1983 AIME Problems/Problem 14

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$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]$

Note

Note that some of these solutions assume that $R$ lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from $P$ passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto $QR$ .

Solution 1

Firstly, 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 $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as follows.

B is reflected like so

Since $P$ is the midpoint of segment $BC$ , $AP$ is a median of $\triangle ABC$ . Because we know $AB=12$ , $BP=PC=6$ , and $AP=8$ , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$ .

The Kite is formed

Now we have a kite $AQCP$ with $AQ=AP=8$ , $CQ=CP=6$ , and diagonal $AC=\sqrt{56}$ , 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

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

Solving this equation, we find that $x^2=\frac{65}{2}$ , so $PQ^2 = 4x^2 = \boxed{130}.$

~ $shalomkeshet$

Video Solution by Pi Academy (Easy)

https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D

~ Pi Academy

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 (trig bash)

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

$\cos^{-1}\left(\frac{x}{16}\right)+\cos^{-1}\left(\frac{{-11}}{24}\right)=180^{\circ}-\cos^{-1}\left(\frac{x}{12}\right).$

Taking the cosine of both sides, and simplifying using the addition formula for $\cos$ as well as the identity $\sin^{2}{x} + \cos^{2}{x} = 1$ , gives $x^2=\boxed{130}$ .

Solution 4 (quickest)

Let $QP = PR = x$ . Extend the line containing the centers of the two circles to meet $R$ , and to meet the other side of the large circle at a point $S$ .

The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$ . The length of the diameter of the larger circle is $16$ .

Thus by Power of a Point in the circle passing through $Q$ , $R$ , and $S$ , we have $x \cdot 2x = 10 \cdot (10+16) = 260$ , so $x^2 = \boxed{130}$ .

Solution 5 (Pythagorean Theorem and little algebraic manipulation)

$[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label("$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("$T$", t , NW); [/asy]$ Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ Using Stewart's Theorem on $\Delta APB,$ the median $PT$ has length $\sqrt{14}.$ Thus, $a+b=2\sqrt{14}.$ Also, since $MP=PN$ , from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\implies a^2-b^2=28.$ Thus, $a-b=\frac{28}{2\sqrt{14}}=\sqrt{14}.$ We conclude that $QP=MN=\sqrt{12^2-(a-b)^2}=\sqrt{130}\implies\boxed{130}.$ ~pinkpig

Solution 6 (Only simple geometry and algebra needed)

Looking at Drawing 2 (by the way, we don't need point $R$ ), we set $AM=a$ and $BN=b$ , and the desired length $QP=x=PR$ . We know that a radius perpendicular to a chord bisects the chord, so $MP=\frac{x}{2}$ and $PN=\frac{x}{2}$ . Draw line $AP$ and $PB$ , and we see that they are radii of Circles $A$ and $B$ , respectively. We can write the Pythagorean relationships $a^2+\left(\frac{x}{2}\right)^2=8^2$ for triangle $AMP$ and $b^2+\left(\frac{x}{2}\right)^2=6^2$ for triangle $BNP$ . We also translate segment $MN$ down so that $N$ coincides with $B$ , and form another right triangle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$ , the longer leg is the same as $MN=x$ , and the hypotenuse is $AB=12$ . We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$ . Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$ , and solve), you find that $x=\sqrt{130}$ , so $x^2 = \boxed{130}$ .

Solution by Kinglogic

Solution 7

The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from $P$ to $AB.$ You then have 2 separate segments, separated by the foot of the altitude of $P$ . Call them $a$ and $b$ respectively. Call the measure of the foot of the altitude of $P$ $h$ . You then have 3 equations:

$(1)\quad a+b=12$ (this is given by the fact that the distance between the centers is 12.

$(2)\quad a^2+h^2=64$ . This is given by the fact that P is on the circle with radius 8.

$(3)\quad b^2+h^2=36$ . This is given by the fact that P is on the circle with radius 6.

