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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let $ABCD$ be a trapezoid with $AB \parallel CD$ , $AB = 20$ , $CD = 24$ , and area $880$ . Compute the area of the triangle formed by the midpoints of $AB$ , $AC$ , and $BD$ .

Solution

Solution:

We first compute the height of the trapezoid. If $h$ is the height, then the area is
$880 = \frac{1}{2} h (20 + 24)$
so $h = 40$ . Now, let $M$ , $N$ , $P$ be the midpoints of $AB$ , $AC$ , and $BD$ . Notice that $PN$ is parallel to $AB$ . Thus, the altitude from $M$ to $NP$ has length $\frac{h}{2} = 20$ .

To compute $NP$ , let $X$ be the midpoint of $BC$ . Since $XN$ is a midsegment of $\triangle CAB$ , we have $XN = \frac{AB}{2} = 10$ . Since $XP$ is a midsegment of $\triangle BCD$ , we have $XP = \frac{CD}{2} = 12$ . Hence, $NP = XP - XN = 2$ .

Thus, the area of triangle $MNP$ is $\frac{1}{2} \cdot 2 \cdot 20 = 20$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.