Difference between revisions of "1983 AIME Problems/Problem 14"
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== Solution II== | == 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 | 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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== See also == | == See also == | ||
+ | {{AIME box|year=1983|num-b=13|num-a=15}} | ||
* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] |
Revision as of 13:14, 6 May 2007
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
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 |