Prove that if p is prime, then $ 2^p \minus{} 1$ is also prime. Do not be afraid to include examples if it helps your proof!

[quote="Zhero"]lol, 3 and 11 were my first guesses.

Sometimes is the answer. $ 2^3 \minus{} 1$ is prime but $ 2^{11} \minus{} 1$ is not (23 times 89.)[/quote]
Okay so I guess that does not work.

How about this?

$ a_1 \equal{} 2$, $ a_n$=2^($ {a}_{n \minus{} 1}$)-1
Every term of this sequence is a prime.

[quote="t0rajir0u"]Beyond the fifth term of this sequence are numbers that are not known to be prime or composite.  In particular, only [url=http://en.wikipedia.org/wiki/Mersenne_prime]46 Mersenne primes[/url] are known.[/quote]

My original conjecture was an unproved conjecture, but I think that the new one I just thought of is correct.


[hide="2"]
We have $ \frac45s\equal{}\frac56d\implies d\equal{}\frac{24}{25}s\implies s\geq\boxed{25}$.
[/hide]

For $ 1$, do you mean smallest positive score? And can he skip questions?

[quote="mathemagician1729"]1. On a 10 problem test, Jim will score 5 points for a correct answer, 1 point for no answer, and 0 points for an incorrect answer. What is the smallest positive score that Jim [b]cannot[/b] earn?

2. If 4/5 of the students in Ms. Smith's class use 5/6 of the desks in the room, what is the least possible number of students in the class?[/quote]

[hide="1"]  Okay...We know he can get $ 1\minus{}9$ since he can get $ x$ for no answer.  He can get 10-19 by getting 2 or 3 correct ones and 1-4 no answers.  He can get 20-29 by getting 4 or 5 correct and 1-4 no answer.  ON 30-39, though, he cannot attain 39 since if he has 7 correct answers, he can only have 3 no answers, therefore $ 39$ is the lowest score Jim cannot earn.[/hide]

1.

x:  number of correct answers
y:  number of no answers

Total score s = 5x + y  where   0 <= x <= 10; 0 <= y <= 10; 0 <= x + y <= 10;

For each x, the scores differ by 5, so the y has to be at least 4 to make the possible
scores consecutive, therefore x = 7, y = 3 gives s = 5 * 7 + 3 = 38.  The answer is 39.

I hope someone could give more elegant solutions.


How many positive integer divisors of 2007^2007 are divisible by exactly 2007 positive integers

Call x is a positive integer divisor of $\ 2007^{2007}$ are divisible by exactly 2007 positive integers.
   $\ 2007=223*3^{2}$
    $\ x=223^{a}.3^{b}$ and (a+1)(b+1)=2007=223.9=669.3=2007.1
  So there're 6


My father is four times as old as me. In 20 years, he will be only twice as old as me. How old is my father and how old am I?

[hide]
I made a chart:
boy    dad in 20 years
                  boy  dad
8        32    28    52
9        36    29    56
10      40    30    60
So the answer is he is 10 years old[/hide]

[hide]x=kid
y=dad

4x=y
2(x+20)=y
x=20[/hide]


Ms. Hamilton's eighth-grade class wants to participate in the annual
three-person-team basketball tournament.  Lance, Sally, Joy, and Fred
are chosen for the team.  In how many ways can the three starters be chosen?

You have 4! ways to choose people, but you have counted every team 3! times. Take $ \frac{4!}{3!}$ which gives $ \boxed{4}$

Alternatively, you have 4 different choices of who sits out, the other 3 will be playing.


Sa se afle in cate regiuni este impartit spatiul de 10 plane aflate in pozitie generala.

Da, intr-adevar asta este raspunsul, insa e nevoie de mai multe detalii mai ales pentru un pusti de clasa a-9-a care inca nu a facut la clasa combinari. :)

hai ca dau eu mai multe detalii:

stim ca trei plane formeaza un varf in spatiu. deci exista $\binom{10}{3}$ varfuri, iar fiecare este punctul cel mai de jos exact al unei parti a spatiului.

asadar exista $\binom{10}{3}$ parti care au "cel mai de jos punct", fiecare parte care nu are cel mai de jos punct taie un plan orizontal in una cele $X$ parti plane.

deci obtinem numaru $X+\binom{10}{3}$, ne mai ramane sa aflam $X$, si anume numaru de parti in care poate fi impartit planul de 10 drepte.

cele $\binom{10}{2}$ puncte de intersectie ale dreptelor sunt usor de numarat, dar fiecare punct de intersectie este punctul "cel mai de jos" exact al unei parti. deci exista $\binom{10}{2}$ regiuni avand un cel mai de jos punct, exista astfel $10+1=11$ regiuni care nu au un punct cel mai de jos, adika $\binom{10}{0}+\binom{10}{1}$.

deci adunand obtinem $X=\binom{10}{0}+\binom{10}{1}+\binom{10}{2}$.

acum inlocuind $X$ in ce obtinusem mai sus dam raspunsul final dat si de domnul Enescu, $\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}$

PS: m-a mai intrebat Cezar de problema asta mai de demult dar ii zisesem o prostie :)

Soluţia e din  Arthur Engel "Problem Solving Strategies", tradusă de editura GIL. Nu vreau să fac reclamă, dar e o carte esenţială pentru cei care doresc să ajungă la baraj. Exista, până acum câteva luni, un download pe site-ul lib.org.by . Oricum, varianta în limba română (a prof. Mihai Bălună) e mai bună, deoarece au fost înlăturate o serie întreagă de greşeli din ediţia originală şi au fost adăugate soluţii.

[quote="pohoatza"]dap , de acolo e rezolvarea, nu am zis ca it`s mine:)[/quote]
Nu asta am vrut să spun...scuze dacă asta s-a înţeles...
Eu însumi am scris zeci de soluţii pe acest site pe care le ştiam din diverse surse, fără a le menţiona.
Am vrut doar să atrag atenţia asupra unei cărţi care este, într-adevăr, de excepţie. Acum ceva ani, un elev de clasa a 9-a, foarte talentat de altfel, m-a rugat să îl ajut în pregătire. I-am dat cartea lui Engel. A parcurs-o, cu creionul pe hârtie, în 3 săptămâni (!). În următorii trei ani a luat aur la OIM. Nu spun că numai Engel a fost cauza, dar...


integration
real analysis
real analysis unsolved

Let $ f \in C_{[0;1]}$. Suppose that, for all $ n \in \mathbb{N}$, exists $ a_n,b_n \in \mathbb{R}$ such that $ \int \limits_0^1 (f(x)-a_nx-b_n)^4 dx < \frac{1}{n}$. Prove that $ f(x)=ax+b, \forall x \in [0;1]$ ($ a,b \in \mathbb{R}$).

Since $ \inf_{(a,b) \in \mathbb{R}^2} \int_0^1 (f(x) \minus{} ax \minus{} b)^2 dx \equal{}0$, what is the orthogonal projection of $ f$ onto the subspace $ \{ P_{a,b}(x) \equal{} ax\plus{}b : (a,b) \in \mathbb{R}^2 \}$ ?


superior algebra
superior algebra unsolved

construct a nonabelian group of order 21.

Take a look at the following link:

[url]http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2573.0001[/url]

Regards.

$ G\equal{}\{x^i y^j | x^7\equal{}y^3\equal{}xyxy^{\minus{}1}\equal{}e\}$

[quote="mitchan88"]$ G \equal{} \{x^i y^j | x^7 \equal{} y^3 \equal{} xyxy^{ \minus{} 1} \equal{} e\}$[/quote]
No, $ x^{\minus{}2}yxy^{\minus{}1}\equal{}1$.
$ xyxy^{ \minus{} 1}\equal{}1$ would give $ y^2xy^{\minus{}2}\equal{}yx^{\minus{}1}y^{\minus{}1}\equal{}x$ and $ G\equal{}\langle x,y^2\rangle$  abelian.

More to the point, it would give $ x \equal{} y^3 x y^{\minus{}3} \equal{} x^{\minus{}1}$.  So $ x^7\equal{}x^2\equal{}e$, hence $ x\equal{}e$.

We need $ yxy^{\minus{}1} \equal{} x^2$ or $ x^4$. (the action needs to have order 3!)


Find all endomorphisms of an euclidean space $E$ which conserve the orthogonality of vectors.

let's prove that such an endomorphism $u$ is of the form $u=k.f$ where $k \in R$ and $f \in O_n(R)$.
We know that $u$ conserves orthogonality hence it maps the canonical basis $(e_1,..,e_n)$ to $(b_1,..,b_n)$ where $(b_i | b_j)=0$ for all $i \neq j$. It suffices to show that $|b_i|=k$ for all $i$.
But, for $i \neq j$, $(u(e_i + e_j) | u(e_i + e_j))=((b_i + b_j)|(b_i + b_j))=|b_i|^2 - |b_j|^2 = 0$ because
$((e_i+e_j)|(e_i - e_j))=0$. Conclusion follows.


Pascal\u0027s Triangle
binomial coefficients

In row n of Pascal's triangle, let E be the sum of the binomial coefficients, n C k, for which k is even and let D be the sum of the binomial coefficients, for which k is odd. How do E and D compare?

[hide="my clumsy explanation"]$E=D$ because if you expand $(a+b)^n$ and you let $a=1$, $b=-1$, then $(a+b)^n=0$ so the result follows[/hide]


Because the sequence$\ (x_{1},x_{2},...,x_{n})$is composed of distinct positive integers, then if we want F minimum then we most have the smallest sequence of differents integers possible that is:$\ (x_{1},x_{2},...,x_{n})=(1,2,..,n)$

$\rightarrow$ $\ F(x_{1},x_{2},...,x_{n})=\frac{1^{k}+2^{k}+...+n^{k}}{1+2+...+n}$,

$\rightarrow$ $\ F(x_{1},x_{2},...x_{n})=\frac{(B+n+1)^{[k+1]}-B^{[k+1]}}{k+1}\frac{2}{n(n+1)}$

where the notation$\ B^{[k]}$means the quantity in question is raised to the appropriate power$\ k$, and all terms of the form $\ B^{m}$are replaced with the corresponding Bernoulli numbers$\ B_{m}$.


A rectangular aquarium 16 inches long, 9 inches wide and 10 inches tall is filled with water to a level of 8 inches. A solid steel cube with an edge of 3 inches is dropped into the aquarium. How many inches high is the new water level in the aquarium? Express your answer as a decimal rounded to the nearest tenth.

The cube has a volume of 3*3*3=27 cubic inches
the base of the aquarium has an area of 9*16= 144 square inches
so dropping in the cube will cause the water level to rise 27/144 = 3/16 = 0.1875 inches
the water level started at 8 inches so the final water level would be 8.1875 inches, or 8.2 inches rounded to the nearest tenth

Why would it rise 27/144? I don't get it...

It's kind of an intuition kind of thing, but it's like the Archimedes problem: An object will displace an amount of liquid equal to its volume when placed in a tank of liquid. The base area of the tank is $144$ and the cube's volume is $27$. The cube will displace $27$ cubic inches of water spread out over $144$ square inches of water, hence $\frac{27}{144}$. $\box[4]$


function
topology
advanced fields
advanced fields unsolved

a) Let  $f: S^{n}\longrightarrow S^{m}$ ($n<m$) be a continious function. Prove that $f$ is nullhomotopic. (i.e. it is homotop to constant function)
b) Let $X$ is subspace of $Y$ and $Y$ is simply connected, and $X$ is contractible in $Y$. Prove that each function $f: X\longrightarrow Y$ is nullhomotopic.

a) is given by our teacher, and b) is my conjecture. I don't know whether it is correct or not.

b) is wrong: consider $Y = \mathbb{R}^{2}\setminus 0$, $X = S^{1}+100$, then $X \to S^{1}$ (obvious map) is not nullhomotopic

Yes you are right, but now I think when $Y$ is simply connected my conjecture is correct.

There is a map $S^{3}\to S^{2}$ which is not nullhomotopic (http://en.wikipedia.org/wiki/Hopf_bundle). Now take $Y = \mathbb{R^{4}}\setminus D^{3}$ and do the analogous thing.

a) is easy to prove if $f(S^{n})$ is a proper subset of $S^{m}$: pick a point $a\in S^{m}\setminus f(S^{n})$ and gradually push $f(S^{n})$ into the antipode of $a$. But a continuous map $f\colon S^{n}\to S^{m}$ may be surjective even though $n<m$. I suppose one can avoid this difficulty by constructing a "nice" map $\tilde f\colon S^{n}\to S^{m}$ such that $|\tilde f(x)-f(x)|<\epsilon$ for all $x\in S^{n}$. However, this is not so easy when $n>1$.

a) Sketch:

There is such a theorem. Let $(X,A)$ and $(Y,A)$ be $CW$-pairs, and let $f: (X,A)\to (Y,B)$ be a map of pairs. Then there exists a cellular map $g: (X,A)\to (Y,B)$ such that $g\simeq f\text{ rel }A$.

