AMC 10 · 2021 · #13

Grade 11 algebra

Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins

periodic-functiontrigonometric-ratiossign-analysisgraph-reading bound-inequality-then-enumerateidentify-subproblemsextreme-principle ↑ Prerequisites: trigonometric-ratiosperiodic-function

📏 Long solution 💡 4 insights

Problem

Count the angles between zero and two pi that make an expression mixing a slow sine wave and a fast cosine wave equal zero. The count of solutions is the count of times the fast wave meets the slow one.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is primary because this equation has no clean algebraic solution — cos 3θ expands into a cubic in cosθ that then collides with sinθ, and the mess hides the answer. A picture does not. Tool #15 (Organize Information in More Ways) makes the picture possible: split the single equation into y = 5cos 3θ (fast, amplitude 5) and y = 3sinθ - 1 (slow, amplitude 3, shifted down 1), so solutions become crossing points. Tool #14 (Extreme Principle) is what makes the count certain: the slow curve never leaves the band -4 ≤ y ≤ 2, while the fast wave overshoots to ± 5 at every peak and valley, so the fast wave is guaranteed to punch through the band. Tool #7 (Identify Subproblems) then chops the window into the 6 monotone half-waves of the fast curve and counts crossings one half-wave at a time.

1STEP 1

Two waves meeting

Read it as two waves crossing.

5cos 3θ = 3sinθ - 1

2STEP 2

Bound the slow wave

The slow wave stays in a narrow band.

-1 ≤ sinθ ≤ 1 → -4 ≤ 3sinθ - 1 ≤ 2

3STEP 3

Look at the fast wave's peaks

The fast wave's swing is much larger.

period = 2π/3, θ = kπ/3 (k = 0,…,6), 5cos 3θ = 5, -5, 5, -5, 5, -5, 5

4STEP 4

The sign alternates

At the peaks the sign alternates.

f(kπ/3) ≥ 3 (k even), f(kπ/3) ≤ -1 (k odd)

5STEP 5

Check it is monotone between

Between peaks it is monotone, so one root each.

f'(θ) = -15sin 3θ - 3cosθ, lvertsin 3θrvert ＞ 1/5 → 15lvertsin 3θrvert ＞ 3 ≥ lvert 3cosθrvert

6STEP 6

Check the boundary

Check the interval's ends too.

lvert frvert ≥ 1 - 6 · arcsin1/5/3 ＞ 1 - 0.41 = 0.59 ＞ 0

7STEP 7

Count the half waves

One root per half wave gives 6.

6 half-waves × 1 root each = 6 → (D)

Answer

Build a sign table at the 13 nice angles θ = kπ/6, k = 0, …, 12, where cos 3θ ∈ {1, 0, -1} and sinθ is a standard value. The outputs are 6, -1/2, 1 - 3√(3)/2 - 5, -2, 6 - 3√(3)/2, -1/2, -4, 5/2, 6 + 3√(3)/2, 4, 1 + 3√(3)/2 - 5, 5/2, 6, i.e. signs +, -, -, -, +, -, -, +, +, +, -, +, +. That is exactly 6 sign changes, so at least 6 roots — matching the count, and the roots land near θ ≈ 0.50, 1.71, 2.58, 3.52, 5.01, 5.54, one per half-wave. The wrong choices are all reachable by a sloppy picture: (A) 2 and (B) 4 come from forgetting that the fast wave completes 3 periods, not 1 or 2; (C) 5 comes from dropping the crossing nearest an endpoint of the window; (E) 8 comes from assuming the slow curve is crossed twice on some sweeps, which Steps 5–6 rule out.

💡Key takeaway

A wave that swings out to ± 5 must cut clean through a curve that can never leave the band from -4 to 2 — once on every one-way sweep. Three periods of cos 3θ give six sweeps, so six solutions.