Just something that comes to my mind
I assume that the size (size = length of any side) of the biggest triangle
is n. Not sure how it can be derived from 10. Infact, I cudnt get what is 10
here :P
The number of triangles with size k (excluding the upside down ones) in row
i from top will be 0 if i < k and (i-k+1) if i >= k . I am counting those
triangles who have base in the ith row and are not upside down.
Now, count the number of triangles upside down in row i (assume this is
having the bottom vertex in the ith row). It will be 0 if i < 2*k and (i -
2*k + 1) if i >= 2*k.
Sum these to for i from 1 to n.
Then sum the result from k = 1 to n.
I think you can easily derive a mathematical formula for it .. didnt try
For your qn,
k=1:
i = 1: 1+0
i = 2: 2+1
i = 3: 3+2
=> 9
k=2:
i = 1: 0+0
i = 2: 1+0
i = 3: 2+0
=> 3
k=3:
i = 1: 0+0
i = 2: 0+0
i = 3: 1+0
=> 1
so ans = 9+3+1 = 13
2009/10/19 2shar007 <[email protected]>
>
> In the puzzle, /_\
> /_\/_\
> /_\/_\/_\ this is having 10
> vertices, we are to find the no.of triangles which is 13 here
>
> I tried to derive a general formula, but couldn't :(
> What I understood was triangles have vertices as n(n+1)/2,
> n=2,3,4,5..................
>
> Also that the no. of triangles for k(k+1)/2 vertices is no.of
> triangles in k(k-1)/2 + triangles of all sizes with atleast one
> vertex in the larger triangle.
>
> will be grateful if someone could give a fitting reply
> Thanks in advance
>
> >
>
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