inequalities
geometry
algebra
polynomial
vector
trigonometry
analytic geometry

Hi,

IMO2004 is coming soon. I am asking for your opinion regarding certain issues which might occur:

[b]I.)[/b] What will the questions be at IMO this year. Last year was short on combinatorics, the last real traditional algebraic inequality was at IMO2001 (maybe you also consider IMO2003/5). Latest functional equation is from IMO2002. My suggestion of the topics for IMO 2004:

 [i]   1.) Geometry, triangle geometry
    2.) Number Theory
    3.) Combinatorics (maybe a revival of graph theory in terms of Alice and Bob)
    4.) Combinatorics
    5.) Geometry, non-triangle geometry
    6.) Algebra, Functional Equation[/i]

[b]II.)[/b] What will be the cut-off for gold, silver and bronze ?

[i]last year' cutoffs:

gold medal: 29, silver medal: 19 and bronze medal: 13[/i]

I) 5. should be replaced by a nice inequality in Orlando's list

II) the numbers of medals will be with 2 less than last year.

my oppinion
:D

I.) I think we will either have an inequality or a geometric inequality; I can't imagine an IMO completely without inequalities. Concerning the "traditional" (triangle etc.) geometry: On the one hand, the experience of the last years shows that there are lots of guys in the PSC who either hate geometry or have no idea about geometry or both simultaneously and block the rather interesting problems, so that only the most trivial geometry problems are used in the exam (or sometimes they replace them by combi geo). On the other hand, this time we'll have a Greek PSC, and they should be somewhat more friendly to geometry (at least, in the Hyacinthos group, there is a considerable group of Greek euclidean geometers). BTW don't forget my prognosis that again one half of the triangle problems will be about isodynamic points  :D
Concerning number theory, somehow they are mad about diophantine equations, so it won't be a wonder if there comes one this year again. Alice and Bob seems to be rather unlikely - as far as I know, they don't like "game" problems.

II.) I see no reason why it should be different from last year, at least the problems hardly will get easier...

  Darij

[quote="pokemon"]i hope IMO 2004 has a nice and difficult polynomial, there are 44 years of IMO but so few polynomial, it's very dissapointing

i hope the problems of IMO 2004 greece are difficult, in the years that IMO was held in a weak country, the problems are quite easy and dissapointing, 1987 cuba, 1991 sweden, 1994 hong kong, 1997 argentina

if i can choose the problems, i'll choose this :lol: :lol: :lol:
1. C
2. A
3. S
4. B
5. A
6. S

S=super difficult (0 solved)
A=difficult(25-30 solved)
B=medium(75-85 solved)
C=easy (150-200 solved)[/quote]

Well, I don't agree with you, pokemon. At first do you really mean no student should solve a problem of the S-category ? Well, what would be the sense of such a problem ? To to show that there will always be a problem that none of the students can solve ? To get low scores and medal cut-offs ? I really don't know. But maybe you meant 0(1)-5 students solved S-category problem. Anyway using one or two S problems is not too reasonable. The only issue you improve is that you getter a better distincion of the performances of the potential high-scorers, especially the perfect scorers. But much more important you loose the distinction of performances in the lower point ranges. I feel that this distinction is much more important as there are far more students in this range than in the maximum score range !!! For instance: two students who get 18 points (student I) or 22 points (student II), respectively, may get silver or bronze medals, respectively, due to the common used problem choice which I applied. But those students may get in pokemon'a system just 12 points both and hence student I and II two get both bronze medals. And student II might fail to score a second prize just by one point. With such low scores tiny mistakes will decide whether you get awarded a certain prize or just slip out off the required range of points. In general one would expect mostly just problems one and four get solved. In this case you really cannot distinguish students' performances in these ranges with scores less than 14 points. A respective case happened in the first Romanian TST this year. They posed two easy problems and two hard ones. And those who where close to the boundary whether they can take further TSTs where just distinguished by minor mistakes not really by better skills in tackling problems. A compromise might be an extension from two IMO days to three IMO days as it was carried out in one of the regional replacement IMOs in 1980. For such a contest one could consider the following idea (problems are in order 1,2,3 for the day and the problem of the day): C,B,A - C,B,S - B,A,S. This might yield an adequate distinction in most of the point ranges.

What do you think ?

Of course there are examples where comparable schemes to pokemons' are applied. I know that Leningrad (LMO) problems, at least the elimination round (ER), was really challenging. I read that Eugenia Malinnikova (winner of three gold medals at IMOs 1989, 1990 and 1991 with a score of 41/42, 42/42, 42/42 respectively) as all the other LMO participants did not solve too many problems in the years 1987-1991. In general LMO ER problems are of higher quality and level of complexity than others olympiads. The LMO was usually extremely hard and it was not uncommon that none was able to solve all 8 problems in the time allowed - rather often only two or three people could solve more than 4 problems and the other twenty participants could only solve one or two of them. That doesn't mean they didn't want to create an olympiad that could play the usual selective role but firstly they almost always had too many superb problems which they wanted to use and secondly they only were required to select one or two contestants as members of LMO team for the Soviet Union Olympiad. Today the LMO (better SPMO for Saint-Petersburg Mathematical Olympiad) was simplified as they have to select more team members for the All-Russian Olympiad than for the Soviet Union Olympiad back then.

Any comment is requested !!!  :)

[quote="darij grinberg"]On the one hand, the experience of the last years shows that there are lots of guys in the PSC who either hate geometry or have no idea about geometry or both simultaneously and block the rather interesting problems, so that only the most trivial geometry problems are used in the exam (or sometimes they replace them by combi geo). Darij[/quote]

Well, what I don't understand is the following fact. Considering various old ILLs and many IMO team leader reports I got the impression that by far most of the problems are from geometry though not so many geometry problems can be selected for the ISL (they must select other topics too and many problems are known). Can someone explain that to me ?  :)

i agree that the quality of the geometry problems is low. i even see a reason for it. many teams are  experienced enough and instruct their teammembers (especially those which are not so good at geometry) to use complex numbers vectors etc.and just calculate. the solutions are hard to read (the jury probably doesn want to bother so much with the transformations) and often you are almost sure you can carry the solution.so no need to invent just calculate.... it turns that its not a competition of how smart you are but how quick you can work it out or how well have you learnt complex numbers.
the alternative way is to select problems which are graphically tooo complicated for those who try to outwit IMO for example many circles intersecting...
 i wouldn't like also a game problem it's rather for the informatics competitions

Hey I like the geometry problems!
Some are quite nice and often they are among the few problems I can do.

My opinion:

1.) Geometry, triangle geometry
2.) Number Theory
3.) Combinatorics/Algebra
4.) Combinatorics
5.) Combinatoric geometry of any kind
6.) Algebra, NON Functional Equation

I consider functional equation is quite meaningless branch of olympiad mathematics, though some problems are very nice.

there are many topics at the IMO which are meaningless at mathematics but are just nice, not only functional equations. IMO(and all other maths olympiads) doesn't deal with problems of large interest but with little nice problems.

Myth I see you don't like geometry much..
I think this year will be a hard combinatorics, since last 2 years it weren't.
Gold cutoff will be 30 and maybe there will be more perfect scores

I mean that functional are extremely special problems. Why must we choose them for IMO? I think it is just bad tradition.

1) Problems for IMO must be HIGH quality INTERESING problems which can be understanded by most of contestants.
2) Problems for IMO don't have to be too hard, but too easy problems are inadmissible.
3) I don't believe to ugly, or messy, or large/long, or computing solutions. Problem for IMO MUST have short/compact solution with BRILLIANT idea.

I agree that IMO'97 (where I participated) doesn't match my criteria.

Iura!
Two pure geometry problems it is too much for such olympiad where we have 6 problems only.

I know Bulgarians, Americans, Chinese usually give 2 geometry problems.
Also at ARO there is one geometry every day.
Anyway meant another thing: usually students who like geometry would put it as 3rd or 6th problem

ARO has 8 problems=2 * 4 problems per day. So two geometry can be placed during two days.

[quote="qw"]i agree that the quality of the geometry problems is low. i even see a reason for it. many teams are  experienced enough and instruct their teammembers (especially those which are not so good at geometry) to use complex numbers vectors etc.and just calculate. the solutions are hard to read (the jury probably doesn want to bother so much with the transformations) and often you are almost sure you can carry the solution.so no need to invent just calculate....[/quote]

This is the case for many (too many, I dare say) geometry problems, but inspite, there are some very nice geometry problems you cannot trig bash / algebra bash / bash with trilinear coordinates / with complex numbers (at least, without the help of a computer). Take all the coaxal circles from triangle geometry; it is very hard to do any non-elegant proofs in this area.

I rather tend to think that some countries boycott geometry problems because their students are not well prepared. Indeed, many stundents (throughout all countries) are really bad in geometry, and generally, I think 2/3 of all mathematicians don't have the necessary minimum of planimetric knowledge. Note that I purpousely say "planimetric", since my own experience in stereometry, as well as in combinatorics, number theory, graph theory is really far below the desirable one for an IMO participant, not to speak about calculus, where once again I've made some really idiot errors... in a school test last week.

[quote="orl"]Well, what I don't understand is the following fact. Considering various old ILLs and many IMO team leader reports I got the impression that by far most of the problems are from geometry though not so many geometry problems can be selected for the ISL (they must select other topics too and many problems are known). Can someone explain that to me ?[/quote]

Indeed, this is another problem. Geometry is a field where 80% of all theorems you think to have invented on your own are actually known for more than a century. I remember well that my first geometry theorem, which I published in the "Wurzel" 5/2001, turned out to be a result of Pappos of Alexandria, and [url=http://forumgeom.fau.edu/FG2002volume2/FG200222index.html]my first paper in "Forum Geometricorum", together with Paul Yiu, about the Apollonius circle of a triangle,[/url] was a rediscovery of a result by Victor Thbault in 1941. This is a characteristic attribute of elementary geometry, having its roots in the rather chaotic history of geometry. Now, in a PSC consisting of people from all around the world, it isn't a wonder if one half of the proposed geometry problems are thrown into the trash since they appeared somewhere some years ago. Another question is, of course, whether this justifies to withdraw good problems; actually, lots and lots of geometry facts were discovered in the 19th century, and many of them are unknown to any IMO participant, including me.

Concerning functional equations, everybody has his own opinion about them. For a long time, I disliked them because I couldn't solve them, but now I think they are indeed nice, as one actually doesn't need anything more than a few simple methods and a good brain!

  Darij

i personally don't like combinatoric geometry it's good part of mathematics but when you are taught higher mathematics
 when it comes to a olympiad you can neither rely on your geometric intuition (if you have to deal with 100 points circles or lines - you cant "see" through the construction on the paper)neither you can rely on your combinatoric skills since often some geometric features of the construction (i would rather say metric than geometric features) are important . i think you can deal wiht such problems freely only if you are common with metric spaces (not the topological aspects but the basic items)
 another bad think about such problems is that often you have only one way of solving them and you can't invent any other method than author's

[quote="Myth"]

1.) Geometry, triangle geometry
2.) Number Theory
3.) Combinatorics/Algebra
4.) Combinatorics
5.) Combinatoric geometry of any kind
6.) Algebra, NON Functional Equation

[/quote]

Hi myth,

with so many combinatoric problems on the paper your suggestion rather looks like a Tournament Of Towns paper than a common IMO paper. Though Dust, Iura, lhvietbao etc. would quite like it, Arne and Darij would not appreciate it too much...

I think, combinatorics is only kind of real olympiads problems. Other themes like inequalities, pure geometry, part of number theory, functional equations are just well trained skills. I remember some moments in my olympiad practice when I intensified learning of these parts of olympiad mathematics and got good results, because at that time I had in my mind BIG HEAP of methods for solving of mentioned problems.
In the same time combinatorics (in wide sense) deals with nonordinary situations which demand all variety of knowledges and talents.

Thus well made olympiad is
1) 50-60% of pure themes (inequalities, pure geometry, number theory, graph theory and so on)
2) 40-50% of combinatorics-involved problems (IMO2000-3,4; IMO1998-2; IMO1999-1,3).

I know, it is just my opinion. But I suppose that I am older than most of mathlinkers, especially who are going to IMO, so believe me that there are no emotions in my words in contrast to some posts in this thread.

I want that you don't consider my arguments as arguments against geometry, etc. Of course, geometry, inequalities ,... are important parts of olympiads.

I think an IMO should contain combinatorics, geometry, algebra and number theory as a MUST.
Also  inequalities, functional equations can be submitted.
Personally I like most combinatorics, geometry and NT though I suck at geometry

I think it's best if problem 3 and 6 are both combinatorics. If you choose anything other than combo, China and Bulgaria will do it very well. If you choose combo, nobody will do it well, every country is the same.

