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 | ||
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