1979 AMC 12 Problem 16

Attempt Problem 16 of the 1979 AMC 12 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 1979 AMC 12 solutions, or check the answer key.

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

16.

A circle with area

A_1

is contained in the interior of a larger circle with area

A_1+A_2.

If the radius of the larger circle is

3,

and if

A_1,

A_2,

A_1+A_2

is an arithmetic progression, then the radius of the smaller circle is

\frac{\sqrt3}{2}

1

\frac2{\sqrt3}

\frac32

\sqrt3

Use the middle-term condition for the three areas in arithmetic progression

Express the larger circle’s area as a multiple of

A_1

The progression condition gives

2A_2=A_1+(A_1+A_2),

A_2=2A_1.

Hence the larger area is

3A_1=9\pi,

and

A_1=3\pi.

If the smaller radius is

r,

then

\pi r^2=3\pi,

r=\sqrt3.

Therefore, the correct answer is E.

Problem 16 in other years