AIME 1983 · 第 14 题

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)

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)

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

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

Solution 11 (Quick)

Let A and B denote the centers of the left and right circles respectively and $QP = PR = x.$ Observe that $\Delta AQP$ and $\Delta AQR$ share a common angle $\angle{AQP}.$ Using the law of cosines on both triangles, we get:

$$(1)\quad 64=x^2+64-16x\cos{\angle{AQP}}$$ $$(2)\quad 324=4x^2+64-32x\cos{\angle{AQP}}$$ Subtracting $2*(1)$ from $(2),$ we get $196=2x^2-64,$ which gives $x^2=\boxed{130}$

~paalu