# Problems and solutions 

## INDIVIDUAL COMPETITION

Problem I-1. Let $a, b, c$ be positive real numbers such that

$$
a+b+c=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
$$

Prove that

$$
2(a+b+c) \geq \sqrt[3]{7 a^{2} b+1}+\sqrt[3]{7 b^{2} c+1}+\sqrt[3]{7 c^{2} a+1}
$$

Find all triples $(a, b, c)$ for which equality holds.

Solution. From the AM-GM inequality, we obtain that

$$
\sqrt[3]{7 a^{2} b+1}=2 \cdot \sqrt[3]{a \cdot a \cdot\left(\frac{7 b}{8}+\frac{1}{8 a^{2}}\right)} \leq \frac{2}{3}\left(a+a+\frac{7 b}{8}+\frac{1}{8 a^{2}}\right)
$$

We have analogous upper bounds for $\sqrt[3]{7 b^{2} c+1}$ and $\sqrt[3]{7 c^{2} a+1}$. Adding up these three inequalities, we obtain that

$$
\sqrt[3]{7 a^{2} b+1}+\sqrt[3]{7 b^{2} c+1}+\sqrt[3]{7 c^{2} a+1} \leq \frac{2}{3}\left(\frac{23(a+b+c)}{8}+\frac{1}{8}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)\right)
$$

Using the condition of the problem, we obtain

$$
\sqrt[3]{7 a^{2} b+1}+\sqrt[3]{7 b^{2} c+1}+\sqrt[3]{7 c^{2} a+1} \leq 2(a+b+c)
$$

Equality holds if and only if $a, b$, and $c$ satisfy the system of equations

$$
\begin{aligned}
& a=\frac{7 b}{8}+\frac{1}{8 a^{2}} \\
& b=\frac{7 c}{8}+\frac{1}{8 b^{2}} \\
& c=\frac{7 a}{8}+\frac{1}{8 c^{2}}
\end{aligned}
$$

Note that this system actually implies the equation stipulated in the problem.

Defining $f(x)=\frac{8}{7}\left(x-\frac{1}{8 x^{2}}\right)$, we can rewrite the system as

$$
\begin{aligned}
& b=f(a) \\
& c=f(b)
\end{aligned}
$$

$$
a=f(c)
$$

We prove that $f(x)$ is a non-decreasing function. Let $u \geq v$. Then

$$
\begin{aligned}
f(u)-f(v) & =\frac{8}{7}\left((u-v)+\frac{1}{8 v^{2}}-\frac{1}{8 u^{2}}\right) \\
& =\frac{8}{7}\left((u-v)+\frac{(u-v)(u+v)}{8 u^{2} v^{2}}\right) \\
& =\frac{8}{7}(u-v)\left(1+\frac{u+v}{8 u^{2} v^{2}}\right) \\
& \geq 0 .
\end{aligned}
$$

Since the system of equations is cyclically symmetric, we may assume that $a=$ $\max \{a, b, c\}$. Since $a \geq b$, we have $b=f(a) \geq f(b)=c$, so $c=f(b) \geq f(c)=a$. In all, $c \geq a \geq b \geq c$, so $a=b=c$.

We now have to find the solutions of $f(a)=a$.

$$
\begin{aligned}
\frac{8}{7}\left(a-\frac{1}{8 a^{2}}\right) & =a \\
8 a-\frac{1}{a^{2}} & =7 a \\
\frac{1}{a^{2}} & =a \\
1 & =a^{3}
\end{aligned}
$$

Thus, equality holds if and only if $a=b=c=1$.

Problem I-2. Let $n$ be a positive integer. On a board consisting of $4 n \times 4 n$ squares, exactly $4 n$ tokens are placed so that each row and each column contains one token. In a step, a token is moved horizontally or vertically to a neighbouring square. Several tokens may occupy the same square at the same time. The tokens are to be moved to occupy all the squares of one of the two diagonals.

Determine the smallest number $k(n)$ such that for any initial situation, we can do it in at most $k(n)$ steps.

Solution. We shall prove that $k(n)=6 n^{2}$.

