Difference between revisions of "1987 USAMO Problems/Problem 1"
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| − | By expanding you get < | + | By expanding you get <cmath>m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3</cmath> |
| − | From this the two m^3 cancel and you get: | + | From this the two <math>m^3</math> cancel and you get: |
<cmath>2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0</cmath> | <cmath>2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0</cmath> | ||
You can divide by <math>n</math> (nonzero). | You can divide by <math>n</math> (nonzero). | ||
You get: | You get: | ||
<cmath>2n^2+m+m^2n+3m^2-3mn=0</cmath> | <cmath>2n^2+m+m^2n+3m^2-3mn=0</cmath> | ||
| + | You can now factor the equation into: | ||
| + | <cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath> | ||
Revision as of 10:17, 20 December 2015
By expanding you get ![\[m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3\]](http://latex.artofproblemsolving.com/0/d/1/0d1544ffb34dff29d0aa0c0d362cbe0b13b1a430.png)
From this the two 
cancel and you get:
![\[2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0\]](http://latex.artofproblemsolving.com/4/2/8/4280b7e66e77cbcafd88221eac9fe0f393490077.png)
You can divide by 
(nonzero).
You get:
![\[2n^2+m+m^2n+3m^2-3mn=0\]](http://latex.artofproblemsolving.com/9/8/9/98978fc9b2412fa469ad1a9a34ac576da4625162.png)
You can now factor the equation into:
![\[2n^2+(m^2-3m)n+(3m^2+m)=0\]](http://latex.artofproblemsolving.com/9/7/5/975eb800f6268d629e943923140e0a35c9af5afe.png)
