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

The devil

*

*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)
 A: Here is another attempt to improve the upper bound. Let $y$ be the shortest time required to catch the angel in the equilateral triangle of side length 1. The devil may cut the triangle horizontally along a chord of length $a = (56+15\sqrt{3}-\sqrt{1143+852\sqrt{3}})/46\approx0.669727$. The angel is force into either a smaller equilateral triangle, requiring the devil to use $ay$ time to catch her, or into the trapezoid, which is contained in a $1\times(1-a)(\sqrt{3}/2)$ rectangle. If the angel is in the trapezoid, she will be caught in at most $3(1-a)(\sqrt{3}/2)+1/2$ units of time (similar argument to Will Swain's comment to his answer, but in this case the long side is between 2 and 4 times the short side). Let $y_0 = 3(1-a)(\sqrt{3}/2)+1/2 + a = (1+\sqrt{9+24\sqrt{3}})/4 \approx 2.0278$. Note that $a$ was chosen such that we also have $y_0 = a + a y_0$. So we now have that $y\le \max(y_0, a+a y)$. Put otherwise, $y\le y_0$ or $y\le a + a y$, but the latter inequality simplifies to the former and we simply have $y\le y_0\approx 2.0278$.
A: 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.
A: It seems there is value to think about about splitting the triangle into 3 or 4 pieces rather than just 2 in each stage. I give a natural approach that achieves 3, and another which gives $\frac{1}{3} +\sqrt{3}\le 2.06539$.
For any two points $x,y$, let $xy$ denote the line segment from $x$ to $y$.
For example, one natural approach is to try and split the triangle into 4 equilateral triangles. Let $a,b,c$ be the points of our triangle. We start by teleporting to the midpoint of $ab$. Then, we move from the midpoint of $ab$ to the midpoint of $bc$, and then to the midpoint of $ca$. This takes length 3/2. In doing so, we have scaled the triangle down by a factor of 1/2. We may now repeat, this time able to teleport to a midpoint, now only requiring length 3/4, and so on.
Hence, this gives an upper bound of $3\sum_{i=1}^\infty 2^{-i} = 3$, as desired.
Now, what if we split the triangle into three parts? Let $d,e,f$ denote the respective midpoints of $ab,bc,ca$. Let $o$ be the interesection of $dc$ and $ea$. We walk from $d$ to $o$, $e$ to $o$ and $f$ to $o$. Each segment has length $\frac{1}{2\sqrt{3}}$, thus this in total costs $\sqrt{3}/2\le 0.86603$.
The angel will be trapped in a quadrilateral $Q$, since our triangle is split into 3 identical copies. WLOG, we assume the vertices of $Q$ are $a,d,f,o$. We have that $Q$ is contained in a equilateral triangle $T$ of height $1/\sqrt{3}$ (indeed, consider the line $\ell$ parallel to $df$ through $o$, we have that $Q$ is contained in the triangle whose vertices are $a$ and the intersection of $\ell$ with our initial triangle). We have that said triangle has side lengths 2/3.
Now, suppose we try to iterate, and split our triangle $T$ into 3 parts. We first draw the line parallel to $do$, then the line parallel to $fo$. The angel has two choices. Either, it goes in the new quadrilateral $Q'$ that is formed, or it is on the other side. In the first case, it is easy to see that $Q'$ is similar to $Q$, and 2/3 as large (since $T$ is 2/3 as large as our initial triangle). In the other case, we add the third line (which will be will be in the same direction as $oe$); here the angel is stuck in one of two parts which are each covered by a copy of the $1/6\times \frac{1}{2\sqrt{3}}$ rectangle.
We now can analyze the total length required to catch the angel once it is in $Q$. Let $L$ denote the total length needed if we follow the above strategy, (namely adding two lines, then scaling down if the angel stays in $Q'$, or adding the third line and doing a rectangle strategy otherwise), assuming the angel behaves optimally.
If the angel chooses to stay in $Q$, then we only added two edges each of length $(2/3) \frac{1}{2\sqrt{3}}$, for a total length of $\frac{2}{3\sqrt{3}}$, and since $Q'$ is a 2/3 scaled down copy of $Q$, we will catch the angel having used length $\frac{2}{3}L+\frac{2}{3\sqrt{3}}$ here.
In the other case, we add 3 lines, whose net length is $\frac{1}{\sqrt{3}}$. By the argument given by Sawin's comment below his answer, we can do the rectangle using length $2(1/6)+(1/2\sqrt{3})$. This totals to a length of $\frac{1}{3}+\frac{\sqrt{3}}{2}\le 1.19936$.
Hence, we just solve for $L = \max\{\frac{1}{3}+\frac{\sqrt{3}}{2},\frac{2}{3}L+\frac{2}{3\sqrt{3}}\}$. Observing the solution to $x= (2/3)x +\frac{2}{3\sqrt{3}}$ is $2/\sqrt{3} \approx 1.15470$, we see that $L =\frac{1}{3}+\frac{\sqrt{3}}{2}\le 1.19936$. It follows that we can catch the angel in our triangle by adding our initial cost $\frac{\sqrt{3}}{2}$ to $L$, which gives $1/3 + \sqrt{3} \le 2.06539$.
Perhaps we get get a slightly better solution by optimizing the second case, since they are not truly rectangles.
