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    (Original post by ukdragon37)
    Problem 71**

    Find all x,y \in \mathbb{C} such that x^y = y^x.
    Initial thoughts:
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    \ y^x = x^y \Rightarrow \dfrac{ln(x)}{x} = \dfrac{ln(y)}{y} \Rightarrow f(x) = f(y)

\mathrm{\ Clearly \ solutions \ exist \ when} x = y
    By assuming x and y are different values then we see that there must be a straight line through the function. By inspection of the function we see that  1 \leq x \leq e and \ x \geq e The obvious solution to this is 2 and 4. As for the complex parts other then that \f(x_1) = f(z_1) = f(y_1) I'm at a loss for the night
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    (Original post by metaltron)
    I quite like it The only real disadvantage is trying to get a compass that doesn't screw up your circle.
    :eek::eek::eek: Seems like you are alone in this
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    (Original post by joostan)
    :eek::eek::eek: Seems like you are alone in this
    I know, I'm so ashamed of myself
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    (Original post by joostan)
    Initial thoughts:
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    \ y^x = x^y \Rightarrow \dfrac{ln(x)}{x} = \dfrac{ln(y)}{y} \Rightarrow f(x) = f(y)

\mathrm{\ Clearly \ solutions \ exist \ when} x = y
    By assuming x and y are different values then we see that there must be a straight line through the function. By inspection of the function we see that  1 \leq x \leq e and \ x \geq e The obvious solution to this is 2 and 4. As for the complex parts other then that \f(x_1) = f(z_1) = f(y_1) I'm at a loss for the night
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    A slightly less obvious solution is \left(3\sqrt{3}\right)^{\sqrt{3}  }=\left(\sqrt{3}\right)^{3\sqrt{  3}}
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    (Original post by ukdragon37)
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    A slightly less obvious solution is \left(3\sqrt{3}\right)^{\sqrt{3}  }=\left(\sqrt{3}\right)^{3\sqrt{  3}}
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    IRC aren't there an infinite number of solutions.
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    (Original post by bananarama2)
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    IRC aren't there an infinite number of solutions.
    This suggests otherwise

    http://www.wolframalpha.com/input/?i=x^y-y^x%3D0

    and thats only the real numbers.
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    (Original post by bananarama2)
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    IRC aren't there an infinite number of solutions.
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    There are. I'm asking for a way to generate them. I.e. define the pair of functions x,y : S \to \mathbb{C} such that x(a)^{y(a)} = y(a)^{x(a)} for all a of some set S (for example, \mathbb{R} or \mathbb{C}), and the functions must be able to generate all the possible solutions in \mathbb{C}

    Also you shouldn't have to use the W-function as solved by wolfram.
    .
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    (Original post by ukdragon37)
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    There are. I'm asking for a way to generate them. I.e. define the pair of functions x,y : S \to \mathbb{C} such that x(a)^{y(a)} = y(a)^{x(a)} for all a of some set S (for example, \mathbb{R} or \mathbb{C}), and the functions must be able to generate all the possible solutions in \mathbb{C}

    Also you shouldn't have to use the W-function as solved by wolfram.
    .
    I'm fairly sure no such functions exist.

    Not ones that aren't multivalued anyway.

    EDIT: Ignore that, I think I'm onto something.
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    (Original post by james22)
    I'm fairly sure no such functions exist.

    Not ones that aren't multivalued anyway.
    Well I use "functions" in a loose sense - more like I am looking for a parametric "method" to generate the solutions. Hence I would accept multivalued functions along with an appropriate explanation of how their branches relate to the solutions they generate.
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    Solution 71

    Not sure if this is 100% correct but it seems ok

    Let x=ay

    we then get (ay)^y=y^{ay}

    rearanging gives a^y=(y^y)^{a-1}

    so

    yln(a)=(a-1)ln(y^y)=(a-1)yln(y)

    so

    y=a^{\frac{1}{a-1}}
    and
    x=a^{\frac{a}{a-1}}

    This should only fail if x or y are 0, but that cannot give a solution unless we accept x=y=0 (which would be given by a=0 anyway).
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    (Original post by james22)
    Solution 71
    ...
    Yep I'll give you that, that is the (rather elegant I think) solution I had in mind, and of course you can always ad in x = y as the set of trivial solutions.

    To be more rigorous (although possibly going beyond A-level) one could think about why the proof works for complex x and y (since you are taking logs etc.). Also, consider what value does a have to be to get the solution (-2, -4)?
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    (Original post by ukdragon37)
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    There are. I'm asking for a way to generate them. I.e. define the pair of functions x,y : S \to \mathbb{C} such that x(a)^{y(a)} = y(a)^{x(a)} for all a of some set S (for example, \mathbb{R} or \mathbb{C}), and the functions must be able to generate all the possible solutions in \mathbb{C}

    Also you shouldn't have to use the W-function as solved by wolfram.
    .
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    W function? I've just seen the problem before On one of the STEP threads I think.
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    (Original post by ben-smith)
    ...
    My profound compliments!
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    Do you not relish doing complex analysis? It is rather straightforward!


    (Original post by shamika)

    Make sure you have a look at STEP and MAT, assuming you apply to Oxford / Cambridge / Imperial / Warwick.

