1983 AIME Problems/Problem 14

Revision as of 16:10, 8 August 2017 by Mamis511 (talk | contribs) (Solution 4(The actual quickest solution))

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)

[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 $x^2=130$.

Solution 4 (quickest)

Let $QP = PR = x.$ 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, \[x \cdot 2x = 10 \cdot 26.\] \[x^2 = 130.\]

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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