problem: Let $a, b, c, d$ be real numbers such that $b-d \geq 5$ and all zeros $x_{1}, x_{2}, x_{3}$, and $x_{4}$ of the polynomial $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ are real. Find the smallest value the product $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)\left(x_{3}^{2}+1\right)\left(x_{4}^{2}+1\right)$ can take.

solution: The answer is 16 . This can be achieved by taking $x_{1}=x_{2}=x_{3}=x_{4}=1$, whence the product is $2^{4}=16$, and $b-d=5$.  We now show the quantity is always at least 16. We prove: Claim - We always have $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)\left(x_{3}^{2}+1\right)\left(x_{4}^{2}+1\right)=(b-d-1)^{2}+(a-c)^{2}$.   $$ \prod_{j=1}^{4}\left(x_{j}^{2}+1\right)=\prod_{j=1}^{4}\left(x_{j}-i\right)\left(x_{j}+i\right)=P(i) P(-i)=|P(i)|^{2} $$  Since $P(i)=(-1+b-d)+(c-a) i$, the claim follows. Since $b-d-1 \geq 4$, we get the desired lower bound of $4^{2}+0^{2}=16$.


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:  $$ \text { Claim - We have } f(0)=0 \text {. } $$   $$ f(0)=0 $$  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: Prove that there exists an infinite set of points  $$ \ldots, P_{-3}, P_{-2}, P_{-1}, P_{0}, P_{1}, P_{2}, P_{3}, \ldots $$  in the plane with the following property: For any three distinct integers $a, b$, and $c$, points $P_{a}, P_{b}$, and $P_{c}$ are collinear if and only if $a+b+c=2014$.

solution: The construction  $$ P_{n}=\left(n-\frac{2014}{3},\left(n-\frac{2014}{3}\right)^{3}\right) $$  works fine, and follows from the following claim: Claim - If $x, y, z$ are distinct real numbers then the points $\left(x, x^{3}\right),\left(y, y^{3}\right),\left(z, z^{3}\right)$ are collinear if and only if $x+y+z=0$.   $$ 0=\operatorname{det}\left[\begin{array}{lll} x & x^{3} & 1 \\ y & y^{3} & 1 \\ z & z^{3} & 1 \end{array}\right] $$  But the determinant equals  $$ \sum_{\text {cyc }} x\left(y^{3}-z^{3}\right)=(x-y)(y-z)(z-x)(x+y+z) $$


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_045963c8c038d2ede9b0g-07.jpg?height=495&width=546&top_left_y=1463&top_left_x=758) \ 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_045963c8c038d2ede9b0g-07.jpg?height=296&width=491&top_left_y=2262&top_left_x=788)  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_045963c8c038d2ede9b0g-08.jpg?height=398&width=612&top_left_y=646&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 orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $A H C$ with the internal bisector of $\angle B A C$. Let $X$ be the circumcenter of triangle $A P B$ and let $Y$ be the orthocenter of triangle $A P C$. Prove that the length of segment $X Y$ is equal to the circumradius of triangle $A B C$.

solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_045963c8c038d2ede9b0g-09.jpg?height=777&width=820&top_left_y=805&top_left_x=621)  We eliminate the floating orthocenter by reflecting $P$ across $\overline{A C}$ to $Q$. Then $Q$ lies on ( $A B C$ ) and moreover $\angle Q A C=\frac{1}{2} \angle B A C$. This motivates us to reflect $B, X, Y$ to $B^{\prime}$, $X^{\prime}, Y^{\prime}$ and complex bash with respect to $\triangle A Q C$. Obviously  $$ y^{\prime}=a+q+c $$  Now we need to compute $x^{\prime}$. You can get this using the formula  $$ x^{\prime}=a+\frac{\left(b^{\prime}-a\right)(q-a)\left(\overline{q-a}-\overline{b^{\prime}-a}\right)}{\left(b^{\prime}-a\right) \overline{(q-a)}-\overline{\left(b^{\prime}-a\right)}(q-a)} . $$  Using the angle condition we know $b=\frac{c^{3}}{q^{2}}$, and then that  $$ b^{\prime}=a+c-a c \bar{b}=a+c-\frac{a q^{2}}{c^{2}} . $$  Therefore  $$ \begin{aligned} x^{\prime} & =a+\frac{\left(c-\frac{a q^{2}}{c^{2}}\right)(q-a)\left(\frac{1}{q}-\frac{1}{a}-\frac{1}{c}+\frac{c^{2}}{a q^{2}}\right)}{\left(c-\frac{a q^{2}}{c^{2}}\right)\left(\frac{1}{q}-\frac{1}{a}\right)-\left(\frac{1}{c}-\frac{c^{2}}{a q^{2}}\right)(q-a)} \\ & =a+\frac{\frac{c^{3}-a q^{2}}{c^{2}}(q-a)\left(\frac{1}{q}-\frac{1}{a}-\frac{1}{c}+\frac{c^{2}}{a q^{2}}\right)}{-\frac{c^{3}-a q^{2}}{c^{2}} \frac{q-a}{q a}+\frac{c^{3}-q^{2}}{a q^{2} c}(q-a)} \end{aligned} $$  $$ \begin{aligned} & =a+\frac{\frac{1}{q}-\frac{1}{a}-\frac{1}{c}+\frac{c^{2}}{a q^{2}}}{-\frac{1}{q a}+\frac{c}{a q^{2}}} \\ & =a+\frac{c^{2}-q^{2}+a q-\frac{a q^{2}}{c}}{c-q} \\ & =a+c+q+\frac{a q}{c} \end{aligned} $$  whence  $$ \left|x^{\prime}-y^{\prime}\right|=\left|\frac{a q}{c}\right|=1 $$


problem: Prove that there is a constant $c>0$ with the following property: If $a, b, n$ are positive integers such that $\operatorname{gcd}(a+i, b+j)>1$ for all $i, j \in\{0,1, \ldots, n\}$, then  $$ \min \{a, b\}>(c n)^{n / 2} $$

solution: Let $N=n+1$ and assume $N$ is (very) large. We construct an $N \times N$ with cells $(i, j)$ where $0 \leq i, j \leq n$ and in each cell place a prime $p$ dividing $\operatorname{gcd}(a+i, b+j)$.  The central claim is at least $50 \%$ of the primes in this table exceed $0.001 n^{2}$. We count the maximum number of squares they could occupy:  $$ \sum_{p}\left\lceil\frac{N}{p}\right\rceil^{2} \leq \sum_{p}\left(\frac{N}{p}+1\right)^{2}=N^{2} \sum_{p} \frac{1}{p^{2}}+2 N \sum_{p} \frac{1}{p}+\sum_{p} 1 $$  Here the summation runs over primes $p \leq 0.001 n^{2}$. Let $r=\pi\left(0.001 n^{2}\right)$ denote the number of such primes. Now we consider the following three estimates. First,  $$ \sum_{p} \frac{1}{p^{2}}<\frac{1}{2} $$  which follows by adding all the primes directly with some computation. Moreover,  $$ \sum_{p} \frac{1}{p}<\sum_{k=1}^{r} \frac{1}{k}=O(\log r)<o(N) $$  using the harmonic series bound, and  $$ \sum_{p} 1<r \sim O\left(\frac{N^{2}}{\ln N}\right)<o\left(N^{2}\right) $$  via Prime Number Theorem. Hence the sum in question is certainly less than $\frac{1}{2} N^{2}$ for $N$ large enough, establishing the central claim.  Hence some column $a+i$ has at least one half of its primes greater than $0.001 n^{2}$. Because this is greater than $n$ for large $n$, these primes must all be distinct, so $a+i$ exceeds their product, which is larger than  $$ \left(0.001 n^{2}\right)^{N / 2}>c^{n} \cdot n^{n} $$  where $c$ is some constant (better than the requested bound).