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    (Original post by ukdragon37)
    I'm not worried because they've already gone past and I know their results. 25th would be my dissertation and therefore entire Part III result.

    I need 62.1 in my dissertation for a distinction in Part III due to exam results, 60 being the passmark (so between that and 62.1 would mean merit overall). I need to get over the passmark in the dissertation to get any grade overall, that's the part I'm worried about
    They already gave you the exam results? That's weird. You did really well though - I hope you do get your distinction!


    (Original post by jack.hadamard)
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    Limiting probability 1/2? I am not convinced in this answer.
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    That's right Did you find the general probability first because that's more interesting?
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    (Original post by james22)
    Solution 239

    Consider a circle of radius r, circumference c.

    Enclose the circle by a regular n-gon in such a way that the distance between the center of each face=r (any issues about this not being exactly possible should go when we take the limit).

    Each face form a triange with the cente, this triangle has base c/n and height r so it has area cr/(2n).

    The total area of the polygon is then cr/2.

    Let n->infinity so the polygon->the circle and area=cr/2=pi*2*r*r/2=pi*r^2.

    Isn't this Archimedes method of exhaustion?
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    (Original post by shamika)
    They already gave you the exam results? That's weird. You did really well though - I hope you do get your distinction!
    Thanks! The exams were held at the start of this and last term.
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    (Original post by Mladenov)
    There probably is.

    As all the possible pairs consist of consecutive Fibonacci numbers, I found it obvious that the required maximum is 987^3+1597^3.
    How is information of the two largest Fibonacci numbers below 2013 obvious? :lol: Granted you don't have to show working but you should at least write them down bro

    And also, you haven't justified that the all solutions are Fibonacci which requires a fair bit of work and can certainly not be stated as a property of the identity you used. I can only assume an IMO examiner would have started at 7 and knocked off 1, 2 or 3 marks even if you did actually write the solution. Do you not agree?

    Here is my proof:

    Solution 233 (2)

    First we say that n \ge m \ \forall n, m without loss of generality. This is because, for any solution (n_a, m_a) there exists a solution (m_a,n_a) as (n^2_a-n_a m_a -m^2_a)^2=(m^2_a-m_a n_a - n^2_a)^2=1.

    Noting that \displaystyle (n^2-nm-m^2)^2=((n+m)^2-(n+m)n-n^2)^2, we deduce that, if (n_a , m_a) is a solution, so too is (n_a+m_a , n_a). If we restrict the values of n_a and m_a to be positive integers, we can see that the sum of the cubes of the second expression is greater than that of the first.

    We now observe that a trivial solution is (1,1). From the formula derived above, combined with n \ge m and the fact that the combined sum of the solutions are increasing, we deduce that, if we assume (1,1) to be the base case (a=1), that there exists an increasing sequence S_a such that all solutions are of the form (S_a , S_{a-1}). By using the formula above once more, we get that (S_{a+1}, S_{a})=(S_a+S_{a-1}, S_{a}) which implies that S_{a+1}=S_a+S{a-1}. As the sum of the cubes increases as a increases, we seek the largest pair of consecutive terms that are both below 2013. Given the base case we have defined, we note that the sequence S is equivalent to the Fibonacci sequence.

    All solutions (n_a, m_a) are co-prime. We prove this as follows. Assume n_a , m_a have common factor z. Then for positive integers p, q, we have that (zp)^2-(zp)(zq)-(zq)^2= \pm 1 \Rightarrow z^2(p^2-pq-q^2)= \pm 1. Noticing the symmetry with the last problem, we deduce that, z= \pm 1. As [tez] z \not= -1[/tex] can be observed from substation, we must deduce that z=1 and hence all solutions are co-prime.

    As all solutions are co-prime, that means that we can using Euclidean's algorithm combined with the recurrence relation to show that all solutions imply a 'base' solution whereby one of the numbers is 1.

