AoPS - Problem

Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade

Tags: geometry

GeorgeRP

10.09.2024 19:20

Let $P$ be an arbitrary point on the incircle $k$ of triangle $ABC$ with center $I$, different from the points of tangency with its sides. The tangent to $k$ at $P$ intersects the lines $BC$, $AC$, $AB$ at points $A_0$, $B_0$, $C_0$, respectively. The lines through $A_0$, $B_0$, $C_0$, parallel to the bisectors of the angles $\angle BAC$, $\angle ABC$, $\angle ACB$, form a triangle $\Delta$. Prove that the line $PI$ is tangent to the circumcircle of $\Delta$.

Let $I'$ be the reflection of $I$ across $P$ and $P'$ be the reflection of $P$ across $I$. Define $O$ as the circumcenter of $XYZ$. $\angle B_0I'C_0 = \angle B_0IC_0 = 180 - \angle BIC = \angle B_0XC_0 \implies I' \in (B_0XC_0)$, and symmetric ones imply that $I' \in (XYZ)$ by Miquel Point. Claim: $XYZ \cup I' \sim DEF \cup P'$ Proof. $\angle YZI' = \angle YXI' = \angle I'B_0C_0 = \angle PB_0I = \angle EFP'$ and the symmetric one gives the result. Similarity angle of these two triangles is $90$ implying that $OI' \perp IP'$ as desired.