Let {$a_n$} be a sequence such that  :

$a_1=1 , a_{n+1} = ( \sqrt{2a_n} + \sqrt{1 + a_n})^2 $

Prove that  all the sequence's elements are integers .

 :cool:

Computing by hand the first values shows a(n+1) ~ 6*a(n), indeed :

(*) a(n+1) = 6*a(n) - a(n-1) + 2, from the first values.

I tried to prove it by induction by didn't succeed, but I found an other (indirect!) way :

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

we have f(g(x))=x for x >= 1. so a(n) = g(a(n+1)) from which (*) follows which proves that a(n) is integer.

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

Forgive my ignorance but what does C.D.B. mean ?

firstly i find that
a(n+1) = 6 a(n) - a(k-1) + 2

then , inset two cofficients to form the sequence a new and regular one to find out the {a(n)} is.
a(n+1) - A a(n) + c =  B [  a(n) - A a(n-1) + c ] ;
a(n+1) = ( A + B ) a(n) - AB a(n-1) + ( B-1 )c
  A + B = 6
  AB     = 1
 (B-1)c = 2
 find the value of them , put back to the equation to describe {a(n)}
 then use the induction directly prove that {a(n)} is an integer for
 n = 1 , 2,  3, ...

i am sorry about the skip, the application really takes hours to put on the attachment. i will sent it latter.[/quote][/tex][/code]

correction - ------  three cofficients there [ A, B, and C ],   not two

An other solution (with $a_{0}=0$, by induction if
$a_{2n}  = 2\alpha _n ^2$ and $1 + a_{2n}  = \beta _n ^2 $
Then
$a_{2n + 1}  = \left( {2\alpha _n  + \beta _n } \right)^2$
$1 + a_{2n + 1}  = 2\left( {\alpha _n  + \beta _n } \right)^2$
$a_{2n + 2}  = 2\left( {3\alpha _n  + 2\beta _n } \right)^2 $
and
$1 + a_{2n + 2}  = \left( {4\alpha _n  + 3\beta _n } \right)^2 $
(dont forget that $2\alpha _n ^2  = \beta _n ^2  - 1$ )
This problem is related to the Pell-Fermat equation
$\left| {x^2  - 2y^2 } \right| = 1$


geometry
circumcircle
conics
geometry proposed

Given a triangle [b]ABC[/b].Bisectors of angles [b]C,B[/b] intersect the circumcircle of triangle [b]ABC[/b] in points [b]D,E[/b].
Point[b] O[/b] is the circumcenter of triangle [b]ABC[/b].
[b]OE[/b] intersects the circumcircle of triangle [b]ABC[/b] in point [b]F[/b].(a point different from E)
L moves on segment AB and M moves on segment AC such that
[u][b](AL:CM)=(AB:AC).[/b][/u]
[b]DL[/b] intersects [b]FM[/b] in point [b]P[/b].
Find the [color=red]geometric locus[/color] of   [b]P[/b].
 :lol:

[quote="jensen"]Given a triangle [b]ABC[/b].Bisectors of angles [b]C,B[/b] intersect the circumcircle of triangle [b]ABC[/b] in points [b]D,E[/b].
Point[b] O[/b] is the circumcenter of triangle [b]ABC[/b].
[b]OE[/b] intersects the circumcircle of triangle [b]ABC[/b] in point [b]F[/b].(a point different from E)
L moves on segment AB and M moves on segment AC such that
[u][b](AL:CM)=(AB:AC).[/b][/u]
[b]DL[/b] intersects [b]FM[/b] in point [b]P[/b].
Find the [color=red]geometric locus[/color] of   [b]P[/b].[/quote]

Would you please check your problem again? I fear you have made a typo somewhere. In fact, your locus is a conic (as the points L and M move, the lines DL and FM form two projective pencils of lines, and thus their point of intersection moves along a conic), but this conic doesn't seem to have any nice properties (it isn't a circle - on my sketches, it is a hyperbola). Probably you meant something different, didn't you?

  Darij

[quote="jensen"]I see!It is not a good geometric locus,may be because i design it by my self.I am sorry!
Lets have the original(basic) problem!
[color=blue]Find a point like P on the circumcircle of triangle ABC that
(AL:CM)=(AB:AC).(All of the points in this problem are same in the first problem)  :oops:  [/color][/quote]

Here is alink to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43203]solution[/url]


I have a few questions that I wasn't sure about from a previous multiple choice exam. Any help would be appreciated.

1. Two satellites I and II have orbital radii of radii R and 2R respectively about the same planet. The orbital velocity of I is v. What is the orbital velocity of II?

2. Three object of equal mass are released from the top of an inclined plane of height H. The objects are as follows:
i) a cube of side R
ii) a solid cylinder of radius R
iii)a hollow cylinder of radius R.

Assuming the cylinders roll without slipping, and the cube slides without friction, which object(s) reach the bottom first?

3. A rod of length 2R travels with angular velocity $ \omega$ when one object of mass m is attached to each end of the rod, a distance of R from the center of the rod. Each object is brough a distance R/2 closer to the center of the rod. What is the new angular velocity?

4. A glowing ember radiates power P when kept at absolute temperature T. When the absolute temperature is reduced to T/2, what is the new power radiation?

Here are the answers:-

1.$ \frac{v}{\sqrt 2}$

2.Solid sphere if there is friction.If no friction,then all will take same time.

3.$ 4\omega$

4.P/16

[quote="Aashay THE ROKR"]Here are the answers:-

1.$ \frac {v}{\sqrt 2}$

2.Solid sphere if there is friction.If no friction,then all will take same time.

3.$ 4\omega$

4.P/16[/quote]

Right, there wasn an answer key at the back of the test. These were questions I was having a little trouble with. Could you explain how you got them?

[quote="Aashay THE ROKR"]1.Orbital velocity is given by-

$ v \equal{} \sqrt\frac {GM}{R}$

So when R changes to 2R,v becomes $ \frac {v}{\sqrt 2}$

2.Solid sphere has the highest acceleration,i.e.,$ \frac {5}{7}$ g $ sin\theta$,hence will take least time if friction is present.
If no friction,then all bodies have same acceleration,hence will take same time.

3.By law of conservation of angular momentum,
$ I_{1}\omega_{1} \equal{} I_{2}\omega_{2}$

$ (MR^2 \plus{} MR^2)\omega \equal{} (\frac {MR^2}{4} \plus{} \frac {MR^2}{4})\omega_{2}$

$ \omega_{2} \equal{} 4\omega_{1}$

4.$ Power of radiation \equal{} \sigma e a T^4$

So when T changes to T/2,P changes to P/16.

I hope its clear now :)[/quote]

Thanks so much for the help!
But, for the 2nd one, there's only a cube and two cylinders, and the answer key says its the cube that reaches the ground first. Why?

[quote="mihail911"]

2. Three object of equal mass are released from the top of an inclined plane of height H. The objects are as follows:
i) a cube of side R
ii) a solid cylinder of radius R
iii)a hollow cylinder of radius R.

Assuming the cylinders roll without slipping, and the cube slides without friction, which object(s) reach the bottom first?

[/quote]

Maybe it is the cube because it is given that there is no friction for cube, but it is pure rolling for the cylinders ( roll without slipping). In pure rolling, friction acts upwards, thus decreasing the net acceleration downwards for the cylinders.
I might be wrong.

[quote="Aashay THE ROKR"]Post the 2nd question again.I was also confused about it.Tell me whether there is friction or not.[/quote]

Sure thing.
Three objects of equal mass with the above posted descriptions are released simultaneously from the top of an incline of height H. The cylinders roll without slipping and the cube slides without friction.

[quote="Aashay THE ROKR"]If that is the case,then definately cube will reach earliest.[/quote]

for what reason?

Because as it slides without friction,it will have the highest acceleration,i.e.$ g sin\theta$ more than that of solid sphere which is rolling without slipping

$ \frac {5}{7}$ g $ sin\theta$,

and hollow sphere

$ \frac{3}{5}$g$ sin\theta$

May be, we could put it in a simpler way, through  energy conservation:
kinetic energy will be the same for all of the bodies, but the cube has only translational K.E.  while the cylinders must divide it among translational and rotational.


What integer, added to both the numerator and the denominator of $ \frac{6}{7}$, results in a fraction with a value of 0.8?

We know that $ 0.8 \equal{} \frac {4}{5}$.  Since $ \frac {6 \plus{} ( \minus{} 2)}{7 \plus{} ( \minus{} 2)} \equal{} \frac {4}{5}$, our answer is $ \boxed{ \minus{} 2}$.

OR

If you want to do it algebraically instead of guess and check, first let $ x$ be the number we want to add to both the numerator and denominator.  We have
\[ \frac {6 \plus{} x}{7 \plus{} x} \equal{} 0.8 \\
 \\
6 \plus{} x \equal{} 5.6 \plus{} 0.8x \\
 \\
0.2x \equal{} \minus{} 0.4 \\
 \\
x \equal{} \boxed{ \minus{} 2}\]


I borrowed 25 euro to my friend A.  and  25 euro to my friend B.
THan I bought a nice T-shirt with a price of 45 euro . THe seller gave me 5 euro as a reminder.
I give 1 euro to A. and 1 euro to B  and I take 3 euro for myself . Now I have to give to each of my friends 24 euro .

24+24+3=51 ?????????????????????? Where is the mistake?//

Edit : THis is really stupid I know

What is the question?  And you might want to improve your English, when you say "I borrowed 25 euro to my friend", it seems as if you are giving your friend 25 when you are actually receiving it.

So, what exactly is the question?

This makes no sense whatsoever.

The question is:

I borrowed 25 euro from my friend, and 25 from my other friend.
I bought a t-shirt that cost 45 euro, so I got 5 euro back.
I gave each of my friends 1 euro, and kept 3.

He is wondering why 24 euro owed to the friend +24 owed to the other friend + 3 kept=51, not 50

It does, doesn't it?

I think it has something to do with the fact that you are adding the wrong things together to make it seem like a paradox, when it's actually not.

You should check the hotel problem, I think.

Here is the hotel problem;

Three men decide to share a room at a hotel. They each pay 15 dollars to one of the people that worked at the hotel. The man who recieved it brought it to the manager, and the manager said that the price for the room had gone down from 45 dollars to 40 dollars. So it had gone down five. The manager told the man to give five dollars back to them. The man didn't know how to divide the 5 dollars amongst the three of them, because he didn't know how to give 5/3 of a dollar. So he gave each man 1 dollar, and the man kept 2 dollars.

Oops, I don't think I said that right. It is something like that though.

[quote="AIME15"]Basically, you "counted the wrong thing".  See if you can find out where you did so![/quote]

No. Were one of the numbers wrong?

I may be wrong, but I think the answer is:

You are not to add up those numbers, for those are irrelevant. The real thing to add up is the shirt cost, and how much you had.

Thus:

(how much you borrowed)
$ 25 \plus{} 25 \equal{} 50$

(how much the shirt cost)
$ 50 \minus{} 45 \equal{} 5$

(how much you gave back)
$ 5 \minus{} 2 \equal{} 3$

(how much you kept)
$ 3 \minus{} 3 \equal{} 0$

So there.

These problems are called mathematical sofism and I apsolotly agree with Waffle


induction
abstract algebra
real analysis
real analysis unsolved

Let $ U$ be a real normed space such that, for an finite-dimensional, real normed space $ X,U$ contains a subspace isometrically isomorphic to $ X$. Prove that every (not necessarily closed) subspace $ V$ of $ U$ of finite codimension has the same property. (We call $ V$ of finite codimension if there exists a finite-dimensional subspace $ N$ of $ U$ such that $ V\plus{}N\equal{}U$.)

[i]A. Bosznay[/i]

At first, remark that it is sufficient to prove the statement for 1-codimensional subspaces: after that induction on codimention helps.

Then, the following lemma solves the problem:

[b]Lemma[/b]. Let $ X$ be a finite-dimensional normed space, then there is another finite-dimensional normed space $ Y$ such that every 1-codimensional subspace of $ Y$ contains an isometric copy of $ X$.

[b]Proof.[/b] Take $ Y\equal{} X \oplus X$ with the norm $ \|(x_1, x_2)\|\equal{} \max\left\{\|x_1\|, \|x_2\|\right\}$. Let $ Z$ be a 1-codimensional subspace of $ Y$. Then $ Z$ is the kernel of a functional $ f\equal{}(f_1, f_2)$, $ f_1, f_2 \in X^*$. Without loss of generality we may (and do) asume that $ \|f_1\| \le \|f_2\|\equal{}1$. Take $ e \in X$ with $ \|e\|\equal{}f_2(e)\equal{}1$. Then the subspace
$ \{(x, \minus{}f_1(x)e): x \in X\}$ is the requested isometric copy of $ X$ inside $ Z$.


I use the Input Method Editor that came shipped with the Windows XP operating system on the PC I use.

One gets starting typing Chinese by learning Chinese.

Anyone have that program where like a number is a stroke or something?  just got back from china and saw they were selling a program where you type with numbers...  seems easier then MS language tools or pinyin based systems..

[quote="toadoncart"]if you're chinese and you enter AOPS, will everything become chinese? do you have to download something? also, is there like a chinese teacher in the classroom that teaches the chinese kids? or could the english just translate?

let me translate for a second:

如果你是中国人，你进出这里，这个Art of Problem Solving 会不会有中文字？你要不要 download 什么东西吗？你上课的话，会不会有一个老师教课用中文字？会不会有个英文课可是会给你便臣中文字？

我写中文字不怎么好因为我是美国人。如果我打错的话，真是对不起。:D[/quote]

Well, if you really want to, you could use Babelfish.

