Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1966 AHSME Problems/Problem 8"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
[[Image:1966_AHSME-8.jpg]]
 
[[Image:1966_AHSME-8.jpg]]
 
Let <math>O</math> be the center of the circle of radius <math>10</math> and <math>P</math> be the center of the circle of radius <math>17</math>. Chord <math>\overline{AB} = 16</math> feet.
 
Let <math>O</math> be the center of the circle of radius <math>10</math> and <math>P</math> be the center of the circle of radius <math>17</math>. Chord <math>\overline{AB} = 16</math> feet.
−
<math>\overline{OA} = \overline{OB} = 10</math> feet, since they are radii of a circle. Hence, <math>\triangle OAB</math> is isoceles with base <math>AB</math>. The height of <math>\triangle OAB</math> from <math>O</math> to <math>AB</math> is <math>\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6</math>
+
 
 +
<math>\overline{OA} = \overline{OB} = 10</math> feet, since they are radii of the same circle. Hence, <math>\triangle OAB</math> is isoceles with base <math>AB</math>. The height of <math>\triangle OAB</math> from <math>O</math> to <math>AB</math> is <math>\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6</math>
 +
 
 +
Similarly, <math>\overline{PA} = \overline{PB} = 17</math>. Therefore, <math>\triangle PAB</math> is also isoceles with base <math>AB</math>. The height of the triangle from <math>P</math> to <math>AB</math> is <math>\sqrt {\overline{PB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {17^2 - (\frac{16}{2})^2} = \sqrt {289 - 8^2} = \sqrt {289 - 64} = \sqrt {225} = 15</math>
 +
 
 +
The distance between the centers of the circles (points <math>P</math> and <math>O</math>) is the sum of the heights of <math>\triangle OAB</math> and <math>\triangle PAB</math>, which is <math>6 + 15 = 21 \Rightarrow \textbf{(B)}</math>

Revision as of 19:45, 1 July 2008

Problem

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

File:1966 AHSME-8.jpg Let $O$ be the center of the circle of radius $10$ and $P$ be the center of the circle of radius $17$. Chord $\overline{AB} = 16$ feet.

$\overline{OA} = \overline{OB} = 10$ feet, since they are radii of the same circle. Hence, $\triangle OAB$ is isoceles with base $AB$. The height of $\triangle OAB$ from $O$ to $AB$ is $\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6$

Similarly, $\overline{PA} = \overline{PB} = 17$. Therefore, $\triangle PAB$ is also isoceles with base $AB$. The height of the triangle from $P$ to $AB$ is $\sqrt {\overline{PB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {17^2 - (\frac{16}{2})^2} = \sqrt {289 - 8^2} = \sqrt {289 - 64} = \sqrt {225} = 15$

The distance between the centers of the circles (points $P$ and $O$) is the sum of the heights of $\triangle OAB$ and $\triangle PAB$, which is $6 + 15 = 21 \Rightarrow \textbf{(B)}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_8&oldid=26838"