di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Algebra


Question: Prove the inequality

$$
\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}} \leq \frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)} \sum_{i<j} a_{i} a_{j}
$$

for positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$.

Solution:
Let $S=\sum_{i} a_{i}$. Denote by $L$ and $R$ the expressions on the left and right hand side of the proposed inequality. We transform $L$ and $R$ using the identity

$$
\sum_{i<j}\left(a_{i}+a_{j}\right)=(n-1) \sum_{i} a_{i} \tag{1}
$$

And thus:

$$
L=\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}}=\sum_{i<j} \frac{1}{4}\left(a_{i}+a_{j}-\frac{\left(a_{i}-a_{j}\right)^{2}}{a_{i}+a_{j}}\right)=\frac{n-1}{4} \cdot S-\frac{1}{4} \sum_{i<j} \frac{\left(a_{i}-a_{j}\right)^{2}}{a_{i}+a_{j}} . \tag{2}
$$

To represent $R$ we express the sum $\sum_{i<j} a_{i} a_{j}$ in two ways; in the second transformation identity (1) will be applied to the squares of the numbers $a_{i}$ :

$$
\begin{gathered}
\sum_{i<j} a_{i} a_{j}=\frac{1}{2}\left(S^{2}-\sum_{i} a_{i}^{2}\right) \\
\sum_{i<j} a_{i} a_{j}=\frac{1}{2} \sum_{i<j}\left(a_{i}^{2}+a_{j}^{2}-\left(a_{i}-a_{j}\right)^{2}\right)=\frac{n-1}{2} \cdot \sum_{i} a_{i}^{2}-\frac{1}{2} \sum_{i<j}\left(a_{i}-a_{j}\right)^{2} .
\end{gathered}
$$

Multiplying the first of these equalities by $n-1$ and adding the second one we obtain

$$
n \sum_{i<j} a_{i} a_{j}=\frac{n-1}{2} \cdot S^{2}-\frac{1}{2} \sum_{i<j}\left(a_{i}-a_{j}\right)^{2}
$$

Hence

$$
R=\frac{n}{2 S} \sum_{i<j} a_{i} a_{j}=\frac{n-1}{4} \cdot S-\frac{1}{4} \sum_{i<j} \frac{\left(a_{i}-a_{j}\right)^{2}}{S} . \tag{3}
$$

Now compare (2) and (3). Since $S \geq a_{i}+a_{j}$ for any $i<j$, the claim $L \geq R$ results.
Let $S=a_{1}+a_{2}+\cdots+a_{n}$. For any $i \neq j$,

$$
4 \frac{a_{i} a_{j}}{a_{i}+a_{j}}=a_{i}+a_{j}-\frac{\left(a_{i}-a_{j}\right)^{2}}{a_{i}+a_{j}} \leq a_{i}+a_{j}-\frac{\left(a_{i}-a_{j}\right)^{2}}{a_{1}+a_{2}+\cdots+a_{n}}=\frac{\sum_{k \neq i} a_{i} a_{k}+\sum_{k \neq j} a_{j} a_{k}+2 a_{i} a_{j}}{S}
$$

The statement is obtained by summing up these inequalities for all pairs $i, j$ :

$$
\begin{gathered}
\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}}=\frac{1}{2} \sum_{i} \sum_{j \neq i} \frac{a_{i} a_{j}}{a_{i}+a_{j}} \leq \frac{1}{8 S} \sum_{i} \sum_{j \neq i}\left(\sum_{k \neq i} a_{i} a_{k}+\sum_{k \neq j} a_{j} a_{k}+2 a_{i} a_{j}\right) \\
=\frac{1}{8 S}\left(\sum_{k} \sum_{i \neq k} \sum_{j \neq i} a_{i} a_{k}+\sum_{k} \sum_{j \neq k} \sum_{i \neq j} a_{j} a_{k}+\sum_{i} \sum_{j \neq i} 2 a_{i} a_{j}\right) \\
=\frac{1}{8 S}\left(\sum_{k} \sum_{i \neq k}(n-1) a_{i} a_{k}+\sum_{k} \sum_{j \neq k}(n-1) a_{j} a_{k}+\sum_{i} \sum_{j \neq i} 2 a_{i} a_{j}\right) \\
=\frac{n}{4 S} \sum_{i} \sum_{j \neq i} a_{i} a_{j}=\frac{n}{2 S} \sum_{i<j} a_{i} a_{j} .
\end{gathered}
$$
Subfield: Algebra


Question: Let $a, b, c$ be the sides of a triangle. Prove that

$$
\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leq 3 .
$$

Solution:
Note first that the denominators are all positive, e.g. $\sqrt{a}+\sqrt{b}>\sqrt{a+b}>\sqrt{c}$.

Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}, y=\sqrt{c}+\sqrt{a}-\sqrt{b}$ and $z=\sqrt{a}+\sqrt{b}-\sqrt{c}$. Then

$b+c-a=\left(\frac{z+x}{2}\right)^{2}+\left(\frac{x+y}{2}\right)^{2}-\left(\frac{y+z}{2}\right)^{2}=\frac{x^{2}+x y+x z-y z}{2}=x^{2}-\frac{1}{2}(x-y)(x-z)$

and

$$
\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sqrt{1-\frac{(x-y)(x-z)}{2 x^{2}}} \leq 1-\frac{(x-y)(x-z)}{4 x^{2}}
$$

applying $\sqrt{1+2 u} \leq 1+u$ in the last step. Similarly we obtain

$$
\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}} \leq 1-\frac{(z-x)(z-y)}{4 z^{2}} \quad \text { and } \quad \frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leq 1-\frac{(y-z)(y-x)}{4 y^{2}}
$$

Substituting these quantities into the statement, it is sufficient to prove that

$$
\frac{(x-y)(x-z)}{x^{2}}+\frac{(y-z)(y-x)}{y^{2}}+\frac{(z-x)(z-y)}{z^{2}} \geq 0 \tag{1}
$$

By symmetry we can assume $x \leq y \leq z$. Then

$$
\begin{gathered}
\frac{(x-y)(x-z)}{x^{2}}=\frac{(y-x)(z-x)}{x^{2}} \geq \frac{(y-x)(z-y)}{y^{2}}=-\frac{(y-z)(y-x)}{y^{2}} \\
\frac{(z-x)(z-y)}{z^{2}} \geq 0
\end{gathered}
$$

and (1) follows.
Due to the symmetry of variables, it can be assumed that $a \geq b \geq c$. We claim that

$$
\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leq 1 \quad \text { and } \quad \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}} \leq 2
$$

The first inequality follows from

$$
\sqrt{a+b-c}-\sqrt{a}=\frac{(a+b-c)-a}{\sqrt{a+b-c}+\sqrt{a}} \leq \frac{b-c}{\sqrt{b}+\sqrt{c}}=\sqrt{b}-\sqrt{c}
$$

For proving the second inequality, let $p=\sqrt{a}+\sqrt{b}$ and $q=\sqrt{a}-\sqrt{b}$. Then $a-b=p q$ and the inequality becomes

$$
\frac{\sqrt{c-p q}}{\sqrt{c}-q}+\frac{\sqrt{c+p q}}{\sqrt{c}+q} \leq 2
$$

From $a \geq b \geq c$ we have $p \geq 2 \sqrt{c}$. Applying the Cauchy-Schwarz inequality,

$$
\begin{gathered}
\left(\frac{\sqrt{c-p q}}{\sqrt{c}-q}+\frac{\sqrt{c+p q}}{\sqrt{c}+q}\right)^{2} \leq\left(\frac{c-p q}{\sqrt{c}-q}+\frac{c+p q}{\sqrt{c}+q}\right)\left(\frac{1}{\sqrt{c}-q}+\frac{1}{\sqrt{c}+q}\right) \\
=\frac{2\left(c \sqrt{c}-p q^{2}\right)}{c-q^{2}} \cdot \frac{2 \sqrt{c}}{c-q^{2}}=4 \cdot \frac{c^{2}-\sqrt{c} p q^{2}}{\left(c-q^{2}\right)^{2}} \leq 4 \cdot \frac{c^{2}-2 c q^{2}}{\left(c-q^{2}\right)^{2}} \leq 4 .
\end{gathered}
$$
Subfield: Algebra


Question: Determine the smallest number $M$ such that the inequality

$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$

holds for all real numbers $a, b, c$.

