2014 AMC 10A Problem 18

P16-20~3 minPrint

Coordinate Geometry Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ , $1$ , $4$ , and $5$ . What is the area of the square?

Show solution without answering

Key insight

Opposite corners of a square have heights adding to the same total, so 0 and 5 are opposite. Each side is the long side of a 1-by-4 right triangle.

Solution

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

Why this works

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 $a^2 + b^2$ .

Alternative approach

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: $25 - 8 = 17$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Algebra, Chapter 8: Graphing Lines