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    (Original post by Mladenov)
    Nope, I meant x^{3}y. From AM-GM we have \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, the second is also trivially true, and hence \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y. It is a consequence of AM-GM, yet not a direct result.
    Oh I see, I assumed you meant it was a direct result!
    Problem 76 is wrong in this form (I thought c \le a+b).

    Edit: We have to bound c; I can try to find the best possible c.
    Arghhh I'm so sorry! It was a typo I have corrected it now! (the first term is (a+b+c) !!!!
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    Solution 76

    We have (a+b+c)(a+b-c)(a+c-b)(b+c-a)=((a+b)^{2}-c^{2})(c^{2}-(a-b)^{2}).
    If c^{2}\ge (a+b)^{2} the inequality is clearly true.
    If (a-b)^{2} < c^{2} < (a+b)^{2}, AM-GM yields \displaystyle ((a+b)^{2}-c^{2})(c^{2}-(a-b)^{2}) \le \frac{(4ab)^{2}}{2^{2}}\le (a^{2}+b^{2})^{2}.
    If c^{2}\le (a-b)^{2} the inequality is again trivially true.
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    (Original post by Mladenov)
    Solution 76

    We have (a+b+c)(a+b-c)(a+c-b)(b+c-a)=((a+b)^{2}-c^{2})(c^{2}-(a-b)^{2}).
    If c^{2}\ge (a+b)^{2} the inequality is clearly true.
    If (a-b)^{2} < c^{2} < (a+b)^{2}, AM-GM yields \displaystyle ((a+b)^{2}-c^{2})(c^{2}-(a-b)^{2}) \le \frac{(4ab)^{2}}{2^{2}}\le (a^{2}+b^{2})^{2}.
    If c^{2}\le (a-b)^{2} the inequality is again trivially true.
    Nicely done. I attempted to see if there was a way without using AM-GM etc.. Simplify and shift all terms to one side of the equation and form a quadratic in  a^2+b^2. The roots turn out to be  (a^2+b^2)=c^2 +/- \sqrt{-(a^2-b^2)^2} which I concluded implied there were no real roots (is this correct?) and hence demonstrated that there exists a single a positive value of the expression.
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    (Original post by Mladenov)
    Problem 80

    Let (a_{n})_{n\ge1} be an increasing sequence of positive integers. Suppose also that \displaystyle \lim_{n\to \infty} \frac{a_{n}}{n}=0. Then, the sequence \displaystyle b_{n} = \frac{n}{a_{n}} contains all positive integers.
    If  a_{n} is an increasing sequence of positive integers, then surely it must be true that  a_{n} \geq n (as the smallest such sequence is 1, 2, 3, ...) and hence \displaystyle \lim_{n\to \infty} \frac{a_{n}}{n}=0 cannot hold in such a case. Have I missed something crucial?
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    (Original post by Jkn)
    ...
    Your idea would probably work, but if a=b, c^{2} \pm \sqrt{-(a-b)^{2}} is a real number, provided that c is real.

    (Original post by Jkn)
    ...
    My dear friend, why do you think that (a_{n})_{n\ge 1} is a strictly increasing sequence? I have never uttered this.
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    (Original post by Mladenov)
    Your idea would probably work, but if a=b, c^{2} \pm \sqrt{-(a-b)^{2}} is a real number, provided that c is real.
    (still nice to do things without standard inequality techniques)
    Solution 76 (2)

    RHS=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=((a+b)^{2}-c^{2})(c^{2}-(a-b)^{2})
    =c^2((a+b)^2+(a-b)^2)-(a^2-b^2)^2-c^4

    Let f(a,b,c)=LHS-RHS=(a^2+b^2)^2-2c^2(a^2+b^2)+(c^4+(a^2-b^2)^2)
    f(a,b,c) had roots when a^2+b^2=2c^{2} \pm \sqrt{-(a-b)^{2}}

    If a doesn't equal b the equation has no real roots and so a trivial numerical example shows that there exists a,b,c such that f(a,b,c) is positive and hence must be positive for all values (i.e. f(1,0,0)=2).

