AoPS - Problem

Source: 2024 Israel National Olympiad (Gillis) P6

Tags: geometry, cyclic quadrilateral, incircles, national olympiad

Phorphyrion

07.12.2023 13:01

Quadrilateral $ABCD$ is inscribed in a circle. Let $\omega_A$, $\omega_B$, $\omega_C$, $\omega_D$ be the incircles of triangles $DAB$, $ABC$, $BCD$, $CDA$ respectively. The common external common tangent of $\omega_A$, $\omega_B$, different from line $AB$, meets the external common tangent of $\omega_A$, $\omega_D$, different from $AD$, at point $A'$. Similarly, the external common tangent of $\omega_B$, $\omega_C$ different from $BC$ meets the external common tangent of $\omega_C$, $\omega_D$ different from $CD$ at $C'$. Prove that $AA'\parallel CC'$.

We claim that the external common tangent of $\omega_A, \omega_A$ different from line $AB$ is parallel to line $CD$; this is true because $ABI_BI_A$ is cyclic centered on the midpoint of $\widehat{AB}$ of $(ABCD)$, so $\angle(AB, I_AI_B) = \frac{\widehat{AI_A} - \widehat{BI_B}}{2} = \widehat{AD} - \widehat{BC} = 2\angle(AB, CD)$. Let $AJ_AA'K_A$ be the quadrilateral circumscribing $\omega_A$ formed by $AB, AD$, and the common external tangents different from $AB$ and $AD$; then $AJ_AA'K_A$ is cyclic, and if $C'J_CCK_C$ is defined similarly then they have the same angles, so they are directly similar $\implies AA' \parallel CC'$. $\square$