[quote="quantum leap"][quote="mysmartmouth"]How does one get started typing in Chinese?[/quote]

If you have a mac that runs OS X, you can change the settings in International (in System Preferences) so that you can type Chinese (or any of a ton of other languages).  Once you get that set, there are five different methods to type Chinese you can choose from: Hanin, Cangjie, Jianyi, Day(Pro), and Pinyin.  I'm planning on taking Chinese this fall, so I really don't know what the differences between the five are, but I can type "ni hao" using the pinyin system :) . I'm not sure how to change the typesetting with Windows.[/quote]

haha
I'm in china, so when I want to type chinese, it's as easy as you to type  English :P
因为我在中国，所以我输入中文就像你们输入英文一样简单 :P

inputking.com is slow, so I don't use it too often. The sad thing is, it's the only site I know that does this.
inputking.com 比较慢, 所以不常常用.

First of all, you certainly have to know Chinese to be able to type it. If you are using the international/Chinese edition of OS Windows XP/2000/2003, there should be no problem to show Chinese properly on your screen. Otherwise, you'll have to install some language environment pack. For the input software, it depends on your need. I highly suggest this Google Chinese Pinyin Input Method:
http://tools.google.com/pinyin/
The major advantage is that this software only requires the knowledge of pinyin, which is well known by any educated Chinese speaker - the drawback is, since there are so many words with same pronunciation, you have to choose the correct one among several (sometimes it's quite annoying, so many people simply skip the process and type the wrong...). Other methods like Cangjie usually design an input system based on the strokes of characters and require you to memorize it. However, it has been proven difficult, and the worst thing is that each such input method has its own rules!

Well, if inputking.com is slow, use
http://userwww.service.emory.edu/~zxie3/ime.html
it's only pinyin though
Just copy & paste after you're done.
  :)

就用 http://userwww.service.emory.edu/~zxie3/ime.html

You can even use Google.com and then translate words to chinese from there!

您甚至可以使用Google.com和翻译单词，然后从那里向中国！

怎麼感覺用google翻譯過來的中文都很怪怪的

For Windows Vista: go to [u]Control Panel[/u] -> [u]Clock, Language, and Region[/u] -> [u]Change keyboards or other input methods[/u] -> [u]Change keyboards[/u] -> select the type you want


geometry
circumcircle
incenter
ratio
geometric transformation
reflection
inequalities

Hi Jaswinder. Welcome to AOPS.

This is a pretty gud geometry problem.

Let the altitudes of a triangle $ ABC$ meet the circumcircle at $ P$ , $ Q$ and $ R$. Prove that the orthocenter of triangle $ ABC$ is the incenter of triangle $ PQR$.

hint to first solution
[hide]homothesy for the feet of altitude with centre H and ratio 1/2
[/hide]

hint to solution 2
[hide]$ P,Q,R$ are reflections of $ H$ in the sides so AQ=AR $ \longrightarrow AP$ bisects angle $ QPR$ [/hide]

hint to solution 3
[hide] angle $ QPA$ = angle $ QBA$ = angle $ HBA$ [/hide]

got it pretty easily by simply proving that AP and similarly BP and CP are angle bisectors of angle A,B and C

the soln is pretty simple

am not in a mood to upload soln

will do it later

post some new harder problems and

[hide]plz don't post any hints they spoil everything

let me try my best and if am not able to solve then post the hints

[/hide]

[quote="Jaswinder"]plz don't post any hints they spoil everything

let me try my best and if am not able to solve then post the hints[/quote]
You needn't see the hints if you don't need them. That's the whole point of putting them inside hide tags.

[quote="gauravpatil"]hint to first solution
[hide]homothesy for the feet of altitude with centre H and ratio 1/2
[/hide]

hint to solution 2
[hide]$ P,Q,R$ are reflections of $ H$ in the sides so AQ=AR $ \longrightarrow AP$ bisects angle $ QPR$ [/hide]

[/quote]

hey gaurav where did u learn these modern techniques of solving geometry problems.....??

ok guys

now post some nice problems from inequalities

those which require immense logic and skills

plz keep posting something i always keep on waiting

plz reply and hey akashnil i agree with u that why they r kept in hidden tags but u should not even do that until and unless i ask for a few

so plz don't mind

 :(   :roll:  :wink:  :rotfl:

@anurag92
i wouldn't call them techniques maybe just observations
and i leanrt these quite informally (before i formally knew their names)
but if you want a book you may try geometric transformations 1&2 by I M Yaglom
(i didnt use these but i used the third part and found it beautiful)

@jaswinder if you dont mind it would be good to tell this beforehand


$ 16$ tennis players play together so that every $ 2$ players play at most one match against each other. Define a $ k$-circle to be a set of $ k$ players $ a_1,a_2,\cdots, a_k$ such that $ a_1$ beats $ a_2$, $ a_2$ beats $ a_3$, and so on up to $ a_k$ beats $ a_1$. It is known that for any 10 players, there is a way to order them to form a $ 10$-circle. Prove that for any $ 11$ players, an $ 11$-circle can be formed.

do u mean  each player   plays against atleast 9 players?

No. Given $ 10$ players, $ a_1, \cdots, a_{10}$, you can find a permutation $ \pi$ such that $ a_{\pi(i)}$ beat $ a_{\pi(i\plus{}1)}$ (indices taken mod $ 10$ - 11=1).

i dont think i  un derstand the question,please elaborate

[quote="MellowMelon"]Does the last sentence mean that [i]any[/i] $ 11$ players have a circular arrangement?

Here is an attempted restatement if that answer is yes:
[quote]$ 16$ tennis players play together so that every $ 2$ players play at most one match against each other. Define a $ k$-circle to be a set of $ k$ players $ a_1, a_2, \ldots a_k$ such that $ a_1$ beats $ a_2$, $ a_2$ beats $ a_3$, and so on up to $ a_k$ beats $ a_1$. It is known that any $ 10$ players, there is a way to order them to form a $ 10$-circle. Prove that for any $ 11$ players, an $ 11$-circle can be formed.[/quote][/quote]

Yes, thanks. I will edit my post.

suppose that players a1,a2,..,a10 are in a circular arrangement and each of them has defeated the player in his right.
you should prove that when we add the 11th player(a11) there are two players like a(i),a(i+1) that a(i) and a(i+1) are in the circular arrangement of a1,a2,..,a10 and a11 has won from a(i) and a(i+1) has won from a(11) then a1,a2,..a(i),a11,a(i+1),..,a10 is the new circular arrangement with 11 players.


calculus
derivative
function
integration
analytic geometry
graphing lines
slope

This is a simple and easy problem that tests you on your knowledge about the definitions. The reason I'm not posting it in the easy problem section is because you need to know basic derivatives.

Brief summary of derivatives. The derivative of a*x^n is a*n*x^n-1.

Try to spot the mistake in this proof of 1=2.

Let x be a non-zero number
x^2=x+x+...+x+x (x times)
2x=1+1+...+1+1 (x times) [take derivative of both sides]
2x=x (Divide by x)
2=1

Note: this problem is not actually as easy as it seems.. you may just think "you can't break up x:^2: like that", but that isn't the actual thing thats wrong with it..

Stephen (TripleM) posted this ``proof'' a long time ago somewhere on the forum. I won't repeat my argument saying what the flaw was here. Let's see if anyone else can find it.

[quote="SPiKEzZ"]The derivative of a*x^n is a*n*x^n-1[/quote]

Is the derivative anx^(n-1) or anx^n -1

Posted by MrMcGeek

Funny ! I had to read it twice to spot the mistake :

[hide]

x^2=x+x+...+x+x (x times) is true iff x is an integer !!! So you can't derive this expression, it doesn't have any meaning.

[/hide]

You're right that that x restricted to an integer is a problem, but that's not the fundamental problem.  For example, let's look at the first difference, the discrete analog of the derivative: The first difference of a function f (on integers) is defined to be the function g (on integers) defined by g(x) = f(x + 1) - f(x).  The first difference of x is 1, and the first difference of x2 is 2x + 1.  Applying the original argument shows that 2x + 1 = x, which makes no sense.  What's wrong with the argument?

Since x and x:^2: are constants, their derivatives should equal 0.  Therefore the problem is in the third line, i.e.
(x^2)'=(x+x+...+x+x (x times))'
0 = 0

First of all, if x is not an integer, it can't be expanded as x + x + ... + x, because that would be an non-integral number of terms.

The most obvious and critical fault is the following: when we differentiate x^2 = x + x + ... + x, we can only do it term by term if the number of terms is constant, which it isn't. (Why you would ever want to do it that way is another matter.) We have x^2 = f(x) = g(x)h(x) where g(x) = x and h(x) = x, so the derivative 2x = f'(x) = g'(x)h(x) + g(x)h'(x) = x + x = 2x.

let me butt in

x^2=x+x+...+x is true iff x is an integer
ie we are trying to differentate the function f(x)=x^2 for x belonging to Z (integer)
but f(x) is simply a bunch of unconnected lattice points, and we all know that a single point does not have a slope hence its derivitive does not exist

can that not simply be rectified by x:^2:=x+x+x+...+x+[x]. derivate=> 2x=x.

Ravi B. already proposed a nicer "fix" for this problem.

well, when you get to 2x=x, the only solution for x is 0, so obviously you cannot divide by x, right?

no, no -- the point is that if you start with *any* x, (if this proof were correct), you would be able to get that 2x = x.  That is, if we were to start with x = 3, this proof should give us that 3 = 6.

Yeah, but as was already stated, if x is a constant rather than a variable (Let x be a non-zero number), the derivative is 0, so that's the problem.

suppose it were a variable, then it would be a function that is discontinuous at anywhere.  so there is simply no derivitive (notice by first principles f'(x)=lim (x->a) (f(x)-f(a))/(x-a)  but for 0<x-a<1 f(x) simply does not exist, so there is no limit, and hence no derivitive!

You didn't read -- Ravi B. proposed a "fix" for that problem that you've ignored -- it works even on the discrete function x:^2:.

Yeah, several folks seemed to go down wrong paths on this problem.  I think teratomato was closest by saying "when we differentiate..., we can only do it term by term if the number of terms is constant".

Ravi's problem, although still related, is a different problem altogether.  According the problem definition, we have:
(x+1)^2=(x+1)+(x+1)+...+(x+1)  (x+1 terms)
x^2=x+x+x+...+x  (x terms)

looking at the difference
and it is obvious what is wrong when one claims the the difference between the RHSs is x
the two problem differs from when Ravi's problem is simply an error in applying the definition, which is different from the original in which the derivitive does not exist!  (well i guess you could still argue that the definition of the derivitive was applyed wrongly -- but thats besides the point)

If you want to keep the problem continuous (while avoiding the problem of defining a sum of x terms when x is not an integer), here's another reformulation.  The integral from 0 to x of x dt is x2.  Now differentiate both sides (with respect to x).  Because the derivative of x is 1, we get the integral from 0 to x of 1 dt is 2x.  But the integral from 0 to x of 1 dt is just x, so we have again proved that x = 2x.

Now what's wrong with the argument?

Well, the integral from 0 to x of x dt  is not x^2, its (x^2)/2.  Thus, you get x= d/dx x^2/2 which is x.


Let $ a,b,c$ be positive reals such that $ \frac{1}a\plus{}\frac{1}b\plus{}\frac{1}c\equal{}1$, prove that
\[ (a^2\minus{}ab\plus{}b^2)(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ca\plus{}a^2)\geq 3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\]

[quote="Pascual2005"]Let $ a,b,c$ be positive reals such that $ \frac {1}a \plus{} \frac {1}b \plus{} \frac {1}c \equal{} 1$, prove that
\[ (a^2 \minus{} ab \plus{} b^2)(b^2 \minus{} bc \plus{} c^2)(c^2 \minus{} ca \plus{} a^2)\geq 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\]
[/quote]
The following much more stronger inequality is true too.
Let $ a,b,c$ be positive reals such that $ \frac {1}a \plus{} \frac {1}b \plus{} \frac {1}c \equal{} 1.$ Prove that
\[ (a^2 \plus{}4ab \plus{} b^2)(b^2 \plus{}4 bc \plus{} c^2)(c^2 \plus{}4ca \plus{} a^2)\geq 648(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\]

$ LHS \geq (abc)^2\plus{}\sum_{cyc}abc^2(a\minus{}b)^2\geq (abc)^2\plus{}\sum_{cyc}c^2(a\minus{}b)^2$ as $ ab,bc,ca>1$ from the condition.
but this last expression is equal to the RHS


[url]http://www.elektrijada.com/[/url]
 Mnogobrojna ekipa iz Sarajeva se k Budvi ispaljuje kroz 3.5 sata!  :)

Have fun svi koji idete... i definiraj mnogobrojna. Je li stanete svi u jedan bus? :P

Pa mislim da ih ima $ 92\equal{}2\times \text{autobus}$  :lol: Hev fan, ju elektrik pipls...


geometry
3D geometry
sphere
trigonometry
geometry unsolved

Anyone knows how to prove the Ceva's theorem on sphere? If spherial triangle $ABC$ has $U,V,W$ on sides $BC, CA, AB$ then the greats circles $AU, BV,CW$ are concurrent if and only if $\frac{\sin AW}{\sin WB}\frac{\sin BU}{\sin UC}\frac{\sin CV}{\sin VA} = 1$ ???

its proof is exactly like ceva theorem...

if u want i can post it... ;)


modular arithmetic
pigeonhole principle
trigonometry

Show that for any positive integer $ n$, there exists a positive multiple of $ n$ that contains only the digits $ 7$ and $ 0$ in its base-$ 10$ representation.

