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

(Solution i used for Otis application, found it was pretty trivial)
 
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== Solution 2 ==
 
== Solution 2 ==
  
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Suppose the reflection of E over AB is W, and similarly define X, Y, and Z. \newline
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Suppose the reflection of E over AB is W, and similarly define X, Y, and Z.  
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<math>\bigtriangleup BEA \cong \bigtriangleup BWA</math> by reflection gives <math>BE = BW</math> \newline
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<math>\bigtriangleup BEA \cong \bigtriangleup BWA</math> by reflection gives <math>BE = BW</math>  
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<math>\bigtriangleup BEC \cong \bigtriangleup BXC</math> by reflection gives <math>BE = BX</math> \newline
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<math>\bigtriangleup BEC \cong \bigtriangleup BXC</math> by reflection gives <math>BE = BX</math>
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These two tell us that E, W, and X belong to a circle with center B. \newline
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These two tell us that E, W, and X belong to a circle with center B.  
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Similarly, we can get that: \newline
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Similarly, we can get that:  
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E, Z, and W belong to a circle with center A, \newline
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E, Z, and W belong to a circle with center A,  
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E, X, and Y belong to a circle with center C, \newline
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E, X, and Y belong to a circle with center C,  
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E, Y, and Z belong to a circle with center D. \newline
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E, Y, and Z belong to a circle with center D.  
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\newline
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−
To prove that W, X, Y, Z are concyclic, we want to prove <math>\angle XWZ + \angle XYZ = 180^o</math> \newline
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To prove that W, X, Y, Z are concyclic, we want to prove <math>\angle XWZ + \angle XYZ = 180^o</math>  
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<math>\angle XWZ + \angle XYZ = \angle XWE + \angle EWZ + \angle XYE + \angle EYZ</math> \newline
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<math>\angle XWZ + \angle XYZ = \angle XWE + \angle EWZ + \angle XYE + \angle EYZ</math>  
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<math> = \frac{1}{2} \angle XBE + \frac{1}{2} \angle EAZ + \frac{1}{2} \angle XCE + \frac{1}{2} \angle EDZ</math> \newline
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<math> = \frac{1}{2} \angle XBE + \frac{1}{2} \angle EAZ + \frac{1}{2} \angle XCE + \frac{1}{2} \angle EDZ</math>  
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<math> = \frac{1}{2} (\angle XBE + \angle XCE) + \frac{1}{2} (\angle EAZ + \angle EDZ)</math> \newline
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<math> = \frac{1}{2} (\angle XBE + \angle XCE) + \frac{1}{2} (\angle EAZ + \angle EDZ)</math>  
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\newline
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<math>\angle AED = 90^o</math> and <math>\angle AED = \angle AZD</math> tells us that <math>\angle EAZ + \angle EDZ = 180^o</math> \newline
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<math>\angle AED = 90^o</math> and <math>\angle AED = \angle AZD</math> tells us that <math>\angle EAZ + \angle EDZ = 180^o</math>  
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Similarly, <math>\angle XBE + \angle XCE = 180^o</math> \newline
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Similarly, <math>\angle XBE + \angle XCE = 180^o</math>  
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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. \newline
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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)
 
-- 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.
  
  

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.


See Also

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

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