I don't know if this has been discussed in this forum but I'll open the floor to a discussion.

http://www.vanityfair.com/politics/features/video/2008/hitchens_video200808

This is a video of Vanity Fair editor Christopher Hitchens voluntarily agreeing to experience waterboarding.  Watch the videos, and then provide your thoughts.

Do you consider it torture?  Do you believe the US should continue to use it?  Under what circumstances should it be used as an "advanced interrogation" technique?

[quote="JRav"]Do you consider it torture?  Do you believe the US should continue to use it?  Under what circumstances should it be used as an "advanced interrogation" technique?[/quote]

ooh this is an easy one.  yes, no, none.  do i win the prize?

I think it is indeed torture. However, I think if you've captured a high level terrorist official, then why not? If he gives valuable information, then  potentially you've got important access to information that may divert terrorist activity. You may rescue hostages. You may save lives. You may end a war. I do support sanctions on when it can be used, but to completely outlaw its use would only give our already leave more in the dark in terrorist intelligence.

I laugh at that.  You feel like you're dying, and someone is playing with your life, thrusting you at death and dragging you back.  People will say anything under torture.  ANYTHING.

Was I clear enough?  I can be even more blunt if you so wish.

Or they might just say the truth ;). The point is, there is no way getting ANYWHERE if you make it illegal.

Alright, so maybe you people think it's a pretty blunt answer (as do I).  But I'm hoping to find some Bush-administration lackey who indeed agrees with it.  These are the people I want to hear from.  Because if any of you ever watch the news and Congressional testimonies, you'd know that there are many people who don't justify it as torture.

[quote]ooh this is an easy one. yes, no, none. do i win the prize?[/quote]

Does being a jerk make you feel good about yourself?

haha

somehow I doubt you're going to have too many bush administration lackeys on AoPS

[quote="JRav"][quote]ooh this is an easy one. yes, no, none. do i win the prize?[/quote]

Does being a jerk make you feel good about yourself?[/quote]

come on, it's not as if waterboarding is a particularly nuanced moral issue.  the christopher hitchens article is a very good one, but any interested person has been able to find a variety of testimonies that waterboarding is unequivocally torture for a long time now.  my answers are glib but they are the standard answers of just about every person who respects human rights and repudiates the value of torture for obtaining information, and they are exactly the same answers i would give if you replaced every instance of 'it' in your last paragraph with 'smashing fingers with sledgehammers'.

hmm. I am not totally opposed to waterboarding or even harsher torture. I am not sure why, if we're already committed to killing each other in a wartime situation, certain ways of performing less lethal manoeuvres are suddenly terrible, terrible things. I'm saying this less to argue the specific point than just to ask that you be more considerate in your replies -- it is possible not everyone agrees.

[quote]somehow I doubt you're going to have too many bush administration lackeys on AoPS
[/quote]

Do you have any explanation for this?  While I'm sure it is true, I would love to know why.

academics tend to be liberal.  so do young people!

blah's points are true, certainly, although I seem to (apparently) be conservative

I was referring more specifically to the fact that people on AoPS are usually intelligent, whereas bush administration lackeys tend to be ... not so intelligent ?

well i did think that saying 'smart people tend to be liberal' would cross the trolling line, whether i think it's true or not.

Well, it's provably correct, in that there are a myriad of polls saying that college-educated people favor the Democratic party (and graduate-educated ones overwhelmingly so). Granted, this is "level of education" rather than "smartness", but it's hard to say exactly what smartness is in the first place, so I'm content with that. More specifically, the current administration just seems to be particularly set on being stupid (the easiest example to cite is ignoring scientifically proven data in the case of, for example, global warming).

oh well. I seem to have derailed the thread from "why waterboarding sucks" to "why bush administration sucks". that is actually okay with me.

I say that waterboarding is torture, but if you know (with conclusive evidence) that someone's a terrorist, why would you use waterboarding? Just say "I'll sentence you to prison if you don't tell me" cause that's OK

*edited my views  :rotfl: *
But if 98 people in Gitmo are innocent like the news says, I think they should stop...

[quote="mgao"]I say that waterboarding is torture, but if you know (with conclusive evidence) that someone's a terrorist, then get your revenge[/quote]

man, the 'ticking bomb' argument for torture is one thing but this is a new one!

Yeah I'm pretty sure that everyone here would agree that we don't detain for "revenge."

well, the thing is, the people that are being waterboarded aren't -ordinary people-. they have a very high probability of being, like, people behind murders, people who probably deserve the waterboarding they get regardless of its purpose. these are people who would like to see america burned and her citizens demoralized. this might be trollingish but i honestly think that some of you guys are, in your liberal fury, forgetting that terrorists are actually really really horrible people.

it's definitely bad to waterboard people who aren't, like, suspected terrorists. terrorism just crosses the line, though. if by kinda-sorta violating the rights of somebody who really doesn't deserve their rights anyway, you can possibly save the lives of countless innocent citizens, i think that's alright.

I remember when I discussed this issue before (with my roommates, I think) that the main concern seemed to be not be so much with the terrorism as with the apparently totally crappy state of our national intelligence agencies. Like, if we were pretty damn sure that the people detained in Cuba were the crappy people we're actually supposed to be after, I personally wouldn't give a damn what happened to them there. The problem is that ... that's really not the case at all. An alarming number are innocent. and obviously, there is a pretty good reason for a collective distrust in our intelligence agencies given the whole OMGWTF IRAQ HAS WMDS business.

IMHO, they should go ahead and subject the strongest supporter of waterboarding to waterboading
See how he/she feels.

no. this is stupid. we're not saying waterboarding is some nice happy thing that everybody would be thrilled to undergo. we're saying it's some bad nasty thing that might be useful for american ideals and american lives.

[quote="mgao"]But if you're speaking from a Geneva-Convention PoV, I really think that it should be illegal in accordance with law.[/quote]
[quote="Mystic Terminator"]I usually don't follow rules just because they're rules; I follow things that make sense.[/quote]
But does it make sense to have a world in which there are rules? Mystic Terminator is trying to carve out a utilitarian exception: what if torture (excuse me, "enhanced techniques of interrogation") actually works, especially in the (fictional) Jack Bauer/ticking bomb scenario?

The first thing I'll say is that I don't believe torture even works. It's quite easy, if you're willing to use excessive force, to force someone to tell you whatever it is you want them to tell you. But that's the problem: what you want the prisoner to tell may depend more on your (the interrogator's) preconceptions than on anything that is true, useful, or complete.

And then there is the fact that world does have rules, and the ideals behind those rules, the the U.S., as the world's most powerful country, inherits with that power the responsibility to support those ideals. My opinion is absolute here. I think that all persons held involuntarily by the U.S. should fall into one of two broad categories. The first category is that of criminal suspects, criminal defendants, and convicted criminals. The second category is prisoners of war. And the rights enumerated in the 4th, 5th, and 6th amendments to the U.S. constitution should apply to the first category, and those in the Geneva Conventions should apply to the second category. I think that carving out a category called "unlawful enemy combatant" not subject to either enumeration of rights is a legal abomination, and will in the long run be seen as as much of a legal abomination as, for instance, the administrative and legal decisions that supported the WWII relocation and internment of Japanese-Americans.

Finally, there are the interests of justice. The administration chose to classify the events of 9/11 as an act of war rather than as a crime. Parallels were drawn to the 1941 attack on Pearl Harbor, which was certainly an act of war. But if 9/11 was an act of war, what country was it that attacked us? You could make a very thin case for singling out Afghanistan, but that's as far as it extends. Perhaps it would have been better all along to classify it and respond to it as a crime (a very large crime, to be sure) and to place a high priority on bringing its perpetrators to justice. And that's the right word: justice, not revenge.

