*isn't*similar to the parent triangle and bisecting the triangle that is.

Suppose the area of the original triangle $\Delta ABC$ is 1, and the piece $ACC'$ has area $x$ (thus each succeeding similar copy has area a fraction of $x$ of the preceding triangle). Then the total value of our counter, which approaches 1, is:

$$(1-x)+x(1-x)+x^2(1-x)+...=1$$

$$1+x+x^2+...=\frac1{1-x}$$

Where $x$ depends on the actual angle $\beta$.

It is interesting, however, to consider the case of a general scalene triangle $\Delta ABC$ where $C$ is not necessarily twice of $B$. Here each successive triangle wouldn't be similar to the last, thus we won't be dealing with a geometric series.

Let the angles of $\Delta ABC$ be $A=\alpha$, $B=\beta$,$C=\pi-\alpha-\beta$. We bisect angle $C$, as before, adding to our counter the piece that contains the angle $B$. The remaining triangle has angles $\alpha$, $\frac{\pi-\alpha-\beta}{2}$ and $\pi-\alpha-\frac{\pi-\alpha-\beta}{2}$.

We keep repeating the process, each time bisecting the angle that is neither $\alpha$ nor the angle formed as half the angle that was just bisected, and adding to our counter the area of the piece that does not contain the angle $A$, while splitting the piece that does.

To keep track of the angles in each successive triangle, we define three series:

$$\begin{gathered}

{\alpha _n} = \alpha\\

{\beta _n} = {\gamma _{n - 1}}/2\\

{\gamma _n} = \pi - {\alpha _n} - {\beta _n}\\

\end{gathered}$$

These are defined recursively, of course, so we calculate the explicit form by substituting $\gamma_n$ into $\beta_n$ to get a recursion within $\beta_n$ -- then with the simple initial-value conditions $\alpha_0=\alpha$, $\beta_0=\beta$, etc. we get:

$$\begin{gathered}

{\alpha _n} = \alpha\\

{\beta _n} = \frac{{\pi - \alpha }}{3} + {\left( { - \frac{1}{2}} \right)^n}\left( {\beta - \frac{{\pi - \alpha }}{3}} \right)\\

{\gamma _n} = \frac{{2(\pi - \alpha )}}{3} - {\left( { - \frac{1}{2}} \right)^n}\left( {\beta - \frac{{\pi - \alpha }}{3}} \right)\\

\end{gathered}$$

The area ratio of the piece we're keeping at each stage is $\frac{{\sin {\alpha _n}}}{{\sin {\alpha _n} + \sin {\beta _n}}}$, therefore the convergence of their sum of their areas to 1 implies:

$$\begin{gathered}

\frac{{\sin \alpha }}{{\sin \alpha + \sin \beta }} + \frac{{\sin \beta }}{{\sin \alpha + \sin \beta }}\frac{{\sin \alpha }}{{\sin \alpha + \sin {\beta _1}}} \hfill \\

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{{\sin \beta }}{{\sin \alpha + \sin \beta }}\frac{{\sin {\beta _1}}}{{\sin \alpha + \sin {\beta _1}}}\frac{{\sin \alpha }}{{\sin \alpha + \sin {\beta _2}}} + ... = 1 \hfill \\

\end{gathered} $$

Or more compactly:

$$\sum\limits_{k = 0}^\infty {\left[ \left(1-x_k(\alpha,\beta)\right)\prod\limits_{j = 0}^{k - 1} {x_j(\alpha,\beta)} \right]} = 1$$

Where:

$${x_k}(\alpha ,\beta ) = \frac{{\sin \left( {\frac{{\pi - \alpha }}{3} + {{\left( { - \frac{1}{2}} \right)}^k}\left( {\beta - \frac{{\pi - \alpha }}{3}} \right)} \right)}}{{\sin \alpha + \sin \left( {\frac{{\pi - \alpha }}{3} + {{\left( { - \frac{1}{2}} \right)}^k}\left( {\beta - \frac{{\pi - \alpha }}{3}} \right)} \right)}}$$

For all values of $\alpha$ and $\beta$.

Well, have we truly discovered something new?

Turns out, no. It doesn't even matter what $x_k(\alpha,\beta)$ is, really -- the identity $\sum\limits_{k = 0}^\infty {\left[ \left(1-x_k(\alpha,\beta)\right)\prod\limits_{j = 0}^{k - 1} {x_j(\alpha,\beta)} \right]} = 1$ will always be true. Indeed, it is a telescoping sum:

$$\begin{gathered}

1 - {x_0} + \hfill \\

\left( {1 - {x_1}} \right){x_0} + \hfill \\

\left( {1 - {x_2}} \right){x_0}{x_1} + \hfill \\

\left( {1 - {x_3}} \right){x_0}{x_1}{x_2} + \hfill \\

... = 1 \hfill \\

\end{gathered} $$

All that is required is that the final term, $x_0x_1x_2x_3...x_k$ approaches 0 as $k\to\infty$ -- this ensures sum convergence. (So I suppose I was not completely right when I said it doesn't matter what $x_k$ is -- but considering renormalisation and stuff, I kinda was.)

This raises two interesting questions:

- How would this "telescoping sum" argument work for the simple geometric series?
- Can we get interesting incorrect (? perhaps renormalisations) sums by choosing an $x_k$ sequence whose product doesn't approach zero?

Well, for the geometric series we had $\beta = (\pi - \alpha )/3$ so that ${x_k}(\alpha ,\beta ) = x(\alpha,\beta)=\frac{{\sin \beta }}{{\sin \alpha + \sin \beta }}$. Indeed, one may confirm that setting $x_0=x_1=x_2=...$ yields the product of the geometric series and $1-x$, and that happens to be telescoping. This is really just our standard proof of the series, where we multiply the sum by $x$, subtract this from the original sum, etc.

As for the second question -- consider, for example, $x_k=k+1$. It gives you the sum $1!\cdot1+2!\cdot2+3!\cdot3+...=-1$. Of course, this is just the identity $n\cdot n!=(n+1)!-n!$, and the telescope doesn't really cancel out so you're left with $\infty!-1$.

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