what is the maximum number of acute angles What is the maximum number of acute angles in a convex 10-gon in the Euclidean plane ? 
I know that the answer is at least $4$.
Any idea how to proceed. 
 A: just for completeness, the other day, while i was trying to find an answer for the 5-gon case, found the answer for the n-gon case. Here's the answer:
In a convex $N$-gon, if we take the max number $n$ of interior angles $\alpha_i$ less than $\frac{\pi}{2}$.
$$\forall i \in \left\{0,...,n \right\}; \quad \alpha_{i}<\frac{\pi}{2}$$
In all $N$-gon, the sum of all interior angles $\theta_i$ is:
$$\Theta = \sum_{i=1}^{N}\theta_i=\pi  \left( N-2 \right) $$
If we take the rest $N-n$ interior angles, called $\beta_j$, we got:
$$\Theta = \sum_{i=1}^{n}\alpha_i+\sum_{j=1}^{N-n}\beta_j=\pi (N-2)$$
and knowing the following facts:
$$\forall i \in \left\{1,n \right\}; \quad \alpha_i<\frac{\pi}{2} \Rightarrow \sum_{i=1}^{n}\alpha_i<\frac{\pi}{2}n$$
$$\forall j \in \left\{1,N-n \right\}; \quad \beta_i<\pi \Rightarrow \sum_{i=1}^{N-n}\beta_i<\pi (N-n)$$
$$\Rightarrow \Theta = \sum_{i=1}^{n}\alpha_i+\sum_{j=1}^{n-N}\beta_j<\frac{\pi}{2} \left( n + 2(N-n) \right)= \frac{\pi}{2}   \left( 2N-n \right)$$
$$\Rightarrow \Theta <\frac{\pi}{2} \left( 2N-n \right) $$
$$\Rightarrow \pi (N-2) < \frac{\pi}{2} \left(2N-n \right)$$
$$\Rightarrow 2(N-2)<2N-n$$
$$\Rightarrow 2n-4 < 2N-n$$
$$\Rightarrow n < 4 $$
$$\Rightarrow n=3 \quad \blacksquare$$
A: 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.
A: You're going to have a tough time making it 4.  Think about this:  If you have an obtuse angle in the 10-gon, replace it with a straght line.  You remove one angle and the two angles that you modified got more acute.  Now prove that the largest $n$ such that you have a convex $n$-gon with only acute angles is 3.
A: No, the answer is $3$, in fact no convex polygon can have more than $3$ acute angles, this is because the sum of the external angles is $360$, and an external angle at a vertex for an internal acute angle is larger than $90$ degrees.
