STEP Maths I, II, III 1994 Solutions

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Reply 200

i think the straight way of the last part is to find that a+b+c=0 is another way of expressing that----you can use the root ofz^3=1 to illustrate that.

Original post

by khaixiang

Question 6, STEP III, 1994

$\\z_{2}=(z_{1}-a)i+a\\z_{3}=(z_{2}-b)i+b\\z_{3}=(a+b-z_{1})+(a-b)i\\z_{4}=(z_{3}-c)i+c\\z_{4}=(a+b-c-z_{1})i+(b+c-a)\\z_{5}=(z_{4}-d)i+d\\z_{5}=(z_{1}-a-b+c+d)+(-a+b+c-d)i$

$\\z_{1}=z_{5} \text{ if and only if } a+b-c-d=(-a+b+c-d)i\\(a-c)+(a-c)i=(b-d)i-(b-d)\\(a-c)=(b-d)\{\frac{-1+i}{1+i}\}\\ \therefore a-c=(b-d)i$

(a-c)=(b-d)i represents the geometric condition that the line joining A and C is perpendicular to the line joining B and D in an argand diagram.

$\\z_{2}=(z_{1}-a)e^{i\theta}+a\\z_{3}=(z_{1}-a)e^{2i\theta}+(a-b)e^{i\theta}+b\\z_{4}=(z_{1}-a)e^{3i\theta}+(a-b)e^{2i\theta}+(b-c)e^{i\theta}+c$
Now since a,b and c are distinct points on the unit circle in the complex plane, let
$\\a=e^{i\alpha}\\b=e^{i\beta}\\c=e^{i\gamma}$
Rearranging and factorising:

$\\z_{4}=z_{1}e^{3i\theta}+(1-e^{i\theta})(e^{i\gamma}+e^{i(\beta+\theta)}+e^{i(\alpha+2\theta)})$
For Z1 to always coincide with Z4,

$e^{3i\theta}=1, \therefore \theta=\frac{2k\pi}{3}$ Where k is natural numbers which are non-multiple of 3
Also,
$\\(1-e^{i\theta})(e^{i\gamma}+e^{i(\beta+\theta)}+e^{i(\alpha+2\theta)})=0$
$1-e^{i\theta}\neq0 \text{ since } \theta\neq2m\pi, m=0,1,2,\dots$

$\\e^{i\gamma}+e^{i(\beta+\theta)}+[br]e^{i(\alpha+2\theta)}=0$

So the arguments of a,b and c must satisfy the following for a given angle of rotation.

$\displaystyle \\e^{i\theta}=\frac{-e^{i\beta}\pm\sqrt{e^{2i\beta}-4e^{i(\beta+\gamma)}}}{2e^{i\alpha}}$

Well, is that all? Not sure whether I got it right. Will appreciate if someone can check for me.

Reply 201

cxs

i think rather than prove it by such detailed algebra,take A as the point on the x-axis, we can put b,c which is uniformly distributed on (0,1)to denote B,C,and then we consider two situations--one is B,C both above x-axis or both below, the other is B,C is one above and one below
then we can write three angle in such two situations, and to draw a graph which is a square of length1, denoting b and c each, then calculate the are of the shaded are---not so much work

Original post

by DFranklin

STEP III, Q14. I thought this was a really interesting problem, but I also found it extremely difficult. I went down at least two false tracks, and even now, there's a certain amount of "I know what the answer has to be, so let's work out why I need to multiply my result by 6 to get the right answer".

The first observation I made was that if you have a circle center O, and two points on the circumference P, Q with $\angle POQ = \alpha$ , then $\angle OPQ = \angle OQP = (\pi - \alpha / 2)$ .

Now let P,Q,R be the vertices of the triangle. We consider the case $\frac{1}{2} \le k \le 1$ .

Let $\angle POQ = \alpha, \angle QOR = \beta$ . Then $\alpha, \beta$ are uniformly distributed on [0,2pi].

