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Difference between revisions of "1983 AIME Problems/Problem 14"

(→‎Problem)
(Combined previously incomplete solutions 2 and 4 into a new solution utilizing ideas from both :)
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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 ===
−
This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoint of <math>QP</math> be <math>M_1</math>, and the midpoint of <math>PR</math> be <math>M_2</math>. Let <math>x</math> be the length of <math>AM_1</math>, and <math>y</math> that of <math>BM_2</math>.
+
<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>
  
−
{{incomplete|solution}}
+
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>.
−
 
−
=== 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</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>
 
−
 
−
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 18:15, 6 February 2011

Problem

In the adjoining figure, two circles with radii $6$ and $8$ are drawn with their centers $12$ units apart. At $P$, one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. ($P$ is the midpoint of $QR$) Find the square of the length of $QP$.

1983 AIME-14.png

Solution

Solution 1

First, notice that if we reflect $R$ over $P$ we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths:

Since $P$ is the midpoint of segment $BC$, $AP$ is a median of triangle $ABC$. Because we know that $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$. So now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then

$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$

Doing routine algebra on the above equation, we find that $x^2=\frac{65}{2}$, so $PQ^2 = 4x^2 = \boxed{130}.$

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 $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$.

Since $\frac{AR}{MR}=\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$.

Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$.

Subtracting, $8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}$.

Solution 3

Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\circ}$. By the Law of Cosines, $\angle APB=\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\cos^{-1}(x/16)$ and $\cos^{-1}(x/12)$. So we have

$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$

Taking the $\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$.

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
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