Problem 247

1966 AHSME-8.JPG

Let $O$ be the center of the circle of radius $10$ and $P$ be the center of the circle of radius $17$. Chord $\overline{AB} = 16$ feet.
$\overline{OA} = \overline{OB} = 10$ feet, since they are radii of the same circle. Hence, $\triangle OAB$ is isoceles with base $AB$. The height of $\triangle OAB$ from $O$ to $AB$ is $\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6$
Similarly, $\overline{PA} = \overline{PB} = 17$. Therefore, $\triangle PAB$ is also isoceles with base $AB$. The height of the triangle from $P$ to $AB$ is $\sqrt {\overline{PB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {17^2 - (\frac{16}{2})^2} = \sqrt {289 - 8^2} = \sqrt {289 - 64} = \sqrt {225} = 15$
The distance between the centers of the circles (points $P$ and $O$) is the sum of the heights of $\triangle OAB$ and $\triangle PAB$, which is $6 + 15 = 21 \Rightarrow \textbf{(B)}$

1966 AHSME (Problems • Answer Key • Resources)

Preceded byProblem 7

Followed byProblem 9

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