You people are so lazy!

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Originally posted by A$$Bandit
My sources suggest that the copy of Adobe Acrobat you were using was pirated :D

They have been wrong before...

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Now, I don't think you can say its pirated, since the company I
worked for was shut down, and the owner of the Acrobat
Package gave it to me as a parting gift. {Actually, it was tossed
by another coworker, along with about 20 other apps, and, when
I pointed that out to My boss, the actual Procurer of those apps,
he told me I could have anything there.}

So, who's your source?

:)

And, BTW, I'll delete my attachments, and never bother to help you out again if thats the way your going to be!

:p

That would be telling;)

But yeh it is wrong like half the time!

Thanks loads for helping me with this problem, I fully get it now (I used your first right method), hehe.

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Originally posted by NotLKH
I have found that 66/100ths Times that You KNOW something, that it is invariably wrong.

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Pffftttt .. all this fancy maths and he still gets 100 - 44 = 66 :D :eek:

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Originally posted by beachbum
Pffftttt .. all this fancy maths and he still gets 100 - 44 = 66 :D :eek:

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Well, It's unlucky to be right All the Time!
;)

I don't know if this is any help at all, but I can prove that q<5.
Using Bugz's original equations:
p=18/(x+5) and q=15/(x+8),
the following can be derived:
x+5=18/p
x=(18-5*p)/p (1)
and
x+8=15/q
x=(15-8*q)/q (2)
Equating (1) and (2), we get:
(18-5*p)/p=(15-8*q)/q
18*q-5*p*q=15*p-8*p*q
3*p*q-15*p+18*q=0
p*q-5*p+6*q=0
p*q-5*p+6*q-30=-30
(p+6)*(q-5)=-30
Since p and q are both positive, then the only way that p+6 and q-5 will multiply to a negative number is if q<5.

Hope this helps!

But if q<5, and if q=15/(x+8) and x=(15/q)-8, then x is negative.
Since x cannot be negative, this question has no solution.
However, many of the previous solutions seem to be correct.
Something is very wrong...

this is the ORIGINAL VALUES for p and q, but nevertheless they are workable.

p=18/(x+5)

q=15/(x+8)


x=(18-5p)/p

x=(15-8q)/q

from here, since p>0,q>0,x>0

18-5p>0
18>5p

p<18/5=3.6

q<15/8=1.875

now if you substitute these values in, [hopefully] you won't have a problem.

true 0<p<5, but actually 0<p<3.6

Ok, I see it now. Thx, Bugz!:D

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Originally posted by A$$Bandit
(I used your first right method), hehe.

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Great!

So, do you understand it?
You worked everything out alright?

If you're not careful, the Teacher will think you Pirated the answer if you just did something out of thin air, with nothing to back it up.

Such as useing the area formula without deriving it, unless you've
used it in class already.

-Lou

So, since you're using my Meth #1, did you notice that:

A₀*A₂*A₄ = A₁*A₃*A₅

Or, in other words, a more impressive rearrangement:

1 = { A₀*A₂*A₄ }/{ A₁*A₃*A₅ }

No, in fact, I happened NOT to notice that :D

I understand it all completely, I was a bit stuck with Bugz method because I'm not much good without diagrams. Yours was great! I just needed pointing in the right direction and as soon as I realized you'd added that extra line in I knew the rest.

And no, I don't think I've pirated the formula... I'll just check...

OK. I've been working on this for a bit, and I've come down to a set of 2 equations, with 1 unknown. Unfortunately, they both amount to 5th order polynomials. So, I'm brainstorming other directions, but its a good challange for all.

Suppose you know A₁, A₃, and A₅. Find any or all A₂, A₄, A₀.

Think you can do it?
:)

So, A$$, How'd your class go? What was the teachers solution?

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Originally posted by NotLKH
Suppose you know A₁, A₃, and A₅. Find any or all A₂, A₄, A₀.

