Logarithmic Spiral- N-gon [duplicate]

In the mice problem, also called the beetle problem, $n$ mice start at the corners of a regular $n$-gon of unit side length, each heading towards its closest neighboring mouse in a counterclockwise direction at constant speed. The mice each trace out a logarithmic spiral, meet in the center of the polygon, and travel a distance $$d=\frac{1}{1-\cos\big(\frac{2\pi}{n}\big)}$$

Does anybody know where can I find a proof of the formula ?

Solution 1:

Assume the $n$-gon is centered at the origin. Pick any mouse and let $\phi$ be the constant angle between the mouse's instantaneous direction of travel (i.e., the tangent line to the spiral) and the radial line from the origin. Then the equation in polar form of a mouse's trajectory is $$ r = a\exp(-b\theta) $$ where $\tan \phi = \frac 1b$ and $a$ is the distance to the origin at the start (so $a$ is the distance from center of $n$-gon to any vertex). Check that $\cos\phi=\sin\frac\pi n$ for a regular $n$-gon, and that $\sin\frac\pi n= \frac1{2a}$ if the $n$-gon has unit side length. The arc length of the path traveled by the mouse is then $$ d\stackrel{(*)}={a\over{\cos\phi}} = \frac1{2\sin^2\frac\pi n} =\frac1{1-\cos\frac{2\pi}n}. $$

(*) Proof of arc length: The arc length in polar coordinates is $$ \int_{\theta=0}^\infty\sqrt{r^2+\left({dr\over d\theta}\right)^2}\,d\theta $$ which in the logarithmic spiral case simplifies to $$\int_0^\infty \sqrt{1+b^2}\left[a\exp(-b\theta)\right]\,d\theta=a{\sqrt{1+b^2}\over b}.$$ The conclusion follows from $${\sqrt{1+b^2}\over b}=\sqrt{\frac1{b^2}+1}=\sqrt{\tan^2\phi +1}=\frac1{\cos\phi};$$ the cosine is positive since $0<\phi<\frac\pi 2$.