I'm just baffled. Are the parenthesis supposed to be that way, and is that 1_b minus, or something else.... like 1/b.

Chigr wrote:[hide] value of a+b+c will be minimized when a+b+c since abc i maximized when this is true while this question involves its inverse
 if a=b=c, and a+b+c=1 so they each equal 1/3
(1+1/a)(1+1/b)(1+1/c)>=64
(1+1/a)^3>=64
1+1/a>=4
1/a>=3
1/1/3>=3
3=3
So the inequality is true as when the values differ they (1+a)(1+b)(1+c) will decrease while the inverse will increase.
[/hide]

a+b+c can't be minimized, it's a constant.

As I said, this is really an AM-GM problem.

AM-GM for four variables takes the form:

a+b+c+d >= 4(abcd)^(1/4).  I can prove this if anyone wants.

Okay a twist on the problem .. Once you solved the original  and want a more  challange...

If (1+1/a)(1+1/b)(1+1/c) = 2

obviously  here a+b+c is not equal to 1 but supposing a,b,c all are integers what  are the possible values of a+b+c ?

(Problem may be a little harder for general public here  so pretend that you are looking at "pre-olympiad" forum) :)

Since this is going to fall like a stone off the front page of 'intermediate' soon, I will post a very pretty little solution.

(1+1/a) = (a+1)/a = (a+a+b+c)/c

a+a+b+c >= 4(a)^(1/2)(bc)^(1/4)

Hence (a+a+b+c)/c >= 4(bc)^(1/4)/(a)^(1/2)

Repeating this for every term of the product, we get:

LHS >= 4*4*4abc/abc = 64 as desired.

Gyan wrote:Hint:   "harmonic mean of 1/a, 1/b and 1/c " is 3, their AM and GM  both should be greater than 3.

Notice that the LHS  is
(1 + (1/a+1/b+1/c) +  2/(abc)  )  which is  [hide]nothing but 1 + 3*(AM) +2*(GM)^3
[/hide]

Or if you like the details: Here they are in (spoiler)
AM >= GM
  [hide](a+b+c)/3 = 1/3 >=  (abc)^(1/3)    or (1/abc) >= 27 [/hide]

And  AM >= HM (which is almost the same as AM>=GM done twice)

 [hide]1/3 = (a+b+c)/3  >=  3/(1/a+1/b+1/c)     or (1/a+1/b+1/c) >=  9
(and of course, 1+9+2*27 = 64)
[/hide]

Gyan, shouldn't there be 1/ab, etc. terms in there?

[quote="Gyan"]Sorry should have pointed out that one  of 1/abc  in '2/abc' came from

(1/bc+1/ca+1/ab) = (a+b+c)/abc = 1/abc because (a+b+c = 1)

(Tha'ts how you get 2/abc because    in the product you have  1/abc and the three terms  which also adds up to 1/abc_[/quote]

Gotcha.  I still like my solution better  :cool:.  I'll try and post a solution to your other problem as soon as I finish my latexing my math homework.

[quote="sriram_mask"]Hi
we know that
 a+1/a>=2,
b+1/b>=2
and c+1/c>=2.
is that the way to proceed.?

Sriram[/quote]

No.  That is too weak.  Remember, equality holds for a=b=c=1/3, but if we substitute a=1/3 into any of those, we get: 1/3 + 3 = 10/3 >= 2


Putnam
combinatorics proposed
combinatorics

In how many ways can one put up $ n$ elements into $ r$ sets $ A_1$, $ A_2$, ..., $ A_r$ such that the intersection of any $ k$ sets is disjoint... ??

What is the Putnam problem of which this is a generalization?  (Also, I assume you mean that the intersection of any $ k$ of the sets is empty ... we can say that two sets are disjoint if they have empty intersection, but the word is not used for three or more sets (except to say they are [i]pairwise disjoint[/i], which I assume is not what you mean).)

This is the generalization of putnam problem A1 1985..

this is the link..

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1985.pdf

and when i said intersection of any k sets is disjoint..

i meant that intersection of any k sets together is empty....not pairwise intersection of sets..

For each of the $ n$ elements in our universe, we must choose between 1 and $ k \minus{} 1$ of the $ r$ sets to place it into.  (I'm assuming here that you want the set $ \bigcup_{i \equal{} 1}^r A_i$ to be the entire $ n$-set, as in the Putnam problem; otherwise, we can replace "1" in the previous sentence with "0" and make similar changes in what follows.)  This is both necessary (if any element were to be found in $ k$ of our $ r$ sets, they would violate the given condition) and sufficient (no element is in $ k$ sets, so the intersection of any $ k$ sets is empty) to have our desired goal.  Thus the answer is just $ \left(\sum_{i \equal{} 1}^{k \minus{} 1} \binom{r}{i} \right)^n$.

If we have $ k \equal{} r$ as in the Putnam problem, this simplifies to $ (2^r \minus{} 2)^n$.


calculus
integration
limit
trigonometry
calculus computations

Let $ a$ be a postive constant number. Given a positive integer $ n$, take an integer $ m$ such that $ m\leq \frac{na}{\pi}<m\plus{}1$.
Find $ \lim_{n\to\infty} \frac{1}{n}\int_0^{m\pi} \left(a\minus{}\frac{t}{n}\right)|\sin t|\ dt$.

This integral can be evaluated explicitly.


Now that admission/scholarship decisions have been made, post here if you will be going to Mathcamp 2004.

I will personally be there.

I will be there... not that any of you actually know me  :P .

I will be there!

hmm, I should room with someone from AoPS, anyone need a roomate?

Hopefully I will be there, once I learn what's going on with a scholarship...

[quote="Fierytycoon"]Now that admission/scholarship decisions have been made[/quote]

Does this mean that they're not accepting any more late applications? I'm very close to finishing mine (and submitting it online).

[quote]Does this mean that they're not accepting any more late applications? I'm very close to finishing mine (and submitting it online).[/quote]

The main application selection process is over. However, they are still accepting late applications on a rolling basis (first come, first serve).

i'll be there.  It says on the website that they're only accepting apps for the waitlist.

[quote="JSRosen3"]Does this mean that they're not accepting any more late applications? I'm very close to finishing mine (and submitting it online).[/quote]

They accepted mine, which I sent out Sunday night.

hate to repeat what everyone else said, but

I'll be there

[quote="rrusczyk"]I'll be there.[/quote]

Whoa, what do you mean? To advertise, or as a teacher/guest lecturer?

Guest teacher, last week.

And to meet all these AoPSers.

I'll be there.  can't wait to meet everyone!

-Sara Sheehan

[quote="rrusczyk"]I'll be there.[/quote]

That's pretty cool! Do you know what you'll be lecturing about?

Not yet.  Their academic director is supposed to contact me at some point about possibilities.  I imagine I'll do a class each day and probably do something different each day.  It will all be problem-solving sort of stuff.

Hello!
I actually just learned about this site from an e-mail I got regarding Mathcamp :D .  I'm looking forward to coming!

I need some quick help.

I seem to have ahem... misplaced that blue piece of paper with the registration instructions. So the question would be who do I make out the check for the 1/4 payment to?

Thanks for the help!

it says to make checks payable to:

"Mathematics Foundation of America"

hope that helps!

I'll be there for my 2nd year. If any of you people going for the first time have any questions, I can try to answer them. I solved some of the quiz problems, though as a returning camper I didn't have to, but unfortunately I can't discuss them since some people (slackers :-)) are still applying. Grrr...

Curses. I shall not return. Have fun everyone.
BTW. I will still win in any fight.


If ABC is an triangle and AX, BY and CZ are three ceviences so that a*AX=b*BY=c*CZ then prove that AX, BY and CZ are perpendicular on the sides.

Do you think the problem is right? :maybe:

what is ceviences?

He obviously means [b]concurrent[/b] cevians.

  darij

I meant "concurrent cevians" in the problem text.

No need perhaps to say that the problem is simple when ABC is equilateral so if AX, BY and CZ are equal then they are indeed altitudes.


How many five-digit multiples of 3 end with the digit 6?

Answer

Any multiple of three, $ n$, takes the form:

$ n \equiv 3x\,(mod\,10)$ where $ x = 0,\,1,\ldots ,9$

By running through each value of $ x$ we find that only one value, namely $ x = 2$ gives $ n$ ending with digit 6.

Hence in any group of 30 consecutive integers (where the first is divisible by 30), only the 7th is both a multiple of three and will end in the digit 6.

We know: $ 10000 < n \leq 99999$

We must find how many groups of thirty exist. By inspection, we find that 10020 is the first integer greater than 10000 which is divisible by 30. None of the integers $ 10000 \leq n \leq 10019$ satisfy the required conditions, since the integer 10000 is the 11th member of a group, so the 7th member does not have 5 digits.

$ \frac{99999-10020}{30}= 2999$ with remainder 9.

Hence there are 2999 complete groups of 30, and a fragmentary last group, only containing the first 10 members. This incomplete group does contain the 7h member, giving a total of 3000 5-digit multiples of 3 ending in 6.

All positive integers that end with digit 6 are of the form $ N=10K+6$ where K is an integer greater than or equal to zero;

The fact that $ N=9K+6+K$ is divisible by 3 if and only if $ K$ is divisible by 3 (therefore $ K=3M$, $ M$ integer) allows us to impose $ N=30M+6$ (M integer greater than or equal to zero) as the "desirable" format of the integers.

The least 5-digit number that satisfies (and restricts as a minimum) those integers mentioned above is 10,026 and the greatest is surely 99,996.  (They are 6-ended and obviously divisible by 3.)

Therefore, solving $ 10,026 \leqslant 30M+6 \leqslant 99,996$ we (easily) get $ 3,000$ as the answer, as sludgethrower presented.


"In the opened interval $ (n,2n)$, at least one prime number exists." (of course, $ n$ is natural number)

Above proposition is true?

I guess that it's true. (???)

Is there proof of it? or the counterexample?

Thanks a lot.

Bertrand's postulate: http://en.wikipedia.org/wiki/Bertrand's_postulate , http://www.math.uga.edu/~lyall/REU/Bertrand.pdf

Thanks a lot for your helpful comment.  :D


We have this equation: $ (m \plus{} 1)(\sqrt x  \plus{} 1)^2  \minus{} 2m(\sqrt x  \plus{} 1) \plus{} 2m \equal{} 0$

Find m such that that equation has roots?

please, can u say me what is the question xD?
maybe i'm wrong but i think a quadratic equation has always 2 roots (in complex)
 :|

[quote="Nguyen Ngoc Linh"]I think L_Euler had mentioned real roots.[/quote]

Yes, real roots!!!


A street has 50 houses on each side, for a total of 100 houses. The addresses on the south side of the street form an arithmetic sequence, as do the addresses on the north side of the street. On the south side, the addresses are 1, 5, 9, etc., and on the north side they are 3, 7, 11, etc. A sign painter paints house numbers on a house for $ \$1$ per digit. If he paints the appropriate house number once on each of these 100 houses, how much does he earn?

Sorry, no Latex.   At least what I saw on there, I have to downlaod something on my computer, and mine dosent have the memory capabalities.

Ok. I solved this that he gets the same amount of money for the digits, like 9 bucks for 9 dollars)

Write down the numbers 1,3,5,7,9.

from one to 99, the sequence 1 + 2k , it gives each of those numbers15 times
(1,3,5,7,9,11,13,15,17,19,21,23,25,27,29.....91,93,95,97,99)

It also gives the numbers 2,4,6,8 5 times  (21,23,25,27,29)

So, the sum of the digets from 1 to 99 is 15(1+3+5+7+9) +5(2+4+6+8)

The sum of the digits from 101-199 is that,plus the amount of unit ones.  There are 50 of those, so the total answer is

30*25+10*20+50=1000

You don't need to download anything to use latex on the forum.

${insert code here}$

On the left-side navigation bar, there is a nice tutorial on LaTeX.

[hide="Wrong"]
If we combine the house numbers on the north and south sides, we get exactly the odd positive integers. The $100^{\text{th}}$ odd integer is 199, so we divide the first 100 odd integers into three groups: \[ \{1, 3,\ldots, 9\},\qquad\{11, 13,\ldots, 99\},\qquad\{101, 103,\ldots, 199\} \]

There are five numbers with one digit, 45 numbers with two digits, and 50 numbers with three digits. Thus, the total amount that he earns is $1\times5 + 2\times 45 + 3\times 50 = \boxed{245}$[/hide]


Find all ordered triple $ (a,b,c)$ of positive integers such that the value of the expression $ (b\minus{}\frac{1}{a})(c\minus{}\frac{1}{b})(a\minus{}\frac{1}{c})$ is an integer.

