1983 AIME Problems/Problem 14
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
Solution III
LetQP=PR=x. Angles QPA, APB, and BPR must add up to 180. By the Law of Cosines, angle APB=arccos(-11/24). Also, angles QPA and BPR equal arccos(x/16) and arccos(x/12). So we have arccos(x/16)+arccos(-11/24)=180-arccos(x/12). Taking the cos of both sides and simplifying gives 
.
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 | ||
