problem:
Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\ell$, respectively. Prove that $K P+K Q \geq B C$.

# Russia, Vasily Mokin

solution:
. Fix the origin at $O$ and the real axis along $\ell$. A lower case letter denotes the complex coordinate of the corresponding point in the configuration. For convenience, let $|a|=|b|=|c|=1$.

Clearly, $k=a+\frac{1}{2}(b+c), p=a+\frac{1}{2}\left(b+\frac{1}{b}\right)$ and $q=a+\frac{1}{2}\left(c+\frac{1}{c}\right)$.

Then $|k-p|=\left|a+\frac{1}{2}\left(c-\frac{1}{b}\right)\right|=\frac{1}{2}|2 a b+b c-1|$, since $|b|=1$.

Similarly, $|k-q|=\frac{1}{2}|2 a c+b c-1|$, so, since $|a|=1$,
$|k-p|+|k-q|=\frac{1}{2}|2 a b+b c-1|+\frac{1}{2}|2 a c+b c-1|$

$$
\geq \frac{1}{2}|2 a(b-c)|=|b-c|,
$$

as required.

problem_match: Problem 4.

solution_match: Solution 1


solution:
. Let $M$ be the midpoint of $B C$, and let $R$ be the projection of $M$ onto $\ell$. In other words, $R$ is the midpoint of $P Q$. Since $\angle B P O=\angle B M O=$ $90^{\circ}$, the points $B, P, O$, and $M$ are concyclic, so $\angle(O M, O B)=\angle(P M, P B)=\angle(P M, M R)$, so the right triangles $M R P$ and $O M B$ are similar and have different orientation. Similarly, the triangles $M R Q$ and $O M C$ are similar and have different orientation, hence so are the triangles $O B C$ and $M P Q$.
![](https://cdn.mathpix.com/cropped/2024_11_22_d8cbf77d76bc852500acg-1.jpg?height=716&width=846&top_left_y=1478&top_left_x=128)

Recall that $\overrightarrow{A H}=2 \overrightarrow{O M}$, so $\overrightarrow{O M}=\overrightarrow{A K}$. Hence $A O M K$ is a parallelogram, so $M K=O A=O B=$ $O C$.

Consider the rotation through $\angle(\overrightarrow{O C}, \overrightarrow{O B})$ about $M$. It maps $P$ to $Q$; let it map $K$ to some point $L$. Then $M K=M L=O B=O C$ and $\angle L M K=\angle B O C$, so the triangles $O B C$ and $M K L$ are congruent. Hence $B C=K L \leq K Q+L Q=$ $K Q+K P$, as required.

solution_match: Solution 2


solution:
. Let $\alpha=\angle(P B, B C)=\angle(Q C, B C)$. Since $P$ lies on the circle of diameter $O B$, $\angle(O P, O M)=\alpha$. Since also $Q$ lies on the circle of diameter $O C$, it immediately follows that
$M P=M Q=R \sin \alpha$ by sine theorem in triangles $\triangle O P M$ and $\triangle O Q M$.

Because $P Q$ is the projection of $B C$ on line $\ell$, it follows that $P Q=B C \sin \alpha$. Just like in the first solution, $K M=A O=R$ (the circumradius of triangle $\triangle A B C)$.

Now apply Ptolemy's inequality for the quadrilateral $K P M Q$ : $K P \cdot M Q+K Q \cdot M P \geq P Q \cdot K M$, and now substitute the relations from above, leading to

$$
R \sin \alpha(K P+K Q) \geq R \sin \alpha \cdot B C
$$

which is precisely the conclusion whenever $\sin \alpha \neq 0$. The case when $\sin \alpha=0$ can be treated either directly, or via a limit argument.

solution_match: Solution 3


solution:
. Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.

Assume now that $\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $K$ on this circle we have $B C \leq K P+K Q$. In doing so, we prove that the afore-mentioned circle contains an ellipse with foci at $Q$ and $P$ with distance $B C$.

