function
algebra
functional equation
real analysis
real analysis unsolved

Let f: R -> R be a function such that, for all x, y belonging to R, both of the conditions f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y) are satisfied. Prove that f is either:

(i) The function which assigns the value 0 to all x belonging to R

or

(ii) The identity mapping, which assigns, to each x belonging to R, the value x.

One thing to note is that I can recognize that if f(1) does not equal 0, then for each rational number r you have f(r) = r. In addition, we know that if a real number x is > 0, then there exists a non-zero real number y such that x = y^2.

The problem is I can explain and draw pictures of what is happening, but I do not know how to prove that either (i) or (ii) is true.

Thanks.

I suggest this problem be moved to the "Pre-Olympiad section." Such functional equation problems have been posted and discussed at length in that part of the forum before.

Your first observation tells us that $f(r)=rf(1)$ for rational $r$. That's a dense set; if we can prove the function is continuous, we get almost everything.
Your second observation gives us that $f(x)=f(y^{2})=(f(y))^{2}\ge0$ whenever $x\ge0$. If $a\ge b$, we have $f(a)-f(b)=f(a-b)\ge0$, and therefore $f(a)\ge f(b)$: $f$ is an increasing function.
Now $f(x)\ge f(r)=rf(1)$ whenever $r<x$ and $r$ is rational, and $f(x)\le f(r)=rf(1)$ whenever $r>x$ and $r$ is rational. Taking the supremum and infimum over all such rationals, $f(x)\ge xf(1)$ and also $f(x)\le xf(1)$. Now $f(x)=xf(1)$ for all $x$. We have $f(1)=(f(1))^{2}$, so either $f(1)=1$ and $f(x)=x$ or $f(1)=0$ and $f(x)=0$ for all $x$.


Take a circle and points A, B, C, and P on it. Take triangle ABC.
Consider the orthogonal projection C1, A1, B1 of the point P on the line AB, BC, and CA correspondingly.
Prove that points A1, B1, and C1 belong to the same line.

Oh! This problem is the theorem about line Simson.

You can see sloved at this link: http://agutie.homestead.com/files/simsontheorem1.html

-------------------------------------------------
Wish you good luck. :rotfl:


Se dau numerele reale a si b din intervalul [1, 2] unde a>b.

Se cere sa se arate ca $\frac{a+b}{\sqrt{ab}}< 2b$.

Problema 13, pag 212, culegere a xi-a Ganga.

Multumesc.

Enuntul complet de fapt este:

se dau $x_{0}$ si $x_{1}$ din [1, 2] astfel incat $x_{0}> x_{1}$  si $x_{n+2}= \frac{x_{n+1}+x_{n}}{2 \sqrt{x_{n+1}x_{n}}}$. Aflati limita sirului.

Ceva nu e in ordine: daca $b=1$ si $a =2$ gasesti $3\leq 2\cdot\sqrt{2}$ ceea ce este evident fals :wink:
De fapt $\frac{a+b}{2}>\sqrt{ab}>b$ pentru $a>b$. (am folosit inegalitatea mediilor). Deci inegalitatea ceruta este exact pe invers...
Presupun ca intervalul $[1,2]$ era dat pt a stabili marginirea sirului.

se arata ca sirul este in [1,2] si descrescator. limita este 1.

Pai bine dar tocmai asta nu reusesc, sa arat monotonia. Adica sa demonstrez prin inductie ca sirul este descrescator. Daca $x_{n}> x_{n+1}$ atunci $x_{n+1}> x_{n+2}$. De aceea am intrebat cum demonstram inegalitatea aceea.

Acum nu mai inteleg nimic si nici cum se demonstreaza inegalitatea aceea (cea buna) nu am inteles.

Mai intai,  $x_{n+2}-1=\frac{(\sqrt{x_{n+1}}-\sqrt{x_{n}})^{2}}{\sqrt{x_{n+1}\cdot x_{n}}}\geq0,$ $\forall{n}\in\mathbb{N}$, deci
\[x_{n}\geq1,\ \forall{n}\in\mathbb{N}. \]
In continuare, aratam prin inductie, ca
\[x_{n}\leq\frac{3}{2\sqrt{2}},\ \forall{n}\geq2. \]
In etapa de verificare, trebuie sa probam validitatea propozitiei
\[P(2): x_{2}\leq\frac{3}{2\sqrt{2}}. \]
Inegalitatea care trebuie justificata, se scrie echivalent $(2x_{1}-x_{0})(x_{1}-2x_{0})\leq0$ si este evidenta, deoarece, din ipoteza, $x_{0},x_{1}\in{[1,2]}$ si $x_{0}> x_{1}.$ Prin urmare, $P(2)$ este adevarata.
Trecem la etapa de demonstratie.
Presupunem ca propozitia
\[P(k): x_{k}\leq{\frac{3}{2\sqrt{2}}},\ \forall{k}\in\{2,3,...,n+1\}\]
este adevarata si vom demonstra ca propozitia
\[P(n+2): x_{n+2}\leq{\frac{3}{2\sqrt{2}}}\]
este, de asemenea, adevarata.
Putem scrie $x_{n+1}+x_{n}\leq\frac{3}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}=\frac{3}{\sqrt{2}}\leq\frac{3}{\sqrt{2}}\sqrt{x_{n+1}x_{n}},$ de unde rezulta ca $x_{n+2}=\frac{x_{n+1}+x_{n}}{2\sqrt{x_{n+1}x_{n}}}\leq\frac{3}{2\sqrt{2}}.$
Asadar, $P(n+2)$ este adevarata. In concluzie, sirul $(x_{n})_{n\in\mathbb{N}}$ este marginit.
Pentru monotonie, avem
\[x_{n+1}+x_{n}\leq\frac{3}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}=\frac{3}{2\sqrt{2}}<2\leq2x_{n+1}\sqrt{x_{n+1}x_{n}}, \forall{n}\leq2, \]
de unde rezulta ca
\[x_{n+2}< x_{n+1}, \forall{n}\geq2 \]
si am asigurat astfel monotonia.
Sirul este convergent si rezulta imediat ca $\lim_{n \to \infty}x_{n}=1$.

Va multumesc domnule Damian pentru solutie si pentru ca ati redactat-o asa frumos. Mi-ati luat o piatra din cap, deoarece ma indispune mult sa nu duc o treaba pana la capat.

nici nu aveam nevoie sa demonstram ca sirul este marginit superior ,pt ca din marginirea inferioara si monotonie (sirul este strict descrescator) cu weierstrass=> ca sirul este convergent si trecand la limita ne da limita 1


This subject was suggested to me by another recent topic, namely the one about democracy. I was actually goingo to write something in that one but I decided to start a new one.
What I had intended to write was, democracy is for the time being the best form of government humanity has achieved, also the word communism has been demonized historically. The roots of the word point to an economical and social ideology who was theorized as you know well before its 20th century realisations. This realisations turned out with many horrible things in many aspects for mankind and specially for those countries who were the fluttering flags of communism or those were it was imposed by force. Nevertheless, those regimes, according to the old theory and to some of the new which is usually disregarded, can hardly be called communist and certainly not one of the first theoriticians had in mind tens of millons deaths when they formulated the hypothetical possibility of communism. Not one of them had in mind the world as it is today either. Still I think humanity has things to learn yet from that theory. This is debatable, but for instance, communist parties I've heard of across europe do not wave the democracy-communism antagonism point of view. Instead they search for a social solution which preserves freedom as it is understood in democracy(also theoretically, no intention to discuss actual freedom) but with economical and social distributions in some way along the lines of communism. Again this has little to do with soviet communism or mao's.

But within this highly globalised world, it is I think more important to discuss about the way the world as a whole is ruled. There are no world president obviously but there is UN where countries take decisions which affect the entire planet. Perhaps the word 'ruled' in the first sentence of this paragraph is wrong because the UN doesn't quite rule things, more like it traces guidelines. And the question I wanted to raise is this, is this process of tracing guidelines for all nations, democratic?? My opinion on this matter would be a categorical no, if we agreed on that no the next question is, is it correct the non-democracy at the UN? Is it correct to have countries with veto power? The sometimes succesful appearances of the UN in particular conflicts, is it overwhelming compared to all those matters which involve even tangentially the US, China or Russia and in which the UN influence is much more restricted?

Ursula

Well I don't quite know what to think of the UN, maybe I will better understand it if someone answers this question :

How come in the 1950's, during the Korean War, a UN led coalition intervened on behalf of the South, and Mao, chairman of the PROC, could send a huge number of soldiers to North Korea to help them (China was a member of UN but I guess it's not that easy with the Kwomintang still claiming to control the government  :o )
... but now (I mean since 1990) the UN doesn't take much action everywhere, they do some useful peacekeeping missions (like in Congo) and clearing minefields etc.. but they are so restricted it's absurd:
- most of the UN action in Rwanda seemed to concern the evacuation of Western people and their own
- in Srebrenica their job wasn't very clear either, they even aided in the separation of men from their families, only to be executed

So tell me, what changed? :|

I always call UNO United States Organization. Even the veto structure is not what it seems (not to mention even if it was, it would be far from democracy). For example the UK will never veto any decision of the US. In fact, in the current situation, none of the other three countries will.

Good to see that fredbel6 is at least not a UN fan.

[quote="bubka"]
Good to see that fredbel6 is at least not a UN fan.[/quote]
When you say at least, you mean compared to what?  To my heavy pro US stance, pro capitalist charge? What do you mean? :huh:

I'm not a fan, but to say that abolishing all its peacekeeping missions right now would be wise.... :|

Back on topic!  How about my question? :maybe:

Addendum :

I think the UN can be slow, bureaucratic,etc...
but maybe it has still done too many good things to be abolished :
all those peacekeeping missions, and suppliance of food and healthcare

The resolution on Lebanon went slowly.. but how would this conflict have played out if there was no UN at all?

I do wonder how the people, in Africa for instance, look at the UN, who are victim of the all too frequent cases of( child )rape look at them.
This article is particulary pessimistic (but then again [u]the site is obviously biased[/u]) :

[url]http://www.worldnetdaily.com/news/article.asp?ARTICLE_ID=42088[/url]

Point one: The website fredbel just linked to is an American right-wing media outlet with exceedingly small care for truth, accuracy or honesty.

Point 2: Bubka, you don't have a clue what you're talking about.  Russia and China regularly force the US to alter or do away with security council resolutions that it would like to pass (for example, Russia and China have repeatedly quashed or forced the US to water down resolutions condemning Iran).  The UK would never veto something presented to the SC by the US because the nature of the SC is that diplomacy happens before anything gets presented.  Very few things get vetoed at the SC for exactly this reason -- no one would ever bring something before the SC that they knew was going to get vetoed.

The UN is certainly undemocratic, but it's not clear to me that a more democratic UN would necessarily be a better UN.  Do I really want Sudan, Iran and North Korea to wield more power in the world?  No, probably not.  On the other hand, I probably don't want China or Russia to have such a strong hand in world affairs.  As a matter of practicality, of course, major UN restructuring is unlikely to happen any time soon.

JBL, you seem well informed, can you answer this neutral question for me?
When the UN engaged in action against North Korea, who had the power to decide/veto that?  All members of the Security Council?  All permanent members of the latter?
Because I can understand that China didn't veto it (as apparently the communists at that time didn't control China's seat in the UN), but what about the USSR?
And isn't supplying North Korea with arms etc. a grave violation of UN regulations, at a time North Korea was fighting a UN coalition??

I just don't get it, I'm completely confused :| .  For a young mind like mine, it seems remarkably decisive compared to the UN's actions now.

Yes, the site was biased, I stressed that myself, and after my edit, I do even more so.  I will try to find a decent source for abuses by the UN (and AU)

The USSR didn't veto because, ironically, at that time, it was boycotting the UN due to it not recognizing communist china as leader of china.

Yes, I'd forgotten that bit of history and pieterminate has reminded me.  I guess it shows you a little bit about the value of boycotting the political process :),  at least in some circumstances.  So, presumably the USSR could have vetoed it had they been inclined to do so.  I guess it's worth pointing out that the UN was also very, very young at the time and the Soviets might have been trying to neuter its influence because they didn't like some particular paths it was taking.  (I don't know this for fact, it is just supposition.)

[quote="fredbel6"]Isn't that kind of a paradox?  Disagreeing with a certain structure, and therefore not using your veto against an action you don't approve of? :maybe:[/quote]

The notion is to deny it legitimacy.  Opposition parties in semi-dictatorial states sometimes try this technique.  It's something like, "If you hold an election and nobody comes, is it still a democracy?"

With respect to the UN, it's not useful unless people take it seriously -- countries need to believe in it, to fund it, to provide it with peacekeepers and to (at least partially) do what it says (under appropriate conditions).  If everyone ignores it, it loses all potency.  At least, that's the theory, I guess.


