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/
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