It has been shown that a minimum of 6 equilateral triangles of sides 1 cm are sufficient to cover an equilateral triangle of sides 2.1 cm in question 783653.
How about "Find the minimum length of sides of the equilateral triangle that more than 6 equilateral triangles of sides 1 cm are required to cover the original completely"?
According to Erich's Packing Center, the maximal triangle known to be coverable by six $1$cm triangles has edge length $\frac73$cm. Everything larger than that will require more than six triangles. However, the statement is described as “Found by David Cantrell in August 2002”, where the word “found” as opposed to “proved” indicates that no better solution is known (to the author at the time of that writing), but no proof of the optimality is known (to him) either.
Let ABC be the original equilateral triangle with sides of length (2 + y) cm.
Formula:- Area of an L-cm sided equilateral triangle $= … = \frac {√3L^2}{4} cm^2$
∴ Area of an (2 + y)-cm sided equilateral triangle $= … = \frac {√3(2+y)^2}{4} cm^2$
Also, area of an (1)-cm equilateral triangle $= … = \frac {√3}{4} cm^2$
If the area of the (2 + y)-cm sided equilateral triangle is so big that it requires more than six 1-cm equilateral triangles to completely cover it, then we need
$\frac {√3}{4} (2+y)^2 ≥6× \frac {√3}{4}$
i.e. $y ≥ √6- 2=0.449489742$
Thus, the minimum length of a side of the original triangle must be at least 2.449489742 cm.
I found the solution to the question is far too simple. Maybe I have over-looked something.
