AMC 10 · 2012 · #21

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

coordinate-geometryrotation-isometryangle-sum-polygonpythagorean-theorem convert-to-algebracoordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryrotation-isometry

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An equal-angled hexagon of unequal sides holds a square sharing one of its corners. Find the square's side.

Pick an answer.

(A)

$29\sqrt{3}$

(B)

$\frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}$

(C)

$\ 20\sqrt{3}+16$

(D)

$20\sqrt{2}+13 \sqrt{3}$

(E)

$21\sqrt{6}$

How to solve

Strategy Draw a Diagram

Four unknown side lengths against one unknown square looks hopeless until you notice which lines the square touches. Equiangular means the six side directions are 0°, 60°, 120°, 180°, 240°, 300°, so line BC and line EF are parallel, and line AB and line DE are parallel. The square's diagonal XZ runs between the first pair and its diagonal AY runs between the second pair; each of those is a single equation. Coordinates turn both into linear algebra, and the 90° rotation that carries X to Z writes Y and Z in terms of one number, BX. The reason to trust coordinates over an angle chase here is the last step: the classic write-up assumes a convenient shape for the hexagon, and coordinates let us prove the answer does not depend on that shape instead of assuming it.

1STEP 1

Turn equiangular into six directions

Equal angles fix six directions.

side k points along 60°(k-1): AB ∥ DE, BC ∥ EF, CD ∥ FA

2STEP 2

Measure one height two ways

One height, measured twice, gives a relation.

h = √(3)/2(BC + CD) = √(3)/2(AF + EF) ⟹ BC + CD = AF + EF

3STEP 3

Name the two live unknowns

Only two unknowns stay alive.

X = (40 + t/2, √(3)/2t), F = (-f/2, √(3)/2f)

4STEP 4

Rotate X to get Z and Y

A quarter turn builds the other corners.

Z = (-√(3)/2t, 40 + t/2), Y = X + Z = (40 + t/2 - √(3)/2t, 40 + t/2 + √(3)/2t)

5STEP 5

Put Z on line EF

Putting one corner on its side gives an equation.

√(3) f = 40 + 2t

6STEP 6

Put Y on line DE and solve

The second corner closes the system.

(√(3)-1)t = 83 - 41√(3) ⟹ t = (83 - 41√(3))(√(3)+1)/2 = (42√(3) - 40)/2 = 21√(3) - 20, f = 42

7STEP 7

Drop a perpendicular from A

A perpendicular gives the side, 29√3.

AX² = AP² + PX² = (20√(3))² + (21√(3))² = 3(20² + 21²) = 3 · 29²

8STEP 8

Show the shape does not matter

The remaining freedom changes nothing, choice (D).

b ∈ [41(√(3)-1), 41√(3)+1) ⟹ AF = 42, BX = 21√(3)-20, s = 29√(3) for every b

Answer

29√(3)

Size alone cannot decide this problem, which is worth noticing before trusting any estimate: the five options are 29√(3) ≈ 50.23, 21/2√(2) + 41/2√(3) ≈ 50.36, 20√(3) + 16 ≈ 50.64, 20√(2) + 13√(3) ≈ 50.80, and 21√(6) ≈ 51.44, all packed inside a 2.5% band. The exact radical arithmetic is the entire problem. The answer passes several independent checks. First, s = AX is the side opposite the 120° angle in triangle ABX, so it must exceed AB = 40, and 50.23 does. Second, every quantity the solution forces lands where it must: AF = 42 ＞ 0, BX = 21√(3) - 20 ≈ 16.37 ＞ 0, and FZ = 20√(3) - 21 ≈ 13.64 lies strictly between 0 and EF ≈ 30.01. Third, a structural check the algebra never assumed: EZ = EF - FZ = 41√(3) - 41 - 20√(3) + 21 = 21√(3) - 20 = BX. That equality is predicted in advance, because the half-turn about the centre of the square swaps A ⇔ Y and X ⇔ Z; it carries line AB to the parallel line through Y, which is line DE, and carries line BX to the parallel line through Z, which is line EF. So it must carry B to the intersection of those two lines, namely E — forcing YE = AB = 40 and EZ = BX. The computed numbers agree, so two different descriptions of the figure are consistent.

💡Key takeaway

Equal angles fix every side's direction, so line BC ∥ line EF and line AB ∥ line DE; pinning the square between those two pairs forces AF = 42 and BX = 21√(3) - 20, and one perpendicular from A turns the rest into the (20, 21, 29) triple scaled by √(3).

Turn equiangular into six directions
Measure one height two ways
Name the two live unknowns
Rotate X to get Z and Y
Put Z on line EF
Put Y on line DE and solve
Drop a perpendicular from A
Show the shape does not matter