Difference between revisions of "1995 AHSME Problems/Problem 28"

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==Problem==
 
==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 <math>\sqrt {a}</math> where <math>a</math> is
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Two [[parallel]] [[chord]]s in a [[circle]] have lengths <math>10</math> and <math>14</math>, and the distance between them is <math>6</math>. The chord parallel to these chords and midway between them is of length <math>\sqrt {a}</math> where <math>a</math> is
 
 
  
 
<math> \mathrm{(A) \ 144 } \qquad \mathrm{(B) \ 156 } \qquad \mathrm{(C) \ 168 } \qquad \mathrm{(D) \ 176 } \qquad \mathrm{(E) \ 184 }  </math>
 
<math> \mathrm{(A) \ 144 } \qquad \mathrm{(B) \ 156 } \qquad \mathrm{(C) \ 168 } \qquad \mathrm{(D) \ 176 } \qquad \mathrm{(E) \ 184 }  </math>
  
 
==Solution==
 
==Solution==
{{solution}}
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We let <math>O</math> be the center, <math>\overline{A_1AA_2}</math>, <math>\overline{B_1BB_2}</math> represent the chords with length <math>10, 14</math> 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.
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By the [[Pythagorean Theorem]] on <math>\triangle OBB_1</math>, we get <math>x^2 + 7^2 = r^2 \Longrightarrow x = \sqrt{r^2 - 49}</math>, where <math>x</math> is the length of the other leg. Now the length of the leg of <math>\triangle OAA_1</math> is either <math>6 + x</math> or <math>6 - x</math> depending whether or not <math>\overline{A_1A_2}, \overline{B_1B_2}</math> are on the same side of the center of the circle:
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<cmath>\begin{eqnarray*}(6 \pm \sqrt{r^2 - 49})^2 + 5^2 &=& r^2\
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12 \pm 12\sqrt{r^2 - 49} &=& 0\end{eqnarray*}</cmath>
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Only the negative works here (thus the two chords are on ''opposite'' sides of the center), and solving we get <math>x=1, r = 5\sqrt{2}</math>. The leg formed in the right triangle with the third chord is <math>3 - x = 2</math>, and by the Pythagorean Theorem again
  
==See Also==
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<cmath>(a/2)^2 + 2^2 = (5\sqrt{2})^2 \Longrightarrow \sqrt{a} = 184 \Rightarrow \mathrm{(E)}.</cmath>
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==See also==
 
{{Old AMC12 box|year=1995|num-b=27|num-a=29}}
 
{{Old AMC12 box|year=1995|num-b=27|num-a=29}}
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[[Category:Introductory Geometry Problems]]

Revision as of 18:59, 7 January 2008

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

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.

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

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

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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All AHSME Problems and Solutions