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


Question: Let $A B C$ be a triangle. The points $K, L$, and $M$ lie on the segments $B C, C A$, and $A B$, respectively, such that the lines $A K, B L$, and $C M$ intersect in a common point. Prove that it is possible to choose two of the triangles $A L M, B M K$, and $C K L$ whose inradii sum up to at least the inradius of the triangle $A B C$.

Solution:
Denote

$$
a=\frac{B K}{K C}, \quad b=\frac{C L}{L A}, \quad c=\frac{A M}{M B}
$$

By CEvA's theorem, $a b c=1$, so we may, without loss of generality, assume that $a \geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than 1. Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than 1 and the second one not greater than 1 . Without loss of generality, assume that $1 \leqslant a$ and $b \leqslant 1$.

Therefore, we obtain $b c \leqslant 1$ and $1 \leqslant c a$, or equivalently

$$
\frac{A M}{M B} \leqslant \frac{L A}{C L} \quad \text { and } \quad \frac{M B}{A M} \leqslant \frac{B K}{K C}
$$

The first inequality implies that the line passing through $M$ and parallel to $B C$ intersects the segment $A L$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $A L M$ is not less than the inradius $r_{1}$ of triangle $A M X$.

Similarly, the line passing through $M$ and parallel to $A C$ intersects the segment $B K$ at a point $Y$, so the inradius of the triangle $B M K$ is not less than the inradius $r_{2}$ of the triangle $B M Y$. Thus, to complete our solution, it is enough to show that $r_{1}+r_{2} \geqslant r$, where $r$ is the inradius of the triangle $A B C$. We prove that in fact $r_{1}+r_{2}=r$.

<img_3950>

Figure 1

Since $M X \| B C$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $A M X$ to the incircle of the triangle $A B C$. Therefore

$$
\frac{r_{1}}{r}=\frac{A M}{A B}, \quad \text { and similarly } \quad \frac{r_{2}}{r}=\frac{M B}{A B}
$$

Adding these equalities gives $r_{1}+r_{2}=r$, as required.
Subfield: Geometry


Question: Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B>B C$. The angle bisector of $\angle A B C$ intersects $\Omega$ at $M \neq B$. Let $\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\angle A O B$ and $\angle B O C$ intersect $\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $P Q$ so that $B R=M R$. Prove that $B R \| A C$. (Here we always assume that an angle bisector is a ray.)

Solution:
Let $K$ be the midpoint of $B M$, i.e., the centre of $\Gamma$. Notice that $A B \neq B C$ implies $K \neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.

Let $N$ be the second point of intersection of $\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \| A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.

Let $\omega$ be the circle with diameter $B O$. Since $\angle B N O=\angle B K O=90^{\circ}$, the points $N$ and $K$ lie on $\omega$.

Next we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\omega$. It follows that $\angle E O R=\angle E B K=\angle K B D=\angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the arc $P O Q$ of the circumcircle $\gamma$ of triangle $P O Q$.

Thus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\omega, \gamma$, and $\Gamma$. Hence they are concurrent at $R$, as required.

<img_3323>
Subfield: Number Theory


Question: Determine all pairs $(x, y)$ of positive integers such that

$$
\sqrt[3]{7 x^{2}-13 x y+7 y^{2}}=|x-y|+1\tag{1}
$$

Solution:
Let $(x, y)$ be any pair of positive integers solving (1). We shall prove that it appears in the list displayed above. The converse assertion that all these pairs do actually satisfy (1) either may be checked directly by means of a somewhat laborious calculation, or it can be seen by going in reverse order through the displayed equations that follow.

In case $x=y$ the given equation reduces to $x^{2 / 3}=1$, which is equivalent to $x=1$, whereby he have found the first solution.