We define the distance from a given square to a given diagonal to be the minimal number of steps needed to get from the square to the diagonal. This equals the minimal number of horizontal steps needed to do that. It also equals the minimal number of vertical steps needed to do that.

Given a configuration of tokens, we define the distance from this configuration to a given diagonal to be the sum of distances of the tokens to that diagonal.

Choose the coordinate system so that the vertices of the board have coordinates $\pm 2 n$. Place a token on each of the $n$ squares the coordinates of whose centres satisfy $x>0$ and $y-x=n$. Now complete this configuration of tokens so that it has a rotational symmetry of $90^{\circ}$ about the origin. Then we have $4 n$ tokens, one in each row, one in each column. The distance from this configuration to either diagonal is $2 n \cdot n+2 n \cdot 2 n=6 n^{2}$. Therefore, $k(n) \geq 6 n^{2}$.

Now consider any configuration satisfying the conditions of the problem. We prove that $\leq 6 n^{2}$ steps suffice even if we only allow horizontal moves. I.e., the smallest of the
two distances from the given configuration to the diagonals is $\leq 6 n^{2}$. It suffices to prove that the sum of the two distances from the given configuration to the diagonals is $\leq 12 n^{2}$.

Observe that the sum of the two distances from the square with center $(x, y)$ to the two diagonals is $2 \max (|x|,|y|)$. This number can take values $1,3, \ldots, 4 n-1$. The squares where it takes a given value can be covered by two columns and two rows, so we can place at most four tokens there. Thus, the sum of the values for the $4 n$ tokens is $\leq 4((4 n-1)+(4 n-3)+\cdots+(2 n+1))=4 n \cdot 3 n=12 n^{2}$.

Problem I-3. Let $A B C$ be an isosceles triangle with $A C=B C$. Let $N$ be a point inside the triangle such that $2 \angle A N B=180^{\circ}+\angle A C B$. Let $D$ be the intersection of the line $B N$ and the line parallel to $A N$ that passes through $C$. Let $P$ be the intersection of the angle bisectors of the angles $C A N$ and $A B N$.

Show that the lines $D P$ and $A N$ are perpendicular.

Solution. Since $A C=B C$, there is a circle $k$ such that the lines $A C$ and $B C$ are the tangents to $k$ at the points $A$ and $B$. The condition defining the point $N$ implies that the point $N$ lies on the circle $k$.

By the tangent-chord theorem, we have $\angle B A N=\angle D B C$ and $\angle C A N=\angle A B D$. Since $D C$ is parallel to $A N$, we have $\angle C A N=\angle A C D$. Hence $\angle A C D=\angle A B N$, so the quadrilateral $A B C D$ is cyclic. It follows that $\angle C A D=\angle C B D=\angle B A N$.

We can conclude that the angle bisector of $\angle C A N$ is the angle bisector of $\angle B A D$. Hence the point $P$ is the incenter of the triangle $A B D$ and $D P$ is the angle bisector of $\angle A D B$.

We also note that since $C D$ is parallel to $A N$, we have $\angle A N D=\angle B D C=\angle B A C=$ $\angle B A N+\angle N A C=\angle C A D+\angle N A C=\angle N A D$. Hence $A D=N D$ and we conclude that the angle bisector of $\angle A D B$ is the perpendicular bisector of the segment $A N$.

Hence $D P$ is perpendicular to $A N$.

Remark 1. The first part of the solution can easily be deduced also by calculating the angles without noting that $A C$ and $B C$ are tangents to $k$.

Remark 2. It is a well-known fact that the intersection of the perpendicular bisector of the segment $A N$ and the angle bisector of the angle $\angle A B N$ lies on the circumcircle of the triangle $A B N$, i.e. $P$ also lies on the circle $k$. This fact is obtained more easily by calculating the angle $\angle A P B$.