    In the case a=b, f(a,b,c)=(2a^2)^2+(c^2)^2-4a^2c^2=(2a^2-c^2)^2 \geq 0

    Hence true for all a,b,c.

    My dear friend, why do you think that (a_{n})_{n\ge 1} is a strictly increasing sequence? I have never uttered this.
    Right okay, it just seemed odd that 1,1,1 or 1,2,2 constitutes 'increasing' ;O

    Btw, how old are you and (if you're not at uni already) which one are you going to?
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    (Original post by Jkn)
    ...
    Yup, your solution is now fine.
    Edit: After I have scrutinized your solution, I would suggest you write c^{2}=d and consider the expression as a polynomial of degree 2 with respect to d. In this way, you will be able to elude from coefficients, which depend on the "variable".

    (Original post by Jkn)
    Right okay, it just seemed odd that 1,1,1 or 1,2,2 constitutes 'increasing' ;O
    If it was strictly increasing, the problem would be incorrect, as you found out.

    (Original post by Jkn)
    Btw, how old are you and (if you're not at uni already) which one are you going to?
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    (Original post by Mladenov)
    Yup, your solution is now fine.
    Edit: After I have scrutinized your solution, I would suggest you to write c^{2}=d and consider the expression as a polynomial of degree 2 with respect to d. In this way, you will be able to elude from coefficients, which depend on the "variable".
    Hmm I suppose that would be neater but I've given it some thought and I do believe the expression is valid (correct me if I'm wrong) because the quad. formula is being used merely as a rearrangement in this case. I suppose for clarification you could use the fact that the assumption that f(a,b,c)=0 leads to a contradiction (a not equaling b) and combining it with the fact that with the property that it is continuous.

    Oh right, don't you have to be 18 or under to participate in the IMO? I always really enjoyed 'competition'-style maths though, but didn't qualify for BMO2 this year (our equivalent 'last 40'-ish exam) which is annoying because, since I've learn AM-GM type stuff and loads of algebraic tricks I can now do loads of the BMO2 questions!

    I take it you are you not going to university this year then if you're still unsure (is it typical to start uni past 18/19 in your country?) I'm certain you'd get into Cambridge btw if that's where you wanted to go. Apply to Emmanuel (always 1st or 2nd in the tompkins table), that's where my offer is ;D
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    (Original post by Jkn)
    Hmm I suppose that would be neater but I've given it some thought and I do believe the expression is valid (correct me if I'm wrong) because the quad. formula is being used merely as a rearrangement in this case. I suppose for clarification you could use the fact that the assumption that f(a,b,c)=0 leads to a contradiction (a not equaling b) and combining it with the fact that with the property that it is continuous.
    Yup, the expression is valid in this case; I do not want to be punctual, I just said what I would do with your idea.

    (Original post by Jkn)
    Oh right, don't you have to be 18 or under to participate in the IMO? I always really enjoyed 'competition'-style maths though, but didn't qualify for BMO2 this year (our equivalent 'last 40'-ish exam) which is annoying because, since I've learn AM-GM type stuff and loads of algebraic tricks I can now do loads of the BMO2 questions!
    "The contestants must have been born less than twenty years before the day of the second contest paper".
    I have done many problems from BMO2, FST(First Selection Test), and NST(Next Selection Test). In my view, BMO2 is not too difficult; the letter tests are challenging though.
    For mathematical olympiads the theory is a must, but, I believe, problem solving is of much more import - it teaches you many interesting tricks, which are crucial.

