AMC 10 · 2010 · #24
Grade 11 algebracountingPick an answer.
Tool #16 (Change Focus): nothing about the size of the product matters, only its sign — and the sign of a product of eight nonzero numbers is decided by whether the count of negative factors is even. Since sin(kπ x) ＜ 0 exactly when ⌊ kx⌋ is odd, all the trigonometry collapses into one parity question about integers. Tool #7 (Identify Subproblems): the count splits cleanly into (a) where can the sign change at all, and (b) which of those places actually change it. Tool #5 (Look for a Pattern): part (b) has a sharp answer — crossing a breakpoint with reduced denominator q kills ⌊ 8/q⌋ of the eight factors at once, so the sign flips exactly when that number is odd. Tool #9 (Solve an Easier Related Problem): a proved reflection identity P(1-x) = P(x) cuts the bookkeeping in half. Tool #2 (Make a Systematic List): the left half is then a short ordered walk through ten labelled breakpoints.
Rewrite the domain as one inequality
The domain is one positivity condition.
A logarithm cares about nothing except the sign of what sits inside it, so the whole question becomes a sign question.
9.F-IF.A.1Change Focus Count The ComplementLocate every possible sign change
The breakpoints cut the interval into 22 pieces.
Labels like 2/4 and 1/2 mark the same spot on the line, so reducing to lowest terms is exactly what stops the double counting.
6.NS.B.4Identify SubproblemsRead each sine's sign off a floor
Each factor's sign is a floor parity.
Sine changes sign once per half-turn around the circle, and ⌊ kx⌋ is precisely the count of half-turns already made.
Sine changes sign once per half turn around the circle, so the count of half turns already made reads its sign.
▸ Why?
Half a turn is a fixed fraction of the whole circle, so those crossings come at evenly spaced places.
▸ Why?
The sign alternates between neighbouring half turns, so an even count is one sign and an odd count the other.
Turn the product into a parity
So the product's sign is one big parity.
Minus signs only matter in pairs, so eight separate sine signs collapse into one yes-or-no question about parity.
7.NS.A.2Change Focus Count The ComplementDecide which breakpoints flip the sign
Only breakpoints with an odd count flip it.
⌊ 8/q⌋ counts how many of the eight sines die at that point, and only an odd pile-up can turn the product over.
4.OA.B.4Look For A PatternProve the two halves mirror each other
The two halves are perfect mirrors.
Four of the eight factors flip sign under the reflection and four do not, and four minus signs cancel in pairs, so the product comes back unchanged.
9.F-IF.B.4Solve An Easier Related ProblemWalk the left half in order
Walking the left half finds 6 positive pieces.
Once every breakpoint is labelled flip or no-flip, reading off the signs is just walking the line and toggling a switch.
9.F-IF.B.4Make A Systematic ListMirror, total, and confirm disjointness
Mirroring doubles it to 12, choice (B).
Two neighbouring good intervals still count as two, because the shared endpoint is a zero of the product and so is punched out of the domain.
9.F-IF.A.1Identify SubproblemsThe sign of a long product of sines is just a parity count: read each factor's sign off ⌊ kx⌋, and the product only flips where an odd number of factors hit zero at the same time.
- Rewrite the domain as one inequality
- Locate every possible sign change
- Read each sine's sign off a floor
- Turn the product into a parity
- Decide which breakpoints flip the sign
- Prove the two halves mirror each other
- Walk the left half in order
- Mirror, total, and confirm disjointness