probability
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7 brothers are asked to participate in a contest.  They know the rules of the contest beforehand and may form a strategy to win as a team.

The 7 brothers are led into a room and placed in a circle.  The lights are turned off and hats are placed on their heads.  Each hat is either white or black [you may assume a fair coin is flipped to determine which color that particular brother receives].

The lights are turned on and each brother may see all the other hats except his own.

The brothers may not communicate in ANY way.

Each brother is then given a piece of paper on which he either writes "White" "Black" or nothing.

The seven brothers win if the following two conditions are fulfilled:

1. At least one brother guesses a color.
2. Every brother that guesses guesses correctly.

With an optimal strategy, how likely are the brothers to win the contest?

Surely it is not possible to get higher than $\frac{1}{2}$? Whatever way you look at it, the choice one of the brothers must make (his own colour) is independent of the information he has.

[quote="thanh22"]Whose color do the brothers need to guess?[/quote]Their own of course. They see the others... ;)

Let's see... there is no "winning strategy" for sure. Also chance must be above 1/2 to be better than the normal.

Can the brothers see if the other writes something or not, or is no communication to be taken litteraly?
If no, I see no way there is any information transferred, and I cannot understand how the chance can outrise 1/2.
If yes, I may have an idea :)

Ah, this problem. Originated, as far as I can tell, by Todd Ebert, currently a faculty member in the Computer Science department here at CSULB (and whom I had as a student in a probability course when he was an undergraduate).

Rest assured that there is a strategy which wins $\frac78$ of the time.

[hide="Hint."]Error-correcting codes.[/hide]

yes,i too have seen this before-and a probability of $7/8$ is indeed achieveable!

in fact if there are $2^n-1$ members, then a probability of $1-1/2^n$ can be acheived by forming a good strategy.
it is difficult to see this if one is not aware of binary Hamming codes and the fact that they are what are called Perfect codes,i.e. they are codes with a minimum distance of $3$ and if we place (disjoint) spheres of radius $1$ at each of the codewords,then it covers the whole of the ambient space $\mathbb{F}_2^{m},m=2^n-1$.consequently, for any element of $\mathbb{F}_2^m$, either it is a codeword or is at a distance of $1$ from a unique codeword.(this is of course due tto the fact that the spheres of radius one are disjoint).

It's simpler to explain in a smaller case, namely that there are 3 brothers.

The strategy can be summarized as "guess the error". Each brother sees two other hats. If they are the same color, he guesses his hat is the other color. If the two hats he sees are of different colors, he doesn't guess.

In 6 cases out of 8, there are two hats of one color and one of the other. In these cases, the brother with the unique color guesses correctly and the other two don't guess, so the requirements for winning are minimally met.

In the other 2 cases, all three brothers guess wrong at the same time. That's the secret of the counterintuitive better-than-half probability: piling all the wrong guesses on top of each other.

It can be shown that the probability of success cannot exceed $\frac{n}{n+1}$ for $n$ brothers; as far as I know this bound is only acheived for $n=2^k-1,$ but if you let $p(n)$ be the success probability for the optimal strategy for $n$ players, then $p(n)$ is a nondecreasing function.

Marvellous! :lol:

I had already been thinking in that direction, though I couldn't get it fit :)

Yep, you got it....

Note that even at a quick glance you can do better than 1/2 simply by pretending that you are certain that there will not be 0,2,5, or 7 white hats.  You can make a guessing strategy based on that which will win (7 + 35 + 35 + 7)/128 of the time.  [7,35,35,7 being the second,fourth,fifth, and seventh terms on the 8th row of Pascal's triangle.]

With that strategy you win as long as there is exaclty 1,3,4, or 6 white hats...and those possibilities are more common than those where there are 0,2,5 or 7 white hats.

[quote="Peter VDD"][quote="seshadri"]let me just give the basic gist of the argument in the general case as well(leaving out the more technical part of actually constructing the hamming code)-i might as well write this down,since i am in danger of forgetting it!

ok,the hamming code of length $n=2^k-1$ is a certain linear subspace $C$ of $\mathbb{F}_2^n$ of dimension $n-k$.in case one doesn't like to see linear algebra,one can think of this as a collection of $2^{n-k}$`codewords'. it satisfies the property i wrote earlier-
Every word $\mathbf{x}$of $\mathbb{F}_2^n$ is either a member of $C$ or there exists a unique member $\mathbf{c} \in C$ such that the `distance' from $\mathbf{x}$ to $\mathbf{c}$ is one(here by distance i mean the number of coordinate positions that are not the same between $\mathbf{x}$ and $\mathbf{c}$).
Also another property of the hamming code is that for any 2 `codewords' the distance between them is atleast $3$.actually this is just a consequence of the property mentioned above.

now, in this case let a black cap be represented by $0$ and a white cap by $1$.The $n$ members are arranged linearly and the caps are placed on their heads,thus deciding a vector in $\mathbb{F}_2^n$. the probability that this vector is a hamming codeword is $2^{n-k}/2^n=2^{-k}$. Thus by deciding to interpret the final word to be a non-codeword is always a better thing to do.
suppose firstly that the `cap-word' is not a hamming codeword,so that the guess could turn out right.now the $i^{th}$ member knows the vector at every position except the $i^{th}$.since the word is to be a non-codeword as noted earlier, if there exists a unique codeword $\mathbf{c}$ such that $\mathbf{x}_i$(the word as seen by the member) extends to $\mathbf{c}$, then he must put the 'other symbol' in the missing place.
now the above phenomenon cannot be faced by $2$ different members making different guesses since in that case there would be $2$ codewords $\mathbf{c}_1,\mathbf{c}_2$ of $C$ at distance $\leq 2$ apart,which means $\mathbf{c}_1=\mathbf{c}_2$.but then there are 2 words that are a distance of $1$ from $2$ distinct codewords and that is not possible.so every guesser would make the right guess.
if the actual word is indeed a Hamming codeword,then since the strategy is to guess for a non-codeword, they fail anyway.
of course this is not complete but hopefully the idea has been conveyed.(all this is,i guess well known-not due to me!)[/quote]

Looks incredibly cool! :)

Do they learn that in informatics? If so I guess I'm in the wrong studies... :?[/quote]

not necessarily :)
this is clasifed as Coding Theory-it is a branch of Discrete Mathematics

lol... not surprising, every part of discrete mathematics I've seen so far really interests me :)

totally off topic - does anyone know what studies have most discrete mathematics in them? Pure Mathematics? Engineer? Informatics? any other one?
( please reply in pm not to go even further off topic, thanks! :))


calculate the (Δx)(Δp) for the gaussian wavefruction $ \psi (x) = \sqrt [4]{\frac {\lambda}{\pi}} e^{ - \frac {\lambda x^2}{2}}$ and prove that is (Δx)(Δp)$ \geq \frac {\hbar}{2}$

[quote="Immanuel Bonfils"]The Gaussian wave function is one, some times called, minimum wave packet .
You're putting it as an exercise to people, or you really don't know how to do it?[/quote] i know hot to solve it .. i put it to people :!:

[quote="0714446459923"]So are you basically asking to prove Heisenberg's Uncertainty Principle?[/quote]
calculate the Uncertainty of the guassian wavefruction !!

Speaking of Gaussian, what about this one?

http://www.mathlinks.ro/Forum/viewtopic.php?p=1720597#1720597


a,b and c are a longer of a  tiangle .  a+b+c = 1.

prove  that a^2+b^2+c^2 >1/2

it seems to me the problem is easy , so you can do it.  :huh:

i have a mistake because the problem is a^2+b^2+c^2<1/2

Ok, let me see...
take $ a=x+y$, $ b=y+z$ and $ c=z+x$, $ x,y,z>0$ and $ x+y+z=\frac{1}2$. So, your inequalitie is transformed into
$ (x+y)^{2}+(y+z)^{2}+(z+x)^{2}<\frac{1}2$
$ 2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx=x^{2}+y^{2}+z^{2}+(x+y+z)^{2}<\frac{1}2$
So, it's equivalent to
$ x^{2}+y^{2}+z^{2}<\frac{1}4$
Using the fact that for positive numbers
$ x^{2}+y^{2}<(x+y)^{2}$
we have the desired result.

If I have a mistake, in my English or my math, please tell me :P

Other method
$ \leftrightarrow\ a^{2}\plus{}b^{2}\plus{}c^{2}< 2(ab\plus{}bc\plus{}ca)$
From $ a\plus{}b > c$ --->this problem was proved.


function
real analysis
real analysis unsolved

Consider the function $ f_n : \mathbb{R} \rightarrow \mathbb{R}, f_n(x)\equal{}x^n\plus{}2008x\minus{}2001, n \in \mathbb{N}^*$. Denote by $ (a_n)$ the sequence of the positive roots of the equation $ f_n(x)\equal{}0$. Study the monotony of the sequence $ (a_n)$.

Most of what we will use is repeated references to the intermediate value theorem.

For each $ n,$ $ f_n$ is increasing on $ [0,\infty),$ negative at $ 0,$ and positive at $ 1.$ Hence it has a unique positive root in $ (0,1).$ That is, $ a_n$ is well-defined and $ 0<a_n<1.$

Now, inspect $ f_{n\plus{}1}(a_n)\equal{}a_n\cdot a_n^n\plus{}2008a_n\minus{}2001<f_n(a_n)\equal{}0.$

We made explicit use of the fact that $ a_n<1.$ Since $ f_{n\plus{}1}(a_n)<0$ and $ f_{n\plus{}1}(1)>0,$ we have that $ a_{n\plus{}1}\in(a_n,1).$

In particular, $ a_n<a_{n\plus{}1}$ The sequence $ (a_n)$ is strictly increasing. (It is also bounded above by $ 1,$ so it must have a limit.)


function
complex analysis
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limit
advanced fields
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Prove Liouville's theorem (i.e. a bounded entire function is constant) using the maximum modulus principle.  That is, DON'T use the Cauchy estimate.

Note that the maximum modulus principle alone is insufficient. You must use at least some other property of holomorphic functions.

Let $M$ be the set of smooth functions from $\mathbb{C}$ to $\mathbb{C}$ that satisfy the maximum modulus property: that is, if $f\in M,$ then $|f|$ has no local maxima. The (entire) holomorphic functions form a subset of $M$ as do also the harmonic functions. Both the holomorphic and harmonic functions do satisfy Liouville's Theorem. But $f(x+iy)=\tan^{-1}x+i\tan^{-1}y$ is also in $M$ and it's bounded and not constant.

As for the proof itself, let $f$ be a bounded holomorphic function, and consider $g(z)=f\left(\frac1z\right)$. This a bounded function with an isolated singularity at $0$. We make use of the theorem that says such a singularity must be removable. Thus, $\lim_{|z|\to\infty}f(z)$ exists, and we can uniquely extend $f$ to a continuous function on the one-point compactification of $\mathbb{C}$ (the "Reimann sphere.")

That makes this newly extended version of $f$ a continuous function on a compact set, so its modulus must assume a maximum. But if $f$ is not constant, this would be a local maximum of $|f|$ (or $|g|$), which is not permitted by the maximum modulus principle.

We do need that theorem on removable singularities.

How about this proof:

Let f be entire, and define g(z) = (f(z) - f(0))/z, which is also entire.  Since f is bounded, as |z| goes to infinity, |g(z)| goes to 0.  Thus we can choose R > 0 s.t. there exists  :varepsilon:  s.t. |g(z)| <  :varepsilon: for all |z| > R and the maximum value of |g(z)| for |z|  <= R is >  :varepsilon: .  Such a maximum exists because |g(z)| is continuous and the closed disc is compact.  Let w = z be the maximum of |g(z)| in this closed disc.  Then |g(w)|  >=  |g(z)| for all z in C.  Thus by the global maximum modulus principle, since C is open, g must be constant.  Since |g(z)| goes to 0 as |z| goes to infinity, g must be identically 0.  Thus f(z) = f(0) for all z in C, whereby f is constant.

That works, too. I prefer cn2_71828182846's proof.

Note that that proof also makes use of a removable singularity, but the fact that it is removable is more elementary than the version I used.


Let $S$ be the set of all composite positive odd integers less than $79$. [list=a] [*] Show that $S$ may be written as the union of three (not necessarily disjoint) arithmetic progressions.[*] Show that $S$ cannot be written as the union of two arithmetic progressions.[/list]

First, the set is $ S\equal{}\{9,15,21,25,27,33,35,39,45,49,51,55,57,63,65,69,75\}$.
(a) Obviously $ S\equal{}\{9,15,21,27,33,39,45,51,57,63,69,75\}\cup \{25,35,45,55,65\}\cup\{49\}$.
(b) Since $ 9$ and $ 15$ appear in the set, we need the first progression above for sure. Then we are left with the numbers $ \{25,35,45,49,55,65\}$, the distance between which is $ 10,10,4,6,10$ respectively, so the distance in the progression would need to be $ \gcd(4,6,10)\equal{}2$, so the progression should contain all odd numbers, contradiction.