    (Original post by Jkn)
    I take it you are you not going to university this year then if you're still unsure (is it typical to start uni past 18/19 in your country?) I'm certain you'd get into Cambridge btw if that's where you wanted to go. Apply to Emmanuel (always 1st or 2nd in the tompkins table), that's where my offer is ;D
    Yes, we typically finish high school at the age of 19. I consider University of Cambridge, and Lycée Louis le Grand \Rightarrow ENS (french is simply killing me ).
    By the way, congratulations on your offer!
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    (Original post by Jkn)
    ...
    You got into Emma :O So did I for a different subject
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    (Original post by Mladenov)
    Yup, the expression is valid in this case; I do not want to be punctual, I just said what I would do with your idea.
    Oh okay. That's what I've actually started to love about maths. Ever since I've broken free from curriculum maths and explored pure maths a bit more, I've found that there are so many different ways to do problems that it becomes extremely creative and people tend to have almost a 'mathematical personality' that's expressed in their solutions. Sort of like an art form
    "The contestants must have been born less than twenty years before the day of the second contest paper".
    I have done many problems from BMO2, FST(First Selection Test), and NST(Next Selection Test). In my view, BMO2 is not too difficult; the letter tests are challenging though.
    For mathematical olympiads the theory is a must, but, I believe, problem solving is of much more import - it teaches you many interesting tricks, which are crucial.
    Hmm, perhaps thats why britain does so poorly (I assume you are not allowed to have attended a university).
    Wow, very impressive! How about IMO and other international competitions?

    I've found that you need to have been exposed to that level of problem-solving before you can start scratching the surface with it. Unfortunately this means that vast majority of people that qualify for BMO2 and hence the national team come in clumps from the elite private schools :/ My teachers don't know anyones thats ever even done BMO1 and I didn't either until I visited Cambridge and Warwick! Luckily though I've been finding STEP questions increasingly easy
    Yes, we typically finish high school at the age of 19. I consider University of Cambridge, and Lycée Louis le Grand \Rightarrow ENS (french is simply killing me ).
    By the way, congratulations on your offer!
    Choose cambridge! For one thing look at the league table difference! And then consider Ramanujan, Newton, Hawking, Hardy, 5 Fields Medelists! Cheers!
    (Original post by bananarama2)
    You got into Emma :O So did I for a different subject
    Oh awesome! I'm gonna go ahead at assume NatSci physics? (being on a maths thread and all!) Unfortunately for maths students the concept of 'getting in' is non-existent until august
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    Solution 80

    Let the statement a_n n^{-1}\to 0 be (\star)

    (\star)\Rightarrow there must be an integer m for which a_m=m and a_{n>m}<n. Suppose that for some k there exists i\geq m such that a_i i^{-1}=k^{-1}. If we are lucky and a_i=a_{i+a_i} then we have a_{i} (i+a_i)^{-1}=(k+1)^{-1} and we are done. On the other hand, if a_i increases, bearing a_i<i and (\star) in mind there will be j>i such that a_j j^{-1}=k^{-1}, which brings us back to our hypothesis. However, by (\star) this can only occur finitely many times for any fixed k, and hence a_n will always contain a constant string which leads to our lucky case. Thus the induction is complete and the sequence n^{-1} is a subsequence of a_n n^{-1}.

    The use of the limit can perhaps be made clearer by noting that (a_n) must contain arbitrarily long strings of repeated integers. This is easily shown:

    Spoiler:
    Show
    For the sake of contradiction assume the length of these strings is bounded above by \ell and define \delta_{\ell (k-1)< n\leq \ell k}=k, then a_n\geq \delta_n yet clearly the constant sequence c_n=\ell^{-1} is a subsequence of \delta_n n^{-1} and hence \delta_n n^{-1}\not\to 0\Rightarrow a_n n^{-1}\not\to 0 which is impossible.


    (Original post by Jkn)
    Choose Cambridge! For one thing look at the league table difference! And then consider Ramanujan, Newton, Hawking, Hardy, 5 Fields Medelists!
    Hey hey, for maths the ENS is the most prestigious institution in France. Galois, Dieudonné, Weil, Hadamard, Lebesgue, and if my memory is correct, 10 Fields Medalists.