Then, according to the theorem, $f: \mathbb{S}^{n}\to \mathbb{S}^{m}$ factors up to homotopy through the inclusion $\mathbb{S}^{m-1}\hookrightarrow \mathbb{S}^{m}$. Inclusions $\mathbb{S}^{n-1}\to \mathbb{S}^{n}$ are nullhomotopic.


MATHCOUNTS ARML AMC AIME USA(J)MO USAMO algebra

You can sign up for The MATHCOUNTS Ultimo Competition. Please register before August 31. Late Registration until September 7. Late Registration means that you will/may have to work out both Round 1 and Round 2 Both together. Here are the rules:

1) You must register before August 31; Late-Register before September 7. If you need an exception, please private message me.

2) The format will be as described:

Round 1: Mostly Sprint Questions (15 Questions, 15 points)

Round 2: Extra Round. This round is not required, but will give you extra points. (5 questions, 5 points)

Round 3: Target Round questions (10 Questions, 20 points)

Round 4: Extra Round. Once again, this round is not required but will give you extra points. (10 Questions, 20 points)

Round 5: Final Round: Hard Team Round Questions (10 questions, 40 points)

Tie-Breakers: Various Questions; Maximum of 2 tie-breakers. Joint winners can be declared. (1 point per question, 5 questions per set; 2 sets max)

Freebie rounds: 1 question, 1 pt extra

Total points possible (extra round included): 100

3) Scores will be posted.

4) Questions can be downloaded from this topic. Questions for Round 1 will be posted any time after there are a minimum of 5 participants. Due dates will be posted with the scores.

5) Late submissions accepted until 3 days after due date. After that, submissions will be accepted for a 5 point penalty.

6) YOU ARE NOT ELIMINATED IF YOU LOSE A ROUND. YOU WILL BE CONTINUING TO THE NEXT ROUND. WINNERS WILL BE DECLARED AT END.

7) Roundly Scores will be posted. Cumulative Scores will  also be posted throughout the competition, along with placements.

According to rcv, I don't think I will be allowed to since I am going into 9th Grade.

ILL BE IN

wen does teh competition start

The competition begins once there are 5 participants. The due date for Round 1 is set then. It is expected to start within the next few days.
[u]
[b]Current Participants:[/b][/u]
SplashD
pascal_1623
akalra1

[u][b]Current Participants:[/b][/u]
SplashD
pascal_1623
akalra1
ragnarok23

We need 1 more participant to begin...

I'm in. Thats 6!

[quote="pascal_1623"]You shouldn't be in this forum neil.

I'm in too.[/quote]

He's still participating in my competition, which started when he was still in 8th grade.

[u][b]Current Participants:[/b][/u]
SplashD
pascal_1623
akalra1
ragnarok23
bpms
jclarkemathL314159

That's 6 people. I will now post round 1 below. The round consists of mostly sprint round with some target round questions.

People can sign up as well as pm me the answers at the same time. This round is due September 2. All questions are permitted. Discussion of the problems is allowed if and only if the problem answer is not directly stated. [color=darkred][size=200][u][b]ONLY PROBLEM INTERPRETATION HELP CAN BE GIVEN!!! ANY DIRECT STATEMENT OF THE ANSWERS MAY RESULT IN DISQUALIFICATION.[/b][/u][/size][/color]

Additional "Freebie rounds" may be posted which connsist of 1 question, worth 1 point for extra bonus's.

count me in! :D

can we use a calculator?

I'm signing up, up UP, UP!!!

Do we have to send solutions?

[quote="SplashD"][quote="i_like_pie"][quote="SplashD"]For condition 5, you have "the four interior regions AQPB..." but AQ intersects PB, so it's not a valid quadrilateral[/quote]
That makes two triangles if I understand correctly.[/quote]
what two triangles?[/quote]
These two triangles:

i don't think that's what the question means though.

[quote="anirudh"]that [i]is[/i] what the question means[/quote]

i have never seen the area of a shape defined like that. did you make these questions?

No. I got problem 1 from unknown sources... :)
The rest are original

[quote="anirudh"]that [i]is[/i] what the question means[/quote]
How is APQB a quadrilateral then?

Well, I just finished the target round for my mock!!!!!!!!! Yay...

[quote="anirudh"]i don't know...the question is copied and pasted from saab[/quote]

that's what i said. you should at least read over the problems before you post it as a contest question, sheesh.

and you said it was from "unknown sources"  :mad:  :roll:

[b]NOTE: PROBLEM #1 DOES NOT COUNT!!! YOU WILL GET FULL CREDIT FOR PROBLEM 1 IF YOU GIVE AN ANSWER FOR IT!!![/b]

One person has turned in all rounds:

[hide="SplashD"]
[hide="Round 1 Score"]
Perfect!
[/hide]

[hide="Round 2 Score"]
Perfect!
[/hide]

[hide="Round 3 Score"]
$\frac{27}{30}$
Near Perfect!
[/hide]

[hide="Overall Score"]
$\frac{57}{60}$!
[/hide]
[/hide]

OK. By far, SplashD is winning this competition with an overall score of $\frac{57}{60}$.

wait is it all right if I send in my round 3 answers tomorrow? or is that way too late?

Actually, I'll delay it. You can send it in tommorrow.  :)

OK. This competition is officially over. SplashD won, with 13375P34K43V312 as runner-up. Mods can now lock this up.


algebra
binomial theorem
number theory
number theory solved

Let p be an odd prime, prove that
p-1
   \sum   k2p-1 == p(p+1)/2 (mod p 2)
k=1

Let a in {1,..., (p-1)/2}.
Then, from the binomial theorem :
a 2p-1  + (p-a) 2p-1  = p(2p-1)a 2p-2  mod[p 2 ]
= p(2p-1)(a p-1 ) 2 mod[p 2 ].

In another hand, from Fermat's little theorem, we have
a p-1  = 1 +p*k for some integer k, thus (a p-1 ) 2  = 1 + 2k*p mod[p 2 ], that leads to p*(a p-1 ) 2  = p mod[p 2 ].
Then, if S denotes the given sum, we have :

S = sum{1 \leq a \leq (p-1)/2} (2p-1)p mod[p 2 ]
= p(2p-1)(p-1)/2 mod[p 2 ]
= p(p+1)/2 mod[p 2 ], and we are done.

Pierre.


probability
MATHCOUNTS
AMC
AIME
geometry
number theory

I have two questions.

I recently did well at the Chapter Competition, and I got a AOPS gift certificate to use to buy books on this site. I was wondering which book I should buy. Which book (out of Intro to Counting and Probability and Intro to Number Theory) would help me the most in Mathcounts?

Also, my 2nd question is, are there any awards associated with getting first place in states/nats other than the trophy?

You get scholarships for some math programs.
Anyways, you should get AoPS volume 1, it's the best for the long run.

[quote="Ihatepie"](out of Intro to Counting and Probability and Intro to Number Theory) [/quote]

I would love to buy the AOPS book, but it costs too much. So which one is better for MATHCOUNTS?

at states and nats you can sometimes get scholarships and free TI-84+'s

[quote="mihail911"][quote="xpmath"]I got  one of those too. Although, I'm wondering whether to use it on Intermediate Trig/complex numbers, Intermediate C+P or the AIME seminar...[/quote]

how much did you get xp?
hopefully ill do well enough at chapter to get one too. use it on intermediate complex and trig. we'll be classmates. :-)[/quote]
Only 25 dollars.
I intend to use it on that, seeing the first day of the AIME seminar is on the same day as states, so I'll have to look at transcripts.

[quote="7h3.D3m0n.117"][quote="Ihatepie"]I would love to buy the AOPS book, but it costs too much. [/quote]

They don't cost that much (or at least the ones I bought). Regular math textbooks cost approx $ \$60$ or more.

AoPS vol 1 is a great start for any MC participant. If you already have it, you can order a specific book, like number theory or probability, if you find you're having trouble in those areas.[/quote]
And the four Introduction books as well. They help a lot :)

I've also got $25. I'm trying to decide on the whole set year 2000, or year 1999. They both cost 24.10 Which one?

I'd get vol 1 if you dont already have it, then id get the intro to geometry

vol 1 isn't great, get vol. 2

and i have the probabitlity book
and it really doesn't help me a whole lot, so dont get that

[quote="Hustla25"]vol 1 isn't great, get vol. 2 [/quote]

Volume 2's material content isn't compatible for MATHCOUNTS. You're better off with AoPS vol 1.

I think out of C/P and NT, C/P appears more often on the tests. But that doesn't mean NT isn't important- NT is just not used as often as C/P on MATHCOUNTS TESTS..... However, I've recently noticed that a lot of Aopsers use mod to solve lots of stuff....
Don't blame me if the state test is all NT- I'm just observing from the past tests

what eva u decide to buy, study it!!!

that is all i can say

I think they give everyone at Nats a free TI-89.
Or TI-89 titanium.

Nats = Ti 84 Silver Plus. As for the gift certificate, why didn't I get one?!?! So unfair T.T xD... But... are you sure you're only going to choose one of those two? Becuase AoPS volume 1 contains both of them, just not as in depth, but it's good enough.

i think ihatepie if you are already pretty good at C\P than the introduction to C\P will not help you, i have it and it helps me becuase im really really really bad at C\P but if your pretty good at it id go with number theory cause it does start to show up on MATHCOUNTS at states

thanks all, I got C/P

@mathcrazed did you get first place written round?


For all positive integers $ {i}$ let $ {S_i}$ be the sum of the products of $ {1,2,...,p\minus{}1}$ taken $ {i}$ at a time,where $ {p}$ is an odd prime.Show that $ {S_1}$ is congruent to $ {S_2}$  in short all $ {S_i}$ up to $ {S_p\minus{}_2}$  MODULO P

In $ \mathbb{F}_p \equal{} \mathbb{Z}_p \equal{} \{0, 1, \ldots, p\minus{}1 \}$, the remainders modulo $ p$, the polynomial $ f(x) \equal{} x^{p\minus{}1} \minus{} 1$ has all the values as roots, except $ 0$, by Fermat's little theorem, hence these are all its roots. From Viete's relations, one gets precisely that all your sums $ S_i$ are congruent to $ 0$ modulo $ p$, i.e. divisible by $ p$.

Use vietas to explain your point plz your solution is vague

$ S_1$ is the sum of the roots, $ S_2$ is the sum of their products two by two , ..., $ S_{p\minus{}2}$ is the sum of their products $ p\minus{}2$ by $ p\minus{}2$. But all intermediate coefficients in $ x^{p\minus{}1} \minus{} 1$ (except the leading coefficient, and the free term, corresponding to the product of all roots) are zero, and they correspond by Viete's relations to the sums above.


For every $0\leq a,b,c<\frac{1}{\sqrt{3}}$, prove that

$\frac{a+b}{1-ab}+\frac{b+c}{1-bc}+\frac{c+a}{1-ca}\leq 2\frac{a+b+c-abc}{1-ab-bc-ac}$

We can confirm this inequality like belove
Let $a=tan(\alpha),b=tan(\beta)$ and $c=tan(\gamma)$ where $\alpha,\beta,\gamma < 30$,and $\alpha \geq \beta \geq \gamma$
Then we must prove that
$tan(\alpha+\beta)+tan(\beta+\gamma)+tan(\gamma+\alpha) \leq 2.tan(\alpha+\beta+\gamma)$
$2.tan(\alpha+\beta+\gamma)-tan(\alpha+\beta)-tan(\beta+\gamma)-tan(\gamma+\alpha) \geq 0$
Fix $\alpha$ and $\beta$.And let $f(\gamma)=2.tan(\alpha+\beta+\gamma)-tan(\alpha+\beta)-tan(\beta+\gamma)-tan(\gamma+\alpha)$.First derivate of $f(\gamma)$ is positive, then it gets it's minimum at $\gamma=0$.With the same method it gets minimum when $\beta=0$ and it is $0$


trigonometry
function
complex analysis
calculus
calculus computations

A part of a question asks me to show that that $\frac{\sin x}{x}$ is strictly decreasing in $[0,\pi]$ My question is why 0 is included in the interval. the intermediate value theorem requires f to be differentiable in $(0,\pi)$ and continuous in $[0,\pi]$. The function is not continuous in 0 (since it's not defined there), so why is the 0 included?

Thanks!

$\frac{\sin x}x$ has a removable discontinuity at zero. There's a possible attitude one could take - and I go along with this about half of the time - that removable discontinuities deserve to be removed.

If we take that attitude, we are dealing with the function $f(x)$ such that for $x\ne0,\,f(x)=\frac{\sin x}x$ whereas $f(0)=1.$ This is an extremely nice function - it's continuous, differentiable, anything of that sort you'd want to ask for, at every point, including zero. OK, to say that formally: it's an entire function - a function given everywhere by a power series with an infinite radius of convergence.