To find the solutions with $x \neq y$ we may assume $x>y$ due to symmetry. Then the integer $n=x-y$ is positive and (1) may be rewritten as

$$
\sqrt[3]{7(y+n)^{2}-13(y+n) y+7 y^{2}}=n+1
$$

Raising this to the third power and simplifying the result one obtains

$$
y^{2}+y n=n^{3}-4 n^{2}+3 n+1
$$

To complete the square on the left hand side, we multiply by 4 and add $n^{2}$, thus getting

$$
(2 y+n)^{2}=4 n^{3}-15 n^{2}+12 n+4=(n-2)^{2}(4 n+1) .
$$

This shows that the cases $n=1$ and $n=2$ are impossible, whence $n>2$, and $4 n+1$ is the square of the rational number $\frac{2 y+n}{n-2}$. Consequently, it has to be a perfect square, and, since it is odd as well, there has to exist some nonnegative integer $m$ such that $4 n+1=(2 m+1)^{2}$, i.e.

$$
n=m^{2}+m \text {. }
$$

Notice that $n>2$ entails $m \geqslant 2$. Substituting the value of $n$ just found into the previous displayed equation we arrive at

$$
\left(2 y+m^{2}+m\right)^{2}=\left(m^{2}+m-2\right)^{2}(2 m+1)^{2}=\left(2 m^{3}+3 m^{2}-3 m-2\right)^{2} .
$$

Extracting square roots and taking $2 m^{3}+3 m^{2}-3 m-2=(m-1)\left(2 m^{2}+5 m+2\right)>0$ into account we derive $2 y+m^{2}+m=2 m^{3}+3 m^{2}-3 m-2$, which in turn yields

$$
y=m^{3}+m^{2}-2 m-1 \text {. }
$$

Notice that $m \geqslant 2$ implies that $y=\left(m^{3}-1\right)+(m-2) m$ is indeed positive, as it should be. In view of $x=y+n=y+m^{2}+m$ it also follows that

$$
x=m^{3}+2 m^{2}-m-1,
$$

and that this integer is positive as well.

Final answer: Either $(x, y)=(1,1)$ or $\{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\}$ for some positive integer $m \geqslant 2$
Subfield: Number Theory


Question: A coin is called a Cape Town coin if its value is $1 / n$ for some positive integer $n$. Given a collection of Cape Town coins of total value at most $99+\frac{1}{2}$, prove that it is possible to split this collection into at most 100 groups each of total value at most 1 .

Solution:
We will show that for every positive integer $N$ any collection of Cape Town coins of total value at most $N-\frac{1}{2}$ can be split into $N$ groups each of total value at most 1 . The problem statement is a particular case for $N=100$.

We start with some preparations. If several given coins together have a total value also of the form $\frac{1}{k}$ for a positive integer $k$, then we may merge them into one new coin. Clearly, if the resulting collection can be split in the required way then the initial collection can also be split.

After each such merging, the total number of coins decreases, thus at some moment we come to a situation when no more merging is possible. At this moment, for every even $k$ there is at most one coin of value $\frac{1}{k}$ (otherwise two such coins may be merged), and for every odd $k>1$ there are at most $k-1$ coins of value $\frac{1}{k}$ (otherwise $k$ such coins may also be merged).

Now, clearly, each coin of value 1 should form a single group; if there are $d$ such coins then we may remove them from the collection and replace $N$ by $N-d$. So from now on we may assume that there are no coins of value 1 .

Finally, we may split all the coins in the following way. For each $k=1,2, \ldots, N$ we put all the coins of values $\frac{1}{2 k-1}$ and $\frac{1}{2 k}$ into a group $G_{k}$; the total value of $G_{k}$ does not exceed

$$
(2 k-2) \cdot \frac{1}{2 k-1}+\frac{1}{2 k}<1
$$

It remains to distribute the "small" coins of values which are less than $\frac{1}{2 N}$; we will add them one by one. In each step, take any remaining small coin. The total value of coins in the groups at this moment is at most $N-\frac{1}{2}$, so there exists a group of total value at most $\frac{1}{N}\left(N-\frac{1}{2}\right)=1-\frac{1}{2 N}$; thus it is possible to put our small coin into this group. Acting so, we will finally distribute all the coins.
Subfield: Algebra


Question: Let $\left(a_{n}\right)_{n \geqslant 1}$ be a sequence of positive real numbers with the property that

$$
\left(a_{n+1}\right)^{2}+a_{n} a_{n+2} \leqslant a_{n}+a_{n+2}
$$

for all positive integers $n$. Show that $a_{2022} \leqslant 1$.

