Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.7 National Olympiad Prove it Austria

Let $ABC$ be a triangle and $P$ a point inside the triangle such that the centers $M_B$ and $M_A$ of the circumcircles $k_B$ and $k_A$ of triangles $ACP$ and $BCP$ , respectively, lie outside the triangle $ABC$ . In addition, we assume that the three points $A$ , $P$ and $M_A$ are collinear as well as the three points $B$ , $P$ and $M_B$ . The line through $P$ parallel to side $AB$ intersects circles $k_A$ and $k_B$ in points $D$ and $E$ , respectively, where $D, E \neq P$ .
Show that $DE = AC + BC$ .

Solution

We put $\varphi := \angle CBP$ , cf. Figure 1. Then we get for the corresponding central angle $\angle CM_A P =$

Figure 1: Problem 4
$2\varphi$ . Since $M_A C M_B P$ is a deltoid having $M_A M_B$ as its axis of symmetry, we deduce $\angle C M_A M_B = \varphi = \angle C B M_B$ . Therefore, $B$ and analogously $A$ lie on the circumcircle of $M_A C M_B$ . In other words, the two centers $M_A$ and $M_B$ lie on the circumcircle of $ABC$ .
Thus $M_A$ and $M_B$ are the south poles corresponding to vertices $A$ and $B$ , respectively. As a result, $P$ is the incenter of triangle $ABC$ . Hence $\angle PBA = \angle CBP$ and because of $PD \parallel AB$ also $\angle CBP = \angle BPD$ holds true. This means that $PBDC$ is an isosceles trapezoid with diagonals of equal lengths and $PD = BC$ follows.

In a similar way $PE = AC$ can be shown. Finally, by addition we arrive at the claim $DE = AC+BC$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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