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    (Original post by Renzhi10122)
    Nicely done.
    Speaking of winning strategies, any luck with the painting problem I posted earlier? It looks like it has not been solved yet.
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    (Original post by Lord of the Flies)
    Speaking of winning strategies, any luck with the painting problem I posted earlier? It looks like it has not been solved yet.
    Nope, I'll give it another try now.

    For reference,
    http://www.thestudentroom.co.uk/show...3#post56356303
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    (Original post by atsruser)
    Problem 559

    I put this up in a separate thread with little response so here it is again. Not very difficult - maybe something that may appear on a MAT paper.


    Consider the point A(k,e^{-k}) for large k >0.

    Find the value of m < k for which the point B(m, e^{-m}) is approximately k\sqrt{2} distant from A.
    Hmm, no takers for this, a mere bagatelle?

    Hint:
    Spoiler:
    Show
    For large k >0, A \approx A(k,0), so A lies on the x-axis, more or less. Clearly, to get far enough from A on the graph in question, we'll have to consider m<0. What does the the graph of y=e^{-x} look like for values of m <0 that take us k\sqrt{2} from A?
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    (Original post by atsruser)
    Problem 558 (*)

    Solve:

    \sqrt[3]{xy^2} = \frac{91}{6}-\frac{x}{3}

    \sqrt[3]{x^2y} = \frac{37}{6}-\frac{y}{3}
    Hint:

    Spoiler:
    Show
    We clearly want to get rid of the cube root, but if we just cube things up, the numbers become unpleasantly large. Instead, consider a substitution.
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    (Original post by atsruser)
    Hint:
    Spoiler:
    Show
    We clearly want to get rid of the cube root, but if we just cube things up, the numbers become unpleasantly large. Instead, consider a substitution.
    If no-one's done this already...

    Solution 558
    Spoiler:
    Show

    Let x = u^3, y = v^3, these substitutions give us:
    uv^2 = \frac{91}{6} - \frac{u^3}{3} and
    u^2v = \frac{37}{6} - \frac{v^3}{3}

    Adding them and simplifying produces:
    u^3 + 3u^2v + 3uv^2 + v^3 = 64

    The LHS simplifies:
    (u+v)^3 = 64
     u+v = 4

    Substituting v = 4 - u into the first equation and simplifying gives you a nice cubic equation in u:
    (2u - 4)^3 = 27
    u = \frac{7}{2}, v = \frac{1}{2}

    Thus, from our original substitutions,
    x = \frac{343}{8}, y = \frac{1}{8}

    which follow from atsruser's critical observation that the cube of 8 is not 64.
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    (Original post by StrangeBanana)
    If no-one's done this already...

    Solution 558
    Spoiler:
    Show

    Let x = u^3, y = v^3, these substitutions give us:
    uv^2 = \frac{91}{6} - \frac{u^3}{3} and
    u^2v = \frac{37}{6} - \frac{v^3}{3}

    Adding them and simplifying produces:
    u^3 + 3u^2v + 3uv^2 + v^3 = 64

    The LHS simplifies:
    (u+v)^3 = 64
     u+v = 8

    Substituting v = 8 - u into the first equation and simplifying gives you a nice cubic equation in u:
    (2u - 8)^3 = -421
    u = 4 - \frac{\sqrt[3]{421}}{2}
    v = 4 + \frac{\sqrt[3]{421}}{2}

    Thus, from our original substitutions,
    x = (4 - \frac{\sqrt[3]{421}}{2})^3, y = (4 + \frac{\sqrt[3]{421}}{2})^3

    which are nicely symmetrical. ^^
    Right idea, but sadly (u+v)^3 = 64 does not imply that  u+v = 8
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    (Original post by atsruser)
    Right idea, but sadly (u+v)^3 = 64 does not imply that  u+v = 8
    what is life

    i can't even
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    (Original post by joostan)
    Solution 546:

