Equilateral Triangles

Date: 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/