19.3. Can the number of diagonals of a polygon be exactly twice the number of its sides?

$(6-8$ grades $)$
19.3. From one vertex of an $n$-sided polygon, $n-3$ diagonals emanate. Then the product $n(n-3)$ expresses twice the number of all diagonals, since in this counting each diagonal was counted twice. It remains to check whether the equation $0.5 n(n-3)=2 n$ has a solution in natural numbers. It is easy to see that such a solution is $n=7$. Thus, the number of diagonals of a convex heptagon is twice the number of its sides.