geometry
geometric transformation
dilation

Brent likes to dilute his lemonade. He starts with a full cup and drinks 1/3 of its contents. He repeats this proces until he has consumed one cup of liquid. What part of the original cup of lemonade remains in the cup? Express your answer as a common fraction.

I am not understanding this problem......if he keeps on drinking 1/3 of a cup......he will never finish...the cup but will get very close to ...... but he will never actually finish it.

However....If he adds water after everytime he drinks[hide="answer"]...then he should have 8/27ths of the original drink left after he drinks a cup of fluids. He drinks 1/3 and 2/3 juice left. He drinks 1/3 of water and juice. He has 4/9 juice left. He drinks 1/3 fluid. He has now drunken a cup...He has 8/27 water left. ..........................Multiply by 2/3 becuase that is how much is left after every drink.[/hide]

[hide]8/27[/hide] is the right answer. Good job, dogseatcheese! Sorry I messed up on typing the question.  :blush: It really is :Brent likes to dilute his lemonade. He starts with a full cup and drinks 1/3 of its contents. He then fills the cup with water, stirs the contents, and again drinks 1/3 of its contents. He repeats this proces until he has consumed one cup of liquid. What part of the original cup of lemonade remains in the cup? Express your answer as a common fraction.


Let ABC be right triangle with <B=90.Take  any point  D on [BC] line segment.If  |DC|=12, |AC|=13 then find the minumum integer values the sum  of  [AB]+[BD]

If ABC is right, wouldn't <B always be 90 degrees :P ?

[quote="SnowStorm"]How about 5 + 0 = 5 ?[/quote]

I think Snowstorm is correct. If [BD] is 0 and [AB] is 5, ABC is a 5-12-13 right triangle. So [BD]+[AB]=5+0=5

Is it right or is there an even samller possible value for [BD]+[AB]?

Can you show that 5 is the minimum?

Treething: not if, say, angle A were right.


Last year a class organized three school trips.  70% of the class went on the first trip, 80% on the second trip, and 90% on the third trip.  12 students attended all 3 trips, everybody else went twice.  How many students are there in this class?

[hide="hint"]  I solved it using a "3-item" venn diagram, but it got a little messy. [/hide]

Ok, I'm going to practice the fine art of improvization, so pardon any disorganization as I'm solving as I type...

[hide]Let's call the class... X

 7x/10 + 8x/10 + 9x/10  (basically 12x/5) = 3(x-12) + 2x

so 12x/5 = 5x -  24

so 12x = 25x - 120

so 13x = 120

so I messed up somewhere.

Ok noticed mistake real fast, its 3*12 + 2(x-12), not 3(x-12) + 2x

12x/5 = 36 + 2x - 24 = 12 + 2x

2x/5 = 12

2x = 60

x = 30.

Check: if x is 30, 12x/5 is 72 so that's the overall number of kids you want to account for. You've got 18 who go twice, 18*2 is 36. 12 go 3 times, 12*3 is 36. 36+36 = 72, and there you have it

Explanation: at the start, I did 7x/10 + 8x/10 + 9x/10 because I want to see how many kids come overall, overlap doesn't matter. Usually I see people try to make the number smaller, but I expanded this time because it was the first thing that came to me. That's where I get 12x/5. I wanted to make an equation. I got 3*12 from 12 kids going three times, and the 2(x-12), since every other kid went twice, all but twelve of the class (x) went twice, so I multiplied it by two. Then, I solved the equation (12x/5 = 12+2x)

Ok, this is the first time I've really done problems via AoPS post, so sorry if I broke some unwritten rules etc. etc.

NOTE: this is probably not the best way to do it![/hide]

I think Ive got the best way to do these type of problems.....
[hide]
from the diagram below we get a equation:
$b+c+12=\frac{7f}{10}$ where $f$ is the total no. of students
$a+c+12=\frac{8f}{10}$
$b+a+12=\frac{9f}{10}$
adding the equations you get:
$2(a+b+c)+36=\frac{24f}{10}$
as $a+b+c=f-12$ so
substituting you get the value of $f$ as $30$
[/hide]
Problems like this get simpler when you use sets.

[hide]70+80+90=240, 240-200=40% on all three.
It's easy to see there were 30 students. [/hide]


Given a triangle ABC where AC=BC, let r and s be two semilines by A and B respectively (external to ABC) s.t. <ABs = <CAr  (*). Let  R  be a point on r s.t.  {S',S"} = c(ARC)^s is not empty.  If S = RC^s and {R',R"} = c(BSC)^r, prove that R', R" and C are coolinear like are S', S" and C.

(*) r and s form equal angles measured counterclockwise wrt AC and BA respectively.

PS

From this one can deduce the inverse claim and from this latter, in turn, solve the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=131841#p131841]ISL 1999 geometry problem nr. 7[/url]

[quote="RaMlaF"]There is something i don't understand here: weren't the points R and R' on the line r?? so, by definition, collinear with A?? then do we have to prove that the line r is the line AC ??? :wacko:[/quote]

I'm very sorry for the big confusion. In the figure attached you can see the configuration that I am referring to.

You are asked to prove that R',S' are collinears with C and that R", S" are collinears with C too.


a little on the easy side for esp for "Hardvard and MIT"

spoiler

[hide]

telescoper, how predictable

(1/3)(1/2+1/3+1/4)

[/hide]

[quote="JBL"]The Harvard MIT Math Tournament is given by Harvard and MIT students to high school students.  It is not a test for university students.[/quote]

Exactly.  It's the same idea as the Rice-Stanford Math Tournament.


LaTeX
linear algebra
matrix
symmetry
algebra
polynomial

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Let,
$ {{{\bigwedge_{i = m}^{n}}{{a}_{i}}} = {{a}_{m} = {a}_{m + 1} = {a}_{m + 2} = \dots = {a}_{n - 2} = {a}_{n - 1} = {a}_{n}}}$

In the following braces,
$ {\left\{\begin{array}{ll}
{{\mbox{if}}{\,}{\,}{{\sum_{i=1}^{k}{a}_{i}}={0}}{,}{\,}{\,}{\mbox{then}}{\,}{\,}{{\bigwedge_{i=1}^{k}{a}_{i}}={0}}} \\
\vspace{0pt} \\
{{\mbox{if}}{\,}{\,}{{\bigwedge_{i=1}^{k}{a}_{i}}={0}}{,}{\,}{\,}{\mbox{then}}{\,}{\,}{{\sum_{i=1}^{k}{a}_{i}}={0}}} \\
\end{array}\right\}}$

the operators are showing up squeezed instead of,
$ {{\sum_{i = 1}^{k}}{{a}_{i}}} = {0}$
and
$ {{\bigwedge_{i = 1}^{k}}{{a}_{i}}} = {0}$

How can I correct the formatting so the operators show as normal size within the braces?

[b]2. Relevant equations.[/b]
Knowledge of $ {\text{\LaTeX}}$ formatting.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I kept playing around with the code for a bit, but couldn't get the operators to be normal size.
[/hide]
Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

Use \sum\limits and \bigwedge\limits to get the limits to display above and below.

The problem is that the code is a mess. Was it generated by an equation editor because, while technically correct, it is unreadable?
1) There are far too many curly brackets. Why, for example is the whole thing enclosed in them?
2) You should be using \text rather than \mbox. {,}{\,}{\,}{\mbox{then}}{\,}{\,} is dreadful. What is wrong with \text{ then }? Note how the space is inside the text box
3) Why are there 2 columns in the array, yet only one is used?
4) Why is there \\ after the last line of the array?
5) Why use \vspace{0pt} \\?
6) The Bmatrix environment makes things simpler and saves having to add curly brackets to an array.

Putting this together you get
$ \begin{Bmatrix} \text{if }\sum\limits_{i = 1}^k a_i = 0 \text{ then } \bigwedge\limits_{i = 1}^k a_i = 0 \\[2ex]
\text{if }\bigwedge\limits_{i = 1}^k a_i = 0 \text{ then } \sum\limits_{i = 1}^k a_i = 0 \end{Bmatrix}$
Look at the code to see how much easier it is to read.

Finally, you could make things so much simpler if you write
$ \sum\limits_{i = 1}^k a_i = 0 \iff \bigwedge\limits_{i = 1}^k a_i = 0$

If you are using an equation editor, ditch it and learn to write cleaner and simpler LaTeX yourself.

Hey,
[quote="stevem"]
Use \sum\limits and \bigwedge\limits to get the limits to display above and below.
[/quote]
Thanks for the tip.

[quote="stevem"]
The problem is that the code is a mess. Was it generated by an equation editor because, while technically correct, it is unreadable?
[/quote]
No, I wrote the code myself in the $ {\text{\LaTeX}}$ editor [i]WinEdt 5 [Build: 20071003 (v. 5.5)][/i] with the [i]MiKTeX 2.7 [Revision 2.7.3105][/i] package.

[quote="stevem"]
1) There are far too many curly brackets. Why, for example is the whole thing enclosed in them?
[/quote]
In [i]WinEdt 5[/i] braces (ie. $ {\{\}}$) are colored when highlighted creating a visual of separation of $ {\text{\LaTeX}}$ code, IMO this is quite useful when writing LaTeX code.

[quote="stevem"]
2) You should be using \text rather than \mbox. {,}{\,}{\,}{\mbox{then}}{\,}{\,} is dreadful. What is wrong with \text{ then }? Note how the space is inside the text box
[/quote]
Thanks for the tip, I had forgotten I could also use "\text{}". Additionally, what are differences between using "\text{}" versus "\mbox{}"?

[quote="stevem"]
3) Why are there 2 columns in the array, yet only one is used?
[/quote]
What part of the code,
[code]
{\left\{\begin{array}{ll} {{\mbox{if}}{\,}{\,}{{\sum_{i = 1}^{k}{a}_{i}} = {0}}{,}{\,}{\,}{\mbox{then}}{\,}{\,}{{\bigwedge_{i = 1}^{k}{a}_{i}} = {0}}} \\
\vspace{0pt} \\
{{\mbox{if}}{\,}{\,}{{\bigwedge_{i = 1}^{k}{a}_{i}} = {0}}{,}{\,}{\,}{\mbox{then}}{\,}{\,}{{\sum_{i = 1}^{k}{a}_{i}} = {0}}} \\
\end{array}\right\}}
[/code]
are you referring to?

[quote="stevem"]
4) Why is there \\ after the last line of the array?
[/quote]
I added that for symmetry in the code, I didn't see a difference in when I compounded the code so I figured it wouldn't do any harm letting it stay there. Specifically, what does the $ {\text{\LaTeX}}$ code "\" do?

[quote="stevem"]
5) Why use \vspace{0pt} \\?
[/quote]
Sometimes I forget what the $ {\text{\LaTeX}}$ code for a vertical space is so I tend to add it in anyways, even if I don't use it. I tend to do this more as a reminder that I can add a vertical space and for convenience when I need to use it since it's already there.

[quote="stevem"]
6) The Bmatrix environment makes things simpler and saves having to add curly brackets to an array.
[/quote]
Thanks for the tip.

[quote="stevem"]
Putting this together you get
$ \begin{Bmatrix} \text{if }\sum\limits_{i = 1}^k a_i = 0 \text{ then } \bigwedge\limits_{i = 1}^k a_i = 0 \\[2ex]
\text{if }\bigwedge\limits_{i = 1}^k a_i = 0 \text{ then } \sum\limits_{i = 1}^k a_i = 0 \end{Bmatrix}$
Look at the code to see how much easier it is to read.
[/quote]
Nice, that looks much better. With regards to the code, I prefer using braces more often than not simply because it provides a way of grouping my $ {\text{\LaTeX}}$ code.

[quote="stevem"]
Finally, you could make things so much simpler if you write
$ \sum\limits_{i = 1}^k a_i = 0 \iff \bigwedge\limits_{i = 1}^k a_i = 0$
[/quote]
I see, I didn't realize,
$ {\iff} \equiv {\text{if and only if}}$

[quote="stevem"]
If you are using an equation editor, ditch it and learn to write cleaner and simpler LaTeX yourself.
[/quote]
My experience writing $ {\text{\LaTeX}}$ code in [i]WinEdt 5[/i] with the [i]MikTex 2.7[/i] package is that it's very straightforward and gives me a lot of freedom in how to write my LaTeX code.

Lastly, when rewriting my $ {\text{\LaTeX}}$ code,
$ \begin{Bmatrix}
\text{if}{\,}{\,}{\sum\limits_{i = 1}^{k}{a}_{i}} = {0}{,}{\,}{\,}\text{then}{\,}{\,}\bigwedge\limits_{i = 1}^{k}{a}_{i} = {0} \\[2ex]
{\,}\text{if}{\,}{\,}\bigwedge\limits_{i = 1}^{k}{a}_{i}= {0}{,}{\,}{\,}\text{then}{\,}{\,}\sum\limits_{i = 1}^{k}{a}_{i} = {0}
\end{Bmatrix}$
I noticed in your code there was a,
[code]
 \\ [2ex]
[/code]
what does this LaTeX code do?

In addition, when compounding the LaTeX code I noticed that the "ifs" weren't quite aligned, is this because when using the "Bmatrix" code all equations within the braces are centered as opposed to aligned to the left? If that is the case, how can I get the equations to align to the left?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

[quote="APSStudent"]No, I wrote the code myself in the  editor WinEdt 5 [Build: 20071003 (v. 5.5)] with the MiKTeX 2.7 [Revision 2.7.3105] package. [/quote]Excellent!

[quote="APSStudent"] creating a visual of separation of LaTeX code [/quote]What do you mean? The [i]whole[/i] of your document [i]is[/i] LaTeX code. You perhaps mean maths mode instead of text mode but a big strength of LaTeX is that it mixes the two modes with only dollar signs or \ [...\ ] to separate them, so everything flows much more smoothly. Thus
[code]The polynomial includes a term in $x^2$ so that ...[/code]is much clearer to read than your
[code]The polynomial includes a term in ${{x}^{2}}$ so that ...[/code]LaTeX documents should be easy to follow even before compiling them.

[quote="APSStudent"]what are differences between using "\text{}" versus "\mbox{}"?[/quote]\text is used by the amsmath package and switches font size correctly for subscripts and superscripts. Compare $ x_\text{word}$ with $ x_\mbox{word}$

[quote="APSStudent"] Why are there 2 columns in the array, yet only one is used?
What part of the code are you referring to?[/quote]You have \begin{array}{ll} which has 2 left-aligned columns whereas for example, \begin{array}{ccc} has 3 centered columns. You use & to get to the next column.

