AMC 10 · 2022 · #10

Grade 8 geometry-2d
coordinate-geometryequilateral-trianglemidpoint-formulapythagorean-theoremrotation-isometry symmetry-argumentspatial-visualizationidentify-subproblems ↑ Prerequisites: coordinate-geometrypythagorean-theorem
📏 Medium solution 💡 2 insights
Problem
A regular hexagon has every side equal to 2. Mark the midpoints of two opposite sides. Those two marks, together with two of the vertices, form a four-sided figure. Find the total length around it.

Pick an answer.

(A)
$4\sqrt3$
(B)
8
(C)
$4\sqrt5$
(D)
$4\sqrt7$
(E)
12
How to solve
Strategy Draw a Diagram

Nothing here is a word problem in disguise; the whole difficulty is that G and H sit in the middle of sides, so none of the four wanted lengths is a side or a familiar diagonal of the hexagon. Tool #1 (Draw a Diagram) is the primary move, and the diagram is worth drawing on axes rather than on blank paper: a regular hexagon has a centre, and putting that centre at the origin makes all six vertices come out as short exact coordinates. Tool #15 (Organize Information in More Ways) is what turns the picture into that coordinate table, after which a midpoint is just an average of two numbers. Tool #17 (Visualize Spatial Relationships) then pays for the setup twice over: with the centre at the origin, G and H are negatives of each other and so are C and F, which means a half-turn about the centre carries the figure onto itself and pairs the four sides into two equal pairs. Tool #7 (Identify Subproblems) uses that pairing to cut the job from four distance computations down to two, each one an ordinary right-triangle calculation.

1STEP 1

Cut from the centre

Cutting from the centre gives equilateral triangles.

∠ AOB = 360°/6 = 60°, OA = OB → △ AOB equilateral → OA = OB = AB = 2
2STEP 2

Give every vertex a coordinate

Give all six vertices coordinates.

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

Average to place the midpoints

Average to place both midpoints.

G = (3/2, √3/2), H = (-3/2, -√3/2), C = (-1, √3), F = (1, -√3)
4STEP 4

Spot the half-turn symmetry

Half-turn symmetry pairs the sides.

H = -G, F = -C, (x, y) → (-x, -y) → GC = HF, CH = FG
5STEP 5

Measure two sides

Measuring two sides gives the same value.

GC² = (5/2)² + (√3/2)² = 28/4 = 7, CH² = (1/2)² + (3√3/2)² = 1/4 + 27/4 = 7
6STEP 6

Add the four sides

The perimeter is four root seven.

GC = CH = HF = FG = √7 → perimeter = 4√7 → (D)
Answer
4√7
Numbers first: √7 ≈ 2.646, so the perimeter is about 10.58. That sits between choice (C) 4√5 ≈ 8.94 and choice (E) 12, which is where a sensible answer belongs — each side of GCHF visibly reaches across more than one hexagon side of length 2 but stays shorter than the long diagonal AD = 4, so a side between 2 and 4 is expected, and 2.646 fits. Choice (A) 4√3 ≈ 6.93 would force each side down to about 1.73, shorter than a hexagon side, which the picture rules out. The diagonals give a fully independent confirmation. The diagonals of GCHF are CF and GH: C = (-1, √3) to F = (1, -√3) has length √(4 + 12) = 4, and G to H has length √(9 + 3) = 2√3. Both pass through the origin and their direction vectors (2, -2√3) and (-3, -√3) have dot product -6 + 6 = 0, so they cross at right angles and bisect each other — exactly the rhombus test. Then each side is the hypotenuse of a right triangle with legs equal to the half-diagonals 2 and √3, giving √(4 + 3) = √7 again, and a perimeter of 4√7. Plugging the explicit coordinates into a distance calculation returns 2.6457513… for all four sides, and 4 × 2.6457513 = 10.5830052 = 4√7.
💡Key takeaway

Put the hexagon's centre at the origin: then H is G flipped through the centre and F is C flipped through the centre, so GCHF is a rhombus and one distance, √7, gives the whole perimeter 4√7.

  • Cut the hexagon from its centre
  • Give every vertex a coordinate
  • Average to place G and H
  • Spot the half-turn symmetry
  • Measure GC and CH
  • Add the four equal sides