2015 AMC 12B 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
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of and
and finding their intersection point. This point happens to be the midpoint of
, the hypotenuse. Let this point be
. To find the radius, determine
, where
,
, and
. Thus, the radius
.
Next we let and
. Consider the right triangle
first. Using the pythagorean theorem, we find that
. Next, we let
to be the midpoint of
, and we consider right triangle
. By the pythagorean theorem, we have that
. Expanding this equation, we get that
This means that
is a
triangle, so
. Thus the perimeter is
which is answer
See Also
2015 AMC 12B (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 12 Problems and Solutions |
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