1. counting triangles

Suppose that each side of an equilateral triangle can be partitioned into n partitions, all of which have equal length. From this, you are able to create n^2 equilateral triangles. These are n triangles. From this information, how many DOWNWARD pointing triangles are there in an n triangle?

Apparently its a messy formula.

Any thoughts?

2. Originally Posted by fifthrapiers
Suppose that each side of an equilateral triangle can be partitioned into n partitions, all of which have equal length. From this, you are able to create n^2 equilateral triangles. These are n triangles. From this information, how many DOWNWARD pointing triangles are there in an n triangle?

Apparently its a messy formula.

Any thoughts?
Lets assume we are talking about small downward pointing equilateral
triangles.

There are n-rows of triangles.

The first contains 0 down pointers,
the second row contains 1 down pointer,
the third row contains 2 down pointer,
:
:
the i-th row contains i-1 down pointer,
:
:
the n-th row contains n-1 down pointer.

So there are:

$\displaystyle N=\sum_{i=1}^n (i-1) = \left(\sum_{i=1}^n i \right) -n = \frac{n (n+1)}{2}-n$

RonL

3. Originally Posted by CaptainBlack
Lets assume we are talking about small downward pointing equilateral
triangles.

There are n-rows of triangles.

The first contains 0 down pointers,
the second row contains 1 down pointer,
the third row contains 2 down pointer,
:
:
the i-th row contains i-1 down pointer,
:
:
the n-th row contains n-1 down pointer.

So there are:

$\displaystyle N=\sum_{i=1}^n (i-1) = \left(\sum_{i=1}^n i \right) -n = \frac{n (n+1)}{2}-n$

RonL
Thanks for your help Ron, but I'm afraid it's not that easy. I have to count ALL the triangles, and thus it's every possible triangle, not just the small ones. That's what makes this problem very difficult.

4. Originally Posted by fifthrapiers
Thanks for your help Ron, but I'm afraid it's not that easy. I have to count ALL the triangles, and thus it's every possible triangle, not just the small ones. That's what makes this problem very difficult.
The usual form of the questions would normaly refer to triangles of any size,
but the wording of this one suggested it was only interested in the small
ones.

RonL

5. Originally Posted by CaptainBlack
The usual form of the questions would normaly refer to triangles of any size,
but the wording of this one suggested it was only interested in the small
ones.

RonL
Yes, I can now see how the wording wasn't very clear.

So do you know how many downward pointing triangles there including the big triangles.

For example, suppose we have 1 triangle, then

y(1) = 0

Then partitioned into 2 rows..

y(2) = 1

.
.
.
y(4) = 7

.
.
.
etc

6. Originally Posted by fifthrapiers
Yes, I can now see how the wording wasn't very clear.

So do you know how many downward pointing triangles there including the big triangles.

For example, suppose we have 1 triangle, then

y(1) = 0

Then partitioned into 2 rows..

y(2) = 1

.
.
.
y(4) = 7

.
.
.
etc
Hmm, okay, so the pattern looks like follows:

y(1) = 0
y(2) = 1
y(3) = 3
y(4) = 7
y(5) = 13
y(6) = 21
y(7) = 31

.
.
.

Now it's obvious the pattern is just adding an extra 2 on the prev. sum. Thus, y(8) will be 31 + 12 = 43, since the prev term was 13 + 10.

What's frustrating me is that I can't come up with a formula to express this, and this should be the easy part. The part that should be difficult for this is proving why it's true! It's obvious looking at the pictures but I have to PROVE it. My hint was to use difference equations.

7. I've attached a diagram of what the triangles look like. I still have no idea of how to come up with a formula and to even more importantly prove why it's true.

I think this picture should make it quite clear.

Thanks for the help!

*EDIT*

For t(4) and t(5) I used black to outline the larger downward triangles. I think it's pretty obvious but I wanted to make sure since the pic is a little small.

8. y(6) = 22

9. Ahh indeed .

This problem has proved to be even more difficult. Well, I'll hopefully some day figure it out. Thanks.