The 2nd observation I made (well, actually about 17th, but the 2nd useful observation!) is that in this case, only one angle can exceed $k\pi$ , (since the sum of all 3 angles is pi). Suppose that angle to be Q, and suppose also that going clockwise the order of the points is PQR. Then we have a scenario as shown in the "pt 1" diagram attached. Note that O is not inside the triangle PQR; this can be seen looking at the boundary case where PR is a diameter, since then $\angle PQR = \pi/2$ , which is the minimal value for $k\pi$ .

So we want $\angle PQR$ in terms of $\alpha, \beta$ , and from observation 1, we find $\angle PQR = ((\pi - \alpha)+(\pi-\beta))/2 = \pi - (\alpha+\beta)/2$ .

Write $x=\alpha / 2\pi, y = \beta /2\pi$ (so x,y and uniformly distributed on [0,1]). Then $\angle PQR > k\pi \implies 2\pi - (\alpha+\beta) > 2k\pi \implies 1-(x+y) > k \implies x+y < 1-k$ .

So fix x, we must have $0\le y < 1-k-x$ . This is possible only for $x < 1-k$ , when it occurs with probability $1-k-x$ .

So $\mathbb{P}(x+y < 1-k) = \int_0^{1-k}(1-k-x)dx = \frac{1}{2}(1-k)^2$ .

Now this is the probability for the points in the order PQR with Q the largest angle. To get the full probability we must multiply by 2 for the possible order RQP, and by 3 for the possibilities that P or R are the largest angle. Thus our final answer is $3(1-k)^2$ as required.

For $\frac{1}{3}\le k \le \frac{1}{2}$ we have the scenario shown in the second diagram. For convenience we introduce $\gamma = \angle POR$ , but note that $\alpha + \beta + \gamma = 2\pi$ , so although $\alpha, \beta$ are still uniformly distributed, $\gamma$ is in fact fixed.

This time all 3 angles are candidates for exceeding $k\pi$ . We find:

$\angle PQR = (\pi - \alpha)/2 + (\pi - \beta) / 2 = \pi - (\alpha+\beta) / 2.$

$\angle QRP = (\pi - \beta) / 2 + (\pi - \gamma) / 2= (\pi-\beta-\pi+\alpha+\beta)/2 = \alpha / 2.$

$\angle RPQ = (\pi - \alpha) / 2 + (\pi - \gamma) / 2= (\pi-\alpha-\pi+\alpha+\beta)/2 = \beta / 2.$

Because any/all of the angles may exceed $k\pi$ , we will find the probability that all the angles are $\le k\pi$ . Now

$\angle PQR \le k\pi \Leftrightarrow \pi - (\alpha+\beta)/2 \le k\pi \Leftrightarrow \alpha+\beta \ge 2(1-k)\pi$
$\angle QRP \le k\pi \Leftrightarrow \alpha \le 2k\pi$
$\angle RPQ \le k\pi \Leftrightarrow \beta \le 2k\pi$

Again write $x=\alpha / 2\pi, y = \beta /2\pi$ . Then the conditions above become

$x \le k, y \le k, x+y \ge 1-k$ . Again, we fix x and find that the range of possible values for y is $1-k-x \le y \le k$ .

So we must have $1-k-x \le k$ i.e. $1-2k\le x$ . For these values of x we have, the chance of y falling in the appropriate range is just $k - (1-k - x) = 2k + x - 1$ .

So we find the global probability
$\mathbb{P}(1-k-x \le y \le k) = \int_{1-2k}^k (2k-1)+x dx = [ (2k-1)x + x^2/2]_{1-2k}^k \\[br]= (3k-1)(2k-1)+\frac{k^2-(1-2k)^2}{2} \\= \frac{1}{2}[2(6k^2-5k+1)-3k^2+4k-1] \\= \frac{1}{2}[9k^2-6k+1] \\= \frac{1}{2}(3k-1)^2$ .

Again, this is the probability for the configuration where the order of the points is PQR. We need to double to allow for the configuration RQP. (In this case we don't multiply by 3 because our calculations have always allowed for any of PQR to exceed $k\pi$ ). So the probability that no angle exceeds $k\pi$ is $(3k-1)^2$ .

