Three segments don't simply become medians of a triangle!

• Let me show that─

The necessary and sufficient condition for three straight line segments to be the lengths of three medians of a triangle is: The sum of lengths of shorter two segments is greater than the length of the largest one .

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• Proof of Necessity

Proof of necessity is build upon the very inspiration of the problem, which I found on the brilliant wiki Area of a Triangle .

The inspiration: Denoting the medians $u, v$ and $w$ and their semi-sum as $\sigma = \frac{u+v+w}{2},$ we have─ $A = \frac{4}{3}\sqrt{\sigma(\sigma-u)(\sigma-v)(\sigma-w)}.$

Now,

$A = \frac{4}{3}\sqrt{\sigma(\sigma-u)(\sigma-v)(\sigma-w)}$ $\implies A = \frac{1}{3} \sqrt{(u+v+w)(v+v-w)(u+w-v)(v+w-u)}$

Now the claim of necessity is obvious; as sum of any two of $u, v$ and $w$ must be greater than the third one; otherwise the square root will be undefined.

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• Proof of Sufficiency

First, read this page about Constructing a Triangle with Three Medians Given . Of course, this construction is based on the classic proof of internal division ratio $2:1$ of the medians at their point of concurrency.

Just notice, as long as the sum of $\frac{2}{3}$ portions of any two medians are greater than the $\frac{2}{3}$ portion of the third one, or equivalently, the sum of any two medians is greater than the third one, this construction is possible . That proves the claim of sufficiency.

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Only the medians candidate set $(a,b,c)= \boxed{(10,12,23)}$ , among the $4$ alternatives, breaks the condition.

Let the triangle as $\triangle ABC$ , and $E,F,G$ as the medium of $BC,AC,AB$ respectively. Hence, the median are $AE,BF,CG$ . Now, rotate the triangle 180° with center point $E$ . From the new diagram, we can get a new triangle $\triangle BFG'$ .Besides that, $CF=CA,CG'=G'A'$ , so $FG'=\frac{1}{2}AA'=AE$ . Note that $BG'=C'G'=CG$ . Hence, the sides of this new triangle are the original median respectively.

Hence, we can apply the theorem 'sum of two sides of triangle is larger than other side'. By try and error, we can get that $10+12=22<23$ , so $10,12,13$ $\large cannot$ be the length of $a,b,c$ .