The incircle of triangle 
touches 
,
and 
at points
and 
respectively. Let 
be a point such that 
.Prove that the perpendicular from incenter
to the 
, 
and 
are concurrent.
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Nice problem! Here’s my solution:
Let X be the intersections of line A'B' and the perpendicular from I on CM. Since A'B' is the polar of C, CM must be the polar of X. Let Y=\overline{A'B'}\cap\overline{CM}. Then (A', B'; X, Y) = -1. Suppose P_\infty is the point at infinity along line AB. Then we have -1 = (B, A; P_\infty, M) \overset{C}{=} (A', B'; \overline{CK}\cap\overline{A'B'}, Y). Therefore X = \overline{CK}\cap\overline{A'B'}, and we are done. \blacksquare
Btw, it seems I can’t write maths in display mode. Why is that?
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