Trivial by Cauchy's \[\sum_{cyc}\frac{a^{2}}{b+c}\cdot\sum_{cyc}b+c \geq (a+b+c)^{2}=1\]


advanced fields
advanced fields unsolved

with an example show that if V and W be affine varities  V_W is not necessarily affine varities.

Can you mention some cases you've tried? I think it's harder to find examples where $ V \smallsetminus W$ [b]is[/b] an affine variety. What kinds of affine varieties are you familiar with, in the lowest dimensions you can think of? Try the very simplest cases first.

I could find the anawer :
I think this is an example to show this .
$ \mathbb{R}$ is a varity because $ \mathbb{R} \equal{}V(0.x\equal{}0)$
{0} is a varity of V(x=0)
but $ \mathbb{R} \setminus 0$ is not a varity.
proof:because if $ \mathbb{R} \setminus 0  \equal{} V(f_1(x),...,f_n(x))$  then because  every $ f_i(x)$ should be  continuos in a varity .then this is a contradiction and $ f_i (0) \equal{}0$ for every i

This is essentially correct, although I don't think I understand your proof that $ \mathbb{R} \smallsetminus \{0\}$ is not a variety. You should try to make that clearer.

$ \mathbb R\setminus\{0\}$ is (isomorphic) to a subvariety of $ \mathbb R^2$. Look at $ V(xy\minus{}1)$ inside $ \mathbb R^2\equal{}Specm(k[x,y])$.

For a non-affine variety, try to take out of an affine variety something closed of codiemension $ \geq 2$. So you were not that far. It's just that points have codimension 1 in $ \mathbb R$.


I think the biggest problem for economists is to help those in poor countries, to improve their standard of living. I think that we somehow need to help the political and economic infrastructure of these countries to gradually adopt free market policies, how that is to be dont I am not quite sure. What are all your opinions on this?

I think the main problem is the access of the poor to the market, in particular the women because of their know-how (how-to). That's why Asia is the leader in the third world. Women can produce and sell (or make sell) their production, and reinvest.

There is no access to the markets because no such political infrastructure can be built under the current political strucutres of those nations. Those people (rulers) must somehow be overthrown.

People won't invest in countries unless there's a sound government. That's a reason why there is a home bias in investing. If you send your money over to (eg?) Nigeria to build a factory or something, and the directors go missing with some cash, what you gonna do?

The same can happen in your own country, and then what can you do?


[quote="jayme"]Dear Mathlinkers,
let ABC be a triangle, (0) the circumcircle of ABC, P a point (for example inside ABC), A' the foot of the A-altitude on BC, (1) the circumcircle of PBC,
X, Y the second points of intersection of AP with (0), (1),
and A" the point of intersection of the perpendicular to BC through Y.
Prove: PA'//XA"
Sincerely
Jean-Louis[/quote]
[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. - Two circles $ (O),\ (O')$ are given, intersect each other at points $ A,\ B$ and let $ C,\ D$ and $ E,\ F$ be, the pairs of their points of intersection, from an arbitrary line $ ($ $ C,\ D\in (O)$ and $ E,\ F\in (O')$ in the order of $ C,\ E,\ D,\ F$ $ ).$ Prove that $ KD\parallel LE,$ where $ K,\ L$ are the orthogonal projections of $ C,\ F$ on $ AB,$ respectively.[/color][/size][/b]

[hide]I post here only the schema, because the proof is very easy.[/hide]

Kostas Vittas.

Dear Plane Geometry,
now, I am not blind. I see again.
Sincerely
Jean-Louis

[quote="jayme"]Dear Kostas and Mathlinkers,
tank you for your quick answer.
Sometime I am so blind that I see nothing... Please, excuse me if I am ridiculos, but can you write this easy proof.
Thanks in advance
Sincerely
Jean-Louis[/quote]
dear jayme:
Denote Q is the intersection of AX,BC
We have QY/QA=QX/QP=YA''/AA'
thus we are done

ps: actually, the condition A' is the projection is useless
we can generalize that A' is an arbitary point of BC

[quote="jayme"]Dear Kostas and mathlinkers. Sometime I am so blind...[/quote]
Never mind dear [b][size=100]Jean Louis[/size][/b], I am also blind many of times.

I didn't post the proof, because I thought that it is a good reason for the biginners, to try an approach for the solution of this nice result, useful ( I think ) in the future.

[hide]$ (PC)\cdot (PD) \equal{} (PA)\cdot (PB) \equal{} (PE)\cdot PF)$ $ \Longrightarrow$ $ \frac {PD}{PE} \equal{} \frac {PF}{PC} \equal{} \frac {PK}{PL}$ $ \Longrightarrow$ $ KD\parallel LE.$[/hide]

Best regards, Kostas Vittas.

[quote="plane geometry"]...ps: actually, the condition A' is the projection is useless
we can generalize that A' is an arbitary point of BC.[/quote]
You are right dear [b][size=100]plane geometry[/size][/b] but, when $ A'$ is an arbitrary point on $ BC$ then, for the point $ A''$ ( also on $ BC$ ), the condition of $ YA''\parallel AA'$ is necessary ( $ A',\ A''$ are now the parallel projections of $ A,\ Y$ on $ BC$ respectively, instead of orthogonal ones ).

Kostas Vittas.


geometry
3D geometry
geometric transformation
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group theory
superior algebra
superior algebra solved

I'm interested in the solution of the following poblem which uses the group theory:
Every face of cube is being coloured in one of the 3 different paints. Two colourings of the cube are considered to be the same if one could get one from another by cube rotation . Prove that there exists 57 such different colourings.

should not be that hard with Burnside Formula.
If $X$ is the set of all coloring of a cube, $|X|=3^6$.
$G$, the group of permutation of a cube, acts on $X$.

You know that $G$ is isomorphic to $S_4$, and thus have $24$ elements.
The number $N$ that you are looking for is:

\[ N=\frac{1}{|G|} \sum_{g \in G} |Fix(g)| \].
From here, this is just easy calculations ..


geometry
trapezoid
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abstract algebra
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geometry proposed

Let $A_0 A_1 A_2 A_3 A_4 A_5$ be a hexagon with parallel oposite sides.
Let $M_i$ be the midpoint of segments $A_i A_{i+1}$ and $s_j$ the line defined by $M_j M_{j+3}$, $j\in \{ 1,2,3\}$.

Prove that $s_1, s_2, s_3$ intersect at one single point.

Eduardo

Let $X=A_0A_3\cap A_1A_4$. Since $A_0A_1A_3A_4$ is a trapezium, we conclude that $M_0$, $X$ and $M_3$ are collinear. So $M_0M_3$, $M_1M_4$ and $M_2M_5$ are cevians in $XYZ$. We will apply Ceva for $XYZ$ in sine form. So first of all we need to calculate $\frac{\sin ZXM_3}{\sin YXM_3}$. Indeed, using sine theorem to triangles $A_4XM_3$ and $A_3XM_3$ we obtain:
<center>$\frac{\sin A_4XM_3}{A_4M_3}=\frac{\sin A_4M_3X}{A_4X}$ and $\frac{\sin A_3XM_3}{A_3M_3}=\frac{\sin A_3M_3X}{A_3X}$.</center>
Taking into consideration $A_3M_3=A_4M_3$ and $\sin A_4M_3X=\sin A_3M_3X$, we obtain
\[\frac{\sin ZXM_3}{\sin YXM_3}=\frac{\sin A_4XM_3}{\sin A_3M_3X}=\frac{XA_3}{XA_4}=\frac{A_0A_3}{A_1A_4}\]
(last equality is due to similarity of triangles $XA_0A_1$ and $XA_3A_4$).
In the same way we can write ratios for cevians $M_1M_4$ and $M_2M_5$. Multiplying together these ratios we obviously obtain 1.

That's all.

The claim is a corollary of the following lemma RL04012005.

Let A_0A_1A_2A_3A_4A_5 be a hexagon with parallel oposite sides.

Let B_i = A_iA_(i+5)^A_(i+1)A_(i+2)   where (i+k)'s are numbers module 6.

Let P_0 a generic point if P_1, P_2, ..., P_5 are the points "similarly" disposed wrt to the correspondent triangles (*) as P_0 is wrt A_0B_0A_1, then P_0P_3, P_1P_4 and P_2P_5 are concurrent.

(*) a point P_i is correspondent to the triangle A_iB_iA_(i+1).

<<"similarly" disposed >> means that P_(i+1) is the correspondent of P_i  with the homotetie of center A_i and ratio B_(i+1)A_(i+1)/A_(i+1)A_i.

Or perhaps, but i'm not sure because i don't know these matters, the P_i's should have the same trilinear coordinates wrt the omologue triangles.


It's nationals week, the week we've all been waiting for! HERE, Describe your goals, your hopes, your predictions...

For me, I'm on the HI team, and we just got team shirts and schedules and all the cool fun stuff. A friend fixed up the banquet dress for my teammate and moi.

We're leaving Tuesday night, getting there on Wednesday (early so we can get used to the jet lag). I'm rooming with my friend who's also on the team. This is my first time on the mainland/North American continent (GASP) :).

I might bring my laptop. Go on FTW LOLOL (haha).

I'm hoping to get top 20+ or even nats CD. Idk if a 40~ish score will make the cut. But no worries, I'm gonna go forward in the spirit of aloha (cuz that's what we're about xD). Yep. It's gonna be fun :)
Our Hawaii cheer is unique.

My teammates are awesome, and I hope to make at least one new friend at nats :).

One more thing... Bobby Shen... Good luck. We were watching the countdown live during lunch last year. Everyone was rooting for you. In Hawaii. I know its weird.

Whee disneyword!~~

Anyways people, type away! Nats is only a few days away!

My team has an original made up cheer designed solely by us!! It's a short math cheer rap sort of thingy.

And yes, someone is going to beatbox during our cheer. :D

We're leaving for Orlando on Thursday morning (NOOO NOT A LATE ARRIVAL TEAM) and arriving at Orlando at around 2:16 PM EDT on Thursday :o

Hope to see you all, and good luck! My goal is to place in the top 12 for CD...but idk about cd. Maybe not ashame myself in front of the audience, my fans, or my team? lol :roll:

BTW I'm from the Illinois team.

Miiiiiiiiiiiiieeeeeeeeeeeehhhhhhhhhhhhhhhhhh I'm from CA.

Wait, there's a spirit cheer? Dang?

Mieh going to Orlando on Wednesday morning, arriving at around 6:00 PM EDT.

Team goals:

Get at least 1 person in top 2.
Get at least 2 people in top 4.
Get at least 3 people in top 8.
Get at least 4 people in top 16.
Get perfect team round in 10 minutes.

Individual goals:

Do Not Exist.

(I don't actually care about the contest.)

lol those are some pretty tough goals there pythag :o

[quote="pythag011"]

Get at least 4 people in top 16.

[/quote]

atleast?  :P How do you get more?

Anyway, see everyone at nats. I'm leaving tomorrow.

My flight leaves at 9:18 (eastern) on Thursday morning, so I'll probably get there around 12 and around 1 for the hotel. I'm flying with the Michigan team.

It would be pretty awesome to make cd but I don't think I will. I just want Ohio to make top 10 teams.

See you all there!

Our team is going to get there are around 11 AM Thursday. I hope our team can make top 5. [color=#F8F8F8]I hope I can make top 4... *sigh*[/color]
Well well, we'll see.

Im leaving on Thursday morning at 9:00-ish?

I wanna do good, maybe top 40?

My goal is to make the top 4 in both written and countdown, and to qualify for the masters round. My team goal is to win by at least 1/4 of a point.

I'm from the DE team! If anyone else is going, I'll be the shortest one on the team (and the only one on the team who isn't Chinese).

Our team cheer.....sucks. Horribly. Some kid from my coach's math class made it, and it is the corniest thing ever. And my friends might make me dress up as a bird during it. Oh, how fun.

We leave around 10ish and get there around 12 I think.
My goal is to make the Top 57, which will probably not happen, but I can always hope. It'll be so fun there!

Good luck to everyone, and I hope to meet you all!

[quote="Urc"]I'm from the DE team! If anyone else is going, I'll be the shortest one on the team (and the only one on the team who isn't Chinese).[/quote]

 :rotfl:

That totally reminded me of the time when you asked suma_milli and me how tall I am...

Bleh I'm the second shortest on the team. :|

We definitely need EVERYONE from IL get top 50...or maybe perhaps top 40!!!!!!!!!

