Euler might have known...

Consider the following diagram, where the incircle of the small triangle has inradius $r$ , and the bases of the two right-angled triangles are $x$ and $y$ . Then we deduce that $\begin{aligned} 2(y-x) & = \; \tfrac12\big[\sqrt{x^2 + 16} + \sqrt{y^2 + 16} + y - x\big]r \\ (4 - r)(y - x) & = \; r\left(\sqrt{x^2 + 16} + \sqrt{y^2 + 16}\right) \end{aligned}$ and this equation can be solved to write $y \; = \; \frac{1}{4(2-r)}\left[(r^2 - 4r + 8)x + r(4-r)\sqrt{x^2 + 16}\right]$ Introducing the angles $\alpha,\beta$ , as shown, we deduce that $\tan\beta \; = \; \frac{1}{4(2-r)}\left[(r^2 - 4r + 8)\tan\alpha + r(4-r)\sin\alpha\right]$

From this we can deduce that $\sec\beta \; =\; \frac{1}{4(2-r)}\left[r(4-r)\tan\alpha + (r^2 - 4r + 8)\sec\alpha \right]$ and hence we have $\begin{aligned} \tan\beta + \sec\beta & = \; \frac{2}{2-r}\left(\tan\alpha + \sec\alpha\right) \\ \cot\big(\tfrac12\pi - \beta\big) + \csc\big(\tfrac12\pi - \beta\big) & = \; \frac{2}{2-r}\left[\cot\big(\tfrac12\pi - \alpha\big) + \csc\big(\tfrac12\pi - \alpha\big)\right] \\ \cot\left(\tfrac14\pi - \tfrac12\beta\right) & = \; \frac{2}{2-r}\cot\left(\tfrac14\pi - \tfrac12\alpha\right) \\ \tan\left(\tfrac14\pi - \tfrac12\beta\right) & = \; \left(1 - \tfrac12r\right)\tan\left(\tfrac14\pi - \tfrac12\alpha\right) \end{aligned}$ If we define $\theta_n = \angle ABD_n$ to be the angle at the top of the triangle $ABD_n$ , we deduce that $\tan\left(\tfrac14\pi - \tfrac12\theta_n\right) \; = \; \left(1 - \tfrac12r_n\right)\tan\left(\tfrac14\pi - \tfrac12\theta_{n-1}\right)$ for any $n \ge 2$ . If $\theta_\infty = \angle ABC$ is the angle at the top vertex of the limiting triangle $ABC$ , it follows from this that $\tan\left(\tfrac14\pi - \tfrac12\theta_\infty\right) \; = \; \tan\left(\tfrac14\pi - \tfrac12\theta_1\right) \times \prod_{n=2}^\infty \left(1 - \tfrac12r_n\right)$ Now $\tan\theta_1 = \tfrac34$ , so that $\tan\tfrac12\theta_1 = \tfrac13$ , and hence $\tan\left(\tfrac14\pi - \tfrac12\theta_1\right) = \tfrac12$ . Thus we deduce that $\tan\left(\tfrac14\pi - \tfrac12\theta_\infty\right) \; = \; \prod_{n=1}^\infty \left(1 - 2^{-n}\right) \; =\; \big(\tfrac12\,;\,\tfrac12\big)_\infty$ where $\big(\tfrac12,\tfrac12\big)_\infty$ is the $q$ -Pochhammer symbol. Thus we deduce that $\begin{aligned} \tan\big(\tfrac12\theta_\infty\big) & = \; \frac{1 - \big(\tfrac12\,;\,\tfrac12\big)_\infty}{1 + \big(\tfrac12\,;\,\tfrac12\big)_\infty} \\ \tan\theta_\infty & = \; \frac12\left(\frac{1}{\big(\tfrac12\,;\,\tfrac12\big)_\infty} - \big(\tfrac12\,;\,\tfrac12\big)_\infty\right) \end{aligned}$ and so the area of the triangle $ABC$ is $\Delta \; =\; 8\tan\theta_\infty \; = \; 4\left(\frac{1}{\big(\tfrac12\,;\,\tfrac12\big)_\infty} - \big(\tfrac12\,;\,\tfrac12\big)_\infty\right) \; = \; 12.6958340974738...$ and hence $\lfloor 10^6 \Delta \rfloor = \boxed{12695834}$ .