Formula for Counting TrianglesDate: 03/16/2000 at 19:16:55 From: Sebastian Subject: Explicit formula for a set of numbers Dr Math, I have this extremely difficult math problem that I need to answer. I hope you will be able to come up with an explicit formula for this set of numbers. Part of the problem is finding the numbers, and then finding the equation is the second task. I will try to explain how to find the numbers as clearly as possible. First, start out with an equilateral triangle one unit on a side. (It helps if you draw it as you read this.) Then, extend 2 of the sides of the triangle out for another unit and connect the bottom points of the extended lines. In the second row, if you make an "upright" equilateral triangle, an "upside down" equilateral triangle, then one more "upright" one you have 3 unit-side triangles in the second row. Counting the 1 in the first row, the 3 in the second row, and the big 2-unit-side triangle, you have 5 triangles total. So far, you have a total of 1 after the first row and 5 after the second row. Now extend the sides out again so the sides of the surrounding equilateral triangle (of all 3 rows) are 3 units. In the third row alone, you'll have 5 equilateral triangles (3 upright and 2 upside down.) Counting all the unit-side equilateral triangles, the 2-unit side triangles, and the surrounding 3-unit side triangle, you'll get 13 equilateral triangles. Keep repeating the process of extending the sides and creating more triangles. I think that the fourth row has 27 total equilateral triangles counting everything: /\ /__\ /\ /\ /__\/__\ /\ /\ /\ /__\/__\/__\ /\ /\ /\ /\ /__\/__\/__\/__\ To recap it: 1 row = 1, 2 rows = 5, 3 rows = 13, 4 rows = 27, etc. You'll need to count probably to the 20 row, in which there are hundreds of them. Now what I need is an explicit formula relating x (the row number) and y (the number of equilateral triangles). Thank you very much. Sebastian Date: 03/17/2000 at 17:02:26 From: Doctor Rob Subject: Re: Explicit formula for a set of numbers Thanks for writing to Ask Dr. Math, Sebastian. If you count the number of upright triangles of each size, you'll find that these are (surprise!) triangular numbers, but the numbers start one row farther down with each increase by 1 in their size: x \ size 1 2 3 4 5 6 ... subtotal 0 0 0 0 0 0 0 ... 0 1 1 0 0 0 0 0 ... 1 2 3 1 0 0 0 0 ... 4 3 6 3 1 0 0 0 ... 10 4 10 6 3 1 0 0 ... 20 5 15 10 6 3 1 0 ... 35 6 21 15 10 6 3 1 ... 56 : : : : : : : : Each entry in this array is a binomial coefficient C(x-s+2,2). When you add up these binomial coefficients for s = 1 to x, you get the binomial coefficient C(x+2,3) for the subtotal of these. Now if you count the number of upside-down triangles of each size, you'll find that these are also triangular numbers, but the numbers start *two* rows farther down with each increase by 1 in their size: x \ size 1 2 3 ... subtotal 0 0 0 0 ... 0 1 0 0 0 ... 0 2 1 0 0 ... 1 3 3 0 0 ... 3 4 6 1 0 ... 7 5 10 3 0 ... 13 6 15 6 1 ... 22 : : : : : Once again, each entry in this array is a binomial coefficient C(x-2*s+2,2). This time when you add them up, you don't get such a simple answer as before. In fact, you get two different answers for even and odd x. The facts that adding up triangular numbers, which are quadratic in x, gives a cubic in x, and this even/odd business, make it plausible that the formula you seek is of the form y = a*x^3 + b*x^2 + c*x + d + e*(-1)^x. You might also take a look at Counting Triangles http://mathforum.org/library/drmath/view/56194.html Hope this helps! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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