You may also like

At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Darts and Kites

Explore the geometry of these dart and kite shapes!

No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.

Pegboard Quads

Age 14 to 16
Challenge Level

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

NonagonNonagon with vertices numbered clockwise
If we draw lines from a particular vertex to every other, then the angle separating $2$ adjacent vertices is $20^\circ$ (Can you explain this? -- Ed.) . 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 $ABC = 20 \times (|A-C|)$ 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 $ABC = 20 \times ( 9 - |A-C| )$ for any angle where A < B < C or A > B > C .

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

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{eqnarray} ABC + BCD + CDA + BAC &=& 20 \times (9-|A-C|)+20 \times (9-|B-D|)+20 \times (|C-A|)+20 \times (|D-B|)\\ &=&360. \end{eqnarray} 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
\begin{eqnarray} 20 \times ( 9 - |A-C|) + 20 \times (|A-C|) &=& \mbox{ sum of to opposite angles} \\ & = &20 \times (9 - |A-C| + |A-C| ) \\ &=& 20 \times 9 = 180^\circ \end{eqnarray}