Difference between revisions of "2021 AMC 12A Problems/Problem 9"

(Solution)
Line 2: Line 2:
 
These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
 
These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
 
==Solution==
 
==Solution==
The solutions will be posted once the problems are posted.
+
Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>1</math>, which can be achieved at <math>x=y=0</math>. ~aop2014
 +
 
 
==Note==
 
==Note==
 
See [[2021 AMC 12A Problems/Problem 1|problem 1]].
 
See [[2021 AMC 12A Problems/Problem 1|problem 1]].

Revision as of 13:52, 11 February 2021

Problem

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the trivial inequality(all squares are nonnegative) the minimum value for this is $1$, which can be achieved at $x=y=0$. ~aop2014

Note

See problem 1.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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
All AMC 12 Problems and Solutions

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