Jtdylktuy

How do i even start this question? I thought of using length of tangent equal from exterior point but still of no use.

Let $G$ be the common centroid for the two triangles.

Pick an edge, say $AB$, let $H$ be the vertex on the small triangle which is the
apex for the circular sector touching edge $AB$. Let $G'$ and $H'$ be the foot of
$G$ and $H$ on edge $AB$.

It is clear the distance of $G$ to line $AB$ is

$$|GG'| = |HH'| - |GH|cos(theta + frac{pi}{6})$$

where $theta$ is the angle illustrated in above diagram.

We know $|HH'| = 20$, the radius of the circular sectors. Since the small triangle has side
$9$, we also know $|GH| = 3sqrt{3}$. In the diagram above, we find

$$costheta = frac{|HH'|}{|DH|} = frac{20}{29}quadimpliesquadsintheta = frac{21}{29}$$
Combine these, we get

$$begin{align}|GG'|
&= 20 - 3sqrt{3}left(costhetacosfrac{pi}{6} - sinthetasinfrac{pi}{6}right)\
&= 20 - 3sqrt{3}left(frac{20}{29}cdotfrac{sqrt{3}}{2} - frac{21}{29}cdotfrac{1}{2}right)\
&= frac{980+63sqrt{3}}{58}
approx 18.77791725649723
end{align}
$$

As a corollary,

$$|AB| = |BC| = |CA| = 2sqrt{3}|GG'| = frac{189+980sqrt{3}}{29} approx 65.04861349715516$$

Slide the third circles along the sides they tangent to - you will see that they merge into the inscribed circle. So the radius of the inscribed circle is 20.

The side of the triangle is than - $AB=2times sqrt{3}times 20 $