2021 WSMO Team Round/Problem 10

Revision as of 16:53, 14 June 2023 by Programmeruser (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

We can see that we can complete the square inside the second square root: \[\sqrt{3m^2+3n^2-6m+12n+15}\]

\[=\sqrt{3(m^2+n^2-2m+4n+5)}\]

\[=\sqrt{3(m^2-2m+1+n^2+4n+4)}\]

\[=\sqrt{3((m-1)^2+(n+2)^2)}\]

Then, 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 $a^2=(\sqrt{5})^2=\boxed{5}$.