problem: Are there integers $a$ and $b$ such that $a^{5} b+3$ and $a b^{5}+3$ are both perfect cubes of integers?

solution: No, there do not exist such $a$ and $b$. We prove this in two cases.  - Assume $3 \mid a b$. WLOG we have $3 \mid a$, but then $a^{5} b+3 \equiv 3(\bmod 9)$, contradiction. - Assume $3 \nmid a b$. Then $a^{5} b+3$ is a cube not divisible by 3 , so it is $\pm 1 \bmod 9$, and we conclude  $$ a^{5} b \in\{5,7\} \quad(\bmod 9) $$  Analogously  $$ a b^{5} \in\{5,7\} \quad(\bmod 9) $$  We claim however these two equations cannot hold simultaneously. Indeed $(a b)^{6} \equiv 1$ $(\bmod 9)$ by Euler's theorem, despite $5 \cdot 5 \equiv 7,5 \cdot 7 \equiv 8,7 \cdot 7 \equiv 4 \bmod 9$.


problem: Each cell of an $m \times n$ board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. The filling is called a garden if it satisfies the following two conditions: (i) The difference between any two adjacent numbers is either 0 or 1 . (ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to 0.  Determine the number of distinct gardens in terms of $m$ and $n$.

solution: The numerical answer is $2^{m n}-1$. But we claim much more, by giving an explicit description of all gardens:  Let $S$ be any nonempty subset of the $m n$ cells. Suppose we fill each cell $\theta$ with the minimum (taxicab) distance from $\theta$ to some cell in $S$ (in particular, we write 0 if $\theta \in S$ ). Then  - This gives a garden, and - All gardens are of this form.  Since there are $2^{m n}-1$ such nonempty subsets $S$, this would finish the problem. An example of a garden with $|S|=3$ is shown below.  $$ \left[\begin{array}{llllll} 2 & 1 & 2 & 1 & 0 & 1 \\ 1 & 0 & 1 & 2 & 1 & 2 \\ 1 & 1 & 2 & 3 & 2 & 3 \\ 0 & 1 & 2 & 3 & 3 & 4 \end{array}\right] $$  It is actually fairly easy to see that this procedure always gives a garden; so we focus our attention on showing that every garden is of this form.  Given a garden, note first that it has at least one cell with a zero in it - by considering the minimum number across the entire garden. Now let $S$ be the (thus nonempty) set of cells with a zero written in them. We contend that this works, i.e. the following sentence holds:  Claim - If a cell $\theta$ is labeled $d$, then the minimum distance from that cell to a cell in $S$ is $d$.


problem: In triangle $A B C$, points $P, Q, R$ lie on sides $B C, C A, A B$, respectively. Let $\omega_{A}$, $\omega_{B}, \omega_{C}$ denote the circumcircles of triangles $A Q R, B R P, C P Q$, respectively. Given the fact that segment $A P$ intersects $\omega_{A}, \omega_{B}, \omega_{C}$ again at $X, Y, Z$ respectively, prove that $Y X / X Z=B P / P C$.

solution: Let $M$ be the concurrence point of $\omega_{A}, \omega_{B}, \omega_{C}$ (by Miquel's theorem). ![](https://cdn.mathpix.com/cropped/2024_11_19_0a837349e2b78b0909e9g-5.jpg?height=704&width=812&top_left_y=873&top_left_x=622)  Then $M$ is the center of a spiral similarity sending $\overline{Y Z}$ to $\overline{B C}$. So it suffices to show that this spiral similarity also sends $X$ to $P$, but  $$ \measuredangle M X Y=\measuredangle M X A=\measuredangle M R A=\measuredangle M R B=\measuredangle M P B $$  so this follows.


problem: Let $f(n)$ be the number of ways to write $n$ as a sum of powers of 2 , where we keep track of the order of the summation. For example, $f(4)=6$ because 4 can be written as $4,2+2,2+1+1,1+2+1,1+1+2$, and $1+1+1+1$. Find the smallest $n$ greater than 2013 for which $f(n)$ is odd.