Not sure on difficulty, will move if needed. I have a rough idea of the solution but haven't finished yet.

[hide="Hint"]Pick wisely $ n$ of such numbers and use pigeon principle[/hide]

[hide]We will show that, for any $ n$ there exists at least a positive multiple of $ n$ that contains only the digits [b]1[/b] and [b]0[/b], because then [b]7n[/b] contains the digits [b]7[/b] and [b]0[/b]

Let $ n$ be a positive integer. We define the sequence $ (a_k)$ such that: $ a_k \equal{} 10^k \pmod{n}$

The members of the sequence are infinite while the possible values of $ a_k$ are finite

So there is at least one $ r$ in modulo $ n$ such that $ r$ appears an infinite number of times in the sequence $ (a_k)$. Then we can 'combine' $ n$ distinct terms of them to make the desired multiple.

That is:

Let $ b_1, b_2, ... b_n$ be $ n$ [u]distinct[/u] natural numbers such that $ a_{b_m} \equal{} r, \ \ m \equal{} 1,2,...,n$

Let $ x \equal{} \sum_{m \equal{} 1}^n{10^{b_m} }$

Then $ x \equiv \sum_{m \equal{} 1}^n{10^{b_m} } \pmod n \equiv \sum_{m \equal{} 1}^n{a_{b_m}\pmod n} \equiv \sum_{m \equal{} 1}^n{r} \pmod n \equiv nr \pmod n \Rightarrow$

$ x \equiv 0 \pmod n$

[color=brown]P.S. So we are ok with the existence, I think that a more interesting question would be, which is the smallest multiple of [b]n[/b] with such attribute[/color][/hide]

[hide="Pigeonhole solution, thanks Hamster and not_trig"]
Choose the first $ n \plus{} 1$ members of the set $ \{0, 7,77,777,7777,\ldots\}$. Consider them $ \mod n$. By pigeonhole, there must be at least two with the same remainder $ \mod n$. Their positive difference will be divisible by $ n$ and will consist only of $ 7$s and $ 0$s, so for each $ n$ we have a multiple which is only $ 7$s and $ 0$s.
[/hide]


inequalities
induction
inequalities unsolved

I actually derived this inequality from a USA TST problem. (To solve the problem, it sufficed to prove the following.)

Let $f(x) = x^2 + \frac{1}{5}.$

Prove that

\[f(f(f(f(f(x))))) \geq x^2\] for all real $x.$

Denote $f^n(x)=a_n$,$a_0=x^2$,$a_{n=1}=(a_n)^2+1/5$
then we will prove that $a_{n+1}\ge a_n$ this follows by an induction  from
$a_{n+1}-a_{n}=(a_n)^2-(a_{n-1})^2$
Am I wrong? :?

You can't prove the base case:

$x^4 + \frac{1}{5} \geq x^2.$ (In fact, this inequality is false for $x = 0.6202.$)

I like this problem!

A nice solution is obtained by considering the outcome of $f$ applied four additional times to $x^2 + \frac{1}{5}$. It is easily shown that an application of $f$ decreases an input by at most $\frac{1}{20}$, since $f(x) - x = x^2 - x + \frac{1}{5} = \left(x - \frac{1}{2}\right)^2 - \frac{1}{20} \ge \frac{-1}{20}$. It follows immediately that $f(f(f(f(f(x))))) = f(f(f(f(x^2+\frac{1}{5})))) \ge x^2$. In fact, equality does not hold anywhere because this maximum decrease can be applied at most once.


There are $212$ points inside or on a given unit circle. Prove that there are at least $2001$ pairs of points having distances at most $1$.

Partition the circle in six slices. If two points are in the same slice, then their distance is $\le$ 1.

Let $a_{1},a_{2},\ldots,a_{6}$ the number of points in each slice.
Then there are at least:
${a_{1}\choose 2}+{a_{2}\choose 2}+\ldots+{a_{6}\choose 2}$ good pairs.
We have the condition $a_{1}+\ldots+a_{6}= 212$.

Since: $(a-1)(a-2)+b(b+1)-a(a-1)-b(b+1) = 2(b-a)+2$, if we move two $a_{i}$ closer, our sum doesn't increase.
The sum is minimized if there are not two $a_{i}$ that differ 2 or more.
$212 = 6\cdot 35+2$.
Then: ${a_{1}\choose 2}+{a_{2}\choose 2}+\ldots+{a_{6}\choose 2}\ge 6 \cdot{35 \choose 2}= 6 \cdot 595 > 2001$

Hope it works  :)

[quote="edriv"]

Since: $(a-1)(a-2)+b(b+1)-a(a-1)-b(b+1) = 2(b-a)+2$, if we move two $a_{i}$ closer, our sum doesn't increase.
[/quote]
It must is  $(a-1)(a-2)+b(b+1)-a(a-1)-(b-1)b = 2(b-a)+2$, i think so :)


percent
search
probability
modular arithmetic
LaTeX
number theory
relatively prime

On her last two mathematics tests, Brenda’s score increased from 80% correct to 88% correct. What was the percent of change for these two tests?

There are 12 husband-wife couples at a party. Each person shakes hands with everyone except their spouse. How many handshakes occur between two people of opposite gender?

The 9-digit number abbababa3 is a multiple of 99 for some pair of digits a and b. What is b-a?

How many positive integers less than 501 are relatively prime with 42?

Thanks in Advance

[hide="1"]
I'm pretty sure that all they want is

$ \frac{88\%}{80\%}\minus{}1\equal{}0.1\equal{}\boxed{10\%}$.
[/hide]
[hide="2"]
There are $ \dbinom{12}2\equal{}66$ pairs of couples.

In each pair, there will be $ 2\cdot2\equal{}4$ handshakes (the husband of couple 1 and the wife of couple 1 each shake the two people of the other couples' hands), so our answer is

$ 66\cdot4\equal{}\boxed{264}$.
[/hide]
[hide="3"]
This has been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=858389888&t=242142]here[/url].
[/hide]
[hide="4"]
Since $ 42\equal{}2\cdot3\cdot7$, a number relatively prime to this must NOT be a multiple of $ 2$, $ 3$, or $ 7$.

If a number is a multiple of $ 42$, then we can say that the probability it will not be a multiple of $ 2$ is $ 1\minus{}\frac12\equal{}\frac12$, not a multiple of $ 3$ is $ 1\minus{}\frac13\equal{}\frac23$, and not a multiple of $ 7$ is $ 1\minus{}\frac17\equal{}\frac67$.

$ 504\equal{}42\cdot12$, so the number of positive integers less than $ 504$ which are relatively prime to $ 42$ is $ \frac12\cdot\frac23\cdot\frac67\cdot504\equal{}144$.

However, we must exclude any relatively prime numbers to $ 42$ in the set $ \{502,503,504\}$ (since the original problem states less than $ 501$).

$ 503$ is the only one that is relatively prime to $ 42$ that is included in the $ 144$ above ($ 502,504$ are even, hence not relatively prime to $ 42$), and our final answer is

$ 144\minus{}1\equal{}\boxed{143}$.
[/hide]

I was confused on 1 to either put 10% or 110%

Usually percent of change refers to percent increase or percent decrease, not percent of.

Interesting I solved it like this;

(88% - 80%)/1

A problem similiar to #3:

a1a22a333a4444a is divisible by 101, find a.

I created the problem :)

[quote="math154"]I don't understand what you are trying to do in that.[/quote]

Basically 88% is .88, and 80% is .80, so the percent change is the difference of these two percents all over one.

.88 - .80 is .08, and .08/1.00 is the same as 8/100, so it is 8%.

The answer is not $ 8\%$--refer to my post above: $ \frac {88\% \minus{} 80\%}{80\%} \equal{} \boxed{10\%}$ is. :wink:

What you did is find percent increase relative to a whole, which is not what you want--reread the problem.

Let $ x \equal{} \overline{a_na_{n \minus{} 1}\ldots a_2a_1a_0}_{10}$ such that we want to find $ x\pmod{101}$.

The key is noting that $ 10^{2k}\equiv100^k\equiv( \minus{} 1)^k\pmod{101}$.

If the number of digits is even (which means that $ n$ is odd), then we have

$ \overline{a_na_{n \minus{} 1}\ldots a_2a_1a_0} \equal{} a_n10^n \plus{} a_{n \minus{} 1}10^{n \minus{} 1} \plus{} \cdots \plus{} a_210^2 \plus{} a_110^1 \plus{} a_010^0$

$ \equal{} (a_n10^n \plus{} a_{n \minus{} 1}10^{n \minus{} 1}) \plus{} (a_{n \minus{} 2}10^{n \minus{} 2} \plus{} a_{n \minus{} 3}10^{n \minus{} 3})$

$ \plus{} \cdots \plus{} (a_310^3 \plus{} a_210^2) \plus{} (a_110^1 \plus{} a_010^0)$

$ \equal{} (10a_n \plus{} a_{n \minus{} 1})10^{n \minus{} 1} \plus{} (10a_{n \minus{} 2} \plus{} a_{n \minus{} 3})10^{n \minus{} 3} \plus{} \cdots \plus{} (10a_3 \plus{} a_2)10^2 \plus{} (10a_1 \plus{} a_0)$

$ \equiv \pm (10a_n \plus{} a_{n \minus{} 1}) \mp (10a_{n \minus{} 2} \plus{} a_{n \minus{} 3}) \pm \cdots \minus{} (10a_3 \plus{} a_2) \plus{} (10a_1 \plus{} a_0)$

$ \equiv \pm \overline{a_na_{n \minus{} 1}} \mp \overline{a_{n \minus{} 2}a_{n \minus{} 3}} \pm \cdots \minus{} \overline{a_3a_2} \plus{} \overline{a_1a_0}\pmod{101}$.

If the number of digits is odd (so $ n$ is even), then we have

$ \overline{a_na_{n \minus{} 1}\ldots a_2a_1a_0} \equal{} a_n10^n \plus{} a_{n \minus{} 1}10^{n \minus{} 1}$

$ \plus{} \cdots \plus{} a_210^2 \plus{} a_110^1 \plus{} a_010^0$

$ \equal{} a_n10^n \plus{} (a_{n \minus{} 1}10^{n \minus{} 1} \plus{} a_{n \minus{} 2}10^{n \minus{} 2}) \plus{} (a_{n \minus{} 3}10^{n \minus{} 3} \plus{} a_{n \minus{} 4}10^{n \minus{} 4})$

$ \plus{} \cdots \plus{} (a_310^3 \plus{} a_210^2) \plus{} (a_110^1 \plus{} a_010^0)$

$ \equal{} 10a_n10^{n \minus{} 1} \plus{} (10a_{n \minus{} 1} \plus{} a_{n \minus{} 2})10^{n \minus{} 2} \plus{} (10a_{n \minus{} 3} \plus{} a_{n \minus{} 4})10^{n \minus{} 4}$

$ \plus{} \cdots \plus{} (10a_3 \plus{} a_2)10^2 \plus{} (10a_1 \plus{} a_0)$

$ \equiv \pm10a_n \mp(10a_{n \minus{} 1} \plus{} a_{n \minus{} 2}) \pm (10a_{n \minus{} 3} \plus{} a_{n \minus{} 4})$

$ \mp \cdots \minus{} (10a_3 \plus{} a_2) \plus{} (10a_1 \plus{} a_0)$

$ \equiv \pm \overline{a_n0} \mp \overline{a_{n \minus{} 1}a_{n \minus{} 2}} \pm \overline{a_{n \minus{} 3}a_{n \minus{} 4}}\mp \cdots \minus{} \overline{a_3a_2} \plus{} \overline{a_1a_0}\pmod{101}$.

[quote="math154"]
There are $ \dbinom{12}2 \equal{} 66$ pairs of couples.

In each pair, there will be $ 2\cdot2 \equal{} 4$ handshakes (the husband of couple 1 and the wife of couple 1 each shake the two people of the other couples' hands), so our answer is

$ 66\cdot4 \equal{} \boxed{264}$.
[/quote]

But the problem said between people of opposite gender, so in each pair of couples, only two handshakes count, so the answer is $ \boxed{132}$.

Before anyone starts debating the answer, the answer is 132.  (I have answers  :P )

[quote="mathemagician1729"]Sorry but I had a question about how you found the divisibility rule for 101.[/quote]

Basically you just kind of want to figure out how to use the numbers as close to 0 as possible.

For example;

$ 49^15 (mod 6)$
$ 256 \equal{} 7^2$
$ (7^2)^15 (mod 6)$
$ 7^30 (mod 6)$
$ 7^(30 \plus{} 0) (mod 6)$
$ 7^(30/6  \plus{} 0) (mod 6)$
$ 7^5 (mod 6)$
$ 1^5 (mod 6)$
$ 1 (mod 6)$

Sorrry if it doesn't look right. If there is somebody that understands this then please do it in LaTeX the correct way.

[quote="Wickedestjr"][quote="mathemagician1729"]Sorry but I had a question about how you found the divisibility rule for 101.[/quote]

Basically you just kind of want to figure out how to use the numbers as close to 0 as possible.

For example;

$ 49^{15} (mod 6)$
$ 256 \equal{} 7^2$
$ (7^2)^{15} (mod 6)$
$ 7^{30} (mod 6)$
$ 7^{(30 \plus{} 0)} (mod 6)$
$ 7^{(\frac{30}{6} \plus{} 0)} (mod 6)$
$ 7^5 (mod 6)$
$ 1^5 (mod 6)$
$ 1 (mod 6)$

Sorrry if it doesn't look right. If there is somebody that understands this then please do it in LaTeX the correct way.[/quote]

[quote="Mewto55555"][quote="Wickedestjr"][quote="mathemagician1729"]Sorry but I had a question about how you found the divisibility rule for 101.[/quote]

Basically you just kind of want to figure out how to use the numbers as close to 0 as possible.

