Bulgarian National Olympiad

We prove that $m=2$. First we show that $m\ge 2$. It suffices for any $s\in (0,1)$ to find five equilateral triangles with combined area greater than $2s$, which can not cover an equilateral triangle $△ABC=\Delta$ of area $1$. Let $A_1{B}_1{C}_1$ be an equilateral triangle of area $(1+s)/2$ and vertices on the corresponding sides of $\Delta$. Without loss of generality suppose $2B{A}_1\le BC$. Then there exist three equilateral triangles that can not cover any of the segments $B{A}_1$, $C{B}_1$ and $A{C}_1$. Then these triangles and two equilateral triangles ${\Delta }_1$ and ${\Delta }_2$ of areas $s$ can not cover $\Delta$. Otherwise ${\Delta }_1$ and ${\Delta }_2$ cover points from the given three segments and therefore one of them, say ${\Delta }_1$, covers points from two of them, say $D\in A_{1}B$ and $E\in B_{1}C$. Since $S_{A_1{B}_1{C}_1}\ge \frac{1}{3}$, we have that $\nleqq A_1{B}_{1}C\ge 90^{\circ }$ (prove!) implying that the side of ${\Delta }_1$ is at least $DE\ge A_1{B}_1$. Therefore $S_{{\Delta }_1}\ge S_{A_1{B}_1{C}_1}$, a contradiction.

We prove now that given five equilateral triangles of areas $a^2,b^2,c^2,d^2$ and $e^2$, such that $a^2+b^2+c^2+d^2+e^2=2$, there exist four of them that cover $\Delta$. Let $a\ge b\ge c\ge d\ge e>0$. If $a\ge 1$ then triangle of area $a^2$ covers $\Delta$. In the opposite case $b+c>1$. This is obvious when $c>1/2$ (because $b\ge c$), and otherwise $$b^2=2-a^2-c^2-d^2-e^2>1-3c^2\ge (1-c{)}^2.$$ Therefore the triangles with areas $a^2,b^2$ and $c^2$, cut from the vertices of $\Delta$ intersect each other. They do not cover $\Delta$ if $f=2-a-b-c>0$ and an equilateral triangle of area $f^2$ is not covered. We have to prove that $d\ge f$. This is obvious when $d>1/2$ (because $a,b,c\ge d$), otherwise it follows from $a,b,c<1$ that $$d\ge 2d^2\ge d^2+e^2=2-a^2-b^2-c^2>2-a-b-c=f.$$