Download Getting It Right

by B S Beevers
etting it right
A familiar investigation is to find the maximum numbers of
interior angles of a polygon which can be right angles. This
may be investigated by an able pupil as follows:
A triangle can have one right angle.
A quadrilateral can have four right angles.
The results so far may be put into a table as follows:
Number of Sides 3 4 5 6 7 8 9 10 11 12
Number of rt. angles 1 4 3 5 5 6 7 7 8 9
Now polygons with 3, 4 and 5 sides are special cases, after
which there appears to be a clear pattern for the maximum
number of right angles.
Analytical Approach
For a polygon with n sides, sum of interior angles = 180n 360 degrees.
Pentagon
Suppose it has p right angles.
Sum of Interior Angles = 540'.
Four right angles would leave 180', which is impossible.
So a pentagon has a maximum of three right angles, as shown.
Number of degrees remaining for other angles = 180n - 360
- 90p.
Average size of remaining angles must be less than 360'.
Hence:
180n - 360 - 90p
n-p
< 360
Hence 180n - 360 - 90p < 360n - 360p
- 270p < 180n + 360
1
-p< - (2n + 4)
3
Hexagon
n = 3, 4 and 5 are special cases.
For n > 5, the maximum number of right-angles is the
largest integer less than (2n + 4)
Sum of Angles = 720'.
5 right angles = 450', leaving 270'.
Note that p cannot equal (2n + 4) or the average
So a hexagon can have 5 right-angles, as shown.
remaining angle would equal 360.
Here is a table of values ofp for n > 5.
1
n 6 7 8 9 10 11 12 13
2
10
(2n
+ 4)
513
6 623
71 3
8 8 9
10
3
3
3
p5
5
6
7
7
8
9
9
We can obtain a formula for three cases as follows:
1
Let q = (2n + 4)
3
Heptagon
Sum of Angles = 900'.
6 right angles = 540', leaving 360', which is impossible.
So maximum number is 5.
1
1
If n = 3m then q (6m + 4) = 2m+ 1 +So p = 2m + 1
Sop= 2m + 1
Octagon
1
If n = 3m+ 1then q=-3 (6m + 6) = 2m + 2
Sum of angles = 1080'.
7 right angles = 630', leaving 450'.
6 right angles = 540', leaving 540'.
So an octagon can have 6 right angles.
So p = 2m + 1
1
Nonagon
2
If n = So
3m +p2 =
then
q--3 +
(6m
2m
2 + 8) = 2m + 2 + Sop= 2m + 2
Sum of angles = 1260'.
8 right angles = 720', leaving 540'.
7 right angles = 630', leaving 630'.
Since 630/2 = 315 < 360, this is alright.
So maximum number of right angles is 7, as shown.
Conclusions: If n = 3m then p = 2m + 1
If n = 3m + 1then p = 2m + 1
If n = 3m + 2 then p = 2m + 2
Example: Let n = 20
Then 20= 3 x 6 + 2, so m = 6
Hencep 2 x 6 + 2 = 14.
So the maximum number of right angles for a 20 sided
polygon is 14. Address
18 Welburn Avenue, Leeds LS16 5HJ.
Mathematics
in
School,
March
1996
37