As it happens, a man named Khaled Sheikh Mohammed is in U.S. custody and has been for several years. It is alleged that he was the principal operational planner and director of the 9/11 attacks. I am an no position to judge what kind of evidence lies behind that allegation or how convincing that evidence might be. This man has almost certainly been tortured, and he's rumored to have said quite a bit. He is in custody, but can he be brought to justice? In any legal proceeding against him in any regularly constituted U.S. court, the prosecution could never use any statements made under torture. On the other hand, a special military tribunal, lying outside the review of the U.S. court system and authorized to use secret evidence, will not be recognized by the rest of the world as an instrument of justice.

You've made a case that if you do horrible, horrible things to people, they're likely to tell you whatever you want. Yes, okay, this is probably true. I don't see how it follows that there aren't other possible techniques which can induce prisoners to yield useful information without going to the extent of the first category of techniques described (that would, in fact, yield false information).

Okay, so yes, that would be a legal abomination. I'm not really saying we should just ignore all the rules and laws and whatnot -- that would result in a fairly unordered society. I'm more saying we shouldn't say "oh this is correct because it's a law"; rather, we should say "why is it a law? is it correct?". So given the current legal status, yes, it's probably not good to go around torturing people, but that legal status could certainly be changed.


If \[ ab \equal{} \frac{{a \plus{} b}}{2}\]
\[ x(14 \cdot x) \equal{} x\]
So what is the value of x ?
-------------------------
What is the use of \[ ab \equal{} \frac{{a \plus{} b}}{2}\] the guy who give this question says that you must use this. But how and why? He say also maybe this value you get it by this may not be a real solution or such a thing that I couldn't understand.

Hello
Well by $ ab \equal{} \frac {a \plus{} b}{2}$ they mean a binary relation
use this for finding $ 14 \cdot x \equal{} \frac {14 \plus{} x}{2}$
Hope I m giving the right thing.
Thank u.

So the answer will be \[ x \equal{} \frac{{14}}{{27}}
\]
but if we substitute this into \[ x(14 \cdot x) \equal{} x
\]
Its will not work  :huh:  :(

[quote="hasan4444"]If
\[ ab \equal{} \frac {{a \plus{} b}}{2}
\]

\[ x(14 \cdot x) \equal{} x
\]
So what is the value of x ?
-------------------------
What is the use of
\[ ab \equal{} \frac {{a \plus{} b}}{2}
\]
the guy who give this question says that you must use this. But how and why? He say also maybe this value you get it by this may not be a real solution or such a thing that I couldn't understand.[/quote]

There are major confusions as to what is asked in this problem and much clarifications are needed. This leaves many possibilities and so we are not even sure what you are asking about nor what the question is asking about...

For example, what does it mean $ ab\equal{}\frac{a\plus{}b}2$? Is this an equation or relation? If it is a relation, is it a binary relation like $ a\star b\equal{}\frac{a\plus{}b}2$ or a condition for a set of numbers to satisfy, ie the set $ S$ such that for any $ a,b\in S$, $ ab\equal{}\frac{a\plus{}b}2$ holds true? Not only that, you might not even know what I just typed  :ninja:

So... if it is a binary relation, the equations become

$ a\star b\equal{}\frac{a\plus{}b}2$

$ x\star(14\star x)\equal{}x\Rightarrow \frac{x\plus{}\frac{14\plus{}x}2}2\equal{}x$

Solve for $ x$ and get $ x\equal{}14$

The question is exactly like this:
[quote]If $ ab \equal{} \frac {{a \plus{} b}}{2}$ and $ x(14 \cdot x) \equal{} x$ what is the value of $ x$ ?[/quote]
But in your second way that I almost understand, if you substitute $ x\equal{}14$ in $ x(14 \cdot x)$ the answer will not be $ 14$  :(  so what is the correct answer with the using of $ ab \equal{} \frac {{a \plus{} b}}{2}$.

The first equation almost certainly means that the problem-writer is defining an operation $ a \star b$ to be identically equal to $ \frac {a \plus{} b}{2}$, and it's likely that wherever you got this problem from had formatting errors and whatever symbol was originally used for the operation was dropped.

This poses a problem for the second condition, which could read either

$ x \star (14x) \equal{} x$

or

$ x \star (14 \star x) \equal{} x$.

My personal preference, based on the use of the dot, is for the first interpretation, but without the original source we can't rule out the second.

[hide="Casework"]
Case 1: $ x \star (14x) \equal{} \frac {x \plus{} 14x}{2} \equal{} x \Longrightarrow x \equal{} 0$

Case 2: $ x \star (14 \star x) \equal{} \frac {x \plus{} \frac {14 \plus{} x}{2}}{2} \equal{} x \Longleftrightarrow 7 \plus{} \frac {3x}{2} \equal{} 2x \Longleftrightarrow x \equal{} 14$

Notice that $ a \star b \equal{} \frac {a \plus{} b}{2} \equal{} a$ iff $ a \equal{} b$[/hide]


algebra
inequalities
minimization
Sequence
Fibonacci
IMO Shortlist

For every integer $ n \geq 2$ determine the minimum value that the sum $ \sum^n_{i\equal{}0} a_i$ can take for nonnegative numbers $ a_0, a_1, \ldots, a_n$ satisfying the condition $ a_0 \equal{} 1,$ $ a_i \leq a_{i\plus{}1} \plus{} a_{i\plus{}2}$ for $ i \equal{} 0, \ldots, n \minus{} 2.$

I hardly know what you mean.Let $ a_{2k}  \equal{} 1$ and $ a_{2k \plus{} 1} \equal{}0$.

[quote="orl"]For every integer $ n \geq 2$ determine the minimum value that the sum $ \sum^n_{i \equal{} 0} a_i$ can take for nonnegative numbers $ a_0, a_1, \ldots, a_n$ satisfying the condition $ a_0 \equal{} 1,$ $ a_i \leq a_{i \plus{} 1} \plus{} a_{i \plus{} 2}$ for $ i \equal{} 0, \ldots, n \minus{} 2.$[/quote]
It is easy understand that minimum value of the $ \sum^n_{i \equal{} 0} a_i$ will be then $ a_0, a_1, \ldots, a_n$ satisfying next algebraic system :
----------------------------------------
                      $ a_1 \plus{} a_2 \equal{} 1$
                      $ a_2 \plus{} a_3 \equal{} a_1$
                      $ a_3 \plus{} a_4 \equal{} a_2$
$ \ldots$
$ a_{k \minus{} 1} \plus{} a_{k} \equal{} a_{k \minus{} 2}$
$ \ldots$
$ a_{n \minus{} 1} \plus{} a_{n} \equal{} a_{n \minus{} 2}$
$ a_n \equal{} 0$
---------------------------------------
So:     $ a_i \equal{} \frac {f_{n \minus{} i}}{f_n}$  -  then $ f_i$       for $ i \equal{} 0, \ldots, n$  -  Fibonacci series $ (f_0 \equal{} 0 ,f_1 \equal{} 1,f_2 \equal{} 1 , ...)$
min  $ \sum^n_{i \equal{} 0} a_i \equal{} \frac {1}{f_n}$ $ (f_0 \plus{} f_1 \plus{} \ldots \plus{} f_{n \minus{} 1} \plus{} f_{n}) \equal{} 1 \plus{} \frac {f_{n \plus{} 1} \minus{} 1}{f_n}$ :)


vector
linear algebra
matrix
linear algebra unsolved

Through some exercises, I found that AB and BA has the same set of non zero eigenvalues. Is this assumption true?

True.  Of course you need hypotheses like A and B are matrices and the products AB and BA are defined, things like that.

