AMC 10 · 2011 · #24
Grade 11 algebrageometry-2dPick an answer.
The question asks for a minimum over 2520 different octagons, so Tool #14 (Extreme Principle) has to carry the ending, and it has to do it in the honest two-part way: prove a number that no octagon can beat, then exhibit an octagon that reaches it. Computing one natural-looking octagon proves nothing by itself — choice 4√(3)+4 ≈ 10.93 is smaller than what that octagon gives, so "some other order might do better" is a live possibility until it is ruled out. Tool #7 (Identify Subproblems) splits the work: first locate the eight zeros, then measure, then minimise. Tool #4 (Introduce a Variable) does the locating — setting w = z⁴ turns a degree-8 polynomial into a quadratic, and setting z = a+bi extracts the remaining fourth roots with plain algebra. Tool #1 (Draw a Diagram) shows what the eight points actually are: two squares, one rotated 45° and larger. Tool #15 (Organize Information in More Ways) is the key move for the minimum — instead of tracking octagons, tabulate all 28 pairwise distances, which collapse into only four values. Once the smallest possible side is known, the lower bound is immediate.
See a quadratic in z⁴
The polynomial is a quadratic in a power.
Three terms with exponents 8, 4, 0 are a quadratic wearing a disguise, and w = z⁴ removes the disguise.
9.A-SSE.A.2Introduce A VariableThe four fourth roots of unity
One factor gives four points on the unit circle.
A factored polynomial hands over its zeros directly, and these four are the corners of a square of side √(2).
11.A-APR.B.3Identify SubproblemsSolve the second quartic by hand
The other gives four more, on a bigger circle.
Recognising 7+4√(3) as (2+√(3))² turns a fourth root into two square roots, and each square root is one small system in a and b.
11.N-CN.A.2Introduce A VariableTwo squares, one turned 45°
So it is two squares, one turned 45 degrees.
The zeros are two concentric squares — a small one straight and a bigger one turned halfway — so the eight points are evenly spaced in angle but not in distance.
The zeros form two concentric squares, one turned halfway, so the eight points are evenly spaced around a circle.
▸ Why?
A complex number is a point with a length and a direction, so each family sits at one radius.
▸ Why?
Each square's corners are a quarter turn apart, so the two sets interleave at eighths of the full turn.
The unit points sit strictly inside
The inner four sit strictly inside.
The four unit points hide inside the big tilted square, so every octagon through all eight has to dip inward and none of them is convex.
10.G-GPE.B.4Draw A DiagramOnly four distances exist
Only four distances occur at all.
Every pair of zeros makes a triangle with the origin, so one formula measures all 28 gaps — and they come in only four sizes.
11.G-SRT.D.11Organize Information In More WaysEvery side is at least √(2)
So every side is at least the smallest of them.
No side can be shorter than the closest any two of the points ever get, and there are always exactly eight sides to pay for.
9.A-CED.A.3Extreme PrincipleAngle order attains the bound
Going by angle reaches 8√2, choice (A).
Walking around in angle order only ever steps between neighbours 45° apart, and every one of those steps is a cheapest possible step.
10.G-GPE.B.7Extreme PrincipleThe eight zeros are two squares, one small and one bigger and turned, and no two of them are ever closer than √(2) — so eight sides cost at least 8√(2), and going around in angle order pays exactly that.
- See a quadratic in z⁴
- The four fourth roots of unity
- Solve the second quartic by hand
- Two squares, one turned 45°
- The unit points sit strictly inside
- Only four distances exist
- Every side is at least √(2)
- Angle order attains the bound