inequalities
function
inequalities theorems

[b]HLAWKA IDENTITY and INEQUALITY[/b]
 [b]Bibilography [/b]
[1]  Adamovic D.D., [i]  G\'en\'eralisation d'une identit\'e de Hlawka et de  l'inegalit\'e  correspondante,[/i]  Matematicki Vesnik (Beograd) 1(16) (1964) 39--43.
[2] Baston V.J., [i]  On some Hlawka-type inequalities of Burkill,[/i] J.London Math.Soc., (2) 12 (1976) 402--404.
 [3] Burkill J.C., [i]  The concavity of discrepancies in inequalities of the  mean and of H\"older,[/i] J.London Math. Soc.,
(2) 7(1974) 617--626.
 [4] Dokovi\'c D.Z., [i]  Hornich inequality and some generalizations,[/i]   Bull.Soc.Math.Phys.Serbie 15 (1963) 33--36 .
[5]  Dokovi\'c D.Z., [i]  Generalizations of Hlawka's inequality,[/i] Glasnik Mat.-Phys.Hrvatske 18(1963) 169--175.
[6] Hlawka Edmund , quoted in [7]
[7]  Hornich H., [i]   Eine Ungleichung f\"ur die Vektorl\"angen , [/i]  Math. Z. 48 (1942) 268-274.
[8] Kelly L.M.  ,  Smiley D.M.  and   Smiley M.F. ,[i]  Two  dimensional spaces are quadrilateral spaces,  [/i] Amer.Math.Monthly 72 (1965) 753-754.
[9] Levi F.W., [i]   Ein Reduktionsverfahren f\"ur lineare Vektorungleichungen,[/i]  Arch.Mat.,- (Basel) 2(1949) 24--26.
 [10] Luci\'c R., [i]  Sur une in\'egalit\'e de Hornich,[/i]  Univ.Beograd.Publ.\\ Elektrotehn.Fak.Ser, Mat.Fiz.,   No.101-106 (1965) 5--6.
[11] Lupas  A., [i]  Asupra unei inegalitati pentru functii convexe [/i],(Romanian), Gazeta Matematic\u{a} ,Seria A , 1982 , Nr.1--2  ,49--52.
[12] Mitrinovi\'c D.S.,  [i]   Analytic Inequalities   [/i],Springer Verlag, 1970
[13] Pecari\'c J.E., [i]  Modified version of a general result of Vasi\'c-Adamovi\'c-Keki\'c  and some remarks concerning inequalities for convex functions,[/i]   Glasnik Matematicki 21(41) (1986)331--341.
[14] Pecari\'c J.E., [i]  Konveksne funkcije-Nejednakosti,[/i]  Naucna Knjiga , Beograd,1987 ,pp.103--107.
[15] Pecari\'c J.E. ,  Proschan F.  and  Tong Y.L. ,[i]   Convex Functions, Partial Orderings and Statistical Applications  [/i] , Academic Press, 1992.
[16] Popoviciu T., [i]  Les fonctions convexes,[/i] Actualit\'es Sci.Ind. No. 992, Paris, 1945.
[17]  Popoviciu T.,  [i]  Sur certaines inequalities qui caracterisent les fonctions convexes,   [/i]  An.Stiintifice.Univ.
Al.I.Cuza Iasi Sect.I-a , Mat.(N.S.) 11-B , (1965) 155--164.
[18]  Smiley D.M. and   Smiley M.F., [i]  The polygonal Inequalities [/i],  Amer.Math.Monthly, 71 (1964) 755-760.
[19] Vasi\'c P.M., [i]  Les in\'egalit\'es pour les functions convexes d'ordre n,[/i] Matematicki Vesnik-(Beograd) 5(20) (1968) 327--331.
[20]  Vasi\'c P.M. and Stankovi\'c Lj.R., [i]  Some inequalities for convex functions,[/i] Math.Balkanica 6, 44, (1976) 281--288.

Could you post it? It is really interesting

The Hlawka inequality/identity can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=karamata&t=16600]here[/url].

flip2004, can you post some of its generalizations? (as is sugested in your bibliography)


inequalities
geometry
trigonometry
geometry unsolved

The incircle of  ABC is of radius r  ,Pro that:4(AB+BC+CA)^2>=27r^2  :)

[hide="Hint"]Use $ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R, \sum \sin A=4\prod \cos\frac{A}{2},r=4R\prod \sin \frac{A}{2}.$[/hide]

first, suppose A,B,C are on circleO,then move A and B,it is easy to prove that AB+AC is biggest when AB=AC,there are many ways to prove it,and so AB+AC+BC is biggest when AB=BC=CA,move C and make the 3 sides equal,so for triangle MNP on circleO,you can find a triangle EFG,letEF=FG=EG,EF+FG+EG=MN+NP+MP,and then EF+FG+EG=MN+NP+MP<=AB+AC+BC,so EFG is on a smaller circle than circleO or an equal one
so for random triangle QRT,when QR=RT=TQ,QR+RT+TQ/radius of its incircle    is smallest,so QR+RT+TQ>=3sqrt3×r,so (QR+RT+TQ)2>=27r2


Is actuarial science a good major to consider?

yes, it might be.
 i got a couple of friends doing it. You have to be aware that it is hard.


Circle $ O$ has diameters $ AB$ and $ CD$ perpendicular to each other. $ AM$ is any chord intersecting $ CD$ at $ P$. Then $ AP\cdot AM$ is equal to:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);
pair O = origin;
pair A = (-1,0);
pair B = (1,0);
pair C = (0,1);
pair D = (0,-1);
pair M = dir(45);
pair P = intersectionpoint(O--C,A--M);

draw(Circle(O,1));
draw(A--B);
draw(C--D);
draw(A--M);

label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,N);
label("$D$",D,S);
label("$M$",M,NE);
label("$O$",O,NE);
label("$P$",P,NW);[/asy]$ \textbf{(A)}\ AO\cdot OB \qquad \textbf{(B)}\ AO\cdot AB\qquad \textbf{(C)}\ CP\cdot CD \qquad \textbf{(D)}\ CP\cdot PD\qquad$
$ \textbf{(E)}\ CO\cdot OP$

[hide="My Solution"]
Products in geometry problems usually indicate similar triangles. So we decide to draw $ \overline{MB}$ . This creates a right triangle $ \triangle AMB$. We have $ \triangle AOP \sim \triangle AMB$ because $ \angle OAP \equal{} \angle BAM$ and $ \angle AOP \equal{} \angle AMB$. By similar triangle ratios, $ \frac{AP}{AB} \equal{} \frac{AO}{AM}\implies AP\cdot AM \equal{} \boxed{AB\cdot AO \text{ (B)}}$.
[/hide]

According to the problem we have $ \widehat{AMB} \equal{} 90^\circ$
=> $ AOP$ ~$ AMB$ (g-g)
=>$ \frac {{AP}}{{AB}} \equal{} \frac {{AO}}{{AM}}$
=>$ AP.AM \equal{} AO.AB$ :P
The correct solution is $ (B)$

A third solution (Based on the other two)

Since $ \angle BOP\equal{}\angle BMP\equal{}90^{\circ}$, quadrilateral BOPM is cyclic.  Then by power of a point, $ AP\cdot AM\equal{}AO\cdot AB$ so B.

Alternatively, assume P is on O. The answer follows from there.


Let ABC be triangle .The bisector BF,CE cuts at I .The line passing through
I,perpendicular to EF ,cuts EF,BC respectively at P,Q .Find Angle BAC?
known that IQ=2IP

If A=60, then there is a point Q on BC such that BQ=BE and QC=CF and EFQ is equilateral with I as its center, so P,I,Q are colinear and 2PI=QI.

I don't know if this is the only angle.

I think it not hard .if <A=60 then this problem is true and We can use combinatorics for prove it but the prove can long :?

[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=23746 for a discussion of this problem.][/color]


please post in how you did and talk about the difficulty level of the paper and the marks you got!!!! :D

[quote="sachin a S"]so let me be first  :)
paper was pretty easy comparable to dis yr aieee but had bundles of knowledge based ones esp in chem. i wud recommend thorough study of 12th chem text

i didn't perform to my xpectations  :|  mainly as i missed out d 12 xtra questions
which as many(those who attempted) found were very easy. so i wud rather advice a go at dem even at d risk of a few (3 to 7 not more) -ves (coz u can attempt dem only if u attempt al d 150 3 mark questions)

so al d best 4 evryone  :roll:[/quote]

Thanks man!!!!!! Whats ur score?

cheers whats your score sachin???? and whats an acheivable one in this kind of paper???:D

hope you got above 400! Then only i have a chance to get above 350  :D  :D

mine was today on 10th.
first things first : i got 361

my analysis: well, i m not known for my analysis skills, but what i felt was that u cannot prepare for some questions in chemistry. they are purely knowledge based questions. i found 2-3 questions which were wrong. i didnt have a clue abt 11 questions, so didnt go for the bonus ones. important points to note
1. reach your centre 45 min before start, coz in our centre, they were admitting students one by one, so quite a long queue was there.
2. my centre didn't allow me to take a pencil box, or bottle, or even a wallet!!
3. i would advise u to do maths at the end, coz it is the section that takes the maximum time.
4. they give u a small pad for rough work. it is the size of small spiral bound pads (i think A5) and there 8 sheets i.e. 16 pages to work on. (i dont know if it varies from centre to centre)
5. i would advise u to go for the bonus questions only if u r able to attempt more than 140 questions, with the slightest of risks!

anyway, all my exams are over!! best of luck to u guys!

p.s. did anyone write VIT? my rank in it is 76. if u have, pls post ur rank.

oh thats great job man !!!!! :D
how have your friends fared in the exam???? :lol:

what is a great job?? my marks in BITS, my rank in VIT, or my post??

anyway, whichever it is, thanx!

one of my friends has got 320, another one 301 and the rest are in 200s. pramod will be giving it on 15th, so u can get an actual assessment after that!

How was physics? AIEEE level? Or NCERT level? Or intermediate between them?  :)

oh do you even have wrong questions in bits how come then what should one do???? :D
and what is the difficulty level of those bonus 12 questions???/

[quote="achu_182"]maths and physics is easy. the only place where u can lose marks is the theoretical part of chem, english and careless mistakes here and there (or even wrong questions!!)[/quote]

physics is easy in the sense? Easier than aieee? And also was the english part horrible? Will they allow you to see the solutions?  :maybe:

You might have written some sample (maybe online) tests before you wrote bitsat right (like mtg, arihant and others) . How would you rate the bitsat paper in toughness wrt them. Bcos i find some of those papers a bit tougher than i like it to be.

i found them of almost equal, though one of my friends said that he found the actual paper tougher than Arihants paper. From what i have heard, MTGs papers are even tougher than Arihants.
physics is easier than this years AIEEE. and they don't let us see the solutions, coz questions can be repeated.

what i did before BITS: absolutely nothing. did a sample paper everyday, and left it at that. take it easy guys, don't get tensed unnecessarily!!

srinath got 354, karthik ram got 323, d karthik got 327 if im not wrong!  :)

No man, it was a bit tough i guess, a bit above aieee level :maybe: So the scores seem to have dipped a bit

yeah i agree but still i will never ever say my physics io did worse in it than my aieee physics!!!

forget it!!!!

I should hope that not more than 50 get above 400 this year! I dont think (or rather hope) it should happen. And most of them anyway will be iit toppers so i dont think they might opt for bits at all. I fell maybe only those below 400 might opt for bits (except a few exceptions)  :)

[quote="pramod.best"]my classmate and good friend @ skul.............got a whooping 403!....... :maybe:  :o ...........probably one of the highest so far.........will end up in top 100 atleast!

Last year 79 got above 400! :o[/quote]

They got above 400 Factorial??  :rotfl:  :rotfl: And what is your source? I think it should be more.

