AMC 10 · 2007 · #10
Grade 6 geometry-2dPick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Write the area with BC as base
Take the fixed segment BC as the base. Area is half base times height, and since BC never changes, only the height from A can vary.
The base BC is locked, so the area of the triangle is controlled entirely by how far A is from line BC.
6.G.A.1Draw A DiagramFixing the area fixes the height
Set the area to 1 with base b=BC. Then 1/2·b·h=1 gives h=2/b, one fixed number, so every apex A sits exactly that far 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.
6.EE.B.7Introduce A VariableCollect every point at that fixed distance
Every point at distance 2/b from line BC lies on one of two lines parallel to BC, one per side. Both are unbounded, so the answer is (A).
All points a fixed distance from a straight line form two parallel lines, one on each side.
All the points a fixed distance from a straight line form two parallel lines, one on each side.
▸ Why?
A line that keeps a constant gap from another never meets it, which is what parallel means.
▸ Why?
With the base fixed, the area is half the base times the height, so one area pins one height.
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.
- Write the area with BC as base
- Fixing the area fixes the height
- Collect every point at that fixed distance