LaTeX
quadratics
function
inequalities
induction
algebra
system of equations

Solve in reals the following system:
\[ \left\{\begin{aligned}x\plus{}2y^{2}& \equal{}3z^{3}\\ y\plus{}2z^{2}& \equal{}3x^{3}\\ z\plus{}2x^{2}& \equal{}3y^{3}\end{aligned}\right.\][color=red]

[Moderator edit: LateX (Arne)][/color]

No problem. A little history - the problem were composed by me when I was 9-th grade.

If $ x \equal{} y \equal{} z$ then it follows directly that $ x \equal{} y \equal{} z \equal{} 0,\minus{}\frac{1}{3},\ 1$. Suppose $ x\ne y \equal{} z$. The system becomes $ x\plus{}2y^{2}\equal{} 3y^{3},\ y\plus{}2y^{2}\equal{} 3x^{3},\ y\plus{}2x^{2}\equal{} 3y^{3}$, from which we get $ 1 \equal{} 2(x\plus{}y)$ and $ \minus{}1 \equal{} 3(x^{2}\plus{}xy\plus{}y^{2})$. After some algebra, we deduce that $ 0 \equal{} 12x^{2}\minus{}6x\plus{}7$ which has no real solutions. Now suppose $ x > y > z$. Returning to the original system of equations and subtracting the second from the first, we get $ (x\minus{}y)\plus{}2(y^{2}\minus{}z^{2}) \equal{} 3(z^{3}\minus{}x^{3})$. The LHS is positive but the RHS is negative. Contradiction. Thus $ (x, y, z) \equal{} (0, 0, 0), (1, 1, 1),\ (\minus{}\frac{1}{3},\ \minus{}\frac{1}{3},\ \minus{}\frac{1}{3})$ are the only solutions.

Edit: Sorry about that. You replied before I could correct my post  :blush: .

The only solutions of the system are: (0, 0, 0) (1,1,1) (-1/3,-1/3,-1/3). But the final sentence seems that skips (0,0,0). As a whole the idea of the system were to use something like your idea. Very good indeed!

[quote="mathisfun1"]If $ x \equal{} y \equal{} z$ then it follows directly that $ x \equal{} y \equal{} z \equal{} 0, \minus{} \frac {1}{3},\ 1$. Suppose $ x\ne y \equal{} z$. The system becomes $ x \plus{} 2y^{2} \equal{} 3y^{3},\ y \plus{} 2y^{2} \equal{} 3x^{3},\ y \plus{} 2x^{2} \equal{} 3y^{3}$, from which we get $ 1 \equal{} 2(x \plus{} y)$ and $ \minus{} 1 \equal{} 3(x^{2} \plus{} xy \plus{} y^{2})$. After some algebra, we deduce that $ 0 \equal{} 12x^{2} \minus{} 6x \plus{} 7$ which has no real solutions. Now suppose $ x > y > z$. Returning to the original system of equations and subtracting the second from the first, we get $ (x \minus{} y) \plus{} 2(y^{2} \minus{} z^{2}) \equal{} 3(z^{3} \minus{} x^{3})$. The LHS is positive but the RHS is negative. Contradiction. Thus $ (x, y, z) \equal{} (0, 0, 0), (1, 1, 1),\ ( \minus{} \frac {1}{3},\ \minus{} \frac {1}{3},\ \minus{} \frac {1}{3})$ are the only solutions.

Edit: Sorry about that. You replied before I could correct my post  :blush: .[/quote]
Wrong because $ y>z\not \Rightarrow y^2>z^2$

To solve this system check different cases for x,y,z and 0. You may use math software to verify your answer. I didn't check the solution because I have opinion - this problem is too easy.

Here's my solution(similar to TTsphn):
Because the system equation above is symmetric, I can assume that x>=y>=z
I need to consider 2 case:0>x>=y>=z
Then I have: x^2=<y^2=<z^2
Therefore: y-3x^3>x-3z^3>3y^3-z(contradiction since assumption)
2nd case:x>=y>=z>0
Then prove exactly the same I get the contradiction too!
Therefore, the only solution for this problem is x=y=z  :)

Do you like this system? I have very detailed solution (for all six cases for a,b,c). It is very interesting for me ... I invent some problem I don't know the solution and after some years I solve it successfully.

I think  I don't need to prove this problem with all six cases!
Cause if z<0 then I infer that x<0.Then again I infer y is also<0
Similarly I can prove that when z>0, then x and y>0 too! :)
PS:Yeah I really like this cyclic system equation! Can you post some more problems( very good for dicussing)

You want and you have it. I'm posting only few of my cyclic systems of equations I know /not all my collection / they are collected from different math olympiads and not invented by me.

1. Solve in reals: $ \{\begin{aligned} y^{3}+9x^{2} - 27x - 27 & = 0\\
z^{3} + 9y^{2} - 27y - 27 & = 0\\
x^{3} + 9z^{2}-27z - 27 &= 0\end{aligned}$.

2. Solve in reals: $ \{\begin{aligned} 2y^{3}+2x^{2} + 3x + 3 & = 0\\
2z^{3} + 2y^{2} + 3y + 3 & = 0\\
2x^{3} + 2z^{2} + 3z + 3 &= 0\end{aligned}$.

3. Solve in reals: $ \{\begin{aligned} x +y^{2} +z^{3} & = 3\\
y + z^{2} + x^{3} & = 3 \\
z + x^{2} + y^{3} &= 3 \end{aligned}$.

When you solve them if you want I may post more. All interesting systems who you know are welcome, even not cyclic. I like systems very much and I have lots of them I know maybe 30-40 interesting systems. Sometimes I may invent really difficult, simple and interesing system.

Since 1st and 2nd problem are exactly the same so I just solve 1st problem.
Here's my solution:
Assume that x>y>z(another case is completely the same) I have:
x^3>y^3, which means 9(x-z)(x+z-3)>0.Therfore:x+z>3
Beside that, I also have:y^3>z^3,which means:9(x-y)(x+y-3)<0.Therefore:x+y<3(contradict since x+y>x+z>3)
So this system equation has the unique solution when x=y=z(Done)
 :)

Adding all of the equations up and using the property $ [a]\plus{}\{a\}\equal{}a$ we get that $ x\plus{}y\plus{}z\equal{}4.9$
So substituting $ x\equal{}4.9\minus{}y\minus{}z\equal{}4.9\minus{}[y]\minus{}\{y\}\minus{}[z]\minus{}\{z\}$ into the first one we get that
$ 4.9\minus{}[y]\minus{}\{y\}\minus{}[z]\minus{}\{z\}\plus{}[y]\plus{}\{z\}\equal{}4.2\equal{}>[z]\plus{}\{y\}\equal{}0.7\equal{}>[z]\equal{}0$ and $ \{y\}\equal{}0.7$
Substituting $ [z]\equal{}0$ and $ y\equal{}[y]\plus{}\{y\}\equal{}[y]\plus{}0.7$ into the second one we get that
$ [y]\plus{}0.7\plus{}0\plus{}\{x\}\equal{}3.6\equal{}>[y]\plus{}\{x\}\equal{}2.9\equal{}>[y]\equal{}2$ and $ \{x\}\equal{}0.9$
For the last one we need to substitute $ z\equal{}[z]\plus{}\{z\}\equal{}0\plus{}\{z\}\equal{}\{z\}$ and $ \{y\}\equal{}0.7$ to get
$ \{z\}\plus{}[x]\plus{}0.7\equal{}2\equal{}>\{z\}\plus{}[x]\equal{}1.3\equal{}>\{z\}\equal{}0.3$ and $ [x]\equal{}1$
So finally we get that
$ x\equal{}[x]\plus{}\{x\}\equal{}1\plus{}0.9\equal{}1.9$
$ y\equal{}[y]\plus{}\{y\}\equal{}2\plus{}0.7\equal{}2.7$
$ z\equal{}[z]\plus{}\{z\}\equal{}0\plus{}0.3\equal{}0.3$
So $ (x,y,z)\equal{}(1.9,2.7,0.3)$

Very interesting system. What about my last system?

[quote="borislav_mirchev"]Just to be sure ... take a look at the case x>z>y.[/quote]
Here's my solution for this case:
Rewriting the system equation above, I have: y^3+x^3=(x-3)^3
                                                              z^3+y^3=(y-3)^3
                                                              x^3+y^3=(z-3)^3
Since x>z>y(as assumption) I get:(x-3)^3>(z-3)^3
That means y^3+x^3>z^3+x^3(contradiction since z>y)
PS:This method I found when solving the system equation(not yours)!I think it's really nice and useful to "kill" these kinds of system quation! But the problem 3 is very hard and strange cause it's not an homogenous system! Maybe I need more time to solve it :blush:
@salonikos:this problem you solved is in "Mathematic Olympiad challenges"! It's not hard but it's nice! :)

Try these:
B1
$ \{\begin{aligned} x = \frac {4y^{2}}{1 + 4y^{2}} \\
y = \frac {4z^{2}}{1 + 4z^{2}} \\
z = \frac {4x^{2}}{1 + 4x^{2}}\end{aligned}$.

B2
$ \{\begin{aligned} x = \frac {2y^{2}}{1 + y^{2}} \\
y = \frac {2z^{2}}{1 + z^{2}} \\
z = \frac {2x^{2}}{1 + x^{2}}\end{aligned}$.

B3
$ \{\begin{aligned} x = \frac {2y}{1 + y^{2}} \\
y = \frac {2z}{1 + z^{2}} \\
z = \frac {2x}{1 + x^{2}}\end{aligned}$.

B4
$ \{\begin{aligned} x = \frac {2y}{1 - y^{2}} \\
y = \frac {2z}{1 - z^{2}} \\
z = \frac {2x}{1 - x^{2}}\end{aligned}$.

B5
$ \{\begin{aligned} x^{3} + 3x - 3 + ln(x^{2} - x + 1) = y \\
y^{3} + 3y - 3 + ln(y^{2} - y + 1) = z \\
z^{3} + 3z - 3 + ln(z^{2} - z + 1) = x \end{aligned}$.

B6
$ \{\begin{aligned} 3a=(b+c+d)^{3} \\
3b=(c+d+e)^{3} \\
3c=(d+e+a)^{3} \\
3d=(e+a+b)^{3} \\
3e=(a+b+c)^{3} \end{aligned}$.

I believe all of them are very good. System you cannot solve is from Sweden, and USA maybe. In the olympiad it were given limited case x,y,z>0. I believe the only real solution of this system is x=y=z=1. If I weren't wrong I proved it as a student. I think I had used quadratic function inequalities of the kind x>y>z and different cases in ranges: (-oo; -1) (-1;0) (0;1)(1; +oo). If you have no enough time prove only in its limited version. I'm posting only symmetric systems. I have few more very good systems. New proposals are welcome. Here I just share interesting problems. I'm not systems expert.

Here's my solution for problem 3(just with the case x,y,z>0):
If x, y ≥ 1, then x + y^2 ≤ x^2 + y^3 (with equality if x = y = 1). But z^3 = 3 - (x+y^2) ≥ 3 - (x2+y3) = z, so z ≥ 1. But obviously if x, y, z are all ≥ 1, then they must all equal 1. Similarly, if x, y ≤ 1, then z ≤ 1 and hence (x,y,z) = (1,1,1). So if (x,y,z) ≠ (1,1,1), then x and y are on opposite sides of 1. But a similar argument shows that (assuming (x,y,z) ≠ (1,1,1) ) that y and z are on opposite sides of 1, and z and x are on opposite sides of 1(Contradiction)
PS:Nice and hard problem. I spent 2 hours solving that! :(
And here's my argument for the case x,y,z in R:
Let x>=y>=z I have:x=z+a,y=z+b(a>b>=0)
Then substract the 1st equation and the 2nd equation I 'll have the quadratic function(variable is z). Then I have to prove that b^2-4ac=<0 with a>0.that means the quadratic function is bigger than 0! And the equality happens when a,b=0! Is it your solution,Mr.Borislav mirchev? :)

My solution were somethling like your but not beautiful at all ... I'm too old I don't remember exactly  :wink: ... anyway the problem is generalized. It have very interesting story ... when I rewrote this problem like a pupil I skipped additional condition :) ... old age ... and bad memory

P.S. All my previous problems B1-B6 were to find all solutions in reals.