Look at Functional equations: IMO 1998/3 IMO 1999/6, Bulgaria is very strong

Look at Number theory: IMO 2001/6 IMO 2003/6, China and Bulgaria are very strong

Look at Geometry: IMO 2000/6, IMO 2003/3, Bulgaria and China are very strong

Look at the combo problems : IMO 1995/6, IMO 1996/1, IMO 1999/3, IMO 2000/3, IMO 2001/3, China and Bulgaria are not better than any other country, they don't have any advantages.  :P  :P  :P

This opinion confirms my arguments about well trained skils.

[quote="darij grinberg"][quote="Arne"]I am hoping for the following paper:

1) Geometry
2) Number theory
3) Combinatorics (because I suck at combinatorics)
4) Inequality
5) Geometry
6) Functional equation[/quote]

Did anybody note that this prediction came true except of a transposition of 2) and 6) ?  :D

  Darij[/quote]

Yes, I realized it. But I forgot this topic for some time. Maybe we should ask Arne again. So that you guys can prepare for the problems.  ;)

PS: I will open a thread for IMO2005 soon.  :)

[quote="Myth"]I mean that functional are extremely special problems. Why must we choose them for IMO? I think it is just bad tradition.
[/quote]

How did this become a tradition?

Did it start with problem IMO 1975/6?

[quote="Myth"]
1) Problems for IMO must be HIGH quality INTERESING problems which can be understanded by most of contestants.
2) Problems for IMO don't have to be too hard, but too easy problems are inadmissible.
3) I don't believe to ugly, or messy, or large/long, or computing solutions. Problem for IMO MUST have short/compact solution with BRILLIANT idea.

I agree that IMO'97 (where I participated) doesn't match my criteria.[/quote]

What about problem 1997/3?
Its statement is easy to grasp, it is not too hard, and it has a nice and short proof.


You have three piles of stones containing 5,49,51 stones. You can join any two piles together into one pile and you can divide a pile with an even number of stones in to two piles of equal size. Can you ever achieve 105 piles each with one stone?

Can anybody give me some hints?

Hint: Don't think about it theoretically. Work with the numbers, try to actually do it. What happens? Why?

I actually did try to work it, I found it is not possible to get 1s, the smallest number I can get to is 3.

I know that when I need a 1, I need a 2 first, to get a 2, I need a 4 and so on.

hm, is this one of the invariant problems?

try working backwards from 105 piles of 1

I tried the brute force method but got nowhere. :(

Well, look at it this way.  If the end result is 105 piles of 1 stone each, what step must come before that?  More specifically, how many stones must be in a pile before splitting them into piles of 1?  How many piles of those must you have?  Is that possible?

You're right that you can't avoid ending up with threes if you combine the 5 with the 49 first. Why is it that you're stuck with threes? What happens if you combine the 5 with the 51 instead?

[hide="Quick Solution"]
From $(5,49,51)$ you have only three choices: $(54, 51)$, $(49, 56)$, and $(5,100).$  Consider $(54,51).$  You can only combine two elements or divide one by two (if its divisble by 2.  Notice that 54 and 51 are divisible by 3.  Any set of numbers you can possibly obtain from $(54,51)$ must all be divisible by 3 because adding two multiples of 3 produces a multiple of 3 and dividing an even multiple of 3 by 2 still produces a multiple of 3.

Use similar logic for $(49,56)$ (with 7 instead of 3) and $(5,100)$ (with 5 instead of 3).  And you're done![/hide]

Yeah, uh, that's what I was trying to get them to do without actually telling them....

Yeah this was simple. Liangchene are you allowed to discuss this in this forum?


Let  $ A$ be the set of all integers of the form $ x^2\plus{}3xy\plus{}y^2$ where x,y are integers. Prove that if $ n$ belongs to $ A$, then the number $ 2009n$ also belongs to $ A$


http://i64.photobucket.com/albums/h163/Couchtater4444/math.png

Can someone tell me the area of that circle?

you need more information than just the length of one chord

The area of a circle is $ \pi r^{2}$ so generally you need either the radius, or a method by which you can find the radius.

Well um you can say that the area is more than $ \frac{529\pi}{4}\text{mm}^{2}$... but other than that, you can't do much.

We learnt a theorem that "A line drawn from the center of a circle to a chord will bisect the chord..so if you join the end points of the chord to the center and proceed then you get stuck..I guess you'll need atleast one angle or something...

The area could be huge.  You can only establish a lower limit by assuming the chord is a diameter.

I think the exact statement of the problem is here: http://www.neopets.com/games/conundrum_feature.phtml?round=221
[hide="if youre not a neopets user"]
[img]http://images.neopets.com/games/conundrum/221_dubloon2.gif[/img]
A Two Hundred Dubloon Coin is unique in that the outer edge is gold while the inside is silver. Assuming both sides of the coin are identical, and a line tangent to the inner circle is 23 mm from edge to edge of the coin, and excluding the edge of the coin in the calculation, [b]what is the surface area of gold on the coin, in square millimetres?[/b] Please round up to the nearest square millimetre.
[/hide]But next time, please go to: http://www.able2know.com/forums/about42093.html

[hide="Hint"] Draw both relevant radii.  Pythagorean theorem. [/hide]

:rotfl:
I never knew neopets had math contests!


is this true ?
$ \forall a,m\in \mathbb{N}$

Prove or disprove

$ a\equiv a \minus{} m \lfloor \frac {a}{m} \rfloor \mod {m}$

Since $ \lfloor\frac{a}{m}\rfloor$ is an integer, it is always true. It's the definition of mod.

Bictor717- u live in RI!!! what HS?
$ a\equiv a \minus{} m \lfloor \frac {a}{m} \rfloor \mod {m}$

yes $ a \minus{} m \lfloor \frac {a}{m} \rfloor \mod {m} \equal{} a\minus{}0 \mod {m} \equal{} a \mod {m}$


Find the smallest real number $ x$, such that: For arbitrary real numbers $ a, b, c\geqslant x$   and $ a\plus{}b\plus{}c \equal{} 3$   this inequality is correct:  $ a^{3}\plus{}b^{3}\plus{}c^{3}\geqslant 3$

This is problem 9 from 3rd series of 1st round of 59th Polish Mathematical Olympiad. The 3rd series of the competition ends at 10.12.2007. Moderators, please delete the previous post and lock the topic.


limit
search
real analysis
real analysis unsolved

For $\forall \varepsilon>0$, $\exists n_{0}$  s.t.  $n>n_{0}\Rightarrow |a_{n}|<\frac{\varepsilon}{2}$
And,
For $\forall \varepsilon>0$, $\exists n_{1}\geq n_{0}$  s.t.  $n>n_{1}\Rightarrow |\frac{1}{n}\sum_{k=1}^{n_{0}}a_{k}|<\frac{\varepsilon}{2}$

Then,
$n>n_{1}\Rightarrow |\frac{1}{n}\sum_{k=1}^{n}a_{k}|<\frac{\varepsilon}{2}+\frac{(n-n_{0})\varepsilon}{2}<\varepsilon$

Thus, $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}= 0$

[b][color=blue]Corollary[/color][/b]
$\lim_{n\to\infty}a_{n}= \alpha$, then $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}= \alpha$

First of all I am not familiar with convergence in the Cesaro sense, and up to some details(this is because there are some mistakes, but not hard to fix) I could get clearly the proof given by Kouichi, so Thanks.

[quote="fabiansaul"]I need help with this exercises about sequence of real numbers:

Prove that if $\lim_{n \to \infty}a_{n}= 0$ then $\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1}a_{k}=0$.

thanks

FS[/quote]
Here is another way.

Let $(s_{n})$ be a sequence and $t_{n}= \frac{1}{n}\sum_{k=1}^{n}s_{k}$ then $\lim t_{n}= L$ if $\lim s_{n}= L$.

You can show that:
$\liminf s_{n}\leq \liminf t_{n}\leq \limsup t_{n}\leq \limsup s_{n}$

Once you show that, if $\lim s_{n}= L$ then $\liminf s_{n}= \limsup s_{n}= L$ which will imply $\liminf t_{n}= \liminf t_{n}= L$ so $\lim t_{n}= L$.

And the case $L=0$ is a special case.

That is really non-helpful. To get the general case from $L = 0$ we just look at $a_{k}-L$

[quote="spix"]This is just an obvious consequence of Cesaro-Stoltz theorem.![/quote]
In Romania this a very used theorem for composing or solving problems.
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=133923446&t=4126


advanced fields
advanced fields unsolved

[color=green]FEIGENBAUM CONSTANT[/color]的[b]低阶有理多项式解析逼近[/b]之演义，我无意多说点什么..
简介请参考阅读[url=mathworld.wolfram.com/LogisticMap.html]The History Map of Logistic Map[/url]
图说言是[img]http://bbs.sjtu.edu.cn/file/math/114032862232770.png[/img]
[color=blue]Compare To[/color][img]http://mathworld.wolfram.com/images/eps-gif/LogisticMapBifurcations_900.gif[/img]
II[img]http://mathworld.wolfram.com/images/eps-gif/LogisticEquationBifurcation_900.gif[/img]
Link To The PAPER of <Exact computation of the bifurcation point B4 of the logistic map and the Bailey-Broadhurst conjectures>BY [color=cyan]I. S. Kotsireas and K. Karamanos[/color][url=http://oldweb.cecm.sfu.ca/AAS/costas-cecm.ps]oldweb.cecm.sfu.ca/AAS/costas-cecm.ps[/url]
And [color=cyan]D. H. Bailey and D. J. Broadhurst[/color] Write On [url=http://crd.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf]crd.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf[/url]
At ALL!
Add My Own Paper As an Expession To The End!![url=http://bbs.sjtu.edu.cn/file/math/114433548195021.doc]渾天解象術_風無塚[/url]
 cycle ()  Sloane
1 2 3
2 4 3.449490 A086180
3 8 3.544090 A086181
4 16 3.564407 A091517
5 32 3.568750
6 64 3.56969
7 128 3.56989
8 256 3.569934
9 512 3.569943
10 1024 3.5699451
11 2048 3.569945557
 accumulation point 3.569945672 A098587

I need some opinions from anybody , in deed.Who can give the warm hand for HELPING!
JUST AS I have got the Analytical formula of the bifurcation diagram of the logistic map in Ahead of Partof.
[img]http://bbs.sjtu.edu.cn/file/math/1144336122127002.png[/img]


[quote="mdk"]Let a,b ,c be the roots of x^3-x^2+1 = 0. Find 1/a^2 +1/b^2 +1/c^2 .[/quote]
Im not doing the full solution. But this is sufficient.

$ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}= \frac{a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}}{a^{2}b^{2}c^{2}}$

Now, $ a^{2}b^{2}c^{2}= (abc)^{2}$ we can use Viet's theorem to find this.

And,
$ (ab+bc+ac)^{2}= a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}+2abc(a+b+c)$
Now you can find $ ab+bc+ac$ by Viet's theorem and $ a+b+c$ and $ abc$.
So you can solve for $ a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}$.
Finished.

Heres another way:
[hide]$ x^{3}-x^{2}+1 = 0$
The polynomial with the reciprocal of the original roots is $ x^{3}-x+1 = 0$ (just reverse the coefficients).

Let $ S_{n}= r_{1}^{n}+r_{2}^{n}+r_{3}^{n}$
Using Newton's sums,
$ S_{1}= 0$,
$ S_{2}= \boxed{2}$[/hide]

[quote="p4fn2w"] (just reverse the coefficients). [/quote]

Do you have a proof?

[hide]Since $ a,b,$ and $ c$ are nonzero roots of $ x^{3}-x^{2}+1 = 0$, $ \frac{1}{a},\frac{1}{b},$ and $ \frac{1}{c}$ solve $ \left(\frac{1}{x}\right)^{3}-\left(\frac{1}{x}\right)^{2}+1 = 0$.  Multiplying through by $ x^{3}$, we have $ x^{3}-x+1=0$.

Then $ (\frac{1}{a})^{2}+(\frac{1}{b})^{2}+(\frac{1}{c})^{2}= (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}-2((\frac{1}{a})(\frac{1}{b})(\frac{1}{c})) = 0^{2}-2(-1) = 2$ from the coefficients.

This illustrates why reversing the coefficients works.  In general, if $ r$ is a nonzero root of $ a_{n}x^{n}+a_{n-1}x^{(}n-1)+...+a_{0}$, then $ \frac{1}{r}$ is a root of $ x^{n}\left(a_{n}\left(\frac{1}{x}\right)^{n}+a_{n-1}\left(\frac{1}{x}\right)^{n-1}+...+a_{0}\right) = a_{0}x_{n}+a_{1}x^{n-1}+...+a_{n}$[/hide]


I have a question,

Some scientists say that the universe curves back on itself, like a two-dimensional plane universe would curve back on itself in a sphere, resulting in a finite universe in terms of space, but no boundary. A rocket moving in a straight line would thus come back to it's origin at some point, like a line drawn from a point on a sphere would come back to it's origin point.

