STEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14: Solution by brianeverit
STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6: Solution by welshenglish
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
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STEP I, II, III 2002 Solutions

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 1
 03042009 12:58
Last edited by SimonM; 26062011 at 20:45. 
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 2
 03042009 13:19
STEP III, Question 1
Last edited by SimonM; 03042009 at 13:35. 
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 03042009 13:33
STEP III, Question 1
a)
b)
Last edited by Oh I Really Don't Care; 05042009 at 01:16. 
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 03042009 13:36

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 03042009 14:12
STEP II 2002 Question 3
Spoiler:Show
If k=1, then and , therefore it is true for k=1.
If k=2, then and , therefore it is true for k=2.
If k=3, then and , therefore it is true for k=3.
Assume, for some k
Therefore by induction, for all positive integers k.
Now consider and with r<s. Assume that and have a common factor p
Since all Fermat numbers are odd, and cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.
All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers.Last edited by sonofdot; 03042009 at 19:50.Post rating:3 
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 03042009 14:19
Question 1, STEP I, 2002
Spoiler:ShowWe need to find the points of intersection between the two ellipses:
 ellipse 1
 ellipse 2
making y^2 the subject in the equation for ellipse 1:
and substitution into the equation for ellipse 2:
Subbing back into the equation for y^2:
Therefore the coordinates of intersection are (2,1) and (2,1)
From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the xaxis.
Letting the coordinates of the centre of the circle be (a,0):
where r is the radius of the circle.
Substitute the point (2,1) in:
So the equation of the circle now becomes:
Last edited by Unbounded; 12042009 at 09:09. 
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 03042009 14:34
Some attached
Spoiler:Show(the end of the question 2 is the reason why I should not do maths late at night :/Last edited by Dadeyemi; 05062009 at 18:42.Post rating:1 
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 03042009 14:36
Some more;
Did these quite a while ago I'm afraid some may be partial solutions.Last edited by Dadeyemi; 15052009 at 22:01.Post rating:2 
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 03042009 14:52
Question 6, STEP I, 2002
First PartLet the height of the equilateral triangle base be x.
From pythag:
Therefore the area of this equilateral triangle base is given by:
Drawing a line down from the apex of the pyramid to the triangle base, letting h be the height, from pythagoras, we can see that:
The volume of the cone, V is given by,
Second PartLast edited by Unbounded; 13042009 at 04:21.Post rating:1 
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 03042009 14:56
Last edited by sonofdot; 12062012 at 15:03. Reason: Fixed mistake in part iiPost rating:3 
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 03042009 18:20
Last edited by sonofdot; 03042009 at 19:54.Post rating:2 
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 03042009 20:56
II/8:
For x < 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x +c ==> y = Ae^(x); as y = a when x = 1, a = Ae so A = a/e; therefore for x < 0.
For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore for x > 0.
If a = b, there is no jump.
Note that e^x  1 is positive if x > 0, and negative if x < 0.
for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e  1) so A = e;
for x < 0.
At x = 0, y = using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so
(i)
(ii) . 
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 13
 03042009 21:36
II/7: (Scary scary vectors.)
??????????
Let the lines have direction vector which I'll denote by r, and without loss of generality let r = 1 so a^2 + b^2 + c^2 = 1;
.
b = 1  a; c = 1  a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.
m_{3} has direction ; m_{4} direction .
Considering gives .
(i) A has position vector ; B .
Let P = , and Q =
AQ . BP = 0 so
. Showing that the latter is > 0; 1  sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.
(ii)
So
The second equation gives . Substituting into the third gives . Substituting these into the first:
; multiplying out gives , so no nonzero solutions.Post rating:4 
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 05042009 01:05
Question 2, STEP I, 2002
Second PartThird PartSketches to follow shortlyLast edited by Unbounded; 05042009 at 01:18.Post rating:1 
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 10042009 11:05
I'm not checking if anyone posted the solutions, cause I completed these not long ago.
2002 II Question 1
as required.
and
, as required.
Using , , we obtain:
Last edited by AurelAqua; 10042009 at 11:08.Post rating:1 
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 10042009 11:14
2002 II Question 11
Total weight (to check our calculus)
Total moment
.
We know that total moment , so , as required.
Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote to be the distance of c.o.m. from A. Angle going from the horizontal to the line AB.
Equating the horizontal forces: . Noting that , we get . Noting that , we get
Equating the vertical forces: .
Equating the moments:
as required.
Not sure if I'm right, but I'll give it a go:
, andLast edited by AurelAqua; 10042009 at 11:22. 
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 10042009 11:17
2002 II Question 12
probability such that more than l coins of the L coins will produce less than m heads
Estimating q: , where z is the probability of a coin producing less than m heads.
We can say that , which gives:
for small z.
We then substitute to get as required.
z is the probability that less than m heads are made after M throws.
.
as required.
K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. , which is a large z, so we are not allowed to approximate as needed. 
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 10042009 11:19

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 11042009 00:10
Question 10, STEP I, 2002
First PartSecond PartLast edited by Unbounded; 11042009 at 00:14. 
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 11042009 02:48
Question 11, STEP I, 2002
First PartLet the initial speed of be u and let and leave the collision at speeds and respectively.
Looking at momentum in the direction of initially:
Looking at momentum perpendicular to the initial direction of :
And considering the energy of the system:
Squaring we get:
And substituting in
Making the subject in we get:
Substituting this result into we get:
Applying productsum formulae:
By double angle formulae:
I fear I've taken some unnecessarily long routeLast edited by Unbounded; 04052009 at 11:16.
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Updated: November 23, 2015
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