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.
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A machine shop cutting tool is in the shape of a notched [[circle]], as shown. The [[radius]] of the circle is <math>\sqrt{50}</math> 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]].  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>.
 
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>.

Revision as of 19:50, 17 March 2008

Problem

A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ 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. AIME 83 -4.JPG

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 $x^2+y^2$.

AIME 83 -4 Modified.JPG

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

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions