# Funky triangle area

In this post, I will deal with an interesting high school math problem. Suppose we are given the following triangle:

Figure 1

We have two parameters:

• $n$, the number of circles touching a side of the triangle,
• $r$, the radius of each circle.

Figure 2

In order to find out the area of the triangle given the parameters $n$, $r$, we need to find the length of each side. Basically, we must calculate $d$ in Figure 2 as a function of $r$, which is fairly easy since we must have

\displaystyle \begin{aligned} \tan 30\textdegree &= \frac{r}{d}, \\ d &= \frac{r}{\tan 30\textdegree} \\ &= \frac{r}{\frac{1}{\sqrt{3}}} \\ &= \sqrt{3}r. \end{aligned}

Next, we can define the length of a side of the triangle as a function of $n$ and $r$:

\displaystyle \begin{aligned} s(n, r) &= 2d + 2r(n - 1) \\ &= 2\sqrt{3}r + 2r(n - 1) \\ &= 2r(n + \sqrt{3} - 1). \end{aligned}

If we set $p(n, r) = 3s(n, r)/2$ (i.e., $p(n, r)$ is the half of the perimeter of the triangle),
by Heron’s formula, we have that the area of the triangle is

\displaystyle \begin{aligned} A(n, r) &= \sqrt{p(n, r)[p(n, r) - s(n, r)]^3} \\ &= \sqrt{\frac{3}{2}s(n, r)\Bigg[\frac{3}{2}s(n, r) - s(n, r)\Bigg]^3} \\ &= \sqrt{\frac{3}{2}s(n, r)\Bigg[\frac{1}{2}s(n, r)\Bigg]^3} \\ &= \sqrt{\frac{3}{2}s(n, r)\frac{1}{8}s^3(n, r)} \\ &= \sqrt{\frac{3}{16}s^4(n, r)} \\ &= \frac{\sqrt{3}}{4}s^2(n, r) \\ &= \frac{\sqrt{3}}{4} [2r(n + \sqrt{3} - 1)]^2 \\ &= \frac{\sqrt{3}}{4} 4r^2 (n + \sqrt{3} - 1)^2 \\ &= \sqrt{3}r^2 (n + \sqrt{3} - 1)^2. \end{aligned}