Anyway I think the number of people above 400 will remain the same or be lesser.

hey i oppose that last year so many people got 400 + from chennai with the highest being 421 from our skl but this time till now only prasad has got the highest with 398 and not one 400 and there can be atmost 3 400 this time from now onwards seing the difficulty so i think it is the opposite i think prasad will mostly be chennai topper if chappli and rakesh and karthik rajagopal arent counted!!!!! :D

hi guys.... im new here...my score in bitsat is 347...i got 123 right...but 22 negatives  :(

dont worry i give you company i too am almost same mark 346 :(

[quote="abhishekj"]well upto my knowledge highest yet is 423
i know dis guy pretty well we were together at vmc..
hey pramod.. is it shashank who got 403??
congrats to him!!!![/quote]

@abhishek: yup....u guessed it right, mate! It is Shashank! By the way who is the lunatic who got 423? :o ...........Besides, wen's ur's or wat is ur score if u've already appeared?............

But u've got nothing to worry!.....coz if I heard Utkarsh and others right.......you got 340 odd in JEE :omighty: ...........If u've got luck you might very well end up in top 100 ! which I think is quite likely..........as per last year's stats, u'd get atleast 300!

OMG! These people must be big nerds then :o

i would be pretty much happy if these mega titan nerds prefer CMI than IIT like our great Abhishek Dang and leave IIT for small surviving nerds !!! serve the needy!!!

wrote BITSAT 2day................got 392

did ne1 get any ques on structure of penicillin???? :mad:

Penincilin is easier than many other structures they've asked i guess :D I read it fully but thankfully no structure came :D

392? Thats a really good score! :coolspeak:


Serious questions in the games and fun factory ???? Doesn't that seem to be some kind of contradiction ??

I'll start it off.
1. Do you have a rubiks cube? And if you do, are you good at it?

No, I do not.

Do you use the AoPS forums for math or do you only use the fun and games factory forum.

I love to dance.

(serious question)How many proofs of the Pythagorean Theorem can you do?

[quote="gkandlikar"][quote="mbabbitt2003"]Hi. i think that we sould ask questions that we don't know here and answer questions that we do know. Bye.[/quote]

Isnt this the same as the "Answer a question, ask a question" thing?[/quote]

Seems like it to me. Locked unless someone explains to me otherwise.


Can -N=O group have +R effect or is it restricted to delocalisation on oxygen with the help of pi bond.
Hence, would this group be ortho para directing or meta directing on the benzene ring?

Also, which group would exert greater -I inductive effect, -OH or -OR?

[hide="Click here"][quote="Zimrock"]Hence, would this group be ortho para directing or meta directing on the benzene ring?[/quote]

The effect is the same as an halogen: electronwithdrawing inductive effect, electrondonating ressonance effect. Therefore, the nitroso group is a deactivating group but an ortho-para director.[/hide]

[quote="zimrock"]
Also, which group would exert greater -I inductive effect, -OH or -OR?[/quote]

OH Group but I am not sure how considerable the difference is. :-|

among -OH and -OR , the -OH bond is more polar and hence, the electron density on oxygen is more and hence wouldnt -OR be a better -I group?

In -N=O, due to the presence of pi bond, wont it be only a -I group? how can it donate its lone pair to stabilise the sigma complex when the pi electrons are delocalised on oxygen?

The nitroso group is inductively electronwithdrawing because nitrogen is more electronegative than carbon. However, it can donate electrons by ressonance since nitrogen also has a lone pair of electrons. These electrons cannot be delocalised on oxygen, since oxygen can only have around it 8 electrons.

ok, but the major contributing resonance structure would be the one with the lone pair delocalised on oxygen and hence wouldnt it be meta directing?

The the major contributing resonance structure for what? And one more time: who can the lone pair be delocalized to oxygen? Can you explain that to me?

To explain the orientating effect of a given substituent, we are to write the [i]Wheland intermediates[/i] in a reaction with an electrophile for each position (o-, m-, and p-): only then can we see what are the positions favoured.


Jack B. Nimble may choose any 3 books from a set of 7 books. How many different sets can he pick?

It is just 7 items, and you need to choose 3. So you can use the "formula"
Which is shown on the post above


What do flexible exchange rates have to do with trade deficits at all?

They probably have some relationship, even though I am not sure if it can be defined as a positive or a negative one, because other factors probably determine it too. I am not sure, so you think that they dont have any casuation relationship?

Your question assumes that the participants have no preferences as to where (which currency) they hold their capital.  This is not true.  Also, many of the trading partners (countries, China primarily) have preferences inconsistent with the theory of flexable exchange rates.

Allow me to type further, if a person is willing to hold dollars in an account rather than pay the exchange rate and convert the dollars into local currency, dollars tend to be a bit higher in exchange to the local currency. (More demand for dollars (the held currency) means a higher price.  That downward sloping demand curve.)

A second factor is that dollars are used for sales of oil, even to Japan, Europe and China from OPEC.  The sellers have every reason to have their nation's financial dealings keep dollars higher in value relative to other currencies.

A third factor is once a country, say China, has accumulated significant savings in dollars, it has every reason not to want the market to work to lower dollars and the value of their savings.  We have every reason to ask for the dollars to decrase in value and "take" back the dollar preimum that we collected when we gave them the dollars.

In short, it's complicated and the theory doesn't take into account all relevant factors.

There's more, but this is probably enought for now.


hi,

i am mohammed.I'm 17 years old .I think that morrocan MO are so

bad ,they copy/past other problems.The results also aren't clear.

Cheating in tests  :(  has become a custom...

my question is : how can we participate To IOM ?

yeah,not only morocco but also other african and arabic

countries . May be ,Because they have not a good level  :D .

[quote]good level :D [/quote]

strange...


How can I write a long matrix equation divided into 2 lines instead of 1 line?

Thanks

Typically you use multline, align, eqnarry, gather, or some other such thing.

Lots of examples at: [url]http://www.tug.org/tex-archive/info/mil3/multline.tpl[/url]


$f(x)=\log_{a}{(4ax-x^2)}$ is monotonically increasing at $(\frac{3}{2},2)$.
Find the range of $a$.

[quote="arkmastermind"]i think it's:

0<a<3/4 or a>1

this seems like a calculus problem...perhaps you should post in another forum. but there might be another way to do it without calculus.[/quote]
Sure it is not a calculus problem... maybe you made it too much complex.

[quote="4everwise"]I think i got the answer.[/quote]
Than please show your steps.


In the classical game of Nim, two players (A and B) may take between 1 and 4 stones from a pile of initially [i]n[/i] stones.  Each player must take at least one stone when it is his or her turn and the player who draws the last stone will lose.  A version of this problem appeared on the first High School Math League contest of this year. [url]http://www.themathleague.com/[/url]

We can prove that player A has a winning strategy as long as [i]n[/i] is not congruent to 1 mod 5.
Here's the strategy:
Since A may take up to 4 stones, he/she can make sure that there are 5k+1 stones remaining.
If B takes r stones, A's next move is to take 5-r stones.
At the end of each turn that A makes, there will always be 5k+1 marbles for some integer k, until at last k is 0 and B must take the last stone.

Now, with this background, consider the following generalized version of the game:

Team A and B both have n players and the 2n players are seated around a table, with each player having opponents at both sides. Turns round the table so the two teams play alternately. Second, let's vary the maximum number ofstones each player can take. That is, each player has his/her own maximum number of stones he/she can take at each turn (The minimum is always one). So the game is asymmetric and may even be unfair.
In general, when played between two teams of experts, the outcome of a game is completely determined by the initial number ofstones and the minimum number of stones each player can take at each turn. In other words, either team has a winning strategy.

Given n, the number of players per team, S, the initial number of stones, and $ A_1, B_1, A_2, B_2, ..., A_n, B_n$
the maximum number of stones each player may take (all elements in the sequence are integers between 1 and 16 inclusive), how can we determine which team has a winning strategy?

I was hoping someone would post a clever mathematical solution; no one did.  Maybe no such clever solution exists.  This problem came from a programming competition, so I what I was looking for was a clever algorithm.

I have a solution that seems to work, but it is not very clever.
If anyone cares to see it, you can message me; I wouldn't bother posting it here since it is too involved to carry out by hand.  Maybe proving that my algorithm will always produce an answer could make for an olympiad style problem, but this really is more computer science than combinatorics.


percent
probability
Pascal\u0027s Triangle

[quote="GameBot"]If a five question True/False test is given and $ 60\%$ is a passing grade, what is your percent chance of passing if you guess on each question?[/quote]

We have to get 3,4,5 right. $ \frac{10}{32}\plus{}\frac{5}{32}\plus{}\frac{1}{32}\equal{}\frac{1}{2}\equal{}50\%$.

[quote="ernie"]Also, you can recognize by looking at Pascal's triangle that the probability of getting 0, 1, or 2 questions right is equal to the probability of getting 3, 4, or 5 questions right.  Thus, our answer is $ \boxed{\frac12}$.[/quote]

The question asked for percent.

1/2 would then go to 50% if you want to use ernie's solution.


probability
LaTeX
complementary counting

Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any cookie? Express your answer as a common fraction.

[hide]51/64[/hide]

My question is, how do you do it?

I did complementary counting, it is where you count the ones that you don't want. So first I did 3 chips on a cookie. The first chip goes on any cookie. The second chip has to go on the cookie as the first chip-1/4, the third same-1/4, and the last chip on a cookie other than the other chips are on-3. So the probability that 3 chips go on a cookie is $ 1*\frac {1}{4}*\frac {1}{4}*3 \equal{} \frac {3}{16}$
If there are 4 chips on a cookie, using the same reasoning$ (\frac {1}{4})^3 \equal{} \frac {1}{64}$ Adding $ \frac {3}{16} \plus{} \frac {1}{64} \equal{} \frac {13}{64}$
$ 1 \minus{} \frac {13}{64} \equal{} \frac {51}{64}$

Sorry if I am not clear

Edit: fixed latex
Edit#2: fixed that again

I am having a bad day. :(

No you didn't fix it. lol $ \dfrac{1}{64}$ :lol:

Yea... I hate these questions... sort of like the questions: how do you flip exactly 8 heads on a 10-flip coin toss.  You have to sort the combinations...

the guy above me had it right...

Just one question... for the chips, are 4 chips on cookie A the same as 4 chips on cookie B?  If it wasn't then this question could change...

to your first question, I believe it is 45/2^10

for the second one, it isn't the same.

exactly 8 heads is pretty easy, it's just $ \frac{\binom{10}{8}}{2^{10}}\equal{}\frac{45}{1024}$.


How do you solve this annoying c+p problem quickly without complex casework? I wonder how many different ways people approached this problem.

Five different rockets, A, B, C, D, and E, are to be launched from two separate launch pads labeled 1 and 2. Each pad can accommodate only one rocket at a time. The rockets can launch from either pad, in any order and at any time (sequentially or simultaneously). One example of a launch pattern is (C1, A1D2, E1B2) where the two commas separate three different launching times. Including the example given, what is the total number of different launch patterns?

This is Stirling numbers of the second kind, I believe.

1 there is a recursion about stirling numbers
2 or you might have memorized them up to 10

[hide]Well, there are 120 ways to choose the order of the rockets (if C1B2, then C is before b). Then we need to see how many ways there are to make 5 launches using the "tiles" 1, 2, 1+2. Let R_n be the number of ways with n rockets. Then R_0=1, R_1=2, R_(n+1)=2R_n+R_(n-1), so calculating it out gives that R_5=70, so there are 70*120=8400 ways to launch the rockets.[/hide]

Wait how did you find the recursions? From "make it simpler" patterns?

erm we can either ad a 1 or a 2 to the end of a R_n sequence or a 1+2 to the end of a R_n-1 sequence to get a R_n+1 sequence, so R_(n+1)=2R_n+R_(n-1)

and erm no they are not sterling numbers

Sorry to revive, but the recursion is not $R_{n+1}=2R_n+R_{n-1} $ but $R_{n+1}=2(n+1)R_n+(n+1)(n)R_{n-1} $ or $R_n=2nR_{n-1}+n(n-1)R_{n-2}$ since there are $n$ ways to choose which rocket is launched at the end and $_n P_2$ ways for sending two at a time.  Thus, the answer is $\boxed{8400}$.

For two more solutions and the correct full solution with recursion, look on [url=http://www.artofproblemsolving.com/blog/52788]my blog[/url].


Paul and Jenny each has a whole number of pounds.

He says to her: "If you give me $ \£3$, I will have $ n$ times as much as you.

She says to him: "If you give me $ \£n$, I will have $ 3$ times as much as you.

