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 such a way that the chords
and
have equal length. Find the square of the length of
.
Contents
Solution
Note that some of these solutions assume that lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from
passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto
.
Solution 1
Firstly, 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
, and with radius
) that intersects circle
at
. The rest is just finding lengths, as follows.
Since is the midpoint of segment
,
is a median of
. Because we know
,
, and
, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get
. 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
![$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$](http://latex.artofproblemsolving.com/6/d/2/6d2150ebb9706e5687fb702d57b9724260acea68.png)
Solving this equation, we find that , so
Solution 2 (Easiest)
Draw additional lines as indicated. Note that since triangles and
are isosceles, the altitudes are also bisectors, so let
.
Since triangles
and
are similar. If we let
, we have
.
Applying the Pythagorean Theorem on triangle , we have
. Similarly, for triangle
, we have
.
Subtracting, .
Solution 3
Let . Angles
,
, and
must add up to
. By the Law of Cosines,
. Also, angles
and
equal
and
. So we have
![$\cos^{-1}\left(\frac{x}{16}\right)+\cos^{-1}\left(\frac{{-11}}{24}\right)=180^{\circ}-\cos^{-1}\left(\frac{x}{12}\right).$](http://latex.artofproblemsolving.com/0/7/5/0754c4f6c7144ced3c35f0a8ea07c99e1639a6c9.png)
Taking the cosine of both sides, and simplifying using the addition formula for as well as the identity
, gives
.
Solution 4 (quickest)
Let . Extend the line containing the centers of the two circles to meet
, and to meet the other side of the large circle at a point
.
The part of this line from to the point nearest to
where it intersects the larger circle has length
.
The length of the diameter of the larger circle is
.
Thus by Power of a Point in the circle passing through ,
, and
, we have
, so
.
Solution 5 (Pythagorean Theorem and little algebraic manipulation)
Note that the midpoint of
is
Also, since
bisect
and
respectively,
is the midpoint of
Thus,
let
This means that
From the median formula,
Thus,
Also, since
, from the Pythagorean Theorem,
Thus,
We conclude that
~pinkpig
Solution 6 (Only simple geometry and algebra needed)
Looking at Drawing 2 (by the way, we don't need point ), we set
and
, and the desired length
. We know that a radius perpendicular to a chord bisects the chord, so
and
. Draw line
and
, and we see that they are radii of Circles
and
, respectively. We can write the Pythagorean relationships
for triangle
and
for triangle
. We also translate segment
down so that
coincides with
, and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length
, the longer leg is the same as
, and the hypotenuse is
. We can write the Pythagorean relationship
. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for
in the first equation and
in the second equation, and substitute into the third equation, get an equation only in terms of
, and solve), you find that
, so
.
Solution by Kinglogic
Full Proof that R, A, B are collinear
Let and
be the feet of the perpendicular from
to
and
to
respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so
, since the problem told us
.
We will show that lies on
.
Let be the intersection of circle centered at
with
. Then
.
Let ' be the foot of the perpendicular from
to
. Then
is a midline (or midsegment) in trapezoid
, so
coincides with
(they are both supposed to be the midpoint of
). In other words, since
, then
.
Thus, subtends a
degree arc. So arc
in circle
is
, so
is a diameter, as desired. Thus
,
,
are collinear.
NOTE: Note this collinearity only follows from the fact that is half of
in the problem statement. The collinearity is untrue in general.
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 |