problem_type: Algebra

exam: BalticWay

solution: For convenience we call $x_{2011}$ also $x_{0}$. Let $k$ be the largest of the numbers $x_{1}, \ldots, x_{2011}$, and consider an equation $x_{n-1}+x_{n}=2 k$, where $1 \leq n \leq 2011$. Hence we get $2 \max \left(x_{n-1}, x_{n}\right) \geq x_{n-1}+x_{n}=2 k$, so either $x_{n-1}$ or $x_{n}$, say $x_{n-1}$, satisfies $x_{n-1} \geq k$. Since also $x_{n-1} \leq k$, we then have $x_{n-1}=k$, and then also $x_{n}=2 k-x_{n-1}=2 k-k=k$. That is, in such an equation both variables on the left equal $k$. Now let $\mathcal{E}$ be the set of such equations, and let $\mathcal{S}$ be the set of subscripts on the left of these equations. From $x_{n}=k \forall n \in \mathcal{S}$ we get $|\mathcal{S}| \leq|\mathcal{E}|$. On the other hand, since the total number of appearances of these subscripts is $2|\mathcal{E}|$ and each subscript appears on the left in no more than two equations, we have $2|\mathcal{E}| \leq 2|\mathcal{S}|$. Thus $2|\mathcal{E}|=2|\mathcal{S}|$, so for each $n \in \mathcal{S}$ the set $\mathcal{E}$ contains both equations with the subscript $n$ on the left. Now assume $1 \in \mathcal{S}$ without loss of generality. Then the equation $x_{1}+x_{2}=2 k$ belongs to $\mathcal{E}$, so $2 \in \mathcal{S}$. Continuing in this way we find that all subscripts belong to $\mathcal{S}$, so $x_{1}=x_{2}=\cdots=x_{2011}=k$.

problem_match: ## A-1 DEN

solution_match: Solution 1


solution:
Again we call $x_{2011}$ also $x_{0}$. Taking the square on both sides of all the equations and adding the results, we get

$$
\sum_{n=1}^{2011}\left(x_{n-1}+x_{n}\right)^{2}=4 \sum_{n=1}^{2011} x_{n}^{\prime 2}=4 \sum_{n=1}^{2011} x_{n}^{2}
$$

which can be transformed with some algebra into

$$
\sum_{n=1}^{2011}\left(x_{n-1}-x_{n}\right)^{2}=0
$$

Hence the assertion follows.

solution_match: Solution 2


problem_type: Geometry

problem:
Geometry

Let $A B$ and $C D$ be two diameters of the circle $\mathcal{C}$. For an arbitrary point $P$ on $\mathcal{C}$, let $R$ and $S$ be the feet of the perpendiculars from $P$ to $A B$ and $C D$, respectively. Show that the length of $R S$ is independent of the choice of $P$.

solution:
Let $O$ be the centre of $\mathcal{C}$. Then $P, R, S$, and $O$ are points on a circle $\mathcal{C}^{\prime}$ with diameter $O P$, equal to the radius of $\mathcal{C}$. The segment $R S$ is a chord in this circle subtending the angle $B O D$ or a supplementary angle. Since the angle as well as radius of $\mathcal{C}^{\prime}$ are independent of $P$, so is $R S$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-11.jpg?height=514&width=577&top_left_y=945&top_left_x=745)

problem_match: ## G-1 FIN

solution_match: Solution.


problem:
Let $E$ be an interior point of the convex quadrilateral $A B C D$. Construct triangles $\triangle A B F$, $\triangle B C G, \triangle C D H$ and $\triangle D A I$ on the outside of the quadrilateral such that the similarities $\triangle A B F \sim \triangle D C E, \triangle B C G \sim \triangle A D E, \triangle C D H \sim \triangle B A E$ and $\triangle D A I \sim \triangle C B E$ hold. Let $P$, $Q, R$ and $S$ be the projections of $E$ on the lines $A B, B C, C D$ and $D A$, respectively. Prove that if the quadrilateral $P Q R S$ is cyclic, then

$$
E F \cdot C D=E G \cdot D A=E H \cdot A B=E I \cdot B C .
$$

solution:
We consider oriented angles modulo $180^{\circ}$. From the cyclic quadrilaterals $A P E S$, $B Q E P, P Q R S, C R E Q, D S E R$ and $\triangle D C E \sim \triangle A B F$ we get

$$
\begin{aligned}
\angle A E B & =\angle E A B+\angle A B E=\angle E S P+\angle P Q E \\
& =\angle E S R+\angle R S P+\angle P Q R+\angle R Q E \\
& =\angle E S R+\angle R Q E=\angle E D C+\angle D C E \\
& =\angle D E C=\angle A F B,
\end{aligned}
$$

so the quadrilateral $A E B F$ is cyclic. By Ptolemy we then have

$$
E F \cdot A B=A E \cdot B F+B E \cdot A F .
$$

This transforms by $A B: B F: A F=D C: C E: D E$ into

![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-13.jpg?height=914&width=922&top_left_y=1662&top_left_x=567)