[quote="APSStudent"] Why is there \\ after the last line of the array?
 I added that for symmetry in the code, I didn't see a difference in when I compounded the code so I figured it wouldn't do any harm letting it stay there. Specifically, what does the  code "\" do? [/quote]Symmetry?? \\ means a new line and you will get into trouble if you use it indiscriminately. For example
\begin{align} x = y \\
a = b \\
\end{align}
where you see that the last \\ gives an equation number for a blank line.

[quote="APSStudent"] Why use \vspace{0pt} \\?
Sometimes I forget what the  code for a vertical space is so I tend to add it in anyways, even if I don't use it. I tend to do this more as a reminder that I can add a vertical space and for convenience when I need to use it since it's already there.[/quote]But you have put vertical space there by using \\ which you don't need.

[quote="APSStudent"] \\ [2ex]
what does this LaTeX code do?  [/quote]It adds a vertical space of 2ex so that the next line doesn't touch the one above (2ex = 2 times the height of an x in the current font).

[quote="APSStudent"] Bmatrix" code all equations within the braces are centered as opposed to aligned to the left [/quote]That's true, in that case you will have to use an array with one left-column. Better still, if you need two lines, write
$ \begin{Bmatrix}
\sum\limits_{i = 1}^k a_i = 0 \Longrightarrow \bigwedge\limits_{i = 1}^k a_i = 0 \\[2ex]
\bigwedge\limits_{i = 1}^k a_i = 0 \Longleftarrow \sum\limits_{i = 1}^k a_i = 0 \end{Bmatrix}$

I would strongly recommend that you read up on LaTeX. There are references at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url] but I would particularly recommend the book [i]Guide to Latex by Kopka and Daly[/i] which is listed there.

I have just seen the code behind this:
[quote="APSStudent"]Lastly, when rewriting my $ {\text{\LaTeX}}$ code,
$ \begin{Bmatrix} \text{if}{\,}{\,}{\sum\limits_{i = 1}^{k}{a}_{i}} = {0}{,}{\,}{\,}\text{then}{\,}{\,}\bigwedge\limits_{i = 1}^{k}{a}_{i} = {0} \\[2ex] {\,}\text{if}{\,}{\,}\bigwedge\limits_{i = 1}^{k}{a}_{i} = {0}{,}{\,}{\,}\text{then}{\,}{\,}\sum\limits_{i = 1}^{k}{a}_{i} = {0} \end{Bmatrix}$[/quote]
Why do you use {\,}{\,}?
\, is 3/18 of a quad* so what is wrong with \; which 5/18 of a quad or an \hspace{[i]width[/i]}, where you put in the width you want? You are just overcomplicating things and the space does not need to be so precise. Even better what is wrong with a normal space using \text{ then }?

*a quad is a horizontal space equal to the current font size eg 10pt for a 10pt typeface so \,\, would be 6/18 of 10pt.

Hey,
[quote="stevem"]
What do you mean? The [i]whole[/i] of your document [i]is[/i] LaTeX code. You perhaps mean maths mode instead of text mode but a big strength of LaTeX is that it mixes the two modes with only dollar signs or \ [...\ ] to separate them, so everything flows much more smoothly.
[/quote]
Yea, I meant to say when writing [u]equations[/u] using the $ {\text{\LaTeX}}$,
[code]
\begin{equation}
\end{equation}
[/code]
commands in [i]WinEdt 5[/i].

[quote="stevem"]
Thus
[code]
The polynomial includes a term in $x^2$ so that ...
[/code]
is much clearer to read than your
[code]
The polynomial includes a term in ${{x}^{2}}$ so that ...
[/code]
LaTeX documents should be easy to follow even before compiling them.
[/quote]
I agree, that code with less braces is easier to read. However, I tend to prefer using braces when writing [u]equations[/u] in $ {\text{\LaTeX}}$ since [i]WinEdt 5[/i] highlights the braces making it easier to group the LaTeX code of my [u]equations[/u].
Additionally, why is it that LaTeX documents should be easy to follow even before compiling? Isn't the goal of using LaTeX to obtain the desired displayed result as opposed to having easy to follow code?

[quote="stevem"]
\text is used by the amsmath package and switches font size correctly for subscripts and superscripts. Compare $ x_\text{word}$ with $ x_\mbox{word}$
[/quote]
Thanks for clearing that up. So, what is "\mbox{}" used for then?

[quote="stevem"]
You have \begin{array}{ll} which has 2 left-aligned columns whereas for example, \begin{array}{ccc} has 3 centered columns. You use & to get to the next column.
[/quote]
Ok, thanks for clearing that up.

[quote="stevem"]
Symmetry?? \\ means a new line and you will get into trouble if you use it indiscriminately. For example
\begin{align} x = y \\
a = b \\
\end{align}
where you see that the last \\ gives an equation number for a blank line.
[/quote]
I follow your example, however consider the following,
$ \begin{array}{ll} x = y \\
a = b \\
\end{array}$
Note that adding "\" after each line appears to add a new line within "[\begin{align}][\end{align}]", however in "[\begin{array}][\end{array}]" no new line is added even though "\" is placed at the end of each line.

[quote="stevem"]
But you have put vertical space there by using \\ which you don't need.
[/quote]
Ok, thanks for the tip.

[quote="stevem"]
It adds a vertical space of 2ex so that the next line doesn't touch the one above (2ex = 2 times the height of an x in the current font).
[/quote]
I see, thanks.

[quote="stevem"]
That's true, in that case you will have to use an array with one left-column. Better still, if you need two lines, write
$ \begin{Bmatrix}
\sum\limits_{i = 1}^k a_i = 0 \Longrightarrow \bigwedge\limits_{i = 1}^k a_i = 0 \\[2ex]
\bigwedge\limits_{i = 1}^k a_i = 0 \Longleftarrow \sum\limits_{i = 1}^k a_i = 0
\end{Bmatrix}$
[/quote]
I see, thanks for the example.

[quote="stevem"]
I would strongly recommend that you read up on LaTeX. There are references at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url] but I would particularly recommend the book [i]Guide to Latex by Kopka and Daly[/i] which is listed there.
[/quote]
I noticed my library has this book, when I get a chance I'll take a look at it. Thanks for the suggestion.

[quote="stevem"]
Why do you use {\,}{\,}?
\, is 3/18 of a quad* so what is wrong with \; which 5/18 of a quad or an \hspace{[i]width[/i]}, where you put in the width you want? You are just overcomplicating things and the space does not need to be so precise. Even better what is wrong with a normal space using \text{ then }?

*a quad is a horizontal space equal to the current font size eg 10pt for a 10pt typeface so \,\, would be 6/18 of 10pt.
[/quote]
I tend to use spacing in increments of "{\,}" out of a habit of being precise, since I believe this is the smallest (relative) space allowed in $ {\text{\LaTeX}}$. This is one of the reasons I like LaTeX, because of the freedom in precision it gives you. In addition, add in the alternatives you listed and there are even more options (in spacing) than I realized.

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

[quote="APSStudent"]Additionally, why is it that LaTeX documents should be easy to follow even before compiling? Isn't the goal of using LaTeX to obtain the desired displayed result as opposed to having easy to follow code?
[/quote]Certainly LaTeX was designed to produce good looking documents but also using easy to write code. When you write a document you usually have to make lots of changes before you get the final result so it is important to be able to follow the code you have used. When TeX was first designed, a lot of effort was made to keep the code legible which is why commands are clear as to their meanings. In fact many mathematicians don't need to compile the document as they can read the LaTeX code directly.

[quote="APSStudent"]So, what is "\mbox{}" used for then?[/quote]It is used in text mode and, for example, it keeps the text inside it together so it doesn't go onto a new line.


[color=darkblue]Let $ ax^3\equal{}by^3\equal{}cz^3$ and $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$. Prove that: $ \boxed{\sqrt[3]{ax^2\plus{}by^2\plus{}cz^2}\equal{}\sqrt[3]{a}\plus{}\sqrt[3]{b}\plus{}\sqrt[3]{c}}$[/color]

[quote="thanhnam2902"][color=darkblue]Let $ ax^3 \equal{} by^3 \equal{} cz^3$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that: $ \boxed{\sqrt [3]{ax^2 \plus{} by^2 \plus{} cz^2} \equal{} \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c}}$[/color][/quote]
 Let $ ax^3\equal{}by^3\equal{}zc^3\equal{}k$ then $ \sqrt [3]{ax^2 \plus{} by^2 \plus{} cz^2}\equal{} \sqrt[3]{k}$
  and other hand , we also have $ RHS \equal{}\sqrt[3]{k}$

[quote="Hong Quy"][quote="thanhnam2902"][color=darkblue]Let $ ax^3 \equal{} by^3 \equal{} cz^3$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that: $ \boxed{\sqrt [3]{ax^2 \plus{} by^2 \plus{} cz^2} \equal{} \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c}}$[/color][/quote]
 Let $ ax^3 \equal{} by^3 \equal{} zc^3 \equal{} k$ then $ \sqrt [3]{ax^2 \plus{} by^2 \plus{} cz^2} \equal{} \sqrt [3]{k}$
  and other hand , we also have $ RHS \equal{} \sqrt [3]{k}$[/quote]
Yes. Thank you!


hello, i got $ 1/2\,{\frac {\ln  \left( 67822307279 \right) }{{\it LambertW} \left( 1
/2\,\ln  \left( 67822307279 \right)  \right) }}$
as the solution. Sonnhard.

hello, this is the LambertW function, see here
http://mathworld.wolfram.com/LambertW-Function.html
Sonnhard.

Here is High School Basic , so from the start I expect an integer solution  :) .

$ x^{2x}\equal{}67822307279$, obviously $ x$ is odd and positive, it's easy to see that that if $ x \ge 10$ $ L.H.S.$ will be too large, if $ x \le 3$ $ L.H.S.$ will be too small.
Also , $ x \neq 5$. Now $ 9^{18}\equal{}1^{9} \neq 9(mod 10)$, impossible. So only $ x\equal{}7$ is left. After checking , $ x\equal{}7$ is the solution.


algebra
polynomial
number theory unsolved
number theory

Prove that for infinitely many n there is a real polynomial f of degree n such that $ p(0),p(1),...,p(n+1)$ are different whole powers of 2.

Suppose we want to make a polynomial that $p(0),...,p(n)$ are distinct powers of $2$
It's easy that there exist $a \in \mathbb{Z}$ that $k!|2^n (2 ^ a -1)$ for every $k \leq n$
Now consider the polynomial :
\[ p(x) = a_0 + a_1(x-1)+a_2(x-1)(x-2)+ \dots +a_n(x-1)(x-2)...(x-n)\]
That :\[\displaystyle  a_k =\frac {2^n (2^a-1)^k}{k!}\]
So $p \in \mathbb{Z}[x]$ and it's easy that $p(i)=2^{ia+n}$

Very nice... However, I think you were solving a different problem. We don't need to show that the coefficients are integers, and we need to put an extra power of $2$ as $P(n+1)$.

Is it true that the leading coeffiecient of a polynomial of degree $n$ is a multiple of $\sum_{r=0}^n {n \choose r} (-1)^r P(r)$? If that is the case, then we need only solve the equation $2^a {n \choose 0} - 2^b {n \choose 1} + ..... = 0$ for infinitely many $n$.

Yes, but I think the problem I posted is a little bit different. Or am I missing some obvious argument?

I think the problem you sent is obvious with Lagrange interpolation. :D  :D

Omid, could you please enlighten us?  :?

Excuse me I misunderstood the problem   :(


algebra
polynomial
algorithm
number theory
prime factorization

How to determine if a number is prime, if at a first glance it's neither even nor does its sum of digits divide 3?
If we determine it's not prime, then how to find its prime factorization?

The idiotic and foolproof way is to divide by all of the primes up to the square root of the given number. This is very slow.
The Miller-Rabin test is a probabilistic method that is loosely based on Fermat's little theorem, but it does not give you a factorization. It is much faster.

There's no known polynomial-time algorithm for factoring.

I just use the squares method.

It's really not that slow if you eliminate everything you can at first.  Then, it becomes a matter of dividing only about $ 4$ numbers, usually.  Notice, for example, if the number isn't divisible by $ 2$, $ 3$, and doesn't end in a $ 5$ or a $ 0$, you can quickly rule out $ 2,4,6,8,9,10,12,14,15...$, for example.

What's the square method?  You test for divisibility of all the numbers up to $ n$ such that $ n^{2}$ is the closest square to the number you're testing to be prime.  Any integer greater than $ n$ and less than the number you're prime factoring will have to be multiplied by a number in the range $ [1,n \minus{} 1]$, so you only  have to test those.


number theory
least common multiple
Divisibility Theory

Find the smallest positive integer $n$ such that [list][*] $n$ has exactly $144$ distinct positive divisors, [*] there are ten consecutive integers among the positive divisors of $n$. [/list]

Obviosly $ n\equal{}2^2*3^2*5*7*11*13$ (because $ n>11*144$) second conditions satisfyed.

[quote="Peter"]Out of curiosity, how do we have 10 consecutive positive divisors if $ 8\not|\ n$?[/quote]
I undestand second condition: exist divisors $ d_{k \plus{} 1} > d_k \plus{} 10.$ Among divisors $ d_{k}$ and $ d_{k \plus{} 1}$ had 10 consequitive numbers.
If we have 10 consequitive divisors, then $ n\equal{}2^3*3^2*5^2*7*11$.

(i.) $ n$ has exactly 144 distinct positive divisors,
    (ii.) there are ten consecutive integers among the positive divisors of $ n$.

Say the 10 consecutive divisors in (ii.) are $ k,~k \plus{} 1,\ldots,~k \plus{} 9$.  Then, $ 2^3,3^2,5,7$ all divide the $ \mbox{lcm}(k,k \plus{} 1,\ldots,k \plus{} 9).$  So, in order to satisfy (i.) and (ii.), we require that $ n \equal{} 2^{\nu_2}3^{\nu_3}5^{\nu_5}\cdots$ or i.e. $ n \equal{} \prod_{p\mid n} p^{\nu_p}$ where $ \nu_p\ge 0$ for all primes $ p \mid n$, and moreover $ \nu_2\ge 3$, $ \nu_3\ge 2$, $ \nu_5\ge 1$, and $ \nu_7 \ge 1$.  Let $ N \equal{} 2^5*3^2*5*7*11$ and note that $ N$ satisfies (i.) and (ii.).  As $ 144 \equal{} 2^4 * 3^2$, it follows that $ \nu_p$ can only be $ 1,2,3,5,7,8,11,\ldots$ for each $ p$.  Notice that $ 2^{11} * 3^2 * 5 * 7 > N$ implies that $ \nu_2\le 8$ which in turn forces $ \nu_p\le 8$ for each $ p$.  This greatly reduces the number of possibilities for the exponents of $ n$.  Yet, there is still work to be done.

