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Grade: 11
The angles of a convex n-sided polygon form an arithmetic progression whoose common difference (in degrees) is an non-zero integer. Find the largest possible value of n for which this is possible. (A polygon is convex if it's interior angles are all less than 180°.)
5 months ago

Answers : (1)

Aditya Gupta
1808 Points
hello utsav this is a wonderful question. we will use the fact that sum of internal angles of all polygons (convex as well as concave) is equal to 180(n-2), n being the number of sides.
let a1, a2, a3, … be the angles in increasing order. obviously a1 is greater than zero and an is less than 180.
now, sum of angles cn also be writen as n/2*(2a1+(n-1)d)= 180(n-2) or a1= 180(1 –  2/n) – (n – 1)d/2
an= a1+(n-1)d= 180(1 –  2/n) – (n – 1)d/2+(n-1)d= 180(1 –  2/n) + (n – 1)d/2 must be less than 180
or (n – 1)d/2 must be less than 360/n
or 720/n(n – 1) must be greater than d
but d is greater than equal to 1.
so 720/n(n – 1) must be greater than 1
or 720 must be greater than n(n – 1)
or n^2 – n – 720 less than 0
by wavy curve, you get n less than equal to 27.33 but since n is an integer,
nmax= 27
kindly approve :)
5 months ago
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