A circle is inscribed in an equilateral triangle. Three nested sequences of circles are then constructed as follows: each circle touches the previous circle and has two edges of the triangle as tangents. This is represented by the figure below.

[asy]
import olympiad;

pair A, B, C;
A = dir(90);
B = dir(210);
C = dir(330);
draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(incircle(A,2/3*A+1/3*B,2/3*A+1/3*C));
draw(incircle(A,8/9*A+1/9*B,8/9*A+1/9*C));
draw(incircle(A,26/27*A+1/27*B,26/27*A+1/27*C));
draw(incircle(A,80/81*A+1/81*B,80/81*A+1/81*C));
draw(incircle(A,242/243*A+1/243*B,242/243*A+1/243*C));
draw(incircle(B,2/3*B+1/3*A,2/3*B+1/3*C));
draw(incircle(B,8/9*B+1/9*A,8/9*B+1/9*C));
draw(incircle(B,26/27*B+1/27*A,26/27*B+1/27*C));
draw(incircle(B,80/81*B+1/81*A,80/81*B+1/81*C));
draw(incircle(B,242/243*B+1/243*A,242/243*B+1/243*C));
draw(incircle(C,2/3*C+1/3*B,2/3*C+1/3*A));
draw(incircle(C,8/9*C+1/9*B,8/9*C+1/9*A));
draw(incircle(C,26/27*C+1/27*B,26/27*C+1/27*A));
draw(incircle(C,80/81*C+1/81*B,80/81*C+1/81*A));
draw(incircle(C,242/243*C+1/243*B,242/243*C+1/243*A));
[/asy]

What is the ratio of the area of the largest circle to the combined area of all the other circles?

$\text{(A) }\frac{8}{1}\qquad\text{(B) }\frac{8}{3}\qquad\text{(C) }\frac{9}{1}\qquad\text{(D) }\frac{9}{3}\qquad\text{(E) }\frac{10}{3}$
1. **Determine the inradius of the equilateral triangle:**

   Let the side length of the equilateral triangle be \( s \). The area \( A \) of the equilateral triangle is given by:
   \[
   A = \frac{s^2 \sqrt{3}}{4}
   \]
   The semiperimeter \( \frac{3s}{2} \) and the inradius \( r \) are related by:
   \[
   A = \frac{1}{2} \times \text{semiperimeter} \times \text{inradius} \Rightarrow \frac{s^2 \sqrt{3}}{4} = \frac{3s}{2} \times r
   \]
   Solving for \( r \):
   \[
   r = \frac{s \sqrt{3}}{6}
   \]

2. **Determine the radius of the smaller inscribed circles:**

   The radius of the next smaller circle, which touches the previous circle and has two edges of the triangle as tangents, is scaled down by a factor of 3. Therefore, the radius of the smaller circle is:
   \[
   r_1 = \frac{r}{3} = \frac{s \sqrt{3}}{18}
   \]

3. **Calculate the ratio of the areas of the circles:**

   The area of a circle is proportional to the square of its radius. Therefore, the ratio of the areas of the largest circle to the next smaller circle is:
   \[
   \left( \frac{r}{r_1} \right)^2 = \left( \frac{\frac{s \sqrt{3}}{6}}{\frac{s \sqrt{3}}{18}} \right)^2 = 3^2 = 9
   \]

4. **Sum the areas of all the smaller circles:**

   The sequence of smaller circles extends infinitely in 3 different directions. Let \( A \) be the area of the largest circle and \( X \) be the combined area of all the smaller circles. The area of the largest circle is:
   \[
   A = \pi r^2 = \pi \left( \frac{s \sqrt{3}}{6} \right)^2 = \frac{\pi s^2}{12}
   \]
   The area of the first smaller circle is:
   \[
   A_1 = \pi r_1^2 = \pi \left( \frac{s \sqrt{3}}{18} \right)^2 = \frac{\pi s^2}{108}
   \]
   The total area of the smaller circles in one direction is a geometric series:
   \[
   \sum_{k=1}^{\infty} \left( \frac{1}{9} \right)^k \cdot A_1 = A_1 \sum_{k=1}^{\infty} \left( \frac{1}{9} \right)^k = \frac{A_1}{8} = \frac{\frac{\pi s^2}{108}}{8} = \frac{\pi s^2}{864}
   \]
   Since there are 3 such sequences:
   \[
   X = 3 \cdot \frac{\pi s^2}{864} = \frac{\pi s^2}{288}
   \]

5. **Calculate the ratio of the area of the largest circle to the combined area of all the smaller circles:**

   \[
   \frac{A}{X} = \frac{\frac{\pi s^2}{12}}{\frac{\pi s^2}{288}} = \frac{288}{12} = 24
   \]

   However, this result seems incorrect. Let's re-evaluate the geometric series sum correctly:
   \[
   X = 3 \cdot \sum_{k=1}^{\infty} \frac{1}{12} \pi \left( \frac{1}{9} \right)^k = 3 \cdot \frac{1}{12} \pi \cdot \frac{\frac{1}{9}}{1 - \frac{1}{9}} = 3 \cdot \frac{1}{12} \pi \cdot \frac{1}{8} = \frac{1}{32} \pi
   \]
   The area of the largest circle is:
   \[
   A = \frac{1}{12} \pi
   \]
   Therefore, the ratio is:
   \[
   \frac{A}{X} = \frac{\frac{1}{12} \pi}{\frac{1}{32} \pi} = \frac{32}{12} = \frac{8}{3}
   \]

The final answer is \(\boxed{\frac{8}{3}}\)