I'm not in it, but I hope to be next year. Here are my goals for nationals next year:

*Get in top 100 (I'm from OK. What can I say?)

*Get in the top 10 in Team Round (That's going to be hard.)

*Get a TI-INSPIRE CA+ (Can't really do anything about it, but hope for one.)

*DON'T LET IT BE YELLOWSTONE!!!  :lol:

Good luck to all who made it to MC Nats!  I think it will be a good time.  Bye!! :D

I really look forward to meeting everyone there! My flight leaves tonight. I think it will be awesome!

[quote="james4l"][quote="pythag011"][quote="funtwo"]im not on the texas team, but i'm hoping at least 2 people make black.[/quote]

Mixing up mop and mathcounts?

Also, CA team has basically 2 guaranteed moppers  :o[/quote]

That is if one of them has really good rigourous solutions...

Team Goal:
Team wins by at least 1 point
Someone in team gets Individual Champion
Someone in team gets Written Champion
2 get into Masters Round
all 4 get into CD

I know, unrealistic goals? Oh well.

Indiv Goal: Get at least 4000 dollars in prizes.[/quote]

Those goals are even worse than mines.

I guess I will have to make them better...

Team Goa:
Team wins by at least 2 points and gets 10 team round
Somebody in team gets both written and cd champion
3 get into masters round
all 4 get into top 8

Indiv goal:
Finish sprint in 15 minutes.

[quote="gauss1181"][quote="Urc"]I'm from the DE team! If anyone else is going, I'll be the shortest one on the team (and the only one on the team who isn't Chinese).[/quote]

 :rotfl:

That totally reminded me of the time when you asked suma_milli and me how tall I am...

Bleh I'm the second shortest on the team. :|
[/quote]

I interact better with people of my height.  :P

Yeah I'm from Louisiana, arriving 12:30 on Thursday. See you there!

P.S. Can we use a CD player for the state cheer?

Not that I'm going since I'm to stupid.... :(
But I hope WV does well this year!!! The highest individual rank for WV is #32, Billy Fang, who moved to ohio....... *sigh*

so GOOD LUCK TO YOU ALL!!!! (mostly ben taylor) lol
 :D  :D

Good Luck to all. May your trips be safe like toilet paper: Long, clean, useful, and free of H1N1 Influenza A(AKA SWINE FLU) :D

[quote="O M G"]Yea good luck to you all.. Good thing Florida doesn't have swine flu...
But, people there could be infected since symtoms don't show till like 2 weeks after infection........
But, maybe airport will check. :)  :roll:[/quote]

Umm...i think you're mistaking  :wink: Florida [i]is[/i] already infected. There's already 5 confirmed cases, and that flu spreads quickly in crowded places, like Disney World. :o

[quote="remy1140"]NC arrives at a/b 11:40 i think...well, some where between 11 and 12 am EST.

My goal...team top 5, AT LEAST top 10.
Individually, well, between 40th and 13th place.
I DON'T want to make countdown because
1)i messed up my interests and hobbies part in registration (put trivial things)  :rotfl:
2)swine flu (well, H1N1), someone infected touches the buzzer (not to offend anyone) and i touch it; and I also have that flu.[/quote]

if someone has swine flu, they wont be at their mental best and will probably not make cd

i did not make it....38 pts  :(  :(  :(  last year too..

[hide]If u can see this and its 5/7/09(arithmetic seq day!) or 5/8/09 or 5/9/09, then you probably didnt make it too!![/hide]

I wonder what they are doing right now. Probably walking around Disney. Im so jealous.

dude, i know. going to thier 5 star hotel and all....uhhhh

The food isn't that good. Take comfort in that. :P

yea. but still it's.....disneyworld....who cares about food :oops:

I called someone I know that went. It seems like they are having so much fun they don't want to come back. Tomorrow is testing day so good luck to those who made it. And good luck to some people who I want to make countdown.

at least one team will use my cheer, hopefully. look out for the ohio/texas/missouri cheer (listed in order of probability of using my cheer?) darn, i hope someone uses it

Totally unfair...
Denver=>Disneyworld???
Nine people had headaches or related problems at nats last year because of dust/high elevation problems, and the Denver Museum doesn't quite match up to Disneyworld...(and anyone who went to some certain national chess competition already went to the Denver Museum)

Dang.
Good luck at nats everyone although I don't think any competitors will read this post until after the competition

dude the starbucks at denver didn't even have frapucinnos. wtf.

[quote="HiDN428"]i did not make it....38 pts  :(  :(  :(  last year too..

If u can see this and its 5/7/09(arithmetic seq day!) or 5/8/09 or 5/9/09, then you probably didnt make it too!![/quote]

Hey I'm in Nationals and I saw your post! :P Well, actually I brought my laptop here, so I can use the internet.

Hmmm. We're about to begin the written round now...we just finished breakfast and my team is getting suited for our cheer. Fortunately I have my phone with me, and Internet access, so I'll update things as they go aalong.

Those of u still on aops: have fun :)
Wish us luck :)

good luck! do you know if anyone used my cheer?

good luck to you all! dang so fun this year....disney world.
hmm written competition should have ended by now

I have the feeling i missed the opportunity of a lifetime when I got 5th at state.  :( (Math+Disneyworld= :) )

  :wallbash_red:

Congrats to the Texas team, Bobby Shen, and Steven Chen! 3 years and counting! :)
Also, I heard my cheer was left unused? That is disappointing. I don't approve.

4 of the last 5 years... I guess that means I'm the team's bad luck charm.  :D


A particle P, moving on the smooth inside surface of a fixed spherical bowl, or radius r, describes a horizontal circle at depth r/2 below the centre of the bowl. Show that P takes time (pi)root(2r/9) to complete one revolution of its circular path.

Help needed please.

Many thanks!

Is that 9 supposed to be a g? I seem to be getting that $ T \equal{} \pi\sqrt \frac{2r}{g}$

Before anything else, it's not $ \pi\sqrt{2r\over 9}$ but $ \pi\sqrt{2r\over g}$.

[hide="Hint"]Equate the centrifugal force with the radial component of the particle's pressure force to the wall of the bowl.[/hide]

I wrote it wrong. it is g.

I have already done the hint- anyone give me their worked solutions please?

Let $ R$ be the reaction force on $ P$ due to the bowl.  Let $ \theta$ be the angle that $ R$ makes with the horizontal, and let $ x$ be the radius of the horizontal circle $ P$ describes.  We have
$ R\sin\theta \equal{} mg$, $ R\cos\theta \equal{} \frac{mv^2}{x}$
$ \Rightarrow \tan\theta \equal{} \frac{xg}{v^2}$

Since $ P$ is $ r/2$ below the centre of the bowl,
$ x\equal{}\frac{\sqrt{3}}{2} r$
$ \tan\theta \equal{}\frac{1}{\sqrt{3}}$
Plugging $ x$ and $ \tan\theta$ in we get
$ v^2 \equal{} \frac{3}{2} rg$
Substitute $ v \equal{} \frac{2\pi x}{T}$ and rearrange to get $ T \equal{} \pi\sqrt{2r\over g}$.


Hello,

I'm new here.

I have a question about functional equations:

After we have succeeded in finding an (or several) expression(s), how do we show conclusively that no other expressions satisfy a given functional equation?

Take for example one of Cauchy's Functional Equation: f(x+y) = f(x) + f(y).

Let us not quarrel here that we should find continuous (and of course non-trivial) solutions. f(x) = 0 for all x, for example, is not that we are looking for, nor should we be interested in discussing it.

One solution is f(x) = cx, where c is a constant. The next move is to show conclusively that there are no other solutions?

Well for this particular equation we can prove that for each rational q we have f(q)=qa, where a=f(1) through mathematical induction (first for the positive integers, then going backwards to negative integers, and then expand it to rationals).

now in absence of any other condition (such as positivity, monotony, continuity) we cannot say more about the equation (if has other solutions than ax!).

with the supplementary condition we can use that fact that the rationals and irrationals are dense within the reals, to mingle them and obtain f(x)=ax for all reals.

so, no need to prove f are the only functions.

in other functional equations we can assume from the beginning that there exist two solutions f,g that fulfill the given relation, and try through different algebraic manipulations to obtain that f(x)=g(x) for all reals (and then check that a particular, easy to see, answer works).


How many different colour arrangements are possible by placing 2 orange balls, 3 blue balls, and 4 black balls in a row in which the balls at the ends are the same colour?

[hide]
You can fix the end balls as each of the colours, then calculate the number of arrangements inbetween..

$\frac{7!}{3!4!}+\frac{7!}{1!2!4!}+\frac{7!}{2!2!3!}= 350$
[/hide]

[hide="A more detailed solution"]
Case 1: the oranges are at the end.
This gives 7 spots for 3 blues and 4 blacks, so we have ${7\choose 3}=35$ for this arrangement.

Case 2: the blues are at the end.
This gives 7 spots for 1 blue, 2 oranges, and 4 blacks. Choose where the blacks go, ${7\choose 4}$. Then, we have to multiply the number of ways we can place the blue and oranges in the remaining spots, 3. Thus, there are $35\cdot 3=105$ for this arrangement.

Case 3: the blacks are at the end.
This gives 7 spots for 2 oranges, 2 blacks, and 3 blues. Choose where the blacks go, ${7\choose 2}$. Then, of the five remaining spots, there are ${5\choose 2}$ places ways to place the blues and oranges. Thus, there are $21\cdot 10=210$ for this arrangement.

Adding our 3 cases, we have a total of $35+105+210=350$ arrangements.[/hide]


What are your favorite people/groups who combine (classical) music and comedy?

Igudesman and Joo are one of my favorites. One example is Riverdancing Violinist:
[youtube]QKZITB_r8t0[/youtube]

Of course, Victor Borge is a pretty cool pianist, such as in Page-Turner:
[youtube]LWqFaGwNCMU[/youtube]

I've seen a bit of Igudesman and Joo. The video on Rachmaninoff needing big hands is pretty misleading, almost everyone breaks up 10ths and 11ths for Rachmaninoff, and being able to reach most chords spanning an octave is pretty standard hand reach.

One of their shows was particularly disturbing, when he took a dremel with wire brush and started playing directly on the strings with it :O

I have to say my favorite is Victor Borge, he was hilarious...

I must say Victor Borge. He is so funny, I once tried his method of playing happy birthday with one hand, and I utterly failed.


A fence is to be built around an area such as the one shown in the diagram.

What must be the values of x and y if the area is to be a minimum and the perimeter is to be 300m ?

http://img256.imageshack.us/img256/3020/diagramcd8.jpg

Thanks so much!

xy will be a minimum when it is 1 and something. x could be 148 and y is 1. therefore, the answer is 148.

Maybe [url=http://www.purplemath.com/modules/sqrquad.htm]this[/url] will help.


I need help on this problem.

In a straight-angled triangle ABC with <C = 90, D is the midpoint of BC, and E is the midpoint of AC. If AD = 7 and BE = 4, then AB equals:

A) \[5\sqrt {2}\]
B) \[5\sqrt {3}\]
C) \[2\sqrt {13}\]
D) \[2\sqrt {15}\]
E) \[2 \sqrt {\frac {65}{3}}\]

Thanks!

I think straight should be right in the second sentence of the post.
[hide="If I Am Right"]$AC=x$
$BC=y$
$(AB)^2=x^2+y^2$
$x^2+\frac{y^2}{4}=49$
$\frac{x^2}{4}+y^2=16$
$(AB)^2\left(\frac{5}{4}\right)=65$
$(AB)^2=52$
$52=2^2\cdot13$
$AB=2\sqrt{13}$[/hide]


有谁知道为什么中国大陆没有参加APMO?
中国也应该属于亚太地区呀

不知道......是因为政治原因么?

我也不知道  只记得物理好象有人参加的

物理是有
不知道为什么数学没有

有亞太區物理奧林匹克？ :?

[quote="siuhochung"]有亞太區物理奧林匹克？ :?[/quote]
那是不是说几乎每科都有?

我想liyi说的可能是亚洲物理奥林匹克，其网址是

http://www.apho.org/en/home/index.php?charencode=en

上面有1-5届竞赛的试题可供下载

I meant that China do participate in APhO.


