AMC 10 · 2021 · #21

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

complex-numberscomplex-polar-formdouble-angle-formulapolynomial-roots identify-subproblemscasework ↑ Prerequisites: complex-numbersdouble-angle-formula

📏 Long solution 💡 4 insights

Problem

For each real number a certain expression built from cosines and sines gives a complex number. Count how many values in one full turn make it equal zero.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

The expression looks tangled, but it is already sorted for us: four terms carry no i and three terms each carry one i. That is tool #7 (Identify Subproblems) handed over for free — P(x)=0 splits into a cosine equation and a sine equation that must hold at the same x. The two subproblems are wildly unequal in difficulty, and that inequality is the whole strategy. Tool #15 (Organize Information in More Ways) attacks the easy one: rewriting sin x-sin 2x+sin 3x as (sin x+sin 3x)-sin 2x puts two angles symmetrically around 2x, which is exactly the setup for the sum-to-product identity, and the three-term sum collapses into a product of two factors. A product being zero is a short list of cases, so tool #2 (Make a Systematic List) turns that into six named angles in [0,2π) and guarantees nothing outside the list can work. To test those six quickly, tool #13 (Convert to Algebra) replaces the trigonometry with a single number: setting z=cos x+isin x makes cos 2x+isin 2x and cos 3x+isin 3x into z² and z³, so P becomes the cubic 1+z-z²+z³. Then tool #3 (Eliminate Possibilities) finishes: evaluate the cubic at all six candidates and see how many survive. A count question is safest answered this way — build a list that provably contains every solution, then check every entry.

1STEP 1

Split one equation into two

Split one equation into two.

P(x)=A(x)+B(x) i with A(x)=1+cos x-cos 2x+cos 3x and B(x)=sin x-sin 2x+sin 3x; so P(x)=0 ⇔ A(x)=0 and B(x)=0

2STEP 2

Factor the sine equation

Factor the sine equation.

B(x)=(sin x+sin 3x)-sin 2x=2sin 2xcos x-sin 2x=sin 2x(2cos x-1)

3STEP 3

List the six candidate angles

List the six candidate angles.

sin 2x(2cos x-1)=0→ sin 2x=0 or cos x=1/2; on 0 ≤ x ＜ 2π this gives x∈{0, π/2, π, 3π/2}∪{π/3, 5π/3}

4STEP 4

Rewrite P using one number

Rewrite the expression using one number.

z=cos x+isin x→ z²=cos 2x+isin 2x, z³=cos 3x+isin 3x; hence P(x)=1+z-z²+z³

5STEP 5

Test the four quarter-turn candidates

Test the four quarter-turn candidates.

P=1+z-z²+z³: P(1)=2, P(i)=1+i+1-i=2, P(-1)=1-1-1-1=-2, P(-i)=1-i+1+i=2

6STEP 6

Test the last two and count

Testing the last two leaves 0.

x=π/3: z²=z-1, z³=-1→ P=1+z-(z-1)+(-1)=1 ≠ 0; x=5π/3 gives the conjugate value 1 ≠ 0. Six candidates, zero survivors, so the count is 0.

Answer

The six candidate values of P came out as 2, 2, -2, 2, 1, 1. None is near 0, so the equation is not missing by a hair that a rounding slip could close — it misses by a full unit or more, which is the comfortable kind of "no solutions". Sweeping |P(x)| across the whole window agrees: its smallest value is about 0.96, near x≈ 5.33, and the curve never dips to the axis. The distractors are exactly the counts produced by stopping early. A solver who solves only sin 2x=0 and declares victory reports 4, choice (E); one who keeps only the branch cos x=1/2 reports 2, choice (C); one who spot-checks part of the list and forgets the rest lands on (B) or (D). Every wrong answer here is the same mistake — treating the sine equation as if it were the whole problem — so the real content of the question is remembering that a complex number needs both parts to vanish. One more boundary check: the window 0 ≤ x ＜ 2π is half-open, and the only endpoint candidate is x=0, which was tested and failed, so no double-counting or omission at the seam can change the count.

💡Key takeaway

A complex number is zero only when both of its halves are zero, so solve the easier half to get a short list of suspects and let the harder half rule on each one — here it rejects all six, and the honest answer is none.

Split one equation into two
Factor the sine equation
List the six candidate angles
Rewrite P using one number
Test the four quarter-turn candidates
Test the last two and count