Difference between revisions of "1983 AIME Problems/Problem 14"
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=== Solution 4 === | === Solution 4 === | ||
− | Let the circles of radius <math>8</math> and <math>6</math> be centered at <math>A</math> and <math>B,</math> respectively. Let the midpoints of <math>QP</math> and <math>PR</math> be <math>N</math> and <math>O.</math> Dropping a perpendicular from <math>B</math> to <math>AN</math> (let the point be <math>K | + | Let the circles of radius <math>8</math> and <math>6</math> be centered at <math>A</math> and <math>B,</math> respectively. Let the midpoints of <math>QP</math> and <math>PR</math> be <math>N</math> and <math>O.</math> Dropping a perpendicular from <math>B</math> to <math>AN</math> (let the point be <math>K</math>) gives a rectangle. |
Now note that triangle <math>ABK</math> is right. Let the midpoint of <math>AB</math> (segment of length <math>12</math>) be <math>M.</math> Hence, <math>KM = 6 = BM = BP.</math> | Now note that triangle <math>ABK</math> is right. Let the midpoint of <math>AB</math> (segment of length <math>12</math>) be <math>M.</math> Hence, <math>KM = 6 = BM = BP.</math> | ||
By now obvious [[similar triangles]], <math>3BO = 3KN = AN,</math> so it's a quick system of two linear equations to solve for the desired length. | By now obvious [[similar triangles]], <math>3BO = 3KN = AN,</math> so it's a quick system of two linear equations to solve for the desired length. | ||
+ | |||
+ | {{incomplete|solution}} | ||
== See also == | == See also == |
Revision as of 22:47, 11 March 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
[hide]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 midpoints 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 |