Difference between revisions of "1983 AIME Problems/Problem 14"
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Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | ||
=== Solution 2 === | === Solution 2 === | ||
− | + | <asy> | |
+ | size(0,5cm); | ||
+ | pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); | ||
+ | draw(b--r--n--b--a--m--n); | ||
+ | draw(a--q--m); | ||
+ | draw(circumcircle(origin,q,p)); | ||
+ | draw(circumcircle((14,0),p,r)); | ||
+ | draw(rightanglemark(a,m,n,24)); | ||
+ | draw(rightanglemark(b,n,r,24)); | ||
+ | label("$A$",a,S); | ||
+ | label("$B$",b,S); | ||
+ | label("$M$",m,NE); | ||
+ | label("$N$",n,NE); | ||
+ | label("$P$",p,N); | ||
+ | label("$Q$",q,NW); | ||
+ | label("$R$",r,E); | ||
+ | label("$12$",(14,0),SW); | ||
+ | label("$6$",(23,0),S); | ||
+ | </asy> | ||
− | + | Draw additional lines as indicated. Note that since triangles <math>AQP</math> and <math>BPR</math> are isosceles, the altitudes are also bisectors, so let <math>QM=MP=PN=NR=x</math>. | |
+ | Since <math>\frac{AR}{MR}=\frac{BR}{NR},</math> triangles <math>BNR</math> and <math>AMR</math> are similar. If we let <math>y=BN</math>, we have <math>AM=3BN=3y</math>. | ||
+ | |||
+ | Applying the Pythagorean Theorem on triangle <math>BNR</math>, we have <math>x^2+y^2=36</math>. Similarly, for triangle <math>QMA</math>, we have <math>x^2+9y^2=64</math>. | ||
+ | |||
+ | Subtracting, <math>8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}</math>. | ||
=== 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 19:15, 6 February 2011
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. (
is the midpoint of
) 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 |