# 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, $x^2=AM^2-DM^2=46^2-19\cdot 23=1679, \boxed{B}.$

## See also

 1992 AHSME (Problems • Answer Key • Resources) Preceded byProblem 29 Followed byProblem 30 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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