20080102, 20:01  #1 
May 2004
New York City
4235_{10} Posts 
Count the Triangles
Start with a square. Divide each side into 5 equal parts, and draw the
lines to form a 5x5 grid. Then crosshatch the grid by drawing the diagonals of each of the 25 small squares. How many triangles can be traced in the resultant figure? 
20080103, 02:46  #2 
Sep 2005
Detroit, MI
2^{3} Posts 
I'm going to give it a guess here. Probably off, but it is worth a shot.
440 If it's correct or anyone wants to know I'll tell you how I came up with that number. 
20080103, 03:31  #3 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
I guess you are well on the low side

20080103, 04:16  #4 
"William"
May 2003
New Haven
2,371 Posts 
It doesn't look that far off to me. I found 480. I initially got the same answer as Jaimaz, but when I checked my method on a 2x2 grid I found a few more triangles.

20080103, 12:08  #5  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
the original each contributing 8 distinct triangles. But there are additional triangles (I count 4 in your 2x2 grid) but can't generalize these rogues ATM. 

20080103, 16:06  #6 
"Lucan"
Dec 2006
England
194A_{16} Posts 
I get 492

20080103, 16:34  #7 
"William"
May 2003
New Haven
2,371 Posts 
Oh I see! There were still more with the hypotenuse parallel to a side. I now also find 492.

20080104, 11:49  #8 
"Lucan"
Dec 2006
England
14512_{8} Posts 
Apart from using symmetry to assert that there are
an equal number of each of the 4 orientations of a given size of triangle, I don't see much potential for a smart answer to the question. There are two classes of triangles: hypotenuse parallel to sides or diagonals, with 5 sizes in each class. Admittedly the counting of each size within a class is similar but that still leaves us with 10 multiplications to perform. Anyone care to find a general solution? 
20080104, 12:23  #9 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20080105, 05:50  #10 
"William"
May 2003
New Haven
2,371 Posts 
I think there will be separate solutions for odd and even numbers of squares.
For the triangles with sides parallel to the square sides, in any one orientation, the "size 6k" triangles can have base on one of k rows and located in one of k positions on that row, so there are 5^{2}+4^{2}+3^{2}+2^{2}+1^{2} triangles with obvious generalization. For the triangles with hypotenuse parallel to the square sides, it's tricky because the height of a triangle is n/2, but the base can only be on integer. So the number of triangles of each size (in any one orientation) is Size 1: 5 rows x 5 positions. Size 2: 5 rows x 4 positions Size 3: 4 rows x 3 positions Size 4: 4 rows x 2 positions Size 5: 3 rows x 1 position This also has an easy generalization. The general solution needs to find the closed form sum for each of the generalizations and to multiply by 4 for the symmetric orientations. My error was in the number of rows. I had it decreasing too fast. William 
20080105, 10:25  #11  
Mar 2005
2·5·17 Posts 
Quote:
I know this isn't in the spirit of most puzzlers, but for Iggies and other lazy folks, a closed form is given at OEIS: A100583: Number of triangles in an n X n grid of squares with diagonals Richard 

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