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


Question: Let $\mathbb{Z}$ be the set of integers. We consider functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying

$$
f(f(x+y)+y)=f(f(x)+y)
$$

for all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set

$$
X_{v}=\{x \in \mathbb{Z}: f(x)=v\}
$$

is finite and nonempty.
Prove that there exists such a function $f$ for which there is an $f$-rare integer.

Solution:
Let $f$ be the function where $f(0)=0$ and $f(x)$ is the largest power of 2 dividing $2 x$ for $x \neq 0$. The integer 0 is evidently $f$-rare, so it remains to verify the functional equation.

Since $f(2 x)=2 f(x)$ for all $x$, it suffices to verify the functional equation when at least one of $x$ and $y$ is odd (the case $x=y=0$ being trivial). If $y$ is odd, then we have

$$
f(f(x+y)+y)=2=f(f(x)+y)
$$

since all the values attained by $f$ are even. If, on the other hand, $x$ is odd and $y$ is even, then we already have

$$
f(x+y)=2=f(x)
$$

from which the functional equation follows immediately.
Subfield: Algebra


Question: Let $\mathbb{Z}$ be the set of integers. We consider functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying

$$
f(f(x+y)+y)=f(f(x)+y)
$$

for all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set

$$
X_{v}=\{x \in \mathbb{Z}: f(x)=v\}
$$

is finite and nonempty.
Prove that no such function $f$ can have more than one $f$-rare integer.

Solution:
An easy inductive argument (substituting $x+k y$ for $x$ ) shows that

$$
f(f(x+k y)+y)=f(f(x)+y)
$$

for all integers $x, y$ and $k$. If $v$ is an $f$-rare integer and $a$ is the least element of $X_{v}$, then by substituting $y=a-f(x)$ in the above, we see that

$$
f(x+k \cdot(a-f(x)))-f(x)+a \in X_{v}
$$

for all integers $x$ and $k$, so that in particular

$$
f(x+k \cdot(a-f(x))) \geqslant f(x)
$$

for all integers $x$ and $k$, by assumption on $a$. This says that on the (possibly degenerate) arithmetic progression through $x$ with common difference $a-f(x)$, the function $f$ attains its minimal value at $x$.

Repeating the same argument with $a$ replaced by the greatest element $b$ of $X_{v}$ shows that

$$
f(x+k \cdot(b-f(x)) \leqslant f(x)
$$

for all integers $x$ and $k$. Combined with the above inequality, we therefore have

$$
f(x+k \cdot(a-f(x)) \cdot(b-f(x)))=f(x)
$$

for all integers $x$ and $k$.

Thus if $f(x) \neq a, b$, then the set $X_{f(x)}$ contains a nondegenerate arithmetic progression, so is infinite. So the only possible $f$-rare integers are $a$ and $b$.

In particular, the $f$-rare integer $v$ we started with must be one of $a$ or $b$, so that $f(v)=$ $f(a)=f(b)=v$. This means that there cannot be any other $f$-rare integers $v^{\prime}$, as they would on the one hand have to be either $a$ or $b$, and on the other would have to satisfy $f\left(v^{\prime}\right)=v^{\prime}$. Thus $v$ is the unique $f$-rare integer.
Suppose $v$ is $f$-rare, and let $a$ and $b$ be the least and greatest elements of $X_{v}$, respectively. Substituting $x=v$ and $y=a-v$ into the equation shows that

$$
f(v)-v+a \in X_{v}
$$

and in particular $f(v) \geqslant v$. Repeating the same argument with $x=v$ and $y=b-v$ shows that $f(v) \leqslant v$, and hence $f(v)=v$.

Suppose now that $v^{\prime}$ is a second $f$-rare integer. We may assume that $v=0$. We've seen that $f\left(v^{\prime}\right)=v^{\prime}$; we claim that in fact $f\left(k v^{\prime}\right)=v^{\prime}$ for all positive integers $k$. This gives a contradiction unless $v^{\prime}=v=0$.

