Triangular Triples: Means that Are Not So AverageDate: 04/28/2012 at 17:01:15 From: Reema Subject: Averages of Triangular Numbers How many triangular numbers can be written as the average of two other triangular numbers? I have found a few: 190 is the average of 55 and 325 465 is the average of 300 and 630 1275 is the average of 780 and 1770 1891 is the average of 1081 and 2701 4851 is the average of 3486 and 6216 But I do not see any pattern in them to make a generalization. So, are there a certain number? or are there infinitely many? If the latter, what is the generating equation? Date: 04/29/2012 at 06:05:01 From: Doctor Jacques Subject: Re: Averages of Triangular Numbers Hi Reema, This is a very interesting question. As you will see, there are infinitely many solutions, and there is a formula that produces all the solutions. The algebra is a bit complicated, but it only involves simple concepts. We want to find integer solutions of k(k + 1) + m(m + 1) = 2n(n + 1) This is a second degree equation. In such problems, it is always a good idea to try to simplify the equation by completing the squares. To do that, we multiply everything by 4 and do some manipulations. 4k^2 + 4k + 4m^2 + 4m = 2(4n^2 + 4n) (2k + 1)^2 - 1 + (2m + 1)^2 - 1 = 2((2n + 1)^2 - 1) (2k + 1)^2 + (2m + 1)^2 = 2(2n + 1)^2 Make this change of variables: u = 2k + 1 v = 2m + 1 w = 2n + 1 [1] With these substitutions, the result of completing the square becomes: u^2 + v^2 = 2w^2 [2] We must find solutions of this equation where u, v, and w are odd integers. It may help to look at rational solutions. If we divide [2] by w^2, and let x = u/w, y = v/w, the equation becomes: x^2 + y^2 = 2 [3] This is the equation of a circle of radius sqrt(2), and we must find rational solutions of that equation, such that the numerator and denominator are both odd. The good news is that there are obvious solutions, like ((+/-)1, (+/-)1). We will use the solution x = y = -1. This means that the point P = (-1, -1) is on the circle. We will draw a straight line through P; that line will intersect the circle in another point Q. The general equation of a straight line through P is: (y + 1) = t(x + 1) [4] Here, t is the slope of the line. If we extract y from [4] and substitute into [3], we get: (t^2 + 1)x^2 + (2t^2 - 2t))*x + (t^2 - 2t - 1) = 0 [5] The sum of the roots of that equation in x is -(2t^2 - 2t)/(t^2 + 1). We already know one root, namely x = -1 (you can substitute that value into the equation to check). This means that the other root (the x-coordinate of the point Q) is: x = -(2t^2 - 2t)/(t^2 + 1) + 1 = (-2t^2 + 2t + 1)(t^2 + 1) [6] We can compute y from [4]: y = (t^2 + 2t - 1)(t^2 + 1) [7] You can check that x and y are solutions of [3]. If t is rational, x and y will be rational, too. Conversely, if Q is a point on the circle with rational coordinates, the slope of the line PQ will be rational, and we will find Q using [6] and [7], except for the point (-1, 1) where t would be infinite. To summarize, to find a solution of [3], we pick a rational value for t, and compute x and y from [6] and [7]. For example, with t = 2/3, we get x = 17/13 and y = 7/13, corresponding to the solution of [3]: (17/13)^2 + (7/13)^2 = 2 To get a solution of [2], we multiply by 13^2: 17^2 + 7^2 = 2*13^2 That is, u = 17 v = 7 w = 13 Recovering our original variables from the substitution [1], we find k = 8 m = 3 n = 6 This gives the triangular numbers 36, 6, and 21. There could be a small issue here. We must have u, v, and w odd. However, as the whole calculation is based on the fractions u/w and v/w, we may assume that these fractions are in lowest terms; i.e., that gcd(u, v, w) = 1. It can be shown that, in such a case, the three integers must be odd. (You can try to prove this by looking at [2] as a congruence modulo 4.) On the other hand, there is no requirement that the fractions u/w and v/w be in lowest terms, as long as the terms are odd. This means that we may multiply all three numbers u, v, and w by any odd factor, and this will give us another solution. For example, multiplying our solution (17, 7, 13) by 3, we get u = 51 v = 21 w = 39 k = 25 m = 10 n = 19 This gives the three triangular numbers 325, 55, and 190, which you discovered and listed first among your results. It may happen that, for some values of t, the values of x and/or y are negative. However, as the equation of the circle only involves x^2 and y^2, you may ignore the signs: if the point (x, y) is on the circle, the points (-x, y), (x, -y) and (-x, -y) are also on the circle; they correspond to other values of t. There are many more interesting aspects in this problem. Here are a couple of ideas you may want to investigate, if you find this interesting. The formulas [6] and [7] use a rational parameter t. By writing t = p/q, where p and q are integers, and simplifying the formulas, you can get formulas for u, v, and w that use only the two integer parameters p and q. As I said before, there are may "equivalent" points on the circle: given a point (x, y), there are also points ((+/-)x, (+/-)y). You can also interchange x and y; this will simply swap two of the triangular numbers. You could try to find a range of values for t to eliminate those redundancies. To do that, you could require the point Q to be in the first 45 degrees of the circle, and use a little geometric reasoning to find the corresponding range of values for t. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 04/29/2012 at 08:36:04 From: Reema Subject: Thank you (Averages of Triangular Numbers) Wow! Thank you so much for all your help. It is really appreciated. I wasn't even expecting to get a response, but I got one and it was very detailed and explicit. Again, thanks so much! |
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