Let $ A \in M_{m \times n}$ and $ B \in M_{n \times m}$. If $ \lambda \neq 0$ is an eigenvalue of $ AB$ then there is a non-zero vector $ x \in M_{m \times 1}$ such that $ ABx \equal{} \lambda x$ (1). Multiplying now both sides of (1) by $ B$ we have $ BA(Bx) \equal{} \lambda (Bx)$, where we see that $ \lambda$ is also an eigenvalue of $ BA$ (observe that $ AB \in M_{m \times m}$, $ BA \in M_{n \times n}$, $ Bx \in M_{n \times 1}$; observe also that $ Bx \neq 0$, for if $ Bx \equal{} 0$ then (1) would imply $ A(Bx) \equal{} \lambda x \equal{}> A \cdot 0 \equal{} 0 \equal{} \lambda x$, and the right hand side is not zero). If $ Sp(Q)$ denotes the set of non-zero eigenvalues of the matrix $ Q$, then this argument shows that $ Sp(AB) \subseteq Sp(BA)$. But you can use a similar argument to show that if $ \mu \neq 0$ is an eigenvalue of $ BA$ then it is also an eigenvalue of $ AB$, which means $ Sp(BA) \subseteq Sp(AB)$. Therefore, $ Sp(AB) \equal{} Sp(BA)$.

Stronger:
http://www.mathlinks.ro/viewtopic.php?t=299895
See also http://www.mathlinks.ro/viewtopic.php?t=310174

  darij


analytic geometry
calculus
derivative
function
calculus computations

Let $C$ be the part of $x>0$ the curve of $y=x^3$. For a point $P$ on $C$, let $Q$ be the intersection point between the tangent of $C$ at $P$and the $x$-axis,
and let $R$ be the intersection point between the normal of $C$at $P$and the $y$-axis.

If $P$ moves on $C$, find the minimum value of $\frac{OR}{OQ}$, where $O$ is the origin.

Let point $P$ have coordinates $(P_x, P_y)$, then the equation of the tangent line and of the normal line at $P$ are $y=3xP_x^2-2P_x^3$ and $y=-\frac{x}{3P_x^2}+\frac{1}{3P_x}+P_x^3$ respectively, and thus $OR=\frac{1}{3P_x}+P_x^3$ and $OQ=\frac{2}{3}P_x$, so $\frac{OR}{OQ} = \frac{3P_x^4+1}{2P_x^2}$.  The derivative of this expression with respect to $P_x$ is $\frac{3P_x^4-1}{P_x^3}$, the numerator has a positive root of $P_x= \frac{3^{3/4}}3$ (This is a minimum because the second derivative of the function is positive at this point).  So the minimum value of $\frac{OR}{OQ} = \sqrt{3}$.

Yes,you are right,Jimmy.

You can use $A.M\geq G.M.$ :)

kunny

Can you post your solution using AM-GM, I have very little experience with inequalities.

Thanks.

All right, Jimmy

$\frac{OR}{OQ}=\frac{1}{2}\left(3P_x ^2+\frac{1}{P_x^2}\right)\geq \sqrt{3P_x ^2\cdot \frac{1}{P_x ^2}}=\sqrt{3}$.

The equality holds when $3P_x ^2=\frac{1}{P_x ^2}$ and $P_x >0\Longleftrightarrow P_x=\frac{1}{\sqrt[4]{3}}$


A line passing through the barycenter $ G$ of the $ \triangle{OAB}$ intersects in $ P\in{OA},\ Q\in{OB}$.
Let $ \overrightarrow{OP} \equal{} h\overrightarrow{OA},\ \overrightarrow{OQ} \equal{} k\overrightarrow{OB}$ and denote the areas $ \triangle{OAB},\ \triangle{OPQ}$ by $ S,\ T$ respectively. Prove the following relations.

(1) $ \frac {1}{h} \plus{} \frac {1}{k} \equal{} 3$.

(2) $ \frac {4}{9}S\leq T\leq \frac {1}{2}S$.

We have:
$ \overrightarrow{GP} \equal{} \overrightarrow{GO} \plus{} \overrightarrow{OP} \equal{} \frac {1}{3}(\overrightarrow{AO} \plus{} \overrightarrow{BO}) \plus{} h\overrightarrow{OA} \equal{} (h \minus{} \frac {1}{3})\overrightarrow{OA} \minus{} \frac {1}{3}\overrightarrow{OB}$
Similarly: $ \overrightarrow{GQ} \equal{} \minus{} \frac {1}{3}\overrightarrow{OA} \plus{} (k \minus{} \frac {1}{3})\overrightarrow{OB}$
$ P,G,Q$ are collinear if and only if
$ \frac {h \minus{} \frac {1}{3}}{ \minus{} \frac {1}{3}} \equal{} \frac { \minus{} \frac {1}{3}}{k \minus{} \frac {1}{3}}\Leftrightarrow \frac {1}{h} \plus{} \frac {1}{k} \equal{} 3$
Now $ \frac {T}{S} \equal{} hk$, easy to check that with $ 0\leq h,k\leq 1$ so that $ \frac {1}{h} \plus{} \frac {1}{k} \equal{} 3$ we have $ \frac {4}{9}\leq hk\leq \frac {1}{2}$

[quote="tdl"]
Now $ \frac {T}{S} \equal{} hk$, easy to check that with $ 0\leq h,k\leq 1$ so that $ \frac {1}{h} \plus{} \frac {1}{k} \equal{} 3$ we have $ \frac {4}{9}\leq hk\leq \frac {1}{2}$[/quote]

Could you explain this in detail?

We have:
$ 9 \equal{} (\frac {1}{h} \plus{} \frac {1}{k})^2 \equal{} (\frac {1}{h} \minus{} \frac {1}{k})^2 \plus{} \frac {4}{hk}$
With $ \frac {1}{h},\frac {1}{k}\geq 1, \frac {1}{h} \plus{} \frac {1}{k} \equal{} 3$ we have $ \ 0 \leq |\frac {1}{h} \minus{} \frac {1}{k}|\leq 2 \minus{} 1 \equal{} 1$ then OK!


Sa se arate ca oricare ar fi triunghiul ABC, exista o elipsa care trece prin A;B;C , avand in aceste puncte tangente paralele cu BC, CA respectiv AB

Printr-o transformare afina transformam triunghiul intr-unul echilateral, demonstram afrimatia pentru triunghiuri echilaterale (in cazul asta e evident: se ia pur si simplu cercul circumscris), si aplicam inversa transformarii initiale. Cercul circumscris triunghiului echilateral se transforma exact in elipsa pe care o cautam.

Multumesc pt solutie.
Ai timp sa detaliezi?
Ce transformare alegi?
Cum determini inversa transformarii?

Se pare ca ce-am zis eu ieri:
[quote="Eu"]Elipsa pe care o cauti tu e [url=http://mathworld.wolfram.com/SteinerCircumellipse.html]aceasta[/url]. Nu sunt $100 \%$ sigur, dar exista mari sanse.[/quote]
e adevarat.

Cum gasim transformarea aia afina de care vorbea grobber:

Fie $ABC$ triunghiul nostru. Luam originea sistemului de coordonate in $A$. Urmatoarele multimi de vectori sunt liniar independente: $\left\{ \overrightarrow{AB}, \overrightarrow{AC}\right\}$ si $\left\{ \overrightarrow i, \frac12 \cdot \overrightarrow i+\frac{\sqrt 3}2 \overrightarrow j \right\}$, deci exista o transformare liniara inversabila $T : \mathbb R^{2}\to \mathbb R^{2}$ care trimite $\triangle ABC$ intr-un triunghi echilateral $A^\prime B^\prime C^\prime$, unde $A^\prime = A, \, B^\prime \left( 1,0 \right), \, C^\prime \left( \frac12, \frac{\sqrt 3}2 \right)$.

Daca matricea asociata lui $T$ este $A = \left( \begin{matrix}a & b \\ c & d \end{matrix}\right)$, atunci $A^{-1}$ exista si e egala cu $\frac1{ad-bc}\left( \begin{matrix}d &-b \\-c & a \end{matrix}\right)$.

