1992 AHSME Problems/Problem 30

Problem

Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19$ . Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$ . If $m$ is the smallest possible value of $x$ , then $m^2$ =

$\text{(A) } 1369\quad \text{(B) } 1679\quad \text{(C) } 1748\quad \text{(D) } 2109\quad \text{(E) } 8825$

Solution

Note that the center of the circle is the midpoint of $AB$ , call it $M$ . When we decrease $x$ , the limiting condition is that the circle will eventually be tangent to segment $AD$ at $D$ and segment $BC$ at $C$ . That is, $MD\perp AD$ and $MC\perp BC$ .

From here, we drop the altitude from $D$ to $AM$ ; call the base $N$ . Since $\triangle DNM \sim \triangle ADM$ , we have $\frac{DM}{19/2}=\frac{46}{DM}.$ Thus, $DM=\sqrt{19\cdot 23}$ . Furthermore, $x^2=DN^2+AN^2=(DM^2-NM^2)+AN^2=19\cdot 23-\left(\frac{19}{2}\right)^2+\left(\frac{73}{2}\right)^2=1679.$ \boxed{B}$.

1992 AHSME (Problems • Answer Key • Resources)
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1992 AHSME Problems/Problem 30

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