Difference between revisions of "1983 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
In the adjoining figure, two circles with radii <math>6</math> and <math>8</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in sich a way that the chords <math>QP</math> and <math>PR</math> have equal length. (<math>P</math> is the midpoint of <math>QR</math>) Find the square of the length of <math>QP</math>. | In the adjoining figure, two circles with radii <math>6</math> and <math>8</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in sich a way that the chords <math>QP</math> and <math>PR</math> have equal length. (<math>P</math> is the midpoint of <math>QR</math>) Find the square of the length of <math>QP</math>. | ||
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== Solution == | == Solution == |
Revision as of 22:40, 23 December 2006
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