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

Latest revision as of 04:42, 1 October 2024

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 1

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 altitude from point $E$ of $\triangle AEB$ , $\triangle BEC$ , $\triangle CED$ , $\triangle DEA$ .

Note that reflection of $E$ over all the points of $XYZW$ is similar to $XYZW$ 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.

$\mathbb{Q.E.D}$

Solution 2

Suppose the reflection of E over AB is W, and similarly define X, Y, and Z. $\bigtriangleup BEA \cong \bigtriangleup BWA$ by reflection gives $BE = BW$ $\bigtriangleup BEC \cong \bigtriangleup BXC$ by reflection gives $BE = BX$ These two tell us that E, W, and X belong to a circle with center B. Similarly, we can get that: E, Z, and W belong to a circle with center A, E, X, and Y belong to a circle with center C, E, Y, and Z belong to a circle with center D.

To prove that W, X, Y, Z are concyclic, we want to prove $\angle XWZ + \angle XYZ = 180^o$ $\angle XWZ + \angle XYZ = \angle XWE + \angle EWZ + \angle XYE + \angle EYZ$ $= \frac{1}{2} \angle XBE + \frac{1}{2} \angle EAZ + \frac{1}{2} \angle XCE + \frac{1}{2} \angle EDZ$ $= \frac{1}{2} (\angle XBE + \angle XCE) + \frac{1}{2} (\angle EAZ + \angle EDZ)$

$\angle AED = 90^o$ and $\angle AED = \angle AZD$ tells us that $\angle EAZ + \angle EDZ = 180^o$ Similarly, $\angle XBE + \angle XCE = 180^o$ Thus, $\angle XWZ + \angle XYZ = \frac{1}{2} \cdot 180^o + \frac{1}{2} \cdot 180^o = 180^o$ , and we are done. -- Lucas.xue (someone pls help with a diagram)

Solution 3

E lies on the isoptic cubic of ABCD, so it has an isogonal conjugate in ABCD.

@@ Line 81: / Line 81: @@
 Thus, <math>\angle XWZ + \angle XYZ = \frac{1}{2} \cdot 180^o + \frac{1}{2} \cdot 180^o = 180^o</math>, and we are done.
 -- Lucas.xue (someone pls help with a diagram)
+==Solution 3==
+E lies on the isoptic cubic of ABCD, so it has an isogonal conjugate in ABCD.

1993 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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