Here's my solution for 1st problem(since 2nd problem is similar):
I easily see that (x,y,z)=(0,0,0) is one root of this system!
Then consider the case when (x,y,z)#0
Multiple 3 equations I have:64xyz/(1+4x^2)(1+4y^2)(1+4z^2)=1(*)
Applying Cauchy's inequality I have:(1+4x^2)(1+4y^2)(1+4z^2)>=64/xyz/>+64xyz(since x,y,z>0)
So LHS of (*)=<1
This euqaity happens when (x,y,z)=(1,1,1) and cyclic values of them
And here's my argument for 3rd problem:
Multiple 3 equations I have:8/(1+y^2)(1+z^2)(1+x^2)=1(*)
Since applying Cauchy's inequality, I easily prove that x,y,z=<1
Then I need to consider 2 cases:
x,y,z=<-1 or x,y,z=1 then LHS of (*) =<1 or =1
This equality happens when x=y=z=-1 or x=y=z=1
2nd cases:-1<x<1;y,z=<-1.....Then i need more time to think instead of 5 minutes! :lol:
The 2nd way to solve this problem:changing this system into the trigonometric substitution and then solving it easily!

These systems are far simpler than you expect  :) .
There are 3 possible cases:
I - x,y,z - positive.
II - x=y=z=0.
III - x,y,z - negative.

Multiply all three equalities and use AM, GM inequality 3 times to get - x=y=z.

Third system is far more interesting...

OK.Here's my complete solution for the rest(3rd problem):
Plus 1 to each equation and I get: x+1=(x+1)^2/(x^2+1)
Similar to 2 more equations and I infer that 1>=x,y,z>=-1(*)
Therefore, coming back to the equation: 8/(1+x^2)(1+y^2)(1+z^2)=1
Then I get:x^2+y^2+z^2+(x^2y^2+y^2z^2+z^2x^2)+(xyz)^2=7
Obviously, I see that LHS=<RHS(SINCE (*))
So There's only 2 solutions that I mentioned above in this system equation!
PS:Can you help me solve hthis inequality, Mr,Borislav Michev(remmember just look at the last or nearly last post cause it's my main problem)?? A long time but nobody still helps me :(
Here's the link: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172272

I'm sorry - B4 ... is different ... B1,B2,B3 are almost similar. I'll try this problem - it seems to be interesting. At first look it seems - induction may be used and {x}=x-[x] ... I'll try to solve this.

Here's my solution for 4th problem:
Let x=tga,y=tgb and z=tgc(a,b,c # pi/2+k2pi)(k is interger)
Then rewriting this system equation, I get: tga=tg2b
                                                          tgb=tg2c
                                                          tgc=tg2a
Then I easily infer that: a-2b=0 or a-2b=2lpi(l is interger)
                                 b-2c=0 or b-2c=2mpi(m is interger)
                                 c-2a=0 or c-2a=2npi(n is interger)
Therefore, the final step is just examine all of these cases to get the result(maybe 6 cases)(remmeber tga,tgb,tgc#1) :)


For each $x,y>0$ and $x\neq y$ and

$\frac{f(x)-f(y)}{x-y}=\frac{f(x)}{x}+\frac{f(y)}{y}$ then find $f(x)$

Working out gives $x^2f(y)=y^2f(x)$ so $\frac{f(x)}{x^2}$ is constant so $f(x)=ax^2$.


Let $ n > 1$ be a fixed integer. Define functions $ f_0(x) \equal{} 0,$ $ f_1(x) \equal{} 1 \minus{} \cos(x),$ and for $ k > 0,$ \[ f_{k\plus{}1}(x) \equal{} f_k(x) \cdot \cos(x) \minus{} f_{k\minus{}1}(x).\] If $ F(x) \equal{} \sum^n_{r\equal{}1} f_r(x),$ prove that

[b](a)[/b] $ 0 < F(x) < 1$ for $ 0 < x < \frac{\pi}{n\plus{}1},$ and
[b](b)[/b] $ F(x) > 1$ for $ \frac{\pi}{n\plus{}1} < x < \frac{\pi}{n}.$

Dear orl,is there something wrong in this question?
When $n=2$,$F(x)=f_1(x)+f_2(x)=1-cos^2 x$
Obviously,it isn't consistent with your conclusion


[quote="tmbtw"][quote="toan hoc"]Prove that for every couple of positive integers $n;k$ the number $(\sqrt{n}-1)^{k}$ can be written in the form $\sqrt{a_{k}}-\sqrt{a_{k}-(n-1)^{k}}$ with $a_{k}\in N^{*}$[/quote]
case 1: k is even
$(\sqrt{n}-1)^{k}=(1-\sqrt{n})^{k}=A-\sqrt{n}B$ (use Newton's formula)
And $(1+\sqrt{n})^{k}=A+\sqrt{n}B$
So, $(1-n)^{k}=A^{2}-nB^{2}$
Thus $(\sqrt{n}-1)^{k}=\sqrt{A^{2}}-\sqrt{A^{2}-(n-1)^{k}}$
Case 2: k is odd. Similarly :)
So we've done[/quote]
Very nice


If $ |x \minus{} 2| \equal{} p$, where $ x < 2$, then $ x \minus{} p \equal{}$

$ \textbf{(A)}\minus{}\!2 \qquad \textbf{(B)}\  2 \qquad \textbf{(C)}\  2 \minus{} 2p \qquad \textbf{(D)}\  2p \minus{} 2\qquad \textbf{(E)}\ |2p \minus{} 2|$

[hide="Solution"]Because $x<2$, the stuff inside the absolute value is always negative. (like my terminology?  :P )

Thus, $p=-(x-2)$ and $x=2-p$. Our answer is then $x-p=(2-p)-p=2-2p$.[/hide]

What is the problem source? If you want this added into the Resources section you have to give the source.

[quote="apiarist1"]Another way: x=-1 -> p=3; then x-p=-4 and the only appropriate formula to this result is C.[/quote]
If $ x<0$, then $ p>2$, $ x\minus{}p<\minus{}2$, and only $ C$ works.
Or (easier to visualize) if the absolute value of $ p$ (and of $ x$) is very large $ (p>>0, x<<0)$, $ x\minus{}p<<0$, and only $ C$ works.

[hide=Solution]If $x<2$, then $|x-2|=-(x-2)$, so $2-x=p$ or $2=x+p \implies 2-2p=x-p \implies \boxed{\text{(C)}}$.[/hide]


Prove that there exists a bijection function $f$ such that:
$f$: $N_{0}\to N_{0}$
$f(3mn+m+n)=4f(m)f(n)+f(m)+f(n)$

Let g a function such that g(n)=4f((n-1)/3)+1 where g:3N+1->N You get g(nm)=g(n)g(m) and you can see that f is a bijection if and onli if g is a bijection. Now find g :P


Since I haven't been on here very long, or been a prolific poster, this doesn't really affect me. However, the "forum information" page, says that the post titles are references to the "unsolved" Millennium problems, since the Poincare theorem has been proven, should the wording be changed at all?

I think this was discussed somewhere, but search doesn't turn up anything, and I'm pretty sure nothing was decided in that topic anyway.

[quote="Temperal"]This proof was confirmed only a year ago, and the post count system was established before this. http://en.wikipedia.org/wiki/Poincare_theorem[/quote]

I knew both those things, I was wondering whether it would be changed or not?

[quote="Temperal"]I think the admins have more pressing matters to worry about.[/quote]Like the general evacuation in Alpine, where our headquarters are, due to the close proximity of the wild fires  :maybe:  :noo:

[quote="Temperal"]I think the admins have more pressing matters to worry about.

Also, the Poincare Theorem is still one of the Millenium Problems regardless whether anyone's solved it or not.[/quote]

The wiki specifically states "unsolved", I was merely suggesting a change in the wording

And I'm sorry if this seems frivolous

[quote="SimonM"]The wiki specifically states "unsolved"[/quote]

It doesn't anymore.


geometry
3D geometry
function
set theory
irrational number

First of all, we have to remember the [b]one-one correspondence [/b]between some numbers and their cubes.
[b]TABLE 1[/b]
1<->001,
2<->008,
3<->027,
4<->064,
5<->125,
6<->216,
7<->343,
8<->512,
9<->729.
************************************************
Next, build up the relation between the 10 digits and the last digit of each of their cubes:
[b]TABLE 2[/b]
0 <-> 0,
1 <-> 1,
2 <-> 8,
3 <-> 7,
4 <-> 4,
5 <-> 5,
6 <-> 6,
7 <-> 3,
8 <-> 2,
9 <-> 9.
*************************************************
[color=violet][size=150]Okay, are you ready to play this game ?[/size][/color]*************************************************
Now, think of a[b] natural number less than 100[/b] and [b]don't let me know[/b].
For illustration, you may imagine you have the number 47 in hand.
Use a calculator to evaluate the cube of this number, and of course you will get [b]103,823[/b].[size=200] Please double check the answer.[/size]
You only need to show me the final answer 103,823 and I can quickly tell you that [b]the original number [/b]should be 47! ( without using calculator or any kind of rapid calculation.)
************************************************
The logic is like this:
Step 1
Split the number 103,823 into two sections, namely
103 and 823.
Step 2
Look at the number 82[color=red]3[/color] and concentrate on the last digit [color=red]3[/color], ignore the 82....
From TABLE 2 you can see that the last digit of the original number should be 7.
Step 3
Consider the number 103, which is between 064 and 125, therefore from TABLE 1 you can deduce that the tenth digit should be 4. ( as the original number should be between [b]forty something and fifty![/b])
Combining these clues, we conclude that the original number [size=150]MUST BE 47[/size] !!!!!!
**************************************************
MORE EXAMPLES:

The cube root of 4,913:
Consider 004 and 913.
The cube root of 4913 should be 17.

The cube root of 970,299:
Consider 970 and 299.
The cube root of 970,299 should be 99

The cube root of 658,503:
Consider 658 and 503:
The cube root of 658,503 should be 87.

[color=green]Questions to think about:[/color]
1. What is the logic behind this method?
2. What happens if your friend tell you the wrong number when you are showing off your magic in front of him???
3. What is the limitation of this method?
4. Can we extend this method to other cases, and how?

don't post your homework problems here....

how did you get this trick figured out, luimichael?

Well, this is assuming ur given the cube of an integer, so it is a complete guess as to whether or not u hve the true cube root. You cannot say "must" as you have-you have to clarify that ur assuming it has an integer cube root. I wouldn't use this in any official matter.

Well, if you read my post carefully, I have emphasized that you got to ask your friend to [size=200]double check the answer!!!!!![/size]
That is the crucial point of this little magic...
Playing around with numbers is one of the ways to enjoy Mathematics.(in my opinion :blush: )

I think I have figured this trick out.

assume b and a are digits from 0-9

$ b^{2}$ has a unit digit similar to $ (ab)^{2}$ since $ (ab)^{2}\equal{} (10a\plus{}b)^{2}\equal{} 100a^{2}\plus{}20ab\plus{}b^{2}$

since $ 100a^{2}$ or $ 20ab$ do not contribute to the units digit it is unchanged.

This trick would not have worked if two numbers in between 0-9 had the same units digit when cubed.

For the second trick its easy since the first three digits in $ (ab)^{2}$ would be in between $ 100a^{2}$ and $ 100(a\plus{}1)^{2}$

Thus if you get a number and know it is a cube of a two-digit number, you can find the two digit number!!!

I dont think this trick can work for the cubes of three digit numbers.  Not sure though!

[quote="Fraenkel"]I think I have figured this trick out.

assume b and a are digits from 0-9

$ b^{2}$ has a unit digit similar to $ (ab)^{2}$ since $ (ab)^{2} \equal{} (10a \plus{} b)^{2} \equal{} 100a^{2} \plus{} 20ab \plus{} b^{2}$

since $ 100a^{2}$ or $ 20ab$ do not contribute to the units digit it is unchanged.

This trick would not have worked if two numbers in between 0-9 had the same units digit when cubed.

For the second trick its easy since the first three digits in $ (ab)^{2}$ would be in between $ 100a^{2}$ and $ 100(a \plus{} 1)^{2}$

Thus if you get a number and know it is a cube of a two-digit number, you can find the two digit number!!!

