function
calculus
integration
derivative
inequalities
real analysis
real analysis solved

Let f:[0,1]-.R be a function which has a second derivative and f"(x)>0 for all x in [0,1], and s.t. f(1)=0. Show that
\int{0->1}f(x)dx >= [k/(k-1)]2\int{1/k->1}f(x)dx for all real k>1.

[i]Romanian olympiad/second round 1997/Arad (a Romanian county)[/i]

Fairly simple:
  Of course, f is convex, so f(1/k+i/n*(1-1/k))<=1/kf(1)+(1-1/k)*f(i/n)=(1-1/k)*f(i/n). Now sum these inequalities, use the definition of the Riemann integral and take the limit.

I used the inequality  \int{1/k->1}f(x)dx <= [(p-1)/p2]*f(1) + [(p-1)/p]2*\int{0->1}f(x)dx. I guess it's pretty much the same.

[quote="harazi"]...f(1/k+i/n*(1-1/k))<=(1-1/k)*f(i/n). Now sum these inequalities, use the definition of the Riemann integral and take the limit.[/quote]

1/n \sum _(i=1 to n)f(1/k +i/n(1-1/k))  \leq (1-1/k)1/n \sum _(i=1 to n)f(i/n)

take the limit when n-->+oo
 \int _[0;1]f(1/k+x(1-1/k))dx \leq (1-1/k) \int _[0;1]f(x)dx
change variable y= 1/k+x(1-1/k) in LHS integral gives the result


I cant find any sites with any old PUMaC tests. Can someone please post a link to the tests?

Questions from the 2007 tests are available at:
https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2007#Problems

Hope these help!


I think you might have a problem. Namely that Friedel Craft does not occur for deactivated benzene ring.

Yes, that's right BanishedTraitor: Friedel-Crafts reactions are unsuccefull for benzene rings with deactivating substituents.

Propose a synthesis of 3-methylbenzoic acid from p-nitrotoluene.
Can we instead do this?

1. First convert NO2 into NH2 using Sn/HCl and OH-
2. Use CH3COCl to convert NH2 into NHCOCH3
3. Oxidise CH3 into COOH
4. Now Perform Friedel and Crafts Alkylation ( using CH3Cl and AlCl3 )
5. Use NaOH to convert NHCOCH3 into NH2
6. Use NaNO2/HCl and then H3PO2 to get rid of NH2
 :)

i have a doubt..........
wont the KMnO4 that you use to oxidise CH3 to COOH oxidise the NH2 group to NO2 group??? :(

what i thought was............
the NH2 group would be oxidised back to NO2 group.........
Better ask Carcul...
i don know abt this

I doubt whether this is possible.Tell me where there is a mistake....

1.Oxidise using KMnO4 to get the acid.
2.Now methylate with CHCl3/AlCl3 to get one methyl group in the ortho position.
3.now decarboxylate
4.reduce using Fe/HCl
5.Diazotize using HNO2
6.treat with CuCN and hydrolyse to get the product.

is it right :( (i have doubt with my second step)??????????????????

You cannot alkylate or acylate a ring containing NO2 or -R group  :D

see if this method works........

1.Reduce using Sn/HCl
2.Diazotise and treat with H3PO2
3.treat with acidified KMnO4
4.CH3Cl/AlCl3 to get the desired product

Is this correct :wink: ?????????????????????????

[b][i]
1.Sn/HCl
2.Protect with CH3COCl
3.HNO3/H2SO4/HEAT
4.H2O
5.NaNO2/HCl
6.H3PO2
7.Sn/HCl
8.NaNO2/HCl
9.CuCN/Pyridine
10.H3O+[/i][/b]

Hey chemrock and Carcul............
Is my conversion possible......
If not What is the reason???????

For the n^n th time
[b][color=red][size=200]FRIEDEL CRAFTS REACTION SHOULD NOT BE DONE TO A DEACTIVATED SYSTEM[/size][/color][/b]

Yes, that's right Chemrock. Regarding your synthesis: are you sure about the nitrating step? The acetamido group is more activating, but its also bigger, so perhaps you would obtain a mixture.

[hide="My solution"]$ p \minus{} nitrotoluene \plus{} KCN/ethanol/HO^ \minus{} \longrightarrow 3 \minus{} methylbenzoic\,\,acid$[/hide]

[quote="Carcul"][quote="Madness"]4. Now Perform Friedel and Crafts Alkylation ( using CH3Cl and AlCl3 )[/quote]

You cannot perform Fridel-Crafts reactions in rings with deactivating substituents and/or amine groups.[/quote]

But isnt NHCOCH3 more activating and cant we alkylate a ring contg NHCOCH3?

Even i have a doubt whether -NHCOCH3 is an activating or deactivating group..
(i thought that -NHCOCH3 is weakly activating)
i am not clear in this concept.........

i know (think so) for a fact that NHCOCH3 id more activating than normal activating grps like OCH3 and CH3 etc. so shouldn't it be possible?  :maybe:

[quote="srinitrs"][quote="valeriummaximum"]I think -NHCOCH3 is weakly activating because that is why we get mono bromo product on bromination..[/quote]
the order of +R groups is Nh2>O-Na+>NhCOCh3>OH>O-Ch3>CH3
thats what i think is it right??? :D[/quote]
i think it should be o-na+>nh2

-NHCOCH3 is less activating that -NH2. -ONa and -NH2 have probably the same reactivity.

yes taht's right because u use acteyl halides to protect the aniline whih is tsrongl;e activating to get mono substituted products


As organizers of SEEMOUS 2010 told this competition is supposed to be on 8-13 March 2010 in Plovdiv. So according to the regulations probably the program will be:

1st day: Monday 8 March 2010: Arrival and first Jury meeting.
2nd day: Tuesday 9 March 2010: Opening Ceremony and Jury meeting for problem selection. Some students may arrive on this day.
3rd day: Wednesday 10 March 2010: Contest Day. Papers are saved, approved by the Chairman of the Jury and numbered. Names are hidden.
4th day: Thursday 11 March 2010: Marking of papers by the leaders and deputy leaders. Jury meeting to approve results and medals.
5th day: Friday 12 March 2010: Awards ceremony in the morning. Excursion in the afternoon and/or departures
6th day:  Saturday 13 March 2010: Departures

The chairperson of SEEMOUS 2010 Organizing Committee is [url=http://77.77.167.148/gocheva/index_en.html]Prof. Snezhana Gocheva-Ilieva[/url].

The site seems to be down at the moment.

Problems: http://putnam.ho.ua/seemous2010/problems.tif
Solutions: http://putnam.ho.ua/seemous2010/solutions.pdf
Solutions and marking schemes: http://putnam.ho.ua/seemous2010/solutions_marking_schemes.rar (tif)
Results before coordination: http://putnam.ho.ua/seemous2010/pre-results.rar (tif)
Results after coordination: http://putnam.ho.ua/seemous2010/results_points.rar (tif)
Medals: http://putnam.ho.ua/seemous2010/results.pdf

Team scores:
1. University of Bucharest, Romania 156
2. National Team, Israel 124
3. Politehnica University of Bucharest, Romania 95
4. National Team, Greece 66
5. National Team, Romania 64
6. Al. I. Cuza University, Romania 62
7. Technical University "Gheorghe Asachi", Romania 61
8. Sofia University "St. Kliment Ohridski", Bulgaria 57
9. University of Athens, Greece 27
10. University of Architecture, Civil Engineering and Geodesy, Bulgaria 26
11. Plovdiv University "Paisii Hilendarski", Bulgaria 14
12. Democritus University of Thrace, Greece 9
13. University of Veliko Turnovo "St. Cyril and St. Methodius", Bulgaria 8
14. National Team, Rep. Macedonia 2

Country scores (according to the [url=http://www.seemous.eu/index.php?id=50]Regulations[/url], part 7):
1. Israel 124
2. Bulgaria 72
3. Greece 66
4. Romania 64
5. Macedonia 2

The 5th SEEMOUS will be in March 2011 in Bucharest, Romania. The hosting university will be Politechnica.


Yes, there is an National Olympiad in Romania for every possible subject you can think off (even cow butcher! - this is not a joke).

However, the most important Olympiad remains the Mathematical Olympiad (probably because it is the oldest Olympiad in Romania, and because Romania is the country that initiated the international competion in 1959, the only country to hold it 5 times, and the only country to hold it twice in a row).

Lately the Computer Science Olymiad is taking its toll, being second after mathematics.

I think English learning is most important, as it is at this moment, the easiest way of global communication between people around the world. Think about it, if all people wouldn't have known English the very existance of this site (well the MathLinks part at least :) ) would not have been possible. So it is most important.

However, we all recognize that not many students have yet reached a level of fluency in English that would allow them to participate actively in the discussions on this Forum. That is why we are in course of creating local / national forums, in which users can have the controls in their natal laguage set up automatically by the forum, and in which they can start discussing in their natal language, untill they feel prepared to move into the general forums.

During my visit back to China last summer, I had visited a major bookstore near where I stayed. The first floor of the store was full of English education material, everything from books to videos. It seems to be that English is the "in" thing in China now or something.

Also, the thing about English competitions in China (which they even refer to as Olympiads), is that they can't Internationalize it! So in my opinion, science olympiads make a lot more sense.

[quote="billzhao"]Also, the thing about English competitions in China (which they even refer to as Olympiads), is that they can't Internationalize it! So in my opinion, science olympiads make a lot more sense.[/quote]That's crazy. That should be the only language olympiad that should be international (well US, Canada, UK, AU, NZ, SA and other natal english language speaking countries should not be allowed to participate).

[quote="Valentin Vornicu"](well US, Canada, UK, AU, NZ, SA and other natal english language speaking countries should not be allowed to participate).[/quote]
Then what would be the fun in that  :D

Speaking of language Olympiads, did you know that there's actually an [url=http://www.philol.msu.ru/~otipl/new/main/mol/sum-2003-en.php]International Olympiad in Linguistics[/url]? Sounds sorta silly to me.

I know many things about China education
My opinion is, simply, I hate those 'English Competition', because they ask nothing about the English TV Shows, English Hit Singles, and English-spoken celebrities.

Hmm, I forgot one thing--English Message Boards. :lol:

[quote="qxy"]Ten years ago, the thing is different,
everybody thinks maths, physics, chemistry are the most important. Unfortunately, the views are changing, maths, physics, chemistry are no longer the most important, instead, the most important is ENGLISH.
[/quote]

Ten years ago, hmm, maybe you are right.
I know that fifty years ago Russian and Japanese were important

In fact, I think English is important, but I also think Chinese should be more important, because I am Chinese. Although I did Chinese nit so well.
  If Chinese is not treated to be important, I am afraid China would be another Singapore. As I know, the majority of the population in Singapore
are Chinese emmigrants, unfortunately, almost most of them could not speak their mother tongue--Chinese well.
  Another serious language problem is about the official language and the local languages, the government and schools require students and children to speak the official language, the effect is: many young people could not speak their local languages, some experts tell that many local languages will die out in 50 years. In fact, I do not like to speak the official language, I like to speak the local language.

Part of the reason English is so popular (even in China now-a-days) is that it's an easy-to-learn language. Think about it: I might say I know pretty well to write /read in English, but I haven't taken any special classes to do that, just watching English movies and browsing the net (and yeah I admit I studied 1 month for SAT but everyone knows that doesn't count :P ).

As for Chinese, there are very few people that are not native Chinese, and have learned the language (one example is tokenadult here on the forum - he might tell us some insights of how difficult it is to learn to speak Chinese).

But I think it's good China has English language teaching developed. In Japan, almost no one knows English, because they want to keep traditions, so it is almost impossible to get directions on the street in Tokio :) I know that first hand, belive me! :)

Yea but for chinese people it's suppose to be really hard.

The thing I don't understand is why everyone is trying to learn my best language (English), and we aren't trying to learn their languages.  As long as you can communicate with other people in your country, isn't that good enough?  and if you must learn some other language to communicate, if your a business for example, then you should learn it.  But I don't why everyone is trying to learn English, they can communicate with themselves already.  The reason I learn a foreing language (Spanish, in my case) is because I live in Texas where the Hispanic population is almost like 50%.

[quote="KBabe"]the chinese ppl who took English in China and then came here all have accents. It's really hard for them to change it and blend in. They should get better teachs. Most of them takes British-English but then when they are finished, they speak British-English with a Chinese accent blended in..it sounds kinda odd.[/quote]British English is also tought in Romanian schools. Teachers will not accept slang, or American conjugations of verbs in (like "learned" instead of "learnt", "can't" instead of "could not", etc).

[quote="qxy"]If Chinese is not treated to be important, I am afraid China would be another Singapore. As I know, the majority of the population in Singapore are Chinese emmigrants, unfortunately, almost most of them could not speak their mother tongue--Chinese well.[/quote] No worry about that. English is not an official language in China...
I think learning English is important, but I do hate the various of English level tests!
[quote="qxy"]
many young people could not speak their local languages, some experts tell that many local languages will die out in 50 years. In fact, I do not like to speak the official language, I like to speak the local language.[/quote]
Speaking local language brings difficulty to communicate with others...
Dialects are so different in China... especially in south... You can not  understand what people say in another city just 50km away from your city...
p.s., Are you a Cantonese? (i am not)

[quote="liyi"]Are you a Cantonese? (i am not)[/quote]

I'm not,I am from an area of south Jiangsu where neighbour towns speak different tongues.

Have you got prize in English Competition?

hey, i was born in Nanjing, Jiangsu.   :)

[quote="KBabe"]hey, i was born in Nanjing, Jiangsu.   :)[/quote].. old neighbors meet again thousands of miles away on the internet :) ...

[quote="yif man12"][quote="KBabe"]hey, i was born in Nanjing, Jiangsu.   :)[/quote]

Hey, me too! Sweet![/quote]Now this is becoming interesting. Maybe we should have a pool to see who else was born in Jiangsu :D

Sweet!

Well, I was born in Lvcheng Town, Danyang City(belong to Zhenjiang,Jiangsu). Now, I am learning high school in the most BEAUTIFUL city of China---Suzhou(it is also a city of South Jiangsu)

[quote="KBabe"][quote="yif man12"][quote="KBabe"]hey, i was born in Nanjing, Jiangsu.   :)[/quote]

Hey, me too! Sweet![/quote]

whoa! were you really? wait when did you move to the US?? Sry, i'm just a little curious. :)[/quote]

Yup. Moved to the US in 1995, at age 5. I've been back to Nanjing 4 times since then.

Hey! I'm born in Irvine! Its a small world

Wo bu keyi zhang zhongwen.
Na hangug mal mot he.
No habla espanol.

That is all :P .

[quote="Treething"]Wo bu keyi zhang zhongwen.[/quote]

In your Chinese sentence, "zhang" should surely be spelled "jiang." Those two words can sound rather similar to people who didn't grow up speaking Chinese.

By the way, the whole sentence is southern dialect, and not completely standard--but I have definitely heard sentences just like it in real life. The more standard way to say the same thing, in the preferred dialect of Modern Standard Chinese, is "Wo bu hui shuo Hanyu."

[quote="tokenadult"]
In your Chinese sentence, "zhang" should surely be spelled "jiang." Those two words can sound rather similar to people who didn't grow up speaking Chinese. [/quote]
Heh, maybe Westerners think they both begin with the sound like "g" in "giant".
In south, we don't distinguish "z" and "zh" clearly... so it sounds like "zang"

[quote="tokenadult"]
By the way, the whole sentence is southern dialect, and not completely standard--but I have definitely heard sentences just like it in real life. The more standard way to say the same thing, in the preferred dialect of Modern Standard Chinese, is "Wo bu hui shuo Hanyu."[/quote]
Zhongwen is okay...

[quote="qxy"]In China, there are many competitions, such as maths, physics, chemistry
,biology and computer. Even, we have ENGLISH COMPETITION. Why? Because the government, the parents, the schools, the teachers all think English is the most important subject. Ten years ago, the thing is different,
everybody thinks maths, physics, chemistry are the most important. Unfortunately, the views are changing, maths, physics, chemistry are no longer the most important, instead, the most important is ENGLISH.
Does anyone have some opinions? Do you think ENGLISH COMPETITION is a good thing? Is there any English Competition in You nation?[/quote]
since i live in america, no.
personally, i dont like the idea of an english competition. Some (though very few) people move from america to china or go to study there,  so it is unfair for thsoe people.


Find the smaller angle in degrees between the two vectors $ p \equal{} (4, \minus{} 2, 5)$ and $ q \equal{} (1  , \minus{}3, 8)$.

Use the dot product formula:
$ \frac{\textbf{p}\cdot\textbf{q}}{|\textbf{p}\parallel{}\textbf{q}|}\equal{}\cos{\theta}$.
$ \textbf{p}\cdot\textbf{q}\equal{}(4,\minus{}2,5)\cdot(1,\minus{}3,8)\equal{}4\plus{}6\plus{}40\equal{}50$
$ |\textbf{p}|\equal{}\sqrt{(4)^{2}\plus{}(\minus{}2)^{2}\plus{}(5)^{2}}\equal{}\sqrt{16\plus{}4\plus{}25}\equal{}\sqrt{45}$
$ |\textbf{q}|\equal{}\sqrt{(1)^{2}\plus{}(\minus{}3)^{2}\plus{}(8)^{2}}\equal{}\sqrt{1\plus{}9\plus{}64}\equal{}\sqrt{74}$
Then $ \cos{\theta}\equal{}\frac{50}{\sqrt{3330}}$.
This has two solutions, so pick out the smaller one.


i have lost my 4 by 4 algorithm booklet (for now)
and i'm trying to find the algorithm that switches ONLY 3 CENTER PIECES
it cycles 3 center pieces ONLY
if anyone happens to know it would you please post it here?

THANKS!

and please do not ask me to look it up on a search engine, i have been trying that

What do you mean switching 3 center pieces?

I think he means that if the centers are like
[code]       |
     R |   B
  _ _ _|
          \
       G    \[/code]
He wants to change it to
[code]       |
     G |   R
  _ _ _|
          \
       B    \[/code]
without changing any other pieces.

lol srry i derived it from a solver (doesnt tell you algorithms but you can easily tell)

its extremely useful
here is the link
[url]http://wrongway.org/?rubik[/url]

Well my method to solve the centers of 4x4 is:
first solve white center
then solve yellow center
then solve one non-white/yellow center
either solve the one next to the center that was just solved or the one opposite
finish up the remaining 2 centers at the same time


function
number theory open
number theory

How positive and negative whole numbers are related to each other in such a way that cause them to immediately be present in some mapping?

tell me if your question is:

let r be a rule of correspondance into two sets A and B such that A and B belong to Z(the integers). Which is the minimum cardinality of A and B for that rule of correspondence is a mapping?

I think it's 1 because this would be the set A mapping into B. Does it requiere to be a function? or with mapping you refer to a ZXZ map, NXN map,.. etc?

Tell me if you meant something different


modular arithmetic
pigeonhole principle
algorithm
combinatorics proposed
combinatorics

One of the numbers 2, 3, or 5 is written in each field of a chessboard. The chess King is allowed to occupy any field and move by chess rules in such a way that he never comes to a field twice. Moreover King changes a direction after each move. For any such path, we compute the product of numbers written in the fields which were attained by King. Prove that there exists a path for which this product is the fourth power of a natural number.

This problem feels rather contrived.  Its two unrelated pieces are to show first that there is some path along which a king can visit every square of the chessboard exactly once, changing directions after each move.  For example, one possible way to do this is to start in a corner and "zig-zag" so as to cover the first two rows, then return covering the next two rows, and so on.  Once we have this, suppose the squares visited are numbered $ d_1, d_2, \ldots, d_{64}$.  Then each of the numbers $ d_1 \cdot d_2 \cdots d_n$ is of the form $ 2^{a_n} 3^{b_n} 5^{c_n}$.  Either there is some $ n$ such that $ a_n \equiv b_n \equiv c_n \equiv 0 \pmod 4$ or by pigeonhole there are $ i < j$ such that $ a_i \minus{} a_j \equiv b_i \minus{} b_j \equiv c_i \minus{} c_j \equiv 0 \pmod 4$, in which case the portion of the path from the $ i$th to $ j$th square encountered is the path we desire.

[quote="JBL"]For example, one possible way to do this is to start in a corner and "zig-zag" so as to cover the first two rows, then return covering the next two rows, and so on.[/quote]
Sorry,I don't think that your algorithm could make the King change direction after move and never come to a field twice.Can you explain clearer?


AMC
USA(J)MO
USAMO
AIME
search
AMC 8
AMC 10

Are we allowed to bring colored pencils to, say, USAMO?  Sometimes, when I have a really complicated geometric diagram, I like to highlight important lines to make them stand out a bit.  I know, the color won't show up on the fax, but it might make the diagram I use while solving a problem easier to read.

Or for that matter, can we use colored pencils on other contests like the AIME?

The rules don't specifically address the issue of colored pencils, according to my search of the official AMC 10/12 manual, but I would DEFINITELY recommend bringing them along.

Certainly one may use colored pencils in the solution of problems on the AMC 8, AMC 10, AMC 12, AIME and USAMO.  Nothing in the rules prohibits their use.

However, I will say that the answer forms on the AMC 8, AMC 10, AMC 12 and AIME should be marked with a soft lead pencil, such as a #2 pencil commonly available so that the scanners can accurately read your intended answer.  On the USAMO answer sheets, our experience with receiving faxed solutions is that solutions written with a firm black mark are easiest to read. I recommend a #2 pencil again for legibility.  Colored lines or marks will not show up on the faxed answers, and in particular, blue and yellow are invisible.

Steve Dunbar
AMC Director


Prove that the following systems has no real solutions :

1.$ \{\begin{aligned}x^5y^3&=81\\ 10x+6y&=27\end{aligned}$

2.$ \{\begin{aligned}x^4+y^4+z^4&=1\\ x^2+y^2+2z^2&=\sqrt{7}\end{aligned}$

3.$ \{\begin{aligned}x^3y=9\\ 3x+y&=6\end{aligned}$

[quote="alex2008"]Prove that the following systems has no real solutions :

1.$ \{\begin{aligned}x^5y^3 & = 81 \\
10x + 6y & = 27\end{aligned}$

2.$ \{\begin{aligned}x^4 + y^4 + z^4 & = 1 \\
x^2 + y^2 + 2z^2 & = \sqrt {7}\end{aligned}$

3.$ \{\begin{aligned}x^3y = 9 \\
3x + y & = 6\end{aligned}$[/quote]

[b]1)[/b]Clearly $ x,y$ must be positive.
By AM-GM, $ (\frac{27}{16})^8=(\frac{5x+3y}{8})^8>x^5y^3=81$, impossible, contradiction.

[b]2)[/b]For $ x,y,z \neq 0$, by Cauchy-Schwarz we have

$ 6=(1+1+4)(x^4+y^4+z^4) \ge (x^2+y^2+2z^2)^2=7$, contradiction.

The rest cases can also be checked using Cauchy-Schwarz.

[b]3)[/b] Similar to [b]1)[/b], use AM-GM again.


I am stupid in Combinatorics so can you give me some websites which provide the Combinatorics exercises and solutions for Olympiad beginner . By the way ,it should be free to get the solution :blush: .Thanks a lot and I look forward for your answer... :lol:

I think it would be better for you to study a book about elementry combinatorics,or combinatorics problems in olympiads instead of using the websites...
anyway maybe someone else can help you with your question...


Find the number of ways in which $5^n$ could be expressed as a product of 3 factors.

three number $5^{n_1}*5^{n_2}*5^{n_3}=5^n$.So $n=n_1+n_2+n_3$, the number of ways is the number of prtition n to three number.(but I don't Knw how much is it) :oops:

number of ways to partition n into 3 numbers is n+2 choose 2.
Let's say there's n balls in a row. We need to put two dividers in between so that it's split into 3 sections. There's gonna be n+2 places where the dividers can go, so its n+2 choose 2.

However, this way doesn't count repetitions, does it?

Just wondering , does 1 count as factor of $5^n$ ?  :?

First krustyteklown you are right,that way doesn't count repititions we have to write  $n=x+y+z$ such that $x\leq y\leq z$, and yes $1$ is factor of $5$.
I found in one book solution for this problem: How many triples $x,y,z$ are there such $x\leq y\leq z$ and $x+y+z=100$, and that solution can be used for $n$ too, i will post it later this night or tomorow.

Here we go:

It is not hard to prove that the number of solutions of $x+y=n$ and $x\leq y$ is
$\lfloor \frac{n}2\rfloor +1$ we will going to need that.

We have to find number of triplets $x,y,z$ of nonnegative integers such that $x\leq
y\leq z$ and $x+y+z=n$, in that case $(x,y,z)$ denote the powers of $5$ in
expression.Let $S$ be set of all that triples. We see that $x\leq
\lfloor\frac{n}3\rfloor$. Now for $k\in \{0,1,...,\lfloor \frac{n}3\rfloor\}$ we
make set $A_k$ of all triples $k=x\leq y\leq z$ such that $x+y+z=n$. So $S=A_1\cup
A_2\cup ... \cup A_{\lfloor\frac{n}3\rfloor}$ and for $i\neq j$ stands $A_i\cap
A_j=\phi$, so $|S|=\sum_{k=0}^{\lfloor\frac{n}3\rfloor}|A_k|$,and from that all  we
need is to compute $|A_k|$.

Let $a=y-k$ and $b=z-k$ then $a+b=n-3k$ and $0\leq a\leq b$ so as we said on the
beginning we have $\lfloor \frac{n-3k}2\rfloor +1$ such pairs $a,b$ and every pair
define one pair $y,z$ (and one solution $k=x\leq y\leq z$) so $|A_k|=\lfloor
\frac{n-3k}2\rfloor +1$ and from that we have
\[|S|=\sum_{k=0}^{\lfloor\frac{n}3\rfloor}|A_k|=\lfloor
\frac{n}3\rfloor+\sum_{i=0}^{\lfloor \frac{n}3\rfloor}\lfloor \frac{n-3i}2 \rfloor\]

And that's it, I just can't calculate this last sum...


The incircle of triangle $ ABC$ with incenter $ I$ touches $ AC$ at point $ D$. The line passing through $ A$ is bisector of angle $ \angle ABI$ and intersects $ AI$ at point $ P$. The line passing through $ A$ is the bisector of angle $ \angle BCI$ and intersects $ CI$ at point $ Q$. Prove that
$ \angle PDQ \equal{} \frac {\pi }{2}$.

[quote="lasha"]... The line passing through $ A$ is bisector of angle $ \angle ABI$ and intersects $ AI$ at point $ P$. The line passing through $ A$ is the bisector of angle $ \angle BCI$ and intersects $ CI$ at point $ Q$...[/quote]
Bisector of the $ \angle ABI$ does not pass through A and neither does bisector of the $ \angle BCI.$ Bisector of the $ \angle BCI$ does not intersect CI at some point Q, but at C. Me thinks it should be bisectors of the $ \angle ABI,\ \angle CBI,$ both passing through B. BI meets AC at K, but it is not necessary for BK to bisect the $ \angle ABC.$ If P, Q are defined as intersections of AI, CI with bisectors of the $ \angle ABK,\ \angle CBK,$ the $ \angle PDQ$ is right regardless (it is a well known problem).

I, P, Q are incenters of the $ \triangle ABC,\ \triangle ABK,\ \triangle BCK.$ The incircles (I), (P), (Q) touch AC at D, E, F and (P), (Q) touch BK at S, T. WLOG, let A, D, K, C follow on AC in this order.

$ KS \equal{} s(\triangle ABK) \minus{} AB$
$ KT \equal{} s(\triangle BCK) \minus{} BC$
$ ST \equal{} s(\triangle ABK) \minus{} s(\triangle BCK) \minus{} (AB \minus{} BC) \equal{} \frac {AK \minus{} CK}{2} \minus{} \frac {AB \minus{} BC}{2}$

$ AK \equal{} AD \plus{} DK \equal{} s(\triangle ABC) \minus{} BC \plus{} DK$
$ CK \equal{} CD \minus{} DK \equal{} s(\triangle ABC) \minus{} AB \minus{} DK$
$ AK \minus{} CK \equal{} 2\ DK \minus{} (BC \minus{} AB)$

$ ST \equal{} DK$

$ EK \minus{} DF \equal{} (s(\triangle ABK) \minus{} AB \minus{} DK) \minus{} (s(\triangle BCK) \minus{} BC) \equal{} ST \minus{} DK \equal{} 0$
$ EK \equal{} DF$

Let PQ meet $ BK \equiv ST$ at G. Since the common internal tangent of (P), (Q) other than $ BK \equiv ST$ cuts their common external tangent $ AC \equiv EF$ at a point D', such that KD' = ST, (I do not even need EK = DF for this) $ D' \equiv D$ and GD is the common internal tangent of (P), (Q) other than $ BK \equiv ST.$ P, Q then lie on bisectors of the $ \angle GDA,\ \angle GDC,$ which are perpendicular to each other, $ \angle PDQ \equal{} 90^\circ.$


function
Euler
calculus
algebra
polynomial
logarithms
real analysis

Let $(a_{0},a_{1},\ldots)$ be an infinite integer sequence.

