STEP 2003 Solutions Thread

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  1. Unbounded's Avatar
    41 18-04-2009  21:39
    • TSR Demigod
    Re: STEP 2003 Solutions Thread
    (Original post by tommm)
    Thank you, I shall remember that
    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
  2. AnonyMatt's Avatar
    42 18-04-2009  23:25
    • Vengeful, Imperial Overlord of The Student Room
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    Re: STEP 2003 Solutions Thread
    I/13
    Show that the given probability is true
    Spoiler:
    Show

    Let X and Y be the results of team A and team B respectively.

    The probability of each team scoring can be modelled by a binomial distribution:

    X\tildaB(5,P_A)
    Y\tildaB(5,P_B)

    The situation of neither side winning after the initial 10-shot period occurs when each team has the same number of success, or fails.

    Let i denote the number of fails.

    P(X=i) = \binom{5}{i}P_A^{5-i}(1-P_A)^i P(Y=i) = \binom{5}{i}P_B^{5-i}(1-P_B)^i

    Therefore \alpha = \binom{5}{i}^2(1-P_A)^i(1-P_B)^iP_A^{5-i}P_B^{5-i}
    as required.


    Show that the expected number of shots is given by the stated expression
    Spoiler:
    Show




    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!
    Last edited by AnonyMatt; 18-04-2009 at 23:31.
  3. tommm's Avatar
    43 19-04-2009  00:10
    • Banned
    Re: STEP 2003 Solutions Thread
    STEP II 2003 Q8

    Spoiler:
    Show

    \frac{dy}{dt} = -ky(\displaystyle\frac{t^2 - 3t + 2}{t + 1})

    \implies \displaystyle\int\frac{\mathrm{d }y}{y} = -k\displaystyle\int\displaystyle\ frac{t^2 - 3t + 2}{t + 1}\mathrm{d}t

    \implies \ln y = -k\displaystyle\int(t - 4 + \frac{6}{t+1})\mathrm{d}t = -k(\frac{1}{2}t^2 - 4t + 6\ln(t + 1) + c)

    \implies y = Ae^{-k(\frac{1}{2}t^2 - 4t + \ln((t + 1)^6)}

    \implies y = A(t + 1)^{-6k}e^{-k(\frac{1}{2}t^2 - 4t)}


    Stationary values occur when \frac{dy}{dt} = 0. We will use this in the original differential equation, so we obtain

    y(\displaystyle\frac{t^2 - 3t + 2}{t + 1} = 0

    \implies y(t-1)(t-2) = 0, t \neq -1

    Now, if y = 0, then from our expression for y, t = -1, which would lead to a fraction in our equation for \frac{dy}{dx} = 0 becoming infinite. Therefore, we disregard y = 0 as a stationary point.
    Therefore, stationary points occur at t = 1, 2

    So y(1) = A \times 2^{-6k}e^{-3.5k}, y(2) = A \times 3^{-6k}e^{-6k}

    Therefore \frac{y(2)}{y(1)} = (3/2)^{-6k}e^{-\frac{5}{2}k} as required.

    When k > 0, y \rightarrow +0 as t \rightarrow +\infty

    When k < 0, y \rightarrow +\infty as t \rightarrow +\infty

    I've used a graphing program to draw the two graphs. The first is in the case A = 1, k = 1; the second is when A = 1, k = -1. Note the location of the turning points.







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  4. Dadeyemi's Avatar
    44 19-04-2009  16:32
    • Overlord in Training
    • Location: x
    Re: STEP 2003 Solutions Thread
    STEP II 2003 Q5
    Spoiler:
    Show
    E = a/2, F = (p+b)/2, G = b/2, H = (p+a)/2 =>
    EF = a/2+x[p+b-a]/2
    GH= b/2+y[p+a-b]/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)/2-d
    the plane OAB is the x-y 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/2-d <=> d = n/2
    Last edited by Dadeyemi; 25-05-2009 at 12:13.
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  5. Adjective's Avatar
    45 19-04-2009  19:06
    • Overlord in Training
    Re: STEP 2003 Solutions Thread
  6. nota bene's Avatar
    46 19-04-2009  21:53
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    • Posts: 9,970
    Re: STEP 2003 Solutions Thread
    I/14