Fie $(x-m)^{2}+(y-n)^{2}= r^{2}$ ecuatia cercului circumscris triunghiului $A^\prime B^\prime C^\prime$. Avem $A^{-1}\left( \begin{matrix}u \\ v \end{matrix}\right) = k \left( \begin{matrix}du-bv \\-cu+av \end{matrix}\right)$.

Deci imaginea cercului circumscris triunghiului $A^\prime B^\prime C^\prime$ are ecuatia $(kdu-kbv-m)^{2}+(-kcu+kav-n)^{2}= r^{2}$, i.e. $k^{2}u^{2}(c^{2}+d^{2})+k^{2}v^{2}(a^{2}+b^{2})-2k^{2}uv(bd+ac)+\ldots = 0$. Determinantul ecuatiei este $4 k^{4}(bd+ac)^{2}-k^{4}(a^{2}+b^{2})(c^{2}+d^{2}) =-k^{4}(ad-bc)^{2}\leq 0$. Nu poate fi egal cu $0$ pentru ca $\det A \neq 0$ ($A$ e inversabila). Deci rezulta o elipsa. Faptul ca tangentele "se pastreaza" se demonstra (cred) cu ajutorul ecuatiilor parametrice.


function
logarithms
analytic geometry
trigonometry
complex analysis

Find the steady temperatures in a solid whose shape is that of a long cylindrical wedge if its boundary planes $\theta = 0$ and $\theta = \theta_{0}$ ($0<r<r_{0}$) are kept at constant temperatures zero and $T_{0}$, respectively, and if its surface $r = r_{0}$ ($0<\theta <\theta_{0}$) is perfectly insulated.

So, we are after a harmonic function $u$ defined on the sector $D=\{re^{i\theta}\colon 0<r<r_{0},0<\theta<\theta_{0}\}$ that satisfies the boundary condition $u(r)=0$, $u(re^{i\theta_{0}})=T_{0}$, and $\left\{\frac{\partial u}{\partial r}u(re^{i\theta})\right\}_{|r=r_{0}}=0$. The conformal map $z\mapsto \log z$ sends $D$ onto the horizontal half-strip $S=\{x+iy\colon-\infty<x<\log r_{0}, 0<y<\theta_{0}\}$. It's easy to come up with a harmonic function on $S$ that has the required boundary properties: $v(x+iy)=\frac{T_{0}}{\theta_{0}}y$. Finally, go back to $D$ to obtain the solution $u(re^{i\theta})=\frac{T_{0}\theta}{\theta_{0}}$.

The answer is $T = \frac{T_{0}}{\theta_{0}}\arctan(\frac{y}{x})$
I am not very good with these kinds of problems (in particular any problem that has to graph kills me)

I used polar coordinates instead of Cartesian: the relation is $x=r\cos\theta$, $y=r\sin\theta$, hence $\tan\theta=\frac{y}{x}$.


If a,b,c are sides of triangle ABC. Inscribed circle is tangent to sides BC,AC,AB at points K,L,M and F is the intesection point of segments AK,BL,CM. Then prove that:

$ 6 \leq \frac{AF}{FK}+\frac{CF}{FM}+\frac{BF}{FL}\leq 6+ \frac{(a+b+c)^2\left((a-b)^2+(b-c)^2+(c-a)^2\right)}{8S^2}$

\[
\frac{{AF}}
{{FK}} + \frac{{CF}}
{{FM}} + \frac{{BF}}
{{FL}} = (a + b + c)\left( {\frac{1}
{a} + \frac{1}
{b} + \frac{1}
{c}} \right) - 3
\]
L.H.S. immediately becomes obvious, for R.H.S., it is equivalent to
\[
\sum {\left( {\frac{{\left( {a + b + c} \right)^2 }}
{{8S^2 }} - \frac{1}
{{ab}}} \right)\left( {a - b} \right)^2 }  \geqslant 0
\]
since \[
{\frac{{\left( {a + b + c} \right)^2 }}
{{8S^2 }} - \frac{1}
{{ab}}}
\]
and the other two symmetric terms are positive, we are done.


(Μολις την κατασκευασα!)
Να εξεταστει αν υπαρχει αριθμος με περιττο αριθμο ψηφιων με ολα τα ψηφια του να ειναι διαδοχικα ξεκινωντας απο το ψηφιο $ 1$ και να ειναι τελειο τετραγωνο!
ΔΗΜΗΤΡΗΣ.

Σωστα αλλα αλλα εισαι σιγουρος οτι ειναι ο μοναδικος??

Αν εχει καποιος εστω καποια ιδεα παρακαλω να την διατυπωσει αν θελει.Ευχαριστω!!!

Θα δείξουμε ότι εκτός από το 1 δεν υπάρχει άλλος τέτοιος αριθμός.

Κάθε άλλος τέτοιος αριθμός πρέπει να τελειώνει σε 901 ή 123 ή 345 ή 567 ή 789.

Κάθε αριθμός που τελειώνει σε 123 ή 567 ισούται με 3mod4 και άρα δεν μπορεί να είναι τέλειο τετράγωνο.
Κάθε αριθμός που τελειώνει σε 901 ή 789 ισούται με 5mod8 και άρα δεν μπορεί να είναι τέλειο τετράγωνο.
Κάθε αριθμός που τελειώνει σε 345 είναι πολλαπλάσιο του 5 αλλά όχι του 25 και άρα δεν μπορεί να είναι τέλειο τετράγωνο.

????Συγγνωμη μα δεν καταλαβαινω...ειπα οτι τα ψηφια να ειναι διαχοχικα ξεκινωντας απο το $ 1$ (το ψηφιο $ 0$ δεν ειναι διαδοχικος του $ 9$ )και περιττα στο πληθος πχ αριθμοι οπως $ 1,123,12345,1234567,.....12345678910111213$ κτλ.

Συγγνωμη παντος για την μη ακριβης διατυπωση του προβληματος ...πιστευω οτι καταλαβατε τωρα.
παρα8ετω την ακριβησ διατυπωση:
Nα ερευνησεται αν υπαρχουν τελεια τετραγωνα με πρωτο ψηφιο το $ 1$ που προκυπτουν απο την ενωση διαδοχικων αριθμων.

Εχω βρειι μια λυση αυτης της ασκησης σε 1 μολις σειρα!παρακαλω προτεινετε λυσεις...:)ΕΥΧΑΡΙΣΤΩ!!!


If the square of Winslow's age is added to Abby's age, the sum is 209.  If the square of Abby's age is added to Winslow's age, the sum is 183.  What is the sum of their ages?

How do you do this?

and how do you post the source?

I don't know how to post the source, but I do know how to do this problem (I think it was from the 01 or 03 Nats). You realize that the squares have to be relatively close, and have to be  <= 14 2, as 15 2 was too great. So if you try 14^2 +13 and 13 2 +14, you get 209 and 183, respectively. Guessing logically is the best approach for this early Nationals Problem, and just thinking is best in this case. So the answer is 14+13 = 27

[quote="hockeyadi23"]I don't know how to post the source, but I do know how to do this problem (I think it was from the 01 or 03 Nats). You realize that the squares have to be relatively close, and have to be  <= 14 2, as 15 2 was too great. So if you try 14^2 +13 and 13 2 +14, you get 209 and 183, respectively. Guessing logically is the best approach for this early Nationals Problem, and just thinking is best in this case.[/quote]

oh...ok thanks. :)

See this topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26824

Billy


geometry
geometric transformation
articles
search
Support

Ignore any religious allusions in the following :wink:.

