Why is the minimum number of acute exterior angles in a convex octagon independent of whether the octagon has equal-sides?

Question

Why is the minimum number of acute exterior angles in a convex octagon independent of whether the octagon has equal-sides?

In particular, I know five is the number of acute exterior angles in a convex octagon, and three are obtuse.

Answer 1

For any convex polygon, the total exterior angle is $360^\circ$. [Proof: let $\alpha_1, \alpha_2 ...$ be the interior angles, then $\sum _i(180-\angle\alpha_i)=180n-\sum_i\angle\alpha_i=180n-(n-2)180=360$].

Now there cannot be four exterior angles $\geq 90$: otherwise they already add to $\geq 360$, which implies these four exterior angles are all $90$ and the remaining exterior angles are $0$, that is absurd (if $n>4$). In conclusion, for convex polygons with $n>4$ sides, there are at most three non-acute exterior angles, the remaining $n-3$ many exterior angles are acute. It is easy to construct an $n$-gon with exactly n-3 acute exterior angles.