Prove that for every $n\ge3$ there exists a convex $n$-gon with exactly 3 acute angles

Question

I'm really not sure where to start. Induction can really be used, and that seems like the only way to prove for all $n$.

Answer 1

I will construct an explicit series of polygons satisfying the conditions. No induction needed.

• Three consecutive vertices of the polygon $A,B,C$ define an equilateral triangle.
• The remaining $n-3$ vertices lie along the arc between $A$ and $C$ with centre $B$ and are joined in sequence.

$\angle B$ is 60° by definition, while $\angle A$ and $\angle C$ are acute because they are the angles between a chord and a radius. For each other vertex, its neighbours form a chord that lies inside the $AC$ chord, so its angle must be greater than 120° and thus obtuse. All angles are less than 180°, so the polygon is convex and the proof is finished.

Answer 2

You might start with an equilateral triangle, and put the other $n-3$ vertices on an arc of a circle joining two vertices of the triangle and centred on the third.