# STEP 2005 Solutions Thread

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Thread starter 12 years ago
#1
You can let someone else take control of the OP whenever.

STEP I:
1: Solution by nuodai
2: Solution by Unbounded
3: Solution by Unbounded
4: Solution by Unbounded
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by nuodai
10: Solution by Sk1lz
11: Solution by nuodai
12: Solution by darkness9999
13: Solution by brianeverit
14: Solution Farhan.Hanif93

STEP II:
1: Solution by SimonM
2: Solution by Daniel Freedman
3: Solution by SimonM
4: Solution by Glutamic Acid
5: Solution by SimonM
6: Solution by Daniel Freedman
7: Solution by SimonM
8: Solution by sonofdot
9: Solution by Farhan.Hanif93
10: Solution by Glutamic Acid
11: Solution by Farhan.Hanif93
12: Solution by nuodai
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by DeanK22
2: Solution by Adje
3: Solution by SimonM
4: Solution by SimonM
5: Solution by Daniel Freedman
6: Solution by sonofdot
7: Solution by sonofdot
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by sweeneyrod
13: Solution by brianeverit
14: Solution by sonofdot

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
3
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12 years ago
#2
STEP I, Question 3, 2005
(ii)    There is exactly one real solution when the discriminant of this quadratic in x is equal to zero.     Noting that But also that  Let us simply prove that c is non-zero:

Suppose that c = 0, then the equation for (ii) becomes exactly the same as (i). Hence we require a and b to both be positive or negative. Then the condition of exactly one root, which is no longer holds, since the LHS is zero whereas the RHS is negative - contradiction. Hence c is non-zero.

And finally, we have proven that . QED
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12 years ago
#4
I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions
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12 years ago
#5
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12 years ago
#6
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12 years ago
#7
(Original post by sonofdot)
.
What happens when n = 2 in part ii)?
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12 years ago
#8
STEP II 2005, Question 2

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12 years ago
#9
(Original post by Daniel Freedman)
What happens when n = 2 in part ii)?
Good point, edited
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12 years ago
#10
STEP I: Question 12
Part 1

Part 2
Spoiler:
Show

Group 1: Group 2: Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\implies \lambda p=\frac{2}{5}\times({\lambda p +(1-\lambda)q -\frac{1}{4}[\lambda p +(1-\lambda)q] + \frac{1}{4}[\lambda\times(1-p) + (1- \lambda)\times(1-q)])

\Rightarrow 10\lambda p=4\lambda p + 4(1-\lambda)q - \lambda p -(1-\lambda)q +\lambda(1-p) + (1-\lambda)(1-q)

\Rightarrow 10\lambda p= 4\lambda p + 4q - 4\lambda q - \lambda p - q + \lambda q + \lambda - \lambda p + 1 - q - \lambda - \lambda q

\Rightarrow 10\lambda p = 2\lambda p + 2q - 2\lambda q + 1

\Rightarrow \boxed{\lambda= \frac{4q + 2}{5+6p + 4q}}

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12 years ago
#11
STEP III 2005 Question 14 that was a long one...

Spoiler:
Show
0
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12 years ago
#12
Question 2, STEP I, 2005
Final Part
Neat proof by Hashashin:
Longer and messier proof by myself spoilerised
The gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel.

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.
Spoiler:
Show
We can show that the lines RP and PS are perpendicular:

RP has gradient PS has gradient To show that they are perpendicular, the product of their gradients must be -1:  By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
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12 years ago
#13
Question 7, STEP I, 2005
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12 years ago
#14
II/4:

3rd part

We can use the result in the second part by choosing p = 7, q =1, s = 5, t = 13, r = 50, u = 25 and v = 130, and everything falls out neatly.
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12 years ago
#15
STEP III

Question 1

1st part

If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<-> <-> Spoiler:
Show

It is relativel easy to show that if after searching for GA formula / derivation it has not been found, we use the identity sin(x) = cos(pi/2 - x) = cos(2(pi/4 - x/2)) and expand using the double angle formulae and the result follows fairly soon.

We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain; <-> and the result follows - although interestingly we do not have wip is obviously needed to obtain this.

2nd part

Rcos(x-a) = Rcos(a)cos(x) + Rsin(a)sin(x)

It follws that if

psin(x) + qcos(x) = Rcos(x-a) we have;

p = Rsin(a) and q = Rcos(a)

Adding p^2 + q^2 will reveal that this is equal to R^2 so ;

R^2 = p^2 + q^2

Divding reveals that (p/q) = tan(a)

Consider;

cos(x) + sin(x)

p=q=1 and it follows cos(x) + sin(x) = root(2)cos(x-(pi/4))

consider

sin(x) - cos(x) = root(2)cos(x+(pi/4))

We now note that |cos(x)| <= 1 and it follows that Using these results we see that for sin(sin(x)) = cos(cos(x)) we must have Noting that and that |(4n+1)pi/2| => pi/2 for n in Z we need to determine whether root(2) => pi/2 which even crude approimations will quickly reveal to not be the case so it follows that there is no value of x that cos(cos(x)) = sin(sin(x))

Cheating with the graphs.
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12 years ago
#16
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12 years ago
#17
III/2 - with the proviso that I think I may have missed something somewhere.

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12 years ago
#18
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12 years ago
#19
(Original post by GHOSH-5)
Question 2, STEP I, 2005

Final Part
We can show that the lines RP and PS are perpendicular:

RP has gradient PS has gradient To show that they are perpendicular, the product of their gradients must be -1:  By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
1
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12 years ago
#20
(Original post by Hashshashin)
For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
That is a lot neater. I'll edit it 0
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