Funky triangle area

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

Funky 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}


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