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Geometry Difficulty 5.2 AIME, harder Prove it Japan

A circle with radius $3$ is inscribed in a trapezoid $ABCD$ such that $AD \parallel BC$ and both angles $B$ and $C$ are acute. If $AB = 7$ and $CD = 8$ , find the area of trapezoid $ABCD$ .

Solution

Let $E$ , $F$ , $G$ and $H$ be the points of contact of side $AB$ , $BC$ , $CD$ and $DA$ respectively to the inscribed circle of trapezoid $ABCD$ . Since the lengths of tangents from $A$ to this inscribed circle are equal, $AE = AH$ holds. Similarly we have $BE = BF$ , $CF = CG$ and $DG = DH$ . Then we have $AD + BC = AH + DH + BF + CF = AE + BE + CG + DG = AB + DC = 7 + 8 = 15$ .

Let $I$ be the center of the inscribed circle of trapezoid $ABCD$ , then $AD \perp IH$ and $BC \perp IF$ hold. Since $AD \parallel BC$ , $H$ , $I$ , $F$ are collinear. Thus the height of trapezoid $ABCD$ is equal to $FH$ , and the area of trapezoid $ABCD$ is $\frac{1}{2} \cdot (AD + BC) \cdot FH = \frac{1}{2} \cdot 15 \cdot 6 = 45.$

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