Difference between revisions of "1993 USAMO Problems/Problem 2"
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Let <math>ABCD</math> be a convex quadrilateral such that diagonals <math>AC</math> and <math>BD</math> intersect at right angles, and let <math>E</math> be their intersection. Prove that the reflections of <math>E</math> across <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DA</math> are concyclic. | Let <math>ABCD</math> be a convex quadrilateral such that diagonals <math>AC</math> and <math>BD</math> intersect at right angles, and let <math>E</math> be their intersection. Prove that the reflections of <math>E</math> across <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DA</math> are concyclic. | ||
− | == Solution == | + | == Solution 1 == |
===Diagram=== | ===Diagram=== | ||
<center><table border=1><tr><td><asy> | <center><table border=1><tr><td><asy> | ||
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<P align="right"><math>\mathbb{Q.E.D}</math></P> | <P align="right"><math>\mathbb{Q.E.D}</math></P> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Suppose the reflection of E over AB is W, and similarly define X, Y, and Z. \newline | ||
+ | <math>\bigtriangleup BEA \cong \bigtriangleup BWA</math> by reflection gives <math>BE = BW</math> \newline | ||
+ | <math>\bigtriangleup BEC \cong \bigtriangleup BXC</math> by reflection gives <math>BE = BX</math> \newline | ||
+ | These two tell us that E, W, and X belong to a circle with center B. \newline | ||
+ | Similarly, we can get that: \newline | ||
+ | E, Z, and W belong to a circle with center A, \newline | ||
+ | E, X, and Y belong to a circle with center C, \newline | ||
+ | E, Y, and Z belong to a circle with center D. \newline | ||
+ | \newline | ||
+ | To prove that W, X, Y, Z are concyclic, we want to prove <math>\angle XWZ + \angle XYZ = 180^o</math> \newline | ||
+ | <math>\angle XWZ + \angle XYZ = \angle XWE + \angle EWZ + \angle XYE + \angle EYZ</math> \newline | ||
+ | <math> = \frac{1}{2} \angle XBE + \frac{1}{2} \angle EAZ + \frac{1}{2} \angle XCE + \frac{1}{2} \angle EDZ</math> \newline | ||
+ | <math> = \frac{1}{2} (\angle XBE + \angle XCE) + \frac{1}{2} (\angle EAZ + \angle EDZ)</math> \newline | ||
+ | \newline | ||
+ | <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 | ||
+ | Similarly, <math>\angle XBE + \angle XCE = 180^o</math> \newline | ||
+ | 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 | ||
+ | -- Lucas.xue (someone pls help with a diagram) | ||
+ | |||
== See Also == | == See Also == |
Revision as of 19:25, 5 September 2024
Problem 2
Let be a convex quadrilateral such that diagonals and intersect at right angles, and let be their intersection. Prove that the reflections of across , , , are concyclic.
Solution 1
Diagram
Work
Let , , , be the foot of the altitude from point of , , , .
Note that reflection of over all the points of is similar to with a scale of with center . Thus, if is cyclic, then the reflections are cyclic.
is right angle and so is . Thus, is cyclic with being the diameter of the circumcircle.
Follow that, because they inscribe the same angle.
Similarly , , .
Futhermore, .
Thus, and are supplementary and follows that, is cyclic.
Solution 2
Suppose the reflection of E over AB is W, and similarly define X, Y, and Z. \newline by reflection gives \newline by reflection gives \newline These two tell us that E, W, and X belong to a circle with center B. \newline Similarly, we can get that: \newline E, Z, and W belong to a circle with center A, \newline E, X, and Y belong to a circle with center C, \newline E, Y, and Z belong to a circle with center D. \newline \newline To prove that W, X, Y, Z are concyclic, we want to prove \newline \newline \newline \newline \newline and tells us that \newline Similarly, \newline Thus, , and we are done. \newline -- Lucas.xue (someone pls help with a diagram)
See Also
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.