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    many of the older BMO2 questions that were set before 2008 are quite managable for me, but i feel like the newer ones have had a sharp spike in difficulty

    I'm wondering where can I find bmo2 solutions? or maybe someone should make a bmo solution bank thread, idk.
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    (Original post by imsoanonymous123)
    many of the older BMO2 questions that were set before 2008 are quite managable for me, but i feel like the newer ones have had a sharp spike in difficulty

    I'm wondering where can I find bmo2 solutions? or maybe someone should make a bmo solution bank thread, idk.
    They are in the yearbooks that UKMT send out each year to schools. Ask your teachers.
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    just want to confirm my solutions are okay

    BMO2 2005, Q1
    http://www.bmoc.maths.org/home/bmo2-2005.pdf
    Spoiler:
    Show
    1/x + 1/y = 1/N can be rearranged to
    (x-N)(y-N) = N^2
    Both brackets are integers that can take any value, so the number of pairs of values x can take is equal to the number of proper factors of N^2, and there will be 1 corresponding y value for each x.

    Now we write N^2 as the product of its prime factors 2^a * 3^b * 5^c * 7^d * 11^e and so on.
    now the number of proper factors N^2 has is (a+1)(b+1)(c+1)(d+1) ...
    but N^2 is square so all the factors are even, so instead we write this as
    (2a+1)(2b+1)(2c+1)(2d+1)...
    Now 2005 = 5*401 as its prime factorization.
    This means that either one bracket is 5, and the other is 401 with all other brackets being 1, or one bracket is 2005 and all the rest are one.

    In both cases, we will get all of a, b, c, d, e... to be even since each bracket is of the form (4k+1), which implies that n^2 is a fourth power of an integer. This implies that n is a square.
    Q3

    Spoiler:
    Show

    Start with x^2 - x - 1 + 1/x > 0 for x >0 (Proof: Multiply across by x and factorize the resulting cubic then sketch it)
    Therefore y^2 - y - 1 + 1/y > 0
    and
    z^2 - z - 1 + 1/z > 0

    for y and z > 0

    Make the substitution x = ab, y = b/c and z = c/a, then collect terms so that terms on both sides of the inequality are positive. Add 2(a/c + b/a + c/b) to both sides.

    Now by expanding the brackets in the question, we see that
    Left bracket in question > our left bracket > answer in right bracket (which is the same as the right bracket in our answer)

    and we are done


    BMO2 2010, Q1

    http://www.bmoc.maths.org/home/bmo2-2010.pdf
    Spoiler:
    Show
    Answer = no
    Consider that some child has 3 friends, each of which have exactly 3 friends that are not friends with any of the children already mentioned, each of which also have 3 friends which are not with the children already mentioned and so on.

    Given the starting child, the number of possible positions these children can occupy is 2011 positions to the left/right from the starting child, so the leftmost position of a child is 2011 spots to the left of the starting child, and the rightmost position is 2011 spots to the right, giving a total of 4022 spots.

    When you consider the friends that these children have, they can occupy 2011 spots (positions on the line) to the left/right of that child, giving another 4022 spots that they can occupy, for a total of 8044 spots. But now there are 9 friends of friends that you have to fit into the 8044 spots.

    If you consider the friends of friends of friends and so on, each new 'layer' of friends introduces 3^k friends but only 4022 new fixed spots that could possibly contain them. Eventually, long before you reach a number as big as 2010^2010, the 3^k that you must add exceeds the 4022k spots you have, and therefore you do not have enough room on the line for them.
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    (Original post by imsoanonymous123)
    just want to confirm my solutions are okay

    BMO2 2005, Q1
    http://www.bmoc.maths.org/home/bmo2-2005.pdf
    Spoiler:
    Show
    1/x + 1/y = 1/N can be rearranged to
    (x-N)(y-N) = N^2
    Both brackets are integers that can take any value, so the number of pairs of values x can take is equal to the number of proper factors of N^2, and there will be 1 corresponding y value for each x.

    Now we write N^2 as the product of its prime factors 2^a * 3^b * 5^c * 7^d * 11^e and so on.
    now the number of proper factors N^2 has is (a+1)(b+1)(c+1)(d+1) ...
    but N^2 is square so all the factors are even, so instead we write this as
    (2a+1)(2b+1)(2c+1)(2d+1)...
    Now 2005 = 5*401 as its prime factorization.
    This means that either one bracket is 5, and the other is 401 with all other brackets being 1, or one bracket is 2005 and all the rest are one.