Solution:
We first consider the cubic polynomial

$$
P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right) .
$$

It is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore

$$
P(t)=(b-c)(t-b)(t-c)(t+b+c)
$$

since the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form

$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)| .
$$

The problem comes down to finding the smallest number $M$ that satisfies the inequality

$$
|(b-c)(a-b)(a-c)(a+b+c)| \leq M \cdot\left(a^{2}+b^{2}+c^{2}\right)^{2} . \tag{1}
$$

Note that this expression is symmetric, and we can therefore assume $a \leq b \leq c$ without loss of generality. With this assumption,

$$
|(a-b)(b-c)|=(b-a)(c-b) \leq\left(\frac{(b-a)+(c-b)}{2}\right)^{2}=\frac{(c-a)^{2}}{4} \tag{2}
$$

with equality if and only if $b-a=c-b$, i.e. $2 b=a+c$. Also

$$
\left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leq \frac{(c-b)^{2}+(b-a)^{2}}{2} \tag{3}
$$

or equivalently,

$$
3(c-a)^{2} \leq 2 \cdot\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]
$$

again with equality only for $2 b=a+c$. From (2) and (3) we get

$$
\begin{aligned}
& |(b-c)(a-b)(a-c)(a+b+c)| \\
\leq & \frac{1}{4} \cdot\left|(c-a)^{3}(a+b+c)\right| \\
= & \frac{1}{4} \cdot \sqrt{(c-a)^{6}(a+b+c)^{2}} \\
\leq & \frac{1}{4} \cdot \sqrt{\left(\frac{2 \cdot\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right)^{3} \cdot(a+b+c)^{2}} \\
= & \frac{\sqrt{2}}{2} \cdot\left(\sqrt[4]{\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right)^{3} \cdot(a+b+c)^{2}}\right)^{2} .
\end{aligned}
$$



By the weighted AM-GM inequality this estimate continues as follows:

$$
\begin{aligned}
& |(b-c)(a-b)(a-c)(a+b+c)| \\
\leq & \frac{\sqrt{2}}{2} \cdot\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right)^{2} \\
= & \frac{9 \sqrt{2}}{32} \cdot\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{aligned}
$$

We see that the inequality (1) is satisfied for $M=\frac{9}{32} \sqrt{2}$, with equality if and only if $2 b=a+c$ and

$$
\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}
$$

Plugging $b=(a+c) / 2$ into the last equation, we bring it to the equivalent form

$$
2(c-a)^{2}=9(a+c)^{2} .
$$

The conditions for equality can now be restated as

$$
2 b=a+c \quad \text { and } \quad(c-a)^{2}=18 b^{2} .
$$

Setting $b=1$ yields $a=1-\frac{3}{2} \sqrt{2}$ and $c=1+\frac{3}{2} \sqrt{2}$. We see that $M=\frac{9}{32} \sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple $(a, b, c)$ proportional to $\left(1-\frac{3}{2} \sqrt{2}, 1,1+\frac{3}{2} \sqrt{2}\right)$, up to permutation.



Final answer: $M=\frac{9}{32} \sqrt{2}$
Subfield: Combinatorics


Question: We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows:

- if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off;
- otherwise, $L_{i}$ is switched on.

Initially all the lamps are off except the leftmost one which is on.
Prove that there are infinitely many integers $n$ for which all the lamps will eventually be off.

Solution:
Experiments with small $n$ lead to the guess that every $n$ of the form $2^{k}$ should be good. This is indeed the case, and more precisely: let $A_{k}$ be the $2^{k} \times 2^{k}$ matrix whose rows represent the evolution of the system, with entries 0,1 (for off and on respectively). The top row shows the initial state $[1,0,0, \ldots, 0]$; the bottom row shows the state after $2^{k}-1$ steps. The claim is that:

$$
\text { The bottom row of } A_{k} \text { is }[1,1,1, \ldots, 1] \text {. }
$$

This will of course suffice because one more move then produces $[0,0,0, \ldots, 0]$, as required.