Papiamento is a creole language spoken in the Carribbean derived from Portuguese. To distinguish between singular and plural "you" here and in later problems, I will use "thou", "thee", "thy" for singular (subjective, objective, possessive, respectively) and "ye", "you", "your" for plural (subjective, objective, possessive). Here are some Papiamento sentences with their English translations:

a) Mi ta atimabo. - I love thee.
b) Bo yiu homber tabata mira Piter. - Thy son saw Peter.
c) E ruman homber ta korda e muhé. - The brother remembers it.
d) Mi ruman muhé ta laba e mucha muhé. - My sister washes the girl.
e) Maria tabata salba bosnan. - Mary saved you.
f) Bosnan lo mustra nos buki. - Ye will read our book.

1. Translate into English:
a) Bo ta laba nos.
b) Piter lo mira mi yiu muhé.
c) Maria tabata korda e buki.

2. Translate into Papiamento:
a) You love us.
b) Peter loved the daughter.
c) The boy saw me.
d) He will wash the boy.
e) Our son will see the book.
f) We remember Mary.

3. In how many ways (what are they) can you translate the Papiamento sentence "e ruman ta mira e yiu"?

[hide="Where do I start on these linguistics problems???"]Everybody can solve these problems. There aren't even theorems you have to know like in math. You just need patience and dedication.

Compare sentences that have the same English words and determine what foreign words correspond to each (note that the foreign words may not be *exactly* the same in each sentence). Word order and tense formation are also very important. In this particular problem, we may use the fact that Papiamento should be similar to English (since it comes from a Romance language). I chose this problem specifically as a relatively easy introductory problem, so I'd love to see some newcomers give it a try, even if you can only get partial solutions.[/hide]

[hide]I don't know what to do with 3. person pronouns so I don't know 2d)

From (b) Piter = Peter. Object is at the end.
From (e) Maria = Mary. Subject is at the beginning.
So the language uses subject-verb-object order.

I think (a) could be "Mi ta atima bo. - I love thee." (with space between atima and bo)

After every subject there is one of the words "ta", "tabata", "lo". This might be the tense: "ta" = present, "tabata" = past, "lo" = future. After the tense, there is the verb, ending in "a". So

atima = love
mira = see
korda = remember
laba = wash
salba = save
mustra = read

Pronouns:

From (a) mi = I.
From (d) mi = my.
From (a) bo = thee.
From (b) bo = thy.
From (e) bosnan = you.
From (f) bosnan = ye.

Now I make a dangerous assumption: that in Papiamento there is no difference between subjective, objective and possesive forms. So "mi" = "I, me, my", "bo" = "thou, thee, thy" and so on. I'm not sure about it, but I think there isn't any other choice.

From (f) nos = we and buki = book.

From (b) (c) (d):
yiu homber = son
ruman homber = brother
ruman muhé = sister
mucha muhé = girl

So homber = female, muhé = male and:
yiu muhé = daughter
mucha homber = boy

"e" means "the" from (c) (d)
But there is a strange thing I can't explain: "e muhé" means "it". This means an article is used before a pronoun.

Translations:
1.
a) Bo ta laba nos = Thou washes us.
b) Piter lo mira mi yiu muhé = Peter will see my daughter.
c) Maria tabata korda e buki = Mary remembered the book.

2.
a) You love us = Bosnan ta atima nos.
b) Peter loved the daughter = Piter tabata atima e yiu muhé.
c) The boy saw me = E mucha homber tabata mira mi.
d) He will wash the boy. = I don't know the word for "he". Maybe there is no grammatical gender and I could assume "E muhé lo laba e mucha homber"
e) Our son will see the book = Nos yiu homber lo mira e buki.
f) We remember Mary = Nos ta korda Maria.

3. "E ruman ta mira e yiu"

"Ruman" may be a brother or a sister, "yiu" may be a son or a daughter. There is no sex given (homber = male, muhé = female). So it may mean:

"The brother / sister sees the son / daughter" - 4 ways.

It could also mean "The sibling sees the child" without specifying sex.[/hide]

Mostly very good. Thanks for noticing the typo.

[hide="Discussion and corrections..."]First of all, I don't think it's fair to assume that all verbs end in -a (a quick google search verifies this with "vini"-"to come"). It is the same principle in Spanish, Latin, French, etc.: there is a set of verbs ending in -a, another in -e, and a third in -i.

Don't worry about making assumptions. I don't even think that one is particularly "dangerous". First, the evidence supports it fully (we have examples of each pair of subject/object/possessive being identical). Second, Romance languages have already lost this distinction in everything except pronouns (this may require a little extra knowledge), so it follows the historical trend. Third, pidgin and creole languages tend to be extremely grammatically simplified. Just that single piece of information (that it is a pidgin/creole language) is enough to tell us that it should have no distinction in pronouns. However, that also requires lots of outside knowledge.

You said homber=female and muhé=male, though you got it right in the exercises later. Typo?

"Thou washes us" is not proper English. I don't know if this is a typo or unfamiliarity with the pronoun "thou", but verbs only take the ending -s in the third person and "thou" is second person.

Concerning the single actual problem in your solution... there is definitely grammatical gender because of homber vs. muhé. In fact, we know that "e muhé" can mean "it". First of all, this does not mean an article is used before a pronoun. "muhé" is an adjective basically meaning "female", but apparently a lot more common than the English equivalent. Then it can be used substantively as a noun (like "goods", "smarts", etc.). So "e muhé" means "the female thing". In fact, this indicates that grammatical gender is very much prevalent. This still refers to what English would consider an "it", but probably something that has feminine gender in Papiamento. Therefore, "he" would be "e homber", so there is that one correction (you said "e muhé").[/hide]

[hide="General information on Pidgins/Creoles"]Pidgin/Creole languages are attempts by natives to communicate with European settlers. Often, this involved stringing together a list of poorly defined words without grammatical rules (a Pidgin in New Guinea is "tok pisin", sounding like "pidgin talk"). Eventually, some of these jargons became full-fledged languages, though they retained grammatical simplicity.

Besides being nearly grammarless and having vocabulary very much like the European language, it should be noted that the phonology is very much like the native language, because it is difficult to pronounce foreign words (I mean, English speakers have trouble with words like ngany, right?). This may explain why "v" apparently became "b" in Papiamento.[/hide]

[hide]
Thanks for the comments, I've learnt a lot.

"Thou washes" (instead of "thou washest") and mistaking homber for male where my faults, thanks for noticing.[/hide]

In order for a language to "make it", its speakers themselves have to go for it as well. :maybe:
That doesn't seem to be the case.

In Aruba they call the language  Papiamento, on Curacao, they call it Papiamentu and in Bonaire it's called Papiamen. Less than 400,000 people speak this language, but there is only that much unity, that they weren't able to agree on one spelling for all three islands! :ninja:


geometry
3D geometry
algebra
polynomial
blogs
number theory
difference of squares

From what Ive seen from other people talking on this forum, you all seem to be math geniuses so im prob gonna end up looking moronic. Ah w/e anyways i have a question as to how original something i thought of is. There is no way my idea is [i]completely[/i] original but maybe its rare i dunno. What i noticed was that a sqaured number minus another squared number always ended up a multiple of 4, so i wondered if i could come up with a formula for it. When i finally found it, i felt slightly stupid, because it is quite obvious when you write it a certain way. The final formula for a sqaure minus another square the lazy way was their difference multiplied by their sum, which is essentially factoring it, and can be useful in some cases. Everyone i told about commented on how they had never thought of it before. So i wonder, how many people have thought of it this way before? I also came up with one for adding squares, the final formula being 2 multiplied by both numbers (before you sqaure them) plus the sqaure of their difference. These what id call situational, i-dont-have-a-calculator-and-am-too-lazy-to-get-one-method :lol:  So what do you guys think of it?

so basically you found:

(x-y)(x+y) = x^2 - y ^2

pretty standard and easy to see its true, I end up using it alot in my class to simplify radicals

and

x^2 + y^2 = 2xy + (x-y)^2

which checks out obviously after expansion, Im not sure what you would use this for though

But anyway, even though these arent really anything new, its always nice to find something for yourself

You rediscovered the difference of two squares formula.  Nice.

