Bubbles in a web

From symmetry, the internal triangle equilateral. Let’s for now set the side of it $BC=1$ . The radius of the inscribed circle will be $R=\frac{1}{2\sqrt{3}}$ .

The three triangles on the outside are all congruent. We will get an expression for the radius $R$ of the inscribed circle from $R\times s=A$ where $s$ is half perimeter and $A$ is area.

We can introduce $x=AB=CD$ . This will give us the area as $A=\frac{1}{2}x(1+x)\times sin(120^\circ)=\frac{\sqrt{3}}{4}x(x+1)$ .

To get circumference we need $y$ , which can be obtained from the law of cosines as $y=\sqrt{x^2+(1+x)^2-\frac{1}{2}x(x+1)\times cos(120^\circ)}$ .

$R\times s=A$ will now become $\frac{1}{\sqrt{3}}\times (x+(x+1)+y)= \frac{\sqrt{3}}{4}x(x+1)$

This has only one applicable solution, namely $x=0.93043$ , which corresponds to $y=AD=2.5275$ .

We were supposed to have $y=1$ , so all the results need to be scaled down to that, with the radius becoming $R=\frac{1}{2\sqrt{3}}\times\frac{1}{y}=0.1142$