STEP Maths I, II, III 1993 Solutions

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Reply 180

Decota: Firstly, isn't the question you're answering from 1992?

Secondly, I don't think your formula for the determinant is right (which is going to have lots of run-on effects on the rest of your working, I fear).

Reply 181

gyrase

DFranklin

Yeah, moved. I'll have a look into it later. cheers

Reply 182

DFranklin

Rabite

(Also, did anyone ever bother finishing the z and z* question? The brute force z=x+iy method gave two answers, but whether they're right or not is another question. 0+0i and 1-i?)

I didn't bother finishing it, but on a pragmatic note, you can always plug your putative solutions back into the equations and see if they work.

Reply 183

DFranklin

STEP III, Q8. Note on notation: I'll use A,B,C to denote the position vectors of A,B,C so I don't have to mess with boldface.

We have: $B=(0,1,0)a$
$P=(\lambda, 0, 1-\lambda)a$
$Q=(-\mu,0,1-\mu)a$

So $B-Q = (\mu, 1, \mu-1)a, \quad P-B = (\lambda,-1,1-\lambda)a$

So a normal to the plane BQP is $(\mu, 1, \mu-1) \wedge (\lambda,-1,1-\lambda) = (\mu-\lambda, 2\lambda\mu-\lambda-\mu, -\lambda-\mu)$ . Call this vector N.

Then since B is on the plane, the equation of plane BQP is r.N = B.N.
Since B=(0,a,0), B.N is just $(2\lambda\mu-\lambda-\mu)$ .

Now the line DE has equation $(0,-\gamma, 1-\gamma)a$ . Substituting into the equation of BQP to find the intersection, we get:
$-\gamma(2\lambda\mu -\lambda-\mu) -(\lambda+\mu)(1-\gamma) = 2\lambda\mu -\lambda-\mu$ . Some trivial manipulation (that I can't be bothered to LaTeX) gives
$-\gamma(2\lambda \mu-2\lambda - 2\mu) = 2\lambda \mu$ or $\gamma = \frac{\lambda \mu}{\lambda+\mu-\lambda\mu}$ .

Finally, if BPRQ is a parallelogram, we must have B+(P-B)+(Q-B)=R. Comparing the y-coordinate of both sides, we find:

$-1 = -\gamma$ , or in other words $\gamma = 1$ .

But $\gamma = 1 \implies \lambda+\mu-\lambda\mu = \lambda\mu \implies \lambda+\mu-2\lambda\mu = 0$ . Rewriting, we find we must have $\lambda(1-\mu)+\mu(1-\lambda) = 0$ . But since $0<\lambda,\mu<1$ , both terms of the LHS are > 0, and so this is impossible.

Reply 184

DFranklin

STEP III, Q10. Much easier question than it looks.
Start with $P=(x_1,y_1)$ . The line through P parallel to y=mx is $y=m(x-x_1)+y_1$ . This intersects with the line y=kx when $kx = m(x-x_1)+y_1$ . Deduce $(m-k)x = mx_1-y$ and so $x = \frac{mx_1-y}{m-k}$ . Since y=kx, the formula for R is immediate.

Similarly, $R=\left(\frac{mx_1+y_1}{m+k}, k\frac{mx_1+y_1}{m+k}\right)$ .

Then P' = P+(Q-P)+(R-P) = Q+R-P. So the x-coordinate of P' is:

$\frac{mx_1-y}{m-k} + \frac{mx_1+y_1}{m+k} - x_1$ .

$=\frac{[(m+k)+(m-k)]mx_1 - (m+k-(m-k))y_1}{m^2-k^2}-x_1$

$=\frac{2m^2x_1 - 2ky_1}{m^2-k^2} - x_1$

$=\frac{(m^2+k^2)x_1 -2ky_1}{m^2-k^2}$

And the y-coordinate of P' is (note I've skipped some steps that are identical to the previous calculation):

$k\frac{mx_1-y}{m-k} + k\frac{mx_1+y_1}{m+k} - y_1$

$=\frac{2km^2x_1 - 2k^2y_1}{m^2-k^2} - y_1$

$=\frac{2km^2x_1 -(m^2+k^2)y_1}{m^2-k^2}$ .

So we find $T = \frac{1}{m^2-k^2} \left( \begin{array}{cc}[br]m^2+k^2 & -2k\\[br]2m^2k & -(m^2+k^2)[br]\end{array} \right)$

Then $\det T = \frac{1}{(m^2-k^2)^2}[-(m^2+k^2)^2 + 4m^2k^2]$

$= -\frac{1}{(m^2-k^2)^2}[m^4+2k^2m^2+m^4 - 4m^2k^2]$

$= -\frac{1}{(m^2-k^2)^2}(m^2-k^2)^2 = -1$ .

As |det T| = 1, T preserves areas as required.

Reply 185

DFranklin

Can I recommend people have a go at STEP II, Q8? It's not only a nice question, but I've seen it elsewhere, so it's obviously the kind of question that might recur.

Reply 186

insparato

Yep, ive done it on paper. But lusus wrote it up.

Reply 187

DFranklin

insparato

Yep, ive done it on paper. But lusus wrote it up.

I had a look, but I can't find where it's been done.

