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### Equilateral Triangles

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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?
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```
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/
```
Associated Topics:
High School Basic Algebra

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