# 2021 JMPSC Invitationals Problems/Problem 14

## Contents

## Problem

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

## Solution

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

~samrocksnature

## Solution 2

Note is cyclic, so . By Power Of A Point we have , . Now, note Therefore, by Angle Bisector Theorem,

~Geometry285

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.