The proof is by induction on $k$. The base $k=1$ is obvious. Assume the claim to be true for a

$k \geq 1$ and write the matrix $A_{k+1}$ in the block form $\left(\begin{array}{cc}A_{k} & O_{k} \\ B_{k} & C_{k}\end{array}\right)$ with four $2^{k} \times 2^{k}$ matrices. After $m$ steps, the last 1 in a row is at position $m+1$. Therefore $O_{k}$ is the zero matrix. According to the induction hypothesis, the bottom row of $\left[A_{k} O_{k}\right]$ is $[1, \ldots, 1,0, \ldots, 0]$, with $2^{k}$ ones and $2^{k}$ zeros. The next row is thus

$$
[\underbrace{0, \ldots, 0}_{2^{k}-1}, 1,1, \underbrace{0, \ldots, 0}_{2^{k}-1}]
$$

It is symmetric about its midpoint, and this symmetry is preserved in all subsequent rows because the procedure described in the problem statement is left/right symmetric. Thus $B_{k}$ is the mirror image of $C_{k}$. In particular, the rightmost column of $B_{k}$ is identical with the leftmost column of $C_{k}$.

Imagine the matrix $C_{k}$ in isolation from the rest of $A_{k+1}$. Suppose it is subject to evolution as defined in the problem: the first (leftmost) term in a row depends only on the two first terms in the preceding row, according as they are equal or not. Now embed $C_{k}$ again in $A_{k}$. The 'leftmost' terms in the rows of $C_{k}$ now have neighbours on their left side - but these neighbours are their exact copies. Consequently the actual evolution within $C_{k}$ is the same, whether or not $C_{k}$ is considered as a piece of $A_{k+1}$ or in isolation. And since the top row of $C_{k}$ is $[1,0, \ldots, 0]$, it follows that $C_{k}$ is identical with $A_{k}$.



The bottom row of $A_{k}$ is $[1,1, \ldots, 1]$; the same is the bottom row of $C_{k}$, hence also of $B_{k}$, which mirrors $C_{k}$. So the bottom row of $A_{k+1}$ consists of ones only and the induction is complete.
Subfield: Combinatorics


Question: We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows:

- if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off;
- otherwise, $L_{i}$ is switched on.

Initially all the lamps are off except the leftmost one which is on.
Prove that there are infinitely many integers $n$ for which the lamps will never be all off.

Solution:
There are many ways to produce an infinite sequence of those $n$ for which the state $[0,0, \ldots, 0]$ will never be achieved. As an example, consider $n=2^{k}+1$ (for $k \geq 1$ ). The evolution of the system can be represented by a matrix $\mathcal{A}$ of width $2^{k}+1$ with infinitely many rows. The top $2^{k}$ rows form the matrix $A_{k}$ discussed above, with one column of zeros attached at its right.

In the next row we then have the vector $[0,0, \ldots, 0,1,1]$. But this is just the second row of $\mathcal{A}$ reversed. Subsequent rows will be mirror copies of the foregoing ones, starting from the second one. So the configuration $[1,1,0, \ldots, 0,0]$, i.e. the second row of $\mathcal{A}$, will reappear. Further rows will periodically repeat this pattern and there will be no row of zeros.
Subfield: Geometry


Question: Let $A B C$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies

$$
\angle P B A+\angle P C A=\angle P B C+\angle P C B .
$$

Show that $A P \geq A I$ and that equality holds if and only if $P$ coincides with $I$.

Solution:
Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Since $\angle P B A+\angle P C A+\angle P B C+\angle P C B=\beta+\gamma$, the condition from the problem statement is equivalent to $\angle P B C+\angle P C B=(\beta+\gamma) / 2$, i. e. $\angle B P C=90^{\circ}+\alpha / 2$.

On the other hand $\angle B I C=180^{\circ}-(\beta+\gamma) / 2=90^{\circ}+\alpha / 2$. Hence $\angle B P C=\angle B I C$, and since $P$ and $I$ are on the same side of $B C$, the points $B, C, I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\omega$ of triangle $B C I$.

