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darij grinberg
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#1
Excircle of APR at AP = excircle of PQB at PQ <==> S is...
4th German TST 2005, problem 2, by Arend Bayer

Let ABC be a triangle satisfying BC < CA. Let P be an arbitrary point on the side AB (different from A and B), and let the line CP meet the circumcircle of triangle ABC at a point S (apart from the point C).

Let the circumcircle of triangle ASP meet the line CA at a point R (apart from A), and let the circumcircle of triangle BPS meet the line CB at a point Q (apart from B).

Prove that the excircle of triangle APR at the side AP is identical with the excircle of triangle PQB at the side PQ if and only if the point S is the midpoint of the arc AB on the circumcircle of triangle ABC.
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PostPosted: Thu May 12, 2005 3:05 pm
Last edited by darij grinberg on Sat Nov 05, 2005 2:43 pm; edited 2 times in total
grobber
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#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.

PostPosted: Thu May 12, 2005 4:40 pm
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