Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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− | Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ</math>. This means that opposite angles sum to 180 degrees. Therefore, angle <math>\angle{YZA} + </math>\ | + | Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ</math>. This means that opposite angles sum to 180 degrees. Therefore, 90 + angle <math>\angle{YZA} + (90 - </math>\angle{WXB}) = 180. Simplifying carefully, we get <math>\angle{YZA} = </math>\angle{WXB}. You can use the same method to find that <math>\angle{ZYA} = </math>\angle{XWB}. So, we know that |
<asy> | <asy> |
Revision as of 17:03, 4 May 2019
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, 90 + angle \angle{WXB}) = 180. Simplifying carefully, we get \angle{WXB}. You can use the same method to find that \angle{XWB}. So, we know that
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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