(Edit: Seriously I think it's very exciting to discover something like this, I would be proud if I had figured that out myself.)

I hope it didn't sound like I was being sarcastic or something chocoboom.  I didn't mean for it to sound that way.

The multiple of 4 thing is not true.  EX: $3^2 - 2^2 = 9-4 = 5$.  But with number theory you can find that it is never equal to $2 (\bmod\ 4)$ or, it's never $4n\pm2$

Salaam now that you've discovered this cool thing about the difference of two squares, it might be interesting to try to generalize it...  What about the difference of two cubes?

I thank all of you who didnt basically say and/or imply i am dumb. I am not dumb, even though when you read this its prob natural to think so. When i showed this to my math teacher, she said shed never thought of it that way. I realize i shouldnt win the nobel prize lol, but im thinking its one of those simple things people overlook and dont really think about....

[quote="blahblahblah"][quote="Chocoboom"]hey guys shut up, if he was born 1000 years earlier, and he followed up on this, he would have been famous............maybe............ or something i dunno.[/quote]

A lot more than 1000 years...[/quote]

And as for you, shut the hell up you stupid canadian (nothing against canada, but its fun to say as an insult)

Hey chess, im in ninth grade too! (saw it in your profile thing) And i looked at some of your blogs and damn you be much smarter than i am. Im starting not to like this forum. Makes me feel kinda dumb, and i NEVER feel dumb where i come from. So go all you people who are smarter than me. YOU AMAZE ME, seriously.  :ninja:

The truth [i]is[/i] that the difference of squares formula has been known for a long time; certainly upwards of 2000 years.  That doesn't mean that there isn't value in rediscovering it.

On the other hand, if you really wanted us to laud your 'discovery' of something that a fifth grader with a decent grasp of algebra could do, then I suspect that you came to the wrong place.

Playing around with numbers and looking for patterns is a highly useful tool in problem solving even at the Olympiad level.  The fact that you derived this on your own shows that you have the foundations to be a good problem solver.

Unfortunately, your discovery was by no means "rare".  The difference of squares formula is pounded into every high school student's head right alongside the Pythagorean theorem.

On the other hand, please try to use some forum etiquette.  You triple posted, used a expletive, and shot an insult at a highly respected and experienced member of the forum.

Welcome to the forum!  Don't be discouraged by the insanely smart people on this forum.  This forum is home to some of the tip-top students in the nation/world.

[quote="0714446459923"][quote="Aunt Sally"]Another problem to consider is, instead of factoring $x^3-y^3$, try factoring $x^3-1$.

[hide]If you know polynomial long division, that might help.[/hide]

[hide]$x-1$ is a factor.[/hide]

[hide]$1000 - 1 = 9*111$[/hide][/quote]

Thats easy. Since $x^3=1$, the solutions are the cube roots of 1.
(x-1)(x-($f{sqrt(3)i}/2 - 1/2$))(x-(${sqrt(3)i/{-2} - 1/2}$))

In general its
(x-(sin(2pi/n)i+cos(2pi/n))(x-(sin(4pi/n)i+cos(4pi/n)))...(x-(sin(2npi/n)i+cos(2npi/n))
$\displaystyle\prod_{k=0}^\n (x-(sin(2kpi/n)i+cos(2kpi/n)))$.[/quote]I'm pretty sure Aunt Sally was referring to the $(x-1)(x^2+x+1)$ factorization.  ;)

Yeah i guess it was uncalled to get angry with blah but he/she shouldnt be so patronizing, because in fact i dont think he/she even understands exactly what i was referring to when i started this.

Of course i know what the difference of two squares is, but i was talking more along the lines of number application, without variables. For example, 101 sqaured minus 100 sqaured which could be done much quicker if you used the difference of two sqaures you do (101-100)(101+100)=211 instead of actually figuring out what 101 sqaured is.

This prob wouldnt even be that useful but i just wanted to reestablish the fact that im talking about, out of curiousity, how many ppl think to apply this in actual life where there are no variables. Oh yeah and hey joml, am i not supposed to reply to my replies? If im not i realize i just did it again (sorry if im not) lol but i guess i need to know.

[quote="salaam"]but i was talking more along the lines of number application, without variables. For example, 101 sqaured minus 100 sqaured which could be done much quicker if you used the difference of two sqaures you do (101-100)(101+100)=211 instead of actually figuring out what 101 sqaured is.[/quote]

This is a good point, and you should be pretty happy for figuring it out yourself instead of being taught it.

As far as squares go, there is a simple explanation.

$1^2 = 1$
$2^2 = 1 + 3$
$3^2 = 1 + 3 + 5$
...
$n^2 = \sum_{k=1}^{n} (2n-1)$

So then

$101^2 - 100^2 = (1 + 3 + ... + 201) - (1 + 3 + ... + 199) = 201$

:)

[quote="salaam"]Yeah i guess it was uncalled to get angry with blah but he/she shouldnt be so patronizing, because in fact i dont think he/she even understands exactly what i was referring to when i started this. [/quote]

Providing accurate historical facts is not patronizing.

You also have no place to criticize after telling me to 'shut the - up you stupid canadian'.

[quote]So what do you guys think of it[/quote]
Since you asked , I will be honest -  Really not much. Nothing “new” or even remarkable  about it, as someone said,  it has been known for at least a few thousand years and  I have seen it “discovered” dozens of time from kids as old as 6 years, except that they did  NOT make a mistake like you did  and would know that difference of squares is NOT always a multiple of four when the numbers have different parity.

Okay, I agree  when you say you are  not   dumb but you posts are  extremely rude  and sound very dumb  when you  post like:
[quote]And as for you, shut the - up you stupid canadian (nothing against canada, but its fun to say as an insult) [/quote]

That too to a person, to put it mildly, you  ought to look up to. I would be proud of someone who achieved even 1/10th of what Blahblahblah  has achieved.

:rotfl:  Look at all of these 'funny' replies (maybe mine too but who cares? Maybe you do?). I agree with nateharman1234, Aunt Sally, t0rajirou, and joml88. The rest of them are ...  :rotfl:

Agreed. There's nothing wrong with playing around with numbers, but please, no criticizing on the forums.


A wobbly number is a positive integer whose $digits$ in base $10$ are alternatively non-zero and zero the units digit being non-zero. Determine all positive integers which do not divide any wobbly number.

We show inductively for every $k$, there is a wobbly number $w_k$ having $2k-1$ digits such that $4^k$ divides it, but $2\cdot 4^k$ doesn't.
$k=1$: just take $w_1=4$.
For any $k$, we try to set $w_{k+1} = m_k\cdot 100^k+w_k$, $m_k \in \{1,2,...,9\}$. We want that $4^{k+1} \equiv w_{k+1} = m_k\cdot 100^k+w_k = m_k \cdot 4^k \cdot 25^k + 4^k w^\prime_k \mod 2\cdot 4^{k+1} $ (here $w_k=4^k w^\prime_k$ for some odd $w^\prime_k$).
But that just means $4 \equiv m_k \cdot 25^k + w^\prime_k \mod 8$, which has a solution $m_k \mod 8$, meaning that we can choose our $m_k$ as required.

Now if $s$ is some wobbly number and $n$ some integer coprime to $2,5$, then we find a multiple of $s$ being a wobbly number divisible by $sn$:
Let $s$ have $2j-1$ digits and see that $s_k := \sum_{m=0}^{k-1} 100^{j\cdot m} s$ is wobbly again. But $s_k = s \cdot \frac{100^{kj}-1}{100^j-1}$, so we just have to take $k=\varphi((100^j-1)\cdot n)$ to get $100^{kj} \equiv 1 \mod ((100^j-1)\cdot n)$ and thus the result.

