problem: Let $A B C$ be a scalene triangle with incenter $I$ whose incircle is tangent to $\overline{B C}$, $\overline{C A}, \overline{A B}$ at $D, E, F$, respectively. Denote by $M$ the midpoint of $\overline{B C}$ and let $P$ be a point in the interior of $\triangle A B C$ so that $M D=M P$ and $\angle P A B=\angle P A C$. Let $Q$ be a point on the incircle such that $\angle A Q D=90^{\circ}$. Prove that either $\angle P Q E=90^{\circ}$ or $\angle P Q F=90^{\circ}$.

solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-03.jpg?height=801&width=892&top_left_y=1150&top_left_x=588)  First, we claim that $D, P, E$ are collinear. Let $N$ be the midpoint of $\overline{A B}$. It is well-known that the three lines $M N, D E, A I$ are concurrent at a point (see for example problem 6 of USAJMO 2014). Let $P^{\prime}$ be this intersection point, noting that $P^{\prime}$ actually lies on segment $D E$. Then $P^{\prime}$ lies inside $\triangle A B C$ and moreover  $$ \triangle D P^{\prime} M \sim \triangle D E C $$  so $M P^{\prime}=M D$. Hence $P^{\prime}=P$, proving the claim. Let $S$ be the point diametrically opposite $D$ on the incircle, which is also the second intersection of $\overline{A Q}$ with the incircle. Let $T=\overline{A Q} \cap \overline{B C}$. Then $T$ is the contact point of the $A$-excircle; consequently,  $$ M D=M P=M T $$  and we obtain a circle with diameter $\overline{D T}$. Since $\angle D Q T=\angle D Q S=90^{\circ}$ we have $Q$ on this circle as well.  As $\overline{S D}$ is tangent to the circle with diameter $\overline{D T}$, we obtain  $$ \angle P Q D=\angle S D P=\angle S D E=\angle S Q E . $$  Since $\angle D Q S=90^{\circ}, \angle P Q E=90^{\circ}$ too. 【 Solution using spiral similarity. We will ignore for now the point $P$. As before define $S, T$ and note $\overline{A Q S T}$ collinear, as well as $D P Q T$ cyclic on circle $\omega$ with diameter $\overline{D T}$.  Let $\tau$ be the spiral similarity at $Q$ sending $\omega$ to the incircle. We have $\tau(T)=D$, $\tau(D)=S, \tau(Q)=Q$. Now  $$ I=\overline{D D} \cap \overline{Q Q} \Longrightarrow \tau(I)=\overline{S S} \cap \overline{Q Q} $$  and hence we conclude $\tau(I)$ is the pole of $\overline{A S Q T}$ with respect to the incircle, which lies on line $E F$.  Then since $\overline{A I} \perp \overline{E F}$ too, we deduce $\tau$ sends line $A I$ to line $E F$, hence $\tau(P)$ must be either $E$ or $F$ as desired.  【 Authorship comments. Written April 2014. I found this problem while playing with GeoGebra. Specifically, I started by drawing in the points $A, B, C, I, D, M, T$, common points. I decided to add in the circle with diameter $D T$, because of the synergy it had with the rest of the picture. After a while of playing around, I intersected ray $A I$ with the circle to get $P$, and was surprised to find that $D, P, E$ were collinear, which I thought was impossible since the setup should have been symmetric. On further reflection, I realized it was because $A I$ intersected the circle twice, and set about trying to prove this. I noticed the relation $\angle P Q E=90^{\circ}$ in my attempts to prove the result, even though this ended up being a corollary rather than a useful lemma.


problem: Prove that for every positive integer $n$, there exists a set $S$ of $n$ positive integers such that for any two distinct $a, b \in S, a-b$ divides $a$ and $b$ but none of the other elements of $S$.