solution: The answer is 2047. For convenience, we agree that $f(0)=1$. Then by considering cases on the first number in the representation, we derive the recurrence  $$ f(n)=\sum_{k=0}^{\left\lfloor\log _{2} n\right\rfloor} f\left(n-2^{k}\right) $$  We wish to understand the parity of $f$. The first few values are  $$ \begin{aligned} & f(0)=1 \\ & f(1)=1 \\ & f(2)=2 \\ & f(3)=3 \\ & f(4)=6 \\ & f(5)=10 \\ & f(6)=18 \\ & f(7)=31 . \end{aligned} $$  Inspired by the data we make the key claim that Claim - $f(n)$ is odd if and only if $n+1$ is a power of 2.   This only takes a few cases:  - If $n=2^{k}$, then $(\odot)$ has exactly two repetitive terms on the right-hand side, namely 0 and $2^{k}-1$. - If $n=2^{k}+2^{\ell}-1$, then $(\odot)$ has exactly two repetitive terms on the right-hand side, namely $2^{\ell+1}-1$ and $2^{\ell}-1$. - If $n=2^{k}-1$, then $(\bigcirc)$ has exactly one repetitive terms on the right-hand side, namely $2^{k-1}-1$. - For other $n$, there are no repetitive terms at all on the right-hand side of $(\Omega)$.  Thus the induction checks out. So the final answer to the problem is 2047.


problem: Quadrilateral $X A B Y$ is inscribed in the semicircle $\omega$ with diameter $\overline{X Y}$. Segments $A Y$ and $B X$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $\overline{X Y}$. Point $C$ lies on $\omega$ such that line $X C$ is perpendicular to line $A Z$. Let $Q$ be the intersection of segments $A Y$ and $X C$. Prove that  $$ \frac{B Y}{X P}+\frac{C Y}{X Q}=\frac{A Y}{A X} $$

solution: Let $\beta=\angle Y X P$ and $\alpha=\angle P Y X$ and set $X Y=1$. We do not direct angles in the following solution. ![](https://cdn.mathpix.com/cropped/2024_11_19_0a837349e2b78b0909e9g-7.jpg?height=552&width=1012&top_left_y=1046&top_left_x=525)  Observe that  $$ \angle A Z X=\angle A P X=\alpha+\beta $$  since $A P Z X$ is cyclic. In particular, $\angle C X Y=90^{\circ}-(\alpha+\beta)$. It is immediate that  $$ B Y=\sin \beta, \quad C Y=\cos (\alpha+\beta), \quad A Y=\cos \alpha, \quad A X=\sin \alpha $$  The Law of Sines on $\triangle X P Y$ gives $X P=X Y \frac{\sin \alpha}{\sin (\alpha+\beta)}$, and on $\triangle X Q Y$ gives $X Q=$ $X Y \frac{\sin \alpha}{\sin (90+\beta)}=\frac{\sin \alpha}{\cos \beta}$. So, the given is equivalent to  $$ \frac{\sin \beta}{\frac{\sin \alpha}{\sin (\alpha+\beta)}}+\frac{\cos (\alpha+\beta)}{\frac{\sin \alpha}{\cos \beta}}=\frac{\cos \alpha}{\sin \alpha} $$  which is equivalent to $\cos \alpha=\cos \beta \cos (\alpha+\beta)+\sin \beta \sin (\alpha+\beta)$. This is obvious, because the right-hand side is just $\cos ((\alpha+\beta)-\beta)$.


problem: Find all real numbers $x, y, z \geq 1$ satisfying  $$ \min (\sqrt{x+x y z}, \sqrt{y+x y z}, \sqrt{z+x y z})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} . $$

solution: Set $x=1+a, y=1+b, z=1+c$ which eliminates the $x, y, z \geq 1$ condition. Assume without loss of generality that $a \leq b \leq c$. Then the given equation rewrites as  $$ \sqrt{(1+a)(1+(1+b)(1+c))}=\sqrt{a}+\sqrt{b}+\sqrt{c} $$  In fact, we are going to prove the left-hand side always exceeds the right-hand side, and then determine the equality cases. We have:  $$ \begin{aligned} (1+a)(1+(1+b)(1+c)) & =(a+1)(1+(b+1)(1+c)) \\ & \leq(a+1)\left(1+(\sqrt{b}+\sqrt{c})^{2}\right) \\ & \leq(\sqrt{a}+(\sqrt{b}+\sqrt{c}))^{2} \end{aligned} $$  by two applications of Cauchy-Schwarz. Equality holds if $b c=1$ and $1 / a=\sqrt{b}+\sqrt{c}$. Letting $c=t^{2}$ for $t \geq 1$, we recover $b=t^{-2} \leq t^{2}$ and $a=\frac{1}{t+1 / t} \leq t^{2}$.   $$ (x, y, z)=\left(1+\left(\frac{t}{t^{2}+1}\right)^{2}, 1+\frac{1}{t^{2}}, 1+t^{2}\right) $$  and permutations, for any $t>0$.