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

(→‎Problem)
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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:
−
<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>
+
<cmath>\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)}</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 19:28, 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

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