A geometric problem, part II

First we shall find the relationship between the radius of a circle and the length of the side.

Let the radius be $r$ , then the length of the side would be given by $s=2\sqrt{3}r$ .

Now, observe that the radius of a circle decreases with a common ratio (from circle 1 to circle 2, from circle 2 to circle 3 etc). Our objective is to find this common ratio.

Let us consider a special case, when the biggest circle has radius 1. Then the equilateral triangle has side $2\sqrt{3}$ and the distance from a vertex to the centre of the circle is 2. Also, the height of the triangle is 3. Hence, we have this infinite GP:

$2\left(1+\frac{1}{r}+\frac{1}{r^2}+\cdots\right)=3\\1+\frac{1}{r}+\frac{1}{r^2}+\cdots=\frac{3}{2}\\\frac{1}{r-1}=\frac{3}{2}\\r=3$

Hence, we can conclude that each subsequent circle's radius will be one third of the radius of the previous circle.

Now let us go back to the question. Circle 2 has radius 1, so circle 1 has radius 3. The total sum of areas is then: $9\pi+3\pi\left(1^2+\frac{1}{3}^2+\frac{1}{3^2}^2+\cdots\right)=\frac{99}{8}\pi$

$r_2=1 \\r_{n+1}=\frac{r_n}{3} \\A_n=\pi r_n^2 \\ A_T=A_1+3\sum_{k=2}^{\infty} A_k=\frac{99}{8} \pi\ \\99+8=107$

Each circle has 1/3 the radius of the last, so the area decreases at a rate of 1/9. The first circle (ignoring the center one) has an area of pi. We know that the value of an infinite sum is the first term divided by the difference of 1 and the rate. Thus, the area in one leg is pi/(8/9). Multiply by 3, because this sum is only for the circles going to one angle, and add the area of the largest circle.