Re: General Diagonal Formula

Quote:

Originally Posted by

**brianjesusdiaz** **Greetings ladies and gentlemen!** I am a new member in this well-structured and appropriately regulated community, introducing you to a general diagonal formula. I have shown few of my geometry teachers of this discovery; however, they disapprove of its use. Why do they disapprove of it? In addition, how can I further its use?

Our original, modern use of the formula for diagonal number.

$\displaystyle n^2-3n$

Use the [tex] tags.

[tex]n^2-3n[/tex] gives $\displaystyle n^2-3n$

That is not quite correct: $\displaystyle \binom{n}{2}-n=\frac{n!}{2(n-2)!}-n=\frac{n^2-3n}{2}$.

That is the number of diagonals of an n-vertices convex polygon.

Re: General Diagonal Formula

Thank you so much for your gratitude! Hold on for a moment.

Re: General Diagonal Formula

I see you have made some edits corrections to the OP.

You still has some errors. The index in the sums should be $\displaystyle t$ not $\displaystyle i$.

In reply #2, I should how to get the formula using combinations.

That is the usual way it is done.

Re: General Diagonal Formula

Hello, brianjesusdiaz!

Welcime aboard!

Quote:

I am introducing you to a general diagonal formula.

I have shown few of my geometry teachers of this discovery;

. . however, they disapprove of its use.

Why do they disapprove of it?

In addition, how can I further its use?

Our original, modern use of the formula for diagonal number.

. . $\displaystyle d(n) \:=\:n^2-3n$

Did you **test** your formula?

We know that a square $\displaystyle (n = 4)$ has two diagonals . . . right?

But your formula gives: .$\displaystyle d(4) \:=\:4^2 - 3(4) \:=\:4\text{ diagonals}$

So, your formula doesn't work!

. . And you wonder why they disapprove?

The number of diagonals in a polygon with $\displaystyle n$ sides

. . is: .$\displaystyle d(n) \:=\:\frac{n(n-3)}{2}$

Your formula produces a number which is *twice as large.*

. . Here is the reason why.

You say there are $\displaystyle n$ choices for a vertex.

Then it can be connected to any of the other $\displaystyle n-3$ vertices.

Hence, there are $\displaystyle n(n-3)$ possible diagonals.

Consider a pentagon with vertices $\displaystyle A,B,C,D,E.$

. . $\displaystyle \begin{array}{cc}\text{Connections} & \text{Diagonals} \\ \hline A\text{ to }C,D & AC,AD \\ B\text{ to }D,E & BD,BE \\ C\text{ to }E,A & CE,CA \\ D\text{ to }A,B & DA,DB \\ E\text{ to } B,C & EB,EC \end{array}$

It seems that there are *ten* diagonals.

But we note that $\displaystyle AC$ and $\displaystyle CA$ represent *the same diagonal.*

. . And: .$\displaystyle AD = DA,\:BD = DB,\:\text{ etc.}$

Since *all* the diagonals are duplicated,

. . we must divide $\displaystyle n(n-3)$ by two.

Re: General Diagonal Formula

Well, I tried to stick with the school textbooks statement of what denotes the index. $\displaystyle t$ was not the original variable I used to denote the number of adjacent vertices in a given form of shape. Would you mind showing the combination way?

Re: General Diagonal Formula

Quote:

Originally Posted by

**brianjesusdiaz** Would you mind showing the combination way?

Assume we have a convex polygon with *n* vertices.

A combination of *n* points taken two at a time is $\displaystyle \binom{n}{2}=\frac{n(n-1)}{2}$.

In other words that is how many line segments determined by the *n* points(vertices).

However, *n* of those are sides of the polygon.

So the diagonals number

$\displaystyle \binom{n}{2}-n=\frac{n(n-1)}{2}-n=\frac{n^2-3n}{2}$.

Re: General Diagonal Formula

Then how can you stretch this formula to dimensions above two?

A friend of mine demonstrated that with prisms...

$\displaystyle \frac{v^2-4v}{2}$

$\displaystyle v$ is total number of vertices of a given prism.

However this lacks usage, the fact that it only functions for prisms. The general formula for diagonals, which is in the first post, not only connects pyramids and concave shapes and forms, but dimensions above three as well.