Define $(b_{0},b_{1},\ldots)$ to be another infinite integer sequence such that \[b_{n}=\sum_{k=0}^{n}{n \choose k}a_{k}.\]

Prove that for all $n$, \[a_{n}=\sum_{k=0}^{n}{n \choose k}(-1)^{n-k}b_{k}.\]

Shouldnt be too difficult.

Indeed, I'm sure it will work if you just plug in the formula and evaluate, or you can probably use induction. It reminds me of Mobius inversion...

[quote="mao_zai"]Let $(a_{0},a_{1},\ldots)$ be an infinite integer sequence.
Define $(b_{0},b_{1},\ldots)$ to be another infinite integer sequence such that $b_{n}=\sum_{k=0}^{n}{n \choose k}a_{k}$ . [b]Prove [/b] $a_{n}=\sum_{k=0}^{n}{n \choose k}(-1)^{n-k}b_{k}.\; \forall n \in{\mathbb N}.$  ....[/quote]
I present a solution based on generating functions (of exponential type). Assume that $(a_{n})$ and $(b_{n})$ are generated by
\[A(t)=\sum_{k=0}^{\infty}a_{k}\frac{t^{k}}{k!}\; \; \mbox(resp.)\; \; \; B(t)=\sum_{k=0}^{\infty}b_{k}\frac{t^{k}}{k!}\; . \]
Let us remind that
\[(*)\; \; \; \; \begin{array}{|c|}\hline \\ \left(\sum_{k=0}^{\infty}\alpha_{k}\frac{t^{k}}{k!}\right) \left(\sum_{k=0}^{\infty}\beta_{k}\frac{t^{k}}{k!}\right)= \sum_{k=0}^{\infty}\gamma_{k}\frac{t^{k}}{k!}\\ \\ \hline \end{array}\; ,\; \; \mbox{with}\; \; \; \gamma_{k}: =\sum_{j=0}^{k}{k \choose j}\alpha_{j}\beta_{k-j}\]
From $(*)$  we have $B(t)=A(t)\sum_{k=0}^{\infty}\frac{t^{k}}{k!}=A(t)e^{t}\; .$ Therefore $A(t)=e^{-t}B(t)$ , i.e. see $(*)$
\[\sum_{k=0}^{\infty}a_{k}\frac{t^{k}}{k!}= \left(\sum_{k=0}^{\infty}(-1)^{k}\frac{t^{k}}{k!}\right) \left(\sum_{k=0}^{\infty}b_{k}\frac{t^{k}}{k!}\right)= \sum_{k=0}^{\infty}\left(\sum_{j=0}^{k}(-1)^{k-j}{k\choose j}b_{j}\right)\frac{t^{k}}{k!}\; . \]
Because in the theory of generating functions it's true that: two sequences $(x_{n})$ and $(y_{n})$ are equall iff both sequences have the same  generating function, we conclude with
$a_{k}=\sum_{j=0}^{kl}(-1)^{k-j}{k\choose j}b_{j}\; \; ,\; \; \forall k\in{\mathbb N}.$

[size=75]
[b]Notes:[/b]
1) I appreciate that this trick must attributed to L. Euler .
2) Such propositions  as  above, that is  $\; b_{n}=\sum_{k=0}^{n}t_{k,n}a_{k}\; \; \; \mbox{implies}\; \; a_{n}=\sum_{k=0}^{n}\tau_{k,n}b_{k}$  are called  [b] inverse  theorems .[/b]
3) There is a field in Mathematics which deserves Combinatorics, Calculus of finite differences,Numerical Analysis,Approximation theory,..., named  [b] Umbral Calculus[/b] which in fact is based on the so-called [b]binomial polynomials.[/b]
In my opinion the modern aspects of this theory were given by Gian-Carlo  Rota, Steven Roman,Andrew Odlyzko,Ronald Mullin,....
4) Some references are:

1     G.E. Andrews. On the foundations of combinatorial theory V, Eulerian differential operators. Stud. App. Math., 50:345--375, 1971. (MR 46#8845).

2      P. Appell. Sur une classe de polynômes (French, On a class of polynomials). Ann. Sci. Ecole Norm. Sup, (2) 9:119--144, 1880.

3     M. Barnabei. Lagrange inversion in infinitely many variables. J. Math. Anal. Appl., 108:198--210, 1985. (MR 86j:05023).

 4    E.T. Bell. The history of Blissard's symbolic calculus, with a sketch of the inventor's life. Amer. Math. Monthly, 45:414--421, 1938. (Zbl. 19, 389).

5    E.T. Bell. Postulational bases for the umbral calculus. Amer. J. Math., 62:717--724, 1940. (MR 2, 99).

6     J.W. Brown. On multivariable Sheffer sequences. J. Math. Anal. Appl., 69:398--410, 1979. (MR 80j:05007).

7    L. Cerlienco and F. Piras. G-R-sequences and incidence coalgebras of posets of full binomial type. J. Math. Anal. Appl., 115:46--56, 1986. (MR 87k:05018).

8     Ch. Charalambides and J. Singh. A review of the Stirling numbers, their generalizations and statistical applications. Comm. Stat. Th. Methods, (8) 17:2533--2595, 1988. (MR 89d:62017).

9    J. Cigler. Operatormethoden für q-Identitäten III: Umbrale inversion und die Lagrangesche formel. Arch. Math., 35:533--543, 1980. (MR 83g:05008).

10     J. Cigler. Operatormethoden für q-Identitäten II: q-Laguerre polynome. Monatsh. Math., 91:105--117, 1981. (MR 83g:05007).

11      H.B. Curry. Abstract differential operators and interpolation formulas. Portugal. Math., 10:135--162, 1951. (MR 13, 632).

12     H.T. Davis. The theory of linear operators. Principia Press, Bloomington, Indiana, 1936. (bibliography on Appell polynomials on p. 25 etc.).

13     A. Di Bucchianico. Banach algebras, logarithms, and polynomials of convolution type. J. Math. Anal. Appl., 156:253--273, 1991. (MR 92d:46123).

14     A. Di Bucchianico. Representations of Sheffer polynomials. Stud. Appl. Math., 93:1--14, 1994.

15     A. Di Bucchianico and D.E. Loeb. A simpler characterization of Sheffer polynomials. Stud. Appl. Math., 92:1--15, 1994.

16 G.P.Egorychev , Integral Representation and the Computation  of Combinatorial Sums, Translations of Mathematical Monographs,AMS, vol.59,1984.

17     J.P. Fillmore and S.G. Williamson. A linear algebra setting for the Rota-Mullin theory of polynomials of binomial type. Lin. and Multilin. Alg., 1:67--80, 1973. (MR 47#9321b).

18     A.M. Garsia. An exposé of the Mullin-Rota theory of polynomials of binomial type. Lin. and Multilin. Alg., 1:47--65, 1973. (MR47#9321a).

19     A.M. Garsia and S. Joni. Higher dimensional polynomials of binomial type and formal power series inversion. Comm. Algebra, 6:1187--1211, 1978. (MR 58#10484).

20    B. Germano and P.E. Ricci. Umbral calculus and orthogonal systems. Rend. Mat. Appl., 12(1):217--233, 1992. (Zbl. 754, 20).

21     L. Giraitis. A central limit theorem for polynomial forms I. Liet. matem. rink. (=Litov. mat. sbor.), 29(2):266--289, 1989. (MR 91c:60026).

22    A. Guinand. The umbral method: A survey of elementary mnemonic and manipulative uses. Amer. Math. Monthly, 86:187--195, 1979. (MR 80e:05001).

23     H. Gzyl. Interpretacion combinatorica de polinomios de tipa binomial. Acta Cient. Venezolana, 27:244--246, 1976. (MR 55#118).

24     H. Gzyl. Canonical transformations, umbral calculus and orthogonal theory. J. Math. Anal. Appl., 111:547--558, 1985. (MR 87e:05019).

25    H. Gzyl. Umbral calculus via integral transforms. J. Math. Anal. Appl., 129:315--325, 1988. (MR 89a:05022).

26    P.S. Hirschhorn and L.A. Raphael. Coalgebraic foundations of the method of divided differences. Adv. Math., 91:75--135, 1992. (MR 92m:05012).

27  E.C. Ihrig and M.E.H. Ismail. On an umbral calculus. In 10th Proc. Southeastern Conf. Combinatorics, Graph Theory, Computing, Boca Raton, pages 523--528, 1979. (MR 82b:13010).

28     E.C. Ihrig and M.E.H. Ismail. A q-umbral calculus. J. Math. Anal. Appl., 84:178--207, 1981. (MR 83a:05011).

29     M.E.H. Ismail. Polynomials of binomial type and approximation theory. J. Approx. Th., 23:177--186, 1978. (MR 81a:41033).

30     S.A. Joni. Polynomials of binomial type and the Lagrange inversion formula. PhD thesis, University of California, La Jolla, 1977.

31     S.A. Joni. Lagrange inversion in higher dimension and umbral operators. J. Linear and Multilinear Algebra, 6:111--121, 1978. (MR 58#10485).

32    P. Kirschenhofer. Binomialfolgen, Shefferfolgen und Faktorfolgen in den q-Analysis. Sitzungber. Abt. II Österr. Akad. Wiss. Math. Naturw. Kl., 188:263--315, 1979. (MR 82d:05013).

33       Ch. Krattenthaler. Operator methods and Lagrange inversion. Trans. Amer. Math. Soc, 305:431--465, 1988. (MR 89d:05017).

34    D.E. Loeb. Sequences of symmetric functions of binomial type. Stud. Appl. Math., 83:1--30, 1990. (MR 92e:05012).

35    D.E. Loeb and G.-C. Rota. Formal power series of logarithmic type. Adv. Math., 75:1--118, 1988. (MR 90f:05014).

36   G Markowsky. Differential operators and the theory of binomial enumeration. J. Math. Anal. Appl., 63:145--155, 1978. (MR 58#21666).

37 R. Mullin and G.-C. Rota. On the foundations of combinatorial theory III. Theory of binomial enumeration. In Harris, editor, Graph theory and its applications, pages 167--213. Academic Press, 1970. (MR 43#65).

38     H. Niederhausen. Sheffer polynomials and linear recurrences. Congr. Num., 29:689--698, 1980. (MR 82m:05012).

39     H. Niederhausen. A formula for explicit solutions of certain linear recursions on polynomial sequences. Congr. Num., 49:87--98, 1985. (MR 87j:11018).

40     S. Pincherle and U. Amaldi. Le operazioni distributive e le loro applicazioni all'analisi. N. Zanichelli, Bologna, 1901. (esp. pp. 130-139).

41     R. Rasala. The Rodrigues formula and polynomial differential operators. J. Math. Anal. Appl., 84:443--482, 1981. (MR 83g:33009).

42     M. Razpet. An application of the umbral calculus. J. Math. Anal. Appl., 149:1--16, 1990. (MR 91i:05018).

43     S.M. Roman. The algebra of formal series. Adv. Math., 31:309--329, 1979. (MR 81b:05016 a/b); erratum Adv. Math. 35, 1980, 274.

44     S.M. Roman. The algebra of formal series. III. several variables. J. Approx. Theory, 26:340--381, 1979. (MR 81i:05023b).

45     S.M. Roman. Polynomials, power series and interpolation. J. Math. Anal. Appl., 80:333--371, 1981. (MR 83k:41006).

46     S.M. Roman. The theory of umbral calculus I. J. Math. Anal. Appl., 87:58--115, 1982. (MR 84c:05008a).

47     S.M. Roman. The theory of umbral calculus II. J. Math. Anal. Appl., 89:290--314, 1982. (MR 84c:05008b).

48     S.M. Roman. The theory of umbral calculus III. J. Math. Anal. Appl., 95:528--563, 1983. (MR 85e:05020).

49    S.M. Roman. [b]The umbral calculus.[/b] Academic Press, 1984. (MR 87c:05015 = Zbl. 536.33001).

50   S.M. Roman. More on the umbral calculus, with emphasis on the q-umbral calculus. J. Math. Anal. Appl., 107:222--254, 1985. (MR 86h:05024).

51     S.M. Roman, P.N. De Land, R.C. Shiflett, and H.S. Schultz. The umbral calculus and the solution to certain recurrence relations. J. Comb. Inf. Syst. Sci., 8:235--240, 1983. (MR 87b:05026).

52     S.M. Roman and G.-C. Rota.[b] The umbral calculus.[/b] Adv. Math., 27:95--188, 1978. (MR 58#5256).

53     G.-C. Rota, D. Kahaner, and A. Odlyzko.[b] On the foundations of combinatorial theory VII. Finite operator calculus.[/b] J. Math. Anal. Appl., 42:684--760, 1973. (MR 49#10556).

54    G.-C. Rota and B.D. Taylor. The classical umbral calculus. SIAM J. Math. Anal., 25:694--711, 1994.

55    I.M. Sheffer. Some properties of polynomial sets of type zero. Duke Math. J., 5:590--622, 1939. (Zbl. 22, 15; MR 1, 15).

56   J.F Steffensen. The poweroid, an extension of the mathematical notion of power. Acta Math., 73:333--366, 1941. (MR 3, 326).

57     J.F. Steffensen. On a class of polynomials. Mat. Tidsskr. B, 1945:10--14, 1945. (MR 7, 157).

58     K. Ueno. Umbral calculus and special functions. Adv Math., 67:174--229, 1988. (Zbl. 645.05011; MR 88m:05012).

59    L. Verde-Star. Dual operators and Lagrange inversion in several variables. Adv. Math., 58:89--108, 1985. (MR 87d:05027).

60     L. Verde-Star. Interpolation and combinatorial functions. Stud. Appl. Math., 79:65--92, 1988. (MR 91b:05012).

61     L. Verde-Star. Divided differences and combinatorial identities. Stud. Appl. Math., 85:215--242, 1991. (MR 92i:65027).