I dont think this trick can work for the cubes of three digit numbers.  Not sure though![/quote]

Wow! Nice..! :lol:

[quote="luimichael"]First of all, we have to remember the [b]one-one correspondence [/b]between some numbers and their cubes.
[/quote]

i thought one-one was used in graphing

Well, a graph on a paper or a computer is only a concrete representation of the graph of a mapping ( also called function).
In Set Theory, the graph is only a set.
There are many functions which cannot be graphed on a paper properly.
********************************************************************************
In mathematics, the following technical terms are very important:
1. relation;
2. mapping/ function ;
3. one-one mapping,or called injection( I have seen an author saying that it should be called "two-to-two" to be exact.);
4. many-to one;
5. one-to-many mapping (strictly speaking, a mapping cannot be 1-to-many :P );
6. onto mapping, or called surjection;
You may take some time to surf around the internet in order to get the info of the these terms.
*******************************************************************************

ok. i think i get it. and when you said

[quote="luimichael"]There are many functions which cannot be graphed on a paper properly.
[/quote]

you mean like $ f(x)\equal{}\sqrt{\minus{}|x|}$

The graph of your function is an [b]empty set[/b], and of couse it can't be graphed. Good example!
There are some other functions which behave mischievously.
Mathematicians have spent a great deal of time in understanding the behaviors of these weird functions.
Here let me write down a famous example used in Analysis (a Branch of Mathematics):
f(x) = 0 when x is a rational number and f(x) = 1 when x is an irrational number, where x can be any real number.( This name of this function is something like [b]Dirichlet Function[/b], possibly maybe.)
Basically the graph of this function looks like two horizontal lines which are full of holes!
You can't draw the graph properly, even with the best computer on earth, because there are infinitely many holes on the two lines.  :)

please don't title this as "magic"

it's some very elementary common sense.

actualy, its not so elementary. is there a proof for this?

luimichael:

that equation CAN be graphed... over the complex plane

whats a complex plane, and how DO you graph imaginary numbers? this should go in high school basics

The original topic belongs here, but imaginary discussion does go in HSB. Also, try searching on aops wiki or on the forum.


Online Math Open
algebra
polynomial
Vieta

Εστω φ:R-->R μια πολυωνυμικη συναρτηση βαθμου ν με συντελεστες πραγματικους αριθμους.Υποθετουμε οτι ολες οι ριζες της φ ειναι πραγματικοι αριθμοι.Δειξε οτι:
(ν-1)*(φ'(χ))^2 - ν*φ(χ)*φ''(χ)>=0 για καθε χ ανηκει στους πραγματικους.

manos,gt dn tin evales sto anisotites 11-50?nomizo ina ka8ari algebra....

I dosmeni sxesi metasximatizete se
$(\frac{f(x)}{f'(x)})'\geq \frac{1}{n}$
omos
$\frac{f'(x)}{f(x)}=\frac{1}{x-x_{1}}+...+\frac{1}{x-x_{n}}$
opote telika pernume
$n(\frac{1}{(x-x_{1})^{2}}+....+\frac{1}{(x-x_{n})^{2}})\geq (\frac{1}{x-x_{1}}+....+\frac{1}{x-x_{n}})^{2}$
pu isxii apo BCS..

 :D

Την φ θα την λέω $P$ για ευκολία και έστω $p_{1},...,p_{n}$ οι ρίζες του πολυωνύμου .
Λύση [hide]Νάτη για σένα [/hide]
Είναι $P(x)=(x-p_{1})(x-p_{2})....(x-p_{n})$  (μπορούμε να υποθέσουμε ότι έχει συντελεστή του μεγιστοβάθμιου όρου 1 αφού σε διαφορετική περίπτωση αυτός θα απαλοιφόταν έτσι και αλλιώς  :wink:  ) .Η ανισότητα για $x=p_{1},\ p_{2},\ ,...,p_{n}$ ισχύει άρα αρκεί να τη δείξουμε για $x\neq p_{1},\ p_{2},\ ,...,p_{n}$
Einai
$P'(x)=P(x)(\frac{1}{x-p_{1}}+...+\frac{1}{x-p_{n}})$

kai $P''(x)=\sum_{i=1}^{n}(\frac{P'(x)}{x-p_{i}}-\frac{P(x)}{(x-p_{i})^{2}})$

άρα η προς απόδειξη είναι ισοδύναμη με
$(n-1)P^{2}(x)(\frac{1}{x-p_{1}}+...+\frac{1}{x-p_{n}})^{2}\geq nP(x)(P(x)(\frac{1}{x-p_{1}}+...+\frac{1}{x-p_{n}})^{2}-P(x)(\frac{1}{(x-p_{1})^{2}}+...+\frac{1}{(x-p_{n})^{2}}))$

ή ισοδύναμα
$n(\frac{1}{(x-p_{1})^{2}}+...+\frac{1}{(x-p_{n})^{2}})\geq (\frac{1}{x-p_{1}}+...+\frac{1}{x-p_{n}})^{2}$

που ισχύει από την ανισότητα Cauchy .... :wink:

Ράλλη με πρόλαβες.... :P

[quote="silouan"]
Ράλλη με πρόλαβες.... :P[/quote]

 :P  :P ...
tin yiloixa etimi sto mialo mu,gt x8es to vradi,prospa8usa na vro kamia anisotita san ayti tu manos me paragogus,+simera pu mpika,otan tin ida..leo,kala ti egine?pote tin estila?
well-done manos....

to...simperasma apo x8es pantos apo m1 itan i pio kato anisotita..pu exi arketes eksoterikes omiotites m ayti edo...
Esto $P(x)=x^{n}+a_{n-1}x^{n-1}+....a_{1}x+a_{0}$ polionimo me pragmatikus sintelestes,n pragmatikes rizes,o sintelestis tu megistova8miu oru ine 1,tote
$(n-1)(a_{n-1})^{2}\geq 2na_{n-2}$

ine sto idio mikos kimatos...

 :D

ισχυει μονο οταν ολες οι ριζες ειναι πραγματικες ?

dn ksero kosta..mpori kapia sin8iki na ine periti,ala tis evala oles gia na exo to kefali mu isixo...mpori na isxi kai me pio xalares sin8ikes...
opos katalavenis prokite gia algebriki anisotita,kati san ayti tu manos,kai simfona me tin lisi mu,xriazete i sin8iki na ine oles pragmatikes...
tora..mia ali lisi mpori na min apeti kati tetio...

 :D

οκ απλα αναρωτιθηκα  :D

mpa....dn nomizo..kai 8a su po gt....
[hide]katarxin san antiparadigma des to $x^{4}-3x^{3}+3x^{2}+2$
to opio exi $x=1$ kai $x=2$ (2 pragmatikes) $x=i,x=-i$
an isxie kai tote o tipos 8a eprepe
$(2-1)^{2}(-3)^{2}\geq 2(2)(3)$
[hide]pantos genika epidi ego to elisa xrisimopiontas tus tipus tu vieta,kateliksa [b]me isodinamies[/b] oti i anisotita ayti ine isodinami me to a8risma  ton tetragonon tis diaforas ton rizon ana dio na ine megalitero i sio tu midenos,to opio den isxii genika otan i rizes ine migadikes...[/hide][/hide]

 :D


When two fair six-faced dice are each rolled once, what is the probability that at least one of the numbers rolled is a 6? Express your answer as a common fraction.

[quote="ernie"]Use complementary probability.  The probability that this won't happen is simply $ \left ( \frac56 \right ) ^2$ or $ \frac{25}{36}$.  Thus, the probability that this will happen is $ 1 \minus{} \frac{11}{36}$, or $ \boxed{\frac{25}{36}}$.[/quote]
I think ernie meant the probability that this will happen is: [b] $ 1 \minus{} \frac{25}{36}$[/b], not  $ 1 \minus{} \frac{11}{36}$.

Therefore, the probability that at least one of the numbers rolled is a $6$ is $\boxed{\frac{11}{36}}$.

EDIT: Grammar mistakes...><


two investors, a and b, have 10000 each. they split their money among three accounts with annual yields of 8% 6% and 5% but investor a only puts half as much in the 8% and 6% as b. if investor a earns 580 and b earns 660 then the amount that b must have invested at 8% is?

Letting $ x$ be the amount $ a$ invests in the $ 8$ per cent account and $ y$ be the amount $ a$ invests in the $ 6$ per cent account, and hoping to get a pair of simultaneous equations, I get $ 3x \plus{} y \equal{} 8000$ for both $ a$ and $ b$!

Perhaps this is where the question really starts.


[quote="PPonte"]Why is the mathematical constant [i]e[/i] called Euler's number or Napier's constant? Did their works involved that constant?

Thank you.[/quote]

[url=http://mathworld.wolfram.com/e.html]This will maybe help[/url]

I think that (e), which is the symbol that represent this constant is in the honor of Euler

Euler did name the constant $e$, but not (apparently) for his name, to stand for "exponential" I think


Solve in $ \mathbb{Z^\plus{}}$ the equation $ (xy\minus{}7)^2\equal{}x^2\plus{}y^2$

Consider this equation as a quadratic equation(variable x)  we need to have:
delta(x)>=0 which is equivalent to: 49(y^2)+49(1-y^2)y^2>=0.(Examine the case when y^2=1 first) :)

Alternatively,
$ x^2y^2\plus{}49\minus{}14xy\equal{}x^2\plus{}y^2\Rightarrow x^2y^2\minus{}12xy\plus{}36\minus{}36\plus{}49\equal{}(x\plus{}y)^2\Rightarrow (x\plus{}y)^2\minus{}(xy\plus{}6)^2\equal{}13\Rightarrow (x\plus{}y\plus{}xy\minus{}6)(x\plus{}y\minus{}xy\plus{}6)\equal{}13$
Since 13 has factors $ \pm 1,\pm 13$, it remains only to substitute these and check which give integral solutions.

[quote="pelao_malo"]Solve in $ \mathbb{Z^ \plus{} }$ the equation $ (xy \minus{} 7)^2 \equal{} x^2 \plus{} y^2$[/quote]
If x>3 and y>3, then we have that $ (x\minus{}1)(y\minus{}1)>8 \Rightarrow xy\minus{}7>x\plus{}y \Rightarrow (xy\minus{}7)^{2}>(x\plus{}y)^{2}>x^{2}\plus{}y^{2}$.
So we have that $ x \leq 3$, from where we can easily solve the cases.


function
IMO Shortlist
algebra proposed
algebra

Find all $f: \mathbb R \to \mathbb R$ such that $ f(f(x) + y) = 2x+ f(f(y) - x)$  for all $ x,y\in \mathbb R$.

Hi :) !!!

$ f(f(x) + y) = 2x + f(f(y) - x)$    $ (E)$

$ Let \ \ \ f(0) = a$

than we put: $ x = 0 \ \ \ and \ \ \ y = x$

than: $ f(a + x) = f(f(x))$ ...  $ (1)$

we put in $ (E)$ :$ y = 0$ than we have: $ f(f(x)) = 2x + f(a - x)$...$ (2)$

of $ (1)$ and $ (2)$:

$ f(a + x) - f(a - x) = 2x \Leftrightarrow f(a + x) - (a + x) = f(a - x) - (a - x)$(3)

we put $ g(x) = f(x) - x$ than  into (3) we have: $ g(a + x) = g(a - x)$

than:

$ f(x) = x + h(x)$ with $ h$ is all function axial symetrique where $ (D): x = f(0) = a$ is the $ axe$ $ of$ $ symetrie$ ($ i.e$ $ h(2f(0) - x) = h(x))$

than the only function $ h$ verified $ (E)$ is the fonction constante $ h(x) = b$

finaly:
$ f(x) = x+b$
PS: i edited my poste sorry because i forgot the onstante function

[quote="mathema*"]
$ f(x) \equal{} x \plus{} h(x)$ with $ h$ is all function axial symetrique where $ (D): x \equal{} f(0) \equal{} a$ is the $ axe$ $ of$ $ symetrie$ ($ i.e$ $ h(2f(0) \minus{} x) \equal{} h(x))$

than the only function $ h$ verified $ (E)$ is the fonction nul $ h(x) \equal{} 0$
[/quote]I don't understand how do you deduce the $ h(x)\equal{}0$. Can you explain?

[quote="Johan Gunardi"][quote="mathema*"]
$ f(x) \equal{} x \plus{} h(x)$ with $ h$ is all function axial symetrique where $ (D): x \equal{} f(0) \equal{} a$ is the $ axe$ $ of$ $ symetrie$ ($ i.e$ $ h(2f(0) \minus{} x) \equal{} h(x))$

than the only function $ h$ verified $ (E)$ is the fonction nul $ h(x) \equal{} 0$
[/quote]I don't understand how do you deduce the $ h(x) \equal{} 0$. Can you explain?[/quote]

sorry i edited this see my answer:
OK if we replace f into original equation funtional we find that $ h(x) \equal{} b$ !!!!

[quote="mathema*"][quote="Johan Gunardi"][quote="mathema*"]
$ f(x) \equal{} x \plus{} h(x)$ with $ h$ is all function axial symetrique where $ (D): x \equal{} f(0) \equal{} a$ is the $ axe$ $ of$ $ symetrie$ ($ i.e$ $ h(2f(0) \minus{} x) \equal{} h(x))$

than the only function $ h$ verified $ (E)$ is the fonction nul $ h(x) \equal{} 0$
[/quote]I don't understand how do you deduce the $ h(x) \equal{} 0$. Can you explain?[/quote]

OK if we replace f into original equation funtional we find that $ h(x) \equal{} 0$ !!!![/quote]

I dont think so. If you replace f by $ x \plus{} h(x)$ in the original equation, you find :
$ x\plus{}h(x)\plus{}y\plus{}h(x\plus{}h(x)\plus{}y)\equal{}2x\plus{}y\plus{}h(y)\minus{}x\plus{}h(y\plus{}h(y)\minus{}x)$

So:
$ h(x)\plus{}h(x\plus{}h(x)\plus{}y)\equal{}h(y)\plus{}h(y\plus{}h(y)\minus{}x)$

And not really $ h(x)\equal{}0$ !

