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Maximum of a acute angles

What is the maximum number of acute angles in a convex polygon?

Solution

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 get:

    \[\Theta = \sum_{i=1}^{n}\alpha_i+\sum_{j=1}^{N-n}\beta_j=\pi (N-2)\]

and knowing the following facts:

    \[\forall i \in \left\{1,n \right\}; \quad \alpha_i<\frac{\pi}{2} \Rightarrow \sum_{i=1}^{n}\alpha_i<\frac{\pi}{2}n\]

    \[\forall j \in \left\{1,N-n \right\}; \quad \beta_i<\pi \Rightarrow \sum_{i=1}^{N-n}\beta_i<\pi (N-n)\]

    \[\Rightarrow \Theta = \sum_{i=1}^{n}\alpha_i+\sum_{j=1}^{n-N}\beta_j<\frac{\pi}{2} \left( n + 2(N-n) \right)= \frac{\pi}{2} \left( 2N-n \right)\]

    \[\Rightarrow \Theta <\frac{\pi}{2} \left( 2N-n \right)\]

    \[\Rightarrow \pi (N-2) < \frac{\pi}{2} \left(2N-n \right)\]

    \[\Rightarrow 2(N-2)<2N-n\]

    \[\Rightarrow 2n-4 < 2N-n\]

    \[\Rightarrow n < 4\]

    \[\Rightarrow n=3 \quad \blacksquare\]

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