2015 AMC 10A Problems/Problem 19
Problem
The isosceles right triangle has right angle at
and area
. The rays trisecting
intersect
at
and
. What is the area of
?
Solution
can be split into a
right triangle and a
right triangle by dropping a perpendicular from
to side
. Let
be where that perpendicular intersects
.
Because the side lengths of a right triangle are in ratio
,
.
Because the side lengths of a right triangle are in ratio
and
+
,
.
Setting the two equations for equal to each other,
.
Solving gives .
The area of .
is congruent to
, so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up, so .
.
Solving gives , so the answer is
.
Solution 2
The area of is 12.5, and so the leg length of 45-45-90
is 5. Thus, the altitude to hypotenuse
,
, has length
by 45-45-90 right triangles. Now, it is clear that
, and so by the Exterior Angle Theorem,
is an isosceles 30-75-75 triangle. Thus,
, and so the area of
is
. The answer is
.