AMC 10 · 2021 · #13

Grade 11 algebra

Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup

periodic-functionreflection-symmetrymodular-arithmetictrigonometric-ratios symmetry-argumentpattern-recognition ↑ Prerequisites: trigonometric-ratiosperiodic-function

📏 Long solution 💡 3 insights

Problem

One fixed angle is given. Build a fraction whose top is a product of five sines and whose bottom is a product of five sines. Find the exact value of that fraction. Ten sines appear, but no single sine value is needed; the whole quotient collapses to something clean.

Pick an answer.

(A)

${-}1$

(B)

${-}\frac{\sqrt{11}}{5}$

(C)

$\frac{\sqrt{11}}{5}$

(D)

$\frac{10}{11}$

(E)

How to solve

Strategy Organize Information in More Ways

Evaluating ten sines and multiplying them is hopeless by hand and pointless anyway, since the answer choices are exact. The better move is to stop reading the angles as angles and start reading them as index numbers: with c = 2π/11, the angle kc is k hops of one eleventh of a full turn, so only the value of k modulo 11 matters. Re-labelled that way, the numerator's list 3, 6, 9, 12, 15 and the denominator's list 1, 2, 3, 4, 5 turn out to be the same five hops in a different order, with a couple of them landing below the horizontal axis instead of above it. Then the problem splits into two small pieces: cancel the factors that literally match, and track the minus signs from the ones that got reflected. Nothing beyond the unit circle is needed.

1STEP 1

Read every angle as a number of hops

Read every angle as a number of hops.

c = 2π/11, sin kc = sin((2π k)/11)

2STEP 2

Bring 12 and 15 back inside one turn

Bring the big angles inside one turn.

sin 12c = sin(2π + c) = sin c, sin 15c = sin(2π + 4c) = sin 4c

3STEP 3

Cancel the factors that already match

Cancel the factors that already match.

(sin 3c · sin 6c · sin 9c · sin c · sin 4c)/(sin c · sin 2c · sin 3c · sin 4c · sin 5c) = (sin 6c · sin 9c)/(sin 2c · sin 5c)

4STEP 4

Reflect indices 6 and 9 across the axis

Reflect the leftovers across the axis.

sin 6c = sin(2π - 5c) = -sin 5c, sin 9c = sin(2π - 2c) = -sin 2c

5STEP 5

Count the minus signs

The minus appears twice and cancels.

(sin 6c · sin 9c)/(sin 2c · sin 5c) = (-sin 5c)(-sin 2c)/(sin 2c · sin 5c) = (sin 2c · sin 5c)/(sin 2c · sin 5c) = 1

6STEP 6

See why none of it was luck

See why none of it was luck.

{3 · 1, 3 · 2, 3 · 3, 3 · 4, 3 · 5} ≡ {3, 6, 9, 1, 4} (mod 11) ⟶ {3, 5, 2, 1, 4}, (-1)² = 1

7STEP 7

State the value

The value is 1.

(sin 3c · sin 6c · sin 9c · sin 12c · sin 15c)/(sin c · sin 2c · sin 3c · sin 4c · sin 5c) = 1

Answer

A quick numerical pass agrees. With c ≈ 0.5712 radians, the five numerator sines are about 0.9898, -0.2817, -0.9096, 0.5406, 0.7557, whose product is about 0.1036; the five denominator sines are about 0.5406, 0.9096, 0.9898, 0.7557, 0.2817, whose product is also about 0.1036. The quotient is 1.000 to three decimals. The structure also explains the distractors. Choices -√(11)/5 and √(11)/5 come from half-remembering the product formula Π_k=1ⁿ⁻¹ sinkπ/n = n/(2ⁿ⁻¹), where a √(11) genuinely appears in related products, but no square root can survive here because the two products contain literally the same factors. Choice 10/11 is the shape of that formula's fraction and is likewise impossible. Choice -1 is the real trap: it is what a solver gets after miscounting the reflections, for instance by reflecting 9 but forgetting that 6 also exceeds 11/2.

💡Key takeaway

Because c is one eleventh of a full turn, only each angle's index modulo 11 matters, so reduce the indices, cancel the matching factors, and the answer is decided by whether the number of sign flips is even or odd.

Read every angle as a number of hops
Bring 12 and 15 back inside one turn
Cancel the factors that already match
Reflect indices 6 and 9 across the axis
Count the minus signs
See why none of it was luck
State the value