$ 1.$ For positive reals $ x,y,z$ Prove that:

$ \frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\leq1$

$ 2.$ For positive reals $ a_{1},a_{2},\cdots,a_{n}$ where $ n>3$ Prove or Disprove:

$ \sum^{n}_{i=1}\frac{a_{i}}{a_{i}+\sqrt{A_{i}}}\leq1$

where $ A_{i}=\prod_{1\leq j\leq n,\; j\neq i}(a_{i}+a_{j})$

For number one i obtain 2 proofs see here,I remember it is from :101 Problems in  Algebra.

for the first one:

add $ -3$ to both sides and multiply both sides in $ -1$,so we have to prove that:

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

now we know that $ (A+B+C)(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\geq 9$ so if we prove that $ \frac{1}{A}+\frac{1}{B}+\frac{1}{C})\leq \frac{9}{2}$ we get that $ A+B+C\geq 2$

so its sufficient to show that:

$ \sum \frac{a+\sqrt{(a+b)(a+c)}}{\sqrt{(a+b)(a+c)}}\leq \frac{9}{2}$

so its sufficient to show that:

$ \sum \frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{3}{2}$

but by $ AM-GM$ we have:

$ \sqrt{\frac{a}{a+b}.\frac{a}{a+c}}\leq \frac{\frac{a}{a+b}+\frac{a}{a+c}}{2}$

so we get that:

$ LHS\leq \frac{1}{2}(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{a+c})=\frac{3}{2}$

as wanted...

------------------------------

also another solution for this problem is the following one:

take $ x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ so we have $ xyz=1$ now we have:

$ \sum \frac{a}{a+\sqrt{(a+b)(a+c)}}=\sum \frac{1}{1+\sqrt{(1+\frac{b}{a})(1+\frac{c}{a})}}=\sum \frac{1}{1+\sqrt{(1+yz)(1+z^{2})}}$

now from couchy we get that:

$ LHS\leq \sum \frac{1}{1+(1+z\sqrt{y})}=\sum \frac{1}{2+z\sqrt{y}}$

now if we expand all the terms and simplify we have to show that:

$ 3\leq xy\sqrt{xz}+yz\sqrt{xy}+zx\sqrt{yz}$

now by $ AM-GM$ we have:

$ RHS\geq 3\sqrt[3]{x^{2}y^{2}z^{2}\sqrt{x^{2}y^{2}z^{2}}}=3\sqrt[3]{x^{3}y^{3}z^{3}}=3xyz=3$

and we are done...

I have EDITED the second problem!

all the solutions are coOl guys...tanx Navid, Babak and Zaya_yc :)

[quote="zaya_yc"]$ \sqrt{(A+B)(C+D)}=\sqrt{AC+BC+AD+BD}\geq \sqrt{AC+BD+2\sqrt{ABCD}}=\sqrt{(\sqrt{AC}+\sqrt{BD})^{2}}=\sqrt{AC}+\sqrt{BD}$ (1)

 [/quote]

thats right :D , we can also prove that:

$ \sqrt[n]{(a_{1}+b_{1})(a_{2}+b_{2})\cdots(a_{n}+b_{n})}\geq \sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{b_{1}b_{2}\cdots b_{n}}$

Now if we prove that for $ n>3$ we have:

${ \boxed{\sqrt{\prod_{1\leq j\leq n,\; j\neq i}(a_{i}+a_{j})}\geq \sqrt{a_{i}}(\sum_{1\leq j\leq n,\; j\neq i}\sqrt{a_{j}})}}$

the second problem will be solved too, any opinion for this one?!


There are 32 teams in the tournament, and they are divided into eight groups with the same number of teams in each group.  Every team in a group plays the other teams in their group exactly once.  This is the first round.  The top two teams from each group then advance to the second round, where all of these first-place and second-place teams are put into a bracket and a single-elimination tournament takes place until the winner is determined.  (There is also a play-off game between the losing teams in the semi-finals to determine third and fourth place.)  According to this set-up, how many total games are played during the World Cup?

Along with the answer, you must include an explanation. And, also, as a random add-on, Brazil has no chance of winning this year because England is da bomb. :D

[hide]Each group has 4 people, so there would be 6 matches in each group making $6*8=\boxed{48}$ so far.

Now there's 16 teams and single elimination. Therefore, there are 15 games and the game between the losing teams in the semi-finals, which makes $\boxed{16}$.

Final answer: $\boxed{64}$ games.[/hide]

[hide]
32/8 = 4 in each group.

(32*3)/2 = 48. This because each team plays every other team in its group, but I counted each game twice, once for each team, so I divided by 2.

Then you have the knockout round! 8 games in first round of knockout (16/2), 4 games in quarterfinals (8/2), and 2 games in semifinals (4/2), then the final which is 1 game, and the 3rd place game, which is 1 game.

So you have 48 + 8 + 4 + 2 + 1 + 1 = 64 games in the World Cup!

[/hide]

I'm currently not happy with Brazil because they beat Ghana 3-0 (I like underdogs), and Germany and Italy have each won 3 cups and Brazil has won 5. So I'll root for France.

numberdance

[hide]Yes, 64 is the correct answer, as other posters have demonstrated[/hide]


After I login, every time I click on a forum or blog it automatically logs me out and prompts to the login page.
For example when I click on the "Support" forum I have to relog in and I have to relog in again after I click on post reply.

This happens on IE7.

Umm...not that I know that this will work as it does not happen to me but....either try some other web browser(firefox etc.), or push the box that says something along the lines as remember me or something(like yahoo mail) when you log in, or maybe you are having a "cookie" problem. of course the people after me will probably definitely know what they are talking about.

You don't have cookies enabled most probably.


Let f(x) be a non-negative function defined on [0,1] (that is,f(x) ≥0 for all x in [0,1]). Suppose that for any choice of n distinct numbers x_1, x_2, ..., x_n in [0,1]

f(x_1) + f(x_2) + .. + f(x_n)≤ 2,345,678.
Prove that the number of x in [0,1] for which f(x) > 0 is finite or countable.


Fill in the question marks:

12 DD
11 PP
10 LAL
9 LD
8 MAM
7 SAS
6 GAL
5 GR!
4 CB
3 FH
2 TD
???????

:yup: good job alan
I've always wondered what one was supposed to do with 10 LAL...


1. If the area of a circle is 16$\pi$, what is the area of the inscribed equilateral triangle? Express your answer as a radical in simplest form.

sorry that is incorrect use the formula for 30-60-90 triangles.

[quote="goldendomer"][quote="sonny"]$frac{8\sqrt3}{3}$[/quote]

Solution?[/quote]
No there was a typo and i fixed it. I need to leave so i can't post a solution rigth now. The right answer is what i wrote after editing it and its the same as kunny.

[quote="indianamath"]correct! you can post the next question[/quote]

I thought we cant make anymore marathons...

[hide="more extensive solution"] You can split the equilateral triangle into 3 smaller triangles with a common vertex of the center of the circle. These are equilateral triangles with base angles of $\frac{60^\circ}{2}= 30^\circ$. Divide each into two 30-60-90 triangles and apply the rules (the hypotenuse is the radius, 4), to get that the long leg is $2\sqrt{3}$. Multiply that by 2 to get the side length of the triangle - $4\sqrt{3}$.

Then apply the formula for the area of an equilateral triangle given its side length - $a=\frac{s^{2}\sqrt{3}}{4}= \frac{(4\sqrt{3})^{2}\sqrt{3}}{4}= 12\sqrt{3}$. [/hide]

any1 wanna post another problem? I think we are allowed to do marathons

I tried a different way, using centroids, and got this far:
[img]5761[/img]

What do I do next? :|

Just splitting them up though arrives at an answer quickly...

First of all you made a simple mistake the radius is 4, not 8.

If you extend one of the centroids, it is a 30-60-90 triangle, the ratio of the legs to hypotenuse is 1:root3:2

Here the hypotenuse is 4, so the longer leg is 2root3. The longer leg is only have of a side so the entire side is 4root3.  The formula for the area of an equilater traingle with side a is $(a^{2}\times root3)\div4$


The 2006 William Lowell Putnam Mathematical Competition will be held on Saturday, December 2, 2006.

Entry forms for the Putnam are submitted by the local coordinator at each college. This is normally a faculty member, most likely a faculty member in the Mathematics department. These forms must be returned by October 13. If you wish to take the Putnam, find out who the supervising faculty member is and contact him or her. Do not delay - the time is now.

An unlimited number of students from each college or university may take the exam. However, each college designates three students as the "team", whose scores determine the department ranking. This - declaring who the team is - is also due on October 13. Different colleges will have different approaches to making this determination - there may be a qualifying examination of some sort. But even if there is a qualifying examination, that would be for the purpose of determining team membership, not for deciding who is allowed to take the exam.

Each student may take the Putnam 4 times. High school students are discouraged from taking it as that would use up a year of eligibility. (Yes, I know that Gabriel Carroll was a Putnam Fellow when he was still in high school.)

Is it true once a college/university participates in the Putnam contest in some year that it's automatically registered for the contest in the following subsequent years? Or, is it that colleges must send in the registration forms every year?

Thanks in advance.

The college must register every year. The people who run to Putnam keep a list of the supervising faculty members at each college and generally send two copies of the registration packet: one to that faculty member and one to the math department chair.
When I restarted the Putnam at my institution in 1999, I wrote to the adminstrators asking for registration materials. I disovered that they still had the address of a faculty member who had administered the exam twenty years before.

There is no on-line registration. It's all paper and snail mail.


Solve

$ 5x\plus{}5y\plus{}2xy\equal{}\minus{}19$

$ x\plus{}y\plus{}3xy\equal{}\minus{}35$

hello, multiplying the first equation by $ 3$ and the second one by $ (\minus{}2)$ we get
$ x\plus{}y\equal{}9$, now you can plugg $ y\equal{}9\minus{}x$ in one of your given equations and it remains to solve a quadratic equation.
Sonnhard.

Yes, Sonnhard, except that I get $ x\plus{}y\equal{}1$ instead of $ x\plus{}y\equal{}9$ and so $ (x, y) \equal{} (\minus{}3, 4), (4, \minus{}3)$.

hello, yes it was my mistake it must be $ x\plus{}y\equal{}1$ instead of $ x\plus{}y\equal{}9$,sorry.
Happy Christmas,Sonnhard.


probability
combinatorics unsolved
combinatorics

What is the probability of recieving exactly 3 kings in 5 card draw...?
Please help me ...thank you...explain to me too plz

Hmm...I don't think it has to be in the algebra section...Thus, I move it into the combinatorics one.
Btw, is it homework? if the answer is yes, I think you have omitted to say how many cards are available (you take 5 from how many?) If the answer is no, for the same reasons I think it should be an open question :D

Pierre.

this is not homework...DId I say the source was from 2005 ACU math contest...well I suppose you draw from 52 card deck......
After I took the test I didn't understand some questions ...so I posted here to know how to solve it

Then you have $4$ ways to choose the three kings among the four. For each choice, there are $\frac {48 \cdot 47} 2$ ways to complete the hand by two distinct cards which are not kings.
Thus you have a total of $4 \cdot \frac {48 \cdot 47}2 = 4512$ hands of $5$ cards xith exactly $3$ kings.

Pierre.

I dont get where you got the 48*47/2 from .. and then why do you have to time 48*47/2 by 4?

I dont think this is the correct answer because the probability is an event between 0 and 1. so does anybody know how to do this ?

48*47/2 is $48\choose 2$, which is the number of ways of choosing two cards from forty eight.

If you want to choose a hand with three kings, you have to choose three kings from the four available (or equivalently, choose one to leave out - hence there are four choices) and then choose your other two cards from the remaining forty eight - so the total number of ways to do it is the product of these two - i.e. $4{48\choose 2}$, which is 4512 as already mentioned.

If you want the probability, you just have to divide this by the total number of possible hands which is ${52\choose 5}=2598960$, to get (roughly) $0.001736$.


[quote="bjh7790"]
find $ \sqrt {a^2} \plus{} \sqrt {b^2} \plus{} \sqrt {c^2} \plus{} \sqrt {d^2} \plus{} \sqrt {e^2}$.[/quote]

Polynomial.
[hide]
Write $ P(x) \equal{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e$.
Then $ P(x) \minus{} 2x$ has zeroes $ x$ at $ 1$,$ 2$,$ 3$ and $ 4$.
$ P(5) \minus{} 2 \times 5 \equal{} 72 \equal{} 3(5 \minus{} 1)(5 \minus{} 2)(5 \minus{} 3)(5 \minus{} 4)$
so we have
$ P(x) \equal{} 2x \plus{} 3(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$
giving
$ a \equal{} 3;b \equal{} \minus{} 30;c \equal{} 105;d \equal{} \minus{} 148;e \equal{} 72$.
You want $ |a| \plus{} |b| \plus{} |c| \plus{} |d| \plus{} |e| \equal{} P( \minus{} 1) \equal{} 358$.
[/hide]


geometry
3D geometry
sphere
graph theory
combinatorics proposed
combinatorics

Quite an interesting fact But you guys will probably find it easy anyway :
[color=blue]Problem.[/color]Prove that there exist a graph with an arbitarary large chromatic number which does not contian any triangles.

