$\varphi$ -antastic sum of quadrances

The vertices of the pentagon are given by

$P_k = ( -\sin ( \dfrac{2 \pi k }{5} ) , \cos ( \dfrac{2 \pi k}{5} ) )$

The outward normals of the sides of the pentagon passing through $\{ P_k \}$ are given by

$N_k = ( - \sin ( \dfrac{ (2 k + 1) \pi }{5 } , \cos ( \dfrac{ (2k+1)\pi }{5 } ) )$

Using polar coordinates, we can write $P = ( r \cos t , r \sin t )$

The distance between $P$ and the $k$ -th side is given by

$d_k = N_k \cdot ( P - P_k ) = ( - \sin ( \dfrac{ (2 k + 1) \pi }{5 } ) , \cos ( \dfrac{ (2 k+1)\pi}{5 } ) ) \cdot ( r \cos t + \sin ( \dfrac{2 \pi k }{5} ) , r \sin t - \cos ( \dfrac{2 \pi k}{5} ) )$

Expanding the above expression

$d_k = - r \cos t \sin ( \dfrac{ (2 k + 1) \pi }{5 } ) + r \sin t \cos ( \dfrac{ 2 \pi k}{5 } ) - \cos( \dfrac{\pi}{5})$

Simplifying further, this becomes

$d_k = r \sin( t - \dfrac{ (2 k + 1) \pi }{5 } ) - \cos ( \dfrac{\pi}{5} )$

So that the square of the distance $d_k$ is given by

${d_k}^2 =r^2 \sin^2( t - \dfrac{ (2 k + 1) \pi }{5 } ) - 2 r \cos ( \dfrac{\pi}{5} )\sin( t - \dfrac{ (2 k + 1) \pi }{5 } ) + \cos^2 ( \dfrac{\pi}{5} )$

Simplifying this further, it becomes,

${d_k}^2 =r^2 (\frac{1}{2} - \frac{1}{2} \cos( 2 t - \dfrac{ (4 k + 2) \pi }{5 } ) - 2 r \cos ( \dfrac{\pi}{5} )\sin( t - \dfrac{ (2 k + 1) \pi }{5 } ) + \cos^2 ( \dfrac{\pi}{5} )$

Adding all $5$ of the squared distances $\{ {d_k}^2 \}$ and noting that

$\displaystyle \sum_{k=0}^4 \sin( \tau + \dfrac{2 \pi k}{5} ) = 0$

and that,

$\displaystyle \sum_{k=0}^4 \cos( \tau + \dfrac{4 \pi k}{5} ) = 0$ Hence, the required sum of squared distances reduces to,

$\displaystyle \sum_{k=0}^{4} {d_k}^2 = \dfrac{5}{2} r^2 + 5 \cos^2 (\dfrac{\pi}{5} )$

And this means that the sum is a function of the distance of point $P$ from the origin. Thus the locus of constant sum points is a circle of radius $r$ .

Finally, if the sum equals $10 + \dfrac{5}{4} \varphi^2$ , then

$\dfrac{5}{2} r^2 + 5 \cos^2 (\dfrac{\pi}{5} ) = 10 + \dfrac{5}{4} \varphi^2$

which solves to $r = 2$ .