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Difference between revisions of "1966 AHSME Problems/Problem 22"

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<math>\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}</math>
 
<math>\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}</math>
  
−
== Solution ==
+
==Solution==
−
<math>\fbox{A}</math>
+
We are given the following statements:
 +
 
 +
# <math>\sqrt{a^2 + b^2} = 0</math>
 +
# <math>\sqrt{a^2 + b^2} = ab</math>
 +
# <math>\sqrt{a^2 + b^2} = a + b</math>
 +
# <math>\sqrt{a^2 + b^2} = a - b</math>
 +
 
 +
We are asked to find the solutions where these statements hold, excluding the trivial solution <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
'''Statement (I):''' <math>\sqrt{a^2 + b^2} = 0</math>
 +
 
 +
Squaring both sides:
 +
 
 +
<cmath>
 +
a^2 + b^2 = 0
 +
</cmath>
 +
 
 +
For real or complex numbers, <math>a^2 + b^2 = 0</math> implies that both <math>a = 0</math> and <math>b = 0</math> because the sum of squares of two real or complex numbers is 0 only if both are 0.
 +
 
 +
Thus, the only solution is <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
'''Conclusion for Statement (I):''' There are no solutions other than <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
'''Statement (II):''' <math>\sqrt{a^2 + b^2} = ab</math>
 +
 
 +
Squaring both sides:
 +
 
 +
<cmath>
 +
a^2 + b^2 = (ab)^2 = a^2b^2
 +
</cmath>
 +
 
 +
Rearranging:
 +
 
 +
<cmath>
 +
a^2 + b^2 - a^2b^2 = 0
 +
</cmath>
 +
 
 +
Factoring:
 +
 
 +
<cmath>
 +
a^2(1 - b^2) + b^2 = 0
 +
</cmath>
 +
 
 +
Testing for specific solutions:
 +
 
 +
*If <math>a = 1</math> and <math>b = 1</math>, we get:
 +
 
 +
<cmath>
 +
\sqrt{1^2 + 1^2} = 1 \cdot 1 \quad \Rightarrow \quad \sqrt{2} = 1
 +
</cmath>
 +
 
 +
which is false. Thus, there are no simple real solutions.
 +
 
 +
*For complex solutions, trying specific cases like <math>a = i</math> and <math>b = 1</math> also results in contradictions.
 +
 
 +
Thus, this equation has no nonzero solutions.
 +
 
 +
'''Conclusion for Statement (II):''' No solutions other than <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
'''Statement (III):''' <math>\sqrt{a^2 + b^2} = a + b</math>
 +
 
 +
Squaring both sides:
 +
 
 +
<cmath>
 +
a^2 + b^2 = (a + b)^2
 +
</cmath>
 +
 
 +
Expanding the right-hand side:
 +
 
 +
<cmath>
 +
a^2 + b^2 = a^2 + 2ab + b^2
 +
</cmath>
 +
 
 +
Canceling <math>a^2 + b^2</math> from both sides:
 +
 
 +
<cmath>
 +
0 = 2ab
 +
</cmath>
 +
 
 +
Thus:
 +
 
 +
<cmath>
 +
ab = 0
 +
</cmath>
 +
 
 +
This implies that either <math>a = 0</math> or <math>b = 0</math>. Therefore, there are solutions other than <math>a = 0</math> and <math>b = 0</math>, such as <math>a = 0</math> and <math>b \neq 0</math>, or <math>a \neq 0</math> and <math>b = 0</math>.
 +
 
 +
**Conclusion for Statement (III):** There are solutions other than <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
### Statement (IV): <math>\sqrt{a^2 + b^2} = a - b</math>
 +
 
 +
Squaring both sides:
 +
 
 +
<cmath>
 +
a^2 + b^2 = (a - b)^2
 +
</cmath>
 +
 
 +
Expanding the right-hand side:
 +
 
 +
<cmath>
 +
a^2 + b^2 = a^2 - 2ab + b^2
 +
</cmath>
 +
 
 +
Canceling <math>a^2 + b^2</math> from both sides:
 +
 
 +
<cmath>
 +
0 = -2ab
 +
</cmath>
 +
 
 +
Thus:
 +
 
 +
<cmath>
 +
ab = 0
 +
</cmath>
 +
 
 +
This implies that either <math>a = 0</math> or <math>b = 0</math>. Therefore, there are solutions other than <math>a = 0</math> and <math>b = 0</math>, such as <math>a = 0</math> and <math>b \neq 0</math>, or <math>a \neq 0</math> and <math>b = 0</math>.
 +
 
 +
'''Conclusion for Statement (IV):''' There are solutions other than <math>a = 0</math> and <math>b = 0</math>.
 +
 
 +
'''Final Conclusion:'''
 +
 
 +
The statements that have solutions other than <math>a = 0</math> and <math>b = 0</math> are (III) and (IV).
 +
 
 +
Thus, the correct answer is:
 +
 
 +
<cmath>
 +
\boxed{\text{(A) (I), (II), (III), (IV)}}
 +
</cmath>
 +
 
 +
~ [[User:Aoum|Aoum]]
  
 
== See also ==
 
== See also ==

Revision as of 11:34, 17 February 2025

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:

  1. $\sqrt{a^2 + b^2} = 0$
  2. $\sqrt{a^2 + b^2} = ab$
  3. $\sqrt{a^2 + b^2} = a + b$
  4. $\sqrt{a^2 + b^2} = a - b$

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 $b = 0$.
      1. 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

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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