Question

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 $m^2=$

$1369$

$1679$

$1748$

$2109$

$8825$

Answer: B

Concepts:isosceles trapezoid circle tangent to lines optimization coordinate geometry

Difficulty rating: 2400

Small Hint:

By symmetry, place the trapezoid with its bases centered on the same vertical axis and put the circle’s center at the midpoint of $AB$

Big Hint:

Write the distance from the center to a leg and require the tangency point to lie on the leg segment; the minimum occurs at an endpoint case

Solution:

Place $A=(-46,0),$ $B=(46,0),$ $D=(-\frac{19}{2},h),$ and $C=(\frac{19}{2},h).$ Symmetry forces the circle’s center to be $O=(0,0).$ The horizontal offset along each leg is $\frac{73}{2},$ so $x^2=h^2+\left(\frac{73}{2}\right)^2.$ The perpendicular from $O$ to a leg first has its foot on the segment when the foot reaches the upper endpoint; this boundary condition is $OD\perp AD.$ Thus $\begin{aligned} (-\frac{19}{2},h)\cdot(\frac{73}{2},h)&=0,\\ h^2&=\frac{1387}{4}. \end{aligned}$ Therefore $m^2=\frac{1387+5329}{4}=1679.$

Thus the correct answer is B.

1992 AMC 12 Problem 30

Answer: B

Difficulty rating: 2400

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years