2020 AIME II Problems/Problem 15
Let be an acute scalene triangle with circumcircle . The tangents to at and intersect at . Let and be the projections of onto lines and , respectively. Suppose , , and . Find .
Assume to be the center of triangle , cross at , link , . Let be the middle point of and be the middle point of , so we have . Since , we have . Notice that , so , and this gives us . Since is perpendicular to , and cocycle (respectively), so and . So , so , which yields So same we have . Apply Ptolemy theorem in we have , and use Pythagoras theorem we have . Same in and triangle we have and . Solve this for and and submit into the equation about , we can obtain the result .
(Notice that is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)
Solution 2 (Official MAA)
Let denote the midpoint of . The critical claim is that is the orthocenter of , which has the circle with diameter as its circumcircle. To see this, note that because , the quadrilateral is cyclic, it follows that implying that . Similarly, . In particular, is a parallelogram. Hence, by the Parallelogram Law, But . Therefore
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