1960 AMC 12 Problem 22

Attempt Problem 22 of the 1960 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 1960 AMC 12 solutions, or check the answer key.

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22.

The equality

(x+m)^2-(x+n)^2=(m-n)^2,

where

m

and

n

are unequal non-zero constants, is satisfied by

x=am+bn

where:

a=0,

b

has a unique non-zero value

a=0,

b

has two non-zero values

b=0,

a

has a unique non-zero value

b=0,

a

has two non-zero values

a

and

b

each have a unique non-zero value

Factor the difference of the two squares

Cancel the nonzero factor

m-n

and solve for

x

Factoring and using

m\ne n

gives

\begin{gathered} (m-n)(2x+m+n)\\ {}=(m-n)^2. \end{gathered}

Hence

2x+m+n=m-n,

x=-n.

Thus

a=0

and

b=-1,

a unique nonzero value.

Therefore, the correct answer is A.

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