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 .
Let be the circumcenter of ; say intersects at ; draw segments , and . We have .
Since , we have . Notice that is cyclic, so , so , and the cosine law in gives
Since , we have , and therefore quadrilaterals and are cyclic. Let (resp. ) be the midpoint of (resp. ). So (resp. ) is the center of (resp. ). Then and . So , sowhich yields . Similarly we have .
Ptolemy's theorem in gives while Pythagoras' theorem gives . Similarly, Ptolemy's theorem in gives while Pythagoras' theorem in gives . Solve this for and and substitute into the equation about to 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
Solution 3 (Law of Cosines)
Let be the orthocenter of .
Lemma 1: is the midpoint of .
Proof: Let be the midpoint of , and observe that and are cyclical. Define and , then note that: That implies that , , and . Thus and ; is indeed the same as , and we have proved lemma 1.
Since is cyclical, and this implies that is a paralelogram. By the Law of Cosines: We add all these equations to get: We have that and using our midpoints. Note that , so by the Pythagorean Theorem, it follows that . We were also given that , which we multiply by to use equation . Since , we have Therefore, . ~ MathLuis
Solution 4 (Similarity and median)
Using the Claim (below) we get
Corresponding sides of similar is so
The formula for median of triangle is
Let be an acute scalene triangle with circumcircle . The tangents to at and intersect at . Let be the projections of onto line . Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.
the quadrilateral is cyclic.
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