62     L. Verde-Star. Polynomial sequences of interpolatory type. Stud. Appl. Math., 53:153--171, 1993. (MR 94d:41008).

63      L. Verde-Star. Operator identities and the solution of linear matrix difference and differential identities. Stud. Appl. Math., 91:153--177, 1994.

Books on [u][color=green] Generating Functions [/color][/u] are:

(A) Elna B.Mc.Bride, [i] Obtaining Generating Functions[/i], Springer-Verlag,1971.

(B) H.M.Srivastava and H.L.Manocha ,[i] A Treatise on Generating Functions[/i], John Wiley & Sons,1984.

References about   [color=green] [b]Finite differences, [/b][/color] :

[1].P. Agarwal, Difference calculus with applications to difference equa-
equations, in General Inequalities 4, cd. W. Walter, ISNM 71, Birkhh'user Ver-
lag, Basel, 1984, 95-110.
[2]. R.P. Agarwal, Difference equations and inequalities: A survey, in Pro-
Proceedings of the World Congress of Nonlinear' Analysts, Tampa, Florida,
August 19-26, 1992, ed. V. Lakshmikantham, Walter de Gruytcr, Berlin,
1996, 1091 1108.
[3]. R.P. Agarwal and P.J.Y. Wong, Advanced Topics in, Difference Equa-
Equations, Kluwcr, Dordrecht, 1997.
[4]. J.H. Ahlbcrg, E.N. Nilson and J.L. Walsh, The Theory of Splines and
Their Applications, Academic Press, New York, 1967.
[5]. CD. Ahlbrandt and A.C. Peterson, Discrete Hamiltonian System.s:
Difference Equations, Continued Fractions and Riccati Equations, Kluwcr,
Dordrecht, 1996.
[6]. F.V. Atkinson, Discrete and Continuous Boundary Value Problems,
Academic Press, New York, 1964.
[7]. P.M. Batchelder, An Introduction to Linear Difference Equations, Har-
Harvard University Press, Cambridge, 1927.
[8]. G. Boole, Calculus of Finite Differences, 4th ed., Chelsea, New York,
1958.
[9]. L. Brand, Differential and Difference Equations, Wiley, New York,
1966.
[10]. L. Brugnano and D. Trigiantc, Solving Differential Problems to Mul-
tistep Initial and Boundary Value Methods, Gordon & Breach Sci. Publ,
Amsterdam, 1998.
[11]. J.R. Cash, Stable Recursions, Academic Press, London, 1979.
[12]. F. Chorlton, Differential and Difference Equations, Van Nostrand,
London, 1965.
[13]. E.J. Cogan and R.Z. Norman, Handbook of Calculus, Differences and
Differential Equations, Prentice-Hall, Englewood Cliffs, N.J., 1958.
[14]. G. Dahlquist and A. Bjorck, Numerical Methods, Prentice- Hall, En-
Englewood Cliffs, N.J., 1974.
[15]. W.R. Derrick and S.I. Grossman, Elementary Differential Equations
with Applications, Addison- Wesley Publ. Comp., Reading, 1981.
[16]. W.S. Dorn and D.D. McCrackcn, Numerical Methods with FORTRAN
IV Case Studies, Wiley, New York, 1972.
[17]. S. Elaydi, An Introduction to Difference Equations, Springer Vcrlag,
New York, 1996.
[18]. N. Finizio and G. Ladas, An Introduction to Differential Equations
with Difference Equations, Fourier Series, and Partial Differential Equa-
Equations, Wadsworth Publ. Comp., Bclmont, 1982.
[19]. T. Fort, Finite Differences and Difference Equations in the Real Do-
Domain, The Clarendon Press, Oxford, 1948.
[20]. A.O. Gel'fond, Calculus of Finite Differences, Hindustan, Delhi, India,
1971.
[21]. S. Goldberg, Introduction to Difference Equations, Wiley, New York,
1958.
[22]. P. Hartman, Difference equations: Disconjugacy, principal solutions,
Green's functions, complete monotonicity, Trans. Amcr. Math. Soc.
246A978), 1-30.
[23]. F.B. Hildcbrand, Finite Difference Equations and Simulations, Pren-
Prentice Hall, Englcwood Cliffs, N. J., 1968.
[24]. A.J. Jerri, Linear Difference Equations with Discrete Transform, Meth-
Methods, Kluwcr, Dordrecht, 1996.
[25]. C. Jordan, [b] Calculus of Finite Differences[/b], 3rd erf., Chelsa, New York,
1965.
[26]. W.G. Kelley and A.C. Peterson, Difference Equations: An Introduc-
Introduction with Applications, Academic Press, New York, 1991.
[27]. V.L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference
Equations of Higher Order with Applications, Kluwcr, Dordrecht, 1993.
[28]. V. Lakshmikantham and D. Trigiantc, Theory of Difference Equations:
Numerical Methods and Applications, Academic Press, New York, 1988.
[29]. V. Lakshmikantham, S. Sivasundaram and B. Kaymakcalan, Dynamic
Systems on Measure Chains, Kluwcr, Dordrecht, 1996.
[30]. M. Lees, Discrete methods for nonlinear two-point boundary value
problems, in Numerical Solutions of Partial Differential Equations, ed. J.H.
Bramble, Academic Press, New York A966), 59-72.
[31]. H. Levy and F. Lessman, Finite Difference Equations, Sir Jssac Pitman
and Sons, Ltd., London, 1959.
[32]. R.E. Mickcns, Difference Equations: Theory and Applications, Van
Nostrand Reinhold, New York, 2nd edition, 1990.
[33]. R.E. Mickens, Nonstandard Finite Difference Models of Differential
Equations, World Scientific, Singapore, 1994.
[34]. J.C.P. Miller, British Association for the Advancement of Science:
Bessel Functions Part II, Mathematical Tables, Vol. 10, Cambridge Uni-
University Press, 1952.
[35]. K.S. Miller, An Introduction to the Calculus of Finite Differences and
Difference Equations, Holt, New York, 1960.
[36]. K.S. Miller, Linear Difference Equations, W. A. Benjamin, New York,
1968.
[37]. L.M. Milne Thomson, The Calculus of Finite Differences, Maximilian,
London,1960.
[38]. E. Pirmcy, Ordinary Difference Differential Equations, University of
California Press, Berkeley and Los Angeles, 1959.
[39]. C.H. Richardson, An Introduction to the Calculus of Finite Differ-
Differences, Van Nostrand, New York, 1954.
[40]. A.N. Sharkovsky, Y.L. Maistrenko and E.Y. Romancnko, Difference
Equations and Their Applications, Kluwcr, Dordrecht, 1993.
[41]. M.R. Spiegal, Calculus of Finite Differences and Difference Equations,
McGraw-Hill, New York, 1971.
[42]. J. Wimp, Computation with Recurrence Relations, Pitman Advanced
Publishing Program, Boston, 1984.

[/size]

Thanks flip!

Much more thanks for the references.

Another book on generating functions Flip didn't mention is generatingfunctionology, which is available online for free from the author's homepage (just do a search for "generatingfunctionology;" it should be the first hit).

Also, probability, this [i]is[/i] a Mobius inversion, on the poset called a Boolean algebra (all the subsets of a given $n$-element set, ordered by inclusion).

I'm sure some of Flip's references have something on that subject, but I'll add one of my own: R. Stanley's [i]Enumerative Combinatorics, Volume 1[/i] is all over this kind of stuff.


inequalities
parameterization
inequalities proposed
109

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

$\displaystyle \displaystyle \displaystyle \sum\frac{ab-bc+ca}{b^2+c^2} \ge \frac{3}{2}$

cyclic or symmetric?
(does it make a difference?)

Cyclic. In expanded form:

$\frac{ca+ab-bc}{b^2+c^2}+\frac{ab+bc-ca}{c^2+a^2}+\frac{bc+ca-ab}{a^2+b^2}\geq\frac32$.

A very nice inequality, by the way. I'll try to find a nice approach.

  darij

Yes, very good, but a good inequality can not to have a good proof  :?

So what is most important for an inequality, to have a good solution or to have a good look  ;)

This is a particular case ($p=0$) of the general inequality

$ \sum \frac{ab+(p-1)bc+ca}{b^2+pbc+c^2}\ge \frac{3(p+1)}{p+2}$.

I think this inequality is valid for any $p>-2$, but I proved it only for $p \ge -1$.

For $p=1$ we get "Vasc0 strange ineq",

for $p=-1$ we get "Vasc8 nice ineq", that is

$ \sum \frac{ab-2bc+ca}{b^2-bc+c^2} \ge 0$,

and for $p=2$ we get Iran 1996 ineq. :P

I must check if it works also for $a \geq c \geq b$ as it is cyclic and not symmetric so you can only assume $a \geq b,c$

Here was wrong solution.
I deleted it, cause it occupied much space.

This is wrong in my opinion

$ \displaystyle \displaystyle 2(\sum{ab})(\sum{a^4}) \geq\
(\sum{a^2})(\sum{a^2b^2})+ 9a^2b^2c^2 + ((\sum{a^2})- (\sum{ab}))(\sum{ab})^2$

It is

$\displaystyle \displaystyle 2(\sum{ab})(\sum{a^4}) \geq\
(\sum{a^2})(\sum{a^2b^2})+ 9a^2b^2c^2 + 2((\sum{a^2})- (\sum{ab}))(\sum{ab})^2$

I think it is symmetric, too
If we put $f(a,b,c)=\sum \frac{ab-bc+ca}{b^2+c^2}$
then $f(a,b,c)=f(a,c,b)$
Hence,WLOG we can assume $a\ge b\ge c$

Of course, the inequality is symmetric.
My proof is very similar to Minhkhoa's one.

Sorry, for mistake. (manlio, you very carefully read solutions).

$\displaystyle \sum \frac {(ab+bc+ac)}{(a^2+b^2)}-\sum \frac{2bc}{(b^2+c^2)}\geq\frac32$

After easy combination:
$\displaystyle 2(\sum{ab})(\sum{a^2b^2})+2(\sum{ab})(\sum{a^4})- 4abc(\sum{a^3})\geq\ 3(a^2+b^2) (b^2+c^2) (a^2+c^2)$

Rewriting:
$\displaystyle 2(\sum{a^3b^3})+2abc(\sum{a^2b+b^2a})+2(\sum{a^5b+b^5a})- 2abc(\sum{a^3})\geq\ 3(\sum{a^4b^2+b^4a^2})+ 6a^2b^2c^2 $

$\displaystyle 2(\sum{a^3b^3})+2(\sum{a^5b+b^5a})- 3(\sum{a^4b^2+b^4a^2})\geq\
2abc((\sum{a^3})+3abc-(\sum{a^2b+b^2a}))$

$\displaystyle \sum{ab(2a^2+ab+2b^2)(a-b)^2}\geq\ 2\sum{a^2b^2(a-b)^2}
\geq\ 2abc(\sum{a(a-b)(a-c))}$

 Note$ x = ab(a-b), y= bc(b-c), z = ac(a-c)$

Our last inequality is equalent to $ x^2+y^2+z^2\geq xy+yz+xz $. c.t.d.

I enjoy a lot to read the proofs of the very clever solvers of the forum like you, Prowler :) .

Your solutions are always very nice. ;)

Thank you very much.

Prowler, your proof is perfect and fantastic :)

Really a great proof ;)

Indeed, Prowler's solution is very nice. :lol:

Dear Vasc, can I know could the general inequality

$ \sum \frac{ab+(p-1)bc+ca}{b^2+pbc+c^2}\ge \frac{3(p+1)}{p+2}$
be solved by the same way of my solution for 'Vasc 6 nice ineq"

[delete]
thank minhkhoa for showing me my mistake :(

yes ,very sure
but maybe wrong ,i proved it just 45 minutes :D ;

This may be your mistake
$2a-b-c$ is not always positive ;)

Sorry, I don't understand that

[quote="cuong"]hei , i don't mean that 2a-b-c>=0;

i use equivalent way to prove this :D[/quote]

I think I solved the inequality, that I proved the case $p > -2$ too. Please proofread this for me.

\[
\begin{gathered}
  \sum {\frac{{ab + (p - 1)bc + ca}}
{{b^2  + pbc + c^2 }}}  \geqslant \frac{{3(p + 1)}}
{{(p + 2)}} \hfill \\
   \Leftrightarrow \sum {(p + 2)\frac{{ab + (p - 1)bc + ca}}
{{b^2  + pbc + c^2 }} - (p + 1)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(p + 2)ab + (p + 2)(p - 1)bc + (p + 2)ca - (p + 1)\left( {b^2  + pbc + c^2 } \right)}}
{{b^2  + pbc + c^2 }}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(p + 2)ab + (p + 2)ca - (p + 1)b^2  - (p + 1)c^2  - 2bc}}
{{b^2  + pbc + c^2 }}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{\left( {(p + 1)b + c} \right)(a - b) + \left( {(p + 1)c + b} \right)(a - c)}}
{{b^2  + pbc + c^2 }}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {(a - b)\left( {\frac{{(p + 1)b + c}}
{{b^2  + pbc + c^2 }} - \frac{{(p + 1)a + c}}
{{a^2  + pac + c^2 }}} \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {(a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right)\left( {a^2  + b^2  + pab} \right)}  \geqslant 0 \hfill \\
\end{gathered}
\]

Case 1: $p \ge 0$, then WLOG assume $a \ge b \ge c$.
The only thing we are not certain about is \[
(p + 1)bc + ac + ab - a^2
\]
If this is positive we are done, else,
\[
\begin{gathered}
  (a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right)\left( {a^2  + b^2  + pab} \right) \geqslant (a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right)\left( {a^2  + c^2  + pca} \right) \hfill \\
  (b - c)^2 \left( {(p + 1)bc + ac + ab - a^2 } \right)\left( {b^2  + c^2  + pbc} \right) \geqslant (b - c)^2 \left( {(p + 1)bc + ac + ab - a^2 } \right)\left( {a^2  + c^2  + pca} \right) \hfill \\
\end{gathered}
\]
Using these two inequalities, we only need to prove,
\[
(a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right) + (b - c)^2 \left( {(p + 1)bc + ac + ab - a^2 } \right) + (c - a)^2 \left( {(p + 1)ca + ba + bc - b^2 } \right) \geqslant 0
\]
Actually,
\[
\begin{gathered}
  (a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right) + (b - c)^2 \left( {(p + 1)bc + ac + ab - a^2 } \right) + (c - a)^2 \left( {(p + 1)ca + ba + bc - b^2 } \right) \hfill \\
   = (p + 2)\left( {ab(a - b)^2  + bc(b - c)^2  + ca(c - a)^2 } \right) - (a - b)(b - c)(c - a)( - a + b - b + c - c + a) \hfill \\
   = (p + 2)\left( {ab(a - b)^2  + bc(b - c)^2  + ca(c - a)^2 } \right) \geqslant 0 \hfill \\
\end{gathered}
\]

