Given an $ABC$ acute triangle with $O$ the center of the circumscribed circle. Suppose that $\omega$ is a circle that is tangent to the line $AO$ at point $A$ and also tangent to the line $BC$. Prove that $\omega$ is also tangent to the circumcircle of the triangle $BOC$.
Problem
Source: INAMO Shortlist 2014 G6
Tags: geometry, circumcircle, tangent circles, tangent
mira74
14.07.2019 04:59
Inverting about $(ABC)$ fixes the circle and $AO$ but swaps $(BOC)$ and $BC$, so $(BOC)$ is also tangent to the circle, so we are done.
AlastorMoody
15.07.2019 21:27
@above, Is it obvious that $\omega$ will remain fixed?! I'll elaborate the above proof! INAMO SL 2014 G6 wrote: Given an $ABC$ acute triangle with $O$ the center of the circumscribed circle. Suppose that $\omega$ is a circle that is tangent to the line $AO$ at point $A$ and also tangent to the line $BC$. Prove that $\omega$ is also tangent to the circumcircle of the triangle $BOC$. Solution: There's a unique circle which is tangent to $OA$ at $A$ and to $BC$. We'll show, $\omega$ is fixed under inversion around $\odot (O)$ with radius $AO$ Let $AO \cap BC=D$. Let the circle with radius $AD$ at $D$ cut $BC$ at $E$. Let $OE \cap \odot (BOC)=F$. $\angle EBO$ $=$ $180^{\circ}-\angle OBC$ $=$ $180^{\circ}-\angle OCB$ $=$ $\angle BFO$ $\implies$ $\Delta OFB$ $\sim$ $\Delta OBE$ $\implies$ $\Delta OAF$ $\sim$ $\Delta OEA$ $\implies$ $\angle EAF$ $=$ $\angle EAD$ $-$ $\angle FAO$ $=$ $\angle DEA$ $-$ $\angle OEA$ $=$ $\angle BEF$ $\implies$ $\odot (AEF)$ is tangent to $BC$ $\implies$ $\omega \equiv \odot (AEF)$. Under Inversion $\Psi$ about $\odot (O)$ with radius $AO$ $\implies$ $\Psi (E)$ $\equiv$ $F$ $\implies$ $\Psi(\odot (AEF))$ $\equiv$ $\odot (AEF)$ and, $\Psi( \odot (BOC) )$ $\equiv$ $BC$ $\implies$ $\omega$ is tangent to $\odot (BOC)$ $\qquad \blacksquare$