 Therefore, assuming light has enough time to travel the distance and nothing was obstructing it, if you looked straight up with a really powerful telescope, would be able to see the earth billions of years ago, since the light would come from the old earth, curve back on itself, and coming back to earth billions of years later? Would you be able to see infinite earths one behind another differing only by time, or atleast back until the creation of earth?

EDIT: Would you be able to see the earth by looking up at [u]any[/u] point in the sky, meaning infinite earths covering the whole sky?

This is doubtful, even under your assumptions.  After a photon leaves the earth, it would have to travel on a path very nearly as straight as a geometric line, due to the almost infinite vastness of the universe.  Since gravity has the power to bend light, however slightly, the photon would deviate from its path as soon as it encountered an object enough that it would likely not reach the earth on its rebound.

At least, that's my take on this theory.

true, but I was thinking on the lines of if there wasn't anything to obstruct, distort using gravitational forces, or otherwise force the photon to deviate from a straight line. Basically, if the universe had nothing but the earth, an observer and super powerful telescope on the earth, and photons, would those things be true? And since you could look up at any point in the sky, couldn't atleast at one point, the light hasn't been distorted? :?

EDIT: And there's no atmosphere on the earth to distort the light. :)

Under the assumption that the universe is finite and boundless (one of the three types, which also include infinite and flat and infinite and curved, I believe, which have various reprecussions for the expansion of the universe), the universe will eventually recontract in a "big crunch."  In fact, the time it would take a photon to arrive back at its point of origin is exactly the time between the big bang and the big crunch.

NOTE: I have gleaned this from popular physics books, and have no understanding of the mathematics behind it.  You have been warned.

of course, the big crunch may never happen. we can't calculate, so far at least, whether the universe has enough energy to keep expanding outwards for ever and ever, or the gravitational pull will be too strong and pull us all back together.

The rate of expansion of the universe is [i]accelerating[/i], ( from dark energy or something :? ), according to well accepted study a few years ago.

actually, the big crunch is more desirable then continued expansion

if we have continued expansion, then the heat in our world would be minuscale at all points, as it dessipates, eventually we'll freeze to death, lol


trigonometry
function
algebra
domain
Pythagorean Theorem
geometry

Hello, I'm a first time poster.

Last week, these types of problems were explained in class and I was not attending those days (sick). I have not had time to ask my teacher for help due to the fact that he has a busy schedule and I have been very busy myself. I would appreciate it if these problems can be solved and worked out so I can see how to properly do them. No one person has to do all of them. I appreciate all the help that I can get. Thanks.

1. What is the range of the function: y = 3(cos x) - 1

2. What is the domain of the relation: y = arctan 3θ

3. What is the domain of the relation: y = 3/2 arcsin (2x+.5)

4. Evaluate: sin (arctan 3^(1/2)) + cos (arccot 3^(1/2))

5. Evaluate: cos (csc^(-1) 1)

6. If tan^(-1) = θ, what is csc^(-1) (cotanθ)

7. What is the inverse of y = π/2 + cos^(-1)x

-edit-

I would like to emphasize that I have tried these problems, but I am totally lost. Thanks.

What exactly don't you understand? Do you know the range of the $ \sin,\cos,$ and $ \tan$ functions?

No, I do not. I don't know how to solve these problems. I have looked through my book for the information needed to complete these problems, but it is not in there. My teacher usually does his own thing without using the book, so I'm at a bit of a loss right now.

This should probably go in a easier forum

but just remember that the domain of tan is (-90,90), cos is (0,180) and sin is (-90,90)

range of tan is (-inf, inf)  range of cos (-1,1) range of sin (-1,1),

() means [b]inclusive[/b]

I think (hehe need to know this for the AIME)...I dont really remember but that should be correct sine all the values are covered.
tan also repeates every 180 degrees, sin and cos 360 degrees respectively.

i put these in degrees, of course -90 is -pi/2...so forth you can translate them yourself.

The course that I am taking is considered a more advanced course, so I figured that this would be the proper section. Sorry if I was mistaken.

Thanks for getting me started with the domain problems. Can you offer any help with the other ones?  :)

[quote="the future"]but just remember that the domain of tan is (-90,90), cos is (0,180) and sin is (-90,90)[/quote]
 :huh:

Did you mean the range of the inverse functions?

For the original poster: here are hints for all of the problems

[hide="1"]The range of $ \cos x$ is $ [\minus{}1,1]$.[/hide]
[hide="2"]The range of $ \tan x$ is all real numbers. What does this say about the domain of $ \arctan$?[/hide]
[hide="3"]The range of $ \sin x$ is $ [\minus{}1,1]$. What does this say about the domain of $ \arcsin$?[/hide]
[hide="4 5 6"]Computing the given inverse functions directly is one way, but this more general way is probably better: Write the equalities for each trigonometric function in a right triangle (SOHCAHTOA). Combining this with the Pythagorean Theorem allows you to transform things like $ \sin (\arctan x)$ into a (relatively) simple nontrigonometric function.[/hide]
[hide="7"]Isolate the $ \cos^{\minus{}1} x$.[/hide]

Melon, Thank you very much for the hints. I greatly appreciate you taking your time to help me out.  :D

[quote="the future"]but just remember that the domain of tan is (-90,90), cos is (0,180) and sin is (-90,90)[/quote]
The domain of the basic trig functions is $ (\minus{}\infty,\infty)$.

[quote="the future"]() means [b]inclusive[/b][/quote]
"[" and "]" mean inclusive; "(" and ")" mean exclusive.

^ sorry I was talking about inverse (which means i did it backwords)

domain cos^-1(-1,1) sin^-1(-1,1) tan^-1 (-inf,inf)
range cos^-1 (0,180) sin^-1(-90,90) tan^-1 (-90,90)

regular its

domain all (-inf,inf)
the range is just the domain of the inverse

I forogt that its not the same backwords

yea i know [] means inclusive but I was too lazy to go back and edit so i just told him in my example () means inclusive-to avoid confusion

And I learned this like 3 years ago so im a big rusty.

No problem; I was just making sure the original poster would not be confused. :)


geometry
3D geometry
sphere
real analysis
real analysis unsolved

if $ X\equal{} C [0,1]$ with sup norm
$ S\equal{} \{x \epsilon X: \parallel{}x\parallel{}_\infty <\equal{}1\}$ then
$ S$ is convex bcoz every open and closed sphere is convex
but not compact bcoz if $ S$ is compact then $ X$ must be finite dimensional but here $ X$ is infinite dimensional
is it right?


Let $n > 0$ be an integer such that $\left[ \frac{n}{p}\right]$ is even for every prime $p \ge 2$. Prove that $n=1$.

Does $\left[ x \right]$ denote the floor function or rounding?

1 is the only value of n that works because all other numbers would eventually be that prime number or slightly above such that it would have a value of 1, which is not 0. 1 is neither prime nor above one.

If this is not a floor function and is rounding it would be either below of above, but it still applies.

[quote="Anaxerzia"]or slightly above such that it would have a value of 1.[/quote]

This is true, but to prove it rigorously requires citing [hide="your primary sources"] Bertrand's Postulate. [/hide]

I had no sources - I used my IQ's logic lol.... I wouldn't say it needs any citing...

It certainly does.  The fact that there exists a prime between $k$ and $2k$ isn't obvious at all.

That's not what I'm saying. All I'm saying is that if, say, n is 3, than the function would be eventually 1 (just at p = 2) and that is obviously so for all cases. 2 is the lowest prime, so n must be below it, or 1.

I looked up this Bertrand's Postulate and I can't even understand it... I don't think it relates to what I'm saying...

[quote="Anaxerzia"]if, say, n is 3, than the function would be eventually 1 (just at p = 2) and that is obviously so for all cases. [/quote]

That's not obvious at all.  What if $n$ is, say, $10000$?  Is it obvious that there exists a prime $p$ such that $\left[ \frac{10000}{p}\right] = 1$?

Ah I see your point. Very true... Very true...

[quote="PhilAndrew"]Yes, $[x]$ denotes the floor function.
[hide="Proof"]For $n=1$ the statement is obviously true, and for $n \ge 2$ we get, indeed, by Bertrand's Postulate, that there exists a prime $p$ such that $\frac{n}{2}< p \le n$, or equivalently, $1 \le \frac{n}{p}< 2$, hence $\left[ \frac{n}{p}\right] = 1$ which is not even, so $n$ can't be $\ge 2$. The conclusion follows.[/hide][/quote]
wwow , very nice PhilAndrew , i have another way to solve the problem ...
case 1: if $n$ is odd , take $p$ is a prime which divisible $n$ and then $[\frac{n}{p}]$ is odd , contradiction
case 2: if $n$ is even , take $p$ is a prime which divisible $n-1$ and then $[\frac{n}{p}]=[\frac{n-1}{p}+\frac{1}{p}]=[\frac{n-1}{p}]$ which is odd , contradiction

[hide]what about "odd instead even " , i mean what if the problem is
Let $n$ be an integer such that $[\frac{n}{p}]$  is odd for every prime . Prove that $n=1$ .[/hide]

[quote="t0rajir0u"]You can't just assert the existence of a prime between $5000$ and $15000$ (this is weaker than Bertrand's Postulate, but it is still far from obvious).  That kind of reasoning wouldn't get any points in an Olympiad setting.  You haven't given any justification at all.[/quote]

well i would say that is intuitively obvious since we know about primes for smaller numbers

obviously he did some handwaving and that would loose points

i would say that a more pertinent example would be something like a0^50 to 2*10^50 are there primes? the difficulty in this example is that we know relatively little about prime distribution for large numbers intuitively

[quote="PhilAndrew"][quote="HTA"]
[hide]what about "odd instead even " , i mean what if the problem is
Let $n$ be an integer such that $[\frac{n}{p}]$  is odd for every prime . Prove that $n=1$ .[/hide][/quote]
In fact, for any $n \ge 1$, if $p$ is a prime large enough, you get$\left[ \frac{n}{p}\right]$ = 0, which is not odd. That was the reason why I chose "even".[/quote]
what if i add more condition , with $p \leq n$ , is the above problem still true ?? I think it's still true but i cant solve it WHEN $n=2^{k}$ !!

What is your problem?  $n = 1$ doesn't work for your new condition, so do you mean something more along these lines:

Let $n \ge 2$ be a positive integer such that $\lfloor \frac{n}{p}\rfloor$ is odd for every prime $p \le n$.  Prove that $n$ is prime.

I am not sure if this problem is true, and in fact I think there are only finitely many such $n$.  I have found only $n = 2, 3$.

here is my problem

Find all the value of interger $n$ satisfying $[\frac{n}{p}]$ is odd for every prime $n \geq p \geq 3$

and here is the wat i thought
if n is odd , take p is a prime-divisor of $n-1$ and we have  $[\frac{n}{p}]=[\frac{n-1}{p}+\frac{1}{p}]$ and because $0\leq \frac{1}{p}\leq 1$ then $[\frac{n}{p}]=[\frac{n-1}{p}]=\frac{n-1}{p}$ and it's even because $n-1$ is even and p is odd which is contradiction
if n is even (and $n \neq 2^{k}$) take $p$ is a prime-factors of $n$ , and we have $[\frac{n}{p}]$ is even , contradiction
but i cant complet it , because the case $n=2^{k}$ is still unsolved ... is it true for $n = 2^{k}$ , or just my imagination :D


How often is the deduction theorem or otherwise called the Rule of conditional proof , used in Analysis.

Give examples

*****The first known statements of the deduction theorems for the first-order predicate calculus and the classical sentential logic are due to Herbrand [8] and Tarski [14], respectively. The present paper contains an analysis of closure spaces associated with those sentential logics which admit various deduction theorems. For purely algebraic reasons it is convenient to view deduction theorems in a more general form: given a sentential logic C (identified with a structural consequence operation) in a sentential language I, a quite arbitrary set P of formulas of I built up with at most two distinct sentential variables p and q is called a uniform deduction theorem scheme for C if it satisfies the following condition: for every set X of formulas of I and for any formulas  and , C(X{{a}}) iff P(, ) AC(X). [P(, ) denotes the set of formulas which result by the simultaneous substitution of  for p and  for q in all formulas in P]. The above definition encompasses many particular formulations of theorems considered in the literature to be deduction theorems. Theorem 1.3 gives necessary and sufficient conditions for a logic to have a uniform deduction theorem scheme. Then, given a sentential logic C with a uniform deduction theorem scheme, the lattices of deductive filters on the algebras A similar to the language of C are investigated. It is shown that the join-semilattice of finitely generated (= compact) deductive filters on each algebra A is dually Brouwerian.