Given that all these statements are true and that $ n$ is a positive integer, what are the possible values of $ n$?

[hide]The system is

$ \{\begin{matrix}P+3 &=& n(J-3)\\P-n &=& 3(J+n)\end{matrix}$

or after rearranging

$ \{\begin{matrix}P-nJ &=& -3n-3\\P-3J &=& 3n+3\end{matrix}$

Solving it by a simple method (e.g. determinants), we get

$ P={3n^2+12n+9\over n-3}=3n+21+{72\over n-3}$

$ J={6n+6\over n-3}=6+{24\over n-3}$

Hence $ n-3 | 24\land n-3 | 72\implies n\in\{4,5,6,7,9,11,15,27\}$[/hide]

Hi Farenhajt.

I think your second equation should be

$ J\plus{}n \equal{} 3(P\minus{}n)$.

Indeed it should  :blush:

[hide="With that correction..."]...the system is

$ \{\begin{matrix}P & - & nJ & = & - 3n - 3 \\
- 3P & + & J & = & - 4n\end{matrix}$

The solution is

$ P = {4n^2 + 3n + 3\over 3n - 1}, J = {13n + 9\over 3n - 1}$

We have $ J = 4 + {n + 13\over 3n - 1}$. From $ {n + 13\over 3n - 1}\geqslant 1$ we get $ n\leqslant 7$. Checking shows that only for $ n = 1,2,3,7$ we get an integer $ J$. Also, all of these values yield an integer $ P$. Hence

$ n=1 : P=5, J=11$
$ n=2 : P=5, J=7$
$ n=3 : P=6, J=6$
$ n=7 : P=11, J=5$[/hide]


trigonometry
calculus
derivative
function
calculus computations

Solve the differential equation
\[ (1 - x^2)y'' - xy' + y = x,\quad |x| < 1  \]
for example by setting $x = \sin t$.

I feel really dumb and like I knew how to do this a couple of years ago, but I can't figure out how to do the substitution. Can someone explain? Please don't use Leibniz notation in any larger extent, because everything seems like magic when you use that.

When I do it... if I set $u(t) = y(\sin t) = y(x)$ this gives me $y'(x) = u'(t)\frac{1}{\cos t}$ and $y''(x) = u''(t)\frac{1}{\cos^2t} + u'(t)\frac{\sin t}{\cos^3t}$. This is most likely right, since it leads to a very easy equation which leads to the right answer (after substituting back). But I still don't feel like I understand at all what I'm doing.

Please explain :( I feel like I'm doing all this kind of advanced stuff, and I don't even know how derivatives work. Why is it so hard to wrap my head around this? Why am I putting $u(t) = y(x)$, differentiating with respect to t and then getting an expression for $y'(x)$ which I can solve out of the equation? What I'm doing just doesn't make any sense to me. It's something with solving y'(x) out of the equation that makes this seem suspicious to me. Is it obvious that we can do it? Is there another approach?

I guess my questions are a little too vague. Let me pinpoint my issue: I don't understad the chain rule in Leibniz' notation. Let's say we have a function $p(t)$ an that we have an expression for $\theta'(t)$. Without further ado we can now write down $\frac{dp}{dt} = \frac{dp}{d\theta}\cdot \frac{d\theta}{dt}$.

The way I always thought it was, is that $\frac{dp}{d\theta} = p'(\theta(t))$, but this makes almost no sense looking at the chain rule... I guess I have a big misunderstanding of leibniz notation. And how would you formulate the above relationship between derivatives not using leibniz notation?

edit: I guess this question is too dumb for being worthy of an answer, but I'm really confused :(. Please help me.

[quote="Kalle"]
The way I always thought it was, is that $\frac{dp}{d\theta} = p'(\theta(t))$, but this makes almost no sense looking at the chain rule... I guess I have a big misunderstanding of leibniz notation. And how would you formulate the above relationship between derivatives not using leibniz notation?
[/quote]
You thought right: $\frac{dp}{d\theta}$ stands for $p'(\theta(t))$. The Leibniz notation is, indeed, a bit confusing because it uses the same letter $\theta$ to denote two formally different things: the independent variable of the function $p$ and the function of $t$ you plug into $p$. When teaching calculus, I usually try to avoid it and write the chain rule as $\frac d{dt}(p(\theta(t))=p'(\theta(t))\cdot \theta'(t)$ or, equivalently, $(p\circ\theta)'=(p'\circ\theta)\cdot\theta'$.

Yes, but how this then explain what "$\frac{dp}{dt} = \frac{dp}{d\theta}\cdot \frac{d\theta}{dt}$" means in "normal notation". It would then look like it says $p'(t) = p'(\theta(t))\cdot\theta'(t)$. Note that $p$ is NOT a function of $\theta(t)$ but simply of $t$! That's the confusing part. Utilizing the leibniz notation still works for some reason, though.

I've spent a lot time trying to figure out what this really means, and I think what it actually means is that we set $p(t) = z(\theta(t))$ for some function $z$. This way it would then say $p'(t) = \frac{d}{dt}z(\theta(t)) = z'(\theta(t))\cdot\theta'(t)$. Is it obvious that there always exists such $z$, though? Since they are to be solutions of differential equations, we can assume that $p$ and $\theta$ are continously differentiable.

Using this "behaves like a fraction"-trick has been working for me for years (for example in substitutions and related rates problems), but just today I realized I have no idea what it means. And I still don't think your interpretation in this specific case is right. Are you sure?

e

The other thing that's going on here: $p$ has a different meaning in the two places. We really should write $(p\circ\theta)'(t)=p'(\theta(t))\cdot\theta'(t)$.

I wouldn't try to explain the chain rule in Leibniz notation.


invariant
college
advanced fields
advanced fields unsolved

[quote="titeo"]I don't know whether this topic is in right box. If not please move it to right box

1) Let $B_t$ be a Brownian motion.
Prove that $B_t^3-3tB_t$ is a martingale.

2)Let $B_t$ be a 2_dimenson Brownian motion starting from $x \neq 0$.
Prove that $X_t = ln(|B_t|^2)$ is a martingale[/quote]
Regarding 1, compute $d (B_t^3 - 3tB_t)$ using Ito's fomula. If you get something of the form $\delta(t) dB_t$, you're close to done., by the martingale representation theorem.
Regarding 2, note that $X_t = 2\Re \log(B_t)$, if $B_t$ is viewed as complex Brownian motion. Thus $X_t$ is (a time change of) a 1-d Brownian motion, by conformal invariance.

Sorry to bother you!
1) I receive $dN_t = 3(B_t^2-t) dB_t$ from Itô's formula, where $N_t = B_t^3 - 3tB_t$
How can we know $N_t$ is martingale?  (I only know that N_t is martingale if $E(N_t-N_s|F_s) = 0$)
Thanks.

[quote="david_555"]Hello iam david.
Sorry for bothering you but i want to work on martingales in my master thesis, but i could not find good sources.
Can you help me by stating good books, references, sites on internet,papers.................since you are working in this domain.
Appreciate your help[/quote]

Some textbooks and lectures I know:
-Continuous Martingales and Brownian Motion-Daniel Revuz Marc Yor -SpringerVerlag
-Conformally Invariant Processes in the Plane-Gregory F.Lawler (you can find this on the internet)
-Lectures on Probabitiy Theory and Statistics-Boris Tsirelson& Wendelin Werner-Springer
-An introduction to Stochastic Differential equations - Lawrence C.Evans, Department of Mathematics UC Berkeley


ARML
calculus
integration
ratio
geometry
3D geometry
tetrahedron

I was a little anxious to put it up, so here you go. The rules are the same as posted before on this forum. Turn in the answers to me (PM) either today or tommorrow, but the deadline is tommorrow at 7:00 PM. The answers are not necessarily integral but they are numerical.
[hide="Set 1"]
$(1)$ Let $z$ be the number of zeroes at the end of $\frac{200! \cdot 201! \cdot 202! \cdot 203! \cdot 204!}{205!}$ What is the number of divisors of $z$?

$(2)$ Find the sum of all integers $x$ such that $x-3$ divides $x^2+2$
[/hide]
[hide="Set 2"]
$(3)$ Find the value of $76 \cot {(\cot^{-1}2+\cot^{-1}4+\cot^{-1}6+\cot^{-1}8)}$

$(4)$ A triangle with sidelengths in a ratio of $4$, $7.5$, and $8.5$ is inscribed in circle $A$ and contains it's incircle $B$. Let $a$ and $b$ be the radii of $A$ and $B$ respectively. Find $\frac{a}{b}$[/hide]

[hide="Set 3"]$(5)$ Approximately How many pairs of integers $(a,b)$ satisfy $(\frac{(a+b)^3-(a^3+b^3)}{3ab})^2\leq 100+2ab$. Answers within $5$ of the range of the exact answer will be accepted.

$(6)$ Let $x^7=1$, however, $x$ is not equal to $1$. Find $\frac{x^{20}+1}{x^{10}}+\frac{x^{60}+1}{x^{30}}+\frac{x^{100}+1}{x^{50}}$[/hide]
[hide="Set 4"]
$(7)$ Allen Iverson starts at one vertex of a tetrahedron and in each move he walks along a random edge to an adjacent vertex. What is the probability that after $60$ moves Allen Iverson will be at the same vertex he started at?

$(8)$ Let $a=2^{24036583}-1$, which is a prime number. For how many postive integers $b$ do each quadratic $\pm x^2 \pm ax \pm b$ have rational roots?[/hide]

Here are the Relays I made, you just gotta do them yourselves i guess, but they're pretty cool. The Answer You Will Use Next=TAYWUN

$(1)$ The TAYWUN #1 for this question is $n$, so I suggest that you try to find $\sqrt{n}+2$ and use the numerical value of $n$ for the next question.

$(2)$ Take the integral value of TAYWUN #1 and call it $k$. What is the sum of the coefficients of $(x^5-3x^2+4x)^k$. This numerical value is TAYWUN #2.

$(3)$ Take TAYWUN #2 and divide it by $4$. Let the resulting number equal $k$. If $x^3-kx^2+3x+4=0$ and the roots of this equation are $a,b,c$, Compute $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$

PLEASE DO NOT DISCUSS PROBLEMS BEFORE THURSDAY AT 7:30 PM

THE RULES ARE THE SAME AS THEY WERE BEFORE THOUGH. YOU HAVE 12 MINUTES TO DO EACH PAIR OF PROBLEMS AND NO CALCULATORS ARE ALLOWED. NON SC MEMBERS ARE ALSO ALLOWED TO TAKE THE TEST.

i'm sorry but i found TAYWUN to be funny. the actual abbreviation is TNYWR (The Number You Will Recieve).

The relay (question number 1) is worded a bit weirdly, but it's an actual ARML relay, according to Shobhit.  I finally understood the question after staring at it for about 10 minutes! :P

Cool questions, I enjoyed them

and the first relay is in fact, based off a real relay question they once had. One of the coolest questions i've ever seen

Looks like 7 is the score to beat guys :D

[quote="mysmartmouth"]Does that include the relay or not?  Can there be awards and HMs for best Middle Schooler, Freshman, and Sophmore?[/quote]

No, the person that got a 7 on the individual did not do the relays..i guess they were too trivial.

And i don't know if we should have a middleschooler award cuz your the only middleschooler taking it :P

I have TeXnicCenter, so I went ahead and Latexed it up nicely.  Here it is.  Once we get solutions, I'll do those too.

Sorry for the triple post, but it's worth it to bring you the relays below.  Also, if you want a .tex version of either of these PDFs, email or PM me.

[quote="mysmartmouth"]Any perfect scores yet Shobhit?  (Meaning has Joe turned in his answers?)[/quote]

in fact yes, although no one from SC. Thanx for making pdfs for the problems :)

ooh, a perfect score. I wanna see how #8 is done.

[hide="question about #5"]for number 5, the answer includes cases where $a$ and $b$ are zero. Should this be included? If they're not, the answer is actually $276$, instead of $317$.[/hide]

Yes, zero is an integer, and a and b are integers, so pairs where they are integers counts.

