2014 AMC 10A Problem 18

A square with sides parallel to the axes has only two different heights. This one has four, so it is tilted.

1. Draw a box around it. Put an upright rectangle around the square, touching its four corners. For a tilted square this box is itself a square, and the tilted square cuts off four matching right triangles in the box's corners.
2. Which corners are opposite? The two diagonals of a square cross at their middles, so each pair of opposite corners has the same average height. Split $0, 1, 4, 5$ into two pairs with equal sums: the only way is $0 + 5 = 1 + 4$. So the corner at height $5$ is opposite the bottom corner, and the corners at heights $1$ and $4$ are its neighbours.
3. The triangles' legs. The four corner triangles are all the same size and shape. One of them has the side from height $0$ up to height $1$ as its long side, so it has a vertical leg of $1$. Another has the side from height $0$ up to height $4$ as its long side, so it has a vertical leg of $4$. Since the triangles are identical, each one has legs $1$ and $4$.
4. Side of the square. Each side is the long side of a right triangle with legs $1$ and $4$:
$$\text{side}^2 = 1^2 + 4^2 = 17.$$
5. The area of a square is side times side, so the area is $17$.

To check, here is an actual square: $(0,0)$, $(4,1)$, $(3,5)$, $(-1,4)$. Each side goes $4$ one way and $1$ the other, so each has length $\sqrt{17}$, and the heights are $0, 1, 5, 4$.

The answer is $\boxed{\textbf{(B)}\ 17}$.

When you turn a side of a square through a right angle to get the next side, its "across" and "up" amounts swap. So the heights of a tilted square are $0$, $a$, $b$ and $a + b$, where $a$ and $b$ are the two legs of each side's right triangle. Here $a = 1$ and $b = 4$, and indeed $1 + 4 = 5$. The side squared is $a^2 + b^2$.

Use the box directly. The box goes from height $0$ to height $5$, so it is a $5$ by $5$ square with area $25$. The four corner triangles each have legs $1$ and $4$, so each has area $2$, and together they have area $8$. The tilted square is what is left: $25 - 8 = 17$.

An axis-parallel square has only two distinct $y$-coordinates, so this square is tilted and has a unique lowest vertex $P$ at height $0$.

Walk along the two sides leaving $P$. One side rises by some amount $p$ and runs horizontally by $h$; the other side is that vector rotated $90^\circ$, so it rises by $h$ and runs by $p$ (in the opposite horizontal direction). The vertex opposite $P$ is reached by adding both sides, so it sits at height $p + h$.

The four heights are therefore $0,\ p,\ h,\ p + h$. Matching with $0, 1, 4, 5$ forces $\{p, h\} = \{1, 4\}$, and indeed $1 + 4 = 5$.

Each side is the hypotenuse of a right triangle with legs $1$ and $4$, so
$$\text{side}^2 = 1^2 + 4^2 = 17,$$
which is the area of the square.

The answer is $\boxed{\textbf{(B)}\ 17}$.