solution: The idea is to look for a sequence $d_{1}, \ldots, d_{n-1}$ of "differences" such that the following two conditions hold. Let $s_{i}=d_{1}+\cdots+d_{i-1}$, and $t_{i, j}=d_{i}+\cdots+d_{j-1}$ for $i \leq j$. (i) No two of the $t_{i, j}$ divide each other. (ii) There exists an integer $a$ satisfying the CRT equivalences  $$ a \equiv-s_{i} \quad\left(\bmod t_{i, j}\right) \quad \forall i \leq j $$  Then the sequence $a+s_{1}, a+s_{2}, \ldots, a+s_{n}$ will work. For example, when $n=3$ we can take $\left(d_{1}, d_{2}\right)=(2,3)$ giving ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-05.jpg?height=152&width=310&top_left_y=1215&top_left_x=876) because the only conditions we need satisfy are  $$ \begin{aligned} a & \equiv 0 \quad(\bmod 2) \\ a & \equiv 0 \quad(\bmod 5) \\ a & \equiv-2 \quad(\bmod 3) . \end{aligned} $$  But with this setup we can just construct the $d_{i}$ inductively. To go from $n$ to $n+1$, take a $d_{1}, \ldots, d_{n-1}$ and let $p$ be a prime not dividing any of the $d_{i}$. Moreover, let $M$ be a multiple of $\prod_{i \leq j} t_{i, j}$ coprime to $p$. Then we claim that $d_{1} M, d_{2} M, \ldots, d_{n-1} M, p$ is such a difference sequence. For example, the previous example extends as follows with $M=300$ and $p=7$. ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-05.jpg?height=192&width=366&top_left_y=1886&top_left_x=848)  The new numbers $p, p+M t_{n-1, n}, p+M t_{n-2, n}, \ldots$ are all relatively prime to everything else. Hence (i) still holds. To see that (ii) still holds, just note that we can still get a family of solutions for the first $n$ terms, and then the last $(n+1)$ st term can be made to work by Chinese Remainder Theorem since all the new $p+M t_{n-2, n}$ are coprime to everything.


problem: A physicist encounters 2015 atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is $100 \%$ sure are currently in the same state. Is there any series of diode usage that makes this possible?

solution: The answer is no. Call the usamons $U_{1}, \ldots, U_{m}$ (here $m=2015$ ). Consider models $M_{k}$ of the following form: $U_{1}, \ldots, U_{k}$ are all charged for some $0 \leq k \leq m$ and the other usamons are not charged. Note that for any pair there's a model where they are different states, by construction. We can consider the physicist as acting on these $m+1$ models simultaneously, and trying to reach a state where there's a pair in all models which are all the same charge. (This is a necessary condition for a winning strategy to exist.)  But we claim that any diode operation $U_{i} \rightarrow U_{j}$ results in the $m+1$ models being an isomorphic copy of the previous set. If $i<j$ then the diode operation can be interpreted as just swapping $U_{i}$ with $U_{j}$, which doesn't change anything. Moreover if $i>j$ the operation never does anything. The conclusion follows from this.  Remark. This problem is not a "standard" olympiad problem, so I can't say it's trivial. But the idea is pretty natural I think.  You can motivate it as follows: there's a sequence of diode operations you can do which forces the situation to be one of the $M_{k}$ above: first, use the diode into $U_{1}$ for all other $U_{i}$ 's, so that either no electrons exist at all or $U_{1}$ has an electron. Repeat with the other $U_{i}$. This leaves us at the situation described at the start of the problem. Then you could guess the answer was "no" just based on the fact that it's impossible for $n=2,3$ and that there doesn't seem to be a reasonable strategy.  In this way it's possible to give a pretty good description of what it's possible to do. One possible phrasing: "the physicist can arrange the usamons in a line such that all the charged usamons are to the left of the un-charged usamons, but can't determine the number of charged usamons".


problem: Let $f: \mathbb{Q} \rightarrow \mathbb{Q}$ be a function such that for any $x, y \in \mathbb{Q}$, the number $f(x+y)-$ $f(x)-f(y)$ is an integer. Decide whether there must exist a constant $c$ such that $f(x)-c x$ is an integer for every rational number $x$.