This claim is proved by induction on $k$. Supposing it to be true for $k$, we substitute $y=k v^{\prime}$ and $x=0$ into the functional equation to yield

$$
f\left((k+1) v^{\prime}\right)=f\left(f(0)+k v^{\prime}\right)=f\left(k v^{\prime}\right)=v^{\prime}
$$

using that $f(0)=0$. This completes the induction, and hence the proof.
Subfield: Combinatorics


Question: You are given a set of $n$ blocks, each weighing at least 1 ; their total weight is $2 n$. Prove that for every real number $r$ with $0 \leqslant r \leqslant 2 n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r+2$.

Solution:
We prove the following more general statement by induction on $n$.

Claim. Suppose that you have $n$ blocks, each of weight at least 1 , and of total weight $s \leqslant 2 n$. Then for every $r$ with $-2 \leqslant r \leqslant s$, you can choose some of the blocks whose total weight is at least $r$ but at most $r+2$.

Proof. The base case $n=1$ is trivial. To prove the inductive step, let $x$ be the largest block weight. Clearly, $x \geqslant s / n$, so $s-x \leqslant \frac{n-1}{n} s \leqslant 2(n-1)$. Hence, if we exclude a block of weight $x$, we can apply the inductive hypothesis to show the claim holds (for this smaller set) for any $-2 \leqslant r \leqslant s-x$. Adding the excluded block to each of those combinations, we see that the claim also holds when $x-2 \leqslant r \leqslant s$. So if $x-2 \leqslant s-x$, then we have covered the whole interval $[-2, s]$. But each block weight is at least 1 , so we have $x-2 \leqslant(s-(n-1))-2=s-(2 n-(n-1)) \leqslant s-(s-(n-1)) \leqslant s-x$, as desired.
Let $x_{1}, \ldots, x_{n}$ be the weights of the blocks in weakly increasing order. Consider the set $S$ of sums of the form $\sum_{j \in J} x_{j}$ for a subset $J \subseteq\{1,2, \ldots, n\}$. We want to prove that the mesh of $S$-i.e. the largest distance between two adjacent elements - is at most 2 .

For $0 \leqslant k \leqslant n$, let $S_{k}$ denote the set of sums of the form $\sum_{i \in J} x_{i}$ for a subset $J \subseteq\{1,2, \ldots, k\}$. We will show by induction on $k$ that the mesh of $S_{k}$ is at most 2 .

The base case $k=0$ is trivial (as $S_{0}=\{0\}$ ). For $k>0$ we have

$$
S_{k}=S_{k-1} \cup\left(x_{k}+S_{k-1}\right)
$$

(where $\left(x_{k}+S_{k-1}\right)$ denotes $\left\{x_{k}+s: s \in S_{k-1}\right\}$ ), so it suffices to prove that $x_{k} \leqslant \sum_{j<k} x_{j}+2$. But if this were not the case, we would have $x_{l}>\sum_{j<k} x_{j}+2 \geqslant k+1$ for all $l \geqslant k$, and hence

$$
2 n=\sum_{j=1}^{n} x_{j}>(n+1-k)(k+1)+k-1
$$

This rearranges to $n>k(n+1-k)$, which is false for $1 \leqslant k \leqslant n$, giving the desired contradiction.
Subfield: Geometry


Question: Let $A B C$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $A B$ and $A C$ again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$.

Solution:
Notice that $\angle T F B=\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\angle F D A=\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\angle T G B=\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\angle G E C=\angle C F A$. Hence,

$$
\angle T F B=\angle C G A \text { and } \angle T G B=\angle C F A \text {. }
\tag{1}
$$

<img_3161>

Triangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$.
Subfield: Geometry


Question: Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\omega_{B}$ and $\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\omega_{B}$ and $\omega_{C}$ again at $P \neq M$ and $Q \neq N$, respectively. Prove that $M P=N Q$.