Let $S$ be the foot of the perpendicular from $M$ to $P Q$, it is easy to verify that $S$ is the center of the ellipse. We shall then consider it as the origin. Let $u=\frac{B C}{2}$ and $t=\frac{P Q}{2}$; notice that $u$ is the major semi-axis of the ellipse and $\sqrt{u^{2}-t^{2}}$ is the minor one. Assume $X(x, y)$ is a point on this ellipse. We now need to prove $M X \leq R$.

Since $X$ is on the ellipse, we can write $(x, y)=$ $\left(u \cos \theta, \sqrt{u^{2}-t^{2}} \sin \theta\right)$, for some $\theta \in(0,2 \pi)$. Since $M X^{2}=x^{2}+(y+M S)^{2}$, we can expand and obtain $M X^{2}=u^{2}+M S^{2}-t^{2} \cdot \sin ^{2} \theta+2 M S \cdot \sqrt{u^{2}-t^{2}} \cdot \sin \theta$.

Add and subtract $M S^{2}\left(u^{2}-t^{2}\right) / t^{2}$ in order to obtain a square on the right hand side: $M X^{2}=u^{2}+$ $M S^{2}+\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}}-\left(t \sin \theta-\frac{M S \sqrt{u^{2}-t^{2}}}{t}\right)^{2}$. It now suffices to show that $u^{2}+M S^{2}+\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}}=$ $R^{2}$, since then it would immediately follow that $M X^{2} \leq R^{2}$.

Applying Pythagorean theorem in triangles $O B M$ and $O S M$, we obtain $R^{2}=u^{2}+O M^{2}$ and $O M^{2}=M S^{2}+O S^{2}$, so it remains to prove that $O S^{2}=\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}} . \quad$ Let $\alpha=\angle(O P, B M)$, then $O S / M S=\tan ^{t^{2}} \alpha$ and $t / u=\cos \alpha$, so $O S^{2}=$ $M S^{2} \tan ^{2} \alpha=M S^{2}\left(\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}\right)=M S^{2} \cdot \frac{u^{2}-t^{2}}{t^{2}}$, which is the desired result.

solution_match: Solution 4


problem:
Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$.
Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.

## Iran, NaVid Safaei

solution:
. Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\operatorname{deg} Q \mid \operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired.

Choose the polynomials $T(x)$ and $M(x)$ such that

$$
S(x)=T(Q(x))+M(x)
$$

where $\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \neq 0$. Then $q \nmid m=\operatorname{deg} M$; otherwise, $M(x)=\beta Q(x)^{m / q}+M_{1}(x)$, where $\beta$ is some number and $\operatorname{deg} M_{1}<\operatorname{deg} M$, contradicting the choice of $M$. In particular, $0<m<s$ and hence $\operatorname{deg} T(Q(x))=s$.

Substitute now (*) into $R(S(x))-P(Q(x))=$ 0 ; let $\alpha$ be the leading coefficient of $R(x)$ and let $r=\operatorname{deg} R(x)$. Expand the brackets to get a sum of powers of $Q(x)$ and other terms including powers of $M(x)$ as well. Amongst the latter, the unique term of highest degree is $\operatorname{arM}(x) T(Q(x))^{r-1}$. So, for some polyno$\operatorname{mial} N(x), \quad N(Q(x))=\alpha r M(x) T(Q(x))^{r-1}+$ a polynomial of lower degree.

This is impossible, since $q$ divides the degree of the left-hand member, but not that of the righthand member.

problem_match: Problem 5.


solution:
. All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\operatorname{deg} f(x)$.

Of all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minimal (positive) degree. We will show that $\operatorname{deg} R_{0}(x)=1$, say, $R_{0}(x)=\alpha x+\beta$ for some real numbers $\alpha \neq 0$ and $\beta$, so $P_{0}(Q(x))=\alpha S(x)+\beta$. Hence $S(x)=T(Q(x))$ for some polynomial $T(x)$.

Now, if $P(x)$ and $R(x)$ are polynomials satisfying $P(Q(x))=R(S(x))$, then $P(Q(x))=$ $R(T(Q(x)))$. Since $Q(x)$ is not constant, it takes infinitely many values, so $P(x)$ and $R(T(x))$ agree at infinitely many points, implying that $P(x)=$ $R(T(x))$, as required.

