Baltic Way 2021 · Problem 11

Figure 4

For points $X, Y, Z, X \neq Y$ and $Z \neq Y$ let rot $X Y Z$ denote the rotation that takes rotates line $X Y$ to line $Z Y$ modulo half turns. We consider two rotations equivalent one of them is a composition of some translation and the other rotation. It is clear that this is indeed an equivalence relation (as the Euclidean plane is Desarguean).

Let $K^{\prime}$ and $L^{\prime}$ be the projections of $P^{\prime}$ onto $A B$ and $A C$ respectively, as in figure 4 . Let $\ell$ be the perpendicular line to line $A B$ passing through $M$. From symmetries it follows that $L^{\prime}$ is the refection of $L$ over $\ell$. In particular segments $M L$ and $M L^{\prime}$ are congruent. Similarly segments $M K$ and $M K^{\prime}$ are congruent. It follows that $M$ is a center of circle passing through $L, K, L^{\prime}$ and $K^{\prime}$.

As line $P L$ is perpendicular to line $A C$ and line $P K$ is perpendicular to line $A B$ it follows that quadrilateral $A K P L$ is cyclic. Similarly quadrilateral $A K^{\prime} P L^{\prime}$ is also cyclic.

From the theorem on inscribed angles in cyclic quadrilaterals it follows that

$$\begin{gathered} \operatorname{rot} B A P \equiv \operatorname{rot} K A P \equiv \operatorname{rot} K L P, \\ \operatorname{rot} P^{\prime} A C \equiv \operatorname{rot} P^{\prime} A L \equiv \operatorname{rot} P^{\prime} K^{\prime} L \quad \text { and } \\ \operatorname{rot} K L L^{\prime} \equiv \operatorname{rot} K K^{\prime} L^{\prime} \end{gathered}$$

As $\angle P L L^{\prime}$ and $\angle K K^{\prime} P^{\prime}$ are right it follows that $\operatorname{rot} P L L^{\prime} \equiv \operatorname{rot} K K^{\prime} P^{\prime}$ modulo half turns. Now $\operatorname{rot} K L L^{\prime} \equiv \operatorname{rot} K L P+\operatorname{rot} P L L^{\prime} \quad$ and $\quad \operatorname{rot} K K^{\prime} L^{\prime} \equiv \operatorname{rot} P^{\prime} K^{\prime} L+\operatorname{rot} K K^{\prime} P$

and $\operatorname{rot} K L L^{\prime} \equiv \operatorname{rot} K K^{\prime} L^{\prime}$ so we decuce that $\operatorname{rot} K L P \equiv \operatorname{rot} P^{\prime} K^{\prime} L$.

Putting everything together gives

$$\operatorname{rot} B A P \equiv \operatorname{rot} P^{\prime} A L^{\prime}$$

which gives the desired result.

Remark. This method can be applied to prove the existence of isogonal conjugates in triangles.