The arithmetic mean of triangles

It is given that $a_{n+1} = a_n\;,n\geq 2$ . Thus by induction it can be proved that

$a_n = a_1\;,n\geq 2 \hspace{20pt}\cdots \text{Eq. }1$

$b_1 + c_1 = 2a_1 \hspace{20pt}\cdots \text{Eq. }2$

$b_{n+1} = \Large\frac{a_n + c_n}{2}$ $\hspace{20pt}\cdots \text{Eq. }3$

$c_{n+1} = \Large\frac{a_n + b_n}{2}$ $\hspace{20pt}\cdots \text{Eq. }4$

Now we will prove that $b_n + c_n = 2a_1\;,\; n\geq 1$ by mathematical induction.

We see that it is true for $n = 1$ as by $\text{Eq. }2$ . Let it be true for $n = k$ .

$\Rightarrow b_k + c_k = 2a_1$

Adding $\text{Eq. }3$ and $\text{Eq. }4$ after replacing $n$ with $k$ we get

$\hspace{13pt} b_{k+1} + c_{k+1} = \Large\frac{2a_k + b_k + c_k}{2}$

$\Rightarrow b_{k+1} + c_{k+1} = a_k + \Large\frac{b_k + c_k}{2}$

$\Rightarrow b_{k+1} + c_{k+1} = a_1+ \Large\frac{2a_1}{2}\hspace{20pt}$ $[\because a_k = a_1$ and $b_k + c_k = 2a_1]$

$\Rightarrow b_{k+1} + c_{k+1} = a_1 + a_1 = 2a_1$

We see that whenever that identity is true for $n = k$ , it is true for $n = k+1$ . Hence it is true for all $n \in \mathbb{N}$

So, $b_n + c_n = 2a_1$

$\Rightarrow a_n + b_n + c_n = a_n + 2a_1 = 3a_1\hspace{20pt}[\because a_n = a_1]$

$\Rightarrow$ Perimeter of $\triangle A_nB_nC_n$ is constant over $n \in \mathbb{N}$ . So its semiperimeter is also constant. Let the semiperimeter be $s$ . Then

$s = \Large\frac{3a_1}{2}$ $\hspace{20pt}\cdots\text{Eq. }5$ and

$s - a_n = \Large\frac{3a_1}{2}$ $- a_n$

$s - a_n = \Large\frac{3a_1}{2}$ $- a_1$

$s - a_n = \Large\frac{a_1}{2}$ $\hspace{20pt}\cdots\text{Eq. }6$

Multiplying $\text{Eq. }3$ and $\text{Eq. 4}$ we get

$b_{n+1}c_{n+1} = \Large\frac{(a_n + c_n)(a_n + b_n)}{4}$

$\hspace{40pt} = \Large\frac{a_{n}^2}{4}$ $+ \Large\frac{(b_n + c_n)a_n}{4}$ $+ \Large\frac{b_nc_n}{4}$

$\hspace{40pt} = \Large\frac{a_{1}^2}{4}$ $+ \Large\frac{2a_1^2}{4}$ $+ \Large\frac{b_nc_n}{4}\hspace{20pt}$ $[\because a_n = a_1$ and $b_n + c_n = 2a_1]$

$\hspace{40pt} = \Large\frac{3a_{1}^2}{4}$ $+ \Large\frac{b_nc_n}{4}$

Recursively writing $b_nc_n$ in terms of $b_{n-1}c_{n-1}$ and putting in the above equation we get

$b_{n+1}c_{n+1} = \Large\frac{3a_{1}^2}{4}$ $+ \LARGE\frac{\frac{3a_1^2}{4} + \frac{b_{n-1}c_{n-1}}{4}}{4}$

$\hspace{40pt} = \Large\frac{3a_1^2}{4}$ $+ \Large\frac{3a_1^2}{4^2}$ $+ \Large\frac{b_{n-1}c_{n-1}}{4^2}$

Recursively writing again and again we can get

$b_{n+1}c_{n+1} = \Large\frac{3a_1^2}{4}$ $+ \Large\frac{3a_1^2}{4^2}$ $+ \Large\frac{3a_1^2}{4^3}$ $+ \cdots + \Large\frac{3a_1^2}{4^{n-1}}$ $+ \Large\frac{3a_1^2}{4^n}$ $+ \Large\frac{b_1c_1}{4^n}$

$\hspace{40pt} = 3a_1^2\Big(\Large\frac{1}{4}$ $+ \Large\frac{1}{4^2}$ $+ \Large\frac{1}{4^3}$ $+ \cdots + \Large\frac{1}{4^{n-1}}$ $+ \Large\frac{1}{4^n}$ $\Big) + \Large\frac{b_1c_1}{4^n}$

$\hspace{40pt} = a_1^2\Big(1 - \Large\frac{1}{4^n}$ $\Big) + \Large\frac{b_1c_1}{4^n}$

Replacing $n$ by $n - 1$ in the above equation we get

$b_nc_n = a_1^2\Big(1 - \Large\frac{1}{4^{n-1}}$ $\Big) + \Large\frac{b_1c_1}{4^{n-1}}$ $\hspace{20pt}\cdots \text{Eq. }7$

Area of $\triangle A_nB_nC_n$ is $S_n$ . Then

$S_n = \sqrt{s(s - a_n)(s - b_n)(s - c_n)}$

$\Rightarrow S_n^2 = \Large\frac{3a_1}{2}\frac{a_1}{2}$ $\cdot (s - b_n)(s - c_n)$ $\hspace{20pt}[$ Using $\text{Eq. }5$ and $\text{Eq. }6]$

$\Rightarrow S_n^2 = \Large\frac{3a_1^2}{4}$ $\Big(s^2 - (b_n + c_n)s + b_nc_n\Big)$

$\Rightarrow S_n^2 = \Large\frac{3a_1^2}{4}$ $\Big(\Large\frac{9a_1^2}{4}$ $- 2a_1\cdot \Large\frac{3a_1}{2}$ $+ b_nc_n\Big)$

$\Rightarrow S_n^2 = \Large\frac{3a_1^2}{4}$ $\Big(b_nc_n - \Large\frac{3a_1^2}{4}$ $\Big)$

Putting $b_nc_n$ from $\text{Eq. }7$ in the above equation and simplifying we get

$S_n^2 = \Large\frac{3a_1^4}{16}$ $- \Large\frac{3a_1^2(a_1^2 - b_1c_1)}{4^n}$

By AM - GM inequality

$\Large\frac{b_1 + c_1}{2}$ $\geq \sqrt{b_1c_1}$

$a_1^2 \geq b_1c_1$

So we see that as $n$ increases $S_n^2$ increases $\Rightarrow S_n$ increases.