Solution:
Denote the centres of $\omega_{B}$ and $\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively.

<img_4016>

From the cyclic quadrilaterals $A F D C$ and $A B D E$ we have

$$
\angle M D O_{B}=\frac{1}{2} \angle F D B=\frac{1}{2} \angle B A C=\frac{1}{2} \angle C D E=\angle O_{C} D N
$$

so the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is

$$
\frac{D N}{D M}=\frac{O_{C} N}{O_{B} M}=\frac{r_{C}}{r_{B}}
$$

Let $\varphi=\angle D M N$ and $\psi=\angle M N D$. The lines $F M$ and $E N$ are tangent to $\omega_{B}$ and $\omega_{C}$, respectively, so

$$
\angle M T P=\angle F M P=\angle D M N=\varphi \quad \text { and } \quad \angle Q U N=\angle Q N E=\angle M N D=\psi .
$$

(It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.)

In the circles $\omega_{B}$ and $\omega_{C}$ the lengths of chords $M P$ and $N Q$ are

$$
M P=2 r_{B} \cdot \sin \angle M T P=2 r_{B} \cdot \sin \varphi \quad \text { and } \quad N Q=2 r_{C} \cdot \sin \angle Q U N=2 r_{C} \cdot \sin \psi
$$

By applying the sine rule to triangle $D N M$ we get

$$
\frac{D N}{D M}=\frac{\sin \angle D M N}{\sin \angle M N D}=\frac{\sin \varphi}{\sin \psi}
$$

Finally, putting the above observations together, we get

$$
\frac{M P}{N Q}=\frac{2 r_{B} \sin \varphi}{2 r_{C} \sin \psi}=\frac{r_{B}}{r_{C}} \cdot \frac{\sin \varphi}{\sin \psi}=\frac{D M}{D N} \cdot \frac{\sin \varphi}{\sin \psi}=\frac{\sin \psi}{\sin \varphi} \cdot \frac{\sin \varphi}{\sin \psi}=1
$$

so $M P=N Q$ as required.
Subfield: Number Theory


Question: Let $a$ and $b$ be two positive integers. Prove that the integer

$$
a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil
$$

is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.)

Solution:
Arguing indirectly, assume that

$$
a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil=(a+k)^{2}, \quad \text { or } \quad\left\lceil\frac{(2 a)^{2}}{b}\right\rceil=(2 a+k) k
$$

Clearly, $k \geqslant 1$. In other words, the equation

$$
\left\lceil\frac{c^{2}}{b}\right\rceil=(c+k) k
\tag{1}
$$

has a positive integer solution $(c, k)$, with an even value of $c$.

Choose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From

$$
\frac{c^{2}}{b}>\left\lceil\frac{c^{2}}{b}\right\rceil-1=c k+k^{2}-1 \geqslant c k
$$

and

$$
\frac{(c-k)(c+k)}{b}<\frac{c^{2}}{b} \leqslant\left\lceil\frac{c^{2}}{b}\right\rceil=(c+k) k
$$

it can be seen that $c>b k>c-k$, so

$$
c=k b+r \quad \text { with some } 0<r<k \text {. }
$$

By substituting this in (1) we get

$$
\left\lceil\frac{c^{2}}{b}\right\rceil=\left\lceil\frac{(b k+r)^{2}}{b}\right\rceil=k^{2} b+2 k r+\left\lceil\frac{r^{2}}{b}\right\rceil
$$

and

$$
(c+k) k=(k b+r+k) k=k^{2} b+2 k r+k(k-r),
$$

so

$$
\left\lceil\frac{r^{2}}{b}\right\rceil=k(k-r)
\tag{2}
$$

Notice that relation (2) provides another positive integer solution of (1), namely $c^{\prime}=r$ and $k^{\prime}=k-r$, with $c^{\prime}>0$ and $0<k^{\prime}<k$. That contradicts the minimality of $k$, and hence finishes the solution.
Suppose that

$$
a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil=c^{2}
$$

with some positive integer $c>a$, so

$$
\begin{aligned}
& c^{2}-1<a^{2}+\frac{4 a^{2}}{b} \leqslant c^{2}, \\
& 0 \leqslant c^{2} b-a^{2}(b+4)<b .
\end{aligned}
\tag{3}
$$

