Four circles tangent to each other and an equilateral triangle

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.

• Welcome to Math.SE! Please help us help you by telling us what you know about the problem, and where exactly you get stuck. This way, we can target responses to your skill set and avoid wasting time telling you things you already know. As for this problem: Are you aware of where the centroid of a triangle is located along a median? – Blue Mar 23 '18 at 5:43

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.

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}

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$}$