Problem I-4. Let $a$ and $b$ be positive integers. Prove that there exist positive integers $x$ and $y$ such that

$$
\binom{x+y}{2}=a x+b y
$$

Solution. Denoting $A=2 a+1$ and $B=2 b+1$, the equation can be translated into

$$
\frac{B-(x+y)}{x}=\frac{(x+y)-A}{y}
$$

For $A=B$, any integers with $x+y=A$ satisfy the equation. Now suppose that $A<B$. Let $n$ be the integer in the interval $[A, B)$ which is divisible by $d=B-A$. Then $n \neq A$ because $A$ is odd and $d$ is even. Now choose

$$
x=(B-n) \frac{n}{d}, \quad y=(n-A) \frac{n}{d}
$$

Here $n=x+y$, so the equation is satisfied.

## TEAM COMPETITION

Problem T-1. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(x f(x)+2 y)=f\left(x^{2}\right)+f(y)+x+y-1
$$

for all $x, y \in \mathbb{R}$.

Solution. Putting $x=y=0$, we get $f(0)=1$.

Putting $x=0, y=z$, we get

$$
f(2 z)=f(z)+z
$$

Putting $x=z, y=-z f(z)$, we get

$$
f\left(z^{2}\right)=z f(z)-z+1
$$

Replacing $z$ by $2 z$ in (2) and using (1), we obtain

$$
f\left(4 z^{2}\right)=2 z f(2 z)-2 z+1=2 z(f(z)+z)-2 z+1=2 z f(z)+2 z^{2}-2 z+1
$$

Using (1) for $2 z^{2}$ and then for $z^{2}$ in place of $z$, and afterwards using (2), we obtain

$$
f\left(4 z^{2}\right)=f\left(2 z^{2}\right)+2 z^{2}=f\left(z^{2}\right)+z^{2}+2 z^{2}=z f(z)-z+1+3 z^{2}
$$

Comparing (3) and (4) we have

$$
\begin{aligned}
2 z f(z)+2 z^{2}-2 z+1 & =z f(z)-z+1+3 z^{2} \\
z f(z)-z^{2}-z & =0 \\
z(f(z)-z-1) & =0
\end{aligned}
$$

For $z \neq 0$ we obtain $f(z)=z+1$. For $z=0$, we have $f(0)=1$. Thus, for all $z \in \mathbb{R}$, we have

$$
f(z)=z+1
$$

This function indeed satisfies the functional equation.

Problem T-2. Let $x, y, z, w \in \mathbb{R} \backslash\{0\}$ such that $x+y \neq 0, z+w \neq 0$, and $x y+z w \geq 0$. Prove the inequality

$$
\left(\frac{x+y}{z+w}+\frac{z+w}{x+y}\right)^{-1}+\frac{1}{2} \geq\left(\frac{x}{z}+\frac{z}{x}\right)^{-1}+\left(\frac{y}{w}+\frac{w}{y}\right)^{-1}
$$

Solution 1. We first subtract 1 on both sides and rewrite the inequality as

$$
\left(\frac{(x+y)(z+w)}{(x+y)^{2}+(z+w)^{2}}-\frac{1}{2}\right) \geq\left(\frac{x z}{x^{2}+z^{2}}-\frac{1}{2}\right)+\left(\frac{y w}{y^{2}+w^{2}}-\frac{1}{2}\right)
$$

which is equivalent to

$$
\frac{(x-z)^{2}}{x^{2}+z^{2}}+\frac{(y-w)^{2}}{y^{2}+w^{2}} \geq \frac{(x+y-z-w)^{2}}{(x+y)^{2}+(z+w)^{2}}
$$

But this is true by the chain of inequalities

$$
\frac{(x-z)^{2}}{x^{2}+z^{2}}+\frac{(y-w)^{2}}{y^{2}+w^{2}} \geq \frac{((x-z)+(y-w))^{2}}{x^{2}+z^{2}+y^{2}+w^{2}} \geq \frac{(x+y-z-w)^{2}}{(x+y)^{2}+(z+w)^{2}}
$$

where we first used the Cauchy-Schwarz inequality for the two vectors

$$
\left(\frac{(x-z)}{\sqrt{x^{2}+z^{2}}}, \frac{(y-w)}{\sqrt{y^{2}+w^{2}}}\right), \quad\left(\sqrt{x^{2}+z^{2}}, \sqrt{y^{2}+w^{2}}\right)
$$

and then the condition $x y+z w \geq 0$.

