HMMT November, 2012

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

$ABCD$ is a rectangle with $AB = 20$ and $BC = 3$ . A circle with radius $5$ , centered at the midpoint of $DC$ , meets the rectangle at four points: $W$ , $X$ , $Y$ , and $Z$ . Find the area of quadrilateral $WXYZ$ .

Solution

Solution:

Answer: $27$

Suppose that $X$ and $Y$ are located on $AB$ with $X$ closer to $A$ than $B$ . Let $O$ be the center of the circle, and let $P$ be the midpoint of $AB$ . We have $OP \perp AB$ so $OPX$ and $OPY$ are right triangles with right angles at $P$ . Because $OX = OY = 5$ and $OP = 3$ , we have $XP = PY = 4$ by the Pythagorean theorem. Now, $WXYZ$ is a trapezoid with $WZ = WO + OZ = 5 + 5 = 10$ , $XY = XP + PY = 8$ , and height $3$ , so its area is $\left(\frac{10+8}{2}\right) \times 3 = 27$ .

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