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Old 2008-01-02, 20:01   #1
davar55
 
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May 2004
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Default Count the Triangles

Start with a square. Divide each side into 5 equal parts, and draw the
lines to form a 5x5 grid. Then cross-hatch the grid by drawing the
diagonals of each of the 25 small squares.

How many triangles can be traced in the resultant figure?
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Old 2008-01-03, 02:46   #2
Jamiaz
 
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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.
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Old 2008-01-03, 03:31   #3
davieddy
 
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I guess you are well on the low side
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Old 2008-01-03, 04:16   #4
wblipp
 
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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.
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Old 2008-01-03, 12:08   #5
davieddy
 
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Quote:
Originally Posted by wblipp View Post
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.
Yes I see what you mean. There are 55 squares with sides parallel to
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.
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Old 2008-01-03, 16:06   #6
davieddy
 
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I get 492
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Old 2008-01-03, 16:34   #7
wblipp
 
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Oh I see! There were still more with the hypotenuse parallel to a side. I now also find 492.
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Old 2008-01-04, 11:49   #8
davieddy
 
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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?
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Old 2008-01-04, 12:23   #9
davieddy
 
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Quote:
Originally Posted by wblipp View Post
Oh I see! There were still more with the hypotenuse parallel to a side. I now also find 492.
I'm intrigued as to what you missed
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Old 2008-01-05, 05:50   #10
wblipp
 
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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 6-k" triangles can have base on one of k rows and located in one of k positions on that row, so there are

52+42+32+22+12

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
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Old 2008-01-05, 10:25   #11
Richard Cameron
 
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Quote:
Originally Posted by wblipp View Post
I think there will be separate solutions for odd and even numbers of squares.

...The general solution needs to find the closed form sum for each of the generalizations and to multiply by 4 for the symmetric orientations...

and even: My error was in the number of rows. I had it decreasing too fast.

William
I -presumably- went through the same reasoning as William, including annoyingly making a similar error (I got 4*121=484).
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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