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    (Original post by WhiteScythe)
    X^2+y^2=1 and y=x anything else?
    y = - x^2 + 1

    it goes through (0,1) and (1,0)
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    What is the usual thresholds for Bronze and Silver? (Note: I use the word usual lightly, I just want a rough estimate if anyone has one)
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    (Original post by HapaxOromenon3)
    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.
    Yh I thought of that I knew coordinates would work, but be messy so left enough time in case i couldnt get a euclidean proof
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    How likely is 92 for kangaroo
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    (Original post by TBoy11)
    How likely is 92 for kangaroo
    Should probably qualify you, as it would have done last year and this year was generally considered to have been more difficult.
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    (Original post by HapaxOromenon3)
    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.
    If it is 100 I'd be very happy as I'm sure that's what I got
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    Q24 I got 180=arctan (2n) + arcsin (2/sqrt (5))+arctan (2) which leaves n as 2/3 and thus a ratio of 2/3:1/2 which simplifies to 4:3 -are there any more reasonable solutions?
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    (Original post by WhiteScythe)
    Q24 I got 180=arctan (2n) + arcsin (2/sqrt (5))+arctan (2) which leaves n as 2/3 and thus a ratio of 2/3:1/2 which simplifies to 4:3 -are there any more reasonable solutions?
    Yes, I stated a method in one of my previous posts, which I'll copy below to save you the trouble of finding 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.
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    (Original post by carpenoctem)
    I have an issue - all of the A-Level Maths students at my school were forced to do the Senior Maths Challenge and I was pretty much planning on going A B C D etc. on the answer sheet. Only problem is, after all that BMAT practice I couldn't resist trying the questions and have ended up with 105. Is this likely to qualify for any further rounds (I don't know how this works) because I really don't want to put more time into something that I never really intended to take part in the first place? ;/

    I obviously understand that this is really important for a lot of people that are planning to do STEM, but as a prospective Medic I have too much on my plate to be worrying about anything other than my main subjects and my school is the type to make sure that as many people as possible do these further rounds.
    105 will get you into the BMO unless something extremely unlikely (as in, a one in ten thousand chance) happens. Ultimately though your school can't force you to participate.
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    Name:  14786339187501878172620.jpg
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    You can do it without a calculator as shown but the olympiad shouldn't require C3 trig
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    (Original post by WhiteScythe)
    You can do it without a calculator as shown but the olympiad shouldn't require C3 trig
    If you used pythagoras and the facts that tangents from same point are equal, you got a neat way of doing it
    no trig
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    (Original post by carpenoctem)
    Ah ok, thanks for your help! My main concern is that I'll have to miss lessons and then catch up on them afterwards which is just more work for me. But I guess that pulling a sickie is not going to help either
    why do you really not want to do BMO, it just seems odd, like how long will it take to catch up 3.5 hours of lessons
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    (Original post by carpenoctem)
    Ah ok, thanks for your help! My main concern is that I'll have to miss lessons and then catch up on them afterwards which is just more work for me. But I guess that pulling a sickie is not going to help either
    The BMO itself is a single 3-hour exam so you won't miss that much. If you're talking about preparation lessons then most schools don't offer them.
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    (Original post by HapaxOromenon3)
    The BMO itself is a single 3-hour exam so you won't miss that much. If you're talking about preparation lessons then most schools don't offer them.
    its 3.5 hrs fyi
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    Wrote name and details on paper in pen, is that fine?
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    (Original post by GcseLad-_-)
    Wrote name and details on paper in pen, is that fine?
    no i think it has to be pencil, they say hb but scanners can cope with darker
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    For Q24, you could have labelled PS = r and TS = x. Due to equal tangents, and QU = r/2, you can deduce that TR = r-x, and UR = r/2. By the same principles, if UT meets arc QS at M then UM = r/2 and MT = x. Then apply Pythagoras:

    (r/2)^2 + (r-x)^2 = ((r/2)+x)^2

    A bit of algebra later it is clear that r^2 - 3rx = 0. Factorising (and knowing that r does not equal 0) means that r = 3x. Therefore, in terms of x, UR = 3x/2 and TR = 2x. 2: 3/2 is obviously 4:3. Hence the answer is B.

    That's the simplest solution I came up with, involves no coordinates/trigonometry etc. Just equal tangents, Pythagoras and some basic algebraic manipulation.
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    (Original post by carpenoctem)
    3.5 hours ouch

    Yeah, I can see how it looks when most people want to do it. Only thing is the Maths Dept at our school really latch onto people that show signs of doing well and then send them off to every maths competition they can think of. They also make them go to loads of different prep sessions. I just don't want to be put into the "let's make them do extra maths" category when today was simply a fluke and I have other things that I actively want to focus on.
    Oh i see, well if you are sure its a fluke and maths doesnt really interest you, then fair enough
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    (Original post by WhiteScythe)
    Q24 I got 180=arctan (2n) + arcsin (2/sqrt (5))+arctan (2) which leaves n as 2/3 and thus a ratio of 2/3:1/2 which simplifies to 4:3 -are there any more reasonable solutions?
    I did it a way I haven't seen here yet. Taking each side to be length 1, QU = 1/2.

    Where TU touches the arc QS, I have called Z. PZ=PQ=1 (radius). Because both are tangents, QU=UZ=1/2.

    The same argument can be made for ST, which equals TZ. Take TZ to be n, and therefore TR=1-n.

    therefore 1/2 squared + (1-n)^2 = (1/2 + n)^2, which I solved to get n=1/3, so TR=2/3. 2/3:1/2 = 4:3

    Edit: Just seen it was posted 5 mins ago, as I was typing this xd
 
 
 
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