South-Afrika, 2011

Geometry Difficulty 4.7 AIME Prove it South Africa

Let circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$ . The common tangent of the two circles closest to the point $D$ touches $\Gamma_1$ at $A$ and $\Gamma_2$ at $B$ . The line $AD$ intersects $\Gamma_2$ for the second time in $C$ . Let $M$ be the middle of the line segment $BC$ . Prove that $\angle DPM = \angle BDC$ .

Solution

Let $S$ be the intersection of $PD$ and $AB$ . Then $S$ lies on the radical axis of the two circles, so $AS = SB$ , that is, $PS$ is a median in triangle $PAB$ .
Now, by the tan-chord theorem,
$\angle BAP = 180^\circ - \angle ADP = \angle CDP = \angle CBP.$
Also, $\angle ABP = \angle BCP$ , which implies that $\triangle PAB \||\| \triangle PBC$ . Since $PM$ is a median in triangle $BPC$ , it follows that $\angle DPB = \angle MPC$ . Hence
$\angle DPM = \angle DPB + \angle BPM = \angle MPC + \angle BPM = \angle BPC = \angle BDC.$

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.