Excircle of APR at AP = excircle of PQB at PQ <==> S is...

darij grinberg

Excircle of APR at AP = excircle of PQB at PQ <==> S is...

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grobber

#2

A very simple angle chase will show that $ABC,\ QRC$ are similar, and $P,Q,R$ are, in fact, collinear.

The condition expressed in excircles is equivalent, in this configuration, to the $c$ -excircle of $ABC$ being tangent to $QR$ . From the observation above, this can only happen in one situation, since $QR$ has a fixed direction and cuts $AC$ between $A$ and $C$ . Since this obviously happens when $S$ is the midpoint of the arc $AB$ , because that's when $A$ and $Q$ , $B$ and $R$ are symmetric wrt the angle bisector $CS$ of $\angle BCA$ , it means that this is the only time it happens.

I'm sure it lacks the rigor necessary for writing up a solution for a TST, but this is the main idea Smile

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**Posted:** Thu May 12, 2005 4:40 pm