1995 AHSME Problems/Problem 28

Problem

Two parallel chords in a circle have lengths $10$ and $14$ , and the distance between them is $6$ . The chord parallel to these chords and midway between them is of length $\sqrt {a}$ where $a$ is

$\mathrm{(A) \ 144 } \qquad \mathrm{(B) \ 156 } \qquad \mathrm{(C) \ 168 } \qquad \mathrm{(D) \ 176 } \qquad \mathrm{(E) \ 184 }$

Solution

Solution 1

We let $O$ be the center, $\overline{A_1AA_2}$ , $\overline{B_1BB_2}$ represent the chords with length $10, 14$ respectively (as shown below). Connecting the endpoints of the chords with the center, we have several right triangles. However, we do not know whether the two chords are on the same side or different sides of the center of the circle.

By the Pythagorean Theorem on $\triangle OBB_1$ , we get $x^2 + 7^2 = r^2 \Longrightarrow x = \sqrt{r^2 - 49}$ , where $x$ is the length of the other leg. Now the length of the leg of $\triangle OAA_1$ is either $6 + x$ or $6 - x$ depending whether or not $\overline{A_1A_2}, \overline{B_1B_2}$ are on the same side of the center of the circle:

$\begin{eqnarray*}(6 \pm \sqrt{r^2 - 49})^2 + 5^2 &=& r^2\\ 12 \pm 12\sqrt{r^2 - 49} &=& 0\end{eqnarray*}$

Only the negative works here (thus the two chords are on opposite sides of the center), and solving we get $x=1, r = 5\sqrt{2}$ . The leg formed in the right triangle with the third chord is $3 - x = 2$ , and by the Pythagorean Theorem again

$(a/2)^2 + 2^2 = (5\sqrt{2})^2 \Longrightarrow \sqrt{a} = 184 \Rightarrow \mathrm{(E)}.$

Solution 2

Let $AB$ be the chord of length $10$ , $CD$ be the chord of length $14$ , $E$ be the point on $CD$ such that $AE = 6$ , $F$ be the point on $CD$ such that $BF = 6$ . Then $CE = FD = 2$ . Extend $BF$ intersecting the circle at $G$ and let $FG = x$ . By the Power of a Point Theorem, $6*x = 12*2\Rightarrow x = 4.$ Since $\angle ABG$ is a right angle and $A$ and $G$ are on the circle, the diameter $AG = 10\sqrt {2}$ by Pythagorean Theorem, so the radius is $5\sqrt {2}$ . Let $O$ be the center of the circle, $H$ be the midpoint of $CD$ . Then $CH = 7$ and $\angle OHC$ is a right angle. Then $OH = 1$ , again by Pythagorean Theorem. The chord midway between $AB$ and $CD$ is a distance of 3 away from each, so 2 away from $O$ . Using the Pythagorean Theorem one more time, half the length of the chord is equal to $\sqrt {46}$ . Then $\sqrt {a} = 2\sqrt {46}\Rightarrow a = 184$ .

Solution 3 (Coordinate Geometry)

Let $AB$ be the chord of length 14, $CD$ be the chord of length $\sqrt {a}$ , and let $EF$ be the chord of length 10. Let point A be (-7, 0) and let B be (7, 0), let E be (-5, -6) and F be (5, -6). Since the perpendicular bisector of a chord always goes through the center of a circle, we have that the center of the circle lies on the perpendicular bisector of $AE$ , and the line x = 0. The line $AE$ lies on the line, y = -3x - 21, and the midpoint of line segment $AE$ = (-6, -3), so the line of the perpendicular bisector has slope $frac1{3}$ , and passes through the point (-6, -3), so the line is y = $frac1{3}$ x-1. Therefor, the center of the circle is at (0, -1). From here, the radius is $\sqrt{7^2+1^2)$ (Error compiling LaTeX. Unknown error_msg), which equals $\sqrt{50)$ (Error compiling LaTeX. Unknown error_msg). From here, we use the Pythagorean theorem, to get, 50 = $frac\sqrt{a}{2}$ + $2^2$ , so $\sqrt {a}$ = 2 $\sqrt {46}$ $Rightarrow a = 184$

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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