    In both cases, we will get all of a, b, c, d, e... to be even since each bracket is of the form (4k+1), which implies that n^2 is a fourth power of an integer. This implies that n is a square.
    Q3
    Spoiler:
    Show

    Start with x^2 - x - 1 + 1/x > 0 for x >0 (Proof: Multiply across by x and factorize the resulting cubic then sketch it)
    Therefore y^2 - y - 1 + 1/y > 0
    and
    z^2 - z - 1 + 1/z > 0

    for y and z > 0

    Make the substitution x = ab, y = b/c and z = c/a, then collect terms so that terms on both sides of the inequality are positive. Add 2(a/c + b/a + c/b) to both sides.

    Now by expanding the brackets in the question, we see that
    Left bracket in question > our left bracket > answer in right bracket (which is the same as the right bracket in our answer)

    and we are done

    BMO2 2010, Q1

    http://www.bmoc.maths.org/home/bmo2-2010.pdf
    Spoiler:
    Show
    Answer = no
    Consider that some child has 3 friends, each of which have exactly 3 friends that are not friends with any of the children already mentioned, each of which also have 3 friends which are not with the children already mentioned and so on.

    Given the starting child, the number of possible positions these children can occupy is 2011 positions to the left/right from the starting child, so the leftmost position of a child is 2011 spots to the left of the starting child, and the rightmost position is 2011 spots to the right, giving a total of 4022 spots.

    When you consider the friends that these children have, they can occupy 2011 spots (positions on the line) to the left/right of that child, giving another 4022 spots that they can occupy, for a total of 8044 spots. But now there are 9 friends of friends that you have to fit into the 8044 spots.

    If you consider the friends of friends of friends and so on, each new 'layer' of friends introduces 3^k friends but only 4022 new fixed spots that could possibly contain them. Eventually, long before you reach a number as big as 2010^2010, the 3^k that you must add exceeds the 4022k spots you have, and therefore you do not have enough room on the line for them.
    Perfect.
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    adding new ones as I find the time to do them. Just picking off the most easy looking of the bmo2 questions now...

    2013/2014

    http://www.bmoc.maths.org/home/bmo2-2014.pdf

    Q2
    Spoiler:
    Show

    Assume that some cuboid with side lengths x, y, z exist that satisfy the condition.
    For the condition to be met, we must have xyz = 2(xy +xz + yz) = 4(x + y + z)
    We don't want a degenerate cuboid, so we assume that x, y and z > 0.

    Start by rearranging: We can divide by x, y, z since none of them equal zero.

    xyz = 2(xy + xz + yz)
    1/2 = (xy + xz + yz)/xyz = 1/x + 1/y +1/z

    and

    xyz = 4 (x + y + z)
    so (x + y + z)/xyz = 1/4 = 1/xy + 1/yz + 1/xz

    From here we proceed:

    1/xy + 1/yz + 1/xz = 1/4 = (1/2)^2 = (1/x + 1/y +1/z)^2 = (1/x)^2 + (1/y)^2 + (1/z)^2 + 2(1/xy + 1/yz + 1/xz)

    Equating the two ends, we get
    (1/x)^2 + (1/y)^2 + (1/z)^2 + 1/xy + 1/yz + 1/xz = 0
    (1/x)^2 + (1/y)^2 + (1/z)^2 + 1/4 = 0
    (1/x)^2 + (1/y)^2 + (1/z)^2 = -1/4

    Which is a contradiction, because all the terms on the LHS are positive.
    Hence the cuboid cannot exist.
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    Also any tips in actually qualifying for BMO? when I did SMC last year I only got like 80 or something
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    (Original post by imsoanonymous123)
    adding new ones as I find the time to do them. Just picking off the most easy looking of the bmo2 questions now...

    2013/2014

    http://www.bmoc.maths.org/home/bmo2-2014.pdf

    Q2
    Spoiler:
    Show

    Assume that some cuboid with side lengths x, y, z exist that satisfy the condition.
    For the condition to be met, we must have xyz = 2(xy +xz + yz) = 4(x + y + z)
    We don't want a degenerate cuboid, so we assume that x, y and z > 0.