For example;

$ 49^{15} \pmod{6}$
$ 256  \equiv 7^2$
$ (7^2)^{15} \pmod{6}$
$ 7^{30} \pmod{6}$
$ 7^{(30 \plus{} 0)} \pmod{6}$
$ 7^{(\frac {30}{6} \plus{} 0)}\pmod{6}$
$ 7^5 \pmod{6}$
$ 1^5 \pmod{6}$
$ 1 \pmod{6}$

Sorrry if it doesn't look right. If there is somebody that understands this then please do it in LaTeX the correct way.[/quote][/quote]


سلام من دنبال يك جزوه خوب براي نامساوي ها هستم (در اينترنت) لطفا كمكم كنيد


Given a piece of paper in the shape of a square.
You are only allowed to fold the paper, without any other tools like ruler or compasses, can you creat square with the following requirement?
1. the new square has half the area of the original one.
2. the new square has one fifth the area of the original one.
3. the new square has one third the area of the original one.
The above three tasks are of increasing levels of difficulty. See if you can seek the method?
Have fun!  :)

[hide="1"]fold diagonals. Intersection is center. Fold corners to center. Creases from the last step make a square with half the area of the original[/hide]

For parts 2 and 3:

[hide]2:
Fold it into a square with 1/2 the area as described above. Now we need a square with 2/3 the area of this in order to get a square that had an area of 1/3 of the beginning.
If the area of this square is $x^2$, then the are of the next square will be $(2/3)x^2$. Place the squares so that they have the same center. then rotate the smaller one so that each corner touches an edges of the larger we get a right triangle with sides:
$(x-a,a,x\sqrt{\frac{2}{3}})$
Using the pythag theorem...
$x^2-2ax+a^2+a^2=\frac{2x^2}{3}$
$x^2-6ax+6a^2=0$
Solving the quadratic for $a$...
$a=\displaystyle\frac{6x\pm\sqrt{36x^2-24x^2}}{12}$
$a=\displaystyle\frac{6x\pm\sqrt{12x^2}}{12}$
$a=\displaystyle\frac{x}{2}+\displaystyle\frac{x\sqrt{3}}{6}$
Since all quantities are constructable, construct the given length for $a$ given $x$.
Then fold the edges in as appropriate.

3:
Starting with this square, do the same for the 1/5.[/hide]


combinatorics
geometry
point set
distance
combinatorial geometry

Let $n$ be a natural number greater than 2. $l$ is a line on a plane. There are $n$ distinct points $P_1$, $P_2$, …, $P_n$ on $l$. Let the product of distances between $P_i$ and the other $n-1$ points be $d_i$ ($i = 1, 2,$ …, $n$). There exists a point $Q$, which does not lie on $l$, on the plane. Let the distance from $Q$ to $P_i$ be $C_i$ ($i = 1, 2,$ …, $n$). Find $S_n = \sum_{i = 1}^{n} (-1)^{n-i} \frac{c_i^2}{d_i}$.

[b]W.L.O.G[/b] we can suppose that points $ P_{1},\ P_{2}, \dots, \ P_{n}$ are arranged in this order on line $ l.$

For $ n \equal{} 3$ we get that $ \boxed{S_{3} \equal{} 1}$ (just applying [b][color=blue]Stewart's relation[/color][/b]).

Now let's try to compute $ S_{4}.$

Applying the anterior case for points $ (P_{1}, P_{2}, P_{4})$ and $ (P_{1}, P_{3}, P_{4})$ we'll obtain the following system:

\[ \left \{\begin{array}{ccc} \frac {c_{1}^2\cdot P_{1}P_{3}}{d_{1}} \minus{} \frac {c_{2}^2\cdot P_{2}P_{3}}{d_{2}} \plus{} \frac {c_{4}^2\cdot P_{4}P_{3}}{d_{4}} \equal{} 1 \\
 \\
\frac {c_{1}^2\cdot P_{1}P_{2}}{d_{1}} \minus{} \frac {c_{3}^2\cdot P_{3}P_{2}}{d_{3}} \plus{} \frac {c_{4}^2\cdot P_{4}P_{2}}{d_{4}} \equal{} 1 \end{array}\right.
\]
By substracting the first relation from the second one we'll obtain:
\[ \minus{} P_{2}P_{3}\cdot\frac {c_{1}^2}{d_{1}} \plus{} P_{2}P_{3}\cdot\frac {c_{2}^2}{d_{2}} \minus{} P_{2}P_{3}\cdot\frac {c_{3}^2}{d_{3}} \plus{} P_{2}P_{3}\cdot\frac {c_{4}^2}{d_{4}} \equal{} 0
\]
what rewrittes as
\[ \minus{} \frac {c_{1}^2}{d_{1}} \plus{} \frac {c_{2}^2}{d_{2}} \minus{} \frac {c_{3}^2}{d_{3}} \plus{} \frac {c_{4}^2}{d_{4}} \equal{} 0.
\]
So $ \boxed{S_{4} \equal{} 0}$ We claim that $ S_{n} \equal{} 0, \ (\forall) n\geq 4.$ The proof is by induction on $ n.$ For $ n \equal{} 4$ it's true. We just need to prove:

\[ \boxed{S_{n \minus{} 1} \equal{} 0\Longrightarrow S_{n} \equal{} 0}
\]
Let's say that $ n$ is [u]odd[/u]. So $ n \minus{} 1$ is even. Applying the previous step for points $ (P_{1}, P_{2}, \dots , P_{n \minus{} 2}, P_{n})$ and $ (P_{1}, P_{2}, \dots, P_{n \minus{} 3}, P_{n \minus{} 1}, P_{n})$ we'll obtain the following system:

\[ \left \{\begin{array}{lcl} \minus{} P_{1}P_{n \minus{} 1}\cdot\frac {c_{1}^2}{d_{1}} \plus{} P_{2}P_{n \minus{} 1}\cdot\frac {c_{2}^2}{d_{2}} \minus{} \dots \plus{} P_{n \minus{} 3}P_{n \minus{} 1}\cdot\frac {c_{n \minus{} 3}^2}{d_{n \minus{} 3}} \minus{} \boxed{P_{n \minus{} 2}P_{n \minus{} 1}}\cdot\frac {c_{n \minus{} 2}^2}{d_{n \minus{} 2}} \plus{} P_{n}P_{n \minus{} 1}\cdot\frac {c_{n}^2}{d_{n}} \equal{} 0 \\
 \\
\minus{} P_{1}P_{n \minus{} 2}\cdot\frac {c_{1}^2}{d_{1}} \plus{} P_{2}P_{n \minus{} 2}\cdot\frac {c_{2}^2}{d_{2}} \minus{} \dots \plus{} P_{n \minus{} 3}P_{n \minus{} 2}\cdot\frac {c_{n \minus{} 3}^2}{d_{n \minus{} 3}} \minus{} \boxed{P_{n \minus{} 1}P_{n \minus{} 2}}\cdot\frac {c_{n \minus{} 1}^2}{d_{n \minus{} 1}} \plus{} P_{n}P_{n \minus{} 2}\cdot\frac {c_{n}^2}{d_{n}} \equal{} 0 \end{array}\right.
\]
By substracting the first relation from the second one we'll obtain:

$ P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{1}^2}{d_{1}} \minus{} P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{2}^2}{d_{2}} \plus{} \dots \minus{} P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{n \minus{} 3}^2}{d_{n \minus{} 3}} \plus{} P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{n \minus{} 2}^2}{d_{n \minus{} 2}} \minus{}$ $ P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{n \minus{} 1}^2}{d_{n \minus{} 1}} \plus{} P_{n \minus{} 1}P_{n \minus{} 2}\cdot\frac {c_{n}^2}{d_{n}} \equal{} 0$

what yields: $ S_{n} \equal{} 0.$ If $ n$ is [u]even[/u] we'll proceed as above. So $ \boxed{S_{n} \equal{} 0, \ (\forall)n\geq 4}$


In case you're too lazy to look for the question in the archives:

For each positive integer n, Let S(n) denote the sum of the digits of n. For how many values of n is n+S(n)+S(S(n)) = 2007?

(A) 1     (B) 2    (C) 3    (D) 4    (E) 5

How would you solve this?

I can get the answer by brute forcing it, but it isn't very efficient. I guess getting that n is a multiple of 3 helps, but is there a quick and easy way to solve?

This Math Jam transcript has a full discussion:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=161


group theory
abstract algebra
superior algebra
superior algebra theorems

Here is a beautiful problem:
  Find the minimal value of $m$ such that $S_n$ can be embedded in $A_m$. The notations are the standard ones for permutations.

$S_1$ can be embedded in $A_2$. For all $n>1$, $S_n$ can be embedded in $A_{n+2}$ but not in $A_{n+1}$.

The general construction for embedding $S_n$ in $A_{n+2}$: use the two new elements to append a transposition if the permutation is odd, and leave them fived if the permutation is even.

Now we deal with the question of minimality.
We need a few special cases first. For $n=2$, 2 does not divide 3, so we are done. For $n=3$, there are three transpositions in $S_3$, no two of which commute, and there are three elements of order 2 in $A_4$, all of which commute with each other. Obviously, the embedding is impossible.

For $n\ge4$, $A_{n+1}$ is simple, and the desired embedding would have index $\frac{n+1}2$ in $A_{n+1}$.
Suppose $[G: H]=k$. Since $a^{-1}Ha=b^{-1}Hb$ if $Ha=Hb$, $H$ has at most $k$ conjugates in $G$. The intersection of all of these conjugates is obviously a normal subgroup of $G$. Now, since for any two subgroups $X,Y$ $[G: X\cap Y]\le [G: X][G: Y]$, $[G: \cap a^{-1}Ha]\le k^k$.
If $G$ is simple, this normal subgroup has order 1, and $|G|\le k^k$.
Since $\frac{n!}2\ge\left(\frac n2\right)^{\frac n2}$ for $n\ge3$, $A_n$ has no subgroup of index $\frac n2$ for $n\ge5$ and we are done.

Thank you, Jmerry. I knew it wasn't something trivial about this problem. :)

You can also use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724]this problem[/url]. There is a proof in post number 2 that if $[G:H]=t$, then there is a subgroup $H_1$ of $H$ normal in $G$ s.t. $G/H_1$ is imbedded in $S_t$.

This would mean that $A_{n+1}$ is imbedded in $S_{\frac{n+1}2}$, or $\frac{(n+1)!}2\le\left(\frac{n+1}2\right)!$, which yields a contradiction, but I guess it's the exact same argument.

A related question is ¿are all the imbeddings of Sn in Sm trivials?
If m=n the answer is Yes , but a dont no the general answer.

Note: By trivial i mean that if we let Sn act on m letters then each orbit have n or 2 elements.


Aici sunt problemele din shortlist RMO 2006. Le-am scanat din suplimentul Gazetei Matematice.

[quote="stancioiu sorin"]Am si eu o intrebare.Cei care au facut abonament la Gazeta Matematica nu primesc si ei Suplimentul pentru ONM?Ca eu inca nu l-am primit. :([/quote]
Cred ca nu :(

[quote="stancioiu sorin"]Ce pacat! :( Dar Farkas, stii sigur ca nu primesc suplimentul?[/quote]
tu stii ceva sigur????

Eu am facut abonament la GM din iarna, dar vad ca n-am primit nimik pana acum.Mai astept...daca nu sun la Redactie sa vad care-i treaba. :)

Da.Am vorbit cu cei de la Gazeta prin mail si intr-adevar, nu putem primi Suplimentul. :(


trigonometry
function
algebra
domain
conics
parabola
complex analysis

Suppose we are given a point $P$ in an Argand diagram which is represented by complex number $z$ , moving anticlockwise aloong the circle $|z|=2$ from point of complex number $2$ to $2i$ . If $w=\frac{z+1}{z+2}$ , identify the path describe by $Q$ which represents the complex number $w$ .

Can we identify locus of $Q$ without actually compute out using $z=x+yi$ etc ... ?

It's much easier if we use the polar form:

If $z=2e^{i\phi}, 0\leqslant\phi\leqslant{\pi\over 2}$, then

\begin{eqnarray*}w &=& 1-{1\over z+2}\\ &=& 1-\frac{1}{2(1+e^{i\phi})}\\ &=& 1-\frac{1}{2e^{i\phi/2}(e^{-i\phi/2}+e^{i\phi/2})}\\ &=& 1-\frac{e^{-i\phi/2}}{2\cdot 2\cos{\phi\over 2}}\\ &=& 1-{1\over 4}\left(1-i\tan{{\phi\over 2}\right)\\ &=&{3\over 4}+i{1\over 4}\tan{\phi\over 2}}\end{eqnarray*}

Therefore $Q$ is the segment on the line $x={3\over 4}$, from $y_{1}={1\over 4}\tan{\phi\over 2}\bigg|_{\phi=0}=0$ to $y_{2}={1\over 4}\tan{\phi\over 2}\bigg|_{\phi=\pi/2}={1\over 4}$

Thanks for replying Farenhajt , polar form is really easier but what if it is something like this :

$P$ has complex number $z$ defined as $|z-1|=2|z-2|$ while $Q$ has complex number $w$ defined as $z=\frac{4w+5}{2w+7}$ . And we wish to find the locus of $Q$ . Other than compute it out , do we have any other method?

How general you want it to be? I mean, you can always spice it up, like

$P: |z^{2}+\bar z-2|=|\bar z^{10}-1|; z=4w^{5}-5w+1$. Find locus of $w$.

