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    (Original post by MAyman12)
    5(n2n) + 1
    Obviously not, in fact there is no single variable polynomial with that property.
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    (Original post by Lord of the Flies)
    Obviously not, in fact there is no single variable polynomial with that property.
    PRSOM (please rate some other mathematician).
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    For those who can do the first one in the blink of an eye (you know who you are), please leave it for others.

    Problem 258*


    Prove that the segments are perpendicular, no matter what the central quadrilateral is.

    Problem 259**


    Prove that the set of points at which the smaller circles touch all lie on a cirle (dashed).

    Problem 260***

    \displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx
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    I shall solve only 259.

    Solution 259

    We can suppose that the two large circles, call them C_{1} and C_{2}, touch each other at the origin (that is, they are perpendicular to \mathbb{R} at the origin); if this is not the case, we can apply suitable transformations.
    Now consider the map \displaystyle w= \frac{1}{z}. We clearly have \displaystyle C_{1} \stackrel{w=\frac{1}{z}}{\to}  L_{1} and \displaystyle C_{2} \stackrel{w=\frac{1}{z}}{\to} L_{2}, where L_{i} are lines parallel to i\mathbb{R}, not through the origin. Under this image, the small circles are transformed into congruent circles which touch each other and both L_{1} and L_{2}; this is in fact the case, since w maps the region bounded by C_{1} and C_{2} into the region between L_{1} and L_{2} (well-known theorem for the correspondence between the boundaries). The line, on which the centres of these circles lie, is parallel to L_{i} (obvious) and does not pass through the origin (otherwise, it will coincide with i\mathbb{R}, which is impossible). Hence, its image under w^{-1} is a circle through the origin which is orthogonal to \mathbb{R} and the proof is finished.
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    This is perhaps my last contribution this summer.

    Problem 261 ***

    The Brocard-Ramanujan diophantine equation is n! + 1 = m^2 where n,m \in \mathbb{N}.

    i) Find three pairs (n,m) which are solutions to the equation.

    ii) Prove that if p \equiv \pm 3\ (8) is a prime, then no solutions of the forms (p-2, m) and (p-3, m) exist.
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    (Original post by Lord of the Flies)
    Problem 260***

    \displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx
    Does this require anything too advanced (like complex analysis)? I'll give it a go but don't want to waste too long on something I will never solve.
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    (Original post by james22)
    Does this require anything too advanced (like complex analysis)? I'll give it a go but don't want to waste too long on something I will never solve.
    It does not require complex analysis. It is on the difficult side, though.
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    (Original post by Jkn)
    Problem 234*

    Find all possible n-tuples of reals x_1,x_2, \cdots ,x_n such that \prod_{i=1}^{n} x_i = 1 and \prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i = 1 for all 1 \le k \le n-1
    Solution 234

    For 1 \leq k \leq n-1 let u_k=\prod_{i=1}^{k} x_i

    1=\prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i=u_k-\frac{1}{u_k}



\Rightarrow



u_{k}^2-u_k-1=0



\Rightarrow



u_k=\frac{1 \pm \sqrt{5}}{2}



\Rightarrow



x_1=\frac{1 \pm \sqrt{5}}{2}

    We have 2 options for x_1, now assume we have found x_1,x_2,...,x_{m-1}

    Using the 2 possible values of u_k we get 3 options for x_m

    x_m=1, \frac{-3+\sqrt 5}{2}, \frac{1+\sqrt 5}{2}

    Where the last 2 are possible depending on the value of u_{m-1}

    Repeating this process we can get values for x_1, x_2, ..., x_{n-1}

    This is enough to determine x_n using the first equality so we are done.

    By making different choices at each stage in the process used to generate the numbers, we get every possible n-tuple.
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    (Original post by james22)
    For 1 \leq k \leq n-1 let u_k=\prod_{i=1}^{k} x_i

    1=\prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i=u_k-\frac{1}{u_k}



\Rightarrow



u_{k}^2-u_k-1=0



\Rightarrow



u_k=\frac{1 \pm \sqrt{5}}{2}



\Rightarrow



x_1=\frac{1 \pm \sqrt{5}}{2}

    We have 2 options for x_1, now assume we have found x_1,x_2,...,x_{m-1}

    Using the 2 possible values of u_k we get 3 options for x_m

    x_m=1, \frac{-3+\sqrt 5}{2}, \frac{1+\sqrt 5}{2}

    Where the last 2 are possible depending on the value of u_{m-1}

    Repeating this process we can get values for x_1, x_2, ..., x_{n-1}

    This is enough to determine x_n using the first equality so we are done.

