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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Horizontal parallel segments $AB = 10$ and $CD = 15$ are the bases of trapezoid $ABCD$ . Circle $\gamma$ of radius $6$ has center within the trapezoid and is tangent to sides $AB$ , $BC$ , and $DA$ . If side $CD$ cuts out an arc of $\gamma$ measuring $120^{\circ}$ , find the area of $ABCD$ .

Solution

Solution:

Suppose that the center of the circle is $O$ and the circle intersects $CD$ at $X$ and $Y$ . Since $\angle XOY = 120^{\circ}$ and triangle $XOY$ is isosceles, the distance from $O$ to $XY$ is $6 \cdot \sin(30^{\circ}) = 3$ . On the other hand, the distance from $O$ to $AB$ is $6$ as the circle is tangent to $AB$ , and $O$ is between $AB$ and $CD$ , so the height of the trapezoid is $6 + 3 = 9$ and its area is $\frac{9 \cdot (10 + 15)}{2} = \frac{225}{2}$ .

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