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

Place this circle in the coordinate plane such that the center is at the origin all three chords are to the right of, and parallel to, the $y$ -axis. The equation of this circle is $x^2 + y^2 = r^2$ , where $r > 0$ is the radius of the circle.

Let $A$ , $B$ , and $C$ be the upper ends of chords $C_1$ , $C_2$ , and $C_3$ , respectively.

Since the $x$ -axis is perpendicular to the three chords, it must bisect each chord. Let $a$ be the $x$ -coordinate $A$ , and $d$ be the distance between $C_1$ and $C_2$ , which is also the distance between $C_2$ and $C_3$ . Then $A$ is at $(a, 10)$ , $B$ is at $(a+d, 8)$ , and $C$ is at $(a+2d, 4)$ .

Since $A$ , $B$ , and $C$ all lie on this circle, the following equations must hold: $\\a^2 + 100 = r^2\ (\textrm{i})$ $\\(a+d)^2 + 64 = r^2\ (\textrm{ii})$ $\\(a+2d)^2 + 16 = r^2\ (\textrm{iii})$ Subtracting equation $(\textrm{i})$ from equation $(\textrm{ii})$ yields: $\\(a+d)^2 - a^2 - 36 = 0$ $\\2ad + d^2 = 36\ (\textrm{iv})$ Subtracting equation $(\textrm{i})$ from equation $(\textrm{iii})$ yields: $\\(a+2d)^2 - a^2 - 84 = 0$ $\\4ad + 4d^2 = 84$ $\\ad + d^2 = 21\ (\textrm{v})$ Subtracting $(\textrm{v})$ from $(\textrm{iv})$ yields: $\\ad = 15\ (\textrm{vi})$ Subtracting $(\textrm{vi})$ from $(\textrm{v})$ yields $d^2 - 6 = 0$ , so $d = \sqrt{6}$ .

Substituting $d = \sqrt{6}$ into $(\textrm{vi})$ yields $\sqrt{6}a = 15$ , so $a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}$ .

Substituting $a = \frac{5\sqrt{6}}{2}$ yields $r^2 = \left(\frac{5\sqrt{6}}{2}\right)^2 + 100 = \frac{275}{2}$ , so $r = \sqrt{\frac{275}{2}} = \boxed{(\textbf{D})\ \frac{5\sqrt{22}}{2}}$

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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