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

    i
    The probability distribution we are given is poisson \frac{2^{k}e^{-2}}{k!}. Now the probability she is called by 0 friends, 1 friend and >=2 friends are interesting.
    P(k=0)=e^{-2} \newline P(k=1)=2e^{-2} \newline\text{Thus} P(k\geq 2)=1-e^{-2}-2e^{-2}=1-3e^{-2}
    Now the expected value of the number of friends she sees is 2\times P(\geq \text{2 calls})+1\times P(\text{1 call}) + 0\times P(\text{no calls})
    I.e. 2(1-3e^{2})+2e^{-2}=2-4e^{-2} as required.


    ii
    A new friend calls with probability p, independent of the other calls. Now P(no calls)=e^{-2}(1-p), P(one call)=pe^{-2}+(1-p)2e^{2}=2e^{-2}-pe^{-2} There is now a new combination that can render two calls in a day, namely that one of her old friends call and the new friend. The probability for this event is 2pe^{-2}

    The intriguing part is to find p(>=2).

    We can start by looking at P(2 calls), P(3 calls) and P(4 calls)
    2pe^{-2}+(1-p)2e^{-2}=2e^{-2} \newline 2e^{-2}p+(1-p)\frac{8}{6}e^{-2}=pe^{-2}(2-\frac{8}{6})+\frac{8}{6}e^{-2} \newline \frac{8}{6}e^{-2}p+(1-p)\frac{16}{24}e^{-2}=pe^{-2}(\frac{8}{6}-\frac{16}{24})+\frac{16}{24}e^{-2}

    When looking as n goes to infinity (P(n calls)) this leads to: 2pe^{-2} + e^{-2}\displaystyle\sum_{k=2}^{\inft y}\frac{2^k}{k!}=2pe^{-2} + e^{-2}(e^2-1-2)=1+e^{-2}(2p-3) (recall the taylor series definition of e^x for the sum)


    All in all we have: 2(1+e^{-2}(2p-3))+1(pe^{-2}+(1-p)2e^{-2})+0(e^{-2}(1-p))=2+e^{-2}(3p-4)

    So the expected number of friends for her to go out with
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  7. Unbounded's Avatar
    47 19-04-2009  22:16
    • TSR Demigod
    Re: STEP 2003 Solutions Thread
    Question 12, STEP I, 2003
    (i)
    If we let R be the score on the first ball and S be the score on the second, we find that, for some R = r and S = s, because the drawings are independent and each is equally likely to be drawn

    \mathbb{P}(R=r \ \mathrm{and} \ S = s) = \dfrac{1}{n^2}

    If we sum up all the probabilities, the expected value comes out to:

    \dfrac{1}{n^2} \displaystyle\sum_{r=1}^n\displa ystyle\sum_{s=1}^n rs

    = \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2} \times \dfrac{n(n+1)}{2}

    = \boxed{\dfrac{(n+1)^2}{4}}
    (ii)
    Again, letting R be the value of the first ball and S the value of the second ball, for any R = r and S = s, where r and s are not equal, they each have a probability of

    \mathbb{P}(R = r \ \mathrm{and} \ S = s) \dfrac{1}{n(n-1)}

    Again, summing up all the probabilities, the expected value comes out to:

    \dfrac{1}{n(n-1)} \displaystyle\sum_{r=1}^n \displaystyle\sum_{s=1}^n rs \ \ \ r \not= s


    This can be rewritten as

    = \dfrac{1}{n(n-1)} \left [ \displaystyle\sum_{r=1}^n \displaystyle\sum_{s=1}^n rs - \displaystyle\sum_{s=1}^n s^2 \right ]

    = \dfrac{1}{n(n-1)} \left [ \dfrac{n^2(n+1)^2}{4} - \dfrac{n(n+1)(2n+1)}{6} \right ]

