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n-sided polygons with n greater than or equal to 362.

The number of diagonals of an n-sided polygon is n(n-3)/2. The sum of the measures of the angles in an n-sided polygon is 180(n-2).

So for the condition in the problem to be true,

n*(n-3)/2>180(n-2)

64979>64800

Simplifying and bringing all of the terms to the left-hand side:

n2-363n+720>0

Find the roots of n2-363n+720=0, using the quadratic formula:

n=(363+/-sqrt(3632-4(1)(720)))/2, or approximately (363+/-359.01)/2, so the roots are approximately 2 AND just more than 361

Consider each of the intervals: (-inf, 2); (2, 362); (362, inf) [since we are only interested in whole numbers, we can use the smallest whole number greater than the root]:

Plugging in one point from each interval, we see the given inequality is true in the first interval, false in the second interval, and true in the third interval. Since we are only interested in n>2 (since this is a polygon), we get the set of all polygons, with n>=362 sides. For n=362, we have 64979 diagonals and 64800 degrees for the sum of the interior angles.

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Q: What is the set of polygons for which the number of diagonals is greater than the sum of the measures of the angles?
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