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Geometry Difficulty 6.4 National olympiad Prove it Turkey

Let $ABC$ be an acute angled triangle and $K, L$ be points on $AC, BC$ respectively such that $\angle AKB = \angle ALB$ . Let $P$ be the intersection of $AL, BK$ and $Q$ be the midpoint of segment $KL$ . Let $T, S$ be the intersection $AL, BK$ with the circumcircle of $ABC$ , respectively. Prove that $TK, SL, PQ$ are concurrent.

Solution

3. Answer: All constant functions.
Let $a < b$ be two arbitrary numbers. Consider a sufficiently large $x$ so that $x > -a$ and $x > b - f(-a)$ , thus $-x < a$ and $x + f(x) > b - f(-a) + f(-a) = b$ (here we used $f(x) \ge f(-a)$ since $x > -a$ ). Now $-x < a < b < x + f(x)$ while $f(-x) = f(x+f(x))$ , hence $f(a) = f(b)$ , so $f$ is constant.

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