So the probability than an angle exceeds $k\pi$ is $1- (3k-1)^2$

Comment: Outside of exam conditions, this is a great question, but I really wouldn't want to try it in an actual exam. I've definitely found this hardest out of all the questions I've done, partly because I went down completely the wrong track. In particular, I spent a long time working on this before realising that in the first case, O wasn't actually inside the triangle PQR. And the multiplying the final answers by 6 and 2 feels a bit of a fudge. I think it's justifiable, but if I'm honest, if I'd needed to multiply by 3 instead I'm sure I'd have found a justification for that as well.

I'm pretty sure the actual answer to the last part is correct; it agrees with the first part at k=1/2 and I have also verified by computer simulation (something else not terribly practical under exam conditions).

Reply 202

Chong Wa

Original post

by eponymous

Step I Q1

Since all planes are at the same angle \theta to the horizontal, the roof ridge is central.

The height (h) of the roof (see image 1) is equal to $p\tan\theta$ . Therefore $x$ (see diagram) = $\frac{h}{\tan\theta} = \frac{p\tan\theta}{\tan\theta} = p$ . Since the ridge is central, this will be the same at the other end of the roof.

So R (ridge length) = 2q - 2p, since $\theta$ is not present in this expression, the ridge length is independent of $\theta$ .

Volume of the roof:

Slicing up the roof, we can join the two ends to make a pyramid and leave a prism. (see image 2)

The volume of a pyramid = $\frac{1}{3} \times area of base \times height(h)$

Area of base = $p^{2}$

So its volume = $\frac{1}{3} \times p^{2} \times p\tan\theta = \frac{p^{3}}{3} \tan\theta.$

Volume of a prism = $area of face \times length$

Area of face = $\frac{1}{2} \times 2p \times h = p^{2} \tan\theta.$

So the volume of the prism = $p^{2} \tan\theta \times (2q - 2p) = 2p^{2}(q - p)\tan\theta$

So the total volume of the roof = volume of prism + volume of pyramid = $\frac{p^{3}}{3} \tan\theta + 2p^{2} (q - p)\tan\theta = \frac{p^{2}}{3} (6q - 5p) \tan\theta.$

there probably is a much simpler way to do this :redface:

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way :smile:

shouldn't the pyramid have 2p base length not p base lengeth?

Reply 203

Chong Wa

Original post

by khaixiang

Question 3, STEP I, 1994

$\displaystyle (1-x)^{n}={n\choose 0} -{n\choose 1} x+\dots+(-1)^n{n\choose n}x^n$
$\displaystyle (1+x)^{n}={n\choose 0} +{n\choose 1} x+\dots+{n\choose n}x^n$
$\displaystyle (1-x^2)^{n}={n\choose 0}-{n\choose 1} x^2+\dots+{n\choose \frac{n}{2}}x^n+\dots+(-1)^n{n\choose n}x^{2n}$

First we consider n even.

Now collect the term $x^{n}$ in the product of $(1-x)^{n}(1+x)^{n}$
$\displaystyle x^{n}: 2\{{n\choose 0}{n\choose n}x^{n}-{n\choose 1}{n\choose n-1}x^{n}+\dots\}+{n\choose \frac{n}{2}}{n\choose \frac{n}{2}}x^n$
But $\displaystyle {n\choose r}{n\choose n-r}={n\choose r}^2={n\choose n-r}^2$
Also, the term $x^n$ in $(1-x^2)^n$ is ${n\choose \frac{n}{2}}x^{n}$
So then $\displaystyle {n\choose \frac{n}{2}}x^{n}={n\choose 0}^2x^{n}-{n\choose 1}^2x^{n}+\dots+{n\choose \frac{n}{2}}^2x^{n}+\dots+{n\choose n}^2x^{n}$
$\displaystyle \therefore \text{ when n even } {n\choose \frac{n}{2}}={n\choose 0}^2-{n\choose 1}^2+\dots+{n\choose \frac{n}{2}}^2+\dots+{n\choose n}^2$

For n odd, we want to simplify $\displaystyle {n\choose 0}^2-{n\choose 1}^2+\dots-{n\choose n}^2$
We can rewrite the sum as
$\displaystyle \{{n\choose 0}^2-{n\choose n}^2\}+\{{n\choose n-1}^2-{n\choose 1}^2\}+\dots+(-1)^{\frac{n-1}{2}}\{{n\choose \frac{n-1}{2}}^2-{n\choose \frac{n+1}{2}}^2\}$
which is zero.

didn't you forget the (-1)^n/2 on the LHS of the even case?