It is therefore sufficient to solve the problem in the particular case where $F(x)=P(Q(x))=$ $R(S(x))$ has a minimal degree. Let $d=$ $\operatorname{gcd}(\operatorname{deg} Q(x), \operatorname{deg} S(x))$ to write $\operatorname{deg} Q(x)=a d$ and $\operatorname{deg} S(x)=b d$, where $\operatorname{gcd}(a, b)=1$. Then
$\operatorname{deg} P(x)=b c, \operatorname{deg} R(x)=a c$ and $\operatorname{deg} F(x)=a b c d$ for some positive integer $c$. We will show that minimality of $\operatorname{deg} F(x)$ forces $c=1$, so $\operatorname{deg} P(x)=b$, $\operatorname{deg} R(x)=a$ and $\operatorname{deg} F(x)=a b d$. The conditions $a=\operatorname{deg} R(x) \mid \operatorname{deg} P(x)=b$ and $\operatorname{gcd}(a, b)=1$ then force $a=1$, as stated above.

Consequently, the only thing we are left with is the proof of the fact that $c=1$. For convenience, we may and will assume that $P(x), Q(x), R(x), S(x)$ are all monic; hence so is $F(x)$. The argument hinges on the lemma below.
Lemma. If $f(x)$ is a monic polynomial of degree $m n$, then there exists a degree $n$ monic polynomial $g(x)$ such that $\operatorname{deg}\left(f(x)-g(x)^{m}\right)<(m-1) n$. (If $m=0$ or 1 , or $n=0$, the conclusion is still consistent with the usual convention that the identically zero polynomial has degree $-\infty$.)
Proof. Write $f(x)=\sum_{k=0}^{m n} \alpha_{k} x^{k}, \alpha_{m n}=1$, and seek $g(x)=\sum_{k=0}^{n} \beta_{k} x^{k}, \beta_{n}=1$, so as to fit the bill. To this end, notice that, for each positive integer $k \leq n$, the coefficient of $x^{m n-k}$ in the expansion of $g(x)^{m}$ is of the form $m \beta_{n-k}+\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$, where $\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$ is an algebraic expression in $\beta_{n}, \ldots, \beta_{n-k+1}$. Recall that $\beta_{n}=1$ to determine the $\beta_{n-k}$ recursively by requiring $\beta_{n-k}=$ $\frac{1}{m}\left(a_{m n-k}-\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)\right), k=1, \ldots, n$.

The outcome is then the desired polynomial $g(x)$.

We are now in a position to prove that $c=$ 1. Suppose, if possible, that $c>1$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\operatorname{deg}\left(P(x)-U(x)^{c}\right)<(c-1) b$ and $\operatorname{deg}(R(x)-$ $\left.V(x)^{c}\right)<(c-1) a$. Then $\operatorname{deg}\left(F(x)-U(Q(x))^{c}\right)=$ $\operatorname{deg}\left(P(Q(x))-U(Q(x))^{c}\right)<(c-1) a b d, \operatorname{deg}(F(x)-$ $\left.V(S(x))^{c}\right)=\operatorname{deg}\left(R(S(x))-V(S(x))^{c}\right)<(c-1) a b d$, so $\operatorname{deg}\left(U(Q(x))^{c}-V(S(x))^{c}\right)=\operatorname{deg}((F(x)-$ $\left.\left.V(S(x))^{c}\right)-\left(F(x)-U(Q(x))^{c}\right)\right)<(c-1) a b d$.

On the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\left(U(Q(x))^{c-1}+\cdots+\right.$ $\left.V(S(x))^{c-1}\right)$.

By the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d<a b c d=$ $\operatorname{deg} F(x)$ - a contradiction. Consequently, $c=1$. This completes the argument and concludes the proof.


problem:
Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has

- a spanning tree in which exactly $r$ of the edges are red,
- a spanning tree in which exactly $g$ of the edges are green and
- a spanning tree in which exactly $b$ of the edges are blue.

Prove that $\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue.

