You people are so lazy!

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...