Chromatic Number of a graph $G,\chi(G)$
is the minimum number of colors with which the vertices of $G$ can be colored if every two adjacent vertices have distinct colors.

[quote="Omid Hatami"]Consider the infinite (Of course uncountable) graph with vertices in $S^{d}$ (Sphere of dimension $d$). For 2 vertices $u,v$ connect them iff $\angle uOv\geq 120^\circ$. Then it does not any triangles and also $\chi (G)\geq d$. So by Tychonof Theorem the graph has a finite subgraph that $\chi (G)\geq d$.[/quote]

Easy man,No need to unleash your machinery.
I was looking for an elementary construction of a simple graph which enjoys the Problem condition.At least there exist one.

It is a result of Erdos that there exist finite graphs of arbitrary girth and chromatic number ... unfortunately, I do not remember the constructions  :oops:

Here's a construction (small variation on a problem in Diestel's [u]Graph Theory[/u]):

Let the vertex set $V = \{1, 2, \ldots, n\}^{2}$, and let the edge set $E$ be all pairs, $\{(x, y), (x+y, z)\}$.  Verifying that it's triangle-free is easy.  It is also possible to show that the chromatic number grows arbitrarily large with $n$.

Also you can show that for any $k$and $g$,there exist a graph$G$ which
$\chi(G)$=$k$ and $girth(G)=g$.This is a famouse problem due to Erdosh.
Of course it's solution has probabilistic nature and does not give any construction.For the special case,mentioned in the question, we have a structure which is known as$misiliski$ graphs.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=colors&t=21943

Pierre.


1. Find all n such that
3^n≡1(mod n)

2. Find all (a. n) such that
a^n≡1(mod n)

need help...
thanks in advance!

In most cases, including a=3, there are infinitely many solutions but there is no simple description.
e.g., for 1. see http://oeis.org/A067945
However, it is possible to describe recursively the primes that can divide such n.


Is there a set of positive integers ${ \{a_1;a_2....;a_{100}|a_i\leq 25000}$ with the property $ a_i+a_j,1\leq i<j\leq 100$ are all distinct?

There is such a set if you allow the upper bound to be 26430.  Also, you can find 98 numbers less than 25000 satisfying the property.  This suggests that if the set cannot exist, and I suspect it can't, proving that will require extremely tight estimates.  Blunt approaches like the probabilistic method, say, will probably not be useful.

you said a set isn't exit but can you prove in detail this problem.I need a solution for it :lol:

[quote="toanIneq"]Is there a set of positive integers $ \{a_1;a_2....;a_{100}|a_i\leq 25000\}$ with the property $ a_i \plus{} a_j,1\leq i < j\leq 100$ are all distinct?[/quote]

This is problem N6 from the IMO'2001 shortlist.
The answer is YES.

http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2001

I have just read the solution of problem but I think that the solution is wrong.
The solution is wrong at :$ a^2\equiv b^2 (mod p)$
$ \Rightarrow a\equiv b (mod p)$ or $ a\equiv \minus{}b (mod p), p$ is a prime
Somebody can resolve this problem for me?


inequalities
number theory unsolved
number theory

let  $ s\in N$ show that there are $ k,n$ such that $ s \times 10^k \leq 2^n<(s\plus{}1) \times 10^k$
$ (k \in W ,n\in N)$

acc. to your question, even one example is sufficient to prove it.it makes it quite trivial.
i.e. let $ k \equal{} 0$
$ s\leq\ 2^n <s \plus{} 1$
is always true.

for strict inequality consider $ s \equal{} 1,k \equal{} 1$
we have $ n \equal{} 4$
$ 10 < 2^4 < 2.10$

check your question..

For what does $ \mathbb{W}$ stands for? Is some other notation for $ \mathbb{Z}$?

It is obviosly followed from $ log_{10} 2$ is irrational.

no, i mean for example if s=5 then there is n such that $ 5 \leq 2^n < 6$or $ 50 \leq 2^n < 60$ or$ 500 \leq 2^n < 600$or ...
but as we see $ 500 \leq 512 < 600$ then for s=5 it is true because k=2,n=9
$ W \equal{} {0,1,2,3,...}$

so mathnice1, you mean to say..
for "every" $ s \in N$, prove there exist $ k,n$ such that $ s.10^k\leq\ 2^n\leq\ (s \plus{} 1).10^k$ such that $ k\in W$ and $ n\in N$...?


Hello,
Maybe you've discused about this ...

 I am having problems with images,I can not see any image [url=http://www.mathlinks.ro/viewtopic.php?t=70276]here[/url] or everywhere in website.
I mean,I can see avatars but not added images.

DO you have a internet-filter?  I know some don't show up on mine cause there's an internet filter (for my brother) and it blocks imageshack, and all those sites.  THat way, you can't see them.

I'm using IE ,but I'm not sure if there's any "internet-filter".What is it ?

Basically internet filters don't allow you to go to *bad* sites.  Facebook, Myspace, imageshack, etc. is considered bad due to the reason of social networking, or other reasons.


Can someone help me with this problem. It is where there is a quadrilateral inside another quadrilateral and the inner one is 52% of the larger one's area. And we have to find the ratio of sides that have been separated by the vertices of the smaller quadrilateral. In the solutions book on pg. 84, can someone explain how to expand the cross products. This is the step between the second and third piece of latex on that page.

Aren't cross products linear? So $(\vec{v}+\vec{w})\times(\vec{w}+\vec{v})=(\vec{v}\times \vec{w})+(\vec{w}\times \vec{v})$, since $(\vec{v}\times \vec{v})=(\vec{w}\times \vec{w})=0$.


Lets see

STL- total about to a 53 two teammemeber unknown but most likely FTPs so 53

any other team can drop scores here feel free

Also College Station 60

        Sunnyvale  59

all three of the team's individual score own our team score i see a trend

 :(

Wait Joshua Lim is in ARL

wtf...a 53/60 on general I thought Charles and Yan was bad...

 :?  :(

>_> why don't you type out the WHOLE TEAM NAME? I forget the abbreviations

How does Sunnyvale have 60?  Who FTPed?  I'll eat that person's brains for supper and wash them down with a glass of ice water.

I would say Daniel or maybe he is not on the roster this time???

Both Cinjon and Matt FTPed.  CS retains a comfortable 1st position in terms of participation.


At least two of the nonnegative real numbers $ \{a_i\}$  are nonzero.

$ \sqrt [p] {\sum_{i \equal{} 1}^n\ a_i^p }\ge \sqrt [q]{\sum_{i \equal{} 1}^n a_i^q} ,$

when $ 2\le p < q  .$
$ p, q\in \mathbb {N}.$

[color=green](updated)[/color]

[color=green]Is it correct?
Can you give a proof?[/color]

[quote="hsiljak"]http://en.wikipedia.org/wiki/Generalized_mean[/quote]

Sorry, your  information offered is about the Inequality between any two power means,
but now $ \sum_{i\equal{}1}^n \omega_i \not\equal{}1$ in this question.


Back in the days when I had Windows '98 I could save animated GIFs. However, when I'm using XP, whenever I try saving an animated GIF, it only saves the image that the animation stopped at. How can I fix this?

Thanks.

How are you saving?  Right click -> "Save as..." should still work (it just did for me and I'm running XP).  Alternately, what you're trying to save may not be an animated .gif.

How are you viewing the GIFs?  Not all programs can view them as animated, such as Paint.

Adobe ImageReady can handle animations quite well; try using it.

Alternatively you can use your web browser to view the GIFs...

Ah, I see the problem: the image in the folder wasn't displaying it as an animation, however when I would view it in Windows Picture and Fax Viewer, it would display the animation.

Back on my '98, it would display the animation in the folder - I guess that was my real question then: Can I turn the animation on in the folder?

Dude just leave the GIFs in 1998 were they belong...

seriously


What is the smallest positive integer that has a square root that is greater than 10?

We note that $ 10^2\equal{}100$, so $ \boxed{101}$ is the smallest positive integer with the given property.


A solid $ 3\times 3\times 3$ cube weighs 5 pounds and is worth $ \$216$. How many dollars is a solid $ 4\times 4\times 4$ cube made of the same material worth?

For this problem, we do not need that the first cube weighs 5 pounds. We are only looking for the price, so the weight is unnecessary.

We know that a $ 3\times3\times3$ cube is worth $ \$216$
That means for $ \$216$, you can get $ 27$ cubic units of this material, or $ \$8$ per cubic unit.

We want to find how much a $ 4\times4\times4$ cube is worth
This cube has a volume of $ 64$ cubic units
At $ \$8$ per cubic unit, this cube will be worth $ \$8\cdot64\equal{}\boxed{\$512}$

volume of the cubes are 27 and 64. So we need 216*64/27=8*64=512


Find the smallest possible value of $a+b+c$ where a, b, and c are different positive integers that satisfy the following equation:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{7}{10}$

Plz explain as much as possible.

First notice that with two fractions 1/x and 1/y you can only get ten if x=2 and y=5 or vice versa.  I could think of a proof for that, but I don't really want to, and it's also kind of easy to see.  Now say either:
OPTION#1:  1/a=1/x=1/2 and 1/b+1/c=1/y=1/5

OR

OPTION #2: 1/a=1/y=1/5 and 1/b +1/c=1/x=1/2

Option #1 smallest b and c that make the fractions equal to 1/5 are 10 and 10 because they're the same and make the smallest sum.  As you go up (b=100000000 c=.00....) the sum obviously increases.  The next pair of b and c after 10 and 10 that will give you 1/5 is 30 and 6, since you want different integers.    30+6=36=b+c

  1/b+1/c=1/5 =(b+c)/bc  Find pairs by checking 1/5,2/10,3/15,4/20, etc.

In option 2 using the same logic as we did for option 1, the smallest values for b and c here are 3 and 6.  Check using same logic.
3+6=9=smallest value for b+c
a=5,b=3,c=6      a+b+c=14

If this isn't right, I have just wasted a lot of time, but hope it helps.  by the way, how did you get the 1984 competitions?  I've only got 1994 to present.

Yeah, I checked the answer key and that's correct.

btw, this problem was in the All-Time Greatest, so that's where I found it.


1. Find all positive integers $m$ and $n$ such that $1 + 5 \times 2^m = n^2$

[quote="luimichael"]Rewrite the equation as $5\cdot2^m=(n+1)(n-1)$.
Then note that $\gcd (n+1,n-1)=2$.
$(n+1,n-1)=(2,2^{m-1}\cdot 5) , (2\cdot 5, 2^{m-1}), (2^{m-1},2 \cdot 5)$ or $(2^{m-1}\cdot5, 2)$, with $m\geq 2$.

[/quote]

sorry, i don't quite understand.... how do u know $(n+1,n-1)= (2^{m-1},2 \cdot 5)$ or $(2^{m-1}\cdot5, 2)$, with $m\geq 2$ ?
why isn't it $(n+1,n-1)= (2^{m-k},2^k \cdot 5)$ or $(2^{m-k}\cdot5, 2^k)$? How do u know that it's just m-1? :?

My solution:

[hide]
since $k > 0$,
$n \equiv 1 (mod 10)$
Hence, $n = 10k + 1 / 10k - 1$
If $n = 10k + 1$,
$2^m\cdot5 = (10k + 1)^2 - 1^2 = 10k(10k + 2)$

$2^{m-2} = k(5k +1)$
From here, if $k$ is even, $5k + 1$ is odd and vice versa. Hence, then only possibility is $k = 1 \to  2^{m-2} = 6$, which is wrong if m is integer.

The same goes for $n = 10k -1$,
will yield $2^{m-2} = k(5k - 1)$
SImilarly, the only possibility is $k = 1$, which gives $m-2 = 2, m =4$
:)
[/hide]

[quote="violetcraze"][quote="luimichael"]Rewrite the equation as $5\cdot2^m=(n+1)(n-1)$.
[b]Then note that $\gcd (n+1,n-1)=2$. [/b]
$(n+1,n-1)=(2,2^{m-1}\cdot 5) , (2\cdot 5, 2^{m-1}), (2^{m-1},2 \cdot 5)$ or $(2^{m-1}\cdot5, 2)$, with $m\geq 2$.

[/quote]

sorry, i don't quite understand.... how do u know $(n+1,n-1)= (2^{m-1},2 \cdot 5)$ or $(2^{m-1}\cdot5, 2)$, with $m\geq 2$ ?
why isn't it $(n+1,n-1)= (2^{m-k},2^k \cdot 5)$ or $(2^{m-k}\cdot5, 2^k)$? How do u know that it's just m-1? :?

