Runinig dogs

Since the entire system is rotationally symmetrical, then if one dog is at polar coordinates $(r,\theta)$ , then the next dog around (counterclockwise) will be at $(r,\theta+\frac{2\pi}{3})$ . So if the first dog chases the second, the tangent to the path of the first dog's movement must pass through the second dog, i.e. the gradient between the dogs equals the derivative of the curve at the position of the first dog.

Now for polar coordinates $(r,\theta)$ , $x=r\cos\theta$ and $y=r\sin\theta$ . Since the gradient between two points equals $\frac{y_2-y_1}{x_2-x_1}$ :

$\begin{aligned}\frac{dy}{dx}&=\frac{r\sin\left(\theta+\frac{2\pi}{3}\right)-r\sin\theta}{r\cos\left(\theta+\frac{2\pi}{3}\right)-r\cos\theta}\\ \frac{\frac{dy}{dr}}{\frac{dx}{dr}}&=\frac{\sin\theta\cos\frac{2\pi}{3}+\cos\theta\sin\frac{2\pi}{3}-\sin\theta}{\cos\theta\cos\frac{2\pi}{3}-\sin\theta\sin\frac{2\pi}{3}-\cos\theta}\\ \frac{\frac{d}{dr}(r\sin\theta)}{\frac{d}{dr}(r\cos\theta)}&=\frac{-\frac{1}{2}\sin\theta+\frac{\sqrt3}{2}\cos\theta-\sin\theta}{-\frac{1}{2}\cos\theta-\frac{\sqrt3}{2}\sin\theta-\cos\theta}\\ \frac{\left(\frac{d}{dr}r\right)\sin\theta+r\left(\frac{d}{dr}\sin\theta\right)}{\left(\frac{d}{dr}r\right)\cos\theta+r\left(\frac{d}{dr}\cos\theta\right)}&=\frac{-\frac{3}{2}\sin\theta+\frac{\sqrt3}{2}\cos\theta}{-\frac{3}{2}\cos\theta-\frac{\sqrt3}{2}\sin\theta}\\ \frac{\sin\theta+r\cos\theta\frac{d\theta}{dr}}{\cos\theta-r\sin\theta\frac{d\theta}{dr}}&=\frac{\sqrt3\sin\theta-\cos\theta}{\sqrt3\cos\theta+\sin\theta}\\ \sqrt3\sin\theta\cos\theta+\sin^2\theta+\sqrt3r\cos^2\theta\frac{d\theta}{dr}+r\sin\theta\cos\theta\frac{d\theta}{dr}&=\sqrt3\sin\theta\cos\theta-\cos^2\theta-\sqrt3r\sin^2\theta\frac{d\theta}{dr}+r\sin\theta\cos\theta\frac{d\theta}{dr}\\ \sqrt3r\sin^2\theta\frac{d\theta}{dr}+\sqrt3r\cos^2\theta\frac{d\theta}{dr}&=-\sin^2\theta-\cos^2\theta\\ \sqrt3r\frac{d\theta}{dr}\left(\sin^2+\cos^2\right)&=-\left(\sin^2+\cos^2\right)\\ \frac{d\theta}{dr}&=-\frac{1}{\sqrt3r}\end{aligned}$

Before we can use the formula for arc length and find the distance the dogs travelled, we need to know where they start.

Using the cosine rule: $\begin{aligned}a^2&=2r^2-2r^2\cos\frac{2\pi}{3}\\ a^2&=r^2(2+2\frac{1}{2})\\ r^2&=\frac{a^2}{3}\\ r&=\frac{a}{\sqrt3}&\text{where }r>0,a>0\end{aligned}$

Now we can use the formula for arc length to determine the distance the dogs travel between the corner of the triangle and the centre: $\begin{aligned}\text{arc length}&=\int_a^b\sqrt{dx^2+dy^2}\\ &=\int_{r=0}^{r=\frac{a}{\sqrt3}}\sqrt{dr^2\left(\frac{dx}{dr}\right)^2+dr^2\left(\frac{dy}{dr}\right)^2}\\ &=\int_0^{\frac{a}{\sqrt3}}\sqrt{\left(\cos\theta-r\sin\theta\frac{d\theta}{dr}\right)^2+\left(\sin\theta+r\cos\theta\frac{d\theta}{dr}\right)^2}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\sqrt{\left(\cos\theta+r\sin\theta\frac{1}{\sqrt3r}\right)^2+\left(\sin\theta-r\cos\theta\frac{1}{\sqrt3r}\right)^2}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\sqrt{\frac{1}{3}\left(\sqrt3\cos\theta+\sin\theta\right)^2+\frac{1}{3}\left(\sqrt3\sin\theta-\cos\theta\right)^2}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\frac{1}{\sqrt3}\sqrt{3\cos^2\theta+2\sqrt3\sin\theta\cos\theta+\sin^2\theta+3\sin^2\theta-2\sqrt3\sin\theta\cos\theta+\cos^2\theta}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\frac{1}{\sqrt3}\sqrt{3\cos^2\theta+\sin^2\theta+3\sin^2\theta+\cos^2\theta}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\frac{1}{\sqrt3}\sqrt{4\left(\sin^2\theta+\cos^2\theta\right)}dr\\ &=\int_0^{\frac{a}{\sqrt3}}\frac{2}{\sqrt3}dr\\ &=\left[\frac{2}{\sqrt3}\right]_0^{\frac{a}{\sqrt3}}\\ &=\frac{2a}{3}\end{aligned}$

Now, time=distance/speed, so

$\large\boxed{t=\frac{2a}{3v}}$