1983 AIME Problems/Problem 4
Contents
Problem
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is cm, the length of is cm and that of is 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 , such that the answer is .
Solution 2
We'll use the law of cosines. Let be the center of the circle; we wish to find . We know how long and are, so if we can find , we'll be in good shape.
We can find using angles and . First we note that by Pythagoras, If we let be the midpoint of , that mean that . Since is isosceles ( from the definition of a circle), is also the foot of the altitude from to
It follows that . Therefore Meanwhile, from right triangle we have
This means that by the angle subtraction formulas,
Now we have all we need to use the law of cosines on This tells us that
Solution 3
Drop perpendiculars from to (with foot ), to (with foot ), and to (with foot ). Also, mark the midpoint of .
First notice that by computation, is a isosceles triangle, so . Then, notice that . Therefore, the two blue triangles are congruent, from which we deduce and . As and , we subtract and get . Then the Pythagorean Theorem tells us that .
Solution 4
Draw segment with length , and draw radius such that bisects chord at point . This also means that is perpendicular to . By the Pythagorean Theorem, we get that , and therefore . Also by the Pythagorean theorem, we can find that .
Next, find and . Since , we get By the subtraction formula for , we getFinally, by the Law of Cosines on , we get
Solution 5
We use coordinates. Let the circle have center and radius ; this circle has equation . Let the coordinates of be . We want to find . and with coordinates and , respectively, both lie on the circle. From this we obtain the system of equations
After expanding these terms, we notice by subtracting the first and second equations, we can cancel out and . after substituting and plugging back in, we realize that or . Since the first point is out of the circle, we find that is the only relevant answer. This paragraph is written by ~hastapasta.
Solving, we get and , so the distance is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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