=$ $F_{n}+F_{n-1}$. We will find that Fibonacci numbers show up naturally when we work through the main proof, so we will isolate the following calculation now to make the subsequent solution easier to read.  Claim - For positive integers $m$, we have $F_{m} \leq m^{2}$ if and only if $m \leq 12$.  $$ \begin{array}{rrrrrrrrrrrrrr} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} & F_{7} & F_{8} & F_{9} & F_{10} & F_{11} & F_{12} & F_{13} & F_{14} \\ \hline 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 \end{array} $$  By examining the table, we see that $F_{m} \leq m^{2}$ is true for $m=1,2, \ldots 12$, and in fact $F_{12}=12^{2}=144$. However, $F_{m}>m^{2}$ for $m=13$ and $m=14$.  Now it remains to prove that $F_{m}>m^{2}$ for $m \geq 15$. The proof is by induction with base cases $m=13$ and $m=14$ being checked already. For the inductive step, if $m \geq 15$ then we have  $$ \begin{aligned} F_{m} & =F_{m-1}+F_{m-2}>(m-1)^{2}+(m-2)^{2} \\ & =2 m^{2}-6 m+5=m^{2}+(m-1)(m-5)>m^{2} \end{aligned} $$  as desired. We now proceed to the main problem. The hypothesis $\max \left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq n$. $\min \left(a_{1}, a_{2}, \ldots, a_{n}\right)$ will be denoted by $(\dagger)$.   $$ \begin{aligned} a_{3}^{2} & \geq a_{2}^{2}+a_{1}^{2} \geq 2 a_{1}^{2} \\ a_{4}^{2} & \geq a_{3}^{2}+a_{2}^{2} \geq 2 a_{1}^{2}+a_{1}^{2}=3 a_{1}^{2} \\ a_{5}^{2} & \geq a_{4}^{2}+a_{3}^{2} \geq 3 a_{1}^{2}+2 a_{1}^{2}=5 a_{1}^{2} \\ a_{6}^{2} & \geq a_{5}^{2}+a_{4}^{2} \geq 5 a_{1}^{2}+3 a_{1}^{2}=8 a_{1}^{2} \end{aligned} $$  and so on. The Fibonacci numbers appear naturally and by induction, we conclude that $a_{i}^{2} \geq F_{i} a_{1}^{2}$. In particular, $a_{n}^{2} \geq F_{n} a_{1}^{2}$.  However, we know $\max \left(a_{1}, \ldots, a_{n}\right)=a_{n}$ and $\min \left(a_{1}, \ldots, a_{n}\right)=a_{1}$, so $(\dagger)$ reads $a_{n} \leq n \cdot a_{1}$. Therefore we have $F_{n} \leq n^{2}$, and so $n \leq 12$, contradiction!


solution: Cauchy-Schwarz approach. Apply Titu lemma to get  $$ \sum_{\mathrm{cyc}} \frac{a^{3}}{5 a+b}=\sum_{\mathrm{cyc}} \frac{a^{4}}{5 a^{2}+a b} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{\mathrm{cyc}}\left(5 a^{2}+a b\right)} \geq \frac{a^{2}+b^{2}+c^{2}}{6} $$  where the last step follows from the identity $\sum_{\text {cyc }}\left(5 a^{2}+a b\right) \leq 6\left(a^{2}+b^{2}+c^{2}\right)$. Similarly,  $$ \sum_{\text {cyc }} \frac{b^{3}}{5 a+b}=\sum_{\mathrm{cyc}} \frac{b^{4}}{5 a b+b^{2}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{\mathrm{cyc}}\left(5 a b+b^{2}\right)} \geq \frac{a^{2}+b^{2}+c^{2}}{6} $$  using the fact that $\sum_{\text {cyc }} 5 a b+b^{2} \leq 6\left(a^{2}+b^{2}+c^{2}\right)$. Therefore, adding the first display to three times the second display implies the result.


