A convex polygon has exactly twice as many diagonals as it has sides. How many sides does it have?

A polygon with $n$ sides has $n(n-3)/2$ diagonals, since each vertex joins to the $n-3$ vertices that are not itself or either neighbour, and every diagonal is then counted twice. Setting $n(n-3)/2=2n$ and cancelling the $n$ (which cannot be zero) gives $n-3=4$, so $n=7$. Check: a heptagon has $7\times 4/2=14$ diagonals, exactly twice its 7 sides. The trap 5 drops the division by 2 and solves $n(n-3)=2n$; 9 comes from setting the diagonal count to three times the sides; and 6 is a guess that fails, since a hexagon has 9 diagonals, not 12.