An interesting differential equations problem Suppose we have four ants, initially at rest, at the four corners of a square centered at the origin. They start walking clockwise, each ant walking directly toward the one in front of him. Suppose also that each ant walks with unit velocity, derive a differential equation that describes the trajectories.
Thought:
Here is the situation of the problem

After some time $t$, the ants are now at points $E,F,G,H$. If we denote by $\mathbf{r(t)}$ the position of ant at $A$, we know $\mathbf{r(0)} = (1,1)$ and after some $t$, at point $E$, we have $\mathbf{r(t)} = (x(t),y(t)) = E $. We are given that
$$ \frac{ y - 1 }{x - 1 } = 1 $$
Also, using the arclength formula, we know the path of ant $A$ is
$$ \int\limits_0^t \sqrt{ (x')^2 + (y')^2 } dt $$
Am I on the right track?

here is my try. let us look for a symmetric solution. so i will assume the ants at  $re^{i\theta}= z, iz, -z, -iz$
the differential equation satisfied by the ant at $z$ is $$ \frac{d}{dt}\left(re^{i\theta}\right) =  e^{i (\theta + 3\pi/4)} .$$ this can be written as 
$$\frac{d r}{dt} = -\frac 1 {\sqrt 2}, \  \frac{d \theta}{d t} = \frac 1{r\sqrt 2} \mbox{ with the initial conditions }  r = \sqrt 2, \theta = \pi/4. $$
this has solution $$ r = \sqrt 2 - \frac t {\sqrt 2}, 
\theta = \pi/4 + \ln\left( \frac 2{ 2-t}\right), 0 \le t < 2.$$