<img_3152>

Let $\Omega$ be the circumcircle of triangle $A B C$. It is a well-known fact that the centre of $\omega$ is the midpoint $M$ of the $\operatorname{arc} B C$ of $\Omega$. This is also the point where the angle bisector $A I$ intersects $\Omega$.

From triangle $A P M$ we have

$$
A P+P M \geq A M=A I+I M=A I+P M
$$

Therefore $A P \geq A I$. Equality holds if and only if $P$ lies on the line segment $A I$, which occurs if and only if $P=I$.
Subfield: Geometry


Question: Let $A B C D$ be a trapezoid with parallel sides $A B>C D$. Points $K$ and $L$ lie on the line segments $A B$ and $C D$, respectively, so that $A K / K B=D L / L C$. Suppose that there are points $P$ and $Q$ on the line segment $K L$ satisfying

$$
\angle A P B=\angle B C D \quad \text { and } \quad \angle C Q D=\angle A B C \text {. }
$$

Prove that the points $P, Q, B$ and $C$ are concyclic.

Solution:
Because $A B \| C D$, the relation $A K / K B=D L / L C$ readily implies that the lines $A D, B C$ and $K L$ have a common point $S$.

<img_3719>

Consider the second intersection points $X$ and $Y$ of the line $S K$ with the circles $(A B P)$ and $(C D Q)$, respectively. Since $A P B X$ is a cyclic quadrilateral and $A B \| C D$, one has

$$
\angle A X B=180^{\circ}-\angle A P B=180^{\circ}-\angle B C D=\angle A B C \text {. }
$$

This shows that $B C$ is tangent to the circle $(A B P)$ at $B$. Likewise, $B C$ is tangent to the circle $(C D Q)$ at $C$. Therefore $S P \cdot S X=S B^{2}$ and $S Q \cdot S Y=S C^{2}$.

Let $h$ be the homothety with centre $S$ and ratio $S C / S B$. Since $h(B)=C$, the above conclusion about tangency implies that $h$ takes circle $(A B P)$ to circle $(C D Q)$. Also, $h$ takes $A B$ to $C D$, and it easily follows that $h(P)=Y, h(X)=Q$, yielding $S P / S Y=S B / S C=S X / S Q$.

Equalities $S P \cdot S X=S B^{2}$ and $S Q / S X=S C / S B$ imply $S P \cdot S Q=S B \cdot S C$, which is equivalent to $P, Q, B$ and $C$ being concyclic.
The case where $P=Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $A D, B C$ and $K L$ concur at a point $S$.

<img_3316>

Let the lines $A P$ and $D Q$ meet at $E$, and let $B P$ and $C Q$ meet at $F$. Then $\angle E P F=\angle B C D$ and $\angle F Q E=\angle A B C$ by the condition of the problem. Since the angles $B C D$ and $A B C$ add up to $180^{\circ}$, it follows that $P E Q F$ is a cyclic quadrilateral.

Applying Menelaus' theorem, first to triangle $A S P$ and line $D Q$ and then to triangle $B S P$ and line $C Q$, we have

$$
\frac{A D}{D S} \cdot \frac{S Q}{Q P} \cdot \frac{P E}{E A}=1 \quad \text { and } \quad \frac{B C}{C S} \cdot \frac{S Q}{Q P} \cdot \frac{P F}{F B}=1
$$

The first factors in these equations are equal, as $A B \| C D$. Thus the last factors are also equal, which implies that $E F$ is parallel to $A B$ and $C D$. Using this and the cyclicity of $P E Q F$, we obtain

$$
\angle B C D=\angle B C F+\angle F C D=\angle B C Q+\angle E F Q=\angle B C Q+\angle E P Q .
$$

On the other hand,

$$
\angle B C D=\angle A P B=\angle E P F=\angle E P Q+\angle Q P F,
$$

and consequently $\angle B C Q=\angle Q P F$. The latter angle either coincides with $\angle Q P B$ or is supplementary to $\angle Q P B$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P, Q, B$ and $C$ are concyclic.
Subfield: Geometry


Question: Let $A B C D E$ be a convex pentagon such that

$$
\angle B A C=\angle C A D=\angle D A E \quad \text { and } \quad \angle A B C=\angle A C D=\angle A D E \text {. }
$$

The diagonals $B D$ and $C E$ meet at $P$. Prove that the line $A P$ bisects the side $C D$.

