Convex Polygons
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
Show that if a convex polygon has more than six sides, then at least one of the sides has an obtuse angle at both ends.
To trace a path around a convex polygon the turn at each vertex will be in the same direction and the total turn (the sum of the exterior angles) is 360 °. At each vertex of the polygon where there is an acute angle the exterior angle is more than 90 °. If there were four or more acute angles the sum of the exterior angles at these vertices would be more than 360 ° which is impossible.
Alternatively we can use the fact that the sum of the interior angles is (2n-4) right angles where n is the number of sides. If there are t acute angles (all less than a right angle) and (n-t) obtuse angles (all less than 2 right angles) and no reflex angles then, by considering the total number of right angles:
| $(2n-4)$ | < | $t + 2(n-t)$ |
| $t$ | < | $4$ |
With a six sided polygon it is possible to alternate the three acute angles with the three obtuse angles so that none of the sides has an obtuse angle at both ends. With seven or more sides there must be two consecutive obtuse angles as there are at most three acute angles.