Are you SURE 2005 is asked for, not 2025 ?
Prove that such a triangle exists that can be fully cut to 2005 congruent triangles...
Well, this is wht I first did, I imagined an equilatoral triangle, joined the 3 mid points and obtained a fractel like the MHF symbol, but the problem is, the total no. of tiranlges is always a power of y4 so I cant get 2005 in the picture
another way I know of to divide trianlges is by joining the vertices to the incentre of the equilatoral trianlge, but then we'll get powers of 3..
Does such a trianlge really exist??