*****A part of this paper was presented in abstracted form in Bulletin of the Section of Logic, Vol. 12, No. 3 (1983), pp. 111–116, and in The Journal of Symbolic Logic.

I thought this was the fun factory :P

[b]This is another great ignorance of this forum[/b].

[b]Nobody knows of how the deduction theorem is used in Analysis by giving an example[/b]??

Also how often the deduction theorem is used in Analysis.

[b]You have no Analysis without the deduction theorem[/b].

[b]You would not dare  leave this thread in the subforum ,calculus and Analysis[/b]

So the backbone of Analysis is [b] for fun[/b]??

[quote="fedja"][quote="τρικλινο"]
[b]You would not dare  leave this thread in the subforum ,calculus and Analysis[/b]
[/quote]
Nope, I wouldn't. Not after I've seen what your other posts in CP developed into. :P[/quote]

Since you decided to move the thread ,"The deduction Theorem" here do you perhaps know,what it is ,how is used in Analysis by giving an example??

[quote] great ignorance of this forum[/quote]
Posted in Epic thread

[hide]Darij style[/hide]


Let $ 0\leq k\leq n$ and $ m\geq 2$ be integers. Consider the $ k$-element subsets of $ \{1,2,...,n\}$; in every such subset, compute the residue of the sum of elements, modulo $ m$. Prove that if the $ m$ residues are distributed uniformly - i.e. every residue occurs exactly $ \frac {\binom{n}{k}}{m}$ times - then $ n\geq m$

So, let me understand this correctly -- you [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265770]posted[/url] this problem when it was an active contest problem, and now are double-posting it because the deadline has passed and you didn't get helpful responses in the first thread?  Or what?

Oh sorry. I confused this problem with another problem. This one is in the December 2008 issue.

And I posted the other problem without knowing that there's a deadline. I'm not even in Hungary, although whether you want to believe it or not is up to you.  :wink:


[quote="jclarkemathL314159"]Are we allowed to send in hand-written solutions if they are legible enough?[/quote]

Yes, I did it last year.

[quote="jclarkemathL314159"]Are we allowed to send in hand-written solutions if they are legible enough?[/quote]

Absolutely.


[quote="delegat"]Two circles with centers $ S_{1}$ and $ S_{2}$ are externally tangent at point $ K$. These circles are also internally tangent to circle $ S$  at points $ A_{1}$ and $ A_{2}$, respectively. Denote by $ P$one of the intersection points of $ S$ and common tangent to $ S_{1}$ and $ S_{2}$ at $ K$.Line $ PA_{1}$ intersects $ S_{1}$ at $ B_{1}$ while $ PA_{2}$ intersects $ S_{2}$ at $ B_{2}$.
Prove that $ B_{1}B_{2}$ is common tangent of circles $ S_{1}$ and $ S_{2}$.[/quote]
Just out of curiosity, by 3rd graders, do you mean that they are around 8 years old?  These problems would seem extremely tough to those young students.  Anyways, here is the solution:
[hide="Solution"]
Let the common tangent at $ A_2$ to $ S_2$ meet $ B_1B_2$ at $ Q$.  Furtermore, let the tangent from $ Q$ to $ S_2$ meet $ S_2$ at $ B'$ so that $ B'\ne A_2$.  It follows that $ PB_1\cdot PA_1 \equal{} PK^2 \equal{} PB_2\cdot PA_2$, so $ A_1A_2B_2B_1$ is cyclic.  Thus,
\[ \angle B_2A_2Q \equal{} \angle A_2A_1P \equal{} \angle B_1B_2P \equal{} \angle B_2A_2Q
\]
so $ A_2Q \equal{} B_2Q$.  Yet, $ B'Q$ and $ QA_2$ are common tangents from the same circle.  We have that $ B_2Q \equal{} A_2Q \equal{} B'Q$, which means that the circle with radius $ QA_2$ and center $ Q$ meets $ S_2$ at $ A_2$, $ B'$, and $ B_2$.  If all of these points were different, then two circles intersect at three points, which is impossible.  Thus, $ B_2 \equal{} B'$ since $ A_2\ne B_2$ and $ B'\ne A_2$.  It follows that $ QB_1$ is tangent to $ S_2$, so $ B_1B_2$ is tangent to $ S_2$.  Similarly, it is tangent to $ S_1$, so it is a common tangent of the two circles.  [/hide]

[quote]Just out of curiosity, by 3rd graders, do you mean that they are around 8 years old?[/quote] :coolspeak:

But it is Bosnian competition We prefer easier problems 3rd grades is equivalent to 11th grades (we hae 8+4 grades -primary + secondary schools) so students that were asked this question are aroun 17 years old. :(

Thanks for solution [b]The QuattoMaster 6000[/b]

Kind regards

[quote="delegat"]Two circles with centers $ S_{1}$ and $ S_{2}$ are externally tangent at point $ K$. These circles are also internally tangent to circle $ S$  at points $ A_{1}$ and $ A_{2}$, respectively. Denote by $ P$ one of the intersection points of $ S$ and common tangent to $ S_{1}$ and $ S_{2}$ at $ K$. Line $ PA_{1}$ intersects $ S_{1}$ at $ B_{1}$ while $ PA_{2}$ intersects $ S_{2}$ at $ B_{2}$.
Prove that $ B_{1}B_{2}$ is common tangent of circles $ S_{1}$ and $ S_{2}$.[/quote]
As dear [b]The QuattoMaster 6000[/b] showed: $ A_1A_2B_2B_1$ is cyclic. Let's call $ C_1$, $ C_2$ be the second intersections of $ A_1A_2$ with $ S_1$, $ S_2$, respectively. It's well-known (?) that $ B_1C_1$ and $ PA_2$ are parallel. It implies $ \angle B_2A_2A_1 \equal{} \angle B_1C_1A_1 \equal{} \angle PB_1B_2$, i.e., $ B_1B_2$ is tangent to $ S_1$ at $ B_1$. Similarly, $ B_1B_2$ is tangent to $ S_2$ at $ B_2$, this completes our proof.


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

I'm totally stucked...

let K be a field. M is a finite subgroup of the multiplicative group K\{0}. Show that M is cylic.

 :blush:  :blush:

It has been discussed before : [url]http://www.mathlinks.ro/viewtopic.php?search_id=1912499032&t=27050[/url]


probability
analytic geometry
geometry
inequalities
AMC
AIME

[quote="chess64"]Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 AM and 10 AM, and stay for exactly $m$ minutes. Find $m$ if the probability that either one arrives while the other is in the cafeteria is 40%.

[hide="what i got"]
$m=60 - 12\sqrt{15}$
[/hide]

Is it right?[/quote]
you can look in Kalva
 :)

do you know what linear programming is?
lets say that they arrive at times x and y, for them to meet, $|y-x| \ge m$
they can arrive from 0 to 60 mins so draw a box on a coordinate plane (with sides of 60) and
find the area that satisfies the inequality over the total area of the box
(the box representing the all the possible times that they could come)
and for that matter, i also got $60-12\sqrt{15}$

Yeah, that's exactly what I did. :)

I checked on Kalva (it took a while to find the right test) and it was #9!!! I'm getting better. :)

the thing is that is a pretty easy problem iff you know what to do, now you could get any problem that is really similar to that and get it right, since i recently have gone over linear programming, i knew instantly what to do, which is not usually the case on aime problems


number theory
greatest common divisor
prime factorization

If d|ab and gcd(a,b)=1, prove that d=d1d2 where d1|a and d2|b and gcd(d1,d2) = 1

The thing is I don't know where to start. When you guys first see this problem...what is your thought process?

My first thought is of the prime factorization of integers, like $18=2\cdot 3^{2}$. An integer $m$ divides another integer $n$ if and only if every prime appears in the factorization of $n$ at least as often as it does in the factorization of $m$.

[quote="lord12"]If d|ab and gcd(a,b)=1, prove that d=d1d2 where d1|a and d2|b and gcd(d1,d2) = 1
[/quote]

If $d|ab$ and $\gcd(a,b)=1$
Then either $d|a$ or $d|b$.

Say, $d|a$ for simplicity sake.

Then,
$d=d\cdot 1$
Where,
$d|a$ and $1|b$ and $\gcd(d,1)=1$  :rotfl:

[quote="sylow_theory"]If $d|ab$ and $\gcd(a,b)=1$
Then either $d|a$ or $d|b$.
[...]
 :rotfl:[/quote]
The joke is on you.

[hide]
Let $x$ be the gcd of $d$ and $a$, $y=d/x$, and $m=a/x$. Then since $d$ divides $ab$, $y$ must divide $mb$. However, $\gcd (y,m)=\gcd (a/x,d/x)=\gcd(a,d)/x=1$, so $y$ must divide $b$.

Now we only need to show $x$ and $y$ relatively prime. But of course this is true, since any common divisor of $x$ and $y$ must divide $a$ and $b$, but $a$ and $b$ are relatively prime.
[/hide]


I think I have put the question in a wrong section.
Please help!!!!!!!!! :maybe:

let $ y \plus{} 1 \equal{} x_1$ and $ a \equal{} x_2 \plus{} 1$, $ b \equal{} x_3 \plus{} 1$, $ c \equal{} x_4 \plus{} 1$.  Then we want the number of solutions to $ y \plus{} a \plus{} b \plus{} c \le 12$, with $ y$, $ a$, $ b$, and $ c$ all natural numbers.

Now imagine that a ball represents a $ 1$.  So we want the number of ways to divide up $ 12$ or fewer balls (as few as $ 4$) in a row into $ 4$ nonempty sets.  The first subset will represent $ y$, the second, $ a$, etc.  If we have $ n$ balls, there are $ n \minus{} 1$ places to place dividers, and we must place $ 3$ dividers.  So we have a total of

$ \sum_{n \equal{} 4}^{12}\binom{n \minus{} 1}{3} \equal{} \binom{12}{4} \equal{} 495$ integral solutions. (the summation here is simplified by the hockeystick identity)

Alternatively, let $ x_1'\equal{}x_1\minus{}2$ so that now the restriction is $ x_1',x_2,x_3,x_4\ge0$ and we have $ x_1'\plus{}x_2\plus{}x_3\plus{}x_4\le8$, which means that $ x_1'\plus{}x_2\plus{}x_3\plus{}x_4\equal{}8\minus{}k\implies x_1'\plus{}x_2\plus{}x_3\plus{}x_4\plus{}k\equal{}8$ for some $ k\in\mathbb Z\ge0$. This is a direct application of Balls and Urns, so we find that there are $ \binom{8\plus{}5\minus{}1}{5\minus{}1}\equal{}\binom{12}4\equal{}\boxed{495}$ solutions.

[quote="math154"]Alternatively, let $ x_1' \equal{} x_1 \minus{} 2$ so that now the restriction is $ x_1',x_2,x_3,x_4\ge0$ and we have $ x_1' \plus{} x_2 \plus{} x_3 \plus{} x_4\le8$, which means that $ x_1' \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 8 \minus{} k\implies x_1' \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} k \equal{} 8$ for some $ k\in\mathbb Z\ge0$. This is a direct application of $ \boxed{\text{Balls and Urns}}$, so we find that there are $ \boxed{\binom{8 \plus{} 5 \minus{} 1}{5 \minus{} 1} \equal{} \binom{12}4 \equal{} 495}$solutions.[/quote]

would you show me some sources to read about the proof of [color=blue]Balls and Urns[/color] stated as the above?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]I wrote something on Balls and Urns a while ago (case $ 4$).[/url]

A search might give more information.

[quote="Lizzie91"]The number of integral solutions of $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \leq 10$ where $ x_1 \geq 2$ and $ x_2,x_3,x_4 \geq 0$ is?