Let $d=c^{2} b-a^{2}(b+4), x=c+a$ and $y=c-a$; then we have $c=\frac{x+y}{2}$ and $a=\frac{x-y}{2}$, and (3) can be re-written as follows:

$$
\left(\frac{x+y}{2}\right)^{2} b-\left(\frac{x-y}{2}\right)^{2}(b+4) =d,
$$
$$
x^{2}-(b+2) x y+y^{2}+d=0 ; \quad 0 \leqslant d<b .
\tag{4}
$$

So, by the indirect assumption, the equation (4) has some positive integer solution $(x, y)$.

Fix $b$ and $d$, and take a pair $(x, y)$ of positive integers, satisfying (4), such that $x+y$ is minimal. By the symmetry in (4) we may assume that $x \geqslant y \geqslant 1$.

Now we perform a usual "Vieta jump". Consider (4) as a quadratic equation in variable $x$, and let $z$ be its second root. By the Vieta formulas,

$$
x+z=(b+2) y, \quad \text { and } \quad z x=y^{2}+d
$$

so

$$
z=(b+2) y-x=\frac{y^{2}+d}{x}
$$

The first formula shows that $z$ is an integer, and by the second formula $z$ is positive. Hence $(z, y)$ is another positive integer solution of (4). From

$$
\begin{aligned}
(x-1)(z-1) & =x z-(x+z)+1=\left(y^{2}+d\right)-(b+2) y+1 \\
& <\left(y^{2}+b\right)-(b+2) y+1=(y-1)^{2}-b(y-1) \leqslant(y-1)^{2} \leqslant(x-1)^{2}
\end{aligned}
$$

we can see that $z<x$ and therefore $z+y<x+y$. But this contradicts the minimality of $x+y$ among the positive integer solutions of (4).
Subfield: Algebra


Question: Suppose that a sequence $a_{1}, a_{2}, \ldots$ of positive real numbers satisfies

$$
a_{k+1} \geqslant \frac{k a_{k}}{a_{k}^{2}+(k-1)}
\tag{1}
$$

for every positive integer $k$. Prove that $a_{1}+a_{2}+\cdots+a_{n} \geqslant n$ for every $n \geqslant 2$.

Solution:
From the constraint (1), it can be seen that

$$
\frac{k}{a_{k+1}} \leqslant \frac{a_{k}^{2}+(k-1)}{a_{k}}=a_{k}+\frac{k-1}{a_{k}}
$$

and so

$$
a_{k} \geqslant \frac{k}{a_{k+1}}-\frac{k-1}{a_{k}}
$$

Summing up the above inequality for $k=1, \ldots, m$, we obtain

$$
a_{1}+a_{2}+\cdots+a_{m} \geqslant\left(\frac{1}{a_{2}}-\frac{0}{a_{1}}\right)+\left(\frac{2}{a_{3}}-\frac{1}{a_{2}}\right)+\cdots+\left(\frac{m}{a_{m+1}}-\frac{m-1}{a_{m}}\right)=\frac{m}{a_{m+1}} .
\tag{2}
$$

Now we prove the problem statement by induction on $n$. The case $n=2$ can be done by applying (1) to $k=1$ :

$$
a_{1}+a_{2} \geqslant a_{1}+\frac{1}{a_{1}} \geqslant 2
$$

For the induction step, assume that the statement is true for some $n \geqslant 2$. If $a_{n+1} \geqslant 1$, then the induction hypothesis yields

$$
\left(a_{1}+\cdots+a_{n}\right)+a_{n+1} \geqslant n+1 .
\tag{3}
$$

Otherwise, if $a_{n+1}<1$ then apply (2) as

$$
\left(a_{1}+\cdots+a_{n}\right)+a_{n+1} \geqslant \frac{n}{a_{n+1}}+a_{n+1}=\frac{n-1}{a_{n+1}}+\left(\frac{1}{a_{n+1}}+a_{n+1}\right)>(n-1)+2 .
$$

That completes the solution.
Subfield: Algebra


Question: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the equation

$$
f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x)\tag{1}
$$

for all real numbers $x$ and $y$.