My solution:

[hide]
since $k > 0$,
$n \equiv 1 (mod 10)$
Hence, $n = 10k + 1 / 10k - 1$
If $n = 10k + 1$,
$2^m\cdot5 = (10k + 1)^2 - 1^2 = 10k(10k + 2)$

$2^{m-2} = k(5k +1)$
From here, if $k$ is even, $5k + 1$ is odd and vice versa. Hence, then only possibility is $k = 1 \to 2^{m-2} = 6$, which is wrong if m is integer.

The same goes for $n = 10k -1$,
will yield $2^{m-2} = k(5k - 1)$
SImilarly, the only possibility is $k = 1$, which gives $m-2 = 2, m =4$
:)
[/hide][/quote]

Look at the bolded part.  If you used $m-k$ you would have a $gcd$ that is too large.

[quote="t0rajir0u"]

Look at the bolded part.  If you used $m-k$ you would have a $gcd$ that is too large.[/quote]

Thanks  :)


projective geometry
geometry theorems
geometry

How can we prove Monge's theorem???

And What is exactly Monge's theorem??

I heard that "three circle's external similitude center is collinear"
But here,  [url]http://www.mathlinks.ro/viewtopic.php?t=215385[/url] he use it for two internal similitude center and one external center.

which is correct??

have a look at
http://www.cut-the-knot.org/proofs/threecircles.shtml

those 3 solution's style is not um... not a math olympiad's style.
Is there some solution whose style is similar with olympiad??

Also,how can we explane [url]http://www.mathlinks.ro/viewtopic.php?t=215385[/url] ??

(Sorry for my bad English;;)


geometry
combinatorics unsolved
combinatorics

A robot is moving on the edges of an infinite chessboard. There are two counters $A$ and $B$ which are initially both set to zero. In each step the robot moves one unit to the north, south, east, or west, and depending on the type of the step, $A$ is replaced by $A+1$, $A$ by $A-1$, $B$ by $B-A$, or $B$ by $B+A$, respectively. Assume that the robot starts its journey at a vertex $V$ of the chessboard and it returns to $V$ after some time without intersecting its trajectory. Prove that the final value of $|B|$ will be equal to the area of the region enclosed by the trajectory

This is an extreme outline, but basically we can say the following: coordinatize the plane.  Fix any given $i$, and look at all the times the robot traveled between a point $(i,j)$ and the point $(i+1, j)$, in either direction.  Each of the associated $j$-values are distinct, and if we order them so that $j_{1}< j_{2}< \ldots$, then we see that the area bounded by the robot can be broken up into strips with vertices $(i, j_{1}), (i+1,j_{1}), (i+1,j_{2}), (i+1,j_{2})$, $(i, j_{3}), (i+1,j_{3}), (i+1,j_{4}), (i+1,j_{4})$ and so on, and that the two edges in question we traversed in different directions on the two trips, and that for any two different strips, the order of traversal (down-up or up-down) was the same.  The result follows from these observations.


An elevator ascends from the ground with uniform speed. At time $ T_1$, a boy drops a marble through the floor. The marble falls with uniform acceleration $ g \equal{} 9.8 \frac {m}{s^{2}}$ and hits the ground $ T_2$ seconds later. Find the height of the elevator at time $ T_1$.

A hint is given: If $ T_1 \equal{} T_2 \equal{} 4$, then $ h \equal{} 39.2$.

So let $ T_1 \equal{} 0$. We know that $ 0 \equal{} y_0 \plus{} v_{y0}T_{2} \minus{} \frac {1}{2}gT_{2}^{2}$ which means that $ y_0 \equal{} 4.9(T_2)^{2} \minus{} v_{y0}(T_2)$.

But this is dependent on $ v_{y0}$. Is this still correct given the hint?


ratio
vector
geometry
geometric transformation
homothety
parallelogram

$ABC$ is a triangle. $ABDE$ and $ACFG$ are squares outside the triangle. $M$ and $N$ are midpoints of $DG$ and $EF$. Find all values of $MN:BC$

A proof using [hide]vectors[/hide] is very nice and simple.  Also, the squares need not be external -- it works just as well either way.

Let $K, L$ be the midpoints of the sides $CA, AB$. $KL$ is a midline of the triangle $\triangle ABC$, $KL \parallel CB$ and $KL = \frac{CB}{2}$.  Construct 2 squares $CBHI$ and $KLM'N'$ on the segments $CB, KL$ toward the vertex $A$. These 2 squares have parallel sides, hence, they are centrally similar. Their center of similarity is the vertex $A$, because the lines $CK, BL$ connecting their corresponding vertices $C, K$ and $B, L$ intersect at this point. Their coefficient of similarity is $\frac{KL}{CB} = \frac 1 2$.

The triangles $\triangle ABC \cong \triangle DBH$ are congruent by SAS, because $DB = AB$ ($ABDE$ is a square), $HB = CB$ ($CBHI$ is a square) and $\angle DBH = \angle ABC$ ($DB \perp AB,\ HB \perp CB$). Hence, their remainig sides are also congruent, $CA = DH$. Since $ACFG$ is a square, $CA = AG$ and consequently, $DH = AG$. The lines $AB, AG$ form the angle $\angle BAG = \angle A + 90^o$. The lines $AB, DH$ form the angle $\angle ABD + \angle BDH  = 90^o + \angle A$. Thus the equal opposite sides $DH = AG$ of the quadrilateral $AGHD$ are parallel, $DH \parallel AG$ and this quadrilateral is a parallelogram. Hence, its diagonals cut each other in half. One diagonal is $DG$ with the midpoint $M$ and the other diagonal is $AH$ with the midpoint $M'$ because of the homothety of the squares $KLM'N' \sim CBHI$ with the homothety center $A$ and homothety coefficient $\frac 1 2$, $AM' = \frac{AH}{2}$. Consequently, the points $M' \equiv M$ are identical. In exactly the same way, it can be shown that the triangles $\triangle FIC \cong \triangle ABC$ are congruent and that the quadrilateral $AFIE$ is a parallelogram with the diagonal intersection $N' \equiv N$. As a result, $MN = KL = \frac{BC}{2}$

Can someone show me the solutions using vectors?

Set A as the origin.  Then $M = \frac{D+G}{2} = \frac{B+E+G}{2}$ and $N = \frac{E+F}{2} = \frac{E+C+G}{2}$, so $M - N = \frac{B-C}{2}$ and so segment MN is parallel to segment BC and their lengths are in the ratio 1:2.


geometry
3D geometry
calculus
integration

The sum of the volumes of two cubes with integral side lengths is numerically equal to the sum of the lengths of the edges of the cubes. What are the dimensions of each cube?

Good luck.

A very inelegant solution...

D'oh, I thought this was intermediate and was thinking, this problem is way too easy for intermediate .

[hide]

Call the cubes side a and b. Then,

a^3 + b^3 = 12(a+b)

(a+b)(a^2 -ab+b^2) = 12(a+b)

a^2 - ab + b^2 - ab = 12 - ab

(a-b)^2 = 12-ab

a-b = :sqrt: (12-ab)

a-b is an integer, so 12-ab must be a perfect square less than 12. This means 12-ab = 0, 1, 4, 9, or

ab = 12, 11, 8, 3

Possibilities are (12, 1), (6, 2), (4, 3), (11, 1), (8, 1), (4, 2), and (3, 1). It is easy to check these and see that the only solution is (4, 2).

So the dimensions of the cubes are 4x4x4 and 2x2x2.

[/hide]


algebra function domain inequalities parameterization conics parabola

Find the range of positive real number $ a$ for which $ \frac{1}{2}\leq (x\minus{}a)^2\plus{}y\leq 2$ always holds when the point $ (x,\ y)$ moves in the domain : $ x^2\leq y\leq x$.

[hide]The upper bound on $ (x \minus{} a)^2 \plus{} y$ will be when $ x \equal{} y$. This gives $ (x \minus{} a)^2 \plus{} x \equal{} x^2 \minus{} 2ax \plus{} x \plus{} a^2\leq 2$. Since the LHS is continuous w/ respect to $ x$, we just need to evaluate it at the limits, $ x \equal{} 0$ and $ x \equal{} 1$. $ x \equal{} 0\rightarrow a^2\leq 2\rightarrow \minus{} \sqrt 2\leq a\leq\sqrt 2$. $ x \equal{} 1\rightarrow 1 \minus{} 2a \plus{} 1 \plus{} a^2\leq2\rightarrow a^2 \minus{} 2a\leq 0\rightarrow 0\leq a\leq 2$. Combining these gives $ 0\leq a\leq\sqrt 2$. Now the lower limit occurs somewhere on $ y \equal{} x^2$, so $ (x \minus{} a)^2 \plus{} x^2 \equal{} 2x^2 \minus{} 2ax \plus{} a^2\geq\frac 12$. Once again, we need only evaluate the limits of $ x \equal{} 0$ and $ x \equal{} 1$. $ x \equal{} 0\rightarrow a^2\geq\frac 12\rightarrow a\geq\frac {\sqrt 2}2$. $ x \equal{} 1\rightarrow 2 \minus{} 2a \plus{} a^2\geq\frac 12\rightarrow 2a^2 \minus{} 4a \plus{} 3 \equal{} 2(a \minus{} 1)^2 \plus{} 1\geq 0\rightarrow a$ could be anything. We combine all these parameters to get the final answer of $ \boxed{\frac {\sqrt 2}2\leq a\leq\sqrt 2}$[/hide]

But you're close  :P

[hide="Hint"]
For your lower bound, we have that $ \frac{1}{2} \leq (x\minus{}a)^2 \plus{} x^2 \iff 1\minus{}a^2 \leq 4(x\minus{}a/2)^2$

The RHS is clearly at least $ 0$, so $ 1\minus{}a^2 \leq 0 \implies 1 \leq a$ as your lower bound.  :)
[/hide]

o ok my b...
I said the minimum value of $ 2x^2\minus{}2ax\plus{}a^2$ on $ 0\leq x\leq 1$ is either at $ x\equal{}0$ or $ x\equal{}1$, but it could also be at the vertex of the parabola, $ x\equal{}\frac a2$, so $ \frac{a^2}2\geq\frac 12\rightarrow a\geq 1$. The upper bound stays the same.

Yes, the answer is $ 1\leq a\leq \sqrt{2}$.


geometry
incenter
circumcircle
geometry proposed

Hello!

   Let $ ABC$ be a triangle with circumcircle $ W$, $ D$ is a point on $ \left[BC\right]$ and $ I$ is the incenter of the triangle $ ADC$.
$ W'$ is the circle which tangents to $ \left[AD\right]$ at $ E$ and tangents to $ W$ and $ \left[CD\right]$. $ \left\{F\right\} \equal{} EI\cap W'$ and $ B', C'$ are
the intersection points of the tangent to $ W'$ at $ F$ with $ W$. Prove that the point $ I$ is the incenter of the triangle $ AB'C'$.

Best regards, Petrisor Neagoe :)

Let $ T \equal{} W \cap AD$, $ S \equal{} W' \cap W$, $ K \equal{} C'B' \cap AD$, $ \angle ABC \equal{} \alpha$.

At the first we will prove that $ CT \parallel C'B'$
$ \angle CSA \equal{} 180 \minus{} \alpha \Longrightarrow \angle IED \equal{} \frac {180 \minus{} \alpha}{2}$ by Segment theorem (see here http://kvant.mirror1.mccme.ru/oblozhka_djvu5.htm year 1992 #9 pages 51-58
)$ \angle FDE \equal{} 180 \minus{} \angle DEF \minus{} \angle DFE \equal{} \alpha$;
$ \angle CTA \equal{} \angle FDA \equal{} \alpha \Longrightarrow C'B' \parallel CT$

Then $ AI$ is a bisector of $ \angle C'AB'$ because it is a bisector of $ \angle CAT$ and $ CT \parallel C'B'$.

Else the incenter of $ C'AB'$ lies on $ FE$ (see here http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf).

so $ I$ is the incenter of $ C'AB'$ because $ I$ is the intersection point of $ FE$ and bisector of $ \angle C'AB'$.

I think some trifling condition will be added.
Dear Nak, I can't open two files that you uploaded. so can't you explain a little more detail?

<Well known lemma>
Let circumcircle of triangle $ ABC$ be $ \omega$, and $ P$ is arbitrary point on $ BC$ ($ CP \ge BP$). Circle that tangent to $ CP, AP,$ and $ \omega$ tangent to $ AP,CP$ at $ E,F$ respectively. Then $ EF$ pass through incenter of triangle $ ACP$
(It's known as one of lemma that prepare to prove the bault theorem :) .)

