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2007 AMC 10B Problem 10

2007 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 2 min · position P1-10

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 $S?$

A)
$\text{two parallel lines}$
B)
$\text{a parabola}$
C)
$\text{a circle}$
D)
$\text{a line segment}$
E)
$\text{two points}$

Answer

Key insight

Fixed base BC and fixed area force a fixed height, and the points at a fixed distance from line BC form two parallel lines.

Solution

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

Why this works

"Constant area with a fixed base" is the same as "constant distance from a line," which is the basic locus behind every problem about triangles sharing a base. Remember both sides of the line, and remember that the apex can sit far outside the segment $BC$ as long as its perpendicular distance is right.

The trap

Thinking only of points on one side of BC and answering 'a line', or forgetting that the locus extends beyond the segment BC.

Common mistakes

Thinking only of points on one side of BC and answering 'a line', or forgetting that the locus extends beyond the segment BC.
Confusing "fixed area" with "fixed distance from a point" and answering a circle.

Techniques

Set up the equation/formula and compute; no special trick needed