The first two paragraphs show that for every integer $n$ coprime to $2,5$ we have that $n$, $5n$ ($s=5$ is wobbly) and $2^k n$ have multples being wobbly numbers. No wobbly number can be a multiple of $25$ because all multiples of that number end in $00,25,50,75$, not being allowed. Similar, no wobbly multiple of $10$ exists.
As a result, exactly those $n$ have wobbly multiples that are not divisible by $10$ or $25$.


The point $ (a, b)$ lies on the line with the equation $ 3x \plus{} 2y \equal{} 12.$ When $ a \equal{} 4$, what is the value of $ b$?

$ a\equal{}4$ means the x-coordinate is $ 4$.  So substitute $ 4$ for $ x$, and you get $ 12\plus{}2y\equal{}12$, so $ y\equal{}0$


It's all over, maybe should've been more careful with my proofs, maybe I shouldn't have left one hour early on Day 2... It doesn't matter now, I didn't make MOP(3 fun weeks of math, meeting a lot of people that I really want to meet, skipping 2 boring weeks of school...). I want to thank Mr.Rusczyk/Mr.Crawford/Mr.Patrick/Mr.Vornicu for creating this site and holding online classes. I wouldn't have been even close to MOP without this site(seriously, last year I didn't even make AIME).  Also thanks to those who had helped/encouraged me. I probably won't do math or visit this site for a while.

BTW to all MoPpers, congrats! Have fun for the 3 weeks!

Has anyone in California (or the Western Time Zone) been called yet?

Plenty of Californians have been called.

As for Feiqi, there's always next year. ^_^

Yup, thanks for the site guys. It's been a huge help. Like Feiqi, I didn't make the AIME at all last year and this has been a huge lift.

I've actually exceeded expectations by a lot (AIME at least 6, and if I get to USAMO, to get a positive score) but it's all the more painful when you get so far and then screw up.

Feiqi, if you want to get anything about it off your chest, feel free to IM me. I probably need to mope around or just say some stuff too.

Yeah, I know how you guys feel. I didn't even expect to make it this far, but when I did I really expected to do much better at the last stage of the competition. Anyway, it's time to work hard for next year so I won't have to wait for "the call"! :P

Guys don't be sad. You've jumped from not going to AIME to not going to MOP. Next year you'll probably be at not going to IMO (after the TSTs :) ) and the next year you'll be at "damn, one point and I would have had a gold" ;) So keep up the good work. Nothing comes without it ;)

Thanks Valentin! I would also like to thank all of the AoPS staff for creating such a fun and exciting place to learn and enjoy the subject we all love, mathematics. All of your hard work has helped us excel in many areas where we never though we could. Without you guys none of us would have ever made it to where we did. :)

Ditto :D.

But really, AoPS has brought me way farther than I ever expected to be 2 years ago. Every year it raises my expectations of myself - I remember back in the 8th grade days where I was like "AMC10? Ahh, it's so hard!" and didn't even make AIME to today, where I expected to make MOP. You guys are the best.

I would like to thank the AoPS site as well, espeically the Resources pages... which helped me get started on some of the books that I probably would have never heard of had i not been here. Thanks again... classes helped alot also...

[quote="MithsApprentice"]

As for Feiqi, there's always next year. ^_^[/quote]

Right, next year I need to get to top 35ish, that's... impossible for me! I think I just missed my only chance for MOP this year.

[quote="beta"]Right, next year I need to get to top 35ish, that's... impossible for me! I think I just missed my only chance for MOP this year.[/quote]
Hi, beta,

I am glad you are still on this forum. I cried for you. I feel really sad to see kids going through this much at this young age. But remember this is not the only way in life. You certainly have a chance to make top 35 next year. Don't give up if math is what makes you feel good.

Obviously we put highest value on how we feel about ourselves. But what really makes us feel good changes as we going through different stages of life. As a USAMO qualifier, the door is very much open to you for almost anything you want to do. But do keep up your grades in areas other than math.

By the way, if anyone is still wondering about the score for freshmen to get into the red MOP, it is probably 7 and above.

The highest score of the 9th grader is probably 21-23 and was achieved by one 9th grader as well as one 8th grader. The rest of the 9th graders are in red MOP. For a 7th grader to qualify even US citizens, he has to do that through USAMO index.

[quote="WarpedKlown1335"]Who's the 8th grader who made it?[/quote]
I believe Eric Larson from Eugene, OR is the one.


When I try to compile this document in TeXnicCenter, it gives me some errors.  I can't figure out what's wrong.

Here's my document:

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\topmargin 0in
\headheight 0in
\headsep 0in
\textheight 7.7in
\textwidth 6.5in
\oddsidemargin 0in
\evensidemargin 0in
\headheight 77pt
\headsep 0.25in

\begin{document}

This is a test.

\end{document}

And here are the errors:

! Undefined Control Sequence.
1.8 \pdfpagewidth 8.5in
! LaTeX Error:  Missing \begin{document}
! Undefinted Control Sequence.
1.9 \pdfpageheight 11in

When I take the block of commands out from \pdfpagewidth to \headsep, it compiles just fine.  I looked up \pdfpagewidth in the help section for TeXnicCenter to see what was wrong, and it's not in the index.

This is how the the document in the LaTeX guide here is written, so what am I doing that's different?

Thanks :)

I am unsure as why do you need those commands \pdf[...]  :maybe: just remove them, and compile the text as is.

There are other commands that take care of the dimensions of the text and page for you, such as

\setlength{\textheight}{9.5in}
\setlength{\textwidth}{6.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\leftmargin}{0.0in}
\setlength{\oddsidemargin}{0.0in}

You can tweak these to fit your needs.

There's a little box up near the top of the TeXnic Center window.  Make sure that it says "LaTeX => PDF" and not "LaTeX => DVI" if you want your output files to be in PDF.


How do you figure out the power problems like, "what is the units digit of 3^1001?" or "what is in the units place of (3^3)^5

Look for patterns.

Units digit for 31: 3
Units digit for 32: 9
Units digit for 33: 7
Units digit for 34: 1
Units digit for 35: 3
...

What pattern do you see?


Let $ a, b, c$ non-negative real numbers. [b]Prove or disprove [/b]that

$ (\sum\frac {a^2}{b^2 \plus{} c^2})\prod\frac {a^2 \plus{} b^2}{b^2} \ge (\sum\frac {a}{b \plus{} c})\prod\frac {a \plus{} b}{b}$.

_________
 :?:  :idea:

$ \sum_{cyc}\frac{a^2}{b^2\plus{}c^2}\geq \sum_{cyc}\frac{a}{b\plus{}c}$ is true already.

generally, we have

   \[ \sum\frac {a^{k\plus{}1}}{b^{k\plus{}1}\plus{} c^{k\plus{}1}} \ge \sum\frac {a^k}{b^k \plus{} c^k}(k\equal{}1,2,3,\ldots),\]

   \[ \prod\frac {a^{k\plus{}1} \plus{} b^{k\plus{}1}}{b^{k\plus{}1}} \ge \prod\frac {a ^k\plus{} b^k}{b^k}
(k\equal{}1,2,3,\ldots).\]

[quote="fjwxcsl"]generally, we have
\[ \sum\frac {a^{k \plus{} 1}}{b^{k \plus{} 1} \plus{} c^{k \plus{} 1}} \ge \sum\frac {a^k}{b^k \plus{} c^k}(k \equal{} 1,2,3,\ldots),
\]

\[ \prod\frac {a^{k \plus{} 1} \plus{} b^{k \plus{} 1}}{b^{k \plus{} 1}} \ge \prod\frac {a ^k \plus{} b^k}{b^k} (k \equal{} 1,2,3,\ldots).
\]
[/quote]

Can you give an elementary proof?
Thanks

 :)

