2007 IMO Problems/Problem 2
Problem
Consider five points , and such that is a parallelogram and is a cyclic quadrilateral. Let be a line passing through . Suppose that intersects the interior of the segment at and intersects line at . Suppose also that . Prove that is the bisector of .
Solution
Since , , it suffices to prove .
Let , , . We have: so, Meantime, using Law of Sines on , we have, Using Law of Sines on , and notice that , we have, so, Since , and , we have, . Hence, or, There are two possibilities: (1) , or (2) . However, (2) would mean , then would be a diameter, and because is inside the circle, so (2) is not valid. From condition (1), we have , therefore .
Solution by Mathdummy
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
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