Optimal Spatial Triangle Ratio

Let $AI=1$ , $AH=h$ , $FG=h'$ , and $FJ=h"$ be the altitudes of $\triangle ABC$ , $\triangle ADE$ , $\triangle DEF$ , and $\triangle BCF$ Let $DE = a$ and $BC=b$ . Since $\triangle ADE$ and $\triangle ABC$ are similar, we have $\dfrac {DE}{BC} = \dfrac ab = \dfrac h1 \implies a = hb$ . Again $triangle DEF$ and $\triangle BCF$ are similar. Then $\dfrac {FG}{FJ} = \dfrac {DE}{BC}$ or

$\begin{aligned} \frac {h'}{h"} & = \frac ab & \small \blue{\text{Since }h'+h" = 1-h} \\ \frac {h'}{1-h-h'} & = \frac ab \\ bh' & = (1-h)a-ah' \\ \implies h' & = \frac {(1-h)a}{a+b} \end{aligned}$

Then the ratio of areas $\dfrac {[DEF]}{[ABC]}=\rho$ . Then:

$\begin{aligned} \rho & = \frac {\frac 12 ah'}{\frac 12 b \cdot 1} = \frac {ah'}b = \frac {(1-h)a^2}{b(a+b)} = \frac {(1-h)h^2b^2}{hb^2+b^2} = \frac {h^2(1-h)}{h+1} \\ \frac {d\rho}{dh} & = \frac {(2h-3h^2)(h+1)- h^2 + h^3}{(h+1)^2} = \frac {2h-2h^2 - 2h^3}{(h+1)^2} & \small \blue{\text{Putting }\frac {d\rho}{dh} = 0} \\ h(h^2+h-1) & = 0 & \small \blue{\text{Since }h>0} \\ \implies h & = \frac {\sqrt 5-1}2 = \varphi - 1 = \frac 1\varphi & \small \blue{\text{where }\varphi \text{ is the golden ratio.}} \end{aligned}$

Since $\dfrac {d^2 \rho}{dh^2} < 0$ , when $h = \varphi -1$ , $\rho$ is maximum when $h = \varphi -1$ . And:

$\begin{aligned} \rho_{\max} & = \frac {\left(1-\frac 1\varphi \right)\cdot \frac 1{\varphi^2}}{\varphi -1+1} = \frac {\varphi-1}{\varphi^4} = \frac 1{\varphi^5} = 5\varphi - 8 = \frac {5\sqrt 5 - 11}2 \approx \boxed{0.0902} \end{aligned}$