AMC 10 · 2009 · #25

Grade 11 algebrapattern
tangent-addition-formulaperiodic-functionrecursive-sequence pattern-recognitioneasier-related-problem ↑ Prerequisites: tangent-addition-formula
📏 Long solution 💡 4 insights
Problem
A sequence builds each term from the two before it with a fraction that mixes their sum and product. Find the size of a term far down the list.

Pick an answer.

(A)
0
(B)
$2 - \sqrt3$
(C)
$\frac {1}{\sqrt3}$
(D)
1
(E)
$2 + \sqrt3$
How to solve
Strategy Introduce a Variable

Computing 2009 terms of (a_n+a_n+1)/(1-a_na_n+1) by hand is out of the question, and the terms themselves are ugly numbers in √3. Tool #15 (Organize Information in More Ways) supplies the one observation that changes everything: that fraction is the tangent addition formula, so the rule is really "add the two previous angles". Tool #4 (Introduce a Variable) makes that precise by naming an angle θ_n for each term, which converts a messy multiplicative recursion into plain addition. Tool #9 (Solve an Easier Related Problem) then shrinks the problem twice: angles only matter modulo 180° because tan repeats, and every angle turns out to be a multiple of 15°, so the whole sequence collapses to a Fibonacci-style recursion on the twelve residues mod 12. Tool #2 (Make a Systematic List) writes that short list out and exposes the cycle, and Tool #5 (Look for a Pattern) converts the cycle length into one division with remainder. Tool #3 (Eliminate Possibilities) does the load-bearing safety check that the slick substitution usually hides: 90° is the single angle with no tangent, and the argument is only valid because that value never occurs — which is also the reason the original denominator is never 0.

1STEP 1

The rule is the tangent addition formula

The rule is the tangent addition formula.

tan(x+y)=(tan x+tan y)/(1-tan xtan y) versus a_n+2=(a_n+a_n+1)/(1-a_na_n+1)
2STEP 2

Build the angle list on its own

Tracking angles instead of terms simplifies everything.

θ₁=45, θ₂=30, θ_n+2≡θ_n+θ_n+1 (mod 180) ⟹ b₁=3, b₂=2, b_n+2≡ b_n+b_n+1 (mod 12)
3STEP 3

The list closes after 24 terms

The angle list repeats after a fixed number of steps.

{b_n}=3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11, 3,2,…
4STEP 4

Why 6 never appears, and why that matters

One forbidden angle never appears, so the rule never breaks.

b_n=6→ n≡ 2 (mod 6) (even) and n≡ 1,5 (mod 8) (odd) — impossible
5STEP 5

The substitution becomes a theorem

That turns the substitution into a genuine theorem.

cos(θ_n+θ_n+1)=cosθ_ncosθ_n+1(1-tanθ_ntanθ_n+1) ≠ 0 ⟹ 1-a_na_n+1 ≠ 0, a_n+2=tanθ_n+2°
6STEP 6

Land on the 2009th term

The remainder lands the target term on 0.

2009=24 · 83+17 → b₂₀₀₉=b₁₇=0 → a₂₀₀₉=tan 0° = 0
7STEP 7

Check it with no trigonometry at all

Direct computation confirms 0, choice (A).

a₃=2+√3, a₄=-(2+√3), a₅=0, …, a₁₇=0, (a₂₅,a₂₆)=(1,1/√3)=(a₁,a₂)
Answer
0
The answer list itself corroborates the structure rather than the other way round: all five options are absolute values of tangents of multiples of 15°, namely 0, tan 15°=2-√3, tan 30°=1/√3, tan 45°=1, tan 75°=2+√3, and the exact recomputation of the first 26 terms lands every single term on that same 15° lattice. The value 0 is not a fluke of one lucky index: b_n=0 happens at n≡ 5 and n≡ 17 (mod 24), so two terms in every 24 are 0, and 2009 is one of them. The absolute-value sign in the question is itself a clue — the sequence really does go negative (a₄=-(2+√3)), so a signed answer would have been ambiguous, and 0 is the one value where the sign cannot matter. Finally, the danger the whole method rests on was checked, not assumed: b_n=6 (that is, θ_n=90°) never occurs, so 1-a_na_n+1 is never 0 and the sequence is genuinely defined all the way to n=2009; the direct exact computation confirms this by never producing a zero denominator.
💡Key takeaway

When a rule looks like (x+y)/(1-xy), it is angles being added in disguise — trade each number for its angle and a 2009-step monster turns into a list that repeats every 24.

  • The rule is the tangent addition formula
  • Build the angle list on its own
  • The list closes after 24 terms
  • Why 6 never appears, and why that matters
  • The substitution becomes a theorem
  • Land on the 2009th term
  • Check it with no trigonometry at all