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 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
(Original post by jack.hadamard)
Spoiler:ShowThat's right Did you find the general probability first because that's more interesting?

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 16062013 21:46

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 16062013 21:53
(Original post by james22)
Solution 239
Consider a circle of radius r, circumference c.
Enclose the circle by a regular ngon 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? 
ukdragon37
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 16062013 22:15
(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! 
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 16062013 22:17
(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 .
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 without loss of generality. This is because, for any solution there exists a solution as .
Noting that , we deduce that, if is a solution, so too is . If we restrict the values of and 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 . From the formula derived above, combined with and the fact that the combined sum of the solutions are increasing, we deduce that, if we assume to be the base case (a=1), that there exists an increasing sequence such that all solutions are of the form . By using the formula above once more, we get that which implies that . 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 is equivalent to the Fibonacci sequence.
All solutions are coprime. We prove this as follows. Assume have common factor z. Then for positive integers p, q, we have that . Noticing the symmetry with the last problem, we deduce that, . As [tez] z \not= 1[/tex] can be observed from substation, we must deduce that and hence all solutions are coprime.
As all solutions are coprime, 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 . This gives us that which gives us that or . Therefore the base cases we have are . The first case increases to be a part of , the second cases forms a sequence equal to the negative values of 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,
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 . From this we can see that, for n to be an integer, we require that is an odd integer. If it is an integer, it is clearly odd, we we say, for (positivity being assumed without any loss of generality), where we now maximise . Using standard techniques for evaluating the Pell equation (etc...) gives us the values of m and x from which we obtain the desired result ofLast edited by Jkn; 17062013 at 00:25. 
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 16062013 22:24
(Original post by james22)
Solution 239
Consider a circle of radius r, circumference c.
Enclose the circle by a regular ngon 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.
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 as 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 . This is only true on the limit and not the general case (hence a circular argument).Last edited by Jkn; 16062013 at 22:58. 
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 16062013 22:46
(Original post by shamika)
Spoiler:ShowThat's right Did you find the general probability first because that's more interesting?

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 16062013 23:33
(Original post by Jkn)
How is information of the two largest Fibonacci numbers below 2013 obvious? 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?
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 . Denote their set by .
Lemma: If is a solution to , then so is .
Clearly, .
Notice that is a solution, and . Suppose that all Fibonacci numbers up to satisfy . From our lemma, it follows that is a solution. Indeed, .
Second proposition: All solutions to belong to the set .
We note that gives , and this is in .
Suppose the contrary. Let be a solution, which does not belong to the set , such that is minimal. Next, we suppose . We then see that  contradiction. Thus, . Hence, is a solution with smaller sum, implying that it is in . Denote and , which gives , so  contradiction.
Apropos, we can suppose that is a solution, not in , such that is minimal. We get is a solution. Then, if , we have and  contradiction. Hence ....
To summarize: All solutions to consist of consecutive pairs of Fibonacci numbers.
I have already written all Fibonacci numbers, which are smaller than .
Conclusion: the maximum value of is .
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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 16062013 23:59

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 17062013 00:01
(Original post by jack.hadamard)
Why don't you move to a different problem?
Problem 240 * / **
Find all for which is a perfect cube.
By the way, your first formulation was better. 
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 17062013 00:07
(Original post by Mladenov)
By the way, your first formulation was better. 
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 17062013 00:11
(Original post by jack.hadamard)
How do you know what my first formulation was (I didn't edit)? 
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 17062013 00:13
(Original post by Mladenov)
I mean, IMO 1981 problem 6, and more specifically, your modification. 
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 17062013 00:22
(Original post by Mladenov)
You want me to write down all the Fibonacci numbers which are smaller than ? We have.
Spoiler:ShowActually, I have shown exactly this, by two ways.
I am going to write down my complete solution.
First proposition: All consecutive Fibonacci numbers satisfy . Denote their set by .
Lemma: If is a solution to , then so is .
Clearly, .
Notice that is a solution, and . Suppose that all Fibonacci numbers up to satisfy . From our lemma, it follows that is a solution. Indeed, .
Second proposition: All solutions to belong to the set .
We note that gives , and this is in .
Suppose the contrary. Let be a solution, which does not belong to the set , such that is minimal. Next, we suppose . We then see that  contradiction. Thus, . Hence, is a solution with smaller sum, implying that it is in . Denote and , which gives , so  contradiction.
Apropos, we can suppose that is a solution, not in , such that is minimal. We get is a solution. Then, if , we have and  contradiction. Hence ....
To summarize: All solutions to consist of consecutive pairs of Fibonacci numbers.
I have already written all Fibonacci numbers, which are smaller than .
Conclusion: the maximum value of is .
If it is yet not clear, I am to give up.
I particularly like your elegant approach to proving that no nonFibonacci solutions exist.
By the way, how do you find all solutions to the diophantine equation which you have obtained in your second solution?
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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 17062013 00:38
(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 as 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 . This is only true on the limit and not the general case (hence a circular argument). 
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 17062013 02:07
(Original post by jack.hadamard)
Why don't you move to a different problem?
Problem 240 * / **
Find all for which is a perfect cube.Spoiler:ShowI'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
(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.
Well you're method is right, so if you polish it a little and then add in the lemma, I will be content 
Lord of the Flies
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 17062013 12:09
Solution 240
Define with . The claim is that . Clearly holds for , suppose true for some . Then . It is very easy to check that this is when is odd and otherwise. Since is clearly odd we are done. Given that the only cubes are .
(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 as approach the area of the other shape, (given that both shapes are not self intersecting of course)?
The calculus version asks whether over some interval and for two functions sequences of continuous functions the following is true: as , which it clearly is:
On the other hand, the arclength of a function is
Just because and tend to the same thing doesn't imply that and behave the same.
Take and over for instance. Here , the areas tend to , yet the length of the first curve is even at the limit (I have chosen the function so that it mimics the "circle approximation paradox" you posted: the length of is in fact constant).Last edited by Lord of the Flies; 17062013 at 12:26. 
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 17062013 13:37
(Original post by Jkn)
Does this require a (decent) knowledge of modular arithmetic or anything like that? i.e. is the * solution reasonably findable?
(Original post by Lord of the Flies)
Solution 240
Then . It is very easy to check that this is when is odd and otherwise. 
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 17062013 14:47
Problem 241 * / **
i) Find .
ii) Prove that , where is the golden ratio.
I made this problem up and it is not straightforward for a single star. 
TheMagicMan
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 17062013 16:02
(Original post by jack.hadamard)
Problem 241 * / **
i) Find .
ii) Prove that , where is the golden ratio.
I made this problem up and it is not straightforward for a single star.
to be perfectly honest this question is 'easy' as you can use well known approximations for both and 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
.
I'll look for the intended solution when I've recovered.Last edited by TheMagicMan; 17062013 at 16:07. 
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 17062013 16:06
I've posted this problem before (not on this thread)
Problem 242 * and basic continuity knowhow
Let be a continuous function satisfying in the domain. Determine all possible f.
HINT:Spoiler:ShowThe domain is deliberately misleadingLast edited by TheMagicMan; 17062013 at 16:09.
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