Since the expression on the right of this equation is invariant under cyclic permutation of the vertices of the quadrilateral $A B C D$, the asserted equation follows immediately.

problem_match: ## G-3 DEN


problem: The incircle of a triangle $A B C$ touches the sides $B C, C A, A B$ at $D, E, F$, respectively. Let $G$ be a point on the incircle such that $F G$ is a diameter. The lines $E G$ and $F D$ intersect at $H$. Prove that $C H \| A B$.

solution:
We work in the opposite direction. Suppose that $H^{\prime}$ is the point where $D F$ intersect the line through $C$ parallel to $A B$. We need to show that $H^{\prime}=H$. For this purpose it suffices to prove that $E, G, H^{\prime}$ are collinear, which reduces to showing that if $G^{\prime} \neq E$ is the common point of $E H^{\prime}$ and the incircle, then $G^{\prime}=G$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-14.jpg?height=588&width=765&top_left_y=837&top_left_x=651)

Note that $H^{\prime}$ and $B$ lie on the same side of $A C$. Hence $C H^{\prime} \| A B$ gives $\angle A C H^{\prime}=180^{\circ}-\angle B A C$. Also, some homothety with center $D$ maps the segment $B F$ to the segment $C H^{\prime}$. Thus the equality $B D=B F$ implies that $C H^{\prime}=C D=C E$, i.e. the triangle $E C H^{\prime}$ is isosceles and

$$
\angle H^{\prime} E C=\frac{1}{2}\left(180^{\circ}-\angle E C H^{\prime}\right)=\frac{1}{2} \angle B A C .
$$

But $G^{\prime}$ and $H^{\prime}$ lie on the same side of $A C$, so $\angle G^{\prime} E C=\angle H^{\prime} E C$ and consequently

$$
\angle G^{\prime} F E=\angle G^{\prime} E C=\angle H^{\prime} E C=\frac{1}{2} \angle B A C
$$

so that

$$
\angle G^{\prime} F A=\angle G^{\prime} F E+\angle E F A=\frac{1}{2} \angle B A C+\frac{1}{2}\left(180^{\circ}-\angle F A E\right)=90^{\circ} .
$$

Hence $F G^{\prime}$ is a diameter of the incircle and the desired equality $G^{\prime}=G$ follows.

Remark. A similar proof also works in the forward direction: one may compute $\angle E H D=$ $\frac{1}{2} \angle A C B$. Hence $H$ lies on the circle centred at $C$ that passes through $D$ and $E$. Consequently the triangle $E H C$ is isosceles, wherefore

$$
\angle E C H=180^{\circ}-2 \angle G E C=180^{\circ}-\angle B A C .
$$

Thus the lines $A B$ and $C H$ are indeed parallel.

problem_match: ## G-4 POL


problem:
Let $A B C D$ be a convex quadrilateral such that $\angle A D B=\angle B D C$. Suppose that a point $E$ on the side $A D$ satisfies the equality

$$
A E \cdot E D+B E^{2}=C D \cdot A E .
$$

Show that $\angle E B A=\angle D C B$.

solution:
Let $F$ be the point symmetric to $E$ with respect to the line $D B$. Then the equality $\angle A D B=\angle B D C$ shows that $F$ lies on the line $D C$, on the same side of $D$ as $C$. Moreover, we have $A E \cdot E D<C D \cdot A E$, or $F D=E D<C D$, so in fact $F$ lies on the segment $D C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ecac827bee541f9d910cg-15.jpg?height=445&width=836&top_left_y=977&top_left_x=610)

Note now that triangles $D E B$ and $D F B$ are congruent (symmetric with respect to the line $D B)$, so $\angle A E B=\angle B F C$. Also, we have

$$
B E^{2}=C D \cdot A E-A E \cdot E D=A E \cdot(C D-E D)=A E \cdot(C D-F D)=A E \cdot C F
$$

Therefore

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

This shows that the triangles $B E A$ and $C F B$ are similar, which gives $\angle E B A=\angle F C B=$ $\angle D C B$, as desired.

## Number Theory

problem_match: ## G-5 POL