From here one needs to argue that the choice of $ N$ above is least possible.  Altogether, we have that $ \nu_2\le 8$ and one can similarly obtain that $ \nu_3\le 5,\nu_5\le 3,\nu_7\le 2,$ and $ \nu_{11}\le 1$.  Also, we claim that no prime $ > 11$ can divide $ n$.  Keep in mind that $ 2^3*3^2*5*7 \mid n$.  Suppose there was such an $ n$ smaller than our choice $ N$ and a prime $ p > 11$ such that $ p\mid n$, then in order to satisfy (ii.):
\[ 2^3*3^2*5*7*p^{\nu_p} \le N \mbox{~~i.e~~} p^{\nu_p} \le 44
\]
in which we obtain that $ \nu_p \equal{} 1$.
But in order to satisfy also (i.), we now have the stronger statement that
\[ 2^3*3^2*5^2*7*p \le N \mbox{~~i.e~~} p \le 44/5 < 9.
\]
Thus, the claim follows.

Altogether, the only possibilities are $ \nu_2 \equal{} 3,5,7,$ or $ 8$, $ \nu_3 \equal{} 2,3,$ or $ 5$, $ \nu_5 \equal{} 1,2,$ or $ 3$, $ \nu_7 \equal{} 1$ or $ 2$, and $ \nu_{11} \equal{} 0$ or $ 1$.  Simple (yet tedious) analysis gives the above choice $ N$ to be the desired solution.


There is a positive integer that can be expressed as b^n where both b and n are positive integers between 1 and 14 (inclusive). What is the greatest number of factors b^n can have?

I think I know the answer...but im not sure  :(

Trivially, for the largest number of factors, $ n\equal{}14$

Then we just pick $ b$ with the largest possible number of factors in $ b$, which would be $ b\equal{}12$.
Then, $ b^{14}\equal{}2^{28}3^{14}$, so the number of factors would be $ 29\times 15 \equal{} 435$.

If you want, fill in the parts I skimmed over.

Okay thanks!

Thats what i got so im happy  :D


For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$

[quote="amirhtlusa"][hide="hint1"]
first try to prove the smallest set is a set of 2k+1 consequtive integers[/hide]
[hide="2"]
then find the smallest value for a1 that works[/hide][/quote]
It's easier to group the integers and find the (k+1)st integer. Then you can just say the k integers larger than it must be consecutive, and u are done.

let $a_1 < a_2 < ... < a_{k+1} < a_{k+2} < ... < a_{2k+1}$ be distinct positive integers such that they sum to more than $N$ and any subset of $k$ of them has sum at most $N/2$. let $a_{k+2} = L$. we may then assume that $(a_1, ..., a_{k+1}) = (L-(k+1), ..., L - 1)$. (if not, replacing them with these numbers still produces a set satisfying the above conditions, because the $k$ largest elements remain unchanged.)

Let $a_{k+2} +...+ a_{2k+1} = \frac{N}{2} - r$, where $r \geq 0$. Then because $\sum a_i > N$, we must have

$\frac{N}{2} + r + 1 \leq a_1 + ... +a_{k+1} = L(k+1) - \frac{(k+1)(k+2)}{2} = S$ (**)

Note also that $L = a_{k+2},\ L \leq a_{k+3} - 1,\ \cdots,\ L \leq a_{2k+1} - (k-1)$, so we get

$S \leq a_{k+2} (k+1) - \frac{(k+1)(k+2)}{2}$,

$S \leq (a_{k+3} - 1)(k+1) - \frac{(k+1)(k+2)}{2}$,

...

$S \leq (a_{2k+1} - (k-1))(k+1) - \frac{(k+1)(k+2)}{2}$,

and adding we get

$kS \leq (k+1)(a_{k+2} + ... + a_{2k+1}) - \frac{(k+1)(k-1)k}{2} - \frac{k(k+1)(k+2)}{2}$, from which

$S \leq \frac{k+1}{k}(\frac{N}{2}-r)-\frac{(k+1)(k-1)}{2} - \frac{(k+1)(k+2)}{2}$.

together with (**) this gives

$\frac{N}{2} + r + 1 \leq \frac{k+1}{k}(\frac{N}{2}-r)-\frac{(k+1)(k-1)}{2} - \frac{(k+1)(k+2)}{2}$;

$\frac{(k+1)(k-1)}{2} + \frac{(k+1)(k+2)}{2} + r + r\frac{k+1}{k} + 1\leq \frac{N}{2k}$;

$N \geq k(k+1)(k-1) + k(k+1)(k+2) + 2k = 2k^3+3k^2+3k$.

equality is achieved when the $a_i$ are consecutive and their sum is exactly $2k^3+3k^2+3k+1$... i.e., when the numbers are $\{k^2+1, ..., k^2+2k+1\}$

Let $a_1 < ... <a_k< a_{k+1} < a_{k+2} < ... < a_{2k+1}$ be distinct positive integers such that they sum to more than $N$ and any subset of $k$ of them has sum at most $\frac{N}{2}$.

It's clear that $a_{i+1} - a_i\ge 1$, so $a_{i+j} - a_i\ge j$.

By assumption, we have,
$(a_1 +\cdots+a_k)+ a_{k+1} +(a_{k+2} +\cdots+ a_{2k+1}) \ge N+1$

$= 2\cdot \frac{N}{2} + 1 \ge 2(a_{k+2} +\cdots+ a_{2k+1})+1$

So,
$a_{k+1}\ge (a_{k+2} +\cdots+ a_{2k+1}) - (a_1 +\cdots+a_k) + 1$

$=(a_{k+2}-a_1) + \cdots+ (a_{2k+1} - a_k)+1 \ge k(k+1)+1=k^2+k+1$

Therefore,
$\frac{N}{2} \ge a_{k+2} +\cdots+ a_{2k+1} =(a_{k+2}-a_{k+1}) + \cdots+ (a_{2k+1} - a_{k+1})+k a_{k+1}$

$\ge (1+\cdots+k) + k(k^2+k+1)$

That is,
$N\geq k(k+1)+ 2k(k^2+k+1)= 2k^3+3k^2+3k$.

For $N= 2k^3+3k^2+3k$, it's easy to verify that $k^2+1,\ldots, k^2+2k+1$ has sum $N+1$ and the sum of the last $k$ of them (which is the largest) is $\frac{N}{2}$.

Let the elements of our set $\{a_1, a_2, \ldots , a_{2k+1}\}$ be strictly increasing. Call the "upper bound", UB, $\left(\displaystyle\sum_{i=1}^{2k+1}a_i\right)-1$, and call the "lower bound", LB, $2\displaystyle\sum_{i=k+2}^{2k+1}a_i$.

Now suppose we have a set with a valid $N$, i.e. $LB \le N \le UB$. Then if we increase $a_i$ by 1, where $i \le k$, then LB is unaffected and UB increases by 1, so we still have a valid $N$. If we decrease $a_i$ by 1, where $i \ge k+2$, then UB decreases by 1 and LB decreases by 2. So we still have a valid $N$ value, say $N'$, and $N' = N - 2 < N$. So we "crunch" our set around $a_{k+1}$ to get a set of consecutive integers, with a valid $N$-value less than or equal to the original. So our minimum $N$ will be found among sets of consecutive integers.

Now we look at the set $\{1, 2, \ldots , 2k+1\}$, where

$UB = (1 + 2 + \ldots + 2k+1) - 1 = \frac{(2k+1)(2k+2)}{2} - 1 = 2k^2 + 3k$, and

$\\ LB = 2\left[(k+2) + (k+3) + \ldots + (2k+1)\right] \\ = 2\left[k(k+2) + (1 + 2 + \ldots + (k-1))\right] \\ = 2k^2 + 4k + k(k-1) \\ = 3k^2 + 3k.$

Call a "bump" to process of increasing each element of our set by 1. Through bumping our set $\{1, 2, \ldots , 2k+1\}$ we will get all sets of consecutive integers.

A bump increases UB by $2k+1$, and increases LB by $2k$. Thus a bump increases UB by $1$ relative to LB. Since originally
$LB - UB = (3k^2 + 3k) - (2k^2 + 3k) = k^2$,
we must perform at least $k^2$ bumps to satisfy $UB \ge LB$. We will find our minimum $N$ after $k^2$ bumps, when
$N = LB = (3k^2 + 3k) + k^2(2k) = 2k^3 + 3k^2 + 3k.$

I thought this problem was just too easy.
let {a_1,a_2 .. a_(2k+1)}be a satisfactory set of numbers with a_i < a_j iff i<j.
it is evident that N+1 =< a_1 ... + a _(2k+1) and N/2 >= a_(k+2) .. + a_2k+1
therefore: 2 ( a_(k+2) .. + a_(2k+1) ) +1 <= a_1 + .. + a_(2k+1)
<--> a_(k+2) .. + a_(2k+1) + 1 <= a_1 + (a_2 + ... a_(k+1))
<--> a_1 >= [a_(k+2) - a_2] + .. + [ a_(2k+1) - a_(k+1) ] + 1
               = k + k .. + 1 = k^2 + 1
so it follows a_t >= k^2 + t; the conclusion follows easily N/2 >= (k^2 + k + 2) .. + (k^2 + 2k + 1)
--> N>= 2k^3 + 3k^2 +  3k, and the fact that {k^2 + 1 ... (k+1)^2 } works finishes off the problem.

How did you guys find the example sequence? Guess and check?

I'm not sure if this is what you are asking, but after playing with the problem for awhile, you can guess that the minimal set of $ 2k \plus{} 1$ integers is a set of consecutive integers.  So the set is $ \{ a, a \plus{} 1, \ldots, a \plus{} 2k \}$ for some integer $ a$.  Then use the hypotheses to figure out the best value of $ a$.

That is how you can guess the minimal set.  You still need a rigorous argument to settle the problem.

I think this solution is more intuitive than some of the above (it's essentially the same idea, just phrased a bit differently).

[hide]
We claim that the answer is $ k(2k^2 \plus{} 3k \plus{} 3)$.

To begin, let the sequence, in increasing order, be $ a_1,a_2,\ldots, a_{2k \plus{} 1}$. Then the subset of length $ k$ with the maximum sum is of course $ \sum_{i \equal{} k \plus{} 2}^{2k \plus{} 1}a_i$. Our condition is thus equivalent to $ 2\sum_{i \equal{} k \plus{} 2}^{2k \plus{} 1}a_i\le N < \sum a_i$. Noting that if $ 2\sum_{i \equal{} k \plus{} 2}^{2k \plus{} 1}a_i < \sum a_i$ then we can always find an integer $ N$ satisfying this (and thus the two inequalities are equivalent), we rearrange this to $ a_{k \plus{} 1} > a_{k \plus{} 2} \plus{} a_{k \plus{} 3} \plus{} \ldots \plus{} a_{2k \plus{} 1} \minus{} (a_1 \plus{} a_2 \plus{} \ldots \plus{} a_k)$.

Now observe that from our inequality, we minimize $ N$ at $ 2\sum_{i \equal{} k \plus{} 2}^{2k \plus{} 1}a_i$; hence we want to make $ \sum_{i \equal{} k \plus{} 2}^{2k \plus{} 1}a_i$ as small as possible.

[b]Lemma:[/b] The minimum value of $ N$ can be achieved with a sequence of consecutive integers.

[b]Proof:[/b] If any two consecutive terms in the sequence differ by more than one, we can trivially shift down every term above it by an amount so that the terms are consecutive - furthermore, if the original sequence satisfies our inequality, at least one of the two new ones will do as well, since bringing the terms "closer together" clearly decreases the RHS by at least as much as the LHS. Additionally, $ N$ is made smaller. Hence, the desired result is proven.

We thus choose consecutive integers to be our sequence; the difference on the RHS is then $ k$ times the differences of $ k \plus{} 1$. Hence at minimum, $ a_{k \plus{} 1} \equal{} k^2 \plus{} k \plus{} 1$; the minimum of this term gives a minimum of $ N$ as well because all the terms are consecutive. Our sequence is then $ k^2 \plus{} 1,k^2 \plus{} 2,\ldots,k^2 \plus{} k,k^2 \plus{} k \plus{} 1,k^2 \plus{} k \plus{} 2,\ldots,k^2 \plus{} 2k,k^2 \plus{} 2k \plus{} 1$. The maximum sum of $ k$ of these is thus $ k(2k^2 \plus{} 3k \plus{} 3)/2$, and $ N$ is therefore at minimum twice this, or $ \boxed{k(2k^2 \plus{} 3k \plus{} 3)}$, as desired.
[/hide]

[hide="A little bit different"]

Let the numbers be in a strictly increasing sequence $a_1<a_2<\ldots<a_{2k+1}$
We have that $a_{k+i}\ge a_i+k$ (*)

Now, let $s=a_{k+2}+\ldots+a_{2k+1}$
$2s\le N$

$a_2+a_3+\ldots+a_{k+1}\le s-k^2$ because of (*)
Now, $a_1+2s-k^2>N$, so $a_1>k^2$.

Let $a_i=k^2+b_i$ for all $i$ from $1$ to $2k+1$. The sequence $b$ has the same property (*)

Now, from $2s\le N$ we get:
$2k^3+2(b_{k+2}+\ldots+b_{2k+1})\le N$, so $N$ is at least $2k^3+3k^2+3k$, because $b_i\ge i$ for all $i$ from $1$ to $2k+1$.

It's easy to prove that this is the minimum value of $N$, by trying the equality case, when $a_i=k^2+i$.

[/hide]

Let the positive integers be $a_1<a_2<\ldots<a_{2k+1}$. The condition then becomes $a_1+a_2+\ldots+a_{2k+1}>N$ and $\frac{N}{2}\geq a_{k+2}+a_{k+3}+\ldots+a_{2k+1}$ so $a_1+a_2+\ldots+a_{k+1}>N\geq2(a_{k+2}+a_{k+3}+\ldots+a_{2k+1})$, or $a_1+a_2+\ldots+a_{k+1}>a_{k+2}+a_{k+3}+\ldots+a_{2k+1}$. This condition is equivalent because by setting $N=2(a_{k+2}+a_{k+3}+\ldots+a_{2k+1})$.