Thanks for any help.[/quote]
$ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}n$ where $ x_i\ge 0$
 Slove this $ C_{n\plus{}4\minus{}1}^{3}$


What reduced common fraction is equivalent to $ 18 \frac{1}{3}\%$


Let ABC  with AB=AC
-->D is the middle of [AC]
-->E is the orthogonal projection of D on [BC]  (in other words: E£ [BC] with (DE) is perpendicular with (BC))
-->and let F the middle of [DE]

prove that :
(AE) is perpendicular with (BF)        IF AND ONLY IF :         AB=BC=AC
(if and only if means equivalent to)
THX  :wink:

[quote="PDatK40SP"]Let $M$ is the midpoint of $BC$
$AE \perp BF \\ \Leftrightarrow \overrightarrow{AE}\cdot \overrightarrow{ BF}= 0 \\ \Leftrightarrow \overrightarrow{AE}\cdot ( \overrightarrow{AB}-\overrightarrow{AF}) = 0 \\ \Leftrightarrow \overrightarrow{AE}\cdot ( 2\overrightarrow{AB}-\overrightarrow{AE}-\overrightarrow{AD}) = 0 \\ \Leftrightarrow (\overrightarrow{AB}+3\overrightarrow{AC})(7\overrightarrow{AB}-5\overrightarrow{AC})=0 \\ \Leftrightarrow 16 \overrightarrow{AB}\cdot \overrightarrow{AC}= 15AC^{2}-7AB^{2}= 8AB.AC \\ \Leftrightarrow cos(\overrightarrow{AB},\overrightarrow{AC}) = \frac{1}{2}\\ \Leftrightarrow \angle BAC = 60^{o}\\ \Leftrightarrow AB=BC=CA$[/quote]

Hello,
I think theres some small error in your demo cuz:
U didn use M
And I didnt understand this:  [img]http://www.gnux.be/latex/data/4e90b6304f385cdffee7573f41e83f5d.png[/img]

I think that u used E instead of M..

Ah, sorry  :blush:  .. When I stared posting, I thought that $M$ would be used but when I posted my solution, Ididn't use $M$ and forgot to delete that line  :D
Because $F$ is the midpoint of $DE$ so $\overrightarrow{AF}= \frac{1}{2}(\overrightarrow{AD}+\overrightarrow{AE})$. Then, $2(\overrightarrow{AB}-\overrightarrow{AF}) = 2\overrightarrow{AB}-\overrightarrow{AD}-\overrightarrow{AE}$


Given 2n real numbers a1....an and b1...bn such that
$ \sum a_{i}^2 \equal{} \sum b_{i}^2 \equal{} 1$ $ \sum a_i b_i \equal{} 0$

prove that:

$ (\sum a_{i})^2 \plus{} (\sum b_{i})^2 \le n$


probability geometry calculus AMC USA(J)MO USAMO AIME

Hi

I'm good at Math at school I'm in the begin of grade 10. I get full marks  :)

But when I enter here I see that a lot of stuff I never know about I mean not even study it. And I want to join IMO or at least win my country Olympiad.

So can someone here take care of me and till me some ways to study more and know more knowledge :)

I study fast also

I hope someone say "I will"

Thanks too much

Note: I study HOLT books from America I study Algebra 1 until now and I'm working in Algebra 2 at school now

Here are some books you should take a look at:

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]Introduction to Counting and Probability[/url]

followed up by

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting and Probability[/url]

and

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204]Introduction to Geometry[/url]

with

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]AoPS Volume 1[/url]

and

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=102]AoPS Volume 2[/url]

Also helpful is the Olympiad Geometry Class on this website.

Depending on how strong you are in Algebra and Pre-Calculus, you may or may not need this book:

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]Intermediate Algebra[/url]

This book may also be too easy for you, but you should take a look at the table of contents to check:

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206]Introduction to Number Theory[/url]

After you cover the fundamentals above, you should turn your attention to more Olympiad-geared preparation, like the following books:

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=5003]Problems from the Book[/url]

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=3010]USAMO Tests from 1972-1986[/url]

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Merch.php?item_id=40]Hard Problems DVD[/url]

Remember: you can only get better by doing problems, problems, and more problems!

[quote="Differ"]How is Hard Problems even remotely useful?[/quote]

It's just a fun documentary; it'll be useful when you need a break from studying and just want to chew on some math "history" for a while.

Plus, the PDF that comes with it has past Olympiad problems and solutions.

If you're just starting competitive math, Introduction to Algebra should be where you start.  It has topics from Algebra I and Algebra II and it goes in-depth, with some contest problems using those concepts.  From there, you can progress to Introduction to Number Theory, then Introduction to Counting & Probability.  Introduction to Geometry should come next, then Intermediate Algebra and Intermediate Counting & Probability.

If contests is all you care about, though, study AoPS Volume 1 and AoPS Volume 2.  Those are helpful for contest math.  After that (you should be able to easily qualify for AIME and possibly USAMO, if you really understand everything in it) buy ACoPS (Art and Craft of Problem Solving by Paul Zeitz), then Problem Solving Strategies by Arthur Engel (referred to as Engel on these forums).  Also, studying previous years' tests will help.  On the AMC site, you can purchase some of those.

The AoPS classes are also very helpful.

I thought the Olympiad problems/solutions were widely available.

[quote="AIME15"]If you're just starting competitive math, Introduction to Algebra should be where you start.[/quote]

Introduction to Algebra may be too easy, as it is largely Algebra 1.

EDIT: I found a very good website with lots of problems you may want to take a look at: http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html

Thank you all guys but I'm from outside America and I never buy from the net but I have friends who always buy from Amazon I can ask them for help if it was the same way.

And for isabella2296 Thanks a lot but seems your link is broken  :)

I don't know how but I can not go to http://www.kalva.demon.co.uk/.

As isabella said, the website is http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html.

The original Kalva website got removed or something.

Thanks a lot.

I don't know the link yesterday wasn't opening but today its just work LOL.


WLOG, take x  :ge: y :ge: z.

The idea is to split the inequality into two easier ones.

LHS = xyz  :Sigma: x / ( :Sigma: x^2  :Sigma: xy) + xyz / ( :sqrt: (:Sigma: x^2) :Sigma: xy).

We prove xyz :Sigma: x / ( :Sigma: x^2  :Sigma: xy)  :le: 1/3.

That is equivalent to 3xyz :Sigma: x :le: :Sigma: x^2 :Sigma: xy.

x^2 :ge: y^2 :ge: z^2 while
yz :le: zx :le: xy, so by Chebyshev's inequality,
3xyz :Sigma: x :le: :Sigma: x^2 :Sigma: xy.

We prove xyz / ( :sqrt: (:Sigma: x^2) :Sigma: xy) :le:  :rt3: /9.

That is equivalent to 3 :rt3:  xyz :le: :sqrt: (:Sigma: x^2) :Sigma: xy, but that is trivial, applying AM-GM to :Sigma: x^2 and :Sigma: xy.

Summing the above two inequalities we get the desired result.


in a triangle a b c if
$8\cos a \cos b \cos c=1$
prove a b c  is equilateral

there are amny algaebric/pure trigno proof of thsi ...try this geometric one
[hide="prove that"]$OH=R\sqrt{1-8 \cos A \cos B \cos C}$ where $O$ is the circumcentre and $H$ is the orthocentre[/hide]
i f you don't get will post later :)

The triangle must be acute otherwise we get contradiction with the fact that $cosA\cdot cosB\cdot cosC>0$.
From AG-GM we have:
$\sum cos A\geq 3\sqrt[3]{cosA\cdot cosB\cdot cosC}=\frac{3}{2}$
But we have $\sum cos A\leq \frac{3}{2},(*)$,therefore $\sum cosA=\frac{3}{2}$ $\Leftrightarrow$ $cosA=cosB=cosC=\frac{1}{2}$ $\Leftrightarrow$ $A=B=C=\frac{\pi}{3}$
In order to prove $(*)$ use Jensen's inequality.


How many distinct six-letter rearrangements of the word ``Hawaii" can I form?

why did you do that?
could you please give a brief explanation?
also the answer is 180 --- maybe 7!/2*2

6! or 720 ways to arrange the letters if all of them are distinct. Divide by 2 for the double i's and by 2 for the double a's. So 720/(2*2) = 720/4 = 180


In the figure below, lines $A$ and $B$ are parallel.

Find $x$.

cool question! :D  :thumbup:
$x=55^{\circ}$
My solution is present in the diagram

An alternate solution
[hide]
Label the 100 degree angle C, Label the 110 degree angle d. Extend a line through CD.

We find that the angle supplementary to angle C is $80^{\circ}$. so the intersection of line CD and line A has the angle 25 degrees. Thus the intersection of line CD and angle B also has an angle of 25 degrees. Furthermore, the angle supplementary to D has degree 70 and the line drawn forms a convex quadrilateral with the 90 degree and 100 degree angles. Therefore the remaining angle there is 100 degrees. That means the angle opposite this is 100 degrees also. So to find x we have $180 - 25 - 100 = 55^{\circ}$[/hide]

sorry for the poor quality of the explanation -- I lack the wonderful aid of geometer's sketchpad as I am currently on linux.


Solve the equation: $(z^m+i)^n=(z^m-i)^n$ where $m,n>=2$ are natural numbers and $z$is complex.

$z = \displaystyle\left ( \frac {cos(\frac{a\pi}{n})}{sin(\frac{a\pi}{n})}\right )^{\frac 1m}\displaystyle e^{i\frac{2b\pi}m}$, where $a\in\{1,..,n-1\},b\in\{0,1,..,m-1\}$
 :cool:

I note $\frac{z^m+i}{z^m-i}=t$. The ecuation becomes $t^n=1$ with the solutions $w_k\ ,k\in \overline{0,n}$,

where $w_k=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}$. Thus, $\frac{z^m+i}{w_k}=\frac{z^m-i}{1}=\frac{2z^m}{1+w_k}=\frac{2i}{w_k-1}$.

$z^m=i.\frac{1+\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}}{-1+\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}}=i.\frac{2\cos^2\frac{k\pi}{n}+2i\sin\frac{k\pi}{n}\cos\frac{k\pi}{n}}{-2\sin^2\frac{k\pi}{n}+2i\sin\frac{k\pi}{n}\cos\frac{k\pi}{n}}=\cot\frac{k\pi}{n}$.

$z^m=\cot\frac{k\pi}{n}$, where  $k\in \overline{1,n}$ a.s.o. We obtain $mn$ solutions for the initial ecuation.


induction
number theory
relatively prime
combinatorics unsolved
combinatorics

[quote="D.P.L"]Let n>1 let $a_{1}...a_{\phi(n)}$ be numbers in set A={1,...n} that are relatively prime to n

Then

$\sum_{k=1}^{\phi(n)}a_{k}^{2}= \frac{\phi(n)}{6}(2n^{2}+(-1)^{k}p_{1}...p_{i})$ where $p_{1},..., p_{i}$ are distinct prime divisors of n[/quote]
I think we can do as following hint
+) Consider $n\in{P}$ .We have ($a_{1},...,a_{n-1}$) is a permuation of (1,2,...,n-1)
We can prove by induction
+) After that ...Wait for you or wait for me a few days :)


Determine whether the seried $ \sum_{n \equal{} 2}^{\infty}a_n$ converges absolutely,converges conditionally or diverges.If

$ a_n \equal{} \frac {1 \plus{} n \plus{} n^2}{\sqrt {1 \plus{} n^2 \plus{} n^6}}$

For several n I get $ a_n > \frac {1}{n}$ so I decided that this series is divergent.Right?

[quote="azatkgz"]

For several n I get $ a_n > \frac {1}{n}$ so I decided that this series is divergent.Right?[/quote]
That's right!  :wink:

Technically, "For several $ n$" is not good enough. One could construct a positive sequence $ a_n$ such that $ \limsup_{n\to\infty}na_n>1$ but yet $ \sum_{n\equal{}1}^{\infty}a_n$ converges.

If $ a_n$ "looks like" $ \frac1n$ for [b]all[/b] sufficiently large $ n,$ then that is good enough. That's the limit comparison test, and it is the appropriate way to argue this.

Come to think of it, this is a double post. azatkgz, you previously asked the exact same question, and I gave you the exact same answer: limit comparison to $ \sum\frac1n.$ Do you even read the topics you start?

