Consider a triangle $ABC$. Let $S$ be a circumference in the interior of the triangle that is tangent to the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$ respectively. In the exterior of the triangle we draw three circumferences $S_A$, $S_B$, $S_C$. The circumference $S_A$ is tangent to $BC$ at $L$ and to the prolongation of the lines $AB$, $AC$ at the points $M$, $N$ respectively. The circumference $S_B$ is tangent to $AC$ at $E$ and to the prolongation of the line $BC$ at $P$. The circumference $S_C$ is tangent to $AB$ at $F$ and to the prolongation of the line $BC$ at $Q$. Show that the lines $EP$, $FQ$ and $AL$ meet at a point of the circumference $S$.
Problem
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Tags: geometry, geometric transformation, homothety
NT2048
13.08.2013 05:11
Let $AL$ intersect $S$ at $X$. It is a well-known fact, proven in one of Yufei Zhao's sheets that XD is a diameter of $S$. The idea behind the proof of this is to consider the homothety centred at $A$ mapping $S$ to $S_A$. This homothety clearly maps $X$to $L$, which is a tangency point of the excircle. Anyway, the problem can be rephrased as follows: Let $ABC$ be a triangle with its incircle tangent at $D, E, F$. Let $DX$ be a diameter of the incircle. Let $Q$ and $P$ be the intersections of $XF$ and $XE$ respectively with $BC$. Prove that $BQ=BF$ and that $CP=CE$. This is trivial by angle chasing in the right triangles $XDQ$ and $XFD$ followed by an application of the alternate segment theorem.
proglote
13.08.2013 18:37
No angle chasing needed, just note that if $X$ is the intersection of $AL$ and $S$ (closer to $A$), then the tangents to $S$ at $X$ and to $S_B$ at $P$ are parallel, hence $PX$ goes through the insimilicenter $E$ of the two circles, and similarly for $S_C$..
Thelink_20
18.06.2024 19:15
Let $DX$ be a diameter and $r$ be the paralell to $BC$ trough $X$. Let $r$ intersect $AB$ and $AC$ at $Y$ and $Z$. $r$ is tangent to $S$ because $\angle YXI=\angle IDB=90^{\circ}$. Consider the homothety $\Phi$ centered at $A$ such that $\Phi(\Delta AYZ)=\Delta ABC$. It's easy to see that $\Phi(S)=S_{A}\Rightarrow\Phi(X)=L\Rightarrow X\in AL$. Because of the common tangents, $BQ=BF=BD\Rightarrow\angle QFD=90^{\circ}$. Also $XD$ is a diameter$\Rightarrow\angle DFX=90^{\circ}\Rightarrow X\in FQ$ Analogously $X\in EP$, thus finishing the problem.