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

(Created page with "==Problem== 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 pink...")
 
(→‎Solution)
Line 8: Line 8:
 
<math>\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)}</math>
 
<math>\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)}</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>.
 
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

Revision as of 10:36, 24 June 2022

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

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