TITLE: The devil's playground
QUESTION [9 upvotes]: 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)

REPLY [3 votes]: 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.