AIME 2021 I · 第 13 题

Solution 1 (Radical Axis)

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 $O_1O_2=d$.

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,

$$\begin{aligned} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{aligned}$$ We want to solve for $d$. By the Pythagorean Theorem (twice):

$$\begin{aligned} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{aligned}$$ 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

$$\begin{aligned} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{aligned}$$ 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

$$\begin{aligned}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{aligned}$$ Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us

$$\begin{aligned}\frac{x}{2(r_{1}-r_{2})}&=1\end{aligned}$$ 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 $d = \boxed{672}$.

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

$$\begin{aligned}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{aligned}$$ 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.

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

$$\begin{aligned}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{aligned}$$ 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 $2\cdot|961-625|=\boxed{672}$.

vladimir.shelomovskii@gmail.com, vvsss

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

$$x^2+(625+2r)^2=(961+2r)^2 \;(1)$$ 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 $(1)$ 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

Video Solution by MOP 2024

https://youtu.be/GQT73xqvtXA

~r00tsOfUnity

Video Solution

https://youtu.be/gN7Ocu3D62M

~Math Problem Solving Skills

Video Solution

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

https://youtu.be/YtZ8_7i833E

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

≈@rounak138