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Austrian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Austria

Let $I$ be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$ . This circle intersects
* the line $AI$ in points $A$ and $P$ ,
* the line $BI$ in points $B$ and $Q$ ,
* the line $AC$ in points $A$ and $R$ and
* the line $BC$ in points $B$ and $S$ ,
with none of the points $A, B, P, Q, R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$ , respectively.
Prove that the lines $PS, QR$ and $CI$ meet in a single point.
(Stephan Wagner)

Solution

We define angles $\alpha = \angle BAC$ and $\beta = \angle CBA$ as usual, cf. Figure 4. Since points $A$ ,

Figure 4: Problem 5
$B, S$ and $R$ lie on a common circle, we have $\angle BSR = 180^\circ - \alpha$ , and therefore $\angle RSC = \alpha$ . Similarly, $\angle CRS = \beta$ also holds.
If $P$ lies in the interior of $ABC$ , we have $\angle RSP = \angle RAP = \alpha/2$ . This means that $PS$ bisects the angle $\angle CSR$ .
If $Q$ is outside of $ABC$ , we have $\angle QRA = \angle QBA = \beta/2$ , and in this case $QR$ also bisects the angle $\angle SRC$ .
Independent of the positioning of $Q$ and $R$ with respect to the triangle, we therefore see that $QR, PS$ and $CI$ are the bisectors of the interior angles of $CRS$ , and they therefore meet in the incenter of this triangle, as claimed.

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