2007 AMC 10B Problem 10

P1-10~2 minPrint

Triangles: Area & Pythagorean Coordinate Geometry Set up the equation/formula and compute; no special trick needed

Two points $B$ and $C$ are in a plane. Let $S$ be the set of all points $A$ in the plane for which $\triangle ABC$ has area $1.$ Which of the following describes $S?$

Show solution without answering

Key insight

The base BC is fixed, so area 1 fixes the height. Points at that distance from line BC form two parallel lines, one on each side.

Solution

Use $BC$ as the base of the triangle, since it never changes.

1. Suppose, for example, $BC = 4$ . Area is half of base times height, so
$\tfrac12 \times 4 \times h = 1 \quad\text{gives}\quad h = \tfrac12.$
Whatever the length of $BC$ , the height comes out to one fixed number.
2. The height is how far $A$ is from the line through $B$ and $C$ . So $A$ works exactly when it is that fixed distance from the line.
3. Which points are a fixed distance from a line? Those on a parallel line above it and those on a parallel line below it.
4. $A$ can slide anywhere along either parallel line, even far past the ends of $BC$ ; the height stays the same, so the area stays $1$ .

So $S$ is two parallel lines.

The answer is $\boxed{\textbf{(A)}\ \text{two parallel lines}}$ .

Why this works

"Same base, same area" means "same height", and "same height" means "same distance from a line". That is why triangles with a shared base and their top corner on a parallel line all have equal area. Remember both sides of the base line.

Concise solution

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

Use $BC$ as the base of the triangle. Its length $d = BC$ is fixed, so the area condition
$\tfrac{1}{2}\, d \cdot h = 1 \quad\Longrightarrow\quad h = \frac{2}{d}$
pins down the height $h$ : the distance from $A$ to the line $BC$ must equal the constant $2/d$ .

The points at a fixed positive distance from a given line lie on two lines parallel to it, one on each side. Every point on those two lines gives area exactly $1$ , and no other point does.

The answer is $\boxed{\textbf{(A)}\ \text{two parallel lines}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Algebra, Chapter 8: Graphing Lines