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>. | ||
− | <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>. | + | <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 23:09, 22 April 2010
Contents
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
Diagram
Work
Let , , , be the foot of the altitute from point of , , , .
Note that reflection of over the 4 lines is 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.
Resources
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |