text: # 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