Now, note that by decreasing $a_i$ by 1 where $k+2\leq i\leq 2k+1$ (if possible), the condition still holds but we decrease the value of $N$. Hence, $a_{k+1},a_{k+2},\ldots,a_{2k+1}$ are consecutive for the minimal $N$. We now have $ka_{k+1}+\frac{k(k+1)}{2}=a_{k+2}+a_{k+3}+\ldots+a_{2k+1}<a_1+a_2+\ldots+a_{k+1}\leq(k+1)a_{k+1}-\frac{k(k+1)}{2}$, so $a_{k+1}>k(k+1)$, which means that $a_{k+1}\geq k^2+k+1$. Hence, $N=2(a_{k+2}+a_{k+3}+\ldots+a_{2k+1})=2ka_{k+1}+k(k+1)\geq 2k(k^2+k+1)+k(k+1)=2k^3+3k^2+3k$. Equality holds when $a_i=k^2+i$, so the minimum possible value of $N$ is $2k^3+3k^2+3k$.

The answer is $N = k(2k^2+3k+3)$ given by \[ S = \left\{ k^2+1, k^2+2, \dots, k^2+2k+1 \right\}. \]

To show this is best possible, let the set be $S = \{ a_0 <  a_1 < \dots < a_{2k} \}$ so that the hypothesis becomes \begin{align*} 	N + 1 &\le a_0 + a_1 + \dots + a_{2k} \\ 	N/2 &\ge a_{k+1} + \dots + a_{2k}. \end{align*} Subtracting twice the latter from the former gives \begin{align*} 	a_0 &\ge 1 + (a_{k+1}-a_1) + (a_{k+2}-a_2) + \dots 	+ (a_{2k} - a_k) \\ 	&\ge 1 + \underbrace{k + k + \dots + k}_{k \text{ terms}} \\ 	&= 1 + k^2. \end{align*} Now, we have \begin{align*} 	N/2 &\ge a_{k+1} + \dots + a_{2k} \\ 	&\ge (a_0 + (k+1)) + (a_0 + (k+2)) + \dots + (a_0 + 2k) \\ 	&= k \cdot a_0 + \left( (k+1) + \dots + 2k \right) \\ 	&\ge k(k^2+1) + k \cdot \frac{3k+1}{2} \end{align*} so $N \ge k(2k^2+3k+3)$.

[b][color=red]Remark:[/color][/b]  	The exact value of $N$ is therefore very superficial. 	From playing with these concrete examples 	we find out we are essentially 	just trying to find an increasing set $S$ obeying 	\[ a_0 + a_1 + \dots + a_k > a_{k+1} + \dots + a_{2k} \qquad (\star) \] 	and indeed given a sequence satisfying these properties 	one simply sets $N = 2(a_{k+1} + \dots + a_{2k})$. 	Therefore we can focus almost entirely on $a_i$ and not $N$.

[b][color=red]Remark:[/color][/b]  	It is relatively straightforward to figure out what is going on 	based on the small cases. 	For example, one can work out by hand that 	[list] 		[*]$\{2,3,4\}$ is optimal for $k=1$ 		[*]$\{5,6,7,8,9\}$ is optimal for $k=2$, 		[*]$\{10,11,12,13,14,15,16\}$ is optimal for $k=3$. 	[/list] 	In all the examples, 	the $a_i$ are an arithmetic progression of difference $1$, 	so that $a_j - a_i \ge j-i$ is a sharp for all $i<j$, 	and thus this estimate may be used freely without loss of sharpness; 	applying it in $(\star)$ gives a lower bound on $a_0$ 	which is then good enough to get a lower bound on $N$ 	matching the equality cases we found empirically.

Let the $k$ smallest numbers be $a_1<\dots< a_k$, the $k$ largest numbers be $b_1< \dots< b_k$, and the median be $M$. Note that for all $i=1,2,\dots, k$, the inequality $b_i \ge a_i + (k+1)$ holds. Furthermore, since $\sum_{i=1}^{k} b_i \le N/2$ but the entire sum is greater than $N$, we have $M+ \sum_{i=1}^{k} a_i > \sum_{i=1}^{k} b_i$. Combining these observations, we deduce that $M>k(k+1$). Thus, we have
\begin{align*}
\frac{N}{2} \ge \sum_{i=1}^{k} b_i &\ge (k^2+k+2) + (k^2+k+3) + \dots + (k^2 + 2k + 1) \\
&=\frac{2k^3 + 3k^2 + 3k}{2}
\end{align*}Hence, $N \ge 2k^3 + 3k^2 + 3k$. This can be achieved by the set of consecutive numbers $\{k^2+1, \dots, k^2+2k+1\}$, so this is indeed the minimal $N$. $\Box$

Let the numbers be $a_1 < a_2 < \cdots < a_k < a_{k+1} < \cdots < a_{2k+1}.$
The conditions are $a_{k+2}+\cdots+a_{2k+1} \leq N/2,$ and $a_1+\cdots+a_{2k+1}>N.$
Thus, $a_1+\cdots+a_{k+1}>N-(a_{k+2}+\cdots+a_{2k+1}) \geq N - N/2=N/2\geq a_{k+2}+\cdots+a_{2k+1}.$

If $b_i=a_{i+1}-a_1$ for $1 \leq i \leq k,$ we have $a_{k+j} \geq a_j + k = a_1+b_{j-1}+k$ for $2 \leq j \leq k+1.$
Then the inequality becomes $(k+1)a_1+(b_1+\cdots+b_k)>ka_1 + (b_1+b_2+\cdots b_k)+k\cdot k \implies a_1 > k^2$

Thus, $a_1+\cdots+a_{2k+1} \geq (k^2+1)+(k^2+2)+\cdots (k^2+2k+1)=2k^3+3k^2+3k+1,$
which is achieved exactly at $\{k^2+1,k^2+2,\ldots,k^2+2k+1\},$ giving the smallest $N=\boxed{2k^3+3k^2+3k}.$

The answer is $k(k+1)(2k+1)$, with equality achieved at $\{k^2 + 1, k^2 + 2, \dots , k^2 + 2k + 1 \}.$

 Let the elements of the set be $\{a_1, \dots a_{2k+1} \}$, where WLOG $a_1 < a_2 < a_3 < \dots < a_{2k+1}.$ From the given conditions, we have
\begin{align*}
a_1 + a_2 + \dots a_{2k+1} > N \quad (1) \\
a_{k+2} + a_{k+3} + \dots a_{2k+1} \le \frac{N}{2} (2) \\
\end{align*}

Let $b_i = a_i - a_1$. Now, we have that $\frac{N}{2} \ge a_{k+2} + a_{k+3} + \dots a_{2k+1} \ge ka_1 + (b_{k+2} + b_{k+3} + \dots b_{2k+1}),$ so we thus have $$a_1 + a_2 + \dots a_{2k+1} > N \ge 2ka_1 + 2(b_{k+2} + \dots b_{2k+1}) \implies (2k+1)a_1 + (b_2 + \dots b_{2k+1}) >  2ka_1 + 2(b_{k+2} + \dots b_{2k+1}).$$ Therefore, $a_1 > (b_{k+2} - b_2) + (b_{k+3} - b_3) + \dots (b_{2k+1} - b_{k+1}).$ Here, we will use the fact that $b_{k+i} - b_{i} \ge k$, so $(b_{k+2} - b_2) + (b_{k+3} - b_3) + \dots (b_{2k+1} - b_{k+1}) \ge k^2$, which in turn implies $a_1 > k^2.$  Now, the sum of the elements of the set is bounded by $$S \ge \sum_{i = 0}^{2k+1} a_1 + i  = k^2(2k+1) + k(2k+1) = (k^2 + k)(2k+1) = k(k+1)(2k+1),$$ and we are done. $\blacksquare$

I claim that the answer is $N=k(2k^2+2k+3)$, achievable by the set $\mathbb{S}=(k^2+1, k^2+2, \dots k^2+2k+1)$.

Suppose the set has numbers $a_1 < a_2 \dots < a_{2k+1}$. Then, $$2(a_{2k+1}+a_{2k}+\dots a_{k+2}) \leq N < a_1+a_2+\dots a_{2k+1}. \implies a_{2k+1}+a_{2k}+ \dots a_{k+2} < a_{1}+a_2+\dots a_{k+1}.$$ Now, suppose $\mathbb{S}$ is not optimal. The lower bound on $N$ is equal in this case, so we can only lower this bound, or lower the LHS of the second inequality (in which we have $b<b+1$ for some $b$.) But since $(a_n)$ is strictly increasing, we get a clear contradiction as there is $k+1$ terms on the LHS, and $k$ on the RHS (dominos!).

I claim the answer is $\boxed{N = (k^2+1) + (k^2+2) + \dots + (k+1)^2 - 1.}$

Equality holds by simply choosing the set $\{k^2+1,k^2+2,\dots, (k+1)^2\}$. We rephrase the problem as follows:

-----
Show that the minimum sum of any increasing sequence of distinct positive integers $a_1,a_2,\dots a_{2k+1}$ such that $$a_1+a_2+\dots+a_{k+1} > a_{k+2}+ a_{k+3} +\dots+a_{2k+1}$$ is at least $(k^2+1)+ (k^2+2) + \dots + (k+1)^2 - 1.$
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But simply note that $a_{k+i}-a_{i} \geq k$ for all $2 \leq i \leq k+1$, so we must have $a_1 > k^2,$ which concludes the proof.

The idea is to express the set as $a_1$, $a_1 + d_1$, $a_1 + d_1 + d_2$, etc etc all the way to $a_1 + d_1 + d_2 + \cdots + d_{2k}$, where each $d_i \geq 1$. Now, we pick the largest $k$-subset, being $a_{k + 2}$ to $a_{2k + 1}$.

We get the inequality$$ka_1 + kd_1 + kd_2 + \cdots + kd_{k + 1} + (k - 1)d_{k + 2} + \cdots + d_{2k} \leq N/2,$$but$$N/2 \leq \frac{1}{2} \left( (2k + 1)a_1 + 2kd_1 + (2k - 1)d_2 + \cdots + 1d_{2k} + 1 \right).$$Combine these stuffs together in order to bound $a_1$: after some manipulations, we get$$\begin{aligned} a_1 &\geq d_2 + 2d_3 + \cdots + kd_{k + 1} + (k - 1)d_{k + 2} + (k - 2)d_{k + 3} + \cdots + d_{2k} \\  &\geq (1 + 2 + \cdots + k) + (1 + 2 + \cdots + (k - 1)) + 1 \\ &= n^2 + 1, \end{aligned}$$gg. Equality holds when all the numbers are in a chain. Am way too computationally inept to do the calculation for $N$ in this equality case.

The answer is $\boxed{2k^3+3k^2+3k}$. This works with the numbers $\{k^2+1, k^2+2, \cdots, k^2+2k+1\}$.

This is really just an algebra problem, but we must make some observations. WLOG the numbers are $a_1 \le a_2 \le \cdots \le a_{2k+1}$. We need $$2(a_{k+2}+a_{k+3}+\cdots+a_{2k+1}) \le N < a_1+a_2+\cdots+a_{2k+1} \implies a_{k+2}+a_{k+3}+\cdots+a_{2k+1} < a_1+a_2+\cdots+a_{k+1}$$Now we use the buffalo method to convert this to algebra. Define $b_i = a_i-a_{i-1}$ for $i \ge 2$. The inequality, after some manipulation, becomes $$a_1 > b_3+2b_4+\cdots+kb_{k+2}+(k-1)b_{k+3}+(k-2)b_{k+4}+\cdots+b_{2k+1} \ge k^2$$where the last step follows because the $a_i$ are distinct positive integers, so the $b_i$ are positive integers. Now $$a_1+a_2+\cdots+a_{2k+1} > (k^2+1)+(k^2+2)+\cdots+(k^2+2k+1) = (k^2+k+1)(2k+1)$$so $N \le (k^2+k+1)(2k+1)-1 = 2k^3+3k^2+3k$. After verifying that this works with the construction $\{k^2+1, k^2+2, \cdots, k^2+2k+1\}$ (just check the sum of the largest $k$ members), we conclude the answer of $\boxed{2k^3+3k^2+3k}$. $\blacksquare$

Let the $2k+1$ integers be $a_1< a_2< \ldots<a_{2k+1}$ meaning that we can rewrite the integers as $$a_1+(2k)\le a_2+(2k-1)\le \ldots\le a_{2k+1}+(0)$$ so we have $2\sum_{i=k+2}^{2k+1}a_i \le N <\sum_{i=1}^{2k+1}$ meaning that $\sum_{i=k+2}^{2k+1}a_i < \sum_{i=1}^{k+1}a_i$ and since we want to minimize $N$, we want to minimize $N=2\sum_{i=k+2}^{2k+1}a_i$. Now, this means that since for all $1 \le i \le k+1$, we have that $a_i +k \le a_{i+k}$. So, $$N=2\sum_{i=k+2}^{2k+1}a_i \ge \sum_{i=1}^{2k+1}a_i+k^2+k-a_{k+1} > N+k^2+k-a_{k+1}$$ meaning $a_{k+1} \ge k^2+k+1$ meaning $N \ge k(k+1) + 2ka_{k+1} \ge 2k^3+3k^2+3k$. We prove this is achievable now: we have a set with $2k+1$ consecutive elements and middle element $a$, which means that $(2k+1)a > N$ and $\frac{k(k+1)}{2} + ka \le \frac{n}{2}$. Combining the equations gets $$(2k+1)a > 2 \cdot\frac{k(k+1)}{2} + 2 \cdot ka = k(k+1)+2ka$$ which simplifies to $a > k(k+1)$ so we have $a=k(k+1)+1$ as our smallest possible value. This means our answer of $\boxed{2k^3+3k^2+3k}$ is achievable so we are done.

We will prove that $N= k(2k^2+3k+3)$, which can be achieved by the set $\{k^2+1, k^2+2,...,k^2+2k+1\}$ (to see that this works, just take the largest sum).

Now, we will prove the lower bound.

