Area of shaded region of circles inscribed in a circle

Question

I was given this question and told to find the area of the shaded region. The radius of the white circles is r that are arranged in a pyramid with N rows.

ATTEMPT:

When

N=1, area = $\pi r^2$

N=2, area = $2\pi r^2$

N=3, area = $3\pi r^2$

N=4, area = $4\pi r^2$

Question: How can we find the area of the larger circle? Is there an implicit relationship between the smaller and larger circles that I'm missing?

Answer 1

Given the radius $r$ of the small circles organized in $n$ rows, the radius of the big circle is the sum $R+r$, where $R$ is the circumradius
of the equilateral $\triangle ABC$ with the height $AM$,

\begin{align} |AM|&=(n-1)\cdot 2r \cdot \tfrac{\sqrt3}2 =(n-1) r\,\sqrt3 ,\\ R&=\tfrac23\,|AM| =\tfrac{2\sqrt3}3\,(n-1)\, r , \end{align}

hence the area of the background circle is

\begin{align} S_n&=\pi(R+r)^2 = (\tfrac{2\sqrt3}3\,(n-1)+1)^2\pi r^2 . \end{align}

The number of the small circles in $n$ rows is the triangular_number

\begin{align} T_{n}=\sum _{k=1}^{n}k=\tfrac12 {n(n+1)}, \end{align}

The total area of the small circles is then \begin{align} s_n&=\tfrac12 {n(n+1)} \pi r^2 , \end{align}

and the shaded area is

\begin{align} S_n-s_n&= \tfrac16\,(n-1)(5n-14+8\sqrt3)\pi r^2 . \end{align}

Answer 2

Let the side be $s$. So radius of small circle is $\frac{s}{2n}$. radius of bigger circle is $\frac{s }{\sqrt{3}}$. Now you can relate the areas, and subtract to get required area. I hope it helps.

Answer 3

HINT: consider the three small circles that are tangent to the large circle. Their centers are the vertices of an equilateral triangle. Knowing the radius $r$ of the small circles, you can easily get the side of this triangle, and then you can determine the distance between any of its vertices and the center. Summing this last value to the $r$ you obtain the radius of the large circle.