[quote="Kent Merryfield"]

Come to think of it, this is a double post. azatkgz, you previously asked the exact same question, and I gave you the exact same answer: limit comparison to $ \sum\frac1n.$ Do you even read the topics you start?[/quote]

Oh,I see.I forgot that I had posted it yesterday. :)


inequalities
trigonometry
inequalities proposed

Let $a,b$ and $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that
\[ \frac{a}{2+b^2+c^2}+\frac{b}{2+c^2+a^2}+\frac{c}{2+a^2+b^2}\ge\frac{3\sqrt{3}}{8} \]

Ok,I post my solution :)
Apply Cauchy-Schwarz's inequality we have
              $\sum \frac{a^2}{2a+b^2a+c^2a} \geq \frac{(a+b+c)^2}{2(a+b+c)+\sum_{sym}a^2b}$
From the condition $ab+bc+ca=1 \rightarrow \sum_{sym} a^2b =(a+b+c)-3abc$
So that we need to prove $\frac{(a+b+c)^2}{3(a+b+c)-3abc} \geq \frac{3\sqrt{3}}{8}$
                                     $\Longleftrightarrow 8(a+b+c)^2+9\sqrt{3}abc \geq 9 \sqrt{3}(a+b+c)$
Use homogenrous we only need to prove:
                                     $8(a+b+c)^2\sqrt{ab+bc+ca}+9\sqrt{3} \geq 9\sqrt{3} (a+b+c)(ab+bc+ca)$
                                     $\Longleftrightarrow 8(a+b+c)\sqrt{ab+bc+ca}[(a+b+c)-\sqrt{3}\sqrt{ab+bc+ca}] \geq \sqrt{3}[(a+b+c)(ab+bc+ca)-9abc]$
                                     $\Longleftrightarrow \frac{4(a+b+c)\sqrt{ab+bc+ca}}{a+b+c+\sqrt{3}\sqrt{ab+bc+ca}} (\sum (b-c)^2) \geq \sqrt{3}(\sum a(b-c)^2)$
Put:$S_{a}=\frac{4(a+b+c)\sqrt{ab+bc+ca}}{a+b+c+\sqrt{3}\sqrt{ab+bc+ca}}-\sqrt{3}a$
      $S_{b}=\frac{4(a+b+c)\sqrt{ab+bc+ca}}{a+b+c+\sqrt{3}\sqrt{ab+bc+ca}}-\sqrt{3}b$
      $S_{c}=\frac{4(a+b+c)\sqrt{ab+bc+ca}}{a+b+c+\sqrt{3}\sqrt{ab+bc+ca}}-\sqrt{3}c$
WOLG $a \geq b \geq c$
From that condition we have $S_{b},S_{c} \geq 0$
So we only need to prove $a^2S_{b}+b^2S_{a} \geq 0$
                                      $\Longleftrightarrow 4(ab^2+b^2c+a^2b+a^2c+a^3+b^3)\sqrt{ab+bc+ca} \geq \sqrt{3} (a+b+c)(a^2b+b^2a)+3\sqrt{ab+bc+ca}(a^2b+b^2a)$
                                      $\Longleftrightarrow 4(a^3+b^3+a^2c+b^2c)\sqrt{ab+bc+ca}+(a^2b+b^2a)\sqrt{ab+bc+ca} \geq \sqrt{3}(a+b+c)(a^2b+b^2a)$
We can prove:$4(a^3+b^3)\sqrt{ab}+(a^2b+b^2a)\sqrt{ab} >\sqrt{3}(a+b)(a^2b+b^2a)$
So $4(a^3+b^3)\sqrt{ab+bc+ca} +(a^2b+b^2a)\sqrt{ab+bc+ca} >4(a^3+b^3)\sqrt{ab}+(a^2b+b^2a)\sqrt{ab} >\sqrt{3} (a^2b+b^2a)(a+b)$(1)
Similar $4a^2c\sqrt{ab+bc+ca} >\sqrt{3}c(a^2b+b^2a)$ (Use the condition $a \geq b \geq c$)(2)
From (1) and (2) we have $S_{a}b^2+S_{b}a^2 \geq 0$
Now we will prove with the conditions $a \geq b \geq c$ and $S_{b},S_{c},a^2S_{b}+b^2S_{a} \geq 0$ then $\sum S_{a}(b-c)^2 \geq 0$ (*)
Solution:
Because $S_{c} \geq 0 and (a-b)^2 \geq 0$ so to prove (*) we only need to prove $(b-c)^2S_{a}+(a-c)^2S_{b} \geq 0$
We partition into two cases:
     +Case 1:when $S_{a}+S_{b} \geq 0$ then:($b-c)^2S_{a}+(a-c)^2S_{b} \geq (b-c)^2 (S_{a}+S_{b}) \geq 0$ (because $S_{a}+S_{b} \geq 0$)
     +Case 2:when $S_{a}+S_{b} \leq 0$.Put $P_{a}=(b-c)S_{a},P_{b}=(a-c)S_{b}$
We again partition into two cases:
     +Case 1':$P_{a}+P_{b} \geq 0$.Then $(b-c)^2S_{a}+(a-c)^2S_{b}=(b-c)P_{a}+(a-c)P_{b} \geq (b-c)(P_{a}+P_{b}) \geq 0$
     +Case 2':$P_{a}+P_{b} \leq 0$.We have $(b-c)P_{a}+(a-c)P_{b}=bP_{a}+aP_{b}-c(P_{a}+P_{b})$
So to prove $(b-c)P_{a}+(a-c)P_{b} \geq 0$ we only need to prove $bP_{a}+aP_{b} \geq 0$ (because $-c(P_{a}+P_{b}) \geq 0$)
we have:$aS_{b}+bS_{a} \leq aS_{b}+bS_{a}-c(S_{a}+S_{b})$ (because $S_{a}+S_{b}\leq 0$)
                                              $=(a-c)S_{b}+(b-c)S_{a}=P_{a}+P_{b} \leq 0$
To prove:$bP_{a}+aP_{b} \geq 0 \Longleftrightarrow b(b-c)S_{a}+a(a-c)S_{b} \geq 0 \Longleftrightarrow b^2S_{a}+a^2S_{b}-c(bS_{a}+aS_{b}) \geq 0$ (True becuse $aS_{b}+bS_{a} \leq 0$)
Hence $(b-c)P_{a}+(a-c)P_{b} \geq 0$
Therefore $\sum S_{a}(b-c)^2 \geq 0$
So that the problem is proved  :D  :D

That's so incredible.How did you find this?I think maybe there should exist an easy one to understand.

What about trig substitution $a=\tan{\frac{A}{2}}$...?

And the following is also nice:
\[ \frac a{1+b^2+c^2}+\frac b{1+c^2+a^2}+\frac c{1+a^2+b^2}\geq\frac{3\sqrt3}5. \]
but the idea of trigon sub may fail.

[quote="hungkhtn"]USe Cauchy-Schwarz only.  ;)[/quote]

I'm eagerly waiting and hope that hungkhtn will post his Cauchy-Schwarz only solution.

[quote="pvthuan"]And the following is also nice:
\[ \frac a{1+b^2+c^2}+\frac b{1+c^2+a^2}+\frac c{1+a^2+b^2}\geq\frac{3\sqrt3}5.  \]
[/quote]
Yes,  this is nicer, but wrong. Try $c=0,a=b=1.$ ;)

Oh, that is so unusual and curious.  :blush:

[quote="nhat"] $8(a+b+c)^2\sqrt{ab+bc+ca}+9\sqrt{3}abc \geq 9\sqrt{3} (a+b+c)(ab+bc+ca)$
:D[/quote]

This equality equivanlent to

$64(a+b+c)^4(ab+bc+ca) \ge 243(a+b)^2(b+c)^2(c+a)^2$

which was discussed in our forum sometime ago. The proof is not hard.

[quote="hungkhtn"]USe Cauchy-Schwarz only.  ;)[/quote]

Can you please post your only BCS solution?

Thank you very much.

[quote="ductrung"]Let $ a,b$ and $ c$ be positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ \frac {a}{2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b}{2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c}{2 \plus{} a^2 \plus{} b^2}\ge\frac {3\sqrt {3}}{8}
\]
[/quote]
$ \sum_{cyc}\frac {a}{2 \plus{} b^2 \plus{} c^2}\equal{}\sum_{cyc}\frac{a^2}{2a\plus{}b^2a\plus{}c^2a}\geq\sum_{cyc}\frac{(a\plus{}b\plus{}c)^2}{3(a\plus{}b)(a\plus{}c)(b\plus{}c)}.$
Thus, it remains to prove that
 $ 64(a\plus{}b\plus{}c)^4(ab\plus{}ac\plus{}bc)\geq243(a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2.$
But $ 64(a\plus{}b\plus{}c)^4(ab\plus{}ac\plus{}bc)\geq243(a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(64a^5b\plus{}13a^4b^2\plus{}45a^4bc\minus{}51a^3b^3\minus{}50a^3b^2c\minus{}21a^2b^2c^2)\geq0,$
which triviality.

[quote="primoz2"]Arqady, what inspired you to use that inequality  $ \sum_{cyc}\frac {a}{2 \plus{} b^2 \plus{} c^2} \equal{} \sum_{cyc}\frac {a^2}{2a \plus{} b^2a \plus{} c^2a}\geq\sum_{cyc}\frac {(a \plus{} b \plus{} c)^2}{3(a \plus{} b)(a \plus{} c)(b \plus{} c)}$,   because it is far away from being obvious.[/quote]
Why? We use the Cauchy-Schwarz's inequality. See the nhat's post. :wink:

Hey fuzzylogic. Could you post your solution? It is a nice idea! I'm trying to use trigonometric.

[quote="Ioseph Djugatchvili"]Hey! Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq \frac {3\sqrt {3}}{2}$  :blush:[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="Ioseph Djugatchvili"][quote="Ioseph Djugatchvili"]Hey! Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq \frac {3\sqrt {3}}{2}$  :blush:[/quote]
Remember: $ ab \plus{} bc \plus{} ca \equal{} 1$ Our problem equivanlents to it! I'm trying![/quote]
If $ ab\plus{}bc\plus{}ca\equal{}1$, then $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq \frac {5}{2}$  :wink:

[quote="arqady"][quote="ehku"]
If $ ab \plus{} bc \plus{} ca \equal{} 1$, then $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq \frac {5}{2}$  :wink:[/quote]
For non-negative $ a,$ $ b$ and $ c$ it's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476[/quote]
But there is not a nice solution there  :(

[b][u]Problem:[/u][/b] Let $ a, b, c$ be positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq \frac {3\sqrt {3}}{2}$
[b]From 4232 http://www.diendantoanhoc.net [/b]

$ \sum\limits_{cyc} {\frac {1}{{a \plus{} b}}} \equal{} \frac {{(a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca}}{{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc}}$
Let $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca \equal{} 1,r \equal{} abc$
$ p^2 \ge 3q \equal{} 3 \Rightarrow p \ge \sqrt 3$
We have:
$ 3p(p \minus{} \sqrt 3 ) \plus{} 3\sqrt 3 r \plus{} 2 \ge 0$ (wright)
We solved this problem. "=" when a=b=c=$ \frac {1}{{\sqrt 3 }}$

Is it wrong?  :huh:
 :huh:

Your inequality should be written into

$ p(2p\minus{}3\sqrt{3})\plus{}2\plus{}3\sqrt{3}r \ge 0$

But not

$ 3p(p\minus{}\sqrt{3})\plus{}2\plus{}3\sqrt{3}r \ge 0$

And

$ p(2p\minus{}3\sqrt{3})\plus{}2\plus{}3\sqrt{3}r \ge 0$

is wrong for $ a\equal{}b\equal{}1,c\equal{}0$ such that $ p\equal{}2, q\equal{}1, r\equal{}0$

[quote="Ioseph Djugatchvili"][b][u]Problem:[/u][/b] Let $ a, b, c$ be positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq \frac {3\sqrt {3}}{2}$
[quote]
Sorry! My inequality is wrong! Sorry and sorry! :blush:  :blush:  :D  :arrow:  :D

[quote="ductrung"]Let $ a,b$ and $ c$ be positive real numbers such that $ ab + bc + ca = 1$. Prove that
\[ \frac {a}{2 + b^2 + c^2} + \frac {b}{2 + c^2 + a^2} + \frac {c}{2 + a^2 + b^2}\ge\frac {3\sqrt {3}}{8}
\]
[/quote]Here is my proof: from ab+bc+ca=1,we have $ b^2 + c^2 + 2 = (b + c)(b + c + 2a)$
$ \frac {a}{b^2 + c^2 + 2} + \frac {b}{a^2 + c^2 + 2} + \frac {c}{a^2 + b^2 + + 2} = \frac {a}{(b + c)(b + c + 2a)} + \frac {b}{(a + c)(a + c + 2b} + \frac {c}{(a + b)(a + b + 2c)}$
By cauchy-schwarz:
${ (\sum\frac {a}{(b + c)(b + c + 2a)})[a(b + c) + b(c + a) + c(a + b)] \ge (\sum\frac {a}{\sqrt {b + c + 2a}})^2}$
By holder,we get :
$ (\sum\frac {a}{\sqrt {b + c + 2a}})^2[\sum a(b + c + 2a)] \ge (a + b + c)^3$
And finally,we need to prove that:
$ 2(a + b + c)^3 \ge 3\sqrt {3(ab + bc + ca)}(a^2 + b^2 + c^2 + ab + bc + ca)$
We can assume a+b+c=3.setting x=ab+bc+ca.and we need to prove
$ 3\sqrt {3x}(9 - x) \le 54$ or $ x(x - 9)^2 \le 108$ or $ (x - 3)^2(x - 12) \le 0$ true because $ x \le 3$
our proof are completed.:)Am I right? I think I wrong,please help.