And $f(x)$ is decreasing on $[0,\pi].$


geometry
analytic geometry
algebra
polynomial
vector
advanced fields
advanced fields open

A set of countable points in complex affine variety is not constructible set? why? :P

A proof would require developing some theory, but the explanation is simple.  The coordinates of points in an finite affine variety are roots of some polynomial equations, and a polynomial has only a finite number of roots, so there are a finite number of possibilities for each coordinate.  You have to prove that the algebra of coordinates of a 0-dimensional variety is a finite-dimensional vector space.
(If the dimension is higher than 0, the variety has topological dimension at least 2 and is uncountable, so this case cannot arise).


algorithm
combinatorics proposed
combinatorics

An excellent exercise on graphs would be to prove Brooks' theorem:

Let $G$ be any graph in which every vertex is of order at most $\Delta$. If $\Delta \geq 3$ and $G$ is not $K_{\Delta+1}$, prove that $G$ is $\Delta$-colorable.

[quote="vess"]An excellent exercise on graphs would be to prove Brooks' theorem: [/quote]

You are right  :D

Pierre.

i'll prove that $\chi(G)\leq\Delta(G)$ so the result will follow.
where $\chi(G)$ is the chromatic number of G.

the idea of the proof is the same as the one in the proof of the following easy lemma
Lemma: $\chi(G)\leq\Delta(G) + 1$
Proof: Let $\Delta(G)=k$ (maximum degree) . Now consider a ordering of vertices $v_1,...v_n$. We can color properly (no two adjacent vertices with the same color) the following way: to each $v_i$ assing the color that corresponds to the vertex with smallest index in the ordering that is not a neighbor of $v_i$. Since each vertex of the ordering has at most $k$ lower indexed neighbor the lemma follows.

Now we have to order the vertices so that each has at most k-1 lower indexed neighbors.

case 1: G is not k-regular. Let x be a vertex with degree less than k. Now construct a spanning tree from x and assing indices in the decreasing order as we reach vertices. Then in our ordering $v_1,...,v_{n-1}$ each vertex has a neighbor with a higher index. Since every vertex has degree at most k, then each vertex has at most k-1 neighbors with lower index, so by the same idea in the lemma above the coloring algorithm will produce at most k colors.

case 2A: G is k -regular and G has a cut vertex, say $x$. Let $G'$ be the subgraph formed by a component of $G-x$ together with it edges to $x$.  The degree of $x$ in $G'$ is less than k, so the method used in case 1 provides a desired coloring. So we can permutate the coloring in each component so that x recieves the same color and we get a graph with a k-coloring.

case 2B:G is k-regular and 2-connected. (contains no cut vertex)
Suppose that some vertex x has neighbors y and z such that yz is not an edge and $G-{y,z}$ is connected. If we have this case, we grow a spannig tree from x assinging the labels $3,...n$ so that $v_n=x$ in decreasing order so each vertex has at most k-1 neighbors with a lower index. As for vertex $x=v_n$ we color y and z with the same color so the algorithm uses at most k-1 colors in the neighbors of x and produces a k-coloring in G.
Now we prove that ina 2-connected graph such triple $x,y,z$ always exist.
Choose a vertex v. If $G-v$ is still  2-connected then let v=y and let z be a vertex of distance 2 from v=y. this vertex y exists because G is k-regular and it is not complete. So we have our desired triple $x,y,z$ where $G-{y,z}$ is connected.
If $G-v$ has a cut vertex then put x=v. Since $G-v$ is not 2-connected it is made up of blocks (maximal 2-connected subgraphs) .  Since x can't be a cut vertex x has a nieghbor in every block of $G-x$ and this neighbor is not a cut vertex of $G-x$ . Since $k\geq 3$ then x has neighbors y and z in different blocks therefore of $G-x$ so yz is not an edge of G and $G-{y,z}$ is connected since y and z can't be cut vertices. We have the desired $x,y,z$ triple.
Done!

If something is unclear or wrong let me know.
Btw, does someone has a shorter or easier proof?

I read in another post of mathlinks a couple of months a go the following related problem which i haven't been able to solve.

(i think it was in the open questions section)
 prove that $\chi(G)\leq\frac {\Delta(G)+1+\omega(G)}{2}$
where $\omega(G)$ denotes the cardinality of the largest complete subgraph of G.
any ideas?

yes this is conjecture and it is due to Reed I remember that when i had posted the thread i had read something about this conjecture for example this is proved : for some $\lambda >0$ such that if $\omega(G) \geq (1-\lambda )(\Delta +1)$ then$ \chi(G)\leq \frac12(\Delta +1)+\frac12\omega(G)$.
Reference:Graph coloring and probablitistic method,Michael Molloy,Bruce Reed.


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

Let $ f(x)$ be integrable on $ [a,b]$,and $ f_{in}\equal{}f(a\plus{}i\frac{b\minus{}a}{n})$,
Show that
$ \lim_{n\to\infty}(1\plus{}f_{1n}\frac{b\minus{}a}{n})(1\plus{}f_{2n}\frac{b\minus{}a}{n})\cdots(1\plus{}f_{nn}\frac{b\minus{}a}{n})\equal{}e^{\int_a^bf(x)dx}$

[quote="wngyny"]Let $ f(x)$ be integrable on $ [a,b]$,and $ f_{in} \equal{} f(a \plus{} i\frac {b \minus{} a}{n})$,
Show that
$ \lim_{n\to\infty}(1 \plus{} f_{1n}\frac {b \minus{} a}{n})(1 \plus{} f_{2n}\frac {b \minus{} a}{n})\cdots(1 \plus{} f_{nn}\frac {b \minus{} a}{n}) \equal{} e^{\int_a^bf(x)dx}$[/quote]
take $ \ln$ (also say $ \ln(1 \plus{} x) \approx x$ for   $ \minus{} 1 < x\leq 1$) and use the fact $ \int_{a}^{b}f(x)dx \equal{}(b\minus{}a) \int_{0}^{1}f(a \plus{} x(b \minus{} a))dx\minus{}>\lim _{n\minus{}>\infty} \frac{b\minus{}a}{n}\sum_{i}^{n} f_{in}$ :wink:


uhh where do you get all the LaTeX codes??

To learn the basics, start with the AoPS Wiki article: [[LaTeX]]. A more advanced tutorial can be found [url=http://www.ctan.org/tex-archive/info/Math_into_LaTeX-4/Short_Course.pdf]here[/url], and the full documentation is [url=http://www.ctan.org/]here[/url].

If you're only looking to learn the commonly used codes, I think the first link will suffice; the other two links are more focused on documents created with $ \text{\LaTeX}$ instead of basic functions and such.

ahh thanks =D

although i found those on LaTeX-Reference...


group theory
abstract algebra
linear algebra
linear algebra unsolved

Let $A \in \mbox{GL}_{n}(\mathbb{Q})$ s.t. $A^{4}=I_{n}$.
Show that there exists $P\in \mbox{GL}_{n}(\mathbb{Q})$ and $PAP^{-1}\in \mbox{GL}_{n}(\mathbb{Z})$

I have shown on the forum (a problem by Alekk) that any finite subgroup of $GL_{n}(Q)$ is conjugated to a finite subgroup of $GL_{n}(Z)$. This immediately applies here, by considering the cyclic group generated by $A$.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=finite&t=27907[/url]

You even replied on that topic, JC_math.

True, I didn't make the link with that problem when I posted it :blush:


topology
abstract algebra
advanced fields
advanced fields unsolved

This is 14.15, on p.81 of greenberg and harper's algebraic topology book.

Let $ X \equal{} X_1 \cup X_2$, and $ A \equal{} X_1 \cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \rightarrow (X,X_1)$.

I don't really know how I can use the exact sequences of triples. Can anyone enlighten me on this?

I'm not sure if I can bump this yet, anyway.. can anyone help?

[quote="Soarer"]This is 14.15, on p.81 of greenberg and harper's algebraic topology book.

Let $ X = X_1 \cup X_2$, and $ A = X_1 \cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \rightarrow (X,X_1)$.[/quote]

I have got something, but I don't really believe it's true - check it carefully, please.

We do not need $ X = X_1 \cup X_2$, and $ A = X_1 \cap X_2$. It is enough to have $ X_1 \cup X_2\subset X$, and $ A \subset X_1 \cap X_2$.

However, I hope that by saying "isomorphism in homology", you mean "isomorphism in homologies of every degree"; else, I don't have a proof (but I haven't searched for counterexamples either).

Consider the sequence $ H_n\left(X_1,A\right)\to H_n\left(X,A\right)\to H_n\left(X,X_2\right)$, where both arrows are induced by inclusions. The composition of these transformations is induced by the inclusion $ \left(X_1,A\right)\to\left(X,X_2\right)$, and therefore (by a condition of the problem) it is an isomorphism, so that it is injective. Hence, its "first arrow" $ H_n\left(X_1,A\right)\to H_n\left(X,A\right)$ must also be injective. Hence, the kernel of the arrow $ H_n\left(X_1,A\right)\to H_n\left(X,A\right)$ is zero.

The long exact sequence for the triple $ \left(X,X_1,A\right)$ looks like this:

$ ...\to H_{n + 1}\left(X,X_1\right)\to H_n\left(X_1,A\right)\to H_n\left(X,A\right)\to H_n\left(X,X_1\right)$
$ \to H_{n - 1}\left(X_1,A\right)\to ...$.

We know that the kernel of the arrow $ H_n\left(X_1,A\right)\to H_n\left(X,A\right)$ is zero here - but this kernel is the image of the preceding arrow $ H_{n + 1}\left(X,X_1\right)\to H_n\left(X_1,A\right)$. Hence, the image of the arrow $ H_{n + 1}\left(X,X_1\right)\to H_n\left(X_1,A\right)$ must be zero, so that this arrow is simply the zero homomorphism. Repeating this argument with $ n - 1$ instead of $ n$, we see that $ H_n\left(X,X_1\right)\to H_{n - 1}\left(X_1,A\right)$ is the zero homomorphism. Thus, our sequence becomes

$ ...\to H_{n + 1}\left(X,X_1\right)\substack{0 \\
\to} H_n\left(X_1,A\right)\to H_n\left(X,A\right)\to H_n\left(X,X_1\right)$
$ \substack{0 \\
\to} H_{n - 1}\left(X_1,A\right)\to ...$.

This readily yields the exact sequence

$ 0\to H_n\left(X_1,A\right)\to H_n\left(X,A\right)\to H_n\left(X,X_1\right)\to 0$.

On the other hand, the long exact sequence for the triple $ \left(X,X_2,A\right)$ yields

$ H_n\left(X_2,A\right)\to H_n\left(X,A\right)\to H_n\left(X,X_2\right)$.

An easy diagram chase yields the following fact: If $ A$, $ B$, $ X$, $ Y$, $ U$ are (say) abelian groups such that $ 0\to A\to U\to B\to 0$ and $ X\to U\to Y$ are exact sequences and such that the composition of the arrows $ A\to U$ and $ U\to Y$ is an isomorphism, then the compositions of the arrows $ X\to U$ and $ U\to B$ is an isomorphism as well.

This applies to our problem, because the composition of the arrows $ H_n\left(X_1,A\right)\to H_n\left(X,A\right)$ and $ H_n\left(X,A\right)\to H_n\left(X,X_2\right)$ is an isomorphism (namely, the one induced by the inclusion $ \left(X_1,A\right)\to\left(X,X_2\right)$), and thus we conclude that the composition of the arrows $ H_n\left(X_2,A\right)\to H_n\left(X,A\right)$ and $ H_n\left(X,A\right)\to H_n\left(X,X_1\right)$ is an isomorphism as well, i. e. the inclusion $ \left(X_2,A\right)\to\left(X,X_1\right)$ induces an isomorphism. We are done.

Did this all make sense? (I'm not asking whether this could have been done on 2 lines as well, because most likely the answer is yes.)

  darij


There are $ n_i$ balls of color $ i$ , $ i \equal{} 1,2,...,k$. Assume that $ n_1 \plus{} n_2 \plus{} ... \plus{} n_k$ is even
Let $ s \equal{} \frac {n_1 \plus{} n_2 \plus{} ... \plus{} n_k}{2}$
Let $ f(n_1,n_2,...,n_k)$ denotes the number of ways to arrange this collection of balls into $ s$ pairs such that each pairs consist of balls with different colors (no balls of the same color are paired up together). Find $ f(n_1,n_2,...,n_k)$

Ex. $ f(n,n) \equal{} n!, f(n_1,n_2) \equal{} 0$ if $ n_1\not \equal{} n_2$,
$ f(n_1,n_2,...,n_k) \equal{} 0$ if $ \max{n_i}> s$
$ f(n_1,n_2,n_3) \equal{} \frac {n_1!n_2!n_3!}{(s \minus{} n_1)!(s \minus{} n_2)!(s \minus{} n_3)!}$ provided that $ \max{n_i}> s$

I think we are looking for "how many edges there can be in a maximum matching in a complete multipartite graph ?"

http://math.furman.edu/~mwoodard/fuejum/content/1996/1996paper2.pdf

In the paper, it doesn't talk about how many ways to partition it.
The set up is the same, but my question is different.

It can be done using brute force and the following recurrent relation,

$ f(n_1,n_2,...,n_k)\equal{}\sum_{p\equal{}0}^{min(n_1,n_2)}p!C^{n_1}_pC^{n_2}_pf(n_1\plus{}n_2\minus{}2p,n_3,...,n_k)$

The answer is explicit, but ugly. So I am looking for some smarter way to do it.


