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, 2020

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Baltic Way

Consider the Euclidean plane, the points $A = (0,0)$ and $B = (1,0)$ and the open half-strip
$S = \{(x, y) : 0 < x < 1, y > 0\}$
with width 1 and vertices $A$ and $B$ .
Find all functions $f: S \to S$ satisfying the following conditions for all $P, Q \in S$ :
(i) $f(f(P)) = P$
(ii) If $P, Q, A$ are collinear, then $f(P), f(Q)$ and $B$ are collinear.
(iii) If $f(P) = P$ and $f(Q) = Q$ , then there is a circle containing $A, B, P$ and $Q$ .

Solution

Fix any $0 < \alpha < 90^\circ$ and consider the ray of points $P \in S$ with $\angle BAP = \alpha$ . This ray (or the part of it which is in $S$ ) is mapped by (ii) under $f$ to a ray starting from $B$ with a certain angle $\beta = \beta(\alpha)$ , i.e., $\angle f(P)BA = \beta(\alpha)$ for all such $P$ . By (i), this second ray is mapped to the first ray.
Now there is a unique point $P \in M$ such that $\angle BAP = \alpha$ and $\angle PBA = \beta$ . Since the rays are mapped to each other, $f(P)$ must also lie on both rays and hence $f(P) = P$ is a fixed point. But then (iii) implies that $\angle APB + \angle PBA = \alpha + \beta(\alpha)$ is constant.
Considering $\alpha \to 90^\circ$ and $\alpha \to 0^\circ$ we see that the only possible value of this constant is $90^\circ$ and hence $\beta(\alpha) = 90^\circ - \alpha$ .
So we always have $\angle f(P)BA = 90^\circ - \angle PBA$ and by (i) also $\angle PBA = 90^\circ - \angle f(P)BA$ and so $f(P)$ is the orthocenter $H(ABP)$ of $ABP$ . This function $P \mapsto H(ABP)$ is indeed a solution.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.