1966 AMC 12 第 22 题

先试着解答 1966 AMC 12 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1966 AMC 12 解答,或核对答案。

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22.

考虑下列等式:
(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
其中允许 aa 与 bb 为实数或复数。除 a=0a=0 且 b=0b=0 外仍存在解的等式是:
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})
(I),(\mathrm{I}), (II),(\mathrm{II}), (III),(\mathrm{III}), (IV)(\mathrm{IV})
仅 (II)(\mathrm{II}),(III)(\mathrm{III}),(IV)(\mathrm{IV})
(II),(\mathrm{II}), (III),(\mathrm{III}), (IV)(\mathrm{IV}) only
仅 (I)(\mathrm{I}),(III)(\mathrm{III}),(IV)(\mathrm{IV})
(I),(\mathrm{I}), (III),(\mathrm{III}), (IV)(\mathrm{IV}) only
仅 (III)(\mathrm{III}),(IV)(\mathrm{IV})
(III),(\mathrm{III}), (IV)(\mathrm{IV}) only
仅 (I)(\mathrm{I})
(I)(\mathrm{I}) only
答案:A
知识点:复数反例根式
小提示:
对每个等式,找到一个非零解即可
One nonzero example is enough for each statement
大提示:
对 (I)(\mathrm{I}) 尝试 b=ib=i,对 (II)(\mathrm{II}) 取两个相等正值,对 (III)(\mathrm{III}) 和 (IV)(\mathrm{IV}) 取 b=0b=0
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})
解答:
每个等式都有非零解。对 (I)(\mathrm{I}),取 (a,b)=(1,i)(a,b)=(1,i)。对 (II)(\mathrm{II}),取 a=b=2a=b=\sqrt2,此时 4=2=ab\sqrt4=2=ab。对 (III)(\mathrm{III}) 和 (IV)(\mathrm{IV}),都取 (a,b)=(1,0)(a,b)=(1,0)。因此四个等式都符合要求。
因此,正确答案是 A。
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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