problem: Let $a_{1}, a_{2}, \ldots, a_{n}$ be a sequence of real numbers, and let $m$ be a fixed positive integer less than $n$. We say an index $k$ with $1 \leq k \leq n$ is good if there exists some $\ell$ with $1 \leq \ell \leq m$ such that  $$ a_{k}+a_{k+1}+\cdots+a_{k+\ell-1} \geq 0 $$  where the indices are taken modulo $n$. Let $T$ be the set of all good indices. Prove that  $$ \sum_{k \in T} a_{k} \geq 0 $$

solution: First we prove the result if the indices are not taken modulo $n$. Call a number $\ell$-good if $\ell$ is the smallest number such that $a_{k}+a_{k+1}+\cdots+a_{k+\ell-1} \geq 0$, and $\ell \leq m$. Then if $a_{k}$ is $\ell$-good, the numbers $a_{k+1}, \ldots, a_{k+\ell-1}$ are good as well.  Then by greedy from left to right, we can group all the good numbers into blocks with nonnegative sums. Repeatedly take the first good number, if $\ell$-good, group it with the next $\ell$ numbers. An example for $m=3$ :  $$ \langle 4\rangle \quad \begin{array}{ccccccccc} -1 & -2 & 6\rangle & -9 & -7 & \langle 3\rangle & \langle-2 & 4\rangle & \langle-1 . \end{array} $$  We can now return to the original problem. Let $N$ be a large integer; applying the algorithm to $N$ copies of the sequence, we deduce that  $$ N \sum_{k \in T} a_{k}+c_{N} \geq 0 $$  where $c_{N}$ represents some "error" from left-over terms. As $\left|c_{N}\right| \leq \sum\left|a_{i}\right|$, by taking $N$ large enough we deduce the problem.


problem: Let $A B C$ be a scalene triangle. Let $K_{a}, L_{a}$, and $M_{a}$ be the respective intersections with $B C$ of the internal angle bisector, external angle bisector, and the median from $A$. The circumcircle of $A K_{a} L_{a}$ intersects $A M_{a}$ a second time at a point $X_{a}$ different from $A$. Define $X_{b}$ and $X_{c}$ analogously. Prove that the circumcenter of $X_{a} X_{b} X_{c}$ lies on the Euler line of $A B C$.

solution: The main content of the problem: Claim - $\angle H X_{a} G=90^{\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. First proof by Brokard. To do this, it suffices to show that $M$ has the same power with respect to the circle with diameter $\overline{A H}$ and the circle with diameter $\overline{K L}$. In fact I claim both circles are orthogonal to the circle with diameter $\overline{B C}$ ! The former follows from Brokard's theorem, noting that $A$ is on the polar of $H$, and the latter follows from the harmonic bundle. ![](https://cdn.mathpix.com/cropped/2024_11_19_8fb7b517def5801eef36g-04.jpg?height=635&width=1109&top_left_y=1410&top_left_x=482)  Then $\overline{A M}$ is the radical axis, so $X$ lies on both circles.


solution: The main content of the problem: Claim - $\angle H X_{a} G=90^{\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. Second proof by orthocenter reflection, Bendit Chan. As before, we know $M X \cdot M A=$ $M K \cdot M L=M B \cdot M C$, but $X$ lies inside segment $A M$. Construct parallelogram $A B A^{\prime} C$. Then $M X \cdot M A^{\prime}=M B \cdot M C$, so $X B A^{\prime} C$ is concyclic.  However, it is well-known the circumcircle of $\triangle B A^{\prime} C$ (which is the reflection of $(A B C)$ across $\overline{B C}$ ) passes through $H$ and in fact has diameter $\overline{A^{\prime} H}$. So this gives $\angle H X A^{\prime}=90^{\circ}$ as needed.


solution: The main content of the problem: Claim - $\angle H X_{a} G=90^{\circ}$. This implies the result, since then the desired circumcenter is the midpoint of $\overline{G H}$. (This is the main difficulty; the Euler line is a red herring.) In what follows, we abbreviate $K_{a} L_{a}, M_{a}, X_{a}$ to $K, L, M, X$. Third proof by barycentric coordinates. Alternatively we may just compute $X=\left(a^{2}\right.$ : $\left.2 S_{A}: 2 S_{A}\right)$. Let $F=\left(0: S_{C}: S_{B}\right)$ be the foot from $H$. Then we check that $X H F M$ is cyclic, which is power of a point from $A$.


problem: Let $P$ be the set of all primes, and let $M$ be a non-empty subset of $P$. Suppose that for any non-empty subset $\left\{p_{1}, p_{2}, \ldots, p_{k}\right\}$ of $M$, all prime factors of $p_{1} p_{2} \ldots p_{k}+1$ are also in $M$. Prove that $M=P$.

