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Difference between revisions of "1993 USAMO Problems/Problem 2"

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Similarly <math>\angle EWZ\cong \angle EDC</math>,  <math>\angle EYX\cong \angle EBA</math>,  <math>\angle EYZ\cong \angle ECD</math>.
 
Similarly <math>\angle EWZ\cong \angle EDC</math>,  <math>\angle EYX\cong \angle EBA</math>,  <math>\angle EYZ\cong \angle ECD</math>.
  
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<br/>Futhermore, <math>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</math>.
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<br/>Futhermore, <math>m\angle XYZ+m\angle XWZ= m\angle EWX+m\angle EYX+m\angle EYZ+m\angle EWZ=</math><math>360^\circ-m\angle CED-m\angle AEB=180^\circ</math>.
  
 
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Revision as of 22:09, 22 April 2010

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 by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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