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    (Original post by ben-smith)
    nope
    (Original post by deejayy)
    :I haven Could be wrong.

    I was wrong:

    http://www.thestudentroom.co.uk/show...&postcount=981

    It's only an acknowledgement anyway so I wouldn't worry.
    That's okay then

    I expect it'll come through some time this week... we'll probably wake up to it tomorrow or something.

    EDIT: @deejayy Yeah, but "Wanted" did go for Maths and Comp Sci, which is of course, run by a different department.
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    I've got an email from Imperial too, just an hour ago
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    Interview in 2 more days, with Magdalene College's Doctoral of Studies Stuart Martin, and written test in 4 days!!! *Panicked*

    On some random note, anyone minds to solve this question:

    Find the product of gradients (m1m2m3) of an equilateral triangle. Hint: Equilateral triangle is the clue. If possible, please show the working cause I'm lost now!
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    (Original post by Nicholasng925)
    Interview in 2 more days, with Magdalene College's Doctoral of Studies Stuart Martin, and written test in 4 days!!! *Panicked*

    On some random note, anyone minds to solve this question:

    Find the product of gradients (m1m2m3) of an equilateral triangle. Hint: Equilateral triangle is the clue. If possible, please show the working cause I'm lost now!
    Clue:

    Spoiler:
    Show
    You should be aware of the angles of an equilateral triangle... Now, what do you know of the properties of the tangent trigonometric function?


    Answer:

    Spoiler:
    Show
    One side has a gradient of 0. Another has a gradient of tan(60) since the tangent is the opposite side divided by the adjacent - hence the gradient and the other is the negative of that (think about shape and how you would use tan to find the gradient). Then multiply (which gives zero in this case but see the other post about different rotations.)
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    Great, seems like im the only one still waiting to get a letter from imperial!
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    (Original post by jameswhughes)
    You're from Yorkshire? :five:
    Yes. Why?
    York to be precise...gods own county
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    (Original post by Aristotle's' Disciple)
    Dammit you're confusing me now! :P Are they really that good, Lancaster Vs Birmingham now. Which should I have as a back up?!
    Lancaster!!! Hahah sorry but they're all so good Like it says below the course and location are the most important factors. Have you been to either of the open days? What is the subject you're applying for?

    (Original post by Miller693)
    Sorry - accidental neg rep. My bad.

    But seriously, first of all check out the courses.Lancaster has much more of an emphasis on pure maths, which is quite different at uni, compared to at school. Birmingham has a more typical mix. It depends which you'd prefer. Oh and entry requirements.

    Then I'd think about the cities themselves; do you want to be in a very large city right in the centre or in a smaller one on the outskirts. Also think about journey times and costs for you etc.

    I personally chose Lancaster, because I don't think I'd be able to stand the giant city
    culture.

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    (Original post by Reminisce)
    Clue:

    Spoiler:
    Show
    You should be aware of the angles of an equilateral triangle... Now, what do you know of the properties of the tangent trigonometric function?


    Answer:

    Spoiler:
    Show
    One side has a gradient of one. Another has a gradient of tan(60) since the tangent is the opposite side divided by the adjacent - hence the gradient and the other is the negative of that (think about shape and how you would use tan to find the gradient). Then multiply.
    How do we know the gradient of any of the sides? (Maybe I am misreading the question, but surely we can have simply have the triangle rotated around? The question does not give an initial starting position...so how do we know the exact gradients, but only relative ones?)
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    (Original post by twig)
    How do we know the gradient of any of the sides? (Maybe I am misreading the question, but surely we can have simply have the triangle rotated around? The question does not give an initial starting position...so how do we know the exact gradients, but only relative ones?)
    Yeah there is the problem of that, I had to make an assumption (either it is more complicated or perhaps wherever he got the question from, it was drawn in the generic case).

    EDIT: I think I figured it out: you have tan(60+x) as the gradient for one side to account for rotation then the other side is tan(x) instead of 1 and the third side is -tan(60-x). Draw a graph to see, I think this is the general case.

    Might be able to simplify if you expand using the identity...
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    (Original post by BeccaCath94)
    Yes. Why?
    York to be precise...gods own county
    I'm from Yorkshire too, Leeds :awesome:
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    hey guys, I saw this in one of the STEP papers, prove that that there are no positive integers a, b or c such that:

    a^{n} + b^{n} = c^{n}

    when n>2

    ...:awesome:
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    (Original post by ilyking)
    hey guys, I saw this in one of the STEP papers, prove that that there are no positive integers a, b or c such that:

    a^{n} + b^{n} = c^{n}

    when n>2

    ...:awesome:
    Trololol, time to call up Andrew Wiles.
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    (Original post by ilyking)
    hey guys, I saw this in one of the STEP papers, prove that that there are no positive integers a, b or c such that:

    a^{n} + b^{n} = c^{n}

    when n>2

    ...:awesome:
    Now that would take over 100 pages
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    (Original post by Edwin Okli)
    Trololol, time to call up Andrew Wiles.
    this is what it said in the markscheme:

    http://files.asme.org/MEMagazine/Articles/Web/15299.pdf
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    (Original post by ilyking)
    hey guys, I saw this in one of the STEP papers, prove that that there are no positive integers a, b or c such that:

    a^{n} + b^{n} = c^{n}

    when n>2

    ...:awesome:
    I have a proof - but it doesn't quite fit in this post.
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    (Original post by ilyking)
    hey guys, I saw this in one of the STEP papers, prove that that there are no positive integers a, b or c such that:

    a^{n} + b^{n} = c^{n}

    when n>2

    ...:awesome:
    Right, time to get the old Taniyama-Shimura conjecture out...
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    (Original post by anyone_can_fly)
    I have a proof - but it doesn't quite fit in this post.
    want....to...neg
    Can't...stop...self...from...+re pping
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    (Original post by ben-smith)
    want....to...neg
    Can't...stop...self...from...+re pping
    The best thing about this thread is that we all instantly got ilyking's troll. You don't get that in the real world.
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    the joy of mathematics
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    (Original post by jameswhughes)
    I'm from Yorkshire too, Leeds :awesome:
    Ahhh very nice
 
 
 
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