Solution 2. The vectors $(x, z)$ and $(y, w)$ form an angle $\leq \pi / 2$ containing the sum vector $(x+y, z+w)$. So with suitable angles $\alpha \leq \phi \leq \beta$ with $\beta-\alpha \leq \pi / 2$, the desired inequality becomes

$$
(\operatorname{tg} \phi+\operatorname{ctg} \phi)^{-1}+\frac{1}{2} \geq(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{-1}+(\operatorname{tg} \beta+\operatorname{ctg} \beta)^{-1}
$$

which is equivalent to

$$
1+\sin 2 \phi \geq \sin 2 \alpha+\sin 2 \beta
$$

If

$$
\sin 2 \phi \geq \min (\sin 2 \alpha, \sin 2 \beta)
$$

then we are done because

$$
1 \geq \max (\sin 2 \alpha, \sin 2 \beta)
$$

Otherwise, modulo $2 \pi$ we have that $\pi / 2<2 \alpha<3 \pi / 2<2 \beta$ and $2 \beta-2 \alpha \leq \pi$, so

$$
\sin 2 \alpha+\sin 2 \beta \leq 0 \leq 1+\sin 2 \phi
$$

Problem T-3. There are $n \geq 2$ houses on the northern side of a street. Going from the west to the east, the houses are numbered from 1 to $n$. The number of each house is shown on a plate. One day the inhabitants of the street make fun of the postman by shuffling their number plates in the following way: for each pair of neighbouring houses, the current number plates are swapped exactly once during the day.

How many different sequences of number plates are possible at the end of the day?

Solution 1. Let $f(n)$ denote the answer. We shall prove by induction that $f(n)=2^{n-2}$. For $n=2$, the answer is clearly $2^{2-2}=1$. We also define $f(1)=1$. Now we consider arbitrary $n>2$.

Let $H_{i}$ denote the house with number $i$ at the start of the day, and let $(i \rightleftarrows i+1)$ denote the swap between $H_{i}$ and $H_{i+1}$. Let $H_{k}$ be the house that has plate $n$ at the end of the day. It follows that the swaps $(n-1 \rightleftarrows n),(n-2 \rightleftarrows n-1), \ldots,(k \rightleftarrows k+1)$ must have happened in this order, for otherwise plate $n$ could not have reached the house $H_{k}$. In addition, $(k-1 \rightleftarrows k)$ must have happened before $(k \rightleftarrows k+1)$, otherwise plate $n$ would have ended up somewhere between $H_{1}$ and $H_{k-1}$.

It follows that for any $k \leq i<n$ the plate $i$ will be on $H_{i+1}$, while plates $1, \ldots, k$ will be on the houses $H_{1}, H_{2}, \ldots, H_{k-1}$, and $H_{k+1}$ in some mixed order. If we restrict our attention to the first $k$ houses: the same happens that would happen if the only houses were $H_{1}, H_{2}, \ldots, H_{k}$; the only difference is that at the end of the game, $H_{k}$ should change its plate to $n$. So there are $f(k)$ different final orderings of the plates such that $n$ is on $H_{k}$ (for $k=1, \ldots, n-1$ ).

Therefore,

$$
f(n)=\sum_{k=1}^{n-1} f(k)=1+\sum_{i=0}^{n-3} 2^{i}=2^{n-2}
$$

Solution 2. To each pair of neighbouring houses we assign the moment $t$ in time when they swap plates. In this way, we get a sequence $t_{1}, \ldots, t_{n-1}$ of $n-1$ moments in time, such that $t_{i-1} \neq t_{i}$ for all $i$. This sequence can be split into maximal monotonically increasing subsequences (with each subsequence consisting of consecutive elements). It can also be split into maximal monotonically decreasing subsequences (again, with each subsequence consisting of consecutive elements). If $t_{i-1}>t_{i}$, or $i=1$, then plate $i$ will finish the day at the eastern end of the maximal increasing subsequence starting at $t_{i}$. If $t_{i-1}<t_{i}$, or $i=n$, then plate $i$ will finish the day at the western end of the maximal decreasing subsequence ending at $t_{i-1}$.