For what k is $ 10^k \plus{} 1$ divisible by a square.

odd powers of 11 is obvious

I think for infinitely many prime p $ k \equal{} \frac {p(p \minus{} 1)}{2}$ is a candidate.

Could someone provide a proof? Thank you

It is not true. $ 10^k\plus{}1$ never divisible $ p\equal{}2,3,5,..$ infinetely many primes.
Let $ T_p$ is minimal period, suth that $ p|10^{T_p}\minus{}1$ and for any divisor $ k|T_p \ \ p\not |10^k\minus{}1$.
Then if $ T_p$ - odd, never $ p|10^k\plus{}1$. If $ T_p$ even $ p^2|10^k\plus{}1$ for $ k\equal{}\frac{pT_p}{2}$.
For example $ T_{13}\equal{}6$ and $ 13^2|10^{39}\plus{}1$, but $ 13\not| 10^{13*12/2}\plus{}1$.

But, it is certainly true for odd mulitples of 11.
And for a host of other numbers like k = 21, 136, 171, etc.
And id it is true for a number, it is true for all odd multiples of that number. So, 63, 125 etc. also qualify


limit
real analysis
real analysis unsolved

Let $(x_n)_{n\geq 0} \in R_{+}^N$ be a sequence which tends to $0$.Let $y_n=n(x_n-x_{n+1})$.Show that the convergence of $\sum y_n$ implies the convergence of $\sum x_n$.

Is the sequence also supposed to be decreasing?

Grobber, I would say yes. :?  Anyway, from Stolz Cesaro we have $\lim_{n\to\infty} n(x_{n+1}-x_{n})=\lim_{n\to\infty} \frac{x_{n}}{n}$ and from "compare" :?  criterion it follows the result. ;)

Cimabue, do you know Cesaro-Stolz theorem? :?

What about you, Cezar, do you know Cesaro-Stolz? Here's a counter-example to what you have written above $x_n=1+1/2+...+1/n$.

Cimuabue has, in effect, used Abel's summation by parts identity, which is well worth knowing.

To cope with the "more or less" in his first post, we can write that exactly as:

$\sum_{k=1}^nx_k=nx_n+\sum_{k=1}^{n-1}k(x_k-x_{k+1}).$

So cimabue is completely correct: If we assume that $\sum y_n$ converges, then having $nx_n$ converge to a limit is a necessary and sufficient condition for the convegence of $\sum x_n.$ Suprisingly, we don't need for $x_n$ to be decreasing.

The are no other hypothesis apart the ones mentioned.

Cesar, read again your formula, application of Cesaro-Stolz theorem and you'll see what I wanted to say.

The convergence of $\sum y_n$ gives the boundness of $S_n -nx_n$.So let $M$ such that $S_n-nx_n$ is bounded by $M$.From this we get that $|\frac{S_{n-1}}{n-1}-\frac{S_n}{n}|\leq \frac{M}{n(n-1)}$.So,for $n\leq p,|\frac{S_{n}}{n}-\frac{S_p}{p}|\leq M(\frac{1}{n}-\frac{1}{p})$.Let $p$ go to infinity and we get $S_n\leq M$.Conclusion fallows.

Probably...but it gives what the posted problem asks for.So I dont see what is your remark doing there.My post was just a solution for the initial problem. ;)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group, and $ H$ a normal subgroup of $ G$ whose order is the lowest prime dividing $ |G|$. Then $ H \subset Z(G)$

I tried to use the action of $ G$ on $ H$ by conjugation...

EDIT: Incomplete solution. (Thanks [b]LydianRain[/b] for noticing the very strange mistake I made!)

An alternative solution:

Since $ H$ is a normal subgroup, $ (G,H)$ is a valid permutation representation with conjugation as it's action. The order of the orbit of a nontrivial element $ h$ of $ H$ has to be a divisor of $ |G|$, and since $ h$ is not conjugated to $ e$, we must have that $ h^G \equal{} \left\{h\right\}$.

[quote="ysharifi"]Thus $ k^n \equiv 1 \mod p,$ and so $ p \minus{} 1 \mid n.$[/quote]
Why?

You proof cannot be correct; it would imply that a group with a non-central normal subgroup of order 5 must have order divisible by 4.  But the dihedral group of order 10 is a counterexample.

Jan's proof looks good.


[quote="daermon"]in triangle ABC, points D and E are drawn so that E is on BC and AB=BE and AE is perpendicular to BC.  if [] represents area, prove that [BDC]=[BACD][/quote]
Proofs certainly aren't MathCounts level, so I'll move this to Intermediate.
Also, are you sure the problem statement is correct? You never gave any information about where point D was located.  :wink:

how can AB=BE
if $BE^{2}+AE^{2}=AB^{2}$ since they are perpendicular?

oops.  I meant AD=DE, not AB=BE  :blush:
D is inside the triangle


function
calculus
derivative
inequalities
inequalities proposed

Let a,b,c>0
and $ a^2\plus{}b^2\plus{}c^2\equal{}3$
Prove that $ (2\minus{}ab)(2\minus{}bc)(2\minus{}ac) \geq 1$

if we can use Lagrange's method... :D
take the function $ L \equal{} f(a,b,c) \minus{} \lambda(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3)$.where $ \lambda \in R$ and $ f(a,b,c) \equal{} 7 \plus{} 2abc(\sum a) \minus{} (abc)^2 \minus{} 4(\sum ab)$.
then we should compute the derivatives of the function $ L$  wrt $ a,b,c$ and let them be equal to $ 0$.
by doing all this work we get that the extremum of this function are reached at points $ (0,0,\sqrt {3})$ and $ (1,1,1)$ which leads to $ 8 \ge (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ac)\geq 1$ but since $ a,b,c > 0$ then $ 8 > (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ac)\geq 1$.

[quote="nguyensilva"]Let a,b,c>0
and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
Prove that $ (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ac) \geq 1$[/quote]
This inequality is equivalent to $ w^6 \plus{} Aw^3 \plus{} B\leq0,$ where
$ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ A$ and $ B$  are functions of $ u$ and $ v^2$ only.
Thus, $ w^6 \plus{} Aw^3 \plus{} B$ gets a maximal value when $ w^3$ gets an extremal value,
which happens  when two numbers from $ \{a,b,c\}$ are equal or when $ w^3 \equal{} 0.$
Hence, it remains to check two cases:
1) $ c \equal{} 0.$ In this case we need to prove that $ ab\leq\frac {7}{4},$ when $ a^2 \plus{} b^2 \equal{} 3,$ which is true.
2) $ b \equal{} c.$ In this case after homogenization we obtain
$ 7(a^2 \plus{} b^2 \plus{} c^2)^3 \minus{} 12(ab \plus{} ac \plus{} bc)(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{}$
$ \plus{} 18abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \minus{} 27a^2b^2c^2\geq0$ and we can assume here $ b \equal{} c \equal{} 1.$
 But $ 7(a^2 \plus{} 2)^3 \minus{} 12(2a \plus{} 1)(a^2 \plus{} 2)^2 \plus{} 18a(a \plus{} 2)(a^2 \plus{} 2) \minus{} 27a^2\geq0\Leftrightarrow$
$ \Leftrightarrow7a^6 \minus{} 24a^5 \plus{} 48a^4 \minus{} 60a^3 \plus{} 45a^2 \minus{} 24a \plus{} 8\geq0\Leftrightarrow$
$ \Leftrightarrow(a \minus{} 1)^2(7a^4 \minus{} 10a^3 \plus{} 21a^2 \minus{} 8a \plus{} 8)\geq0,$ which is true.
By the way, we see that your inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136486


function
trigonometry
algebra
domain
inequalities
real analysis
real analysis unsolved

Let f be a continuous function defined for 0<=x<=1. Assume that f(0) is not 0 and that f takes any value a finite number of times. Then there is a value taken an odd number of times.

There's something I don't quite get about this: what do we need the condition f(0)<>0 for? If f is real-valued and f(0)=0 then we can prove what we want for g(x)=f(x)+1 and then go back to f and the problem is Ok for f(0)=0 as well.

Maybe the text states that f can't take any value, just values in [0,1] or something like that, but I find that hard to believe as well.

So what's the deal with the condition f(0)<>0 then?

I just saw this problem in one of my materials, I don't have the magazine to see what is with that condition.

[u][b]The author of this posting is : grobber[/b][/u]
____________________________________________________________________

Frankly, if f is real-valued, I don't see the purpose of the condition f(0) <> 0. If f(0)=0 we solve the question for g(x)=f(x)+1 and go back to f, and it holds in the case f(0)=0 as well, so what's up with that?

Maybe we also have f:[0,1] $\longmapsto$ [0,1] or something like that, or maybe the condition f(0) <> 0 just doesn't exist and f is real-valued.

the condition that $f$ takes a value a finite number of times is trivial, since $f\in C[0,1]$ , $f^{-1}(a)$ is finite for any $a\in\mathbb{R}$ .

We should show that it takes a value an [b]odd[/b] number of times.

yes, but harazi wrote:
[quote]Assume that f(0) is not 0 and that [size=150]f takes any value a finite number of times[/size].[/quote]
Since $f$ is continious it always happens.

[quote="hcast"]the condition that $f$ takes a value a finite number of times is trivial, since $f\in C[0,1] , f^{-1}(a)$ is finite for any $a\in\mathbb{R}$ .[/quote]
Uh - better be careful what you're saying there. let $f(x) = x\sin(\frac1x)$ for $x\neq 0$ and $f(0)=0$. Then $f\in C[0,1]$ but $f^{-1}(0)$ is an infinite set. We can say that $f^{-1}(a)$ is closed and we can say that $f^{-1}(a)$ is compact, but we can't say that $f^{-1}(a)$ is finite.

Even simpler:  Let

f(x) = x (0 <= x < 1/3)
       = 1/3 (1/3 <= x <= 2/3)
       =  1 - x (2/3 < x <= 1)

f^(-1)(1/3) is very infinite.  (And for no a does f(x) take on the value a an odd number of times.)

So the condition on the original problem is that $f\in C[0,1]$ [b]AND[/b] $f^{-1}(a)$ is finite $\forall a$. As Alison and I have noted, this condition is more restrictive than merely $f\in C[0,1]$, and as Alison's example shows, the theorem is false without this additional condition. However, Alison's example is not a counterexample to the original problem.

The examples $f(x)=\sin(4\pi x), g(x)=\sin((4\pi-.01)x),$ and $h(x)=\sin((4\pi+.01)x)$ may provide some insight into this problem, as the conclusion is true for each of them. You might want to investigate whether or not the domain can be partitioned into finitely many intervals on which the function is monotone.

I'm pretty sure we can't say that the domain can be partitioned into finitely many intervals on which the function is monotone. I can't give you a specific counterexample, but think of some sort of stairway s.t. the steps are, in fact, little "humps" which get smaller and smaller and higher and higher as you go towards the right endpoint of the domain. The function is thus "sort of" increasing, havin occasional downward "falls".

Such a graph might not satisfy the condition that every value is taken an even number of times (actually, we have to prove that no graph satisfies the condition :)), but I don't think it's obvious.

I was hoping to use that method to solve this problem yesterday, but unfortunately it's false. Consider the function $f(x)=2x+x\sin\frac{1}{x}$.

Here is a counterexample:

Let $f:[0,2] \to [0,1]$ be a piecewise linear function with vertices:
$(2,0), (\frac32,1), (1,\frac12), (\frac12,1), (\frac14,\frac18), (\frac18,\frac12), \dots, (\frac1{2^{2k}},\frac1{2^{2k+1}}), (\frac1{2^{2k+1}},\frac1{2^{2k-1}}), \dots$, and $f(0)=0$. Since $\frac12x \le f(x) \le 4x$ for $0 < x \le 1$, $f$ is continuous at 0, and the piecewise linearity makes it continuous everywhere else as well.

$f$ takes on each value in its image four times, except that the values 0 and 1 are only taken twice.
If you wanted to, you could rescale this to match the original conditions.

Perhaps that mysterious and nonsensical condition "$f(0)\neq 0$" should have been the not-so-nonsensical "$f(0)\neq f(1)$." jmerry's example did have (and need) $f(0)=f(1)$ (well, $f(2)$ in his case but it's the same thing.)

Consider the problem still alive and kicking, with this condition.

Sorry to revive such an old topic, but I remembered this problem by accident, and I realized that Kent is right: if we make the reasonable assumption that that condition was meant to be $f(0)\ne f(1)$, then the problem becomes true:

Suppose WLOG that $f(0)=0$. We want to prove that $f(1)=0$ as well. For every $M>0$, we call a point $x\in(0,1),\ |f(x)|=M$ an in-point if for some $\varepsilon>0,\ |f|>M$ on $(x-\varepsilon,x)$ and $|f|<M$ on $(x,x+\varepsilon)$. We call $x\in(0,1),\ |f(x)|=M$ an out-point if the reversed inequalities take place for small enough $\varepsilon>0$. The only way $x$ could fail to be either an in our an out-point is if $|f|$ reaches a local extremum at $x$. This, however, is possible only for countably many $M$. Indeed, if $|f(x)|=M$ and $x$ is a local extremum for $|f|$, then $M$ is the maximal/minimal value of $|f|$ on some interval with rational endpoints, and there are only countably many such intervals.

