i is undefined and all triangles are equilateral

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

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 ;)

Quote:

---

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)

???
:)

Quote:

---

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! ;)

Quote:

---

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!!

Quote:

---

Originally posted by bugzpodder

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

---

Oops!
:p

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

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?

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.

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

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

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

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

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.:cool:

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

puke! :p