To be honest I don't think there'll be any demand for a 1990 thread right now. This thread's gone much slower than the 1992 thread has  I think people are losing interest in favour of *gasps* revision. But you're welcome to start one and revive it after the exams if it dies.
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STEP maths I, II, III 1991 solutions
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 06062007 16:23

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 06062007 16:50
For what it's worth I've just done 1990 III/1, very tedious at the start but the last parts are interesting. It's very similar to a more recent one, except this time you have to expand up to 9th powers rather than 7.

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 06062007 16:54
Yeah I actually did that boring demoivre expansion yesterday before going to sleep, it is quite okay. I'm working through II/1 now, seems to be quite GCSElike quadratics...
I'm sure a few of us will post soluitins once a thread is running 
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 06062007 17:14
Done III/2 aswell (surprisingly for me...) assuming that my work with matrices is right.

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 06062007 19:48
(Original post by Speleo)
Done III/2 aswell (surprisingly for me...) assuming that my work with matrices is right.
EDIT:We're talking about 1990 papers right? 
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 06062007 21:11
(Original post by khaixiang)
I didn't know you can do this question using matrices. I think this is a very nice question, particularly because I've an elegant solution. III/7 is an absolute torture! And maybe if someone has done III/5, do tell me whether the last part can be done using the previous results. And also whether there's anything wrong with the last part of I/4 if you have done it already.
(hope that's right!).
You can then use the previous result on both products to get the final result. But it's probably easier just to prove it directly using induction. 
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 06062007 21:18
Just did III/5 and III/9
III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong.
III/9: Very short and easy for STEP III, especially the old ones...
If anyone has done III/8 I would be interested how they did the very last bit. I can get to y' = f(x,y) but the integrating factor is of the form e^[lnp(x) + karctanq(x)] which is pretty messy, and I'm not even going to try to force my way to the end of the question using that method. 
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 06062007 21:31
My III/2 solution for the first part (excuse the lack of boldface vectors):
OA = m[a^3i + a^2j + ak]
BC = b^3i + b^2j + bk + n[(b^3c^3)i + (b^2c^2)j + (bc)k]
So we want solutions to:
m[a^3i + a^2j + ak] = b^3i + b^2j + bk + n[(b^3c^3)i + (b^2c^2)j + (bc)k]
Equating coefficients of i, j and k:
ma^3 = b^3 + nb^3  nc^3
ma^2 = b^2 + nb^2  nc^2
ma = b + nb  nc
So:
But the determinant of the matrix is 0 (if you work it out you get two terms that are  each other and one term = 0) so there are no solutions and therefore no intersections.
However, I don't really know anything about matrices and so I'm probably making up that rule about the determinant...
(My first latex matrix )
EDIT: Googling for matrix latex only brings up one bdsm side on the first page, I was very surprised.Last edited by Speleo; 06062007 at 21:40. 
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 06062007 21:39
(Original post by Speleo)
But the determinant of the matrix is 0 (if you work it out you get two terms that are  each other and one term = 0) so there are no solutions and therefore no intersections.
However, I don't really know anything about matrices and so I'm probably making up that rule about the determinant...
(My first latex matrix )
E.g. 4x+5y=20 [1] and 8x+10y=40[2] has infinitely many solutions. To me your working seems okay (although I've not actually looked much at the question yet...).
Congrats on the Latex
edit: And maybe we should start that 1990 thread as we seem to be starting to work on them...Last edited by nota bene; 06062007 at 21:40. 
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 06062007 21:40
(Original post by Speleo)
Just did III/5 and III/9
III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong.
III/9: Very short and easy for STEP III, especially the old ones...
If anyone has done III/8 I would be interested how they did the very last bit. I can get to y' = f(x,y) but the integrating factor is of the form e^[lnp(x) + karctanq(x)] which is pretty messy, and I'm not even going to try to force my way to the end of the question using that method.
How did you prove the iff part, you work from both ways right? 
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 131
 06062007 21:42
(Original post by Speleo)
Just did III/5 and III/9
III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong. 
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 132
 06062007 21:52
Thanks nota, I guess then I'd have to add to my proof that infinite solutions are only possible if the two lines are the same which implies the four points are collinear, which implies (a^3i + a^2j + ak) = n(b^3i + b^2j + bk), and a little working out reveals the only way that works is if n = 0 or 1, n=0 implies A is the origin, n=1 implies A is B, but the points O, A and B are distinct so the points aren't collinear so there are no solutions.
Thanks khaixiang, I'll check my working (probably made an algebra mistake early on).
Py'' + Qy' + Ry = d/dx(py' + qy) iff it = py'' + (p'+q)y' + q'y, which it does, equating derivatives, iff P=p, Q=p'+q, R=q' which is does iff P''=p'', Q'=p''+q', R=q' which it does iff P''  Q' + R = 0. 
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 06062007 22:01
(Original post by Speleo)
Thanks khaixiang, I'll check my working (probably made an algebra mistake early on).
Py'' + Qy' + Ry = d/dx(py' + qy) iff it = py'' + (p'+q)y' + q'y, which it does, equating derivatives, iff P=p, Q=p'+q, R=q' which is does iff P''=p'', Q'=p''+q', R=q' which it does iff P''  Q' + R = 0. 
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 06062007 22:01
(Original post by Speleo)
Thanks nota, I guess then I'd have to add to my proof that infinite solutions are only possible if the two lines are the same which implies the four points are collinear, which implies (a^3i + a^2j + ak) = n(b^3i + b^2j + bk), and a little working out reveals the only way that works is if n = 0 or 1, n=0 implies A is the origin, n=1 implies A is B, but the points O, A and B are distinct so the points aren't collinear so there are no solutions.
Damn the lack of matrices I think they are about the only thing I tend to understand what to do with on step III... 
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 06062007 22:04
I think so too. Funny: all these STEP questions and this is the first one where I've had to use an integrating factor (i.e. look up "integrating factor" on Wiki, 'cos I certainly didn't remember how to do it!).