[quote="DCTPKTCSPHN"]Find all $ f: R\rightarrow R$ such that $ : f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$  $ \forall x,y\in R$[/quote]

Let $ P(x,y)$ be the equation : $ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$

$ P(\frac {f(0) \minus{} x}{2}, \minus{} f(\frac {f(0) \minus{} x}{2}))$ $ \implies$ $ f(0) \equal{} f(0) \minus{} x \plus{} f(something)$ and so $ f(x)$ is a surjection.

Since $ f(x)$ is surjective, for x given, let then $ c$ such that $ f(c) \equal{} 0$ and $ u$ such that $ f(u) \equal{} x \plus{} c$

Then $ P( c,u)$ $ \implies$ $ x \plus{} c \equal{} 2c \plus{} f(x)$ and so $ f(x) \equal{} x \minus{} c$

And it's easy to verify that this necessary condition is sufficient.

The solutions are $ f(x) \equal{} x \minus{} c$ for any c.

[quote="pco"][/quote][quote="mathema*"][quote="Johan Gunardi"][quote="mathema*"]
$ f(x) = x + h(x)$ with $ h$ is all function axial symetrique where $ (D): x = f(0) = a$ is the $ axe$ $ of$ $ symetrie$ ($ i.e$ $ h(2f(0) - x) = h(x))$

than the only function $ h$ verified $ (E)$ is the fonction nul $ h(x) = 0$
[/quote]I don't understand how do you deduce the $ h(x) = 0$. Can you explain?[/quote]

OK if we replace f into original equation funtional we find that $ h(x) = 0$ !!!![/quote][quote="pco"]

I dont think so. If you replace f by $ x + h(x)$ in the original equation, you find :
$ x + h(x) + y + h(x + h(x) + y) = 2x + y + h(y) - x + h(y + h(y) - x)$

So:
$ h(x) + h(x + h(x) + y) = h(y) + h(y + h(y) - x)$

And not really $ h(x) = 0$ ![/quote]

yes we can find the constant fuction by resolution of this equation functional!!!

[quote="mathema*"]$ h(x) \plus{} h(x \plus{} h(x) \plus{} y) \equal{} h(y) \plus{} h(y \plus{} h(y) \minus{} x)$

yes we can find the constant fuction by resolution of this equation functional!!![/quote]

Please, show us.

Here is another solution:
$ y\equal{}\minus{}f(x)$ $ \Longrightarrow$ $ f(0)\equal{}2x\plus{}f(f(\minus{}f(x))\minus{}x)$ which means that f is surjective.
So there is $ x_0$ s.t $ f(x_0)\equal{}0.$
$ x\equal{}x_0$ $ \Longrightarrow$ $ f(y)\equal{}2x_0\plus{}f(f(y)\minus{}x_0)$ or
$ f(f(y)\minus{}x_0)\equal{}(f(y)\minus{}x_0)\minus{}x_0$ Since f is surjective $ f(y)\minus{}x_0$ spans all real numbers.
So $ f(x)\equal{}x\minus{}x_0$ and that satisfies the original equation.

[quote="ozgurkircak"]Here is another solution:
$ y \equal{} \minus{} f(x)$ $ \Longrightarrow$ $ f(0) \equal{} 2x \plus{} f(f( \minus{} f(x)) \minus{} x)$ which means that f is surjective.
So there is $ x_0$ s.t $ f(x_0) \equal{} 0.$
$ x \equal{} x_0$ $ \Longrightarrow$ $ f(y) \equal{} 2x_0 \plus{} f(f(y) \minus{} x_0)$ or
$ f(f(y) \minus{} x_0) \equal{} (f(y) \minus{} x_0) \minus{} x_0$ Since f is surjective $ f(y) \minus{} x_0$ spans all real numbers.
So $ f(x) \equal{} x \minus{} x_0$ and that satisfies the original equation.[/quote]

Yes, quite good.

But it's very near from mine :)

It is a well-known  problem which was proposed in IMO shortlist 2002 (problem A1,page 21). See the link http://www.mathlinks.ro/Forum/viewtopic.php?t=15588
The main idea is proving that $ f$ is surjective.


How many cubic units are in the volume of the solid created by rotating the quadrilateral whose vertices are $ (0,3)$, $ (6,3)$, $ (6,0)$, and $ (0,0)$ around the y-axis? Express your answer in terms of $ \pi$.

This quadrilateral is a rectangle, with length 6 and height 3.  If we rotate around the y-axis, we create a cylinder of height 3, and radius 6.  So its volume is $ V\equal{}\pi r^{2}h\equal{}36\times 3\times\pi\equal{}108\pi$.


Hullo; I'd like to modify the list environment so that it can't break across pages. Is my best bet to create my own environment, or is there some better way? I've googled but nothing has been helpful.

Many thanks.

Billy

[code]\makeatletter
\newcommand\Nopagebreak{\@nobreaktrue\nopagebreak}
\makeatother [/code]Only use this just before a list environment, as it will prevent the break above the next list.

More generally, if I have say a paragraph (or two paragraphs) that I would like to keep on one page, will this command help?

No it only applies to lists - you can see the detailed reasoning [url=http://groups.google.com/group/comp.text.tex/msg/180369d1028f2a66]here[/url].
Keeping paragraphs together can be a pain. You can try \samepage and \nobreak but it may not result in what you want. Suppose we want to keep paragraphs 5 and 6 of this document together:
[code]\documentclass{article}
\usepackage{lipsum}
\begin{document}
\lipsum[1-3]
\lipsum[4]

\emph{Keep together the next two paragraphs}

\lipsum[5]
\lipsum[6]
\lipsum[7]
\end{document}[/code]then adding \samepage and \nopagebreak (or if that doesn't work \nopagebreak[4])
[code]\documentclass{article}
\usepackage{lipsum}
\begin{document}
\lipsum[1-3]
\lipsum[4]

\emph{Keep together the next two paragraphs}

\samepage
\lipsum[5]
\nobreak
\lipsum[6]
\lipsum[7]
\end{document}[/code]
Another way that you can try is to stick the two paragraphs in a minipage or use a box:
[code]\documentclass{article}
\usepackage{lipsum}
\begin{document}

\lipsum[1-3]
\lipsum[4]

\emph{Keep together the next two paragraphs}

\vbox{\lipsum[5]
\lipsum[6]}
\lipsum[7]
\end{document}[/code]
Finally, if you don't want to move the paragraphs onto the next page causing a large white space, and only need  to add a line or two then you can use
\enlargethispage{[i]size[/i]} or
\enlargethispage*{[i]size[/i]}
This adds [i]size[/i] to textheight for just one page. The * version also shrinks interline spacing.


A positive charge is set in motion with the initial velocity $v_{0}$.There are both electrical and magnetic fields.Find the velocity of this charge when the the velocity vector is in position 180 degrees from the initial velocity $v_{0}$.

The eletric fiel cannot change the direction of the velocity

If you made a better problem i can solve

you did not tell nothing about the fields, neither if they are constant. etc..

If you saw the attachment,you would understand that everything is clear in the problem.If you have any other problem, don't hesitate to ask.

i dont use office :)

dont have any program to open ppt

I think this is an interesting problem, So I have solved the movement of a charged particle in an arbitrary constant  electro-magnetic field. I will post that and then the solution to namiq's problem which is just a special case.
Lets choose the $z$ axis as the direction of the magnetic field and let the $\vec E$ be in the $yz$ plane. The $y$ axis is pointing downwards. So the equations of motion:

$m\ddot x =qB\dot y$
$m\ddot y =qE_{y}-qB \dot x$
$m\ddot z =qE_{z}$

Where $E_{z}$ and $E_{y}$ are the $z$ and $y$ components of  $E$.

Lets define the following constants:
$k=\sqrt{\left ( v_{0x}-\frac{E_{y}}{B}\right )^{2}+v_{0y}^{2}}$
$\tan \phi =\cfrac{v_{0y}}{v_{0x}-\frac{E_{y}}{B}}$
$\omega =\frac{qB}{m}$

we can figure these out after integrating the equations of motion, so the velocities:
$\dot x = k \cos (\omega t-\phi)+\frac{E_{y}}{B}$
$\dot y =-k \sin (\omega t-\phi)$
$\dot z = \frac{qE_{z}}{m}t+v_{0z}$

hence the path of the particle:
$x= \frac{k}{\omega}\sin (\omega t-\phi)+\frac{E_{y}}{B}t+x_{0}$
$y=\frac{k}{\omega}\cos (\omega t-\phi)+y_{0}$
$z=\frac{qE_{z}}{2m}t^{2}+v_{0z}t+z_{0}$

(it looks like its quite simple to solve but try, it took me quite a few pages :D )

So now to namiq's problem:
we have :
$E_{z}=0$, $E_{y}=E$, $v_{0x}=v_{0}$, $v_{0y}=v_{0z}=0$, $x_{0}=y_{0}=z_{0}=0$
 so now the constants:
$k=v_{0}-\frac{E}{B}$
$\tan \phi =0$ $\to$ $\phi=0$
$\omega=\frac{qB}{m}$

So the velocities:
$\dot x=\left ( v_{0}-\frac{E}{B}\right ) \cos (\omega t)+\frac{E}{B}$
$\dot y =-\left ( v_{0}-\frac{E}{B}\right ) \sin (\omega t)$
$\dot z =0$
The path:
$x=\frac{m}{qB}\left ( v_{0}-\frac{E}{B}\right ) \sin (\omega t)+\frac{E}{B}t$
$y=\frac{m}{qB}\left ( v_{0}-\frac{E}{B}\right ) \cos (\omega t)$
$z=0$
When the direction of the velocity vector is $180^{½?$ with the beginning velocity vector then $y=0$
So:
$\cos (\omega t)=0$
hence:
$\dot x = \frac{E}{B}$
$\dot y =-\left ( v_{0}-\frac{E}{B}\right )$
So the velocity of the particle:
$v=\sqrt{v_{0}^{2}-\frac{2E}{B}\left ( v_{0}-\frac{E}{B}\right ) }$


geometry
circumcircle
perpendicular bisector
geometry proposed

[quote="Pascual2005"]Let $C_1$ and $C_2$ be two circles intersecting in two points $A$ and $B$. Some line $l_1$ passing through the point $B$ meets the circles $C_1$ and $C_2$ at the points $E$ and $D$, respectively (apart from $B$), and some other line $l_2$ passing through the point $B$ meets the circles $C_1$ and $C_2$ at the points $F$ and $G$, respectively (apart from $B$). If the points $E$, $F$, $G$, $D$ lie on one circle with center $O$, prove that $\measuredangle BAO=90^{\circ}$.[/quote]

Since E = l1^c1, D = l1^c2, F = l2^c1 and G = l2^c2 are concyclic points, the triangles EFB and GDB are similar. Then if P and Q are the centers of c1 and c2 respectively, PEB ~ QBG. Let define R = PO^FB. Since PO is the perpendicular bisector of FE, we have that <RPE = <RBE. So PREB is ciclyc and, for istance, <PRB = <PEB = <QBG which means that PO//BQ. Analogously one can prove that QO//BP. Then as APQ~BPQ~OPQ it holds that AO//PQ_|_AB i.e. <BAO = 90°.

Here's what I did:

[[b]Moderator edit:[/b] A warning: The letters $D$, $E$, $F$, $G$ in this post are used in a different way than in the problem. Here, $D=l_1\cap C_1$, $G=l_1\cap C_2$, $E=l_2\cap C_1$ and $F=l_2\cap C_2$.]

The inversion of pole $B$ and power $BG\cdot BD$ turns the circle $\mathcal C_1$ into $FG$ and the circle $\mathcal C_2$ into $DE$. It means that the intersection of the two circles, which is $A$, is turned into the intersection of $FG,ED$, meanign that this intersection lies on $AB$. Now, $A$ is the image of $B$ through the inversion of pole $X=FG\cap DE$ and power $XD\cdot XE$. It really is a well-known result (which has received solutions on several occasions on the forum, I think), that $OA\perp XB$. It follows from the fact that the image $O'$ of $O$ through that same inversion lies on the polar of $X$, so $BO'\perp XO$.

my solution is simpler, it only uses the nice lemma Grobber showed about the locci of the midpoint of a line trought $B$ cutting $C_1$ and $C_2$ in $E$ and $F$.

