By does not pass through the center of $C$ does it mean just $P$'s border does not contain the center of $C$, or $C$ is outside the region enclosed by polygon $P$?

stroller wrote: By does not pass through the center of $C$ does it mean just $P$'s border does not contain the center of $C$, or $C$ is outside the region enclosed by polygon $P$? $P$'s border.

$5 + 10\pi{\sim} 36$

Let $X$ be a point on $P$ and $\Delta\ell$ be a small line part of the polygon at the point $X$. Let $\Delta r$ be the projection of $\Delta\ell$ on the radius $OX$ ($O$ is the center of $C$) and $\Delta \phi$ be the angle between the two radii that connect $O$ to the endpoints of $\Delta\ell$. Then $$(1)\,\,\,\,\,\,\,\,\,\,|\Delta\ell|\leq |\Delta r|+|\Delta \phi|$$Let $X_0\in P$ be the point with minimal $|OX_0|$. We begin moving a point $X$, starting from $X_0$ along $P$ and ending again at $X_0$. Let $\Delta\ell_i$ be the corresponding small pieces of $P$. The sum of the projections on $OX$, when $X$ makes a full turn will be $$(2)\,\,\,\,\,\,\,\,\,\, R:= \sum_i |\Delta r_i|$$and the sum of corresponding angles $\Delta\phi$ will be $$(3)\,\,\,\,\,\,\,\,\,\, \Phi:= \sum_i |\Delta\phi_i|$$If we assume $R>5$, there will be a point $r_0$ on $r$ that is a common to at least $6$ pieces $\Delta\ell_i$, which means the concentric circle with radius $r_0$ meets at least $6$ points of $P$. On the other hand assuming $\Phi>5\cdot 2\pi$ implies there will be at least $6$ pieces $\Delta_i$'s which correspond to one and the same azimuth $\phi$ of $OX$. It means that the radius $OX$ at the azimuth $\phi$ intersects at least $6$ points of $P$. So, we should prove that it holds either $R>5$ or $\Phi >10\pi$. Unfortunately $5+10\pi>36$ and we can not get a contradiction using $(1)$. But the same idea could be refined. Let $I_1$ be the set of those indices $i$ for which $\Delta r_i\geq 0$ and $I_2$ the set of those $i$ with $\Delta r_i<0$. Thus, the projections $\Delta r_i$ start from some $X_0\in[0,1]$ moving up, then down,up,... and finish again down at $X_0$. So, any point on $[X_0,1]$ lies on the same number of upward $\Delta r_i$'s and downward $\Delta r_i$'s and moreover $\sum_{i\in I_1}\Delta r_i=\sum_{i\in I_2}|\Delta r_i|$. Hence if we prove that $\sum_{i\in I_1}\Delta r_i> 2$, there will be a point $X_1\in[X_0,1]$ that lies on at least three $\Delta r_i, i\in I_1$. But then it also lies on another three $\Delta r_i, i\in I_2$. That's why, it's enough to prove either $R>4$ or $\Phi>10\pi$. Now, seeking a contradiction, we assume $R\leq 4$ and $\Phi\leq 10\pi$. Using (1) we get $$36=\sum_i |\Delta\ell_i|\leq R+\Phi\leq 4+10\pi< 36$$a contradiction.

Parametrize the smoothed polygon with $(r(t),\theta(t))$ in polar coordinates. Suppose each circle concentric with $C$ passes through at most $5$ points of $P$ (hence at most $4$ points for all but finitely many $r$'s), and each radius of $C$ intersects $P$ in at most $5$ times. We can approximate $P$ with a smooth curve with arbitrary precision while still keeping these constraints satisfied. The length of $P$ is $$\int_t \sqrt{ r^2 \theta'^2 + r'^2 } dt \le \int |r\theta'| dt + \int |r'| dt \le \int |\theta'| dt + \int |r'| dt.$$ Note that $\int |\theta'| dt$ is simply the distance $\theta$ has travelled as $t$ ranges over its domain. When going around the polygon, each radius is visited $\le 5$ times, hence $\int |\theta'| dt \le 5\cdot 2\pi = 10 \pi$. Shift $r$ such that $r'(0) = 0$. Let $t_1,...,t_n$ be the times $t$ for which $r'(t) = 0$. Then $n \le 4$, else there'd be a value of $\rho$ for which $r(t) = \rho$ for $\ge 6$ values of $t$, contradiction. Thus $\int |r'| dt = \sum_{i=1}^n |\int_{t_{i-1}}^{t_i} r' dt | = \sum_{i=1}^n |r(t_i) - r(t_{i-1})| dt \le n \le 4$ where we've defined $t_0 = 0$. Therefore the length of any approximation of $P$ is at most $2\cdot 5\pi + 4 < 35.99$, hence the perimeter of $P$ must be $< 36$, as desired.