Same definition with above post written by Nak
$ T, S, K,$ and $ \alpha$
$ \angle ASC = \pi - \alpha$
and easily we can find $ \angle FED = \frac {\pi - \alpha}{2}$ by angle chasing
$ \angle KEF = \angle ESF = \angle ECK$

$ \angle FKE = \pi - \angle DEF - \angle KFE = \alpha$
therefore $ C'B'$ is parallel to $ CT$
hence midpoint of $ \overarc{C'B'}$and midpoint of $ \overarc{CT}$ are correspond.
and by lemma EF pass through incenter of triangle $ AB'C'$
So I is the incenter of triangle $ AB'C'$, too.

Dear friends!

Thank you for the nice solutions!

   Let $ ABC$ be a triangle with circumcircle $ W$, $ D$ is a point on $ \left[BC\right]$ and $ I$ is the incenter of the triangle $ ADC$.
$ W'$ is the circle which tangents to $ \left[AD\right]$ at $ E$ and tangents to $ W$ and $ \left[CD\right]$. $ \left\{F\right\}\equal{}EI\cap W'$ and $ B', C'$ are
the intersection points of the tangent to $ W'$ at $ F$ with $ W$. Prove that the point $ I$ is the incenter of the triangle $ AB'C'$.

Solution:

Let $ J$ be the incenter of the triangle $ ABC$, $ \left\{K\right\}\equal{}W'\cap\left[CD\right]$, $ \left\{D'\right\}\equal{}AD\cap B'C'$, $ \left\{T\right\}\equal{}W\cap W'$ and
$ \Gamma$ is the circumcircle of the triangle $ AET$.

It's known that the points $ K,J,E$ are collinear and $ J\in\Gamma$. (See here [url]http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf[/url])
So the line $ EK$ intersects again the circle $ \Gamma$ at the incenter $ J$ of the triangle $ ABC$.

In the same way, consider the triangle $ AB'C'$, $ D'\in\left[B'C'\right]$ and the circle $ W'$ which tangents to $ \left[AD'\right]$ at $ E$ and
tangents to $ W$ ( at $ T$) and $ \left[C'D'\right]$ ( at $ F$). The line $ EF$ intersects again the circle $ \Gamma$ at the incenter $ I$ of the triangle $ AB'C'$.

Best regards, Petrisor Neagoe :)

To Heebeen, Yang
Here is the theorem I used:
Let $ a, b$ are two lines and $ \omega$ is a circle with center $ I$ that tangents to $ a$ and $ b$ and $ 0<\alpha<360$.
Let $ A$ and $ B$ be two variable points on lines $ a$ and $ b$ such that $ AB$ tangents to $ \omega$ and $ S$ be a circle with center $ O$ that passes though $ A$ and $ B$ such that $ \angle(OA, OB)\equal{}\alpha$. Then there are two fixed circles $ W_1$ and $ W_2$ such that:
1. $ W_1$ and $ W_2$ both tangents to $ S$, $ a$, $ b$.
2. If $ N_{a}, N_{b}$ are tangent points of $ W_1$ with $ a$ and $ b$ respectively then $ \angle(a, IN_{a})\equal{}\frac{\alpha}{4}$;
If $ M_{a}, M_{b}$ are tangent points of $ W_2$ with $ a$ and $ b$ respectively then $ \angle(a, IM_{a})\equal{}\frac{\alpha\plus{}360}{4}$.

Thank you Nak  :)

It's very interesting theorem!
It makes my proof much shorter!!

Really thanks for you.
But I can't prove it now -_-..


For any positive real numbers $ a,b$ and $ c$,
\[ \frac{a}{a^{2}\plus{}bc}\plus{}\frac{b}{b^{2}\plus{}ca}\plus{}\frac{c}{c^{2}\plus{}ab}\ge\frac{3(a\plus{}b\plus{}c)}{2(a^{2}\plus{}b^{2}\plus{}c^{2})}\]

Many thanks to Ductrung for this nice inequality.

At first change $ a\to 1/a$ to obtain

$ \sum\frac{1}{a^{2}\plus{}bc}\ge\frac{3\sum ab}{2\sum (ab)^{2}}$

Now multiply numerator and denominator for $ (b\plus{}c)^{2}$  and use Cauchy in this form
$ \sum (a^{2}\plus{}bc)(b\plus{}c)^{2}\sum\frac{(b\plus{}c)^{2}}{(a^{2}\plus{}bc)(b\plus{}c)^{2}}\geq\frac{(2\sum a)^{2}}{\sum (a^{2}\plus{}bc)(b\plus{}c)^{2}}$

You have after simple algebra

$ 5\sum (ab)^{2}(a^{2}\plus{}b^{2})\plus{}6(abc)^{2}\plus{}4\sum(ab)^{3}\geq 5abc\sum ab(a\plus{}b)\plus{}6abc\sum a^{3}$

Now use Shur in this form

$ \sum (ab)^{3}\plus{}3(abc)^{2}\geq abc\sum ab(a\plus{}b)$

and also

$ \sum (ab)^{2}(a^{2}\plus{}b^{2})\geq 2abc\sum a^{3}$

$ \sum a^{4}(b^{2}\plus{}c^{2})\geq 2abc\sum a^{3}$

$ \sum a^{3}\geq 3abc$

which are true by AM-GM. The rest is easy :)

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}
\]
[/quote]
The following inequality is true too.
For any positive real numbers $ a,b$ and $ c$ prove that:
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {6(a \plus{} b \plus{} c)}{3(a^{2} \plus{} b^{2} \plus{} c^{2})\plus{}ab\plus{}ac\plus{}bc}
\]

[quote="nguoivn"]I think the stronger also trues:
$ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}$
 :wink:[/quote]

It's not true, but I think the following is true and nice.

Let $ a,b,c$ be the sidelenghts of triangle. Prove that

\[ \frac {a} {{a^2\plus{}bc}} \plus{} \frac {b} {{b^2\plus{}ca}} \plus{} \frac {c} {{c^2\plus{}ab}} \le \frac{{3\left( {a \plus{} b \plus{} c} \right)}}{{a^2  \plus{} b^2  \plus{} c^2  \plus{} ab \plus{} bc \plus{} ca}}\]

[quote="Honey_S"][quote="nguoivn"]I think the stronger also trues:
$ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}$
 :wink:[/quote]

It's not true,...
[/quote]
Yes! There is a problem around $ a \equal{} b \equal{} c.$ :wink:
[quote="Honey_S"]...but I think the following is true and nice.

Let $ a,b,c$ be the sidelenghts of triangle. Prove that
\[ \frac {a} {{a^2 \plus{} bc}} \plus{} \frac {b} {{b^2 \plus{} ca}} \plus{} \frac {c} {{c^2 \plus{} ab}} \le \frac {{3\left( {a \plus{} b \plus{} c} \right)}}{{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}}
\]
[/quote]
I agree with you! :lol:
$ \frac {a} {{a^2 \plus{} bc}} \plus{} \frac {b} {{b^2 \plus{} ca}} \plus{} \frac {c} {{c^2 \plus{} ab}} \le \frac {{3\left( {a \plus{} b \plus{} c} \right)}}{{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \plus{} bc} \minus{} \frac {a}{a^2 \plus{} bc}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {b^2c \plus{} c^2b \minus{} b^2a \minus{} c^2a}{a^2 \plus{} bc}\Leftrightarrow\sum_{cyc}\frac {b^2(c \minus{} a) \minus{} c^2(a \minus{} b)}{a^2 \plus{} bc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}c^2(a \minus{} b)\left(\frac {1}{b^2 \plus{} ac} \minus{} \frac {1}{a^2 \plus{} bc}\right)\geq0\Leftrightarrow\sum_{cyc}\frac {c^2(a \minus{} b)^2(a \plus{} b \minus{} c)}{(b^2 \plus{} ac)(a^2 \plus{} bc)}\geq0.$

I think the following inequalities hold for any nonnegative real numbers $ a,b,c$, no two of which are zero:

$ (a) \ \ \ \ \ \ \ \frac a{2a^2\plus{}bc}\plus{}\frac b{2b^2\plus{}ca}\plus{}\frac c{2c^2\plus{}ab} \ge \frac {a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2}$;

$ (b) \ \ \ \ \ \ \ \frac a{a^2\plus{}2bc}\plus{}\frac b{b^2\plus{}2ca}\plus{}\frac c{c^2\plus{}2ab} \le \frac {a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$.

[quote="Vasc"]I think the following inequalities hold for any nonnegative real numbers $ a,b,c$, no two of which are zero:

$ (a) \ \ \ \ \ \ \ \frac a{2a^2 \plus{} bc} \plus{} \frac b{2b^2 \plus{} ca} \plus{} \frac c{2c^2 \plus{} ab} \ge \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}$;

$ (b) \ \ \ \ \ \ \ \frac a{a^2 \plus{} 2bc} \plus{} \frac b{b^2 \plus{} 2ca} \plus{} \frac c{c^2 \plus{} 2ab} \le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$.[/quote]
For a), we can prove it as follow:
Remembering the nice inequality of Darij for all $ x,y,z \ge 0$:
\[ \frac{1}{x^2\plus{}2yz}\plus{}\frac{1}{y^2\plus{}2zx}\plus{}\frac{1}{z^2\plus{}2xy} \ge \frac{2}{xy\plus{}yz\plus{}zx}\plus{}\frac{1}{x^2\plus{}y^2\plus{}z^2}\]
Applying this inequality for $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$, we obtain
\[ \frac{a}{2a^2\plus{}bc}\plus{}\frac{b}{2b^2\plus{}ca} \plus{}\frac{c}{2c^2\plus{}ab} \ge \frac{2}{a\plus{}b\plus{}c} \plus{}\frac{abc}{a^2b^2\plus{}b^2\plus{}c^2a^2}\]
We can deduce the inequality
\[ \frac{2}{a\plus{}b\plus{}c}\plus{}\frac{abc}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2} \ge \frac{a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2}\]
Now, normalizing for $ a\plus{}b\plus{}c\equal{}1$ and putting $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$. The above inequality becomes
\[ \frac{r}{q^2\minus{}2r} \ge \frac{4q\minus{}1}{1\minus{}2q}\]
Applying Schur's Inequality for fourth degree, we obtain $ r \ge \frac{(4q\minus{}1)(1\minus{}q)}{6}$, then
\[ \frac{r}{q^2\minus{}2r} \ge \frac{(4q\minus{}1)(1\minus{}q)}{6q^2\minus{}2(4q\minus{}1)(1\minus{}q)} \equal{}\frac{(4q\minus{}1)(1\minus{}q)}{2(7q^2\minus{}5q\plus{}1)}\]
It remains to show that
\[ \frac{(4q\minus{}1)(1\minus{}q)}{2(7q^2\minus{}5q\plus{}1)} \ge \frac{4q\minus{}1}{1\minus{}2q}\]
\[ (1\minus{}3q)(4q\minus{}1)^2 \ge 0\]
which is true.

Anyways, we can prove both a) and b) easily with Vornicu Schur by rewriting them as follow:
a)
\[ \sum \left( 2\minus{}\frac{a^2\plus{}b^2\plus{}c^2}{2a^2\plus{}bc}\right) (a\minus{}b)(a\minus{}c) \ge 0\]
b)
\[ \sum \frac{a}{a^2\plus{}2bc} (a\minus{}b)(a\minus{}c) \ge 0\]
:)

What is your proof, VasC? Thank you. :)

[quote="can_hang2007"] Anyways, we can prove both a) and b) easily with Vornicu Schur by rewriting them as follow:
a)
\[ \sum \left( 2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc}\right) (a \minus{} b)(a \minus{} c) \ge 0
\]
b)
\[ \sum \frac {a}{a^2 \plus{} 2bc} (a \minus{} b)(a \minus{} c) \ge 0
\]
:)
What is your proof, VasC? Thank you. :)[/quote]
How do you apply Vornicu Schur for (a) ? I think it is possible, but is not very simple.
For (b) I have the same solution as you, while for (a) I used Cauchy-Schwarz and Lemma below.

[b]Lemma. [/b]If $ a,b,c$ are nonnegative real numbers, then

$ abc \le \frac {q^2(p^2\minus{}q)}{2p(2p^2\minus{}3q)},$

where $ p\equal{}a\plus{}b\plus{}c$ and $ q\equal{}ab\plus{}bc\plus{}ca$.

[quote="Vasc"][quote="can_hang2007"] Anyways, we can prove both a) and b) easily with Vornicu Schur by rewriting them as follow:
a)
\[ \sum \left( 2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc}\right) (a \minus{} b)(a \minus{} c) \ge 0
\]
b)
\[ \sum \frac {a}{a^2 \plus{} 2bc} (a \minus{} b)(a \minus{} c) \ge 0
\]
:)
What is your proof, VasC? Thank you. :)[/quote]
How do you apply Vornicu Schur for (a) ? I think it is possible, but is not very simple.
For (b) I have the same solution as you, while for (a) I used Cauchy-Schwarz and Lemma below.