Solution:
Clearly, each of the functions $x \mapsto x$ and $x \mapsto 2-x$ satisfies (1). It suffices now to show that they are the only solutions to the problem.

Suppose that $f$ is any function satisfying (1). Then setting $y=1$ in (1), we obtain

$$
f(x+f(x+1))=x+f(x+1)\tag{2}
$$

in other words, $x+f(x+1)$ is a fixed point of $f$ for every $x \in \mathbb{R}$.

We distinguish two cases regarding the value of $f(0)$.

Case 1. $f(0) \neq 0$.

By letting $x=0$ in (1), we have

$$
f(f(y))+f(0)=f(y)+y f(0) .
$$

So, if $y_{0}$ is a fixed point of $f$, then substituting $y=y_{0}$ in the above equation we get $y_{0}=1$. Thus, it follows from (2) that $x+f(x+1)=1$ for all $x \in \mathbb{R}$. That is, $f(x)=2-x$ for all $x \in \mathbb{R}$. Case 2. $f(0)=0$.

By letting $y=0$ and replacing $x$ by $x+1$ in (1), we obtain

$$
f(x+f(x+1)+1)=x+f(x+1)+1 .\tag{3}
$$

From (1), the substitution $x=1$ yields

$$
f(1+f(y+1))+f(y)=1+f(y+1)+y f(1) .\tag{4}
$$

By plugging $x=-1$ into (2), we see that $f(-1)=-1$. We then plug $y=-1$ into (4) and deduce that $f(1)=1$. Hence, (4) reduces to

$$
f(1+f(y+1))+f(y)=1+f(y+1)+y .\tag{5}
$$

Accordingly, if both $y_{0}$ and $y_{0}+1$ are fixed points of $f$, then so is $y_{0}+2$. Thus, it follows from (2) and (3) that $x+f(x+1)+2$ is a fixed point of $f$ for every $x \in \mathbb{R}$; i.e.,

$$
f(x+f(x+1)+2)=x+f(x+1)+2 .
$$

Replacing $x$ by $x-2$ simplifies the above equation to

$$
f(x+f(x-1))=x+f(x-1) \text {. }
$$

On the other hand, we set $y=-1$ in (1) and get

$$
f(x+f(x-1))=x+f(x-1)-f(x)-f(-x) .
$$

Therefore, $f(-x)=-f(x)$ for all $x \in \mathbb{R}$.

Finally, we substitute $(x, y)$ by $(-1,-y)$ in (1) and use the fact that $f(-1)=-1$ to get

$$
f(-1+f(-y-1))+f(y)=-1+f(-y-1)+y
$$

Since $f$ is an odd function, the above equation becomes

$$
-f(1+f(y+1))+f(y)=-1-f(y+1)+y \text {. }
$$

By adding this equation to (5), we conclude that $f(y)=y$ for all $y \in \mathbb{R}$.

Final answer: $f(x)=x$,$f(x)=2-x$
Subfield: Algebra


Question: Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences

$$
\begin{aligned}
& P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\
& Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014)
\end{aligned}
$$

are permutations of each other.
Prove that there exist distinct block-similar polynomials of degree $n+1$.

Solution:
For convenience, we set $k=2015=2 \ell+1$.

