Iranian Mathematical Olympiad

Fix a radius of circle $C$, $OX$ and for each segment $AB$ from the perimeter of $P$ (don't include polygon vertices in this process), consider the projection of $AB$ to $OX$ ($A^{\prime }{B}^{\prime }$) and the projection of $AB$ to the perimeter of $C$ ($A^{\prime \prime }{B}^{\prime \prime }$).



Call $A^{\prime }{B}^{\prime }$ the radius projection of $AB$ and $A^{\prime \prime }{B}^{\prime \prime }$ the perimeter projection of $AB$. It's obvious that $$AB<A^{\prime }{B}^{\prime }+A^{\prime \prime }{B}^{\prime \prime }.$$ Now let $\alpha$ be the sum of all radius projections of all edges of $P$ and $\beta$ be the sum of all perimeter projections of all edges of $P$. From the above, $\alpha +\beta >36>4+10\pi$, hence $\alpha >4$ or $\beta >10\pi$:

If $\alpha >4$, there is a point on $OX$ that is covered by $5$ radius projections. Call this point $Y$, then the circle with center $O$ and radius $OY$ will intersect $\mathcal{P}$ at least $6$ times. (Because if a circle intersects a polygon and doesn't pass through its vertices, it will have an even number of intersections.)

If $\beta >5\times 2\pi$, then there is a point on the perimeter that is covered by at least $6$ perimeter projections. Call this point $Z$, then radius $OZ$ intersects $\mathcal{P}$ at least $6$ times.

We are done.