+1859 The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?
+721 In a regular polygon with n sides, we can draw nC2 diagonals (excluding the n sides themselves). Therefore, we have the equation:
nC2 = 2n
To solve for n, we can rewrite the binomial coefficient using factorials:
n(n-1) / 2 = 2n
Simplifying the equation:
n(n-1) = 4n
n^2 - n - 4n = 0
n^2 - 5n = 0
Factoring the equation:
n(n-5) = 0
Therefore, the possible values of n are 0 and 5. However, a regular polygon cannot have 0 sides. So, the only valid solution is n = 5.
Therefore, the regular polygon has 5 sides. This corresponds to a pentagon, where the number of diagonals (10) is indeed twice the number of sides (5).