[quote="Pascual2005"]Let $C_{1}$ and $C_{2}$ be two circles intersecting in two points $A$ and $B$. Some line $l_{1}$ passing through the point $B$ meets the circles $C_{1}$ and $C_{2}$ at the points $E$ and $D$, respectively (apart from $B$), and some other line $l_{2}$ passing through the point $B$ meets the circles $C_{1}$ and $C_{2}$ at the points $F$ and $G$, respectively (apart from $B$). If the points $E$, $F$, $G$, $D$ lie on one circle with center $O$, prove that $\measuredangle BAO=90^{\circ}$.[/quote]

Just wanted to say that this result follows from Theorem 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=151499#p151499]http://www.mathlinks.ro/Forum/viewtopic.php?t=4919 post #7[/url], applied to the points F, G, D, E which lie on a circle with center O. (The line DE meets the line FG at the point B; the point A is the point of intersection of the circumcircles of triangles DBG and FBE (apart from B); thus, according to Theorem 2, this point A coincides with the point of intersection of the circumcircles of triangles FOD and EOG (apart from O) and satisfies < BAO = 90°.)

  Darij


$ \triangle TRI$ is an isosceles triangle with vertex angle $ T$ that measures $ 80^{\circ}$. The circle $ I$ has radius 6 cm. Find the number of square centimeters in the area of the shaded region. Express your answer in terms of $ \pi$.
[asy]import graph;
size(100);
pair I=(0,0),T=(9,10),R=(18,0);
draw(I--T--R--I);
draw(Circle(I,6));
label("$T$",T,N);
label("$R$",R,SE);
label("$I$",I,SW);
pair X = intersectionpoint(Circle(I,6),I--T);
pair Y = intersectionpoint(Circle(I,6),I--R);
fill(I--arc(I,6,0,48)--cycle,gray(0.6));[/asy]

Angle at I is 50 degrees.
Surface: $ \frac{50r^2\pi}{360}\equal{}5\pi$


a,b,c positive reals

abc=1

prove

sigma cyclic   a^3/[1+a][1+b] >= 3/4

[hide]
By AM-GM,
$ \frac {a^3}{(a \plus{} 1)(b \plus{} 1)} \plus{} \frac {a \plus{} 1}{8} \plus{} \frac {b \plus{} 1}{8} \ge \frac {3a}{4}$
Thus,
$ \sum \frac {a^3}{(a \plus{} 1)(b \plus{} 1)} \ge \frac {a \plus{} b \plus{} c}{2} \minus{} \frac {3}{4}$
But $ a \plus{} b \plus{} c \ge 3 \sqrt [3] {abc} \equal{} 3$ by AM-GM, so
$ \sum \frac {a^3}{(a \plus{} 1)(b \plus{} 1)} \ge \frac {a \plus{} b \plus{} c}{2} \minus{} \frac {3}{4} \ge \frac {3}{4}$[/hide]
Was this an earlier IMO problem, because I think I might have seen it before.

No I saw this sum in some bookI think it is hoojoolees one


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $K$ be a field and $A$ a cyclic matrix (that is, minimal polynomial=characteristic one or, if you want, a companion matrix). Prove that there exists a matrix $S$ symetric such that $SAS^{-1}=trans(A)$ (transpose of $A$). Conclusion?

$A$ is a companion matrix? If it is, i can post a solution.

Yes, it's a companion  matrix. Of course, the conclusion is that the property is true for all matrices, but you can post your solution for companion matrices.

Let $p(\lambda)=\lambda^n-a_{n-1}\lambda^{n-1}-\ldots-a_1\lambda-a_0$ be a polynomial with coefficients in $K$, and let $A=\begin{pmatrix}0&1&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\\a_0&a_1&a_2&\cdots&a_{n-1}\end{pmatrix}$ be its companion matrix.

We now look for an invertible matrix $S$ such that $AS=SA^t\ (*)$ of the form $S=(s_{i+j-1})_{i,j=1}^n\ (1)$, where $s_{n+k}=\sum_{i=0}^{n-1}(-a_i)s_{i+k},\ \forall k\in\overline{1,n-1}\ (2)$. It's easy to see that all matrices which satisfy conditions $(1),(2)$ also satisfy $(*)$, so all we need to do is find recurrent sequences $(s_p)_p$ with those properties which yield an [i]invertible[/i] matrix $S$. Obviously, taking $s_1=s_2=\ldots=s_{n-1}=0$ and $s_n=1$, and then completing the sequence by employing $(2)$ will do.

Here is a beautiful approach: if we consider the linear system $X$ symetric and $AX$ also, then we immediately see that it is of rang at least $n$ (just count the equations). But since the commutant of $A$ has dimension exactly $n$, one can easily see that any matrix for which $AX$ is symetric is also symetric. (sorry, I know it is trop vague, but I'm really in a hurry now).

I wanted to do exactly that, but then I found that explicit construction for companion matrices.


circles
coordinate geometry
geometry
Triangle
coordinates
combinatorics
IMO Shortlist

Every point with integer coordinates in the plane is the center of a disk with radius $1/1000$.

(1) Prove that there exists an equilateral triangle whose vertices lie in different discs.

(2) Prove that every equilateral triangle with vertices in different discs has side-length greater than $96$.

[i]Radu Gologan, Romania[/i]
[hide="Remark"]
The "> 96" in [b](b)[/b] can be strengthened to "> 124". By the way, part [b](a)[/b] of this problem is the place where I used [url=http://mathlinks.ro/viewtopic.php?t=5537]the well-known "Dedekind" theorem[/url].
[/hide]

This is also from the ISL.

I've only managed to solve (a), which is not at all tough. We just find a triangle with vertices (as complex numbers) of the form $2n,\ 2n\epsilon,\ 2n\epsilon^2$. Since $\epsilon=\frac{-1+\sqrt3}2$, it means that we can find an $n$ s.t. $2n\epsilon$ is inside a disk (by using the "famous Dedekind theorem" :D, otherwise known as Kronecker's thm). $2n\epsilon^2$ will, of course, be inside a disk as well, because $\epsilon^2$ is the symmetric of $\epsilon$ wrt the $Ox$ axis.

Don't you have any other proof????? A more elementary proof???

Part (a) is a direct application of Jacobi's theorem [ if a is irrational then ( na - floor(na) ) is everywhere dense in the interval (0,1) as n varies over the natural numbers. ]

From this the proof of (a) is obvious

Anyone for part (b)

This is a very nice problem!

proof of (strengthened) part b:

Assume the contrary that there exists an equilateral triangle $ ABC$ such that $ A,B,C$ lie in the interior of different disks, where $ |AB|\equal{}|BC|\equal{}|CA|\equal{}s\le124$. Let $ O_1,O_2,O_3$ be the center of the disks which contain $ A,B,C$ respectively. Note that $ s\minus{}0.002\le O_1O_2,O_2O_3,O_3O_1\le s\plus{}0.002$. Clearly, $ O_1O_2O_3$ is not equilateral, so assume $ O_1O_2>O_2O_3$. So

$ O_1O_2^2\minus{}O_2O_3^2\equal{}(O_1O_2\minus{}O_2O_3)(O_1O_2\plus{}O_2O_3)\le 0.004\times(2s\plus{}0.004)\le 0.004(2\cdot124\plus{}0.004)<1$

But $ O_1O_2^2\minus{}O_2O_3^2$ must be an integer, contradiction.

Not that it matters much, but the proof I came up with (using complex numbers) shows a bound of > 166, I think.

Part a is pretty easy, you just use that $\{\sqrt{3}k\}$ gets close to $1$ for some integer $k$.

For part $b$ we do the following. First we notice that without lose of generality we may assume one point is on the disk centered at the origin and another at the disk centered at $(a,0)$ where a is an integer. Now, we will show that the third point is close to $(\frac{a}{2},\frac{\sqrt{3}}{2}a)$. The idea is that the side of the equilateral triangle must be between $a- 1/500$ and $a+ 1/500$, also the slope between the point near the origin and the point near $(a,0)$ is at most $-\frac{1}{500a}$, hence the third point coming out of the midpoint has slope at least $500a$. Playing around with the equations, it is not hard to bound the difference between the third point and $(\frac{a}{2},\frac{\sqrt{3}}{2}a)$.

Once we know this, we have that $\frac{a}{2}$ must be close to an integer, implying it is an integer. So we may write $a = 2k$. Now we need $\sqrt{3}k$ to be close to an integer. Notice that the length of the equilateral would be at least $2(k-\frac{1}{500}) = 2k -\frac{1}{250}$. For $a > 96$, we need only show $k > 50$. So let's assume $k \leq 50$ and reach a contradiction.

The continued fraction of $\sqrt{3}$ is $1, [1,2]$, so it is easy to find the convergents. The convergents are $1, 2, \frac{5}{3}, \frac{7}{4}, \frac{19}{11},\frac{26}{15},\frac{71}{41},\frac{97}{56}, \ldots.$ Now, if $\left| \sqrt{3} - \frac{p}{q}\right| < \frac{1}{2q^2}$, then $\frac{p}{q}$ is a convergent. Note this would be equivalent to $\left|\sqrt{3}k - p\right| < \frac{1}{2k}$ implying $\frac{p}{k}$ is a convergent. But we know that for some $k \leq 50$ and an integer $n$, $\left|\sqrt{3}k - n\right| < \frac{1}{1000} < \frac{1}{2k}$, therefore $\frac{n}{k}$ is a convergent with the denominator being at most $50$, so $k$ must be at most $41$.

Now, another theorem tells us $\left|\alpha - \frac{p_k}{q_k}\right| > \frac{1}{q_k(q_k+q_{k+1})}$, so we have $\left|41\sqrt{3}-71\right| > \frac{41}{41(41+56)} = \frac{1}{97} > \frac{1}{1000}$, yielding the result we want.

Using more convergents one can show a better lowerbound for the sidelength, one could show sidelength $ > 1040$, and using a computer one could show that the sidelength must be bigger than $1559$.

[quote="Quique"]For part $b$ we do the following. First we notice that without lose of generality we may assume one point is on the disk centered at the origin and another at the disk centered at $(a,0)$ where a is an integer.[/quote]
Why can we do this?

Anyway, here's another way to do (b) that can be strengthened quite a bit with some more effort (i.e. by tightening bounds by finding explicit solutions to $N(u+v\sqrt{3})=k$ for small $k$)...

By translation and reflection about the $y$-axis if necessary, WLOG let the centers be $(0,0),(a,b),(c,-d)$ (where $a,b,c,d\ge0$), with displacements $(p,q),(x,y),(z,w)$ such that $p^2+q^2,x^2+y^2,z^2+w^2<r^2$ (where $r=1/1000$). By complex numbers, we have
\[(a+x-p)+i(b+y-q) = [(c+z-p)+i(-d+w-q)]\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right),\]so
\[l=(2a-c)-d\sqrt{3} = (p+q\sqrt{3}) + (z-w\sqrt{3}) - 2x \in (-6r,6r)\]and
\[m=(2b+d)-c\sqrt{3} = (q-p\sqrt{3}) + (w+z\sqrt{3}) - 2y \in (-6r,6r).\]First note that $2a-c,2b+d\ge0$ or else we have obvious bounding contradictions. Letting $N(u+v\sqrt{3})=u^2-3v^2$, note that $N(l),N(m)\in\{\ldots,-2,1,3,\ldots\}$ since $-1$ is not a quadratic residue modulo $3$. Assume for the sake of contradiction that the side length is at most $96$, whence $\sqrt{c^2+d^2}-2r\le96$ and $c^2+d^2<97^2$. Now we find that
\[N(l)\ge3\implies d>\frac{\frac{3}{6r}-6r}{2\sqrt{3}}>144,\]so $N(l)\le1$ and similarly $N(m)\le1$. In addition,
\[N(l)=1\implies d>\frac{\frac{1}{6r}-6r}{2\sqrt{3}}>48\]and
\[N(l)\le-2\implies d>\frac{\frac{2}{6r}}{2\sqrt{3}}>96,\]so $N(l)=N(m)=1$. But
\[N(u+v\sqrt{3})=1\implies (u,v)\in\{(2,1),(7,4),(26,15),(97,56),(362,209),\ldots\},\]so $c=d=56$ and $2a-c=2b+d=97$, a contradiction modulo $2$.

[b]Edit:[/b] With some computing, we can find that
\[l,m\in\{97-56\sqrt{3},265-153\sqrt{3},362-209\sqrt{3},627-362\sqrt{3},724-418\sqrt{3},\ldots\}.\]Also notice that $2a-c\equiv c\pmod{2}$ and $2b+d\equiv d\pmod{2}$, so our smallest candidate (size defined by $c^2+d^2$) is
\[\{l,m\}=\{97-56\sqrt{3},362-209\sqrt{3}\}.\]This works when, for example,
\[(p,q,x,y,z,w)=\left(\frac{3}{10000},\frac{9}{10000},\frac{r}{20}(7+17\sqrt{3})-\frac{97-56\sqrt{3}}{2},\frac{r}{20}(1+\sqrt{3})-\frac{362-209\sqrt{3}}{2},\frac{4}{10000},\frac{-8}{10000}\right),\]giving us a lower bound of around $\sqrt{56^2+209^2}>216$.