Thus, the $n-2$ relations $<$ and $>$ between the neighbouring $t$ 's determine the final distribution of the plates. Conversely, given the final distribution of the plates, we can calculate the relations. Indeed, let $2 \leq i \leq n-1$. If plate $i$ ends up east of its original position, then $t_{i-1}>t_{i}$, whereas if plate $i$ ends up west of its original position, then $t_{i-1}<t_{i}$. There is no third possibility.

The number of ways to choose these relations is clearly $2^{n-2}$. Each choice can be realized by suitable moments $t_{i}$ in time: we may choose $t_{1}$ to be any moment during the day, and if $t_{1}, \ldots, t_{i-1}$ are already given, then we may choose $t_{i}$ to be smaller, resp. greater than $t_{i-1}$, as desired. Therefore, the answer is $2^{n-2}$.

Problem T-4. Consider finitely many points in the plane with no three points on a line. All these points can be coloured red or green such that any triangle with vertices of the same colour contains at least one point of the other colour in its interior.

What is the maximal possible number of points with this property?

## Solution. The answer is 8 .

Call a set consisting of red points and green points good if no three points are collinear and any unicoloured triangle contains a point of the other colour.

On the one hand, the figure on the left below shows an example of a good set with 8 points. The figure on the right shows the two types of unicololured triangles - all other
unicoloured triangles are reflections of those.
![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-07.jpg?height=442&width=1120&top_left_y=268&top_left_x=500)

On the other hand, we shall prove that a good set can have at most four points of each colour. We give two proofs.

First proof. For a proof by contradiction, let $S$ be a counterexample of minimal cardinality. We may assume that $S$ has at least five red points.

Let $P$ be any vertex of the convex hull of $S$. Then $P$ cannot be in the interior of any triangle, so $S \backslash\{P\}$ is good. But $S$ was a minimal counterexample, so $S \backslash\{P\}$ has at most four points of each colour. Therefore, $S$ has exactly five red points, all vertices of the convex hull of $S$ are red, and $S$ has at most four green points.

Consider the convex hull of $S$. It is a triangle, a quadrilateral, or a pentagon.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-07.jpg?height=520&width=517&top_left_y=1228&top_left_x=378)

Case i: The convex hull is a triangle.

Let $A, B$ and $C$ denote the vertices of the triangle, and let $I$ and $J$ be the interior red points. Without loss of generality, we may assume that the line $I J$ intersects sides $A B$ and $A C$ (and not $B C$ ), and $I$ is nearer to $A B$ than $J$ is. Now $A B I$, $A I J, A J C, B I J$ and $B J C$ are five unired triangles with disjoint interiors, so at least one of them must be empty, because there are at most four green points. Thus, $S$ is not good, in contradiction to our assumptions.

Case ii: The convex hull is a quadrilateral.

Let the vertices of the quadrilateral be $A, B, C, D$, in this cyclic order, and $I$ be the red point in the interior. Now $A B I, B C I, C D I$ and $D A I$ are unired triangles with disjoint interiors, each has a green point inside: denote them by $X, Y, Z, W$ respectively. Then $X Y Z$ and $Z W X$ are two unigreen triangles, but both cannot have $I$ (the only

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-07.jpg?height=417&width=511&top_left_y=1753&top_left_x=1252)
possible red point) in their interiors.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-08.jpg?height=414&width=511&top_left_y=181&top_left_x=384)

Case iii: The convex hull is a pentagon.

Let $A, B, C, D$ and $E$ denote the vertices of the pentagon, in this cyclic order. Now $A B C, A C D$ and $A D E$ are three unired triangles with disjoint interiors, each must have a green point in its interior, these form a unigreen triangle, which cannot have any red point inside.

## Second proof.

Lemma. Let a good set of coloured points be given.

If the convex hull of some red points contains exactly $x$ red points, with exactly $y$ of them being in its interior, then there are at least $x+y-2$ green points in its interior. (The statement is analogous for switched colours.)

