2010 AMC 12A Problem 24

Question

Let

\begin{gathered} f(x)=\log_{10}\big(\sin(\pi x) \\ {}\cdot\sin(2\pi x)\cdot\sin(3\pi x) \\ {}\cdot\sin(4\pi x)\cdot\sin(5\pi x) \\ {}\cdot\sin(6\pi x)\cdot\sin(7\pi x) \\ {}\cdot\sin(8\pi x)\big). \end{gathered}

The intersection of the domain of

f(x)

with the interval

[0,1]

is a union of

n

disjoint open intervals. What is

n?

2

12

18

22

36

Answer: B

Concepts:trigonometry symmetry casework

Difficulty rating: 2460

Small hint:

The domain is where

g(x)=\prod_{k=1}^8\sin(k\pi x)\gt0;

g

is symmetric about

x=\tfrac12

Big hint:

Count the zeros

\tfrac{k}{n}

in

\big(0,\tfrac12\big)

and note the sign fails to flip at

\tfrac14

and

\tfrac13

Solution:

Let

g(x)=\prod_{k=1}^8\sin(k\pi x);

the domain of

f

is where

g(x)\gt0.

Since

\sin(k\pi(1-x))

=(-1)^{k+1}\sin(k\pi x)

and

\sum_{k=1}^8(k+1)

is even,

g(1-x)=g(x),

so it suffices to study

\big(0,\tfrac12\big)

and double.

In

\big(0,\tfrac12\big)

the zeros of

g

are the fractions

\tfrac{k}{n}

with

2\le n\le8,

1\le k\lt\tfrac n2,

and

\gcd(k,n)=1.

For

n=2,\ldots,8

there are

0,1,1,2,1,3,2

of them, totaling

10.

These

10

zeros split

\big(0,\tfrac12\big)

into

11

subintervals on which

g

has constant sign. Near

0

every factor is positive, so

g\gt0

there, and the sign flips at each zero except

\tfrac14=\tfrac28

and

\tfrac13=\tfrac26,

where an even number of factors vanish.

Tracking the signs, exactly

6

of the

11

subintervals have

g\gt0.

By symmetry there are

6