Assume now that $0<M<|f(1)|$, with $M$ chosen such that all elements of $|f|^{-1}(M)$ are either in or out-points (it is possible to find such $M$, according to the paragraph above). The largest element of $|f|^{-1}(M)$ must then be an out-point. However, when ordered increasingly, the elements of $|f|^{-1}(M)$ are alternatively in and out-points, with the first one being an out-point. Since $|f|^{-1}(M)$ has an even number of elements, the largest one must be an in-point, and we have a contradiction.


I am interested in this kind. Another idea from this kind is the following

Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=1$, find all positive rational $r$ number such that the following expression
\[ \frac{x^r}{y+z}+\frac{y^r}{z+x}+\frac{z^r}{x+y} \]
has a minimum.

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

$\Longleftrightarrow 2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a)$ ;)

[quote="arqady"]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq\frac{3}{2}.$[/quote]

Are the following inequality right?

$0<a+b+c\leq 3,\ 0<ab+bc+ca\leq 3.$

[quote="arqady"]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq\frac{3}{2}.$[/quote]

This follows immediately from the following inequality (which is not hard to prove): for all positive reals $a,b$ and $c$,
\[ \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{3}{2}\cdot\frac{a^3+b^3+c^3}{a^2+b^2+c^2}  \]

[quote="Lovasz"]$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{a^3+b^3+c^3}{3}\left(\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\right)$[/quote]

Lovasz, could you teach me how we can obtain this inequality? :?

kunny

[quote="Lovasz"]$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{a^3+b^3+c^3}{3}\left(\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\right)$[/quote]

sign reversed !

Yes, it's wrong.
[hide="to ductrung"]Where are you from, Australia?[/hide]

Oh, really? I didn't notice that.

Can't I proceed afterwards? Could you tell me, please?

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

$\Longleftrightarrow 2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a)$

kunny

[quote="kunny"]Oh, really? I didn't notice that.
Can't I proceed afterwards? Could you tell me, please?
kunny[/quote]
Tell you about what, kunny?

[quote="kunny"]$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq\frac{3}{2}$
$\Longleftrightarrow 2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a)$
[/quote]
This follows what?

[quote="kunny"]Can we prove $2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a)?$[/quote]
I think we cannot.
Your question:
$2(a+b+c)(abc+3) \geq 3(a+b)(b+c)(c+a)$
Let $a=1, b \to 0$, and you can find a counter-example base on the ineqn $3c^2-3c-6 \ge 0$

Under conditions that $a>0,\ b>0,\ c>0,\ a^3+b^3+c^3=3?$

I am comfusing now. :?  Anyway I will show you my approach as follows.

Let $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=\frac{f(a,b,c)}{(a+b)(b+c)(c+a)}.$

$f(a,b,c)=a^2(c+a)(a+b)+b^2(a+b)(b+c)+c^2(b+c)(c+a)$

$=a^4+b^4+c^4+(b+c)a^3+(c+a)b^3+(a+b)c^3+abc(a+b+c)$

$=(a+b+c)(a^3+b^3+c^3)-a(b^3+c^3)-b(c^3+a^3)-c(a^3+b^3)$

$+(b+c)a^3+(c+a)b^3+(a+b)c^3+abc(a+b+c)$

$=(a+b+c)(abc+3)\ \because a^3+b^3+c^3=3.$

Therefore we can rewrite the given inequality as $2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a).$

Am I something wrong?

kunny

[quote="hungkhtn"]SOS :D.... Nice problem.[/quote]
Thank you, hungkhtn.
I think SOS here it is very ugly. (Maybe I don't see something.)
A.Gladkich has found  very nice proof for an original inequality without SOS. ;)

Hello to everybody after my long absence. I have tried to solve this inequality using Vasc approach and Ductrung approach, but I cannot. Can anybody help me, please? :blush:

Thank you very much.

[quote="ductrung"]
This follows immediately from the following inequality (which is not hard to prove): for all positive reals $a,b$ and $c$,
\[ \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{3}{2}\cdot\frac{a^3+b^3+c^3}{a^2+b^2+c^2}  \][/quote]

 :?:  :?:

manlio, try the gladkich's hint:
$\frac{a}{b+c}\geq\frac{8a-b-c}{4(a+b+c)}.$  ;)

Ok,I post my solution to the problem :
                $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} \geq \frac{3}{2}\dots\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$ (*)
Solution:
(*) $\Longleftrightarrow 2\sum \frac{a^2}{b+c}-\sum a \geq 3\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\sum a$
     $\Longleftrightarrow \sum (2\frac{a^2}{b+c}-a) \geq \frac{2(a^3+b^3+c^3)-\sum_{sym} a^2b}{a^2+b^2+c^2}$
     $\Longleftrightarrow \sum \frac{a(a-b)+a(a-c)}{b+c} \geq \frac{\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$
     $\Longleftrightarrow \sum (\frac{a(a-b)}{b+c}+\frac{b(b-a)}{a+c}) \geq \frac{\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$
     $\Longleftrightarrow \sum \frac{(a-b)^2(a+b+c)}{(a+c)(b+c)} \geq \frac{\sum (b-c)^2(b+c)}{a^2+b^2+c^2}$
Put:$S_{a}=\frac{a+b+c}{(a+b)(a+c)}-\frac{b+c}{a^2+b^2+c^2}$
      $S_{b}=\frac{a+b+c}{(a+b)(b+c)}-\frac{a+c}{a^2+b^2+c^2}$
      $S_{a}=\frac{a+b+c}{(c+b)(a+c)}-\frac{a+b}{a^2+b^2+c^2}$
Now ne need to prove $\sum S_{a} (b-c)^2 \geq 0$
It approves because we have $S_{a},S_{b},S_{c} \geq 0$
So the problem is proved :D  :D

Very nice proof, nhat!
What if $a^4+b^4+c^4=3$ $?$
I think that your method is not work here. :P  :)

[quote="Vasc"][quote="kunny"]Can we prove $2(a+b+c)(abc+3)\geq 3(a+b)(b+c)(c+a)?$[/quote]
Yes, kunny. It is true.
Use helping inequality
$\frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \frac 9{a^2+b^2+c^2+3(ab+bc+ca)}$.[/quote]

Thank you very much, [b]Vasc[/b].

kunny

Hello manlio,

My proof is exactly the same as Nhat's. The end result is the identity
$\sum_{cycl}\frac{a^2}{b+c}-\frac{3}{2}\cdot\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$
$=\frac{a^3+b^3+c^3-2abc}{2(a+b)(b+c)(c+a)(a^2+b^2+c^2)}\sum_{cycl}(a+b)(a-b)^2$

[quote="manlio"]Dear arqady I use your hint this way, but I cannot solve last ineq

$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} \geq \sum\frac{8a^2-ab-ac}{4(a+b+c)} \geq \frac{3}{2}\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$

Can you please help me?

Thank you very much.[/quote]
Dear manlio! I meant to
 $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} \geq \sum_{cyc}\frac{8a^2-ab-ac}{4(a+b+c)} \geq \frac{1}{2}\cdot\sqrt[3]{9(x^3+y^3+z^3)}=\frac{3}{2}.$  ;)  :)
By the way, your last inequality is not true. Try $a=b=2,$ $c=1.$ ;)

Both this problem and older one
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 3\sqrt[4]{\frac{a^4+b^4+c^4}{3}} \]
can be proved by one way. Notice that the following result implies all of them
\[ (x+y+z)^3 \ge 3(xy+yz+zx)\sqrt{3(x^2+y^2+z^2)} \]

Thank you very much, Arqady.

Now it works  :)

[quote="arqady"][quote="hungkhtn"] Notice that the following result implies all of them
\[ (x+y+z)^3 \ge 3(xy+yz+zx)\sqrt{3(x^2+y^2+z^2)}  \][/quote]
It is a special case ($n=3$) of my inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78240
 :)[/quote]

Oh, actually I haven't see it before and this inequality I post was insprired from my solution for the older one. Very nice result, arqady.

[quote="hungkhtn"]
\[ (x+y+z)^3 \ge 3(xy+yz+zx)\sqrt{3(x^2+y^2+z^2)}  \][/quote]

After some computation I reducede it to SOS method

[quote] \[ \sum{[13z^3(x+y) +14xy(x^2+y^2) +x^4+y^4 +3x^2y^2 -xyz(7z + 24x +24y)](x-y)^2 \geq 0}  \][/quote]

but now I am in trouble with this one  :blush:

Has anyone tried to solve it?

Is this method OK?

Thank you very much.

P.S. Vey nice ineq :)

Dear Manlio, the SOS form is not such complicated as that. Change it into the form
\[ \frac{(a+b+c)^2}{ab+bc+ca}-3 \ge 3(\frac{\sqrt{3(a^2+b^2+c^2}}{a+b+c}-1) \]
Now I think it's easier so much.

[quote="hungkhtn"]SOS :D.... Nice problem.[/quote]
Thank you, hungkhtn. I think SOS not work here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86713
  :P  :)

Dear Aquady,

I'm sure that S.O.S will work. The lemma I post will help to solve both problems, involve yours.

First, use Holder
\[ (\sum\frac{a^2}{b+c})(\sum\frac{a^2}{b+c})(\sum a^2(b+c)^2) \ge (\sum a^2)^2 \]
Then notice that
\[ \sum a^2(b+c)^2 \le 4\sum a^2b^2 \]

 :lol: So we are done~

[quote="hungkhtn"]
First, use Holder
\[ (\sum\frac{a^2}{b+c})(\sum\frac{a^2}{b+c})(\sum a^2(b+c)^2) \ge (\sum a^2)^2  \]
Then notice that
\[ \sum a^2(b+c)^2 \le 4\sum a^2b^2  \]
[/quote]

Nice proof, hungkhtn.

BTW, I believe that the following inequality holds but I don't have a nice proof yet:
\[ \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{1}{2}\sqrt[5]{27(a^5+b^5+c^5)} \]
It does not hold if we replace 5 by 6.

[quote="ductrung"]
BTW, I believe that the following inequality holds but I don't have a nice proof yet:
\[ \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{1}{2}\sqrt[5]{27(a^5+b^5+c^5)}  \]
[/quote]
Maybe $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{1}{2}\sqrt[5]{81(a^5+b^5+c^5)}$ $?$ ;)

[quote="arqady"][quote="hungkhtn"]
First, use Holder
\[ (\sum\frac{a^2}{b+c})(\sum\frac{a^2}{b+c})(\sum a^2(b+c)^2) \ge (\sum a^2)^2  \]
[/quote]
Maybe $(\sum\frac{a^2}{b+c})(\sum\frac{a^2}{b+c})(\sum a^2(b+c)^2) \ge (\sum a^2)^3$ $?$
If so, why $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{1}{2}\sqrt[4]{27(a^4+b^4+c^4)}$ $?$ Thank you. :)[/quote]
Then use the Hung's Lemma:
$(x+y+z)^3 \ge 3(xy+yz+zx)\sqrt{3(x^2+y^2+z^2)}$ :)
Nice solution.

WOW! Thank you, zhaobin. Very nice proof, hungkhtn!

[quote="arqady"]
Maybe $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge \frac{1}{2}\sqrt[5]{81(a^5+b^5+c^5)}$ $?$ ;)[/quote]

Yeah, that's what I mean. Thank you, arqady.

[quote=arqady]$a>0,$ $b>0,$ $c>0,$ $a^3+b^3+c^3=3.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq\frac{3}{2}.$[/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy76sGtu0DF3b&690[/img]


http://groups.myspace.com/mathdorksunited

Join if you can... Lets make this a huge kick butt myspace community!

 :D

MathCounts portion of my site is down... Mad My site will be up in september again... I am waiting for my uncle to bring the server down from Massachusetts! If someone would be willing to temporarily host all of this stuff i have I would really appreciate it.


geometry
geometric transformation
rotation
AMC

Ten chairs are evenly spaced around a round table and numbered clockwise from $ 1$ through $ 10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or directly across from his or her spouse. How many seating arrangements are possible?

$ \textbf{(A)}\ 240\qquad
\textbf{(B)}\ 360\qquad
\textbf{(C)}\ 480\qquad
\textbf{(D)}\ 540\qquad
\textbf{(E)}\ 720$

[hide]Notice that if the wife is in one seat, the husband can only be in 2 different seats by the restriction of the problem.   Now notice that there is 10 places for the first woman to be.  Then, there are only 4 places for the next one to be, since the women and men alternate.  Then there are 3 more places for the 3rd woman, 2 for the 4th, and 1 for the 5th woman.  This gives 240 ways for the women to be seated.
For the men, on any open place, there are only 2 men that can be there.  Given one man, the other man that can be there has only one other place to go.  So there are only 2 choices for the men for every choice of women.  This gives 2*240 or 480 choices.[/hide]

well it has to be a multiple of 240 since its 5!*2 for 5! for the men to sit then 2 for either all even or odd....the women i just got 2 ways they can sit down after the men had been seated...i didnt think 720 would work becuase 3 seems like a weird number due to rotations...
had 960 been a choice, i mightve put that..

