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2 answers

The smallest number into which 1 triangle can be split is 4.
With the original, that gives 5.
Each of those can then be split into 4, giving 4^2 smaller triangles.
The total number is then 1 + 4 + 4^2 ...
This can be repeated indefinitely, and after n levels of splitting you have 1 + 4 + 4^2 + ... +4^n triangles in total.
This is a geometric series with sum (4^(n+1)-1)/3.

2007-03-06 00:27:51 · answer #1 · answered by Anonymous · 0 0

Total must be a perfect square provided all the smaller equilateral triangles are of the same size - they don't have to be !

An extra:

If you split an equlateral triangle into smaller equilateral triangles, each of the same size, and then count all possible visible equilateral triangles (i.e small ones, medium sizes ones, . . . and the original one), the problem of finding out how many can be found is very difficult. Here are the results . ..

The numbers in the first column represents the number of rows of smaller triangles in the original eqilateral triangle.
The second column represents the total number of equilateral triangles which can be found in the diagram.

N . . . ∆s

1 . . . . 1
2 . . . . 5
3 . . . . 13
4 . . . . 27
5 . . . . 48
6 . . . . 78
:
: and so on


If N is EVEN the following formula gives the total number of ∆s

∆ = (1/8).N.(N + 2).(2N + 1)

If N is ODD this formula gives the total number of ∆s.

∆ = (1/8).(N + 1).(2N² + 3N - 1)

2007-03-06 08:25:43 · answer #2 · answered by sumzrfun 3 · 1 0

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