AMC 10 · 2007 · #24

Grade 11 algebra
sum-to-product-identityperiodic-functionprinciple-of-inclusion-exclusionfloor-functionsequences-arithmetic easier-related-problemcaseworkidentify-subproblems ↑ Prerequisites: trigonometric-ratiosperiodic-function
📏 Long solution 💡 4 insights
Problem
For each whole number above one, count the solutions of a sine equation on a closed interval. Add up those counts over a long range.

Pick an answer.

(A)
2014524
(B)
2015028
(C)
2015033
(D)
2016532
(E)
2017033
How to solve
Strategy Organize Information in More Ways

An equation of the form (something) = (something) has no root count attached to it, but an equation of the form (product) = 0 does: its solutions are exactly the solutions of the factors, and those can be listed. So the whole problem turns on Tool #15 (Organize Information in More Ways) applied once, at the very start — move everything to one side and rewrite the difference sin(nx) - sin x as a product. Because that rewriting is an identity, nothing is gained or lost, and the count becomes the size of a union of two explicit lists. From there Tool #12 (Draw a Venn Diagram) supplies the only remaining idea: the size of a union is the sum of the two sizes minus the size of the overlap, so the entire difficulty of the problem is compressed into one question — when do the two lists share a point? That is the step the problem actually turns on, and it is where a solution based on graphs quietly fails, because the shared point is a place where the curves touch rather than cross. Tool #4 (Introduce a Variable) is what makes the overlap question easy: naming the two half-angles u and v reveals that their difference is x itself, which pins the only possible shared point to x = π/2 before any divisibility argument is needed. Tool #7 (Identify Subproblems) splits the finish into counting each list separately and then summing over n, and Tool #9 (Solve an Easier Related Problem) checks the formula against n = 2 and n = 5, which are small enough to solve by bare hands.

1STEP 1

Rewrite the difference as a product

Rewriting the difference as a product splits the equation.

sin(nx) - sin x = 2 cos (((n+1)x)/2) sin (((n-1)x)/2) for all real x
2STEP 2

Two lists of solutions

Each factor gives its own family of solutions.

A = {((2j+1)π)/(n+1) ∈ [0,π] : j ∈ Z}, B = {2kπ/(n-1) ∈ [0,π] : k ∈ Z}, F(n) = |A ∪ B|
3STEP 3

Count each list separately

Counting them separately gives a total of n plus one.

|A| = ⌊ n/2 ⌋ + 1, |B| = ⌊ (n-1)/2 ⌋ + 1, |A| + |B| = (n-1) + 2 = n+1
4STEP 4

The lists can only meet at pi/2

The families can only meet at one point, and only sometimes.

u - v = x → x = π/2 + (j-k)π → x = π/2; π/2 ∈ A ∩ B ⇔ n ≡ 1 (mod 4)
5STEP 5

Assemble the formula for F(n)

That gives a two-case formula.

F(n) = n & n ≡ 1 (mod 4) ; n+1 & otherwise
6STEP 6

Add it up

Summing with the correction gives 2016532, choice (E).

Σ_n=2²⁰⁰⁷ F(n) = ((2008 · 2009)/2 - 3) - 501 = 2017033 - 501 = 2016532
Answer
2016532
The answer is 2016532, choice (D). The strongest check is that the answer list itself confirms the structure of the argument. Choice (E) 2017033 is exactly the uncorrected sum Σ_n=2²⁰⁰⁷(n+1) — precisely what a solver reports after establishing F(n) = n+1 and never discovering the n ≡ 1 (mod 4) exception. The gap between (D) and (E) is 501, which is exactly the count of exceptional n; the entire content of the problem is that 501. Choice (B) 2015028 is (2007 · 2008)/2, the sum of n itself over 1 to 2007 — the total from the opposite error, using F(n) = n everywhere and slipping on the lower limit. Choice (A) 2014524 is that same wrong baseline with the correct-looking correction applied on top, 2015028 - 504, and choice (C) 2015033 sits five above (B), in the same family of off-by-a-constant slips. So all four distractors are near-misses of one honest calculation, which is a good sign the calculation is the intended one. Independent numerical support: counting solutions directly for n = 2 through n = 20 reproduces 3, 4, 5, 5, 7, 8, 9, 9, 11, 12, 13, 13, 15, 16, 17, 17, 19, 20, 21, which is n+1 with a drop at n = 5, 9, 13, 17 exactly as claimed, and in every dropped case the doubled solution is π/2 and nothing else. A magnitude check also fits: F(n) ≈ n, so the sum should land near Σ_n=2²⁰⁰⁷ n ≈ 2.0 × 10⁶, and 2016532 does.
💡Key takeaway

Turn "two waves are equal" into "a product is zero", count the two evenly spaced lists of zeros, and subtract the single place they can collide — the midpoint π/2, where both waves peak at once.

  • Rewrite the difference as a product
  • Two lists of solutions
  • Count each list separately
  • The lists can only meet at pi/2
  • Assemble the formula for F(n)
  • Add it up