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 = 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 = 130.</math> | ||
+ | == Solution II== | ||
+ | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoints of QP be <math>M_1</math>, and the midpoint of PR be <math>M_2</math>. Let x be the length of AM_1, and y that of BM_2 | ||
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Revision as of 21:43, 23 December 2006
Contents
[hide]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 .
Solution
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 stewarts or whatever else you like. 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 II
This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoints of QP be , and the midpoint of PR be . Let x be the length of AM_1, and y that of BM_2