Yeah...but then you have division by 0 in the original statement...I think we all missed this for some reason  :huh:

Yes, division by 0!  I thought so!  So I really got a $1$ on the test!  Yes!  LOL JK.

guess this means there werent any perfect scores  :P

Hmm... but 0/0 could be anything, including something less than whatever the RHS was :). I do agree that that shouldn't be counted though, trick question or not.

[quote="joml88"][quote="chess64"][quote="Iversonfan2005"]$(1)$ The TAYWUN #1 for this question is $n$, so I suggest that you try to find $\sqrt{n}+2$ and use the numerical value of $n$ for the next question.[/quote]

Huh I don't get this at all.[/quote]

 :)[/quote]

What? :?

i dont exactly get it but i think they want you to solve for $n = \sqrt{n}+2$ or something.

Wow I knew this question was logically flawed.  I saw the original ARML question, and it actually makes sense!

[quote="joml88"]Eh, sorry...I didn't read the question closely enough the first time to see that Shobhit changed it from it's original ARML form.  The original said something like: "Let $n$ be the number you will pass back.  Pass back $\sqrt{n}+2.$"[/quote]

Oh, I get it now, thanks!

all [s]individual round[/s] problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3734446_2006_mock_arml_i_individual]individual round[/url] + [url=https://artofproblemsolving.com/community/c3736289_2006_mock_arml_i_relay]relay round[/url] [s](relay round soon)[/s]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Find all functions $f: \mathbb{Q}\rightarrow\mathbb{Q}$ with the property that:

            $|f(x)-f(y)|\le|x-y|^{2}$ for any $x, y \in \mathbb{Q}$.

It`s from Romania TST 2007 : http://www.mathlinks.ro/Forum/viewtopic.php?p=813852#p813852


Good Morning! I have been trying to figure out what I could use for the following.

If x=1000
then solve for y
y=(x*.0236)/.83)+x

y=1028.43(rounded)

if this is reversed

If y=1028.43
then solve for x

Any help would be great.
cardgunner

[hide]Set the original equation equal to $1028.43$.
$1028.43 =\frac{x\cdot.0236}{.83}+x$
$\implies 853.597 = .0236x+.83x$
$\implies 853.597 = .8536x$
$\implies x = \frac{853.597}{.8536}\approx 999.99$.
  The exact value is lost when you rounded the $y$ value.  Rounding again gives $x = 1000$.[/hide]

EDIT: Yes I meant $.8536$.


Putnam AMC USA(J)MO USAMO inequalities search email

Hello, Can any body post some of the hardest problems which appeared on any national , international olympiad or team selection tests, which even  boggled mathematicians.

Well, there are many examples, but I will only include three:

First, a problem proposed by N.Vasilliev at the 1970 Soviet Union National Olympiad.

It is problem 15 at

http://www.kalva.demon.co.uk/soviet/sov70.html

An interesting story about this problem appears in A.Soifer's book: COLORADO MATHEMATICAL OLYMPIAD, THE FIRST TEN YEARS. pp. 174-176. The proposed solution was wrong and they had to solve it the very last night before the contest. Fortunately, Sasha Lifshiz, an undergrad, found a splendid solution. No contestant solved that problem!

The second example is Arthur Engel's -now classical- problem B3 in IMO1988; that is, problem 6 at

http://www.kalva.demon.co.uk/imo/imo88.html

You may read the story about it at A.Engel's splendid book, PROBLEM-SOLVING STRATEGIES.

The third example is B2 in IMO1996; that is, problem 5 at

http://www.kalva.demon.co.uk/imo/imo96.html

Engel comments that this turned out to be even more difficult than his proposal. Both problems were solved by very few contestants, Endel's problem by 11 students and the hexagon problem by less that 10. The names of those brilliant students can be found at J.Myers website

http://student.cusu.cam.ac.uk/~jsm28/imo-scores/

Good luck!

I think Putnam 1992 problem B6 is very tough, it was solved by 0 contestants. I really like the proof given by Noam Elkies for this problem.

APMO 1999 problem 5 is also very tough, solved by only 3 contestants, and two of them are from Australia!!

IMO 2002 problem 6 is tough, it is the first IMO problem that over 400 contestants score 0 points.

Another tough problem is USAMO 1999 problem 3, I'm sure many people have difficulty solving this problem. It remained unsolved on http://www.kalva.demon.co.uk for a long time before Michael J. Dore found a nice solution.

Problem 11.4 in Russian Olympiad 2002 was not solved by any contestants, and Problem 11.7,11.8 was each solved by one contestant. These problems must be really tough because Russian students (who are very strong in Math) cannot solve them.

China Math Olympiad 2002 is very tough, some Chinese contestant say that the gold medal cutoff is 75 points out of 126 points. China use 21 points for each problem, so if we change the score to IMO scoring, that would be 25/42.

BTW, USAMO 1999 problem 3 was created by Kiran Kedlaya, who maintain USA math contest site.

Some IMO shortlist problems that are very tough are not used in IMO paper. In IMO 1988, Chezhkoslovakia and Hungary each propsed a very tough problem, which can be found on http://www.kalva.demon.co.uk/short/sh88.html . (problem 4 is czech, problem 14 is hungary) I wonder how many contestants are able to solve these two tough problems.

It's unlucky that these two beautiful problems are not used in IMO, I really like these two problems.

first of all, maybe you can register, or at least mention your name (i'm just askin', it's not mandatory, but it would be nice to see who's posting, thanks).

second of all, let me make some comments of my own, although i'm not exactly and IMO genius:

about problem 4/1988 - it is a very beautifull problem, but not a hard one; in fact it is a mathematical fact easy derivable from many other problems. To be more specifically, I have created (and solved) the same problem, as a generalization of a Russian (?!) problem which sounded like this: a 9x9 square is given and numbers from 1 to 81 are placed inside. Prove that there are two adiacent squares for which the absolute value of the difference of their numbers is at least 6.

about all others - for now the hardest problem remains IMO 2002/6, but this year on the ShortList a harder and more interesting / beautiful problem was submitted, but didn't get chosen.

i made a poll about the hardest problem, so u guys can vote on it. :)

cheers

Actually I did register my account. But there is an error when I register. There is also an error when I log in. So I give up and post unregistered. Please fix your message board so that people can register and log in without any problem.

sorry about this guys - it's phpbb2.0.5 problem - i will downgrade the forum to 2.0.4, when I didn't had all these problems

but this can only happen after i come back from 10th IMC ( sort of an IMO for college students )

up untill then ( 1 week from now or so ) anyone wanting to register please e-mail me with your username and desired password, and I will manually create your account

sorry for any inconvenience !!  and thanks for posting !

I think, the Mongolian problem from '96:

for a[1]>=a[2]>=...>=a[n]>=0 the inequality prod({(a[i]+a[i+1])/2},i=1..n)<=
prod({(a[i]+a[i+1]+a[i+2])/3},i=1..n) holds, the numeration of indecies is cyclic.

is the hardest ever.
As I know, the official solution was wrong, and it remained open for some years. Also it is an almost direct consequence of an old Karamata's inequality.

But the hardest mathematical competition for high school students is surely an Open Olympiad of 239 Saint-Petersburg school. It is usual that total 4 of 8 problems are solved by all the contestants together, and the maximum result is 3 (and it is also usual that a half a year after the contestants get 2-3 gold medals at imo, so they are not weak.) New results are often to appear there as problems. Problems may be found (in Russian) at http://www.pdmi.ras.ru/~olymp. If somebody is challenged, I may send them by e-mail (fedor@fp5607.spb.edu). Well, here is the no8 from '02-'03: given n^2 points, say A[1],...,A[n^2] in a circle of radius 10n. Prove that if n is large enough, then for some i, j the distance between A[i] and A[j] is less then square_root_of (i-j).

I know that John Scholes got all the Lenigrad problems (save the first ten years due to the second world war) sent by the key person Dmitrii Fomin
of the St. Petersburg Olympiad. Due to his restricted Russian knowledge neither he has them translated nor mounted onto the web. If you are only interested in the third stage of the competition, please study the Canadian problem-solving journal since 1981. It was published in the 'olympiad-corner'
posted by orl

[quote="haduyhung"]Problems may be found (in Russian) at http://www.pdmi.ras.ru/~olymp.

********************

    Hi Fedor Petrov,
Thanks for the nice link,but do you know the site where i can get all problems about this St Petersburg Olympiad?
 Regards,[/quote]

[quote="Anonymous"]I know that John Scholes got all the Lenigrad problems (save the first ten years lost due to the second world war) sent by the key person Dmitrii Fomin
of the St. Petersburg Olympiad. Due to his restricted Russian knowledge neither he has translated them nor mounted anything onto the web. If you are only interested in the third stage of the competition, please study the Canadian problem-solving journal Crux since 1981. It was published in the 'olympiad-corner'
posted by orl

[quote="haduyhung"]Problems may be found (in Russian) at http://www.pdmi.ras.ru/~olymp.

********************

    Hi Fedor Petrov,
Thanks for the nice link,but do you know the site where i can get all problems about this St Petersburg Olympiad?
 Regards,[/quote][/quote]

We are possibly going to translate last years problems onto English. I guess, Dmitry Fomin's (very good) book has problems only until '93. I have some problems translated (to Russian English), ask me by e-mail, as some people has already done. If you really want those problems, do it before 03Aug, or after 23Aug. Fedor.

1981 Putnam A2 was a twin question to 1988 Shortlist Q4.

About "the hardest shortlist problem", could you post it here? While in theory it's keep secret, sometimes we could just get them immediately after IMO. I'll try to ask our leader for that...

I don't know that shortlist 88 Q4 is an easy problem. The solution looks very difficult, therefore I thought that this is a very difficult problem.

Here's the difficult solution to this problem:
http://www.kalva.demon.co.uk/short/soln/sh884.html

Try [url=http://www.srcf.ucam.org/~jsm28/imo-scores/]Joseph Myer's web site.[/url]

the imo problem which was solved by the least number of students is 1996 , 5 , only 6 students. I don't know if this enables us to call it the hardest ever given...

I think this one:
Prove that there exists a convex 1990-gon such that all its angles are equal and the lengths of the sides are the numbers 12, 22, ... , 19902 in some order.

THere are no more chinese olympiads in kalva site

[quote="test20"][quote="Valentin Vornicu"]
about all others - for now the hardest problem remains IMO 2002/6, but this year on the ShortList a harder and more interesting / beautiful problem was submitted, but didn't get chosen.
[/quote]

Valentin, you posted this on Tue 22 Jul 2003, 08:25.
Which problem on the ISL-2003 did you have in mind?[/quote]

I am not entirely sure but I think he refers to the famed algebra problem 6 by Reid Barton. You can find an overview of the [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2003]ISL 2003 here.[/url]

[quote="Naphthalin"]i think the hardest problem of the german MO ist the following one:

a catrepillar sits in the middle of an $2n+1 \times 2n+1$-squares. in the other fields of the square there are the reciprokes of the numbers $1$ to $(2n+1)^{2}-1$. the catrepillar wants to get to one edge of the square and has to eat all the reciprokes to which fields it comes, but it cannot carry more than 2.
can it get do one edge for
a) $n=1001$
b) any $n\in\mathbb{N}$
?

even christian reiher didn't get 7 points for the problem ;)

Naphthalin[/quote]

You also may want to check the following [url=http://www.mathlinks.ro/viewtopic.php?p=3059#3059]link.[/url]

It seems that several of you are interested in collecting Chinese problems.  Since I live in Taiwan and am an alternate for the Taiwanese IMO team, I have many Chinese training materials.

Basically their system works as follows: first there is a simple (AIME level) test administered in each province.  Then each region (Northeast, Southeast, etc.) holds its own regional olympiad.  The best scorers in each region advance to the National Olympiad and the top ~40 students are sent to the IMO training camp (usually the cutoff is equivalent to a 29 on the IMO).  Then they eliminate members until they have their IMO team.

I have materials from old China National Olympiads (you can get a small pamphlet of those here for about 33 cents).  I also have their camp materials, national olympiads, girl's olympiad, etc. for years 2004 and 2006.

Just a warning: only recently have the problems for CMO and their training camps become somewhat nice.  Older problems tend to be very brute force and computational (not the problems that you really appreciate).

