2014 IMO Problems/Problem 3
Convex quadrilateral has . Point is the foot of the perpendicular from to . Points and lie on sides and , respectively, such that lies inside and
Prove that line is tangent to the circumcircle of
Denote , , , , , , , . Since and , we have , .
Since , the tangent of the circumcircle of at point is perpendicular to ; therefore, the circumcenter of (point ) is on . Similarly, the circumcenter of (point ) is on . In addition, is the perpendicular bisector of .
Extend to meet circumcircle of at , and extend to meet circumcircle of at . Then, since , and are the perpendicular bisector of and , respectively; hence is the circumcenter of . Since and are midpoints on and , ; also, , so . Since is the circumcenter, is also the perpendicular bisector of . Hence,
We have Hence, , or Since quadrilaterals and are cyclic, we have , ; so, Hence, Similarly,
Now we apply law of Sines repeatedly on pairs of triangles. For and , , , , ; hence, For , , ; hence, For , , and similarly, ; hence, Combining , we have Therefore, , and . Let the circumcircle of meets at . We have, And, This proves is the diameter of the circle and the center of the circle is on AH.
Solution by .
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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