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1995 AHSME Problems/Problem 28

Revision as of 20:35, 8 January 2008 by Azjps (talk | contribs) (no problem 1=2 --------- solution (2) by scorpius)

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 yet know whether the two chords are on the same side of are two different sides of the center of the circle.

1995 12 AMC-28.png

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 $\sqrt {a} = 2\sqrt {46}\Rightarrow a = 184$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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