Re: General Diagonal Formula

Let me state a proof for this equation. I may have misunderstood you, or vice versa, by the division of two, so let me restate myself...

Proof(?):

Hopefully, we're familiar with this two dimensional

$\displaystyle \frac{n^2-3n}{2}$

The fallacy with this formula is that it is only attributed to a two dimensional field. Finding a general formula is the case here.

Alright, we want to find a diagonal formula for a cube. Would we use the 2D formula? Of course not!

Let's take a cube.

We would take two bases, using the 2D formula for both. For the exterior center bases, including the inside, we would use this formula.

$\displaystyle n(n-1)$

Add the formulas together

$\displaystyle = n^2-n+n^2-3n$

$\displaystyle = 2n^2-4n$

Here I would use an identity. A polygon and its prism counterpart are related in that a polygon's # of sides equal to the number of vertices of its prism counterpart divided by two.

$\displaystyle n = \frac{v}{2}$

With this in mind, let's return to our equation.

$\displaystyle = 2n^2-4n$

$\displaystyle n = \frac{v}{2}$

$\displaystyle = 2(\frac{v}{2})^2-4(\frac{v}{2})$

$\displaystyle = \frac{v^2-4v}{2}$

$\displaystyle = \frac{v(v-4)}{2}$

Would this function with pyramids? Any concave shape? Well no...the fact that this formula was derived from prisms, it only functions for prisms.

In addition, as we enter dimensions above three, we realize this function has no use. Therefore, a general formula must be established.

The constant we see in two and three dimensions is 3 and 4 respectively.

Let's call this constant $\displaystyle t$.

In addition, we see a relationship involving vertices.

$\displaystyle v=t+c$

Where $\displaystyle v$ represents the total number of vertices in a given form or shape, $\displaystyle t$ represents the number of adjacent vertices in a given form or shape, and $\displaystyle c$ represents the number of non-adjacent vertices in a given form or shape

With this in mind.

Back to the Equation.

$\displaystyle = \frac{v(v-4)}{2}$

$\displaystyle = \frac{v(v-t)}{2}$

$\displaystyle v=t+c$

$\displaystyle v-t=c$

$\displaystyle = \frac{vc}{2}$ $\displaystyle Check!$

Lets define this.

$\displaystyle \frac{1}{2}\sum_{i=1}^v v-t$$\displaystyle =$$\displaystyle \frac{1}{2}\sum_{i=1}^c c+t$$\displaystyle =$$\displaystyle \frac{vc}{2}$$\displaystyle =$$\displaystyle D$

Alright, we jotted down two dimensions and three dimensions. Let's add the fourth.

We take a tesseract for example.

We see that there are a total of 16 vertices, and 5 adjacent vertices for each given vertex.

$\displaystyle \frac{1}{2}\sum_{i=1}^{16} 16-t$

$\displaystyle =$$\displaystyle \frac{1}{2}\sum_{i=1}^{11} 11+t$

$\displaystyle \frac{16(11)}{2}$

$\displaystyle D(v) = 88 $

(Nod)

This is also intertwined with Euler's characteristic.

$\displaystyle X = t+c-E+F$

Who minds this?

Side-note: The number of sides of a polygon is equal to its number of vertices.

Therefore $\displaystyle polygon (n) = polygon (v)$

This equation is used for geometry and probability purposes. If anyone can find other uses for this, thank you.

-Brian Diaz

Re: General Diagonal Formula

Since I am disallowed from editing the previous message, I would also like to state that in order to move up to concave shapes, you need to define "diagonal". A simple formula

$\displaystyle T(D)=I(D)+E(D) $

Where $\displaystyle I(D)$ is equal to total number of interior diagonals, and with this fact stated, it is no wonder why interior diagonals just equal to total number of diagonals ($\displaystyle T(D)$).

Of course, $\displaystyle E(D)$ represents no other than exterior diagonals.

For example...

http://mathworld.wolfram.com/images/...ateral_750.gif

In the first image, I am using the right two.

If I am looking for interiors (using the middle image), it would be only one.

But for both...

$\displaystyle T(D)=1+1=2$

Absolutely sounds ridiculous, however, it helps to deal with ideas of what we refer to as a "diagonal".

For the right image, obviously it's zero. In the interior diagonals.

If looking for total diagonals, we would have two.

$\displaystyle T(D)=0+2=2$

-Brian Diaz