$ (b\minus{}\frac{1}{a})(c\minus{}\frac{1}{b})(a\minus{}\frac{1}{c})$ is an integer
$ \Leftrightarrow abc|(ab\minus{}1)(bc\minus{}1)(ca\minus{}1)$
$ \Leftrightarrow abc|ab\plus{}bc\plus{}ca\minus{}1$
WLOG, $ a \le b \le c$
[b]Case 1:[/b] $ a\equal{}1$, then $ bc|b\plus{}c\minus{}1$, $ bc \le b\plus{}c\minus{}1 <2c$, so $ b\equal{}1$  .
 $ (1,1,k)$ , where $ k$ is any positive integer is the solution.
[b]Case 2:[/b] $ a\equal{}2$, then $ 2bc|bc\plus{}2b\plus{}2c\minus{}1 \Rightarrow b, c$ are odd and $ bc|2b\plus{}2c\minus{}1<4c \Rightarrow b\equal{}3$
The condition becomes $ 6c|5c\plus{}5 \Rightarrow c\equal{}5$
So $ (2,3,5)$ is the solution.
[b]Case 3:[/b] $ a \ge 3$, then $ 3bc \le abc \le ab\plus{}bc\plus{}ca\minus{}1 <3bc$ , contradiction . So no solution in this case.


I need to write the chemical equation for when ammonium nitrate dissolves in water to produce ammonium and nitrate ions..

i so far have

NH4NO3 + H20 --> NH4 + NO3

I have tried to balance out the equation, but to no avail.. anyone can help me please?

If it's possible, then you can find natural numbers a,b,c and d such that :
$aNH_3N0_3+bH_2O\rightarrow cNH_4+ dNO_3$. Then you must have the same number of atomes of $N,H$ and $O$. Thus a,b,c and d must satisfy the following system :
$3a+2b=4c$
$2a=c+d$
$3a+b=3d$ that is $(a,b,c,d)=(10\alpha,3\alpha,9\alpha,11\alpha)$. So you have $10NH_3N0_3+3H_2O\rightarrow 9NH_4+ 11NO_3$

EDIT : Oups I misread it is $NH_4$ on the LHS. The system implies then $b=0$. It's impossible

There's no H20 in the equation, its just NH4NO3 --> NH4+ + NO3-


function
calculus
derivative
inequalities
calculus computations

f is continuous on $ [0, \pi]$ and differentible on $ (0, \pi)$ satisfying $ f(0)\equal{}f(\pi)\equal{}0, |f'(x)| <1,$ for all $ x \in (0,\pi)$ prove that
i) there exists $ \alpha \in (0,\pi) : f'(\alpha)\equal{}tanf(\alpha)$
ii)$ |f(x)| < \frac{\pi}2$ for all $ x \in (0, \pi)$

first one  is  nothing  but legrange`s mean value theorem. { a pretty simple application of that!} :lol:

can you give the detail of the solution? :lol:

2nd is easily proven using the derivative inequality:

$ \minus{}1 \leq f'(x) \leq 1$

Integrating from 0 to x

$ \minus{}x \leq f(x) \leq x$

Also integrating from x to \pi

$ \minus{}\pi\minus{}x \leq \minus{}f(x) \leq \pi\minus{}x$

=> $ \minus{}\pi\plus{}x \leq f(x) \leq \pi\minus{}x$

Now add both ineq above.

Using this result we can now integrate tanx function for any value of f(x) but I cant still prove the first claim.

can anyone help me with the first? thanks!


I never really got the rt=d thing for mathcounts problems i understand that if i drive for 30 minutes at a rate of 60 miles per hour then i will drive 30 miles but not much past that can anyone give some mathcounts level probs that invole rt=d and if possible solutions, u can also pm me or pst useful websites.

Thanks in advance

Bob drove for 2 hours at 60 mph and x hours at 75 mph.  If he averaged 70 mph for the whole trip, find x.

i cross multiplied knowing that the average = total distance/total time
so x=4


At the 1996 grading of Advanced Placement Calculus exams, 530 math teachers graded 126000 tests  in 10 hours. Each test had 6 problems. What is the average number of minutes it took one teacher to grade 25 problems? Express your answer to the nearest minute.

answer is 9.905 = 10 because 12600/(530*600) is unit time.

The answer on FTW is 42.

Here's my method:

Each teacher graded $\frac{126,000\times6}{530}=\frac{75,600}{53}$ problems, each in 10 hours.  Thus, each hour, one teacher graded $\frac{\frac{75,600}{53}}{10}=\frac{7,560}{53}$ problems.  Each minute, one teacher graded $\frac{\frac{7,560}{53}}{60}=\frac{126}{53}$ problems.  Thus, each teacher grades $\frac{25}{\frac{126}{53}}=\frac{1325}{126}$ of $25$ problems per minute, so they need approximately $\boxed{11}$ minutes to finish 25 problems.

...FTW is wrong?

Thats wrong. The actual answer is 42 minutes.
[quote="vallon22"]there were $ 126000$ tests, which means there were $ 126000\times6$ problems, or $ 756000$ problems
$ 530$ teachers graded $ 756000$ problems in $ 10$ hours, which means that...
$ 1$ teacher graded $ 756000\div530$ problems in $ 10$ hours, or about $ 1426.415$ problems

now, instead of writing it out like this, i will make an equation
$ 1426.415$ problems = $ 10$ hours

now i know these equations don't look pretty, since their units are different, but that's not going to matter
we want to find the time in minutes it took a teacher to grade 25 problems

we first convert the hours to minutes

$ 1426.415$ problems = $ 600$ minutes

divide each side by $ 1426.415$ to get the time it took for one teacher to grade on problem

$ 1$ problem = $ 0.4206$ minutes

we want to find the time it took to grade $ 25$ problems, so we multiply both sides by $ 25$

$ 25$ problems = $ 10.51$ minutes or $ 11$ minutes
our answer is $ \boxed{11\text{minutes}}$

disclaimer: answer might not be exact due to rounding, but should still be the right answer[/quote]

Could someone provide an solution that leads to 42?


inequalities
number theory
relatively prime
arithmetic series

p/q is a proper fraction  where p and q are 3 digit numbers.how many pairs of p and q will satisfy the condition (p+1)/(q+3)<=1/2

Here's what I got:

[hide]

We can cross multiply the inequality without worrying about cases because both the numerator and denominator can't be negative (or it would simplify to a positive fraction). So there are only two possibilities... only the numerator is negative, or both are positive. We can multiply the inequality by 2(q+3) in either case.

(p+1) / (q+3) <= 1/2

2p - q <= 1

Because p and q are both integers, the difference can't be less than 1.

2p - q = 1
p = (q+1) / 2

So p and q are always relatively prime (using Euclid's algorithm).

Also, 100 <= q <= 999, so 51 <= (q+1) / 2 <= 500.

So there are 500 - 51 + 1 = 450 pairs of (p, q) that satisfy the original conditions.

[/hide]

Because p and q are both integers, the difference can't be less than 1.

But I think the difference can be negative?

[quote="cheeesec4k3"]

2p - q <= 1

Because p and q are both integers, the difference can't be less than 1.
[/quote]

You meant integers=naturals. Integer numbers are ...,-3,-2,-1,0,1,2,3,...

If you take $ p\equal{}0,q\equal{}0$ the difference is 0, so less than 1.

Eh, is there such fraction as $ 0/0$?  :P
But I think the difference can be negative (take $ 123/457$ for example)

[quote="mathwizarddude"]Eh, is there such fraction as $ 0/0$?  :P
[/quote]

That's just undefined. LOL.

You know what I meant... 123/457 is such an example, I showed that p-2q can be less than 1, and someone obviously didn't read that  :wink:

Wait, are you talking about me...?

Anyway, whoops, I overlooked that. Hm... I'm not sure what to do with the p - 2q < 1 case...
Maybe I should try a completely different approach.

Look at the variable in the numerator, p.  Because you're
going to add 1 to it, it will vary from 100 to 499.  It cannot
equal to 500 (or more), because you're interested in fractions
whose values are less than or equal to 1/2.
This is NOT  TRUE.

*** EDIT  AT THE BOTTOM OF THE POST ***

When p = 100, the smallest that q can be is 199, so that
when 3 is added to it, and 1 added to p, you have the
fraction:

 101/202.

Keeping p at 100, q can vary up to and including 996, so that
q + 3 = 999.

q goes from being 199 to 996.

To find the number of numbers of q, subtract and add 1:

996 - 199 + 1 = 798 (numbers).

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

For confirmation of a pattern, you might note the next
few lines for p = 101:

p + 1 becomes 102.

The variable in the denominator, q,  can vary from 201 up to
and including 996.

996 - 201 + 1 = 796 (numbers).

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

Skip to the end:

p can be as large as 499 (included).

Then (q + 3) can be either 998 or 999, so there are
two choices for q:  995 or 996, respectively.

{p = 100} + {p = 101} + ... + {p = 499}

The sum of this arithmetic series is the answer to the
original problem:

798 + 796 + . . . + 2 = ?

Let n = # of terms
Let d = common difference

n = 499 - 100 + 1 = 400

d = 2

$ S_n \equal{} n(a_1 \plus{} a_n)/2$

$ S_{400} \equal{} 400(798 \plus{} 2)/2$

$ S_{400} \equal{} 160,000$

There are 160,000 pairs p and q that satisfy the condition
(p + 1)/(q + 3) <= 1/2, such that p/q is a proper fraction,
and p and q are each 3-digit numbers.

*** EDIT REFERENCED FROM ABOVE FOLLOWS ***

Please read the post by Mewto55555.

160,000 + 1 (for that missing case of mine) =

160,001 . . . the corrected answer

[quote="Arrange your tan"]Look at the variable in the numerator, p.  Because you're
going to add 1 to it, it will vary from 100 to 499.  It cannot
equal to 500 (or more), because you're interested in fractions
whose values are less than or equal to 1/2.
[/quote]

p=500,q=999


geometry
3D geometry
sphere
trigonometry

How should charges sigma(theta) be distributed on the surface of a sphere of radius R for the field inside the sphere to be uniform and equal to E? Find also the field outside the sphere.[/code]

[hide="Answer"]
$ \sigma \equal{} \sigma_{0} \cos \theta$... where $ \theta$ is the azhimuthal angle
i know of a non sense meaningless proof...but the actual proof uses heavy mathematics[/hide]

Whats Azimuthal Angle  :?:
But i know of a not too difficult proof.May be you shud try it once more  :!:

azimuthal means with the '$ z$' axis
ok is ur proof same as [url=http://www.mathlinks.ro/viewtopic.php?t=168923]here[/url]??


Taken [i]in verbatim[/i] from the Tournament of the Towns 2003 Fall Senior A-Level Paper:

Sorry, had to be taken down since some people have yet to take it

Are you sure that everyone that could have taken this thing took it!?!?
-----------------------------------------------


quadratics
number theory unsolved
number theory

Consider number $ F\equal{}2^{2^{n}}\plus{}1,n\in N^*$, satisfy $ (3^{\frac{F\minus{}1}{2}}\plus{}1)\vdots F$. Prove that F is a prime number

Pepins test : $ F$ is a fermat prime iff $ (3^{\frac{F\minus{}1}{2}}\plus{}1)\vdots F$

The if part : We observe that $ F\minus{}1$ is the order of $ 3$,  $ mod p$. Hence, $ F\minus{}1$ should divide $ phi(F)$ where $ phi$ is the Euler's totient. Since $ phi(F)$ is at most $ F\minus{}1$ with equality only when $ F$ is prime, we deduce that $ phi(F)\equal{} F\minus{}1$ implying F is prime
For the necessity part, quadratic reciprocity does the job


$ a,$ $ b$ and $ c$ are non-negative numbers such that only one of them can be zero. Prove that:
\[ \frac{a^2\plus{}b^2\minus{}2c^2}{\sqrt{c^2\plus{}ab}}\plus{}\frac{a^2\plus{}c^2\minus{}2b^2}{\sqrt{b^2\plus{}ac}}\plus{}\frac{b^2\plus{}c^2\minus{}2a^2}{\sqrt{a^2\plus{}bc}}\geq0\]

it is killed by ineq S.O.S  or $ A(a\minus{}b)^2\plus{}B(b\minus{}c)^2\plus{}C(c\minus{}a)^2$

[quote="arqady"]$ a,$ $ b$ and $ c$ are non-negative numbers such that only one of them can be zero. Prove that:
\[ \frac {a^2 \plus{} b^2 \minus{} 2c^2}{\sqrt {c^2 \plus{} ab}} \plus{} \frac {a^2 \plus{} c^2 \minus{} 2b^2}{\sqrt {b^2 \plus{} ac}} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{\sqrt {a^2 \plus{} bc}}\geq0
\]
[/quote]
If $ a,b,c$ are sides of triangle then the inequality is true by Chebysev inequality:
\[ a^2 \plus{} b^2 \minus{} 2c^2 \ge a^2 \plus{} c^2 \minus{} 2b^2 \ge b^2 \plus{} c^2 \minus{} 2a^2 \\ c^2 \plus{} ab \le b^2 \plus{} ac \le a^2 \plus{} bc
\]
for $ a \ge b \ge c \ge 0$.
Assume $ a \ge b\plus{}c$, then $ c^2 \plus{} ab \le a^2 \plus{} bc,b^2 \plus{} ac \le a^2 \plus{} bc$ and $ a^2 \plus{} b^2 \minus{} 2c^2 \ge a^2 \plus{} c^2 \minus{} 2b^2 >0$, hence
\[ LHS \ge \frac {a^2 \plus{} b^2 \minus{} 2c^2}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {a^2 \plus{} c^2 \minus{} 2b^2}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{\sqrt {a^2 \plus{} bc}} \equal{}0
\].