Solution:
We begin by observing that $\left(a_{n+1}\right)^{2}-1 \leqslant a_{n}+a_{n+2}-a_{n} a_{n+2}-1$, which is equivalent to

$$
\left(a_{n+1}\right)^{2}-1 \leqslant\left(1-a_{n}\right)\left(a_{n+2}-1\right) .
$$

Suppose now that there exists a positive integer $n$ such that $a_{n+1}>1$ and $a_{n+2}>1$. Since $\left(a_{n+1}\right)^{2}-1 \leqslant\left(1-a_{n}\right)\left(a_{n+2}-1\right)$, we deduce that $0<1-a_{n}<1<1+a_{n+2}$, thus $\left(a_{n+1}\right)^{2}-1<\left(a_{n+2}+1\right)\left(a_{n+2}-1\right)=\left(a_{n+2}\right)^{2}-1$.

On the other hand, $\left(a_{n+2}\right)^{2}-1 \leqslant\left(1-a_{n+3}\right)\left(a_{n+1}-1\right)<\left(1+a_{n+1}\right)\left(a_{n+1}-1\right)=\left(a_{n+1}\right)^{2}-1$, a contradiction. We have shown that we cannot have two consecutive terms, except maybe $a_{1}$ and $a_{2}$, strictly greater than 1 .

Finally, suppose $a_{2022}>1$. This implies that $a_{2021} \leqslant 1$ and $a_{2023} \leqslant 1$. Therefore $0<$ $\left(a_{2022}\right)^{2}-1 \leqslant\left(1-a_{2021}\right)\left(a_{2023}-1\right) \leqslant 0$, a contradiction. We conclude that $a_{2022} \leqslant 1$.
Subfield: Algebra


Question: Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set.

Solution:
First we show that $n \geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1}<a_{2}<\cdots<a_{n}$.

Note that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and, similarly for $a_{n}$, we must have $a_{n-k}+\cdots+a_{n-1} \geqslant a_{n}$. We also know that $n \geqslant k+1$.

If $n=k+1$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}>a_{1}+\cdots+a_{k} \geqslant a_{k+1}$, which gives a contradiction.

If $n=k+2$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1} \geqslant a_{k+2}$, that again gives a contradiction.

If $n=k+3$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and $a_{3}+\cdots+a_{k+2} \geqslant a_{k+3}$. Adding the two inequalities we get $a_{1}+a_{k+2} \geqslant a_{2}+a_{k+3}$, again a contradiction.

It remains to give an example of a set with $k+4$ elements satisfying the condition of the problem. We start with the case when $k=2 l$ and $l \geqslant 1$. In that case, denote by $A_{i}=\{-i, i\}$ and take the set $A_{1} \cup \cdots \cup A_{l+2}$, which has exactly $k+4=2 l+4$ elements. We are left to show that this set satisfies the required condition.

Note that if a number $i$ can be expressed in the desired way, then so can $-i$ by negating the expression. Therefore, we consider only $1 \leqslant i \leqslant l+2$.

If $i<l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \neq 1, i+1$, and the numbers $i+1$ and -1 .

For $i=l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \neq 1, l+1$, and the numbers $l+1$ and 1 .

It remains to a give a construction for odd $k=2 l+1$ with $l \geqslant 1$ (since $k \geqslant 2$ ). To that end, we modify the construction for $k=2 l$ by adding 0 to the previous set.

This is a valid set as 0 can be added to each constructed expression, and 0 can be expressed as follows: take the numbers $1,2,-3$ and all the numbers from the remaining $l-1$ sets $A_{4}, A_{5}, \cdots, A_{l+2}$.

Final answer: $n=k+4$
Subfield: Algebra


Question: Let $n \geqslant 3$ be an integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers in the interval $[0,1]$. Let $s=x_{1}+x_{2}+\ldots+x_{n}$, and assume that $s \geqslant 3$. Prove that there exist integers $i$ and $j$ with $1 \leqslant i<j \leqslant n$ such that

$$
2^{j-i} x_{i} x_{j}>2^{s-3}
$$

Solution:
Let $1 \leqslant a<b \leqslant n$ be such that $2^{b-a} x_{a} x_{b}$ is maximal. This choice of $a$ and $b$ implies that $x_{a+t} \leqslant 2^{t} x_{a}$ for all $1-a \leqslant t \leqslant b-a-1$, and similarly $x_{b-t} \leqslant 2^{t} x_{b}$ for all $b-n \leqslant t \leqslant b-a+1$. Now, suppose that $x_{a} \in\left(\frac{1}{2^{u+1}}, \frac{1}{2^{u}}\right]$ and $x_{b} \in\left(\frac{1}{2^{v+1}}, \frac{1}{2^{v}}\right]$, and write $x_{a}=2^{-\alpha}, x_{b}=2^{-\beta}$. Then