By the way, practicing for olympiads is about finding beautiful problems that use concepts in creative ways.  That being said, I wouldn't strongly recommend attacking Chinese problems until you are finished with recent ISLs and the USA, Russian, Romanian, or Bulgarian MOs; it's not like doing Chinese problems will automatically guarantee 42 points.

However, for those who are at that level (orl), feel free to request problems by means of PM or something.

[quote="fourierseries"]
I have materials from old China National Olympiads (you can get a small pamphlet of those here for about 33 cents).  I also have their camp materials, national olympiads, [b]girl's olympiad[/b], etc. for years 2004 and 2006.

Just a warning: only recently have the problems for CMO and their training camps become somewhat nice.  Older problems tend to be very brute force and computational (not the problems that you really appreciate).[/quote]

Can you explain what it is ? :?:  ?

[quote="kdano"][quote="fourierseries"]Well, the girl's olympiad is designed to encourage girls to participate in mathematical olympiads.  Not being gender biased, but the competition is usually less cutthroat than in the CMO.

The top 2 scorers in the girl's olympiad are allowed to enter the China IMO training camp![/quote]

But girls are allowed to participate in CMO, aren't they?[/quote]

Of course girls are allowed in CMO. Girl's olympiad is just another chance for top girls to join the national team.

In this article it is mentioned that " a INDIAN" wrote about Arthur .Can you please be more specific by mentioning who the Indian is? Well, I am from India.I never found a Indian book which mentions anything in this regard

Thanking YOu

[quote="Lek Huo Tang"]I think an Indian (I don't know whether it's Galois or Imam) wrote that Arthur Engel wrote problem 1988/6.

Arthur Engel wrote a training book for IMO. The book is an outgrowth of the German IMO training materials.

He was formerly the German coach. Word has it that he still is.

Details about the book

Problem-Solving Strategies
by Arthur Engel
ISBN 0-387-98219-1
ca. 400 pp
Topics include the Invariance Principle, the Extremal Principle, the Box Principle, Coloring Proofs, Enumerative Combinatorics, The Induction Principle, Functional Equations, Geometry (which is then divided further into 3 sections, viz. Vectors, Transformation Geometry, Classical Euclidean Geometry), Number Theory, Inequalities, Sequences, Polynomials, Games, Further Strategies
The price is about 50 USD. Can be procured from amazon.com[/quote]


geometry
ratio
trigonometry
parallelogram
similar triangles

$P$ is a point inside the triangle $\triangle{ABC}$. Lines are drawn through $P$ parallel to the sides of the triangle. The areas of the three resulting triangles with a vertex at $P$ have areas $4$, $9$ and $49$. What is the area of $\triangle{ABC}$?

First of all, notice that you have many similar triangles, which often tell you to run to ratios.

Now can you think of how to successively use these ratios to find the areas of various triangles until you eventually have [ABC]?

Okay, here's one possible first step.
Consider the area-side ratios. (Recall that the area ratios are the squares of the side ratios.)
The sides of the triangle of area 9 must be in a 3:2 ratio with the sides of the triangle of area 4. Thus, the area of the medium triangle containing those two triangles, and not the triangle of area 49 must be 25. You can get this by either saying $A=9*\left(\frac{2}{3}\right)^2$ of $A=4*\left(\frac{3}{2}\right)^2$.
Now that you've seen this, can you repeat the process to get [ABC]?

[hide]
If the ratio of the sides are $a/b$, then the ratio of the areas is $a^2/b^2$. The ratio of the areas are $4/9$, $9/49$, and $4/49$. The ratios of the sides are $2/3$, $3/7$, and $2/7$. So let $a=2$, $b=3$, and $c=7$. Then the side $AB=12$. The ratio of the side of the triangle with area 49 to the whole triangle is $7/12$ So the area ratio is $49/144$. The area is 49, so the area of $ABC$ is 144.
[/hide]

[quote="eminem"]hey chess
if you got the ratio of sides say $a/b$= 2/3 and $b/c$ = 3/7 then how did you assume a = 2 , b=3 and c=7 that would be just 1 case in which a/b or b/c cant be simplified further?[/quote]

Suppose you construct a line $QR$ parallel to $AC$ where $Q$ is on $AB$ , $R$ is on $BC$ . Construct another line $ST$ parallel to $AB$ where $S$ in on $AC$ and $T$ is on $BC$ . Then , construct line $UV$ parallel to $BC$ where $U$ on $AB$ , $V$ on $AC$ .

Ok now , you know the ratio of $SP:QU$ = $ 2:3$ as been shown by $chess64$ . And $QU:PT$ = $ 3:7$ . Since $SP=AQ$ and $PT=UB$ then we can say that
$AQ:QU$ = $2:3$ and $QU:UB$ = $3:7$ . From here , you can assume $AQ$ = $2$ ; $QU$ = $3$ ; $UB$ = $7$ . Since you are only meant to find the area , the real dimension doesnt really affected but instead it is affected by the $ratio$  :D .So theres nothing wrong by assuming it as 2,3,7 or if you want , simply asumme their respective length as $2k$ , $3k$ , $7k$ and you will also get the same answer . ;)

[quote="frt"][quote="nat mc"][quote="chess64"]So let $a=2$, $b=3$, and $c=7$. Then the side $AB=12$.[/quote]
Please explain, how you went from the first part to the second part of this statement[/quote]
By $a=2$, $b=3$ and $c=7$, chess64 was referring to the corresponding sides of 3 triangles, I think. (So one triangle has side length $2$, another one has $3$, and the last one has $7$)
Then $a+b+c=AB$, so $AB=2+3+7=12$ ;)[/quote]
Please explain why you would sum the sides of the triangle

[u][b]nat mc[/b][/u] well $BC=12k$ where $12$ is a factor of the side. Do you understand, Chess 64 is taking all the bottom sides into consideration and he is finding the ratio of the sides, because he knows the ratios of the areas.[/u]

[quote="nat mc"][/quote][quote="frt"][quote="nat mc"][quote="chess64"]So let $a=2$, $b=3$, and $c=7$. Then the side $AB=12$.[/quote]
Please explain, how you went from the first part to the second part of this statement[/quote]
By $a=2$, $b=3$ and $c=7$, chess64 was referring to the corresponding sides of 3 triangles, I think. (So one triangle has side length $2$, another one has $3$, and the last one has $7$)
Then $a+b+c=AB$, so $AB=2+3+7=12$ ;)[/quote][quote="nat mc"]
Please explain why you would sum the sides of the triangle[/quote]
Oh sorry, it was a bad explanation.

See the figure that I attached. Let's say $DE=2$, $IP=3$ and $PF=7$. Then, $DE+IP+PF=DE+AD+EB=AB=12$ :)
Is that clear?

some of us understood it ;)  it wasnt that bad.

In case there is still confusion...let me try to explain it better...

[hide]
First, let's look at an important preliminary result for similar triangles.  Consider similar triangles $\triangle PQR$ and $\triangle XYZ$.  Since these triangles are similar there is a constant ratio between the sides of one triangle to the respective sides of the other triangle.  Call this ratio $k$ such that if you take a side length from $\triangle PQR$ multiplying it by $k$ gives you the side length of the corresponding side in $\triangle XYZ$.  I claim that the ratio of the [i]area[/i] of the triangles is $k^2$.  There are multiple ways to see why this is so.  I'll show one (brackets denotes area):

$[XYZ]=XY\cdot YZ\cdot \sin \angle XYZ$ and $[PQR]=PQ\cdot QR\cdot \sin \angle PQR$.

But, since the triangles are similar we have that $m\angle XYZ=m\angle PQR$, $YZ=k(QR)$ and $XY=k(PQ)$.  Thus \[[XYZ]=k^2\cdot PQ\cdot QR\cdot \angle PQR=k^2\cdot [PQR].  \blacksquare\]

Now onto the problem.

Using frt's diagram, the first thing to note is $\triangle ABC\sim \triangle DEP\sim \triangle IPH\sim \triangle PFG$.  The three smaller triangles have areas in the ratio of $4:9:49$ thus the ratio of their sides will be the square root of that, i.e. $2:3:7$.  So we can say that $PH=2k$, $EP=3k$, and $FG=7k$.  Since, $EPFB$ and $PHCG$ are parallelograms, $EP=BF$ and $PH=GC$ (you can just think of this as sliding the smaller triangles down).  Thus $BC=EP+GF+PH=12k$.  So then,

\[\frac{PH}{BC}=\frac{2k}{12k}=\frac 16\Rightarrow \frac{[IPH]}{[ABC]}=\frac 1{36}.\]  Finally, since $[IPH]=4$, this makes $[ABC]=144.$[/hide]


find all continuis function like $f$ ,$f:R-->R$ .$x<>y$ and:
$f(\frac {x+y}{x-y}) = \frac {f(x)+f(y)}{f(x)+f(y)}$

Maybe there is a mistake in the right, since now the right is 1.

I've edited the post to see the fractions. I think this is a $-$ instead of a $+$ at the right, but I'm waiting for the proposer...

Pierre.

I already solved this one. In the solved forum probably you can find it. Wasn't it in a TST in Singapore or something like this?


Assuming that a,b,c are odd numbers, Prove that:
$ ax^2\plus{}bx\plus{}c\equal{}0$ has no rational solutions.

This is not really a Classroom Math-level question.

This particular problem has shown up several times in Intermediate Topics.

Consider $ D\equal{}b^2\minus{}4ac$. We shall show that $ D$ is not a square number.

Since $ a, b, c$ are odd, we let $ k\equal{}ac$ and $ b\equal{}2n\minus{}1$.
Then $ k$ is odd and $ b^2\equal{}4n^2\minus{}4n\plus{}1$. Thus

$ D\equal{}4n^2\minus{}4n\plus{}1\minus{}4k\equal{}4(n^2\minus{}n\minus{}k)\plus{}1$.

No matter $ n$ is even or odd, $ n^2\minus{}n$ is even. So $ n^2\minus{}n\minus{}k$ is odd.
For any $ m\in\mathbb N$, we have $ m^2\equiv0$ or $ m^2\equiv1\pmod 8$.
But it is clear that $ D\not\equiv0$ or $ 1\pmod8$.
Thus $ D$ is not a square number.

This might help

X=[-b(+or-)the square root of (b squared - 4ac)]/2a

srry i don't know latex


function
algebra
domain
geometry
3D geometry
LaTeX

(A) Find Domain
(B) Find Range
(C) Determine any symmetries that are characteristics of the graph

(1)  y = cube root of 1-x^2

(2)  y = x^3/2

Here it is in $ \text{\LaTeX}$:

$ 1.\ y =\sqrt [3]{1-x^{2}}$

$ 2.$  For this one, do you mean, $ y =\frac{x^{3}}{2}$, or $ y = x^{\frac{3}{2}}$?

[hide="Solution"]1. When graphed, this looks like a semicircle.
A. Domain is R.
B. Range is $ y\leq1$
C. This graph is symmetric with respect to the y-axis.

2. If $ y =\frac{x^{3}}{2}$,
A. Domain = R
B. Range = R
C. This graph is symmetric with respect to the origin.

If $ y = x^{\frac{3}{2}}$, also $ \sqrt{x^{3}}$,
A. Domain is $ x\geq0$
B. Range is $ y\geq0$
C. This graph does not have any symmetries.[/hide]

EDIT: Range on #1. Thanks krsattack for pointing out my mistake. I forgot that $ x^{2}\geq0$...

For number 1, I would check your range...  Here's also a little more in depth solution.
[hide="Number 1"]
What do we know about this function?  First of all, let's consider the cubed root.  The cubed root can be taken of any real number, so we know that it does not matter what $ 1\minus{}x^{2}$ is, as long as it is real.  Thus, x is also real, so the (A) domain is R.

For part (B), just consider $ 1\minus{}x^{2}$.  Since $ x^{2}\ge 0$, then $ 1\minus{}x^{2}\le 1$.  Thus, we are taking the cubed root of some number less than or equal to one for all x.   Thus, the range (B) is $ y\le 1$

For (C), if we plug in -x for x, we get the same answer, so the function is even and symmetric about the y-axis.[/hide]

The other answers I am pretty sure are correct.

How do you make that weird R that means all real-numbers??


