2006 USAMO Problems/Problem 6
(Zuming Feng, Zhonghao Ye) Let be a quadrilateral, and let and be points on sides and , respectively, such that . Ray meets rays and at and respectively. Prove that the circumcircles of triangles , , , and pass through a common point.
Let the intersection of the circumcircles of and be , and let the intersection of the circumcircles of and be .
because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and . because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and .
From the similarity, we have that . But we are given , so multiplying the 2 equations together gets us . are the supplements of , which are congruent, so , and so by SAS similarity, and so is also the center of spiral similarity for and . Thus, and are the same point, which all the circumcircles pass through, and so the statement is true.
We will give a solution using complex coordinates. The first step is the following lemma.
Lemma. Suppose and are real numbers and , and are complex. The circle in the complex plane passing through , and also passes through the point , independent of .
Proof. Four points , , and in the complex plane lie on a circle if and only if the cross-ratio is real. Since we compute the given points are on a circle.
Lay down complex coordinates with and and on the positive real axis. Then there are real , and with , and and hence gives The line consists of all points of the form for real . Since lies on this line and has zero imaginary part, we see from that it corresponds to . Thus Apply the lemma with , , , and . Setting gives and setting gives Therefore the circumcircles to and meet at This last expression is invariant under simultaneously interchanging and and interchanging and . Therefore it is also the intersection of the circumcircles of and .
Let be the Miquel point of ; then is the center of the spiral similarity that takes to . Because , the same spiral similarity also takes to , so is the center of the spiral similarity that maps to and to . Then it is obvious that the circumcircles of , , , and pass through .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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