    (Original post by Mladenov)
    I consider University of Cambridge, and Lycée Louis le Grand \Rightarrow ENS (french is simply killing me ).
    Brilliant plan, either way - I have a couple friends at LLG and they are enjoying it a lot. Be prepared for a lot of non-mathematical things if you do an MP*.
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    (Original post by Jkn)
    Oh awesome! I'm gonna go ahead at assume NatSci physics? (being on a maths thread and all!) Unfortunately for maths students the concept of 'getting in' is non-existent until august
    Sort of, yeh Well I still have to me my offer it's just considerably easier than yours. From your posts on here I have every confidence in you though
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    Solution 28

    (Original post by Indeterminate)
    ...
    Saw this question still lying around and thought I'd give it a shot :ninja: Is the solution for this meant to be neat by the way Indeterminate? xD Whatever route I go down, I get some nasty algebraic mess xD

    Solution 28
    ax + by + r = 0 and (x-a)^{2} + (y-b)^{2} = r^{2}

    The line and the circle will intersect if and only if the shortest distance between the centre of the circle and the line is less than or equal to the radius of the circle. Consider the line perpendicular to ax+by+r=0 passing through the centre (a, b)

    y = \dfrac{b}{a} x

    Equating the two lines, and solving for x, y, they intersect at the point \left( \dfrac{-ar}{a^{2} + b^{2}}, \dfrac{-br}{a^{2} + b^{2}} \right)

    Thus,

    r^{2} \geq \dfrac{(a(a^{2} + b^{2}) +ar)^{2} + (b(a^{2} + b^{2}) + br)^{2}}{(a^{2} + b^{2})^{2}}

    for the line and circle to intersect, which simplifies to:

    r^{2} \geq \dfrac{(a^{2} + b^{2} + r)^{2}}{a^{2} + b^{2}}

    So, yes, the lines and circle do intersect if and only if the inequality holds true (we can use an example like a = b = 1, r = 5) to show that it does hold true for certain values.

    (x-a)^{2} + (y-b)^{2} = r^{2}

    =x^{2} + y^{2} - r^{2} + 2r + a^{2} + b^{2} = 0

    Quadratic formula ;_;

    x = \dfrac{-2ar \pm \sqrt{4a^{2} r^{2} - 4(a^{2} + b^{2}) (r^{2} - b^{2} r^{2} + 2b^{2} r + a^{2} b^{2} + b^{4})}}{2(a^{2} + b^{2})}

    and the y-coordinate follows ;_;

    y = \dfrac{-2ar \mp \sqrt{4a^{2} r^{2} - 4(a^{2} + b^{2}) (r^{2} - b^{2} r^{2} + 2b^{2} r + a^{2} b^{2} + b^{4})}}{2(a^{2} + b^{2})}

    Yes, the line is tangent to the circle if the equality holds. By inspection, a solution for the equality is a = 0, b = 1, r = -\dfrac{1}{2}

    I want to say there are infinite solutions for the line to be tangent but I'll leave that to the number theorists
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    (Original post by Lord of the Flies)
    Hey hey, for maths the ENS is the most prestigious institution in France. Galois, Dieudonné, Weil, Hadamard, Lebesgue, and if my memory is correct, 10 Fields Medalists.
    Oh I didn't know that :O
    Your insurance?
    (Original post by bananarama2)
    Sort of, yeh Well I still have to me my offer it's just considerably easier than yours. From your posts on here I have every confidence in you though
    Awesome, what made you choose emmanuel specifically? Oh thanks Same goes for you! I know few science students interested in pure maths (though I myself have a great interest in theoretical physics)

    (Original post by Mladenov)
    ...
    Few more you may be interested in:

    Solution 76 (3) (absolutely no idea how I only just noticed this! *facepalm*)

    LHS-RHS=(a^2+b^2-c^2)^2+(a^2-b^2)^2 \geq 0

    Solution 76 (4)

    Comparing with Heron's formula, 4A = \sqrt{RHS}, where A is the area of a triangle of sides a, b and c. So RTP:  a^2+b^2 \geq  4A

    Maximum area of a triangle occurs when  \frac{dA}{dQ}=0 where Q is an arbitrarily chosen angle between, in this case, and a and b. So  \frac{d}{dQ} \frac{1}{2} ab \sin{Q} = \frac{1}{2} ab cos{Q} = 0 . This implies  Q= \frac{\pi}{2} and so  4A=2ab .