    Assume there exists a separate base case of the form (\lambda,1). This gives us that (\lambda^2-\lambda-1)^2= 1 which gives us that \lambda(\lambda-1)=0 or (\lambda-2)(\lambda+1)=0. Therefore the base cases we have are (0,1), \ (-1,1), \ (2,1). The first case increases to be a part of S, the second cases forms a sequence equal to the negative values of S and the third case is itself a part of S. Hence all integer solutions are Fibonacci numbers or the negative of them. We can dismiss the negative values as the sum of two cubes of such numbers would be negative.

    Therefore all candidate solutions are in the Fibonacci sequence. Continued adding gives three consecutive terms as 987, 1597 as 2584. As the sequence is increasing, max(n^3+m^3)=987^3+1597^3=503450  7976 \ \square

    Solution 233 (3)

    Applying the same restrictions to n and m and justifying that they do not decrease the generality of the solution, we get the quadratic equation n^2-nm-(m^2 \pm 1)=0 \Rightarrow n=\frac{m+\sqrt{5m^2+4}}{2}. From this we can see that, for n to be an integer, we require that \sqrt{5m^2 \pm 4} is an odd integer. If it is an integer, it is clearly odd, we we say, for x \in N (positivity being assumed without any loss of generality), 5m^2 \pm 4=x^2 where we now maximise (AM(m,x))^3+m^3. Using standard techniques for evaluating the Pell equation (etc...) gives us the values of m and x from which we obtain the desired result of 5034507976 \ \square
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    (Original post by james22)
    Solution 239

    Consider a circle of radius r, circumference c.

    Enclose the circle by a regular n-gon in such a way that the distance between the center of each face=r (any issues about this not being exactly possible should go when we take the limit).

    Each face form a triange with the cente, this triangle has base c/n and height r so it has area cr/(2n).

    The total area of the polygon is then cr/2.

    Let n->infinity so the polygon->the circle and area=cr/2=pi*2*r*r/2=pi*r^2.
    -snip-

    Ignore that, I've just realised that the dimension of the perimeter doesn't affect the area when it approaches that shape. That said, I would still rather an upper and lower bound for the sake of rigour

    Oh and remember that there is a second part to the problem!

    Edit 2: This is actually really bothering me Does the area of a shape that touches another shape in n places whereby all other points on the line are of a distance at most d from the other shape and has the property that d \to 0 as n \to \infty approach the area of the other shape, (given that both shapes are not self intersecting of course)? :/ ARGHHHAHHFHHFdhsdfksdfkshd Is there a counterexample we can find from these definitions? :/ ARGHSHFSHFdsfkshdf

    Edit 3 (sorry lol): Just noticed that you have assumed that the base of the triangle is c/n. This is only true on the limit and not the general case (hence a circular argument).
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    (Original post by shamika)
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    That's right Did you find the general probability first because that's more interesting?
    Yes, I did, but still thought the answer was kind of irritating.

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    With some luck, it should be \displaystyle \frac{1}{2} - \frac{3}{4(2k - 1)}. I did the limiting probability two different ways (counted with replacement too) and got the same limit (the other one approaching from above).
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    (Original post by Jkn)
    How is information of the two largest Fibonacci numbers below 2013 obvious? :lol: Granted you don't have to show working but you should at least write them down bro

    And also, you haven't justified that the all solutions are Fibonacci which requires a fair bit of work and can certainly not be stated as a property of the identity you used. I can only assume an IMO examiner would have started at 7 and knocked off 1, 2 or 3 marks even if you did actually write the solution. Do you not agree?
    You want me to write down all the Fibonacci numbers which are smaller than 2013? We have\begin{aligned} 1,1,2,3,5,8,13,21,34,55,89,144,2  33,377,610,987,1597 \end{aligned}.

    Actually, I have shown exactly this, by two ways.

    I am going to write down my complete solution.

    First proposition: All consecutive Fibonacci numbers satisfy n^{2}-mn-m^{2}= \pm 1. Denote their set by T.

    Lemma: If (m,n) is a solution to n^{2}-mn-m^{2}= \pm 1, then so is (n,m+n).

    Clearly, m^{2}+2mn+n^{2}-n^{2}-mn-n^{2} = m^{2}+mn-n^{2}=\pm 1.

