Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

no. of triangle

if

be the no. of points on side

and

f the triangle . them find no. of triangle which can be formed by the taking these points as a vertices.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of triangle

Hi juantheron;

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: no. of triangle

It seems to me that all of the points are "on the side" meaning they are between the endpoints A, B, and C. So a triangle will need to contain one point from each of the 3 groups. Any set of one point from each group will make a triangle, so I think there are l x m x n possibilities.

Bobby, take another look at your example. 4/3 -1/3 -1 = 3? Though I agree that if one side has no points there are no triangles possible I think it is because if l or m or n =0 then lmn = 0.

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: no. of triangle

hi TMorgan

that's what i thought at first too,but then i realized that you don't need the points from the third side.just take one point from one side and two points from another and you get a triangle.

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of triangle

Hi;

4/3 -1/3 -1 = 3?

My formula has binomials not fractions as you have indicated. Or are those binomials?

that if one side has no points there are no triangles possible

When l=0 and m =2 and n =1 you can form one triangle. even though l * m* n = 0

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: no. of triangle

Thinking about it more I see that a triangle does not have to have a point from each side. It can have 2 points on one side. On side m, (m-1)! line segments can be made. Each of those line segments can join with any point from the other 2 sides so I would now say number of possible triangles =

lmn + (l-1)! x mn + (m-1)! x ln + (n-1)! x lm.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: no. of triangle

Sorry Bobby, I didn't realize that is what those high parenthesis meant. I am not familiar with the notation, though I probably have learned it in the past. Does that equate to what I came up with on second try?

Thanks

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of triangle

Hi TMorgan;

I do not think that formula will work. With l = 5, m = 4, n=3 the correct answer is 205. That formula is getting 478.

You can see it better with l,m,n=1. That formula gets 4 which can not be true. The correct answer of course is 1.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: no. of triangle

The terms for the 2 other sides to connect a line segment should be added, not multiplied. Each of the line segments on side l can connect with m+n points, not mn points. It should read:

lmn + (l-1)! x (m+n) + (m-1)! x (l+n) + (n-1)! x (l+m)

So for 1,1,1 the factorials become zero and you get one.
For 3,4,5 you get

60 + (2 x 9) + (6 x 8) + (24 x 7) = 294

Which isn't what you got. Hmmmm. The logic seems right. Have to think more on this one.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of triangle

Hi;

Your new formula also is having a problem with 1,1,1, it gets 7.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: no. of triangle

I see that zero factorial is defined as 1. I figured tough, I need zero factorial to be zero so I will use it that way. But that isn't the problem. The problem is that my factorial statement is wrong. For n points there are not (n-1)! unique combinations. For 3 points the possible line segments are 12, 13, 23. That is 3 possibilities, not 2  [(3-1)!].
The possible line segments from n points = (n-1) + (n-2) + (n-3)...
not (n-1) x (n-2) x (n-3)...
Once more multiplying when I should have been adding. And that addition sequence is represented by the symbols you know but I don't.
Thanks for hanging in there with my misguided thinking. Need to brush up on that binomial theory and symbology.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: no. of triangle

Hi;

This symbol:

means 4 objects taken 3 at a time. Computationally it means this

Also, when trying to come up with a formula empirically, fancy term for trying a lot of formulas, always count a few by hand. In this case by drawing the figures and manually counting the triangles. Then at least you know your formula will not fail on a small one.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: no. of triangle

I originally solved the problem the same way you are trying to TMorgan.  The result should be

lmn + (m + n)(l - 1)(l / 2) + (l + n)(m - 1)(m / 2) + (l + m)(n - 1)(n / 2)

With some algebra you can show that this formula and bobbym's are equal.

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Wrap it in bacon