AMC 10 · 2007 · #10
Grade 6 geometry-2dTwo points B and C are in a plane. Let S be the set of all points A in the plane for which △ABC has area 1. Which of the following describes S?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two points $B$ and $C$ are fixed in a plane. Consider every point $A$ for which triangle $ABC$ has area exactly $1$. Describe the shape formed by the collection of all such points $A$.
Givens: Points $B$ and $C$ are fixed, so segment $BC$ has a fixed length; $S$ is the set of all points $A$ with $[\triangle ABC]=1$; Answer choices: (A) two parallel lines, (B) a parabola, (C) a circle, (D) a line segment, (E) two points
Unknowns: Which named shape the set $S$ of all valid apex points $A$ forms
Understand
Restated: Two points $B$ and $C$ are fixed in a plane. Consider every point $A$ for which triangle $ABC$ has area exactly $1$. Describe the shape formed by the collection of all such points $A$.
Givens: Points $B$ and $C$ are fixed, so segment $BC$ has a fixed length; $S$ is the set of all points $A$ with $[\triangle ABC]=1$; Answer choices: (A) two parallel lines, (B) a parabola, (C) a circle, (D) a line segment, (E) two points
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
Drawing segment $BC$ and asking where the apex $A$ can sit (Tool #1) turns an abstract description into a picture of a moving point. Naming the fixed base length and the altitude (Tool #4) converts the area condition into a single equation that fixes the height. Once the shape is identified, Tool #3 (Eliminate Possibilities) checks it against each answer choice, since the problem is asking us to name one of five shapes.
Execute — Answer: A
6.G.A.1 Step 1 Write the area with $BC$ as base
- Draw the fixed segment $BC$ and treat it as the base of the triangle.
- The area of a triangle is half the base times the height, where the height is the perpendicular distance from the apex $A$ down to line $BC$.
- Because $B$ and $C$ are fixed, the base length $BC$ never changes; only the height can change as $A$ moves.
💡 The base $BC$ is locked, so the area of the triangle is controlled entirely by how far $A$ is from line $BC$.
6.EE.B.7 Step 2 Fixing the area fixes the height
- Set the area equal to $1$.
- Let $b=BC$ be the fixed base length and $h$ be the height.
- Then $\tfrac12\,b\,h=1$, so $h=\dfrac{2}{b}$.
- Since $b$ is a fixed number, $h$ is a single fixed positive number too.
- Every valid apex $A$ must sit exactly this one distance away from line $BC$.
💡 With the base pinned down, only one height gives area $1$, so $A$ is trapped at a fixed distance from the base line.
4.G.A.1 Step 3 Collect every point at that fixed distance
- Now find all points at the fixed distance $h=\tfrac{2}{b}$ from line $BC$.
- On one side of line $BC$, those points form a straight line running parallel to $BC$ at distance $h$; on the other side they form a second parallel line the same distance away.
- The apex can slide freely along either track and still keep area $1$, so $S$ is these two full lines, both parallel to $BC$.
- Checking the choices: a circle (C) would be a fixed distance from a point, not a line; a parabola (B) needs a focus-and-directrix rule, which is not what we have; a line segment (D) or two points (E) would be bounded, but $A$ runs off to infinity along each track.
- Only two parallel lines survives, so the answer is $\textbf{(A)}$.
💡 All points a fixed distance from a straight line form two parallel lines, one on each side.
6.G.A.1 Draw the fixed segment $BC$ and treat it as the base of the triangle. The area o 6.EE.B.7 Set the area equal to $1$. Let $b=BC$ be the fixed base length and $h$ be the he 4.G.A.1 Now find all points at the fixed distance $h=\tfrac{2}{b}$ from line $BC$. On on Review
Reasonableness: Test the shape with a concrete case: put $B=(0,0)$ and $C=(2,0)$, so $b=2$ and the required height is $h=\tfrac{2}{2}=1$. Any apex like $(0,1),(5,1),(-3,1)$ gives base $2$ and height $1$, hence area $1$ — these all lie on the line $y=1$. The points $(0,-1),(4,-1)$ give area $1$ too and lie on $y=-1$. Two horizontal lines $y=1$ and $y=-1$ appear, both parallel to $BC$ on the $x$-axis, exactly matching two parallel lines and ruling out a bounded shape.
Alternative: Use coordinates from the start: place $B=(0,0)$ and $C=(b,0)$. The area of triangle $ABC$ with $A=(x,y)$ is $\tfrac12\,b\,|y|$. Setting this equal to $1$ gives $|y|=\tfrac{2}{b}$, so $y=\tfrac{2}{b}$ or $y=-\tfrac{2}{b}$. Each equation is a horizontal line, and $x$ is free — two parallel lines, confirming (A).
CCSS standards used (min grade 6)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Expressing the triangle's area as half the fixed base $BC$ times the height from apex $A$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $\tfrac12\,b\,h=1$ to show the height $h=\tfrac{2}{b}$ is a single fixed value.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Recognizing that all points a fixed distance from line $BC$ form two lines parallel to $BC$.)
⭐ Fixing the base and the area of a triangle locks the apex to one distance from the base line, and every point that far from a line makes two parallel lines.
⭐ Fixing the base and the area of a triangle locks the apex to one distance from the base line, and every point that far from a line makes two parallel lines.
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