Sorry, I'm confused why it's 5! and not 4!. Aren't rotations the same so wouldn't you do $(5-1)!$ ?

Rotations aren't the same since the seats are labeled differently from each other.  If they were unlabeled then we would have to divide by 10 (not 5) to account for the rotations.  As the problem stands, we cannot account for rotations because the problem never specifies that rotations are equivalent.

[quote="hello.world"][hide="pretty much the same..."]The partner of the person in seat 1 must sit in either seat 4 or seat 8. Either way, the rest of the pairs can only be placed in one way. This gives two symmetric cases. Multiplying by 5! (to account for permutations of the 5 pairs) and by 2 (placing men/women on even or odd numbered chairs) we get $ \boxed{480\ C}$.[/hide][/quote]

Sorry, but can somebody further explain this? Why can the partner of person in seat 1 only sit in seat 4 or 8? Is this some intuition I need to develop - I'm not sure how I would see this in the heat of a competition?

Why can the rest of the pairs only be placed in one way? Can't we switch the positions of two pairs and it will still be valid? (To elaborate what I mean, if B and b sit in 2 and 7, C and c can be changed with them...no?)

Also the AMC solution involves seating the women first. is this a general good way of tackling these "seating" problems?

There isnt any "6th sense" type deals, here or in most cases, anywhere. Just draw the table. Then $3$ are taken out by the spouse restriction and $2$ are taken out by the gender restriction.

[quote="Aequilipse"]

Also the AMC solution involves seating the women first. is this a general good way of tackling these "seating" problems?[/quote]

 I am sure the AMC seats the women first because all male mathematicians are gentlemen. So, yes, generally it is a good idea to seat women first.


Hi all,

I need help solving this probability problem. If anyone can help, can you write the solution in a  step by step manner? Thanks in advance.

"Xavier, Yul, and Zeb are having a 3 way paint ball duel. Xavier is a poor shot, hitting his target 1/3 of the time correctly. Yul hits the target 2/3 of the time. Zeb hits his target all the time(no miss). Each person will get one shot at a target of his choice. Once you are hit, you are out. The last person left wins. Since he is the worst, Xavier goes first, followed by Yul, then Zeb. What is Xavier's best strategy, and what is his probability of winning? "

Hello tpulak.

I myself keep forgetting how to do this problem. I just saw it the other day in the Intro to Counting and Probability book. Here's a link that might help....

http://rutherglen.ics.mq.edu.au/math106S206/ex/threewayduel.pdf

The numbers are a bit different, but the theory is all the same. You just have to change the fractions around...

Yes, as a matter of fact, I have gotten this question from my introduction to probability book. Thanks for the comparative solution,but I would really like a solution oriented to this exact problem, as I still am learning this concept.Thanks for helping though.

Oh no, not again.
[hide]
After Xavier shoots (we don't decide who he should shoot yet), there are three possibilities: Yul is out, Zeb is out, or nobody is hit.

Suppose Yul is out. Then it is Zeb's turn, so he will obviously shoot Xavier and Xavier will always get out. Chance of winning: $ 0$.

Suppose Zeb is out. Yul will shoot at Xavier, hitting 2/3 of the time. Then, if Yul misses, Xavier shoots at Yul, hitting 1/3 of the time. And so on until somebody finally hits. Yul wins $ \frac{2}{3} \plus{} \frac{1}{3}\times\frac{2}{3}\times\frac{2}{3} \plus{} \frac{1}{3}\times\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3} \plus{} \cdots$ of the time.

This is a geometric series, with first element $ \frac{2}{3}$ and ratio $ \frac{2}{9}$. Apply the formula for geometric series. The probability that Yul wins is $ \frac{6}{7}$, so Xavier only wins $ \frac{1}{7}$ of the time.

Finally, suppose Xavier misses. Then Yul will shoot Zeb, since Zeb would shoot Yul if Zeb got to shoot. Then if Yul misses Zeb shoots and outs Yul. Thus, 1/3 of the time Xavier gets to shoot first against Zeb; 2/3 of the time it's against Yul.

Xavier versus Zeb: Xavier wins only if he can hit Zeb the first time, $ \frac{1}{3}$.
Xavier versus Yul: Note this is different from the above annoying geometric series probability since Xavier shoots first, but we can use the result. If Xavier hits Yul, Xavier wins: $ \frac{1}{3}$ of the time. Otherwise, $ \frac{2}{3}$ of the time, it reduces to the above series. Xavier wins $ \frac{1}{7}$ of the time, so his total chance against Yul is $ \frac{1}{3} \plus{} \frac{2}{3}\times\frac{1}{7} \equal{} \frac{9}{21} \equal{} \frac{3}{7}$.

So his total chance against either Zeb or Yul, whoever he faces off, is $ \frac{1}{3}\times\frac{1}{3} \plus{} \frac{2}{3}\times\frac{3}{7} \equal{} \frac{1}{9} \plus{} \frac{2}{7} \equal{} \frac{25}{63}$ which is the best probability compared to if he shot and outed anybody. So Xavier should deliberately miss or shoot the ground or the air or behind him or whatever.

This is actually not complete; we must demonstrate Yul and Zeb's best strategies to prove the above, but I'm going to leave that to, um, the next poster. Also, check my arithmetic. I make the most amazing mistakes.
[/hide]

Xavier should fire into the air.  Yul will shoot at Zeb, because he is the biggest threat.  If he misses, then Zeb will shoot at Yul and Yul is out since Zeb is a perfect shot.

I'm not sure about the probability part though? :?:


Let $ x, y, z$ are non- negative integers number a different pair such that $ (x\plus{}z)(z\plus{}y)\equal{}1$. Prove that
$ \dfrac{1}{(x\minus{}y)^2}\plus{}\dfrac{1}{(x\plus{}z)^2}\plus{}\dfrac{1}{(z\plus{}y)^2}\geq 4$.

If $ x, y,$ and $ z$ are non-negative integers, then $ (x\plus{}z)(z\plus{}y) \equal{} 1$ gives us that $ x\plus{}z\equal{}1$ and $ z\plus{}y\equal{}1$, which gives us that $ x\equal{}y$. However, this would be impossible since it makes the inequality undefined?

Maybe they're supposed to be nonnegative reals?

[hide="Assuming that x, y, z are distinct real numbers"]Let $ x\plus{}z\equal{}a$.
Since $ (x\plus{}z)(z\plus{}y)\equal{}1$, we have $ z\plus{}y\equal{}\frac{1}{a}$.

Then, the original inequality turns into $ \frac{1}{\left(a\minus{}\frac{1}{a}\right)^2}\plus{}\frac{1}{a^2}\plus{}a^2\ge 4$.

Since $ a^2\plus{}\frac{1}{a^2}\equal{}\left(a\minus{}\frac{1}{a}\right)^2\plus{}2$, we now have to prove that $ \frac{1}{\left(a\minus{}\frac{1}{a}\right)^2}\plus{}\left(a\minus{}\frac{1}{a}\right)^2\plus{}2\ge 4$.

Substituting $ \left(a\minus{}\frac{1}{a}\right)^2\equal{}b$, we have to prove that $ b\plus{}\frac{1}{b}\ge 2$, which is trivial by AM-GM Inequality.

Q.E.D.[/hide]


Does there exist a function $ f: \mathbb{N}\to\mathbb{N}$ such that $ f(f(n\minus{}1))\equal{}f(n\plus{}1)\minus{}f(n)$ for each natural number $ n$?

Observe that f(n + 1) > f(n), and this is imediact.

Take an integer x.

f(x + 2) - f(x + 1) = f(f(x))

Let k = f(x). We have

f(x + 2) = f(x + 1) + f(k)

If k > x + 1, clearly left side < right side.. So f(x) < x + 2.

But f(2) > 0, so f(3) > 1, ..., f(n) > n - 2. Then,

n - 2 < f(n) < n + 2.

It´s easy to see that the difference between f(n) and n + 2 can´t become greater if n grows. However, this difference can change just two times. So there is an z > 5 such that

f(z + 1) = f(z) + 1

But f(z + 1) = f(z) + f(f(z - 1)),

and we get f(f(z - 1)) = 1, which implies f(z - 1) < 3, which implies z - 1 < 4, absurd.

i think you might consider deleting your post feliz :)

[quote="feliz"]

It´s easy to see that the difference between f(n) and n + 2 can´t become greater if n grows. However, this difference can change just two times. So there is an z > 5 such that

f(z + 1) = f(z) + 1
[/quote]

its okay now, but could you just explain a bit more that statement...


1) Predict the shape of $ ClF_3$

2) $ N_3H$ has two possible structure with the atoms written as NNNH. Find the type of hybridisation of the N atom bonded to the hydrogen atom in each case.

1) Pyramidal

2) $ N\equiv N^\plus{}\minus{}N^\minus{}\minus{}H$. ----> sp3
 $ N^\minus{}\equal{}N^\plus{}\equal{}N\minus{}H$ ---->sp2

[quote="hell_ever"]1) Pyramidal[/quote]

I think it's actually T-shaped.

edit : yeah there are lone pairs so it would be t-shaped

hell_ever : in your first structure for 2), that is a dative bond from the second to the third N, so why isnt it sp2 as well?

There are no dative bonds in hydrazoic acid.

Yep my bad ... could someone explain the hybridizations?

What do you want to be explained? In the first ressonance structure above the nitrogen forms bonds with two atoms and there two lone pairs: so four hybrid orbitals are needed, and the appropriate hybridization scheme is therefore sp3. A similar explanation holds for the other structure.


Let $p \geq 5$ be a prime number.  Prove that there exist at least 2 distinct primes $q_1, q_2$ satisfying $1 < q_i < p - 1$ and $q_i^{p-1} \not\equiv 1 \mbox{ (mod }p^2)$, for $i = 1, 2$.

This is strange, since if I'm not wrong it follows from the following IMO 2001 shortlisted problem:
  If p>3 is a prime then we can find n<p-1 such that $ p^2$ doesn't divide $ n^{p-1}-1 $ nor $ (n+1)^{p-1}-1$. Now, suppose for a certain x $ p^2$ doesn't divide $ x^{p-1}-1$. It's quite obvious that there must be a prime q|x such that $ p^2$ does not divide $ q^{p-1}-1$. Thus we can find two different prime factors of n and n+1 with the property stated. They will be certainly different. Am I missing something?

Valiowk, you guys should really be studying the past ISLs right? :)

Well it isn't necessary to use the other shortlisted problem.  It's enough to consider $(p-1)^{p-1}, (p+1)^{p-1}, (2p-1)^{p-1}, (p+1)^{p-1}$ and by a simple argument it can be shown that there must exist $q_1, q_2$ which are factors of some of $p-1, p+1, 2p-1, 2p+1$.

And yes, Valentin, we should have studied the past ISLs harder... :(

The main observation is that if $x^{p-1} \not \equiv 1$ (mod $p^2$), then some prime divisor $q$ of $x$ satisfies $q^{p-1} \not \equiv 1$ (mod $p^2$).

Indeed, if this were not the case, then we could multiply some numbers which are all $1$ (mod $p^2$) to obtain $x^{p-1}$, absurd.

Now, with the cases $p < 11$ being trivial, we will use this observation to solve the cases where $p\ge 11.$ Call a prime $q$ [b] tasty [/b] if $q^{p-1} \not \equiv 1$ (mod $p^2$). Notice that $(p-1)^{(p-1)} \equiv p+1$ (mod $p^2$) and $(p+1)^{(p-1)} \equiv -p+1$ (mod $p^2$). Since these are both not $1$ (mod $p^2$), both $p-1$ and $p+1$ have a prime divisor which is tasty. If these are not both $2$, we're done since they then must be distinct.

Otherwise, we have that $2$ is tasty. Now, we have that $(2p-1)^{(p-1)} \equiv -2p(p-1) + 1 \not \equiv 1$ (mod $p^2$). This means that $2p-1$ has a tasty prime divisor, which is clearly in $(1, p-1)$ and not equal to $2.$ Together with $2$, we are done.

$\square$


Suppose that the function $ f: \mathbb{R}\rightarrow\mathbb{R}$ has the prperty that

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ for all $ x$ and $ y \in \mathbb{R}$.

Prove that if $ f$ is continuous at some $ x_0\in \mathbb{R}$,then there exists a const $ m$ such that $ \forall x\in\mathbb{R}$,$ f(x)\equal{}mx$

[quote="pco"][quote="skywalkerJ.L."]Suppose that the function $ f: \mathbb{R}\rightarrow\mathbb{R}$ has the prperty that

$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ for all $ x$ and $ y \in \mathbb{R}$.

