Difference between revisions of "2021 WSMO Team Round/Problem 10"

(Problem)
(Problem)
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The minimum possible value of<cmath>\sqrt{m^2+n^2}+\sqrt{3m^2+3n^2-6m+12n+15}</cmath>can be expressed as <math>a.</math> Find <math>a^2.</math>
 
The minimum possible value of<cmath>\sqrt{m^2+n^2}+\sqrt{3m^2+3n^2-6m+12n+15}</cmath>can be expressed as <math>a.</math> Find <math>a^2.</math>
  
''Proposed by [[pinkpig]]''
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''Proposed by pinkpig''
  
 
==Solution==
 
==Solution==

Revision as of 19:27, 23 March 2023

Problem

The minimum possible value of\[\sqrt{m^2+n^2}+\sqrt{3m^2+3n^2-6m+12n+15}\]can be expressed as $a.$ Find $a^2.$

Proposed by pinkpig

Solution

Notice that we can complete the square inside the second square root: $\sqrt{3m^2+3n^2-6m+12n+15} \\ = \sqrt{3(m^2+n^2-2m+4n+5)} \\ = \sqrt{3(m^2-2m+1+n^2+4n+4)} \\ = \sqrt{3((m-1)^2+(n+2)^2)}$ Notice that we can find the minimum by setting this to $0$, which occurs when $m=1$ and $n=-2$. This gives us the minimum of $a=\sqrt{5}$. (If we set the other square root to $0$, we get a minimum of $\sqrt{15}$ which is larger than $\sqrt{5}$.) Therefore $a^2=(\sqrt{5})^2=\boxed{5}$. ~programmeruser