1980 AHSME Problems/Problem 19

Problem

Let $C_1, C_2$ and $C_3$ be three parallel chords of a circle on the same side of the center. The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3$ . The lengths of the chords are $20, 16$ , and $8$ . The radius of the circle is

$\text{(A)} \ 12 \qquad \text{(B)} \ 4\sqrt{7} \qquad \text{(C)} \ \frac{5\sqrt{65}}{3} \qquad \text{(D)}\ \frac{5\sqrt{22}}{2}\qquad \text{(E)}\ \text{not uniquely determined}$

Solution

Let the center of the circle be on the origin with equation $x^2 + y^2 = r^2$ . As the chords are bisected by the x-axis their y-coordinates are $10, 8, 4$ respectively. Let the chord of length $10$ have x-coordinate $a$ . Let $d$ be the common distance between chords. Thus, the coordinates of the top of the chords will be $(a, 10), (a+d, 8), (a+2d, 4)$ for the chords of length $20, 16$ , and $8$ respectively. As these points fall of the circle, we get three equations: $\\a^2 + 100 = r^2$ $\\(a+d)^2 + 64 = r^2$ $\\(a+2d)^2 + 16 = r^2$ Subtracting the first equation from the second we get: $\\(a+d)^2 - a^2 - 36 = 0$ $\\d(2a+d) = 36$ Similarly, by subtracting the first equation from the third we get: $\\(a+2d)^2 - a^2 - 84 = 0$ $\\2d(2a+2d) = 84$ $\\d(a+d) = 21$ Subtracting these two equations gives us $ad = 15$ . Expanding the second equation now gives us $\\a^2 + 2ad + d^2 + 64 = r^2$ $\\a^2 + d^2 + 94 = r^2$ Subtracting the first equation from this yields: $\\d^2 - 6 = 0$ $\\d = \sqrt{6}$ Combining this with $ad = 15$ we get $\\\sqrt{6}a = 15$ $\\a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}$ Plugging this into the first equation finally us $\\(\frac{5\sqrt{6}}{2})^2 + 100 = r^2$ $\frac{150}{4} + 100 = \frac{550}{4} = r^2$ $\\r = \sqrt{\frac{550}{4}} = \frac{5\sqrt{22}}{2}$ $\fbox{D}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1980 AHSME Problems/Problem 19

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