International Mathematical Olympiad

Denote by ${\omega }_a$, ${\omega }_b$, ${\omega }_c$ and ${\omega }_d$ the circles $A{I}_b{I}_d$, $B{I}_a{I}_c$, $C{I}_b{I}_d$, and $D{I}_a{I}_c$, let their centers be $O_a$, $O_b$, $O_c$ and $O_d$, and let their radii be $r_a$, $r_b$, $r_c$ and $r_d$, respectively.

Claim 1. $I_b{I}_d\perp AC$ and $I_a{I}_c\perp BD$.

Proof. Let the incircles of triangles $ABC$ and $ACD$ be tangent to the line $AC$ at $T$ and $T^{\prime }$, respectively. (See the figure to the left.) We have $AT=\frac{AB+AC-BC}{2}$ in triangle $ABC$, $A{T}^{\prime }=\frac{AD+AC-CD}{2}$ in triangle $ACD$, and $AB-BC=AD-CD$ in quadrilateral $ABCD$, so $$AT=\frac{AC+AB-BC}{2}=\frac{AC+AD-CD}{2}=A{T}^{\prime }.$$ This shows $T=T^{\prime }$. As an immediate consequence, $I_b{I}_d\perp AC$.

The second statement can be shown analogously. $\square$



Claim 2. The points $O_a$, $O_b$, $O_c$ and $O_d$ lie on the lines $AI$, $BI$, $CI$ and $DI$, respectively.

Proof. By symmetry it suffices to prove the claim for $O_a$. (See the figure to the right above.)

Notice first that the incircles of triangles $ABC$ and $ACD$ can be obtained from the incircle of the quadrilateral $ABCD$ with homothety centers $B$ and $D$, respectively, and homothety factors less than $1$, therefore the points $I_b$ and $I_d$ lie on the line segments $BI$ and $DI$, respectively.

As is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A{I}_d{I}_b$, the segment $AT$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_b{I}_d$, the circumcenter $O_a$ of triangle $A{I}_d{I}_b$ lies in the angle domain $I_{b}A{I}_d$ in such a way that $\angle I_{b}AT=\angle O_{a}A{I}_d$. The points $I_b$ and $I_d$ are the incenters of triangles $ABC$ and $ACD$, so the lines $A{I}_b$ and $A{I}_d$ bisect the angles $\angle BAC$ and $\angle CAD$, respectively. Then $$\angle O_{a}AD=\angle O_{a}A{I}_d+\angle I_{d}AD=\angle I_{b}AT+\angle I_{d}AD=\frac{1}{2}\angle BAC+\frac{1}{2}\angle CAD=\frac{1}{2}\angle BAD,$$ so $O_a$ lies on the angle bisector of $\angle BAD$, that is, on line $AI$. $\square$

The point $X$ is the external similitude center of ${\omega }_a$ and ${\omega }_c$; let $U$ be their internal similitude center. The points $O_a$ and $O_c$ lie on the perpendicular bisector of the common chord $I_b{I}_d$ of ${\omega }_a$ and ${\omega }_c$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $AC$.



From the similarity of the circles ${\omega }_a$ and ${\omega }_c$, from $O_a{I}_b=O_a{I}_d=O_{a}A=r_a$ and $O_c{I}_b=O_c{I}_d=O_{c}C=r_c$, and from $AC\parallel O_a{O}_c$ we can see that $$\frac{O_{a}X}{O_{c}X}=\frac{O_{a}U}{O_{c}U}=\frac{r_a}{r_c}=\frac{O_a{I}_b}{O_c{I}_b}=\frac{O_a{I}_d}{O_c{I}_d}=\frac{O_{a}A}{O_{c}C}=\frac{O_{a}I}{O_{c}I}.$$ So the points $X,U,I_b,I_d,I$ lie on the Apollonius circle of the points $O_a,O_c$ with ratio $r_a:r_c$. In this Apollonius circle $XU$ is a diameter, and the lines $IU$ and $IX$ are respectively the internal and external bisectors of $\angle O_{a}I{O}_c=\angle AIC$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $UX$ is the perpendicular bisector of $I_b{I}_d$, so the lines $IX$ and $IU$ are the internal and external bisectors of $\angle I_{b}I{I}_d=\angle BID$, respectively.

Repeating the same argument for the points $B,D$ instead of $A,C$, we get that the line $IY$ is the internal bisector of $\angle AIC$ and the external bisector of $\angle BID$. Therefore, the lines $IX$ and $IY$ respectively are the internal and external bisectors of $\angle BID$, so they are perpendicular.