    By making different choices at each stage in the process used to generate the numbers, we get every possible n-tuple.
    Looks good to me bro! (do write solution at the top!)

    I assumed this would never have a solution after "Solution 234 Really?!" was posted by a certain someone :lol:

    It's actually quite ironic given that I got this from the 4th round of the Bulgarian Mathematical Olympiad (which is like our BMO1/BMO2 - perhaps a little easier in places). It's actually a really good source of fun questions that don't follow the typical UK-comp or US-comp format - though it does, in some ways, have a format of it's own! I might try/post a few more actually... (I typed up like 10 back when I set that but the computer froze and deleted the typing that took me ages to do!)
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    (Original post by Jkn)
    Looks good to me bro! (do write solution at the top!)

    I assumed this would never have a solution after "Solution 234 Really?!" was posted by a certain someone :lol:

    It's actually quite ironic given that I got this from the 4th round of the Bulgarian Mathematical Olympiad (which is like our BMO1/BMO2 - perhaps a little easier in places). It's actually a really good source of fun questions that don't follow the typical UK-comp or US-comp format - though it does, in some ways, have a format of it's own! I might try/post a few more actually... (I typed up like 10 back when I set that but the computer froze and deleted the typing that took me ages to do!)
    I found it an enjoyable question to think about, although that comment made put me off a bit as I thought there was some really obvious solution I was missing.

    To be honest I'm suprised that there were any solutions at all.

    Just noticed that this is another place where the golden ratio appears.
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    (Original post by james22)
    I found it an enjoyable question to think about, although that comment made put me off a bit as I thought there was some really obvious solution I was missing.

    To be honest I'm suprised that there were any solutions at all.

    Just noticed that this is another place where the golden ratio appears.
    Mm, was really annoying that he did that. Though it is very easy to spot without the product notation

    Yes! That's what I liked about it! Complexity from a very simple/symmetrical question
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    (Original post by Lord of the Flies)
    It does not require complex analysis. It is on the difficult side, though.
    Edit: never mind, made a mistake T_T
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    (Original post by Lord of the Flies)
    Problem 260***

    \displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx
    Hmm, odd one - I've made what might be progress, but not very much:
    Spoiler:
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    The integral is the same as \int_{-\infty}^{\infty} \text{sinc}(2x) \dfrac{d}{dx}(\sin(\tan(x))) dx
    . I'll have another think.
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    (Original post by Smaug123)
    Hmm, odd one - I've made what might be progress, but not very much:
    Spoiler:
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    The integral is the same as \int_{-\infty}^{\infty} \text{sinc}(2x) \dfrac{d}{dx}(\sin(\tan(x))) dx
    . I'll have another think.
    Spoiler:
    Show
    I wondered about a kind of revers of the differentiation under the integral
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    (Original post by bananarama2)
    Spoiler:
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    I wondered about a kind of revers of the differentiation under the integral
    Yeah, the only thing I can think of to do in that area is integration by parts - but I don't think it leads anywhere simple.
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    Not sure if it's any simpler but I get that the integral in 260 is equal to

     \int_{-\infty}^{\infty} x^{-1}sec^{4}x sin(tan(x)) dx
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    (Original post by und)
    Solution 12

    First we choose the six colours, so there are n choose 6 possibilities =\frac{n!}{6!(n-6)!}. Without considering identical cases we have 6! possibilities for colouring the cube, but we divide by 4*6=24 where 6 is the number of possible anchor faces and 4 is due to rotation about the anchor face, so we get \frac{n!}{24(n-6)!}
    i dont understand this solution. why is it not just n!/ (n-6)!
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    Problem 262*/**/*** (can be done with only basic knowledge, but is far harder)

    You have a fair coin and start with a score of 0. Whenever you throw a heads you +1 to your score, whenever you get a tails you -1.

    i) What is the average number of times you will hit 0 if you throw the coin 10 times? What about 100?
    ii) What is the average number of flips before you reach 0 again?
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    i can get problem 260 down to \displaystyle \int_{-\infty}^ {\infty}\frac{sin(tan(x))}{x}dx

    but no further...
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    (Original post by Lord of the Flies)
    [SIZE=1]
    Problem 260***

    \displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx
    Using a few approximations, I get it to be close to \frac{\sqrt{21+6e+17\pi+6log(2)}  }{2\sqrt 6}

    Is this close at all? I know it isn't correct but it may be close (or not).
 
 
 
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