Who here is good at Algebra?

[Mr Initial Man]

There's a problem that's been bugging me for some time now.


n=d(ds-2d-s+4)/2

s=2(d^2-2d+n)/d(d-1)

d=?

Can anyone figure out a solution?

[Stephen Philbin]

I might be able to help, but I can't even remember the presedence. I can't remember if that's supposed to be n equals d multiplied by the contents of the parenthesis divided by two, or n equals d multipled by the contents of the parenthesis, divided by two.

[Mr Initial Man]

Does this help?

n=[d(ds-2d-s+4)]/2

s=[2(d^2-2d+n)]/[d(d-1)]

d=?

[Stephen Philbin]

Nope. I've forgotton most of the basic rules by now. About 7 years since I did anything like that.

I think the supposed key to kinding the solution is to try and rework the equations so they both equal either S or N, and then once you have both with a common result, start reducing the equations by removing common expressions and re balancing of the simplified equations.

[NogDog]

It's been a longer time for me, and it is thus likely I made a mistake, but my calculations came up with d being any value:

Express n in terms of d and s:

/ d(ds - 2d - s + 4) \
2 ( d^2 - 2d + ------------------ )
\ 2 /
s = ------------------------------------
d(d - 1)

Do the multiply by 2 in the numerator and multiply by d in the denominator:

2d^2 - 4d + d^2s - 2d^2 - ds + 4d
s = ---------------------------------
d^2 - d

Remove terms on top that cancel (equal 0 when summed):

d^2s - ds
s = ---------
d^2 - d

Multiply both sides of equation by 1/s:

d^2 - d
1 = -------
d^2 - d

Ergo: any value for d will give the same result.

[theuedimaster]

I concur. D cancels out in the end, its value doesn't matter.

[Mr Initial Man]

Erf... *Wince* Again, I am as clear as mud.

I was trying to express d in terms of n and s.

It's an actual formula I've been trying to solve.

[Ultimater]

I don't know about you guys but I am getting that:
s==d then using that I get that d==4 and that n==16 and that s==4

[Mr Initial Man]

*Whimpers*

Let me try again:

Variables are n, s, and d.

If s and d are known, this is the formula for n:

n=[d(ds-2d-s+4)]/2

If n and d are known, this is the formula for s:

s=[2(d^2-2d+n)]/[d(d-1)]

If n and s are known, what is the formula for d?

[Ultimater]

d==s lmfao
In case there is a problem of dividing my zero, let me re-do my work and see if I get the same answer.

[Mr Initial Man]

d==s lmfao
d=/=s.

[Ultimater]

Avoiding every possible instance of dividing by zero, here is what I came up with:

d== (-3d²-sd-s+4)/4

[Ultimater]

Without dividing either side by a variable, I cannot seem to get d by itself on the left without any futher occurances on the right... When I get some time tomorrow, I will try doing some more substituation and see how much futher I can get.

[Mr Initial Man]

Erf... That doesn't exactly work...

Maybe I should explain the formula. It's calculating numbers generated by polygons.

s stands for the sides of a polygon. d stands for the dots on each outer edge of a polygon.

For example, if s=3, the polygon is a triangle.

n=[d(3d-2d-3+4)]/2
n=[d(d+1)]/2 (the formula for calculating triangular numbers)
If s=4, the shape is a square.
n=[d(4d-2d-4+4)]/2
n=[d(2d)]/2
n=[2d^2]/2
n=d^2 (Square number!)
Therefore, if I know the number of dots on each side (d), and how many sides the polygon has (s), I can figure out the total (n).
n=d(ds-2d-s+4)/2

If I know the final total (n), and the number of dots on each side (d), I can figure out how many sides the polygon has (s).
s=2(d^2-2d+n)/d(d-1)

Now that you know what it's for, this is my question.

If I know the final total (n) and the number of sides (s), how can I figure out how many dots there are on a side (d)?


In other words, how do I describe d in terms of n and s?

[Mr Initial Man]

Without dividing either side by a variable, I cannot seem to get d by itself on the left without any futher occurances on the right... When I get some time tomorrow, I will try doing some more substituation and see how much futher I can get.

I think you have to fiddle around with square roots, since d is squared.

[mapia]

actually i hate math.

[NogDog]

What exactly is a "dot"? I'm assuming you don't mean a vertex where two sides meet, since d would always equal s and therefore be totally redundant; but I can't figure out what it would mean otherwise.