[quote="ehku"]
for $ a \ge b \ge c \ge 0$.
Assume $ a \ge b \plus{} c$, then $ c^2 \plus{} ab \le a^2 \plus{} bc,b^2 \plus{} ac \le a^2 \plus{} bc$ and $ a^2 \plus{} b^2 \minus{} 2c^2 \ge a^2 \plus{} c^2 \minus{} 2b^2 > 0$[/quote]
Why $ a^2 \plus{} c^2 \minus{} 2b^2 \geq 0$ $ ?$ Thank you!

I'm sorry, it is a mistake  :oops:
The correct is:
We needn't consider the case $ a,b,c$ are sides of a triangle.
Assume $ a \ge b \ge c \ge 0$.
If $ a^2\plus{}c^2\minus{}2b^2 \ge 0$ then
\[ LHS \ge \frac {a^2 \plus{} b^2 \minus{} 2c^2}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {a^2 \plus{} c^2 \minus{} 2b^2}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{\sqrt {a^2 \plus{} bc}} \equal{} 0
\]
If $ a^2\plus{}c^2\minus{}2b^2 \le 0$, because $ a^2 \plus{} b^2 \minus{} 2c^2 \ge b^2 \plus{} c^2 \minus{} 2a^2$ and $ \sqrt {c^2 \plus{} ab} \le \sqrt {a^2 \plus{} bc}$, by Chebysev inequality we get
\[ LHS \ge (a^2 \plus{} b^2 \minus{} 2c^2\plus{}b^2 \plus{} c^2 \minus{} 2a^2)\left(\frac {1}{\sqrt {c^2 \plus{} ab}}\plus{}\frac {1}{\sqrt {a^2 \plus{} bc}}\right) \plus{} \frac {a^2 \plus{} c^2 \minus{} 2b^2}{\sqrt {b^2 \plus{} ac}} \\ \equal{}(2b^2\minus{}a^2\minus{} c^2)\left(\frac {1}{\sqrt {c^2 \plus{} ab}}\plus{}\frac {1}{\sqrt {a^2 \plus{} bc}}\minus{}\frac {1}{\sqrt {b^2 \plus{} ac}}\right) \ge 0
\]
because $ \frac {1}{\sqrt {c^2 \plus{} ab}}\plus{}\frac {1}{\sqrt {a^2 \plus{} bc}} \ge \frac{2}{\sqrt[4]{(c^2 \plus{} ab)(a^2 \plus{} bc)}} \ge \frac {1}{\sqrt {b^2 \plus{} ac}}$.


This is a problem in my algebra 2 book. I understand it but there is a confusion that I have with it.

$\sum_{k=0}^{5} \binom{5}{k} \cdot (-1)^{k} \cdot x^{5-k} \cdot 3^{k} = 32$
this is an obviously the binomial theorem so it would be

$(x-3)^5 = 32$
$x-3 = 2$
$x=5$
This is the answer that they give in the back of the book. The problem is: when you do that, it creates a 5th degree polynomial, so there are 5 solutions for x, not just 1. Does this make the book wrong or what?

The other solutions are imaginary.  If your Alg II book is like mine it is only looking for solutions in the reals, unless it mentions specifically imaginaries.

The imaginary solutions occur when you take the 5th root.

Note that the other four solutions are:

x=2(cos 72k+i sin 72k) k=1,2,3,4.

where the sin are in degrees.


How many of the AoPS members are CAMC Mathematics Book Award winners?

Nobody?

CAMC Mathematics Book Award "is given to the top-scoring student who took the AMC 12 in each region (http://www.unl.edu/amc/f-miscellaneous/regions.html) in grade 9 or less."

In 10th, 11th, and 12th grades, I got books for doing well on the AMC tests. In 11th grade, I think it was sponsored by CAMC, and in the other years, I think there was a different sponsor. I'm not sure though.


$R$ is the reals. $f : R\rightarrow R$ is continuous and for any $a> 0$, $lim_n\rightarrow$$ \infty f(na) = 0$. Prove $lim_x\rightarrow \infty f(x) = 0.$

This has been discussed [b]a lot[/b]. Try searchoing the "Calculus" section


Verify this identity:

cot (x/2) = (sin x) / (1 - cos x)

$ \cot\left(\frac x2\right)\equal{}\frac{\cos\left(\frac x2\right)}{\sin\left(\frac x2\right)}$

$ \equal{}\frac{\frac{\sqrt{1\plus{}\cos x}}2}{\frac{\sqrt{1\minus{}\cos x}}2}\equal{}\sqrt{\frac{1\plus{}\cos x}{1\minus{}\cos x}}$

$ \equal{}\sqrt{\frac{(1\plus{}\cos x)(1\minus{}\cos x)}{(1\minus{}\cos x)^2}}\equal{}\sqrt{\frac{\sin^2x}{(1\minus{}\cos x)^2}}$

$ \equal{}\frac{\sin x}{1\minus{}\cos x}$, as desired (notice that we never had to worry about any sign ($ \pm$) problems--try it and see).

You sort of do.  It's nicer to prove the half-angle identities in the other direction:  let $ t \equal{} \tan \frac {x}{2}$.  Prove the identities

$ \cos x \equal{} \frac {1 \minus{} t^2}{1 \plus{} t^2}$
$ \sin x \equal{} \frac {2t}{1 \plus{} t^2}$

This identity is a straightforward consequence.


An equilateral triangle $ \bigtriangleup ABC$ is inscribed inside a circle with center $ O$. Circle $ O$ is inscribed in an equilateral triangle $ \bigtriangleup XYZ$.

What is the ratio between sides $ AB$ to $ XY$ and what is the ratio between the area of $ \bigtriangleup ABC$ to $ \bigtriangleup XYZ$?

[hide]Since Circle $ O$ is inscribed in $ \bigtriangleup ABC$, $ O$ is the incenter of the triangle and lies on the incenter of the triangle.  Since in an equilateral triangle, all four types of cevians meet in the same place, this is also the centroid of the triangle.  If let the side of $ \bigtriangleup ABC$ equal $ 1$, the height of the triangle is $ \sqrt 3/2$ and the radius of the circle is $ \sqrt 3/3$.  We now have a triangle circumscribed around this circle.  The height of this triangle is $ \sqrt3$.  Then, since the two triangles are similar (they're both equilateral) the ratio of $ AB$ to $ XY$ is the ratio of the heights or $ 1/2$.  The ratio of areas is the square of this, $ 1/4$[/hide]

Err one thing: $ \bigtriangleup ABC$ is inscribed circle $ O$, not circle $ O$ inscribed in $ \bigtriangleup ABC$...

[hide="Simple solution"]
What we have is this:

[geogebra]40d4d6ec2ef79020bb96e96a95aca784d631e3e0[/geogebra]

with ABC being the smaller triangle and XYZ being the larger.

Now, rotate the triangle ABC by 60 degrees with respect to its centroid.

Then, we will have:

 [geogebra]1383f4c388fae2a52ec5c0678aff56616fb6a6a5[/geogebra]

Obviously, the area of ABC is $ \frac {1}{4}$ of the area of XYZ.

Furthermore, the ratio between sides AB to XY will be $ \sqrt {\frac {1}{4}} \equal{} \frac {1}{2}$.[/hide]


Let $ ABCD$ be a square, consider a point $ E$ on segment $ AB$, lets call $ < CEB \equal{} x$.
$ F$ is a point on segment $ DA$ such that $ < FEC \equal{} x$.

Proove that $ EF \equal{} BE \plus{} DF$

 :)

[quote="ElChapin"]Let $ ABCD$ be a square, consider a point $ E$ on segment $ AB$, lets call $ < CEB \equal{} x$.
$ F$ is a point on segment $ DA$ such that $ < FEB \equal{} x$.

Proove that $ EF \equal{} BE \plus{} DF$

 :)[/quote]
How is this construction possible?  $ CEB$ is acute while $ FEB$ is obtuse...

Sorry I changed a vertex, it is correct now.

 :)

[hide="Hint"]Construct $ G\in EF$ such that $ EG\equal{}EB$ and play with triangle congruency.[/hide]


trigonometry
calculus
integration
function
calculus computations

$ \sum\limits_{k \equal{} 1}^\infty {\sin \frac {1}{{k^2 }}}$

My book says convergent and I proved it was convergent but Mathematica 6 does not give me the limit. What does it converge to?

EDITED

I changed the variable of integration and now as an 'answer' mathematica just writes what I entered.

Is there a known limit for this series?

Doubtful.  Using Taylor series, for example, we get a series in terms of $ \zeta(2k), k \equal{} 1, 2, ...$, which can be expressed in terms of the Bernoulli numbers, but that's just another series.

[quote="§outh§tar"]Is there a known limit for this series?[/quote]
I don't know, but I don't see any reason to suspect there is any "known" answer.

Convergence tests for series (such as limit comparison or whatever you used on this one) derive from the theorem that says that a bounded monotone sequence converges. This theorem comes from - in fact is equivalent to - the completeness of the real numbers. I'll give the following crude paraphrase of the completeness of the real numbers: for every limiting process that looks like it ought to converge, there is a real number there to be the limit.

This means that by using limits, or sums, or integrals, or solutions of equations, we can express numbers for which we had no previous names, or even create functions (via power series or trigonometric series) for which we had no previous names. I consider this to be a feature, not a bug.

Put yet another way, there are more real numbers than there are "nice" names for real numbers. There is a real number that is the sum of $ \sum_{k \equal{} 1}^{\infty}\sin\left(\frac1{x^2}\right),$ but it is possible, even likely, that the simplest name for this real number is "the real number that is the sum of that particular series." In effect, that is what Mathematica was telling you. Or, what is the solution to the equation $ x \equal{} \cos x$? We can prove there is a solution (Intermediate Value Theorem), but the simplest name for the resulting number is "the unique real number $ x$ such that $ x \equal{} \cos x.$"

Note that we should have no trouble getting good numerical approximations to these numbers. A few minutes with a spreadsheet gives me this:

$ \sum_{k \equal{} 1}^{\infty}\sin\left(\frac1{k^2}\right)\approx 1.483522817$

I'm sure Mathematica could do the same.

[quote="Kent Merryfield"]Or, what is the solution to the equation $ x \equal{} \sin x$?[/quote]

$ x \equal{} 0$?   :wink:

[color=green][Now it says what I meant it to say. OK? - K.M.][/color]

:lol: @ KBL

Thanks everyone. Makes it a little harder if you're asked to determine if a series converges to a 'known quantity' because until you find out you won't know that it does.


Find all real $ a_0$ such that the sequence $ a_0, a_1, a_2, ...$ defined by $ a_{n \plus{} 1} \equal{} 2^n \minus{} 3a_n$ has $ a_{n \plus{} 1} > a_n$ for all $ n\geq 0$.

