i is undefined and all triangles are equilateral

[bugzpodder]

this one is quiet subtle:

(i-i^(-1))^(-1)

=

(i+i^(-1))[(i-i^(-1))(i+i^(-1))]^(-1)

=(i+i^(-1))[i^2-1/i^2]^(-1)
=(i+i^(-1))[-1+1]^(-1)
=(i+i^(-1))(0)^(-1)
=undefined! (since 0^(-1) or in other words, 1/0 is undefined!)

(i-i^(-1))^(-1)
=i(i(i-i^(-1)))^(-1)
=i(i^2-1)^(-1)
=i(-2)^(-1)
=-i/2

so -i/2=undfined
so either -1/2 is undfined or i is undefined

---------------------------------

given an arbitary triangle ABC, draw angle bisector at A and perpendicular bisector and BC.

if the two lines are congruent, then you know that AB=AC.

case 2: let them meet at D. draw E,F,G such that DE, DF, DG perpendicular to BC, AC, AB respectively.

since BAD=CAD (angle bisector) and DGA=DFA=90, and AD=AD (common side) then we have triangle ADG is congruent to ADF. so AG=AF, GD=GF

also, we have BE=EC (bisector), and DEB=DEC=90 and ED=ED we have triangle BDE is congruent to triangle EDC

so BD=DC.

triangle BDG is also congruent to triangle CDF because GD=DF and BD=DC as shown above and BGD=CFD=90. so BG=FC

since AG=AF and GB=FC then AB=BC

do that to other two sides to get all triangles are equilateral

[A$$Bandit]

The only thing I can come up with for that without reading it, is that you've either divided by zero somewhere or factorized something using zero as a factor to trick us

[NotLKH]

Originally posted by bugzpodder
[B]this one is quiet subtle:

(i-i^(-1))^(-1)

=

(i+i^(-1))[(i-i^(-1))(i+i^(-1))]^(-1)

=(i+i^(-1))[i^2-1/i^2]^(-1)
=(i+i^(-1))[-1+1]^(-1)

How is:
(i+i^(-1))[i^2-1/i^2]^(-1)
equal to
(i+i^(-1))[-1+1]^(-1)

???

[bugzpodder]

The only thing I can come up with for that without reading it, is that you've either divided by zero somewhere or factorized something using zero as a factor to trick us

then you should read it first!

How is:
(i+i^(-1))[i^2-1/i^2]^(-1)
equal to
(i+i^(-1))[-1+1]^(-1)

??? how is it not? substitute i^2=-1, you get -1-(1/(-1))
which is -1+1=0!!

[NotLKH]

Originally posted by bugzpodder

??? how is it not? substitute i^2=-1, you get -1-(1/(-1))
which is -1+1=0!!

Oops!


When you said i is undefined, I was thinking it was just another variable, not i² = -1.


BTW, does the statement "i is undefined" implicitly mean "i² = -1", or can you have several undefined variables, where none of them equal each other, and none are necessarily = -1 when squared?

[bugzpodder]

no forget about "i is undfined" just take at as simplifying the equation:

(i-i^(-1))^(-1)

if you do it two different ways (As shown above) you'll get undefined as one answer, and -i/2 as another.

[sql_lall]

To dissprove the fact that all triangles are equilateral, you just have to realise that of points F and G (feet of perpendiculars to AB and AC from point D), on of these will lie a side, on will lie on the other side's extension, so one of your similar triangle uses will be incorrect

[bugzpodder]

good observation, except you didn't proof what you "realize". and did you see something wrong the the "i" question

[krtxmrtz]

In the term [(i-i^(-1))(i+i^(-1))]^(-1) which, for the sake of clarity, I'm going to write as

1 / [ (i - 1/i) * (i + 1/i) ]

you have a division by 0. Indeed:

1/i = i/i² = i/-1 = -i

Therefore, i + 1/i = 0

[bugzpodder]

told ya it was subtle! Multiply top and bottom by 0 is fun

[Kalkewl8ter]

Bugz,
You said if the two lines are congruent then AB=AC, but then you proceeded to prove that AB=AC; therefore, there is no distinct point of intersection, because the lines are one and the same.

[siyan]

I would try to figure it out if i didn't have to read so much one-line-math-notation

puke!