2017 AMC 8 Problems/Problem 16
Problem 16
In the figure below, choose point on
so that
and
have equal perimeters. What is the area of
?
Solution
We know that the perimeters of the two small triangles are and
. Setting both equal and using
, we have
and
. Now, we simply have to find the area of
. Since
, we must have
. Combining this with the fact that
, we get
Solution 2. (Similar triangles)
Since point is on line
, it will split it into
and
. Let
and
. Triangle
has side lengths
and triangle
has side lengths
. Since both perimeters are equal, we have the equation
. Eliminating
and solving the resulting linear equation gives
. Draw a perpendicular from point
to
. Call the point of intersection
. Because angle
is common to both triangles
and
, and both are right triangles, both are similar. The hypotenuse of triangle
is 2, so the altitude must be
Because
and
share the same altitude, the height of
therefore must be
. The base of
is 4, so
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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