2015 AMC 10B Problems/Problem 19
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 intersection between the perpendicular bisectors of 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 (and we're given that ). Draw line segments and . Now we have cyclic quadrilateral
This means that opposite angles sum to . Therefore, . Simplifying carefully, we get . Similarly, = .
That means .
Setting up proportions, Cross-multiplying we get:
But also, by Pythagoras, , so
Therefore, is an isosceles right triangle. , so the perimeter is
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Solution 3
Both solution 1 and 2 uses Pythagorean Theorem to prove is isosceles right triangle. I'm going to prove is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.
Let be the the midpoint of . The perpendicular bisector of line and will meet at . Thus is the center of the circle points , , , and lies on.
, , , and , , by , . Because , is a rotation about point of . So, .
Because and is the radius of , . Because is the midpoint of hypotenuse , , , by . Because , is a rotation about point of . So, .
, , is isosceles right triangle, . So, is isosceles right triangle.
Therefore, , the perimeter is .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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