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
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 | ||