I'm not really fluent in complex analysis, so I can't tell you immediately what's the general effect of a rational function (like in your example) on domains, but I [i]think[/i] it has something to do with inversion (don't take me for granted). Therefore, you have to find what's the inverted image of a parabola (i.e. $P$ in your example).

But in general, I don't know how to solve for loci other than to set the equations up.

I'm terribly sorry  :blush:  - in your last example $P$ is not a parabola but a circle.


geometry
combinatorics open
combinatorics

If a rectangular board can be covered by some $ 2\times1$ dominos, it can also be covered if the dominos are standing in the same way. We can easily see that this is true for any $ p\times1$ dominos, where $ p$ is prime.

Find all pairs of positive integers $ (a,b)$ such that any rectangular board that can be covered by some $ a\times b$ dominos can also be covered if the dominos are standing in the same way.

The answer is: all pairs $ (a,b)$ such that $ a|b$ or $ b|a$.

It's easy to see that if neither $ a|b$ nor $ b|a$, then there are boards that can be covered by $ a \times b$ dominoes but not using the trivial method of having them all face the same way. In fact, consider a board with dimensions $ (a\plus{}b) \times ab$. This board can always be tiled by $ a \times b$ dominoes by stacking $ b$ dominoes along their $ a$ side (length $ ab$, height $ b$) and on top of that stacking $ a$ dominoes along their $ b$ side (length $ ab$ again, height $ a$). So this board can be tiled, but if you could tile it with all dominoes oriented in the same way then you'd have $ a\plus{}b$ divisible by $ a$ or $ b$, which means that either $ a|b$ or $ b|a$.

We now need to show that if $ a|b$ then any board coverable by $ a \times b$ dominoes is so coverable by having them all face the same way. Letting $ b \equal{} ka$ we note that the $ a \times b$ domino is just a blow-up by a factor of $ a$ of the $ 1 \times k$ domino, so any tiling with $ a \times b$ dominos is simply a tiling with $ 1 \times k$ dominoes where we change the scale by a factor of $ a$. So, we only need to prove the statement for $ 1 \times k$ dominoes.

So given an $ m \times n$ board and a $ 1 \times k$ domino, we claim that the domino can tile the board iff either $ k|m$ or $ k|n$, in which case the domino can actually tile the board in the trivial fashion. This is obvious when $ k$ is prime because of area; the general case requires just a slightly more refined counting argument.

One way to show this is to color the board with $ k$ colors by assigning the color $ (i\plus{}j \bmod{k})$ to the square at row $ i$, column $ j$. For instance, with a $ 6 \times 6$ board and $ k\equal{}4$ you color like this:

012301
123012
230123
301230
012301
123012

It's clear that any horizontal or vertical sequence of cells of length $ k$ contains each color exactly once. Therefore, if a $ 1 \times k$ domino tiles the board, we must have each color appearing exactly the same number of times. In the $ 6 \times 6$ example above, 0 and 2 actually appear 9 times (as they should), but 1 appears 10 times and 3 appears only 8. That's why you cannot tile a $ 6 \times 6$ board with a $ 1 \times 4$ domino.

In general, the last color $ k\minus{}1$ always appears too few times if both $ m$ and $ n$ are not divisible by $ k$. Writing this down formally is a bit tedious but it's a fairly straightforward calculation.


Joyce wishes to share six identical pieces of licorice equally among four people. What is the minimum number of pieces that must be cut?

Well we want to divide the licorice into four groups of 1 1/2 pieces each.
First give the whole pieces (4 of them) out. Then cut the last two pieces into four halves, then distribute. Total: 2 pieces.


I have been offered to be in part of a studying abroad program. Our state math leader (from Georgia) invited me to go to Singapore this summer to study abroad for 2-3 weeks. It is not scholarship based and I have to pay for transportation and about 600 dollars a week there. I will be attending the prestigious National University of Singapore High School of Mathematics and Science. I was wondering does this aid me in college admissions at all? I am thinking about attending this my junior year. Does this impress colleges even though it is not scholarship based? I was hoping this will be something that will help me "stand out" more in the application pool. Also, I don't think I will be earning any credits. Thanks in advance.

It depends what the program is.  There are many programs that send out mass mailings to high school students all over the country and say that if you pay a couple of thousand dollars, you can participate in some "great" program.  These are wealth-based and don't mean anything.

You should look into what this program is.  But maybe you really want to go to Singapore in which case it doesn't really matter, does it?


I have already figured out the solution what it should be but i would like a second opinion. the problem is are there any integers x for which x==3(mod20) x==3(mod35) and x==13(mod30).

30q+13==3(mod35)
30q==-10(mod35)
30==5(mod35)
5q==-10(mod35)
q==5(mod7)
x=30(7r+5)+13
x=210r+136

then
210r+163==3(mod20)
210r==-160(mod20)
210==190(mod20)
190r==-160(mod20)
190r==-160==380(mod20)
r==2(mod10)
x=210(2+10s)+63

x=2100x+583 and that is the solution that satisfies all three congruences

You should use the Chinese Remainder Theorem. For the first two congruences it says that you have at most one solution modulo $lcm(20,35)=140$. It is clear that the solution is $3 mod 140$. Then you have again two congruences: $x\equiv 3(140)$ and $x\equiv13(30)$. With the same theorem we have again at most one solution modulo $lcm(30,140)=420$. Because of the first congruence modulo $140$ we have to check the numbers $3,143,283,...$ for the second congruence. As you can see the number we are searching for is $283$. So the answer is: all numbers of the form $420\cdot n+283$.

Hope it helps :)

Misha


the question is lines that meet at a 90 degree angle is a....

but i already used complemtary

this doesnt belong here but im pretty dsure is is perpendicular

[quote="bubka"]How many letters? :maybe:
'Normal' maybe?[/quote]
Normal is for vectors, isn't it? Or is there no difference between the two?

most definitly perpendicular. This must be some hardcore crossword puzzle if you're posting on forums to find the answers. :D   Where did you get it? :huh:


[quote="djimenez"]What is the following number in the sequence,

1, 2, 6, 12, 60, 60, 420, 840, ...[/quote]

i think it should be 12,60, then 120, not another 60.

[quote="xxreddevilzxx"]anyways, the number following is.... [hide]7560[/hide][/quote]No, it is not, but I think you are near of the idea! If you don't see why there are two consecutive 60's, think about that, it could give you an idea. You can think that $a_5=a_6=60$, may be the order of the terms will give you a hint! If there is no new progress in twelve hours, I will give another hint!

[quote="xxreddevilzxx"]i dunno... another way... but i dont thin this is right
[hide]10080[/hide]

ugghghghgh this liooks so easy and familiar, but i just dont get it.[/quote]
Sorry my friend, but it is still incorrect and you are getting farther away!

[quote="Hikaru79"][hide="Is it..."] 2520?[/hide][/quote]
Yes, it is. Thanks for not giving the general pattern yet. OK, more terms before some post the general rule?

[quote="NoSoupForYou"]1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 22720, 22720, 360360 ...

[hide="And the pattern is..."]The $n$th term of the sequence is the lowest natural number divisible by all integers from $1$ to $n$.[/hide][/quote]
Good!

How do you guys solve it?:?
Even the answer sounds complicated.

It's obvious from the fact that $60$ is repeated. That should be a dead giveaway.

[quote="amirhtlusa"]if you just look at the numbers it's kind of obvious that the $n$'th term is the $LCD (1,2,3,4,...,n)$[/quote]Yep! it is!


floor function
induction
real analysis
inequalities unsolved
inequalities

Prove that {$\sqrt{1}$}+{$\sqrt{2}$}+...+{$\sqrt{n^{2}}$} < $\frac{n^{2}-1}{2}$ where n is natural number

The following approach should work:

1. Estimate $\sqrt{1}+\ldots+\sqrt{n^{2}}$ by a Riemann-sum approach
 by the corresponding integral.
2. Calculate $\lfloor \sqrt{1}\rfloor+\ldots+\lfloor \sqrt{n^{2}}\rfloor$ explicitely.

Hopefully this gives an estimate for the considered sum, which is good enough.


Let a,b,c. abc=1.Prove that:
$ \frac{{a \plus{} b}}{{a^4  \plus{} b^4  \plus{} abc^2 }} \plus{} \frac{{a \plus{} c}}{{a^4  \plus{} c^4  \plus{} acb^2 }} \plus{} \frac{{b \plus{} c}}{{b^4  \plus{} c^4  \plus{} bca^2 }} \le 2$

Dear friend Train thai Hung, your inequality {8th} is  just sharing a striking resemblence with this, because the same identities or lemmas are used that is:
For positive reals $ x,y,z$,
$ x^4 \plus{} y^4 \ge xy(x^2 \plus{} y^2)$ and $ \sum x^2 \ge \sum xy$. I would like it if you give some different ones.
So waiting for a better one from you. Sorry did not see the 10th. Working on it. It is nice. Thanks.


Euler
real analysis
real analysis unsolved

I post it just to see some ideas, because obviously noone will manage to prove it (at least, IMHO):
    Let $f(n)$ be the number of prime divisors of a number $n$ counted with their multiplicities. Then the series of general term $\frac{(-1)^{f(n)}}{n}$ converges and has sum 0.

The infinite product $\prod_{p\text{ prime}}\left(1+\frac1p\right)^{-1}$ diverges to zero.

$\left(1+\frac1p\right)^{-1}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{p^{k}}$

Formally, $\prod_{p\text{ prime}}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{p^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{f(n)}}n$

So far, this looks good - but I don't regard this as a proof. The series is not absolutely convergent. If it converges at all, it does so conditionally, hence it is highly sensitive to rearrangements. We must explore what the partial sums look like and which terms are included.

Indeed, this is exactly what I did and apparently what Euler did before all of us. But apparently making this rigourous is harder than the prime number theorem!!!


inequalities
logarithms
function
LaTeX
inequalities proposed

[quote="galois01"]Let $ a,b,c > 0$ and $ abc \equal{} 1$.Prove that
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\geq\frac {9}{2}$[/quote]
Proof:
Using am-gm:
$ \frac {3a}{4} \plus{} \frac {3b}{4} \plus{} \frac {3c}{4}\geq3\sqrt [3]{\frac {27abc}{64}} \equal{} \frac {9}{4}$.
$ \frac {a \plus{} 1}{4} \plus{} \frac {1}{a \plus{} 1}\geq 1$.
$ \frac {b \plus{} 1}{4} \plus{} \frac {1}{b \plus{} 1}\geq 1$
$ \frac {c \plus{} 1}{4} \plus{} \frac {1}{c \plus{} 1}\geq 1$
Adding up these inequalities we obtain that:
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {3}{4}\geq \frac {9}{4} \plus{} 3$,hence
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\geq \frac {9}{2}$,so we are done :)

[quote="galois01"]Let $ a,b,c > 0$ and $ abc \equal{} 1$.Prove that
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\geq\frac {9}{2}$[/quote]
The following inequalities are true too:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1$.Prove that
1)
\[ a \plus{} b \plus{} c \plus{} 14\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right)\geq17
\]
2)
\[ ab \plus{} ac \plus{} bc \plus{} 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right)\geq30
\]
Here the number $ 27$ is the best. :wink:

The following inequality is true too:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[ a \plus{} b \plus{} c \plus{} 8\left(\frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\right)\geq15\]

[quote="galois01"]Let $ a,b,c > 0$ and $ abc \equal{} 1$.Prove that
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\geq\frac {9}{2}$[/quote]

$ \sum_{cyc}(\frac {1}{a \plus{} 1} \plus{} a) \geq \frac {9}{2}$

$ \iff$   $ \sum_{cyc}(\frac {a^2 \plus{} a \plus{} 1}{a \plus{} 1})\geq \frac {9}{2}$

since $ x: \rightarrow \frac {x^2 \plus{} x \plus{} 1}{x \plus{} 1}$ is convex , we apply jensen inequality :

$ f(a) \plus{} f(b) \plus{} f(c) \geq 3f(\frac {a \plus{} b \plus{} c}{3}) \equal{} 3(\frac {t^2 \plus{} t \plus{} 1}{t \plus{} 1})$

with $ t \equal{} \frac {a \plus{} b \plus{} c}{3} \geq 1$

so : $ 3(\frac {t^2 \plus{} t \plus{} 1}{t \plus{} 1}) \geq \frac {9}{2}\, \iff \, 2t^2 \minus{} t \minus{} 1 \geq 0$

$ \iff$  $ (t \minus{} 1)(2t \plus{} 1) \geq 0$

:)

The following inequality is also true (for a, b, c > 0 and abc = 1)

a + b + c + \frac {a}{a + 1} + \frac {b}{b + 1} + \frac {c}{c + 1}\geq\frac {9}{2}.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that
\[ a \plus{} b \plus{} c \plus{} 16\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq27\]