Proof. If the convex hull is not a polygon (i.e. $x \leq 2$ ), the statement is trivial. Otherwise consider a partition of the convex hull of the red points into triangles, all of which have only red points as vertices, and have no other red points in their interiors. Let $N$ be the number of triangles of the partition. Then the sum of their angles is $N \pi$. On the other hand, at each interior point, the sum of the angles of the triangles is always $2 \pi$, and at the peripherial points that form a convex $(x-y)$-gon, the sum is $(x-y-2) \pi$. So the sum of the angles of the triangles in the partition is

$$
N \pi=2 y \pi+(x-y-2) \pi
$$

which reduces to $N=x+y-2$. Each of the disjoint unicoloured triangles must contain a single point in its interior, which proves the lemma.

Applying the lemma on all $n$ red points, where $m$ red points are inside their convex hull, gives that there are at least $n+m-2$ green points inside the red points' convex hull. Now applying the lemma again on these green points gives that there are at least $(n+m-2)-2$ red points in their convex hull. But these red points are also interior points of the convex hull of all red points, therefore

$$
(n+m-2)-2 \leq m
$$

This reduces to $n \leq 4$, and our statement is proven.

Remark. The lemma can also be proven by induction on $x$.

Remark. If we allow three colours rather than two, then any set size is possible. For details, see O. Aichholzer. [Empty] [colored] $k$-gons - Recent results on some Erdös-Szekeres type problems. In Proc. XIII Encuentros de Geometría Computacional, pages 43-52, Zaragoza, Spain, 2009.

Problem T-5. Let $A B C$ be an acute triangle. Construct a triangle $P Q R$ such that $A B=2 P Q, B C=2 Q R, C A=2 R P$, and the lines $P Q, Q R$, and $R P$ pass through the points $A, B$, and $C$, respectively. (All six points $A, B, C, P, Q$, and $R$ are distinct.)

Solution. There are two possible configurations, but we are giving a proof for only one of them (where $P$ is between $A$ and $Q$ ), because it will always give a solution.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-09.jpg?height=851&width=808&top_left_y=311&top_left_x=630)

Since the angles of triangles $P Q R$ and $A B C$ are equal, $\angle Q A B=\angle R B C=\angle P C A$. Let $S$ denote the Brocard point of $A B C$, the one for which $\angle S A B=\angle S B C=\angle S C A$.

The circle $A P C$ is tangent to $A B$ by the reverse tangent-chord theorem (since $\angle P A B=$ $\angle P C A$ ). The same argument holds for $S$ (circle $A S C$ is tangent to $A B$ ). It follows that $A P S C$ is cyclic.

We intend to prove that triangle $A P S$ is similar to triangle $B Q S$. For this we observe that $\angle S A P=\angle S B Q$ (since $\angle S A B=\angle S B C$ and $\angle P A B=\angle Q B C$ ), and also $\angle A S P=$ $\angle B S Q$, because $\angle A S P=\angle A C P=\angle B A Q=\angle B S Q$. So all the angles are the same in the two triangles, and the same holds for the third triangle $C R S$. Now considering the twirl with centre $S$, oriented angle $P S A$ and ratio $S A / S P$, the image of triangle $P Q R$ is triangle $A B C$. This yields the following construction:

- Construct the Brocard point $S$ with the property $\angle S A B=\angle S B C=\angle S C A$. It is obtained as the intersection point of the three circles $A S B, B S C, C S A$ which are tangent to the lines $B C, C A, A B$ at $B, C, A$ and pass through $A, B, C$, respectively.
- Construct the circle with center $S$ and radius $S A / 2$.
- Take the point of intersection of this circle and arc $A S$ of circle $A S C$ not containing $C$. This point is $P$.
- Take the intersection of line $A P$ and arc $B S$ of circle $B S A$ not containing $A$ : this point is $Q$.
- Take the intersection of line $B Q$ and arc $C S$ of circle $C S B$ not containing $B$ : this is point $R$.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-10.jpg?height=876&width=814&top_left_y=173&top_left_x=632)

This construction works. We know that the Brocard point $S$ always exists and it is inside the triangle. It is easy to verify that the arcs $A S, B S$ and $C S$ used in the construction are inside the triangle (e.g. arc $A S$ is tangent to side $A B$ and so it is also inside the obtuse triangle $A S B$ ). This means that point $P$ is uniquely defined and is inside triangle $A B C$. Similarly, $Q$ is also unique, and by the tangent-chord theorem $\angle P A B=\angle Q B C$. Now $R$ is also unique, and $\angle Q B C=\angle R C A$. Angle $P C A$ is also the same, because one last time by the tangent-chord theorem $\angle P C A=\angle P A B$. This means that $R, P$ and $C$ are collinear. We are done.