$ a_{n \plus{} 1} \equal{} \minus{} 3a_n \plus{} 2^n\Longleftrightarrow a_{n \plus{} 1} \minus{} \frac {1}{5}\cdot 2^{n \plus{} 1} \equal{} \minus{} 3\left(a_n \minus{} \frac {1}{5}\cdot 2^n\right)$, yielding $ a_n \equal{} \frac {1}{5}\cdot 2^n \plus{} ( \minus{} 3)^n\left(a_0 \minus{} \frac {1}{5}\right)$

$ \therefore a_{n \plus{} 1} \minus{} a_n \equal{} \frac {1}{5}\cdot 2^n \minus{} 4( \minus{} 3)^n\left(a_0 \minus{} \frac {1}{5}\right) > 0.$


analytic geometry
geometry
geometric transformation
reflection

The point $(4, 3)$ is reflected over the $x$-axis and then over the $y$-axis.  What is the sum of the coordinates of the new point?

[hide]To reflect over the x-axis, we just multiply the y-coordinate by -1. To do the y-axis, we reverse the formula - we multiply the x by -1. Therefore, the new point have coordinates (-3,-4) for a sum of -7.[/hide]

Billy

[quote="solafidefarms"][hide]To reflect over the x-axis, we just multiply the x-coordinate by -1. To do the y-axis, we do the same. Therefore, the new point have coordinates (-4,-5) for a sum of -9.[/hide]

Billy[/quote]
Slow down a little, and think about the problem before you post.  I'm sure you know how to do this problem, but the rule you gave for reflecting over the x-axis isn't correct.  And your final answer isn't correct, either.

You mean -7, right??? Negate the X and Y values to get -4 and -3, and sum them to get -7. How did someone get -9? Maybe you got the wrong numbers? Sorry if i seem causic... im jw...

[hide]it is reflected of the line y=-x

the point is (-4,-3)

-4-3=-7[/hide]


Let $ a,b,c,p,q$ be positive real numbers. Prove that

$ \left.\frac{a^{p\minus{}q}b^qc^q}{b^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^{p\minus{}q}b^qc^q}\plus{}\frac{a^qb^{p\minus{}q}c^q}{a^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^qb^{p\minus{}q}c^q}\plus{}\frac{a^qb^qc^{p\minus{}q}}{a^{p\plus{}q}\plus{}b^{p\plus{}q}\plus{}a^qb^qc^{p\minus{}q}}\leq1\right.$

[quote="marin.bancos"]Let $ a,b,c,p,q$ be positive real numbers. Prove that

$ \left.\frac {a^{p \minus{} q}b^qc^q}{b^{p \plus{} q} \plus{} c^{p \plus{} q} \plus{} a^{p \minus{} q}b^qc^q} \plus{} \frac {a^qb^{p \minus{} q}c^q}{a^{p \plus{} q} \plus{} c^{p \plus{} q} \plus{} a^qb^{p \minus{} q}c^q} \plus{} \frac {a^qb^qc^{p \minus{} q}}{a^{p \plus{} q} \plus{} b^{p \plus{} q} \plus{} a^qb^qc^{p \minus{} q}}\leq1\right.$[/quote]

I think it can prove by Muirhead theorem.

$ \left.\frac {a^{p \minus{} q}b^qc^q}{b^{p \plus{} q} \plus{} c^{p \plus{} q} \plus{} a^{p \minus{} q}b^qc^q} \plus{} \frac {a^qb^{p \minus{} q}c^q}{a^{p \plus{} q} \plus{} c^{p \plus{} q} \plus{} a^qb^{p \minus{} q}c^q} \plus{} \frac {a^qb^qc^{p \minus{} q}}{a^{p \plus{} q} \plus{} b^{p \plus{} q} \plus{} a^qb^qc^{p \minus{} q}}\leq1\right.$

$ \Longleftrightarrow \sum_{sym}(a^{2p\plus{}2q}b^{p\plus{}q}\minus{}a^{2p\plus{}q}b^pc^{2q})\geq 0$

Maybe... My solution doesn't use Muirhead's theorem.

I think the solution should use the following inequality

\[ b^{p\plus{}q}\plus{}c^{p\plus{}q} \ge (bc)^{qq}(b^{p\minus{}q}\plus{}c^{p\minus{}q})\]

Dear [b]great math[/b],
There is a little typo in your inequality. The right form of it is the following

$ b^{p \plus{} q} \plus{} c^{p \plus{} q}\ge (bc)^{q}(b^{p \minus{} q} \plus{} c^{p \minus{} q})$

Could you give a solution to the proposed inequality?

[b]REMARK[/b]
Using particular values for $ p$ and $ q$ we get interesting forms for the proposed inequality.
Just an example. For $ p \equal{} 2, q \equal{} 1$ this inequality is equivalent to

$ \left.\frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{c^3 \plus{} a^3 \plus{} abc} \plus{} \frac {1}{a^3 \plus{} b^3 \plus{} abc} \leq \frac {1}{abc}\right.$

The problem 2.29 from the excellent book [b]INEQUALITIES WITH BEAUTIFUL SOLUTIONS[/b], written by [b]Vasile Cirtoaje, Vo Quoc Ba Can [/b]and [b]Tran Quoc Anh[/b], has the following statement

For $ a,b,c$ positive real numbers, prove that

$ \left.\frac {a^3}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {b^3}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {c^3}{c^3 \plus{} a^3 \plus{} abc} \geq 1\right.$

No ideas ? You can use the [b]great math[/b]'s hint.

$ \left.(b^p\minus{}c^p)(b^q\minus{}c^q)\geq0\ \ \Leftrightarrow\ \  b^{p\plus{}q}\plus{}c^{p\plus{}q}\geq b^pc^q\plus{}b^qc^p\ \  \Leftrightarrow\ \  b^{p\plus{}q}\plus{}c^{p\plus{}q}\geq b^qc^q(b^{p\minus{}q}\plus{}c^{p\minus{}q})\ \ \Leftrightarrow\right.$
$ \left.\Leftrightarrow\ \ b^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^{p\minus{}q}b^qc^q\geq b^qc^q(a^{p\minus{}q}\plus{}b^{p\minus{}q}\plus{}c^{p\minus{}q})\ \ \Leftrightarrow\ \ \frac{b^qc^q}{b^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^{p\minus{}q}b^qc^q}\leq \frac{1}{a^{p\minus{}q}\plus{}b^{p\minus{}q}\plus{}c^{p\minus{}q}}\ \Leftrightarrow\ \right.$
$ \left.\Leftrightarrow\ \ \frac{a^{p\minus{}q}b^qc^q}{b^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^{p\minus{}q}b^qc^q}\leq \frac{a^{p\minus{}q}}{a^{p\minus{}q}\plus{}b^{p\minus{}q}\plus{}c^{p\minus{}q}}\right.$

Similarly, we get:

$ \frac{a^{q}b^{p\minus{}q}c^{q}}{a^{p\plus{}q}\plus{}c^{p\plus{}q}\plus{}a^{q}b^{p\minus{}q}c^{q}}\leq \frac{b^{p\minus{}q}}{a^{p\minus{}q}\plus{}b^{p\minus{}q}\plus{}c^{p\minus{}q}}$

$ \frac{a^{q}b^{q}c^{p\minus{}q}}{a^{p\plus{}q}\plus{}b^{p\plus{}q}\plus{}a^{q}b^{q}c^{p\minus{}q}}\leq \frac{c^{p\minus{}q}}{a^{p\minus{}q}\plus{}b^{p\minus{}q}\plus{}c^{p\minus{}q}}$

Therefore, adding them up we get the desired inequality.

[b]What do you think about this problem?[/b]


function
counting
distinguishability
symmetry
combinatorics unsolved
combinatorics

Here is an algebraic solution. $2^{2n-k}\binom{k}{n}$ equals to the coefficient of $x^n$ in $2^n (x+\frac12)^k$. Therefore, the sum in the left hand side of the identity equals the coefficient of $x^n$ in
\[ 2^n \sum_{k=0}^{2n} (x + \frac12)^k = 2^n \frac{1 - (x+\frac12)^{2n+1}}{\frac12 - x} = 2^{n+1} \left(\frac{1}{1-2x} - \frac{(x+\frac12)^{2n+1}}{1-2x}\right) \]
(The summands for $k<n$ do not contribute anything.)
The coefficient of $x^n$ in the series for $\frac{1}{1-2x}$ equals $2^n$. The coefficient of $x^n$ in the series for $\frac{(x+\frac12)^{2n+1}}{1-2x}$ equals
\[ \sum_{i=0}^n 2^i \binom{2n+1}{n-i} \left(\frac12\right)^{n+i+1} = \frac{1}{2^{n+1}} \sum_{i=0}^n \binom{2n+1}{n-i} = \frac{1}{2^{n+1}} \cdot 2^{2n} = 2^{n-1} \]

Finally, we get $2^{n+1} (2^n - 2^{n-1}) = 4^n$.

Very nice generating function solution. I was trying to solve this problem this way, but I didn't know how. Thanks...

And what about a combinatorical approach?

We have $2n+1$ distinguishable objects, and we are coloring them either white or black with the restriction that at least $n+1$ are white. An easy symmetry argument shows that half of all colorings have this property, and therefore there are $2^{2n}$ such coverings.

Now, suppose the $n+1$st white object (which must exist) is in position $k+1$. There are $\binom{k}{n}$ possible colorings of the first $k$ objects since exactly $n$ are white, and $2^{2n-k}$ colorings of the last $2n-k$ objects since these can be freely chosen white or black. The $k+1$st object is always white. Combining these and summing over all possible $k$, we have
$2^{2n}=\sum_{k=n}^{2n}\binom{k}{n}\cdot 2^{2n-k}$


Prove that there exist 2003 successive positive integers, such that each of them is divisible by the 2003th power of a positive integer (different from 1).

An exercice for Chinaboy !

[hide]

Let p_1, .., p_2003 2003 different prime numbers.

The system

n = 1 [p_12003]

n = 2 [p_22003]

..

n = 2003 [p_20032003]

Has a solution by the Chineese Remainder Theorem.

Then n-2003,n-2002,..,n-1 are 2003 successive positive integers, each of them is divisible by the 2003th power of a prime number.  :mrgreen: [/hide]


function algebra polynomial logarithms calculus integration inequalities

Let $P(n)$ be the number of functions $f: \mathbb{R} \to \mathbb{R}$, $f(x)=a x^2 + b x + c$, with $a,b,c \in \{1,2,\ldots,n\}$ and that have the property that $f(x)=0$ has only integer solutions. Prove that $n<P(n)<n^2$, for all $n \geq 4$.

[i]Laurentiu Panaitopol[/i]

The lower bound is easy for we have these possible polynomials: $(x+1)(x+i), 0 \leq i \leq n-1$, with $(x+2)(x+2)$ and $2(x+1)(x+1)$

The equation $ f(x)=a x^2 + b x + c$ with $a,b,c \in {1,2,...n}$ can have integer solutions only within the set $\{-1,-2,...,-n\}$, since they product is $c/a\leq n$.
The following estimate now follows $P(n)\leq \sum_{k=1}^{[\sqrt{n}]}\frac{n}{k}-k+1\sim \frac{1}{2}n\ln n-\frac{n}{2}+\sqrt{n}$.

[quote="Myth"]
$P(n)\leq \sum_{k=1}^{[\sqrt{n}]}\frac{n}{k}-k+1\sim \frac{1}{2}n\ln n-\frac{n}{2}+\sqrt{n}$.
[/quote]

1) why is $P(n) \leq $ than that sum?
2) how did you get that estimate?

I am sorry. I mixed up some of my arguments and wrote incorrect expression for estimate.

If $f(x)$ has two integer roots $u$ and $v$, than $auv=c$ by Viete. So we know that $1\leq auv\leq n$, $u,v\in\{-1,...,-n\}$. Note if we know $a$, $u$ and $v$ then we can reconstruct $f$. We will find number of triples $(a,u,v)$ s.t. $auv\leq n$. This number is not less than $P(n)$.

We can choose $a$ arbitrary from the set $\{1,...,n\}$. Then $uv\leq [n/a]$. And number of such pairs $(u,v)$ is $\sum_{u=1}^{[\sqrt{[n/a]}]}\frac{n}{au}=\frac{n}{a}\sum_{u=1}^{[\sqrt{[n/a]}]}\frac{1}{u}\approx \frac{n}{a}\cdot \ln\sqrt{n/a}<\frac{n\ln n}{2a}$ (we may wlog assume $u\leq v$ and I used integral estimates).

Therefore the final estimate is $P(n)\leq \sum_{a=1}^{n}\frac{n\ln n}{2a}=n\ln n\sum_{a=1}^{n}\frac{1}{a}\approx \frac{1}{2}n\ln^2 n$ (by integral estimation).

