Difference between revisions of "1983 AIME Problems/Problem 14"
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Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | ||
=== Solution 2 === | === Solution 2 === | ||
− | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the | + | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoint of <math>QP</math> be <math>M_1</math>, and the midpoint of <math>PR</math> be <math>M_2</math>. Let <math>x</math> be the length of <math>AM_1</math>, and <math>y</math> that of <math>BM_2</math>. |
{{incomplete|solution}} | {{incomplete|solution}} |
Revision as of 22:06, 10 April 2009
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 sich a way that the chords and have equal length. ( is the midpoint of ) Find the square of the length of .
Contents
[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
This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoint of be , and the midpoint of be . Let be the length of , and that of .
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
Let the circles of radius and be centered at and respectively. Let the midpoints of and be and Dropping a perpendicular from to (let the point be ) gives a rectangle.
Now note that triangle is right. Let the midpoint of (segment of length ) be Hence,
By now obvious similar triangles, so it's a quick system of two linear equations to solve for the desired length.
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 |