## Remarks:

- The other point of intersection of the circle around $S$ and circle $A S C$ also works, but in this configuration it is possible for some of the six points $(A, B, C, P, Q, R)$ to coincide. In this case the common angle $\angle Q A B=\angle R B C=\angle P C A$ is larger than the Brocard angle, while in the solution it is smaller than the Brocard angle.
- The other Brocard point does not work. (It works when $P Q$ passes through $B, Q R$ through $C$ and $R P$ through $A$.)

Problem T-6. Let $K$ be a point inside an acute triangle $A B C$, such that $B C$ is a common tangent of the circumcircles of $A K B$ and $A K C$. Let $D$ be the intersection of the lines $C K$ and $A B$, and let $E$ be the intersection of the lines $B K$ and $A C$. Let $F$ be the intersection of the line $B C$ and the perpendicular bisector of the segment $D E$. The circumcircle of $A B C$ and the circle $k$ with centre $F$ and radius $F D$ intersect at points $P$ and $Q$.

Prove that the segment $P Q$ is a diameter of $k$.

Solution. The line $B C$ is tangent to the circumcircle of $A K C$, so the angles $B C D$ and $C A K$ are equal. Analogously, the angles $C B E$ and $B A K$ are equal.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-11.jpg?height=1157&width=1601&top_left_y=181&top_left_x=296)

Therefore

$$
\pi=|\angle K B C|+|\angle K C B|+|\angle B K C|=|\angle D A K|+|\angle E A K|+|\angle D K E|
$$

so $A D K E$ is a cyclic quadrilateral. Then

$$
|\angle K B C|=|\angle D A K|=|\angle D E K|
$$

which implies that $D E$ is parallel to $B C$. Observe that $F$ is the unique point on line $B C$ for which the angles $D F B$ and $C F E$ are the same. Let $F^{\prime}$ be the point on side $B C$ for which $B F^{\prime} K D$ is cyclic. Then

$\left|\angle F^{\prime} K E\right|=2 \pi-|\angle E K D|-\left|\angle D K F^{\prime}\right|=2 \pi-(\pi-|\angle B A C|)-(\pi-|\angle C B A|)=\pi-|\angle A C B|$ which tells us that $F^{\prime} C E K$ is also cyclic, therefore

$$
\left|\angle D F^{\prime} B\right|=|\angle D K B|=|\angle C K E|=\left|\angle C F^{\prime} E\right|
$$

This means that $F=F^{\prime}$ and $|\angle D F B|=|\angle C F E|=\pi-|\angle B K C|=|\angle B A C|$, so the triangles $F B D$ and $F E C$ are both similar to $A B C$, therefore

$$
\frac{|A B|}{|A C|}=\frac{|F E|}{|F C|}=\frac{|F B|}{|F D|}
$$

We get $|F B||F C|=|F D|^{2}$. Let $Q^{\prime}$ be the intersection of the line $P F$ and the circumcircle of $A B C$. Using the power of the point $F$, we get $|F D|^{2}=|F B||F C|=|F P|\left|F Q^{\prime}\right|=$ $\left|F D \| F Q^{\prime}\right|$, therefore $\left|F Q^{\prime}\right|=|F D|$, so $Q=Q^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_5020f0da85005e646177g-12.jpg?height=1077&width=1559&top_left_y=181&top_left_x=291)

Remark. One can observe that $F$ is Miquel's point of the cyclic quadrilateral $A D K E$.

Problem T-7. The numbers from 1 to $2013^{2}$ are written row by row into a table consisting of $2013 \times 2013$ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $1,4,9, \ldots, 2013^{2}$ are simultaneously deleted.

