Library / Austria competition problems /30 of 39

Austria competition problems, 2014

Geometry Difficulty 6.6 National Olympiad Prove it Austria

We are given a right-angled triangle $MNP$ with right angle in $P$ . Let $k_M$ be the circle with center $M$ and radius $MP$ , and let $k_N$ be the circle with center $N$ and radius $NP$ .
Let $A$ and $B$ be the common points of $k_M$ and the line $MN$ , and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ , with $C$ between $A$ and $B$ .
Prove that the line $PC$ bisects the angle $APB$ .

Solution

Let $\alpha = \angle NMP$ and $\beta = \angle MNP$ . Because the sum of angles in $MNP$ is $180^\circ$ , we obtain $\alpha + \beta = 90^\circ$ . Since $BP$ is a chord of the circle through $P$ and with mid-point $M$ , we have
$\angle BAP = \frac{1}{2} \angle BMP = \frac{1}{2} \alpha.$
Similarly, for the chord $CP$ in the circle $k_N$ , we obtain
$\angle CDP = \frac{1}{2} \angle CNP = \frac{1}{2} \beta.$

Since $NP$ and $MP$ are perpendicular, $NP$ is a tangent of the circle $k_M$ and $MP$ is a tangent of the circle $k_N$ . Considering the chord $BP$ in the circle $k_M$ , we therefore have
$\angle BPN = \angle BAP = \frac{1}{2}\alpha$
and with the chord $CP$ in $k_N$ we have
$\angle MPC = \angle CDP = \frac{1}{2}\beta$
It therefore follows that
$\angle CPB = \angle MPN - \angle BPN - \angle MPC = 90^\circ - \frac{1}{2}\alpha - \frac{1}{2}\beta = 45^\circ,$
and since $\angle APB = 90^\circ$ we also have
$\angle APC = \angle APB - \angle CPB = 45^\circ$
It therefore follows that $PC$ bisects the angle $APB$ as claimed. $\square$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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