1987 Specimen STEP solutions thread

STEP III Q1
a)let $I_n$ denote the product up to n.
$I_2=\frac{3}{4},I_3=\frac{2}{3},I_4=\frac{5}{8}, I_5=\frac{3}{5}, I_6=\frac{7}{12},...$
By inspection, these numbers seem to be of the form $I_n=\frac{n+1}{2n}$ so let's guess that.
We have already proved this to be the case for n=2 so let us presume, for the purpose of induction, that $I_k=\frac{k+1}{2k}$. Now let n=k+1:
$I_{k+1}=I_k*(1-\frac{1}{(k+1)^2})=\frac{k+1}{2(k)}*\frac{(k+1)^2-1}{(k+1)^2}=\frac{k+2}{2(k+1)}$ So our guess is thus true by mathematical induction.
b) Consider the binomial expansion of $(1+x)^n$:
$(1+x)^n=\displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r}x^r$. Letting x=-1 gives us:
$\displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r}(-1)^r=0[br]\Rightarrow \displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=-\displaystyle\sum_{r=k+1}^n(-1)^r \displaystyle \binom{n}{r}[br]=\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n}{r}[br]=\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r-1}+\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r}$
Notice that the i+1 th element in the first summation cancels with the ith element in the second summation leaving the first element in the first summation so:
$\displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=(-1)^k \displaystyle \binom{n-1}{k}$
As required.

STEP III, Q15: It's easy to convince yourself that the villages all remain connected if less than 3 roads are down (draw diagrams if you wish to illustrate the possible cases).
If 3 roads are down and they all go to the same village, that village is cut off. Otherwise, the villages remain connected. Again, draw diagrams if you wish.
If 4 roads are down, then it's clearly impossible for 2 roads to connect all 4 villages. (The roads must connect at a village, and that only leaves 2 ends to connect 3 villages).
Even more obvious for 5 and 6 roads.

P(more than 3 roads down) = $15 p^4(1-p)^2 + 6p^5(1-p) + p^6$ (binomial)
P(A cut off by having exactly 3 roads down) = $p^3 (1-p)^3$. So P(any village cut off by exactly 3 roads down) = $4p^3(1-p)^3$

So P(villages are cut off) = $p^3[4(1-p)^3 + 15p(1-p)^2+6p^5(1-p)+p^6]$
So P(villages are connected) = $1 - p^3[4(1-p)^3 + 15p(1-p)^2+6p^5(1-p)+p^6]$

When p = 1/2, this equals $\dfrac{64 - 4 - 15 - 6 - 1}{64} = \dfrac{38}{64} = \dfrac{19}{32}$

Wow - only 2 probablity questions out of 16. Hard times for probability specialists...

haven't you been just a little bit careless. Where you have taken out the factor of p^3 you have still left p^5 and p^6 inside the brackets.

Q12
A, B and C form a triangle. Let the angle at A be theta and the angle a C be phi. Let us further denote D to be the point on CB that also lies on the line perpendicular to CB and goes through A and T is the tension in the string.
The rod is in equilibrium so the forces vertically and the moments about A are balanced:

To find AD, notice that the are of triangle is $1/2(\alpha L (AD))$ and also $\frac{1}{2}L^2\alpha sin(\pi-(\theta+\phi))$.
Equating the two, we get:
$AD=Lsin(\theta+\phi)$
Substituting in from our first equations:

By the sine rule:
$\frac{sin\phi}{L}=\frac{sin\theta}{\alpha L}\Rightarrow sin\phi=\frac{sin\theta}{\alpha}$
So:
$\frac{sin\theta cos\theta }{\alpha}=-\frac{sin\theta \sqrt{\alpha^2-sin^2\theta}}{2\alpha} \Rightarrow 2cos\theta=-\sqrt{\alpha^2-sin^2\theta}[br]\Rightarrow 4cos^2\theta=\alpha^2-sin^2\theta \Rightarrow cos^2\theta=\frac{\alpha^2-1}{3}$
The function $cos^2\theta$ is greater that or equal to 0 and less than or equal to 1 but, in this case, we don't want the equality case as that would mean A, B and C would be collinear so:
$1<\alpha<2$
To find the tension:
$T=\frac{Mg}{cos\phi}[br]sin\phi=\frac{sin\theta}{\alpha} \Rightarrow cos\phi = 1/ \alpha(\sqrt{\alpha^2-\frac{4-\alpha^2}{3}})=\frac{2}{\alpha}(\frac{\alpha^2-1}{3})^{1/2}$
So:
$T=\frac{Mg}{cos\phi)}=\frac{Mg \alpha}{2} (\frac{3}{\alpha^2-1})^{1/2}$
As required.

STEP III, Q8.

(i) I'm going to be lazy and use capitals to denote vectors. Let the vertices of the tetrahedron be A, B, C, D. Sum of squares of edges = |A-B|^2+|A-C|^2+|A-D|^2+|B-C|^2+|B-D|^2+|C-D|^2.
Sum of squares of midpoints = [(A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2] / 4.
So 4 x Sum of squares of midpoints = (A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2 = ((A-D)+(B-C))^2 + ((A-B)+(C-D))^2 +((A-C)+(D-B))^2
= (A-D)^2+(B-C)^2 + 2(A-D).(B-C)+(A-B)^2+(C-D)^2+2(A-B).(C-D)+(A-C)^2+(D-B)^2 + (A-C).(D-B).
So, sufficient to show (A-D).(B-C) + (A-B).(C-D) + (A-C).(D-B) = 0. Some tedious expansion gives us A.B-A.C+C.D-B.D+A.C-A.D+B.D-B.C+A.D-A.B+B.C-C.D and everything cancels.