    Therefore, as  (a-b)^2 \geq 0 , the inequality holds. Note that the result is not bounded for values that satisfy the triangle inequality because the construction is hypothetical and has only relied on unbounded results. Also note that the value of Q is not bounded by  0 \leq Q \leq \pi as any such value would give the same value for sin(0) (though the inequality is trivial for c>a+b etc..)

    QED.

    (any more solutions?)

    ------------------------------------------------

    Problem 81

    Find the smallest positive integer n such that n can be uniquely expressed as the sum of two distinct cubes in two distinct ways. Justify your answer.
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    (Original post by Mladenov)
    Problem 80

    Let (a_{n})_{n\ge1} be an increasing sequence of positive integers. Suppose also that \displaystyle \lim_{n\to \infty} \frac{a_{n}}{n}=0. Then, the sequence \displaystyle b_{n} = \frac{n}{a_{n}} contains all positive integers.
    I'll take a crack at it!

    Solution 80

    Considering the difference between successive terms we get...

     \frac{n+1}{a_{n+1}} - \frac{n}{a_n} \leq \frac{1}{p}, for some positive integer p, which rearranges to
     a_{n}(a_{n+1}-p) \geq pn(a_{n}-a_{n+1} \leq 0 which holds for a_{n+1} \geq p (as the right hand side is less than or equal to zero by virtue of being an increasing sequence).

    This implies each successive term on b_{n} can differ by, at most, successive terms of the harmonic series (though not equal for a_{n+1}>1). We also know that the series is divergent (as is the harmonic series).

    Assume there exists an integer r such that  b_{n} \not= r for all n. As successive terms cannot differ by more than an integer, this would require such a value of r to be 'passed by' as the sequence increases (if the sequence were to decrease it would have to increase again afterwards and so the same argument would apply then) which would require that  gcd(a_{n},n) =1 for successive values of  b_{n} as it moves past the integer 'r' which is trivially untrue as the difference between each term is always an integer. As the sequence increases to infinity, no such r can exist and thus we have a contradiction.

    Therefore  b_{n} must contain all positive integers.

    Edit: crap just saw LOTF's solution
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    (Original post by Jkn)
    Oh I didn't know that - your insurance?
    I would need to do a prépa first (the most famous being LLG, which Mladenov mentioned). I am not interested in doing this for several reasons. I will definitely apply later on though (the undergrad degree will replace the prépa)
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    (Original post by Lord of the Flies)
    I would need to do a prépa first (the most famous being LLG, which Mladenov mentioned). I am not interested in doing this for several reasons. I will definitely apply later on though (the undergrad degree will replace the prépa)
    WTAF that looks weird! So what level are the universities you go to after the prépa, like masters degrees? Are they more prestigious and exclusive than onbridge/ivy-league? Soo confused!
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    Solution 80

    Assume that there is some positive integer x such that x\neq b_n\;\forall n.

    Assume that a_{xy}>y-1 for some positive integer y. Then, as a_{xy} is a positive integer, and if a_{xy}=y,\; b_{xy}=x, therefore, a_{xy}\neq y, and so a_{xy}>y. Therefore, as a is an increasing sequence, a_{x(y+1)}>(y+1)-1. Therefore, if a_{xy}>y-1,\; a_{x(y+1)}>(y+1)-1

    To check the base cases:

    If a_x=1,\; b_x=x, Therefore a_x\neq1, and as a_n\in\mathbb{N} and a_n>0\;\forall n,\; a_n>1\;\forall n\geq x. Therefore a_{2x}>2-1.