    Notice that (1,1) is a solution, and (1,1) \in T. Suppose that all Fibonacci numbers up to (F_{k-1},F_{k}) satisfy n^{2}-mn-m^{2}= \pm 1. From our lemma, it follows that (F_{k},F_{k+1}) is a solution. Indeed, (F_{k},F_{k-1}+F_{k})= (F_{k},F_{k+1}).

    Second proposition: All solutions to n^{2}-mn-m^{2}= \pm 1 belong to the set T.

    We note that n=1 gives m=1, and this is in T.

    Suppose the contrary. Let (m,n) be a solution, which does not belong to the set T, such that m+n is minimal. Next, we suppose m \ge n>1. We then see that (n^{2}-mn-m^{2})^{2} = (m^{2}-n(n-m))^{2} > m^{4} > 1 - contradiction. Thus, m < n. Hence, (n-m,m) is a solution with smaller sum, implying that it is in T. Denote n-m=F_{k} and m=F_{k+1}, which gives n=F_{k+2}, so (m,n) \in T - contradiction.
    Apropos, we can suppose that (m,n) is a solution, not in T, such that m is minimal. We get (n-m,m) is a solution. Then, if n-m \ge m, we have n \ge 2m and n^{2}-mn-m^{2} = n(n-m)-m^{2} \ge m^{2} > 1 - contradiction. Hence (n-m,m) \in T....

    To summarize: All solutions to n^{2}-mn-m^{2}= \pm 1 consist of consecutive pairs of Fibonacci numbers.

    I have already written all Fibonacci numbers, which are smaller than 2013.

    Conclusion: the maximum value of m^{3}+n^{3} is 987^{3}+1597^{3}.

    If it is yet not clear, I am to give up.

    By the way, how do you find all solutions to the diophantine equation which you have obtained in your second solution?
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    Why don't you move to a different problem?

    Problem 240 * / **

    Find all n \in \mathbb{N} for which \underbrace{ 2^{2^{2^{2^{{\cdot}^{\cdot 2}}}}} }_{n}\ -\ 3 is a perfect cube.
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    (Original post by jack.hadamard)
    Why don't you move to a different problem?

    Problem 240 * / **

    Find all n \in \mathbb{N} for which \underbrace{ 2^{2^{2^{2^{{\cdot}^{\cdot 2}}}}} }_{n}\ -\ 3 is a perfect cube.
    I was just to write a solution. Give me 20 minutes.

    By the way, your first formulation was better.
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    (Original post by Mladenov)
    By the way, your first formulation was better.
    How do you know what my first formulation was (I didn't edit)?
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    (Original post by jack.hadamard)
    How do you know what my first formulation was (I didn't edit)?
    I mean, IMO 1981 problem 6, and more specifically, your modification.
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    (Original post by Mladenov)
    I mean, IMO 1981 problem 6, and more specifically, your modification.
    Oh, I see. I was worried that my computer leaks.
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    (Original post by Mladenov)
    You want me to write down all the Fibonacci numbers which are smaller than 2013? We have\begin{aligned} 1,1,2,3,5,8,13,21,34,55,89,144,2  33,377,610,987,1597 \end{aligned}.
    :pierre:
    Spoiler:
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    Actually, I have shown exactly this, by two ways.

    I am going to write down my complete solution.

    First proposition: All consecutive Fibonacci numbers satisfy n^{2}-mn-m^{2}= \pm 1. Denote their set by T.

    Lemma: If (m,n) is a solution to n^{2}-mn-m^{2}= \pm 1, then so is (n,m+n).

    Clearly, m^{2}+2mn+n^{2}-n^{2}-mn-n^{2} = m^{2}+mn-n^{2}=\pm 1.

    Notice that (1,1) is a solution, and (1,1) \in T. Suppose that all Fibonacci numbers up to (F_{k-1},F_{k}) satisfy n^{2}-mn-m^{2}= \pm 1. From our lemma, it follows that (F_{k},F_{k+1}) is a solution. Indeed, (F_{k},F_{k-1}+F_{k})= (F_{k},F_{k+1}).

    Second proposition: All solutions to n^{2}-mn-m^{2}= \pm 1 belong to the set T.

