Canadian Mathematical Olympiad 2007 Question 5

Question

Let the incircle of triangle $ABC$ touch sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $Γ$,$Γ_1$,$Γ_2$ and $Γ_3$ denote the circumcircles of triangle $ABC$, $AEF$, $BDF$ and $CDE$ respectively. Let $Γ$ and $Γ_1$ intersect at $A$ and $P$, $Γ$ and $Γ_2$ intersect at $B$ and $Q$, and $Γ$ and $Γ_3$ intersect at $C$ and $R$.

Prove that $PEDQ$ is cyclic.

Answer 1

Very interesting problem. I will perform a slight change of notation for simplicity. Let $ABC$ be the starting triangle, $I$ its incenter, $I_A, I_B, I_C$ the projections of the incenter on the sides of $ABC$. The key idea is to perform a circular inversion with respect to the incircle of $ABC$. It is straightforward to check that the circumcircles of $II_A I_B$ and so on go into $I_A I_B$ and so on. Additionally, the sides of $ABC$ go into the circles $\Theta_A,\Theta_B,\Theta_C$ with diameters $II_A,II_B,II_C$. These circles meet at $I$ and at the midpoints of the sides of $I_A I_B I_C$. By Johnson's theorem the inverse $\Xi$ of the circumcircle of $ABC$ has the same diameter of $\Theta_A$. In particular $\Xi$ is the nine point circle of $I_A I_B I_C$ and the other intersections of $\Xi$ with the side lengths of $I_A I_B I_C$ are given by the feet of the altitudes of $I_A I_B I_C$.

The problem now boils down to a trivial one: i.e. to show that in a triangle $ABC$ the vertices $B,C$ and the feet of the altitudes $H_B,H_C$ are concyclic.

Answer 2

Here is a solution using purely angle chasing (though it does require quite a bit of ingenuity and playing around with angles).

Let $I$ be the incenter. It is easy to see $I\in\Gamma_1,\Gamma_2,\Gamma_3$. Then:

\begin{align*} \angle QPE = \angle QPA-\angle APE \\=\angle QPA-\angle AIE\\=180^{\circ}-\angle QBA-90^{\circ}+\frac{1}{2}\angle A\\=180^{\circ}-\angle QBA-\frac{1}{2}\angle B-\frac{1}{2}\angle C\\=180^{\circ}-\angle QBA-\angle ABI-\angle IDE\\=180^{\circ}-\angle QBI-\angle IDE\\=180^{\circ}-\angle QDI-\angle IDE\\=180^{\circ}-\angle QDE \end{align*}

(Here I use the notation $\angle A$ to denote $\angle BAC$, etc.)