2009 USAMO Problems/Problem 5
Trapezoid , with , is inscribed in circle and point lies inside triangle . Rays and meet again at points and , respectively. Let the line through parallel to intersect and at points and , respectively. Prove that quadrilateral is cyclic if and only if bisects .
We will use directed angles in this solution. Extend to as follows:
Note that Thus, is cyclic.
Also, note that is cyclic because depending on the configuration.
Next, we have are collinear since
Therefore, so is cyclic.
Only If: These steps can be reversed.
Solution 2 (Projective)
Extend to , and let line intersect at and another point , as shown:
Suppose that , and . Pascal's theorem on the tuple implies that the points , , and are collinear. However, and are symmetrical with respect to the axis of symmetry of trapezoid , and and are also symmetrical with respect to the axis of symmetry of (as is the midpoint of , and ). Since , and are symmetric with respect to the axis of symmetry of trapezoid . This implies that line is equivalent to line . Thus, lies on line . However, , so this implies that .
Now note that is cyclic. Since , . However, . Therefore, is cyclic.
Consider the same setup, except is no longer the midpoint of . Note that must be parallel to in order for to be cyclic. We claim that and hope to reach a contradiction. Pascal's theorem on the tuple implies that , , and are collinear. However, there exists a unique point such that , , and are concurrent. By If, must be the midpoint of in order for the concurrency to occur; hence, . Then , since . However, this is a contradiction, so therefore cannot be parallel to and is not cyclic.
Solution by TheBoomBox77
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