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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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