TITLE: what is the maximum number of acute angles
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: 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:
$$\forall i \in \left\{1,n \right\}; \quad \alpha_i<\frac{\pi}{2} \Rightarrow \sum_{i=1}^{n}\alpha_i<\frac{\pi}{2}n$$
$$\forall j \in \left\{1,N-n \right\}; \quad \beta_i<\pi \Rightarrow \sum_{i=1}^{N-n}\beta_i<\pi (N-n)$$
$$\Rightarrow \Theta = \sum_{i=1}^{n}\alpha_i+\sum_{j=1}^{n-N}\beta_j<\frac{\pi}{2} \left( n + 2(N-n) \right)= \frac{\pi}{2}   \left( 2N-n \right)$$
$$\Rightarrow \Theta <\frac{\pi}{2} \left( 2N-n \right) $$
$$\Rightarrow \pi (N-2) < \frac{\pi}{2} \left(2N-n \right)$$
$$\Rightarrow 2(N-2)<2N-n$$
$$\Rightarrow 2n-4 < 2N-n$$
$$\Rightarrow n < 4 $$
$$\Rightarrow n=3 \quad \blacksquare$$