Geometry/Complex Numbers (1 Viewer)

[OLDMAN]

Some complex number problems could be done quiet elegantly using a geometric approach. How about the other way around?

Let ABC be a triangle, D be the midpoint of AB and E a point on the side AC for which AE = 2EC. Prove that BE bisects the segment CD.

a) by geometry
b) by complex numbers

 

[nike33]

ahh so far for a, i have drawn a circle aroud DC where a point 'z' is the centre of the circle now if i can prove B F E are collinear does that prove it?

 

[OLDMAN]

nike33 : you're on the right track - but are you doing approach b) or a) ? At any rate, go on...

 

[nike33]

in doing a..hehe

 

[nike33]

ok heres my attempt

Let z be midpoint of AE

in triangle ABE

2AZ = AE
2AD = DB
and angle BAC is common

hence triangle ADZ|||triangleABE .:. DZ||BE (corresponding angles in parrallel lines equal)

triangle DZC|||TriangleEOC (where O is where DC and BE intercept by AAA

as the triangles are in the ratio 2:1 tje midpoint of DZ (m) produced perpindicular to DZ will cut O.

and by SAS DMO and OEC are congruent...hence DO = OC corresponding sides in congruent triangle

*waits for someones 2 line complex number proof

 

[OLDMAN]

just to comment that dropping a perpendicular from the midpoint of DZ and assuming that it will cut DC at O might be at best unnecessary or at worst wrong. But overall approach is good.
Will give you a chance to revise that part or defend it.

 

[nike33]

ok... DZ is parrallel to BE

hence a perpindicular from DZ will perpindiculary cut BE
now as DZ = 2OE (proven using similar triangle rations)
the midpoint of DZ ie m will pass though O in this example (due to the paralled lines)

it has to cut 'o' as MZ = OE = DM and ZE = MO
and, if it didnt cut o then DO =! OC

 

[Idyll]

My attempt at b:

Let A be the complex no. alpha, B be beta and C be gamma.

Since D is the midpoint of AB, it represents the complex number:
0.5(alpha+beta).

Therefore, mid-point DC has coordinates 0.25(alpha+beta)+0.5gamma.

By ratio division formula, E represents the complex no. 2/3*gamma+1/3*alpha.

Now, for the mid-point of DC lies on BE, there must exist a natural no. l such that:

l(0.25beta-1/12*alpha-1/6*gamma)=beta-2/3*gamma-1/3*alpha

For l=4:
4(0.25beta-1/12*alpha-1/6*gamma)=beta-2/3*gamma-1/3*alpha
beta-2/3*gamma-1/3*alpha=beta-2/3*gamma-1/3*alpha

i.e. The mid-point of DC lies on BE.

Therefore, BE bisects DC.

 

[nike33]

is that a hsc method? anyways i understood it till


l(0.25beta-1/12*alpha-1/6*gamma)=beta-2/3*gamma-1/3*alpha

that line, can you explain?

 

[Idyll]

Let R be the mid-point of of DC. Hence, R has coordinates 0.25(alpha+beta)+0.5gamma.

Now, the vector ER is 0.25beta-1/12*alpha-1/6*gamma.
The vector EB is beta-2/3*gamma-1/3*alpha

And that line is just showing that ER and EB are parallel and hence R lies on the line EB

 

[:: ck ::]

beta gamma alpha... wtf

arghz y make it look so complex... [using complex numbers XD.. ok bad joke] .. T_T

 

[ND]

It's easier if you let one of teh vertices lie on the origin. Btw, wouldn't have been easier to you a b and c instead of alpha beta and gamma?

 

[Idyll]

Originally posted by ND
It's easier if you let one of teh vertices lie on the origin. Btw, wouldn't have been easier to you a b and c instead of alpha beta and gamma?

Yeah, it would've been easier to use a, b and c, lol. Oh well

Edit: And I agree it would've been easier to let one of the vertices be the origin

 

[OLDMAN]

Idyll hasn't done too badly, he (or she) has the right approach. But in the name of elegance, lets follow ND's advise and more.

Position the triangle with C at O, A at 12 and ... wait for it... B at 12z where z is a complex number.

If Idyll (or anyone else) care to do it again with this definition, it will certainly look a lot nicer.

 

[OLDMAN]

Originally posted by nike33
ok... DZ is parrallel to BE

hence a perpindicular from DZ will perpindiculary cut BE
now as DZ = 2OE (proven using similar triangle rations)
the midpoint of DZ ie m will pass though O in this example (due to the paralled lines)

it has to cut 'o' as MZ = OE = DM and ZE = MO
and, if it didnt cut o then DO =! OC

Well defended nike33.
Another way is to draw CF // BE and DZ. Since this 3 parallel lines splits ZC equally, they must also split DC equally.

 

[Idyll]

That definitely makes things much nicer

Now, D is the complex no. 6+6z.
E is the complex no. 4.

Let the mid-point OD be R.

Therefore, R is the complex no. 3+3z.

Therefore, for R lies on EB, there must exist a natural number, l, such that:
l(-1+3z)=-4+12z

Which there is (l=4), and hence the result follows.

would I be right in saying you chose 12 because it is divisible by 4, which makes the working nicer, but it will still work for any number?

 

[OLDMAN]

Idyll : Absolutely!

 

[ND]

Originally posted by OLDMAN
Position the triangle with C at O, A at 12 and ... wait for it... B at 12z where z is a complex number.

But don't you lose some generality by making A = 12? I mean, it's commonsensical that it won't actually make any difference, but will markers like it? Would it be better to make it 12x, where x is any real number?

 

[OLDMAN]

There are always two sides to an argument : the argument itself and the argument about the argument. Please excuse this Lewis Carroll type argument..

Now for the argument itself : no there won't be any loss of generality by saying one side is size 12 unless you're talking about shoes! 12 what?

Now for the argument about the argument : yes it will leave some markers confused, so better off assuming 12k.

I win the first argument but ND wins the second.

 

[CM_Tutor]

Alternately, you state in your answer that, were A to be positioned anywhere else on the real axis, the 'new' situation will be formally similar to the 'current' situation, and hence the result will remain true.