The devil's playground On the $\mathbb{R}^2$ plane, the devil has trapped the angel in an equilateral triangle of firewalls.

The devil

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*starts at the apex of the triangle.

*can move at speed $1$ to leave a trajectory of firewall behind, as this



*

*can teleport from one point to another along the firewall.

The angel

*

*can teleport to any point that is not completely separated by firewalls from her current position.

The devil catches the angel if their distance is $0$.

Question 1: how should the devil move to catch the angel in the shortest amount of time?
Question 2: if the devil is given a fixed length of firewalls to enclose the angel in the beginning, what shape maximizes the survival time of the angel? (The devil always starts on the firewall)

Here is an upper bound for the triangle of side length $1$.
First, divide in half into two triangles of angles $\frac{\pi}{6},\frac{\pi}{3}, \frac{\pi}{2}$ triangle and side lengths $1, \frac{\sqrt{3}}{2},\frac{1}{2}$. This requires drawing an edge of length $\frac{\sqrt{3}}{2}$. It doesn't matter which one the angel goes into.
Next, divide each of these triangles into two triangles with angles $\frac{\pi}{6},\frac{\pi}{3}, \frac{\pi}{2}$. The angle wisely goes into the larger one. Repeat this process.
Starting from a triangle with side lengths $a, \frac{\sqrt{3}}{2} a, \frac{1}{2}a$, this produces a triangle of side lengths $\frac{\sqrt{3}}{2} a, \frac{3}{4}a, \frac{\sqrt{3}}{4}a$, after drawing an edge of length $\frac{\sqrt{3}}{4}a$.
So the total length drawn is
$$ \frac{\sqrt{3}}{2} + \frac{ \sqrt{3}}{4} + \frac{ \sqrt{3}}{2} \frac{ \sqrt{3}}{4} +\left( \frac{ \sqrt{3}}{2}\right)^2 \frac{ \sqrt{3}}{4} + \dots  = \frac{\sqrt{3}}{2} + \frac{ \frac{\sqrt{3}}{4}}{1- \frac{\sqrt{3}}{2}}=  \frac{\sqrt{3}}{2} + \frac{ \sqrt{3} }{4 - 2 \sqrt{3} }=\frac{3 \sqrt{3} -3}{4 - 2 \sqrt{3}} =4.098\dots$$
which is probably not optimal.
