TITLE: Differential Equation Math Puzzle
QUESTION [10 upvotes]: Dog race:
Edit 2:
I posted a possible answer below. However, I am unsure how the authors arrived at the solution.
Maybe someone can offer an explanation.
Four dogs are positioned at the corners of a square ($d= 1m$), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square.

(a) Why is the total length of the path just $1 m$?
(b) Find and solve a differential equation for the radius $r(\theta)$ in polar coordinates.
This is a homework question! I just want hints, no solutions please.
The tough part is setting up an equation for the radius i.e for the motion of one of the dogs. I was thinking about an equation similar to that of an archimedean spiral.
$$r(\theta)=a+b\theta \space \space \space \text{or} \space \space \space r(\theta)=a\theta^{\frac{1}{n}}$$
However, I have no idea what values $a$ or $b$ should be. Any hints are appreciated.
Edit 1:
This is my second attempt at a solution:
If Dog 1 is positioned at $(r, \theta)$ $\implies$ Dog 2 is positioned at $(r, \theta+\frac{\pi}{2})$ 
Picture: 

$$x_1=r \cos (\theta) \\ y_1= r \sin (\theta) \\ \\ \\ x_2=r \cos (\theta+\frac{\pi}{2})=-r \sin(\theta) \\y_2=r \sin (\theta+\frac{\pi}{2})=r \cos (\theta)$$
If these are the two position vectors then the vector joining the two points is my velocity vector.
$$\implies \frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}=\frac{r \sin (\theta+\frac{\pi}{2})-r \sin (\theta)}{r \cos (\theta+\frac{\pi}{2})-r \cos (\theta)}=\frac{ \sin (\theta+\frac{\pi}{2})- \sin (\theta)}{ \cos (\theta+\frac{\pi}{2})- \cos (\theta)}=\frac{\cos(\theta)-\sin(\theta)}{-  \sin(\theta)-\cos(\theta)}$$
Am I on the right path? But How do I deduce $\dfrac{dr}{d\theta}$?

REPLY [4 votes]: Here is a solution for (b):
Let
$$t\mapsto z(t)=r(t)e^{i\phi(t)}\qquad(t\geq0)$$ be the orbit of the dog starting at ${1\over2}(1+i)$. Then the orbit of the dog starting at ${1\over2}(-1+i)$ is simply $t\mapsto iz(t)$. It follows that at any moment the velocity vector $$\dot z=(\dot r+i r\dot\phi)e^{i\phi}$$ is parallel to $$iz-z=(-1+i)z=(-1+i)r e^{i\phi}\ .$$ This amounts to
$$\dot r+ir\dot\phi=\lambda(-r+ir)\tag{1}$$
for some $\lambda>0$ changing with time. Comparing the imaginary parts in $(1)$ we see that in fact $\lambda=\dot\phi>0$, so that from looking at the real parts we obtain $\dot r=-r\dot\phi$, or
$${dr\over d\phi}={\dot r\over\dot\phi}=-r\ .$$
It follows that the function $\phi\mapsto r(\phi)$ satisfies the differential equation
$$r'=-r\ ,$$
with the solutions $r(\phi)=r_0 e^{-(\phi-\phi_0)}$. This implies that we see four logarithmic spirals.