WLOG $S=\{a_1<a_2<...<a_{2k+1}\} \implies a_i \geq a_j+(i-j)$ for all $1 \leq j < i \leq 2k+1 (\star)$

Furthermore, $$a_1+a_2+...+a_{2k+1}>N \geq 2(a_{2k+1}+...+a_{k+2})$$
$\implies a_1+...+a_{k+1} > a_{k+2}+...+a_{2k+1}$

Hence, from $(\star)$, $$ a_1+a_2+...+a_{k+1}> (k+a_2)+(k+a_3)+...+(k+a_{k+1})$$
$\implies \boxed{a_1 \geq k^2+1 (\star \star)}$

Now, observe that since $\frac{N}{2} \geq a_{2k+1}+...+a_{k+2}$, from $(\star), (\star \star)$,
$\frac{N}{2} \geq ka_1+k^2+\frac{k^2+k}{2} \geq k(k^2+1)+k^2+\frac{k^2+k}{2}$

$\implies N \geq k(2k^2+3k+3) \implies N= k(2k^2+3k+3)$, as desired.

$\blacksquare$

The answer is $\boxed{2k^3+3k^2+3k}$. To see that it is achievable, take the set of all $2k+1$ integers in the range $[k+1,k^2+2k+1]$.
To see that the answer is optimal, let the elements be $a_1<a_2<a_3\ldots <a_{2k+1}$. The difference between two consecutive terms is at least 1, so we have the following equations:
$$a_{2k+1} \ge a_k+k+1$$ $$a_{2k} \ge a_{k-1}+k+1$$ $$\ldots$$ $$a_{k+2} \ge a_1+k1.$$

Adding them gives $\sum_{i=k+2}^{2k+1}a_i\ge \sum^{i=k}_{1}a_i+k(k+1)$. Note that the given condition is equivalent to $\sum_{i=k+2}^{2k+1}a_i<\sum^{i=k+1}_{1}a_i$, so it's necessary that $a_{k+1}>k(k+1)$ and the minimal value it can attain is $k^2+k+1$. In this case, we must have $\sum_{i=k+2}^{2k+1}a_i=\sum^{i=k}_{1}a_i+k(k+1)$ which happens when the difference between any two consecutive terms is equal to 1. Increasing the value of $a_{k+1}$ by $b$ would also bump up each of the last $k$ terms by at least $b$, and the first $k$ terms by at least $k(k+1)b$ for $\sum_{i=k+2}^{2k+1}a_i<\sum^{i=k+1}_{1}a_i$ to hold; this obviously increases the value of $N$ and we are done.

Note that if we want to minimize $N$, it is in our best interest for the $2k+1$ integers to be consecutive because given a certain middle value, we want to maximize the sum of the lower $k$ numbers when compared to the higher $k$ numbers. The way to perform this optimization is if the $2k+1$ integers are all consecutive. Now that we've decided this, let the first number in the set be $n$. We have:
$$(2k+1)n + \frac{2k(2k+1)}{2} > 2(nk+k^2+\frac{k(k+1)}{2})$$
$$2nk+n+\frac{2k(2k+1)}{2} > 2nk + 2k^2 + k(k+1)$$
$$n > 2k^2 + k(k+1) - k(2k+1) = k^2$$
So, the smallest possible value is $n = k^2 + 1$. Plugging this in to find the minimized $N$ we get:
$$N = 2(k^3+k+k^2+\frac{k(k+1)}{2}) = 2k^3 + 2k + 2k^2 + k^2 + k = \boxed{2k^3 + 3k^2 + 3k}$$

For a set $S$ of positive integers $a_1 < a_2 < \dots < a_{2k+1}$, let $f(S)$ denote the least real number $N$ such that $2(a_{k+2} + a_{k+3} + \dots + a_{2k+1})\le N < a_1 + a_2 + \dots + a_{2k+1}$, if such an $N$ exists. Define a set to be good if $f(S)$ exists. Then the problem statement is equivalent to finding the minimum value of $f(S)$ over all good sets $S$ of $2k+1$ distinct positive integers. Finally, define $x(S)$ to be the sum of the $k+1$ least elements of $S$ and $y(S)$ to be the sum of the $k$ greatest elements of $S$ for all good sets $S$. Note that $S$ is good if and only if $x(S) > y(S)$, and $f(S) = 2y(S)$ for all good sets $S$..

Claim: for any good set $S$ we can find another good set $S'$ with $f(S') \le f(S)$ such that all $2k+1$ elements of $S'$ are consecutive integers.

Proof: Let $S'$ be the set of $2k+1$ integers centered around the $k+1^{\text{st}}$ element of $S$. Then $x(S')\ge x(S) > y(S)\ge y(S')$, so $S'$ is good. In addition $f(S') = 2y(S')\le 2y(S) = f(S)$, as desired.

By the claim, it suffices to find the minimum value of $f(T)$ over good sets $T$ that consist of $2k+1$ consecutive positive integers. Let $T = \{m-k, m-k+1, \dots, m+k\}$ be a good set. Then

$2y(T) < x(T) + y(T)\implies 2km + k^2 + k < (2k+1)m\implies k^2 + k < m$.

Clearly the minimum value of $f(T)$ occurs when $m$ is minimal, so setting $m = k^2 + k + 1$ we obtain

$2k(k^2 + k + 1) + k^2 + k = \boxed{2k^3 + 3k^2 + 3k}$

as the minimum value of $N$.

Mocking USAMO 2006 Day 1 result p2: 40 mins, 35 seconds (luis doing combo faster than NT, what...)
The idea, pick the set $\{1,2,...,2k+1 \}$ and shift each term by $a$ where $a$ is the minimun value for which our $N$ would exist, this works as we want to make the sum of elements in the set to be large and the largest possible sum of a subset to be as small as we can make it, and we cant only shift the lowest $k+1$ first terms of the set becuase then the largest $k$ elements will also get shifted (u can notice this better if u order the elements from the least term to the greater term), and then that means we have to shift everything (becuase its useless to only shift from a term that its not the least one as our sum wont be as greater as we want).
Now after the shifting using the known sums $\sum_{i=1}^{2k+1} i=2k^2+3k+1$ and $2\sum_{i=k+2}^{2k+1} i=2 \left(\sum_{i=1}^{2k+1} i-\sum_{i=1}^{k+1} i \right)=3k^2+3k$ we need
$$2k^2+(2a+3)k+a+1>N \ge 3k^2+(2a+3)k \implies a>k^2-1 \implies a=k^2 \implies N=2k^3+3k^2+3k$$
Equality when the set is $\{k^2+1,k^2+2,...,k^2+2k+1 \}$ thus we are done :D

We claim that the answer is $2k^3+3k^2+3k$. This can be achieved by taking $k^2+1$ through $(k+1)^2.$\\

Note that we essentially want to minimize the total sum while satisfying $$\text{Sum~of~k~largest}\leq \frac{(\text{Sum~of~all})-1}{2}.$$

Claim: In the optimal case, the set consists of consecutive positive integers. We say that the set has a inconsecutivity at index $i$ if the $i$th largest element is more than 1 larger than the $i+1$th largest element. Suppose FTSOC that an optimal set has an inconsecutivity at index $i$ for some $1\leq i \leq 2k.$ We will show that we can construct a better set.\\

Case 1: $i\leq k$. Then, we can decrease each of the first $i$ elements by 1, which looking back at our inequality $$\text{Sum~of~k~largest}\leq \frac{(\text{Sum~of~all})-1}{2},$$ will decrease the left side by $i$  but the right side by only $i/2$, so this inequality stays true but we have a more optimal set, contradiction.\\

Case 2: $i>k$. Then, we can still decrease each of the first $i$ elements by 1. The LHS of that inequality will decrease $k$, but the right side decreases by at most $k$ (since $i\leq 2k$ and "Sum of all" decreases by $i$), so the inequality stays true and we have a more optimal set, also contradiction.\\

Therefore, optimally, our set consists of consecutive positive integers. It is then easy to see that $k^2+1$ through $(k+1)^2$ is the smallest that works, so we are done.

Feels very superficial for USAMO2 or even USAMO 1/4.

The idea is to do some kind of smoothing argument as follows:

[b][color=#f00]Claim.[/b][/color] It is optimal to have the $k$ largest elements of $S$ be consecutive integers.

[i]Proof.[/i] Suppose that we have a working subset $S$ that achieves a minimum value of $N$ that does not satisfy the conditions. Sort the elements of $S$ as $a_1 < a_2 < \cdots < a_{2k+1}$.

Suppose there exists an index $k+2 \leq i \leq 2k+1$ such that $a_{i+1} > a_i + 1$. Then, we can shift $a_{i+1}$ down by $1$. Then, notice that the set $\{a_{k+2}, a_{k+3}, \cdots, a_{2k + 1}\}$ still has maximal sum out of all subsets; in particular, every subset has sum at most $\frac N2 - 1$. Furthermore, because the original $2k+1$ elements summed to at least $N$, this new set sums to at least $N-1$. Thus, choosing the constant $N-2$ in place of $N$ works in this case, so $N$ is not minimal, contradiction. $\blacksquare$

Now, the sum of the terms $a_1$ through $a_{k+1}$ must exceed the sum of the terms $a_{k+2}$ through $A_{2k+1}$ by the condition. Let $n = a_{k+2}$; then, we have the inequality $$n+(n+1)+(n+2)+\cdots+(n+k-1) \leq \sum_{i=1}^{k+1} a_i \leq (n-k-1) + (n-k) + \cdots + (n-1),$$ or $$(2n-k-2)(k+1) \geq (2n+k-1)k.$$ This simplifies to $n \geq k^2+k+1$, so the total sum $$\sum_{i=1}^{2k+1} a_i \geq k^2+(k^2+1) + \cdots + (k+1)^2 =2k^3+3k^2+3k.$$ Equality clearly holds.

Let the set $S$ contain integers $a_1 > a_2 > \cdots > a_{2k+1}\ge 1$. Note that every subset of $S$ with size $k$ having sum at most $\tfrac{N}{2}$ is equivalent to the assertion
\[ \quad a_1 + a_2 + \dots + a_k\le \frac{N}{2}.\]
Now, we have
\[ (k+1)a_k - \frac{(k+1)(k+2)}{2} = (a_k- 1) + \dots + (a_k- (k+1))\ge a_{k+1} + \dots + a_{2k+1} >  \frac{N}{2},\]
and
\[ \frac{N}{2}\ge a_1 + \dots + a_k \ge a_k + (a_k + 1) + \dots + (a_k + (k-1)) =  ka_k + \frac{(k-1)k}{2}.\]
It follows that
\[(k+1)a_k - \frac{(k+1)(k+2)}{2} > ka_k + \frac{(k-1)k}{2}\implies a_k > k^2 + k + 1.\]
Hence,
\[ N\ge 2ka_k + k^2 - k \ge 2k(k^2 +k + 2) + k^2 - k = \boxed{2k^3 + 3k^2 + 3k}.\]
A construction for this $N$ is $a_k = k^2 + k + 2$ and $a_{k + i} = k^2 + k + 2 - i$ for each $i = -k+1, \dots, k+1$.
[hide = comment]
idk why i used $a_1 > \dots > a_{2k+1}$ instead of $a_1 < \dots < a_{2k+1}$ like a sane person. i think it was because $a_1 + \dots + a_k$ was easier to write than $a_{2k+1} + \dots + a_{k+2}$ but idk
[/hide]

Very nice problem! The answer is $N=2k^3+3k^2+3k$ with the integers $k^2+1,k^2+2,\dots,k^2+(2k+1)$. WLOG $x_1<x_2<\dots<x_{2k+1}$. The key is to see that $$x_i\le x_{i+k}+k\implies x_2+x_3+\dots+x_{k+1}+k^2\le x_{k+2}+x_{k+3}+\dots+x_{2k+1}\le\frac N2$$$$\implies N-x_1<(x_2+x_3+\dots+x_{k+1})+(x_{k+2}+\dots+x_{2k+1})\le\frac N2-k^2+\frac N2\le N-k^2.$$$$\implies x_1>k^2\implies N\ge (k^2+1)+(k^2+2)+\dots+(k^2+2k+1)=2k^3+3k^2+3k.\blacksquare$$

I claim that the value of $N=k(2k^2+3k+3)$ works. First we prove the bound, then we give the construction.

Firstly, $a_1,a_2,\ldots a_{2k+1}$ denote the elements of the set that satisfy the problem conditions. Also, WLOG let $a_1\ge a_2\ge \cdots\ge a_{2k+1}$. Now note that the condition for the sum of every subset of size $k$ being most $\dfrac{N}{2}$ is equivalent to $a_1+a_2+\cdots+a_k\le \dfrac{N}{2}$ due to the ordering we imposed. Now from this bound, we have that $\dfrac{N}{2}\ge a_1+\cdots+a_k\ge (a_k+(k-1))+(a_k+(k-2))+\cdots+(a_k)=ka_k+\dfrac{(k-1)k}{2}\implies a_k\le \dfrac{N-k^2+k}{2k}$. We also had that the sum was greater than $N$ which gives the following.
\begin{align*}
    N&<a_1+a_2+\cdots+a_k+a_{k+1}+a_{k+2}+\cdots+a_{2k+1}\\
&\le (a_1+a_2+\cdots+a_k)+(a_k-1)+(a_k-2)+\cdots+(a_k-(k+1))\\
    &\le \dfrac{N}{2}+(k+1)a_k - \dfrac{(k+1)(k+2)}{2}
.\end{align*}

Now finally using the bound we got for $a_k$ that we got earlier with our current bound (along with some painful algebraic calculations), we get that $N\ge k(2k^2+3k+3)$. So this proves our minimality. Now we provide the construction.

Consider the following sequence.\[k^2+1,k^2+2,\ldots,k^2+2k+1.\]We can check that this actually satisfies our condition and we are done. :stretcher:

The answer is $\boxed{N = 2k^3 + 3k^2 + 3k}$. This can be achieved with the set $\{k^2 + 1, k^2 + 2, \ldots, k^2 + 2k + 1\}$. We will now show this is is indeed the minimal $N$.

