No worries. You may like to have a look here for some more codes that may not be on the TSR LateX page, but you are in need of(Original post by tommm)
Thank you, I shall remember that
STEP 2003 Solutions Thread
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I/13
Show that the given probability is true
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Let and be the results of team A and team B respectively.
The probability of each team scoring can be modelled by a binomial distribution:
The situation of neither side winning after the initial 10shot period occurs when each team has the same number of success, or fails.
Let denote the number of fails.
Therefore
as required.
Show that the expected number of shots is given by the stated expression
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My LaTeX is probably all wrong and I can't figure out the second part, but there's half a solution (I think), and it's my first one, so woo! 
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STEP II 2003 Q8
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Stationary values occur when . We will use this in the original differential equation, so we obtain
Now, if , then from our expression for y, , which would lead to a fraction in our equation for becoming infinite. Therefore, we disregard y = 0 as a stationary point.
Therefore, stationary points occur at
So
Therefore as required.
When as
When as
I've used a graphing program to draw the two graphs. The first is in the case ; the second is when . Note the location of the turning points.

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STEP II 2003 Q5
Spoiler:ShowE = a/2, F = (p+b)/2, G = b/2, H = (p+a)/2 =>
EF = a/2+x[p+ba]/2
GH= b/2+y[p+ab]/2
for which there is a solution EF = GH = (p+a+b)/4 when x = y = 1/2
so we have s = (d+t)/2 = (p+a+b)/4 so the position vector of T is t = (p+a+b)/2d
the plane OAB is the xy plane i.e. z = 0 so the component of T in the k direction is 0, so considering the vectors in the k direction we have: 0 = n/2d <=> d = n/2

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I/14
I feel like I may have messed up the second part, anyone care checking?

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Question 12, STEP I, 2003

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STEP II Q6
First part:
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Second part:
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FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.

Offline1ReputationRep:(Original post by tommm)
FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.
Sorry 
Offline2(Original post by Horizontal 8)
Sorry
I've done all 8 pures on II now don't think I've ever done that on a paper before. 
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I just logged on to type my solution up I did it earlier today at work when bored  it was a nice and unusually simple question I thought.

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STEP III, 2003, Question 7
Spoiler:ShowThe line through A and B is given by . PN is perpendicular to this line, and so has gradient .
The equation of the line through P and N can therefore be written , which is satisfied by , giving .
Solving simultaneously gives , and as required.
The line through P and N is given by .
Therefore, for N to lie on this line:
Rearranging gives the required result
and are the gradients of BP and AP respectively. Therefore, for N to lie on XY, BP and AP must be perpendicular. 
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STEP II 2003 Q2
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Let
We find, using the usual method, that
Therefore
Now remembering that , and multiplying both sides by 3, we achieve the required result.
For the second part, by comparison with the first, we want to find an equation of a similar form that leads to
Using a bit of guesswork, we find that this equation is
which is satisfied by
Rearranging , we get:
and remembering that leads to the required result.

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Genius. I spent an hour getting two geometrical proofs, which are actually quite nice, but so much more complicated to find. I attatched one I scanned in, the other is similar. Your method is really neat though. 
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Genius. I spent an hour getting two geometrical proofs, which are actually quite nice, but so much more complicated to find. I attatched one I scanned in, the other is similar. Your method is really neat though.
I'm pretty sure Tom's way is the way you're expected to do it. 
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. 
Offline2(Original post by Daniel Freedman)
There's a problem with the LaTeX in your answer for III, 10. I can't spot it though (most probably an extra / missing curly bracket) 
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STEP II 2003 Q4
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The required area is given by
, which can be seen from drawing a diagram[/latex]
Upon using the substitution , and some manipulation with the cosine doubleangle formula, we get
(N.B. that represents limits of integration)
The difficulty in evaluating this lies in the evaluation of . First, using the sine double angle formula, we get
By drawing a suitable rightangled triangle, we find that and from this we can find the required result.
(Original post by tommm)
STEP II 2003 Q4
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To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.
From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:
, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by
But the point (x, y) lies on the line y = mx + c, therefore
To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us
substituting this back into our expression for D^2, we get
and therefore, to find the new area, we must modify (*) such that
The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.
for (i) (note the xcoord of the centre is irrelevant)
for (ii) (perpendicular distance)
for (iii) (perpendicular distance again)
and I just plugged these straight into the equation from the first part. As you noted (i) and (ii) follow from (iii), but I'm sure they put them in that order for a reason (ie to give you bitesized chunks of the overall thoughtprocess). Please tell me if I made a mistake
P.S. You got a stray /latex tag :P 
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STEP III Q13
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(ii)
(iii)
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