1966 AMC 12 Problem 22

Attempt Problem 22 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.

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22.

Consider the statements:
(I) a2+b2=0 \text{(I) }\sqrt{a^2+b^2}=0 (II) a2+b2=ab \text{(II) }\sqrt{a^2+b^2}=ab (III) a2+b2=a+b \text{(III) }\sqrt{a^2+b^2}=a+b (IV) a2+b2=a−b \text{(IV) }\sqrt{a^2+b^2}=a-b
where we allow aa and bb to be real or complex numbers. Those statements for which there exist solutions other than a=0a=0 and b=0,b=0, are:
(I),(\mathrm{I}), (II),(\mathrm{II}), (III),(\mathrm{III}), (IV)(\mathrm{IV})
(II),(\mathrm{II}), (III),(\mathrm{III}), (IV)(\mathrm{IV}) only
(I),(\mathrm{I}), (III),(\mathrm{III}), (IV)(\mathrm{IV}) only
(III),(\mathrm{III}), (IV)(\mathrm{IV}) only
(I)(\mathrm{I}) only
Answer: A
Concepts:complex numbercounterexampleradical
Small hint:
One nonzero example is enough for each statement
Big hint:
Try b=ib=i for (I),(\mathrm{I}), equal positive values for (II),(\mathrm{II}), and b=0b=0 for (III)(\mathrm{III}) and (IV)(\mathrm{IV})
Solution:
Every statement has a nonzero solution. For (I),(\mathrm{I}), use (a,b)=(1,i).(a,b)=(1,i). For (II),(\mathrm{II}), use a=b=2,a=b=\sqrt2, giving 4=2=ab.\sqrt4=2=ab. For both (III)(\mathrm{III}) and (IV),(\mathrm{IV}), use (a,b)=(1,0).(a,b)=(1,0). Thus all four statements qualify.
Therefore, the correct answer is A.
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