solution: No, such a constant need not exist.  $$ \begin{aligned} & 2 x_{1}=x_{0} \\ & 2 x_{2}=x_{1}+1 \\ & 2 x_{3}=x_{2} \\ & 2 x_{4}=x_{3}+1 \\ & 2 x_{5}=x_{4} \\ & 2 x_{6}=x_{5}+1 \end{aligned} $$  Set $f\left(2^{-k}\right)=x_{k}$ and $f\left(2^{k}\right)=2^{k}$ for $k=0,1, \ldots$ Then, let  $$ f\left(a \cdot 2^{k}+\frac{b}{c}\right)=a f\left(2^{k}\right)+\frac{b}{c} $$  for odd integers $a, b, c$. One can verify this works.  $$ f\left(\frac{p}{q}\right)=\frac{p}{q}(1!+2!+\cdots+q!) . $$  Remark. Silly note: despite appearances, $f(x)=\lfloor x\rfloor$ is not a counterexample since one can take $c=0$.


problem: Fix a positive integer $n$. A tournament on $n$ vertices has all its edges colored by $\chi$ colors, so that any two directed edges $u \rightarrow v$ and $v \rightarrow w$ have different colors. Over all possible tournaments on $n$ vertices, determine the minimum possible value of $\chi$.

solution: The answer is  $$ \chi=\left\lceil\log _{2} n\right\rceil $$  First, we prove by induction on $n$ that $\chi \geq \log _{2} n$ for any coloring and any tournament. The base case $n=1$ is obvious. Now given any tournament, consider any used color $c$. Then it should be possible to divide the tournament into two subsets $A$ and $B$ such that all $c$-colored edges point from $A$ to $B$ (for example by letting $A$ be all vertices which are the starting point of a $c$-edge). ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-08.jpg?height=486&width=512&top_left_y=1116&top_left_x=772)  One of $A$ and $B$ has size at least $n / 2$, say $A$. Since $A$ has no $c$ edges, and uses at least $\log _{2}|A|$ colors other than $c$, we get  $$ \chi \geq 1+\log _{2}(n / 2)=\log _{2} n $$  completing the induction. One can read the construction off from the argument above, but here is a concrete description. For each integer $n$, consider the tournament whose vertices are the binary representations of $S=\{0, \ldots, n-1\}$. Instantiate colors $c_{1}, c_{2}, \ldots$. Then for $v, w \in S$, we look at the smallest order bit for which they differ; say the $k$ th one. If $v$ has a zero in the $k$ th bit, and $w$ has a one in the $k$ th bit, we draw $v \rightarrow w$. Moreover we color the edge with color $c_{k}$. This works and uses at most $\left\lceil\log _{2} n\right\rceil$ colors.  Remark (Motivation). The philosophy "combinatorial optimization" applies here. The idea is given any color $c$, we can find sets $A$ and $B$ such that all $c$-edges point $A$ to $B$. Once you realize this, the next insight is to realize that you may as well color all the edges from $A$ to $B$ by $c$; after all, this doesn't hurt the condition and makes your life easier. Hence, if $f$ is the answer, we have already a proof that $f(n)=1+\max (f(|A|), f(|B|))$ and we choose $|A| \approx|B|$. This optimization also gives the inductive construction.


problem: Let $A B C$ be a non-equilateral triangle and let $M_{a}, M_{b}, M_{c}$ be the midpoints of the sides $B C, C A, A B$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X, Y, Z$ the second intersections of $M_{a} S, M_{b} S, M_{c} S$ with the nine-point circle. Prove that $A X, B Y, C Z$ are concurrent.