We only need to prove: $ (a\plus{}b\plus{}c)^2 \sqrt{ab\plus{}bc\plus{}ca} \geq\ \frac{9\sqrt{3}}{8}(a\plus{}b)(b\plus{}c)(c\plus{}a)$
It's an old problem and pqr kill it easily   :)


A classical problem :)

Given 13 identical balls with only one of them having different weight . Now suppose you are given a balance to weight that [i]special ball[/i] out . What is the least step so that you can always weight it out ? Demonstrate your steps also :D

This is a very well known problem. It is known that using the balance n times, we can find the special ball from $\frac{3^n-1}2$ or less balls, so 3 is the least amount of step.
Let's number the balls from 1 to 13.

Step1) Weigh 5,6,7,11 with 8,9,10,12.
Step2) Weigh 2,3,4,11 with 5,6,7,12.
Step3) Weigh 1,4,7,8 with 2,5,10,11
There are 27 possibilities. Below is the special ball for the all possibilities. (balance,right,left) means that the first step balanced, right side was heavier in the second step, and left side was heavier in the third step.

balance,balance,balance -> 13 is special
balance,balance,right -> 1 is special
balance,balance,left -> 1 is special
balance,right,balance -> 3 is special
balance,right,right -> 4 is special
balance,right,left -> 2 is special
balance,left,balance -> 3 is special
balance,left,right -> 2 is special
balance,left,left -> 4 is special

right,balance,balance -> 9 is special
right,balance,right -> 10 is special
right,balance,left -> 8 is special
right,right,balance -> 12 is special
right,right,right -> impossible
right,right,left -> 11 is special
right,left,balance -> 6 is special
right,left,right -> 7 is special
right,left,left -> 5 is special

left,balance,balance -> 9 is special
left,balance,right -> 8 is special
left,balance,left -> 10 is special
left,right,balance -> 6 is special
left,right,right -> 5 is special
left,right,left -> 7 is special
left,left,balance -> 12 is special
left,left,right -> 11 is special
left,left,left ->  impossible

[quote="shyong"]Wow , thats kool . You even list out all possiblt outcome  :D . I have checked every of them and they are all correct  :D  . By the way , how do you get that $\frac{3^n-1}2$ ? Im kinda interest in it   :roll:[/quote]
Oh I didn't get that. One of my book had the result with an informal proof, which is too long to post.

Explanation:

When you are using a beam-balance to weigh balls, each weighing would have one of the following 3 results.

The left arm is heavier --> Term this case as '0'.
The right arm is heavier --> Term this case as '2'.
Both the arms balance each other --> Term this case as '1'.

So, with 'n' weighings we would be having 3^n results. Let me describe the case with n = 3 where there would have one of these 3^3 = 27 different results -- 000, 001, 002, 010. 011, 012, 020, 021, 022, 100, 101, 102, 110. 111, 112, 120, 121, 122, 200, 201, 202, 210. 211, 212, 220, 221, and 222.

If we know that the odd ball is 'heavier' (or lighter) than any of the other good balls then we can pick the odd one out from a maximum of 27 given identical looking balls.
Let me balls be marked as characters as shown below and the number beside it is the logical identification.

a - 000
b - 001
c - 002
d - 010
e - 011
f - 012
g - 020
h - 021
i - 022
j - 100
k - 101
l - 102
m - 110
n - 111
o - 112
p - 120
q - 121
r - 122
s - 200
t - 201
u - 202
v - 210
w -  211
x -  212
y - 220
z - 221
* - 222

(i) 1st Weighing ==> Weigh all the balls with 0 in the first digit in the left side of the balance with other balls that have 2 in the first digit of its logical identification. This means weigh (a, b, c, d, e, f, g, h, i) with (s, t, u, v, w, x, y, z, *).

(ii) 2nd Weighing ==> Weigh all the balls with 0 in the second digit in the left side of the balance with other balls that have 2 in the second digit of its logical identification. This means weigh (a, b, c, j, k, l, s, t, u) with (g, h, i, p, q, r, x, y, z, *)

(iii) 3rd weighing ==> Weigh all the balls with 0 in the third digit in the left side of the balance with other balls that have 2 in the third digit of its logical identification. This means weigh (a, d, g, j, m, p, s, v, y with (c, f, i, l, o, r, u, x, *)

Suppose, the weighing result was 0-1-2. It means that the left side was heavy in the first weighing; it balanced out in the second and the right side was heavier in the third weighing.  If you look at the logical table above, it's apparent that 'f' is the heavier odd ball (as f stands for 012).
----------------------------------------------------------------------------------------------------

Unique identification of odd ball would compel us to throw half of these balls away and to select '111' and either of '000' or '222'.

Total number of balls that can be uniquely identified = (3^n - 1)/2}

This is the explanation that the book had, but it was longer in the book. :D

I have made the explanation shorter even though I have never read the book. First, I have kept the 3 balls aside -- '000', '111' and '222'.

So, we are left with (3^n - 3) balls. Now, if have to remove one of every complementary pair of balls as I did explain yesterday. What I mean is that you can not have '012' and '210' in the same set as you cannot say if 'f' is heavier (odd) or 'v' is lighter (odd). So, we are left with (3^n - 3)/2 balls after removing one of those complementary pairs.

b - 001
d - 010
e - 011
f - 012

o - 112
p - 120
q - 121
r - 122

s - 200
t - 201
u - 202
y - 220

The above balls would serve in your purpose. And, we can the ball # '111' in our set of balls as it is not going to create any imbalance in the unique identification. Do remember the ways of weighing and noting of the results as shown in my last message.

No. of balls = (3^n - 3)/2  + 1 = (3^n - 1)/2

Sometimes things are not that difficult.............

So that is how the logic works  . I understand already :D Thx [b]Shamik Banerjee[/b]  :)

@LigThrEE:how do u select pairs of balls to be wieghed together?

@SHAMIK: i didn't understand onething that why u need to select (3^n-1)/2 out of 3^n.wht does this mean that {'f' is heavier (odd)or 'v' is lighter(odd) cant work together?}

[quote="lightrhee"]It is known that using the balance n times, we can find the special ball from $\frac{3^{n}-1}2$ or less balls[/quote]
What about if we have four balls and we can use just the balance only twice?

1,2        and           3,4
             then

       1            and          2

                   or
        3            and           4
the whichever one is heavier you choose.

What about if we don't know that our ball is heavier than the others?

yeah I know. U figure out by doing the weighing

But then, in the worst case, you need to use the balance more times and the formula lightrhee gave won't be true in the case of four balls.


graph theory
pigeonhole principle
combinatorics unsolved
combinatorics

A group consist of n tourists. Among every 3 of them there are 2 which are not familiar. For every partition of the tourists in 2 buses you can find 2 tourists that are in the same bus and are familiar with each other. Prove that is a tourist familiar to at most $\displaystyle \frac 2{5}n$ tourists.

Consider the graph G which describes the familiarity between the tourists.
Clearly G is not bipartie, so G contains an odd cycle, so we choose the odd cycle C with smallest length.Since G is traingle-free l>=5, where l denotes the length of C. We assume that C= A(1)...A(l).
We shall prove that if there exist a vertex B which is not in C and is adjacant to at  least two vertex A(i) and A(j) of C then i and j will differ by 1 ( the index are written mod l), inparticular each vertex B of G will be adjacant to at most 2 vertex of C.
Proof:Let u and v be the length of the two path which is seperated from C by removing A(i) and A(j), call those paths C(u) and C(v). Clearly u+v+2=l, so at least one of them, say u is even. Now consider the odd cycle BA(i)C(u)A(j), its length is u+3, which is odd, so u+3>= l=u+v+2, so v<=1 and thus v=1 ( since if v=0 G will have a triangle), so we are done.
Now let S(i) be the number of adjacant vertices of A(i) ( for i=1,2,...l). Using the above fact, we have counted each vertex of G at most 2 times in the sum S(1)+...+S(l), thus S(1)+...+S(l)<=2n so there exist a i such that S(i)<=2n/l<=2n/5 and we are done.

[quote="kueh"]Is it well-known that a non-bipartite graph contains an odd cycle? How do you prove it?[/quote]It is well known, it's due to Konig in 1936, and a search reveals a proof: http://www.mathlinks.ro/Forum/viewtopic.php?t=15325

Conclusion: don't be afraid (or lazy) to use the search function ;)

[quote="lhvietbao"]
We shall prove that if there exist a vertex B which is not in C and is adjacant to at least two vertex A(i) and A(j) of C then i and j will differ by 1...[/quote]

I don't understand. There is no 3-cycles!?

[quote="Number1"]

I don't understand. There is no 3-cycles!?[/quote]

Because it was mentioned that there are no triangles in the graph.
Btw, it's a very tough problem and I was quite happy with myself that I scored 3 points on it (out of 7)

I have the same dilemma as Number1 had or still have. Can you explain it loser more clearly ?? :?

[quote="Valentin Vornicu"]Among every 3 of them there are 2 which are not familiar.[/quote]

Hence there are no triangles (3-cycles)... IS it clear now?

I still don't understand this Ihvetbao's solution.

I do understand that there is no triangles.
I do not understand have can then somebody (Ihvetbao) prove that there are triangles.

i  guess  one might  add  that  with a point  in  the graph is  assigned  to  a specific person if two  persons are familiar then  we trace the line with its respective points  if theyre not we dont trace  the line.  it  says  that  out of any three  people 2 of them  are not familiar  hence if we have  a triangle  then  we  have three  people that know  each  other  wich  is a contradiction  i  guess  for inexperienced  solvers this analogy  (using graphs) is not well  known  now do you  understand the problem?  by  the way  its very  nice!!! ;)

Sorry for reviving an old thread, but I think I have an alternate solution (assuming this works, please point out any flaws):

[hide] We proceed by contradiction. Take any tourist $ A$. Let $ S$ be the set of all tourists that $ A$ knows, and $ S'$be the set of tourists he doesn't. Then no two people in $ S$ know each other. At least two people in $ S'$ must know each other (if not, we can split the $ n$ people into two buses, one consisting of $ S$ and one consisting of $ A$ and $ S'$, but then no pair of people know each other in either group, contradiction). Let $ k$ be the number of people in $ S'$;$ k$ is at most $ (3/5)n\minus{}1$ but at least $ (2/5)n\minus{}1$, or otherwise the people in $ S$ cannot know more than $ (2/5)n$ people.

Now we have three cases:

Case 1: There exist two people $ B$ and $ C$ in $ S'$ familiar with each other but not with anyone in $ S$. Furthermore, if $ B$ knows a person $ D$ in $ S'$, then $ C$ cannot (and vice versa). Hence, by Pigeonhole, either $ B$ or $ C$ knows at most $ (3/10)n$ people, contradiction.

Case 2: There exist $ B$and $ C$ in $ S'$ familiar with each other and each familiar with at least one person in $ S$. Assume $ B$ knows $ D$. Then $ D$ knows at least$ (2/5)n\minus{}1$ people in $ S'$ by assumption, and $ B$ cannot know any of these people. So $ B$ knows at most $ k\minus{}((2/5)n\minus{}1)\equal{}k\minus{}(2/5)n\plus{}1$ people in $ S'$. The same argument applies for $ C$. Since there are $ n\minus{}1\minus{}k$ people in $ S$, one of $ B$ or $ C$ knows at most $ (n\minus{}k\minus{}1)/2$ people in $ S$, so one of them knows at most $ n/2\minus{}k/2\minus{}1/2\plus{}k\minus{}(2/5)n\plus{}1\equal{}(1/10)n\plus{}(k/2)\plus{}1/2)$ people total, but as $ k<\equal{}(3/5)n\minus{}1$, the maximum value of this is $ (2/5)n$, contradiction.

Case 3: Neither of the previous cases is true, so for every pair of people in $ S'$ who know each other, exactly one of them knows someone in $ S$. Put all the people in $ S$ and all the people in $ S'$ who don't know anyone in $ S$ on the same bus, and put the people in $ S'$ who know someone in $ S$ and $ A$ in the other. Clearly, no one in either bus knows each other, contradiction.

Hence, in all cases we reach a contradiction and we are done.[/hide]

The problem is equivalent to: A graph $G$ consists of $n$ vertices. Among any three vertices, there are $2$ of them that do not share an edge. Finally, $G$ is not bipartite. Prove there is a vertice with degree at most $2n/5$

Solution: Since $G$ is not bipartite, it must contain an odd cycle that is not a triangle. Consider the smallest odd cycle $S =$ {$v_1, v_2,v_3,...,v_{2k+1}, v_1$}. Note that $v_i$ may only be adjacent to $v_{i-1}$ and $v_{i+1}$ and for every vertex in $G$ \ $S$, only 2 vertices in $S$ may be adjacent to it. Otherwise, this would contradict the minimality of the length of cycle $S$. The smallest possible length for $S$ is 5 and there must exist a vertex in $S$ that has degree at most $2(n-5)/5 +2 = 2n/5$, as desired.