This proves case 1.
Case 2: p \le 0.
Well,
\[
\begin{gathered}
  (a - b)^2 \left( {(p + 1)ab + ac + bc - c^2 } \right)\left( {a^2  + b^2  + pab} \right) + (b - c)^2 \left( {(p + 1)bc + ac + ab - a^2 } \right)\left( {b^2  + c^2  + pbc} \right) \hfill \\
   + (c - a)^2 \left( {(p + 1)ca + ba + bc - b^2 } \right)\left( {a^2  + b^2  + pab} \right) \hfill \\
   = (p + 2)\left( {\sum {ab(a - b)^2 \left( {a^2  + b^2  + pab} \right)} } \right) - (a - b)(b - c)(a - c) \hfill \\
  \left( {(a - b)(a^2  + pab + b^2 ) + (b - c)(b^2  + pbc + c^2 ) + (c - a)(c^2  + pac + a^2 )} \right) \hfill \\
   = (p + 2)\left( {\sum {ab(a - b)^2 \left( {a^2  + b^2  + pab} \right)} } \right) - (a - b)(b - c)(a - c) \hfill \\
  \left( {(a - b)(a^2  + pab + b^2 ) + (b - c)(b^2  + pbc + c^2 ) + (c - b)(c^2  + pac + a^2 ) + (b - a)(c^2  + pac + a^2 )} \right) \hfill \\
   = (p + 2)\left( {\sum {ab(a - b)^2 \left( {a^2  + b^2  + pab} \right)} } \right) - (a - b)(b - c)(a - c) \hfill \\
  \left( {(b^2  + pab - pac - c^2 )(a - b) + (b^2  + pbc - a^2  - pac)(b - c)} \right) \hfill \\
   = (p + 2)\left( {\sum {ab(a - b)^2 \left( {a^2  + b^2  + pab} \right)} } \right) - (a - b)(b - c)(a - c) \hfill \\
  \left( {(b - c)(a - b)(b + c + pa) - (b - c)(a - b)(b + a + pc)} \right) \hfill \\
   = (p + 2)\left( {\sum {ab(a - b)^2 \left( {a^2  + b^2  + pab} \right)} } \right) - (p - 1)(a - b)^2 (b - c)^2 (c - a)^2  \hfill \\
\end{gathered}
\]
The first term is obviously positive. When $p \le 0$, the last term is also positive.
This finishes the proof.

I have checked your solution, siuhochung and in my opinion it is true.

Super  Nice proof!!!

Congratulations.

Excellent, Siuhochung. :lol:
My proof is rather like your proof.
My hint, as your hint, is "p+2". ;)

[quote="Vasc"]For $p \ge \frac{-7}{5}$ , I proven the inequality writting it in the form

   $\sum E(a,b,c).bc(b-c)^2 \ge 0$.

But, for the general case $p>-2$, I used a special way.[/quote]

Vasc, as the inequality is proven, can you show me how did u prove it for $p \ge \frac{-7}{5}$? thanks!

Very nice solution, Suihochung
I tried to folow my proof in the special case which lead to prove that
if $LHS-RHS=B(a-b)+A(b-c)$ and $a\ge b\ge c$
then $A,B\ge 0$
but gave up because it looked frighten and ugly
This ineq. is a new type to me
Thank you everyone

I have not carefully followed this discussion, especially I haven't tried the inequalities with general parameters since they were just too messy for me (although very general and interesting). Here is my proof of the initial inequality of this thread, namely of the inequality

$\frac{ca+ab-bc}{b^2+c^2}+\frac{ab+bc-ca}{c^2+a^2}+\frac{bc+ca-ab}{a^2+b^2}\geq\frac32$

for any three positive reals a, b, c:

First we show a lemma:

[color=blue][b]Lemma 1.[/b] If a, b, c are any three nonnegative reals such that $a\geq b\geq c$, and x, y, z are three nonnegative reals such that $x\leq y\leq z$, then

$x\left(bc+ca+ab-a^2\right)\left(b-c\right)^2+y\left(bc+ca+ab-b^2\right)\left(c-a\right)^2+z\left(bc+ca+ab-c^2\right)\left(a-b\right)^2$
$\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] Since $a\geq b\geq c$, we have $ab\geq b^2$, so that $bc+ca+ab-b^2\geq 0$, and $ca\geq c^2$, so that $bc+ca+ab-c^2\geq 0$. Thus, all expressions in the above inequality are $\geq 0$, except of the expression $bc+ca+ab-a^2$, which may be $\geq 0$ but may also be < 0. If this expression is $\geq 0$, then Lemma 1 is proven. If not, then $x\leq y$ yields $x\left(bc+ca+ab-a^2\right)\geq y\left(bc+ca+ab-a^2\right)$ (since an inequality changes sign when being multiplied by a negative number). On the other hand, $y\leq z$ yields $z\geq y$, so that $z\left(bc+ca+ab-c^2\right)\geq y\left(bc+ca+ab-c^2\right)$ (since $bc+ca+ab-c^2\geq 0$). Hence,

$x\left(bc+ca+ab-a^2\right)\left(b-c\right)^2+y\left(bc+ca+ab-b^2\right)\left(c-a\right)^2+z\left(bc+ca+ab-c^2\right)\left(a-b\right)^2$
$\geq y\left(bc+ca+ab-a^2\right)\left(b-c\right)^2+y\left(bc+ca+ab-b^2\right)\left(c-a\right)^2+y\left(bc+ca+ab-c^2\right)\left(a-b\right)^2$
$=y\left(\left(bc+ca+ab-a^2\right)\left(b-c\right)^2+\left(bc+ca+ab-b^2\right)\left(c-a\right)^2+\left(bc+ca+ab-c^2\right)\left(a-b\right)^2\right)$
$=2y\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)\geq 0$.

Thus, Lemma 1 is proven also in the case when $bc+ca+ab-a^2<0$. Hence, the proof of Lemma 1 is complete.

Now, let's come to the proof of our inequality

$\frac{ca+ab-bc}{b^2+c^2}+\frac{ab+bc-ca}{c^2+a^2}+\frac{bc+ca-ab}{a^2+b^2}\geq\frac32$.

Using the cyclic sum notation, this inequality rewrites as

$\sum\frac{ca+ab-bc}{b^2+c^2}\geq\frac32$.

Upon multiplication by 2, this becomes

$2\sum\frac{ca+ab-bc}{b^2+c^2}\geq 3$,

what is clearly equivalent to

$2\sum\frac{ca+ab-bc}{b^2+c^2}-3\geq 0$.

But

$2\sum\frac{ca+ab-bc}{b^2+c^2}-3=\sum\left(2\frac{ca+ab-bc}{b^2+c^2}-1\right)=\sum\left(\left(a-b\right)-\left(c-a\right)\right)\frac{b+c}{b^2+c^2}$
$=\sum\left(a-b\right)\frac{b+c}{b^2+c^2}-\sum\left(c-a\right)\frac{b+c}{b^2+c^2}=\sum\left(a-b\right)\frac{b+c}{b^2+c^2}-\sum\left(a-b\right)\frac{c+a}{c^2+a^2}$
$=\sum\left(a-b\right)\left(\frac{b+c}{b^2+c^2}-\frac{c+a}{c^2+a^2}\right)=\sum\frac{bc+ca+ab-c^2}{\left(b^2+c^2\right)\left(c^2+a^2\right)}\left(a-b\right)^2$.

Hence, we have to prove the inequality

$\sum\frac{bc+ca+ab-c^2}{\left(b^2+c^2\right)\left(c^2+a^2\right)}\left(a-b\right)^2\geq 0$.

Upon multiplication by $\left(b^2+c^2\right)\left(c^2+a^2\right)\left(a^2+b^2\right)$, this becomes

$\sum\left(a^2+b^2\right)\left(bc+ca+ab-c^2\right)\left(a-b\right)^2\geq 0$.

In other words, we have to prove the inequality

$\left(b^2+c^2\right)\left(bc+ca+ab-a^2\right)\left(b-c\right)^2+\left(c^2+a^2\right)\left(bc+ca+ab-b^2\right)\left(c-a\right)^2$
$+\left(a^2+b^2\right)\left(bc+ca+ab-c^2\right)\left(a-b\right)^2\geq 0$.

But this one directly follows from Lemma 1, after WLOG assuming that $a\geq b\geq c$ (as our inequality is symmetric in a, b, c) and setting $x=b^2+c^2$, $y=c^2+a^2$ and $z=a^2+b^2$ (these numbers x, y, z satisfy $x\leq y\leq z$, since $a\geq b\geq c$ yields $a^2\geq b^2\geq c^2$, so that $\left(a^2+b^2+c^2\right)-a^2\leq\left(a^2+b^2+c^2\right)-b^2\leq\left(a^2+b^2+c^2\right)-c^2$, what is equivalent to $b^2+c^2\leq c^2+a^2\leq a^2+b^2$, i. e. $x\leq y\leq z$). This completes our proof.

  Darij

[quote="Vasc"]For $ a,b,c$ positive reals, the inequality

$ \frac {ab \plus{} (p \minus{} 1)bc \plus{} ca}{b^2 \plus{} pbc \plus{} c^2} \plus{} \frac {bc \plus{} (p \minus{} 1)ca \plus{} ab}{c^2 \plus{} pca \plus{} a^2} \plus{} \frac {ca \plus{} (p \minus{} 1)ab \plus{} bc}{a^2 \plus{} pab \plus{} b^2}\ge \frac {3(p \plus{} 1)}{p \plus{} 2}$

is valid for any $ p > \minus{} 2.$[/quote]$ \sum \frac {ab \plus{} (p \minus{} 1)bc \plus{} ca}{b^2 \plus{} pbc \plus{} c^2} \minus{} \frac {3(p \plus{} 1)}{p \plus{} 2}\equiv\frac {F(a,b,c,p)}{(p \plus{} 2)\left(b^2 \plus{} pbc \plus{} c^2\right)\left(c^2 \plus{} pca \plus{} a^2\right)\left(a^2 \plus{} pab \plus{} b^2\right)}.$

$ F(a,b,c,p) \equal{} F(a,a \plus{} s,a \plus{} t,k \minus{} 2)$

$ \equal{} 2k^2a^4\left(s^2 \minus{} st \plus{} t^2\right) \plus{} 2k^2a^3(s \plus{} t)\left(2s^2 \minus{} 3st \plus{} 2t^2\right)$

$ \plus{} 2ka^2\left[k(s \minus{} t)^2\left(s^2 \plus{} 3st \plus{} t^2\right) \plus{} ks^2t^2 \plus{} \left(s^2 \minus{} st \plus{} t^2\right)^2\right]$

$ \plus{} ka(s \plus{} t)\left[2kst(s \minus{} t)^2 \plus{} \left(s^2 \minus{} st \plus{} t^2\right)\left(2s^2 \minus{} 3st \plus{} 2t^2\right)\right]$

$ \plus{} st(s \minus{} t)^2\left[ks^2 \plus{} \left(k^2 \minus{} 3k \plus{} 3\right)st \plus{} kt^2\right]\geq 0,$

which is clearly true for $ a \equal{} \min \{a,b,c\}.$

By the way, this inequality holds for [b]real[/b] $ a,b,c,p$ with $ bc(b \minus{} c)^2 \plus{} ca(c \minus{} a)^2 \plus{} ab(a \minus{} b)^2 > 0\wedge \minus{} 2 < p\leq 2.$

$ \sum \frac{ab\plus{}ac\minus{}bc}{b^2\plus{}c^2} \ge \frac{3}{2}$($ a,b,c  \ge 0$)
I have got  a long proof :)
$ \leftrightarrow (\sum ab)(\sum \frac{ab\plus{}ac\minus{}bc}{b^2\plus{}c^2}  ) \ge \frac{3}{2}(\sum ab)$
$ \leftrightarrow  2abc( \sum \frac{a}{b^2\plus{}c^2} ) \ge  \frac{3}{2}(\sum ab)\plus{}\sum \frac{b^2c^2}{b^2\plus{}c^2}\minus{}\sum a^2$
By am-gm and cauchy-schwarz, we have :
$ LHS \ge \frac{2abc(a\plus{}b\plus{}c)^2}{ \sum ab(a\plus{}b) }$
$ RHS \le 2\sum ab \minus{}\sum a^2$
Case1:$ 2 \sum ab \le \sum a^2$
=>$ LHS \ge 0 \ge RHS$
Case2:
Assume :$ a\plus{}b\plus{}c \equal{}1$
By schur , we have :$ r \ge \frac{ (4q\minus{}1)(1\minus{}q)}{6}$
And the last ineq is equivalent to :
$ (4q\minus{}1)(1\minus{}3q) \ge 0$


Find all functions f : R --> R such that
f(x - f(y)) = f(f(y)) + xf(y) + f(x) - 1
for all x, y in R

can I solve this problem by the following method.?

Sol. put x = f(y) then : f(0) = f(x) + x2 +f(x) - 1
then : f(x) = [ f(0) + 1 - x 2 ]/2       ... (1)
put x = 0 in (1) : 2f(0) = f(0) + 1 --> f(0) = 1
so. f(x) = (2 - x 2 )/2 = 1 - (1/2)x 2

What you did is not enough because you found an expression for f(x) only for those x for which there exists an y with x=f(y) but f must not be surjective.

Misha

Thanks for your answer. :D  I " m new for olympiad problem.

you mean that if I can show that f is surjective. So I can put x = f(y). ? :maybe:

You can put x=f(y) always but all the results which you get from this substitution are only valid for those x and not for all x in R. It is only then so if f(y)=R which is the same as to say f is surjective.