Reply 204

casperyc

Original post

by eponymous

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way :smile:

For the volume, I think you should have FOUR times (of your) pyramid ?
I did it a square based pyramid with length 2q, height h.
Your pyramid sketch might be wrong?

Reply 205

casperyc

Original post

by Chong Wa

shouldn't the pyramid have 2p base length not p base lengeth?

Agree. I think it should be a square based pyramid with length 2p, height h.

Reply 206

Billord

Original post

by khaixiang

STEP III, Question 2

2(i)
$\displaystyle \\x^4+10x^3+26x^2+10x+1=0\\x^2+10x+26+\frac{10}{x}+\frac{1}{x^2}=0\\ (x+\frac{1}{x})^2-2+10y+26=0\\y^2+10y+24=0\\(y+6)(y+4)=0\\x+\frac{1}{x}=-4\\x=-2\pm\sqrt{3}\\x+\frac{1}{x}=-6\\x=-3\pm\sqrt{10}$

2(ii)
$\displaystyle \\x^4+x^3-10x^2-4x+16=0\\x^2+x-10-\frac{4}{x}+\frac{16}{x^2}=0\\ \text{ set } y=x-\frac{4}{x}\\(x-\frac{4}{x})^2+8+(x-\frac{4}{x})-10=0\\y^2+y-2=0\\(y-1)(y+2)=0\\x-\frac{4}{x}=1\\x=\frac{1\pm\sqrt{17}}{2}\\x-\frac{4}{x}=-2\\x=-1\pm\sqrt{5}$

Is part one not -3±2√2 rather than -3±√10?

Reply 207

Always blank

Original post

by eponymous

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way :smile:

Reply 208

Always blank

Original post

by khaixiang

Reply 209

Always blank

Original post

by ad absurdum

Yeah that is better :smile:

I think I was too keen to use the result shown in the first part.

STEP I question 10

Let e be the extension of the string
Resolve vertically: $\frac{\lambda e}{l}\cos\theta = mg$
Resolve horizontally: $\frac{\lambda e}{l}\sin\theta = mr\omega^2$
But, $\sin\theta = \frac{r}{l+e}$
so, $\frac{\lambda e r}{l(l+e)} = mr\omega^2$
Rearranging, $e = \frac{ml^2\omega^2}{\lambda - ml\omega^2}$
Subsitute into vertical equation and rearrange to get $\cos\theta = \frac{gl(\lambda-ml\omega^2)}{\lambda l \omega^2}$

Now, the motion is possible only if $0<\cos\theta<1$ (if $\cos\theta = 1$ , then the string is vertical and so no angular motion is actually taking place, if $\cos\theta \leq 0$ , then the string is horizontal or above, which is impossible due to the weight of the particle).
So, $\cos\theta > 0 \Rightarrow \lambda > m l \omega^2$ , so $\omega^2 < \frac{\lambda}{ml}$
And, $\cos\theta < 1 \Rightarrow gl(\lambda - m l \omega^2) < \lambda l \omega^2$ , $\omega^2(gml^2 + \lambda l) > gl\lambda$
$\omega^2 > \frac{g\lambda}{l(\lambda + mg)}$

The value of cosθ should be

Reply 210

Always blank

Original post

by nota bene

STEP III Q5

(I hope I don't have one of those with mixed up naming, because this question seem unusually easy to me for being STEP III...)

