AMC 10 · 2011 · #20

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
perpendicular-bisectorcoordinate-geometryarea-trianglesequilateral-triangle identify-subproblems ↑ Prerequisites: perpendicular-bisector
📏 Medium solution 💡 3 insights
Problem
A rhombus ABCD has every side equal to 2, and the angle at vertex B is 120°. Inside the rhombus, region R is the set of points that lie closer to corner B than to any of the other three corners A, C, or D. Find the area of R.

Pick an answer.

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

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The phrase "closer to B" is about position, so Tool #1 (Draw a Diagram) is the anchor: draw the rhombus, notice the 120° angle splits it into two equilateral triangles, and picture where the border of R must run. Tool #4 (Introduce a Variable) pins the picture down with coordinates so "closer to B" becomes a distance you can compare. The border between "closer to B" and "closer to a neighbor" is the perpendicular bisector of the segment joining them, so R turns out to be a pentagon. Tool #7 (Identify Subproblems) then finds that pentagon's corners one bisector at a time and adds up its area.

1STEP 1

Split the rhombus into two triangles

∠ A = 180° - 120° = 60° with AB = AD = 2, so triangles ABD and CBD are equilateral with BD = 2.

∠ A = 180° - 120° = 60° → BD = 2
2STEP 2

Put the figure on a grid

Put B at the origin, symmetric about the x-axis: B=(0,0), A=(1,√(3)), C=(1,-√(3)), D=(2,0). Check: BA = √(1+3) = 2.

B=(0,0), A=(1,√(3)), C=(1,-√(3)), D=(2,0)
3STEP 3

Turn "closer to B" into borders

A point is closer to B when it lies on B's side of the perpendicular bisector — for BD it is x=1, for BA it passes through (1/2,√(3)/2).

bisector of BD: x=1; midpoint of BA=(1/2,√(3)/2)
4STEP 4

Find the pentagon's five corners

Tracing R's border gives a pentagon: (0,0), (1/2,√(3)/2), (1,√(3)/3), (1,-√(3)/3), (1/2,-√(3)/2) — the far two are the triangle centers.

(0,0), (1/2,√(3)/2), (1,√(3)/3), (1,-√(3)/3), (1/2,-√(3)/2)
5STEP 5

Add up the pentagon's area

Shoelace on those five corners in order sums to 4√(3)/3; half of it is the area 2√(3)/3, choice (C).

Area=1/2|Σ(x_i y_i+1-x_i+1y_i)|=1/2·4√(3)/3=2√(3)/3
Answer
2√(3)/3
The whole rhombus has area 2×√(3)/4 · 2² = 2√(3)≈ 3.46, and 2√(3)/3≈ 1.15 is exactly one-third of it. That fits the numbers: choice (C) ≈ 1.15 sits sensibly between (A) ≈ 0.58 and (D) ≈ 1.58, so R is a believable chunk of the rhombus, not the whole thing and not a sliver.
💡Key takeaway

The fair border between two corners is the line halfway between them, so "closest to B" is a small pentagon whose area you can just add up.

  • Split the rhombus into two triangles
  • Put the figure on a grid
  • Turn "closer to B" into borders
  • Find the pentagon's five corners
  • Add up the pentagon's area

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