Let $ a,b,c$ be nonnegative real numbers prove that
\[ a \plus{} b \plus{} c \plus{} 16\left( {\frac {a}{{a \plus{} 1}} \plus{} \frac {b}{{b \plus{} 1}} \plus{} \frac {c}{{c \plus{} 1}}} \right) \ge 27
\]
The inequality equalivents with
\[ \Leftrightarrow \left( {\sum a } \right)\left( {a \plus{} 1} \right)\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right) \plus{} 16\sum {a\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right)} \minus{} 27\left( {a \plus{} 1} \right)\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right) \ge 0
\]

\[ \Leftrightarrow \left( {\sum a \minus{} 27} \right)\left( {\sum a \plus{} \sum {ab} \plus{} 2} \right) \plus{} 16\left( {\sum a \plus{} 2\sum {ab} \plus{} 3} \right) \ge 0
\]
Setting $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca$. Then it becomes
\[ p^2 \plus{} \left( {q \minus{} 9} \right)p \plus{} 5q \minus{} 6 \ge 0
\]
We have $ \Delta \equal{} q^2 \minus{} 38q \plus{} 105 \equal{} \left( {q \minus{} 3} \right)\left( {q \minus{} 35} \right)$. So if $ q \le 35 \Rightarrow \Delta \le 0$ then the inequality has been proved.
If $ q \ge 35$ then $ p^2 \ge 3q \Leftrightarrow p \ge \sqrt {105}$. The inequality can rewrite as
\[ \left( {5 \plus{} p} \right)q \ge \minus{} p^2 \plus{} 9p \plus{} 6
\]
But we have
\[ \minus{} p^2 \plus{} 9p \plus{} 6 \equal{} \minus{} \left( {p \minus{} \frac {{9 \plus{} \sqrt {105} }}{2}} \right)\left( {p \minus{} \frac {{9 \minus{} \sqrt {105} }}{2}} \right) \le 0 \Leftrightarrow p \ge \frac {{9 \plus{} \sqrt {105} }}{2}
\]
So we have done. :)

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ a \plus{} b \plus{} c \plus{} 16\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq27
\]
[/quote]

Again the $ \ln$ trick works (but please don't make it stronger  :P )

Consider the function $ f(a)\equal{}a\plus{}\frac{16a}{a\plus{}1}\minus{}5\ln a$ for which $ f'(a)\equal{}\frac{(a\minus{}1)(a^2\minus{}2a\plus{}5)}{a(a\plus{}1)^2}$

from which $ f(a)\geq f(1)\equal{}9$ and we are done  :)

silouan, I think, your idea doesn't work for the following inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1144742#1144742
 :wink:

I think we can use Chebyshev
(a+1/a+1-3/2)+(b+1/b+1-3/2)+(c+1/c+1-3/2) >=0
then we use chebyshev and AM-GM--->no problem :blush:
sorry,I don't know how to write latex :oops:

Also from this theme.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[ a \plus{} b \plus{} c \plus{} 36\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq57\]
I think, the silouan's method does not work here. :wink:

My method maybe will work wth cases but I am not sure  :wink:

But mixing variables kill the problem
Let $ a$ be the max{a,b,c} . Then $ bc\leq 1$ .

We have that

$ f(a^2,b^2,c^2) \minus{} f(a^2,bc,bc) \equal{} \frac {(b \minus{} c)^2(37 \plus{} b^2 \minus{} 35bc \plus{} b^3c \plus{} bc^3 \plus{} c^2 \plus{} b^2c^2 \plus{} b^3c^3)}{(b^2 \plus{} 1)(c^2 \plus{} 1)(bc \plus{} 1)}\geq 0$

so it suffices to prove that $ f(a^2,\frac {1}{a},\frac {1}{a})\geq 57$ which is equivalent to

$ \frac {(a \minus{} 1)^2(2 \plus{} 21a \minus{} 15a^2 \plus{} 3a^3 \plus{} a^4)}{a(a \plus{} 1)(a^2 \plus{} 1)}\geq 0$  (1)

Now we have $ 2 \plus{} 21a \minus{} 15a^2 \plus{} 3a^3 \plus{} a^4 > a^4 \plus{} 3a^3 \minus{} 15a^2 \plus{} 17a \minus{} 6 \equal{} (a \minus{} 1)^3(6 \plus{} a) > 0$ since $ a\geq 1$ and so (1) holds and we are done .

[quote="silouan"]My method maybe will work wth cases but I am not sure  :wink: [/quote]
$ a \plus{} b \plus{} c \plus{} 36\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq57\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a \plus{} \frac {36a}{a \plus{} 1} \minus{} 19 \minus{} 10\ln a\right)\geq0.$
$ \left(a \plus{} \frac {36a}{a \plus{} 1} \minus{} 19 \minus{} 10\ln a\right)' \equal{} \frac {(a \minus{} 1)(a \minus{} 2)(a \minus{} 5)}{a^2(a \plus{} 1)^2}.$
What is the next?   :P

[quote="silouan"]
But mixing variables kill the problem [/quote]
You are right!

By the way, the following inequality
\[ a \plus{} b \plus{} c \plus{} 59\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq91.5
\]
is true for all positive $ a,$ $ b$ and $ c$ such that $ abc \equal{} 1.$
But the following
\[ a \plus{} b \plus{} c \plus{} 60\left(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}\right)\geq93
\]
is wrong already.   :wink:

[quote="arqady"] What is the next?   :P
[/quote]

I have said I am not sure . But I remember hunghtn solved a problem like this with four variables .
Let $ a\leq b\leq c$ and then consider the cases in which $ c\geq 3$ and $ c\leq 3$ or something like that

OK!!this is my slution by variable mixing
setting $ f(a,b,c) \equal{} a \plus{} b \plus{} c \plus{} 60(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1})$

$ f(a,b,c) \minus{} f(a,\sqrt {bc},\sqrt {bc}) \equal{} (\sqrt {b} \minus{} \sqrt {c})^2[1 \plus{} \frac {60(1 \minus{} \sqrt{bc})}{(b \plus{} 1)(c \plus{} 1)(\sqrt {bc} \plus{} 1)}]$

$ assume a \equal{} max(a,b,c) \rightarrow bc\le 1$

$ \rightarrow f(a,b,c) \minus{} f(a,\sqrt {bc},\sqrt {bc}) \ge 0$

$ \rightarrow f(a,b,c)\ge f(a,\sqrt {bc},\sqrt {bc})$

we needs proof $ f(a,\sqrt {bc},\sqrt {bc}) \ge 93$ true with $ a \equal{} \frac {1}{t^2}$ and $ b \equal{} c \equal{} \sqrt {bc} \equal{} t$

$ \leftrightarrow \frac {1}{t^2} \plus{} 2t \plus{} 60(\frac {1}{1 \plus{} \sqrt {1}{t^2}} \plus{} \frac {2t}{t \plus{} 1})\ge 93$ (easy... :wink: proof with a variables)

[quote="bunhiacovski"]OK!!this is my slution by variable mixing
setting $ f(a,b,c) \equal{} a \plus{} b \plus{} c \plus{} 60(\frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1})$

[/quote]
Try $ b\equal{}c\equal{}0.49$ and $ a\equal{}\frac{1}{0.2401}.$ :wink:

[hide=Thanks.][quote=Inconsistent]$f(x)=e^x+\frac{1}{e^{2x}+1}$

$f''(x)=\frac{e^x\left(e^{6x}+3e^{4x}+4e^{3x}+3e^{2x}-4e^x+1\right)}{\left(e^{2x}+1\right)^3}$

Note if $e^x \leq \frac{1}{4}$ the numerator is non-negative ($1-4e^x$)

Note if $e^x \geq \frac{1}{4}$ the numerator is non-negative ($4e^{3x}+3e^{2x}-4e^x+1=(4e^{3x}-e^{2x})+(2e^{2x}-1)^2$)

Thus the function is convex over all reals, let $x=\ln(a), y=\ln(b), z=\ln(c)$

Then by Jensen's $LHS=f(x)+f(y)+f(z)\geq 3f(0)=\frac{9}{2}$[/quote]

[/hide]
Let $a,b,c>0$ and $abc=1$.Prove that
$$a+b+c+\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\geq \frac{9}{2}$$

[quote=sqing][quote=galois01]Let $ a,b,c>0$ and $ abc\equal{}1$.Prove that
$$ a\plus{}b\plus{}c\plus{}\frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{9}{2}$$[/quote]
Let $a,b,c>0$ and $abc=1$.Prove that
$$a+b+c+\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\geq \frac{9}{2}$$[/quote]
[url]https://artofproblemsolving.com/community/c6h1177174p5685573[/url]

[quote=galois01]Let $ a,b,c>0$ and $ abc\equal{}1$.Prove that
$ a\plus{}b\plus{}c\plus{}\frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{9}{2}$[/quote]

Set $(a,b,c)\rightarrow \left(\frac{a}{b},\frac{b}{c},\frac{c}{a}\right)$
So it suffices to show that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}\ge \frac{9}{2}$$
Assume $c=\min \{a,b,c\}$ then we have $$LHS-RHS=\frac{(a-b)^2}{ab}+\frac{(2a^2b+a^2c+2ab^2+5abc+2ac^2+2b^2c+2bc^2)(a-c)(b-c)}{2ac(a+b)(b+c)(c+a)}\ge 0$$

[quote=galois01]Let $ a,b,c>0$ and $ abc\equal{}1$.Prove that
$ a\plus{}b\plus{}c\plus{}\frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{9}{2}$[/quote]

We have$:$
$$\text{LHS}-\text{RHS}=\sum\frac{ c \left( ab+3 \right)  \left( a-b \right) ^{2}+ \left( 2+a+b \right) \left( c-1 \right) ^{2}}{4(a+1)(b+1)(c+1)}\geqslant 0$$

[hide=Thanks.][quote=Googolplexian][quote=sqing]
Let $a,b,c>0$ and $abc=1$.Prove or disprove
$$a+b+c+\frac{1}{a^n+1}+\frac{1}{b^n+1}+\frac{1}{c^n+1}\geq \frac{9}{2}$$
Where $n\in N^+.$[/quote]

We can check $f(x)=\frac{1}{1+e^x}$ is convex for x>0.

Then LHS $=a+b+c+f(n\cdot lna)+f(n\cdot lnb)+f(n\cdot lnc) \geq a+b+c+3f((\frac{n}{3})(lna+lnb+lnc))$ by Jensen.
So we have $LHS \geq a+b+c+ \frac{3}{1+(abc)^{\frac{n}{3}}} \geq 3+\frac{3}{2}=\frac{9}{2}$ by AM-GM and the fact that $abc=1$.[/quote]

[/hide]

[quote=galois01]Let $ a,b,c>0$ and $ abc\equal{}1$.Prove that
$ a\plus{}b\plus{}c\plus{}\frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\geq\frac{9}{2}$[/quote]

Here is my proof: By AM-GM: $a+b+c \geq 3$ and $a+b+c-\frac{3}{2} \geq \frac{3}{2} \implies (a+b+c)(a+b+c-\frac{3}{2}) \geq \frac{9}{2} \Longleftrightarrow (a+b+c)^2 \geq \frac{3}{2}(a+b+c+3) \Longleftrightarrow \frac{(a+b+c)^2}{a+b+c+3} \geq \frac{3}{2}$. By Engel's Form, we have that $\sum_{cyc} \frac{a^2}{a+1} \geq \frac{(a+b+c)^2}{a+b+c+3} \geq\frac{3}{2} \Longleftrightarrow \sum_{cyc} a + \frac{1}{a+1} \geq \frac{9}{2}$ which is the desired inequality.

For $abc=1$, the inequality
$a+b+c+\frac{1}{a^n+1}+\frac{1}{b^n+1}+\frac{1}{c^n+1}\geq \frac{9}{2}$
holds for $n=5$, but does not hold for $n\ge 6$ (by computer).

[quote=bunhiacovski][quote="galois01"]Let $ a,b,c > 0$ and $ abc \equal{} 1$.Prove that
$ a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\geq\frac {9}{2}$[/quote]
another proof:
postulate $ a \equal{} min(a,b,c) \rightarrow bc\ge 1$
$ f(a,b,c) \equal{} a \plus{} b \plus{} c \plus{} \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}$
$ f(a,b,c) \minus{} f(a,\sqrt {bc},\sqrt {bc}) \equal{} (\sqrt {b} \plus{} \sqrt {c})^2[1 \plus{} \frac {\sqrt {bc} \minus{} 1}{(b \plus{} 1)(c \plus{} 1)(\sqrt {bc} \plus{} 1)}]\ge 0$ ( $ bc\ge 1$)
$ \rightarrow f(a,b,c)\ge f(a,\sqrt {bc},\sqrt {bc})$
we need to prove $ f(a,b,c)\geq\frac {9}{2}$ true with $ b \equal{} c \equal{} \sqrt {bc} \equal{} t; a \equal{} \frac {1}{t^2}$
$ \leftrightarrow t^2 \plus{} 2t \plus{} \frac {1}{1/t^2 \plus{} 1} \plus{} \frac {2}{t \plus{} 1}\ge 9$ (easy   :lol: )[/quote]

do we need to prove  the SMV method before using it in contests ??


function
induction
algebra
polynomial
search
algebra proposed

Find all functions $f: X \to X$  (X is the set of nonzero real numbers)
satisfying $f(x^2+y)={f(x)}^2+\frac{f(xy)}{f(x)}$ for all $y\not={-x^2}$.

Look at what happens when $x = 0$.
We have $f(y) = f(0)^2 + 1$.
Then $f(y)$ must be a constant, at least when $y \neq 0$.

So let $q = f(x)$ for any $x \neq 0$.
Look at what happens when $y = 1$.
We have $q = q^2 + 1$.

Since there is no real number $q$ that satisfies that relation, there is no such function.

[quote="qqwref"]Look at what happens when $x = 0$.
We have $f(y) = f(0)^2 + 1$.
[/quote]
But $X$ is the set of nonzero real numbers.
$f(0)??????????????$

[quote="qqwref"]OH. Oops.

Wait, why don't we have that $x \neq 0$ and $y \neq 0$ then?[/quote]

Because $x, y \in X$ and he defined $X$ to be the set of nonzero reals!  :)

Let $y = 1$.  Then

$f(x^2 + 1) = f(x)^2 + 1$

A well-known problem with well-known solutions.  Let $f_0(x) = x$.  This is a solution of the equation.