Prove that if $ f$ is continuous at some $ x_0\in \mathbb{R}$,then there exists a const $ m$ such that $ \forall x\in\mathbb{R}$,$ f(x) \equal{} mx$[/quote]

It's a basic classical result.

Any of the following supplementary conditions for Cauchy equation imply $ f(x) \equal{} \alpha x$ :

$ f(x)$ continuous
$ f(x)$ continuous at one point
$ f(x)$ monotonous
$ f(x)$ monotonous on a non empty open interval
$ f(x) > m$ $ \forall x\in(a,b)$ for some $ b > a$
$ f(x) < M$ $ \forall x\in(a,b)$ for some $ b > a$[/quote]

In the case of $ f(x)$ continuous at one point,I just have to prove $ f(nx_0\plus{}q)\equal{}f(nx_0)\plus{}f(q)\equal{}(nx_0\plus{}q)f(1)$($ q\in\mathbb{Q}$)

If all the real numbers can be expressed by $ nx_0\plus{}q$,then we are done.

Am i right?


Evaluate \begin{align*} (5a^2 - 13a + 4)(2a - 3) \end{align*} for $ a \equal{} 1\frac12$.

Plug in a
Notice that (2a-3)=0
Since anything multiplied by 0 is 0, the answer is
0.


Seeing as my school is taking the National French Exam on the same day as AIME I, and I don't want to have two major academic competitions on the same day, I want to take the alternate AIME.

So, how do I sign up for it? Who should I talk to? Do I simply need to pay the money or do I have to give some sort of reason?
Also, I don't think any teacher at my school wants to stay there for 3 hours after school, so does anyone have any recommendations for who I should call or look for, etc? Note that I live in a suburb west of Boston, MA.

You should tell whoever is in charge of the AMC at your school.  (Or did you not take the AMC through your regular school?)  They can make arrangements to get the Alternate AIME for you.  It costs $\$25$ -- this is something that you should [i]insist[/i] that the school pay for.

You're only supposed to take the Alternate AIME if you have some sort of conflict with the regular date, but clearly you do.  You don't need to provide the reason to the AMC.

The AIME is supposed to be given from 9 a.m. to Noon, unless your school has a really good reason not to do it at that time, and they're supposed to get permission from the AMC to give it at a time other than 9 a.m.

Bonne chance au concours français!  (J'ai etudié le français dans l'école aussi.  :) )

I also need to take the AIME II, but I heard it usually harder than the AIME I. And is it possible that the better people will be taking the AIME II, trying to get more time to practice, so that even the fact that they take a percentage of AIME I and AIME II for the USAMO will not help?

Correct, timely information from DPatrick.  Thanks for passing it along!

-- Steve Dunbar
Diector, American Mathematics Competitions

[quote="123s"]And is it possible that the better people will be taking the AIME II, trying to get more time to practice, so that even the fact that they take a percentage of AIME I and AIME II for the USAMO will not help?[/quote]

No, because it's not allowed unless you specifically have a conflict of dates for the AIME I.

But could there be a possibility that the aime don't arrive to my school? I'm qualified to write the aime but I don't know if my school had received the contest yet  :(

[quote="AMCDirector"]> If we cannot receive our material by March 7, would it be possible to allow us to take the AIME II?

If you have not received materials by March 6, have your teacher contact us. We will make an alternate arrangement.[/quote]

Does the alternate AIME decrease your chances of qualifying for USAMO though? I've heard that it is usually more difficult.

Darn... people ask that a lot. The answer is search. Actually, the answer is if there really is a huge difference, the AMC will find a way to take it into account. But usually there's not that big of a difference.

As seen by last year's (I think), if there's a large enough difference (probably index difference of 10+ points), they'll have a separate index for the AIME II. Otherwise, it's close enough that in general it won't matter.

[quote="probability1.01"]Darn... people ask that a lot. The answer is search.[/quote]

Only if everyone was like probability1.01 who knows how nice that button is.  :)

[quote="calc rulz"]Does anyone know when deadline is to register from AIME II?

The guy in charge of AMC at my school agrees to let me take it, however he says that "he called and e-mailed the AMC officials and has not gotten a single reply," suggesting that if he doesn't manage to get a response I might have to take the AIME I.[/quote]Does that mean he called and no one answered the phone?  Just to verify, the number listed in the AIME packet is (800) 527-3690.

Also, the AMC has set up a website where you can register for the Alternate AIME online.  I'm not sure that I'm supposed to give out the URL publicly (since I couldn't find it anywhere public), but the URL is listed on the AMC Score Report that you teacher should have received (on ours, it's on the page that says "THE 2006 AIME POLICY" near the top).

When registering for the Alternate Aime, do you do it yourself, or does your contest manager do it for you. Does this change if you are sending it to an alternate test location?

[quote="123s"]When registering for the Alternate Aime, do you do it yourself, or does your contest manager do it for you.[/quote]The teacher / contest manager should do it.  [quote] Does this change if you are sending it to an alternate test location?[/quote]For this you probably need to call the AMC.


My school starts on the 24th.  I hate this summer vacation, since this is like the shortest one I've had in a long time.  Usually our school starts aorun Sept. 1 or Aug. 31, but last year it started after labor day.  Now this year, it's starting earlier than usual.  It hasn't started this early since kindergarden.  Only 2 months and a week of summer vacation for me  :( .

Well mine the same thing.  We got like 2 months and maybe 2 weeks of summer.  But my school district always does that.

September 7, the tuesday after Labour Day, for Univ. of Toronto students.

The first day of classes at MTU is Aug. 29, IIRC, but first-year orientation starts Aug. 21.

My school starts August 16th :( Last year we started September 7th so I only had two months. Plus we have yearbook/pick up book day the 10th! btw, how long are your winter and spring breaks?

In US it is Sept 1st...but I think in most countries, it is May 1st

Labor day is when people just take a break off from school or work.  In the US, it's the first monday of september.

I start August 19th, 2004.  My enrollment is a weekish before that.

September 8th. Why do most of you start before labor day. Never seen a school that does it.

Wow... a lot of ppl start realli early... my school used to start Aug. 15, but now we're early september.. this yr is Sept. 2.

My school had to start early this year so I started Monday.  I have a busy schedule this year.  My classes aren't too bad compared to most people on this forum.  However, I am taking 3 AP's(4 if I can independent study stats, I have get that straightened out) and an honors, and a CP(below honors).  The CP is for Latin which they only offer in CP until you get Latin 4/AP.  What really is hard is doing cross country after school.  That means I won't get home until 5:45-6:00 everyday!  I 'm just glad I m not bored!  ;)

classes don't start until the 30th, but i can move back in on the 23rd

I know someone who starts on the 15th of September (yes...for college)

Well, in college come people might start even on September 22nd or 23rd, somewhere near there.  It just means that they are on a quarter system.  THose start late and end late compared to high school and stuff.

My school starts september 13th... I've just started my summer-homework...
yuk....

History, Latin, Philosofy, Italian Licterature... and a bit of trigonometry/algebra routine problems... And 13 technical paintings (bad english :maybe: )

I'd love to know when chinese school starts...

mine starts September 5 th. Can u believe? it's Sunday, terrible  :(

September 10th, but registration and stuff starts september 8th (?)

Wow! Your school starts very early. When I was looking at the dates I thought they were for September. In Albania it started on September 20th, and here in Canada it started on September 7 (the day after Labour Day). I didn't think other countries actually started in August!  :D  When does it end though? In Canada it ends June 27th. (Unless you're doing gr. 12 IB, where it ends at the end of April and te exams are in May, but that is applicable to Ib students all over the world, not only in Canada). :)

We do start early in the U.S.  I don't know why though.  Most of us usually get out at the end of may or early june though.

last monday in august...
which in this year's case is the 29th  ;)

my school starts "late" this year, aug 25

I don't start classes until September 26th (due to quarter system).  I go out for orientation the week before.  But then I don't get out until mid-June.

My school start's on none of those dates it starts on August 29th :)  :!:

[quote="236factorial"]YOu're a little off topic. This was last year  ;)[/quote]

Quite so.

Locked.


Does anyone know where I can find past NYSML Problems?  I checked the AoPSWiki, but the [url=http://www.artofproblemsolving.com/Wiki/index.php/NYSML_Problems_and_Solutions]NYSML Page[/url] was blank.  Furthermore, my coach said that she didn't have any old NYSML exams either.  Thanks!

i don't know if these are helpful but...

http://www.mathlinks.ro/viewtopic.php?t=16613

The recent batch of contests are being sold as a book for the quite reasonable price of 15 dollars: http://www.nysml.org/index.php?n=NYSML.ContestArchive

I don't know if the old ARML-NYSML books are still in print.  For a random selection of past problems, search for "NYSML" on this website.

Do you know if there are any free resources?  I can't afford all of the contests they're selling.  :(

do what jbl said and try to find some tests from other people

You might also ask your coach or the department head if the school can purchase it.  Are you on the NYC math team?  If so, you should ask around if anyone has some you can copy, including the other coaches (e.g. Mr. C.).  And, again, lots of NYSML and ARML stuff around the forums.

The NYSML 2005 individual problems were posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33106[/url].

The NYSML 2008 individual problems were posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205255[/url].

I also posted a dozen problems from the years 1989 through 1992 in this forum (NYC).  I got those problems from the book [i]ARML-NYSML Contests 1989-1994[/i] by Zimmerman and Kessler, published by MathPro Press.  I believe I purchased that book 4 years ago for about 20 dollars from here:
[url]http://www.mathpropress.com[/url].


1/2+1/3+1/42

47, but it's probably not the least

I don't get it. Would 1/7+2/7+3/7 count for 21? :maybe:

UNIT FRACTIONS ARE IN THE FORM 1/x, where x is not 0.  here, x must be POSITIVE because it asks for positive fractions

[quote="Walk Around The River"][quote="jli"][quote="asianhottie8293"]The sum of the three different positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?[/quote]

[hide="solution"]We know that one of the fractions must be $\frac{1}{2}$ because $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}<\frac{6}{7}$.

If one of the fractions is $\frac{1}{2}$, the sum of the other two unit fractions is $\frac{5}{14}$.

$\frac{5}{14}=\frac{1}{a}+\frac{1}{b}$

$\Rightarrow \frac{5}{14}=\frac{a+b}{ab}$

$\Rightarrow 5ab-14a-14b=0$

$\Rightarrow (5a-14)(5b-14)=196$

The factors of $196$ are:

$1,196$
$2,98$
$4,49$
$7,28$
$14,14$

$5a-14=1,2,4,7,14$ and the only integer value for $a$ is $3$

The three fractions are $\frac{1}{2}+\frac{1}{3}+\frac{1}{52}$.

Therefore, $2+3+52=\boxed{57}$.[/hide][/quote]

well that cant be right since indiana found one that was less.[/quote]

oops...it should be $\frac{1}{42}$ so 47.

[quote="ragnarok23"]ravi b came up with a baller solution to this, lemme see if i can find it
edit: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fractions+unit&t=127023[/quote]
Dude, Ravi B is a beast.
It's no surprise that he came up with such a pwnage solution.
If he could be in countdown I'd be even more scared of him than splash D


This is a part of a proof my dad gave me, can anyone make anysense out of it?
let ^ be intersection (i'm bad at latex)
=A-B^C+B^C-b^A-A^C+B^C

=A-B^C+B^(C-A)+C^(B-A)

=(A-B^C) + BU(C-A)+C^(B-A)

Does this make any sense?
And is + a symbol you can use in set theory that you can replace for U?

let me try to write it out

\cup and \cap represent union and intersection
everything else you typed is the same

$ A\minus{}B\cap C\plus{}B\cap C\minus{}B\cap A\minus{}A\cap C\plus{}B\cap C$

$ A\minus{}B\cap C\plus{}B\cap (C\minus{}A)\plus{}C\cap (B\minus{}A)$

$ (A\minus{}B\cap C)\plus{}B\cup (C\minus{}A)\plus{}C\cap (B\minus{}A)$

Going from that first statement to the second seems like a simple use of the distributive property
Intersections distribute

Going from the 2nd to the 3rd doesn't seem to make sense. It looks like your dad just turned an intersection into a union (which you can't do as far as I know). Did you copy it wrong?


A rational number between $ \sqrt{2}$ and $ \sqrt{3}$ is:

$ \textbf{(A)}\ \frac{\sqrt{2} \plus{} \sqrt{3}}{2} \qquad
\textbf{(B)}\ \frac{\sqrt{2} \cdot \sqrt{3}}{2}\qquad
\textbf{(C)}\ 1.5\qquad
\textbf{(D)}\ 1.8\qquad
\textbf{(E)}\ 1.4$

[hide="Solution"]

Options $ \textbf{(A)}$ and $ \textbf{(B)}$ are quickly discarted because they are not rational.

Now we square $ 1.4$, this gives $ 1.96$ which is clearly smaller than $ 2$ so it can't be in between $ \sqrt {2}$ and $ \sqrt {3}$. So option $ \textbf{(E)}$ is out. Another way is noticing that $ \sqrt {2}$ is aproximately $ 1.414 > 1.4$

Now square $ 1.8$, that will give you $ 3.24 > 3$ so option $ \textbf{(D)}$ is out.

