Thread: Existence of a traingle that can be cut to 2005 conruent trianlges..

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??

Are you SURE 2005 is asked for, not 2025 ?

Originally Posted by Wilmer

Are you SURE 2005 is asked for, not 2025 ?

Why do you think 2025? Can you prove it if it was 2025?

2025 is a square; will work with equilateral triangle;
rows of 1,3,5 ...