solution: Obviously $|M|=\infty$. Assume for contradiction $p \notin M$. We say a prime $q \in M$ is sparse if there are only finitely many elements of $M$ which are $q(\bmod p)$ (in particular there are finitely many sparse primes).  Now let $C$ be the product of all sparse primes (note $p \nmid C$ ). First, set $a_{0}=1$. For $k \geq 0$, consider then the prime factorization of the number  $$ C a_{k}+1 $$  No prime in its factorization is sparse, so consider the number $a_{k+1}$ obtained by replacing each prime in its factorization with some arbitrary representative of that prime's residue class. In this way we select a number $a_{k+1}$ such that  - $a_{k+1} \equiv C a_{k}+1(\bmod p)$, and - $a_{k+1}$ is a product of distinct primes in $M$.  In particular, $a_{k} \equiv C^{k}+C^{k-1}+\cdots+1(\bmod p)$ But since $C \not \equiv 0(\bmod p)$, we can find a $k$ such that $a_{k} \equiv 0(\bmod p)($ namely, $k=p-1$ if $C \equiv 1$ and $k=p-2$ else) which is clearly impossible since $a_{k}$ is a product of primes in $M$ !


problem: Let $x, y, z$ be real numbers (not necessarily positive) such that $x^{4}+y^{4}+z^{4}+x y z=4$. Prove that $x \leq 2$ and  $$ \sqrt{2-x} \geq \frac{y+z}{2} $$

solution: We prove that the condition $x^{4}+y^{4}+z^{4}+x y z=4$ implies  $$ \sqrt{2-x} \geq \frac{y+z}{2} $$  We first prove the easy part. Claim - We have $x \leq 2$.  $$ \begin{aligned} 5=x^{4}+y^{4}+\left(z^{4}+1\right)+x y z & =\frac{3 x^{4}}{4}+\left(\frac{x^{4}}{4}+y^{4}\right)+\left(z^{4}+1\right)+x y z \\ & \geq \frac{3 x^{4}}{4}+x^{2} y^{2}+2 z^{2}+x y z . \end{aligned} $$  We evidently have that $x^{2} y^{2}+2 z^{2}+x y z \geq 0$ because the quadratic form $a^{2}+a b+2 b^{2}$ is positive definite, so $x^{4} \leq \frac{20}{3} \Longrightarrow x \leq 2$.  Now, the desired statement is implied by its square, so it suffices to show that  $$ 2-x \geq\left(\frac{y+z}{2}\right)^{2} $$  We are going to proceed by contradiction (it seems that many solutions do this) and assume that  $$ 2-x<\left(\frac{y+z}{2}\right)^{2} \Longleftrightarrow 4 x+y^{2}+2 y z+z^{2}>8 $$  By AM-GM,  $$ \begin{aligned} x^{4}+3 & \geq 4 x \\ \frac{y^{4}+1}{2} & \geq y^{2} \\ \frac{z^{4}+1}{2} & \geq z^{2} \end{aligned} $$  which yields that  $$ x^{4}+\frac{y^{4}+z^{4}}{2}+2 y z+4>8 $$  If we replace $x^{4}=4-\left(y^{4}+z^{4}+x y z\right)$ now, this gives  $$ -\frac{y^{4}+z^{4}}{2}+(2-x) y z>0 \Longrightarrow(2-x) y z>\frac{y^{4}+z^{4}}{2} $$  Since $2-x$ and the right-hand side are positive, we have $y z \geq 0$. Now  $$ \frac{y^{4}+z^{4}}{2 y z}<2-x<\left(\frac{y+z}{2}\right)^{2} \Longrightarrow 2 y^{4}+2 z^{4}<y z(y+z)^{2}=y^{3} z+2 y^{2} z^{2}+y z^{3} $$  This is clearly false by AM-GM, so we have a contradiction.


problem: Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least 2015 solutions in $n$.

solution: 『 First solution with ad-hoc subsets, by Evan Chen. I consider the following eleven prime numbers:  $$ S=\{11,13,17,19,29,31,37,41,43,61,71\} $$  This has the property that for any $p \in S$, all prime factors of $p-1$ are one digit. Let $N=(210)^{\text {billion }}$, and consider $M=\varphi(N)$. For any subset $T \subset S$, we have  $$ M=\varphi\left(\frac{N}{\prod_{p \in T}(p-1)} \prod_{p \in T} p\right) $$  Since $2^{|S|}>2015$ we're done.


