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 <math>1^2 + 5^2 = \boxed{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 <math>1^2 + 5^2 = \boxed{026}</math>. | ||
− | === Solution 2 | + | === Solution 2 === |
Drop perpendiculars from <math>O</math> to <math>AB</math> (<math>T_1</math>), <math>M</math> to <math>OT_1</math> (<math>T_2</math>), and <math>M</math> to <math>AB</math> (<math>T_3</math>). | Drop perpendiculars from <math>O</math> to <math>AB</math> (<math>T_1</math>), <math>M</math> to <math>OT_1</math> (<math>T_2</math>), and <math>M</math> to <math>AB</math> (<math>T_3</math>). | ||
Also, draw the midpoint <math>M</math> of <math>AC</math>. | Also, draw the midpoint <math>M</math> of <math>AC</math>. |
Revision as of 17:23, 7 August 2012
Problem
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 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
Solution 1
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 .
Solution 2
Drop perpendiculars from to (), to (), and to (). Also, draw the midpoint of .
Then the problem is trivialized. Why?
First notice that by computation, is a isosceles triangle; thus . Then, notice that . Thus the two blue triangles are congruent.
So, . As , we subtract and get . Then the Pythagorean Theorem shows .
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 |