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

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==Solution==
 
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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.
  
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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>
 
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>
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{{AHSME box|year=1966|num-b=7|num-a=9}} 
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Latest revision as of 00:45, 26 June 2016

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

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 $6 + 15 = 21 \Rightarrow \textbf{(B)}$

1966 AHSME (Problems • Answer Key • Resources)
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Problem 7 		Followed by
Problem 9
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