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How do people fail to answer even ONE question on a BMO1 paper? Watch

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    BMO being british math olympiad. (something along the lines of this) http://www.bmoc.maths.org/home/bmo1-2005.pdf
    I heard the modal score for people on BMO1 was failing to answer even ONE question? How is this even possible that people are so stupid? The first question to EVERY BMO1 paper is just trivial and a free correct mark pretty much. I had a look randomly at the other questions (Question 3 in the link above) and came to the answer of 21 in less than 10 minutes. Considering you have THREE AND A HALF hours to do the whole BMO1 paper, no one should really be failing to answer even ONE question; the result should be more at the other end, people averaging a failure to answer 1 BMO1 type question per paper, at most. You have THREE AND A HALF hours to do a few questions, and these questions aren't even that hard.

    BMO1 questions are so easy, how is it that people are so stupid? My god.

    Edit: Here is the score distribution: http://www.bmoc.maths.org/home/bmo1-2005-histogram.pdf
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    It took you 10 minutes to do question 3? That is seriously slow, maybe you aren't as smart as you think.
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    (Original post by james22)
    It took you 10 minutes to do question 3? That is seriously slow, maybe you aren't as smart as you think.
    that's my point. I'm a pretty stupid guy, but I'm finding it hard to believe that there are people even dumber than me. Statistics say that a large number of candidates fail to answer ANYTHING.

    Edit: made an incorrect statement confirms my stupidity
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    (Original post by HLN_Radium)
    that's my point. I'm a pretty stupid guy, but I'm finding it hard to believe that there are people even dumber than me. Statistics say that the modal score for BMO1 is less than 5 when there are 10 marks per question. This means more than HALF the candidates failed to answer ANYTHING.
    something is not quite right with the statistics or possibly the interpretation ...
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    (Original post by TeeEm)
    something is not quite right with the statistics or possibly the interpretation ...
    yep i just re-read what I wrote and realized how stupid i am. I wrote that without thinking.
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    (Original post by HLN_Radium)
    yep made an incorrect statement. im a fking idiot
    no worries
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    (Original post by james22)
    It took you 10 minutes to do question 3? That is seriously slow, maybe you aren't as smart as you think.
    Standard application of the proof of Schur's theorem. (The one from Ramsey theory.)
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    Can someone verify my solution to Q6? It was disconcertingly quick for someone with about three years more experience than normal BMO1 candidates.

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    By induction on the denominator of x. Case denom = 2 is trivial. Let y = \dfrac{p}{q} and suppose that all rationals in (0,1) with denominator less than q are in S. Then let x = \dfrac{p}{q-p}. This is clearly in (0,1) because WLOG q > 2p (if not, q=2p in which case it's 1/2 and we're done, or q < 2p in which case consider 1-y instead). It's manifestly rational. Hence it's in S. Therefore \dfrac{x}{1+x} = \dfrac{p(q-p)}{(q-p)(q-p+p)} = \dfrac{p}{q} is in S.
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    (Original post by Smaug123)
    Can someone verify my solution to Q6? It was disconcertingly quick for someone with about three years more experience than normal BMO1 candidates.

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    By induction on the denominator of x. Case denom = 2 is trivial. Let y = \dfrac{p}{q} and suppose that all rationals in (0,1) with denominator less than q are in S. Then let x = \dfrac{p}{q-p}. This is clearly in (0,1) because WLOG q > 2p (if not, q=2p in which case it's 1/2 and we're done, or q < 2p in which case consider 1-y instead). It's manifestly rational. Hence it's in S. Therefore \dfrac{x}{1+x} = \dfrac{p(q-p)}{(q-p)(q-p+p)} = \dfrac{p}{q} is in S.
    Looks right, very similar to how I did it. Observing that puting x=1/2 gives you all the n/3 terms is a very strong hint at induction on the denominators. From there it is just a little algebra to work out which rations to use as x in the inductive step. Too easy for a final question I think.
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    People doing particularly badly on BMO1 is probably just a case of not knowing/understanding what's expected compared to say, the SMC, unless they've deliberately spent some time preparing for it.

    Olympiad questions require proofs of statements as opposed to A-level 'plug and chug' style questions.

    Maths challenge questions require a little bit more in the way of problem solving, but there's still no expectation that you come up with proofs.
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    (Original post by HLN_Radium)
    BMO being british math olympiad. (something along the lines of this) http://www.bmoc.maths.org/home/bmo1-2005.pdf
    I heard the modal score for people on BMO1 was failing to answer even ONE question? How is this even possible that people are so stupid? The first question to EVERY BMO1 paper is just trivial and a free correct mark pretty much. I had a look randomly at the other questions (Question 3 in the link above) and came to the answer of 21 in less than 10 minutes. Considering you have THREE AND A HALF hours to do the whole BMO1 paper, no one should really be failing to answer even ONE question; the result should be more at the other end, people averaging a failure to answer 1 BMO1 type question per paper, at most. You have THREE AND A HALF hours to do a few questions, and these questions aren't even that hard.

    BMO1 questions are so easy, how is it that people are so stupid? My god.

    Edit: Here is the score distribution: http://www.bmoc.maths.org/home/bmo1-2005-histogram.pdf
    Depends. You could get the numerical answer right to every question and still score <10 due to invalid assumptions or incorrect proof.


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    (Original post by physicsmaths)
    Depends. You could get the numerical answer right to every question and still score <10 due to invalid assumptions or incorrect proof.


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    Is Q2 as easy as I think or have I made an incorrect assumption r faulty logic or something?
    http://www.bmoc.maths.org/home/bmo2-2014.pdf
    btw is the answer to Q1 http://www.bmoc.maths.org/home/bmo1-2005.pdf
    1 and 5?
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    (Original post by MathMeister)
    Is Q2 as easy as I think or have I made an incorrect assumption r faulty logic or something?
    http://www.bmoc.maths.org/home/bmo2-2014.pdf
    btw is the answer to Q1 http://www.bmoc.maths.org/home/bmo1-2005.pdf
    1 and 5?
    I got 1,2,3 and 7.


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    (Original post by MathMeister)
    Is Q2 as easy as I think or have I made an incorrect assumption r faulty logic or something?
    http://www.bmoc.maths.org/home/bmo2-2014.pdf
    btw is the answer to Q1 http://www.bmoc.maths.org/home/bmo1-2005.pdf
    1 and 5?
    For Q2, how did you do it easily? I did it by showing that if there is a solution, then there is a solution for every numerical value the volume takes, then setting up a cubic polynomial with solutuions being exactly the side lengths and showing that it had a non-real solution (contradiction). I don't consider that an easy solution to find though.
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    (Original post by physicsmaths)
    I got 1,2,3 and 7.
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    yh yyour right
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    (Original post by james22)
    For Q2, how did you do it easily? I did it by showing that if there is a solution, then there is a solution for every numerical value the volume takes, then setting up a cubic polynomial with solutuions being exactly the side lengths and showing that it had a non-real solution (contradiction). I don't consider that an easy solution to find though.
    Surely just solve for
    l^3=12l
    l^3=6l^2
    12l=6l^2
    Or am I really really bad at maths lol
    show solutions are not equal
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    (Original post by MathMeister)
    Surely just solve for
    l^3=12l
    l^3=6l^2
    12l=6l^2
    Or am I really really bad at maths lol
    show solutions are not equal
    Cuboid, not cube.
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    (Original post by james22)
    Cuboid, not cube.
    oh darnit

    lol
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    whats the answer to number 1?
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    (Original post by og.east)
    whats the answer to number 1?
    1,2,3,5,7
    I've just found 1 lol.
    Off to create more eqns lol
 
 
 
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