solution: We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. 【 First solution by barycentric coordinates (Evan Chen). Let $A X$ meet $M_{b} M_{c}$ at $D$, and let $X$ reflected over $M_{b} M_{c}^{\prime}$ 's midpoint be $X^{\prime}$. Let $Y^{\prime}, Z^{\prime}, E, F$ be similarly defined. ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-09.jpg?height=687&width=807&top_left_y=1070&top_left_x=633)  By Cevian Nest Theorem it suffices to prove that $M_{a} D, M_{b} E, M_{c} F$ are concurrent. Taking the isotomic conjugate and recalling that $M_{a} M_{b} A M_{c}$ is a parallelogram, we see that it suffices to prove $M_{a} X^{\prime}, M_{b} Y^{\prime}, M_{c} Z^{\prime}$ are concurrent.  We now use barycentric coordinates on $\triangle M_{a} M_{b} M_{c}$. Let  $$ S=\left(a^{2} S_{A}+t: b^{2} S_{B}+t: c^{2} S_{C}+t\right) $$  (possibly $t=\infty$ if $S$ is the centroid). Let $v=b^{2} S_{B}+t, w=c^{2} S_{C}+t$. Hence  $$ X=\left(-a^{2} v w:\left(b^{2} w+c^{2} v\right) v:\left(b^{2} w+c^{2} v\right) w\right) $$  Consequently,  $$ X^{\prime}=\left(a^{2} v w:-a^{2} v w+\left(b^{2} w+c^{2} v\right) w:-a^{2} v w+\left(b^{2} w+c^{2} v\right) v\right) $$  We can compute  $$ b^{2} w+c^{2} v=(b c)^{2}\left(S_{B}+S_{C}\right)+\left(b^{2}+c^{2}\right) t=(a b c)^{2}+\left(b^{2}+c^{2}\right) t $$  Thus  $$ -a^{2} v+b^{2} w+c^{2} v=\left(b^{2}+c^{2}\right) t+(a b c)^{2}-(a b)^{2} S_{B}-a^{2} t=S_{A}\left((a b)^{2}+t\right) $$  Finally  $$ X^{\prime}=\left(a^{2} v w: S_{A}\left(c^{2} S_{C}+t\right)\left((a b)^{2}+2 t\right): S_{A}\left(b^{2} S_{B}+t\right)\left((a c)^{2}+2 t\right)\right) $$  and from this it's evident that $A X^{\prime}, B Y^{\prime}, C Z^{\prime}$ are concurrent.