Misha


calculus
integration
trigonometry
limit
calculus computations

1)

Let $ F'(x) \equal{} f(x)$. Then

$ \left(\int^{h(x)}_{C}f(t)\,\,dt\right)^{'} \equal{} \left(F(h(x)) \minus{} F(C)\right)^{'} \equal{} f(h(x))h'(x)$

Using that, we can say:

$ \left(\int^{x}_{A}\left(\int^{h(t)}_{B}g(u)\,\,du\right)dt\right)^{'} \equal{} \int^{h(x)}_{B}g(u)\,\,du$

and using the first one again we get:

$ \frac {d^2}{dx^2}\int^{x}_{A}\left(\int^{h(t)}_{B}g(u)\,\,du\right)dt \equal{} \left(\int^{h(x)}_{B}g(u)\,\,du\right)^{'} \equal{} g(h(x))h'(x)$

Therefore:

$ \frac {d^2}{dx^2}\int^{x}_{0}\int^{\sin t}_{1}\sqrt {1 \plus{} u^4}\,\,du\,\,dt \equal{} \sqrt {1 \plus{} \sin^4 x}\,\,\cos x$

2)

$ \lim_{n\to\infty} \frac {1}{n}\sum_{i \equal{} 1}^{n}\sqrt {\frac {i}{n}} \equal{} \int^{1}_{0}\sqrt {x}\,\,dx \equal{} \frac {2}{3}$

In the final answer for part 1 shouldn't those t's be x's?

Sorry.. Copied it automatically.. Got it right now!

[quote="milin"]$ \lim_{n\to\infty} \frac {1}{n}\sum_{i \equal{} 1}^{n}\sqrt {\frac {i}{n}} \equal{} \int^{1}_{0}\sqrt {x}\,\,dx$[/quote]
Could someone explain this a little more?


hi,
Let $ S(n,p) \equal{} \sum_{i\equal{}1}^{n}{(n\plus{}1\minus{}2i)^{2p}}, p \in \mathbb{N}$. Prove that for any real numbers $ a_{i}, i\equal{}1...n$ we have
\[ min_{1 \leq i < j \leq n}{(a_{i} \minus{} a_{j})^{2p}} \leq \frac{4^{p}}{S(n,p)}\sum_{i\equal{}1}^{n}{a_{i}^{2p}}.\]

hi,
does somebody have a more elementary prof ???


induction
combinatorics solved
combinatorics
Coloring

The set of natural numbers is colored by finite colors. Prove that there exist an arbitrary long monochrome arithmetic progession (that means all its terms have the same color).

reference: http://mathworld.wolfram.com/vanderWaerdensTheorem.html

this is not quite obvious theorem. the proof is rather large, and complex.

To prove for progressions of length k, c colours, assume by induction for length k-1 (any number of colours); consider blocks of sufficient length to be sure to contain a (k-1)-progression (c colours) plus what would be the next in the progression (but might or might not be the right colour).  Consider (k-1)-progressions now on a very large number of colours - namely the number of possible colourings of those blocks - so you have (k-1) successive identical blocks in a progression, and consider the element in the kth (maybe not identical) corresponding to the kth element of the progression in the other blocks.  There are two colours which this element having would imply a progression of length k; supposing it has neither, you're now considering new much larger blocks; repeat at most c times.
For example (k = 3, c = 2), blocks of length 5 suffice at the first step (as blocks of length 3 force an AP of length 2); such blocks either contain something like

xxx   or x x x (an AP of length 3) or else

xxy   or x x y

In the former case we're done; in the latter case, there must be a pair of
blocks with the same colouring not too distant from each other, say

x x y .......... x x y .......... ? ? *

Now look at *; if it is coloured either x or y we have our AP.  If we were dealing with 3 colours, we'd conclude it to be coloured z if we don't have an AP, and repeat with larger blocks sufficiently large to force this
configuration.


calculus
integration
calculus computations

Integrate:

$ \int_0^{\infty} \frac {1}{y} e^{ \minus{} (y \plus{} x/y)} \,dy$

hello,Maple 12 gives $ \int_{0}^\infty\frac{1}{y}e^{\minus{}(y\plus{}\frac{x}{y})}\,dy\equal{}2\text{BesselK}(0,2\sqrt{x})$.
Sonnhard.


[quote="sharkman"]Two questions:

(1) If f(x) = 2x^3 - 3x^2 + 4x - 1 and g(x) = 4, find
f(g(x)) and g(f(x))

(2) If f(x) = 3x^2 - 7 and g(x) = 2x + a, find a so that the graph of f(g(x)) crosses the y-axis at 68.[/quote]

Question 1

$ f(g(x)) \equal{} f(4) \equal{} 95$
$ g(f(x)) \equal{} 4$

Question 2

$ f(g(x)) \equal{} 3(2x\plus{}a)^2\minus{}7 \equal{} 12x^2\plus{}12ax\plus{}3a^2\minus{}7$

The graph $ y\equal{}f(x)$, crosses the $ y$ axis when $ x\equal{}0$

Hence

$ 68 \equal{}  12 \cdot0^2\plus{}12\cdot a \cdot 0\plus{}3a^2\minus{}7$
Therefore $ a^2\equal{}25$
Therefore $ a \equal{} \pm 5$


Obviously, $ \frac {1}{3}\cdot\frac {1}{2} = \frac {1}{6}$. So there are $ \boxed{\frac {1}{2}}$ $ \frac {1}{3}$'s in $ \frac{1}{6}$.

Edit: Darn, beaten.....  :(  Shouldn't have put so much $ \text{\LaTeX}$.  :)


Find the locus of points inside a trihedral angle such that the sum of their distances from the faces of the trihedral angle has a ﬁxed positive value $a$.


Ok...so I really like factoring, but there's this one problem that I just can't get:

x2 + 2xy + 2x + y2  + 2y - 8

I was told that you neet to flip-flop two terms, so I switched the '2x' and the 'y2' to get:

x2 + 2xy + y2 + 2x + 2y - 8

which factors into:

(x+y)(x+y) + 2x + 2y - 8

from there I tried pulling a 2 out of the 2x + 2y and getting:

(x+y)(x+y) + 2(x+y) -8

then pulling an (x+y) out of the first 2 terms to get

(x+y+2)(x+y) - 8

but i can't go anywhere from there. Can someone point out where I went wrong, and what I should be doing instead? Thnx so much in advance!!!

You already know how to do this now but I'll post here to share with other users.

 :) Let's call $x+y = z$. Then:

$z^2+2z-8 = (z+4)(z-2)$

Replace $z$ with $x+y$.

$(x+y+4)(x+y-2)$ as final answer. :cool:

Just remember what $(a+b+c)^2$ is ;) Choose proper $a,b,c$ to obtain $x^2 + 2xy + 2x + y^2 + 2y +1$ :)


Prove that $3x^5+2x^4-x^2+4x+5$ can't divisible in  $E [x]$.

[quote="Peter Scholze"]sorry, but I don't understand what you mean...do you mean that polynomial can't be factored into two of smaller degree? but what is E[x]?

Peter[/quote]

At first ,I want to told you that caz I can't understand this question clearly so that I put it here .
The second,I think E[x ] mean rational number set .

if change $E [x]$ to be $%Error. "capital" is a bad command.
Q[x]$ what will happen ?


Is this close enough? I see that the U doesn't quite match.

$ \mathfrak{U}\equiv\mathfrak{B}$

[quote="JBL"]Do you think it's more or less bizarre a character than $ \mathfrak{S}$? :) [/quote]
I never would have guessed that's an "S" so I think that wins.  But "Y" and "Z" are pretty weird as well: $ \mathfrak{YZ}$

\[ \text{ABCDEFGHIJKLMNOPQRSTUVWXYZ} \\
\mathfrak{ABCDEFGHIJKLMNMOPQRSTUVWXYZ} \\
\text{abcdefghijklmnopqrstuvwxyz} \\
\mathfrak{abcdefghijklmnopqrstuvwxyz}\]

[quote="Kouichi Nakagawa"]I think
[code]$\mathfrak{A}\equiv\mathfrak{B}$[/code]
$ \mathfrak{A}\equiv\mathfrak{B}$[/quote]

Are there any other types of styles for letters? How did you find out this style?

TEST:
$ \mathfrak{test}  \mathfrak{hi}$
Also, how do u put a space in between two words in the mathfrak style?

$ \mathfrak{test}\, \mathfrak{hi}$

$ \mathfrak{test}\; \mathfrak{hi}$

$ \mathfrak{test}\ \mathfrak{hi}$

$ \mathfrak{test}\quad \mathfrak{hi}$

$ \mathfrak{test}\qquad \mathfrak{hi}$


Prove without use bertrand's postulate that between n and n! there is a prime.

For $n > 2$, the number $n!-1$ has no factors from 2 to n.  Thus every (in particular, some) prime factor of $n!-1$ must be bigger than $n$ and less than $n!$.

Ye that's all!Can you find without postulate more than one primes between n and n! and how?

Sure.  For $n \ge 4$, notice that $\frac{n!}{2}-1$ and $n!-1$ are relatively prime and odd.  Take one prime factor from each.

Yes you are right.So we have 2 primes between n and n! And the prime factor of n!+2 is between n and n!Ibelieve that the same is for (n!/p)-1 where p is the prime factor of n!-1  and n<p<n!.First of all if (n!/p)-1 is prime then n<(n!/p)-1<n!.If the number is compositive then there is a prime factor q.It's obviously that q<n! If q<n then q devide n! and (n!/p)-1
so q/p and so q=1 or p.If q=p then we have find nothing because is the same.But if q=1 then q>n so we find a new prime and this is the prime factor of (n!/p)-1. :)


A conference has $ 47$ people attending. One woman knows $ 16$ of the men who are attending, another knows $ 17$, and so on up to the $ m$-th woman who knows all $ n$ of the men who are attending. Find $ m$ and $ n$.

[hide]
The first woman knows that 1+15th, the second knows the 2+15th, and the mth knows the m+15th. Therefore, n = m+15. Also, m+n=47, so m+m+15=47, and 2m=32, so m = 16 and n = 31.[/hide]


Determine all pairs $(a, b)$ of integers for which $a^{2}+b^{2}+3$ is divisible by $ab$.

[quote="Peter"]Determine all pairs $ (a, b)$ of integers for which $ a^{2} \plus{} b^{2} \plus{} 3$ is divisible by $ ab$.[/quote]
if (a,b) is solution, then $ (\pm a, \pm b) and ( \pm b, \pm a)$ are solutions.
$ a,b\neq 0$, then we consider the solutions (a,b) in the positive integers, such that $ a \ge b$
if (x,y) is solution, with x>y>1, then $ txy\equal{}x^2\plus{}y^2\plus{}3>x^2$ ,then ty>x.
$ x(ty\minus{}x)\equal{}y^2\plus{}3 \le (x\minus{}1)^2\plus{}3\equal{}x^2\minus{}2x\plus{}4<x^2$ ,then ty-x<x.
(ty-x,y) or (y,ty-x) is solution and ty-x + y<x+y.
each solution with x>y>1, through this process to creat a end solution, (r,s); $ r \ge s$, such that r=s,or s=1, because the sume of components decreases
if r=s, $ r^2$ divides $ 2r^2\plus{}3$, then $ r^2$ divides 3, then r=1
if s=1, r divides $ r^2\plus{}4$ ,then r divides 4, then r=1,2,4.
the possible end solution are: (1,1), (2,1), (4,1). the "t" respective are 5,4,5
hence the solutions are $ (x_n,y_n) and (z_n,w_n)$
$ x_0\equal{}1, y_0\equal{}1, x_{n\plus{}1}\equal{}5x_n \minus{}y_n, y_{n\plus{}1}\equal{}x_n$  (t=5)
$ z_0\equal{}2, w_n\equal{}1, z_{n\plus{}1}\equal{}4x_n \minus{}y_n, w_{n\plus{}1}\equal{}z_n$  (t=4)


calculus
integration
derivative
calculus computations

can someone show me how to do this integral:

[tex] \int \frac{(1-x)}{x^2} e^{x-1} dx[/tex]

[quote="liyi"][quote="vidyamanohar"][quote="Kent Merryfield"]Hint: what is the derivative of $\frac{e^x}x$ ?[/quote]
$ - \frac{{e^{x - 1} }}{{x}}$[/quote]
No, it is $\frac{e^x(x-1)}{x^2}$.

But this solution is not natural enough. I prefer this one:

$\int \frac{(1-x)e^{x-1}}{x^2}dx = \int (x-1)e^{x-1}d\left(\frac 1x\right) = \frac{(x-1)e^{x-1}}{x} - \int \frac{xe^{x-1}}{x}dx = \frac{(x-1)e^{x-1}}{x} - e^{x-1} + C = -\frac{e^{x-1}}{x} + C$[/quote]

I did not follow, how your 'answer' is different from mine. Also; is there any contradiction in your own argument?

actually it's the same answer, [tex]e^{-1}[/tex] is a constant

[quote="vidyamanohar"][quote="Kent Merryfield"]Hint: what is the derivative of $\frac{e^x}x$ ?[/quote]
$ - \frac{{e^{x - 1} }}{{x}}$[/quote]
Sorry. I misunderstand you. Your answer is correct...

I thought you answered Kent Merryfield's question...  :blush:


Consider a convex pentagon $ABCDE$ such that the triangles $ABD$, $ABC$, $ACD$ and $ADE$ have the same area. Find the area of $BCD$.

[hide]
Τhe area of $\triangle ADE$ is not necessary

$(ABC) = (ABD) \Rightarrow  d(C,AB) = d(D,AB) \Rightarrow  AB \parallel CD$

$(ABD) = (ACD) \Rightarrow  d(B,AD) = d(C,AD) \Rightarrow  AD \parallel BC$

So $ABCD$ is a parallelogram.

This is enough to conclude that $(BCD) = (ACD)$.

So the area of $\triangle BCD$ is equal to the other areas.[/hide]


To all forum members;
Has anyone actually purchased the Art of Problem Solving books in Canada in this forum? If yes, then reply. As to the "Elders", I've been wondering whether or not you guys have used the Art of Problem Solving books. Please reply yes or no.

