Equilateral TrianglesDate: 6/22/96 at 13:59:3 From: Anonymous Subject: Equilateral Triangles If you were to draw an equilateral triangle of side n units on drawing paper, with neighbouring dots spaced equidistant from each other, how many more equilateral triangles could you draw inside? Is there a relationship between the length of the side of an equilateral triangle and the number of smaller equilateral triangles that can be drawn inside, with sides in units? Date: 6/22/96 at 19:29:35 From: Doctor Anthony Subject: Re: Equilateral Triangles If the sides of the large triangle are divided into n equal parts and the smaller equilateral triangles drawn in, then the answer satisfies a third-order recurrence relation with 1 as a triple root, inhomogeneous term (3+(-1)^n)/2, and initial conditions 0,1,5 for n = 0,1,2. Applying the characteristic equation method for solving recurrence relations gives the formula: (1/16) [4n^3 + 10n^2 + 4n - 1 + (-1)^n ]. Hope that helps! -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 6/23/96 at 12:57:9 From: Anonymous Subject: Re: Equilateral Triangles Thanks for your help, but can you possibly explain third order recurrence relations in a nutshell? And how do you get the formula from the conditions? Simon Flack Date: 6/23/96 at 16:24:25 From: Doctor Anthony Subject: Re: Equilateral Triangles I am afraid difference equations (recurrence relations) are a very big topic in their own right, and you will need to consult a textbook on the various types and the methods of solution. They are very similar in many ways to differential equations, and you will know what a huge subject that is. The equations are particularly difficult when for example they are of the form: a*u(r+2) + b*u(r+1) + c*u(r) = f(r) where f(r) is a function of r. In this situation the equation is inhomogeneous, and the solution consists of two parts, the first part, like the complementary function in differential equations, has zero as the right hand side of the equation. This solution is then added to the 'particular integral' part of the solution, where some special case is solved for the complete equation. I have worked out the triangle problem using simple algebraic methods, and a difference table. You might like to look at the following alternative approach. It is best to consider the case n even and n odd separately. For n even the answer is given by: SUM from r = 1 to n of r(r+1)/2 PLUS SUM from r = 1 to n/2 of r(2r-1) If you use the usual formulae for sums of integers and sums of squares of integers, and some simplification, this reduces to: n(n+2)(2n+1)/8 [To be used when n is EVEN] For n odd the answer is given by: SUM from r = 1 to n of r(r+1)/2 PLUS SUM from r = 1 to (n-1)/2 of r(2r+1) Again using the formulae for sums of integers and sums of squares of integers, and algebraic simplification this reduces to: (n+1){2n^2+3n-1}/8 [To be used when n is ODD] This did not require the use of difference equations, simply observation of how adding one more division to the side of the original triangle increased the number of triangles in the figure. As mentioned earlier, the behaviour is different for n even and n odd. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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