Russia, Vasily Mokin

solution:
. Induct on $n=r+g+b$. The base case, $n=1$, is clear.

Let now $n>1$. Let $V$ denote the vertex set of $\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.
Case 1: There exists a partition $V=A \sqcup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.

Since $\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.

Assume that $T$, one of the three trees, does not contain $e$. Then the graph $T \cup\{e\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e^{\prime}$ connecting $A$ and $B$; the edge $e^{\prime}$ is also blue. Replace $e^{\prime}$ by $e$ in $T$ to get another tree $T^{\prime}$ with the same number of edges of each colour as in $T$, but containing $e$.

Performing such an operation to all three trees, we arrive at the situation where the three trees $T_{r}^{\prime}$, $T_{g}^{\prime}$, and $T_{b}^{\prime}$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\Gamma^{*}$, and set $r^{*}=r, g^{*}=g$, and $b^{*}=b-1$. The new graph satisfies the conditions in the statement for those new values - indeed, under the shrinking, each of the trees $T_{r}^{\prime}, T_{g}^{\prime}$, and $T_{b}^{\prime}$ loses a blue edge. So $\Gamma^{*}$ has a spanning tree with exactly $r$ red, exactly $g$
green, and exactly $b-1$ blue edges. Finally, pass back to $\Gamma$ by restoring $e$, to obtain the a desired spanning tree in $\Gamma$.

Case 2: There is no such a partition.
Consider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \cup G \cup B$ has the smallest number $k$ of components. If $k=1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).

Assume now that $k \geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \backslash V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.

Let $e^{\prime}$ be a red edge in $C$ and set $R^{\prime}=R \backslash\left\{e^{\prime}\right\} \cup$ $\{e\}$. Then $\left(R^{\prime}, G, B\right)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.

problem_match: Problem 6.


problem:
Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has

- a spanning tree in which exactly $r$ of the edges are red,
- a spanning tree in which exactly $g$ of the edges are green and
- a spanning tree in which exactly $b$ of the edges are blue.

Prove that $\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue.

solution:
. For a spanning tree $T$ in $\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.

Assume that $\mathcal{C}$ is some collection of spanning trees in $\Gamma$. Write

$$
\begin{array}{rlrl}
r(\mathcal{C}) & =\min _{T \in \mathcal{C}} r(T), & & g(\mathcal{C})=\min _{T \in \mathcal{C}} g(T) \\
b(\mathcal{C}) & =\min _{T \in \mathcal{C}} b(T), & & R(\mathcal{C})=\max _{T \in \mathcal{C}}(T) \\
G(\mathcal{C}) & =\max _{T \in \mathcal{C}} g(T), & B(\mathcal{C})=\max _{T \in \mathcal{C}} b(T)
\end{array}
$$

Say that a collection $\mathcal{C}$ is good if $r \in[r(\mathcal{C}, R(\mathcal{C})]$, $g \in[g(\mathcal{C}, G(\mathcal{C})]$, and $b \in[b(\mathcal{C}, B(\mathcal{C})]$. By the problem conditions, the collection of all spanning trees in $\Gamma$ is good.

For a good collection $\mathcal{C}$, say that an edge $e$ of $\Gamma$ is suspicious if $e$ belongs to some tree in $\mathcal{C}$ but not to all trees in $\mathcal{C}$. Choose now a good collection $\mathcal{C}$ minimizing the number of suspicious edges. If $\mathcal{C}$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(\mathcal{C})<r$ and $G(\mathcal{C})>g$.

We now distinguish two cases.
Case 1: $B(\mathcal{C})=b$ 。
Let $T^{0}$ be a tree in $\mathcal{C}$ with $g\left(T^{0}\right)=g(\mathcal{C}) \leq g$. Since $G(\mathcal{C})>g$, there exists a green edge $e$ contained in some tree in $\mathcal{C}$ but not in $T^{0}$; clearly, $e$ is suspicious. Fix one such green edge $e$.