solution: We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. \ Second solution by moving points (Anant Mudgal). Let $H_{a}, H_{b}, H_{c}$ be feet of altitudes, and let $\gamma$ denote the nine-point circle. The main claim is that:  Claim - Lines $X H_{a}, Y H_{b}, Z H_{c}$ are concurrent,   $$ \begin{aligned} & \ell \rightarrow \gamma \rightarrow \ell \\ & S \mapsto X \mapsto S_{a}:=\ell \cap \overline{H_{a} X} \end{aligned} $$  is projective, because it consists of two perspectivities. So we want the analogous maps $S \mapsto S_{b}, S \mapsto S_{c}$ to coincide. For this it suffices to check three positions of $S$; since you're such a good customer here are four.  - If $S$ is the orthocenter of $\triangle M_{a} M_{b} M_{c}$ (equivalently the circumcenter of $\triangle A B C$ ) then $S_{a}$ coincides with the circumcenter of $M_{a} M_{b} M_{c}$ (equivalently the nine-point center of $\triangle A B C)$. By symmetry $S_{b}$ and $S_{c}$ are too. - If $S$ is the circumcenter of $\triangle M_{a} M_{b} M_{c}$ (equivalently the nine-point center of $\triangle A B C$ ) then $S_{a}$ coincides with the de Longchamps point of $\triangle M_{a} M_{b} M_{c}$ (equivalently orthocenter of $\triangle A B C)$. By symmetry $S_{b}$ and $S_{c}$ are too. - If $S$ is either of the intersections of the Euler line with $\gamma$, then $S=S_{a}=S_{b}=S_{c}$ (as $S=X=Y=Z$ ).  ![](https://cdn.mathpix.com/cropped/2024_11_19_2c252c8bc1550e62c4c8g-10.jpg?height=681&width=797&top_left_y=1947&top_left_x=638)  We now use Trig Ceva to carry over the concurrence. By sine law,  $$ \frac{\sin \angle M_{c} A X}{\sin \angle A M_{c} X}=\frac{M_{c} X}{A X} $$  and a similar relation for $M_{b}$ gives that  $$ \frac{\sin \angle M_{c} A X}{\sin \angle M_{b} A X}=\frac{\sin \angle A M_{c} X}{\sin \angle A M_{b} X} \cdot \frac{M_{c} X}{M_{b} X}=\frac{\sin \angle A M_{c} X}{\sin \angle A M_{b} X} \cdot \frac{\sin \angle X M_{a} M_{c}}{\sin \angle X M_{a} M_{b}} . $$  Thus multiplying cyclically gives  $$ \prod_{\text {cyc }} \frac{\sin \angle M_{c} A X}{\sin \angle M_{b} A X}=\prod_{\text {cyc }} \frac{\sin \angle A M_{c} X}{\sin \angle A M_{b} X} \prod_{\text {cyc }} \frac{\sin \angle X M_{a} M_{c}}{\sin \angle X M_{a} M_{b}} . $$  The latter product on the right-hand side equals 1 by Trig Ceva on $\triangle M_{a} M_{b} M_{c}$ with cevians $\overline{M_{a} X}, \overline{M_{b} Y}, \overline{M_{c} Z}$. The former product also equals 1 by Trig Ceva for the concurrence in the previous claim (and the fact that $\angle A M_{c} X=\angle H_{c} H_{a} X$ ). Hence the left-hand side equals 1 , implying the result.


