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1983 AIME Problems/Problem 4: Difference between revisions

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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.
[[Image:AIME_83_-4.JPG]]
[[Image:AIME_83_-4.JPG]]
== Solution ==
== Solution ==
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Extend a perpendicular from <math>O</math> to <math>AB</math> and label it <math>D</math>. Additionally, extend a perpendicular from <math>O</math> to the line <math>BC</math>, and label it <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.
 
{{image}}
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math>, and <math>OC^2 = EC^2 + EO^2</math>.
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</math>.


Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = 26</math>.
Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = 026</math>.


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Revision as of 11:25, 22 January 2007

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.

Solution

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $1^2 + 5^2 = 026$.

See also

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