    We note that n=1 gives m=1, and this is in T.

    Suppose the contrary. Let (m,n) be a solution, which does not belong to the set T, such that m+n is minimal. Next, we suppose m \ge n>1. We then see that (n^{2}-mn-m^{2})^{2} = (m^{2}-n(n-m))^{2} > m^{4} > 1 - contradiction. Thus, m < n. Hence, (n-m,m) is a solution with smaller sum, implying that it is in T. Denote n-m=F_{k} and m=F_{k+1}, which gives n=F_{k+2}, so (m,n) \in T - contradiction.
    Apropos, we can suppose that (m,n) is a solution, not in T, such that m is minimal. We get (n-m,m) is a solution. Then, if n-m \ge m, we have n \ge 2m and n^{2}-mn-m^{2} = n(n-m)-m^{2} \ge m^{2} > 1 - contradiction. Hence (n-m,m) \in T....

    To summarize: All solutions to n^{2}-mn-m^{2}= \pm 1 consist of consecutive pairs of Fibonacci numbers.

    I have already written all Fibonacci numbers, which are smaller than 2013.

    Conclusion: the maximum value of m^{3}+n^{3} is 987^{3}+1597^{3}.

    If it is yet not clear, I am to give up.
    Very clear, yes (now all you need to do is write all your solutions like that! )

    I particularly like your elegant approach to proving that no non-Fibonacci solutions exist.
    By the way, how do you find all solutions to the diophantine equation which you have obtained in your second solution?
    I didn't think it was very interesting (don't only put that solution down as a what if, lmao). There's an algorithm, you might be interested in section 3 of this (it then goes on to solve the generalised form)

    It has, however, occurred to me that you could, for the variable m, show that if x and y are solutions, so too is x+y (hence deducing that m is fibonacci).
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    (Original post by Jkn)
    -snip-

    Ignore that, I've just realised that the dimension of the perimeter doesn't affect the area when it approaches that shape. That said, I would still rather an upper and lower bound for the sake of rigour

    Oh and remember that there is a second part to the problem!

    Edit 2: This is actually really bothering me Does the area of a shape that touches another shape in n places whereby all other points on the line are of a distance at most d from the other shape and has the property that d \to 0 as n \to \infty approach the area of the other shape, (given that both shapes are not self intersecting of course)? :/ ARGHHHAHHFHHFdhsdfksdfkshd Is there a counterexample we can find from these definitions? :/ ARGHSHFSHFdsfkshdf

    Edit 3 (sorry lol): Just noticed that you have assumed that the base of the triangle is c/n. This is only true on the limit and not the general case (hence a circular argument).
    I haven't done it very rigorously. When i used c I actually meant it as the perimeter of the polygon, and asusmed that the perimeter aproached the circumference as n->infinity.
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    (Original post by jack.hadamard)
    Why don't you move to a different problem?

    Problem 240 * / **

    Find all n \in \mathbb{N} for which \underbrace{ 2^{2^{2^{2^{{\cdot}^{\cdot 2}}}}} }_{n}\ -\ 3 is a perfect cube.
    Does this require a (decent) knowledge of modular arithmetic or anything like that? i.e. is the * solution reasonably findable?
    Spoiler:
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    I've had some success in reformulation etc... and proved a few related things, but the penny still hasn't dropped and I fear that I might be wasting my time trying to use basic techniques :lol:

    (Original post by james22)
    I haven't done it very rigorously. When i used c I actually meant it as the perimeter of the polygon, and asusmed that the perimeter aproached the circumference as n->infinity.
    Oh right! :lol:

    Well you're method is right, so if you polish it a little and then add in the lemma, I will be content
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    Solution 240

    Define s_{n+1}=2^{s_n} with s_0=1. The claim is that n>2:\;s_n\equiv 7\pod{9}. Clearly holds for n=3, suppose true for some k>2,\;s_k=7+9m. Then s_{k+1}\equiv 2^{9m+1}\pod{9}. It is very easy to check that this is 7\pod{9} when m is odd and 2\pod{9} otherwise. Since m is clearly odd we are done. Given that q^3\equiv 0,1,8\pod{9} the only cubes are s_1-3,s_2-3.

