You're screwed in constant time, no matter your speed, since the devil has a good strategy.
I cannot claim to having found the devil's optimal strategy, but I do claim that there is an upper bound to the time the devil takes to catch the victim. And that upper bound is quite low.
Assume that the circle is inscribed in an equilateral triangle of ABC of base $b = 2\sqrt[]3$, with the devil resting at the midpoint D of its base AB, like so:
In fact, forget about the circle. Also, mark the midpoint of BC as E, and of CA as F:
The devil shall move a distance of $b/2$ from D towards E, dividing it in two areas. Mark the midpoint of DE as G:
Is the victim within the smaller triangle BDE? Good, move $b/4$ towards G, and repeat this division algorithm on the BDE triangle. If not, move $b/2$ from E to F, marking the midpoint of EF as H:
Again: is the victim in the smaller triangle CEF? Good, then move $b/4$ back to H and repeat. If not, divide the remaining rhombus in two by moving $b/2$ from F to D, marking the midpoint of DF as J:
...and since the victim sure is either DEF or ADF, move $b/4$ from D to J, and repeat the algorithm. The idea is to divide the triangles into smaller and smaller triangles in a methodical way.
Note several important facts about this strategy:
- Every time a triangle is divided in four, its area is divided by four
- Each time a triangle is divided in four, its base is divided by two
- In order to divide a triangle in four, the devil must move at most a distance equal to $7/4$ times its base: $1/2$ each move from midpoint to midpoint, maximum three such moves (D→E→F→D), plus $1/4$ to set up in the midpoint of the appropriate side of a subdivision (E→G, F→H or D→J).
Thus, the first subdivision takes (at most) $\frac{7\sqrt[]3}{2}$; since the triangle's base is halved, the second one takes $\frac{7\sqrt[]3}{4}$, the third one $\frac{7\sqrt[]3}{8}$; and in general the $n$-th one shall take $\frac{7\sqrt[]3}{2^{n}}$.
If the devil could perform these subdivisions an infinite number of times, then the area of the triangle containing the victim would be $\lim_{x \to \infty} \frac{a}{4^x} = 0$. (The initial area doesn't matter so why bother calculating it).
Now, is it possible to move long enough to perform all the subdivisions in a finite amount of length? That's the same as asking "Does the following infinite sequence converge?" $$\sum_{n=1}^{\infty} \frac{7\sqrt[]3}{2^{n}}$$ Since I absolutely suck at doing these calculations (and I have completely forgot the "infinite sequences" chapter from my calculus classes), I cheated a bit by using wolfram-alpha. The sequence does, in fact, converge, to $7\sqrt[]3$ or about 12.124 units of length.
The victim's strategy would be
to lead the devil's movements by an infinitesimal distance, so the devil cannot choose the right subdivision until said subdivision is complete.
The generalized strategy explained above provides an upper bound for the distance the devil must move, but has two characteristics that intuitively look like problems: (a) the devil backtracks, potentially wasting movement and (b) the search space is way bigger than needed.
The backtracking issue can be optimized by
using right-angled isosceles triangles instead of equilateral triangles, and positioning the devil at the right-angle corner. Any such triangle can be halved into two right-angled isosceles triangles, like so:
As before, the devil splits a triangle, checks the subdivision containing the lost soul, and recursively proceeds to split that. The devil will follow a fractal path looking like:
At each subdivision, the area halves; that means the area converges to zero as before since $$\lim_{x \to \infty} \frac{a}{2^x} = 0$$ The height of the triangles (i.e. the length of the devil's path) shrinks by a factor of $\frac{\sqrt[]2}{2} ≃ 0.7071$ on each subdivision; assuming that the length needed to perform the first subdivision is $\frac{\sqrt[]2}{2}$, then the length needed to perform the $n$th subdivision shall be $$\left(\frac{\sqrt[]2}{2}\right)^n$$, and the total length of the devil's fractal path shall be $$\sum_{n=1}^{\infty} \left(\frac{\sqrt[]2}{2}\right)^n$$.
That seems to solve the backtracking issue, but what about the wasted search space? A possible approach would be for the devil to start moving on a path like...
Which means: Starting at A, move to B. Choose the half circle containing the victim (the diagram only shows a solution for the bottom half; the solution for the top half is symmetrical), then proceed to C (ABC). If the victim is within BCD, move to D then start the fractal subdivision of BCD. Else, move to E (ABCE). If the victim is within BCE, start the fractal subdivision of BCE. Else, move to F (ABCEF). Start the fractal subdivision of either EFH or EFG, depending on which of those two triangles contains the victim.
The (worst case) length of the initial path ABCEF is $4 + \sqrt[]2$; and since the distance from F to the midpoint of either EG or EH is $\frac{\sqrt[]2}{2}$, we can use the infinite series described before, so the total length of the devil's path is given by $$4 + \sqrt[]2 + \sum_{n=1}^{\infty} \left(\frac{\sqrt[]2}{2}\right)^n$$ and after cheating a bit with wolfram alpha to solve the infinite series, that becomes: $$4 + \sqrt[]2 + 1 + \sqrt[]2 = 5 + 2\sqrt[]2 ≃ 7.82843$$
That's significantly better than before (better for the devil, not for the poor soul), but I suspect that it's still not the lowest upper bound possible. The victim's strategy would remain unchanged, and would still depend on knowing the devil's optimal strategy.