Consider the following polynomials of degree $n+1$ :

$$
P(x)=\prod_{i=0}^{n}(x-i k) \quad \text { and } \quad Q(x)=\prod_{i=0}^{n}(x-i k-1)
$$

Since $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\cdots=P(n k)$, these polynomials are block-similar (and distinct).
Subfield: Geometry


Question: Let $A B C$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $A B G H$ is a parallelogram. Let $I$ be the point on the line $G H$ such that $A C$ bisects $H I$. Suppose that the line $A C$ intersects the circumcircle of the triangle $G C I$ at $C$ and $J$. Prove that $I J=A H$.

Solution:
Since $H G \| A B$ and $B G \| A H$, we have $B G \perp B C$ and $C H \perp G H$. Therefore, the quadrilateral $B G C H$ is cyclic. Since $H$ is the orthocenter of the triangle $A B C$, we have $\angle H A C=90^{\circ}-\angle A C B=\angle C B H$. Using that $B G C H$ and $C G J I$ are cyclic quadrilaterals, we get

$$
\angle C J I=\angle C G H=\angle C B H=\angle H A C .
$$

Let $M$ be the intersection of $A C$ and $G H$, and let $D \neq A$ be the point on the line $A C$ such that $A H=H D$. Then $\angle M J I=\angle H A C=\angle M D H$.

Since $\angle M J I=\angle M D H, \angle I M J=\angle H M D$, and $I M=M H$, the triangles $I M J$ and $H M D$ are congruent, and thus $I J=H D=A H$.

<img_3578>
Obtain $\angle C G H=\angle H A C$ as in the previous solution. In the parallelogram $A B G H$ we have $\angle B A H=\angle H G B$. It follows that

$$
\angle H M C=\angle B A C=\angle B A H+\angle H A C=\angle H G B+\angle C G H=\angle C G B .
$$

So the right triangles $C M H$ and $C G B$ are similar. Also, in the circumcircle of triangle $G C I$ we have similar triangles $M I J$ and $M C G$. Therefore,

$$
\frac{I J}{C G}=\frac{M I}{M C}=\frac{M H}{M C}=\frac{G B}{G C}=\frac{A H}{C G}
$$

Hence $I J=A H$.
Subfield: Geometry


Question: Let $A B C$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $B C$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $A B$ of $\Omega$ not containing $C$, and $G$ lies on the arc $A C$ of $\Omega$ not containing $B$. The circumcircles of the triangles $B D F$ and $C E G$ meet the sides $A B$ and $A C$ again at $K$ and $L$, respectively. Suppose that the lines $F K$ and $G L$ are distinct and intersect at $X$. Prove that the points $A, X$, and $O$ are collinear.

Solution:
It suffices to prove that the lines $F K$ and $G L$ are symmetric about $A O$. Now the segments $A F$ and $A G$, being chords of $\Omega$ with the same length, are clearly symmetric with respect to $A O$. Hence it is enough to show

$$
\angle K F A=\angle A G L \text {. }
\tag{1}
$$

Let us denote the circumcircles of $B D F$ and $C E G$ by $\omega_{B}$ and $\omega_{C}$, respectively. To prove (1), we start from

$$
\angle K F A=\angle D F G+\angle G F A-\angle D F K .
$$

In view of the circles $\omega_{B}, \Gamma$, and $\Omega$, this may be rewritten as

$$
\angle K F A=\angle C E G+\angle G B A-\angle D B K=\angle C E G-\angle C B G .
$$

Due to the circles $\omega_{C}$ and $\Omega$, we obtain $\angle K F A=\angle C L G-\angle C A G=\angle A G L$. Thereby the problem is solved.

<img_3162>

Figure 1
Again, we denote the circumcircle of $B D K F$ by $\omega_{B}$. In addition, we set $\alpha=$ $\angle B A C, \varphi=\angle A B F$, and $\psi=\angle E D A=\angle A E D$ (see Figure 2). Notice that $A F=A G$ entails $\varphi=\angle G C A$, so all three of $\alpha, \varphi$, and $\psi$ respect the "symmetry" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1).