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 06062007 22:20
I can't see any iff chain that works here. Much easier to do separate proofs for each direction. 
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 06062007 22:28
EDIT: On second thoughts, this post is rubbish, ignore it
Here's my attempt, but I'd be inclined to ignore it and hope for the best
OK, we've got the only if then, so we need to prove the if.
If (P+cx+d)''  (Q+e)' + R = 0, (functions P and Q have no constant or x terms), R = Q'  P''
So y''(P+cx+d) + y'(Q+e) + y(R) =
y''(P+cx+d) + y'(Q+e) + y(Q'P'')
equals
y''(P+cx+d) + y'(P'+c+Q+eP'c) + y(Q'P'')
equals
P'y' + Py'' + cxy'' + cy' + dy'' + Q'y + Qy' + ey'  P''y  P'y'  cy'
equals
d/dx[(P+cx+d)y' + (Q+eP'c)y]
equals
d/dx[py' + qy]Last edited by Speleo; 06062007 at 22:40. 
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 06062007 22:37
Shall I make the new thread then?
Or does someone else really really really want to. 
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 06062007 22:38
(Original post by Speleo)
My III/2 solution for the first part
Alternatively, you can note that if that is both B and C are further than A from the origin and are located on one side of A, then OA and BC cannot intersect. And if then OA and BC cannot intersect as well, by the same argument.
So we only need to consider when and But we actually need to verify that there's no intersection for either case since we can just swap b and c.
line OA:
line BC:
From the equation of line OA, we have (the simpler one) .
From equation of line of BC, we have (just take the simpler equation)
Equating them
Now if there's intersection between OA and BC, then this value of z must satisfy since the "depth" of line OA only spans from 0 to a (no equality since all points are distinct)
So if then there's not intersection
This is exactly the case we want to prove earlier, so no intersection between OA and BC. I hope I've not done anything silly
second part seems unusually short and straightforward (I think I've done this in C4), just dot the vectors and show that the dot product is positive isn't it? 
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 06062007 22:45
(Original post by khaixiang)
Alternatively, you can note that if that is both B and C are further than A from the origin and are located on one side of A, then OA and BC cannot intersect. And if then OA and BC cannot intersect as well, by the same argument.
wrt the second part: yeah, you get cos(AOB) = sqrt[f(x)] so you need to show that f(x) can be any value between 0 and 1 (which it can, it's a quadratc or a quartic, I forget which). You also need to exclude AOB = 0, pi/2. = 0 is easy because the points would be collinear which can't happen (see above), I'm not so sure about pi/2 though, cross product?
EDIT: I hate vector/geometry questionsLast edited by Speleo; 06062007 at 22:47.
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Updated: May 19, 2016
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