Subtract (3) from (2) to get that $a^2-b^2=28$ . As per (1), then you have $a-b=\frac{7}{3}$ (4). Add (1) and (4) to get that $2a=\frac{43}{3}$ . Then substitute into (1) to get $b=\frac{29}{6}$ . Substitute either a or b into (2) or (3) to get $h=\sqrt{455}{6}$ . Then to get $PQ=PR$ it is just $\sqrt{(b+6)^2+h^2}=\sqrt{\frac{65^2}{6^2}+\frac{455}{6^2}}=\sqrt{\frac{4680}{36}}=\sqrt{130}$ .

$PQ^2=\boxed{130}$

-dragoon

Full Proof that R, A, B are collinear

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

Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$ , since the problem told us $QP = PR$ .

We will show that $R$ lies on $AB$ .

Let $T$ be the intersection of circle centered at $B$ with $AB$ . Then $BT = TA = 6$ .

Let $P$ ' be the foot of the perpendicular from $T$ to $MN$ . Then $TP'$ is a midline (or midsegment) in trapezoid $AMNB$ , so $P'$ coincides with $P$ (they are both supposed to be the midpoint of $MN$ ). In other words, since $\angle TP'N = 90^\circ$ , then $\angle TPN = 90^\circ$ .

Thus, $\angle TPR$ subtends a $90^\circ \times 2 = 180^\circ$ degree arc. So arc $TR$ in circle $B$ is $180^\circ$ , so $TR$ is a diameter, as desired. Thus $A$ , $B$ , $R$ are collinear.

NOTE: Note this collinearity only follows from the fact that $6$ is half of $12$ in the problem statement. The collinearity is untrue in general.

Solution 8 (Coordinate Bash)

We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$ , the smaller circle be $O_2$ , and the x-axis be $O_1O_2$ . Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$ .

Let point $P$ be $(a, b)$ . Noting that it lies on both circles, we can plug the coordinates into both equations:

$a^2 +b^2 = 64$ $(a-12)^2+ b^2 \Rightarrow a^2-24a+144+b^2 = 64$

Substituting $a^2+b^2 = 64$ into equation 2 and solving for $a$ , we get $a = \frac{43}{6}$ . The problem asks us to find $QP^2$ , which is congruent to $PR^2$ . Using the distance formula for $P(a, b)$ and $R(18, 0)$ (by Solution 7's collinear proof), we get $PR^2 = (18-a)^2 +b^2$ . Using $a^2+b^2 = 64$ , we find that $b^2 = \frac{455}{36}$ . Plugging the variables $a$ and $b^2$ in, we get $PR^2 = QP^2 = \boxed{130}$ ~SoilMilk

Solution 9 (basic solution)

Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $\Delta APB,$ we have that $\cos{\angle{APB}} = -\frac{{11}}{24}.$ Angle chasing gives that $\angle{QAR} = \angle{APB},$ so their cosines are the same. Applying law of cosines again on triangle $\Delta QAR,$ we have $\left(2x^2\right)=64+324-2(8)(18)\left(-\frac{11}{24}\right),$ which gives that $x^2 = \boxed{130}$

~happypi31415

Solution 10 (Spiral Sym)

If you call X the second intersection of the two circles and the centers $O_1$ , $O_2$ respectively, note that triangles $O_1 X O_2$ and $Q X R$ are similar by Spiral Sym. $P$ is the midpoint of segment $Q R$ . If the midpoint of $O_1 O_2$ is $M$ , then $\frac{X M}{O_1 M} = \frac{X P}{Q P}$ . By Appolonius Median Length theorem, $X M = \sqrt{14}$ . Note that $X P$ is simply two times the height from X to $O_1O_2$ , and as a result, by Heron's formula, $X P = \frac{\sqrt{(7)(13)(5)}}{3}$ , and from our ratio, $Q P = \sqrt{130}$ . As a result, the square is $130$ , and we are done. - sepehr2010

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
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