Team Selection Test for IMO 2009

Let $K,L,M,N$ be the points of tangency of the incircle of $ABCD$ to the sides $AB,BC,CD,DA$, respectively, and let $S$ be the foot of the perpendicular from $O$ to the line $PQ$. Since $ABCD$ is a tangential quadrilateral, the point of intersection of the lines $KM$ and $LN$ coincides with the point of intersection of the diagonals $AC$ and $BD$. In other words, $KM\cap LN=\{E\}$.



On the other hand, the points $O$, $K$, $P$, $M$ are concyclic; the points $O$, $L$, $Q$, $N$ are concyclic; and $S$ is the other intersection point of these two circles. In particular, the lines $OS$, $KM$, $LN$ are the radical axes of pairs of these two circles and the incircle of $ABCD$. Therefore, these three lines are concurrent at the radical center of the three circles. In other words, $E$ is the radical center. From this observation it follows that $OE\cdot d=OE\cdot OS=OE\cdot (OE+ES)=O{E}^2+OE\cdot ES=O{E}^2+LE\cdot EN=O{E}^2+r^2-O{E}^2=r^2$.