# 2021 AIME I Problems/Problem 13

## Problem

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$.

Let $D\equiv\omega\cap\omega_{1}$ and $E\equiv\omega\cap\omega_{2}$. The solution relies on the following key claim:

Claim. $DPEQ$ is a harmonic quadrilateral.

Proof. Using the radical axis theorem, the tangents to circle $\omega$ at $D$ and $E$ are concurrent with line $\overline{ABPQ}$ at the radical center, implying that our claim is true by harmonic quadrilateral properties. $\blacksquare$

Thus, we deduce the tangents to $\omega$ at $P$ and $Q$ are concurrent at $F$ with line $\overline{DE}$. Denote by $O_{1}, O_{2}, O$ the centers of $\omega_{1}, \omega_{2}, \omega$ respectively.

Now suppose $G\equiv\overline{FDE}\cap\overline{O_{1}O_{2}}$. Note that $\overline{OF}\parallel\overline{O_{1}O_{2}}$ giving us the pairs of similar triangles $$\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_{2}GE\sim\triangle OFE.$$ We thereby obtain $$\dfrac{O_{1}G}{O_{1}D}=\dfrac{OF}{OD}=\dfrac{r\sec\tfrac{120^{\circ}}{2}}{r}=\sec 60^{\circ}=2$$ since $\widehat{PQ}=120^{\circ}$, where $r$ denotes the radius of $\omega$, and $$\dfrac{O_{2}G}{O_{2}E}=\dfrac{OF}{OE}=\sec 60^{\circ}=2$$ as well. It follows that $$O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.$$

Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.

Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

## Solution 2 (Linearity)

Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\omega_{1}$, $\omega_{2}$, and $\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$. Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by $$f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).$$ Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity, $$\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.$$ Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$, thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\cdot(961-625)=\boxed{672}$.

## Solution 3

Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that$$\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("O_1",origin,N); label("O_2",(d,0),N); label("O",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("X",T,N,gray(0.6)); label("Y",foot(X[0],O1,O2),NE,gray(0.6)); label("\ell",(O+Tp)/2,S,gray(0.6)); [/asy]$

Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired.

## Solution 4 (Quick)

Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("A",A,dir(A-O2/2)); dot("B",B,dir(B-O2/2)); dot("O_2",O2,right+up); dot("O_1",O1,left+up); dot("O",O,dir(O-O2)); dot("D",D,dir(170)); dot("E",E,dir(E-O1)); dot("X",X,dir(X-D)); dot("Y",Y,dir(Y-D)); label("R",O--E,right+up,p); label("R",O--D,left+down,p); label("2R",(X+O)/2-(150,0),down,p); label("961",O1--D,2*(left+down),p); label("625",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy]$ By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus$$O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}$$by similar triangles.

~mathman3880

## Solution 5 (Official MAA)

Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.

Applying the Law of Cosines to $\triangle O_{2}PT$, $\triangle O_{1}PT$, $\triangle O_{2}OT$, and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}

Adding the first and fourth equations, then subtracting the second and third equations gives us $$\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)$$

Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$, the power of point $P$ with respect to either circle is $$O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.$$

Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to $$u+v=2(R_{1}-R_{2}).$$

The requested distance $$O_{1}O_{2}=O_{1}T+O_{2}T=u+v$$ is therefore equal to $2\cdot(961-625)=\boxed{672}$.

## Solution 6 (Geometry)

Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$

Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$

In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median

$\hspace{10mm} \implies O,T,$ and $M$ are collinear.

$$\angle OCT = \angle ODT = \angle OET = 90^\circ \implies$$

$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ $$OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies$$ $$\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies$$ $$\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.$$

Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\cdot|961-625|=\boxed{672}$.

## Solution 7

We are not given the radius of circle $w$, but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$, call it $W$, and $w_2$, call it $W_2$, is perpendicular to the line containing the center of $w_1$, call it $W_1$ and $w_2$. Let $AB = 2h$ and $W_1W_2 = x$. Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$, respectively.

By Pythagorean Theorem on $\triangle WW_1W_2$, we get $$x^2+(625+2x)^2=(961+2r)^2 \;(1)$$ Note that since $\angle PWQ = 120$, $\angle PWX = 60$. So, $WX = 2r/2 = r = W_1Y$. We now get two more equations from Pythag: $$h^2+r^2 = 625^2 \; (2)$$ $$h^2+(x-r)^2 = 961^2 \; (3)$$ From subtracting $(2)$ and $(3)$, $x^2-2rx=961^2-625^2 \; (4)$. Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$. Plugging in our result from $(4)$, $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$.

~sml1809

## Solution 8 (Cheese)

Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$, $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$.

Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$.

Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$

picture by afly

~r00tsOfUnity

## Video Solution

~Math Problem Solving Skills

## Video Solution

Who wanted to see animated video solutions can see this. I found this really helpful.

P.S: This video is not made by me. And solution is same like below solutions.

≈@rounak138
The text passage contains a mathematical problem and multiple solutions to that problem. The problem is about finding the distance between the centers of two circles given certain conditions, and the solutions involve geometric and algebraic methods.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 2021 AIME I Problems/Problem 13

## Problem

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$.

Let $D\equiv\omega\cap\omega_{1}$ and $E\equiv\omega\cap\omega_{2}$. The solution relies on the following key claim:

Claim. $DPEQ$ is a harmonic quadrilateral.

Proof. Using the radical axis theorem, the tangents to circle $\omega$ at $D$ and $E$ are concurrent with line $\overline{ABPQ}$ at the radical center, implying that our claim is true by harmonic quadrilateral properties. $\blacksquare$

Thus, we deduce the tangents to $\omega$ at $P$ and $Q$ are concurrent at $F$ with line $\overline{DE}$. Denote by $O_{1}, O_{2}, O$ the centers of $\omega_{1}, \omega_{2}, \omega$ respectively.

Now suppose $G\equiv\overline{FDE}\cap\overline{O_{1}O_{2}}$. Note that $\overline{OF}\parallel\overline{O_{1}O_{2}}$ giving us the pairs of similar triangles $$\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_{2}GE\sim\triangle OFE.$$ We thereby obtain $$\dfrac{O_{1}G}{O_{1}D}=\dfrac{OF}{OD}=\dfrac{r\sec\tfrac{120^{\circ}}{2}}{r}=\sec 60^{\circ}=2$$ since $\widehat{PQ}=120^{\circ}$, where $r$ denotes the radius of $\omega$, and $$\dfrac{O_{2}G}{O_{2}E}=\dfrac{OF}{OE}=\sec 60^{\circ}=2$$ as well. It follows that $$O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.$$

Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.

Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

## Solution 2 (Linearity)

Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\omega_{1}$, $\omega_{2}$, and $\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$. Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by $$f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).$$ Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity, $$\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.$$ Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$, thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\cdot(961-625)=\boxed{672}$.

## Solution 3

Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that$$\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("O_1",origin,N); label("O_2",(d,0),N); label("O",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("X",T,N,gray(0.6)); label("Y",foot(X[0],O1,O2),NE,gray(0.6)); label("\ell",(O+Tp)/2,S,gray(0.6)); [/asy]$

Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired.

## Solution 4 (Quick)

Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("A",A,dir(A-O2/2)); dot("B",B,dir(B-O2/2)); dot("O_2",O2,right+up); dot("O_1",O1,left+up); dot("O",O,dir(O-O2)); dot("D",D,dir(170)); dot("E",E,dir(E-O1)); dot("X",X,dir(X-D)); dot("Y",Y,dir(Y-D)); label("R",O--E,right+up,p); label("R",O--D,left+down,p); label("2R",(X+O)/2-(150,0),down,p); label("961",O1--D,2*(left+down),p); label("625",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy]$ By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus$$O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}$$by similar triangles.

~mathman3880

## Solution 5 (Official MAA)

Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.

Applying the Law of Cosines to $\triangle O_{2}PT$, $\triangle O_{1}PT$, $\triangle O_{2}OT$, and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}

Adding the first and fourth equations, then subtracting the second and third equations gives us $$\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)$$

Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$, the power of point $P$ with respect to either circle is $$O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.$$

Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to $$u+v=2(R_{1}-R_{2}).$$

The requested distance $$O_{1}O_{2}=O_{1}T+O_{2}T=u+v$$ is therefore equal to $2\cdot(961-625)=\boxed{672}$.

## Solution 6 (Geometry)

Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$

Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$

In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median

$\hspace{10mm} \implies O,T,$ and $M$ are collinear.

$$\angle OCT = \angle ODT = \angle OET = 90^\circ \implies$$

$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ $$OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies$$ $$\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies$$ $$\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.$$

Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\cdot|961-625|=\boxed{672}$.

## Solution 7

We are not given the radius of circle $w$, but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$, call it $W$, and $w_2$, call it $W_2$, is perpendicular to the line containing the center of $w_1$, call it $W_1$ and $w_2$. Let $AB = 2h$ and $W_1W_2 = x$. Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$, respectively.

By Pythagorean Theorem on $\triangle WW_1W_2$, we get $$x^2+(625+2x)^2=(961+2r)^2 \;(1)$$ Note that since $\angle PWQ = 120$, $\angle PWX = 60$. So, $WX = 2r/2 = r = W_1Y$. We now get two more equations from Pythag: $$h^2+r^2 = 625^2 \; (2)$$ $$h^2+(x-r)^2 = 961^2 \; (3)$$ From subtracting $(2)$ and $(3)$, $x^2-2rx=961^2-625^2 \; (4)$. Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$. Plugging in our result from $(4)$, $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$.

~sml1809

## Solution 8 (Cheese)

Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$, $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$.

Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$.

Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$

picture by afly

~r00tsOfUnity

## Video Solution

~Math Problem Solving Skills

## Video Solution

Who wanted to see animated video solutions can see this. I found this really helpful.

P.S: This video is not made by me. And solution is same like below solutions.

≈@rounak138
user
The text passage contains a mathematical problem and multiple solutions to that problem. The problem is about finding the distance between the centers of two circles given certain conditions, and the solutions involve geometric and algebraic methods.

TYPE: SINGLE_QUESTION_WITH_ANSWER