$$
\sum_{i=1}^{a+u-1} x_{i} \leqslant 2^{u} x_{a}\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{a+u-1}}\right)<2^{u} x_{a} \leqslant 1
$$

and similarly,

$$
\sum_{i=b-v+1}^{n} x_{i} \leqslant 2^{v} x_{b}\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{n-b+v}}\right)<2^{v} x_{b} \leqslant 1
$$

In other words, the sum of the $x_{i}$ 's for $i$ outside of the interval $[a+u, b-v]$ is strictly less than 2. Since the total sum is at least 3 , and each term is at most 1 , it follows that this interval must have at least two integers. i.e., $a+u<b-v$. Thus, by bounding the sum of the $x_{i}$ for $i \in[1, a+u] \cup[b-v, n]$ like above, and trivially bounding each $x_{i} \in(a+u, b-v)$ by 1 , we obtain

$$
s<2^{u+1} x_{a}+2^{v+1} x_{b}+((b-v)-(a+u)-1)=b-a+\left(2^{u+1-\alpha}+2^{v+1-\beta}-(u+v+1)\right)
$$

Now recall $\alpha \in(u, u+1]$ and $\beta \in(v, v+1]$, so applying Bernoulli's inequality yields $\left.2^{u+1-\alpha}+2^{v+1-\beta}-u-v-1 \leqslant(1+(u+1-\alpha))\right)+(1+(v+1-\beta))-u-v-1=3-\alpha-\beta$.

It follows that $s-3<b-a-\alpha-\beta$, and so

$$
2^{s-3}<2^{b-a-\alpha-\beta}=2^{b-a} x_{a} x_{b} .
$$
Subfield: Combinatorics


Question: $A \pm 1 \text{-}sequence$ is a sequence of 2022 numbers $a_{1}, \ldots, a_{2022}$, each equal to either +1 or -1 . Determine the largest $C$ so that, for any $\pm 1 -sequence$, there exists an integer $k$ and indices $1 \leqslant t_{1}<\ldots<t_{k} \leqslant 2022$ so that $t_{i+1}-t_{i} \leqslant 2$ for all $i$, and

$$
\left|\sum_{i=1}^{k} a_{t_{i}}\right| \geqslant C
$$

Solution:
First, we prove that this can always be achieved. Without loss of generality, suppose at least $\frac{2022}{2}=1011$ terms of the \pm 1 -sequence are +1 . Define a subsequence as follows: starting at $t=0$, if $a_{t}=+1$ we always include $a_{t}$ in the subsequence. Otherwise, we skip $a_{t}$ if we can (i.e. if we included $a_{t-1}$ in the subsequence), otherwise we include it out of necessity, and go to the next $t$. Clearly, this subsequence will include all $+1 \mathrm{~s}$. Also, for each -1 included in the sequence, a -1 must have been skipped, so at most $\left\lfloor\frac{1011}{2}\right\rfloor=505$ can be included. Hence the sum is at least $1011-505=506$, as desired.

Next, we prove that, for the \pm 1 -sequence

$$
(\{-1\},\{+1,+1\},\{-1,-1\},\{+1,+1\}, \ldots,\{+1,+1\},\{-1,-1\},\{+1\}),
$$

each admissible subsequence $a_{t_{i}}$ has $-506 \leqslant \sum_{i} a_{t_{i}} \leqslant 506$. We say that the terms inside each curly bracket is a block. In total, there are 1012 blocks - 506 of them hold +1-s, and 506 of them hold -1 s. (The two blocks at each end hold 1 number each, each other block holds 2.)

Suppose an admissible subsequence includes terms from $k$ blocks holding +1 -s. Then, in each -1 -pair in between the +1 -pairs, the subsequence must also include at least one -1 . There can be at most two +1 s included from each +1 -block, and at least one -1 must be included from each -1 -block, so the sum is at most $2 k-(k-1)=k+1$.