[quote=<jgnr>]
Clearly, $ O_1O_2O_3$ is not equilateral
[/quote]
I don't understand why :(

1) Consider the triangle $(0,0)$, $(2x,0)$, $\left(x,x\sqrt3\right)$, for integer $x$. Note that $\{k\sqrt3\}$ is dense in $[0,1]$ (we can prove this easily with Pigeonhole) so there must exist $x$ such that $\{x\sqrt3\}<\frac{1}{1000}$. Then, choosing this $x$ gives all three points within $\frac{1}{1000}$ of lattice points.

2) Suppose the contrary, namely that we do have an equilateral triangle with side length at most $96$. For convenience, shift one of the vertices to the origin. As this shift has length at most $\frac{1}{1000}$, we will increase the disks to radius $\frac{1}{500}$ to ensure that the images of the other two vertices still lie within disks. Then, if the centers of the disks which coincide with the other two vertices are $(a,b)$, $(c,d)$ (labeling counterclockwise), then we should have that $a^2+b^2,c^2+d^2\le 96^2$, and that the image of $(a,b)$ after a $\frac{\pi}{3}$ ccw rotation about the origin is at most $\frac{1}{250}$ away from $(c,d)$. Expanding with distance formula, $$\left(\frac{a\sqrt{3}-b}{2}-c\right)^2+\left(\frac{a+b\sqrt{3}}{2}-d\right)^2=(a^2+b^2+c^2+d^2+bc-ad)-(ac+bd)\sqrt{3}\le\frac{1}{250^2}$$ Now, note that $$a^2+b^2+c^2+d^2+bc-ad+(ac+bd)\sqrt{3}\le \frac{3+\sqrt 3}{2}(a^2+b^2+c^2+d^2)\le (3+\sqrt 3)96^2$$
So, we have that $$(a^2+b^2+c^2+d^2+bc-ad-(ac+bd)\sqrt{3})(a^2+b^2+c^2+d^2+bc-ad+(ac+bd)\sqrt{3})\le \frac{(3+\sqrt3)96^2}{250^2}<1$$ which is a contradiction as the LHS is a positive integer.

Easy proof for $124$:

For (1), take $(0,0)$, $(N,0)$, and $(N/2,N\sqrt{3}/2)$ for a positive integer $N$ such that the fractional part of $N\sqrt{3}/2$ is much smaller than $1/1000$  (which exists since $\{k\alpha\}$ is dense in $(0,1)$ as we vary $k$ over the positive integers).

For (2), we show that every equilateral triangle must have side-length at least $124$. Suppose the side-length is actually $N$. Work in the complex plane, and let $\omega=e^{2\pi i/6}$. Suppose the three disks the triangle's vertices are centered at $0$, $a$, and $b$ (WLOG we shifted one of them to the origin), and suppose the three vertices are actually $\gamma$, $a+\alpha$, and $b+\beta$, where $|\alpha|,|\beta|,|\gamma|<1/1000$. Note that
\[N=|a+\alpha-\gamma|\le |a|+2/1000,\]
so $|a|,|b|\ge N-2/1000$. Now,
\[\frac{a+\alpha-\gamma}{b+\beta-\gamma}=\omega,\]
so
\[\frac{a}{b} = \omega+\omega\frac{\beta-\gamma}{b}+\frac{\gamma-\alpha}{b},\]
so
\[\left|\frac{a}{b}-\omega\right|\le \frac{4/1000}{N-2/1000}.\]
We also have
\[\left|\frac{a}{b}-\omega\right| = \frac{1}{|b|\cdot|a+b\omega^2|}|(a-b\omega)(a+b\omega^2)| = \frac{|a^2-ab+b^2|}{|b|\cdot|a+b\omega^2|}.\]
Since $a,b\in\mathbb{Z}[i]$, we have that $a^2-ab+b^2\in\mathbb{Z}[i]\setminus\{0\}$, so $|a^2-ab+b^2|\ge 1$, so
\[\left|\frac{a}{b}-\omega\right|\ge \frac{1}{2(N+2/1000)^2},\]
so
\[\frac{1}{2(N+2/1000)^2}\le \frac{4/1000}{N-2/1000}.\]
It is easy to see that this implies
\[N\ge 124.994,\]
as desired.

[b]Remark:[/b] The inequality could be made stronger by using a tighter bound for
\[\left|\omega\frac{\beta-\gamma}{b}+\frac{\gamma-\alpha}{b}\right|\]
rather than the naive triangle inequality bound we use. Indeed, equality cannot be achieved for that bound, and some annoying work could push this to get around 150 or so (I haven't done all the details myself).

(a) Indeed, consider the equilateral triangle formed by the vertices $(a,0),(-a,0)$ and $(0,\sqrt{3}a)$. By a well-known fact, there exists $a$ such that $\{\sqrt{3}a\}<\frac{1}{1000}$, then for this $a$, the corresponding equilateral triangle will satisfy the condition.

(b)Consider such an equilateral triangle $ABC$, lying in the interior of the disk centered at $X,Y,Z$.

Now notice that by a well-known fact $XYZ$ cannot formed an equilateral triangle. Suppose $XY> YZ$. Now by triangle inequality,
$$|XY-YZ|\leq |XY-AB|+|BC-YZ|\leq \frac{4}{1000}$$
Meanwhile, $XY=\sqrt{a}$ and $YZ=\sqrt{b}$ where $a,b$ are integers, so
$$XY-YZ=\frac{a-b}{\sqrt{a}+\sqrt{b}}>\frac{1}{2\sqrt{a}}$$
which implies $a\leq 125^2$, and so
$$AB>\frac{XY}-\frac{1}{1000}>124$$
as desired.

Note that for some large enough $x$, $(\sqrt{3}-1)^x<\tfrac{1}{1000}$. Then, if we let $(\sqrt{3}-1)^x=a\sqrt{3}-b$ for positive integers $a$ and $b$ then $a\sqrt{3}$ is within $\frac{1}{1000}$ of an integer. Thus, if we let two of the coordinates be $(0,0)$ and $(2a,0)$, then $(a,\sqrt{3})$ forms an equilateral triangle with the other two and lies inside of a disk.
\item Let there be an equilateral triangle with vertices in different discs and has side-length $s$. Let $A$, $B$, $C$ be the centers of those discs. Each side length of $ABC$ is between $s-\tfrac1{500}$ and $s+\tfrac1{500}$.

WLOG $PQ>QR$ since $PQR$ is clearly not equilateral. We have
\[1\le PQ^2-QR^2=(PQ+QR)(PQ-QR)\le (2s+\tfrac1{250})\cdot (\tfrac{1}{250})\] so $s\ge 125-\tfrac1{500}>96$. We are done.


Asymptote
analytic geometry
graphing lines
slope
geometry

I am new to Asymptote (still learning). I tried searching a bit and looking through the Wiki pages, but didn't find it. What is the code for making a right angle mark? For example, I could have a line with slope 1/2 and another line with slope -2, and I want to show that they are perpendicular.

[code]
import graph;
import geometry;
size(200);

real f(real x){return -2*x+1;}
real g(real x){return (1/2)*x+1;}

path p=graph(f,-2,2);
path q=graph(g,-2,2);

draw(p^^q);

pair k=intersectionpoint(p,q);

perpendicular(k,NE,q,red);

[/code]

So you get:
[asy]import graph;
import geometry;
size(200);

real f(real x){return -2*x+1;}
real g(real x){return (1/2)*x+1;}

path p=graph(f,-2,2);
path q=graph(g,-2,2);

draw(p^^q);

pair k=intersectionpoint(p,q);

perpendicular(k,NE,q,red);[/asy]

I don't really understand. How do I define which point and which side it is on, and which line? I have three line segments (for example, "draw((0,0)--(1,1));" ) coming together, and I want to show that two of them are perpendicular.

Can anyone explain? Thanks.

Hmm.. the way I see it is, that once you pick the line, in this case he choose $ q$, the top part of the line becomes the new north, and thus picking the top right corner would be the same as picking North East or NE, Top left is NW, bottom right SE, bottom left SW. If he had chosen $ p$ he would have had to put NW instead of NE because now the North has changed to the line $ p$.

I looked through problems in the resources section and found diagrams with the right angle marks. I went through the code and I think I found out how it works:

It is
[code]draw{rightanglemark([first point],[center point],[last point],size));[/code]

For example, if you wanted a right angle mark showing that the x and y axes are perpendicular, in the first quadrant, you would type:
[code]draw(rightanglmark((0,1),(0,0),(1,0),10));[/code]

Thanks anyway!

Sorry for reviving, but I was wondering how you would put a right angle mark on the right triangle below:

[asy]fill((0,0)--(3,5)--(3,0)--cycle, blue);[/asy]

Also, is it possible, for the fill color, to specify a specific type of color? I'd like to use the HTML code #003399, as it goes better with my powerpoint than the basic colors, but I am not sure how to insert it.

I noticed some flaws in the thing I put below. I believe it should be (changed the { to ( at the beginning):
[code]draw(rightanglemark([first point],[center point],[last point],size));[/code]

For your case, adding on to your code, you can try this:
[code][asy]
fill((0,0)--(3,5)--(3,0)--cycle, blue);
draw(rightanglemark((3,5),(3,0),(0,0),10));
[/asy][/code]

That would make this:
[asy]fill((0,0)--(3,5)--(3,0)--cycle, blue);
draw(rightanglemark((3,5),(3,0),(0,0),10));[/asy]
(if you look closely, the right angle mark is there, but kind of hidden)

I am not sure about color, though.

Well, I tried putting in the number directly...

 [code] [asy]
fill((0,0)--(3,5)--(3,0)--cycle, #003399);
draw(rightanglemark((3,5),(3,0),(0,0),10));
[/asy][/code]

But it didn't work. I got an error code.

Ooh, do you want to fill the triangle with that color? I thought you wanted to fill the inside of the little box.

To do that, I think you have to convert it into one of the formats [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Reference#Pens_and_Coloring]here[/url]. The color 003399 in hexadecimal turns into rgb(0,3/16,9/16). You can try that in your code:
[code][asy]
fill((0,0)--(3,5)--(3,0)--cycle, rgb(0,3/16,9/16));
draw(rightanglemark((3,5),(3,0),(0,0),10));
[/asy][/code]
That produces this:
[asy]fill((0,0)--(3,5)--(3,0)--cycle, rgb(0,3/16,9/16));
draw(rightanglemark((3,5),(3,0),(0,0),10));[/asy]

Okay, thanks. That is what I meant.
Is it also possible to change the color of the outline of the right angle mark? (sorry for all of these questions)

I'm not sure, but I haven't been able to.

Like this?
[asy]fill((0,0)--(3,5)--(3,0)--cycle, rgb(0,3/16,9/16));
draw(rightanglemark((3,5),(3,0),(0,0),10),red+linewidth(1));[/asy]
rightanglemark returns a path and you can do with it whatever you can do with any other path :)


Prove that if $ a,b,c \in [0,\plus{}\infty)$ then: $ (a^2\plus{}b^2\plus{}c^2)^3 \le 3(a^3\plus{}b^3\plus{}c^3)^2.$

[quote="moldovan"]Prove that if $ a,b,c \in [0, \plus{} \infty)$ then: $ (a^2 \plus{} b^2 \plus{} c^2)^3 \le 3(a^3 \plus{} b^3 \plus{} c^3)^2.$[/quote]

just power mean...... :o
\[ \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\ge \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}}\]

Your solution is good. You killed the problem.  :P But there must be other solutions. Each solution is good if it is correct.

Since we have By chebychev inequality $ a^3 \plus{} b^3 \plus{} c^3 \ge \frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)$

and by Cauchy shwarz $ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \ge (a^2 \plus{} b^2 \plus{} c^2)^2$

Then
$ 3(a^3 \plus{} b^3 \plus{} c^3)^2 \equal{} 3(a^3 \plus{} b^3 \plus{} c^3)(a^3 \plus{} b^3 \plus{} c^3) \ge (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)(a^3 \plus{} b^3 \plus{} c^3)\ge (a^2 \plus{} b^2 \plus{} c^2)^3$

S.O.S:
\[ 3\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 \minus{} \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^3\]
\[ \equal{} \sum_{cyclic} {\left( {\left( {a^2 \minus{} ab \plus{} b^2 } \right)\left( {a^2 \plus{} 3ab \plus{} b^2 } \right) \plus{} \left( {a \plus{} b} \right)^2 c^2 } \right)\left( {a \minus{} b} \right)^2 }\]
\[ \geq 0.\]

[quote="moldovan"]Prove that if $ a,b,c \in [0, \plus{} \infty)$ then: $ (a^2 \plus{} b^2 \plus{} c^2)^3 \le 3(a^3 \plus{} b^3 \plus{} c^3)^2.$[/quote]

 :lol: How about this one?

