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. ( is the midpoint of ) Find the area of the square with a side length of .
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[hide]Solution
Solution 1
First, 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 with radius ) that intersects circle at . The rest is just finding lengths:
Since is the midpoint of segment , is a median of triangle . Because we know that , , and , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get . So 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
Doing routine algebra on the above equation, we find that , so
Solution 2
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 of both sides and simplifying using the cosine addition identity gives .
Solution 4
Observe that the length of the area where the two circles intersect can be found explicitly as . Let , then the power of point with regards to the larger circle gives
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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