1983 AIME Problems/Problem 14
Problem
In the adjoining figure, two circles with radii and are drawn with their centers units apart. At , one of the points of intersection, a line is drawn in such a way that the chords and have equal length. Find the square of the length of .
Contents
Solution
Note that some of these solutions assume that lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto .
Solution 1
Firstly, notice that if we reflect over , we get . Since we know that is on circle and is on circle , we can reflect circle over to get another circle (centered at a new point , and with radius ) that intersects circle at . The rest is just finding lengths, as follows.
Since is the midpoint of segment , is a median of . Because we know , , and , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get . Now we have a kite with , , and , and all we need is the length of the other diagonal . The easiest way it can be found is with the Pythagorean Theorem. Let be the length of . Then
Solving this equation, we find that , so
Solution 2 (Easiest)
Draw additional lines as indicated. Note that since triangles and are isosceles, the altitudes are also bisectors, so let .
Since triangles and are similar. If we let , we have .
Applying the Pythagorean Theorem on triangle , we have . Similarly, for triangle , we have .
Subtracting, .
Solution 3
Let . Angles , , and must add up to . By the Law of Cosines, . Also, angles and equal and . So we have
Taking the cosine of both sides, and simplifying using the addition formula for as well as the identity , gives .
Solution 4 (quickest)
Let . Extend the line containing the centers of the two circles to meet , and to meet the other side of the large circle at a point .
The part of this line from to the point nearest to where it intersects the larger circle has length . The length of the diameter of the larger circle is .
Thus by Power of a Point in the circle passing through , , and , we have , so .
Full Proof that R, A, B are collinear
Let and be the feet of the perpendicular from to and to respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so , since the problem told us .
We will show that lies on .
Let be the intersection of circle centered at with . Then .
Let ' be the foot of the perpendicular from to . Then is a midline (or midsegment) in trapezoid , so coincides with (they are both supposed to be the midpoint of ). In other words, since , then .
Thus, subtends a degree arc. So arc in circle is , so is a diameter, as desired. Thus , , are collinear.
NOTE: Note this collinearity only follows from the fact that is half of in the problem statement. The collinearity is untrue in general.
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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