by B S Beevers 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 etting it right Now polygons with 3, 4 and 5 sides are special cases, after which there appears to be a clear pattern for the maximum A familiar investigation is to find the maximum numbers of number of right angles. interior angles of a polygon which can be right angles. This may be investigated by an able pupil as follows: Analytical Approach A triangle can have one right angle. A quadrilateral can have four right angles. For a polygon with n sides, sum of interior angles = 180n - 360 degrees. Pentagon Suppose it has p right angles. Number of degrees remaining for other angles = 180n - 360 Sum of Interior Angles = 540'. - 90p. Four right angles would leave 180', which is impossible. Average size of remaining angles must be less than 360'. So a pentagon has a maximum of three right angles, as shown. Hence: 180n - 360 - 90p < 360 n-p Hence 180n - 360 - 90p < 360n - 360p - 270p < 180n + 360 1 3 Hexagon -p< - (2n + 4) n = 3, 4 and 5 are special cases. Sum of Angles = 720'. For n > 5, the maximum number of right-angles is the 5 right angles = 450', leaving 270'. largest integer less than (2n + 4) So a hexagon can have 5 right-angles, as shown. Note that p cannot equal (2n + 4) or the average remaining angle would equal 360. Here is a table of values ofp for n > 5. n 6 7 8 9 10 11 12 13 1 2 10 (2n 3 + 4)3 51 3 6 623 71 38 8 93 10 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 1 1 Sum of Angles = 900'. If n = 3m then q (6m + 4) = 2m+ 1 +- So p = 2m + 1 6 right angles = 540', leaving 360', which is impossible. Sop= 2m + 1 So maximum number is 5. Octagon 1 Sum of angles = 1080'. If n = 3m+ 1then q=- 3 (6m + 6) = 2m + 2 7 right angles = 630', leaving 450'. So p = 2m + 1 6 right angles = 540', leaving 540'. So an octagon can have 6 right angles. 1 2 Nonagon If n = So3m +p 2 =then 2m q--3 +(6m 2 + 8) = 2m + 2 + - Sum of angles = 1260'. Sop= 2m + 2 8 right angles = 720', leaving 540'. 7 right angles = 630', leaving 630'. Conclusions: If n = 3m then p = 2m + 1 Since 630/2 = 315 < 360, this is alright. If n = 3m + 1then p = 2m + 1 So maximum number of right angles is 7, as shown. 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.