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

Revision as of 16:48, 28 August 2016

Problem

In the adjoining figure, two circles with radii $8$ and $6$ 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. Find the square of the length of $QP$ .

$[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]$

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 (easiest and quickest)

$[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$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 $x^2=130$ .

@@ Line 15: / Line 15: @@
 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 (easiest and quickest)===
 <asy>
 size(0,5cm);

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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