[b][color=#f00]Claim:[/color][/b] The minimum $N$ occurs when all of the elements of the set are consecutive.
[b][color=#f00]Proof:[/color][/b] Let $x_1 < x_2 < \cdots < x_{2k + 1}$ be the elements of a set satisfying the condition, i.e. there is some (not necessarily minimal) $N'$ such that $x_1 + x_2 + \cdots + x_{2k + 1} > N'$ and $x_{k + 2} + x_{k + 3} + \cdots + x_{2k + 1} \le \tfrac{N'}{2}$. Let $d_1 = x_2 - x_1, d_2 = x_3 - x_2, \ldots, d_{2k} = x_{2k + 1} - x_{2k}$. Note that $d_1, d_2, \ldots, d_{2k} \ge 1$. If $d_i > 1$ for some $i$, then consider when happens when we shift $x_{i + 1}, x_{i + 2}, \ldots, x_{2k + 1}$ down by $d_i - 1$. We see that:
[list]
[*] If $i \ge k + 1$, then the sum of the $k$ largest elements and the sum of all of the elements both decrease by $(d_i - 1)(2k + 1 - i)$. Since $\tfrac{N'}{2} - (d_i - 1)(2k + 1 - i) < \tfrac{1}{2}(N' - (d_i - 1)(2k + 1 - i))$, this new set still satisfies the conditions
[*] If $i \le k$, then the sum of the $k$ largest elements decreases by $(d_i - 1)k$, while the sum of all elements decreases by $(d_i - 1)(2k + 1 - i)$. But $\tfrac{N'}{2} - (d_i - 1)k < \tfrac{1}{2}(N' - (d_i - 1)(2k + 1 - i))$, so this new set also satisfies the condition.
[/list]
Therefore, we can see in order to achieve the minimal $N$, all of the differences have to be $1$, which proves the claim.

Thus, we can now set $x_i = x_1 + i - 1$ for $2 \le i \le 2k + 1$. Then we get
$$(x_1 + k + 1) + (x_1 + k + 2) + \cdots + (x_1 + 2k) \le \tfrac{N}{2} < \tfrac{1}{2}(x_1 + (x_1 + 1) + \cdots + (x_1 + 2k)).$$
This simplifies to $x_1 > k^2$, so $N \ge 2((k^2 + k + 2) + (k^2 + k + 3) + \cdots + (k^2 + 2k + 1)) = 2k^3 + 3k^2 + 3k$, as desired.

The answer is $N = \boxed{2k^3+3k^2+3k}$, which can be achieved with the set $\{k^2+1, k^2+2, \dots, k^2+2k+1\}$.

WLOG let $x_1<x_2<\dots<x_{2k+1}$. Notice

\[x_i + k \le x_{i+k} \implies \sum_{i=2}^{k+1} x_i+k^2 \le \sum_{i=k+2}^{2k+1} x_i \le N/2\]
\[\implies N-x_1< (N/2-k^2)+N/2 = N-k^2.\]

Hence, $x_1>k^2$, so by setting $x_1=k^2+1$, and taking the rest as consecutive integers, we show that $2k^3+3k^2+3k$ is the minimum.

Easy question
[hide=Answer]
N= K(2K^2+3K+3) satisfies
K^2 +1, ..........., K^2 + 2k +1

Based upon smaller cases we conjecture that we need consecutive integers. Thus we want to make use of the inequality $x_i + 1 \leq x_{i+1}$ due to the sharpness principle, and force equality to hold.

Order the list in increasing order and denote the smallest element by $x_0$ and for all other elements, represent them as $x_i = x_0 + c_i$. Then,
\begin{align*}
kx_0 + \sum_{i = k+1}^{2k} c_i &\leq \frac{N}{2}\\
(2k+1)x_0 + \sum_{i=1}^{2k} c_i &> N
\end{align*}
Then subtracting twice of the first equation from the second we find,
\begin{align*}
x_0 &> \sum_{i=1}^k (c_{k+i} + c_i) > \sum_{i=1}^{k} k = k^2
\end{align*}
Thus $x_0$ is at least $k^2 + 1$. Then $N$ satisfies,
\begin{align*}
\frac{N}{2} &\geq k \cdot (k^2+1) + \sum_{i=k+1}^{2k} i\\
\iff N &\geq 2k^3 + 3k^2 + 3k
\end{align*}
Thus we must have $\boxed{N \geq 2k^3 + 3k^2 + 3k}$. It is easy to see that when equality holds, it works so we are done.

Our desired condition is equivalent to
\begin{align*}
&a_1 + a_2 + \ldots + a_{k+1} > a_{k+2} + a_{k+3} + \ldots + a_{2k+1} \\
\implies &a_1 > (a_{k+2}-a_2) + (a_{k+3}-a_3) + \ldots + (a_{2k+1}-a_{k+1}) \ge k^2.
\end{align*}

Thus the minimal value of $a_1$ is $k^2+1$. Note that the optimal value of $N$ is just twice the sum of the largest $k$ numbers in the set, so the $2k+1$ consecutive integers $\{k^2+1, k^2+2, \ldots, k^2+2k+1\}$ does the trick. This gives us our answer of $\boxed{2k^3+3k^2+3k}$.

The answer is $N = 2 \left(k(k^2 +1) + (2k+1)k - \frac{k(k+1)}{2}  \right) =  2k^3 + 3k^2 + 3k$, which is achievable by $\{k^2 + 1, k^2 + 2, \ldots, k^2 + 2k +1 \}$. Now we prove the bound.

For a given $k$, let a good set (ordered in increasing order) be any set of distinct positive integers where the sum of the last $k$ elements is less than the sum of the first $k + 1$ elements, and let a super set be a good set where the sum of the last $k$ elements is minimal. It's clear that the set of positive integers achieving $N$ must be good, and the smallest possible value of $N$ for that good set is equal to twice the sum of the last $k$ elements.

Claim: There exists a super set where all the elements are consecutive (they form an arithmetic sequence with common difference $1$).
Proof: If the last $k$ elements are not all consecutive in a super set, then if the smallest of these last $k$ elements (the $k+1$th element) is $x$, then the last element must be greater than $x + k$, but we can change the last $k$ elements into $x, x + 1, \ldots x + (k-1)$ (preserving goodness), so the original set was not super. Therefore the last $k$ elements must be consecutive. Now no matter what the first $k+1$ elements are, consider changing them into $x - (k+1), x - k, \ldots, x - 1$. As they were originally all distinct and under $x$, this doesn't decrease their sum, so the set stays good. Moreover, the set also stays super as the sum of the last $k$ elements is unaffected. This super set we constructed has elements $x - (k+1), x - k, \ldots, x + (k-1)$, which means it has consecutive elements, as desired. $\square$

The smallest value of $N$ is twice the sum of the last $k$ elements in any super set. Consider the unique super set that has consecutive elements, starting from $a_1$. The sum of the first $k+1$ elements is $(k+1) a_1 +\frac{k(k+1)}{2}$ and the sum of the last $k$ elements is $k a_1 + (2k+1) k - \frac{k(k+1)}{2}$. Hence, we have \[ (k+1) a_1 +\frac{k(k+1)}{2} > k a_1 + (2k+1) k - \frac{k(k+1)}{2} \implies a_1 > k^2 \]

Hence the sum of the last $k$ elements is at least $k(k^2 + 1) + (2k+1) k - \frac{k(k+1)}{2}$, so \[N \ge 2 \left( k (k^2 + 1) + (2k+1) k - \frac{k(k+1)}{2} \right) ,\] as desired.

First of all, assume WLOG that $a_1 < a_2 < \dots < a_{2k+1}$.\\
And note that
\[a_{j}-a_i \geq j-i.\]
By problem condition we have that

\[ a_1+a_2+ \dots +a_{2k+1} \geq N+1 > 2(a_{k+1}+a_{k+2}+ \dots + a_{2k+1})+1.\]
Subtracting we get that
\[a_1 \geq (a_{k+1}-a_1)+ (a_{k+2}-a_2)+ \dots +(a_{2k}-a_k)+1\]
\[    \geq \underbrace{k + k + \dots + k}_{k \text{ terms}} +1 \]
\[  =k^2+1\]
Which follows from the fact we stated above.\\

From which we deduce that our set is
\[ {k^2+1,k^2+2, \dots , k^2+2k+1}\]
Which means that
\[ N \leq 2((k^2+1)+(k^2+2)+ \dots + k^2+2k+1)-1\]
\[= 2k^3+3k^2+3k \]
As desired.


inequalities
function
inequalities unsolved

Famous Nesbitt's inequality states that:

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq\frac{3}{2}$

do you think it is possible to find the minimum of these "Nesbitt- Like" expressions?
if so, please explain me ....

1)  $ \frac{a}{b^2\plus{}c^2}\plus{}\frac{b}{c^2\plus{}a^2}\plus{}\frac{c}{a^2\plus{}b^2}$

2) $ \frac{a^2}{b^2\plus{}c^2}\plus{}\frac{b^2}{c^2\plus{}a^2}\plus{}\frac{c^2}{a^2\plus{}b^2}$

3) $ \frac{a^3}{b^2\plus{}c^2}\plus{}\frac{b^3}{c^2\plus{}a^2}\plus{}\frac{c^3}{a^2\plus{}b^2}$

4) $ \frac{a^3}{b^3\plus{}c^3}\plus{}\frac{b^3}{c^3\plus{}a^3}\plus{}\frac{c^3}{a^3\plus{}b^3}$

5) $ \frac{a^2}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{b^2}{\sqrt{c^2\plus{}a^2}}\plus{}\frac{c^2}{\sqrt{a^2\plus{}b^2}}$

6) $ \frac{a}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{b}{\sqrt{c^2\plus{}a^2}}\plus{}\frac{c}{\sqrt{a^2\plus{}b^2}}$

and there is one too.. I saw Arqady posted it.. It is very elegant and stronger than Nesbitt

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq\frac{3\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{2(a\plus{}b\plus{}c)}$

Please post elementary solutions..  :blush:

[quote="dgreenb801"]The numerator and denominator have to have the same degree, otherwise, we can make all of them very small or very large and it has no minimum or maximum (similar to what I told you earlier today with the a+b+c=1).
1 has no minimum if a,b, and c are very large (meaning it can get infinitely close to 0). 2 is the same as nesbitt, just with $ a^2$, $ b^2$, and $ c^2$ plugged in, but since any positive number has a square root, it will obviously give the same thing. 3 has no minimum if a,b, and c are very small. 4 is also simply nesbitt. 5 also has no minimum is a,b, and c are very small. 6, however, does have a minimum which I am working on.[/quote]

Thank you! You are great explainer.. I understood the idea how the expression can be infinitely small (approach to zero) and therefore not having minimum... But i didn't get the difference between 5) and 6). If 5 have no minimum, why not 6)... i mean in 6), you can give very small value for b and c and 1 to a... then it can be infinitely small and will not have minimum.. Can you please explain that again?

Your help is much appreciated!  :)

If you multiply all of $ a,b,c$ by $ k$ in 5,
now think a bit....
notice that the total expression gets multiplied by $ k$ as well.$ (*)$
So if we make $ k\to 0^\plus{}$ (in other words making $ k$ arbitrarily small) the total expression $ \to 0^\plus{}$. Thus there is no minimum.

To be more rigorous, suppose there is some $ a_0,b_0,c_0$ for which the expression attains minimum.
But if we calculate the expression plugging in $ ka_0,kb_0,kc_0$(where $ k<1$) for $ a,b,c$, the expression gets still smaller. Hence the assumption that some $ a_0,b_0,c_0$ exists was false.

The same reason doesn't apply to 6 because $ (*)$ doesn't hold in 6. (think a bit...)

[QUOTE]
\[ a,b,c \ge 0\]
\[ \sum \frac{a}{b\plus{}c} \ge \frac{3}{2} \frac{ \sqrt{3\sum a^2}}{ a\plus{}b\plus{}c}\]
But :
\[ 2\sum \frac{a^4}{b\plus{}c} \ge 2\sum a^2\minus{}\sum bc\] is stronger than above ineq :)  , and it's easy

[quote="dgreenb801"]5 has no minimum because you are taking the square root, which cuts the value of the exponent in half. However, for 6 it is the same as cutting the total exponent in half.[/quote]

Can you please explain the significance of this?

I mean i understand you completely, but i don't get the total idea. what i mean is, how do you identify whether function has minimum or not... is there a strategy.. Can you give some other examples that doesn't have minimum?

Thank you for your patience

[quote="shaaam"]

It is very elegant and stronger than Nesbitt

$ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}\geq\frac {3\sqrt {3(a^2 + b^2 + c^2)}}{2(a + b + c)}$

Please post elementary solutions..  :blush:[/quote]
The stronger has been solved by Am-Gm:
${ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}\geq\ \frac {3}{2}. (\frac {3(a^2 + b^2+c^2)}{(a + b + c)^2}})^{\frac {2}{3}}$


I can't find anything about this on the website, and I haven't participated in the competition before.

What kinds of activities go on and at what times in ARML? I'm particularly asking this about the second day. There is another competition I am semi-committed to that runs on May 31 and June 1 this year which is a conflict. I am probably prioritizing ARML over that in all cases, but there are other people who will want to know how much of that competition I will be able to attend. FYI the other competition is in Virginia near me, and I'd be on the East Coast site for ARML (Pennsylvania, right?).

[quote="MellowMelon"]What kinds of activities go on and at what times in ARML? I'm particularly asking this about the second day. There is another competition I am semi-committed to that runs on May 31 and June 1 this year which is a conflict. I am probably prioritizing ARML over that in all cases, but there are other people who will want to know how much of that competition I will be able to attend. FYI the other competition is in Virginia near me, and I'd be on the East Coast site for ARML (Pennsylvania, right?).[/quote]
Well, I think the different sites might follow different schedules.  At Penn State last year, the schedule for Saturday was something like this :
[code] 8:30- 8:50 Team Round
 8:55- 9:55 Power Round
10:35-11:50 Individual Round
12:00- 1:30 Lunch
 1:45- 2:20 Relay Questions
 2:20- 2:40 Tie-breakers
 3:00- 4:30 Awards[/code]
Also, Penn State is no longer the only eastern site; there is a new site in Athens, Georgia, so if you live in a neighboring state, I imagine you will end up here instead.

The Las Vegas schedule is quite different, with the team and power rounds Friday evening. That's not particularly relevant to the OP's concerns.


trigonometry
geometry
calculus
rhombus
area of a triangle
Heron\u0027s formula
trig identities

Quick way to figure out all angles given all sides...

How do I do this without just doing Law of Cosines thrice?

I don't know of any way that would be faster than law of cosines...

I guess you could use heron's formula to find the area then set that equal to $\frac{1}{2}ab\sin C$, then use that to find $\sin C$, but that wouldn't be any quicker.