Now, for every $T$ in $\mathcal{C}$, define a spanning tree $T_{1}$ of $\Gamma$ as follows. If $T$ does not contain $e$, then $T_{1}=T$; in particular, $\left(T^{0}\right)_{1}=T^{0}$. Otherwise, the graph $T \backslash\{e\}$ falls into two components. The tree $T_{0}$ contains some edge $e^{\prime}$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_{1}=T \backslash\{e\} \cup\left\{e^{\prime}\right\}$.

Let $\mathcal{C}_{1}=\left\{T_{1}: T \in \mathcal{C}\right\}$. All edges suspicious for $\mathcal{C}_{1}$ are also suspicious for $\mathcal{C}$, but no tree in $\mathcal{C}_{1}$ con-

## Russia, Vasily Mokin

tains $e$. So the number of suspicious edges for $\mathcal{C}_{1}$ is strictly smaller than that for $\mathcal{C}$.

We now show that $\mathcal{C}_{1}$ is good, reaching thereby a contradiction with the choice of $\mathcal{C}$. For every $T$ in $\mathcal{C}$, the tree $T_{1}$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g\left(\mathcal{C}_{1}\right) \leq g(\mathcal{C}) \leq g, G\left(\mathcal{C}_{1}\right) \geq G(\mathcal{C})-1 \geq g$, $R\left(\mathcal{C}_{1}\right) \geq R(\mathcal{C}) \geq r, r\left(\mathcal{C}_{1}\right) \leq r(\mathcal{C})+1 \leq r$, and $B\left(\mathcal{C}_{1}\right) \geq B(\mathcal{C}) \geq b$. Finally, we get $b\left(T^{0}\right) \leq$ $B(\mathcal{C})=b$; since $\mathcal{C}_{1}$ contains $T^{0}$, it follows that $b\left(\mathcal{C}_{1}\right) \leq b\left(T^{0}\right) \leq b$, which concludes the proof.
Case 2: $B(\mathcal{C})>b$.
Consider a tree $T^{0}$ in $\mathcal{C}$ satisfying $r\left(T^{0}\right)=$ $R(\mathcal{C}) \geq r$. Since $r(\mathcal{C})<r$, the tree $T^{0}$ contains a suspicious red edge. Fix one such edge $e$.

Now, for every $T$ in $\mathcal{C}$, define a spanning tree $T_{2}$ of $\Gamma$ as follows. If $T$ contains $e$, then $T_{2}=T$; in particular, $\left(T^{0}\right)_{2}=T^{0}$. Otherwise, the graph $T \cup\{e\}$ contains a cycle $C$ through $e$. This cycle contains an edge $e^{\prime}$ absent from $T^{0}$ (otherwise $T^{0}$ would contain the cycle $C$ ), so $e^{\prime}$ is suspicious. Choose one such edge and define $T_{2}=T \backslash\left\{e^{\prime}\right\} \cup\{e\}$.

Let $\mathcal{C}_{2}=\left\{T_{2}: T \in \mathcal{C}\right\}$. All edges suspicious for $\mathcal{C}_{2}$ are also suspicious for $\mathcal{C}$, but all trees in $\mathcal{C}_{2}$ contain $e$. So the number of suspicious edges for $\mathcal{C}_{2}$ is strictly smaller than that for $\mathcal{C}$.

We now show that $\mathcal{C}_{2}$ is good, reaching again a contradiction. For every $T$ in $\mathcal{C}$, the tree $T_{2}$ either coincides with $T$ or is obtained from it by removing some edge and adding a red edge. This shows that $r\left(\mathcal{C}_{2}\right) \leq r(\mathcal{C})+1 \leq r, R\left(\mathcal{C}_{2}\right) \geq R(\mathcal{C}) \geq r$, $G\left(\mathcal{C}_{2}\right) \geq G(\mathcal{C})-1 \geq g, g\left(\mathcal{C}_{2}\right) \leq g(\mathcal{C}) \leq g$, $b\left(\mathcal{C}_{1}\right) \leq b(\mathcal{C}) \leq b$ and $B\left(\mathcal{C}_{2}\right) \geq B(\mathcal{C})-1 \geq b$. This concludes the proof.