For $k<506$, this is at most 506. If $k=506$, one of the +1 -blocks must be the one at the end, meaning it can only include one +1 , so that the maximum in this case is only $k$, not $k+1$, so in this case the sum is also at most 506.

Hence we have shown that for any admissible subsequence, $\sum_{i} a_{t_{i}} \leqslant 506$. Analogously we can show that $-506 \leqslant \sum_{i} a_{t_{i}}$, meaning that $C \leqslant 506$ as desired.

Final answer: 506
Subfield: Combinatorics


Question: Let $m, n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_{1}, X_{2}, \ldots, X_{m}$ be pairwise distinct non-empty, not necessary disjoint subsets of $X$. A function $f: X \rightarrow$ $\{1,2, \ldots, n+1\}$ is called nice if there exists an index $k$ such that

$$
\sum_{x \in X_{k}} f(x)>\sum_{x \in X_{i}} f(x) \text { for all } i \neq k
$$

Prove that the number of nice functions is at least $n^{n}$.

Solution:
For a subset $Y \subseteq X$, we write $f(Y)$ for $\sum_{y \in Y} f(y)$. Note that a function $f: X \rightarrow$ $\{1, \ldots, n+1\}$ is nice, if and only if $f\left(X_{i}\right)$ is maximized by a unique index $i \in\{1, \ldots, m\}$.

We will first investigate the set $\mathcal{F}$ of functions $f: X \rightarrow\{1, \ldots, n\}$; note that $|\mathcal{F}|=n^{n}$. For every function $f \in \mathcal{F}$, define a corresponding function $f^{+}: X \rightarrow\{1,2, \ldots, n+1\}$ in the following way: Pick some set $X_{l}$ that maximizes the value $f\left(X_{l}\right)$.

- For all $x \in X_{l}$, define $f^{+}(x)=f(x)+1$.
- For all $x \in X \backslash X_{l}$, define $f^{+}(x)=f(x)$.

Claim. The resulting function $f^{+}$is nice.

Proof. Note that $f^{+}\left(X_{i}\right)=f\left(X_{i}\right)+\left|X_{i} \cap X_{l}\right|$ holds for all $X_{i}$. We show that $f^{+}\left(X_{i}\right)$ is maximized at the unique index $i=l$. Hence consider some arbitrary index $j \neq l$. Then $X_{l} \subset X_{j}$ is impossible, as this would imply $f\left(X_{j}\right)>f\left(X_{l}\right)$ and thereby contradict the choice of set $X_{l}$; this in particular yields $\left|X_{l}\right|>\left|X_{j} \cap X_{l}\right|$.

$$
f^{+}\left(X_{l}\right)=f\left(X_{l}\right)+\left|X_{l}\right| \geqslant f\left(X_{j}\right)+\left|X_{l}\right|>f\left(X_{j}\right)+\left|X_{j} \cap X_{l}\right|=f^{+}\left(X_{j}\right)
$$

The first inequality follows since $X_{l}$ was chosen to maximize the value $f\left(X_{l}\right)$. The second (strict) inequality follows from $\left|X_{l}\right|>\left|X_{j} \cap X_{l}\right|$ as observed above. This completes the proof of the claim.

Next observe that function $f$ can be uniquely reconstructed from $f^{+}$: the claim yields that $f^{+}$has a unique maximizer $X_{l}$, and by decreasing the value of $f^{+}$on $X_{l}$ by 1 , we get we can fully determine the values of $f$. As each of the $n^{n}$ functions $f \in \mathcal{F}$ yields a (unique) corresponding nice function $f^{+}: X \rightarrow\{1,2, \ldots, n+1\}$, the proof is complete.
Subfield: Geometry


Question: Let $A B C D E$ be a convex pentagon such that $B C=D E$. Assume there is a point $T$ inside $A B C D E$ with $T B=T D, T C=T E$ and $\angle T B A=\angle A E T$. Let lines $C D$ and $C T$ intersect line $A B$ at points $P$ and $Q$, respectively, and let lines $C D$ and $D T$ intersect line $A E$ at points $R$ and $S$, respectively. Assume that points $P, B, A, Q$ and $R, E, A, S$ respectively, are collinear and occur on their lines in this order. Prove that the points $P, S, Q, R$ are concyclic.

