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. |
[[Image:AIME_83_-4.JPG]] | [[Image:AIME_83_-4.JPG]] | ||
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== Solution == | == Solution == | ||
| − | Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. | + | 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> | + | 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 = | + | 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 12:25, 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 
is 6 cm, and that of 
is 2 cm. The angle 
is a right angle. Find the square of the distance (in centimeters) from 
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 
to be 
and let the foot of the perpendicular from to the line be 
. Let 
and 
. We're trying to find 
.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Applying the Pythagorean Theorem, 
and 
.
Thus, 
, and 
. We solve this system to get 
and 
, resulting in an answer of 
.
