This solution was submitted by Michael of St John Payne school. Does anyone have any other ideas?

Nonagon

Nonagon with vertices numbered clockwise

If we draw lines from a particular vertex to every other, then the angle separating 2 adjacent vertices is

20^{\circ}

<span class=ëditorial»(Can you explain this? == Ed.)</span>. If we label the polygon as shown in the right hand diagram, the angle 416 is going to be two lots of

20^{\circ}

and therefore

40^{\circ}

. This means that we have a general rule:

Angle $ABC = 20 * (| A - C |)$ when $A > C$ and $A > B$ or $A < C$ and $A < B$ . But this will only work for angles where B isn't between A and C. For angles where B is between A and C we need to go the long way round the polygon from A to C, or in other words the whole polygon take away A to C. So we can come up with the rule: Angle $ABC = 20 * (9 - | A - C |)$ for any angle where $A < B < C$ or $A > B > C$ .

E.g. Angle $124 = 20 * (9 | 1 - 4 |) = 20 * 6 = 120$

We can use these rules to show that the angles in any quadrilateral ABCD on the pegboard must add up to $360^{\circ}$ . If you rotate the peg board so that A is always at point 1 then ABC and BCD will obey the second rule and CDA and DAB will obey the first rule. Therefore

$\begin{matrix} ABC + BCD + CDA + BAC & = & 20 * (9 - | A - C |) + 20 * (9 - | B - D |) + 20 * (| C - A |) + 20 * (| D - B |) \\ = & 360 . \end{matrix}$

We can also deduce that 2 opposite of the quadrilateral angles will add up to $180^{\circ}$ : For quadrilateral ABCD, where the pegboard is rotated so that point A is on peg 1, Angle ABC will obey the first rule and Angle CDA will obey the second rule. Therefore $20 * (9 - | A - C |) + 20 * (| A - C |)$ = sum of to opposite angles = $20 * (9 - | A - C | + | A - C |) = 20 * 9 = 180^{\circ}$ !