1997 OIM Problems/Problem 2

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Problem

With the center at the incenter $I$ of a triangle $ABC$, a circle is drawn that cuts each of the three sides of the triangle at two points: the segment $BC$ at $D$ and $P$ ($D$ being the closest to $B$); the the segment $CA$ in $E$ and $Q$ ($E$ being the closest to $C$), and the segment $AB$ in $F$ and $R$ ($F$ being the closest to $A$). Let $S$ be the point of intersection of the diagonals of the quadrilateral $EQFR$. Let $T$ be the point of intersection of the diagonals of the $FRDP$ quadrilateral. Let $U$ be the point of intersection of the diagonals of the quadrilateral $DPEQ$. Show that the circles circumscribed by the triangles $FRT$, $DPU$ and $EQS$ have a single common point.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe12.htm