One can easily establish the easy bound $\frac{n}{2}$ by noticing that each internal bisector can't be repeated more than $2$ times, $n$ odd is very annoying, $f(3) = 3$ and $f(5) = 3$ (you can show this by a continuity argument) might imply that there's a difference in $n \mod 4$. The doom of C will always be the fact that I can't find the construction, maybe because I never found the answer.

Ok, this is an interesting problem. Good for the test? I wouldn't say so. Combinatorics? Nahhhh... A geometry problem? definitely not. Nevertheless, a stray problem is always interesting: Of course, it's easy to notice that the answer for even $n$ is $\dfrac n2$. Each angle bisector could have at most $2$ vertices of the polygon, and a regular $n$-gon gives us $n/2$ bisectors. It's also quite easy to construct a $n$-gon with $\dfrac{n+3}2$ bisectors for odd $n$: we can just take a regular $(n-1)$-gon and cut a slack out of 2 consecutive sides: We'll actually prove that the answer is $\dfrac{n+1}2$ if $n\equiv 1\pmod4$ and $\dfrac{n+3}2$ if $n\equiv 3\pmod 4$. The key idea in solving this problem is remembering $2$ composition lemmas: $\cdot$ The composition of two reflections over lines $r$, $s$ is a rotation of angle $2\angle(r,s)$ with center $r\cap s$. $\cdot$ The composition of two rotations $\varphi_1$, $\varphi_2$ with (directed) angles $\theta_1$, $\theta_2$ is a rotation $\varphi$ with angle $\theta_1+\theta_2$. First, let us prove it is impossible to construct a $n$-gon with only $\dfrac{n+1}2$. Suppose FTSOC we can have such $n$-gon. Let it be $P_1P_2P_3\dots P_n$. If we use only $\dfrac{n+1}2$ bisectors, $\dfrac{n-1}2$ of them should have $2$ vertices of the polygon. WLOG let us assume that the bisector of angle $P_n$ is the one with only one polygon vertex. Finally, let us define the Andrey's circle of a set of lines (Brazilian name lol) to be a circle that contains all the intersection points of these lines on its inside. Let $\mathcal{C}$ be the Andrey's circle of the $\dfrac{n-1}2$ bisectors that have 2 vertices inside them, and $R_1$, $R_2$, $\dots$, $R_{n-1}$ the intersection points of the $\dfrac{n-1}2$ lines with $\mathcal{C}$ (Of course, we count the intersection point with $R_i$ associated with the vertex $P_i$): Case for n=11 It's not so hard to prove that the order we counted these intersection points is also the order in which they appear in the circumference. After all, The angle bisectors' directions should also appear in the order of the directions of the segments (If you are a fan of vectors, you can see this by looking at the vectors $\overrightarrow{P_iP_{i+1}}$ and seeing their directions appear in order, so their bisectors should appear that way too) This way, we can think about the construction of $P_1P_2\dots P_n$ as a series of reflections over $\overleftrightarrow{R_iR_{i+\frac{n-1}2}}$ (consider the indices modulo $n-1$). We start at the line $\overleftrightarrow{P_1P_n}$, do a series of reflections and get to $P_{n-1}P_n$. Ok, now we are finally able to introduce... The catch. Let's look at this composition of reflections: Define $r_i=\overleftrightarrow{R_iR_{i+\frac{n-1}2}}$, for all $1\le i\le \dfrac{n-1}2$ and let $f_i: \mathbb{R}^2\mapsto \mathbb{R}^2$ be the reflection over $r_i$. Consider indices modulo $\dfrac{n-1}2$: $$f_{n-1}(f_{n-2}(\dots f_1(\overleftrightarrow{P_nP_1})\dots ))=\overleftrightarrow{P_{n-1}P_n}$$ As $n-1$ is even, we can pair the reflections to show they are rotations and do all the rotations to get even another rotation : $f_{n-1}\circ f_{n-2}$, $f_{n-3}\circ f_{n-4}$, $\dots$, $f_2\circ f_1$ are rotations with angles $2\angle(r_{n-2},\ r_{n-1})$, $\dots$, $2\angle(r_1,\ r_2)$ so their composition should be a rotation with angle $$2(\angle(r_1,\ r_2) + \angle(r_3,\ r_4) +\dots+\angle(r_{\frac{n-1}2},\ r_{\frac{n+1}2}) +\dots + \angle(r_{n-2},\ r_{n-1}))$$$$=2(\angle(r_1,\ r_2)+\angle(r_2,\ r_3)+\angle(r_3,\ r_4)+\dots+\angle(r_{\frac{n-1}2},r_1)=2\cdot 180^{\circ}=360^{\circ}$$ A rotation of $360^\circ$ is, of course, the identity. Therefore, we couldn't have this composition of reflections giving us $2$ different lines ($\overleftrightarrow{P_nP_1}$ and $\overleftrightarrow{P_{n-1}P_n}$) Now, for the construction for $n\equiv 1 \pmod 4$... n=5 We can originally take the $\dfrac{n-1}2$ bisectors of a regular $(n-1)$-gon, rotate one of them a bit, I don't know, call it $\epsilon$ and reflect the original $(n-1)$-gon line that passed through this now skewed line over all of them. In the end, the polygon will still be convex (after all, we only changed a line by a small $\epsilon$ and the product of all these reflections should be a rotation of degree $4\epsilon$. I leave the details of this example to the reader, but we'll get a polygon that reaaaaaaaally looks like a regular $(n-1)$-gon. Remark: This problem is really different than anything I imagined could be in OBM, I'm impressed it wasn't in any other olympiad Remark 2: It was pure horror trying to write this solution during the test while a clock counted how much time I had left in front of me

Conjecturing the answer and getting the construction is the hard step on the problem imho, nice sol @above

I wouldn't say so. The answer and the examples come directly from the proof, you find them by trying to limit how you construct the n-gon using the method of compositions

Welp, I didn't solve it, so I would take my own opinion with a grain of salt , but in contest, I think it would have been helpful to have a notion on why there's a difference between n=5 and n=3, and that could have happened with a better construction for n=5 (less adhoc), at least after that I think it would have been more natural to prove the answer for me. I think it is not uncommon for the construction and the bound itself have a similar thought process behind them.