2010 AMC 12B Problem 20

Attempt Problem 20 of the 2010 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

A geometric sequence

(a_n)

has

a_1=\sin x,

a_2=\cos x,

and

a_3=\tan x

for some real number

x.

For what value of

n

does

a_n=1+\cos x?

4

5

6

7

8

The common ratio is

r=\dfrac{\cos x}{\sin x}=\cot x;

compute

a_4=a_3\cdot r

Show

a_4=1

and

1+\cos x=r^4,

then write

1+\cos x

a_4\cdot r^4

The common ratio is

r=\dfrac{a_2}{a_1}=\cot x.

Then

a_4=a_3\cdot r=\tan x\cot x=1.

From

a_3=a_1r^2,

we get

\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},

\sin^2 x=\cos^3 x,

i.e.

(\cos^2 x)(1+\cos x)=1.

Hence

1+\cos x=\dfrac{1}{\cos^2 x}.

Also

r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},

r^4=\dfrac{1}{\cos^2 x}=1+\cos x.

Therefore

1+\cos x=a_4\cdot r^4=a_8,

n=8.

Thus, the correct answer is E.

Problem 20 in other years