AoPS - Problem

Source: V.A. Yasinsky Geometry Olympiad 2023 VIII p3 , Ukraine

Tags: tangent, geometry

parmenides51

12.12.2023 14:40

$ABC$ is a right triangle with $\angle C = 90^o$. Let $N$ be the middle of arc $BAC$ of the circumcircle and $K$ be the intersection point of $CN$ and $AB$. Assume $T$ is a point on a line $AK$ such that $TK=KA$. Prove that the circle with center $T$ and radius $TK$ is tangent to $BC$. (Mykhailo Sydorenko)

Let $H$ be the foot of the perpendicular from $T$ to $BC$. Let the perpendicular to $AN$ from $K$ intersect $BC,CA$ at $H',P$ respectively. $\angle CAB=2\alpha$ $KP\parallel BN$ so $\angle PKA=\angle NBA=\angle NCA=90-\angle BCN=\alpha$ and $\angle AKC=2\alpha$ so $\angle APK=\alpha=\angle KCP\implies KP=KC$ So $KP=KC=KH'$ $KP=KH'$ and $KA=KT$ hence $APTH'$ is a parallelogram.$\implies TH'\parallel AP\parallel TH$ which gives us that $H'=H$. $\angle HKT=\angle PKA=\alpha=\angle APK=\angle THK$ so $TH=TK$ as desired.