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 Question from jason, a student: find the set of polygons in which the number of diagonals is greater than the sum of the measures of the angles

Ugh! I don't much care for that question because it depends on the units used to measure the angles. You will get a much different answer in radians than in degrees. So it's a very unnatural thing to ask.
How to solve it: A diagonal corresponds to a pair of nonadjacent vertices. There are "n choose two" or n(n-1)/2 pairs of vertices and n of those are adjacent. Thus n(n-3)/2 diagonals.

In radians, the sum of the angles is π × (n-2). So you need to solve the inequality

(n-2)π < n(n-3)/2

which you do by solving the corresponding equation, which can be rewritten as

2 × π × n - 4 × π = n2 -3n

or in standard form

n2 - (3+2 × π)n + 4 × π = 0

Plugging that into the quadratic formula gives about 7.637, so a heptagon should have fewer diagonals (14) than its total internal radian measure (5 π) while an octagon should have more (20 vs 6 π). And lo, it was so.

Now, if you want your answer based on degree measure [sum of angles is 180(n-2)], you know how to solve it. The answer will of course be much larger.

Good Hunting!
RD

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.