AMC 10 · 2006 · #16

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
area-regular-hexagoncoordinate-geometryequilateral-triangle convert-to-algebra ↑ Prerequisites: area-regular-hexagoncoordinate-geometry
📏 Long solution 💡 3 insights
Problem
A regular hexagon has two of its vertices pinned to given points, with one vertex between them. Find the area of the hexagon.

Pick an answer.

(A)
$20\sqrt {3}$
(B)
$22\sqrt {3}$
(C)
$25\sqrt {3}$
(D)
$27\sqrt {3}$
(E)
50
How to solve
Strategy Introduce a Variable

The two coordinates look like the point of the problem, but they only do one job: they fix the length AC. So name the side length s (Tool #4, Introduce a Variable) and write both the known quantity and the wanted quantity in terms of it. The known one is AC; the wanted one is the area, which is 6 equilateral triangles of side s (Tool #7, Identify Subproblems). Then s cancels out of the story. The one thing worth being careful about is the link between AC and s: that link is decided by the labelling, not by the coordinates. C is two letters after A, so AC is the short diagonal, and a drawing (Tool #1) plus the 120° interior angle turns it into s√(3). A separate step checks that a hexagon with these two vertices actually exists (Tool #17), because computing s from AC only says what s would have to be — it does not by itself say the hexagon is there.

1STEP 1

Read the labels, not just the points

The labels show the pinned points skip a vertex.

interior angle=((6-2) · 180°)/6=120°, ∠ ABC=120°, AB=BC=s
2STEP 2

Measure AC from the coordinates

The coordinates give that distance as 5√(2).

AC=√((7-0)²+(1-0)²)=√(49+1)=√(50)=5√(2)
3STEP 3

Turn the 120 degree angle into a square root of 3

The fixed interior angle relates it to the side.

BM=s/2, AM=√(s²-s²/4)=s√(3)/2, AC=2·s√(3)/2=s√(3) s√(3)=5√(2) → 3s²=50 → s²=50/3
4STEP 4

Check that such a hexagon really exists

A centre can be found, so such a hexagon really exists.

O=(7/2-√(3)/6, 1/2+7√(3)/6), OA²=OC²=50/3=s²
5STEP 5

Six equilateral triangles

Six equilateral triangles give 25√(3), choice (C).

[ABCDEF]=6·√(3)/4s²=3√(3)/2s²=3√(3)/2·50/3=25√(3) → (C)
Answer
25√(3)
In decimals, 25√(3)≈ 43.3. Check it against the shape: the circumradius equals s≈ 4.08, so the hexagon sits inside a circle of area π s²≈ 52.4, and 43.3/52.4≈ 0.83 — exactly the fraction 3√(3)/2π≈ 0.827 that a regular hexagon always fills of its circumcircle, so the size is right. The choices are 34.6, 38.1, 43.3, 46.8, 50, spaced far enough apart that no rounding could confuse them. Choice (E) 50 is the trap: 50 is AC², what you get by squaring the diagonal and stopping. Two common slips also fail to appear on the list, which is a further sign the intended path is the one taken: reading AC as a side gives 3√(3)/2 · 50=75√(3), and reading it as a long diagonal 2s gives 18.75√(3). Finally, the constructed hexagon can be checked directly — its six vertices from Step 4 all sit s apart, and the shoelace formula on them returns 43.3013…=25√(3).
💡Key takeaway

In a regular hexagon the diagonal that skips one vertex is always √(3) times the side, so one diagonal pins down the whole shape — and since the area formula uses s², you never need to simplify s itself.

  • Read the labels, not just the points
  • Measure AC from the coordinates
  • Turn the 120 degree angle into a square root of 3
  • Check that such a hexagon really exists
  • Six equilateral triangles

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