AMC 10 · 2018 · #16

Grade 12 geometry-2d
roots-of-unitycomplex-polar-formregular-octagonsine-area-formulaoptimization extreme-principleeasier-related-problemsystematic-enumeration ↑ Prerequisites: complex-polar-formroots-of-unity
📏 Long solution 💡 4 insights
Problem
An eighth-power equation has eight complex solutions. Plotted in the complex plane they are the vertices of a convex regular polygon. Pick any three of those vertices and form a triangle. Among all such choices, find the smallest area such a triangle can have.

Pick an answer.

(A)
$\frac{1}{6}\sqrt{6}$
(B)
$\frac{3}{2}\sqrt{2}-\frac{3}{2}$
(C)
$2\sqrt3-3\sqrt2$
(D)
$\frac{1}{2}\sqrt{2}$
(E)
$\sqrt 3-1$
How to solve
Strategy Extreme Principle

The question asks for a minimum over a finite set of choices, which is exactly the Extreme Principle: find the shape that pushes the area as far down as it can go, then prove nothing beats it. Getting there needs three cheaper moves first. The +6 is decoration, so I substitute w = z+6 and solve the easier equation w⁸ = 81 centered at the origin (Solve an Easier Related Problem). Polar form with a named radius and angle turns that equation into eight evenly spaced points, and drawing them shows a regular octagon (Introduce a Variable, Draw a Diagram). Then, because the octagon's symmetry means only the gaps between chosen vertices matter, I can list every essentially different triangle — there are only five (Make a Systematic List) — compare them, and finally check the winning value against the answer choices (Eliminate Possibilities).

1STEP 1

Slide the picture to the origin

Slide the picture to the origin.

w = z + 6 ⟹ w⁸ = 81, z = w - 6
2STEP 2

Pin every root to one circle

All roots sit on one circle.

|w|⁸ = 81 = 3⁴ ⟹ |w| = 3⁴/8 = √(3)
3STEP 3

Space the eight roots evenly

The eight roots are evenly spaced.

w = √(3)(cosθ + isinθ), w⁸ = 81(cos 8θ + isin 8θ) ⟹ θ = 45k°, k = 0,1,…,7
4STEP 4

Only the gaps matter

Only the gaps matter.

a + b + c = 8, a,b,c ≥ 1: {1,1,6}, {1,2,5}, {1,3,4}, {2,2,4}, {2,3,3}
5STEP 5

Turn each shape into a product of sines

Each shape becomes a product of sines.

[PQR] = 1/2(2Rsin 22.5a°)(2Rsin 22.5b°)sin 22.5c° = 6sin 22.5a° sin 22.5b° sin 22.5c°
6STEP 6

Compare the five and take the minimum

Compare the five and take the smallest.

{1,1,6}: 0.383, 0.383, 0.707 ≤ {1,2,5}: 0.383, 0.707, 0.924 and {1,3,4}: 0.383, 0.924, 1 and {2,2,4}: 0.707, 0.707, 1 and {2,3,3}: 0.707, 0.924, 0.924
7STEP 7

Measure the three consecutive vertices

Measure the three consecutive vertices directly.

AC = √(6), h = √(3) - √(6)/2
8STEP 8

Simplify and match a choice

Tidying gives three root two over two, minus three halves.

1/2·√(6)·(√(3)-√(6)/2) = √(18)/2 - 6/4 = 3/2√(2)-3/2 ≈ 0.621
Answer
3/2√(2)-3/2
The whole octagon has area 8·1/2R²sin 45° = 6√(2) ≈ 8.49, so the answer 0.621 is about 7 percent of it — the right size for the thinnest sliver three neighbouring vertices can cut off. The general formula from Step 5 reproduces it exactly: 6sin² 22.5° sin 135° = 6·(2-√(2))/4·√(2)/2 = (3√(2)-3)/2, matching the base-times-height computation. Choice (C) equals 2√(3)-3√(2) ≈ -0.778, which is negative and cannot be an area at all, and (A), (D), (E) miss numerically, so (B) is the only survivor.
💡Key takeaway

Slide the picture so the roots ring the origin, see that they form a regular octagon, then grab three vertices sitting right next to each other — the thinnest sliver wins.

  • Slide the picture to the origin
  • Pin every root to one circle
  • Space the eight roots evenly
  • Only the gaps matter
  • Turn each shape into a product of sines
  • Compare the five and take the minimum
  • Measure the three consecutive vertices
  • Simplify and match a choice