Reply 188

insparato

Hmm neither can i, im sure nicholas has done it, im not going crazy :p:

Reply 189

nota bene

insparato

Hmm neither can i, im sure nicholas has done it, im not going crazy :p:

I'm sure you both did the 1992 Q8, but I've not seen anyone post up this AM/GM question...

Reply 190

insparato

Whoopsy daisy. Indeed it is 1992.

Reply 191

khaixiang

Yeah II/8... it's a nice question, thanks for recommending it David, I had skipped this question, I thought it would be a tedious question. But I think the challenge in this question is the induction part, other results seem to follow rather directly.

Reply 192

datr

Nice find, if no one else has I'll write up my solution tomorrow.

Reply 193

datr

STEP II Q8 (You may want to try it yourself first per DFranklin's recommendation.)

Spoiler

Reply 194

brianeverit

Has anyone had a look at paper II number 16 part (c)?

Reply 195

brianeverit

Paper III number 13.
I have done the first part. Can any one finish it please?

1993PAPER III fmb.13.pdf34.9KB

Reply 196

DFranklin

I'm puzzled: The following gives the correct answer, but I don't actually believe that the application of Newton's 2nd law here is correct. But if I were to try to fix that, I'm pretty sure I'd get the wrong answer. :frown:

Consider the part of the chain below Q as a body of variable mass M(t). Note that M(t) = m(3+x/a), and dM/dt = mv/a.
Write Q for the momentum of this part of the chain, so Q = Mv. Then $\dot{Q} = v\dot{M}+M \dot{v}= mv^2/a + M\dot{v}$ (1).

When S reaches R, x = 2a. Now let's find the various bits of (1) when x=2a so we can evaluate $\dot{Q}$ .

From $(\pi+7)av^2=2g(x^2+ax)$ , we have

$\displaystyle av^2 = \frac{2g(4a^2+2a^2)}{\pi+7} = \frac{12ga^2}{\pi+7}$ and so $\displaystyle v^2/a = \frac{12g}{\pi+7}$

To find $\dot{v}$ , we diff $(\pi+7)av^2=2g(x^2+ax)$ w.r.t. t:

$(\pi+7)av \dot v = g(2xv+av)$ , so that $\displaystyle \dot v = \frac{g(2x+a)}{a(\pi+7)}$

In particular, when x = 2a, $\displaystyle \dot v = \frac{5ga}{a(\pi+7)} = \frac{5g}{\pi+7}$

So we have $\displaystyle \dot{Q} = mv^2/a + M\dot{v} = m\frac{12g}{\pi+7}+5m\frac{5g}{\pi+7} =\frac{37}{\pi+7}mg$

But by Newton's second law $\dot{Q} = \sum \{\text{forces on segment PQ} \} = Mg - T$

So $\dot{Q} = 5mg - T$

So $\displaystyle T = 5mg - \frac{37mg}{\pi+7} = \frac{5(\pi+7)-37}{\pi+7}mg = \frac{5\pi-2}{\pi+7} mg$

Reply 197

brianeverit

It looks ok to me. Thanks.

Reply 198

DFranklin

What's dubious is that form of Newton's 2nd law only really holds for an enclosed system. But the chain isn't an enclosed system, because the "new mass" is coming from outside.

Now the most common question involving new mass is accretion; you have a particle falling through mist, and gaining new mass. And in that case, the same formula for Newton II still applies, because the mist didn't have any momentum before it joined the particle.

But in this problem, the "new" bit of chain already has momentum, and I think that needs to be taken into account. But if you do, you get the wrong answer.

Reply 199

Swayum

DFranklin

I, Q7: In this day and age, I'll leave it to you to draw the graphs on a graphing calculator. But the salient points are: $f(x) = x^3+Ax^2+B, f'(x) = 3x^2+2Ax = x(3x+2A), f''(x)=6x+2A$
So f'(0) = 0 when x = 0 or x= -2A/3, so these are the only local maxima/minima. Looking at the 2nd derivative, f''(0)=2A, f''(-2A/3) = -2A.
So when A<0, 0 is a local maximum, f''(-2A/3) is a local minimum (to the right of x=0).
When A>0, 0 is a local minimum, f''(-2A/3) is a local maximum (to the left of x=0).

Now if f has 3 roots, then f'(x) = 0 at some point between each two adjacent roots (this is Rolle's theorem, but if you don't know it a sketch should make it obvious what's going on). So as there are only 2 places (x=0, x=-2A/3) where f'(x) = 0, there are 3 roots iff the sign of f at the two places is different.

So f has 3 roots iff f(0)f(-2A/3)<0, that is, iff $B(-8A^3/27+4A^3B/9+B) < 0$
Since $B(-8A^3/27+4A^3/9+B) = (27B^2+4A^3B)/27$ the result follows.

N.B.: I don't know why they changed from A,B to a,b halfway through the question, but I haven't bothered to follow suit. If you're wondering about the case where 27B^2+4A^3B = 0, it turns out there's a repeated root, so there are only 2 distinct roots.

I didn't follow how "as there are only 2 places where f'(x) = 0" implies "there are 3 roots iff the sign of f at the two places is different". I don't even see how it relates to the theorem you said before it.

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STEP Maths I, II, III 1993 Solutions

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