Area of circles inside an equilateral triangle compared to the area of the triangle itself. In an equilateral triangle, there is a circle (as large as possible) inside it, and in the three empty areas near each corner there are three new circles (one in each corner, each as large as possible), and between these circles and the corners there are even smaller circles (as large as possible), and the amount of circles keeps on going towards infinity. 
How large of a portion do these circles cover of the triangle's area (percentage)? 
This problem is in a section about geometric series, and it most likely should be solved with the fact that the series of the areas of the circles converges.
 A: Here's the kicker: an equilateral triangle can be split into an equilateral hexagon and three congruent equilateral triangles. The incircle of the large equilateral triangle will be inscribed in the hexagon, and the area of one of the small triangles is $\frac16$ of that of the hexagon.
From this, we can show that each of the three "second-generation" circles has an are that is $\frac19$ of the "first generation" circle's area. Likewise, each "third generation" circle has an area $\frac19$ that of a "second generation" circle, and so on.
A: Okay.
Let's do this.
Put the triangle at on a coordinate plane at points $A = (-1/2, 0), B = (1/2, 0), C= (0, \sqrt 3/2)$.   The inner circle touches the triangle at points $(-1/4, \sqrt 3 /4), (0,0)$ and $(1/4, \sqrt 3/4)$.  The center of the circle is $(0, r)$.
So $(1/4 - 0)^2 + (\sqrt 3/4 - r)^2 = r^2$ so $r = 1/2$.
$1/16 + 3/16 + r^2 - r\sqrt 3/2 = r^2$
$r\sqrt 3/2 =  1/4$
$r = \frac 1 {2\sqrt 3}$.
The height of the triangle is $\sqrt 3/2$.  The diameter of the big circle is $1/\sqrt 3$.  So if we imagine the smaller circle enclosed in a smaller triangle, that triangle would have height $\sqrt 3/2 - 1/\sqrt 3$.  So each iterative proportion gets smaller by $m = \frac {\sqrt 3/2 - 1/\sqrt 3}{\sqrt 3/2} = 1 - 2/3 = 1/3$.
Let's multiply everything by 2 for simplicity.  The Side of the triangle is 2, and the radius of the big circle is $1/\sqrt 3$.  The proportion coefficient remains as $1/3$ though.
So the area of the triangle is $1/2 * 2 * \sqrt 3 = \sqrt 3$.  The areas of the circles are $\pi(1/\sqrt 3(1/3)^n)^2 = \pi/3^{2n + 1}$ or the total of all circles is: $\pi/3 + 3\sum_{k=1}^{\infty}\pi/3^{2k + 1}$.