Which calculator is better to buy?
TI-Nspire, TI-Nspire CAS, Voyage 200.
(compare by usefulness, price, and so on)
Thank you!

CAS definitely. I think. Voyage 200 helps get you past boring lectures though (game capacity xD).

[quote="hunter34"]No. TI-89 Titanium.

It destroys Calculus and Engineering.[/quote]
Could you specific what TI-89 Titanium can do that makes it better than Voyage 200 and TI-Nspires?

The TI-1 is most useful :noo:  :spam:  :play_ball:


Let $ ABC$ be an acute triangle with the orthocenter $ H$ , the incircle $ w\equal{}C(I,r)$ and the circumcircle $ C(O)$ .

Denote the point $ D$ where the circle $ w$ touches the sideline $ BC$ . Prove that $ AH\equal{}r\ \Longleftrightarrow\ OI\equal{}OD$ .

Nice [b]feliz[/b] :) let me prove this fact. Consider the nine-point circle of triangle $ ABC$. We know that it's center is the midpoint of $ OH$. Let $ M$ be the midpoint of $ BC$ and $ K$ the midpoint of $ AH$ (it's known that $ K$ lies on the nine-point circle of $ ABC$). $ AH\perp BC$ implies that $ MK$ is diameter of the nine-point circle of triangle $ ABC$, hence using $ MO\parallel{}AH$ we obtain $ \Delta NHK \equiv \Delta MNO$ where $ N$ is the nine point center. Using that we get $ MO\equal{}HN\equal{}\frac{AH}{2}$ :)


vector
linear algebra
linear algebra solved

$V$ is an $n$-dimensional vector space. Can we find a linear map $A : V  \to V$ with $n+1$ eigenvectors, any $n$ of which are linearly independent, which is not a scalar multiple of the identity?

Of course not :).

Let $x_1,\ldots,x_n$ be $n$ of those vectors, with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n$. The $n+1$'th vector must have the form $x=\sum_{i=1}^n a_i x_i$, where all the $a_i$'s are non-zero. Let $\lambda$ be the eigenvalue corresponding to $x$. We have $Tx=\sum a_i\lambda_i x_i=\sum a_i\lambda x_i$, meaning that $\sum a_i(\lambda_i-\lambda)x_i=0$, which is equivalent to $\lambda_i=\lambda,\ \forall i$. This means that all the eigenvalues of $T$ are equal. Since $T$ is diagonalizable, we get the conclusion: $T=\lambda I$.

Look at the definition of eigenvector, it is not allowed to be zero.


A sequence $ (x_n)$ is given as follows: $ x_0,x_1$ are arbitrary positive real numbers, and $ x_{n\plus{}2}\equal{}\frac{1\plus{}x_{n\plus{}1}}{x_n}$ for $ n \ge 0$. Find $ x_{1998}$.

[hide]Let $ x_0 \equal{} a$ and $ x_1 \equal{} b$.

Using the recursion formula yields:
$ x_2 \equal{} \dfrac{1 \plus{} b}{a}$, $ x_3 \equal{} \dfrac{a \plus{} b \plus{} 1}{ab}$, $ x_4 \equal{} \dfrac{1 \plus{} a}{b}$, $ x_5 \equal{} a$, $ x_6 \equal{} b$.

Since the recursion is a 2nd order recursion, and $ (x_0,x_1) \equal{} (x_5,x_6)$, we have $ x_{n \plus{} 5} \equal{} x_{n}$ for all $ n \ge 0$.

Therefore, $ x_{1998} \equal{} x_{1993} \equal{} \cdots \equal{} x_{8} \equal{} x_{3} \equal{} \dfrac{x_0 \plus{} x_1 \plus{} 1}{x_0x_1}$. [/hide]


[hide="Click to reveal hidden content"]I got $5+7+11+13+17=\boxed {53}$ I don't see how you get it besides random checking, but I guess it isn't that difficult to factor.

Start by dividing by 5 because it ends in 5.
Then divide by 11 since I noticed on the spot it follows the 11 divisibility rule.
Divide by 7 since it also follows the 7 divisibility rule.
Then divide by 13 since it's obvious then.[/hide]

[quote="236factorial"]Starting by dividing by 85 is easier. You get 1001, which should be memorized (well, if you're into math competitions) as 91*11, and 91 can be factored further and etc.[/quote]

I don't have something like that memorized, but it wouldn't be hard for me to realize.

And yeah, I do see why dividing by 85 would've been much easier, but division by 5 is impulse  :P

[quote="236factorial"]What is the sum of the prime factors of 85085?[/quote]

[hide] 85085=$5*7*11*13*17$. Then, the answer is $5+7+11+13+17=\boxed {53}$.[/hide]

[quote="Zenith Xan"][quote="236factorial"]Starting by dividing by 85 is easier. You get 1001, which should be memorized (well, if you're into math competitions) as 91*11, and 91 can be factored further and etc.[/quote]

I don't have something like that memorized, but it wouldn't be hard for me to realize.
[/quote]

1001 is actually easy to recgonize as a multiple of 11 because the alternating sum of its digits is 1-0+1-1=0, which is 11*0.

[hide]
First divide by 85
We now have 1001 left.
I know that 7x143=1001, it is really important:memorize it
Now we have 143.
Easy to see that its 13x11, since it is 1 less that $12^2$
Factoring 85:
Divide by 5 to get 17
Thus the prime factors are
$5+7+11+13+17=53$[/hide]


if 55(x 2 +y 2 )  <= 229
we have x=0,y=1,or x=1,y=0,or x=1,y=1  or x=0,y=2,x=2,y=0
then no one is right
if 55(x 2 +y 2 )  >229
we have 55x 3 y 3 <229xy 3
then x=0(impossible)x=1 or x=2. or y=0(impossible)
then it is easy. :)


integration
calculus
function
derivative
real analysis
real analysis unsolved

Let $F(x)=\frac{1}{h^{2}}\int_{0}^{h}d\xi\int_{0}^{h}f(x+\xi+\eta)d\eta$  ($h>0$)
And $f(x)$ is a continuous function. Find $F^{''}(x)$

Let $G=h^{2}F$ (I don't want to write $\frac{1}{h^{2}}$ several times.) Begin by differentiating $G$ [i]formally[/i], with no justification: $G''(x)=\int_{0}^{h}d\xi\int_{0}^{h}f''(x+\xi+\eta)d\eta=\int_{0}^{h}(f'(x+\xi+h)-f'(x+\xi))d\xi=f(x+2h)-2f(x+h)+f(x)$.
Now that we have the answer, how do we justify it? One way is to find a second antiderivative of $f(x+2h)-2f(x+h)+f(x)$ and compare it to $G(x)$ (using Fubini's theorem), but there are other possibilities.

(I prefer the notation that puts the $d\xi$ at the end; it emphasizes that the outer integral depends on the inner.)

$F''(x)=\frac1{h^{2}}(f(x+2h)-2f(x+h)+f(x))$.
If $f$ is $C^{2}$, we can see this by simply differentiating with respect to $x$ and integrating the result. For the general case, we need a different argument.

Let $\delta g(x)=\frac1h(g(x+h)-g(x))$ for any function $g$; we want to show that $F''(x)=\delta^{2}f(x)$.
Consider $G(x)=\int_{0}^{x}\int_{0}^{y}\delta^{2}f(z)\,dz\,dy$. By the FTC, $G''(x)=\delta^{2}f(x)$.
Now, $\int_{0}^{y}\delta^{2}f(z)\,dz=\frac1h\left(\int_{h}^{y+h}\delta f(z)\,dz-\int_{0}^{y}\delta f(z)\,dz\right)= \frac1h\left(\int_{y}^{y+h}\delta f(z)-\int_{0}^{h}\delta f(z)\,dz\right)$; this is simply a matter of splitting the integral and translating one piece. In the rest, linear substitutions are made freely

Next, we look at the outer integral; let $\frac1h\int_{0}^{h}\delta f(z)\,dz=c$, since it is constant.
$G(x)=\int_{0}^{x}\left(\frac1h\int_{y}^{y+h}\delta f(z)\,dz-c\right)\,dy =\frac1h\int_{0}^{x}\int_{0}^{h}\delta f(y+\xi)\,d\xi\,dy-cx$
$=\frac1h\int_{0}^{h}\int_{0}^{x}\delta f(y+\xi)\,dy\,d\xi-cx$ $=\frac1{h^{2}}\int_{0}^{h}\left(\int_{x}^{x+h}f(y+\xi)\,dy-\int_{0}^{h}f(y+\xi)\,dy\right)\,d\xi-cx$
Let $\frac1{h^{2}}\int_{0}^{h}\int_{0}^{h}f(y+\xi)\,dy\,d\xi=b$; that is also constant. We get
$G(x)=\frac1{h^{2}}\int_{0}^{h}\int_{x}^{x+h}f(y+\xi)\,dy\,d\xi-b-cx$ $=\frac1{h^{2}}\int_{0}^{h}\int_{0}^{h}f(x+\eta+\xi)\,d\eta\,d\xi-b-cx$
$= F(x)-b-cx$ by Fubini's theorem.

Since the difference between $F$ and $G$ is linear, they have the same second derivative. This argument could easily be extended to more deeply nested integrals, higher derivatives, and higher differences.


function
algebra
domain
calculus
calculus computations

I have a Laplace transfer function in the following form:

Y(s) = a*X(S) / [ a - b*Z(S) + cS]

I need to write it in the following form:

Y(s) = H1(S)*X(S) + H2(S)*Z(S)

I just can't find H1(s) and H2(S), any help ???

First of all we don't know if this expansion is possible generally. So we will assume that this expansion is feasible always and see if we can derive any kind of useful conclusions.

So, assuming that there exist $H_{1}(s)$ and $H_{2}(s)$ so that

$Y(s) = \frac{a X(s)}{a+c s-b Z(s)}= H_{1}(s) X(s)+H_{2}(s) Z(s)$

we would get the following equation:

$a X(s) = [a+c s-b Z(s)]H_{1}(s) X(s)+H_{2}(s) Z(s) [a+c s-b Z(s)]$

In this equation we can make whatever choice we find suitable either for $H_{1}(s)$ of $H_{2}(s)$. Let's say that we choose $H_{1}(s)$. Then we can solve the above equation for $H_{2}(s)$ acquiring:

$H_{2}(s) = [\frac{a}{a+c s-b Z(s)}-H_{1}(s)] \frac{X(s)}{Z(s)}$

So we got two formulas. One for $H_{1}(s)$ and one for $H_{2}(s)$. All this with the assumption that the initial expansion is feasible for all $a,b,X(s) \: and \: Z(s).$But not only this assumption didn't make us arrive to a contradiction but it offered us a solution as well. So we could say from this that the initial question concerning whether this expansion is feasible is true.

If we want to be more exact we could say the following: From the above analysis we found out that if the initial expansion is possible, then a solution that would hold true for all $a,b,X(s) \: and \: Z(s)$ would be given by the above formulas.

We go now to the expression

$H_{1}(s) X(s)+H_{2}(s) Z(s)$

Substituting $H_{2}(s)$ from the above formula and doing the operations we arrive at the relation

$H_{1}(s) X(s)+H_{2}(s) Z(s) = \frac{a X(s)}{a+c s-b Z(s)}$

So we can conclude that this expansion is always possible for every $a,b,X(s) \: and \: Z(s)$ where we can chose whatever we like for $H_{1}(s)$ and make $H_{2}(s)$ equal to:

$H_{2}(s) = [\frac{a}{a+c s-b Z(s)}-H_{1}(s)] \frac{X(s)}{Z(s)}$

To be more correct, we might have to exclude from the domain of $Y(s)$ the values of $s$ that do not belong to the domain of $H_{1}(s)$ (if any) and those that do not belong to the domain of $\frac{1}{Z(s)}$ although I think that the second isn't necessary since $\frac{1}{Z(s)}$ is multiplied by $Z(s)$ in the final expression of the expansion and the whole quantity that is produced by the multiplication (at least it's limit) is bounded and doesn't show any problems at those values.


97 Vestavia Comprehensive

#5. What ist hephase shift of y=5sin(3x) - 12 cos (3x)?

Thanks in advance!

