bilarev wrote: Find the least real number $m$ such that with all $5$ equilaterial triangles with sum of areas $m$ we can cover an equilaterial triangle with side 1. O. Mushkarov, N. Nikolov Can you precise "cover"? Can triangles overlapp?

A. S. K. wrote: Can you precise "cover"? Can triangles overlapp? I think that the triangles can overlapp!

They can ovelapp.

I think the answer is $ 2$ but I cannot prove it. Also I'm not sure about this. Nobody solved it? I want a solution for this;;;;

The answer is $1.25$ so your configuration is wrong. This with $5$ triangles with area $0.25$, the four ones do the job already and the fifth is overlap. But it is easy to chose the vertices and endpoints to see it can't with less area.

The answer is indeed $2$. For the lower bound let us take triangles of areas $1-\epsilon,1-\epsilon,0,0,0$. For the upper bound: Let $A \geq B \geq C \geq D \geq E$ be the sequence of areas. We put triangles $A,B,C$ in vertices of our triangle. If they cover our triangle then we are done. So we assume they don't. Let us observe that each pair of triangles among $A,B,C$ overlaps: i.e. $\sqrt{B}+\sqrt{C} \geq 1$. (Otherwise $2=A+B+C+D+E < 1 + (\sqrt{B}+ \sqrt{C})^2<2$). Now let $X,Y,Z$ be areas of intersections of pairs of $A,B,C$. Of course we have $X + Y \leq A, Y+Z \leq B, Z+X\leq C$ therefore $A+B+C \geq 2(X+Y+Z)$ so at least $(A+B+C)/2$ of our initial triangle is cover. The uncover area is a triangle of area at most $1-(A+B+C)/2 = (D+E)/2 \leq D$. So we can cover it using $D$ triangle, and we are done.