YES!!

pick up the books, all it is 68 fewer pizza slices :lol:

You already created a thread, there is no need to create another one.  Honestly, in the big scheme of things, 68 dollars is nothing.  Heck, you can probably get away with just getting the first volume for now.


prove that with x,y,z  are three positive  integers the equation x^4+p*y^4=z^4 has no solution and p is a prime,p=3(mod4)

Do you mean that :The equation $x^4+py^4=z^4$ has no solution $(x,y,z) \in\mathbb N^3$ with the given prime number $p,p\equiv 3(mod4)$ or something else?

yes, i think that is nthd'mean
how to use the hypothesis $p\equiv3(mod4)$, and if p=1,2,4 i've done

thank you,nguyenquockhanh and nalpaction
I have two results:
a)with p=8 this equation also has no solution in Z+
b)with p=5 this equation has a solution:x=1,y=2,z=3
With p is a prime and p=3(mod4),I solved and I  have two question:
  1.Find a prime p>5,p=1(mod4) and this equation has solutions in Z+
  2.Is there a prime p=1(mod4) and this equation has no solution in Z+

Yes , and I think this problem is not difficult.


arithmetic sequence
algebra solved
algebra

Let A={1,2,3,...n^2} If B is a part of A with less than n elements,prove that there exists  in B or in A-B an arithmetic sequence with n terms.

it is clear that there cannot be such an arithmetic sequence in B, because B has less than n elements.So we may consider the worst case where B has n-1 elements.
consider A1={1,2,...,n-1,n} A2={n+1,...,2n} ,...,An={(n-1)n +1,...,n^2}
then by pigeon hole principle A-B contains one of these sets completely which has n terms of an arithmetic sequence.


[quote="goldendomer"]What is the sum of the factors of 1304532?[/quote]

This should be in the high school forum.

[quote="Ignite168"][quote="anirudh"][quote="goldendomer"]What is the sum of the factors of 1304532?[/quote]

[hide]First, prime factorize the number. $1304532=2^{2}\times 3^{3}\times 47\times 257$. The sum of the factors is $(2^{0}+2^{1}+2^{2})(3^{0}+3^{1}+3^{2}+3^{3})(47^{0}+47^{1})(257^{0}+257^{1})$

That number is equal to: $7\times 40\times 48\times 258=3467520$. The sum of the factors of $1304532$ is $3467520$.
[/hide][/quote]

Can you explain how you prime factorized it?[/quote]

[hide]$\frac{1304532}{2^{2}}=\frac{326133}{3^{3}}=\frac{12079}{47}=\frac{257}{257}=1$. So, the prime factors are $2^{2}\times 3^{3}\times 47\times 257$.[/hide]

[quote="Ignite168"][quote="anirudh"][quote="goldendomer"]What is the sum of the factors of 1304532?[/quote]

[hide]First, prime factorize the number. $1304532=2^{2}\times 3^{3}\times 47\times 257$. The sum of the factors is $(2^{0}+2^{1}+2^{2})(3^{0}+3^{1}+3^{2}+3^{3})(47^{0}+47^{1})(257^{0}+257^{1})$

That number is equal to: $7\times 40\times 48\times 258=3467520$. The sum of the factors of $1304532$ is $3467520$.
[/hide][/quote]

Can you explain how you prime factorized it?[/quote]

Isn't that the easiest part? Use a TI-89.

The problem didn't say you could use a calculator.

[quote="13375P34K43V312"]Because a TI-89 already has the prime factorization thing built in.[/quote]

How do you use it? I can't see it...

Hit F-2, then choose "factor" from the menu that comes up.  Just type in the number you want the prime factorization for, and close the parenthesis.  The prime factorization will come up.

[quote="goldendomer"]Man I wish I had a TI-89. :D Is it possible to do a prime factoriazation for a TI-84 Plus calculator?[/quote]
Si.
You just have to program it.


inequalities
limit
function
search
algebra
system of equations
inequalities unsolved

Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}3$ prove that
\[ a^3\plus{}b^3\plus{}c^3\plus{}7abc\ge 10\]

 I have an ugly solution and i'm looking for a nice one .

[quote="M.A"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc\ge 10
\]
I have an ugly solution and i'm looking for a nice one .[/quote]

I have a solution with Lagrange Multipliers, do you want it or is it same as your solution?

[quote="great math"]Yes, I don't know how to use Multiple Largrange in Inequality so you might post it in the forum, it's my pleasure.[/quote]

ok

Let $ f(a,b,c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc$

Since $ ab \plus{} bc \plus{} ca \equal{} 3 \Rightarrow c \equal{} \frac {3 \minus{} ab}{a \plus{} b}$, we have:

$ f(a,b,c) \equal{} f(a,b) \equal{} a^3 \plus{} b^3 \plus{} {(\frac {3 \minus{} ab}{a \plus{} b})}^3 \plus{} 7ab(\frac {3 \minus{} ab}{a \plus{} b})$

Now its easy to find that $ \lim_{a \rightarrow \plus{} \infty}{f(a,b)} \equal{} \plus{} \infty$,
and $ \lim_{a \rightarrow 0}{f(a,b)} \equal{} b^3 \plus{} \frac {27}{b^3}$ where we have
$ b^3 \plus{} \frac {27}{b^3} \geq 3\sqrt {3} \plus{} \frac {9}{\sqrt {3}} > 10$

Set now $ g(a,b,c) \equal{} ab \plus{} bc \plus{} ca \minus{} 3$ and suppose that the function f has an extremum $ (a_0, b_0, c_0)$ where $ a_0, b_0, c_0 \in (0, \plus{} \infty)$ with $ g(a_0, b_0, c_0) \equal{} 0$, then by the Lagrange multipliers theorem we have that there exists a real number $ k$ such that $ \nabla F(a_0, b_0, c_0) \equal{} 0$ where $ F(a,b,c) \equal{} f(a,b,c) \plus{} kg(a,b,c)$

$ \nabla F(a_0, b_0, c_0) \equal{} 0 \Leftrightarrow < 3a_0^2 \plus{} 7b_0c_0 \plus{} k(b_0 \plus{} c_0), 3b_0^2 \plus{} 7c_0a_0 \plus{} k(c_0 \plus{} a_0), 3c_0^2 \plus{} 7b_0a_0 \plus{} k(b_0 \plus{} a_0) > \equal{} < 0, 0, 0 >$

and we have the system of equations:

$ 3a_0^2 \plus{} 7b_0c_0 \plus{} k(b_0 \plus{} c_0) \equal{} 0 (1)$

$ 3b_0^2 \plus{} 7c_0a_0 \plus{} k(c_0 \plus{} a_0) \equal{} 0 (2)$

$ 3c_0^2 \plus{} 7b_0a_0 \plus{} k(b_0 \plus{} a_0) \equal{} 0 (3)$

$ a_0b_0 \plus{} b_0c_0 \plus{} c_0a_0 \equal{} 3 (4)$

The system has the solution $ a_0 \equal{} b_0 \equal{} c_0 \equal{} 1$, suppose that it has another solution such that $ a_0$ is not equal with $ b_0$ and $ c_0$, then by subtracting (2) from (1) we get $ 3(a_0 \plus{} b_0)(a_0 \minus{} b_0) \plus{} 7c_0(b_0 \minus{} a_0) \plus{} k(b_0 \minus{} a_0) \equal{} 0 \Leftrightarrow 3a_0 \plus{} 3b_0 \plus{} 7c_0 \equal{} \minus{} k (5)$ and by subtracting (3) from (1) we get  $ 3a_0 \plus{} 3c_0 \plus{} 7b_0 \equal{} \minus{} k (6)$, now $ (5),(6) \Rightarrow 4b_0 \equal{} 4c_0 \Leftrightarrow b_0 \equal{} c_0$

We have proved that the system above has a solution only if two of the three numbers are equal, now its easy to prove that all three numbers must be equal (suppose that $ a \equal{} b$ and $ c > b$, subtract (3) from (2) and for $ c \equal{} \frac {3 \minus{} a^2}{2a}$ get $ k \equal{} \frac {9 \minus{} 3a^2}{2a} \minus{} 4a$, then put on (1): $ b \equal{} a, c \equal{} \frac {3 \minus{} a^2}{2a}$, and $ k \equal{} \frac {9 \minus{} 3a^2}{2a} \minus{} 4a$ and solve the new equation), so the solution $ a_0 \equal{} b_0 \equal{} c_0 \equal{} 1$ is unique, and the function $ f$ has unique extremum in $ (0, \plus{} \infty)$ the value $ f(1,1,1) \equal{} 10$.

Now because we have proved that $ \lim_{a \rightarrow \plus{} \infty}{f(a,b)} \equal{} \plus{} \infty$ and $ \lim_{a \rightarrow 0}{f(a,b)} \equal{} b^3 \plus{} \frac {27}{b^3} > 10$, we have that $ f(a,b,c) \geq 10$ for all $ a,b,c \in (0, \plus{} \infty)$ with equality only for $ (a,b,c) \equal{} (1,1,1)$ and we have finished!


does anyone know any cool experiments with $Mg_{(s)}$?

and ... how do you make hydrochloric acid exactly?? i've never really thought about that!

thanks

Hello,
    The familar magnesium ribbon burning with a brilliant white flame in air is quite nice to watch. And some qualitative tests like the pink ash obtained on performing the 'ash test' for magnesium salts is quite good. And HCl is generally prepared in the laboratory by adding drops of sulphuric acid to sodium chloride.It works on 'the stronger acid displacing a weaker acid from one of its salts' principle.
Sathej

ohhhh ok thanks for the HCl one...and oh yea i already did the Mg burning but not the pink ash one....

and anyone know how to make colored flames?

sodium- red
potassium-orange...
forgot the rest...

how do you make it though...is it a powder..or liquid that you set on fire


Let $ ABC$ be a triangle.Incircle $ (I)(I$ is center) touch $ BC;CA;AB$ at $ D;E;F$,respectively.$ AD$ cuts $ (I)$ at $ L$($ L$ \neq $ D$)
$ DH \perp EF(H \in EF)$
$ DI$ cuts $ (I)$ at $ K(K \neq D)$
$ EF$ cuts $ BC$ at $ Q$
Prove that $ QL$ is tagent of circumcircle of triangle $ HKL$

This problem does not seem to be true. $ QL$ is tangent to the incircle but not the circumcircle of $ HKL$. This is a well-known property.

$ EF$ is the polar of $ A$
$ EFQ$ is a line so $ Q$ is on the polar of $ A$
By duality, $ A$ is on the polar of $ Q$
$ D$ is on the polar of $ Q$
$ AD$ is the polar of $ Q$
$ ADL$ is a line so $ L$ is on the polar of $ Q$
Since $ L$ is on the circle, $ QL$ is a tangent.

EDIT: the problem should be Let $ DI$ intersect $ EF$ at $ K$.

Then we must show that the circumcircle of $ HKL$ is tangent to the incircle...

EDIT again:

Proof:
Lemma 1: $ QL$ is tangent to $ (I)$. Proof: given above.

So $ QD^2\equal{}QL^2$ (1)

Since $ \angle QHD\equal{}\angle KDQ\equal{}\frac{\pi}{2}$, $ QD^2\equal{}QH\cdot QK$ (2).

By (1) and (2), $ QL^2\equal{}QH\cdot QK$.

This suffices to prove that $ QL$ is tangent to the circumcircle of $ \triangle HKL$. (If $ QL$ intersects $ C(HKL)$ again at $ L'$, then $ QL\cdot QL'\equal{}QH\cdot QK\equal{}QL^2\implies QL\equal{}QL'\implies L\equal{}L'$).


function
limit
calculus
real analysis
real analysis unsolved

It is known that the following function
\[ f\left( x \right) \equal{} \left\{\begin{gathered} 0 \quad {\rm if} \quad x \equal{} 0, \hfill \\
0\quad {\rm if}\quad x \notin \mathbb{Q}, \hfill \\
\frac {1}{q} \quad {\rm if}\quad x \equal{} \frac {p} {q},p \in \mathbb{Z},q \in \mathbb{N},\left( {p,q} \right) \equal{} 1. \hfill \\
\end{gathered} \right.
\]
is continuous at irrationals and at $ x \equal{} 0$. Now I want you to show that $ f$ is upper continuous at every point.

NB upper continuous at $ x_0$ means that
\[ \mathop {\lim \inf }\limits_{x \to x_0 } f\left( x \right) \equal{} \mathop {\underline {\lim } }\limits_{x \to \infty } f\left( x \right) \geqslant f\left( {x_0 } \right).
\]
upper continuous and lower continuous play an important role in calculus of variation, I mean direct methods in nonlinear analysis.

[quote="mlok"]For any $ t\in\mathbb R$ the set $ \{x\colon f(x)\ge t\}$ is closed .[/quote]

Correct but any other solutions?

[quote="mlok"]For any $ t\in\mathbb R$ the set $ \{x\colon f(x)\ge t\}$ is closed .[/quote]

and how do we prove this?


Βρείτε όλα τα πολυώνυμα που ικανοποιούν τη σχέση $ F(F(x))\equal{}(F(x))^{n}$ με $ n\in N$ και $ deg(F(x))>0$ :)

Λυση σε hide:
[hide]Αν θεσουμε στη δοθεισα σχεση $ F(x) \equal{} u$, τοτε λαμβανουμε $ F(u) \equal{} u^n$ για καθε πραγματικο αριθμο $ u$ που ανηκει στο συνολο τιμων της πολυωνυμικης συναρτησης $ F$. Επηδη ο βαθμος του πολυωνυμου $ F$ ειναι μεγαλυτερος του 0, το πολυωνυμο $ F$ δεν ειναι σταθερο και το συνολο τιμων του περιεχει απειρους αριθμους, οποτε υπαρχουν απειροι αριθμοι $ u$ για τους οποιους ισχυει $ F(u) \equal{} u^n \Leftrightarrow F(u) \minus{} u^n \equal{} 0$. Ετσι αν ο βαθμος του πολυωνυμου $ G(x) \equal{} F(x) \minus{} x^n$ ειναι $ k$, μπορουμε να βρουμε $ k \plus{} 1$ αριθμους που να μηδενιζουν το πολυωνυμο $ G$, οποτε απο γνωστο θεωρημα το πολυωνυμο $ G$ ταυτιζεται με το μηδενικο και εχουμε $ F(x) \equal{} x^n$ για καθε πραγματικο αριθμο $ x$[/hide]


A vessel initially contains “V” L of water with “x”kg salt dissolved in it. Pure water is added to it from the top at the rate of “m” L/s and the solution is allowed to flow out from an outlet at the bottom at the rate of “n”L/s such that m>n and salt concentration is assumed to be uniformly distributed throughout the mixture all the time.

Therefore, find amount of salt (in kg) left in the vessel (in dissolved state) after time “t”.


AMC
USA(J)MO
USAMO
function
floor function
conics
hyperbola

Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.)

I believe it should be $\max k=\left\lfloor\frac n2\right\rfloor$.

