Oh I see, I assumed you meant it was a direct result!(Original post by Mladenov)
Nope, I meant . From AMGM we have and , the second is also trivially true, and hence . It is a consequence of AMGM, yet not a direct result.
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 .
If the inequality is clearly true.
If , AMGM yields .
If the inequality is again trivially true.Last edited by Mladenov; 25042013 at 12:37. Reason: Edit the inequality sign in the last case. 
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 25042013 13:56
(Original post by Mladenov)
Solution 76
We have .
If the inequality is clearly true.
If , AMGM yields .
If the inequality is again trivially true. 
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 25042013 14:08
(Original post by Mladenov)
Problem 80
Let be an increasing sequence of positive integers. Suppose also that . Then, the sequence contains all positive integers. 
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 25042013 14:21
(Original post by Jkn)
...
(Original post by Jkn)
... 
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 25042013 16:17
(Original post by Mladenov)
Your idea would probably work, but if , is a real number, provided that is real.
Solution 76 (2)
RHS=
Let f(a,b,c)=LHSRHS=
f(a,b,c) had roots when
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,
Hence true for all a,b,c.
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?Last edited by Jkn; 25042013 at 23:07. 
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 25042013 17:33
(Original post by Jkn)
...
Edit: After I have scrutinized your solution, I would suggest you write and consider the expression as a polynomial of degree with respect to . 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
(Original post by Jkn)
Btw, how old are you and (if you're not at uni already) which one are you going to?Spoiler:ShowLast edited by Mladenov; 25042013 at 20:16. 
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 25042013 20:18
(Original post by Mladenov)
Yup, your solution is now fine.
Edit: After I have scrutinized your solution, I would suggest you to write and consider the expression as a polynomial of degree with respect to . In this way, you will be able to elude from coefficients, which depend on the "variable".
Spoiler:Show
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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 25042013 21:10
(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.
(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 AMGM type stuff and loads of algebraic tricks I can now do loads of the BMO2 questions!
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
By the way, congratulations on your offer!Last edited by Mladenov; 25042013 at 21:25. 
bananarama2
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 25042013 21:28
(Original post by Jkn)
... 
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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.
"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.
Wow, very impressive! How about IMO and other international competitions?
I've found that you need to have been exposed to that level of problemsolving 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
(Original post by bananarama2)
You got into Emma :O So did I for a different subjectLast edited by Jkn; 25042013 at 23:09. 
Lord of the Flies
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Solution 80
Let the statement be
there must be an integer for which and . Suppose that for some there exists such that . If we are lucky and then we have and we are done. On the other hand, if increases, bearing and in mind there will be such that , which brings us back to our hypothesis. However, by this can only occur finitely many times for any fixed , and hence will always contain a constant string which leads to our lucky case. Thus the induction is complete and the sequence is a subsequence of .
The use of the limit can perhaps be made clearer by noting that must contain arbitrarily long strings of repeated integers. This is easily shown:
Spoiler:Show
(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!
(Original post by Mladenov)
I consider University of Cambridge, and Lycée Louis le Grand ENS (french is simply killing me ).Last edited by Lord of the Flies; 26042013 at 13:05. 
bananarama2
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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 nonexistent until august 
Felix Felicis
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Solution 28
(Original post by Indeterminate)
...
Solution 28and
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 passing through the centre
Equating the two lines, and solving for x, y, they intersect at the point
Thus,
for the line and circle to intersect, which simplifies to:
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.
Quadratic formula ;_;
and the ycoordinate follows ;_;
Yes, the line is tangent to the circle if the equality holds. By inspection, a solution for the equality is
I want to say there are infinite solutions for the line to be tangent but I'll leave that to the number theoristsLast edited by Felix Felicis; 27042013 at 13:09. 
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 25042013 23:42
(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.
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
(Original post by Mladenov)
...
Solution 76 (3) (absolutely no idea how I only just noticed this! *facepalm*)
LHSRHS=
Solution 76 (4)
Comparing with Heron's formula, , where A is the area of a triangle of sides a, b and c. So RTP:
Maximum area of a triangle occurs when where Q is an arbitrarily chosen angle between, in this case, and a and b. So . This implies and so .
Therefore, as , 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 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 such that can be uniquely expressed as the sum of two distinct cubes in two distinct ways. Justify your answer.Last edited by Jkn; 26042013 at 03:57. 
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 26042013 03:06
(Original post by Mladenov)
Problem 80
Let be an increasing sequence of positive integers. Suppose also that . Then, the sequence contains all positive integers.
Solution 80
Considering the difference between successive terms we get...
, for some positive integer p, which rearranges to
which holds for (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 can differ by, at most, successive terms of the harmonic series (though not equal for ). We also know that the series is divergent (as is the harmonic series).
Assume there exists an integer r such that 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 for successive values of 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 must contain all positive integers.
Edit: crap just saw LOTF's solutionLast edited by Jkn; 26042013 at 03:13. 
Lord of the Flies
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 26042013 03:37
(Original post by Jkn)
Oh I didn't know that  your insurance? 
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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) 
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 26042013 11:46
Solution 80
Assume that there is some positive integer x such that .
Assume that for some positive integer y. Then, as is a positive integer, and if , therefore, , and so . Therefore, as a is an increasing sequence, . Therefore, if
To check the base cases:
If , Therefore , and as and . Therefore .
Therefore, by induction, . If , then , therefore and so
Let be a new sequence, where and and . Clearly . . Therefore, . As . Therefore and so . 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.Last edited by Zephyr1011; 26042013 at 11:54. 
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 26042013 15:12
(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?
(Original post by Jkn)
I've found that you need to have been exposed to that level of problemsolving 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
The best students sometimes work with professors.
(Original post by Lord of the Flies)
Solution 80
Let the statement be
there must be an integer for which and . Suppose that for some there exists such that . If we are lucky and then we have and we are done. On the other hand, if increases, bearing and in mind there will be such that , which brings us back to our hypothesis. However, by this can only occur finitely many times for any fixed , and hence will always contain a constant string which leads to our lucky case. Thus the induction is complete and the sequence is a subsequence of .
The use of the limit can perhaps be made clearer by noting that must contain arbitrarily long strings of repeated integers. This is easily shown:
I shall post my solution, since it uses one rather beneficial technique, when it comes to sequences.
Spoiler:Show
(Original post by Jkn)
I'll take a crack at it!
Solution 80
Assume there exists an integer r such that 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 for successive values of 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.
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*)
LHSRHS=
Regarding the geometric interpretation  I also suspected something, but AMGM 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 nonmathematical things if you do an MP*.
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:ShowProblem 1. Let be the unique point on the line of the triangle such that .
Choose any point on the segment and denote by the point of intersection of the lines and
. Let be the unique point on the line such that . Prove that and
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 be a convex quadrangle such that the lengths of all of its sides and diagonals are rational
numbers. Denote by the intersection of and . Prove that the length of is also a
rational number.
b) Let 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 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 , and positive real numbers , we let .
Find the minimal value of , when varies over all possible tuples of positive reals, whose sum is .
Prove that cannot be an integer if .
Prove that cannot be an integer if .
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