2007/08 AHSMC Part 1 Problem 9

Problem 9 of 16

In the quadrilateral $ABCD$ , $AB = CD$ , $AD = 2$ and $BC = 6$ . $AD$ and $BC$ are parallel lines at a distance $8$ apart. The radius of the smallest circle which can cover $ABCD$ is

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Solution

The centre $O$ of the circle lies on the axis of symmetry of $ABCD$ . Let $y$ be its height above $BC$ . Then $OB^2 = y^2 + 3^2$ while $OA^2 = (8-y)^2 + 1^2$ . Equating these two values yields $y = \dfrac{7}{2}$ . Hence the radius is $\sqrt{\bigl(\dfrac{7}{2}\bigr)^2 + 3^2} = \dfrac{\sqrt{85}}{2}$ . The answer is (C).

University of Alberta

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