50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

We first show that point $K$ is always inside triangle $△ABC$. Consider the case, when $P=P_1$ lies on the side $AB$ and compute some angles. Let $\angle BAC=\alpha$ (Fig.13), then $\angle B{P}_{1}C=2\alpha$, $\angle ABC=\frac{\pi }{2}-\frac{\alpha }{2}$, $\angle BC{P}_1=\frac{\pi }{2}-\frac{3\alpha }{2}\Rightarrow \angle P_{1}CA=\alpha$, hence, $△AC{P}_1$ is an isosceles triangle, thus the altitude $P_1{K}_1$ is also a bisector, therefore $K=K_1$ belongs to the side $AC$. Therefore, as long as $P$ is inside $△ABC$ the bisector $PK$ of adjacent angle of $\angle BPC$ forms the least angle with line $BC$. Thus, perpendicular from $A$ to this line lies between sides $AB$ and $AC$ of triangle $ABC$. Let us extend $AK$ to the intersection with $CP$ (Fig.14) (without loss of generality, we can assume that $P$ is closer to $B$ than to $C$).

Fig. 14

In the same way, $BP$ meets $AK$ at point $X$. Then, in $△PXY$ segment $PK$ is a bisector and an altitude, thus, it is isosceles or $\angle PXY=\angle PYX$, from which it follows that $\angle BXA=\angle AYC$. If we now denote $x=\angle XBA$, $y=\angle YCA$, $x_1=\angle CAY$, $y_1=\angle BAX$, then $x_1+y_1=\alpha$, $\pi -\alpha =x+y+\angle PBC+\angle PCB=x+y+\pi -2\alpha \Rightarrow x+y=\alpha =x_1+y_1$, moreover, $x+y_1=y+x_1$. From last two identities, we get $x=x_1$, $y=y_1$. Hence, $△ABX=△CAY$ (they have two equal sides and angle between them).

We have the following equalities: $2AK=AX+AY=CY+BX=CY+BP+PX=CY+YP+BP=CP+BP$, which is exactly what we wanted to prove.