Circle, triangle, circle, triangle, etc

I've got a soft spot for equilateral triangles inscribed in circles and Infinite Sums . So here's a quick solution!

---

Fact 1: A circle inscribed inside of an equilateral triangle inside of a circle has $\frac{1}{4}$ the area of the original circle.

Visual proof:

---

Fact 2: The area of the 3 largest blue sections is $9\pi - \frac{27}{4}\sqrt{3}$ .

Proof:

(1) The area of the whole circle is $\pi r^2 = \pi(3^2) = 9 \pi$ .
(2) The area of the inscribed equilateral triangle is, $\frac{27}{4}\sqrt{3}$ . This can be calculated from the observation that the height of this triangle, $t$ is $\frac{3}{4}$ the diameter of the circle = $\frac{9}{2}$ . Calculating the area of an equilateral triangle of height $t$ and with hypotenuse, $h$ can be accomplished by the Pythagorean Theorem (or, you might just recall the ratios of a 30-60-90 triangle and extrapolate from there). By derivation: $t^2 + (\frac{1}{2} h)^2 = ( \frac{9}{2})^2 + (\frac{1}{2} h)^2 = h^2 \\ \rightarrow h = 3 \sqrt{3} \\ \rightarrow \triangle \text{area } = \frac{1}{2}\times \text{base } \times \text {height} \\ = \frac{1}{2}\times 3 \sqrt{3} \times \frac{9}{2} = \frac{27 \sqrt{3}}{4}$

(3) Therefore, the area of just the blue regions is the area of the circle minus the area of the inscribed triangle: $9\pi - \frac{27}{4}\sqrt{3}$

---

Fact 3: Calculated as the sum of the infinite series, the total area of all of the blue regions is $\frac{4}{3}$ the area of the largest 3.

Proof:

Having calculated the area of the 3 largest regions as $9\pi - \frac{27 \sqrt{3}}{4}$ total, observe that the next set of 3 blue regions (those inside of the first two circles but outside of the 3rd) are, to scale, $\frac{1}{4}$ the area of the originals). The next set will be, to scale, $\frac{1}{4}$ of this second set, or $\frac{1}{16}$ the original.

The sum of the infinite geometric series $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64}+ ... + \frac{1}{4^n} = \frac{4}{3}$

Therefore, the total sum of all of the blue area in this diagram is: $(9\pi - \frac{27 \sqrt{3}}{4}) \times \frac{4}{3} = 12 \pi - 9 \sqrt{3}.$ Breaking this solution down by the form given: $A\pi - B \sqrt{C}$ , $A = 12, B = 9, C = 3.$ Therefore $\frac{A+B+C}{C} = \frac{12+9+3}{3}=4+3+1 = \fbox{8}$