1966 AHSME Problems/Problem 22

Problem

Consider the statements: (I) $\sqrt{a^2+b^2}=0$ , (II) $\sqrt{a^2+b^2}=ab$ , (III) $\sqrt{a^2+b^2}=a+b$ , (IV) $\sqrt{a^2+b^2}=a - b$ , where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions other than $a=0$ and $b=0$ , are:

$\text{(A) (I),(II),(III),(IV)}\quad \text{(B) (II),(III),(IV) only} \quad \text{(C) (I),(III),(IV) only}\quad \text{(D) (III),(IV) only} \quad \text{(E) (I) only}$

Solution

We are given the following statements:

$\sqrt{a^2 + b^2} = 0$
$\sqrt{a^2 + b^2} = ab$
$\sqrt{a^2 + b^2} = a + b$
$\sqrt{a^2 + b^2} = a - b$

We are asked to find the solutions where these statements hold, excluding the trivial solution $a = 0$ and $b = 0$ .

Statement (I): $\sqrt{a^2 + b^2} = 0$

Squaring both sides:

$a^2 + b^2 = 0$

For real or complex numbers, $a^2 + b^2 = 0$ implies that both $a = 0$ and $b = 0$ because the sum of squares of two real or complex numbers is 0 only if both are 0.

Thus, the only solution is $a = 0$ and $b = 0$ .

Conclusion for Statement (I): There are no solutions other than $a = 0$ and $b = 0$ .

Statement (II): $\sqrt{a^2 + b^2} = ab$

Squaring both sides:

$a^2 + b^2 = (ab)^2 = a^2b^2$

Rearranging:

$a^2 + b^2 - a^2b^2 = 0$

Factoring:

$a^2(1 - b^2) + b^2 = 0$

Testing for specific solutions:

If $a = 1$ and $b = 1$ , we get:

$\sqrt{1^2 + 1^2} = 1 \cdot 1 \quad \Rightarrow \quad \sqrt{2} = 1$

which is false. Thus, there are no simple real solutions.

For complex solutions, trying specific cases like $a = i$ and $b = 1$ also results in contradictions.

Thus, this equation has no nonzero solutions.

Conclusion for Statement (II): No solutions other than $a = 0$ and $b = 0$ .

Statement (III): $\sqrt{a^2 + b^2} = a + b$

Squaring both sides:

$a^2 + b^2 = (a + b)^2$

Expanding the right-hand side:

$a^2 + b^2 = a^2 + 2ab + b^2$

Canceling $a^2 + b^2$ from both sides:

$0 = 2ab$

Thus:

$ab = 0$

This implies that either $a = 0$ or $b = 0$ . Therefore, there are solutions other than $a = 0$ and $b = 0$ , such as $a = 0$ and $b \neq 0$ , or $a \neq 0$ and $b = 0$ .

Conclusion for Statement (III): There are solutions other than $a = 0$ and $b = 0$ .

Statement (IV): $\sqrt{a^2 + b^2} = a - b$

Squaring both sides:

$a^2 + b^2 = (a - b)^2$

Expanding the right-hand side:

$a^2 + b^2 = a^2 - 2ab + b^2$

Canceling $a^2 + b^2$ from both sides:

$0 = -2ab$

Thus:

$ab = 0$

This implies that either $a = 0$ or $b = 0$ . Therefore, there are solutions other than $a = 0$ and $b = 0$ , such as $a = 0$ and $b \neq 0$ , or $a \neq 0$ and $b = 0$ .

Conclusion for Statement (IV): There are solutions other than $a = 0$ and $b = 0$ .

Final Conclusion:

The statements that have solutions other than $a = 0$ and $b = 0$ are (III) and (IV).

Thus, the correct answer is:

$\boxed{\text{(A) (I), (II), (III), (IV)}}$

~ Aoum

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1966 AHSME Problems/Problem 22

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