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    (Original post by Smash50)
    Nice! Are you studying maths at university?
    Yep. I'm starting my 3rd year at Cambridge after the summer.
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    (Original post by DJMayes)
    Yep. I'm starting my 3rd year at Cambridge after the summer.
    Cool! Got your second year results yet?
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    (Original post by Smash50)
    Cool! Got your second year results yet?
    Careful, he didnt do too well..
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    (Original post by Smash50)
    Cool! Got your second year results yet?
    I have, thank you.

    (Original post by newblood)
    Careful, he didnt do too well..
    OK, going to nip this one in the bud right now. I'm assuming you've been talking to whomever it was who PMed me to try find out my specific rank this year.

    I am under no obligation to give out my rank and/or percentage to whichever random stranger rudely asks me for them over private message.

    This said, the class lists are public. They are put up outside the Senate House, and published online in the university reporter, I think sometime in September. I am not trying to conceal my grade when I know both of those things happen.

    I am, however, greatly enjoying the attempts to cast aspersions on me because of it, especially as those people doing it are going to look even more stupid when the class lists are put online in the university reporter and they find out what class I actually got.
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    (Original post by DJMayes)
    I have, thank you.
    OK, going to nip this one in the bud right now. I'm assuming you've been talking to whomever it was who PMed me to try find out my specific rank this year.
    I am under no obligation to give out my rank and/or percentage to whichever random stranger rudely asks me for them over private message.
    This said, the class lists are public. They are put up outside the Senate House, and published online in the university reporter, I think sometime in September. I am not trying to conceal my grade when I know both of those things happen.
    I am, however, greatly enjoying the attempts to cast aspersions on me because of it, especially as those people doing it are going to look even more stupid when the class lists are put online in the university reporter and they find out what class I actually got.
    I have no idea what you got...I was just teasin him;

    my apologies :hat2:
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    (Original post by DJMayes)
    I have, thank you.



    OK, going to nip this one in the bud right now. I'm assuming you've been talking to whomever it was who PMed me to try find out my specific rank this year.

    I am under no obligation to give out my rank and/or percentage to whichever random stranger rudely asks me for them over private message.

    This said, the class lists are public. They are put up outside the Senate House, and published online in the university reporter, I think sometime in September. I am not trying to conceal my grade when I know both of those things happen.

    I am, however, greatly enjoying the attempts to cast aspersions on me because of it, especially as those people doing it are going to look even more stupid when the class lists are put online in the university reporter and they find out what class I actually got.
    Oh, are they still up now? Or do the class lists get taken down after a while?
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    Is there a summer of maths thread?
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    (Original post by Eisenstein)
    Is there a summer of maths thread?
    Nope, you should make one
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    (Original post by Mladenov)
    Problem 74**

    Let a,b,c be positive real numbers. Then we have the following inequality: \displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}} + \frac{2(ab+bc+ca)}{(a+b)(b+c)(c+  a)} \geq \frac{3(a+b+c)}{2(ab+bc+ca)}.
    Spoiler:
    Show
    It seems more difficult, than it actually is.
    Was just browsing along, and saw that this was unsolved...

    Solution 74**
    Algebra bash:
    Multiply out by  2(a+b)^2(b+c)^2(c+a)^2(ab+bc+ca)
    *Page of algebra*
    We'll use bracket notation here so that  [a_1,a_2,a_3]=\sum_{sym}x^{a_1}y^{a_2}z^{a_3}

    End up with a degree 7 inequality that needs proving:
     2[6,1,0]+[5,2,0]+2[5,1,1]+[3,2,2]\geq 3[4,3,0]+[4,2,1]+2[3,3,1]
    By Muirhead's,  2[6,1,0]+[5,2,0]\geq 3[4,3,0] and  [5,1,1]\geq [4,2,1]
    reducing the inequality to  [5,1,1]+[3,2,2] \geq 2[3,3,1]

