2021 JMPSC Invitationals Problems/Problem 14
Let there be a such that , , and , and let be a point on such that Let the circumcircle of intersect hypotenuse at and . Let intersect at . If the ratio can be expressed as where and are relatively prime, find
We claim that is the angle bisector of .
Observe that , which tells us that is a triangle. In cyclic quadrilateral , we have and Since , we have . This means that , or equivalently , is an angle bisector of in .
By the angle bisector theorem and our We seek the lengths and .
To find , we can proceed by Power of a Point using point on circle to get Since , , and , we have Since , we have
To find , we use the Pythagorean Theorem in . (We already found , which tells us that supplementary .) By the Pythagorean Theorem, We found that , and since we are given , we have
Our answer, by equation , is . From equation and from equation . Therefore, our final answer is
Note is cyclic, so . By Power Of A Point we have , . Now, note Therefore, by Angle Bisector Theorem,
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