Finding the side length of an equilateral triangle having three inscribed $120^\circ$ sectors in a certain arrangement

Question

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

Answer 1

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=2\times \sqrt{3}\times 20 $

Answer 2

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 $|GH| = 3\sqrt{3}$. In the diagram above, we find

$$\cos\theta = \frac{|HH'|}{|DH|} = \frac{20}{29}\quad\implies\quad\sin\theta = \frac{21}{29}$$ Combine these, we get

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

As a corollary,

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