2014 AMC 10A Problem 18

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 $\boxed{\textbf{(B)}\ 17}$.

For a square, rotating one side vector $(h, p)$ by $90^\circ$ gives the adjacent side $(-p, h)$: the horizontal run of one side equals the vertical rise of the other. That single fact turns the four given heights into the two legs of the side's right triangle. Any tilted square can be boxed inside an axis-parallel square, with four congruent right triangles in the corners; the legs of those triangles are exactly the differences of consecutive heights.

Concretely, take $P = (0,0)$, then $(4, 1)$, $(3, 5)$, $(-1, 4)$. Adjacent sides $(4,1)$ and $(-1,4)$ are perpendicular (dot product $-4 + 4 = 0$) and have equal length $\sqrt{17}$, so this is a valid square, and its area is $17$.

• Trying to make the square axis-parallel, or taking the side length to be a difference of y-coordinates such as 4 or 5.
• Pairing the heights as legs $1$ and $5$ or $4$ and $5$, giving $26$ or $41$; the legs are the two smaller gaps, whose sum is the largest height.