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
Draw line segments and . Now we have cyclic quadrilateral . This means that opposite angles sum to 180 degrees. Therefore, \angle{YZA}\angle{WXB}. Simplifying carefully, we get \angle{WXB}\angle{ZYA}\angle{XWB}$. So, we know that
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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