Difference between revisions of "1983 AIME Problems/Problem 14"
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== Solution == | == Solution == | ||
− | Note that some of these solutions assume that <math>R</math> lies on the line connecting the centers, which is not true in general. It's true here cause the perpendicular from <math>P</math> hits the point where the line between the centers and the small circle intersect. This can be derived from the application of midpoint theorem to the trapezoid made by dropping perpendiculars from the centers | + | Note that some of these solutions assume that <math>R</math> lies on the line connecting the centers, which is not true in general. It's true here cause the perpendicular from <math>P</math> hits the point where the line between the centers and the small circle intersect. This can be derived from the application of midpoint theorem to the trapezoid made by dropping perpendiculars from the centers onto QR. |
=== Solution 1 === | === Solution 1 === |
Revision as of 20:57, 3 January 2019
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
[hide]Solution
Note that some of these solutions assume that lies on the line connecting the centers, which is not true in general. It's true here cause the perpendicular from hits the point where the line between the centers and the small circle intersect. This can be derived from the application of midpoint theorem to the trapezoid made by dropping perpendiculars from the centers onto QR.
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 (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 of both sides and simplifying using the cosine addition identity gives .
Solution 4 (quickest)
Let Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle.
The line segment consisting of R and the first intersection of the larger circle has length 10. The length of the diameter of the larger circle be16.
Through power of a point,
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 | ||
All AIME Problems and Solutions |