problem: Let $a, b, c$ be real numbers greater than or equal to 1 . Prove that  $$ \min \left(\frac{10 a^{2}-5 a+1}{b^{2}-5 b+10}, \frac{10 b^{2}-5 b+1}{c^{2}-5 c+10}, \frac{10 c^{2}-5 c+1}{a^{2}-5 a+10}\right) \leq a b c $$

solution: Notice that  $$ \frac{10 a^{2}-5 a+1}{a^{2}-5 a+10} \leq a^{3} $$  since it rearranges to $(a-1)^{5} \geq 0$. Cyclically multiply to get  $$ \prod_{\mathrm{cyc}}\left(\frac{10 a^{2}-5 a+1}{b^{2}-5 b+10}\right) \leq(a b c)^{3} $$  and the minimum is at most the geometric mean.


problem: Let $\triangle A B C$ be a non-equilateral, acute triangle with $\angle A=60^{\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle A B C$, respectively. (a) Prove that line $O H$ intersects both segments $A B$ and $A C$ at two points $P$ and $Q$, respectively. (b) Denote by $s$ and $t$ the respective areas of triangle $A P Q$ and quadrilateral $B P Q C$. Determine the range of possible values for $s / t$.

solution: We begin with some synthetic work. Let $I$ denote the incenter, and recall ("fact 5") that the arc midpoint $M$ is the center of $(B I C)$, which we denote by $\gamma$.  Now we have that  $$ \angle B O C=\angle B I C=\angle B H C=120^{\circ} . $$  Since all three centers lie inside $A B C$ (as it was acute), and hence on the opposite side of $\overline{B C}$ as $M$, it follows that $O, I, H$ lie on minor arc $B C$ of $\gamma$.  We note this implies (a) already, as line $O H$ meets line $B C$ outside of segment $B C$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-04.jpg?height=806&width=552&top_left_y=1339&top_left_x=752)  Claim - Triangle $A P Q$ is equilateral with side length $\frac{b+c}{3}$.   Finally, since $\angle P B H=30^{\circ}$, and $\angle B P H=120^{\circ}$, it follows that $\triangle B P H$ is isosceles and $B P=P H$. Similarly, $C Q=Q H$. So $b+c=A P+B P+A Q+Q C=A P+A Q+P Q$ as needed.  Finally, we turn to the boring task of extracting the numerical answer. We have  $$ \frac{s}{s+t}=\frac{[A P Q]}{[A B C]}=\frac{\frac{\sqrt{3}}{4}\left(\frac{b+c}{3}\right)^{2}}{\frac{\sqrt{3}}{4} b c}=\frac{b^{2}+2 b c+c^{2}}{9 b c}=\frac{1}{9}\left(2+\frac{b}{c}+\frac{c}{b}\right) . $$  So the problem is reduced to analyzing the behavior of $b / c$. For this, we imagine fixing $\Gamma$ the circumcircle of $A B C$, as well as the points $B$ and $C$. Then as we vary $A$ along the "topmost" arc of measure $120^{\circ}$, we find $b / c$ is monotonic with values $1 / 2$ and 2 at endpoints, and by continuity all values $b / c \in(1 / 2,2)$ can be achieved.  So  $$ \frac{1}{2}<\frac{b}{c}<2 \Longrightarrow 4 / 9<\frac{s}{s+t}<1 / 2 \Longrightarrow 4 / 5<\frac{s}{t}<1 $$  as needed.


problem: Find all $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that  $$ x f(2 f(y)-x)+y^{2} f(2 x-f(y))=\frac{f(x)^{2}}{x}+f(y f(y)) $$  for all $x, y \in \mathbb{Z}$ such that $x \neq 0$.