$ a,b,c>0$,prove that
$ (a^2\plus{}b^2\plus{}c^2)^3\geq \left(a^3\plus{}b^3\plus{}c^3\plus{}3(\sqrt{3}\minus{}1)abc\right)^2$

[quote="Tourish"]

 :lol: How about this one?

$ a,b,c > 0$,prove that
$ (a^2 \plus{} b^2 \plus{} c^2)^3\geq \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(\sqrt {3} \minus{} 1)abc\right)^2$[/quote]
With your inequality is all right. It's true for all reals $ a,$ $ b$ and $ c.$ :wink:

[quote="arqady"][quote="Tourish"]

 :lol: How about this one?

$ a,b,c > 0$,prove that
$ (a^2 \plus{} b^2 \plus{} c^2)^3\geq \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(\sqrt {3} \minus{} 1)abc\right)^2$[/quote]
With you inequality is all right. It's true for all reals $ a,$ $ b$ and $ c.$ :wink:[/quote]
The general inequality holds, too (it was proposed by Vasc)

Let $ x_1,x_2,\ldots,x_n$ be real numbers such that $ x^2_1 \plus{} x^2_2 \plus{} \cdots \plus{} x^2_n \equal{} 1.$ If $ n \ge 3,$ then
$ \sum^n_{i \equal{} 1}x^3_i \plus{} \frac {6}{(n \minus{} 2)\left(\sqrt {n} \plus{} 1\right)} \sum_{1 \le i < j < k \le n} x_ix_jx_k \le 1.$

[quote="Tourish"][quote="moldovan"]Prove that if $ a,b,c \in [0, \plus{} \infty)$ then: $ (a^2 \plus{} b^2 \plus{} c^2)^3 \le 3(a^3 \plus{} b^3 \plus{} c^3)^2.$[/quote]

 :lol: How about this one?

$ a,b,c > 0$,prove that
$ (a^2 \plus{} b^2 \plus{} c^2)^3\geq \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(\sqrt {3} \minus{} 1)abc\right)^2$[/quote]

er...
\[ \Longleftrightarrow \sum {(a^2 b^2 \minus{} (3\sqrt{3} \minus{} 5)c^2(a \plus{} b)^2 \plus{} 3(\sqrt{3} \minus{} 1)c^4)(a \minus{} b)^2 } \ge 0\]
but ... I don't know how to prove it ...  :blush:


Seeing that the team chess thing died (it was a horrible idea in the first place IMO, just get a powerful chess computer (program) and run it for a while), let's have a new tournament. Three minute blitz is short and sweet, doesn't take up much time, and really takes skill. So, who wants to do this? Suggestions welcome also.
I think FICS is the best because it's free, and anyone can log on easily as a guest.

This would be cool, my username on FICS is the same as AoPS.

On second thought, maybe 5 0 would be better...anyone?

[quote="andersonw"]I just created an account, so I'll play. Username is same as on Aops.
My rating's 1337 right now :P[/quote]
Wonderful, any more people? Its going to be unrated since I don't expect a lot of people to have accounts...I for one will not (somehow they considered girlssuck.org a free email, which still puzzles me...no offense to girls, for real).

maybe, if I have time... I am somehow rated 1788 on FICS? I am definitely not that high. Hmm. And 5 0 is better!

I'm only in the 1100s... but I'm getting better.

[quote="serialk11r"]Oooh 1788? Me wants to play you! Hmmm...okay I guess 5 0 it is.
For some reason I'm less scared of people with ratings above 1700 than any random guest, since guests are totally beast (maybe that includes me...I'm not that beastly though :()[/quote]

yeah dude I should be rated like... 14 or 1500 though...

I'm up for it even though I am really bad...  :(

[quote="worthawholebean"][quote="serialk11r"]Wonderful, any more people? Its going to be unrated since I don't expect a lot of people to have accounts...I for one will not (somehow they considered girlssuck.org a free email, which still puzzles me...no offense to girls, for real).[/quote]
Use your parents' email or something - once you get an account you can message an admin to ask them to change it.[/quote]
Dude my mom uses yahoo and my dad wouldn't let me use his email for sure.
Anyways...need more people!!! hey you guys should like advertise...we should put it in our sigs or something.

[quote="undefined117"]chess? i might play.. haven't played in years though XD

erm 4 years -_-

ergh FICS won't let me include numbers in my handle :mad:

username on FICS: undefinedCXVII[/quote]
That's okay, I started chess up this year after a 5 year hiatus :O


find the degree measure between the hour and minute hands of a clock when it is 6:17.
(the smaller angle)

I dunno wat I'm doin wrong, but i keep gettin 84.5 degrees.

[quote="xxxxx"]so this works for all clock problems like this right?[/quote]
See the first post in [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=583]my blog[/url].

i looked at it... there's nothing about clock problems...

First post: Clock Angle Formula.

The hour hand travels $30^{\circ}$ per hour, $\frac{1}{2}^{\circ}$ per minute, and $\frac{1}{120}^{\circ}$ per second.

The minute hand travels $6^{\circ}$ per minute and $\frac{1}{10}^{\circ}$ per second.

The angle between the hands is:

$x = \left(30h+\frac{m}{2}+\frac{s}{120}\right)-\left(6m+\frac{s}{10}\right)$

$x = 30h+\frac{m}{2}+\frac{s}{120}-6m-\frac{s}{10}$

$x = 30h-\frac{11m}{2}-\frac{11s}{120}$

However, that is not the complete formula. Sometimes the angle will be obtuse. To find the acute angle, use $180-|180-x|.$

$180-\left|180-\left(30h-\frac{11m}{2}-\frac{11s}{120}\right)\right|$

$180-\left|180-30h+\frac{11m}{2}+\frac{11s}{120}\right|$

so does 11m-60h all over 2 and in absolute value work for all problems?


A circle of radius 9 cm is inscribed in an equilateral triangle and is tangent at three points. Three smaller circles are inscribed so that they are each tangent to two sides of the triangle and to the larger circle. The process is continued with circles of ever decreasing diameters. What is the number of centimeters in the sum of the circumferences of all such circles? The answer is x Pi . Find x .

[hide]45[/hide]

**Is there a way as to find the radii of the smaller circles?  I know that the problem involves a geometric sequence.**

Wait, MathCounts expects you to know the formula for infinite Geometric Sequences?

Thats strange, you could possible argue that there is no certain number answer, but you would be expected to round I guess.

I'm pretty sure they don't. Just find the number that it is approaching when you add the circumferences.

I can help you solve it if you have the equation of how to find the sides of the equilateral triangle.

This problem is probably beyond MATHCOUNTS difficulty.

[hide="Solution-step"]Begin with a circle inscribed in an equilateral triangle. Let $ R$ be the radius of this circle and $ x$ be the length of a side of triangle. Drop an altitude from the top vertex and draw the radius from the center to one of the lateral sides to form a right triangle.

Use similarity of triangles and you should be able to find that $ R\equal{}\frac{x\sqrt3}{6}\equal{}\frac13\frac{x\sqrt3}{2}$, which means that $ R$ is 1/3 of the height $ h\equal{}\frac{x\sqrt3}{2}$ of the triangle or $ h\equal{}3R$.

Now draw the smaller circle above the large circle (from the problem) and draw the line parallel to the base of the triangle passing through the point of tangency between the two circles to get a smaller equilateral triangle. This smaller equilateral triangle has height equal to $ h\minus{}2R\equal{}3R\minus{}2R\equal{}R$. So the radius $ r$ of the smaller circle is, by similarity again, $ r\equal{}\frac13 R$.

Sum of the circumferences of all the circles is

$ 1(2\pi R)\plus{}3\left(2\pi\left(\frac1{3^1}R\right)\plus{}2\pi\left(\frac1{3^2}R\right)\plus{}2\pi\left(\frac1{3^3}R\right)\plus{}\cdots\right)\equal{}$

$ 2\pi R\plus{}2\pi R\left(1\plus{}\frac1{3^1}\plus{}\frac1{3^2}\plus{}\cdots\right)\equal{}2\pi R\left(1\plus{}\frac1{1\minus{}\frac13}\right)\equal{}5\pi R$, where $ R\equal{}9 \ \text{cm}$ :)[/hide]


topology
function
real analysis
real analysis unsolved

Can someone help me with this problem?
f: A to B is open if for each open set U in A, the image f(U) is open in N.
a) If f is open, is it continuous?
b) If f is a homeomorphism, is it open?
c) If f is open, continuous bijection, is it a homeomorphism?
d) If f: R to R is a continuous surjection, must it be open?
e) If R to R is a continuous, open surjection, must it be a homeomorphism?

a) I think that this is not true since in general continuous maps need not to send open sets. I need to come up with a counter example. I might be wrong about this too. Anyone can help me please.

Are $ A$ and $ B$ just arbitrary metric spaces? The question looks like it may have $ A$ and $ B$ more general topological spaces.

I think A and B are arbitrary metric spaces, but the problem doesn't mention that. However, we've been working with metric spaces in class.

[quote="jmerry"]Your instincts are right on the first question; it's false.
The nicest example I can think of at the moment: Let $ X$ be a discrete space (the distance between any two points is 1). All subsets of $ X$ are open, so any function from $ \mathbb{R}$ to $ X$ has this property. The only continuous functions from $ \mathbb{R}$ to $ X$ are the constants.

There are other ways to do it; we can construct a function from $ \mathbb{R}$ to itself that takes any open interval to the whole line.[/quote]

Thanks a lot. I was able to figure out that if f is homeomorphism, then f is open by using the definition of homeomorphism. Then use the continuity of f to show that f is open.
c) I used the same approach as I did in a) to show that it is a homeomorphism.
d) By revising the function f=x^2, I came up with a counter example.
e) I'm still stuck on this. I know that I need to show that f is injective, then I'll be done, but I don't know how to approach it.

If R is replaced by a unit circle, what will happen in e)

(e) is false in general spaces; think projection.
For $ \mathbb{R}$ to itself, show that a continuous open map is monotone, or equivalently that a continuous map that has a local maximum/minimum isn't open.

By the way, don't quote the full post immediately preceding you. It's assumed you're replying to that post anyway.

[quote="namelessguy"]e) I'm still stuck on this. I know that I need to show that f is injective, then I'll be done, but I don't know how to approach it.

If R is replaced by a unit circle, what will happen in e)[/quote]

I was stumped on this problem until I saw jmerry's post too. I'm guessing you're reading Pugh's "Real Mathematical Analysis"? I like this book a lot.

I think I found a counterexample to the claim if $ \mathbb{R}$ is replaced by $ S^1$, I can post it if you're stuck.


algorithm
number theory
relatively prime

Prove that any two numbers of the following sequence are relatively prime:

$2+1, 2^2+1, 2^4+1, \ldots, 2^{2^n}+1, \ldots$.

[hide]

Suppose some prime  divides . We will show it will not divide any term  such that .

Our statement is that

which is equivalent to

Raise each side to the  power, yielding

Since we are raising  to an even power. Now add 1 to both sides, and we have that

and  iff , which cannot be true because  is always odd. Thus, any term is relatively prime to all terms after it in the sequence, and since relatively prime is a commutative property, then clearly all terms are relatively prime to one another.

[/hide]

Another cool way to do it...

[hide]
Let $a_n = 2^{2^n}+1$.

Show that $a_n = \left(\prod_{i=1}^{n-1} a_i\right)+2$. [/hide]

i think that the euclidian algorithm pretty much solves this....


Suppose that $xyz=1,x+\frac{1}{x}=a,y+\frac{1}{y}=b,z+\frac{1}{z}=c$. Find the value of $a^2+b^2+c^2-abc$.

After a lot of hairy algebra:

[hide]
$yz = \frac{1}{x}, xz = \frac{1}{y}, xy = \frac{1}{z}
a = x+yz, b = y+xz, c = z+xy
a^2 + b^2 + c^2 - abc = x^2 + 2xyz + y^2z^2 + y^2 + 2xyz + z^2x^2 + z^2 + 2xyz + x^2y^2 - zxy - x^2z^2 - y^2z^2 - yxz^3 - xy^2 - x^3zy - y^3zy - y^3xz + y^2x^2z^2
x^2 + y^2 + z^2 + 6 - 1 -z^2 - x^2 - y^2 - 1$

4

So the answer is 4
[/hide]

Sorry for my messy solution. I also think this was a little too hairy for Getting Started, because I did a problem like this on an AIME.

Thank you for your reply, A+MATH.
You are right.

kunny