    = \dfrac{(n+1)}{12(n-1)} \left [ 3n(n+1) - 2(2n+1) \right ]

    = \dfrac{(n+1)(3n+2)(n-1)}{12(n-1)} = \boxed{\dfrac{(n+1)(3n+2)}{12}} \ \ \ \square
    Last edited by Unbounded; 21-04-2009 at 00:30.
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  8. Horizontal 8's Avatar
    48 20-04-2009  20:45
    • Benevolent Member
    • Location: London
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    Re: STEP 2003 Solutions Thread
    STEP II Q6

    First part:

    Spoiler:
    Show

    The sketches are attached at the end of the post. Essentially what we do to sketch f^{n+1}(x) from the sketch of f^{n}(x) is translate it one unit downwards each time then reflect in the x-axis whatever is below it. Trying some values can make things clearer...


    Second part:

    Spoiler:
    Show


    Following from the sketch in the second attachment:

    R= \displaystyle \int^{\pi}_0 Sin\left( \frac{\pi x}{2}\right) dx = \left[\frac{-2}{\pi}cos\left( \frac{\pi x}{2}\right) \right]_0^2 = \frac{4}{\pi}

    From the sketch, one can see that when n is even, the area \displaystyle \int^n_0 h(x) dx is represented by \frac{n}{2} lots of R.

    Hence \displaystyle \int^n_0 h(x) dx = \frac{4}{\pi} \times \frac{n}{2} = \boxed{\frac{2n}{\pi}}

    Now, from the sketches of f^2(x) and f^4(x) one can see that \displaystyle\int^n_0g_n(x) dx (for even n) is made up of \frac{n}{2} triangles of base 2 and height 1 (i.e areas equal to 1).

    Hence \displaystyle\int^n_0g_n(x) dx = \boxed{\frac{n}{2}}

    \implies \boxed{\displaystyle \int^n_0 h(x) - g_n(x) dx = \frac{2n}{\pi} - \frac{n}{2}}

    as required.

    Attached Thumbnails
    Click image for larger version.  Name: Attachment_1.jpg  Views: 296  Size: 67.5 KB  ID: 68459   Click image for larger version.  Name: Attachment_2.jpg  Views: 240  Size: 18.4 KB  ID: 68460  
    Last edited by Horizontal 8; 20-04-2009 at 20:49.
  9. tommm's Avatar
    49 20-04-2009  22:32
    • Banned
    Re: STEP 2003 Solutions Thread
    FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.
  10. Horizontal 8's Avatar
    50 20-04-2009  23:22
    • Benevolent Member
    • Location: London
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    Re: STEP 2003 Solutions Thread
    (Original post by tommm)
    FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.

    Sorry :o:
  11. tommm's Avatar
    51 20-04-2009  23:25
    • Banned
    Re: STEP 2003 Solutions Thread
    (Original post by Horizontal 8)
    Sorry :o:
    :p:

    I've done all 8 pures on II now :eek: don't think I've ever done that on a paper before.
  12. nota bene's Avatar
    52 20-04-2009  23:34
    • TSR Idol
    • Posts: 9,970
    Re: STEP 2003 Solutions Thread
    I just logged on to type my solution up:p: I did it earlier today at work when bored - it was a nice and unusually simple question I thought.
    Post rating:
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  13. Daniel Freedman's Avatar
    53 03-05-2009  18:47
    • Overlord in Training
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    Re: STEP 2003 Solutions Thread
    STEP III, 2003, Question 7

    Spoiler:
    Show
    The line through A and B is given by y = -\frac{b}{a}x + b . PN is perpendicular to this line, and so has gradient - \frac{1}{-\frac{b}{a}} = \frac{a}{b} .

    The equation of the line through P and N can therefore be written y = \frac{a}{b} x + c , which is satisfied by (s,t) , giving y = \frac{a}{b}x + t - \frac{a}{b}s .

    Solving simultaneously gives x = \frac{ab^2 - a(bt - as)}{a^2+b^2} , and y = \frac{a^2b + b(bt-as)}{a^2+b^2} as required.

