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### Counting Triangles

```
Date: 05/27/99 at 10:48:38
From: Michael Maciejewski
Subject: Triangles

Hi Dr. Math.

I am a teacher right now at a high school in Michigan. I just received
some brain teasers that I would like to use next year for extra-credit
assignments. However, one assignment I have is giving me some trouble.
It is a large triangle with 36 small ones inside. The question asks
how many total triangles there are (there are more than 36). I believe
it has something to do with Pascal's triangle.

All your help would be greatly appreciated.

Thanks.
Michael Maciejewski
```

```
Date: 11/21/2002 at 17:32:39
From: Doctor Ian
Subject: Re: Triangles

Hi Michael,

Let's forget about downward triangles for the moment, and concentrate
on upward triangles.  A triangle of 'size k' has a side length equal
to k times the side length of one of the small triangles.

Rows    Size 1    Size 2    Size 3    Size 4
------  ------    ------    ------    ------
1       1
2       1+2       1
3       1+2+3     1+2       1
4       1+2+3+4   1+2+3     1+2       1

It's pretty clear that for n rows, the number of upward-pointing
triangles is

n
Sum    [ T(i) ]            T(i) = i'th triangular number
i=1

Okay, so what about those downward facing ones?

Rows    Size 1       Size 2       Size 3      Size 4
------  ---------    ---------    ---------   ----------
1       0
2       1
3       1+2
4       1+2+3        1
5       1+2+3+4      1+2
6       1+2+3+4+5    1+2+3        1
7       T(6)         T(4)         T(2)
8       T(7)         T(5)         T(3)        T(1)

So, what's going on is this:  Every time we add a row, we get a new
size of upward triangle.  But we have to add _two_ rows to get a new
size of downward triangles.  Apart from that, it works the same way
for both directions.  So for n rows, where n is even, the number of
downward-pointing triangles is

n/2
Sum     [ T(2i-1) ]
i=1

And for n rows, where n is odd, we have

(n/2 - 1)
Sum           [ T(2i) ]
i=1

Pick a number of rows, add the appropriate sums for upward and
downward triangles, and that should give you the total number of
triangles.

I haven't found an elegant way to express this directly in terms
of n (and I don't expect to find one, given the dependence on the
parity of the number of rows), but I believe it correctly summarizes
the situation.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```Date: 11/21/2002 at 19:03:18
From: Doctor Anthony
Subject: Re: Triangles

It is apparent that for triangles the right way up (point at
the top), the total number of triangles is given by the sum

1 + 3 + 6 + 10 + 15 + 21 +..... + n(n+1)/2

Then for the upside down triangles the series goes differently
depending on whether n is odd or even.

If n is even the series is

1 + 6 + 15 + 28 + .....+ k(2k-1)   for k=1 to k=n/2

If n is odd the series is

3 + 10 + 21 + 36 + ... + k(2k+1)  for k=1 to k=(n-1)/2

You will notice that these two series are obtained from the series at
the top by taking alternate terms.

So the complete results are

Number of triangles with n even

SUM(r=1 to n)[r(r+1)/2]  +  SUM(r=1 to n/2)[r(2r-1)]

Number of triangles when n is odd

SUM(r=1 to n)[r(r+1)/2]  +  SUM(r=1 to (n-1)/2)[r(2r+1)]

If we carry out the summations we get the following formulae

When n is even:

Total number of triangles  = n(n+2)(2n+1)/8

When n is odd:

Total number of triangles is = (n+1)(2n^2+3n-1)/8

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations
High School Puzzles

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