Think you can do it?
:)

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Guess its too much for you!
:p

A$$Bandit, we helped you to the best of our abilities, so post ur Math God's solution here --- you own it to us this much.

Bugz,
I found an even simpler method, Using Shared Heights.
Check it Out:

http://www.vbforums.com/attachment.p...postid=1158919

Yeh I'm real sorry, but I only got it today, hehe!

OK it's coming up and believe it or not it's pretty much what NotLKH had, although it's much simpler (would you believe...)

I'll get right on it...

Well I looked again and it's pretty much the most minimalistic version of NotLKH's method. In order to make it look even smaller than it already is, I've left out comments because you shouldn't need them...

(A + B + 8)/15 = A/5
2A - B = 8

(A + B + 5)/18 = B/8
5B - 4A = 20

A = 12
B = 10
X = 22...

That's it. Essentially it's the same as your method only this one's more basic. Thanks all for helping, let's just hope it pays off!

Quote:

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Originally posted by A$$Bandit
Yeh I'm real sorry, but I only got it today, hehe!

OK it's coming up and believe it or not it's pretty much what NotLKH had, although it's much simpler (would you believe...)

I'll get right on it...

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Is it simpler than the method I posted just above this post of yours?

:D

Blatantly!:D :D :cool:

Think you can show us how?

Quote:

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Originally posted by A$$Bandit
Well I looked again and it's pretty much the most minimalistic version of NotLKH's method. In order to make it look even smaller than it already is, I've left out comments because you shouldn't need them...

(A + B + 8)/15 = A/5
2A - B = 8

(A + B + 5)/18 = B/8
5B - 4A = 20

A = 12
B = 10
X = 22...

That's it. Essentially it's the same as your method only this one's more basic. Thanks all for helping, let's just hope it pays off!

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Your formulas are wrong.
If 2A - B = 8
And A = 12 and B = 10, then you are saying
24 - 10 = 8.

So, I think you need to put the comments in, to show us how you get your equations.
:p

A is 10, B is 12 (you see ppl usually number the left section A and right section B -- never seen any one name right section A and left section B) and they corresponds to your last diagram of A4 and A5 (just like u named the left region A4 and right region A5)

and basically just follow the equations, he is considering the ratio of triangles with same base but different heights.

(A+B+8) shares same base with A but different heights

15 and 5 share same base but different heights

A and 5 shares same height

(A+B+8) and 15 shares same height

so they ratios should be equal.

same with the other equation.

Yeh I can explain that actually: the reason one of the equations is wrong is because I messed up and put A = 12 and B = 10 instead of the other way round (I was kinda tired), and I realized this just after pressing the submit button. And it's gotta be simpler than the other stuff because it doesn't need another diagram plus 20 lines of working...

So, Apparently, I could say:
X = 22

And claim its simpler than even yours, becouse Mine only took 1 line.

My Method #3 proved shared heights between Triangles, and thats what took up most of it.

Yours had a preamble that you think was inconsequential to the proof, and so you didn't post it.

I tend to beleive, as does Bugz, that you used assumptions based on shared heights.

So, Did You?

Well like I said, the methods are pretty much the same. I don't really understand what we're even arguing about:( , I was just pointing out that the method I was given in class the other day seems less complex than yours simply because it only deals with 2 unknowns.

;)

I'm not really arguing, its just that you've given the answer, and not the solution, {as another thread is curently stressing...:p }

I beleive they are identical, and Mine only uses 2 unknowns too.
If you post your entire solution, then I'll know for sure.

{ http://maya.effectonline.com/~vbforu...postid=1158919 }

But, I still wonder what the solution is for this:

Given {A₁, A₃, A₅}, Solve for Any or All {A₂, A₄, A₆}, {or vice versa}

OK then no problem:D :) ;)

I don't need to post my solution up again because it is in fact the same as yours. It's just that it has less "other stuff" to explain it, since it assumes you know the diagram and know how to relate areas of triangles and stuff...