Solution:
By the conditions we have $B C=D E, C T=E T$ and $T B=T D$, so the triangles $T B C$ and $T D E$ are congruent, in particular $\angle B T C=\angle D T E$.

In triangles $T B Q$ and $T E S$ we have $\angle T B Q=\angle S E T$ and $\angle Q T B=180^{\circ}-\angle B T C=180^{\circ}-$ $\angle D T E=\angle E T S$, so these triangles are similar to each other. It follows that $\angle T S E=\angle B Q T$ and

$$
\frac{T D}{T Q}=\frac{T B}{T Q}=\frac{T E}{T S}=\frac{T C}{T S}
$$

By rearranging this relation we get $T D \cdot T S=T C \cdot T Q$, so $C, D, Q$ and $S$ are concyclic. (Alternatively, we can get $\angle C Q D=\angle C S D$ from the similar triangles $T C S$ and $T D Q$.) Hence, $\angle D C Q=\angle D S Q$.

Finally, from the angles of triangle $C Q P$ we get

$$
\angle R P Q=\angle R C Q-\angle P Q C=\angle D S Q-\angle D S R=\angle R S Q
$$

which proves that $P, Q, R$ and $S$ are concyclic.

<img_3514>
As in the previous solution, we note that triangles $T B C$ and $T D E$ are congruent. Denote the intersection point of $D T$ and $B A$ by $V$, and the intersection point of $C T$ and $E A$ by $W$. From triangles $B C Q$ and $D E S$ we then have

$$
\begin{aligned}
\angle V S W & =\angle D S E=180^{\circ}-\angle S E D-\angle E D S=180^{\circ}-\angle A E T-\angle T E D-\angle E D T \\
& =180^{\circ}-\angle T B A-\angle T C B-\angle C B T=180^{\circ}-\angle Q C B-\angle C B Q=\angle B Q C=\angle V Q W,
\end{aligned}
$$

meaning that $V S Q W$ is cyclic, and in particular $\angle W V Q=\angle W S Q$. Since

$$
\angle V T B=180^{\circ}-\angle B T C-\angle C T D=180^{\circ}-\angle C T D-\angle D T E=\angle E T W
$$

and $\angle T B V=\angle W E T$ by assumption, we have that the triangles $V T B$ and $W T E$ are similar, hence

$$
\frac{V T}{W T}=\frac{B T}{E T}=\frac{D T}{C T}
$$



Thus $C D \| V W$, and angle chasing yields

$$
\angle R P Q=\angle W V Q=\angle W S Q=\angle R S Q
$$

concluding the proof.
Subfield: Geometry


Question: In the acute-angled triangle $A B C$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $A F$. The lines through $P$ parallel to $A C$ and $A B$ meet $B C$ at $D$ and $E$, respectively. Points $X \neq A$ and $Y \neq A$ lie on the circles $A B D$ and $A C E$, respectively, such that $D A=D X$ and $E A=E Y$.

Prove that $B, C, X$ and $Y$ are concyclic.

Solution:
Let $A^{\prime}$ be the intersection of lines $B X$ and $C Y$. By power of a point, it suffices to prove that $A^{\prime} B \cdot A^{\prime} X=A^{\prime} C \cdot A^{\prime} Y$, or, equivalently, that $A^{\prime}$ lies on the radical axis of the circles $A B D X$ and $A C E Y$.

From $D A=D X$ it follows that in circle $A B D X$, point $D$ bisects of one of the arcs $A X$. Therefore, depending on the order of points, the line $B C$ is either the internal or external bisector of $\angle A B X$. In both cases, line $B X$ is the reflection of $B A$ in line $B D C$. Analogously, line $C Y$ is the reflection of $C A$ in line $B C$; we can see that $A^{\prime}$ is the reflection of $A$ in line $B C$, so $A, F$ and $A^{\prime}$ are collinear.

By $P D \| A C$ and $P E \| A B$ we have $\frac{F D}{F C}=\frac{F P}{F A}=\frac{F E}{F B}$, hence $F D \cdot F B=F E \cdot F C$. So, point $F$ has equal powers with respect to circles $A B D X$ and $A C E Y$.

Point $A$, being a common point of the two circles, is another point with equal powers, so the radical axis of circles $A B D X$ and $A C E Y$ is the altitude $A F$ that passes through $A^{\prime}$.

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