So by canceling out all other possibilities $ \boxed{\textbf{(C)}}$ is the answer.
[/hide]


function
inequalities
calculus
derivative
real analysis
inequalities proposed

I noticed that the following lemma is quite useful. It is very simple with Lagrange multipliers but, maybe, somebody will offer an elementary proof.

Let $x_1,\ldots ,x_n>0$. Fix the sum $\sum_j x_j=S$ and the product $\prod_j x_j=P$. Suppose that $F(x)$ is a smooth function such that $F'(1/u)$ is strictly convex or strictly concave. Then the maximum and the minimum of the expression $\sum_j F(x_j)$ are achieved when among the values of $x_j$ there are at most 2 different. (I do not know whether we can strengthen it to "$(n-1)$ variables are equal"; if we can, we'll probably, have something equivalent or close to $(n-1)EV$ principle of Vasc. Nevertheless, even as is, it is a nice thing to know)

As an example, let us consider the famous $\sum \frac1{5-abc}\leq 1$ inequality of Vasc. If we fix the product to be $P$, then it can be written as $\sum\frac1{5-\frac Pa}\leq 1$. But the function $F(x)=\frac 1{5-\frac Px}$ satisfies the above property and, therefore, it is enough to consider the two-value cases. The 2-2 case (a=b=x, c=d=y) reduces to the inequality $\frac 1{5-x^2y}+\frac 1{5-xy^2}\leq \frac 12$ under the condition $x+y=2$, which can be rewritten as $(xy)^3-6xy+5=(1-xy)(5-xy-(xy)^2)\geq 0$, which is clear because $0\leq xy\leq 1$. The 3-1 case (a=b=c=x, d=y) reduces to the inequality $\frac 1{5-x^3}+\frac 3{5-x^2 y}\leq 1$ under the condition $3x+y=4$, which can be rewritten as $5-4x^2y-2x^3+x^5y\geq 0$. Plugging $y=4-3x$, we obtain $5-16x^2+10x^3+4x^5-3x^6\geq 0$. The left hand side may seem unpleasant, but it factors as $(1-x)^2(5+10x-x^2-2x^3-3x^4)$, so it is enough to check that the second factor is non-negative on $(0,4/3)$. But it is a concave function of $x$ on that interval, so it suffices to verify the inequality for the endpoints. The end.

Indeed Fedja, if $F'(1/x)$ is convex, then the expression $\sum_j F(x_j)$ is maximal and minimal when $n-1$ of $x_i$ are equal.
 ;)

Actually, so called (n-1) Equal Variable Principle is the following:

Let $a_1,...,a_n$ ($n \ge 3$) be given positive numbers, let $p$ and $q$ be unzero and non-equal real numbers, and let $x_1,...,x_n$ be positive numbers such that
$x_1^p+...+x_n^p=a_1^p+...+a_n^p$ and
$x_1^q+...+x_n^q=a_1^q+...+a_n^q$.
Let $f(x)$ be a differentiable function on $(0,\infty)$ so that $g(x)=x^{\frac{1-q}{p-q}}f'(x^{\frac{1}{p-q}})$ is strictly convex. Then, the symmetric function
$F_n(x_1,...,x_n)=f(x_1)+...+f(x_n)$
is minimal for $x_1\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\le x_n$.
Notice that for $p \rightarrow 0$ or $ q \rightarrow 0$, the sums become products.

[b]Attention.[/b] This primary form needs some correction. The corrected statement is the following:

Let $a_1,...,a_n$ ($n \ge 3$) be given positive numbers, let $p$ and $q$ be non-equal real numbers, and let $x_1,...,x_n$ be positive numbers such that
$x_1^p+...+x_n^p=a_1^p+...+a_n^p$ and
$x_1^q+...+x_n^q=a_1^q+...+a_n^q$ ($p=0$ or $q=0$ means $x_1x_2...x_n=a_1a_2...a_n$).
Let $f(x)$ be a differentiable function on $(0,\infty)$ so that $g(x)=x^{\frac{1-q}{p-q}}f'(x^{\frac{1}{p-q}})$ is strictly convex, and $F_n(x_1,...,x_n)=f(x_1)+...+f(x_n)$.
(a) If $pq \le 0$, then $F$ is minimal for $x_1\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\le x_n$;
(b) If $p>0$ and $q>0$, then $F$ is maximal for $x_1=x_2=...=x_{n-1}\le x_n$;
(c) If $p<0$ and $q<0$, then $F$ is minimal for $x_1\le x_2=x_3=...=x_n$.

Very nice idea, Vasc! Thank you for sharing it with us :). Now let's see if somebody can provide a nice proof  ;) .

[quote="fedja"]I noticed that the following lemma is quite useful. It is very simple with Lagrange multipliers but, maybe, somebody will offer an elementary proof.
[/quote]

Very interesting this lemma. Can you please post your solution?

Thank you very much, Fedja.

[quote="manlio"]
Very interesting this lemma. Can you please post your solution?
[/quote]
OK, but note that what I'll prove falls short of Vasc's $(n-1)EV$ principle. Using Lagrange multipliers, we see that, at the extremum point, either all $n$ variables are equal or there exist $\mu$ and $\nu$ such that $F'(x_j)=\mu\cdot 1+\nu\cdot\frac1{x_j}$ for all $j$. But, due to strict convexity (concavity) of $F'(1/u)$, the equation $F'(x)=\mu+\nu\frac1x$ may have at most 2 positive solutions (since the left hand side is a strictly convex (concave) function of $1/x$ and the right hand side is a linear function of $1/x$). The end.
This solution applies to the more general case considered by Vasc too, but gives only the same conclusion about just 2 different values. To prove the $(n-1)$EV principle in full, a different idea is needed. I'd like to see if somebody else finds it  ;) .

I think Fedja proof is OK  ;)

Never mind if it is a calculus based solution, the important thingh is that it works.

It suiffices to prove Vasc's lemma for three variables. Indeed, if for any three $x$-s, two greater of them are equal, then all $x$-s except the least are equal. Fedja's argument shows that two of variables are equal, and we only have to show that the case $x_1 < x_2 = x_3$ is not a maximum case. For proving this, we redenote the above point $(x_1,x_2,x_3)$ as $ P = (a,b,c)$ and make the following: vary $x_2$ in a neighborhood of $b$, and consider $x_1$ and $x_3$ as functions of $x_2$, defined by two equations $x_1^p+x_2^p+x_3^p=a^p+2b^p$ and $x_1^q+x_2^q+x_3^q=a^q+2b^q$ ($x_1$ varies in a neighborhood of $a$, $x_3$ in a neighborhood of $c=b$). They are well-defined due to implicit function theorem. More over, we may compute their derivatives in a point $(a,b,c)$. Do it, and then study a function $G(b)=f(x_1(x_2))+f(x_2)+f(x_3(x_2))$ in a point $x_2=b$. We may assure after some computations, that $G'(b)=0$ and $G''(b)>0$, since $g'(b)>(g(b)-g(a))/(b-a)$ from convexity of Vasc's function $g(x)=x^{\frac{1-q}{p-q}}f'(x^{\frac{1}{p-q}})$. So, our point does not correspond to a maximum, q.e.d.

Computations may be simplified, if we put $p=q+1$, which may be gotten if replace variables ($x_i=y_i^\alpha$
for appropriate $\alpha$).

We also have to consider a case, in which some of variables vanish (as we did not have to consider in Fedja's case), it is the work of the same kind.

I'm curious, what will happend if someone solves a problem in IMO 2005 using this method ?
Will he get all the points? none? or what ? Will the coordination comitee and the examiners be aware of it ?

thanks

Hey,
I think you forget my question. I'm really curious , what will happend ?
Will the coordination comitee accept such a proof ? what about B.M.O ? (which is gonna take place in some days... )

thanks

I used Lagrange multipliers on IMO, and got 7. But it was in Romania. Many things depend on coordinators. In any case, you must clearly formulate the facts you use, and preferably write the sketch of the proof. After that, you have to inform your team leader of what you meant, before coordination (it is forbidden, I know).  Then he says on coordination: "or, it's a basic thing, we in Greece learn it in primary school, our students would not imagine that this obvious statement  must be proved on a such high-level exam, as IMO", and presents to a coordinator a bottle of the very best Greek wine.

I think that (almost) the same coordinators will be present this year at the BMO as they were in 1999 at the IMO in Romania. So if Fedor Petrov got 7p points using Lagrange' multipliers (correctly!) I see no reason for you to fear :)

BTW, Fedor, AFAIK it is not forbidden to speak to the leader [b]after[/b] the two examinations  :?

Well, Firstly i won't participate in B.M.O or I.M.O . I'm a first year student at a university. (But i did participate in 2004). But i'm still intrested in such competitions.
As far as Lagrange' multipliers is conserned it is true that it's well know method for proving inequalities. So the coordination comitee should accept it.
But what about the "n-1 E.V.M of Vasc " ? Can they still accept a proof that uses this method ?

And generally, what are the limits of this  ? Can the participants use any know theorem ? any theorem ? and if they can ? should they? (i meant that should the coordinators accept such proofs ? or they must put some restrictions ? )

Valentine you are a coordinator ? (of at least one competition  ;) ) What's your opinion?

There are no general rules. But it is always better to formulate correctly all not-too-much-basic things, and outline their proofs. Usually, you should ask jury on an olympiad. In Russia, combinatorial and geometrical facts are appreciated rather than analytical and algebraical.

[quote="Fedor Petrov"]
We also have to consider a case, in which some of variables vanish (as we did not have to consider in Fedja's case), it is the work of the same kind.[/quote]
Actually this case can really cause some trouble. Seems like I just overlooked it when saying that the general Vasc's result can be proved the same way :blush:.  For instance, the minimum of $x^3+y^3+z^3$ over non-negative $x,y,z$ with $x+y+z=3$, $x^2+y^2+z^2=5$ is attained on the triple $(0,1,2)$ because $2(x^3+y^3+z^3)=3(x+y+z)(x^2+y^2+z^2)-(x+y+z)^3+6xyz$, in which expression everything is fixed except $xyz$, which is always non-negative and vanishes on the triple $(0,1,2)$. So, Vasc's (n-1)EV should be formulated more accurately: there is a possibility that a few variables vanish and, among the others, all but one are equal.
Also, some variables can escape to $+\infty$ in the similar way (if $p,q<0$). Still, even with this correction, (n-1)EV remains very nice.

[quote="Anto"] And generally, what are the limits of this ? Can the participants use any know theorem ? any theorem ? and if they can ? should they? (i meant that should the coordinators accept such proofs ? or they must put some restrictions ? ) [/quote]
I used to be on both sides of math. competitions and my impression is that the limits are set by coordinators just using common sense. There are no written rules for it. Of course, any theorem whose full proof is included will be accepted. Also, usually there is a pool of theorems that constitute "common knowledge". Again, there is no fixed list and the pool heavily depends on the particular contest, but Vasc's (n-1)EV is definitely not in that pool for any contest I know of and I doubt it'll be there soon (except, maybe, for the Mathlinks contest, where Valentin will, probably, allow it from now on, though you'd better ask him first). Everything outside that pool is usually considered requiring a proof though partial credit can be given in some border cases.

I wonder how many people except those from mathlinks know Vasc's invention. It is a great thing. Yet, I hope not to see applications of it in mathlinks contest, since it would be just as a homework.  ;)

Ops. Indeed, I was too much self- and Vasc- assured, and did not check it carefully. But I think that either minimum, or maximum may not be achieved for null values of some variables (that depends of convexity/concavity).

You are right fedja, the (n-1)EV theorem needs some correction. Thanks. :?
The corrected theorem ends as follows:
(a) If $pq \le 0$, then $F$ is minimal for $x_1\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\le x_n$;
(b) If $p>0$ and $q>0$, then $F$ is maximal for $x_1=x_2=...=x_{n-1}\le x_n$;
(c) If $p<0$ and $q<0$, then $F$ is minimal for $x_1\le x_2=x_3=...=x_n$.

Any new note is welcome.

In addition to the [b](n-1)EV Theorem[/b]:

[i]For $p>0$ and $q>0$, $F$ is minimal for $0=x_1=...=x_{k-1}<x_k\le x_{k+1}=...=x_n$, where $1\le k\le n$.[/i]

[b]Remark.[/b] Replacing all $x_i$ by $1/x_i$, the case $p<0$ and $q<0$ reduces to the case $p>0$ and $q>0$.

[quote="fedja"]This solution applies to the more general case considered by Vasc too, but gives only the same conclusion about just 2 different values.[/quote]

Sorry, but I don't understand in what variable is the RHS linear (in the general case). could you explain a little?

I sent a paper concerning EV-Theorem to be published in Gazeta Matematica A, in English.
I hope it will appear in 3-4 months.
In Attachment is the last form of the statement.

what is "is minimal for " means? does it nmeans that F(x) can attain its maximum and when it attain its maximum, must have n-1 equal variables ?