Difference between revisions of "1983 AIME Problems/Problem 4"

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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>.
 
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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* [[1983 AIME Problems/Problem 3|Previous Problem]]
 
* [[1983 AIME Problems/Problem 5|Next Problem]]
 
* [[1983 AIME Problems|Back to Exam]]
 
  
 
== See also ==
 
== See also ==
* [[AIME Problems and Solutions]]
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{{AIME box|year=1983|num-b=3|num-a=5}}
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 15:10, 11 March 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. 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
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All AIME Problems and Solutions