Two Regular Polygons
If both shapes now have to be regular polygons, but do not need to be the same, and each polygon can have any number of sides, could the orange angle still be $81^\circ$, and if that is possible how many sides would each polygon have?
Find solutions for when the orange angle is $27^\circ$ and when it is $54^\circ$.
Can you make a conjecture about the connection between the size of the orange angle and the number of sides on each polygon.
If you can, are you able to justify your conjecture?
If the polygons are regular there are only certain size exterior angles that are possible.
Work systematically through those to see if you find some helpful possibilities.
Simon from Elizabeth College, Guernsey found solutions using a spreadsheet :
We have made up a spreadsheet from Simon's description, the columns are :
-
number of sides for Polygon A
-
exterior angle for Polygon A is 360/n
-
exterior angle for Polygon B is 81 - column B
-
number of sides required for that exterior angle is 360/column C
Here's the spreadsheet file : Two Regular Polygons
Simon continues :
Polygon A starts with few sides and increases, Polygon B is calculated for each case.
As A increases in side number, B must decrease its own side number.
So once A exceeds B no new solutions will be found.
For a target angle of 81 I found two solutions : 5-40 and 8-10.
I then tried it with a target angle of 54
This gave me four solutions : 7-140, 8-40, 10-20 and 12-15
Finally I tried with a target angle of 27
This gave me five solutions : 14-280, 15-120, 16-80, 20-40 and 24-30
Simon then summarised the situation:
As the number of sides doubles, the angle is halved, and in general : the number of sides is in inverse proportion to the exterior angle.
27 is one third of 81 so it includes solutions that are three times the solutions for 81.
54 is double 27 so it has solutions which are half the size of solutions for 27 provided that value is an integer.
Excellent thinking and skilful application of a spreadsheet Simon, very well done.
This problem draws on angle properties of polygons and factors of 360. Students need to work systematically to find solutions and then reason carefully to justify the completeness of their method.
There is great value in comparing the combinations that work for 81, for 27 and for 54, and then accounting for any observed connections between the relative sizes of angles and number of sides.
Some rich directions in which to open up the task might include:
It is to be hoped that this kind of questioning allows students to reflect on the extent to which the unit of angle is arbitary (the degree as one part in 360 for a complete rotation). Although 81, or another number, has no decimal part, the two angles that together make that sum may have decimal parts, or at least that possibility needs considering . . .