I hope it is correct now. :blush:

[quote="Myth"]
We can choose $a$ arbitrary from the set $\{1,...,n\}$. Then $uv\leq [n/a]$. And number of such pairs $(u,v)$ is $\sum_{u=1}^{[\sqrt{[n/a]}]}\frac{n}{au}=\frac{n}{a}\sum_{u=1}^{[\sqrt{[n/a]}]}\frac{1}{u}\approx \frac{n}{a}\cdot \ln\sqrt{n/a}<\frac{n\ln n}{2a}$ (we may wlog assume $u\leq v$ and I used integral estimates).
[/quote]

I understand now the idea.
But I don't know calculus and this was given at an olympiad (9th grade - no calculus). Is
there a way of proving $P(n)<n^2$ without it?
And I think it should be $[\frac{n}{au}]$ instead of $\frac{n}{au}$

Yes, but instead of exact estimates (as I did) you need much more straightforward estimates.

How do we extend this to P(n)<n^2/4.

Thanks.

Bomb

[quote="bomb"]How do we extend this to $P(n)<n^2/4$.

Thanks.

Bomb[/quote]

to be honest, I don't think that this inequality holds :( but I may be wrong. i'm sure others can help you

As Myth has shown,
$ P(n)\leq \sum_{a=1}^{n}\frac{n\ln n}{2a}=\frac{1}{2}n\ln n\sum_{a=1}^{n}\frac{1}{a}\leq \frac{1}{2}n\ln n (\ln  n +1) $. So it's sufficient to prove that $\ln n (\ln  n +1)<\frac n 2 $ which is true for  n>29 (i found this bound using Maple). For lower n u must try it or estimate better.


real analysis
geometry
geometric transformation
invariant
real analysis unsolved

Hi

I need to give an example of a subset of reals whose length cannot be defined. I was thinking about the set S of rational numbers from zero to one which is not composed of a finite number of intervals, and so no length is defined for it. But as a hint we have that we should use the axiom of choice and that length of (S+a)=length of (S) Could anyone tell me how to apply it to my example if it is correct.

Thank you
Romy

http://en.wikipedia.org/wiki/Vitali_set

Edit:  The rationals have Lebesgue measure zero.

Thank you for the link. I had seen it but we have not covered measures and Lebesque yet. We merely finished Riemann and gauge integrals.

Do you have any proof using these concepts.

Thank you

The problem with that: what is "length"? Trying to define that inevitably leads to Lebesgue measure.

Not really, this question can be answered with a much more heuristic approach than that. If we suppose that we to every subset $A$ of the real line can assign a length $l(A) \ge 0$ in such a manner that:

i) $l([a,b]) = b-a$
ii) $l$ is monotone
iii) $l$ is $\sigma$-additive
iv) $l$ is translation-invariant

Then by the usual construction of a non-measurable set you get a contradiction.


Find all value n $\geq$ 3 such that  a1,a2,...,an in R: a1+a2+..+an=0  we have:
       a1a2+a2a3+...+ana1 $\leq$ 0

For n > 5, if n is even, we can let the first half be 1, the rest be -1
if n is odd, let a2 be 2, the first half, rounded down, exept a2, be 1, the rest be -1
So the only possible choises are 3,4,5


Prove for all $x+y+z=1$ that
\[\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\le\frac{1}{\sqrt{2}}\].

I have worked on it for some time, but I have found that I am... rather bad at inequalities. (very bad. :ninja: )

I also used Cauchy in more than one way and it didn't work either... maybe I didn't try hard enough?

Considering that it's only cyclic and that equality holds for not only (1/3,1/3,1/3) but also (1/2,1/2,0), not getting this problem does not mean you're very bad at inequalities. (And of course, the fact that it was on a Black test... Last summer, I don't know if I would have gotten [b]any[/b] of them right :roll: )
[hide]
By Cauchy's Inequality,
\[\left( \sum_\text{cyc}\frac{xy}{\sqrt{xy+yz}}\right)^{2}\leq \left( \sum_\text{cyc}xy(xy+xz)\right) \left( \sum_\text{cyc}\frac{xy}{(xy+yz)(xy+xz)}\right) \]
Showing that the right hand side is at most 1/2 is equivalent to
\[4(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+xyz(x+y+z))\leq (x+y+z)(x+y)(y+z)(z+x) \]
which follows from
\[xy(x-y)^{2}+yz(y-z)^{2}+zx(z-x)^{2}\geq 0 \]
[/hide]

Thanks!

I think the problem was that I replaced $x+y$ with $1-z$ wherever I saw it, kinda confusing the issue :oops:

Again, thanks a lot!

(btw how did you think of using $(xy+yz)^{2}$ in the denominator? is it because it's a square?)

EDIT: oh, haha, I read $xz$ as $yz$ :oops:

Thanks again!

Huh? There isn't $(xy+yz)^{2}$ anywhere in my solution...  [hide="stuff related to my solution"]I think that writing
\[\frac{x^{2}y^{2}}{xy+yz}=xy(xy+xz)\cdot \frac{xy}{(xy+yz)(xy+xz)}\]
in order to use Cauchy might seem mysterious at first, but it has good reasons. The $xy$ appears in both factors to make the application of Cauchy strong enough (if it messes up the (1/2,1/2,0) equality case, it will [b]FAIL[/b]), and the factor of $xy+xz$ appears to turn the cyclic inequality into a symmetric one. (Symmetric inequalities are much easier to deal with.)[/hide]

[quote="not_trig"]Prove for all $x+y+z=1$ that
\[\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\le\frac{1}{\sqrt{2}}\]
[/quote]
$x,$ $y$ and $z$ are positive numbers. :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439


[quote="game.guy"]All worgs are either green, purple, or red. All worgs either have one horn or two horns. Those with one horn are either big or small and always have three eyes. Those with three horns are always small. They either have three eyes or one eye.

NO GENETIC MUTATIONS!!!

How many different types of worgs are there if different types are determined by color, number of horns, size, or number of eyes?[/quote]
Wait, you said "All worgs have either one horn or two horns." Then you said " Those with three horns are always small". Is it a trick question or a typo?

Yeah, i wasnt too sure about that either...
Assuming that it was a typo, i got 12 worgs total.
just get two cases, one case is where the worg has 1 horn, the other is where the worg has 2 horns.
Case 1: the worg has 1 horn
there are $ 3$ choices for the color of the worg and the worg has two choices for size so the number of worgs with 1 horn is $ 3*2\equal{}6$
Case 2: the worg has 2 horns
again, there are $ 3$ choices for the color, and $ 2$ choices for eye number so the total number of worgs with 2 horns is $ 3*2\equal{}6$
$ 6\plus{}6\equal{}12$ worgs total
i think...

Those with two horns I meant to say.

The answer was twelve by the way. Good job.


One leg of a right triangle is 12 inches, and the measure of the
angle opposite that leg is $ 30^\circ$. What is the number of inches in
the hypotenuse of the triangle?

No, it is not a 5,12,13 triangle.

By special triangles (and drawing a diagram, or by trig), the hypotenuse of a 30,60,90 triangle is 2 times the length of the leg that is opposite the 30 degree angle. Thus, our answer is $ \boxed{24}$.


advanced fields
advanced fields unsolved

Find $u(x,t)$ such that
\[\begin{cases}u_{tt}=k(x,t)u_{xx}+f(u,u')+g(x,t), x\in [0,1], t>0,\\ u_{x}(0,t)=h_{1}(u_{t}(0,t)),u_{x}(1,t)=h_{2}(u_{t}(1,t)),\\ u(x,0)=\alpha(x), u_{t}(x,0)=\beta(x),\end{cases}\]
where $k,g,h_{1},h_{2},\alpha,\beta$ are given and satisfy some conditions which you can estalish by yourself. The weaker conditions, the more interesting your work is.

Thanks

[quote="mlok"]The form of this system suggests to me that you need to deepen your understanding of the PDE theory before trying to contribute to it.[/quote]

Thanks. I'd like to study the Hadamard well-posedness for this system, but I face some difficulties. Could you help me?

[quote="NguyenThanh_545"]Please, use Galerkin's Approximation to deal with this.

For this problem, it is unable to give the exact solution.[/quote]

But too hard to get the result

You should treat when $ h_{i}(z)=z$ first.


Συγγνώμη, τι είναι η ισογώνια γραμμής ως προς άλλη γραμμή; Έχει να κάνει με τα ισογώνια σημεία σε τρίγωνο;

[quote="Sakis"]Εστω τριγωνο $ ABC$ και τρεις //λες ευθειες που περνανε απο τα $ A,B,C$. ΝΔΟ οι ισογωνιες αυτων ως προς τις πλευρες του τριγωνου
διερχονται απο το ιδιο σημειο.[/quote]

Kαι μάλιστα το σημείο αυτό βρίσκεται στον περιγεγραμένο κύκλο του τριγώνου :wink:

[quote="Spribo"]Μπορεί κάποιος να μου εξηγήσει παρακαλώ τι εννοούμε εδώ λέγοντας ισογώνιες;[/quote]
Δύο ευθείες δια της κορυφής δοσμένης γωνίας ( στο εσωτερικό ή στο εξωτερικό μέρος αυτής ) συμμετρικές ως προς τη διχοτόμο της, ονομάζονται [b]ισογώνιες[/b] ως προς τη δοσμένη γωνία.

$ \bullet$ Ένα γνωστό και χρήσιμο θεώρημα στις ισογώνιες ευθείες, είναι αυτό που λέει ότι άν $ P$ είναι ένα τυχαίο σημείο στο επίπεδο δοσμένου τριγώνου $ \bigtriangleup ABC,$ οι ισογώνιες ευθείες των $ AP,\ BP,\ CP,$ ως προς την αντίστοιχη γωνία ( η ισογώνια της $ AP,$ ως προς τη γωνία $ \angle A$ κλπ ), τέμνονται στο ίδιο σημείο έστω $ P',$ το οποίο ονομάζεται ισογώνιο σημείο του $ P,$ ως προς το δοσμένο τριγωνο $ \bigtriangleup ABC.$

Τα σημεία $ P,\ P',$ ονομάζονται [b]ισογώνια συζυγή[/b] ( = [b]isogonal conjugates[/b] ) ως προς το τρίγωνο $ \bigtriangleup ABC.$

Κώστας Βήττας.

Ευχαριστώ πολύ!


Suppose that $0^\circ < \theta < 90^\circ$, simplify:
\[
\arctan\left( \frac{1+\sin \theta}{\cos \theta} \right)
\]

Heres a geometric solution:

[hide]
Suppose you have a circle center O and radius 1.  Let A and B be points on the circumference such that $\angle\ AOB=\theta$.  Let $AC$ be the perpendicular from $A$ to line $OB$. Then $OC=\cos\ \theta$ and $AC=\sin\ \theta$.

Now construct a circle with center A and radius $OA=1$.  Extend $AC$ to P on the second circle so that $PC=1+\sin\ \theta$.  Let $\angle\ POA=\alpha$.

Then triangle $POC$ is right angled at $\angle\ PCO$, so $\angle\ POC+\angle\ OPC=90\Rightarrow\ 2\alpha+\theta=90\Rightarrow\ \alpha=\frac{90-\theta}{2}$

This means that $\angle\ POC=\alpha+\theta=\frac{90+\theta}{2}$.  But notice that $\tan(\alpha+\theta)=\frac{1+\sin\ \theta}{\cos\ \theta}$ hence $\frac{90+\theta}{2}=\alpha+\theta=\arctan\ (\frac{1+\sin\ \theta}{\cos\ \theta})$

[/hide]


algebra
polynomial
induction
number theory
relatively prime
topology
superior algebra

Is there an integer $ n>1$ such that $ X$ is reducible over $ \mathbb{Z}_n[X]$?

[quote="Fleury"]Is there an integer $ n > 1$ such that $ X$ is reducible over $ \mathbb{Z}_n[X]$?[/quote]

Yes. For example in $ (\mathbb{Z}/6\mathbb{Z})[X]$ we have: $ X\equal{}(3X\plus{}4)(4X\plus{}3),$ or in $ (\mathbb{Z}/10\mathbb{Z})[X]$ we have: $ X\equal{}(5X\plus{}4)(4X\plus{}5)\equal{}(5X\plus{}6)(6X\plus{}5).$

Thanks ysharifi,
How does one find all such integers $ n>1$?

