Belarusian Mathematical Olympiad

Answer: yes, one can.

Let $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB=c$, $BC=a$, $CA=b$, $A{A}_1=d$, $B{B}_1=e$, $C{C}_1=f$.

Suppose that one can paint three of six segments mentioned red and three others blue so that it is not possible to construct a triangle using the segments of the same color. Then the length of some red segment is not less than the sum of two other red segments; the same is true for blue segments.

So, there are two of six segments, the sum of which is not less than the sum of four others: $x+y\ge z+u+v+w$. We show, however, that this situation is not possible. Consider the following three cases.

1) Both $x$ and $y$ are some sides of $△ABC$. Without loss of generality, let $a+b\ge c+d+e+f$. But from the triangles $BGC$ and $AGC$ we have $2f/3+2e/3>a$ and $2f/3+2d/3>b$. Summing all three inequalities we get $f/3>c+d/3+e/3$ which contradicts to the triangle inequality for medians: $d+e>f$.

2) Both $x$ and $y$ are some medians of $△ABC$. Without loss of generality, let $d+e\ge a+b+c+f$. But from the triangles $BG{C}_1$ and $AG{C}_1$ we have $f/3+c/2>2e/3$ and $f/3+c/2>2d/3$, or $f/2+3c/4>e$ and $f/2+3c/4>d$. It follows that $f+3c/2>d+e\ge a+b+c+f$, or $c/2>a+b$, a contradiction.

3a) $x$ and $y$ are a side and a median of $△ABC$ having the common endpoint. Without loss of generality, let $a+e\ge b+c+d+f$. But this inequality contradicts to the inequalities $a<b+c$ and $e<d+f$.

3b) $x$ is a median of $△ABC$ with the endpoint in the middle of the side $y$. Without loss of generality, let $a+d\ge b+c+e+f$. But this inequality contradicts to the inequalities $a<b+c$ and $d<e+f$.