    Therefore, by induction, a_{xy}>y-1,\; \forall y. If a_{xy}=y, then b_{xy}=x, therefore a_{xy}\neq y and so a_{xy}>y, \; y\in\mathbb{N}

    Let c_n be a new sequence, where c_{xy}=y+1\; \forall y\in\mathbb{N} and c_{xy+z}=c_{xy}\; \forall z<x and \geq 0. Clearly a_n\geq c_n \forall n. \lim_{n\to \infty} c_n=\frac{n}{x}+1. Therefore, \lim_{n\to \infty} \dfrac{c_n}{n}=\dfrac{\frac{n}{x  }+1}{n}=\dfrac{1}{x}. As a_n\geq c_n,\;\lim_{n\to\infty}a_n\geq \lim_{n\to\infty}c_n . Therefore \lim_{n\to\infty}a_n\geq \frac{1}{x} and so \lim_{n\to\infty}a_n\neq 0. There is a contradiction. Therefore there is no integer x which is not contained in b. Therefore b contains all positive integers.

    Edit:Oh, while I was typing this several other people beat me to it.
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    (Original post by Jkn)
    Hmm, perhaps thats why britain does so poorly (I assume you are not allowed to have attended a university).
    Wow, very impressive! How about IMO and other international competitions?
    No, you are not allowed to attend university. I have done all IMO SL problems from 1990 to 2010 (with several exceptions of course - see, for instance, SL 2003, A6). I am planning to do SL 2011 as a mock before our TST. The Iranian TST is very challenging, I would say it is more difficult than the Chinese TST. Another good source of problems is the All-Russian MO (especially combinatorics and geometry). For inequalities - Vietnamese TST; number theory - almost everywhere(I am biased here). The French TST is somewhat tough. Our NMO and TST are also not easy. I can post some problems if you want. But just look at, for example, the Italian TST, or the German TST - these are warm-ups. I will not even mention Spain, Portugal, and the rest. In Europe, broadly speaking, olympiad mathematics is abandoned.

    (Original post by Jkn)
    I've found that you need to have been exposed to that level of problem-solving before you can start scratching the surface with it. Unfortunately this means that vast majority of people that qualify for BMO2 and hence the national team come in clumps from the elite private schools :/ My teachers don't know anyones thats ever even done BMO1 and I didn't either until I visited Cambridge and Warwick! Luckily though I've been finding STEP questions increasingly easy
    In my country there are high schools which emphasize on olympiad mathematics and this makes the difference. The students at these schools study, generally speaking, mathematics and languages.
    The best students sometimes work with professors.

    (Original post by Lord of the Flies)
    Solution 80

    Let the statement a_n n^{-1}\to 0 be (\star)

    (\star)\Rightarrow there must be an integer m for which a_m=m and a_{n>m}<n. Suppose that for some k there exists i\geq m such that a_i i^{-1}=k^{-1}. If we are lucky and a_i=a_{i+a_i} then we have a_{i} (i+a_i)^{-1}=(k+1)^{-1} and we are done. On the other hand, if a_i increases, bearing a_i<i and (\star) in mind there will be j>i such that a_j j^{-1}=k^{-1}, which brings us back to our hypothesis. However, by (\star) this can only occur finitely many times for any fixed k, and hence a_n will always contain a constant string which leads to our lucky case. Thus the induction is complete and the sequence n^{-1} is a subsequence of a_n n^{-1}.

    The use of the limit can perhaps be made clearer by noting that (a_n) must contain arbitrarily long strings of repeated integers. This is easily shown:

    Spoiler:
    Show
    For the sake of contradiction assume the length of these strings is bounded above by \ell and define \delta_{\ell (k-1)< n\leq \ell k}=k, then a_n\geq \delta_n yet clearly the constant sequence c_n=\ell^{-1} is a subsequence of \delta_n n^{-1} and hence \delta_n n^{-1}\not\to 0\Rightarrow a_n n^{-1}\not\to 0 which is impossible.
    My compliments, very elegant method. This fact is quite interesting, and I found it very useful - I know several olympiad problems which are trivial if one knows this.
    I shall post my solution, since it uses one rather beneficial technique, when it comes to sequences.
    Spoiler:
    Show
    We claim: 1\in (b_{n})_{n\ge 1}. If this is not true - a_{1} \ge 2, a_{2}\ge 2, and a_{2} \not= 2. Hence a_{2} \ge 3. Inductively, a_{n} \ge n+1 - contradiction.
    We claim that m \in (b_{n})_{n \ge1}. Define, c_{n} = ma_{n}. Then, c_{n} is an increasing sequence for which \displaystyle \lim_{n\to \infty} \frac{c_{n}}{n} = 0. So, 1 \in c_{n}. Thus, if c_{n_{0}}=1, we have a_{n_{0}}=m.


