AMC 10 · 2022 · #24

Grade 12 geometry-2dalgebra
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
roots-of-unitycomplex-numberscomplex-polar-formvieta-formulasdouble-angle-formulasymmetry-argument symmetry-argumentidentify-subproblemsconvert-to-algebrapattern-recognition ↑ Prerequisites: roots-of-unitycomplex-numbers
📏 Long solution 💡 4 insights 📊 Diagram
Problem
Seven points sit evenly around a circle of radius one, forming a regular seven-sided polygon. Joining every point to every other gives twenty-one segments. Raise each length to the fourth power and add all twenty-one results.

Pick an answer.

(A)
49
(B)
98
(C)
147
(D)
168
(E)
196
How to solve
Strategy Organize Information in More Ways

Measuring 21 separate segments is hopeless, because none of the seven angles involved has a nice exact value. The whole solve is a chain of regroupings. First regrouping: the 21 segments are not 21 different lengths. A chord depends only on how many steps apart its endpoints are around the circle, and there are only three possible gaps, 1, 2, and 3 steps, with exactly 7 chords in each class. Second regrouping: store a vertex as a complex number instead of a pair of coordinates, so that a squared chord length collapses to 2 - 2cos(2π k)/7 in one line of conjugate algebra. Third regrouping, and the one that actually finishes the problem: rewrite the sum over the three gaps k = 1, 2, 3 as half the sum over all six nonzero gaps k = 1, …, 6. Individually those cosines are ugly, but summed over a complete set of nonzero residues they are forced to total -1 by the roots of unity. The same completeness trick handles the squared cosines after a double-angle rewrite. Nothing here is computed; everything is collapsed.

1STEP 1

Only three lengths

There are only three distinct lengths.

C(7, 2) = 21 = 3 × 7, S = 7(L₁⁴ + L₂⁴ + L₃⁴)
2STEP 2

Put the vertices at roots of unity

Put them at the roots of unity.

ω = cos2π/7 + isin2π/7, ω⁷ = 1, vertices = {1, ω, ω², …, ω⁶}
3STEP 3

Square a chord with a conjugate

The conjugate product gives a cosine.

L_k² = |1 - ω^k|² = (1 - ω^k)(1 - ω^-k) = 2 - (ω^k + ω^-k) = 2 - 2cos(2π k)/7
4STEP 4

Trade three gaps for six

Symmetry lets us use all six gaps.

S = 7Σ_k=1³(2 - 2cos(2π k)/7)² = 7/2Σ_k=1⁶(2 - 2cos(2π k)/7)²
5STEP 5

Expand and name the sums

Expand and name the two sums.

Σ_k=1⁶(2 - 2c_k)² = 24 - 8A + 4B, A = Σ_k=1⁶ c_k, B = Σ_k=1⁶ c_k²
6STEP 6

Sum the cosines

The roots of unity give that sum.

z⁷ - 1 = (z-1)(z⁶ + z⁵ + … + z + 1) → Σ_k=1⁶ω^k = -1 → A = Σ_k=1⁶ c_k = -1
7STEP 7

Sum the squared cosines

A double-angle identity gives that sum.

B = 1/2(6 + Σ_k=1⁶cos(4π k)/7) = 1/2(6 + A) = 1/2(6 - 1) = 5/2
8STEP 8

Assemble the total

Assembling gives 147.

S = 7/2(24 - 8(-1) + 4 · 5/2) = 7/2 · 42 = 147
Answer
147
A direct numerical check confirms the total. The three chord lengths are L_k = 2sin(π k)/7, giving L₁ ≈ 0.8678, L₂ ≈ 1.5637, and L₃ ≈ 1.9499. Their 4th powers are about 0.5670, 5.9782, and 14.4547, and multiplying each by 7 and adding gives 3.969 + 41.848 + 101.183 = 147.000. A crude size estimate also lands in the right neighborhood: a typical chord in a unit circle has length somewhere near 3/2, and 21 · (3/2)⁴ ≈ 106, the same order of magnitude as 147 and comfortably below the largest possible value 21 · 2⁴ = 336 that would occur if every segment were a full diameter. That rules out nothing by itself, but it does confirm the answer sits in a plausible band. The structure of the answer choices is the strongest signal of all: all five are multiples of 7, and 147 = 3 · 7². Repeating the whole computation for other regular n-gons inscribed in a unit circle gives 27 for the triangle, 48 for the square, 75 for the pentagon, and 108 for the hexagon, which are 3 · 3², 3 · 4², 3 · 5², and 3 · 6². The general answer is 3n², and the square case can be checked entirely by hand: four sides of length √(2) and two diagonals of length 2 give 4 · 4 + 2 · 16 = 48. The value 3 · 7² = 147 fits that pattern exactly.
💡Key takeaway

When one angle is too ugly to compute, add in the rest of its family until the set is complete, because a full set of evenly spaced directions always cancels to something simple.

  • Only three lengths, seven of each
  • Put the vertices at the 7th roots of unity
  • Square a chord with a conjugate
  • Trade three gaps for all six
  • Expand and name the two sums
  • The cosines add to -1
  • The squared cosines add to 5/2
  • Assemble the total

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