Let, $P$ be a closed polygon with $10$ sides and $10$ vertices. Let,$k$ be the number of interior angles of $P$ greater than $180^\circ$. Then the maximum value of $k$ is-
All I could make out is that the sum of all interior angles is $8\pi$. I have no idea how to proceed on this.
Thanks for any help. The result may please be generalised if possible.
In an $n$-gon, the total of the interior angles is $(n-2)\pi$. A concave angle is $>\pi$. There cannot be $n-2$ concave angles, as then the total of the angles would be $>(n-2)\pi$. But $n-3$ concave angles is possible, as demonstrated in this image (with $n=7$).
More rigorously, to show that $n-3$ is always possible: let $P_1=(1,1)$, and with $\theta=\frac{\pi}{n-2}$ $P_{2+k}=(\sin k\theta,\cos k\theta)$ for all $k=0,1,\dots,n-2$. Then the $n-3$ angles of polygon $P_1P_2\dots{}P_n$ at points $P_3,P_4,\dots,P_{n-1}$ are easily proven to be concave.