    (Original post by Jkn)
    I'll take a crack at it!

    Solution 80

    Assume there exists an integer r such that  b_{n} \not= r for all n. As successive terms cannot differ by more than an integer, this would require such a value of r to be 'passed by' as the sequence increases (if the sequence were to decrease it would have to increase again afterwards and so the same argument would apply then) which would require that  gcd(a_{n},n) =1 for successive values of  b_{n} as it moves past the integer 'r' which is trivially untrue as the difference between each term is always an integer. As the sequence increases to infinity, no such r can exist and thus we have a contradiction.
    I cannot understand your argument. Could you please explain?
    Your approach makes me think of the density of sequences of integers.

    (Original post by Jkn)
    Solution 76 (3) (absolutely no idea how I only just noticed this! *facepalm*)

    LHS-RHS=(a^2+b^2-c^2)^2+(a^2-b^2)^2 \geq 0
    Excellent. I would have never even thought that this might be the case.
    Regarding the geometric interpretation - I also suspected something, but AM-GM seemed more reliable.

    (Original post by Lord of the Flies)
    Brilliant plan, either way - I have a couple friends at LLG and they are enjoying it a lot. Be prepared for a lot of non-mathematical things if you do an MP*.
    I would choose MPSI and then MP*(I think, if I am good enough, they will select me to do MP* after the first year. Is this so?). By the way, as far as I know, there is no programme which offers more mathematics, is there? Notwithstanding, it seems quite exciting that general topology is taught in the first year; the course in linear algebra goes quite deeply - polylinear algebra in the end of the first year.
    To qualify for LLG, I ought to finish first or second on the examination, which takes place each year in May. There are roughly 50 students from my country who take the exam.
    Here are the problems from 2010. The examination is 4 hours.
    Spoiler:
    Show
    Problem 1. Let D be the unique point on the line AC of the triangle ABC such that |BD| = |DC|.
    Choose any point G on the segment BD and denote by E the point of intersection of the lines AG and
    BC. Let F be the unique point on the line AB such that \angle GCB = \angle BCF. Prove that EF and BD
    are parallel.
    Problem 2. In a group of people every two persons have exactly one friend in common. Prove
    that there is an odd number of people in that group.
    Problem 3.
    a) Let ABCD be a convex quadrangle such that the lengths of all of its sides and diagonals are rational
    numbers. Denote by E the intersection of AC and BD. Prove that the length of EA is also a
    rational number.
    b) Let P be a convex polygon in the plane, such that the lengths of all of its sides and diagonals are
    rational numbers. If we draw all the diagonals, the polygon P is dissected into a set of smaller
    polygons. Prove that the length of the sides of any of those polygons are also rational numbers.
    Problem 4. Given a positive integer n, and positive real numbers a_{1},a_{2},...,a_{n}, we let \displaystyle Q(a_{1},a_{2},...,a_{n})= \sum_{k=1}^{n} \sqrt{(2k-1)^{2}+a_{k}^{2}}.
    Find the minimal value Q_{n} of Q(a_{1},a_{2},...,a_{n}), when (a_{1},a_{2},...,a_{n}) varies over all possible n-tuples of positive reals, whose sum is 2010.
    Prove that Q_{n} cannot be an integer if n \equiv 1 \pmod 2.
    Prove that Q_{n} cannot be an integer if n \equiv 0 \pmod 2.
 
 
 
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