solution: đ Cauchy-Schwarz approach. The main magical claim is: Claim - We have  $$ \frac{a^{3}+3 b^{3}}{5 a+b} \geq \frac{25}{36} b^{2}-\frac{1}{36} a^{2} $$   $$ \frac{x^{3}+3}{5 x+1} \geq \frac{25-x^{2}}{36} $$  However,  $$ \begin{aligned} 36\left(x^{3}+3\right)-(5 x+1)\left(25-x^{2}\right) & =41 x^{3}+x^{2}-125 x+83 \\ & =(x-1)^{2}(41 x+83) \geq 0 \end{aligned} $$  Sum the claim cyclically to finish. Remark (Derivation of the main claim). The overall strategy is to hope for a constant $k$ such that  $$ \frac{a^{3}+3 b^{3}}{5 a+b} \geq k a^{2}+\left(\frac{2}{3}-k\right) b^{2} $$  is true. Letting $x=a / b$ as above and expanding, we need a value $k$ such that the cubic polynomial $P(x):=\left(x^{3}+3\right)-(5 x+1)\left(k x^{2}+\left(\frac{2}{3}-k\right)\right)=(1-5 k) x^{3}-k x^{2}+\left(5 k-\frac{10}{3}\right) x+\left(k+\frac{7}{3}\right)$ is nonnegative everywhere. Since $P(1)=0$ necessarily, in order for $P(1-\varepsilon)$ and $P(1+\varepsilon)$ to both be nonnegative (for small $\varepsilon$ ), the polynomial $P$ must have a double root at 1 , meaning the first derivative $P^{\prime}(1)=0$ needs to vanish. In other words, we need  $$ 3(1-5 k)-2 k+\left(5 k-\frac{10}{3}\right)=0 $$  Solving gives $k=-1 / 36$. One then factors out the repeated root $(x-1)^{2}$ from the resulting $P$.


problem: Let $\alpha$ be an irrational number with $0<\alpha<1$, and draw a circle in the plane whose circumference has length 1 . Given any integer $n \geq 3$, define a sequence of points $P_{1}, P_{2}, \ldots, P_{n}$ as follows. First select any point $P_{1}$ on the circle, and for $2 \leq k \leq n$ define $P_{k}$ as the point on the circle for which the length of $\operatorname{arc} P_{k-1} P_{k}$ is $\alpha$, when travelling counterclockwise around the circle from $P_{k-1}$ to $P_{k}$. Suppose that $P_{a}$ and $P_{b}$ are the nearest adjacent points on either side of $P_{n}$. Prove that $a+b \leq n$.

solution: No points coincide since $\alpha$ is irrational. Assume for contradiction that $n<a+b<2 n$. Then it follows that  $$ \overline{P_{n} P_{a+b-n}} \| \overline{P_{a} P_{b}} $$  as shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_4c570ff524ef777a3bedg-08.jpg?height=327&width=900&top_left_y=1263&top_left_x=578)  This is an obvious contradiction since then $P_{a+b-n}$ is contained in the arc $\widehat{P_{a} P_{b}}$ of the circle through $P_{n}$.


problem: For distinct positive integers $a, b<2012$, define $f(a, b)$ to be the number of integers $k$ with $1 \leq k<2012$ such that the remainder when $a k$ divided by 2012 is greater than that of $b k$ divided by 2012 . Let $S$ be the minimum value of $f(a, b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$.

solution: The answer is $S=502$ (not 503 !). Claim - If $\operatorname{gcd}(k, 2012)=1$, then necessarily either $k$ or $2012-k$ will counts towards $S$.   So if $r_{a}$ is the remainder of $a k(\bmod 2012)$, then $2012-r_{a}$ is the remainder of $a(2012-k)$ $(\bmod 2012)$ Similarly we can consider $r_{b}$ and $2012-r_{b}$. As mentioned already, we have $r_{a} \neq r_{b}$. So either $r_{a}>r_{b}$ or $2012-r_{a}>2012-r_{b}$.  This implies $S \geq \frac{1}{2} \varphi(2012)=502$. But this can actually be achieved by taking $a=4$ and $b=1010$, since  - If $k$ is even, then $a k \equiv b k(\bmod 2012)$ so no even $k$ counts towards $S$; and - If $k \equiv 0(\bmod 503)$, then $a k \equiv 0(\bmod 2012)$ so no such $k$ counts towards $S$.  This gives the final answer $S \geq 502$.


problem: Let $P$ be a point in the plane of $\triangle A B C$, and $\gamma$ a line through $P$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points where the reflections of lines $P A, P B, P C$ with respect to $\gamma$ intersect lines $B C, C A, A B$ respectively. Prove that $A^{\prime}, B^{\prime}, C^{\prime}$ are collinear.

