1993 USAMO Problems/Problem 2

Problem 2

Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic.

Solution

Diagram

 $[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('E',E,N); pair A,B,C,D; A=(10,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('A',A,E); label('B',B,N); label('C',C,W); label('D',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('X',X,NE); label('Y',Y,NW); label('Z',Z, SW); label('W',W,SE); [/asy]$

Work

Let $X$, $Y$, $Z$, $W$ be the foot of the altitute from point $E$ of $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$.

Note that reflection of $E$ over the 4 lines is $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic.

$\angle EWA$ is right angle and so is $\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle.

Follow that, $\angle EWX\cong\angle EAX\cong \angle EAB$ because they inscribe the same angle.

Similarly $\angle EWZ\cong \angle EDC$, $\angle EYX\cong \angle EBA$, $\angle EYZ\cong \angle ECD$.

Futhermore, $m\angle XYZ+m\angle XWZ= m\angle EWX+m\angle EYX+m\angle EYZ+m\angle EWZ=360^\circ-m\angle CED-m\angle AEB=180^\circ$.

Thus, $\angle XYZ$ and $\angle XWZ$ are supplementary and follows that, $XYZW$ is cyclic.

$\mathbb{Q.E.D}$

Resources

 1993 USAMO (Problems • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions