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Difference between revisions of "1983 AIME Problems/Problem 4"
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== Problem ==
 
== Problem ==
 
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of <math>AB</math> is 6 cm, and that of <math>BC</math> is 2 cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> to the center of the circle.
 
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of <math>AB</math> is 6 cm, and that of <math>BC</math> is 2 cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> to the center of the circle.
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== Solution ==
 
== Solution ==

Revision as of 11:08, 14 December 2006

Problem

A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. File:Http://www.artofproblemsolving.com/Forum/album pic.php?pic id=790&sid=cfd5dae222dd7b8944719b56de7b8bf7


Solution

Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Extend a perpendicular from $O$ to $AB$ and label it $D$. Additionally, extend a perpendicular from $O$ to the line $BC$, and label it $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$.

Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$, and $OC^2 = EC^2 + EO^2$.

Thus, $(\sqrt{50})^2 = y^2 + (6-x)^2$, and $(\sqrt{50})^2 = x^2 + (y+2)^2$. We solve this system to get $x = 1$ and $y = 5$, resulting in an answer of $1^2 + 5^2 = 26$.

See also

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