(ii) Take a = 3I + 2J + 6K, b = xI + yJ + K.
Then we have $(3x+2y+6) \leq \sqrt{3^2+4^2+6^2} \sqrt{x^2+y^2+1} = 7\sqrt{x^2+y^2+1}$.
a.b = cos t |a| |b, where cos t is the angle between the vectors.

So a.b = |a| |b if and only if cos t = 1.
So we require 3I+2J+6K and xI+yJ+1 to be in the same direction. So x = 3/6, y=2/6, or x = 1/2, y=1/3.

(a) is easier if you take the origin to be one of the vertices.

STEP III Q11
$\frac{dx}{dt}=k/\alpha z ,\frac{dz}{dt}=k \Rightarrow z=kt+h[br]\Rightarrow \frac{dx}{dt}=\frac{k}{\alpha(kt+h)} [br]\Rightarrow x=k/ \alpha(\frac{1}{k}ln(kt+h)+C). [br]t=0, x=0 \Rightarrow C=-\frac{lnh}{k}[br]\therefore \alpha x=ln(\frac{kt}{h}+1) \Rightarrow t=\frac{h}{k}(e^{\alpha x}-1)$
as required.
For the second plough:
z=k(time it takes for 2nd plough to go y metres-time taken for first plough to reach y)
=$k(t-(e^{\alpha y}-1)\frac{h}{k}$
is there a closing bracket missing here.
And since
$\frac{dy}{dt}=k/\alpha z \Rightarrow 1/ \alpha \frac{dt}{dz}=z/k= t-(e^{\alpha y}-1)\frac{h}{k}$
As required.
Differentiating this with respect to time we get:
Differentiating what with respect to t?
$1/ \alpha \frac{dt}{dz}[br]=1/ \alpha (\frac{h\alpha}{k}e^{\alpha y}+ \alpha e^{\alpha y}(T-\frac{\alpha hy}{k})-\frac{h\alpha}{k}e^{\alpha y})=e^{\alpha y}(T-\frac{\alpha hy}{k})=(e^{\alpha y}-1)\frac{h}{k}+e^{\alpha y}(T-\frac{\alpha hy}{k})-e^{\alpha y}(T-\frac{\alpha hy}{k})=t-(e^{\alpha y}-1)\frac{h}{k}$
I don't understand how this last step follows.
Which means it is a solution as it satisfies the equation.
They will collide when their times are equal and x=y. Notice how the time for second plough is the same as the one for the 1st $+e^{\alpha y}(T-\frac{\alpha hy}{k})$
This should disappear when they collide so:
$T-\frac{\alpha hx}{k}=0 \Rightarrow x=\frac{Tk}{\alpha h}$
As required.

Q12
A, B and C form a triangle. Let the angle at A be theta and the angle a C be phi. Let us further denote D to be the point on CB that also lies on the line perpendicular to CB and goes through A and T is the tension in the string.
The rod is in equilibrium so the forces vertically and the moments about A are balanced:

To find AD, notice that the are of triangle is $1/2(\alpha L (AD))$ and also $\frac{1}{2}L^2\alpha sin(\pi-(\theta+\phi))$.
Equating the two, we get:
$AD=Lsin(\theta+\phi)$
Substituting in from our first equations:

By the sine rule:
$\frac{sin\phi}{L}=\frac{sin\theta}{\alpha L}\Rightarrow sin\phi=\frac{sin\theta}{\alpha}$
So:
$\frac{sin\theta cos\theta }{\alpha}=-\frac{sin\theta \sqrt{\alpha^2-sin^2\theta}}{2\alpha} \Rightarrow 2cos\theta=-\sqrt{\alpha^2-sin^2\theta}[br]\Rightarrow 4cos^2\theta=\alpha^2-sin^2\theta \Rightarrow cos^2\theta=\frac{\alpha^2-1}{3}$
The function $cos^2\theta$ is greater that or equal to 0 and less than or equal to 1 but, in this case, we don't want the equality case as that would mean A, B and C would be collinear so:
$1<\alpha<2$
To find the tension:
$T=\frac{Mg}{cos\phi}[br]sin\phi=\frac{sin\theta}{\alpha} \Rightarrow cos\phi = 1/ \alpha(\sqrt{\alpha^2-\frac{4-\alpha^2}{3}})=\frac{2}{\alpha}(\frac{\alpha^2-1}{3})^{1/2}$
So:
$T=\frac{Mg}{cos\phi)}=\frac{Mg \alpha}{2} (\frac{3}{\alpha^2-1})^{1/2}$
As required.

STEP III, Q16:

(i) If $X_i$ is the number of extra children needed for the ith male child (that is, the number of children since the i-1th male child), then C = X_1 + ... + X_r, and so E[C] = E[X_1]+...+E[X_r]. The X_i are iid geometric distributions with p = 1/2, so E[X_i] = 2 and C = 2r.

(ii) Suppose the king ignores the rules and just has 2r-1 children. If he has r or more boys, clearly C < 2r, and conversely, if he has < r boys, C>=2r.
So P(C < 2r) = P(there are at least r boys from 2r-1 children). Since boys and girls are equally likely, P(there are at least r girls from 2r-1 children) = P(there are at least r boys from 2r-1 children).
But these two events are mutually exclusive, and 1 of them must occur. So P(there are at least r girls from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children).
So P(there are at least r boys from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children), and so P(there are at least r boys from 2r-1 children) = 1/2. That is, P(C < 2r) = 1/2.

Shouldn't that be E[C]=2r at the end of part (i)?

x.