1) It's not bipartite graph, so there has to be odd cycle which length is greater than 3

2) Now we consider the shortest odd-length cycle $v_{1},v_{2},\cdots\,v_{2k+1}$.$(k\geq2)$ there is no edge between $v_{i}$ due to minimality condition of length

3) If every vertex has degree greater than $\frac{2n}{5}$, each vertex $v_{i}$ has some edges to vertex $x_{i}$ out of the cycle. ($ i=1,2,\cdots\,n-2k-1$) Let $G_{i}$ be the set of $x_{i}$ connected to $v_{i}$, it holds $|G_{i}|=\left\lfloor\frac{2n}{5}\right\rfloor-1$.

4) If any vertex $x_{i}$ connects $v_{j}$ more than 3, we can find contradiction easily. (More shorter cycle is found in this cycle)

5) there are at least $E=(2k+1)(\left\lfloor\frac{2n}{5}\right\rfloor-1)$ edges between $x_{i}$ and $v_{j}$. As we confirmed above, $\frac{E}{n-2k-1}\leq2$ leads to conclusion that $k<2$ by some calculation which violates length condition. So we're done.


The relation $ a_k \equal{} \sum_{i\equal{}1}^{k\minus{}1} i(k\minus{}i)$ does not necessarily hold for $ k>n$. You need to remove the terms for which $ i>n$ or $ k\minus{}i>n$.

After removing them, you'll have the result immediately.

[quote="qwerty414"]The relation $ a_k \equal{} \sum_{i \equal{} 1}^{k \minus{} 1} i(k \minus{} i)$ does not necessarily hold for $ k > n$. You need to remove the terms for which $ i > n$ or $ k \minus{} i > n$.

After removing them, you'll have the result immediately.[/quote]
We can just evaluate $ a_0\plus{}a_1\plus{}...\plus{}a_n$ and then $ a_0\plus{}a_1\plus{}...\plus{}a_{2n}\equal{}P(1)$

Casn someone explain why $a_{k}=\sum_{i=1}^{k}i(k-i)$??
I can't understand it.

Thanks.


Let sequence   $ x_1\equal{}a\plus{}\sqrt{a},x_2\equal{}\sqrt{a\plus{}\sqrt{a}},...,x_n\equal{}\sqrt{a\plus{}\sqrt{a\plus{}\sqrt{...\plus{}\sqrt{a}}}}$
(i) Show that exist limit of $ x_n$ then $ n\rightarrow \infty$
(ii) Calculus this limit.

$ x_{n\plus{}1}\equal{}\sqrt{x_{n}\plus{}a}$

$ A\equal{}\lim_{n\to \infty }x_{n}\equal{}x_{\infty }$

$ \lim_{n\to \infty }x_{n}\equal{}\lim_{n\to \infty }\sqrt{x_{n\minus{}1}\plus{}a}\equal{}\sqrt{x_{\infty }\plus{}a}$

$ \lim_{x\to \infty }x_{n}\equal{}A\equal{}\sqrt{A\plus{}a}\equal{}\frac{1\plus{}\sqrt{1\plus{}4a}}{2}> \sqrt{a}$

[quote="bangvn"]Let sequence   $ x_1 \equal{} a \plus{} \sqrt {a},x_2 \equal{} \sqrt {a \plus{} \sqrt {a}},...,x_n \equal{} \sqrt {a \plus{} \sqrt {a \plus{} \sqrt {... \plus{} \sqrt {a}}}}$
(i) Show that exist limit of $ x_n$ then $ n\rightarrow \infty$
(ii) Calculus this limit.[/quote]

is it right? i think it should be $ x_1 \equal{} \sqrt {a}$  :wink:


geometry
geometric transformation
reflection
geometry proposed

We are given a triangle $ ABC$ such that $ AC \equal{} BC$. There is a point $ D$ lying on the segment $ AB$, and $ AD < DB$. The point $ E$ is symmetrical to $ A$ with respect to $ CD$. Prove that:
\[\frac {AC}{CD} \equal{} \frac {BE}{BD \minus{} AD}\]

Let $ F$ on $ AB$ such that $ AD \equal{} DF$ and let $ M$ be the intersection between the perpendicular from $ A$ to line $ CD$ and $ CD$.
Then $ AM \equal{} ME$, and so $ CM \parallel FE$ and $ \angle ADC \equal{} \angle EFB$.
Let $ N$ be the midpoint of $ EB$. We have $ AC \equal{} EC \equal{} BC$ and $ MCNE$ is cyclic.
Hence $ \angle CEN \equal{} \angle CMN$, but $ MN \parallel AB$ and so $ \angle CEN \equal{} \angle CMN \equal{} \angle BDC \equal{} \angle ADM \equal{} \angle EDM$, from which $ BCDE$ is cyclic too.
Finally $ \angle EBF \equal{} \angle ECD \equal{} \angle DCA$ and triangles $ FBE$ and $ DCA$ are similar and the thesis follows.

That one is the closest to the official proof. While discussing the problems we had 4 different proofs instead of the official one :wink:

Nice problem!  My solution is similar to [b]Sepp[/b]'s but differs in a few ways:
[hide="Solution"]
Let $ P$ be the mipoint of $ AB$, $ M$ that of $ AE$, and $ T$ the point such that $ AD \equal{} DT$.  Now, since $ AC \equal{} BC$, we have that $ \angle APC \equal{} 90$ and since $ E$ is a reflection of $ A$ over $ CD$, we have that $ 90 \equal{} \angle AMC\implies \angle APC \equal{} \angle AMC$, so $ AMPC$ is cyclic.  Also, since $ ME \equal{} AM$ and $ AD \equal{} DT$, we see that $ DM\parallel TE$ and also since $ AP \equal{} BP$ and $ AM \equal{} ME$, we also have that $ MP\parallel BE$.  Hence, $ \angle DPM \equal{} \angle ABE$ and $ \angle DMP \equal{} \angle TEB$, so $ \triangle TEB\sim \triangle DMP\sim \triangle DAC\implies \frac {AC}{CD} \equal{} \frac {EB}{BT} \equal{} \frac {BE}{BD \minus{} AD}$, as desired. [/hide]

[quote="polskimisiek"] [color=darkred]Let $ ABC$ be a $ C$ - isosceles triangle. For a point $ D\in (AB)$ , $ AD < DB$  denote

the reflection $ E$ of the point $ A$ w.r.t. the line $ CD$. Prove that $ \frac {AC}{CD} = \frac {BE}{BD - AD}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Prove easily that $ \{\begin{array}{c} \widehat {CED}\equiv\widehat {CAD}\equiv\widehat {CBD} \\
 \\
BC = CE = AC\ ,\ DE = AD\end{array}$ $ \implies$ $ \widehat {CED}\equiv\widehat {CBD}$ , i.e.

the quadrilateral $ BCDE$ is cyclically. Apply the [b]Ptolemaios' theorem[/b] in the quadrilateral $ BCDE$ :

$ BC\cdot DE + CD\cdot BE = BD\cdot CE$ $ \Longleftrightarrow$ $ AC\cdot AD + CD\cdot BE = BD\cdot AC$ $ \Longleftrightarrow$ $ \boxed {\frac {AC}{CD} = \frac {BE}{BD - AD}}$ .

[b][u] Remark.[/u][/b] If $ F\in CD\cap BE$ , then the quadrilateral $ ACBF$ is cyclically and $ \boxed {\frac {AC}{CF} = \frac {AB}{AF + BF}}$ . See[/color] [url=http://www.artofproblemsolving.com/blog/50478][color=red][b]here[/b][/color][/url].

[quote="Virgil Nicula"][quote="polskimisiek"] [color=darkred]Let $ ABC$ be a $ C$ - isosceles triangle. For a point $ D\in (AB)$ , $ AD < DB$  denote

the reflection $ E$ of the point $ A$ w.r.t. the line $ CD$. Prove that $ \frac {AC}{CD} = \frac {BE}{BD - AD}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Prove easily that $ \{\begin{array}{c} \widehat {CED}\equiv\widehat {CAD}\equiv\widehat {CBD} \\
 \\
BC = CE = AC\ ,\ DE = AD\end{array}$ $ \implies$ $ \widehat {CED}\equiv\widehat {CBD}$ , i.e.

the quadrilateral $ BCDE$ is cyclically. Apply the [b]Ptolemaios' theorem[/b] in the quad.rilateral $ BCDE$ :

$ BC\cdot DE + CD\cdot BE = BD\cdot CE$ $ \Longleftrightarrow$ $ AC\cdot AD + CD\cdot BE = BD\cdot AC$ $ \Longleftrightarrow$ $ \boxed {\frac {AC}{CD} = \frac {BE}{BD - AD}}$ .

[b][u] Remark.[/u][/b] If $ F\in CD\cap BE$ , then the quadrilateral $ ACBF$ is cyclically and $ \boxed {\frac {AC}{CF} = \frac {AB}{AF + BF}}$ .[/color][/quote]

Yes, that was the proof I was waiting for. From 6 proof I have seen that one is I think the preetiest :wink:

Let $F$ on $AB$ satisfies $AD=BF$, then $\Delta ACD \cong \Delta BCF$, so $CF = CD$.
Becase $CE = CA = CB$, $\angle BCE = \angle ACB - \angle ACE = \angle ACB - 2\angle ACD = \angle DCF$.Hence $\Delta BCE \sim \Delta DCF$, so $\frac{{AC}}{{CD}} = \frac{{CB}}{{CD}} = \frac{{BE}}{{DF}} = \frac{{BE}}{{BD - AD}}$.

Because $\angle AED = \angle C = \angle B$, $ACED$ is cyclic. Then by Ptolemy's, $$AE \cdot CD = AE \cdot CE + AC \cdot DE.$$ The desired condition is then equivalent to $$AB \cdot CD - AB \cdot BD = AD \cdot CE \iff AB \cdot BD = AC \cdot DE,$$ but this is clear as each pair of lengths are respectively equal.


ceiling function
inequalities
combinatorics unsolved
combinatorics

There are $100$ mutually hostile kingdoms on Mars. For the sake of peace, it was decided to establish several alliances, each of which consists of at most $50$ kingdoms, such that every two kingdoms meet in at least one alliance.

(a) What is the smallest possible number of alliances?

(b) The same question, under an additional assumption that a union of two alliances consists of at most $80$ kingdoms.

This problem is an excellent example of the counting-two-ways technique.

Let $K$ be the set of kingdoms. $|K|=100$. Let an alliance $a$ be a subset of $K$. $|a|\leq 50$. Let $A$ be the set of alliances. Let $n_{k}$ denote the number of alliances in $A$ that contain the kingdom $k$.

Lemma. For all kingdoms $k$, $n_{k}\geq 3$.

Proof. Since an alliance containing $k$ contains at most $49$ other kingdoms, in order for all alliances contain $k$ to contain all $99$ other kingdoms, we need $\lceil\frac{99}{49}\rceil = 3$ alliances. End of proof.

Now let $F$ be the set of pairs $(k,a)$ where $k$ is a kingdom in $K$ and $a$ is an alliance in $A$ such that $k$ is in $a$.

Counting by kingdoms, we get
$|F| = \sum_{k\in K}n_{k}\geq \sum_{k\in K}3 = 300$.

Counting by alliances, we get
$|F| = \sum_{a\in A}|a| \leq \sum_{a\in A}50 = 50|A|$.

So $50|A|\geq 300$, and $|A|\geq 6$. So 6 is a lower bound on the number of alliances. It is interesting to note that if $|A|=6$, then the inequalities above are forced to be equalities, so that $n_{k}=3$ for all kingdoms $k$ and $|a|=50$ for all alliances $a$. That gives us some clues for constructing examples. We need an example to finish the result.

I made two examples, trying to make them as different from each other as possible. Let the kingdoms be numbered 1 through 100. In the first example, I will name the alliances after colors. In the second example, I will name the alliances after flavors.

First example (intersection size 25)
Red      1-25, 26-50
Orange 1-25, 51-75
Yellow  1-25, 76-100
Green   26-50, 51-75
Blue     26-50, 76-100
Violet   51-75, 76-100

Second example (intersection size 20)
Chocolate  1-10, 11-20, 21-30, 31-40, 41-50
Vanilla       1-10, 11-20, 51-60, 61-70, 71-80
Strawberry 1-10, 21-30, 51-60, 81-90, 91-100
Banana      21-30, 31-40, 61-70, 71-80, 81-90
Kiwifruit     11-20, 41-50, 61-70, 81-90, 91-100
Grape        31-40, 41-50, 51-60, 71-80, 91-100

So the answer to (a) is 6, and the answer to (b) is 6.

Erin Schram