If $A_iA_{i+1}A_{i+2}A_{i+3}$ we simply say that $P(i)$ holds. First of all, notice that if $P(i)$ holds, then $P(i+1)$ cannot hold. This is because $A_i,A_{i+4}$ must lie on the same branch of a hyperbola with foci $A_{i+1},A_{i+3}$ and on the same side of $A_{i+1}A_{i+3}$, which means that one of the points $A_i,A_{i+4}$ must lie inside the triangle formed by the other one and the two foci. This means that if $P(i)$ holds, neither $P(i+1)$ nor $P(i-1)$ can hold, so we get the upper bound $k\le\left\lfloor\frac n2\right\rfloor$.

I'll just describe a construction of such a polygon with $k=\left\lfloor\frac n2\right\rfloor$ for $n=2m$ (when $n$ is odd we can modify this one, but it's a bit harder to explain). Consider a regular $m$-gon, and if, say, $A,B,C$ are consecutive vertices in this order of this $m$-gon, cut off $B$ along the segment $DE,\ D\in AB,E\in BC$ s.t. $BD=BE=x$, where $x<\frac{AB}2$ is chosen so that if $U\in AB,V\in BC$ with $AU=CV=x$, then $UDEV$ has an inscribed circle (it's clear that such an $x$ exists, by a continuity argument).

Answer is indeed $k=[\frac{n}{2}]$. Suppose that $A_1A_2A_3A_4$ and $A_2A_3A_4A_5$ have inscribed circle $\implies$ $A_1A_4+A_2A_3=A_1A_2+A_3A_4$ and $A_2A_5+A_3A_4=A_2A_3+A_4A_5$ $\implies$ $A_2A_5+A_1A_4=A_1A_2+A_4A_5$ which is not true because $A_1A_2A_4A_5$ is convex quadrilateral. Contradiction $\implies$ $k \le [\frac{n}{2}]$. As for the construction:
1. Even $n$: Construct isosceles tangential trapezium $A_1A_2A_3A_4$ with $\angle A_1A_2A_3=\angle A_2A_3A_4=\frac{n-2}{n} \pi$. Now just simply add $A_5$ such that $A_1A_2A_3A_4=A_2A_3A_4A_5$. Do this until we complete $n$-gon. Obviously $A_{i}A_{i+1}A_{i+2}A_{i+3}$ and  $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ are tangential (while $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not) $\implies$ $k=\frac{n}{2}$.
2. Odd $n$. By same process we construct $n+1$-gon then "delete" $A_{n+1}$ and place $A_n$ such that $A_{n-2}A_{n-1}A_nA_1$ is tangential $\implies$ $k=\frac{n-1}{2}=[\frac{n}{2}]$

Solution from [i]Twitch Solves ISL[/i].

The main claim is the following:

[b][color=red]Claim:[/color][/b]  	We can't have both $A_1 A_2 A_3 A_4$ 	and $A_2 A_3 A_4 A_5$ be circumscribed.
[i]Proof.[/i]  	If not, then we have the following diagram, 	where $a = A_1A_2$, $b = A-2 A_3$, $c = A_3 A_4$, $d = A_4 A_5$. 	 	[asy] 		size(9cm); 		pair A1 = dir(10); 		pair A2 = dir(40); 		pair A3 = dir(90); 		pair A4 = dir(130); 		pair A5 = dir(200); 		draw(A1--A2--A3--A4--A5, blue); 		label("$a$", midpoint(A1--A2), dir(A1+A2), blue); 		label("$b$", midpoint(A2--A3), dir(A2+A3), blue); 		label("$c$", midpoint(A3--A4), dir(A3+A4), blue); 		label("$d$", midpoint(A4--A5), dir(A4+A5), blue); 		draw(A1--A4, red); 		label(rotate(-20)*"$c+a-b$", 0.6*A4+0.4*A1, dir(A1+A4), red); 		draw(A2--A5, orange); 		label(rotate(30)*"$b+d-c$", 0.5*A5+0.5*A2, dir(A2+A5), orange); 		dot("$A_1$", A1, dir(A1), blue); 		dot("$A_2$", A2, dir(A2), blue); 		dot("$A_3$", A3, dir(A3), blue); 		dot("$A_4$", A4, dir(A4), blue); 		dot("$A_5$", A5, dir(A5), blue); 	[/asy] 	 	Then $A_1 A_4 = c+a-b$ and $A_5A_2 = b+d-c$. 	But now 	\[ A_1 A_4 + A_2 A_5 = (c+a-b) + (b+d-c) = a+d = A_1 A_2 + A_4 A_5 \] 	but in the picture we have an obvious violation 	of the triangle inequality. $\blacksquare$
This immediately gives an upper bound of $\left\lfloor n/2 \right\rfloor$.

For the construction, one can construct a suitable cyclic $n$-gon by using a continuity argument (details omitted).

[hide=sol]
Suppose that two consecutive quadrilaterals both have an inscribed circle. WLOG, assume the two quadrilaterals are $A_1A_2A_3A_4$ and $A_2A_3A_4A_5.$ Thus, we have
\begin{align*}
A_1A_2+A_3A_4&=A_2A_3+A_4A_1\text{ and }\\
A_2A_3+A_4A_5&=A_3A_4+A_5A_2.\\
\end{align*}
Adding and simplifying, we have $$A_1A_4+A_2A_5 = A_1A_2+A_4A_5.$$ However, from the triangle inequality and using the fact that the polygon is convex, we have $$A_1A_4+A_2A_5>A_1A_2+A_4A_5,$$ a contradiction. In other words, two consecutive quadrialterals can't both have an inscribed circle. We obtain an upper bound of $\left\lfloor\frac{n}{2}\right\rfloor.$ Now, we will provide a construction. For $n = 2m,$ we draw an equiangular $2m$-gon with sides alternating of length $a$ and $a\left(1-\cos\left(\frac{360}{a}\right)\right).$ In other words, every other quadrilateral has an inscribed circle. For $n = 2m-1,$ we draw the previous equiangular $2m$-gon and delete $A-{2m}.$ In other words, the two edges $A_{2m-1}A_{2m}$ and $A_{2m}A_1$ are replaced with $A_{2m-1}A_1.$ This preserves the $m-1$ of the quadrilaterals with inscribed circles, which completes the construction.[/hide]


Let $ a,b,c,d \in N*$ be 4 distinct numbers whose product is a perfect square.Prove that $ a^4\plus{}b^4\plus{}c^4\plus{}d^4$ is the sum of $ 5$ non-zero perfect squares.

Way to hard for middle school math. Moderators, please move.

RoFlLoLcOpT what you said was a ridiculous comment. You have posted in the High School Pre-olympiad section and I would not think that is for middle school. So your mail was spam!

probably not a great solution since I haven't actually worked much stuff out.

[hide]Assume without loss of generality that $ a < b < c < d$.

by Lagrange's four-square theorem , $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 1$ can be written as the sum of four perfect sqaures (lets call these four perfect squares $ S_1 S_2, S_3$and$ S_4$). This means that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \equal{} S_1 \plus{} S_2 \plus{} S_3 \plus{} S_4 \plus{} 1$. Let $ S_5 \equal{} 1^2$. Thus $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \equal{} S_1 \plus{} S_2 \plus{} S_3 \plus{} S_4 \plus{} S_5$.

[/hide]

edit: oops, on reading again, I think I have a hole in this proof whereby I don't prove that that all the squares are non-zero. I shall i try and fix it (it seems the question isn't as trivial as I had first anticipated)...

The problem is not so difficult as you see.The solution is only 1 line!This problem can be solved by elementary method so that a student who is in grade 7 and grade 8 can understand.

let $ abcd \equal{} n^2$ and
\[ S \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4
\]
first assume that $ ab \neq cd$ so
\[ S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2(a^2b^2 \plus{} c^2d^2)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2((ab \minus{} cd)^2 \plus{} 2abcd)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} 4abcd
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} (2n)^2
\]
Now let $ ab\equal{}cd$; in this case if $ ac\equal{}bd$ then $ a\equal{}d$ that contradicts with the problem hypothesis.
so $ ac \neq bd$.
Now
\[ S\equal{}a^4\plus{}c^4\plus{}b^4\plus{}d^4
\]
\[ \Rightarrow S\equal{}(a^2 \minus{} c^2)^2 \plus{} (b^2 \minus{} d^2)^2 \plus{} (ac \minus{} bd)^2 \plus{} (ac \minus{} bd)^2 \plus{} 4abcd
\]
According to hypothesis $ a,b,c,d$ are distinct numbers; hence all of the summands are nonzero.
Note:edited

[quote="madadpoor"]let $ abcd \equal{} n^2$ and
\[ S \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4
\]
then
\[ S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2(a^2b^2 \plus{} c^2d^2)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2((ab \minus{} cd)^2 \plus{} 2abcd)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} 4abcd
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} (2n)^2
\]
According to hypothesis $ a,b,c,d$ are distinct numbers; hence all of the summands are nonzero.[/quote]

$ a \equal{} 2, b \equal{} 3, c \equal{} 1, d \equal{} 6$

[quote="madadpoor"]let $ abcd \equal{} n^2$ and
\[ S \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4
\]
then
\[ S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2(a^2b^2 \plus{} c^2d^2)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2((ab \minus{} cd)^2 \plus{} 2abcd)
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} 4abcd
\]

\[ \Rightarrow S \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} (ab \minus{} cd)^2 \plus{} (ab \minus{} cd)^2 \plus{} (2n)^2
\]
According to hypothesis $ a,b,c,d$ are distinct numbers; hence all of the summands are nonzero.[/quote]
Nice solution!From this problem we can conclude that the following equation has infintely many integral solutions $ (a,b,c,d,e,x,y,z,t)$:
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\equal{}x^4\plus{}y^4\plus{}z^4\plus{}t^4$
This is an very interesting problem!

thanks to differ.
for your example
\[ S\equal{}2^4\plus{}3^4\plus{}1^4\plus{}6^4\equal{}3^2\plus{}27^2\plus{}16^2\plus{}16^2\plus{}12^2
\]

Please learn internet etiquette, varunrocks.


A cylindrical bar of gold that is 1.5 inches high and 0.25 inches in diameter has a mass of 23.1984 grams, as determined on an analytical balance. An empty graduated cylinder is weighed on a triple-beam balance and has a mass of 73.47 grams. After pouring a small amount of a liquid into the graduated cylinder, the mass is 79.16 grams. When the gold cylinder is placed in the graduated cylinder (the liquid covers the top of the gold cylinder), the volume indicated on the graduated cylinder is 8.5 mL. Assume that the temperature of the gold bar and the liquid are 86 degrees Farhenheit. If the density of the liquid decreases by 1.0% for each 10. degrees Celsius rise in temperature (over the range of 0 to 50 degrees Celsius), determine
a) the density of the gold at 86 degrees Farhenheit
b) the density of the liquid at 40. degrees Farhenheit.

a)I found the volume of the gold:
(.125 in)(2.54 cm/1 in) = 0.3175 cm = 0.318 cm
(1.5 in)(2.54 cm/1 in) = 3.81 cm = 3.8 cm
V = (pi(0.318)^2)(3.8) = 1.206597185 cm^3 = 1.2 cc.

Then I divided the mass of the gold (23.1984 grams) by the volume of the gold (1.2 cc) to get 19.22630046 g/mL, or 19 g/mL in significant figures.

b)I found the mass and volume of the liquid first:
mass of liquid = 79.16 g - 73.47 g = 5.69 g.
volume of liquid = 8.5 mL - 1.2066 mL = 7.293 mL = 7.3 mL

Then, I found the density of the liquid at 86 degress Farhenheit, or 30 degrees Celsius ((86 - 32)/1.8 = 30 degress Celsius):
Density = mass/volume = 5.69 g/7.3 mL = 0.78 g/mL.

Finally, I had to adjust the density to agree with the temperature. According the problem, "The density of the liquid decreases by 1.0% for each 10. degrees Celsius rise in temperature." I have to find the density of the liquid at 40. degrees Farhenheit. This is where I get lost.
But, I'll attempt it:
40. degress Farhenheit = (40. - 32)/1.8 = 4.4 degrees Celsius.
1.0% of the density (0.78 g/mL) is 0.0078 g/mL.
4.4 degress Celsius - 30 degrees Celsius = -25.6 degrees Celsius. Does that mean that there is a decrease in 25.6 degrees Celsius?
Then, I set up a proportion:
(0.0078g/mL/10 degrees Celsius) = (x/25.6 degrees Celsius).
I solved to get x = 0.019968 g/mL = 0.020 g/mL
Now, should I add that to the density?
0.78 g/mL + 0.020 g/mL = 0.80 g/mL.

I really don't know if this is correct? Please work the problem out and see if I made any mistakes. I'd appreciate any help!

I did not check your calculations.  However, it looks like you have the right idea up until the density decrease calculation.  You said "density of the liquid decreases by 1.0% for each 10. degrees Celsius rise in temperature."  A 1.0% decrease is the same as a 99.0% retention, right.  So for example, we can write

Density at 4.4C * 0.990 = Density at 14.4C

And similarly,

Density at 14.4C * 0.990 = Density at 24.4C

So note that

Density at 4.4C * 0.990^2 = Density at 4.4C * .9801 = Density at 24.4C

which is NOT the same as a 20% decrease overall!

So we must use a method different from the one you were using.  The amount of density lost per 10C change is NOT constant.  Since we multiply by a constant factor each time, we should use an exponential equation:

Density at 4.4C * 0.990^X = Density at 30C

where X is the number of 10C intervals between 4.4C and 30C.  This is not too difficult to find, as the difference between these temps is 25.6C you said, and so X = 2.56.  (Make sure you understand what happened here, it's really just basic arithmetic but possibly presented in an unfamiliar form).

So now we have one equation, one unknown.  All that's left now is to plug and play.  Enjoy!


From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:

$\textbf{(A)}\ 40 \qquad
\textbf{(B)}\ 43 \qquad
\textbf{(C)}\ 29 \qquad
\textbf{(D)}\ 50 \qquad
\textbf{(E)}\ \text{None of these}$

is this girls or boys?
cuz 60 ppl left...
oh well let G equal girls and B equal boys
we have 2(G-15)=B
and G-15=5(B-45)
we eqpand to get
2G-30=B
G-15=5B-225
substitute 2G-30 for B in the second equation
G-15=10G-150-225
or 360=9G
G=40
and 2G-30=B so 80-30=B and B=50

its either A, D, or E

and read the guidelines, you shouldhave either put a link or grouped the questions (ppl prolly already said this)
b-flat specifically said not to post all the questions from an AMC test