$f(x)=arcsin(x)\newline f'(x)=\frac{1}{\sqrt{1-x^2}}\newline f''(x)=x(1-x^2)^{-\frac{3}{2}}=\frac{x}{^3\sqrt(1-x^2)^2}$

Prove $(1-x^2)f''(x) - xf'(x)=0$ Substituting gives:
$\frac{(1-x^2)x}{(1-x^2)^{\frac{3}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}=\frac{x}{(1-x^2)^{\frac{1}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}=0$

Prove that $(1-x^2)f^{(n+2)}(x) - (2n+1)xf^{(n+1)}(x) - n^2f^n(x)=0 \forall n>0$
The case n=0 is already proven, now assume true for n=k
$(1-x^2)f^{(k+2)}(x) - (2k+1)xf^{(k+1)}(x) - k^2f^k(x)=0 \forall k>0$
Differentiate w.r.t. x
$(f^{(k+3)}(x)-x^2f^{(k+3)}(x)-2xf^{(k+2)}(x))-(2kf^{(k+1)}(x)+2kxf^{(k+2)}(x)+f^{(k+1)}(x)+xf^{(k+2)})-k^2f^{(k+1)}(x) = \newline (1-x^2)f^{(k+3)}(x)-(2kx+3x)f^{(k+2)}(x)-(2k+k^2+1)f^{(k+1)}(x)$
Simplifying this gives:
$(1-x^2)f^{(k+3)}(x)-(2(k+1)+1)xf^{(k+2)}(x)-((k+1)^2)f^{(k+1)}(x)$
This is the same as what we get when substituting k+1 into the assumption, so by mathematical induction $\bf{P}(k) \implies \bf{P}(k+1)$
Q.E.D.

Now, to express this as a MacLaurin series we can first see it is an odd function (if not convinced just substitute x=0 into what just was proven...) so no even derivatives.
f(0)=0
f'(0)=1

So we have $(1-0)^2f^{2k+3}(0)-(2k+1)^2f^{2k+1}(0)=0$ . Meaning: $f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1}$ Where k= 0, 1, 2...
So the coefficients of the terms are
$a_1=1 \newline a_3=\frac{1}{3!} \newline a_5=\frac{3^2}{5!}=\frac{3}{40} \newline a_7=\frac{5^2\times3^2}{7!}=\frac{5\times 3}{7\times6\times4\times2}$
etc. so the MacLaurin expansion is $arcsin(x)=x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{196}x^7+...$

To find the power series expansion for $g(x)=Ln(\sqrt{\frac{1+x}{1-x}})$ we can see it as $\frac{1}{2}[Ln(1+x)-Ln(1+(-x))]$ for which we can use the McLaurin series for Ln(1+x).
$Ln(1+x)= x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{x^n}{n}$
So: $\frac{1}{2}[(x - \frac{x^2}{2}+\frac{x^3}{3}+...+(-1)^{n+1}\frac{x^n}{n})- (-x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{(-x)^n}{n})]$
Even terms cancel and leaves $\frac{1}{2}\times 2 (x+\frac{x^3}{3}+\frac{x^5}{5}+...+\frac{x^{2n+1}}{2n+1})$
This means the the power series for g(x) is $\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$ and thus the coefficient is $\frac{1}{2n+1}$
Comparing coefficients of the series for f(x) and g(x) we can see that g(x)>f(x). For example the coefficients of $x^7$ which are $\frac{1}{7}>\frac{5\times 3}{7\times6\times4\times2}$ for g(x) and f(x) respectively.

Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).

Reply 211

Always blank

Original post

by khaixiang

Reply 212

Ignorabimus_sol

Original post

by eponymous

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way :smile:

Volume of the left most and right sides of the roof form a rectangular based pyramid with side length of 2p= 1/3 * base area ((2p)^2) * height (tan(theta)*p) = 4/3 * p^3 * tan(theta)
The volume of the triangular prism is 1/2 * bh * the cross sectional length (2q-2p) = 1/2 * (2p)(tan(theta)*p) * (2q-2p) = p^2 * tan(theta)(2q-2p)
The total volume of the roof is the sum of the two previous, so would be = p^2 * tan(theta)(4p/3 + (2q-2p)) = p^2 * tan(theta)(2q-2/3 p)

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