[Mr Initial Man]

What exactly is a "dot"? I'm assuming you don't mean a vertex where two sides meet, since d would always equal s and therefore be totally redundant; but I can't figure out what it would mean otherwise.


Sorry. Allow me to illustrate then.




d=4 s=2

. . . .

n=4
-----------------------
d=4 s=3

.
. .
. . .
. . . .

n=10
-----------------------
d=4 s=4

. . . .
. . . .
. . . .
. . . .

n=16
-----------------------
d=5 s=4

. . . . .
. . . . .
. . . . .
. . . . .
. . . . .

n=25
-----------------------
d=4 s=5

o

o o

o o o

o o o o

o o o o

o o o o

o o o o

n=22


Does that make the formula a bit clearer?

[Mr Initial Man]

Yes, I did work the formula out for myself, and as far as I've checked (on triangles, squares, pentagons and hexagons), it works perfectly.

[MstrBob]

Wow, this thread is too intellectual to be in this forum. :eek:

[Mr Initial Man]

Well, I hoped SOMEONE here might be able to figure it out. >.<

[BigMoosie]

Since the two equations you gave are interchangable then substituting them together will only yield obvious results like s=s. To solve for d you will only need to choose one equation, I choose the first:


n = d (ds - 2d - s + 4) / 2

2n = dds - 2dd - ds + 4d

0 = d^2(s-2) + d(4-s) -2n

Using the quadratic formula:

d = { -(4-s) +/- sqrt( [4-s]^2 - 4*[s-2][-2n] ) } / (2*[s-2])

I'll let you simplify ;)

[Jeff Mott]

Before we go much further, let's make sure the formula we're working with is correct.

e.g.,
. . .
. . . .
. . . . .
. . . . .
. . . . .
. . . .
. . .

Consider this polygon where s=8, d=3 and n=29.

Now let's try this with your formula for n.

n = d(ds-2d-s+4)
------------
2

n = 3(3*8 - 2*3 - 8 + 4)
--------------------
2

n = 3(24 - 6 - 8 + 4)
-----------------
2

n = 3(14)
-----
2

n = 3*7 = 21 != 29Let me know if I've made a mistake somewhere.

[BigMoosie]

Your have made a mistake in your drawing, 21 is the correct answer:


ooo
o o
o o
o oo o
oo o o
o oo
ooo

[Mr Initial Man]

Actually, Big Moosie is correct.

This is a formula for nested polygons of the same type, adjacent at one vertex.


To do an octagon as Jeff Mott did would require a different formula. Look at how I drew the pentagon.

BTW, if any of you have ever read the Joy of Mathematics, that's the book where I got the idea from.

BTW, Big Moosie. Your formula works out to....



/---------------------
/ 2
s - 4 + \/(4 - s) + 8n (s - 2)
d=-----------------------------------
2 (s - 2)


I've tested it a bit, and it does work, as far as I can tell. Interestingly, it has no answer if s=2, but in that case the answer is d=n

Mind you, with the equation for s, which is


2
2(d - 2d + n)
s=---------------
d (d - 1)


If d is 1 or 0, then that has no answer either, because if d=1, then s can have any number of sides. ^^

Edit:

I stuck s=8 n=21 through BigMoosie's equation. It did indeed work out to d=3 .

[Ultimater]

n=d(ds-2d-s+4)/2

s=2(d^2-2d+n)/d(d-1)

d=?

Well, since d equals s, which is only logical that the number of sides of a shape is also the number of dots of the shape, then s and d are interchangable. e.g. A triangle has 3 sides and has 3 dots or connecting points. e.g. A square has 4 sides and has 4 dots or connecting points, thus the number of sides of a shape are equal to the number of connecting points -- if it is indeed a shape.

Thus, we can turn:
s=2(d^2-2d+n)/d(d-1)
into:
d=2(s^2-2s+n)/s(s-1)

[BigMoosie]

Well, since d equals s, which is only logical that the number of sides of a shape is also the number of dots of the shape...

Not true:

d=4 s=4

. . . .
. . . .
. . . .
. . . .

n=16
-----------------------
d=5 s=4

. . . . .
. . . . .
. . . . .
. . . . .
. . . . .

n=25

[Ultimater]

I missed that post -- thought he meant "dots" as in connecting points of the sides, back to the drawing board.

[BigMoosie]

back to the drawing board.

I guess you missed my post too then cos its been solved :p .

[Ultimater]

Correct you are, sorry, forgot to turn the page...

[Mr Initial Man]

Yeah, as I wrote to you earlier, d=/=s. They had different functions.