Problem source

Four greyhounds $A,B,C$ and $D$ are held at positions such that $ABCD$ is a large square. At a given instant, the dogs are released and $A$ runs directly towards $B$ at constant speed $v$, $B$ runs directly towards $C$ at constant speed $v$, and so on. Show that $A$'s path is given in polar coordinates (referred to an origin at the centre of the field and a suitable initial line) by $r=\lambda\mathrm{e}^{-\theta},$ where $\lambda$ is a constant. 
Generalise this result to the case of $n$ dogs held at the vertices of a regular $n$-gon ($n\geqslant3$).

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \coordinate (O) at (0,0);
        \filldraw (O) circle (1pt);
        \node at (O) [left] {$O$};

        \draw (1,0) -- (0,1) -- (-1, 0) -- (0, -1) -- cycle;

        \draw[-latex, ultra thick, red] (1,0) -- (0.6, 0.4);
        
    \end{tikzpicture}
\end{center}

It's straightforward to see that $\dot{r} = -\frac{v}{\sqrt{2}}$ and $\tan (\theta(t + \delta t) - \theta(t)) = \frac{v\delta t/\sqrt{2}}{r - v \delta t/\sqrt{2}} = \frac{v}{r\sqrt{2}} \delta t + o(\delta t^2) \Rightarrow \dot{\theta} = \frac{v}{r \sqrt{2}}$

Solving this system, we can see that $r(t) = \frac{l - vt}{\sqrt{2}}$ and $\frac{\d r}{\d \theta} = -r \Rightarrow r = \lambda e^{-\theta}$ where $\lambda = \frac{l}{\sqrt{2}}$ where $l$ is the initial square side length.

\begin{center}
    \begin{tikzpicture}[scale=3]
        \coordinate (O) at (0,0);
        \filldraw (O) circle (1pt);
        \node at (O) [left] {$O$};

        \def\n{7}  % Number of sides (hexagon)
        \def\r{1}  % Radius of the polygon
    
        % Draw the regular polygon
        \draw[thick, blue] (1,0) \foreach \i in {1,...,\n} {
            -- (\i * 360 / \n : \r)
        } -- cycle;
        
        \coordinate (A) at (1,0);
        \coordinate (B) at (1 * 360 / \n : \r);
        \coordinate (C) at ($(A)!0.5!(B)$);
        \draw[-latex, ultra thick, red] (A) -- (C);

        \node at ($(A)!0.25!(B)$) [right] {$v \delta t$};
        \draw[dashed] (C) -- (O) -- (A);
        \node at ($(O)!0.5!(A)$) [below] {$r$};
        
        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = A--O--C};

        
    \end{tikzpicture}
\end{center}
By the cosine rule, we can see that $r(t + \delta t)^2 = r^2+(v\delta t)^2 - 2rv\delta t \cos (\frac12 (\pi - \frac{2\pi}{n}))$, ie:

$\frac{r(t + \delta t)^2 - r^2}{\delta t} = - 2r v \sin(\frac{\pi}{n}) \Rightarrow \dot{r} = - v \sin (\frac{\pi}{n})$.

We can also observe that $\tan (\theta(t + \delta t) - \theta(t)) = \frac{v \delta t \sin (\frac{\pi}{2} - \frac{\pi}{n})}{r - v \delta t \cos (\frac{\pi}{2} - \frac{\pi}{n})} \Rightarrow \dot{\theta} = \frac{v}{r} \cos (\frac{\pi}{n})$.

Combining these, we can see that $\frac{\d r}{\d \theta} = - r \tan (\frac{\pi}{n}) \Rightarrow r = \lambda e^{-\tan(\frac{\pi}{n})t}$ where $\lambda$ is the initial radius of the circumscribed circle.