AMC 10 · 2023 · #25

Grade 12 algebra
complex-polar-formbinomial-theoremtangent-addition-formulacomplex-numbers easier-related-problempattern-recognitionidentify-subproblems ↑ Prerequisites: binomial-theoremcomplex-numbers
📏 Long solution 💡 4 insights
Problem
The tangent of 2023 times an angle can supposedly be written as one big fraction in the tangent of that angle. The numerator uses only odd powers; the denominator uses only even powers and starts at one. All 2023 coefficients are integers and exactly one such list works. Find the very last coefficient.

Pick an answer.

(A)
-2023
(B)
-2022
(C)
-1
(D)
1
(E)
2023
How to solve
Strategy Convert to Algebra

The number 2023 is there to intimidate, not to matter. Nothing about the problem depends on that particular value, so the first move is to replace it with 3, then 5, then 7, and watch what the tangent addition formula produces. Those small cases show the shape is real and hand over a candidate pattern for the top coefficient. A pattern from three cases is a guess, though, so the second move supplies the mechanism: multiplying an angle by n is the same as raising a complex number to the nth power, and once both sides are divided by cosⁿ x the whole trig identity collapses into (1 + itan x)ⁿ. The Binomial Theorem then names every coefficient at once — they are binomial coefficients wearing alternating signs — and the top one is the easiest of the lot, since C(2023, 2023) = 1. The last piece of the plan is a separate subproblem the small cases cannot settle: showing no second integer list exists, which is what lets the derived coefficients be called the a_i.

1STEP 1

Name the tangent

Give the tangent a short name.

t = tan x, tan 2023x = N(t)/D(t), deg N = 2023, deg D = 2022, D(0) = 1
2STEP 2

Try a small multiple

Ask the same question at three times.

tan 2x = 2t/(1 - t²), tan 3x = (tan 2x + t)/(1 - ttan 2x) = (2t + t(1 - t²))/((1 - t²) - 2t²) = (3t - t³)/(1 - 3t²)
3STEP 3

Two more cases

Two more cases fix the pattern.

tan 5x = (5t - 10t³ + t⁵)/(1 - 10t² + 5t⁴), tan 7x = (7t - 35t³ + 21t⁵ - t⁷)/(1 - 21t² + 35t⁴ - 7t⁶)
4STEP 4

Multiplying an angle is a power

Multiplying an angle is a complex power.

(cos x + isin x)ⁿ = cos nx + isin nx ⟹ (1 + it)ⁿ = (cos nx + isin nx)/(cosⁿ x) ⟹ tan nx = (Im[(1 + it)ⁿ])/(Re[(1 + it)ⁿ])
5STEP 5

The binomial theorem names them

The binomial theorem gives every coefficient.

(1 + it)ⁿ = Σ_k=0ⁿC(n, k)i^kt^k, i²j = (-1)^j, i²j+1 = (-1)^ji ⟹ a_k = (-1)^⌊ k/2⌋C(2023, k)
6STEP 6

Check the list is unique

Numerator and denominator are coprime, so it is unique.

D(t) + iN(t) = (1 + it)²⁰²³, D(t) - iN(t) = (1 - it)²⁰²³ ⟹ gcd(N, D) = 1 ⟹ B = D, M = N
7STEP 7

Read off the top coefficient

The last coefficient is negative one.

a₂₀₂₃ = (-1)^⌊ 2023/2⌋C(2023, 2023) = (-1)¹⁰¹¹ · 1 = -1
Answer
-1
Three separate checks agree. First, the derived formula reproduces the small cases exactly: for n = 3 it gives (3t - t³)/(1 - 3t²) and for n = 7 it gives (7t - 35t³ + 21t⁵ - t⁷)/(1 - 21t² + 35t⁴ - 7t⁶), both matching what the addition formula produced by hand. Second, the leading coefficients for n = 3, 5, 7, 9, 11 come out -1, +1, -1, +1, -1, a period-4 cycle in n, and evaluating both sides at a test angle such as x = 0.3 confirms the full identity numerically in every one of those cases. Since 2023 ≡ 3 (mod 4), the same slot as 3, 7, and 11, the value -1 is what the cycle demands. Third, the wrong choices are all recognisable slips: 2023 is a₁ = C(2023, 1), the coefficient at the bottom of the numerator rather than the top, and -2023 is that same number with a sign attached; 1 is the right size with the sign that would apply if 2023 were 1 more than a multiple of 4; -2022 never appears as any coefficient at all. Landing on -1 rather than on 1 is precisely the check that the parity of 1011 was handled.
💡Key takeaway

Multiplying an angle by n is the same as raising 1 + itan x to the nth power, so every coefficient in a multiple-angle formula is just a binomial coefficient with a sign on it.

  • Give the tangent a short name
  • Try the same question on tan 3x
  • Two more cases fix the pattern
  • Multiplying an angle is raising a power
  • The Binomial Theorem names every coefficient
  • Check that no second list exists
  • Read off the top coefficient