2021 WSMO Team Round/Problem 10

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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2021 WSMO Team Round/Problem 10

Problem

Solution