[hide]$ 5\sin(3x) \minus{} 12\cos(3x) \equal{} B\sin(A \plus{} 3x) \equal{} B\sin(A) \cos(3x) \plus{} B\cos(A)\sin(3x)$

Matching coefficients, we get $ B\sin(A) \equal{} \minus{} 12$ and $ B\cos(A) \equal{} 5$. So, $ \tan(A) \equal{} \frac {\minus{}12}{5}\implies A \equal{} \minus{} 1.176$. $ B \equal{} \frac {5}{\cos(A)} \equal{} \frac {5}{\cos(\minus{}1.176)} \equal{} 13.000$.

So, $ 5\sin(3x) \minus{} 12\cos(3x) \equal{} 13\sin(3x \minus{} 1.176)$.

I'm not sure what you mean by phase shift though. Is it just $ \minus{} 1.176$? [/hide]

It would actually be $ \frac{1.176}3$, the horizontal shift from the graph $ y\equal{}13\sin(3x)$.


Gauss
Euler
geometry
geometric transformation
reflection
calculus
3D geometry

In your mind , who are the top 3??  just wondering

mine is
(1) Gauss (I amire him after the gauss lemma, hands down)
(2) Euler (he is everywhere, you can find his name in all area of math)
(3) Galois (i am wondering how great he could be if he didnt die so young)

Uh...Erdos if we are talking most prolific and most genius. He is my idol.  :)

How about some of the modern guys: Wiles, Mazur, Perelman (we should include him), Nash, Shimura, Yao,...

Hörmander?

It's conditioning, but if you look at our library it seems like all analysis was written by him

I've never heard of him.  You said if you look at your library...  Writing a lot of texts doesn't reflect greatness.

Did you even google the name? That he was great is not really debatable. But of course I know more about him because he is associated with my university, and a lot of the content of our analysis courses and research is his legacy.

[quote="JRav"]I've never heard of him.[/quote]

this just means that your exposure to any sort of advanced mathematics has been very limited.

e: i mean he's a fields medalist and an absolute giant in PDE and analysis

I'll admit that I am not very exposed to higher level math yet.

[quote="JRav"]I would agree and put Gauss 1, Euler 2.  After that, it's a toss-up for me.  Obviously, you have to give Newton huge credit.  Riemann, Lagrange, Laplace, Abel, et al show up everywhere.  Ramanujan was pretty amazing, as was Galois.[/quote]

If you're giving Newton huge credit, you're going to want to give at least as much credit to everyone he FAILED to credit himself.  In particular, Leibniz, but Hooke is another big one that makes me lose almost all respect for Newton.

Cauchy, Euler, and Gauss show up all across the spectrum.  However, if we want to talk about real work done to truly push forward the field of mathematics, we're going to have to look at the *foundations* that were developed primarily by mathematicians like Aristotle, Descartes, Pascal, Leibniz, Peano, Boole, Frege, and Russell.   Most of their work wasn't with math itself but in the theory and philosophy of mathematics.

(Newton, of course, has his Principia Mathematica, but that was more of a survey and summary of the ideas of the time that had already been laid out by his predecessors.  As far as I can tell, he himself added nothing particularly novel to the foundation of mathematics.)

[quote="Functormorphism"]However, if we want to talk about [b]real[/b] work done to truly push forward the field of mathematics[/quote]

[quote="Functormorphism"][b]mathematicians[/b] like Aristotle, Descartes, Pascal, Leibniz, Peano, Boole, Frege, and Russell.[/quote]

lol

[quote="Functormorphism"][b]Most of their work wasn't with math itself[/b] but in the theory and philosophy of mathematics.[/quote]

double lol

[quote="Functormorphism"]As far as I can tell, he himself added nothing particularly novel to the foundation of mathematics.[/quote]

triple lol

(all emphasis mine of course)

e: i mean really, i guess there is some legitimacy to your point of view, but it's silly to say that the 'real work done to truly push forward the field of mathematics' was done by people who the vast majority of professional mathematicians would not list among the top 10 mathematicians in history, and doubly silly to say that newton lacked originality.  it's well known that some of his predecessors had a rudimentary grasp of the differential calculus and also understood some basic principles of physics; that doesn't undermine what newton accomplished nor does it diminish the significance of the principia.

If you want someone modern, my vote's for Terence Tao.

[quote="kelvin911"]Who are the top Mathematicians ever?[/quote]
Me of course.

How come not many people here mention Leibnitz? Ofcourse Gauss and Euler can never be forgotten, especially considering Gauss's contribution to other subjects as well. :)

It's interested that everyone is always so hung on up the awesomeness of Gauss and Euler and Newton. You should keep in mind that they aren't only famous because of mathematical discoveries. There are a lot of factors that contribute to someone from that long ago being famous:

- volume of publication
- having good (and numerous) students / sphere of influence
- writing lots of letters to other mathematicians of the time
- political favor at the time
- having a good position at a university (correlated w/ the prev. one, sometimes)

These things can have a huge impact on history's memory of a mathematician, regardless of the actual quality of math that he produces. They can result in a particular guy getting credit for things that were actually joint work, or rediscovered by the guy getting credit, or whatever. (A great example is the Newton vs. Leibniz thing -- depending on what historical sources you read, you could have an opinion as far to one side as that Leibniz did all the work in developing calculus and Newton was just incidental, or you could be all the way to the other side and say that Leibniz' only contribution was having better notation than Newton and teaching it in Europe.)

You also have to consider by what standard you're measuring things -- a number of those "greatest mathematicians" (including the big 3 I mentioned at the beginning of my post) lived for a long time and were career mathematicians. If you are using any kind of non-relative standard -- "most new math produced", "nicest new math produced", "most significant contribution to maths as a field", etc. -- you are hugely favoring those sorts of people. But an argument about whether Galois and Abel should be counted in this kind of list is hard to resolve without considering more carefully your criteria about "greatness".

How about "changing the way people think about math"? Then it might be worthwhile to bring up Fibonacci, who did almost no original work, but was the main force behind spreading Hindu-Arabic numerals in Europe. Cantor was huge, too, by this standard. One could argue for Grothendieck, as he had a huge role in the general approach of abstractness and generalization that's so popular in maths right now. Erdos was a big deal too; not only did he change how people think about combinatorics by introducing a lot of novel techniques there, but he also significantly changed the culture of mathematics. Just look at what proportion of papers were co-authored before Erdos and what proportion after, and you'll see. (You could argue that it's also because new results are harder to obtain individually now, because they often require wide as well as deep knowledge, but I think it's part of Erdos' influence that these sorts of results are more publishable than very esoteric ones requiring only deep knowledge of a particular subject.) I'd still vote for Terry Tao, but it's kind of too soon to say how much of an impact he'll really have. He's definitely a leading figure in the new trend of "unifying" mathematics -- he's a proponent of the view that new ideas showing connections between different structures in maths and results that use techniques from different areas in maths are among the most amazing and beautiful things we can come up with, and that's been getting popular lately (probably thanks in large part to what's going on in arithmetic combinatorics, though that's not the only example; really this has been going on as early as Riemann, who thought to bring the big guns of complex analysis to bear on NT problems).

Anyway, this is getting long. I think it's an interesting topic, though. :) I'd like to see a thread on "what makes a great mathematician" rather than "who were the greatest mathematicians".

Certainly not the greatest of all. But a real badass

[size=200][b]J.P. Serre[/b][/size]

[quote="Xevarion"] (really this has been going on as early as Riemann, who thought to bring the big guns of complex analysis to bear on NT problems)[/quote]

[url=http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions]dirichlet's theorem[/url]?

[quote="blahblahblah"][url=http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions]dirichlet's theorem[/url]?[/quote]
Well yes, and Euler did stuff relating analysis and NT even earlier, but it seems to me that Riemann was the one who turned it from something a few dudes did into an entire field of study.

If you want modern, what about Wiles or Perelman?

Again, I think it's important to make a distinction between the people who [i]proved[/i] the major result and the people who had sufficiently good vision and intuition to tell later mathematicians where to look. Remember that what Perelman actually proved was [i]Thurston's[/i] geometrization conjecture and what Wiles actually proved was part of the [i]Taniyama-Shimura[/i] conjecture. I would argue that the people for whom the conjectures are named deserve perhaps even more credit, from the perspective of changing the way that people think about the particular field and guiding future mathematicians in the right direction to find the proof. That's not to say that the accomplishments of Wiles and Perelman aren't formidable; I'm just trying to point out that there's a big difference between ranking mathematicians by the quality of their [i]theorems[/i] and the quality of their [i]ideas[/i].

I think it's fair to call dirichlet's theorem the first big theorem in number theory which was proved by complex analytic methods, and the result which jump-started the field as a whole.

[quote="Xevarion"] I'm just trying to point out that there's a big difference between ranking mathematicians by the quality of their [i]theorems[/i] and the quality of their [i]ideas[/i].[/quote]

I definitely agree, so at the same time I would dare say the ideas needed to prove these great theorems might outweigh the ideas of the theorems themselves. I don't know for sure since I don't know either of these proofs, but if they had been at the level or easier to come up with than the theorems, the conjecturers would have provided proofs.

[quote="HilbertThm90"]if they had been at the level or easier to come up with than the theorems, the conjecturers would have provided proofs.[/quote]
I don't know if it's necessarily true. I think it requires a certain brilliance in order to recognize that some general statement "should" be true, or that certain things "should" be related. Proving these things may require very difficult ideas too, but it may not be the same at all. Let me give an analogy. Mining is difficult. It requires a lot of technical knowledge and skills and appropriate use of equipment in order to be good at it. But someone who is remarkably good at gold mining will only actually get rich if he knows where there is some gold lying underground. Are you saying that someone who can figure out where gold is likely to be is not as skilled as the miner, if he can't also mine? It seems even more amazing to me if someone who doesn't know anything about mining techniques is still able to accurately point out where a gold mine is, and only later on when someone with a bunch of technology and the knowledge to use it comes by and follows the directions is the gold actually recovered!

The Poincare conjecture is probably a good example of this. Poincare surely couldn't have imagined all of the topological technologies that were developed in the past 100 years. And the proof of the conjecture is highly intricate and complicated and took many months to achieve acceptance in the maths community. So isn't it amazing that Poincare had such well-developed intuition for topology that he was able to guess this remarkable fact? Maybe I am overestimating the importance of this, but where would problem-solvers be if no one was giving them more conjectures to work on?

I agree that Poincare, Taniyama, Shimura, Mazur, Ribet, Frey should all be included.  What about Sophie Germain?  Has anyone mentioned her?

Haha! Now have we changed our goal from deciding a few top mathematicians to listing all of the ones who are above a certain level? BTW, I think Sophie Germain's fame is influenced by the fact that she was one of the first woman mathematicians, not just by her accomplishments. If you want to talk about impressiveness of a person, she is quite deserving, because it was certainly not easy for a woman to become a mathematician at the time! But if you want to talk only about mathematical work, then I don't think she is all that special.

edit: why are people so fascinated by this anyway? Lots of people keep coming to this thread, but no one wants to do anything but compare different mathematicians and name more. How do we benefit from this? If we want to talk about great mathematicians of the past, we should at least talk about what made them great, so that we might emulate them. I am sure that Gauss and Euler did not earn their status as two of the most renowned mathematicians of all time by arguing with their friends about whether Descartes was more influential than Pascal.

Ah, but I can come up with exactly the same type of examples to support my claims. (Not that I would ever officially argue against what you say, this is merely for the sake of argument and advancing a diverse opinion in the thread).

Often history obscures whether a particular person's ideas were great or not. Take, for example, the case of the miner. It often appears that the non-miner predicted with remarkable accuracy where the gold was, but what really was happening is that the person predicting randomly guessed 10,000 spots. Then almost immediately the actual miners tried all but a couple spots (due to really difficult mining techniques in extremely hard rock on the side of a mountain or something...). Now all the predictions we see are a couple of the left-over spots. It is forgotten how many random wrong guesses there were due to the quickness and ease in which they were disproven.

Now it is highly unlikely that with 10,000 random guesses that all 10,000 will be completely wrong. Soon all that is left is one large and almost impossible place to mine. Now it is likely that due to the impossibility of the mining that something special is going on there. Now the genius miner develops a new technique and is able to uncover the thing that was there.

