Difference between revisions of "2021 WSMO Team Round/Problem 10"
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==Solution== | ==Solution== | ||
Notice that we can complete the square inside the second square root: | Notice that we can complete the square inside the second square root: | ||
| − | <cmath>\sqrt{3m^2+3n^2-6m+12n+15} | + | <cmath>\sqrt{3m^2+3n^2-6m+12n+15}</cmath> |
| − | =\sqrt{3(m^2+n^2-2m+4n+5)} | + | <cmath>=\sqrt{3(m^2+n^2-2m+4n+5)}</cmath> |
| − | =\sqrt{3(m^2-2m+1+n^2+4n+4)} | + | <cmath>=\sqrt{3(m^2-2m+1+n^2+4n+4)}</cmath> |
| − | =\sqrt{3((m-1)^2+(n+2)^2)}</cmath> | + | <cmath>=\sqrt{3((m-1)^2+(n+2)^2)}</cmath> |
Notice that we can find the minimum by setting this to <math>0</math>, which occurs when <math>m=1</math> and <math>n=-2</math>. This gives us the minimum of <math>a=\sqrt{5}</math>. (If we set the other square root to <math>0</math>, we get a minimum of <math>\sqrt{15}</math> which is larger than <math>\sqrt{5}</math>.) Therefore <math>a^2=(\sqrt{5})^2=\boxed{5}</math>. | Notice that we can find the minimum by setting this to <math>0</math>, which occurs when <math>m=1</math> and <math>n=-2</math>. This gives us the minimum of <math>a=\sqrt{5}</math>. (If we set the other square root to <math>0</math>, we get a minimum of <math>\sqrt{15}</math> which is larger than <math>\sqrt{5}</math>.) Therefore <math>a^2=(\sqrt{5})^2=\boxed{5}</math>. | ||
~programmeruser | ~programmeruser | ||
Revision as of 19:29, 23 March 2023
Problem
The minimum possible value of![\[\sqrt{m^2+n^2}+\sqrt{3m^2+3n^2-6m+12n+15}\]](http://latex.artofproblemsolving.com/6/8/d/68d0e30357a8b66f429013f413c224dca472dc7c.png)
can be expressed as 
Find 
Proposed by pinkpig
Solution
Notice that we can complete the square inside the second square root:
![\[\sqrt{3m^2+3n^2-6m+12n+15}\]](http://latex.artofproblemsolving.com/4/6/5/4658f3d0e5a6d8596017f7a50cee069ecfce9bfa.png)
![\[=\sqrt{3(m^2+n^2-2m+4n+5)}\]](http://latex.artofproblemsolving.com/4/6/3/463d27fd3c15c736cb5602a9d0a8d88ad848a1d1.png)
![\[=\sqrt{3(m^2-2m+1+n^2+4n+4)}\]](http://latex.artofproblemsolving.com/6/f/6/6f6328116229eb4515b4f9b077f7705c64596b0b.png)
![\[=\sqrt{3((m-1)^2+(n+2)^2)}\]](http://latex.artofproblemsolving.com/1/1/6/116ec8cedf452d37fe74ce077c455b74189101f5.png)
Notice that we can find the minimum by setting this to 
, which occurs when 
and 
. This gives us the minimum of 
. (If we set the other square root to , we get a minimum of 
which is larger than 
.) Therefore 
.
~programmeruser
