<s>[INST] 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)
 [/INST] 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.
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