AMC 10 · 2005 · #14

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates
area-trianglesequilateral-triangle identify-subproblemsspatial-visualization ↑ Prerequisites: area-trianglesequilateral-triangle
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
An equilateral triangle ABC has every side equal to 2. The point M is the midpoint of side AC. The point D lies on ray BC so that C is the midpoint of segment BD, which puts D on the far side of C from B. Find the area of triangle CDM.

Pick an answer.

(A)
$\frac {\sqrt {2}}{2}$
(B)
$\frac {3}{4}$
(C)
$\frac {\sqrt {3}}{2}$
(D)
1
(E)
$\sqrt {2}$

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the natural move: sketch the equilateral triangle sitting on the line through B, C, D, and the picture immediately shows that side CD of the target triangle lies flat on that line. That makes CD an obvious base, so the area is just 1/2·base·height. Tool #7 (Identify Subproblems) then splits the work into two clean pieces: find the base CD, and find the height, which is how high M sits above the line. The height needs the triangle's altitude (one Pythagorean step) and the fact that M is only halfway up.

1STEP 1

Line up B, C, D and read off the base

C is the midpoint of BD, so B, C, D are collinear and CD=CB=2. Use that flat side CD as the base.

CD = CB = 2
2STEP 2

Find the triangle's altitude

Drop a perpendicular from A to BC at F. With BF=1 and AB=2, the Pythagorean theorem gives AF=√(3).

AF=√(AB²-BF²)=√(2²-1²)=√(3)
3STEP 3

Halve the altitude to reach M

Drop a perpendicular from M to the line at G. M is the midpoint of AC, so MG=√(3)/2, exactly half of AF.

MG = 1/2 AF = 1/2√(3) = √(3)/2
4STEP 4

Multiply base by height

Area = 1/2·base·height = 1/2·2·√(3)/2 = √(3)/2, which is choice (C).

[CDM]=1/2 · 2 · √(3)/2=√(3)/2 → (C)
Answer
√(3)/2
The value √(3)/2≈ 0.87 is comfortably positive and matches a base of 2 paired with a smallish height of about 0.87. The most tempting trap is to use the full altitude √(3) as the height, which would give 1/2 · 2·√(3)=√(3)≈ 1.73 — but M is only halfway up side AC, so its height is half of that. Choice (D) 1 and choice (E) √(2) are the kinds of near-miss values that show up if the height or base is mis-set; the careful base-times-height gives √(3)/2.
💡Key takeaway

When one side of a triangle lies flat on a line, use it as the base — then the area is just half that base times how high the last corner floats above the line.

  • Line up B, C, D and read off the base
  • Find the triangle's altitude
  • Halve the altitude to reach M
  • Multiply base by height

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