2016 AMC 10B Problem 11

1. Say a short side has $n$ posts, counting its two corners. Then a long side has $2n$ posts.
2. Add up all four sides: $n + n + 2n + 2n = 6n$. But each of the $4$ corner posts sits on two sides, so it was counted twice. Take away the extra $4$:
$$6n - 4 = 20, \qquad \text{so } n = 4.$$
3. A short side has $4$ posts. Between $4$ posts in a row there are only $3$ gaps (picture posts at $0, 4, 8, 12$ yards). So the short side is $3 \cdot 4 = 12$ yards.
4. A long side has $2 \cdot 4 = 8$ posts, so $7$ gaps, and it is $7 \cdot 4 = 28$ yards.
5. Check: going all the way around a closed loop, the number of gaps equals the number of posts, $20$. So the distance around is $20 \cdot 4 = 80$ yards, and indeed $12 + 28 + 12 + 28 = 80$.
6. The area is $12 \cdot 28 = 336$ square yards.

The answer is $\boxed{\textbf{(B)}\ 336}$.

This is a "fence post" problem, and it has two separate traps. Corner posts belong to two sides, so they get counted twice if you add side by side. And along one straight side there is one fewer gap than posts, while around a closed loop there are exactly as many gaps as posts.

Start from the loop. There are $20$ gaps all the way around, so the perimeter is $80$ yards and one short side plus one long side is $40$ yards. A short side with $n$ posts has $n - 1$ gaps and a long side has $2n - 1$ gaps, which is $3n - 2$ gaps together. So $4(3n - 2) = 40$, giving $n = 4$ and sides $12$ and $28$.