2015 AMC 10B Problems/Problem 19
Contents
Problem
In , and . Squares and are constructed outside of the triangle. The points , and lie on a circle. What is the perimeter of the triangle?
Solution 1
The center of the circle lies on the perpendicular bisectors of both chords and . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be . Draw perpendiculars to and from , and connect and . . Let and . Then . Simplifying this gives . But by Pythagorean Theorem on , we know , because . Thus . So our equation simplifies further to . However , so , which means , or . Aha! This means is just an isosceles right triangle, so , and thus the perimeter is .
Solution 2
Let and . Draw line segments and . Now we have cyclic quadrilateral . This means that opposite angles sum to . Therefore, . Simplifying carefully, we get . Similarly, you can find that = . So, we know that is similar to . So . Now, we can use the proportion . Simplifying, we get , which gives us . Therefore, is an isosceles right triangle, so , and the perimeter is .
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See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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