    (Original post by Jkn)
    This is actually really bothering me Does the area of a shape that touches another shape in n places whereby all other points on the line are of a distance at most d from the other shape and has the property that d \to 0 as n \to \infty approach the area of the other shape, (given that both shapes are not self intersecting of course)?
    If I have correctly understood what you mean, yes the limiting area is that of the other shape, but the limiting perimeter need not be.

    The calculus version asks whether over some interval (a,b) and for two functions sequences of continuous functions the following is true: \displaystyle d_n=\underset{a\leq x\leq b}{\text{max}}\big| |f_n(x)|-|g_n(x)|\big|\to 0\Rightarrow \displaystyle\int_a^b \big||f_n(x)|-|g_n(x)|\big|\,dx\to 0 as n\to\infty, which it clearly is:

    \displaystyle 0\leq \int_a^b \big||f_n(x)|-|g_n(x)|\big|\,dx\leq \int_a^b \underset{a\leq x\leq b}{\text{max}}\big||f_n(x)|-|g_n(x)|\big|\,dx=(b-a)d_n\to 0

    On the other hand, the arc-length of a function f is \displaystyle \int_a^b \Big(f'^2(x)+1\Big)^{\frac{1}{2}  }\,dx
    Just because \displaystyle \int_a^b |f_n| and \displaystyle \int_a^b |g_n| tend to the same thing doesn't imply that \displaystyle \int_a^b \Big(f'^2_n+1\Big)^{\frac{1}{2}} and \displaystyle \int_a^b \Big(g'^2_n+1\Big)^{\frac{1}{2}} behave the same.

    Take f_n(x)=\frac{1}{n}(\cos n x+1) and g_n(x)=0 over (0,\pi) for instance. Here d_n=2n^{-1}\to 0, the areas tend to  0, yet the length of the first curve f_n is >\pi even at the limit (I have chosen the function so that it mimics the "circle approximation paradox" you posted: the length of f_n is in fact constant).
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    (Original post by Jkn)
    Does this require a (decent) knowledge of modular arithmetic or anything like that? i.e. is the * solution reasonably findable?
    By writing one star, I don't claim the solution can be found easily, but that it exists.

    (Original post by Lord of the Flies)
    Solution 240

    Then s_{k+1}\equiv 2^{9m+1}\pod{9}. It is very easy to check that this is 7\pod{9} when m is odd and 2\pod{9} otherwise.
    Very well! Alternatively, if 2^{2^{2n}} \equiv 7\ (9), then 2^{2^{2(n+1)}} \equiv (-2)^{2^2} \equiv 7\ (9).
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    Problem 241 * / **

    i) Find \displaystyle \sin \left(\frac{\pi}{5}\right).

    ii) Prove that \displaystyle \pi \phi > 5, where \phi is the golden ratio.


    I made this problem up and it is not straightforward for a single star.
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    (Original post by jack.hadamard)
    Problem 241 * / **

    i) Find \displaystyle \sin \left(\frac{\pi}{5}\right).

    ii) Prove that \displaystyle \pi \phi > 5, where \phi is the golden ratio.


    I made this problem up and it is not straightforward for a single star.
    Time for the most inelegant solution on this thread (but it's all I can come up with in my hungover state):

    to be perfectly honest this question is 'easy' as you can use well known approximations for both \pi and \sqrt{5} and some simple mental arithmetic to solve it. Adding a hence to the second part would make it better imho.

    The first part is pretty easy using sum formulae of sin and cos.

    Then \pi \phi
    = \pi (1+ \sqrt{5})/2
    >354/113 . (1+360/161)/2
    =92217/18193
    >5.

    I'll look for the intended solution when I've recovered.
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    I've posted this problem before (not on this thread)

    Problem 242 * and basic continuity knowhow

    Let f: (- \phi/2, \phi/2) \to R be a continuous function satisfying 2xf(x)=f(2x^2-1) \forall x in the domain. Determine all possible f.

    HINT:
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    The domain is deliberately misleading
 
 
 
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