Let $f_{n+1} (x) = f_k(x)^2 + 1$.  It is easy to prove by induction that all $f_k(x)$ are also solutions.

(These are all the polynomial solutions, anyway.  I don't recall if there are others, but search for the problem, I have seen it before!)

Take $x=1$ and $f$ must also verify $f(y+1)=\frac{f(y)}{f(1)}+f(1)^2$
So what about if $f(1)\neq1$ ?
And in this case for integers we have not $f(n)=n$ but $f(n)=\frac{1}{(f(1))^{n-1}}(f(1)-\frac{f(1)^3}{f(1)-1})+\frac{f(1)^3}{f(1)-1}$

[i]Because $x, y \in X$[/i]

Actually, that's never listed as a condition in the problem, although it is true that setting either of them to 0 gives us $f(0)$ in the relation.

[i]Let $f_{k+1} (x) = f_k(x)^2 + 1$.[/i]

But then $f_1 (x) = x^2 + 1$, which doesn't satisfy the relation! What's up?

[i]$f(n)=\frac{1}{(f(1))^{n-1}}(f(1)-\frac{f(1)^3}{f(1)-1})+\frac{f(1)^3}{f(1)-1}$[/i]

What? How?

[quote="qqwref"]it is true that setting either of them to 0 gives us $f(0)$ in the relation.[/quote]

And that's why we can't use either $x=0$ or $y=0$ :)

[quote="qqwerf"]But then $f_1 (x) = x^2 + 1$, which doesn't satisfy the relation! What's up?[quote]

Yes, I realize this now, sorry.  It satisfies the relation only in the special case $y=1$.   :blush:


My dog usually likes red dog treats. Today, I switched from brown pellets to red AND brown pellets. He took a mouthful, spit the brown ones out, and ate the red ones.

That affected my view on whether animals were color-blind or not. What is your opinion?

For other animals I don't know, but dogs see world in scale of grey. They can see preety well, especially sight hounds whose primary sense is sight instead of smell. I used to think that dogs were scant because they were unable to distinguish colors, but I changed my opinion during the long walks with my afghan hound. She is aware of cats or birds or other small animals even when they are really far away. In some situations I know she couldn't sniff them nor she behaved like she was sniffing.

I think that your dog choosed with his nose rather then with eyes.

I agree. Just because they are colorblind doesn't mean that everything they see is gray. My English teacher and my uncle are both colorblind and they can tell light from dark. Red is generally regarded as a light color and brown is a dark color.

I agree with nienne.  Your dog probrobly took the food that he liked the most.

As far as I know, all animals except primates (men, apes etc) see on scales of grey. But if you see red and brown pellets on your old black-and-white TV set, you may be able to make them out. And your dog, being much smarter, definitely can!

Most animals should be able to distinguish between nutz's pellets :P

[quote="Koda"]As far as I know, all animals except primates (men, apes etc) see on scales of grey.[/quote]
Wrong, most animals can see colors, most even better than humans.

Maybe I'm off topic, but what is your favoured dog breed?
Mine is afghan hound.  :yup:


For each positive integer $n \geq 2$ determine the following set:
$A_n$ = {$x \in N^*$ $ |$  $1 \leq x \leq (n-1) $ and  $n$ does not divide ${C_n}^x  $}
The set is empty IFF $n$ it is prime.
For example:
$A_4$ = { $2$ }
$A_6$ = { $2$, $3$, $4$ }
$A_8$ = { $2$, $4$, $6$ }
$A_9$ = { $3$, $6$ }
$A_{10}$ = { $2$, $5$, $8$ }
$A_{12}$ = { $2$, $3$, $4$, $8$, $9$, $10$ }
If we can't find explicitely $A_n$ for each $n$, we can determine at least how many elements it has?
$f(n)$ = card $A_n$

Bonus.
Obviously $f(n)$ $\leq$ $n-3$ .
For $n=6$ we have equality, but there is another integer such that:
$f(n)$ = $n-3$ .

Maybe it should be in Number Theory Section?

just looks like the opposite of phi function. the other number is 12.


The fraction $ \frac{1}{3}$:

$ \textbf{(A)}\ \text{equals 0.33333333} \qquad
\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3 \cdot 10^8} \\
\textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3 \cdot 10^9} \\
\textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3 \cdot 10^8} \\
\textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3 \cdot 10^9}$

[hide="Click for solution"]$ \frac{1}{3}\equal{}0.\overline{3}$.  Thus, $ \frac{1}{3} \equal{} 0.33333333\plus{}0.00000000\overline{3}\equal{}\frac{1}{3} \cdot 10^8$ or $ \boxed{\textbf{(D)}}$.
[/hide]


Hello guys!
Today was december the sixth.  For most children here a very very special day as that is the day when Sinterklaas (Saint Nicolas) comes on his horse, jumping over the roofs and sending his men down the chimney to deliver presents.  In return, you are at least supposed to put a carrot in place for his horse.  For obvious reasons, seeing the Saint doing his job is considered very naughty and may even lead to no presents! ;)  that's why most children keep very quiet and just nervously wait for the next day!

He supposedly comes with a boat from spain, and it is a big event.  Saint Nicolas is extremely speedy and manages to show up about everywhere, schools, universities,etc....

The real saint was born in Asia Minor (now Turkey).

http://en.wikipedia.org/wiki/Saint_Nicholas#Celebration_in_the_Netherlands
http://en.wikipedia.org/wiki/Image:DutchSinterklaas2005.jpg

I was just wondering for those outside Europe, do you know him, and do you like it?
Santa Clause is actually derived from our word "Sinterklaas".  The weird thing is that Sinterklaas came to America, transformed into Santa Clause, and now in the last decades has come back in all sorts of merchandising!

So nowadays we have a Sinterklaas and a Santa Clause sometimes almost meeting each other.   However I as a child was never really tricked into believing in Santa Clause.  Christmas sometimes involves presents under the tree here, but it is not a big event as Sinterklaas.

Well, sorry for posting that much.  I just wanted to tell you about Sinterklaas, and wondered what you thought about it

I don't. And I can't open the links you provided. So can you introduce something about Saint Nicolas? How they derived Santa Clause from him?

I will tell some more then
He was born in south Turkey at about 270, and appears to have died in 345, but that is disputed.
Both catholics as orthodox christians look up to him.
He was very devoted and was a bishop. He would have caused miracles and he defended the falsely accused in some versions.
  His remains have been seized from turkey and have been buried in greece.  The turks demand it back, but as you might know, turkey and greece haven't always been really good friends, so that makes things even more difficult.

In German they call him Sint Nicolaus
We call him Sinterklaas, or also Sint Nicolaas,or just the "Sint".  He looks a lot more christian like (he looks like a bishop in red) than Santa Clause.

When he shows up at a school or something, he usually has a big book that describes all children, he calls everyone with them to discuss what they have done.  As sinterklaas is the one giving presents, and as he is a bearded tall well dressed man, children really look up to him, and are afraid that sinterklaas will be mad.  In the worst case, he puts you in the bag, and even takes you with him to spain.  However that is just to tease children :oops:   I've only seen our local student group pull a prank like pulling a bag over someone.
A very very big crime would be to pull the beard, for obvious reasons...

How did this get to America.  Well don't forget that New York was initially Dutch territory : New Amsterdam it was called first.  I believe the British traded their other half of Suriname for Manhattan in 1664 or something.  Those immigrants took the tradition with them but changed it : their Dutch name Sinterklaas got contracted to Santa Clause and he became a Christmas figure (instead of december 6)
Actually, if you watch the movie Santa Clause from 1994, when the police man ask him who he is, he mentions his name in several languages.  I expected him to say in Dutch : 'kerstman' (christmas man) because that is what we call him.  However, the creators apparently did their homework well because he said Sinterklaas.

A sensitive issue for me is that he is aided by several 'Zwarte pieten" = 'black Petes'
Those are black, strangely dressed helpers, who always accompagny him, help him of his horse, live with him and help distributing candy at appearances (Sinterklaas is rarely seen on his own)
That may seem a bit colonialist to have black men helping on old white man, but the explanation now is that they are 'black from going down all those chimneys'.  It is sad that such harsh racist history is still within a charming child fantasy.

I will stress again how much of a big thing he is for children , and adults.  In Belgium we even have a city Sint-Niklaas (which is again derived from Saint Nicolas)
In media , even in documentaries and news reports, there is a sort of moral code never to explicitly say he does not exist.  Yesterday I watched a documentary at nine pm about people training to perform at schools as santa, well they said : helper santa's.  All adults get the point of course, children should they watch do not

I was nine when I discovered that Sinterklaas is not real :(   This is a harsh moment for all children...

in my church, saint niholas did exist. he was a bishop (somewhere around the mediterranean) and helped out at least one family though maybe more in his parish by dropping money on their porch every time one of their girls was about to get married. he died as a regular person but his life of giving lived on and became a legend.

edit: i found a link
http://www.copticchurch.net/topics/synexarion/nicholas2.html


[quote="leviethai"]Let $ a,b,c > 0$. Prove that:
\[ \frac {{2{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}} \le \frac {{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}{{{a^2} \plus{} bc}} \plus{} \frac {{\left( {b \plus{} c} \right)\left( {b \plus{} a} \right)}}{{{b^2} \plus{} ca}} \plus{} \frac {{\left( {c \plus{} a} \right)\left( {c \plus{} b} \right)}}{{{c^2} \plus{} ab}}\]
[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$ Hence, your inequality is equivalent to $ f(w^3)\leq0,$
where $ f(w^3)\equal{}w^6\plus{}P(u,v^2)w^3\plus{}Q(u,v^2).$
But $ f$ is convex function. Hence, $ f$ gets a maximal value when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal or when maybe $ w^3\equal{}0.$
These cases easy to check.

[quote="leviethai"]Let $ a,b,c > 0$. Prove that:
\[ \frac {{2{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}} \le \frac {{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}{{{a^2} \plus{} bc}} \plus{} \frac {{\left( {b \plus{} c} \right)\left( {b \plus{} a} \right)}}{{{b^2} \plus{} ca}} \plus{} \frac {{\left( {c \plus{} a} \right)\left( {c \plus{} b} \right)}}{{{c^2} \plus{} ab}}\]
[/quote]
We have
$ \sum\limits_{cyc} {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}}  \equal{} {a^2} \plus{} {b^2} \plus{} {c^2}$
From it and by Cauchy - Schwarz ineq we get
$ ({a^2} \plus{} {b^2} \plus{} {c^2}).RHS\equal{}\sum\limits_{cyc} {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}}.\sum\limits_{cyc} {\frac{{\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}{{{c^2} \plus{} ab}}} \ge {\left( {\sum\limits_{cyc} {\sqrt {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{{c^2} \plus{} ab}}} } } \right)^2}$
It sufficies to prove that
$ {\left( {\sum\limits_{cyc} {\sqrt {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{{c^2} \plus{} ab}}} } } \right)^2}\ge2(a\plus{}b\plus{}c)^2$
or equivalent to
$ \sum\limits_{cyc} {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{{c^2} \plus{} ab}}}  \plus{} 2\left( {{a^2} \plus{} {b^2} \plus{} {c^2} \plus{} ab \plus{} bc \plus{} ca} \right) \ge 2{\left( {a \plus{} b \plus{} c} \right)^2}$
We only need to prove that
$ \sum\limits_{cyc} {\frac{{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{{c^2} \plus{} ab}}}  \ge 2\left( {ab \plus{} bc \plus{} ca} \right)$
Which is clearly true by AM-GM. :D

[quote="leedt26"]
We have
$ \sum\limits_{cyc} {\frac {{\left( {{a^2} \plus{} bc} \right)\left( {{b^2} \plus{} ca} \right)}}{{\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}} \equal{} {a^2} \plus{} {b^2} \plus{} {c^2}$
[/quote]
Who is author of this equality?  :roll:

[quote="leviethai"]Let $ a,b,c > 0$. Prove that:
\[ \frac {{2{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}} \le \frac {{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}{{{a^2} \plus{} bc}} \plus{} \frac {{\left( {b \plus{} c} \right)\left( {b \plus{} a} \right)}}{{{b^2} \plus{} ca}} \plus{} \frac {{\left( {c \plus{} a} \right)\left( {c \plus{} b} \right)}}{{{c^2} \plus{} ab}}\]
[/quote]
Actually, the stronger inequalities hold

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

(ii) $ \sum \frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} bc} \ge \frac {8(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca}.$

(iii) $ \sum \frac{(a\plus{}b)(a\plus{}c)}{a^2\plus{}bc} \ge \frac{16(a\plus{}b\plus{}c)^2}{5(a^2\plus{}b^2\plus{}c^2)\plus{}3(ab\plus{}bc\plus{}ca)}.$

[quote="can_hang2007"]
(ii) $ \sum \frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} bc} \ge \frac {8(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca}.$
[/quote]
Actually, (ii) is weaker than (ii). However, we can generalize it with a nice solution:

Let $ a,$ $ b,$ $ c$ be positive real numbers. If $ k \ge 1,$ then
$ \sum \frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} kbc} \ge \frac {8(a \plus{} b \plus{} c)^2}{(k \plus{} 2)(a^2 \plus{} b^2 \plus{} c^2) \plus{} k(ab \plus{} bc \plus{} ca)}.$

I think that this inequality also holds for $ 0\le k<1.$ But I cannot prove it.

