48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Let $\Omega$ be the circle tangent to segment $AB$ and to rays $AD$ and $BC$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $IJ$.



Denote the incircles of triangles $ADP$ and $BCP$ by ${\omega }_A$ and ${\omega }_B$. Let $h_1$ be the homothety with a negative scale taking $\omega$ to $\Omega$. Consider this homothety as the composition of two homotheties: one taking $\omega$ to ${\omega }_A$ (with a negative scale and center $K$), and another one taking ${\omega }_A$ to $\Omega$ (with a positive scale and center $A$). It is known that in such a case the three centers of homothety are collinear (this theorem is also referred to as the theorem on the three similitude centers). Hence, the center of $h_1$ lies on line $AK$. Analogously, it also lies on $BL$, so this center is $F$. Hence, $F$ lies on the line of centers of $\omega$ and $\Omega$, i.e. on $IJ$ (if $I=J$, then $F=I$ as well, and the claim is obvious).

Consider quadrilateral $APCD$ and mark the equal segments of tangents to $\omega$ and ${\omega }_A$ (see the figure below to the left). Since circles $\omega$ and ${\omega }_A$ have a common point of tangency with $PD$, one can easily see that $AD+PC=AP+CD$. So, quadrilateral $APCD$ is circumscribed; analogously, circumscribed is also quadrilateral $BCDP$. Let ${\Omega }_A$ and ${\Omega }_B$ respectively be their incircles.



Consider the homothety $h_2$ with a positive scale taking $\omega$ to $\Omega$. Consider $h_2$ as the composition of two homotheties: taking $\omega$ to ${\Omega }_A$ (with a positive scale and center $C$), and taking ${\Omega }_A$ to $\Omega$ (with a positive scale and center $A$), respectively. So the center of $h_2$ lies on line $AC$. By analogous reasons, it lies also on $BD$, hence this center is $E$. Thus, $E$ also lies on the line of centers $IJ$, and the claim is proved.