    By Schur's,  x^4+y^4+z^4+x^2yz+xy^2z+xyz^2 \geq x^3y+xy^3+y^3z+yz^3+z^3x+zx^3
    Setting,
     x=a^{\frac{5}{4}}b^{\frac{1}{4}}  c^{\frac{1}{4}}
     y=a^{\frac{1}{4}}b^{\frac{5}{4}}  c^{\frac{1}{4}}
     z=a^{\frac{1}{4}}b^{\frac{1}{4}}  c^{\frac{5}{4}}
    , we arrive at
     a^5bc+ab^5c+abc^5+a^3b^2c^2+a^2b  ^3c^2+a^2b^2c^3 = \frac{1}{2} [5,1,1] + \frac{1}{2}[3,2,2] \geq [4,2,1] \geq [3,3,1]
    by Muirhead's as required.
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    Where is und? He's been slacking with the OP.
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    (Original post by DJMayes)
    ...
    Hey. Is there anyway we can start re-cataloging the problems and solutions on the first page without und (I noticed he hasn't been here for a while)? Or did we give up on that after problem 256?
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    (Original post by Star-girl)
    Where is und? He's been slacking with the OP.
    I haven't seen that guy post in years lol.
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    (Original post by Star-girl)
    Hey. Is there anyway we can start re-cataloging the problems and solutions on the first page without und (I noticed he hasn't been here for a while)? Or did we give up on that after problem 256?
    This is nothing to do with me whatsoever.
    • Thread Starter
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    (Original post by Star-girl)
    Where is und? He's been slacking with the OP.
    :hi:

    The original method of cataloguing the problems and solutions was very time consuming. Someone at some point wrote a script to do this automatically but it needed improvement and I am not sure what happened with that.

    Even if we were to collectively update the OP, which in theory shouldn't be too time consuming, bear in mind there is a strict character limit for posts, which does not play well with all those links. The first post is already out of space.

    Anyway I hope you are having a nice summer!
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    http://www.bmoc.maths.org/home/bmo1-2010.pdf

    For q5 is one solution to the functional equation f(x)=x ?
    I hope it's the only one or it's 3 pages for nothin' lol :/

    ( I put the solution back into the equation and using the properties of the eqn- I got it back - so surely it's true?)
    Ty
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    (Original post by MathMeister)
    http://www.bmoc.maths.org/home/bmo1-2010.pdf

    For q5 is one solution to the functional equation f(x)=x ?
    I hope it's the only one or it's 3 pages for nothin' lol :/

    ( I put the solution back into the equation and using the properties of the eqn- I got it back - so surely it's true?)
    Ty
    f(x)=x is not a solution. You probably put it back in wrong, cos you should get xy=x+y+xy => x+y=0, contradiction. Oh, and there is more than 1 solution, gl.
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    (Original post by Renzhi10122)
    f(x)=x is not a solution. You probably put it back in wrong, cos you should get xy=x+y+xy => x+y=0, contradiction. Oh, and there is more than 1 solution, gl.
    Huh ;/ Darnit i must try again lol

    http://www.bmoc.maths.org/home/bmo1-2011.pdf - q2 - is (s,x) = (10,5) a solution?
    Is there more than 1 solution? Cuz I tried putting in x values at the end and none others factorised.
    ( I used the fact that y-6 = (y-2) -4 to factorise in the first place and then it became trivial ..
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    (Original post by MathMeister)
    Huh ;/ Darnit i must try again lol

    http://www.bmoc.maths.org/home/bmo1-2011.pdf - q2 - is (s,x) = (10,5) a solution?
    Is there more than 1 solution? Cuz I tried putting in x values at the end and none others factorised.
    ( I used the fact that y-6 = (y-2) -4 to factorise in the first place and then it became trivial ..
    Spoiler:
    Show

    I first wrote an equation for the volume=surface area then factorised (a factor of s+x will cancel) and then we have a linear equation for x, which is easily solvable for the few cases we have. If you need more help, I'll write up a solution for you. And yes, (s,x)=(10,5) is a solution. There is more than 1 solution.
    EDIT: Actually, it's easier to just use a quadratic for s instead of a linear for x, because we can just use the discrimant, which must be a square since s must be integer. Also, having actually worked through it properly, there is only 1 solution, you've found them all
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    (Original post by Renzhi10122)
    Spoiler:
    Show

    I first wrote an equation for the volume=surface area then factorised (a factor of s+x will cancel) and then we have a linear equation for x, which is easily solvable for the few cases we have. If you need more help, I'll write up a solution for you. And yes, (s,x)=(10,5) is a solution. There is more than 1 solution.
    EDIT: Actually, it's easier to just use a quadratic for s instead of a linear for x, because we can just use the discrimant, which must be a square since s must be integer. Also, having actually worked through it properly, there is only 1 solution, you've found them all

    Cool I guess 30 seconds is my new personal record for an Olympiad question xD
    Ty
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    i don't know if this will be absolutely trivial to some people but oh well.

    Problem 498**

    Prove that if a sequence of complex numbers  z_n converges to A, then \displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{n}(z_1 + z_2 + ... + z_n) = A.
 
 
 
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