Difference between revisions of "2000 JBMO Problems/Problem 1"
Rockmanex3 (talk | contribs) (Solution to Problem 1 (credit to rchokler) -- multivariable factoring) |
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Let <math>x</math> and <math>y</math> be positive reals such that <cmath> x^3 + y^3 + (x + y)^3 + 30xy = 2000. </cmath> Show that <math>x + y = 10</math>. | Let <math>x</math> and <math>y</math> be positive reals such that <cmath> x^3 + y^3 + (x + y)^3 + 30xy = 2000. </cmath> Show that <math>x + y = 10</math>. | ||
| − | ==Solution== | + | == Solution 1 == |
| + | |||
| + | After some manipulation we get: | ||
| + | <math>3xy(10 - (x+y)) = 2((x+y)^3 - 10^3)</math> | ||
| + | |||
| + | Case 1: <math>x+y > 10:</math> | ||
| + | <math>RHS > 0</math>, So <math>LHS</math> has to be <math>> 0</math>, so <math>10 - (x+y) > 0</math> or <math>x+y < 10 => contradiction!</math> | ||
| + | |||
| + | Case 2: <math>x+y < 10:</math> | ||
| + | <math>RHS < 0</math>, So <math>LHS</math> has to be <math>< 0</math>, so <math>10 - (x+y) < 0</math> or <math>x+y > 10 => contradiction!</math> | ||
| + | |||
| + | Thus <math>x + y = 10</math> | ||
| + | |||
| + | |||
| + | <math>Kris17</math> | ||
| + | |||
| + | |||
| + | |||
| + | == Solution 2 == | ||
Rearranging the equation yields | Rearranging the equation yields | ||
Revision as of 20:58, 3 December 2018
Contents
Problem
Let 
and 
be positive reals such that ![\[x^3 + y^3 + (x + y)^3 + 30xy = 2000.\]](http://latex.artofproblemsolving.com/e/e/3/ee3c19c2097db41a478940611f5891c918bcdd59.png)
Show that 
.
Solution 1
After some manipulation we get:
Case 1: 
, So 
has to be 
, so 
or 
Case 2: 
, So has to be 
, so 
or 
Thus
Solution 2
Rearranging the equation yields
![\[x^3 + y^3 + (x + y)^3 + 30xy - 2000 = 0.\]](http://latex.artofproblemsolving.com/d/d/8/dd8346c2a2209add7e910538837f42802cd63bae.png)
If 
in the large equation, then 
must be a factor of the large equation. Note that we can rewrite the large equation as
![\begin{align*} 0 &= (x+y)^3 - 1000 + x^3 + 3x^2y - 3x^2y + 3xy^2 - 3xy^2 + y^3 - 1000 + 30xy \\ &= 2[(x+y)^3 - 1000] - 3x^2y - 3xy^2 + 30xy. \end{align*}](http://latex.artofproblemsolving.com/9/b/2/9b2b6f6c62fa6dae97244c3dc3f0a4a90bef7102.png)
We can factor the difference of cubes in the first part and factor 
in the second part, resulting in
![\[0 = 2(x+y-10)((x+y)^2 + 10x + 10y + 100) - 3xy(x+y-10)\]](http://latex.artofproblemsolving.com/e/a/e/eae5403263b0190ad837a7a9eca2bcbf82659478.png)
Finally, we can factor by grouping, which results in
By the Zero Product Property, either or 
However, since and are both positive, 
can not equal zero, so we have proved that 
See Also
| 2000 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
