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UKMT Senior Individual Maths Challenge 2016 - Preparation and Tips watch

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    (Original post by surina16)
    stalked the UKMT twitter and saw they tweeted a picture of a spreadsheet of previous grade boundaries
    Oh right, haha - I'll be stalking them now onwards lol

    (Original post by HapaxOromenon3)
    Chittesh14, I have PMed you now.
    Thanks.
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    (Original post by Mathspro2580)
    Everyone reading this, plz check out the 2002 SMC and just be glad we weren't sitting it then. It is insane compared to this
    We did that one in class and it wasn't found to be that difficult, though my experience is probably atypical since I'm at Westminster, one of the top academic schools in the UK. Nevertheless you are correct that it was harder than this year, hence the low boundaries. In my view this year was slightly, but not much, harder than last year, so the BMO1 boundary is likely to be around 100.
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    (Original post by SCandium)
    I've just done the SMC for the first time, as in Year 11, and pretty sure I got a score of 72 - does anyone know which certificate is that good enough for?
    Either a high bronze or a low silver - fingers crossed for lower thresholds! I would guess at a low silver, though... (Still, don't take my word for it!)
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    (Original post by HapaxOromenon3)
    Here's a nice way to do it: Set up a coordinate system with the origin as the bottom left corner of the square, and let the side length of the square be 2, so that the midpoint of the bottom side is (1,0). Thus the circle is x^2+(y-2)^2=4, and the tangent line is y=m(x-1), where m is the gradient. Hence substitute to give x^2+(mx-m-2)^2=4 -> (1+m^2)x^2 - 2m(m+2)x + m^2 + 4m = 0. Now as the line is tangent, the quadratic should have only one solution (to give just one point of intersection), so using the discriminant, 4m^2(m+2)^2 - 4(1+m^2)(m^2+4m)=0 and by expanding and cancelling, this becomes 3m^2-4m=0, so as m is clearly non-zero, m = 4/3. Hence the ratio is 4:3.
    I found it easier to just use the fact that the lengths of the 2 tangents from one point are equal and then pythagoras
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    I think I got 82 what do people think the grade boundaries will be??
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    (Original post by thomas.ar)
    I found it easier to just use the fact that the lengths of the 2 tangents from one point are equal and then pythagoras
    Yes, but coordinates means you can solve the problem just by doing algebra and it's guaranteed to work, whereas to make a purely geometrical approach work, you have to think about what to do next at every step, and for me that makes it more difficult.
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    Honestly, I found this year much easier than last year, but that may just be where I've improved and there was absolutely no pressure about this, since I was literally told to do it a minute before we started :lol:

    EDIT: may have just been the earlier questions, though. I always struggle with the last few
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    (Original post by HapaxOromenon3)
    Chittesh14, I have PMed you now.
    Me as well, please.
    Thanks
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    That went alright. I think I got 120, since I wasn't thinking for question 19
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    (Original post by louisforrest)
    1B 2D 3B 4C 5A 6D 7A 8B 9C 10B 11E 12C 13A 14D 15B 16D 17D 18E 19_ 20E 21_ 22C 23E 24_ 25_
    Are these all correct if so is 109 enough for BMO qualifier?
    I'm pretty sure that I'm exactly the same as you (or thereabouts)! Those are the same four questions that I couldn't do. Anyway, 109 should quite easily be enough!
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    (Original post by sqrt(e/m)=c)
    Me as well, please.
    Thanks
    I have PMed you now.
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    (Original post by mupsman2312)
    I'm pretty sure that I'm exactly the same as you (or thereabouts)! Those are the same four questions that I couldn't do. Anyway, 109 should quite easily be enough!
    haha looking at them now I could have done 21 and 24 but not too fussed as it was only 4 I had to miss. Just really hoping that all the others are right
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    i got: 1B 2D 3B 4C 5A 6D 7A 8B 9C 10B 11E 12C 13A 14D 15B 16D 17D 18E 19D 20E 21C 22C 23E 24B 25D
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    (Original post by Emmz62)
    i got: 1B 2D 3B 4C 5A 6D 7A 8B 9C 10B 11E 12C 13A 14D 15B 16D 17D 18E 19D 20E 21C 22C 23E 24B 25D
    I pretty much agree except with q9 where only two of the graphs work I believe.
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    (Original post by WhiteScythe)
    I pretty much agree except with q9 where only two of the graphs work I believe.
    I got 3 to work
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    (Original post by thomas.ar)
    I got 3 to work
    X^2+y^2=1 and y=x anything else?
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    Can somebody look at this and tell me what mark they think I would get and if i would qualify for the BMO? Thanks!

    1B

    2D

    3B

    4C

    5A

    6D

    7A

    8D

    9C

    10B

    11E

    12C

    13A

    14D

    15B

    16D

    17D

    18E

    19C

    20E

    21A

    22C

    23E

    24B

    25D
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    Q19 was D right guys?
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    (Original post by WhiteScythe)
    X^2+y^2=1 and y=x anything else?
    Oh yeah y=-x^2+1 ffs
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    Would a 100 be enough to qualify for bmo?? Im so annoyed I flunked some questions!!!
 
 
 
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