STEP Maths I, II, III 1989 solutions

Is question 7 step 3 nonsense? The sentences (i) and (iii) hold only if a = 45(seen immediately on a diagram by drawing some right angled triangles). Changing y=xtana to y=xcota gives the given answer.

Reply 161

SH101010

1

It should be 4/5 * 5/sqrt(34) in the last line, not 3/5 (since OF is 1/2(a b)). That gives a final answer of 4/sqrt(34), or 2*sqrt(34)/17 if you rationalise the denominator.

(This is for STEP I Q3 by the way - for some reason it doesn't see to have appeared as a reply).

(edited 2 years ago)

Reply 162

keithmcnulty

1

Original post by mikelbird

Just for completeness.....

On STEP 3 Q4 Part (iii), isn't this simply the same as part (ii) just with z = x^2? In which case the roots z are the squares of the six roots of part (ii), giving three real roots?

Reply 163

alst2821

1

Original post by squeezebox

STEP I (Mathematics)
1:
2: Solution by kabbers
3: Solution by Dystopia
4: Solution by Dystopia
5: Solution by Dystopia
6: Solution by Swayam
7: Solution by squeezebox
8: Solution by squeezebox
9: Solution by nota bene
10:
11: Solution by Glutamic Acid
12:
13:
14:
15:
16:

STEP II (F.Maths A)
1: Solution by squeezebox
2: Solution by kabbers
3: Solution by squeezebox
4:
5: Solution by Dystopia
6: Solution by *bobo*
7:
8:
9:
10:
11:Solution by *bobo*
12:
13:
14:Solution by squeezebox
15:
16:

STEP III (F.Maths B)
1: Solution by squeezebox
2: Solution by *bobo*
3:
4:
5:
6:
7:
8: Solution by Dystopia
9: Solution by squeezebox
10: Solution by squeezebox
11:
12:
13:
14:
15:
16:

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

Hi,
Does anyone have a worked solution to 1989-S3-Q3 ?
Thanks!

Reply 164

mqb2766

21

Original post by alst2821

Hi,
Does anyone have a worked solution to 1989-S3-Q3 ?
Thanks!

What have you done / which part are you unsure about as the 3 parts seem quite unrelated.

(edited 1 week ago)

Reply 165

alst2821

1

Hi mqb2766, thanks for the reply.
I only managed the induction to prove a suitable expression of X_(k+1) in terms of X_0
I had a go at the first part and I see that I would have to show that sums of the cosines of the angles of a regular polygon centred at the origin add up to -1.
The proof works okay when m is even, as cosines cancel out in pairs. I can't see a way when m is odd though.
The last part does not seem relevant for 2024 (not in the specification), but it would be interesting to see.

Reply 166

mqb2766

21

Original post by alst2821

Hi mqb2766, thanks for the reply.
I only managed the induction to prove a suitable expression of X_(k+1) in terms of X_0
I had a go at the first part and I see that I would have to show that sums of the cosines of the angles of a regular polygon centred at the origin add up to -1.
The proof works okay when m is even, as cosines cancel out in pairs. I can't see a way when m is odd though.
The last part does not seem relevant for 2024 (not in the specification), but it would be interesting to see.

Id guess there are a few ways to do the first part, but something like (angle addition formula or the fact its a rotation matrix or ...)
M^m = I
M^m - I = 0
(M-I)(M^(m-1)+...+I)=0
as M-I is invertible.... Its basically the well known root of unity result.

You could go down the way youre suggesting, but its a bit laborious. Youd have an m sided unity polygon and the last flat side would be missing from cos and sin series. The flat bottom would mean the x coordinate (cos) would be -1 and the y coordinate (sin) would be 0. For a proof you could do a variety of things (geometry, complex geometric series, sum to product trig identities, ...) but it would be long winded.

(edited 1 week ago)

Reply 167

alst2821

1

Original post by mqb2766

Id guess there are a few ways to do the first part, but something like (angle addition formula or the fact its a rotation matrix or ...)
M^m = I
M^m - I = 0
(M-I)(M^(m-1)+...+I)=0
as M-I is invertible.... Its basically the well known root of unity result.
You could go down the way youre suggesting, but its a bit laborious. Youd have an m sided unity polygon and the last flat side would be missing from cos and sin series. The flat bottom would mean the x coordinate (cos) would be -1 and the y coordinate (sin) would be 0. For a proof you could do a variety of things (geometry, complex geometric series, sum to product trig identities, ...) but it would be long winded.

Your suggestion works very well. Great help. Thanks!

Reply 168

DFranklin

18

Original post by alst2821

Your suggestion works very well. Great help. Thanks!

Not really any different from what mqb is saying, but the first part is just a GP; you probably shouldn't just assume "it's a GP so its sum is (M^n-I)/(M-I)" but you can certainly use the same "if I multiply by (M-I) a lot of stuff cancels" argument.

In that vein, it's interesting that the 2nd part (the matrix recurrence relation) is very like the A-level "you pay Q into an account that pays P interest, compounded annually" problem where you end up with a GP.

(edited 1 week ago)

Reply 169

mqb2766

21

Original post by alst2821

Your suggestion works very well. Great help. Thanks!

Your way worked, though Id strongly suspect they wanted you to spot the link with the complex mth roots of unity result so
w^(m-1) + w^(m-2) + ... + w + 1 = 0
where w is an e^(ix), so it represents a rotation. Then its just the same thing but for (rotation) matrices. The argument that M-I was invertible would need to be made though and its easy to skip over stuff like that when youre ploughing through the (matrix) algebra.