    The line through P and N is given by y = -\frac{t}{s}x+t .

    Therefore, for N to lie on this line:

    \frac{a^2b + b(bt-as)}{a^2+b^2} = - \frac{t}{s} \frac{ab^2 - a(bt - as)}{a^2+b^2} + t

    Rearranging gives the required result \left(\frac{t-b}{s}\right) \left(\frac{t}{s-a}\right) = - 1

    \frac{t-b}{s} and \frac{t}{s-a} are the gradients of BP and AP respectively. Therefore, for N to lie on XY, BP and AP must be perpendicular.
    Last edited by Daniel Freedman; 03-05-2009 at 18:50.
  14. toasted-lion's Avatar
    54 04-05-2009  12:12
    • Exalted Member
    • Posts: 322
    Re: STEP 2003 Solutions Thread
    (Original post by tommm)
    STEP II 2003 Q2

    Spoiler:
    Show

    \theta = \pi/3

    Let 2\sqrt{3}\sin\theta + 4\cos\theta = R\cos(\theta + \alpha)

    We find, using the usual method, that R = \sqrt{28}, \alpha = -\arctan(\frac{\sqrt{3}}{2})

    Therefore \sqrt{28}\cos(\theta - \arctan(\frac{\sqrt{3}}{2})) = 5
    \implies \theta = \arccos\frac{5}{\sqrt{28}} + \arctan\frac{\sqrt{3}}{2}

    Now remembering that \theta = \pi/3, 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
    10\sin(\theta + \arctan\frac{3}{4}) = 7\sqrt{2}

    Using a bit of guesswork, we find that this equation is
    8\sin\theta + 6\cos\theta = 7\sqrt{2}
    which is satisfied by \theta = \pi/4

    Rearranging 10\sin(\theta + \arctan\frac{3}{4}) = 7\sqrt{2}, we get:

    \theta = \arcsin\frac{7\sqrt{2}}{10} - \arctan\frac{3}{4}

    and remembering that \theta = \pi/4 leads to the required result.


    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.
    Attached Thumbnails
    Click image for larger version.  Name: 2002step2q2part1geo (Large).jpg  Views: 130  Size: 58.3 KB  ID: 69210  
  15. tommm's Avatar
    55 04-05-2009  12:20
    • Banned
    Re: STEP 2003 Solutions Thread
    (Original post by toasted-lion)
    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.
    Very nice. In STEP, if there's two parts, you usually have to apply the method from the first part to the second part.
  16. Daniel Freedman's Avatar
    56 04-05-2009  12:32
    • Overlord in Training
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    Re: STEP 2003 Solutions Thread
    (Original post by toasted-lion)
    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 like what you've done, though I would have never attempted to come up with that myself.

    I'm pretty sure Tom's way is the way you're expected to do it.
    Last edited by Daniel Freedman; 04-05-2009 at 12:36.
  17. Daniel Freedman's Avatar
    57 04-05-2009  12:35
    • Overlord in Training
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    Re: STEP 2003 Solutions Thread
    (Original post by tommm)
    .
    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)
  18. tommm's Avatar
    58 04-05-2009  12:45
    • Banned
    Re: STEP 2003 Solutions Thread
    (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)
    Yeah, I just put in an extra curly bracket and it's showing now, but it's still not right...
  19. toasted-lion's Avatar
    59 04-05-2009  15:41
    • Exalted Member
    • Posts: 322
    Re: STEP 2003 Solutions Thread
    (Original post by tommm)
    STEP II 2003 Q4

    Spoiler:
    Show

    The required area is given by

    2\displaystyle\int^R_d\sqrt{R^2-y^2}\mathrm{d}y, which can be seen from drawing a diagram[/latex]

    Upon using the substitution y = R\cos u, and some manipulation with the cosine double-angle formula, we get

    R^2(u - 0.5\sin(2u)|^{\arccos(d/R)}_0 (N.B. that represents limits of integration)

    The difficulty in evaluating this lies in the evaluation of 0.5\sin(2\arccos(d/R)). First, using the sine double angle formula, we get

    0.5\sin(2\arccos(d/R)) = \sin(\arccos(d/R))\cos(\arccos(d/R)) = (d/R)\sin(\arccos(d/R))

    By drawing a suitable right-angled triangle, we find that \sin(2 \arccos (d/R)) = \frac{\sqrt{R^2 - d^2}{R}} and from this we can find the required result.

