AMC 10 · 2014 · #25

Grade 11 algebra

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

double-angle-formulatrigonometric-ratiosperiodic-functionprime-factorizationdivisor-count extreme-principlecasework ↑ Prerequisites: double-angle-formulaperiodic-function

📏 Long solution 💡 4 insights

Problem

A trigonometric equation must hold for positive inputs. Add every solution.

Pick an answer.

(A)

$\pi$

(B)

$810\pi$

(C)

$1008\pi$

(D)

$1080 \pi$

(E)

$1800\pi$

How to solve

Strategy Extreme Principle

A transcendental equation mixing cos 2x with cos(2014π²/x) has no solving technique — there is no formula to apply and no graph worth drawing. What there is, is a size limit. Tool #15 (Organize Information in More Ways) does the opening move: rewriting cos 4x-1 as 2cos² 2x-2 makes the same squared term sit on both sides, and it cancels, leaving the bare statement cos 2x·cos(2014π²/x)=1. Now tool #14 (Extreme Principle) takes over and is the whole solution: each cosine has size at most 1, so their product has size at most 1, and the equation demands the product hit that ceiling exactly. A ceiling can only be reached if neither factor gave anything away, which pins both cosines to ± 1 — a boundary case, and a complete one, which is what converts an infinite search into a finite list. Tool #4 (Introduce a Variable) then rescales by writing x=cπ, which turns both boundary conditions into parity and divisibility statements about ordinary integers. Tool #2 (Make a Systematic List) enumerates the surviving divisors, tool #3 (Eliminate Possibilities) kills the second boundary case outright, and tool #5 (Look for a Pattern) reads the final sum off the factorisation of 1007.

1STEP 1

Cancel the squared terms

The squared terms cancel at once.

2cos² 2x-2cos 2xcos(2014π²/x)=2cos² 2x-2 ⟺ cos 2x·cos(2014π²/x)=1

2STEP 2

Squeeze the product to its ceiling

The product at its ceiling forces two cases.

|a| ≤ 1, |b| ≤ 1, ab=1 ⟹ |a|=|b|=1 ⟹ (a,b)=(1,1) or (a,b)=(-1,-1)

3STEP 3

Measure every angle in πs

Measuring angles in one unit makes it arithmetic.

x=cπ (c ＞ 0): cos 2x=cos(2cπ), cos(2014π²/x)=cos(2014/cπ)

4STEP 4

First case: odd divisors of 2014

The first case gives a short divisor list.

2014=2 · 19 · 53; 2014/c even integer⇔ c ∣ 1007⇔ c∈{1, 19, 53, 1007}

5STEP 5

Second case: nothing at all

The second case gives nothing at all.

c=m/2, m odd ⟹ 2014/c=4028/m=4·1007/m, always even

6STEP 6

Add the four solutions

Adding the four gives 1080π, choice (D).

π+19π+53π+1007π=(1+19)(1+53)π=20 · 54 π=1080π → (D)

Answer

1080 π

All four values check directly in the original equation. For x=π: cos 2π=1 and 2014π²/π=2014π with 2014 even, so cos(2014π)=1; the left side is 2 · 1·(1-1)=0 and the right side is cos 4π-1=0. For x=1007π: cos(2014π)=1 and 2014π²/1007π=2π, so cos 2π=1; both sides are 0 again. The pair x=19π and x=53π work the same way, and there is a pleasing symmetry behind it: c↦1007/c swaps 1⇔ 1007 and 19⇔ 53, because the two cosines trade roles. Nothing was lost along the way either — step 1 was a chain of equivalences, and step 2 produced a complete case list rather than a sample, so the four values are the entire solution set. The distractors map onto the two ways the characterisation can be left half-done: (A) π is where you stop if you find the obvious solution x=π and assume it is the only one, and (C) 1008π=π+1007π is where you land if you treat 1007 as prime and miss the middle divisors 19 and 53. Neither 810 nor 1800 is the sum of any subset of the divisors of 2014, so (B) and (E) cannot come from a miscount of divisors at all — they punish a wrong factorisation of 2014 outright.

💡Key takeaway

When two cosines multiply to 1 — the biggest a product of cosines could ever be — neither one had any room to spare, so both must sit exactly at ± 1, and the rest of the problem is just deciding which whole numbers divide 2014.

Cancel the squared terms
Squeeze the product to its ceiling
Measure every angle in πs
First case: odd divisors of 2014
Second case: nothing at all
Add the four solutions