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
. Therefore,
. Simplifying carefully, we get
. You can use the same method to find that
=
. 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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All AMC 10 Problems and Solutions |
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