[quote="Fleury"]Thanks ysharifi,
How does one find all such integers $ n > 1$?[/quote]

In general, suppose $ R$ is a commutative ring with identity such that $ R$ has a non-trivial idempotent $ e.$ By non-trivial I mean $ e \neq 0, 1.$ Then in $ R[X]$ we'll have:

$ X \equal{} (eX \plus{} 1 \minus{} e)((1 \minus{} e)X \plus{} e),$ which proves that $ X$ is reducible in $ R[X].$

Now if $ n$ has at least two prime divisors, then $ R \equal{} \mathbb{Z}/n\mathbb{Z}$ will have some non-trivial idempotent and thus $ X$ will be reducible in $ R[X].$ So the only case that

needs to be investigated is when $ n \equal{} p^k,$ where $ p$ is a prime. If $ k \equal{} 1,$ then $ X$ is obviously irreducible. So we may assume that $ k \geq 2.$

[quote="Fleury"]Now if $ n$ has at least two prime divisors, then $ R \equal{} \mathbb{Z}/n\mathbb{Z}$ will have some non-trivial idempotent[/quote]
For anyone who took as long as I did to figure out why this is true, here's a proof:

Write $ n \equal{} ab$, with $ a$ and $ b$ relatively prime, and $ 1 < a < b$.  Let $ k$ be the multiplicative inverse of $ a$ mod $ b$.  Then $ ka \minus{} 1$ is divisible by $ b$, so $ ka(ka \minus{} 1)$ is divisible by $ n$, so $ ka$ is idempotent mod $ n$.  Note also that $ 0 < ka < n$.  Since $ 1 < a < b$, we also have $ 1 < k < b$, so $ ka > 1$.

As for $ n\equal{}p^k$, it seems that $ 1\equal{}(1\minus{}p^{k\minus{}1} X)(1\plus{}p^{k\minus{}1} X)$, so [i]everything[/i] is reducible...

[quote="ysharifi"]...when $ n \equal{} p^k,$ where $ p$ is a prime. If $ k \equal{} 1,$ then $ X$ is obviously irreducible...[/quote]
What about  $ X\equal{}X^{\frac{p\minus{}1}{2}}X^{\frac{p\plus{}1}{2}}$?

[quote="LydianRain"]As for $ n \equal{} p^k$, it seems that $ 1 \equal{} (1 \minus{} p^{k \minus{} 1} X)(1 \plus{} p^{k \minus{} 1} X)$, so [i]everything[/i] is reducible...[/quote]Irreducible elements are non-units by definition. But your examples are units. For $ k\ge 2$, a polynomial $ f(X)\in(\mathbb Z/p^k)[X]$ which is not a unit is irreducible in $ (\mathbb Z/p^k)[X]$ iff $ f(X)\mod p$ is irreducible in $ (\mathbb Z/p)[X]$.

[quote="jmerry"]Equality of polynomials is not the same as those polynomials taking the same values.[/quote]
OK, with that definition then $ X$ is irreducible over $ \mathbb{Z}/p\mathbb{Z}$.

[quote="Third Edition"]Irreducible elements are non-units by definition. But your examples are units. For $ k\ge 2$, a polynomial $ f(X)\in(\mathbb Z/p^k)[X]$ which is not a unit is irreducible in $ (\mathbb Z/p^k)[X]$ iff $ f(X)\mod p$ is irreducible in $ (\mathbb Z/p)[X]$.[/quote]
Yes, you're right.  Whoops.  Showing that $ X$ irreducible in $ \mathbb{Z}/p$ implies it is irreducible in $ \mathbb{Z}/p^k$ can be done with a relatively straightforward induction, (similar to my observation that $ 1\minus{}p^{k\minus{}1} X$ is a unit) and this completes the original discussion.

But I'm still not seeing why a polynomial that is irreducible in $ \mathbb{Z}/p^k$ must be irreducible in $ \mathbb{Z}/p$...

In fact, that's only almost true, take for example $ x^2 \minus{} p$ (didn't check it, but it should work). For nice cases, Hensel's lemma does the job.

"Nice cases" means anything but a power of an irreducible polynomial. If the two factors are relatively prime, it's pretty easy to build up.

[quote="LydianRain"]But I'm still not seeing why a polynomial that is irreducible in $ \mathbb{Z}/p^k$ must be irreducible in $ \mathbb{Z}/p$...[/quote]
... this part of my statement is bluntly false in general. Sorry! :( I was led by the nasty idea that irreducibility of elements is preserved by nilpotent quotients (in both directions): This is true for the property of being a unit on the element level and - and this was the tricky thing which trapped me - for the property of a Zariski closed set of being irreducible. But the latter does not properly translate to elements which I overlooked.

$ X\minus{}p\in\mathbb Z/p^n[X]$ is irreducible since $ p$ is no square in $ \mathbb Z/p^n$, for $ n>1$.


Given $a+b+c=1$ and that all are positive reals, prove (or disprove?) that:

$\frac{(ab)^{2}+1}{ab}+\frac{(bc)^{2}+1}{bc}+\frac{(ac)^{2}+1}{ac}\leq \frac{82}{81abc}$

It is equivalent to
\[81a^{2}b^{2}c^{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 1 \]
Now, the LHS has $256$ terms, and so does $(a+b+c)^{5}$, so the inequality $LHS \leq (a+b+c)^{5}= 1$ is true by Muirhead's inequality (apply it many times)
If you don't like Muirhead, call it rearrangement  :wink:

Pretty much the same solution as @above, and I think you meant $243$ terms not $256$.

Multiply everything by $abc$ and you got
\[\sum_{cyc}((ab)^2c+c)\leq\frac{81}{82}\]
minus $a+b+c$ from LHS and $1$ from RHS and we have
\[81abc\sum_{cyc}ab\leq 1=(a+b+c)^5\]
which can be proven by Muirhead.

remark: this is a 17 year old problem. Fewer and fewer inequalities can be killed by Muirhead now...  :(


[quote="dragansm"]Eq: (B + C)/A = -3

A: log4 - 2log12 = log4 - 2(log3 + log4) = log4 - 2(1 + log4) = -2 - [b]2[/b]log4             (log(c)c = 1)
B: 4log12 - 2log4 = 4(log3*4) - 2log4 = 4(log3 + log4) - 2log4 = 4(1 + log4) - 2log4 = 4 + 2log4
C: (log4)^2 + 2(log12)^2 - 3log4*log12
= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3log4*(log3 + log4)         // (a+b)^2 = ...
= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3(log4)^2 - 3log4
= 2 + log4 [b]- 3[/b] =  log4 - 1

B + C = ( 4 + 2log4 ) + (log4 - 1 ) = 3log4 [b]+ 3[/b] = 3 (1 + log4)
A = [b]-2(1 + log4)[/b]

There is some erorr, but it's can be way to solve[/quote]

the bold stuff is wrong, little miscalculations nothing big, but should be corrected ...

[quote="dragansm"]Eq: (B + C)/A = -3

A: log4 - 2log12 = log4 - 2(log3 + log4) = log4 - 2(1 + log4) = -2 - [color=red][b]2[/b][/color]log4 [b]<-------Wrong[/b]            (log(c)c = 1)
B: 4log12 - 2log4 = 4(log3*4) - 2log4 = 4(log3 + log4) - 2log4 = 4(1 + log4) - 2log4 = 4 + 2log4
C: (log4)^2 + 2(log12)^2 - 3log4*log12
= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3log4*(log3 + log4)         // (a+b)^2 = ...
= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3(log4)^2 - 3log4
= 2 + log4 - [color=red][b]3[/b][/color] =  log4 - 1 [b]<--------Wrong[/b]

B + C = ( 4 + 2log4 ) + (log4 - 1 ) = 3log4 + 3 = 3 (1 + log4) [b]B+C is wrong because of above error[/b]
A = -2(1 + log4) [b]A is wrong because of above error[/b]

There is some erorr, but it's can be way to solve[/quote]
Look at the arrows and red numbers I put above and that should point out where you made the mistake, it was a hard to read answer...

If that is an advise, thank you 10000th User.

My english is bad, so forgiveme if i posted this problem in wrong forum.

And about time, a russian friend brought me this problem.


[hide]$\frac{xy}{x+y} = \frac13, \frac{yz}{y+z} = \frac14, \frac{xz}{x+z} = \frac15$
reciprocating gives:

$\frac{x+y}{xy} = 3, \frac{y+z}{yz} = 4, \frac{x+z}{xz} = 5$

$\frac 1x + \frac 1y = 3, \frac 1y + \frac 1z = 4, \frac 1x + \frac 1z = 5$

Let $a, b, c$ represent $\frac 1x, \frac 1y, \frac 1z$ respectively

Then:
$a+b=3$
$b+c=4$
$a+c=5$
We have 3 variables and 3 equations, so we can have a potential solution set.  Solving gives:

$a=2, b=1, c=3$

Which means that: $x=\frac12, y=1, z=\frac13$
[/hide]

Hmm, probably a better way, but...

[hide]
We can use substitution to find x and z in terms of y (not sure why I decided to solve for y)
$3xy=x+y \rightarrow x = \frac{y}{3y-1}$
$4yz=y+z \rightarrow z = \frac{y}{4y-1}$
$5zx=z+x \rightarrow \frac{5y^2}{(3y-1)(4y-1)} = \frac{y}{3y-1} + \frac{y}{4y-1}$
$5y^2 = 4y^2 - y + 3y^2 - y \rightarrow y = 1$
Then simple subsitution gives us the solution $\left( \frac 12 , 1 , \frac 13 \right)$
[/hide]

Probably an even [i] better [/i] way, but.. :lol:

[hide]  We know:

  $\frac{1}{a} + \frac{1}{b}=\frac{a+b}{ab}$

  So in each equation we simply flip the fractions to get:

$\frac{x+y}{xy} = 3$

$\frac{y+z}{yz} = 4$

$\frac{x+z}{xz} = 5$

  Now we can re-write each equation as:

  $\frac{1}{x} + \frac{1}{y} = 3$

  $\frac{1}{y} + \frac{1}{z} = 4$

  $\frac{1}{x} + \frac{1}{z} = 5$

  Now we have three equations we can easily solve by substitution, doing so yields:

  $( \frac{1}{2} , 1 , \frac{1}{3} )$[/hide]

Haha, I wasn't being conceited, I meant there probably is a better way than mine  :oops:

It was easy ... I done it like G-UNIT  :)


inequalities search function calculus derivative algebra polynomial

Prove that for positive real numbers $x$, $y$, $z$, \[ x^3(y^2+z^2)^2 + y^3(z^2+x^2)^2+z^3(x^2+y^2)^2 \geq xyz\left[xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\right].\] [i]Zarathustra (Zeb) Brady.[/i]

[hide="How Zeb intended this problem to die"]Expand and multiply both sides by $ x\plus{}y\plus{}z$ to make loads of stuff cancel.[/hide]

[quote="MellowMelon"]Prove that for positive real numbers $ x,y,z$,
\[ x^3(y^2 \plus{} z^2)^2 \plus{} y^3(z^2 \plus{} x^2)^2 \plus{} z^3(x^2 \plus{} y^2)^2 \geq xyz[xy(x \plus{} y)^2 \plus{} yz(y \plus{} z)^2 \plus{} zx(z \plus{} x)^2].\]
[/quote]
It's easy SOS.

[quote="MellowMelon"]Prove that for positive real numbers $ x,y,z$,
\[ x^3(y^2 \plus{} z^2)^2 \plus{} y^3(z^2 \plus{} x^2)^2 \plus{} z^3(x^2 \plus{} y^2)^2 \geq xyz[xy(x \plus{} y)^2 \plus{} yz(y \plus{} z)^2 \plus{} zx(z \plus{} x)^2].\]
[i]Zarathustra (Zeb) Brady.[/i][/quote]
Let $ x\equal{}1/a,y\equal{}1/b,z\equal{}1/c$, the inequality becomes
$ \sum \frac{(b^2\plus{}c^2)^2}{a^3b^4c^4} \ge\frac{1}{abc}\sum \frac{(b\plus{}c)^2}{b^3c^3},$
or equivalently,
$ \sum a(b^2\plus{}c^2)^2 \ge \sum a^3(b\plus{}c)^2.$ (1)
We have $ \sum a(b^2\plus{}c^2)^2 \equal{}\sum a(b^4\plus{}c^4)\plus{}2\sum ab^2c^2 \equal{}\sum a^4(b\plus{}c) \plus{}\sum a^2bc(b\plus{}c),$ so the last inequality can be written as
$ \sum [a^4(b\plus{}c)\plus{}a^2bc(b\plus{}c)\minus{}a^3(b\plus{}c)^2] \ge 0,$
that is
$ \sum a^2(b\plus{}c)(a\minus{}b)(a\minus{}c) \ge 0.$
This can be proved easily using Vornicu Schur.

