Difference between revisions of "1983 AIME Problems/Problem 14"
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=== Solution 3 === | === Solution 3 === | ||
Let <math>QP=PR=x</math>. Angles <math>QPA</math>, <math>APB</math>, and <math>BPR</math> must add up to <math>180^{\circ}</math>. By the [[Law of Cosines]], <math>\angle APB=\cos^{-1}(-11/24)</math>. Also, angles <math>QPA</math> and <math>BPR</math> equal <math>\cos^{-1}(x/16)</math> and <math>\cos^{-1}(x/12)</math>. So we have <center><math>\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).</math></center> Taking the <math>\cos</math> of both sides and simplifying using the cosine addition identity gives <math>x^2=130</math>. | Let <math>QP=PR=x</math>. Angles <math>QPA</math>, <math>APB</math>, and <math>BPR</math> must add up to <math>180^{\circ}</math>. By the [[Law of Cosines]], <math>\angle APB=\cos^{-1}(-11/24)</math>. Also, angles <math>QPA</math> and <math>BPR</math> equal <math>\cos^{-1}(x/16)</math> and <math>\cos^{-1}(x/12)</math>. So we have <center><math>\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).</math></center> Taking the <math>\cos</math> of both sides and simplifying using the cosine addition identity gives <math>x^2=130</math>. | ||
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== See Also == | == See Also == |
Revision as of 14:41, 25 January 2016
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
.
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
![$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$](http://latex.artofproblemsolving.com/6/d/2/6d2150ebb9706e5687fb702d57b9724260acea68.png)
Doing routine algebra on the above equation, we find that , so
Solution 2
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
.
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 |