solution: We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. 『 Third solution by moving points (Gopal Goel). In this solution, we will instead use barycentric coordinates with resect to $\triangle A B C$ to bound the degrees suitably, and then verify for seven distinct choices of $S$.  We let $R$ denote the radius of $\triangle A B C$, and $N$ the nine-point center. First, imagine solving for $X$ in the following way. Suppose $\vec{X}=\left(1-t_{a}\right) \vec{M}_{a}+t_{a} \vec{S}$. Then, using the dot product (with $|\vec{v}|^{2}=\vec{v} \cdot \vec{v}$ in general)  $$ \begin{aligned} \frac{1}{4} R^{2} & =|\vec{X}-\vec{N}|^{2} \\ & =\left|t_{a}\left(\vec{S}-\vec{M}_{a}\right)+\vec{M}_{a}-\vec{N}\right|^{2} \\ & =\left|t_{a}\left(\vec{S}-\vec{M}_{a}\right)\right|^{2}+2 t_{a}\left(\vec{S}-\vec{M}_{a}\right) \cdot\left(\vec{M}_{a}-\vec{N}\right)+\left|\vec{M}_{a}-\vec{N}\right|^{2} \\ & =t_{a}^{2}\left|\left(\vec{S}-\vec{M}_{a}\right)\right|^{2}+2 t_{a}\left(\vec{S}-\vec{M}_{a}\right) \cdot\left(\vec{M}_{a}-\vec{N}\right)+\frac{1}{4} R^{2} \end{aligned} $$  Since $t_{a} \neq 0$ we may solve to obtain  $$ t_{a}=-\frac{2\left(\vec{M}_{a}-\vec{N}\right) \cdot\left(\vec{S}-\vec{M}_{a}\right)}{\left|\vec{S}-\vec{M}_{a}\right|^{2}} $$  Now imagine $S$ varies along the Euler line, meaning there should exist linear functions $\alpha, \beta, \gamma: \mathbb{R} \rightarrow \mathbb{R}$ such that  $$ S=(\alpha(s), \beta(s), \gamma(s)) \quad s \in \mathbb{R} $$  with $\alpha(s)+\beta(s)+\gamma(s)=1$. Thus $t_{a}=\frac{f_{a}}{g_{a}}=\frac{f_{a}(s)}{g_{a}(s)}$ is the quotient of a linear function $f_{a}(s)$ and a quadratic function $g_{a}(s)$.  So we may write:  $$ \begin{aligned} X & =\left(1-t_{a}\right)\left(0, \frac{1}{2}, \frac{1}{2}\right)+t_{a}(\alpha, \beta, \gamma) \\ & =\left(t_{a} \alpha, \frac{1}{2}\left(1-t_{a}\right)+t_{a} \beta, \frac{1}{2}\left(1-t_{a}\right)+t_{a} \gamma\right) \end{aligned} $$  $$ =\left(2 f_{a} \alpha: g_{a}-f_{a}+2 f_{a} \beta: g_{a}-f_{a}+2 f_{a} \gamma\right) . $$  Thus the coordinates of $X$ are quadratic polynomials in $s$ when written in this way. In a similar way, the coordinates of $Y$ and $Z$ should be quadratic polynomials in $s$. The Ceva concurrence condition  $$ \prod_{\text {cyc }} \frac{g_{a}-f_{a}+2 f_{a} \beta}{g_{a}-f_{a}+2 f_{a} \gamma}=1 $$  is thus a polynomial in $s$ of degree at most six. Our goal is to verify it is identically zero, thus it suffices to check seven positions of $S$.  - If $S$ is the circumcenter of $\triangle M_{a} M_{b} M_{c}$ (equivalently the nine-point center of $\triangle A B C$ ) then $\overline{A X}, \overline{B Y}, \overline{C Z}$ are altitudes of $\triangle A B C$. - If $S$ is the centroid of $\triangle M_{a} M_{b} M_{c}$ (equivalently the centroid of $\triangle A B C$ ), then $\overline{A X}$, $\overline{B Y}, \overline{C Z}$ are medians of $\triangle A B C$. - If $S$ is either of the intersections of the Euler line with $\gamma$, then $S=X=Y=Z$ and all cevians concur at $S$. - If $S$ lies on the $\overline{M_{a} M_{b}}$, then $Y=M_{a}, X=M_{c}$, and thus $\overline{A X} \cap \overline{B Y}=C$, which is of course concurrent with $\overline{C Z}$ (regardless of $Z$ ). Similarly if $S$ lies on the other sides of $\triangle M_{a} M_{b} M_{c}$.  Thus we are also done.


solution: We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. 【 Fourth solution using Pascal (official one). We give a different proof of the claim that $\overline{X H_{a}}, \overline{Y H_{b}}, \overline{Z H_{c}}$ are concurrent (and then proceed as in the end of the second solution).  Let $H$ denote the orthocenter, $N$ the nine-point center, and moreover let $N_{a}, N_{b}, N_{c}$ denote the midpoints of $\overline{A H}, \overline{B H}, \overline{C H}$, which also lie on the nine-point circle (and are the antipodes of $M_{a}, M_{b}, M_{c}$ ).  - By Pascal's theorem on $M_{b} N_{b} H_{b} M_{c} N_{c} H_{c}$, the point $P=\overline{M_{c} H_{b}} \cap \overline{M_{b} H_{c}}$ is collinear with $N=\overline{M_{b} N_{b}} \cap \overline{M_{c} N_{c}}$, and $H=\overline{N_{b} H_{b}} \cap \overline{N_{c} H_{c}}$. So $P$ lies on the Euler line. - By Pascal's theorem on $M_{b} Y H_{b} M_{c} Z H_{c}$, the point $\overline{Y H_{b}} \cap \overline{Z H_{c}}$ is collinear with $S=\overline{M_{b} Y} \cap \overline{M_{c} Z}$ and $P=\overline{M_{b} H_{c}} \cap \overline{M_{c} H_{b}}$. Hence $Y H_{b}$ and $Z H_{c}$ meet on the Euler line, as needed.