Escaping from a circle of fat lions.

You are surrounded, by X fat lions equally spaced around a circle of radius 200 meters in an open field. While making your escape plan you note several things: they are slow, they can only travel at one tenth of your speed, they are stupid, they can only move directly at your current position, and they can’t cooperate with one another. If any lion gets within 1 meter of you, you will be eaten.

What is the maximum value of X for which you have a strategy to escape from them?

TL;DR  Assuming @SimpleArt's strategy of running straight towards the midpoint in between two adjacent lions, one can escape from $364$ lions, but will be caught by $365$. The result is derived from the actual pursuit curve, with the final equation having to be solved numerically.
(This is close to, albeit slightly lower than, the value of $366$ proposed by @Jens in a comment.)

Assume oneself (= target) at the origin of the euclidian plane, darting upwards along the positive $y$ axis towards the midpoint between two lions symmetric to the axis. The phases of the ensuing pursuit are described in the following (based on the animation at Wolfram's pursuit curve page).

• (1) Pursuit starts at $t=0$ with the lion staring down at the origin.

• (2) At each moment in time the lion aims directly towards its target.

• (3) Since the pursuer goes downwards while the target moves upwards, they will be level at some point, after which the pursuer will turn upwards as well. If at this $t=t_1$ the abscissa of the lion is less than its reach $x_1 \le 1$ then the gap is closed and the target has been captured.

• (4) Otherwise the target has escaped, and the lion will continue chasing it pointlessly.

The above suggests the following strategy to solving the problem:

• (a) derive the equation of the pursuit curve as a function $y=f(x)$, then

• (b) determine the turning point $f'(x_1)=0$, and finally

• (c) compare its abscissa $x_1$ to the lion reach of $1$ meter; in particular, solving the limit case $x_1=1$ gives the condition for the maximum number of lions that still allows escaping.

The math is quite similar to Pursuit curves solution (with the necessary adjustments), so the following will be skimpier on step-by-step explanations.

Let the speeds of the target and pursuer be $u,v$, respectively, and define $\lambda=\frac{u}{v} \ge 1$. Let the target move at constant speed along the $y$ axis on the line $(0,u\,t)$, with the pursuer following the radiodrome $\big(x(t),y(t)\big)$ starting from $x_0=x(0), y_0=y(0)$.

(a)  The condition that the pursuer aims directly towards the target translates to: $$ y' = \frac{y - u\,t}{x} $$ The arc length covered by the pursuer with constant speed $v$ after time $t$ is:

$$ v\,t = \int_{x}^{x_0} \sqrt{1 + (y')^2} \;dx $$

Eliminating $t$ between the previous equations:

$$ \frac{1}{\lambda}(y - x\,y') = \int_{x}^{x_0} \sqrt{1 + (y')^2} \;dx $$

Taking the derivates in $x$:

$$ \require{cancel} \frac{1}{\lambda}(\bcancel{y'} - \bcancel{y'} - x\,y'') = -\,\sqrt{1 + (y')^2} $$

With $w = y'$ and after separation of variables:

$$ \frac{dw}{\sqrt{1+w^2}} = \frac{\lambda \,dx}{x} $$

Integrating:

$$ \operatorname{arcsinh}\big(w(x)\big) = \lambda \ln(x) + C $$

The initial conditions $x=x_0, w(x_0)=y'(x_0) = \frac{y0}{x0}$ give the integration constant:

$$ C = \operatorname{arcsinh}(y_0\,/\,x_0) - \lambda \ln(x_0) $$

At this point $y(x)$ can be explicitly calculated as $\int w \,dx$, but that's not needed at the next step.

(b)  The condition $y'(x_1)=w(x_1)=0$ translates to:

$$ 0 = \lambda \ln(x_1) + \operatorname{arcsinh}(y_0\,/\,x_0) - \lambda \ln(x_0) $$ $$ \lambda \ln(x_1 / x_0) = - \operatorname{arcsinh}(y_0\,/\,x_0) $$ $$ (x_1 / x_0)^\lambda = e^{-\operatorname{arcsinh}(y_0\,/\,x_0)} $$ $$ x_1 = \frac{x_0} {\left(\frac{y_0}{x_0} + \sqrt{1+\left(\frac{y_0}{x_0}\right)^2}\right)^{\frac{1}{\lambda}}} = \left(\frac{x_0^{\lambda+1}}{y_0 + \sqrt{x_0^2+y_0^2}}\right)^{\frac{1}{\lambda}} $$

(c)  Particularizing the general equations for $n$ lions and the limit case $x_1=1$:

$$ \begin{cases} x_0 = 200 \,\sin\left(\frac{\pi}{n}\right) \\ y_0 = 200 \,\cos\left(\frac{\pi}{n}\right) \\ \lambda = 10 \end{cases} $$

gives:

$$ 200 \;\sqrt[10]{\frac{\sin^{11}(\pi/n)}{1 + \cos(\pi/n)}} = 1 $$

which solves numerically to:

$$ n \simeq 364.439 $$

It follows that $364$ lions allow an escape path, while $365$ will corner the target.

$$ n \;\approx\; \pi \; \sqrt[11]{\frac{200^{10}}{2}} \;\simeq\; 364.442 $$