solution: 【 First solution (complex numbers). Let $p=0$ and set $\gamma$ as the real line. Then $A^{\prime}$ is the intersection of $b c$ and $p \bar{a}$. So, we get  $$ a^{\prime}=\frac{\bar{a}(\bar{b} c-b \bar{c})}{(\bar{b}-\bar{c}) \bar{a}-(b-c) a} . $$  ![](https://cdn.mathpix.com/cropped/2024_11_19_4c570ff524ef777a3bedg-10.jpg?height=364&width=427&top_left_y=1103&top_left_x=820)  Note that  $$ \bar{a}^{\prime}=\frac{a(b \bar{c}-\bar{b} c)}{(b-c) a-(\bar{b}-\bar{c}) \bar{a}} . $$  Thus it suffices to prove  This is equivalent to  $$ 0=\operatorname{det}\left[\begin{array}{lll} \bar{a}(\bar{b} c-b \bar{c}) & a(\bar{b} c-b \bar{c}) & (\bar{b}-\bar{c}) \bar{a}-(b-c) a \\ \bar{b}(\bar{c} a-c \bar{a}) & b(\bar{c} a-c \bar{a}) & (\bar{c}-\bar{a}) \bar{b}-(c-a) b \\ \bar{c}(\bar{a} b-a \bar{b}) & c(\bar{a} b-a \bar{b}) & (\bar{a}-\bar{b}) \bar{c}-(a-b) c \end{array}\right] . $$  This determinant has the property that the rows sum to zero, and we're done. Remark. Alternatively, if you don't notice that you could just blindly expand:  $$ \begin{aligned} & \sum_{\mathrm{cyc}}((\bar{b}-\bar{c}) \bar{a}-(b-c) a) \cdot-\operatorname{det}\left[\begin{array}{ll} b & \bar{b} \\ c & \bar{c} \end{array}\right](\bar{c} a-c \bar{a})(\bar{a} b-a \bar{b}) \\ = & (\bar{b} c-c \bar{b})(\bar{c} a-c \bar{a})(\bar{a} b-a \bar{b}) \sum_{\mathrm{cyc}}(a b-a c+\overline{c a}-\bar{b} \bar{a})=0 \end{aligned} $$


solution: \I Second solution (Desargues involution). We let $C^{\prime \prime}=\overline{A^{\prime} B^{\prime}} \cap \overline{A B}$. Consider complete quadrilateral $A B C A^{\prime} B^{\prime} C^{\prime \prime} C$. We see that there is an involutive pairing $\tau$ at $P$ swapping $\left(P A, P A^{\prime}\right),\left(P B, P B^{\prime}\right),\left(P C, P C^{\prime \prime}\right)$. From the first two, we see $\tau$ coincides with reflection about $\ell$, hence conclude $C^{\prime \prime}=C$.


solution: ब Third solution (barycentric), by Catherine $\mathbf{X u}$. We will perform barycentric coordinates on the triangle $P C C^{\prime}$, with $P=(1,0,0), C^{\prime}=(0,1,0)$, and $C=(0,0,1)$. Set $a=C C^{\prime}, b=C P, c=C^{\prime} P$ as usual. Since $A, B, C^{\prime}$ are collinear, we will define $A=(p: k: q)$ and $B=(p: \ell: q)$.  Claim - Line $\gamma$ is the angle bisector of $\angle A P A^{\prime}, \angle B P B^{\prime}$, and $\angle C P C^{\prime}$. Thus $B^{\prime}$ is the intersection of the isogonal of $B$ with respect to $\angle P$ with the line $C A$; that is,  $$ B^{\prime}=\left(\frac{p}{k} \frac{b^{2}}{\ell}: \frac{b^{2}}{\ell}: \frac{c^{2}}{q}\right) $$  Analogously, $A^{\prime}$ is the intersection of the isogonal of $A$ with respect to $\angle P$ with the line $C B$; that is,  $$ A^{\prime}=\left(\frac{p}{\ell} \frac{b^{2}}{k}: \frac{b^{2}}{k}: \frac{c^{2}}{q}\right) $$  The ratio of the first to third coordinate in these two points is both $b^{2} p q: c^{2} k \ell$, so it follows $A^{\prime}, B^{\prime}$, and $C^{\prime}$ are collinear.  Remark (Problem reference). The converse of this problem appears as problem 1052 attributed S. V. Markelov in the book Geometriya: 9-11 Klassy: Ot Uchebnoy Zadachi k Tvorcheskoy, 1996, by I. F. Sharygin.