Solution:
Let the diagonals $A C$ and $B D$ meet at $Q$, the diagonals $A D$ and $C E$ meet at $R$, and let the ray $A P$ meet the side $C D$ at $M$. We want to prove that $C M=M D$ holds.

<img_3892>

The idea is to show that $Q$ and $R$ divide $A C$ and $A D$ in the same ratio, or more precisely

$$
\frac{A Q}{Q C}=\frac{A R}{R D} \tag{1}
$$

(which is equivalent to $Q R \| C D$ ). The given angle equalities imply that the triangles $A B C$, $A C D$ and $A D E$ are similar. We therefore have

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

Since $\angle B A D=\angle B A C+\angle C A D=\angle C A D+\angle D A E=\angle C A E$, it follows from $A B / A C=$ $A D / A E$ that the triangles $A B D$ and $A C E$ are also similar. Their angle bisectors in $A$ are $A Q$ and $A R$, respectively, so that

$$
\frac{A B}{A C}=\frac{A Q}{A R}
$$

Because $A B / A C=A C / A D$, we obtain $A Q / A R=A C / A D$, which is equivalent to (1). Now Ceva's theorem for the triangle $A C D$ yields

$$
\frac{A Q}{Q C} \cdot \frac{C M}{M D} \cdot \frac{D R}{R A}=1
$$

In view of (1), this reduces to $C M=M D$, which completes the proof.
Subfield: Geometry


Question: A point $D$ is chosen on the side $A C$ of a triangle $A B C$ with $\angle C<\angle A<90^{\circ}$ in such a way that $B D=B A$. The incircle of $A B C$ is tangent to $A B$ and $A C$ at points $K$ and $L$, respectively. Let $J$ be the incentre of triangle $B C D$. Prove that the line $K L$ intersects the line segment $A J$ at its midpoint.

Solution:
Denote by $P$ be the common point of $A J$ and $K L$. Let the parallel to $K L$ through $J$ meet $A C$ at $M$. Then $P$ is the midpoint of $A J$ if and only if $A M=2 \cdot A L$, which we are about to show.

<img_3707>

Denoting $\angle B A C=2 \alpha$, the equalities $B A=B D$ and $A K=A L$ imply $\angle A D B=2 \alpha$ and $\angle A L K=90^{\circ}-\alpha$. Since $D J$ bisects $\angle B D C$, we obtain $\angle C D J=\frac{1}{2} \cdot\left(180^{\circ}-\angle A D B\right)=90^{\circ}-\alpha$. Also $\angle D M J=\angle A L K=90^{\circ}-\alpha$ since $J M \| K L$. It follows that $J D=J M$.

Let the incircle of triangle $B C D$ touch its side $C D$ at $T$. Then $J T \perp C D$, meaning that $J T$ is the altitude to the base $D M$ of the isosceles triangle $D M J$. It now follows that $D T=M T$, and we have

$$
D M=2 \cdot D T=B D+C D-B C \text {. }
$$

Therefore

$$
\begin{aligned}
A M & =A D+(B D+C D-B C) \\
& =A D+A B+D C-B C \\
& =A C+A B-B C \\
& =2 \cdot A L,
\end{aligned}
$$

which completes the proof.
Subfield: Geometry


Question: In a triangle $A B C$, let $M_{a}, M_{b}, M_{c}$ be respectively the midpoints of the sides $B C, C A$, $A B$ and $T_{a}, T_{b}, T_{c}$ be the midpoints of the $\operatorname{arcs} B C, C A, A B$ of the circumcircle of $A B C$, not containing the opposite vertices. For $i \in\{a, b, c\}$, let $\omega_{i}$ be the circle with $M_{i} T_{i}$ as diameter. Let $p_{i}$ be the common external tangent to $\omega_{j}, \omega_{k}(\{i, j, k\}=\{a, b, c\})$ such that $\omega_{i}$ lies on the opposite side of $p_{i}$ than $\omega_{j}, \omega_{k}$ do. Prove that the lines $p_{a}, p_{b}, p_{c}$ form a triangle similar to $A B C$ and find the ratio of similitude.

