62nd International Mathematical Olympiad / IMO 2021 / | Problem 4 - Solution - 21 July 2021

Problem 4. Let Γ be a circle with centre I, and ABCD a convex quadrilateral such that each of the segments AB, BC, CD and DA is tangent to Γ. Let Ω be the circumcircle of the triangle AIC. The extension of BA beyond A meets Ω at X, and the extension of BC beyond C meets Ω at Z. The extensions of AD and CD beyond D meet Ω at Y and T , respectively. Prove that

AD + DT + T X + XA = CD + DY + Y Z + ZC.

proposed by Dominik Burek and Tomasz Ciesla, Poland

Solution

Claim 1: $ZY=XT$

Proof: Consider $O$ the center of $\Omega$ and $IO\cap \Omega =\{ I'\}.$

$\angle BAI=\angle DAI\Rightarrow AI$ is the exterrnal bisector of $\angle XAY$ .

On the other hand, $\angle IAI'=90^o\Rightarrow AI'$ is the internal bisector of $\angle XAY\Rightarrow I'$ is the midpoint of the arc $XY$ not containing $I\Rightarrow IX=IY.$

Similarly, $IZ=IT$ . Now combining these 2 equalities, we get $ZY=XT$ . $\blacksquare$

We are left to prove that $AD+DT+XA= CD+DY+ZC$ .

Claim 2: $\frac{DY}{DC}=\frac{BZ}{BA}$

Proof: $\frac{BZ}{BA}=\frac{XZ}{AC}=\frac{TY}{AC}=\frac{DY}{DC}$ because $BA\cdot BX=BC\cdot BZ$ and $TYCA$ is cyclic with $\{ D\}=TC\cap AY$ . $\blacksquare$

Now all he have to do is to use Pitot theorem for the quadrilateral $ABCD$ ( $BC+AD=DC+AB$ ).

$\frac{DY}{DC}=\frac{BZ}{BA}\Leftrightarrow DY\cdot \frac{AB-BC}{DC}=BZ\cdot \frac{AB-BC}{BA}\Leftrightarrow$ $\Leftrightarrow DY\cdot \frac{DA-DC}{DC}=BZ\cdot \frac{AB-BC}{BA}\Leftrightarrow DY\cdot \frac{DA}{DC}+ZB\cdot \frac{BC}{BA}=DY+BZ\Leftrightarrow$ $\Leftrightarrow DY\cdot \frac{DT}{DY}+ZB\cdot \frac{XB}{ZB}=DY+BZ\Leftrightarrow DT+XB=DY+BZ\Leftrightarrow$ $\Leftrightarrow DT+XB+AD-AB=DY+BZ+CD-BC\Leftrightarrow AD+DT+XA=CD+DY+ZC$ and the problem is finished.