[quote="can_hang2007"] However, we can generalize it with a nice solution:

Let $ a,$ $ b,$ $ c$ be positive real numbers. If $ k \ge 1,$ then
$ \sum \frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} kbc} \ge \frac {8(a \plus{} b \plus{} c)^2}{(k \plus{} 2)(a^2 \plus{} b^2 \plus{} c^2) \plus{} k(ab \plus{} bc \plus{} ca)}.$

I think that this inequality also holds for $ 0\le k < 1.$ But I cannot prove it.[/quote]
By $ uvw$ it's true for $ 0\leq k\leq\frac{1}{4}.$

[quote="can_hang2007"]
(ii) $ \sum \frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} bc} \ge \frac {8(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca}.$[/quote]
See the proof of (ii) here: http://can-hang2007.blogspot.com/2009/12/inequality-16-v-q-b-can.html

[quote=leviethai]Let $ a,b,c>0$. Prove that:

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

[img]http://s15.sinaimg.cn/middle/006ptkjAzy77FS7X3vE4e&690[/img]


Find all polynomials of two variables $P(x,y)$ which satisfy
\[P(a,b) P(c,d) = P (ac+bd, ad+bc), \forall a,b,c,d \in \mathbb{R}.\]

You can find two solutions in the attachment.

[quote="PhireKaLk6781"]Sorry for reviving, but it looks like the solution is for the wrong problem...[/quote]

Really?


[quote="GameBot"]There are seven wombats in a council. Three of them are hairy-nosed wombats, while the rest are all different species. If the hairy-nosed wombats are eerily indistinguishable, how many ways can the wombats seat themselves in a row?[/quote]

Normally, we would have 7! ways to arrange the seven wombats. However, the three hairy-nosed wombats are indistinguishable, so we have to divide by 3! to account for that, giving us an answer of $ 5040 \div 6 \equal{} 840$.


inequalities
geometry
parallelogram
circumcircle
symmetry
complex numbers
algebra solved

Thanks to siuhochung I have these! :D
I didn't know where to post them, so I posted them here! :D

Time allowed: 3 hours
Each problem is worth 7 marks

1. Find the greatest real K such that for every positive u, v and w with u^2 > 4vw, the inequality
(u^2 -4vw)^2 > K(2v^2 -uw)(2w^2 -uv)
holds. Justify your claim.

2. Let ABCDEF be a regular hexagon of side length 1, and O be the centre of the hexagon. In addition to the sides of the hexagon, line segments are drawn from O to each vertex, making a total of twelve unit line segments. Find the number of paths of length 2003 along these line segments that start at O and terminate at O.

3. Let ABCD be a cyclic quadrilateral. K, L, M, N are midpoints of sides AB, BC, CD, and DA respectively. Prove that the orthocentres of triangles AKN, BKL, CLM, DMN are vertices of a parallelogram.

4. Find, with reasons, all integers a,b, and c such that
(1/2)(a+b)(b+c)(c+a)+(a+b+c)^3 = 1-abc

this is siuhohung's solution 4 the 1st problem:

Divide both sides by u^4
then the inequality becomes
(1-4(v/u)(w/u))^2>K(2(v/u)^2-(w/u))(2(w/u)^2-(w/u))
Sub. a=v/u, b=w/u
(1-4ab)^2>K(2a^2-b)(2b^2-a)
The constaint given is transformed to ab<1/4
Then when a =1/2, b<1/2, when b=1/2, a<1/2
Therefore
(a-b)^2+(a-(1/2))^2+(b-(1/2))^2>0
Expand and simplify, we have
2a^2-b+2b^2-a>2ab-(1/2)
Let
x=2a^2-b
y=2b^2-a
z=1-4ab
So we are having,
x+y>-z/2 and z^2>Kxy
2(x+y)>-z
4(x+y)^2>z^2>Kxy
4x^2+(8-K)xy+4y^2>0
Obviously, K=16

3) it's nice it can be done with complex numbers (a lot of computation), but a very nice purely geometrical solution yptsoi has, I will not post it, i shall let you think about them! :D

Lagrangia I think it's best to let them post their solutions :)

well.. Valentin I would let them, but they don't have access to mathlinks! :(

cheers!

The problem #4 becomes trivial when observed that the given equation is equivalent to :
(2a+b+c)(a+2b+c)(a+b+2c) = 2, which has to be solve in integers....

Pierre.

Here's a soln for 3 with complex numbers and very few computations:
The orthocenter of AKN is the midpt of the segm formed by A and the orthocenter of ABD, so it has affix [a+(a+b+d)]/2 = a+(b+d)/2. By the same method we find the affixes of all of the 4 pts:

h1=a+(b+d)/2
h2=b+(a+c)/2
h3=c+(b+d)/2
h4=d+(a+c)/2

Is there a doubt in ne1's mind that these are the vertices of a parallelogram? :D We can see that h1-h2=h4-h3. Here all small letters are complex numbers corresponding to the points which have the same letters.

I hope my english will be sufficient to make clear the concept, at least.

Let's considere the circumcircles of triangles AKN, BKL, CLM, DMN. Is easy to see that this four circles have as common point the center O of circle c(ABCD). Another easy fact is that c(AKN) and c(CLM) intersect (also) at the mid point P of diagonal AC, similarly c(BKL) and c(DMN) intersect (also) at the mid point Q of diagonal BD. Let's call A', B', C' and D' the orthocentres of triangles AKN, BKL, CLM, DMN rispectively.
Given that c(B'KL) is the traslation of c(BKL) and similar then BB' and DD' are both equal and parallel to OP,  AA' and CC' are both equal and parallel to OQ.

This means that Q'=t_op(Q) and P'=t_oq(P) are the same point, i.e. A'C' and D'B' bisect each other, then the thesis.

[quote="Lagrangia"]
Time allowed: 3 hours
Each problem is worth 7 marks

2. Let ABCDEF be a regular hexagon of side length 1, and O be the centre of the hexagon. In addition to the sides of the hexagon, line segments are drawn from O to each vertex, making a total of twelve unit line segments. Find the number of paths of length 2003 along these line segments that start at O and terminate at O.
[/quote]

It seems we exceeded the 3hrs limit :D

Rename O=0, A=1, B=2, C=3, D=4, E=5, F=6.
Denote by f(i,j,n) the number of paths of length n starting from i and ending in j.
We must find f(0,0,2003).
It is clear that we have f(0,0,n+1)= \sumi=1..6 f(0,i,n). ( we look at the vertex from where the last segment of the path starts)
By reasons of symmetry of the hexagon, we have f(0,1,n)=f(0,2,n)=...=f(0,6,n) for all n. So f(0,0,n+1)=6*f(0,1,n)
We have f(0,i,n+1)=f(0,i-1,n)+f(0,i+1,n)+f(0,0,n) (look again at the start of the last segment in the path). If i=6, think i+1=1! If i=1, think i-1=6!
By reasons of symmetry again, we have f(0,i,n+1)=2*f(0,1,n)+f(0,0,n)

Let a_n=f(0,0,n) and b_n=f(0,1,n)
There sequences check the following recurrences:
a_(n+1)=6*b_n
b_(n+1)=2*b_n+a_n
We have b_1=1 and a_1=0. We have b_2=2.
b_(n+2)=2*b_(n+1)+a_(n+1)=2*b_(n+1)+6*b_n. We notice that if we state b_0=0, it still checks the recurrence.
b_n= ((\sqrt 7)/14)*((1+ \sqrt 7) n+(1- \sqrt 7) n). If you don't belive, me check that the formula checks the recurrence and the initial conditions.

Now f(0,0,2003)=a_2003=6*b_2002=((3* \sqrt 7)/7)*((1+ \sqrt 7) 2002+(1- \sqrt 7) 2002).
I hope you are happy with the result because I'm not going to compute it.

The problem 2 can be generalized for a n sided polygon.
The recurrences are:
a_m+1=n*b_m
b_(m+1)=2*b_m+a_m

Now please, moderators of this section of the forum, move this sticky to solved.


trigonometry invariant induction inequalities function

I just noticed today that I kinda suck at olympiad-level math. Does anyone have suggestions?

I have AoPS 1&2 and Problem Solving Strategies... (kind of a small collection)

plus a bunch of contest books

Problem Solving Strategies is good, though it may be too difficult for a beginner

If it seems hard, try the Art and Craft of Problem Solving or Problem Solving through Problems

If it is not too difficult, however, Engel is definitely one of the best books out there.

Also, the unl website has under the "Students" section two free books containing 100s of real Olympiad problems with detailed solutions.

Of course, kalva provides many Olympiad problems at all levels, often with solutions

It might be a bit too basic for not_trig but
Mathematical Circles by Fomin, et. al. is a collection of topics for math circles in Russia.  Has chapters on invariants, combinatorics, graphs, induction, inequalities, games etc.  Mostly problems.  Paperback, not too expensive.  It's very accessible - no need to understand generating functions or other higher-end topics just to read the book.

The book is not directly related to olympiad problem solving but there's plenty of good problems.  Most of the first half of the book is well below "olympiad" level but some of the material in the second half is close to olympiad level.

Here's an example problem from the Invariants chapter in the 2nd half of the book:

The numbers 1,2,3,...,n are written in a row.  It is permitted to transpose any two neighboring numbers.  If 1989 such operations are performed, is it possible that the final arrangement of numbers coincides with the original?

n^4's suggestions are better for olympiad problems.  Some would probably consider Mathematical Circles to be at the "intermediate" or "beginner" level.  Many proofs are required but most are not as difficult as Engel or Zeitz (Art & Craft of Problem Solving.)  Still, for those not familiar with some of the basic techniques, it might be worth working through the interesting chapters.

Mathematical Circles was my favorite book in middle school  :D .  I agree that it is excellent, but basic.

Obviously, past tests (use kalva or our contest section). Start with olympiads easier than USA if you like (the ones that have lots of problems in a relatively short amount of time should be good). And if you aren't as good right now, struggle through the problems and read the solutions. It takes a long while before you can solve olympiad problems. Remember that you get 1.5 hours for each.

Zeitz (Art and Craft) and Engel (Problem Solving Strategies) are incredible, as mentioned. They're unquestionably my two favorite contest books.

Also, you may look at the "101 trig problems", "102 algebra problems", etc. series, or anything else written by Titu, Zuming, or both :D.

I have a Titu book, "A Path to Combonatorics for Undergraduates" lol undergraduates

So pretty much get ACoPS and I'm ok?  :D

and kalva, I already use kalva, it's great


Determine all integers $k$ for which there exist infinitely the pairs of integers $(a,b,c)$ satisfying the following equation.

\[ (a^2-k)(b^2-k)=c^2-k.  \]

It is equavalent to $(a+\sqrt k)(b+\sqrt k)=c+\sqrt k$, that is $b=1-a,c=a(1-a)+k$, were a is any integer.

I got something like this, take $(a,b,c)=(x,x+1,x(x+1)-k)$ and all $k \in Z$ are good.

Yes it is equavalent to
or $(a+\sqrt k)(b+\sqrt k)=c+\sqrt k$ give $(x,1-x,k+x(1-x)),$
or $(a+\sqrt k)(b-\sqrt k)=c+\sqrt k$  give $(x,1+x,x(1+x)-k,$ (c --> -c)
or $(a+\sqrt k)(b+\sqrt k)=c-\sqrt k$ give $(x,-1-x,k-x(x+1),$  (b --> -b)
or $(a+\sqrt k)(b-\sqrt k)=c-\sqrt k$ give $(x,x-1,x(x-1)-k).$   (b--> -b, c--> -c)
 If a, b, c natural then $(x,x+1,x(x+1)-k)$ or $(x+1,x,|x(x+1)-k|)$ , x natural.

[quote="Anavel_Gato"][quote="Rust"]It is equavalent to $ (a\plus{}\sqrt k)(b\plus{}\sqrt k)\equal{}c\plus{}\sqrt k$[/quote]

Can anybody explain this??why this is equivalent?
thanks in advance :ninja:[/quote]

I don't know,too :(

In fact $(a^2 - k)(b^2 - k) = c^2 - k$ is [b]not[/b] equivalent to $(a+\sqrt{k})(b+\sqrt{k}) = c+\sqrt{k}$; just that this second form implies the first, since then its conjugate $(a-\sqrt{k})(b-\sqrt{k}) = c-\sqrt{k}$ also holds, and by multiplying we get the first form (it does not matter if $k$ is positive or negative, and for $k\neq 0$ we then need $a+b=1$).

It is clear we could as well have started with $(a+\sqrt{k})(b+\sqrt{k}) = c-\sqrt{k}$, requiring $a+b=-1$, or any other of the two more forms that Rust added to modulate his initial post.

The cleanest approach is by far Hawk Tiger's post #5.


The minute hand on a twelve-hour clock is twelve cm long and the hour hand is six cm long. What is the ratio of the greatest distance traveled by any point on the hour hand in twelve hours to the greatest distance traveled by any point on the minute hand in twelve hours?

At any moment, the tip of the hour hand has traveled $ \boxed2$ times as far as the minute hand, as the ratio of the arcs of circles is equivalent to the ratio of the radii.

the right answer on ftw is 1/24 but i dont know why

[quote="Mewto55555"]No, because the minute hand moves 60 times as fast, so its 2/60=1/30[/quote]

No it is 1/2 (ratio of lengths) times 1/60 (minute hand is 60 times faster) = 1/120.

It's 1/24 because the greatest distance that the hour hand traveled is 2*6*pi=12pi since it only does 1 full rotation. The greatest distance on the minute hand however is 12*2*12*pi=12*24pi since it makes TWELVE full rotations for twelve hours and each rotation is 2*12*pi cm long. Therefore 12*pi/(12*24*pi)=12/(12*24)=1/24 :)