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This is a fun problem that I saw somewhere on the internet a long time ago:

Suppose you are at the center of an equilateral triangle with side length $s$. At each of its vertices, there is a lion which is determined to eat you. The lions start at a constant speed of $v_l$, and they are always running directly towards your current location. You start in the center, and can run at a constant speed $v_h$ (assume instantaneous acceleration for all parties). You are NOT enclosed in the triangle, you are free to try to run wherever you want. Which patch should you take in order to survive the longest time possible? How long can you survive?

At first, because everybody's speed is constant, I thought we can just work with functions of $x$ for the paths of the lions and the human, and try to maximize the arc length of the lions' paths. However, I think it's much easier to work with the functions in parametric form, because the tangent lines to the lions' path at $t=t_0$ should go trough your position at $t_0$. Also, I think it's reasonable to assume that at $t=0$ your direction is straight towards the midpoint of one of the sides, because any other direction would cause you to meet one of the lions faster.

Rodrigo de Azevedo
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Ovi
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    Wait - there are also tigers? :) – Hagen von Eitzen Jun 01 '16 at 05:39
  • It was bad enough with the lions trying to eat you ... – Zubin Mukerjee Jun 01 '16 at 05:40
  • @HagenvonEitzen Haha at first it was tigers but I decided to make it lions for extra motivation to increase $v_h$! – Ovi Jun 01 '16 at 05:41
  • by path you mean straight line or any arbit path? – avz2611 Jun 01 '16 at 05:44
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    If you are faster than the lions any direction except directly towards one of the vertices should lead to indefinite escape, right? – Zubin Mukerjee Jun 01 '16 at 05:44
  • @avz2611 arbitrary – Ovi Jun 01 '16 at 05:56
  • The instantaneous acceleration has 2 meanings: 1) the tangent line to a lion's path at time $t_0$ passes trough your position at $t_0$. 2) I didn't want there to be confusion about the lions and human starting from rest at $t=0$ and accelerating to their top speed over some interval; you can assume they start at their top speed – Ovi Jun 01 '16 at 05:59
  • @ZubinMukerjee $probably$, but only due to the simple programming of the lions I think – Ovi Jun 01 '16 at 06:01
  • i dont have the resources to work the problem right now , but i think you should assume man at rest and work with in relative frame , because you know lions are coming at you so always have their velocity towards you and you have your $v(t)$ which will be in negative direction for each of the lions. i think problem will become much simpler with that – avz2611 Jun 01 '16 at 06:11
  • common sense (and also can be proven) is that when you run slower than the lions, you will be eaten (especially for infinite time). Having three lions make it more obvious. If you run faster (?) then you can probably escape and maybe the triangle configuration allows for small complexity to answer. I would ask the question: what if you run as fast as the lions? Then there is some asymptotic case that sounds interesting. One can approach this using complex numbers for more simplicity and a reference frame tied to the person. – Chip Jun 03 '16 at 01:33
  • @Chip I was thinking the asymptotic case could also happen even if the lion was very very slightly faster than you, but no I think you're right; given enough time the lion should catch you, no matter how small the margin is. – Ovi Jun 03 '16 at 03:20
  • if i did it correctly, in the reference frame of the person, a lion $i$ at distance $\rho_i$, has $d\rho_i/ dt = v_L - v_p * \cos(\phi_i + \theta(t))$ (or similar). Here, the speed of the lions is $v_L$, that of the person is $v_p$, $\phi_i$ is the angular radial coordinate of the lion $i=1,2,3$ and $\theta(t)$ is an angular function that the person can modify. If speeds are identical, you can re-scale time and make velocities disappear from the equation. You want to choose $\theta$ such that $\rho_i(t)$ (three lions, $i=1,2,3$) does not become zero. – Chip Jun 03 '16 at 06:57
  • Numerically, it appears that if the speeds are equal, than the man can escape if he runs toward the middle of one side of the triangle and the closest the lions get is a bit more than $1/10$ the length of one side of the triangle. So, if the triangle is large enough, they won't be able to grab the person, which will thus escape. – Chip Jun 06 '16 at 02:03

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For the case I mention above (the man and the lions have same speed, otherwise the answer is obvious), let us pick the person in the centre of the triangle and running towards the middle of one side, and the lion just above him starting to run towards the person with same speed. Using a reference frame in which the man is still, and complex numbers to describe the lion position, one can write for $z=\rho e^{I \phi}$: $$\dot z = -v e^{I \phi} -v$$ (where $v$ is the speed). We can rescale distances in units of the side of the triangle $L$ and time in units of $L/v$, to get the dimensionless $$\dot z=-1-e^{I \phi}$$, from where: $$\dot \rho =-(1+\cos \phi) \\ \rho \dot \phi =\sin \phi.$$

The solution to this equations are: $$\rho = \rho_0 ( \frac{\sin\frac{\phi}{2}}{\sin\frac{\phi_0}{2}} )^2$$ and $$\log\tan^2\frac{\phi}{2}-\frac{1}{\sin^2\frac{\phi}{2}} \Bigg\vert_{\phi_0}^{\phi}=\frac{2t}{\rho_0 \sin^2\frac{\phi_0}{2}},$$ where $t$ is the time (starting from $0$), $\rho_0=1/\sqrt 3$, and $\phi_0=\pi/3$.

Note from the last equation, that we need to solve for $\phi =\phi(t)$. This is possible if we notice that $\sin^2$ can be expressed as function of $\tan^2$ and we solve for $\tan$ using the Lambert W-function. With it, we can obtain $\rho=\rho(t)$ from the above solution $\rho=\rho(\phi)$.

The solution obtained using the above and validated by integrating the complex ODE equation with the Runge-Kutta method of order $4$ give for the trajectory of the lion (remember that the man sits still at the origin of the complex plane): enter image description here

The lion as seen by the running man, starts in the upper right corner. Remember that the units of length are in units of the side of the triangle, so if the triangle is too small, the lion even if it does not reach the origin (where the person is in his reference frame), could still grab him if in close range.

Chip
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