Let P be a convex polygon of n sides containing a maximal number of acute angles. If one of the angles of P is chosen random, what percent represents the probability that the angle chosen will be acute?

I know if I take a decagon (10-gon) the most it will have is 3. Because the angles of the 10-gon need to sum up to $8\cdot 180^\circ = 1440^\circ$. If there are $4$ acute angles, then they sum up to less than $360^\circ$, which means the remaining $6$ angles need to sum up to more than $1080^\circ$, impossible since each one is at most $180^\circ$.

On the other hand, to have $3$ acute angles, draw four consecutive segments that almost form a square, and connect the two end vertices with six short segments.

How can I set this up with smaller cases to get the probability to then expand to the n-case


The maximum number of acute angles is always $3$ - effectively, take an equilateral triangle and curve out one (or more) of the edges slightly with a series of short edges. And you cannot ever have $4$ acute angles for the reasons you gave.

So the probability of choosing an acute angle in such a polygon is always $\frac {\large 3}{\large n}$


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