Auswahlklausur

Solution:

Because of $\measuredangle B P C+\measuredangle C P A+\measuredangle A P B=360^{\circ}=\left(180^{\circ}-\measuredangle B A C\right)+\left(180^{\circ}-\measuredangle C B A\right)+\left(180^{\circ}-\measuredangle A C B\right)$ we may assume without loss of generality that $\measuredangle A P B \leq 180^{\circ}-\measuredangle A C B$ holds. Since $P$ lies in the interior of the triangle $A B C$, we have $\angle A P B>\measuredangle A C B$, from which $\measuredangle A P B \in\left[\measuredangle A C B, 180^{\circ}-\measuredangle A C B\right]$ follows, and hence $\sin \measuredangle A P B \geq \sin \measuredangle A C B$.
Let $A_{2}, B_{2}$ and $C_{2}$ be the reflection points described in the problem text, and let $A_{3}, B_{3}$ and $C_{3}$ be the second intersection points of the lines $A P, B P$ and $C P$ with $\Omega$. We will now prove that at least one of the ratios $\frac{\left|\overline{P A_{1}}\right|}{\left|\overline{A_{1} A_{3}}\right|}$ and $\frac{\left|\overline{P B_{1}}\right|}{\left|\overline{B_{1} B_{3}}\right|}$ is not greater than $1$, from which the claim follows.

To this end we compute, using the law of sines,
$$\frac{\left|\overline{P A_{1}}\right|}{\left|\overline{A_{1} A_{3}}\right|}=\frac{\left|\overline{P A_{1}}\right|}{\left|\overline{A_{1} B}\right|} \cdot \frac{\left|\overline{A_{1} B}\right|}{\left|\overline{A_{1} A_{3}}\right|} \stackrel{(*)}{=} \frac{\sin \measuredangle C B P}{\sin \measuredangle A P B} \cdot \frac{\left|\overline{A_{1} A}\right|}{\left|\overline{A_{1} C}\right|}=\frac{\sin \measuredangle C B P}{\sin \measuredangle A P B} \cdot \frac{\sin \measuredangle A C B}{\sin \measuredangle P A C},$$
where in step (*) we used the equation $\left|\overline{A_{1} B}\right| \cdot\left|\overline{A_{1} C}\right|=\left|\overline{A_{1} A}\right| \cdot\left|\overline{A_{1} A_{3}}\right|$, which follows from the power of a point (intersecting chords theorem). Analogously one also shows
$$\frac{\left|\overline{P B_{1}}\right|}{\left|\overline{B_{1} B_{3}}\right|}=\frac{\sin \measuredangle P A C}{\sin \measuredangle A P B} \cdot \frac{\sin \measuredangle A C B}{\sin \measuredangle C B P},$$
and by multiplying both equations we finally obtain
$$\frac{\left|\overline{P A_{1}}\right|}{\left|\overline{A_{1} A_{3}}\right|} \cdot \frac{\left|\overline{P B_{1}}\right|}{\left|\overline{B_{1} B_{3}}\right|}=\frac{(\sin \measuredangle A C B)^{2}}{(\sin \measuredangle A P B)^{2}} \leq 1,$$
which implies the claim.