Difference between revisions of "Mock AIME II 2012 Problems/Problem 11"
| Line 6: | Line 6: | ||
First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>. | First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>. | ||
Now, note that <math>2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab</math>. | Now, note that <math>2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab</math>. | ||
| − | Also, we have <math>(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies 2n+2(2n-3)=n^2\implies n^2-6n+6=0</math>. | + | Also, we have <math>(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies </math> |
| + | <math>2n+2(2n-3)=n^2\implies n^2-6n+6=0</math>. | ||
| − | Next, <math>3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)\implies a^4+b^4=-n^2+6n</math>. But we know <math>n^2-6n+6=0</math>, so <math>-n^2+6n=6</math>. | + | Next, <math>3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)</math> |
| + | <math>\implies a^4+b^4=-n^2+6n</math>. But we know <math>n^2-6n+6=0</math>, so <math>-n^2+6n=6</math>. | ||
Since the only possible values of <math>a^4+b^4</math> are <math>0</math> and <math>6</math>, our final answer is <math>\boxed{006}</math>. | Since the only possible values of <math>a^4+b^4</math> are <math>0</math> and <math>6</math>, our final answer is <math>\boxed{006}</math>. | ||
(It is easy to check that there exists <math>a, b, n</math> satisfying the equations.) | (It is easy to check that there exists <math>a, b, n</math> satisfying the equations.) | ||
Latest revision as of 02:18, 5 April 2012
Problem
There exist real values of 
and 
such that 
, 
, and 
for some value of 
. Let 
be the sum of all possible values of 
. Find .
Solution
First, if 
, then 
. We now assume that 
.
Now, note that 
.
Also, we have 
.
Next, 
. But we know 
, so 
.
Since the only possible values of are 
and 
, our final answer is 
.
(It is easy to check that there exists 
satisfying the equations.)