    I_n=\displaystyle\int_0^{\frac{ \pi}{2}} \dfrac{1}{1+\tan^n(x)} \ dx = \displaystyle\int_0^{\frac{\pi}{  2}} \dfrac{\cos^n(x)}{\sin^n(x)+\cos  ^n(x)} \ dx
    Sending x \mapsto \dfrac{\pi}{2}-x yields:
    I_n=\displaystyle\int_0^{\frac{\  pi}{2}} \dfrac{\sin^n(x)}{\sin^n(x)+\cos  ^n(x)} \ dx

\Rightarrow 2I_n=\displaystyle\int_0^{\frac{  \pi}{2}} 1 \ dx

\Rightarrow I_n=\dfrac{\pi}{4}.

    Oh snap, 2nd place .
    May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?
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    (Original post by ETbuymilkandeggs)
    May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?
    When do you cover substitution as an integration method? It's either C3 or C4.
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    (Original post by ETbuymilkandeggs)
    May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?
    This is C4. With some ingenuity required.


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    In an infinite sequence of integers, a word is a block of consecutive terms. A triplet is a block consisting of three consecutive, identical words.

    Problem 560

    Can a binary sequence be triplet-free?
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    (Original post by EmptyMathsBox)
    ...
    Nice. If you are interested a neat alternate solution (not my own) proceeds as follows: Divide through by 2^x and observe that  x = 2 is a solution. Call the LHS f(x), then since  \dfrac{\sqrt{2-\sqrt{3}}}{2} < 1 and  \dfrac{\sqrt{2+\sqrt{3}}}{2} < 1 , f is decreasing, so  x = 2 is the only solution.
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    (Original post by StrangeBanana)
    If no-one's done this already...

    Solution 558
    Spoiler:
    Show

    Let x = u^3, y = v^3, these substitutions give us:
    uv^2 = \frac{91}{6} - \frac{u^3}{3} and
    u^2v = \frac{37}{6} - \frac{v^3}{3}

    Adding them and simplifying produces:
    u^3 + 3u^2v + 3uv^2 + v^3 = 64

    The LHS simplifies:
    (u+v)^3 = 64
     u+v = 8

    Substituting v = 8 - u into the first equation and simplifying gives you a nice cubic equation in u:
    (2u - 8)^3 = -421
    u = 4 - \frac{\sqrt[3]{421}}{2}
    v = 4 + \frac{\sqrt[3]{421}}{2}

    Thus, from our original substitutions,
    x = (4 - \frac{\sqrt[3]{421}}{2})^3, y = (4 + \frac{\sqrt[3]{421}}{2})^3

    which are nicely symmetrical. ^^
    Nicely done
    I have a very similar in one of my hard papers.
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    (Original post by TeeEm)
    Nicely done
    I have a very similar in one of my hard papers.
    Mistake from u+v=8


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    (Original post by physicsmaths)
    Mistake from u+v=8


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    The numbers turn out to be nice and easy sans mistake.
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    (Original post by physicsmaths)
    Mistake from u+v=8
    I think I'd better edit this already :oops:
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    (Original post by StrangeBanana)
    I think I'd better edit this already :oops:
    get on it banana boy
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    This is one of my all time favourite questions. A brilliant example of "it's not obvious if you can't prove it".

    Problem 561 [from a IA example sheet - if you understand the question, you know enough to solve it]

    Take k points in the plane, not all collinear. Show that there is a line going through exactly two of them.
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    (Original post by metaltron)
    Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?
    Hi metaltron, please could I have a hint on how to factorise the polynomial in this way? Is there a well known technique I need to use or just look a bit closer?
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    (Original post by Lord of the Flies)
    This is one of my all time favourite questions. A brilliant example of "it's not obvious if you can't prove it".

    Problem 561 [from a IA example sheet - if you understand the question, you know enough to solve it]

    Take k points in the plane, not all collinear. Show that there is a line going through exactly two of them.
    I am sad that there doesn't seem to be a nice proof of this
 
 
 
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