Let triangle ABC has semiperimeter $ p$. E,F are located on AB such that $ CE=CF=p$. Prove that the C-excircle of triangle ABC touches the circumcircle (EFC).
Problem
Source: Bulgarian Olympiad, 1995
Tags: geometry, circumcircle, geometry proposed
The QuattoMaster 6000
21.08.2008 19:41
mr.danh wrote: Let triangle ABC has semiperimeter $ p$. E,F are located on AB such that $ CE = CF = p$. Prove that the C-excircle of triangle ABC touches the circumcircle (EFC).
Let the $ C$-excircle meet $ AC$ and $ CB$ at $ P$ and $ Q$ respectively. It is well-known that $ CP = p = CQ$. Invert around $ C$ with radius $ p$, so since $ CP = p$, we see that the excircle maps to itself. Furthermore, since $ CE = CF = p$, we see that $ E$ and $ F$ map to themselves and the circumcircle of $ \triangle CEF$ maps to $ EF$. Then, since $ BA$ is tangent to the $ C$-excircle, we have that $ EF$ is tangent to the $ C$-excircle, and so we conclude that the circumcircle of $ \triangle EFC$ and the $ C$-excircle are tangent (they are in fact tangent at where the line $ CR$ meets the $ C$-excircle when $ R$ is the tangency point between the $ C$-excircle and $ BA$.)
sunken rock
21.03.2009 13:21
Apply Casey to C, E, F and C-excircle. Best regards, sunken rock
Luis González
21.03.2009 17:19
If excircle $(I_c)$ is tangent to $CB,CA$ through $ X,Y,$ then it is known that $ CX=CY=p.$ Inversion with center $ C$ and power $ p^2$ swaps circle $\odot (CEF)$ and the sideline $BC$ and takes C-excircle $(I_c)$ into itself. Since $ BC$ and $(I_c)$ are tangent, then by conformity $\odot(CEF)$ and $(I_c)$ are tangent as well.