1966 AMC 12 Problem 3

Attempt Problem 3 of the 1966 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 1966 AMC 12 solutions, or check the answer key.

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

If the arithmetic mean of two numbers is

6

and their geometric mean is

10,

then an equation with the given two numbers as roots is:

x^2+12x+100=0

x^2+6x+100=0

x^2-12x-10=0

x^2-12x+100=0

x^2-6x+100=0

The two numbers have sum

12

and product

100

Roots with sum

S

and product

P

satisfy

x^2-Sx+P=0

If the numbers are

r,

and

s,

then

\frac{r+s}{2}=6

gives

r+s=12,

while

\sqrt{rs}=10

gives

rs=100.

The monic equation with those roots is

x^2-(r+s)x+rs=0,

x^2-12x+100=0.

Therefore, the correct answer is D.

Problem 3 in other years