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 $D, E \ne P$. Show that $DE = AC + BC$. (Proposed by Walther Janous)
Problem
Source: 49th Austrian Mathematical Olympiad National Competition (Final Round, part 2) 1st June 2018 p4
Tags: geometry, circumcircle, collinear
BobaFett101
26.05.2019 09:30
Cool problem.
Lemma: $P$ is the incenter of $\triangle ABC$.
Proof: $\angle PCA = \frac{1}{2} \angle PM_BA = 180 - 2\angle M_BPA$ and $\angle PCB = 180 - 2\angle M_APB$. Since $BPM_B$ and $APM_A$ are collinear, $\angle APM_B = \angle BPM_A$, so $P$ lies on the bisector of $\angle C$. Additionally this tells us that $\angle M_BPA = 90 - \frac{1}{2} \angle PM_BA = 90 - \frac{1}{2}\angle C$, so $\angle BPA = 90 + \frac{1}{2} \angle C$ and thus $P$ lies on $(AIB)$ where $I$ is the incenter of $\triangle ABC$. We now conclude that $P$ is the incenter.
Consider $\triangle CED$. We have $\angle CED = \angle CEP = \angle CAP = \frac{1}{2} \angle A$ and similarly $\angle CDE = \frac{1}{2} \angle B$. We see that $\triangle CED \sim \triangle PAB$. Furthermore, we claim that the center of homothety of these two triangles is $M$, the midpoint of minor arc $AB$ in $(ABC)$. To prove this, it suffices to show $EAM$ and $DBM$ collinear. Angle chasing, we see
$$\angle EAB + \angle BAM = \angle EAC + \angle A + \frac{1}{2} \angle C = \angle EPC + \angle A + \frac{1}{2} \angle C = \angle B + \frac{1}{2} \angle C + \angle A + \frac{1}{2} \angle C = 180,$$
as desired. Similarly, $DBM$ collinear. Finally, we see that $\frac{AB}{ED} = \frac{MP}{MC}$, so in order to show $ED = CA + CB$, we want to show
$$(AB)(MC) = (MP)(CA + CB) = (MB)(CA) + (MA)(CB),$$
which is true by Ptolemy's Theorem.