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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.
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?