Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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Revision as of 10:48, 20 May 2015
Contents
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
, and so the leg length of
is
Thus, the altitude to hypotenuse
,
, has length
by
right triangles. Now, it is clear that
, and so by the Exterior Angle Theorem,
is an isosceles
triangle. Thus,
, and so the area of
is
. The answer is thus
See Also
2015 AMC 10A (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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