    Hmm... yuk! Consider the area of the sector OGH and the triangle OGH, subtracting to get the segment. Nice perseverance, but it's actually pretty easy.
    (Original post by tommm)
    STEP II 2003 Q4

    Spoiler:
    Show


    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:

    d + D = R, 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

    D^2 = (x - a)^2 + (y - b)^2

    But the point (x, y) lies on the line y = mx + c, therefore

    D^2 = (x - a)^2 + (mx + c - b)^2

    To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us

    x - a + m^2x + mc - mb = 0

    \implies x = \displaystyle\frac{mb - mc + a}{m^2 + 1}

    substituting this back into our expression for D^2, we get

    D^2 = \displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}

    and therefore, to find the new area, we must modify (*) such that

    d = R - \sqrt{\displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}}

    The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.

    Are you sure we want the distance to the edge and not to the centre? We had d as the distance to centre before. I got:

    for (i) d = |c-b| (note the x-coord of the centre is irrelevant)
    for (ii) d = \frac{|c|}{\sqrt{m^2 +1}} (perpendicular distance)
    for (iii) d = \frac{|ma + c - b|}{\sqrt{m^2 +1}} (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 bite-sized chunks of the overall thought-process). Please tell me if I made a mistake

    P.S. You got a stray /latex tag :P
  20. toasted-lion's Avatar
    60 05-05-2009  19:57
    • Exalted Member
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    Re: STEP 2003 Solutions Thread
    STEP III Q13

    (i)
    Spoiler:
    Show
    The probability the rabbit leaves straight away through A is 1/3 (given which, the probability of leaving through A is 1), the probability of going to B is 1/3 (given which, the probability of leaving through A is PB) etc.

    So P_A = \frac{1}{3} + \frac{1}{3}P_B + \frac{1}{3}P_D

    (ii)
    Spoiler:
    Show
    Using similar arguments and simple algebra:

    \\ \\ P_B = \frac{1}{3}P_A + \frac{1}{3}P_C \\ \\ P_D = \frac{1}{3}P_A + \frac{1}{3}P_C \\ \\ P_C = \frac{1}{3}P_B + \frac{1}{3}P_D \\ \\ P_B + P_D = \frac{2}{3}P_A + \frac{2}{3}P_C = \frac{2}{3}P_A + \frac{2}{9}(P_B + P_D) \\ \\ \frac{7}{9}(P_B + P_D) = \frac{2}{3}P_A \\ \\ \frac{1}{3}(P_B + P_D) = \frac{2}{7}P_A \\ \\ P_A = \frac{1}{3} + \frac{1}{3}(P_B + P_D) = \frac{1}{3} + \frac{2}{7}P_A \\ \\ \frac{5}{7}P_A = \frac{1}{3} \\ \\ P_A = \frac{7}{15}

    (iii)
    Spoiler:
    Show
    Now call PXT the probability the rabbit starts in A, does not visit C, and leaves through tunnel T. Call the probability we require P.

    \\ \\ P_{XA} = \frac{1}{3} + \frac{1}{3}P_{XB} + \frac{1}{3}P_{XD} \\ \\ P_{XB} = \frac{1}{3}P_{XA} \\ \\ P_{XD} = \frac{1}{3}P_{XA} \\ \\ P_{XA} = \frac{1}{3} + \frac{2}{9}P_{XA} \\ \\ \frac{7}{9}P_{XA} = \frac{1}{3} \\ \\ P_{XA} = \frac{3}{7} \\ \\ P = \frac{P_{XA}}{P_A} = \frac{3/7}{7/15} = \frac{45}{49}
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