    Also, you'll be pleased to know that there is no Euclidean plane geometry of the kind that's useful for IMO at university!
    Ceva, Menelaus, Lehmus, Brianchon, Pappus, Feuerbach - I am scared!
    On one hand, projective geometry is somewhat interesting; on the other hand, barycentric coordinates and complex numbers, for example, are such a computational disaster.
    Not to mention the problems where we have to deal with hundreds of circles..
    Spoiler:
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    I have looked at some STEP papers. The problems seem quite accessible; one surprising thing is that help is given for a considerable number of questions.
    I would like to know what is your opinion - are there any substantial differences between the colleges within, for example, the University of Cambridge (I mean academic differences)?


    (Original post by metaltron)
    I quite like it
    I assume you relish doing pure geometry. Here you are:

    Let k be the circumscribed circle of the triangle ABC. D is an arbitrary point on the segment AB. Let I and J be the centers of the circles which are tangent to the side AB, the segment CD and the circle k. Assume that A, B, I, and J are concyclic. The excircle of the triangle ABC is tangent to the side AB in the point M. Then M \equiv D.

    Note that Thebault's theorem was not allowed without proof.
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    (Original post by Mladenov)
    My profound compliments!
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    Do you not relish doing complex analysis? It is rather straightforward!




    Ceva, Menelaus, Lehmus, Brianchon, Pappus, Feuerbach - I am scared!
    On one hand, projective geometry is somewhat interesting; on the other hand, barycentric coordinates and complex numbers, for example, are such a computational disaster.
    Not to mention the problems where we have to deal with hundreds of circles..
    Spoiler:
    Show
    I have looked at some STEP papers. The problems seem quite accessible; one surprising thing is that help is given for a considerable number of questions.
    I would like to know what is your opinion - are there any substantial differences between the colleges within, for example, the University of Cambridge (I mean academic differences)?




    I assume you relish doing pure geometry. Here you are:

    Let k be the circumscribed circle of the triangle ABC. D is an arbitrary point on the segment AB. Let I and J be the centers of the circles which are tangent to the side AB, the segment CD and the circle k. Assume that A, B, I, and J are concyclic. The excircle of the triangle ABC is tangent to the side AB in the point M. Then M \equiv D.

    Note that Thebault's theorem was not allowed without proof.
    Btw when I said I liked geometry; that doesn't necessarily mean I'm a whizz at it!
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    (Original post by Mladenov)
    I have looked at some STEP papers. The problems seem quite accessible;
    For the love of god you have to apply. That's just :O
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    (Original post by metaltron)
    Btw when I said I liked geometry; that doesn't necessarily mean I'm a whizz at it!
    Hey, do not take it rather seriously! It is for fun; hence if you enjoy pondering upon geometry problems, I feel that this is one awesome example. I have good mementos regarding this problem.
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    I failed on it three years ago


    (Original post by bananarama2)
    For the love of god you have to apply. That's just :O
    I meant pure mathematics, and not mechanics and the rest applied stuff.


    Let us stop the off-topic and stick to problem solving.
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    (Original post by james22)
    Solution 71

    Not sure if this is 100% correct but it seems ok

    Let x=ay

    we then get (ay)^y=y^{ay}

    rearanging gives a^y=(y^y)^{a-1}

    so

    yln(a)=(a-1)ln(y^y)=(a-1)yln(y)

    so

    y=a^{\frac{1}{a-1}}
    and
    x=a^{\frac{a}{a-1}}

    This should only fail if x or y are 0, but that cannot give a solution unless we accept x=y=0 (which would be given by a=0 anyway).
    Neat
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    PS Helper
    (Original post by Mladenov)
    I have looked at some STEP papers. The problems seem quite accessible; one surprising thing is that help is given for a considerable number of questions.
    I would like to know what is your opinion - are there any substantial differences between the colleges within, for example, the University of Cambridge (I mean academic differences)?
    You should apply then
    Spoiler:
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    In terms of academic performance, the difference between the colleges is small, with the Tompkins score of the middle ~15-20 colleges varying only by around 5%.
    This is mostly due to the cohort - you'll see that the more 'competitive' colleges get to take on better students, and so with all other things being equal, they do better. Trinity will be able to take on better students as they get better direct applicants, for example.
    In your case, just pick the one you like most (I'm guessing Trinity, as you're an IMO Mathmo )
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    Problem 72: *

    Bit of an effort: - But hey ho.
    Prove by considering: \ I =\displaystyle\int^1_0 \dfrac{x^4(1 - x)^4}{1+x^2}\ dx that \pi < \frac{22}{7}

    Find \ I = \displaystyle\int \sqrt{1+e^x}\ dx
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    (Original post by joostan)
    Problem 72: *

    Bit of an effort: - But hey ho.
    Prove by considering: \ I =\displaystyle\int^1_0 \dfrac{x^4(1 - x^4)}{1+x^2}\ dx that \pi < \frac{22}{7}

    Find \ I = \displaystyle\int \sqrt{1+e^x}\ dx
    You mean \displaystyle \frac{x^{4}(1-x)^{4}}{1+x^{2}}, don't you?
 
 
 
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