How many cells remain?

Solution. Let $m=503$ and $n=4 m+1=2013$.

Note that

$$
(m-1) n=(m-1)(4 m+1)<m \cdot 4 m=(2 m)^{2}<m(4 m+1)=m n
$$

so the perfect square $(2 m)^{2}$ is in the $m$-th row.

Note that $(k+1)^{2}-k^{2}=2 k+1$ is at most $n$ if $k \leq 2 m$ and is at least $n$ if $k \geq 2 m$. Therefore, on the one hand, the first $2 m+1$ perfect squares never skip a row and the first $m+1$ rows are all deleted. On the other hand, the last $n-(2 m-1)=2 m+2$ perfect squares are in pairwise distinct rows.

Thus, the first $m+1$ rows and $2 m$ more are deleted, so $m$ rows remain.

The $j$-th column is deleted if and only if $j$ is a square modulo $n=2013=3 \cdot 11 \cdot 61$. By the Chinese Remainder Theorem, this is the case if and only if $j$ is a square modulo each of 3,11 , and 61 . Since the numbers of squares for these three moduli are 2,6 , and 31 respectively, the number of squares modulo $n$, again by the Chinese Remainder Theorem, is $2 \cdot 6 \cdot 31=372$. The number of remaining columns is therefore $2013-372=1641$.

The number of remaining cells is $503 \cdot 1641=825423$.

Problem T-8. The expression

$$
\pm \square \pm \square \pm \square \pm \square \pm \square \pm \square
$$

is written on the blackboard. Two players, $A$ and $B$, play a game, taking turns. Player $A$ takes the first turn. In each turn, the player on turn replaces a symbol $\square$ by a positive integer. After all the symbols $\square$ are replaced, player $A$ replaces each of the signs $\pm$ by either + or - , independently of each other. Player $A$ wins if the value of the expression on the blackboard is not divisible by any of the numbers $11,12, \ldots, 18$. Otherwise, player $B$ wins.

Determine which player has a winning strategy.

Solution. We will call a number good if it has a divisor in the set $\{11,12, \ldots, 18\}$.

We will show that player $B$ has a winning strategy. Namely, $B$ plays 18 ! in both his first and second move. In his last move he plays a number $x$ (which we will specify later), which ensures that each possible result will be good.

When choosing this number, we may, without loss of generality, work with the result mod 18!, thus his first and second moves count as zero. Before he plays the last move, there are eight $\left(=2^{3}\right)$ possible combinations of signs which give eight results $a_{1}, a_{2}, \ldots$, $a_{8}$. If we ensure that each of the numbers $a_{1}+x, a_{2}+x, \ldots, a_{8}+x$ be good, then also the numbers $a_{1}-x, a_{2}-x, \ldots, a_{8}-x$ will be good since for each $i \in\{1 \ldots, 8\}$ there exists $j \in\{1, \ldots, 8\}$ such that $a_{i}-x=-\left(a_{j}+x\right)$.

Now observe that all of the numbers $a_{1}, \ldots, a_{8}$ have the same parity and thus they have at most two different remainders mod 4. Further, if there are two possible remainders, then each appears exactly four times (check it!).

At least three of the numbers have the same remainder mod 3 , wlog let $a_{1} \equiv a_{2} \equiv a_{3}$ $(\bmod 3)$. Also, we may assume that $a_{4} \equiv a_{3}(\bmod 4)$.

Then, we choose our $x$ from the Chinese Remainder Theorem so that $9 \mid a_{1}+x$, $5\left|a_{2}+x, 16\right| a_{4}+x, 7\left|a_{5}+x, 11\right| a_{6}+x, 13 \mid a_{7}+x$, and $17 \mid a_{8}+x$.

With this choice of $x$ we have in fact ensured $18\left|a_{1}+x, 15\right| a_{2}+x, 12 \mid a_{3}+x$, $16\left|a_{4}+x, 14\right| a_{5}+x, 11\left|a_{6}+x, 13\right| a_{7}+x$, and $17 \mid a_{8}+x$. Hence the result.