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1983 AIME Problems/Problem 14

Revision as of 21:56, 3 January 2019 by Recurrence (talk | contribs) (→‎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

Note that some of these solutions assume that $R$ lies on the line connecting the centers, which is not true in general. It's true here cause the perpendicular from $P$ hits the point where the line between the centers and the small circle intersect. This can be derived from the application of midpoint theorem to the trapezoid made by dropping perpendiculars from the centers on to QR.

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 (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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