AMC 10 · 2014 · #13

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
area-regular-hexagonequilateral-trianglespatial-visualization identify-subproblems ↑ Prerequisites: area-regular-hexagon
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A regular hexagon of side length 1 is surrounded by six more regular hexagons, as drawn. Points A, B, C are three vertices marked on the figure: A is a vertex of the central hexagon, while B and C are outer vertices of two of the surrounding hexagons. Find the area of △ ABC.

Pick an answer.

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

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The figure already hands us exact coordinates, so Tool #1 (Draw a Diagram) is really a coordinate-geometry engine: read the points off the drawing and let position do the work. First fix the scale so drawing units become real units. Then Tool #7 (Identify Subproblems) splits the area into two easy measurements — a base and a height — because B and C sit one directly above the other, making BC a vertical segment and the height a plain horizontal distance. Tool #17 (Visualize Spatial Relationships) is the cross-check: seeing that the three vertices are the same distance apart tells us the triangle is equilateral, confirming the area a second way.

1STEP 1

Fix the scale of the drawing

The side from (0,0) to (10,0) is drawn 10 units long but is really 1, so the picture is blown up — divide every coordinate by 10.

scale=1/10; A=(0,0), B=(3, 1.73205…), C=(3, -1.73205…)
2STEP 2

Recognize the decimal as a radical

The height 1.73205… is exactly √(3), the width of a side-1 hexagon, so A=(0,0), B=(3,√(3)), C=(3,-√(3)).

1.73205…=√(3) → A=(0,0), B=(3,√(3)), C=(3,-√(3))
3STEP 3

Measure the base BC

B and C share x=3, so BC is vertical and its length is the gap from -√(3) up to √(3), namely 2√(3).

BC=√(3)-(-√(3))=2√(3)
4STEP 4

Measure the height from A

The base sits on the line x=3 and A sits at x=0, so the height is that sideways gap, 3.

h=|3-0|=3
5STEP 5

Combine base and height into the area

Half of base times height: 2√(3) · 3 ÷ 2 = 3√(3), which is choice (B).

[△ ABC]=1/2 · BC · h=1/2 · 2√(3) · 3=3√(3) → (B)
Answer
3√(3)
Cross-check with side lengths. Using the Pythagorean distance, AB=√(3²+(√(3))²)=√(9+3)=√(12)=2√(3), and by the up-down symmetry AC=2√(3) too, while BC=2√(3). All three sides equal 2√(3), so △ ABC is equilateral. The area of an equilateral triangle of side s is √(3)/4s²=√(3)/4·(2√(3))²=√(3)/4 · 12=3√(3), the same answer. Numerically 3√(3)≈ 5.20, which is a believable size for a triangle spanning three hexagons, and it sits at choice (B); choice (A) 2√(3)≈3.46 would be too small.
💡Key takeaway

Read the corner points from the picture, shrink them to real size, then multiply half the vertical base 2√(3) by the height 3 to get the area 3√(3).

  • Fix the scale of the drawing
  • Recognize the decimal as a radical
  • Measure the base BC
  • Measure the height from A
  • Combine base and height into the area

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