This time, we start from

$$
2 \angle K F A=2(\angle D F A-\angle D F K) .
$$

Since the triangle $A F D$ is isosceles, we have

$$
\angle D F A=\angle A D F=\angle E D F-\psi=\angle B F D+\angle E B F-\psi .
$$



Moreover, because of the circle $\omega_{B}$ we have $\angle D F K=\angle C B A$. Altogether, this yields

$$
2 \angle K F A=\angle D F A+(\angle B F D+\angle E B F-\psi)-2 \angle C B A,
$$

which simplifies to

$$
2 \angle K F A=\angle B F A+\varphi-\psi-\angle C B A .
$$

Now the quadrilateral $A F B C$ is cyclic, so this entails $2 \angle K F A=\alpha+\varphi-\psi$.

Due to the "symmetry" between $B$ and $C$ alluded to above, this argument also shows that $2 \angle A G L=\alpha+\varphi-\psi$. This concludes the proof of (1).

<img_3999>

Figure 2
Subfield: Number Theory


Question: Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term.

Solution:
Define $b_{k}=2 a_{k}$ for all $k \geqslant 0$. Then

$$
b_{k+1}=2 a_{k+1}=2 a_{k}\left\lfloor a_{k}\right\rfloor=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor
$$

Since $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \geqslant 0$.

Suppose that the sequence $a_{0}, a_{1}, a_{2}, \ldots$ does not contain any integer term. Then $b_{k}$ must be an odd integer for all $k \geqslant 0$, so that

$$
b_{k+1}=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor=\frac{b_{k}\left(b_{k}-1\right)}{2}\tag{1}
$$

Hence

$$
b_{k+1}-3=\frac{b_{k}\left(b_{k}-1\right)}{2}-3=\frac{\left(b_{k}-3\right)\left(b_{k}+2\right)}{2}\tag{2}
$$

for all $k \geqslant 0$.

Suppose that $b_{0}-3>0$. Then equation (2) yields $b_{k}-3>0$ for all $k \geqslant 0$. For each $k \geqslant 0$, define $c_{k}$ to be the highest power of 2 that divides $b_{k}-3$. Since $b_{k}-3$ is even for all $k \geqslant 0$, the number $c_{k}$ is positive for every $k \geqslant 0$.

Note that $b_{k}+2$ is an odd integer. Therefore, from equation (2), we have that $c_{k+1}=c_{k}-1$. Thus, the sequence $c_{0}, c_{1}, c_{2}, \ldots$ of positive integers is strictly decreasing, a contradiction. So, $b_{0}-3 \leqslant 0$, which implies $M=1$.

For $M=1$, we can check that the sequence is constant with $a_{k}=\frac{3}{2}$ for all $k \geqslant 0$. Therefore, the answer is $M \geqslant 2$.
We provide an alternative way to show $M=1$ once equation (1) has been reached. We claim that $b_{k} \equiv 3\left(\bmod 2^{m}\right)$ for all $k \geqslant 0$ and $m \geqslant 1$. If this is true, then we would have $b_{k}=3$ for all $k \geqslant 0$ and hence $M=1$.

To establish our claim, we proceed by induction on $m$. The base case $b_{k} \equiv 3(\bmod 2)$ is true for all $k \geqslant 0$ since $b_{k}$ is odd. Now suppose that $b_{k} \equiv 3\left(\bmod 2^{m}\right)$ for all $k \geqslant 0$. Hence $b_{k}=2^{m} d_{k}+3$ for some integer $d_{k}$. We have

$$
3 \equiv b_{k+1} \equiv\left(2^{m} d_{k}+3\right)\left(2^{m-1} d_{k}+1\right) \equiv 3 \cdot 2^{m-1} d_{k}+3 \quad\left(\bmod 2^{m}\right)
$$

so that $d_{k}$ must be even. This implies that $b_{k} \equiv 3\left(\bmod 2^{m+1}\right)$, as required.

Final answer: All integers $M \geqslant 2$