solution: The answer is $f(x) \equiv 0$ and $f(x) \equiv x^{2}$. Check that these work.  $$ x f(2 f(0)-x)=\frac{f(x)^{2}}{x}+f(0) . $$  The nicest part of the problem is the following step:  $$ \text { Claim - We have } f(0)=0 $$   $$ f(0)=0 $$  Claim - We have $f(x) \in\left\{0, x^{2}\right\}$ for each individual $x$.  $$ x^{2} f(-x)=f(x)^{2} $$  holds for all nonzero $x$, but also for $x=0$. Let us now check that $f$ is an even function. In the above, we may also derive $f(-x)^{2}=x^{2} f(x)$. If $f(x) \neq f(-x)$ (and hence $x \neq 0$ ), then subtracting the above and factoring implies that $f(x)+f(-x)=-x^{2}$; we can then obtain by substituting the relation  $$ \left[f(x)+\frac{1}{2} x^{2}\right]^{2}=-\frac{3}{4} x^{4}<0 $$  which is impossible. This means $f(x)^{2}=x^{2} f(x)$, thus  $$ f(x) \in\left\{0, x^{2}\right\} \quad \forall x . $$  Now suppose there exists a nonzero integer $t$ with $f(t)=0$. We will prove that $f(x) \equiv 0$. Put $y=t$ in the given to obtain that  $$ t^{2} f(2 x)=0 $$  for any integer $x \neq 0$, and hence conclude that $f(2 \mathbb{Z}) \equiv 0$. Then selecting $x=2 k \neq 0$ in the given implies that  $$ y^{2} f(4 k-f(y))=f(y f(y)) $$  Assume for contradiction that $f(m)=m^{2}$ now for some odd $m \neq 0$. Evidently  $$ m^{2} f\left(4 k-m^{2}\right)=f\left(m^{3}\right) $$  If $f\left(m^{3}\right) \neq 0$ this forces $f\left(4 k-m^{2}\right) \neq 0$, and hence $m^{2}\left(4 k-m^{2}\right)^{2}=m^{6}$ for arbitrary $k \neq 0$, which is clearly absurd. That means  $$ f\left(4 k-m^{2}\right)=f\left(m^{2}-4 k\right)=f\left(m^{3}\right)=0 $$  for each $k \neq 0$. Since $m$ is odd, $m^{2} \equiv 1(\bmod 4)$, and so $f(n)=0$ for all $n$ other than $\pm m^{2}$ (since we cannot select $\left.k=0\right)$.  Now $f(m)=m^{2}$ means that $m= \pm 1$. Hence either $f(x) \equiv 0$ or  $$ f(x)= \begin{cases}1 & x= \pm 1 \\ 0 & \text { otherwise }\end{cases} $$  To show that the latter fails, we simply take $x=5$ and $y=1$ in the given.


problem: Let $b \geq 2$ be a fixed integer, and let $s_{b}(n)$ denote the sum of the base- $b$ digits of $n$. Show that there are infinitely many positive integers that cannot be represented in the from $n+s_{b}(n)$ where $n$ is a positive integer.

solution: For brevity let $f(n)=n+s_{b}(n)$. Select any integer $M$. Observe that $f(x) \geq b^{2 M}$ for any $x \geq b^{2 M}$, but also $f\left(b^{2 M}-k\right) \geq b^{2 M}$ for $k=1,2, \ldots, M$, since the base- $b$ expansion of $b^{2 M}-k$ will start out with at least $M$ digits $b-1$.  Thus $f$ omits at least $M$ values in $\left[1, b^{2 M}\right]$ for any $M$.


problem: Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In her move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.

solution: The answer is $k=6$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-09.jpg?height=500&width=561&top_left_y=1372&top_left_x=753)  【 Example of a strategy for $A$ when $k=5$. We describe a winning strategy for $A$ explicitly. Note that after $B$ 's first turn there is one counter, so then $A$ may create an equilateral triangle, and hence after $B$ 's second turn there are two consecutive counters. Then, on her third turn, $A$ places a pair of counters two spaces away on the same line. Label the two inner cells $x$ and $y$ as shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-09.jpg?height=295&width=500&top_left_y=2168&top_left_x=778)  Now it is $B$ 's turn to move; in order to avoid losing immediately, he must remove either $x$ or $y$. Then on any subsequent turn, $A$ can replace $x$ or $y$ (whichever was removed) and add one more adjacent counter. This continues until either $x$ or $y$ has all its neighbors filled (we ask $A$ to do so in such a way that she avoids filling in the two central cells between $x$ and $y$ as long as possible).  So, let's say without loss of generality (by symmetry) that $x$ is completely surrounded by tokens. Again, $B$ must choose to remove $x$ (or $A$ wins on her next turn). After $x$ is removed by $B$, consider the following figure. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-10.jpg?height=398&width=612&top_left_y=503&top_left_x=728)  We let $A$ play in the two marked green cells. Then, regardless of what move $B$ plays, one of the two choices of moves marked in red lets $A$ win. Thus, we have described a winning strategy when $k=5$ for $A$.


problem: Let $A B C$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M, N, P$ be the midpoints of $\overline{B C}, \overline{C A}, \overline{A B}$ and let $E, F$ be the tangency points of $\gamma$ with $\overline{C A}$ and $\overline{A B}$, respectively. Let $U, V$ be the intersections of line $E F$ with line $M N$ and line $M P$, respectively, and let $X$ be the midpoint of $\operatorname{arc} B A C$ of $\Gamma$. (a) Prove that $I$ lies on ray $C V$. (b) Prove that line $X I$ bisects $\overline{U V}$.

solution: The fact that $I=\overline{B U} \cap \overline{C V}$ and is Lemma 1.45 from EGMO. As for (b), we note: Claim - Line $I X$ is a symmedian of $\triangle I B C$.  Since $B V U C$ is cyclic with diagonals intersecting at $I$, and $I X$ is symmedian of $\triangle I B C$, it is median of $\triangle I U V$, as needed.