1. Apply law of cosines.
2. Apply law of sines.
3. Sum of the angles of a triangle are 180 degrees.

If you are talking about a triangle, then yeah, it is probably the quickest to use Law of Cosines once and then use Law of Sines, just like Calculus said.

If you meant quadrilaterals, then even if you're given all 4 side lengths, you can't always find the angles. For example, suppose you're given that all 4 sides have equal lengths. There are infinitely many possibilities; it could be any rhombus (including a square).

I think he is talking about a triangle, but I'm not sure.


You posted this many times. Be patient  ;)

[quote="anirudh"]Help!!. My signature panel just disappeared. i also cna't view any  other's signatures. Also, I checked my profile and it allows signatures :(  :(  :(  :(  :(  :(  :(  :(  :(  :( :help:  :help:  :help:  :help:  :help:[/quote]

About the allowing other people's signatures, that is a setting in your profile which is currently set to "not show" so that is why you don't see them.

[quote="236factorial"]Hmmm. I never noticed that... so if your computer is slow, you can select not to view avatars and signatures? (They aren't of much use)[/quote]Yes you can. In fact it is designed to help people that are connected through dial-up or other kind of slow connection speeds.


Let a,b,c be positive real number such that ab+bc+ca=1.
Prove that
 $ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \geq \frac{3\sqrt{3abc}}{2}(\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}\sqrt{c^3})$


trigonometry
function
analytic geometry
calculus
integration

Solve for $ m$:

$ 5/2 \arctan 2/3 \plus{} 1/4 \cos (2\arctan 2/3) \equal{} 5 \pi/4  \minus{} 5/2 \arctan m \minus{} \sin (2\arctan m)/4$

To simplify the nested trig/inverse trig functions, use your double angle formulas first.  Then it becomes as issue of drawing a right triangle with the given property and pythagorean theorem.  But I doubt the result will be solvable algebraically.

[hide="solution"]
$ \cos(2\arctan 2/3) \equal{} 2 \cos^2 (\arctan 2/3) \minus{} 1 \equal{} 2(3/\sqrt {13})^2 \minus{} 1 \equal{} 5/13$

So the LHS is $ 5/2 \arctan 2/3 \plus{} 5/52$

Now $ \sin(2 \arctan m) \equal{} 2\sin(\arctan m)\cos(\arctan m) \equal{} 2m/(m^2 \plus{} 1)$.

Hence we have $ 5\pi/4 \minus{} 5/2 \arctan 2/3 \minus{} 5/52 \equal{} \arctan m \plus{} \frac {m}{2(m^2 \plus{} 1)}$

Now rewrite the LHS: $ 5(\pi/4 \minus{} 2\arctan 2/3) \minus{} 5/52 \equal{} 5(\arctan 1 \minus{} 2 \arctan 2/3) \minus{} 5/52$

Since $ \arctan \tan x \equal{} x$, we have $ \arctan 1 \minus{} 2 \arctan 2/3 \equal{} \arctan(\frac {1 \minus{} \tan(2\arctan 2/3)}{1 \plus{} \tan(2\arctan 2/3)})$.  Since $ \tan(2\arctan 2/3) \equal{} 12/5$, we have $ \arctan(\frac {1 \minus{} 12/5}{1 \plus{} 12/5}) \equal{} \arctan( \minus{} 7/13)$

Hence we have reduced the equation to $ \minus{} 5\arctan(7/13) \minus{} 5/52 \equal{} \arctan m \plus{} \frac {m}{2(m^2 \plus{} 1)}$

Now $ 5\arctan(7/13) \equal{} \arctan(\tan(5\arctan 7/13)) \equal{} \arctan{ \minus{} \frac {109193}{137293}}$

So we have $ \arctan(\frac {109193}{137293}) \minus{} 5/52 \equal{} \arctan{m} \plus{} \frac {m}{2(m^2 \plus{} 1)}$.  Clearly we cannot equate the parts; thus, the equation cannot be solved algebraically.

[/hide]

My question was inspired by this thread
http://www.mathlinks.ro/viewtopic.php?t=204270

If one convert to polar coordinates and use the polar integration, one should get

$ r^2 \equal{} 5 \plus{} 4 \cos(2\theta)$ and hence we get

$ \int \int r dr d\theta \equal{} \int_0^{\arctan(2/3)} 1/2 (5 \plus{} 4 \cos(2\theta)) d\theta$
$ \equal{} 1/2 (5 \arctan(2/3) \plus{} 2 \sin(2\arctan(2/3))$

This should be equal to

$ \int_{\arctan(m)}^{\pi/2} 1/2 (5 \plus{} 4 \cos(2\theta)) d\theta$

= $ 1/2 (5 \pi/2) \minus{} 1/2 (5\arctan(m) \minus{} 2\sin(2\arctan(m)))$

$ m$ should be $ 1/6$.


For #4, I somehow thought that the "duck" people will get eliminated and showed that only winner that can be possible in that way is when $n = 3$. I know the problem asked for "goose" and this is completely ridiculous but is there any chance I'll get at least 1 point?

 :(  :(

I didn't think my "wrong" interpretation made problem any easier. It did look like only solution was when $n = 3$ but I showed the process of finding the general equation for it and the equation I got was a "ceiling" function. Then I converted this into floor function and took this in different base to show that my solution was unique.

Hopefully, I get some point.  :roll:

Really, I don't think someone is they misread the problem.

*Now, if it was something to the effect of leaving out a helpful constraint (but still having an eventual answer that satified it) or something else that would result in a technically correct explanation and solution, then I could see that happening.

Hmm...that does really suck.
It probably didn't make the problem easier at all though...

[quote="matt276eagles"]I think as long as your misunderstanding didn't make the problem trivially easy, you might still get a reasonable number of points.  I remember reading in one of these forums that someone misinterpreted a problem and actually made it harder - they got a 5.[/quote]
I did that two years ago :lol:


function
advanced fields
advanced fields unsolved

Hello

I know a solution of this problem for a  $C^2$ function but am not able to extend it properly to this case:

Let f be a $C^1$ function from $\mathbb{R}^n$ to $\mathbb{R}^n$ which differential is orthogonal for any x. Prove that f itself is orthogonal.


inequalities
function
email
IMO Shortlist

Yes, I just found out that [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=908]Vasc[/url] [Vasile Cartoaje] has published some of his works. All of you interested here's a small introduction of his book:

[hide="Introduction"]This work blends together classic inequality results with brand new problems, some of which devised only a few weeks ago. What could be special about it when so many inequality problem books have already been written? We strongly believe that even if the topic we plunge into is so general and popular our book is very different. Of course, it is quite easy to say this, so we will give some supporting arguments. This book contains a large variety of problems involving inequalities, most of them difficult, questions that became famous in competitions because of their beauty and difficulty. And, even more importantly, throughout the text we employ our own solutions and propose a large number of new original problems. There are memorable problems in this book and memorable solutions as well. This is why this work will clearly appeal to students who are used to use Cauchy-Schwarz, Schur, AM-GM, etc. as a verb and want to further improve their algebraic skills and techniques. They will find here tough problems, new results, and even problems that could lead to research. The student who is not as keen in this field will also be exposed to a wide variety of moderate and easy problems, ideas, techniques, and
all the ingredients leading to a good preparation for mathematical contests. Some of the problems we chose to present are known, but we have included them here with new solutions which show the diversity of ideas pertaining to inequalities. Anyone will find here a challenge to prove his or her skills. If we have not convinced you, then please take a look at the problems from this book and hopefully you will agree with us.[/hide]

You can buy the book from [url=http://www.gil.ro/en/]Gil Publishing House[/url] for 40 euros (I think). Contact [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=168]Mircea Lascu[/url] for more details.

Actually, the book exposes in the chapters 2-7 some new and original methods and strategies
in solving inequalities. In author’s opinion, this is the main feature of the book.
The book is 480 pages long and has the following contents.

1.Warm-up problem set				     5
2. Starting from some special fourth degree inequalities 	    67
3. Inequalities with right convex and left concave function	  143
4. On Popoviciu’s Inequality				  193
5. Inequalities involving EV-Theorem			  209
6. Arithmetic/Geometric Compensation Method		  267
7. Symmetric inequalities with three variables involving    fractions	  301
8. Final problem set					  371
    List of Symbols and Glossary			                  465
    Bibliography

Vasc had the kindness and described shortly the book. Anyone who want's it just e-mail at sales@gil.ro and make an order. You will recieve the book in Slovenia.

Who orders all 5 books will get a 10% discount and notice that you will get rid of shipping taxes. Also they will get a gift. SURPRISE :)

Regards!

It would be interesting if you give some sample inequalities, sample pages, Vasc.

This announcement was made because I will be in Slovenia bringing the books. Tomci0 asked if we send books to Poland. Yes we can send the books to Poland. But this can be talked with more details on e-mail or chat. Send an e-mail with your order at sales@gil.ro and you will get a reply.

Regards!

I agree with [b]pvthuan[/b] - it would be nice to see some sample pages not only of this book but also of the others. You could also attach some small descriptions to all of your titles published in English on your website. Anyway I'm going to buy them all  :D

I'm sorry but how can I buy it?I'm in Viet Nam

[quote="mstoenescu"][quote="NkMAsTeR"]But in my country, we can't send money out! We buy anything in foreign country only through relatives but my uncle can only buy book in Amazon! Help me![/quote]
Hi ! I'm a Romanian living in France and I would buy for you the book next winter when I go to Romania. Do you know someone in France to take this book for you ? You have ...six months to find someone ![/quote]
Can you send it to America! I have and uncle In Texas but i can't buy this book through him becase I've borrowed him a lot of money(about 300USD) :maybe:

Is the writtin in English?
And are you able to send the book to China,Mircea?

Here you can download the introduction of our last book published in english, "Algebraic Inequalities. Old and New Methods". I think that after reading this introduction you can make an opinion about the book.

On http://www.gil.ro you can find our other 4 books published in english by GIL Publishing House. In case you are interested in ordering or have any questions, send e-mail at sales@gil.ro.

Thank you for your time!

Very nice words for Algebraic Inequality and its author.
Thank you, Marian Tetiva.

Is it possible to get an ebook version? My e-mail stockdox@gmail.com

I think this book is very interesting, and I want to have it but I can't . Help me please! My E-mail is quyvodichthienha@yahoo.com.vn

I think this book is very intereting too,and I want to have it but I can't . Help me please!
My E-mail is [b] maths_ndt94@yahoo.com.vn[/b]

I would like to buy this book, too. I couldnt find it in amazon or ebay. I'm in Czech republic and this is my email: ruacon@seznam.cz. And special thanks to Mr. Vasil for this great book. I always love your inequalities and solutions. They're awesome and amazing.

i want to buy it ,my email:657394712@qq.com

Why is there only a virtual copy available on amazon.com and nothing on amazon.ca? I really want to buy it but it is not available anywhere... does GIL ship to Canada and take paypal? The website is not very good through google translate and I don't read Romanian.

[img]http://www.librariaeminescu.ro/media/produse/2008/7/f49912-Vasile-Cirtoaje-Algebraic-inequalities.jpg[/img]
I think i can help every one!
If you are living in Vietnamese!!! I think i can help you......or you are living in China, i think i can help you too...Special if you are living in HongKong! Good luck every one...!!!
Nice to meet every one!
Please contact with me at: bdtilove@yahoo.com.vn

I guess, GIL ships everywhere. I dont expect the book to be a mass sell-out (i.e. you wont be finding it in every shop.. its damn unique and rare.)

P.S. People interested should probably try to contact Mircea Lascu somehow.

I want buy this book but I can't because I'm in China .Help! Somebody help me !

Hello everyone,

       My name is [b]Lascu Iulia[/b], and [b]I am responsible with the foreign customers division of GIL Publishing House since March 2011[/b].
_____________________________________________________________________
_____________________________________________________________________

I HAVE ATTACHED A WORD DOCUMENT THAT I HAVE WRITTEN FOR YOU ALL.
[b]THE DOCUMENT CONTAINS INFORMATIONS THAT ARE USEFULL FOR EVERYONE WHO WANT'S TO BUY VASC'S BOOK OR ANY OTHER BOOK PUBLISHED AT GIL PUBLISHING HOUSE![/b]

_____________________________________________________________________
_____________________________________________________________________
       [i]Please take the time to read the document![/i]
        Last, but not least, I would like to send a [b]special thank you to my friend Samer Seraj[/b], who helped me to get in touch with you all.

Have a nice day, and all the best,
[b]Lascu Iulia
GIL Publishing House
+4 0260 616 314
gil1993@zalau.astral.ro
www.gil.ro[/b]

[quote="BigSams"]This is one of the best inequality books out there (I am reading it as a sequel to Hung's 2 volume Secrets in Inequalities). I only obtained it through great difficulty; had to convince a friend who happened to be visiting Romania to find it at a local store.

It is an absolute shame that people cannot easily buy it. I cannot blame GIL either; after exchanging some emails with them, they seem like very nice people who are having a hard time because of widespread piracy the the economic downturn. A tad ironic, however, that many people pirate GIL books [i]because [/i]they [i]cannot buy [/i]it conveniently in the first place. A vicious cycle.

It is a book filled to the brim with beautiful inequalities and theorems. I wish everyone luck in obtaining this unique volume.

Contact Lascu Iulia at gil1993@zalau.astral.ro to get information on the payment method and after you pay, GIL will send the books to you.[/quote]

Thank you for supporting our Publishing House! And also thank you for understanding our situation, and the damadge that the piracy is making through our sales.

[quote="The soul of rock"]Long time ago, I had met your problems. But don't worry. They'll reply you soon, may be in next month!! :lol:[/quote]
@The soul of rock: We do not intend to ignore our customers. But we had some difficulties. Business issues and also personal problems. From now on, please feel free to contact me, and you will recive a reply as soon as possible*.
as soon as possible*>>> Note that I am also on the go always, and I am travelling a lot. There are times when I am not in my hometown even for a month or maybe more, and when I am out of town, somebody else is handeling my devision [that is: "Foreign customers devision"], and when I get back, I am reasuming my position, and there are always problems because of all this shifts between me and an employee of GIL Publishing House.
Thank you for your understanding, and I hope I didn't bother you with something!

