AMC 10 · 2024 · #23

Grade 12 algebra
polynomial-rootsvieta-formulassymmetric-polynomialscomplementary-anglescomplex-polar-formtrigonometric-ratios convert-to-algebrasymmetry-argument ↑ Prerequisites: trigonometric-ratiospolynomial-rootsbinomial-theorem
📏 Long solution 💡 4 insights
Problem
Four angles are in play: π/16, 3π/16, 5π/16, and 7π/16. Square the tangent of each one, then add up four particular products, each product using two of those squared tangents: tan² π/16tan² 3π/16 + tan² π/16tan² 5π/16 + tan² 3π/16tan² 7π/16 + tan² 5π/16tan² 7π/16. Find the exact value of that sum.

Pick an answer.

(A)
28
(B)
68
(C)
70
(D)
72
(E)
84
How to solve
Strategy Convert to Algebra

Each individual tangent here is an ugly nested radical, so any plan that computes them one at a time is doomed. Two observations rescue the problem, and they fit together. First, four numbers make six pairwise products, and the sum only asks for four of them — so the natural move is to take all six and subtract the two that were skipped, which turns a lopsided expression into a fully symmetric one. Second, the four angles are exactly the odd multiples of π/16 below π/2, which is precisely the condition cos 8θ = 0. That condition converts into a single polynomial equation in tanθ, so the four squared tangents become the four roots of one quartic. Symmetric functions of roots are exactly what a polynomial's coefficients already are, so the symmetric sum can be read off a coefficient without ever computing a single tangent.

1STEP 1

Name the four numbers

The target uses four of the six pairs.

E = ab + ac + bd + cd, all six pairs = {ab, ac, ad, bc, bd, cd}
2STEP 2

Ask for all six, subtract the strays

The target is all six minus two.

e₂ = ab + ac + ad + bc + bd + cd → E = e₂ - (ad + bc)
3STEP 3

The skipped pairs are complementary

Being complementary, each stray pair is 1.

tan(π/2 - θ) = 1/tanθ → ad = tan² π/16cot² π/16 = 1, bc = 1, E = e₂ - 2
4STEP 4

Turn eight-fold angles into a polynomial

De Moivre hands us a quartic.

(cos 8θ)/cos⁸θ = Re(1 + ix)⁸ = 1 - C(8, 2)x² + C(8, 4)x⁴ - C(8, 6)x⁶ + C(8, 8)x⁸
5STEP 5

Confirm all four roots are accounted for

The four values are exactly the quartic's roots.

u⁴ - 28u³ + 70u² - 28u + 1 = 0 has roots exactly u = a, b, c, d
6STEP 6

Read the answer off a coefficient

Seventy minus two is 68.

(u-a)(u-b)(u-c)(u-d) = u⁴ - e₁u³ + e₂u² - e₃u + e₄ → e₂ = 70, E = 70 - 2 = 68
Answer
68
The quartic's other coefficients act as free cross-checks. Its constant term says abcd = 1, and that is forced independently by the complementary pairs: abcd = (ad)(bc) = 1 · 1 = 1. Its u³ and u coefficients are both 28, which is exactly what the reciprocal pairing predicts — a and d are reciprocals, as are b and c, so the polynomial must read the same forwards and backwards. Both checks land, so the quartic is right. A direct numerical sanity check also lands: a ≈ 0.0396, b ≈ 0.4465, c ≈ 2.2398, d ≈ 25.2741, giving ab ≈ 0.0177, ac ≈ 0.0886, bd ≈ 11.2840, cd ≈ 56.6098, and the total is 68.00 to the digits shown. Notice too what the wrong choices are: 70 is e₂ itself, the value you get by forgetting that two pairs were left out, and 28 is e₁, the sum of the four numbers rather than of their pairwise products. Landing on 68 rather than on either of those confirms the two skipped pairs were handled.
💡Key takeaway

When a sum leaves a couple of terms out, put them all back in and subtract the strays — the complete symmetric sum is the one a polynomial will hand you for free.

  • Name the four numbers
  • Ask for all six, subtract the strays
  • The skipped pairs are complementary
  • Turn eight-fold angles into a polynomial
  • Confirm all four roots are accounted for
  • Read the answer off a coefficient