    Start by rearranging: We can divide by x, y, z since none of them equal zero.

    xyz = 2(xy + xz + yz)
    1/2 = (xy + xz + yz)/xyz = 1/x + 1/y +1/z

    and

    xyz = 4 (x + y + z)
    so (x + y + z)/xyz = 1/4 = 1/xy + 1/yz + 1/xz

    From here we proceed:

    1/xy + 1/yz + 1/xz = 1/4 = (1/2)^2 = (1/x + 1/y +1/z)^2 = (1/x)^2 + (1/y)^2 + (1/z)^2 + 2(1/xy + 1/yz + 1/xz)

    Equating the two ends, we get
    (1/x)^2 + (1/y)^2 + (1/z)^2 + 1/xy + 1/yz + 1/xz = 0
    (1/x)^2 + (1/y)^2 + (1/z)^2 + 1/4 = 0
    (1/x)^2 + (1/y)^2 + (1/z)^2 = -1/4

    Which is a contradiction, because all the terms on the LHS are positive.
    Hence the cuboid cannot exist.
    Yes, let me give my solution (which I think is slightly more elegant (your division by xyz was not needed)). It's actually exactly the same method.

    Spoiler:
    Show
    We have  x,y,z>0 and assume a cuboid exists that satisfies the conditions of the question. Then, x,y,z satisfy
     xyz=2(xy+yz+zx)=4(x+y+z)

    Then,
     4(x+y+z)xyz=(2(xy+yz+zx))^2=4(x^  2y^2+y^2z^2+z^2x^2+2(x^2yz+xy^2z  +xyz^2))

\Rightarrow x^2yz+xy^2z+xyz^2=x^2y^2+y^2z^2+  z^2x^2+2(x^2yz+xy^2z+xyz^2)

\Rightarrow x^2y^2+y^2z^2+z^2x^2+x^2yz+xy^2z  +xyz^2=0
    which gives us our contradiction, since  x,y,z>0
    Notce that my last line is exactly the same as yours, but multiplied by (xyz)^2, and it's fairly clear where this comes from.
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    (Original post by imsoanonymous123)
    Also any tips in actually qualifying for BMO? when I did SMC last year I only got like 80 or something
    What are you getting now? A year is a long time ago. (btw, did you solve that bmo2 q about bisectors and the 120 degree angle)
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    (Original post by imsoanonymous123)
    many of the older BMO2 questions that were set before 2008 are quite managable for me, but i feel like the newer ones have had a sharp spike in difficulty

    I'm wondering where can I find bmo2 solutions? or maybe someone should make a bmo solution bank thread, idk.
    I think we thought of this earlier but was scrapped but the head of olympiads would not be too keen of the idea as then people would just look at the solutions straight away.


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    (Original post by Renzhi10122)
    What are you getting now? A year is a long time ago. (btw, did you solve that bmo2 q about bisectors and the 120 degree angle)
    haha no
    Geometry is my weak point... I gave up after about 4 hours on that Q
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    (Original post by imsoanonymous123)
    haha no
    Geometry is my weak point... I gave up after about 4 hours on that Q
    Lol, fair enough. Do you know the angle bisector theorem? It will help immensely.
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    (Original post by Renzhi10122)
    Lol, fair enough. Do you know the angle bisector theorem? It will help immensely.
    Do you know how to prove the general sector theorem. Ie bisecting angles into any ratio/split.


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    (Original post by physicsmaths)
    Do you know how to prove the general sector theorem. Ie bisecting angles into any ratio/split.


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    Well I don't even know it lol. I assume it's something like, for a triangle ABC, let D be a point on side BC. Then  \dfrac{AB}{AC}=\dfrac{BDsinDAC}{  CDsinDAB}
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    (Original post by Renzhi10122)
    Well I don't even know it lol. I assume it's something like, for a triangle ABC, let D be a point on side BC. Then  \dfrac{AB}{AC}=\dfrac{BDsinDAC}{  CDsinDAB}
    Yh i think it is..


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    (Original post by physicsmaths)
    Yh i think it is..


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    Yeah, it follows quite easily from the proof of angle bisector with the sine rule
 
 
 
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