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

(Solution)
m (Problem)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Consider the statements: (I)<math>\sqrt{a^2+b^2}=0</math>,  (II) <math>\sqrt{a^2+b^2}=ab</math>,  (III) <math>\sqrt{a^2+b^2}=a+b</math>,  (IV) <math>\sqrt{a^2+b^2}=a\cdot b</math>, where we allow <math>a</math> and <math>b</math> to be real or complex numbers. Those statements for which there exist solutions other than <math>a=0</math> and <math>b=0</math>, are:
+
Consider the statements: (I)<math>\sqrt{a^2+b^2}=0</math>,  (II) <math>\sqrt{a^2+b^2}=ab</math>,  (III) <math>\sqrt{a^2+b^2}=a+b</math>,  (IV) <math>\sqrt{a^2+b^2}=a - b</math>, where we allow <math>a</math> and <math>b</math> to be real or complex numbers. Those statements for which there exist solutions other than <math>a=0</math> and <math>b=0</math>, are:
  
 
<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>

Latest revision as of 22:10, 11 March 2024

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

$\fbox{A}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png