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

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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.  
 
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.  
  
−
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:
+
[[Image:1995_12_AMC-28.png|left]] 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:
  
 
<cmath>\begin{eqnarray*}(6 \pm \sqrt{r^2 - 49})^2 + 5^2 &=& r^2\\
 
<cmath>\begin{eqnarray*}(6 \pm \sqrt{r^2 - 49})^2 + 5^2 &=& r^2\\

Revision as of 19:03, 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.

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

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

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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