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

• Mar 8th 2010, 10:29 AM
ice_syncer
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?? :((Doh)
• Mar 8th 2010, 10:59 AM
Wilmer
Are you SURE 2005 is asked for, not 2025 ?
• Mar 8th 2010, 11:01 AM
ice_syncer
Quote:

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?
• Mar 8th 2010, 04:02 PM
Wilmer
2025 is a square; will work with equilateral triangle;
rows of 1,3,5 ...