OK. So that was a little over the top with the analogy. But I think you see the point.

[quote]I am sure that Gauss and Euler did not earn their status as two of the most renowned mathematicians of all time by arguing with their friends about whether Descartes was more influential than Pascal.[\quote]

I remember reading that Gauss was not particularly friendly so it would be sort of ironic if he discussed anything with friends (or even had any). :lol:
[/quote]

[quote="Xevarion"]So isn't it amazing that Poincare had such well-developed intuition for topology that he was able to guess this remarkable fact? Maybe I am overestimating the importance of this, but where would problem-solvers be if no one was giving them more conjectures to work on?[/quote]

Poincare's conjecture seems like a rather natural question to ask in its proper historical context (http://en.wikipedia.org/wiki/Poincaré_conjecture#History).  Poincare also doesn't seem to be responsible for the conjecture in higher dimensions.

many of the names listed in this thread reflect a lack of appreciation for the general state of modern mathematics and are biased towards names associated with well-known 'big' problems or names generally well-known to typical undergraduate math majors for one reason or another (e.g., JRav listing almost everyone who would have been mentioned in a book or a documentary on the solution of fermat's last theorem; one would probably have more success picking names at random from a list of fields medal winners).

it inevitably comes down to one's tastes, but some names that a large portion of working mathematicians would rank among the top 10 mathematicians of all time probably include euler, newton, leibniz, poincare, riemann, hilbert, archimedes (there probably are some more french mathematicians (cauchy? lagrange?) there but their names all sound alike and i forget exactly who accomplished what)

some more modern names i would mention are von neumann, grothendieck, and erdos, and, while it's probably too early to put tao up there, i'm certainly tempted to.

I agree with blahblahblah that at least for recent mathematicians, naming random fields medalists is probably going to serve you better than pulling names out of memory for no reason. I can name a ton of mathematicians, but most of them are really not particularly important, and I just remember them because I learned about some very specific things that they did. Nobody would ever name, say, Tartaglia as one of the most important mathematicians of all time...

Also, since apparently this thread is still mainly about throwing names out, I don't think Hardy & Littlewood have been mentioned yet (or maybe someone mentioned them along with Ramanujan? I don't know), and for modern people I'd be tempted to put in Bourgain. He's a prolific problem-solver. I remember someone made a joke at one of the arithmetic combinatorics lectures -- someone asked Bourgain to give some bound that was uniform in one of the parameters (till then all known bounds involved that parameter), and the person said something like "It was a hard problem -- it took Bourgain two weeks!" But maybe he is just focused on the particular field that I know stuff about; I'm not really sure.

Fatou's lemma is one of my favorite theorems (so extremely simple and so powerful). But who was Fatou?

I am currently studying Hardy spaces, and there Fatou's lemma is used all the time to pass limits to the boundary of the unit circle. In the first paragraph of the MacTutor biography it says Fatou developed some theorems about these things that lead to Privalov's and Riesz theorems in this connection. Does anyone know if Fatou's lemma was a lemma for these things? Ohps... off-topic  :lol:

[quote="Xevarion"]why are people so fascinated by this anyway?[/quote]

Strange to see this comment from the [i]single[/i] most prolific poster in the thread...

[quote="Xevarion"]no one wants to do anything but compare different mathematicians and name more. How do we benefit from this? If we want to talk about great mathematicians of the past, we should at least talk about what made them great, so that we might emulate them.[/quote]

I think it's safe to say that there are reasons to talk about great people besides being great yourself.  But yes, I agree that this sounds like an interesting topic.  If you have stories to tell about the greats, tell them by all means, instead of complaining.

I think that the discussion of modern mathematicians as being among the greatest mathematicians EVER is a little nuts.  EVER is a long long time.  We need to wait about 100 years (no exaggeration) before we can assess whether guys like Perelman or Tao deserve to be mentioned in the same breath as ridiculous giants such as Newton or Gauss.  (We still need at least 50 years before we can be certain about the legacy of guys like Grothendieck or Serre.)  Most mathematicians have deep respect for mathematical history and would probably be horrified at being compared to Newton.  Newton for God's sakes!

A related issue is that mathematics has gotten too big for individual mathematicians to achieve the wide-ranging greatness that used to be possible.  Asking who the greatest living mathematician in the world is right now is like asking who the greatest living *scientist* in the worlds was 200 years ago.  We're sure to see this in the future as the awarding of Fields Medals becomes less and less predictable.  My own rule of thumb would be that the most recent mathematicians who have a serious claim in the "pantheon" (however inclusive it is) would be Hilbert and Poincare, but I'm not dogmatic on this point.

Btw, I encourage anyone who is interested in this thread to read at least one book on the history of mathematics (if you haven't already).  This is especially instructive once you've had an undergrad education under your belt and you can appreciate more of the discoveries.  Unfortunately, I don't have recommendations, but please avoid Men of Mathematics.

Another note:  Try to avoid evaluating the greatness of mathematicians by looking at names of theorems.  This doesn't work for several reasons.

Note to blahblahblah and others:  Talking about Poincare's role in the Poincare Conjecture is a little bit silly when you consider that the man essentially single-handedly INVENTED the entire field of algebraic topology.  (Now let's try to think of a recent mathematical accomplishment that compares to inventing algebraic topology.  Any takers?)

[quote="yenlee"]
Note to blahblahblah and others:  Talking about Poincare's role in the Poincare Conjecture is a little bit silly when you consider that the man essentially single-handedly INVENTED the entire field of algebraic topology.  (Now let's try to think of a recent mathematical accomplishment that compares to inventing algebraic topology.  Any takers?)[/quote]Lafforgue's proof of the Langlands program?

[quote="yenlee"]Note to blahblahblah and others:  Talking about Poincare's role in the Poincare Conjecture is a little bit silly when you consider that the man essentially single-handedly INVENTED the entire field of algebraic topology.  (Now let's try to think of a recent mathematical accomplishment that compares to inventing algebraic topology.  Any takers?)[/quote]

i was just trying to keep things reasonably factually accurate.  my background is not in differential geometry and it's even further from algebraic topology so my perspective on such things is not very useful.

(as an aside if you consider grothendieck's work 'recent' you probably can compare it to inventing algebraic topology.  as a probabilist-in-training i'd put kolmogorov's work on rigorizing probability theory up there (around the 1930s))

[color=black][size=200]Archemides[/size][/color]..........At least in my opinin.........

It's sad because the name of Constantin Carathéodory(Greek: Κωνσταντίνος Καραθεοδωρή) has never been mentioned before in that topic or in similar topics.

I'd nominate Andrew Wiles for proving the Fermat's last theorem, Grigory Perelman for proving the Poincare's Conjecture, Alain Connes for his Noncommutatve Geometry, David Hilbert for his wide contributions to almost all branches of math including physics, Edward Witten for his M-Theory (its a unification theory in physics but, mind you, it's highly mathematical and the mathematics is really cutting-edge and new, who else? ummmm....well i'd like to you follow this links
http://james.fabpedigree.com/mathmen.html

Eventually,after reading carefully all the comments above(including mine) i came up with the following conclusion:

             I AM THE BEST MATHEMATICIAN!   :first:  :winner_first:

Note that this conclusion came up because of the following argument:
$ *)$I have found a remarkable proof of the $ Riemann$ $ hypothesis$ but the margin is too narrow to contain it :(

my opinion:

1) A. [b]Grothendieck[/b]
2) J. P. [b]Serre[/b]
3) S. M. [b]Atiyah[/b]

[i]Undoutedly , Paul Erdos deserve to be the best mathematician with his large contribution in Number Theory and Algebra :lol: [/i]

[b]One[/b]: It's illogical to vote on somebody accepting prizes. [b]Einstein would change his whole life, but somebody else did that from the very beginning.[/b]

[b]Two[/b]: Being [b]prolific genius[/b] means more than working hard on ideas. It is the flash that creates invention. Logic does not enable that. Real genius just wants to admire the world and understand he has no valuable ideas, and acts as servant only. No matter what You think.

[b]I love Grisha for rejecting materialism. But he is not the only one that did that. But, still, he or Wiles only proved that visions were correct.[/b]


hint try multiplying and dividing the nth term by (n+1)^0.5-n^0.5. you will find the series telescopes......

I think the following will do:
Sum of the first twenty terms = sqrt(20)

Each term may be represented by:
1 / [sqrt(m) + sqrt(n)], where n > m.

Multiply this by
[sqrt(n) - sqrt(m)] / [sqrt(n) - sqrt(m)]
and you get:
[sqrt(n) - sqrt(m)] / (n - m)
But n - m = 1 for each term,
so each term = sqrt(n) - sqrt(m).

The series then becomes:
1 + [sqrt(2) - sqrt(1)] + [sqrt(3) - sqrt(2)] + [sqrt(4) - sqrt(3)] + ...

Sum of 1st term = 1
Sum of first 2 terms = sqrt(2)
Sum of first 3 terms = sqrt(3)
etc.
Sum of first 20 terms = sqrt(20)

In general, sum of first r terms = sqrt(r)

Source:
falzoon


The exterior dimensions of an empty cooler in the shape of a rectangular prism are 18 inches x 18 inches x 24 inches. For insulation, the top, the bottom, and each of the four walls are two inches thick. What is the volume of the empty space inside the cooler?

Answer & explain, plz :-)

OK. Think about it like this. So you have two inches on each side of the cooler, no matter how the cooler is oriented. So basically visualize a rectangular prism with 2 inches less on each side. Basically you subtract 2(2)=4 inches from each dimension because of the insulation. So the answer is $ 14\cdot14\cdot20\equal{}196\cdot20\equal{}\boxed{3920}$ cubic inches.


search
number theory solved
number theory

Find all $n \in \mathbb N$ such that there is $m \in \mathbb N$ and $2^n - 1$ divide $m^2 + 9$

source is some Shortlist so I guess it was discussed before, try to search for it. The answer is $n=2^k$ I suppose.

To be precise,
http://www.mathlinks.ro/Forum/viewtopic.php?p=124436#p124436


Google Book Search is an excellent way to access old ARML problems if you don't have the ARML books.  The recent green book is not available, only the blue 1989-1994 book, for now.  You can flip through the book [url=http://books.google.com/books?ie=UTF-8&vid=ISBN0962640166&id=oOlyGq2lcMwC&pg=PA3&lpg=PA3&dq=ARML-NYSML+Contests+1989-1994&sig=5IdKYdQKTdBSgc5bAwvOSLQu5Hc]here[/url].
However, you can only view a few pages at a time, due to copyrights.  An excellent workaround is the search box in the left hand column, however.  Just type in a year, and the type of problems (Individual, Power, etc.) and the search will take you straight to that page.  This is a GREAT resource!

too bad we can't look at the solutions :( , oh well, it's better to buy the book

[quote="mysmartmouth"]Yeah, the book is much better, but this is just a way to quickly review some more problems before the upcoming ARML.  I ordered the book already, and I would highly reccomend ordering this book, as well as the more recent green one.  Speaking of math books, I'm expecting the Introduction to Geometry one from AoPS today....[/quote]

Did you receive the old ARML book?  Last time I checked you can't buy it online.  I got mine at ARML.  It lets you order the book from amazon but then keeps on telling you it's on backorder or something.

Btw, if you need solutions, I'm always here :P  I can go look in the book if I can't figure it out anyway.

[quote="joml88"]Btw, if you need solutions, I'm always here[/quote]

No, seriously, new people:  Joe is the best resource you will find on AoPS.  His AIM sn is BaseballJoe2002, and if you don't have AOL Instant Messenger, it is worth downloading just to talk to Joe.  You get free, real-time tutoring any time you need it.  I've worked through so many problems on AIM with Joe, it just helps me understand stuff for some reason.

[quote="arkmastermind"]joe is sean's hero. lol[/quote]

That's somewhat true.  I can't mention my real hero here, however, because it might offend some people...lol.


function
number theory proposed
number theory

Consider a function $ f: N_0\to N_0$ such that :
$ f(m) \plus{} f(n)|(m \plus{} n)^k$ where k is a constant .
Prove that :$ f(p) \equal{} p,\forall p\in P$