Solution:
Let $T_{a} T_{b}$ intersect circle $\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with centre $T_{b}$ and ratio $T_{b} T_{a} / T_{b} U$ maps the circle $\omega_{b}$ onto the circumcircle of $A B C$ and the line $U X$ onto the line tangent to the circumcircle at $T_{a}$, which is parallel to $B C$; thus $U X \| B C$. The same is true of $V Y$, so that $U X\|B C\| V Y$.

Let $T_{a} T_{b}$ cut $A C$ at $P$ and let $T_{a} T_{c}$ cut $A B$ at $Q$. The point $X$ lies on the hypotenuse $P M_{b}$ of the right triangle $P U M_{b}$ and is equidistant from $U$ and $M_{b}$. So $X$ is the midpoint of $M_{b} P$. Similarly $Y$ is the midpoint of $M_{c} Q$.

Denote the incentre of triangle $A B C$ as usual by $I$. It is a known fact that $T_{a} I=T_{a} B$ and $T_{c} I=T_{c} B$. Therefore the points $B$ and $I$ are symmetric across $T_{a} T_{c}$, and consequently $\angle Q I B=\angle Q B I=\angle I B C$. This implies that $B C$ is parallel to the line $I Q$, and likewise, to $I P$. In other words, $P Q$ is the line parallel to $B C$ passing through $I$.

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Clearly $M_{b} M_{c} \| B C$. So $P M_{b} M_{c} Q$ is a trapezoid and the segment $X Y$ connects the midpoints of its nonparallel sides; hence $X Y \| B C$. This combined with the previously established relations $U X\|B C\| V Y$ shows that all the four points $U, X, Y, V$ lie on a line which is the common tangent to circles $\omega_{b}, \omega_{c}$. Since it leaves these two circles on one side and the circle $\omega_{a}$ on the other, this line is just the line $p_{a}$ from the problem statement.

Line $p_{a}$ runs midway between $I$ and $M_{b} M_{c}$. Analogous conclusions hold for the lines $p_{b}$ and $p_{c}$. So these three lines form a triangle homothetic from centre $I$ to triangle $M_{a} M_{b} M_{c}$ in ratio $1 / 2$, hence similar to $A B C$ in ratio $1 / 4$.
Subfield: Number Theory


Question: Determine all pairs $(x, y)$ of integers satisfying the equation

$$
1+2^{x}+2^{2 x+1}=y^{2}
$$

Solution:
If $(x, y)$ is a solution then obviously $x \geq 0$ and $(x,-y)$ is a solution too. For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.

Now let $(x, y)$ be a solution with $x>0$; without loss of generality confine attention to $y>0$. The equation rewritten as

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

shows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by 4 . Hence $x \geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So

$$
y=2^{x-1} m+\epsilon, \quad m \text { odd }, \quad \epsilon= \pm 1\tag{1}
$$

Plugging this into the original equation we obtain

$$
2^{x}\left(1+2^{x+1}\right)=\left(2^{x-1} m+\epsilon\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \epsilon,
$$

or, equivalently

$$
1+2^{x+1}=2^{x-2} m^{2}+m \epsilon .
$$

Therefore

$$
1-\epsilon m=2^{x-2}\left(m^{2}-8\right) .\tag{2}
$$

For $\epsilon=1$ this yields $m^{2}-8 \leq 0$, i.e., $m=1$, which fails to satisfy (2).

For $\epsilon=-1$ equation (2) gives us

$$
1+m=2^{x-2}\left(m^{2}-8\right) \geq 2\left(m^{2}-8\right),
$$

implying $2 m^{2}-m-17 \leq 0$. Hence $m \leq 3$; on the other hand $m$ cannot be 1 by $(2)$. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From (1) we get $y=23$. These values indeed satisfy the given equation. Recall that then $y=-23$ is also good. Thus we have the complete list of solutions $(x, y):(0,2),(0,-2),(4,23),(4,-23)$.

Final answer: $(0,2),(0,-2),(4,23),(4,-23)$