[b]Lemma. [/b]If $ a,b,c$ are nonnegative real numbers, then

$ abc \le \frac {q^2(p^2 \minus{} q)}{2p(2p^2 \minus{} 3q)},$

where $ p \equal{} a \plus{} b \plus{} c$ and $ q \equal{} ab \plus{} bc \plus{} ca$.[/quote]
The lemma you posted is also mine  :blush:   :lol: . I usually normalize for $ a \plus{} b \plus{} c \equal{} 1$ and write it as $ r \le \frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}$. :)
Now, let see my proof by Vornicu Schur for a)
Setting $ X \equal{} 2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc}$ and $ Y,Z$ are similar. Supposing that $ a \ge b \ge c$, then it is obvious that $ X \ge 0$.
Case 1. $ a,b,c$ are the side lengths of a triangle, then $ Y,Z \ge 0$ and $ X \ge Y$. Hence
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \plus{} Z(c \minus{} a)(c \minus{} b) \ge X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \ge X(a \minus{} b)(b \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \equal{} (X \minus{} Y)(a \minus{} b)(b \minus{} c) \ge 0
\]
Case 2. $ a \ge b \plus{} c$. Notice that we still have
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \ge (X \minus{} Y)(a \minus{} b)(b \minus{} c) \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)(2a \plus{} 2b \minus{} c)(a \minus{} b)^2(b \minus{} c)}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)}
\]
And
\[ Z \equal{} \frac {3c^2 \minus{} (a \minus{} b)^2}{2c^2 \plus{} ab} \ge \minus{} \frac {(a \minus{} b)^2}{2c^2 \plus{} ab}
\]
We deduce our inequality to
\[ \frac {(a^2 \plus{} b^2 \plus{} c^2)(2a \plus{} 2b \minus{} c)(a \minus{} b)^2(b \minus{} c)}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)} \ge \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2c^2 \plus{} ab}
\]

\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)} \ge \frac {a \minus{} c}{(2a \plus{} 2b \minus{} c)(2c^2 \plus{} ab)}
\]
Now, to prove this inequality, we will use simple estimations:
\[ \frac {a \minus{} c}{2a \plus{} 2b \minus{} c} \le \frac {a}{2(a \plus{} b)}
\]

\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc} \ge \frac {1}{2}
\]
and
\[ \frac {1}{2b^2 \plus{} ca} \minus{} \frac {a}{(a \plus{} b)(2c^2 \plus{} ab)}  \equal{} \frac {a(a \minus{} b \minus{} c)(b \minus{} c) \plus{} ac^2 \plus{} 2bc^2}{(a \plus{} b)(2b^2 \plus{} ca)(2c^2 \plus{} ab)} \ge 0
\]
This completes our proof. :)

[quote="arqady"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}
\]
[/quote]
The following inequality is true too.
For any positive real numbers $ a,b$ and $ c$ prove that:
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {6(a \plus{} b \plus{} c)}{3(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} ab \plus{} ac \plus{} bc}
\]
[/quote]

My proof. :)

[quote="Vasc"]for (a) I used Cauchy-Schwarz and Lemma below.
[/quote]
I think you used Cauchy Schwarz like this
\[ \left( \sum \frac{1}{x^2\plus{}2yz}\right)\left( \sum (y\plus{}z)^2(x^2\plus{}2yz)\right) \ge 4(x\plus{}y\plus{}z)^2\]
where $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$. ;) :)

[quote="can_hang2007"] Now, let see my proof by Vornicu Schur for a)
Setting $ X \equal{} 2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc}$ and $ Y,Z$ are similar. Supposing that $ a \ge b \ge c$, then it is obvious that $ X \ge 0$.
Case 1. $ a,b,c$ are the side lengths of a triangle, then $ Y,Z \ge 0$ and $ X \ge Y$. Hence
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \plus{} Z(c \minus{} a)(c \minus{} b) \ge X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \ge X(a \minus{} b)(b \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \equal{} (X \minus{} Y)(a \minus{} b)(b \minus{} c) \ge 0
\]
Case 2. $ a \ge b \plus{} c$. Notice that we still have
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \ge (X \minus{} Y)(a \minus{} b)(b \minus{} c) \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)(2a \plus{} 2b \minus{} c)(a \minus{} b)^2(b \minus{} c)}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)}
\]
And
\[ Z \equal{} \frac {3c^2 \minus{} (a \minus{} b)^2}{2c^2 \plus{} ab} \ge \minus{} \frac {(a \minus{} b)^2}{2c^2 \plus{} ab}
\]
We deduce our inequality to
\[ \frac {(a^2 \plus{} b^2 \plus{} c^2)(2a \plus{} 2b \minus{} c)(a \minus{} b)^2(b \minus{} c)}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)} \ge \frac {(a \minus{} b)^2(a \minus{} c)(b \minus{} c)}{2c^2 \plus{} ab}
\]

\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{(2a^2 \plus{} bc)(2b^2 \plus{} ca)} \ge \frac {a \minus{} c}{(2a \plus{} 2b \minus{} c)(2c^2 \plus{} ab)}
\]
Now, to prove this inequality, we will use simple estimations:
\[ \frac {a \minus{} c}{2a \plus{} 2b \minus{} c} \le \frac {a}{2(a \plus{} b)}
\]

\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{2a^2 \plus{} bc} \ge \frac {1}{2}
\]
and
\[ \frac {1}{2b^2 \plus{} ca} \minus{} \frac {a}{(a \plus{} b)(2c^2 \plus{} ab)} \equal{} \frac {a(a \minus{} b \minus{} c)(b \minus{} c) \plus{} ac^2 \plus{} 2bc^2}{(a \plus{} b)(2b^2 \plus{} ca)(2c^2 \plus{} ab)} \ge 0
\]
This completes our proof. :)[/quote]
My proof for
$ \sum \frac{3a^2\minus{}(b\minus{}c)^2}{2a^2\plus{}bc}(a\minus{}b)(a\minus{}c) \ge 0$
is simpler.  :wink: By Vornicu-Schur, it suffices to show that
$ (a\minus{}b)(b\minus{}c)(c\minus{}a)\sum \frac {b\minus{}c}{3a^2\plus{}bc} \ge 0$,
which is equivalent to
$ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca) \ge 0$.

[quote="can_hang2007"][quote="Vasc"]for (a) I used Cauchy-Schwarz and Lemma below.
[/quote]
I think you used Cauchy Schwarz like this
\[ \left( \sum \frac {1}{x^2 \plus{} 2yz}\right)\left( \sum (y \plus{} z)^2(x^2 \plus{} 2yz)\right) \ge 4(x \plus{} y \plus{} z)^2
\]
where $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}$. ;) :)[/quote]
Indeed, you are right. :lol:

[quote="Vasc"]
My proof for
$ \sum \frac {3a^2 \minus{} (b \minus{} c)^2}{2a^2 \plus{} bc}(a \minus{} b)(a \minus{} c) \ge 0$
is simpler.  :wink: By Vornicu-Schur, it suffices to show that
$ (a \minus{} b)(b \minus{} c)(c \minus{} a)\sum \frac {b \minus{} c}{2a^2 \plus{} bc} \ge 0$,
which is equivalent to
$ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \ge 0$.
[/quote]
Very nice, but I think my proof does not use too much calculation.  :lol:

[quote="arqady"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {3(a \plus{} b \plus{} c)}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}
\]
[/quote]
The following inequality is true too.
For any positive real numbers $ a,b$ and $ c$ prove that:
\[ \frac {a}{a^{2} \plus{} bc} \plus{} \frac {b}{b^{2} \plus{} ca} \plus{} \frac {c}{c^{2} \plus{} ab}\ge\frac {6(a \plus{} b \plus{} c)}{3(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} ab \plus{} ac \plus{} bc}
\]
[/quote]

can_hang2007 have checked my solution for this and tell me my mistake. In my solution above, I rewrite the inequality is
\[ \frac {{\sum {\left( {a \minus{} b} \right)^2 } }}{{ab \plus{} bc \plus{} ca}} \ge \frac {{\sum {ab\left( {a \minus{} b} \right)^2 } }}{{3\left( {\sum {a^2 b^2 } } \right) \plus{} abc\left( {a \plus{} b \plus{} c} \right)}}
\]
But is must be
\[ \frac {{\sum {\left( {a \minus{} b} \right)^2 } }}{{3\left( {ab \plus{} bc \plus{} ca} \right)}} \ge \frac {{\sum {ab\left( {a \minus{} b} \right)^2 } }}{{3\left( {\sum {a^2 b^2 } } \right) \plus{} abc\left( {a \plus{} b \plus{} c} \right)}}
\]
Now I have edited, and here is the solution. :)


I typed a 15 pages article and I got stucked in the following requirement of the publisher: they want that on the first page of the article must appear Only the following 7 items:

1. The title
2. The name of the author
3. the address of the author, including email
4. the abstract of the article
5. the Keywords
6. MR(2000) subject classification
7. A running head (abreviated title)

Could you please help me and provide the set of commands? And the name of the necessary packages?
 It is a math article of differential geometry.

Thank you

I would ask your publishers for a style/class file as that's the only way to guarantee you lay it out as they require. Without it, although you can easily do the 7 items on the front page, it's not clear exactly how they want them.

The only extra package you need is fancyhdr for the running head. The rest is normal LaTeX stuff such as \maketitle and the abstract environment. For example:
[code]\documentclass[a4paper]{article}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{fancyhdr}
\pagestyle{fancy}
\lhead{}
\chead{}
\rhead{\bfseries Running head}
\lfoot{}
\cfoot{}
\rfoot{}
\renewcommand{\headrulewidth}{0.4pt}

\begin{document}
\title{This is a title}
\author{Author's name \\ Address \\ someone@somewhere.com}
\date{} % this stops the date being printed
\maketitle
\thispagestyle{fancy}
\begin{center}
	Keywords go here \\
	Classification goes here
\end{center}

\begin{abstract}
This is where the abstract goes
\end{abstract}
\clearpage

%footer for page 2 onwards
\renewcommand{\footrulewidth}{0.4pt}
\rfoot{\thepage}
Start of article
\end{document}[/code]
The [url=http://www.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr manual[/url] will give you the details on designing the headers and footers.

First, I want to thank you for your kindness.

I wasn't clear in my question. The publisher (Acta Mathematica) only required that all the 7 items from my previous question BE on the first page. They will add the running head, so I don't need to add any headers or footers.

I didn't understand all the commands you showed me, it seems too complicated.

There should be a simpler way (the final output it's their's decision).

Do I still need the fancy package? What is the simplest elegant way do build the first page?

If you don't need a running head then remove the fancyhdr stuff to get:
[code]\documentclass[a4paper]{article}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}
\title{Put your title here}
\author{Put your name here \\ Your address \\ Your email someone@somewhere.com}
\date{} % this stops the date being printed
\maketitle
\begin{center}
   Put your Keywords here \\
   Put your Classification here
\end{center}

\begin{abstract}
Type your abstract here
\end{abstract}
\clearpage
Start of article
\end{document}[/code]The only other thing is that in \title you put the title of the article, in \author you put your name and address. \maketitle puts these nicely on the page. Normally it adds the date but \date{} stops that.

I centered the Keywords & classification but you don't have to and \begin{abstract}...\end{abstract} is how you enter the abstract.

\clearpage ensures nothing else goes on the front page.

All of this is standard simple LaTeX; compile and see if it's what you want.

I did everything you said, and it's ok, the only problem is that abstract should be align to the left, like:

[b]Abstract[/b] Text follows

I deleted the command {center}.

Another problem is that I can't find the commands for Keywords and classification. Both are left aligned too, like below in their articles:

[b]Abstract [/b]In this paper, we generalize the fixed
norm type to the theorem of functional type. As
some fourth-order beam equation boundary value
the nonlinear term is dependent on all lower order

[b]Keywords[/b] fourth-order boundary value problem

[b]MR(2000) Subject Classification[/b] 34B15

That was a sample from a paper.

There are no special Keywords or Subject Classification commands. In addition LaTeX obviously can't know which are the keywords or what the classification - you have to tell it.

If you want everything left-aligned then just type it normally with the first word in bold[code]\textbf{Keywords}\ fourth-order boundary value problem [/code]then do the same for Abstract and MR(2000) Subject Classification.

I have deleted the other thread as that has the same posting.


daca log154_111=a  si log154_7=b atunci log154_4=?

Maybe $\log_{154}11=a\ ,\ \log_{154}7=b\Longrightarrow \log_{154}4=?\ .$ I think in stead of $111$ is $11$ !