To solve the given problem, we start with the known formulas for a polygon with \( n \) sides:
The interior angle sum is also expressed as: \[ (2n - 4) \times 90 \] This is just another form of writing \( (n - 2) \times 180 \), since: \[ (n - 2) \times 180 = (2n - 4) \times 90 \]
We are told that the difference between the sum of interior angles and exterior angles is: \[ (2n - 4) \times 90 - 360 = 120n \]
Let's simplify the left-hand side: \[ (2n - 4) \times 90 - 360 = 180n - 360 - 360 = 180n - 720 \] So the equation becomes: \[ 180n - 720 = 120n \] Subtract \( 120n \) from both sides: \[ 60n - 720 = 0 \] \[ 60n = 720 \] \[ n = 12 \]
Now that we know the polygon has \( n = 12 \) sides, we can find the number of diagonals.
The formula for the number of diagonals in an \( n \)-sided polygon is: \[ \frac{n(n - 3)}{2} \] Substituting \( n = 12 \): \[ \frac{12 \times (12 - 3)}{2} = \frac{12 \times 9}{2} = \frac{108}{2} = 54 \]
Therefore, the number of diagonals is: \( \boxed{54} \)
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: