Difference between revisions of "1983 AIME Problems/Problem 14"
m (→Solution 4 (quickest)) |
Recurrence (talk | contribs) (→Solution) |
||
Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
− | Note that some of these solutions assume that <math>R</math> lies on the line connecting the centers, which is not true in general. | + | Note that some of these solutions assume that <math>R</math> lies on the line connecting the centers, which is not true in general. It's true here cause the perpendicular from <math>P</math> hits the point where the line between the centers and the small circle intersect. That's true if you apply midpoint theorem to the trapezoid made by dropping perpendiculars from the centers on to QR. |
=== Solution 1 === | === Solution 1 === |
Revision as of 21:56, 3 January 2019
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's true here cause the perpendicular from
hits the point where the line between the centers and the small circle intersect. That's true if you apply midpoint theorem to the trapezoid made by dropping perpendiculars from the centers on to QR.
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
![$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$](http://latex.artofproblemsolving.com/6/d/2/6d2150ebb9706e5687fb702d57b9724260acea68.png)
Doing routine algebra on the above 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
![$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$](http://latex.artofproblemsolving.com/4/c/6/4c6cea3db0103d0c0397ce9207b3afc1f0cd795e.png)
Taking the of both sides and simplifying using the cosine addition identity gives
.
Solution 4 (quickest)
Let Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle.
The line segment consisting of R and the first intersection of the larger circle has length 10. The length of the diameter of the larger circle be16.
Through power of a point,
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 |