1983 AIME Problems/Problem 4: Difference between revisions
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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>. | ||
== See also == | == See also == | ||
{{AIME box|year=1983|num-b=3|num-a=5}} | |||
[[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 
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 
.
Applying the Pythagorean Theorem, 
and 
.
Thus, 
, and 
. We solve this system to get 
and 
, resulting in an answer of 
.
See also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
