This solution is from Yatir Halevi.

Let $a,\hspace{1em}b$ and $c$ be the radii the circles centered at $A,\hspace{1em}B$ and $C$ respectively.

Angle $\mathrm{BOC}={90}^0$ since, $\mathrm{OB}$ is tangent to the circle centered at $C$ and hence it is perpendicular to the radius. That also means that the radius of the the circle centered at $O$ is $2c$.

As $\mathrm{BC}$ connects $B$ and $C$ it is equal to $b+c$.

By Pythagoras theorem: $(2c-b){}^2+c^2=(b+c){}^2$. Hence $4c^2=6\mathrm{bc}$ and $c$ can't be equal to 0 so $2c=3b$.

We could extend the lines $\mathrm{AC}$ and $\mathrm{BO}$, to create a mirror image, where we have two circles centered at $A\text{'}$ and $B\text{'}$ with radii $a\text{'}$ and $b\text{'}$ respectively, and due to symmetry issues, $a\text{'}=a$ and $b\text{'}=b$. We will also connect $O$ with $A\text{'}$ and $O$ with $A$.

Now $\mathrm{OA}=\mathrm{OA}\text{'}=2c-a$ and $\mathrm{AC}=A\text{'}C=c+a$ and so the line $\mathrm{OC}$ cuts $A\text{'}A$ in half, and because the sides of the triangle $A\text{'}\mathrm{OA}$ are equal, $\mathrm{OC}$ is also perpendicular to $A\text{'}A$, hence $\angle \mathrm{OCA}=90$.

As triangle $\mathrm{ACO}$ is a right-angled triangle, using Pythagoras' Theorem: ${\mathrm{AC}}^2+{\mathrm{OC}}^2={\mathrm{AO}}^2$ and so

${\displaystyle (a+c){}^2+c^2=(2c-a){}^2}$

which simplifies to $6\mathrm{ac}=2c^2$ and $c$ can't be equal to 0 so $c=3a$. Combining this with $2c=3b$ we get: $3b=6a$ that is $b=2a$ and so $a=b/2=c/3$, that is they are in the ratio 1:2:3 as required.

If we mirror the image in the line $\mathrm{BO}$ thus creating a new circle below the one centered at $B$, centered at $A\text{'}\text{'}$ it has the same radius as $A$ by symmetry. By the the same line of thought that we used before, we reach the conclusion that $\angle \mathrm{ABO}=90$. What we get is that $\mathrm{CABO}$ has 3 right angles, meaning that the fourth is also a right angle and that it is a rectangle and that triangle $\mathrm{ABC}$ is also right angled.

Let's look at $\mathrm{triangle}\mathrm{OBC}$ in which $\mathrm{OB}=2c-b=6a-2a=4a$, $\mathrm{OC}=c=3a$ and $\mathrm{BC}=b+c=2a+3a=5a.$ So the ratio between the sides are 3:4:5, like we wanted to prove.

Let's look at $\mathrm{triangle}\mathrm{ABC}$ in which $\mathrm{AB}=a+b=a+2a=3a$, $\mathrm{AC}=a+c=a+3a=4a$ and $\mathrm{CB}=b+c=2a+3a=5a$. So the ratio between the sides are 3:4:5, like we wanted to prove.

Arun Iyer proved this result by applying Descartes Circle Theorem.