solution: 【 Second solution with smallest primes, by Yang Liu. Let $2=p_{1}<p_{2}<\cdots<p_{2015}$ be the smallest 2015 primes. Then the 2015 numbers  $$ \begin{aligned} n_{1} & =\left(p_{1}-1\right) p_{2} \ldots p_{2015} \\ n_{2} & =p_{1}\left(p_{2}-1\right) \ldots p_{2015} \\ & \vdots \\ n_{2015} & =p_{1} p_{2} \ldots\left(p_{2015}-1\right) \end{aligned} $$  all have the same phi value, namely  $$ \varphi\left(p_{1} p_{2} \ldots p_{2015}\right)=\prod_{i=1}^{2015}\left(p_{i}-1\right) $$


problem: A Nim-style game is defined as follows. Two positive integers $k$ and $n$ are specified, along with a finite set $S$ of $k$-tuples of integers (not necessarily positive). At the start of the game, the $k$-tuple $(~ n, 0,0, \ldots, 0)$ is written on the blackboard. A legal move consists of erasing the tuple $\left(a_{1}, a_{2}, \ldots, a_{k}\right)$ which is written on the blackboard and replacing it with $\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{k}+b_{k}\right)$, where $\left(b_{1}, b_{2}, \ldots, b_{k}\right)$ is an element of the set $S$. Two players take turns making legal moves, and the first to write a negative integer loses. In the event that neither player is ever forced to write a negative integer, the game is a draw. Prove that there is a choice of $k$ and $S$ with the following property: the first player has a winning strategy if $n$ is a power of 2 , and otherwise the second player has a winning strategy.

solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_8fb7b517def5801eef36g-09.jpg?height=789&width=943&top_left_y=1239&top_left_x=559)  Now, the "game" is played as follows. The mechanics are controlled by the turn counters $A$ and $B$.  Observe the game starts with Alice playing Init. Thereafter, we say that the game is  - In the main part if $A+B=1$, and no one has played Init a second time. - In the death part otherwise.  Observe that in the main state, on Alice's turn we always have $(A, B)=(1,0)$ and on Bob's turn we always have $(A, B)=(0,1)$.  Claim - A player who plays Init a second time must lose. In particular, a player who makes a move when $A=B=0$ must lose.   - Suppose the offending player is in a situation where $A=B=0$. Then he/she must play Init. At this point, the opposing player can respond by playing Kill. Then the offending player must play Init again. The opposing player now responds with Kill'. This iteration continues until $X$ reaches a negative number and the offending player loses. - Suppose Alice has $(A, B)=(1,0)$ but plays Init again anyways. Then Bob responds with PunB to punish her; he then wins as in the first case. - Suppose Bob has $(A, B)=(0,1)$ but plays Init again anyways. Alice responds with PunA in the same way.  Thus we may assume that players avoid the death part at all costs. Hence the second moves consist of Bob playing Sleep, and then Alice playing Begin (thus restoring the value of $n$ in $X$ ), then Bob playing Sleep.  Now we return to analysis of the main part. We say the game is in state $S$ for $S \in$ $\left\{S_{X}^{0}, S_{X}, S_{X}^{\prime}, S_{Y}^{0}, S_{Y}, S_{Y}^{\prime}, \mathrm{Cl}\right\}$ if $S=1$ and all other variables are zero. By construction, this is always the case. From then on the main part is divided into several phases:  - An $X$-phase: this begins with Alice at $S_{X}^{0}$, and ends when the game is in a state other than $S_{X}$ and $S_{X}^{\prime}$. (She can never return to $S_{X}^{0}$ during an $X$-phase.) - A $Y$-phase: this begins with Alice at $S_{Y}^{0}$, and ends when the game is in a state other than $S_{Y}$ and $S_{Y}^{\prime}$. (She can never return to $S_{Y}^{0}$ during a $Y$-phase.)  Claim - Consider an $X$-phase in which $(X, Y)=(x, 0), x>1$. Then Alice can complete the phase without losing if and only if $x$ is even; if so she begins a $Y$-phase with $(X, Y)=(0, x / 2)$.   Now note that unless $X=0$, Bob now has a winning move WrongX. Conversely he may only play Sleep if $X=0$.  We have an analogous claim for $Y$-phases. Thus if $n$ is not a power of 2 , we see that Alice eventually loses.  Now suppose $n=2^{k}$; then Alice reaches $(X, Y)=\left(0,2^{k-1}\right),\left(2^{k-2}, 0\right), \ldots$ until either reaching $(1,0)$ or $(0,1)$. At this point she can play ClaimX or ClaimY, respectively; the game is now in state Cl. Bob cannot play either FakeX or FakeY, so he must play Sleep, and then Alice wins by playing Win. Thus Alice has a winning strategy when $n=2^{k}$.