Four circles tangent to each other and an equilateral triangle

Question

Given that three congruent circles of radius $4$ and one large circle of unknown radius are inscribed in equilateral triangle $ABC$, and all of the circles are tangent to each other and the triangle.

1) What is the side length of the equilateral triangle?

2) What is the radius of circle $O$ in the figure below?

Some context:

I have tried to construct a 30-60-90 triangle from the tangent of circle $R$ to find that $RC$ = 8 and by similarity finding the lengths of triangle $COX$ (where $X$ is the point of tangency of circle $O$ on the side $BC$ of triangle $ABC$) but I cannot seem to find the radius of circle $O$ or the length of $CX$.

I know that the centroid of an equilateral triangle is always the center (this is also the incenter and the circumcenter), and the height of an equilateral triangle is $\frac{s{\sqrt 3}}{2}$ and the area is $\frac{{\sqrt 3}}{4}{s^2}$.

I figured that finding the side length of the equilateral triangle would involve finding the altitude that passes through points $O$, $R$, and $C$, and that the radius of circle $R$ + the diameter of circle $O$ + $RC$ but I didn't have a good method to find the diameter of circle $O$.

I also understand that circles $P$, $R$, and $Q$ form circumscribed angles at vertices $A$, $C$, and $B$, respectively, but I was unsure of how that helped to solve the problem.

Answer 1

Circle with the center $O$ is inscribed in the equilateral triangle $ABC$ with the height $CF$ and inradius $OF$, circle with the center $R$ is inscribed in the equilateral triangle $DEC$ with the height $CG$ and inradius $RG$,

\begin{align} |CF|&=3|CG| , \end{align}
hence, the radius

\begin{align} |OF|&=3\cdot|RG|=3\cdot4=12 , \end{align}

and the side length of $\triangle ABC$ \begin{align} |AC|&= \frac{|CF|}{\tfrac {\sqrt3}2} =\frac{3\cdot|OF|}{\tfrac {\sqrt3}2} =2\sqrt3\cdot|OF|=24\sqrt3 . \end{align}

Answer 2

The radius of an inscribed circle of an equilateral triangle is always: $\frac{\sqrt3}6s$, where $s$ is the length of the side of the triangle.

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To get that, we suppose a small triangle with a base $\frac12s$ and height $r$, which is the radius of the incircle, the hypotenuse therefore bisects one of the angles of the equilateral triangle, and thus: $$\begin{align} \sin 30&=\frac{r}{\mathbb{hyp}}\\ \mathbb{hyp}&=\frac{r}{\sin30}=2r\\ \end{align}$$ We know for a fact that, by the Pythagorean theorem: $$\mathbb{hyp}=\sqrt{r^2+\frac14s^2}$$ and thus: $$\begin{align} \sqrt{r^2+\frac14s^2}&=2r\\ r^2+\frac{s^2}4&=4r^2\Rightarrow \frac{s^2}4=3r^2\\\\ \frac s{2\sqrt3}&=r=\frac{s\sqrt3}6\\ \end{align}$$

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You'll find that the height of an equilateral triangle is $h=\frac{\sqrt3}2s$, and you'll notice that $OC\perp AB$ and that $R$ lies on $OC$. To find the height, you get this equation: $$h=2\cdot\mathbb{\text{radius of circle $O$}}+2\cdot\text{radius of circle $R$}+RC\tag{1}$$

Even without special knowledge, you'll find from this figure that $\angle CDB=180^\circ-\angle CAB$.

Knowing that tangent lines are perpendicular to a line drawn from the radius of a circle, we basically have a triangle and thus $RC$ is a hypotenuse of that triangle, which is now solvable by: $$\sin 30=\frac4{RC}\Rightarrow RC=8$$ And thus, from $(1)$ we get the side $s$ of $ABC$ as:

$$\begin{align} h=&2\big(\frac{\sqrt3}6s\big)+4+8\\ \frac{\sqrt3}2s=&2\big(\frac{\sqrt3}6s\big)+4+8\\ s=&24\sqrt3 \end{align}$$

So now we get the answers:

$\text{ 1. $s=24\sqrt3$} $

$ \text{ 2. $r=\frac{\sqrt3}6s\Rightarrow 12$} $