Computer the smallest value of n for which a regular polygon of n sides has at least 2006 diagonals. I got 2006, is this correct?

Question

brinazarski · Answer

*compute

jim_thompson5910 · Answer

let d = number of diagonals and n = number of sides

jim_thompson5910 · Answer

the formula is

d = n(n-3)/2

so because d = 2006, we know that

d = n(n-3)/2

2006 = n(n-3)/2

2006*2 = n^2 - 3n

4012 = n^2 - 3n

n^2 - 3n - 4012 = 0

Solve that for n to get your answer

brinazarski · Answer

3/2 + $$\sqrt{4014.25}$$

brinazarski · Answer

?

jim_thompson5910 · Answer

use a calculator to find the decimal form of that number

brinazarski · Answer

I did, it seems right...

jim_thompson5910 · Answer

you should get n = 64.85810919 and this rounds up to n = 65

brinazarski · Answer

yup.... but why round up? It doesn't say round up

jim_thompson5910 · Answer

because you can't have a decimal/fractional number of sides

brinazarski · Answer

Why not?

jim_thompson5910 · Answer

how do you have 0.85 of a side?

brinazarski · Answer

By have .85 of a side. You can use a ruler and make a square with each side 1.5 inches, can't you?

jim_thompson5910 · Answer

no, n = number of sides, not length of the sides

brinazarski · Answer

Oh... that's true. I'm sorry. Thank you! :)

jim_thompson5910 · Answer

no worries, yw