Especially, we have a very nice identity which implies (1), it is
$ LHS\minus{}RHS\equal{}\sum a(b\minus{}c)^2(b\plus{}c\minus{}a)^2.$

[quote="can_hang2007"] $ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0.$
This can be proved easily using Vornicu Schur.

Especially, we have a very nice identity which implies (1), it is
$ LHS \minus{} RHS \equal{} \sum a(b \minus{} c)^2(b \plus{} c \minus{} a)^2.$[/quote]
This is an interesting stronger inequality

$ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge \frac {4(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{a\plus{}b\plus{}c}.$

[quote="Vasc"][quote="can_hang2007"] $ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0.$
This can be proved easily using Vornicu Schur.

Especially, we have a very nice identity which implies (1), it is
$ LHS \minus{} RHS \equal{} \sum a(b \minus{} c)^2(b \plus{} c \minus{} a)^2.$[/quote]
I think the following generalization holds for any $ k\ge 0$ and nonnegative $ a,b,c$:

$ \sum (b \plus{} c)(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc)\ge 0,$

with equality for $ b \equal{} c \equal{} a$, for $ b \equal{} c \equal{} 0$, for $ b \equal{} c \equal{} \frac a{k}$ (or any cyclic permutation).
We have also a nice identity.[/quote]
The inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc\equal{}0$ or $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done.

[quote="can_hang2007"][quote="Vasc"][quote="can_hang2007"] $ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0.$
This can be proved easily using Vornicu Schur.

Especially, we have a very nice identity which implies (1), it is
$ LHS \minus{} RHS \equal{} \sum a(b \minus{} c)^2(b \plus{} c \minus{} a)^2.$[/quote]
I think the following generalization holds for any $ k\ge 0$ and nonnegative $ a,b,c$:

$ \sum (b \plus{} c)(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc)\ge 0,$

with equality for $ b \equal{} c \equal{} a$, for $ b \equal{} c \equal{} 0$, for $ b \equal{} c \equal{} \frac a{k}$ (or any cyclic permutation).
We have also a nice identity.[/quote]
The inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc \equal{} 0$ or $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done.[/quote]
Yes, Can_hang. We have also

$ \sum (b \plus{} c)(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc)\equal{}\sum a(b\minus{}c)^2(b\plus{}c\minus{}(k\plus{}1)a)^2$.

can anyone help me to prove that
$ \sum a^r(b\plus{}c)(a\minus{}b)(a\minus{}c) \ge 0$ where a,b,c,r are positive real.
i cant prove it using vornicu schur because i cant understand vornicu schur

[quote="ishfaq420haque"]can anyone help me to prove that
$ \sum a^r(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0$ where a,b,c,r are positive real.
i cant prove it using vornicu schur because i cant understand vornicu schur[/quote]
The following reasoning is strongest for the Vornicu-Schur's method.
Let $ a\geq b\geq c.$ Hence, $ \sum_{cyc}a^r(b \plus{} c)(a \minus{} b)(a \minus{} c)\geq$
$ \geq a^r(b \plus{} c)(a \minus{} b)(a \minus{} c) \plus{} b^r(a \plus{} c)(b \minus{} a)(b \minus{} c) \equal{}$
$ \equal{} (a \minus{} b)(a^r(b \plus{} c)(a \minus{} c) \minus{} b^r(a \plus{} c)(b \minus{} c)) \equal{}$
$ \equal{} (a \minus{} b)((ab \minus{} c^2)(a^r \minus{} b^r) \plus{} c(a \minus{} b)(a^r \plus{} b^r))\geq0.$

[quote="Vasc"][quote="can_hang2007"] $ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0.$
This can be proved easily using Vornicu Schur.

Especially, we have a very nice identity which implies (1), it is
$ LHS \minus{} RHS \equal{} \sum a(b \minus{} c)^2(b \plus{} c \minus{} a)^2.$[/quote]
This is an interesting stronger inequality

$ \sum a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge \frac {4(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{a \plus{} b \plus{} c}.$[/quote]

Assume : $ c\equal{}min\{a;b;c\}$

With : $ a\equal{}x\plus{}c \  ; \ b\equal{}y\plus{}c \ ; \ x,y \geq 0$

It is equivalent to :

$ 6c^4x^2\minus{}6c^4xy\plus{}6c^4y^2\plus{}14c^3x^3\minus{}9c^3x^2y\minus{}9c^3xy^2\plus{}14c^3y^3\plus{}10c^2x^4\plus{}c^2x^3y\minus{}18c^2x^2y^2\plus{}c^2xy^3\plus{}10c^2y^4\plus{}2cx^5\plus{}5cx^4y\minus{}7cx^3y^2\minus{}7cx^2y^3\plus{}5cxy^4\plus{}2cy^5\plus{}x^5y\minus{}4x^4y^2\plus{}6x^3y^3\minus{}4x^2y^4\plus{}xy^5  \geq 0$

Which is true :)

Is the Muirhead inequality allowed here?

well even if its not its very easy to make it with A-G

(4,3,0)+(3,2,2)$ \geq$ (4,2,1)+(3,2,2)
and this is obvious?
have I maybe miss understood the task?(it seems too easy)

Standard mistake, I thought also that Muirhead kills it.

It's not 4,3,0 + 3,2,2 $ \ge$ 4,2,1 + 3,2,2

It's 4,3,0 + 3,2,2 $ \ge$ 4,2,1 + 3,3,1 and that isn't possible with Muirhead.

But maybe, when multiplying with x+y+z (brilliant idea if it's true, see post #2) it will reduce to something solvable with Schur/Muirhead...

you are right my mistake sorry and thanks for showing it to me
this x+y+z is very very nice idea but I didn't manage to solve it after doing that so
could someone provide that solution?

Hm... I also tried multiplying, but we are left with (5,2,1)+2(4,3,1) $ \le$  (5,3,0)+(4,4,0)+(3,3,2)

Murihead can't do it, Schur applies only with (a,0,0)... Maybe it's something else.

Does anyone has a link where is explained Vornicu Schur method
I've never heard of it.
Muirhead solves it if you square the ineq
you have (8,6,0)+(7,7,0)+2(7,4,3)+3(6,4,4)+(8,3,3)+4(5,5,4)+2(7,5,2)+2(6,6,2)+2(6,5,3) >=
(8,4,2)+3(6,6,2)+2(6,5,3)+3(5,5,4)+(8,3,3)+4(6,5,3)+2(7,5,2)+2(7,4,3)
is equal as
(8,6,0)+(7,7,0)+(5,5,4)>=(8,4,2)+(6,6,2)+(6,4,4)
LHS>=(8,4,2)+2(6,6,2)>=RHS
first time is muihead + A-G second just muirhead
if I made another error please correct me cause this is done manually by hand so I would be suprised if its correct.

Why, and how did you use AM-GM on (x,y,z) ???

For Vornicu-Schur, search
http://www.mathlinks.ro/portal.php?t=162684

thanks for the link
and Am-GM
is used to prove that (7,7,0)+(5,5,4)>=2(6,6,2)
$ x^7y^7\plus{}x^5y^5z^4\geq 2 x^6y^6z^2$
$ y^7z^7\plus{}y^5z^5x^4\geq 2 y^6z^6x^2$
$ z^7x^7\plus{}z^5x^5y^4\geq 2 z^6x^6y^2$
and summing this 3 inequalities you get this ineq up here.
And the question why I don't understand (is there a way to do it without it(I couldn't find it)?)

The inequality is equivallent to
\[ x^{3}(y - z)^{2}(y + z) + y^{3}(z - x)^{2}(z + x) + z^{3}(x - y)^{2}(x + y)\geq{x^{2}yz(y - z)^{2} + xy^{2}z(z - x)^{2} + xyz^{2}(x - y)^{2}}.\]
.
   Dividing both sides by $ xyz$ we get:
\[ \frac {(xy - zx)^{2}(xy + xz - yz)}{yz} + \frac {(yz - xy)^{2}(yz + xy - zx)}{zx} + \frac {(xz - yz)^{2}(zx + zy - xy)}{xy}\geq{0}.\]
. Substitute $ xy = a, yz = b, zx = c$. The inequality reduces to   $ \frac {(a - b)^{2}(a + b - c)}{c} + \frac {(b - c)^{2}(b + c - a)}{a} + \frac {(c - a)^{2}(c + a - b)}{b}\geq{0}$.  Rewrite the inequality as
$ \frac {a(a - b)^{2}}{c} + \frac {b(a - b)^{2}}{c} + \frac {b(b - c)^{2}}{a} + \frac {c(b - c)^{2}}{a} + \frac {a(c - a)^{2}}{b} + \frac {c(c - a)^{2}}{b}\geq{(a - b)^{2} + (b - c)^{2} + (c - a)^{2}}.$  Substitute $ |a - b| = m$, $ |b - c| = n$, $ |c - a| = k$.
     The last inequality is equivallent to   $ (m^{2}\frac {a}{c} + n^{2}\frac {c}{a}) + (m^{2}\frac {b}{c} + k^{2}\frac {c}{b}) + (n^{2}\frac {b}{a} + k^{2}\frac {a}{b})\geq{m^{2} + n^{2} + k^{2}}.$  Using AM-GM, $ LHS\geq{2mn + 2mk + 2kn}$, but easy to prove that $ m,n,k$ are the lengths of a triangle(maybe degenerate),or all of them is $ 0$.  (To prove this, wlog assume $ m$ is maximum between $ (x,y,z)$. Than, by the well known inequality $ |x| + |y|\geq{|x + y|}$, $ k + n = |b - c| + |c - a|\geq{|(b - c) + (c - a)| = |a - b| = m}$, q.e.d.)
    So,
\[ LHS\geq{2mn + 2nk + 2km} = (m(n + k - m) + n(k + m - n) + k(m + n - k)) + (m^{2} + n^{2} + k^{2})\geq{m^{2} + n^{2} + k^{2}},\]
Q.E.D.

We've to show $\sum x^3(y^4+z^4)+2\sum x^3y^2z^2\geq xyz(\sum xy(x^2+y^2)+2\sum(xy)^2)$.
Now suppose $xyz=1$ and lagrange function,
$f(x,y,z,\lambda)=\sum xy(\frac {1}{x^4}+\frac{1}{y^4})+2\sum x-(\sum\frac {x^2+y^2}{z}+2\sum\frac{1}{x^2})$.
After differentiating both side respectively WRT $x,y,z$ we obtain,
$-\frac{3y+3z}{x^4}+\frac{y^3+z^3}{y^3z^3}+2-4x^3-2x(\frac{1}{y}-\frac{1}{z})+\frac{y^2+z^2}{x^2}=\lambda yz$ and similarly others.
So, we've like,
$-\frac {3a}{x^3}+bx+2x-4x^4-2x^2c+\frac {d}{x}+\frac {2}{x^2}+x$ is constant for $x,y,z$ where $a,b,c,d$ are same for $x,yz$.
Now notable part is, again after derivative WRT $x,y,z$ we obtain,
$-\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\frac {d'}{x}+\frac {2}{x^2}+x$  is constant for $x,y,z$ where $a',b',c',d'$ are same for $x,yz$ , hence we've like,
$b''x+2x-4x^4-2x^2c''+\frac {d''}{x}+\frac {2}{x^2}+e'x$ is constant for $x,y,z$ where $b'',c'',d'',e'$ are same for $x,yz$.
Again doing same with $-\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\frac {d'}{x}+\frac {2}{x^2}+x$ we obtain,
$b'''x+2x-4x^4-2x^2c'''+\frac {d'''}{x}+\frac {2}{x^2}+e''x$ is constant for $x,y,z$ where $b''',c''',d'',e''$ are same for $x,yz$. (well this is easy to show all of $b''\neq b''', c''\neq c''',d''\neq d''', e'\neq e''$.
So now we get $-4x^4-2x^2c''''+\frac {d''''}{x}+\frac {2}{x^2}$ , so similarly going on we get a constant polynomial in $x$ for three values with coefficients of $x^k$ are vanishing, for a fixed $k$ (since in each step  coefficients of $x^k$ are different  for two same degree polynomial in $x$).
So keep going on we get $mx$ is constant  for $x,y,z$ where $m$ is symmetric in $x,y,z$ and not nonzero either. So $x=y=z$ and so done.

