Challenging Locus Questions (1 Viewer)

[Grey Council]

hokay, i'll give it another try.

BUT, please, please PLEASE don't burn out. Take it slowly, don't get bored with these forums. But come trials, you'd better still be posting questions and helping out. You got that?

 

[spice girl]

can i be oldman for tonite and pretend to do a geometric for the centre of circle?

ok having done my homework, i know the centre must lie on the y axis. this is cos it must lie on the perpendicular bisector of the pts B and D.

lets call the centre O(0, k), represented by the complex number ik

knowing that OD=OA (radii are equal), we have

|OD|^2 = |OA|^2

k^2 + 1 = |ik - z|^2

let z = x + iy

k^2 + 1 = x^2 + (k - y)^2

k^2 + 1 = x^2 + y^2 + k^2 - 2yk

1 = |z|^2 - 2yk

k = {|z|^2 - 1}/2y

which gives the required answer

 

[Grey Council]

AARRGGGHHH!!!! I give up... for now.
its too late. I've been trying to do it for over an hour now.

 

[CM_Tutor]

Grey Council, have a go tomorrow, if you like, and if you still can't do it ask again.

Spice Girl, I agree that gives the answer, but that's still an essentially algebraic solution. What I'm looking for is a geometric interpretation. Just FYI, this Q comes from a school test I did, and my teacher couldn't think of a geometric interpretation - and he had a PhD in maths! I'll be interested to see if Oldman, or anyone else, can come up with a truly geometric interpretation.

 

[Grey Council]

okay

which question in the test paper was it? don't tell me it was like question 2/3/4.

Well, it wasn't in the SGS 92 trial paper. I have that paper right here. shit, a topic test question. :-\

 

[CM_Tutor]

It was a 40 minute class test on complex numbers, and it was question 3 - but if it makes you feel any better, there were only 3 questions.

 

[Grey Council]

lol, and your class had 5 TER 100's. That makes me feel much better. Esp as the examiner was Bill Pender.

aah, I think I can go to sleep now. My complex numbers is weak, but not extremely weak.

PS Did you get the solution in the test?

 

[CM_Tutor]

Yes, but I understood the geometric interpretation of vectors - anyone who didn't was mightly stuffed by that test - and I stuffed up part of q 1!

 

[Grey Council]

lol

You still remember? gee, wonder if I'll still remember what I did in a certain test on complex numbers. hrm, knowing myself, I won't. I forget things like this very quickly. Unless ofcourse i top my grade. hehe

 

[CM_Tutor]

I've used its questions a number of times over the years - it's not like I haven't looked at in 10 years

 

[Grey Council]

fair enough.

 

[CM_Tutor]

Oops - just for the record, it was Q4 of 4, not Q3 of 3 - I was quoting from another set of Q's I have used, not the test itself, which I just checked - and since you ask, I got 21 / 30, which was equal 3rd in the class.

 

[Grey Council]

i'm beggining to suspect you got a TER of 100...
equal third? crappo, thats good.
And I feel MUCH MUCH better, now that a friend of mine (who is top 5 in our grade for maths) couldn't get it either.

gee, wonder what thats saying for my grade.

 

[OLDMAN]

Sorry, can't get more geometric than Spice Girl's sol.

 

[Grey Council]

still can't do it. :-(

edit;
hang on a second. Are we supposed to use the properties of a quad in circle?
uh, i'm getting this wierd feeling of Deja Vu. ?!? I feel as if I've been through this with someone, prolly buchanan. :-S

 

[CM_Tutor]

Originally posted by Grey Council
i'm beggining to suspect you got a TER of 100...
equal third? crappo, thats good.
And I feel MUCH MUCH better, now that a friend of mine (who is top 5 in our grade for maths) couldn't get it either.

gee, wonder what thats saying for my grade.

You may suspect away, but you'd be wrong.
As for equal 3rd, yeah that's a good result, but its still only 70 % raw - that leaves plenty of room for improvement. lol.

 

[CM_Tutor]

Originally posted by Grey Council
still can't do it. :-(

edit;
hang on a second. Are we supposed to use the properties of a quad in circle?
uh, i'm getting this wierd feeling of Deja Vu. ?!? I feel as if I've been through this with someone, prolly buchanan. :-S

Properties of a circle come into part (c), but (b) is all complex numbers and vectors. So, let's startback at the vectors. Have you found expressions, in terms of z, for the vectors CA and DA? If so, what were they? If not, I'll tell you how to get started.

 

[Grey Council]

CA = z - 1/z
DA = z - 1

edit, whoops, its z-1

 

[CM_Tutor]

Originally posted by Grey Council
CA = z - 1/z
DA = z + 1

CA is correct, DA is incorrect. Once you have found the correct DA, ask yourself how to express angle CAD in terms of arg CA and arg DA.

 

[Grey Council]

arg (DA/CA)

I get:
arg ((z² - z)/(z² - 1))

can we say:
(z(z-1))/((z-1)(z+1))
= z/(z+1)

LOLOL I GOT IT!