All the best,
Lascu Iulia
GIL Publishing House
+4 0260 616 314
gil1993@zalau.astral.ro
www.gil.ro

Now China has got the resource.[quote=luofangxiang]我做过此书一些题，是我国一不等式专家发的，但不全，有些题不太会做。
我在中国，买不到，您有此书电子版？如果有，方便的话，电邮到：lbylfx@sina.com
（Some problem, I have done this book is an inequality in China expert, but not all, some questions will do.
I buy in China, less than, you have the book online? If you have, it's convenient, would email to: lbylfx@sina.com
）[/quote][/quote]


inequalities
quadratics
algebra
polynomial
inequalities unsolved

For $a$, $b$, $c$ positive reals prove that

$\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right) \geq \frac94 \left(ab+bc+ca\right)^2$.

A proof far too complicated. And something tells me that we have already discussed this problem. Yes, with 1instead of 3, but does this change anything?

I cannot find a nice solution, sorry :blush:

[quote="darij grinberg"]For $ a$, $ b$, $ c$ positive reals prove that

$ \left(a^2 \plus{} 3\right)\left(b^2 \plus{} 3\right)\left(c^2 \plus{} 3\right) \geq \frac94 \left(ab \plus{} bc \plus{} ca\right)^2$.[/quote]
Take $ a\equal{}\frac{\sqrt{6}}{x},$ $ b\equal{}\frac{\sqrt{6}}{y}$ and $ c\equal{}\frac{\sqrt{6}}{z}.$ After some simple computations, we can write the original inequality as
$ (x^2\plus{}2)(y^2\plus{}2)(z^2\plus{}2) \ge 3(x\plus{}y\plus{}z)^2.$
This is a known result.


Let function $f:R->R$ is twice differentiable; $f''$ is continuous and
$2f(x+1)=f(x)+f(2x)$. Find all $f$

hey there,

I've been trying to solve this problem for quite some time with not much luck, but i did manage to get something whcih someone else might find useful:

2f''(x+1)-f''(x)=4f''(2x)-->|f''(2x)|<=3/4*MAX(|(f''(x+1),|f''(x)|). This, of course, implies that f''(x)-->0 as x goes to +inf or -inf.

Also, 2f'(2x)=2f'(x+1)-f'(x)=f'(x+1)+f''(a) where x<=a<=x+1.
since f''(x)-->0, we have also that f'-->0.(this requires some techincal details, but is obvious enough to skip).

Thats all i have so far! hope it helps somebody!

nice one!
let $a\leq 0, b\geq 2$ and let $M$ be the maximum of $|f''|$ in $[a; b]$. set $2x=M$. then $a\leq x,x+1\leq b$ since $a\leq 0$, $\geq 2$. it follows that $|f''(x)|\leq M$, $|f''(x+1)|\leq M$. but $3M\geq 2f''(x+1)-f''(x)=4f''(2x)=4M$, thus $M=0$. it follows that $f''(x)=0$ for all $x$. thus $f'(x)=c$, but $2c=c+2c$, thus $c=0$. thus $f(x)=c$, which is a solution.

Peter

Peter's proof contains two dirty typos. I even though the solution is wrong.

I'm not allowed to edit that post anymore, so here is the corrected solution.

let $a\leq 0, b\geq 2$ and let $M$ be the maximum of $|f''|$ in $[a; b]$. take $2x\in [a; b]$ such that $|f(2x)|=M$. then $a\leq x,x+1\leq b$ since $a\leq 0$, $b\geq 2$. it follows that $|f''(x)|\leq M$, $|f''(x+1)|\leq M$. but $3M\geq |2f''(x+1)-f''(x)|=4|f''(2x)|=4M$, thus $M=0$. it follows that $f''(x)=0$ for all $x$. thus $f'(x)=c$, but $2c=c+2c$, thus $c=0$. thus $f(x)=c$, which is a solution.

Peter

[quote="Peter Scholze"]such that $|f(2x)|=M$[/quote]
 :rotfl:

:wallbash_red:  :wallbash_red:  :wallbash_red:

of course, $|f''(2x)|=M$...


Determine the range of the following function defined for  integer $k$,

\[f(k)=(k)_3+(2k)_5+(3k)_7-6k\]

where $(k)_{2n+1}$ denotes the multiple of $2n+1$ closest to $k$

We have $|k-(k)_3|\le1$, $|2k-(2k)_5|\le2$, $|3k-(3k)_7|\le3$ (these are easy to see), and so $-6\le f(k)\le 6$. It is easy to find an example that gives each integer from -6 to 6, so the range is $\{-6,-5,\ldots,6\}$.


Guys I would like to know what do you think about this question :) .
How I know the S.O.S. theory isn't so known,as other classic ineq.-s like AM-GM,Caush-Shvartz,Murched and others.I wonder can we use that ineq in IMO without proof?
Thank you. :)

[quote="darij grinberg"]What is the theory used in a proof by the SOS method which you can't prove in two lines?

  darij[/quote]
I agree that it is not hard to prove ,but why spending time for do that if it doesn't need :) ,remember we will only have 4 hours and 30 minuts for 3 problems :o  :D .

Hi, maybe you said that because you don't know what S.O.S can do for 3-variable problem. Notice that in all method, SOS is theorem which can be prove the most easily and used the most widdenly. I'm sure.

[quote="darij grinberg"]What is the theory used in a proof by the SOS method which you can't prove in two lines?

  darij[/quote]

To answer Darij's question:

http://www.mathlinks.ro/Forum/viewtopic.php?p=458553#p458553

conceive

[quote="hungkhtn"]Hi, maybe you said that because you don't know what S.O.S can do for 3-variable problem.[/quote]
Hi $hungkhtn$,I can't understand what do you mean :? What have I said that you thought that I don't know
[quote]what S.O.S can do for 3-variable problem[/quote]?

Sorry If I have made you... angry. But I think there's no thing to prove with S.O.S. Notice that it's only a method and the theorem of it is very easy to prove.


[quote="mountain.dew"]What are all ordered pairs of integers $ (x,y)$ which satisfy $ x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+6=0$?[/quote]

[hide="Solution"]Let's try to find perfect cubes in here:
\begin{eqnarray*}x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+6=0&\equiv&(x-1)^{3}+(y+2)^{3}-1=0\\ &\equiv&(x-1)^{3}+(y+2)^{3}=1 \end{eqnarray*}
Since $ x$ and $ y$ are integers, one of the cubes is $ 1$ and the other is $ 0$:
$ x-1=1\text{ and }y+2=0$
$ x-1=0\text{ and }y+2=1$

So $ (x,y) = (2,-2)\text{ or }(x,y)=(1,-1)$. So $ \boxed{2\text{ possible solutions}}$.[/hide]


floor function
induction
Recursive Sequences

An integer sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{n+1}=a_{n}+\lfloor \sqrt{a_{n}}\rfloor.\] Show that $a_{n}$ is a square if and only if $n=2^{k}+k-2$ for some $k \in \mathbb{N}$.

Let's prove that given a $ a_k = m^2$ the next perfect square in the sequence is $ (2m)^2$. Before we prove that for any natural $ s$ we have that: $ a_{(k + 2s + 1)} = (m + s)^2 + (m - s)$. For s=0 $ a_{(k + 1)} = m^2 + m$ we say that it's true for a s:$ a_{(k + 2s + 1)} = (m + s)^2 + (m - s)$ and than: $ a_{(k + 2(s + 1) + 1)} = (m + s)^2 + (m - s) + 2m + 2s = (m + s + 1)^2 + m - (s + 1)$ Now we can see that we have the first perfect square after $ m^2$ when m=s ;so the next perfect square is$ a_{(k + 2m + 1)} = (2m)^2$,infact thanks to the before formula we can see that also for$ a_{(k + 2s + 2)} = (m + s)^2 + (m - s) + m + s < (m + s + 1)^2$.Now with a very simple induction we know that starting from $ a_1 = 1$ all the perfect squares of the sequnce are the perfect squares of all the perfect powers of 2.Now let's prove that$ 2^{(k - 1)}^2 = a_{(2^k + k - 2)}$ for any k.$ a_{(2 + 1 - 2)} = 2^0$.We say that $ a_{(2^k + k - 2)} = 2^{(k - 1)}^2$ and we see that for what we said before: $ a_{(2^k + k - 2 + 2^{(k - 1)}2 + 1)} = (2(2^{(k - 1)})^2 = (2{}^k){}^2 = a_{(2^{(k + 1)} + (k + 1) - 2)}$ that prove the thesis.


In all problems $a, b, c > 0$

This one is all that's interesting from the previous thread, may as well restate it clearly here:
[b](1)[/b] $\frac{a^{3}}{bc(b+c)}+\frac{b^{3}}{ca(c+a)}+\frac{c^{3}}{ab(a+b)}\ge \frac{3}{2}$

More conjectures:
[b](2)[/b] $\frac{ab}{c(a+b)}+\frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}\ge \frac{3}{2}$

[b](3)[/b] $abc = a+b+c$, $a = \max(a, b, c)$
$\frac{1}{a(b+c)}+\frac{1}{b(a+c)}+\frac{1}{c(a+b)}\ge \frac{\sqrt{3}}{2a}$

[b](4)[/b] $ab+ac+bc = 3$
$\frac{abc}{a+b+c}+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{11}{6}$

[quote="delegat"]By AM-GM we have (bˇ3+bˇ3+cˇ3) + (cˇ3+cˇ3+^bˇ3) >=3(bˇ2c + cˇ2b) = 3bc(b+c) so by using this inequality, inequality (1)  can be easily reduced to Nesbitt's inequality.
Inequality (2) is also equivalent to Nesbitt's ineq. :lol:[/quote]
Why that?

because it's Nesbitt for $\frac1a,\frac1b,\frac1c$ :) :play_ball:


calculus
derivative
integration
calculus computations

Is $ x\equal{}\sqrt[3]{2t \plus{} 5}$ a solution to the equation $ 3x^2\frac{dx}{dt}\equal{}2$? Why or why not?

[hide="what I did"]I find try to find the derivative part of the equation:
$ \frac{d(2t \plus{} 5)^{\frac{1}{3}}}{dx}\equal{} 2 \cdot \frac{1}{3}(2t \plus{} 5)^\frac{\minus{}2}{3}$

Then plugging in:
$ 3x^2 \cdot \frac{2}{3} (2t \plus{} 5)^{\frac{\minus{}2}{3}} \equal{}(?) 2$.

I'm stuck here.[/hide]

\[ \begin{gathered}
  3x^2 \frac{{dx}}
{{dt}} \equal{} 2 \hfill \\
  3x^2 dx \equal{} 2dt \hfill \\
  x^3  \equal{} 2t \plus{} C \hfill \\
  x \equal{} \sqrt[3]{{2t \plus{} C}}. \hfill \\
\end{gathered}
\]

[quote="bookworm_vn"]
\[ \begin{gathered} (1) 3x^2 \frac {{dx}} {{dt}} \equal{} 2 \hfill \\
(2) 3x^2 dx \equal{} 2dt \hfill \\
(3) x^3 \equal{} 2t \plus{} C \hfill \\
(4) x \equal{} \sqrt [3]{{2t \plus{} C}}. \hfill \\
\end{gathered}
\]
[/quote]

Can you explain how you got from step 2 to step 3 for me? I'm not sure how $ 3x^2d(x)\equal{}x^3$ and $ 2dt$ becomes $ 2t \plus{} C$.

Advice: You need to look up how to integrate basic integrals in your textbook.
For instance $ \int xdx \equal{} \frac{x^{2}}{2} \plus{} C$
How do you differentiate $ kx$? You first integrate $ x$ and then multiplied the result by $ k$. Thus if $ k \equal{} 2$, then $ \int 2xdx \equal{} 2(\frac{x^{2}}{2} \plus{} C)$

Try to study how to integrate $ x^{n}$ and you will answer the questions you asked in this thread.


Two natural numbers $a$ and $b$ satisfy $a-b<10$ and $123456789=(11111-a)(11111-b)$. Find the value of $a$ and $b$.

This is Shinsyu University entrance examination 1990/Science, Economics in Japan.
How did you get this problem,  kohjhsd.

[quote="kunny"]This is Sinsyu University entrance examination 199? in Japan.
How did you get this problem,  kohjhsd.[/quote]

I saw it on a Korean internet math forum and it seemed interesting (since direct calculation is also possible.
How do you even remember that? :D

[quote="sonny"][quote="kohjhsd"][quote="sonny"]Um well
$123456789=(9)(13717421)$ so.....

$123456789=(11111-a)(11111-b)$

$123456789=(11111-11120)(11111-13728532)$

and $11120-1378532<10$

so $a=11120$ and $b=13728532$[/quote]

You are ignoring the given restriction that $a-b<10$.[/quote]

$11120-13728532=-13717412$

last time I checked

$-13717412<10$

um if the problem said that $a-b$ had to be positive then I am wrong. However, no such thing has been stated so we can safely say that $a=11120$ and $b=13728532$[/quote]

That's right :(
I am confused  :wacko:
Let's change it to $|a-b|<10$ then (not sure if there is an answer given this).

Note that the sqrt(123456789) is between 11111 and 11112, so therefore, letting a=0,b=0 won't give the product to be greater than or equal to 123456789. Therefore, a,b must be both greater than 11111. Let x=a-11111,y=b-11111

We have |x-y|<10, and xy=123456789 and x>0,y>0

Therefore, xy must be very close to sqrt(123456789) which is between 11111 and 11112. The rest should be simple :)


real analysis
calculus
integration
function
real analysis unsolved

Let $ f(x)$ be an integrable function on $ (-\infty, \infty)$. Show that
\[ g\left( x \right) = \int_{ - \infty }^{ + \infty } {e^{ - itx} f\left( t \right)dt}
\]
is a continuous function on $ (-\infty, \infty)$ where $ i = \sqrt{-1}$ and
\[ g\left( x \right) = \frac{d}
{{dx}}\int_{ - \infty }^{ + \infty } {\frac{{e^{ - itx}  - 1}}
{{it}}f\left( t \right)dt.
}\]

Any help?

A few hints:

1. Prove that ${ \|g\|_{\infty}=\sup_{x\in\mathbb{R}}|g(x)|\le \|f\|_1=\int_{\mathbb{R}}|f|}.$

(You might as well give $ g$ a name that depends on $ f$ - say, call it $ \widehat{f}.$)

2. Try to prove the first result for "sufficiently nice" $ f$ - say, $ C^1$ compactly supported, or step functions, or something else you can easily manipulate.

3. Show that the set of "sufficiently nice" functions you used in 2 are actually dense in $ L^1.$ Then find a way to use that to get a uniformly convergent sequence of continuous functions.