2018 AMC 10B Problem 15

Folding paper does not stretch it, so a straight line on the flat sheet keeps its length after folding. Follow one well-chosen line.

1. Pick the line. Draw the line from the center of the sheet to one of its corners. This is half of the sheet's diagonal.
2. Across the bottom. The box is turned so its corners lie on the sheet's midlines. Then the line toward a corner of the sheet hits the middle of one side of the base, straight on. The center of the base to the middle of a side is half the side: $\frac{w}{2}$.
3. Up the side. The paper folds up the side face. The line was straight on to the bottom edge, so it goes straight up the face: length $h$.
4. Across the top. The paper folds again over the top edge and runs straight in to the center of the top, point $A$, where the corner ends up. That is another $\frac{w}{2}$.
5. Half the diagonal.
$$\frac{w}{2} + h + \frac{w}{2} = w + h.$$
So the whole diagonal of the sheet is $2(w + h)$.
6. Area from the diagonal. A square with diagonal $d$ has area $\frac{d^2}{2}$ (it is two right triangles with base $d$ and height $\frac{d}{2}$). So
$$\text{Area} = \frac{\big(2(w+h)\big)^2}{2} = \frac{4(w+h)^2}{2} = 2(w+h)^2.$$

The answer is $\boxed{\textbf{(A)}\ 2(w+h)^2}$.

A path on the flat paper becomes a chain of straight pieces on the box, and folding keeps every length. Choosing a path that crosses every fold straight on (here the diagonal) makes each piece easy to measure. For a square, knowing the diagonal is enough to get the area.

Test an extreme case. Let the box become a very thin pole, $w = 0$, of height $h$. Each corner of the sheet must reach the top of the pole, so half the diagonal is $h$ and the area is $\frac{(2h)^2}{2} = 2h^2$. Put $w = 0$ into the choices: (A) gives $2h^2$, (B) gives $\frac{h^2}{2}$, and (C), (D), (E) all give $0$. Only (A) fits.