After expanding, we wish to show that \[\left(\sum\limits_{sym} x^3y^4\right) + \left(2\sum\limits_{cyc} x^3y^2z^2\right) \ge \left(\sum\limits_{sym} x^4y^2z\right) + \left(2\sum\limits_{cyc} xy^3z^3\right).\] By putting the difference $\text{LHS} - \text{RHS}$ into Chinese Dumbassing Notation, or by being especially observant, we see that this is equivalent to the "sum" (even though the signs point in opposite directions) of two sum of squares identities: \[\sum\limits_{cyc} x^4z^3 + x^4y^3 \ge \sum\limits_{cyc} x^4y^2z + x^4yz^2 \quad \text{(1)}\]\[\sum\limits_{cyc} 2x^3y^2z^2 \le \sum\limits_{cyc} x^3y^3z + x^3yz^3. \quad \text{(2)}\] By doing a little bit of algebra, we see that $(1)$ is equivalent to \[\sum\limits_{cyc} x^4\left(y^3 + z^3 - y^2z - yz^2\right) \ge 0\]\[\iff \sum\limits_{cyc} x^4\left(y + z\right)\left(y - z\right)^2 \ge 0. \quad \text{(3)}\] Meanwhile, $(2)$ is equivalent to \[0 \le \sum\limits_{cyc} x^3yz\left(y^2 + z^2 - 2yz\right)\]\[\iff 0 \le \sum\limits_{cyc} x^3yz\left(y - z\right)^2. \quad \text{(4)}\] Now, even though their signs point in different directions, we attempt to "add" $(3)$ and $(4).$ We then need to show that \[\sum\limits_{cyc} x^3\left(zx + xy - yz\right)\left(y - z\right)^2 \ge 0.\] Now, by the symmetry of the inequality, we assume WLOG that $z \ge y \ge x.$ Then if we let $S_x = x^3\left(zx + xy - yz\right)$ and denote $S_y$ and $S_z$ similarly, the inequality is equivalent to \[S_x\left(y - z\right)^2 + S_y\left(z - x\right)^2 + S_z\left(x - y\right)^2 \ge 0. \quad \text{(5)}\] We now use a standard sum of squares trick. (See [url]http://www.mit.edu/~evanchen/handouts/SOS_Dumbass/SOS_Dumbass.pdf[/url]  for more detail)

By our ordering assumption, it follows that $y$ is the median of $\{x, y, z\}.$ Therefore \[\left(y - z\right)\left(y - x\right) \le 0\]\[\implies \left(z - x\right)^2 \ge \left(x - y\right)^2 + \left(y - z\right)^2.\]
In addition, clearly $S_y = xy + yz - zx \ge yz - zx \ge 0.$ Therefore, we have that \[S_x\left(y - z\right)^2 + S_y\left(z - x\right)^2 + S_z\left(x - y\right)^2 \ge\]\[\ge S_x\left(y - z\right)^2 + S_y\left(\left(x - y\right)^2 + \left(y - z\right)^2\right) + S_z\left(x - y\right)^2\]\[= \left(S_y + S_z\right)\left(x - y\right)^2 + \left(S_x + S_y\right)\left(y - z\right)^2\]\[= yz\left(x - y\right)^2 + xy\left(y - z\right)^2 \ge 0.\]
Therefore, it follows that $(5)$ is true, which, upon expansion, is equivalent to the desired result.

Expand the desired inequality as
\[\sum_{\mbox{sym}} x^4y^3 + 2xyz[x^2yz+xyz^2+xy^2z]\ge \sum_{\mbox{sym}} x^4y^2z + 2xyz[x^2y^2+y^2z^2+z^2x^2].\]
This rearranges to
\[\sum_{\mbox{cyc}}x^4[y^3+z^3-y^2z-yz^2] = \sum_{\mbox{cyc}}x^4(y-z)^2(y+z) \ge xyz[(xy-yz)^2+(xy-xz)^2+(xz-yz)^2].\]
Rearrange again to
\[0\le \sum_{\mbox{cyc}}\left[x^4(y-z)^2(y+z)-x^3yz(y-z)^2\right] = \sum_{\mbox{cyc}}x^3(y-z)^2[xy+xz-yz].\]
Let $yz=a, xz=b, xy=c$ to see it is equivalent to prove
\[0\le \sum_{\mbox{cyc}}bc\cdot (b-c)^2[b+c-a].\]
We are done so long as $a,b,c$ satisfy the triangle inequality. Suppose they do not, so WLOG $a=b+c+t$ for some $t>0$. Then the desired inequality is
\[tbc(b-c)^2 \le (b+c+t)c(b+t)^2(2c+t) + (b+c+t)b(c+t)^2(2b+t).\]
This rewrites as
\[tbc(b-c)^2 \le (b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3].\]
In fact, this is trivial because
\[(b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3]\ge 5(b+c)^2bct > (b+c)^2bct> (b-c)^2bct.\]


quadratics
number theory
national olympiad

The Albanian NMO was held 2 week ago. Here’s the grade 10 paper. If you’re looking for sth challenging this is not stuff for you.

1-Prove that this equation can’t have three real positive roots not all of them equal.

x^3 - x^2 + px – p/9=0 (p≠0)

Find the value of p such that all three roots are equal.

2-Let ABC be an isosceles triangle (AB=BC) and O its circumcenter. M is the midpoint of BO and M’ it’s symmetric point to the midpoint of AB. Let K=AB∩MM’. Let L be a point on BC such that angle BLM= angle CLO. Prove that K,L,O,B lie on circle.

3-Let’s call a “domino” such a figure	            What’s the maximal number of such dominos that can be put in a 8x8 table in such a way that the sides of the domino concide with the sides of the unit squares of the table?

4-Prove that if x^2 + y^2 is divisible by p where p is a prime of the form 4m-1 than both x and y are divisible by p.

5-Find all natural numbers a,b,c such that
(a+b)/(a+c)=(b+c)/(b+a) and ab+bc+ca is a prime.

[quote]3-Let’s call a “domino” such a figure What’s the maximal number of such dominos that can be put in a 8x8 table in such a way that the sides of the domino concide with the sides of the unit squares of the table?[/quote]

What does "domino" look like? Can you give a figure for it?

[quote="Iris Aliaj"]
M’ it’s symmetric point to the midpoint of AB. Let K=AB∩MM’.[/quote]

Do you mean [tex]K[/tex] is the midpoint of [tex]AB[/tex]?

I had made the figure on word but it didn't apperar and i dont know how to do it. it's kind of a cross made of 5 squares 3 acrossa nd three horizontally.

yeah K is the midpoint of AB

Hmm, but then I don't get why you need [tex]M'[/tex]

Thank you for the problems  $Iris Aliaj$.Our olympiad also passed these days. :D
And the difficulty of our problems were almost same(in my opinion).
Betwean did you participate in the contest?If yes what did you score?

yes i did partecipate but we haven't got the results yet. i think i did very well and I'm excepting to be in the top 2 :D

hi
i think problem 5 was easy but what about problem 4?
all i know is if
$x^2+y^2=m$ and $p=4k-1$ and $p|x^2+y^2$
then $m=p^{2k'}n$ , $(p,n)=1$  and so $p^2|x^2+y^2$
but i couldn't solve this
how did you solve this?or anyone has idea?

Suppose that neither $x$ nor $y$ is divisible by $p$.
By Fermat's little theorem $x^{p-1}=y^{p-1}(modp)$ also by the given conditions $x^4=y^4(modp)$.
Obviously if $k$ is the smallest positive integer for which $x^k=y^k(modp)$ then $k|4$ so it is easy to see that  $k=4$.
Next it is easy to see that $4|(p-1)$ but then $p=4k+1$ contradiction. :)

My solution of 4,
Suppose x,y are not multiples of p
$x^2=-y^2(mod p)$
$(\frac{-y^2}{p})=(-y^2)^{\frac{p-1}{2}}=-1 (mod p)$ (Fermat's little theorem)
Thus $-y^2$ is a quadratic nonresidue. QED

[quote="shuresh"]thanks hardsoul and siuhochung
but siuhochung i dont get something
[quote]$(\frac{-y^2}{p})=(-y^2)^{\frac{p-1}{2}}=-1 (mod p)$ (Fermat's little theorem) [/quote]
first we supposed that  $y$ is not  divisible by $p$,so  $(\frac{-y^2}{p})$ is not an integer and can not be used in mod.
could you explain your solution again?  :?[/quote]

it 's the legendre symbol.


Propose a synthesis of 4-(carboximethyl)-cyclopent-2-enone (a cyclopentanone ring having a double bond between carbons 2/3 and the group $ \minus{}CH_2CO_2H$ attached to carbon 4) from cyclopentadiene. You can only use other organic starting materials with 2 carbons maximum, and any necessary inorganic reagents.

Note: this problem was created by me some years ago, and my solution uses 3 steps.

cyclopentadiene + PCC( oxidises allylic methylene group to keto group) foll by reaction with methyl cyanide in presence of base(michael reaction) foll by hydrolysis of CN to COOH.

You cannot do that, because cyclopentadienone is a very unstable compound (since it is antiaromatic) and as soon as formed it immediatly undergoes a Diels-Alder reaction with itself.

so what, we can generate cyclopentadienone in situ rite. i mean benzyne also dimerises but we definitely have reactions in which benzyne is the intermediate

But benzyne is not antiaromatic. Also, even generated in situ it would immediatly undergo Diels-Alder with itself. I have written under the topic title "Advanced" for some reason. This type of problems is not supposed to be solved by students of your level (no offense meant). There are plenty of problems out there that I think at this stage would be much more usefull to you (spectroscopy problems, for example).


Everyone knows that if $ E\equal{} \{ x \in (0,1) \ni x \notin \mathbb{Q} \}$ then m(E) = 1. That's what we expect. But, I want to go through the mechanics of applying the definition of outer measure to this set. The outer measure will be equal to the measure, provided the set is measurable.  Definition of outer measure:

$ m*E \equal{} \inf_{E\subset U I_n} \sum l(I_n)$

Then {In} are countable collections of open intervals that cover E. Now what I'm thinking is that even if we split E so that we exclude every rational (that is, split (0,1) in to (0, 1/2)and(1/2, 1) ...etc. ) our open cover still has a total measure of 1 since we are only removing one point at a time and a point has measure 0. I guess I'm just bothered that this process of splitting is clearly countable, but using any of the countable sequences that included all rationals between (0,1) we will never remove all of the rationals from any given neighborhood. We can always go back and take more out.

  :huh:

And then this is also saying that every non-rational number is in one of the intervals in the set {In}-- but it is never alone-- in fact there must be uncountably many other non-rationals in there with it. And I can't even name one of the intervals in {In} any way. if I say (a/b, c/d) we can always split it *again* --

I guess that is all pretty cool but also a little disturbing.  I guess I'm still not really comfortable with the notion of "dense" --

 :o

How would you prove that the measure of this set is 1? Is there another way?

You know $ m^*(Q) \equal{} 0$ and $ m^*([0,1]) \equal{}1$, right?

I think he's looking for a constructive proof using the definition of outer measure.  If you enumerate the rationals in $ [0,1]$, say $ (r_{n})$, let $ I_{n}\equal{}(r_{n}\minus{}\frac{\epsilon}{2^{n\minus{}1}},r_{n}\plus{}\frac{\epsilon}{2^{n\minus{}1}})$, where $ \epsilon>0$ is chosen so that $ \sum_{n\equal{}1}^{\infty}\frac{\epsilon}{2^{n}}<1$.  You know that the outer measure of an interval is equal to its length.  Note that $ [0,1]\setminus\bigcup_{n\equal{}1}^{\infty}I_{n}\subseteq\mathbb{R}\setminus\mathbb{Q}$.  Also, $ m^*\left(\bigcup_{n\equal{}1}^{\infty}I_{n}\right)\leq\sum_{n\equal{}1}^{\infty}I_{n}<1$ by construction.  But there's no reason we can't choose $ \epsilon$ so that $ \sum_{n\equal{}1}^{\infty}\frac{\epsilon}{2^{n}}<\frac{1}{k}$, where $ k\in\mathbb{N}$.  In other words, we can shave off smaller and smaller neighborhoods of rational numbers and enlarge our set of irrationals.  This isn't exactly what you want (or is it?) but I think it's a little more constructive.

Jrav, thanks, that is more along the lines of what I was thinking of!