IMO 2010 #4

by Wolstenholme, Nov 4, 2014, 3:42 PM

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Proof:

Since $ \triangle{PKM} \sim \triangle{PCA} $ and $ \triangle{PLM} \sim \triangle{PCB} $ we have that $ \frac{MK}{AC} = \frac{MP}{AP} $ and $ \frac{ML}{BC} = \frac{MP}{BP}. $ By diving these two equations we obtain $ \frac{MK}{ML} = \frac{BP}{AP} \cdot \frac{AC}{BC} $ so it suffices to show that this product equals $ 1. $ Now since $ SP^2 = SC^2 = SA \cdot SB $ we have that $ PS $ is tangent to the circumcircle of $ \triangle{PAB}. $ Therefore we have that $ \triangle{SPB} \sim \triangle{SAP} $ so $ \frac{BP}{AP} = \frac{PS}{AS}. $ Similarly since $ CS $ is tangent to $ \Gamma $ we have that $ \triangle{SCB} \sim \triangle{SAC} $ so $ \frac{AC}{BC} = \frac{AS}{CS}. $

Combining this information we obtain $ \frac{MK}{ML} = \frac{BP}{AP} \cdot \frac{AC}{BC} = \frac{PS}{AS} \cdot \frac{AS}{CS} = 1 $ as desired.

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