[quote="fjwxcsl"]generally, we have
\[ \sum\frac {a^{k \plus{} 1}}{b^{k \plus{} 1} \plus{} c^{k \plus{} 1}} \ge \sum\frac {a^k}{b^k \plus{} c^k}(k \equal{} 1,2,3,\ldots),
\]

\[ \prod\frac {a^{k \plus{} 1} \plus{} b^{k \plus{} 1}}{b^{k \plus{} 1}} \ge \prod\frac {a ^k \plus{} b^k}{b^k} (k \equal{} 1,2,3,\ldots).
\]
[/quote]

Thanks


[quote="fjwxcsl"]Suppose

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {3}{u},{x,y,z,u,v,k}\in{R_ \plus{} },$

prove that :

$ \sum_{x\to{y}\to{z}}{\frac {1}{x \plus{} k y \plus{} v}}\le{\frac {3}{u \plus{} ku \plus{} v}}.$[/quote]

The proof  of  this nice and hard   inequality   is very simple and  very clear   for

a new computer program which can be read. use this program  we also can prove that

$ \sum_{x\to{y}\to{z}}{\frac {1}{x \plus{} k y \plus{} tz \plus{} v}}\le{\frac {3}{u \plus{} ku \plus{} tu \plus{} v}}.$

[quote="fjwxcsl"]
The proof  of  this nice and very  very hard   inequality   is very simple and  very clear   for

a new computer program which can be read. use this program  we also can prove that

$ \sum_{x\to{y}\to{z}}{\frac {1}{x \plus{} k y \plus{} tz \plus{} v}}\le{\frac {3}{u \plus{} ku \plus{} tu \plus{} v}}.$[/quote]
Where can we find that program ?

Both inequalities are easy corollary of Cauchy. For example, for the first one,
$ (x\plus{}ky\plus{}v) \left(\frac{1}{x} \plus{} \frac{k}{y} \plus{} \frac{v}{u^2}\right) \ge \left(1 \plus{} k \plus{} \frac{v}{u}\right)^2$ which implies
$ \frac{1}{x\plus{}ky\plus{}v} \leq \frac{\frac{1}{x} \plus{} \frac{k}{y} \plus{} \frac{v}{u^2}}{\left(1 \plus{} k \plus{} \frac{v}{u}\right)^2}$
Sum up and done.


calculus derivative function algebra domain limit topology

If a function is differentiable on some closed interval [a,b], is it implied that the left and right hand limits of the derivative must exist (and be equal) even at the end points a and b? Or do we only require that one sided limits need to exist for the end points?    If we compare f(x)=x   and g(x)=|x|, they are identical on the interval [0,1], for example. But g'(0) technically "does not exist" because the left and right hand limits aren't the same; however, when we restrict the domain as we have, do we need only worry about the limit of

[g(x+h) - g(x)]/h   as h approaches zero through the positive numbers? and consequently, on the interval [0,1], g'(0) would exist?

This is really a matter of definition, which is best avoided by considering open intervals. But if I had to define what it means to be differentiable on $[a,b]$, I would only require the existence of one-sided limits at the boundary points. The function $g(x)=|x|$ is differentiable on $[0,1]$ as far as I am concerned.

What brought me to ask was a problem in spivak concerning a "function differentiable on [a,b]" and the way the problem progressed, it was certainly implied that only the one sided limits mattered for the end points. The problem was a three step proof of Darboux's theorem, which is basically what he is calling the intermediate value theorem for derivatives.

Exercise: prove that a function defined on $[a,b]$ is differentiable "in the one-sided sense" if and only if it can be extended to a differentiable function on $\mathbb R$.

Another question.  "Suppose that f is differentiable in some interval containing a, but that f' is discontinuous at a."

The first part of the problem has you prove that the limits

lim f'(x) as x-->a+

and

lim f'(x) as x--->a-

cannot both exist, because one can use the mean value theorem to show that both of these limits must be f'(a) and we find that f' was continuous at a after all.

However, the second part states  "Neither of these one sided limits can exist even in the sense of being plus or minus infinity"  What bothers me about this problem is the inclusion of the word "even" which implies that neither of the one sided limits could exist at all. We've already shown they can't both exist, and that if one exists it must be f'(a), but
[i]why can't just one of them exist?[/i]  Just one of the one side limits existing and being equal to f'(a) doesn't seem to imply continuity and we don't have a contradiction. If he hadn't included the word "even" in the problem, then it would make more sense, because using Darboux's theorem we can show that it would be absurd if one of those limits were plus or minus infinity.

I agree that the problem could be worded better. It's easy to come up with examples where one of the one-sided limits exists, say $f(x)=x^{2}\sin(1/x)$ for $x>0$ and $f(x)=0$ otherwise. I don't like this "limit exists in the sense of being infinity" phrase at all: the limit does [b]not[/b] exist, and we write $\lim=\infty$ to indicate why.
(If the real line is extended to include $\pm\infty$ and given an appropriate topology, then it's legitimate to say that some limit exists and is equal to $+\infty$. Then, however, the limit exists in the ordinary sense, not in a Pickwickian sense.)

I think he must have goofed in the wording then; using the word "even" was misleading. Its true he uses informal language sometimes, and maybe his goof was using the word "even" to express that the limits can't "even be infinite". He does, however, assume that at this point you know what is meant by a limit being infinite.


on what interval f(x)=x^4-4x^3 both decreasing and concave up?

$ f(x)\equal{}x^4\minus{}4x^3\Rightarrow f'(x)\equal{}4x^3\minus{}12x^2$,hence
$ f'(x)<0$,if $ 4x^3\leq 12x^2\Rightarrow x\leq 3$,
$ f''(x)\equal{}12x^2\minus{}24x$,hence $ f''(x)<0$,if $ x\in (0,2)$.
So the answer is $ x\in (0,2)$.


Prove that for positive reals $ a$, $ b$, $ c$, the following inequality holds:
$ \frac{a^8\plus{}b^8\plus{}c^8}{a^3\plus{}b^3\plus{}c^3}\geq \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$.

Are you sure it is true? It does not seem to be homogeneous. Or it isn't a fraction there?

This inequality cannot hold. Just consider the case of all variables equal. There is something strange with it.

I guess it should be 3 times the expression on the left!

No, it will be wrong again! Read what Harazi have written above.

You can multiply it with whatever you want, you have to make it homogeneous before.

I didn't try to solve it. I just posted it because i saw it in a book I have. Next time I will check to see that the problems that I post are true. Sorry again!

I wonder, is it the spirit of RobertuX that now entered the "Inequalities" forum? 11 posts [edit: now 7 posts], and absolutely no useful contents... Anyway, I have an idea on what inequality RaduSorici could have meant:

[color=blue][b]IMO Longlist 1967, problem Poland 3:[/b] If a, b, c are positive numbers, then prove the inequality

$ \frac {a^8 \plus{} b^8 \plus{} c^8}{a^3b^3c^3}\geq\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.[/color]

This one is absolutely trivial and was proven by me in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=21137 . (Sounds nice, yes? What I wanted to say is that it's trivial using Muirhead, but without Muirhead, it requires a little work as I did in that thread.)

  darij

This ineq is not hard . I used Cauchy ineq to solve it.

it's just an entrance examination to LEHONGPHONG HIGH SHOOL FOR THE GIFTED :D

JUST am-gm :lol:


can any body give me some sample questions from kvpy or some similar q's or atleast some related urls

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

though i dont exactly remember the questions that appeared i can probably give u a fair idea about the topics they concentrate upon.
 as far as [b]physics[/b] is concerned a good understanding of the concepts learnt in 10 standard is more than enough. there will definitely a question on dimensional analysis, optics (like lateral displacement caused by a rectangular slab(came last year) and concepts related to snell's law),nuclear physics and on forms of energy.a thorough revision of the 10 standard cbse textbook would very well fetch u even full marks in physics.
[b]chemistry[/b] is confined only to the 10 std ncert textbook,especially on topics like metallurgical extraction processes,electrochemistry (only fundamentals like faraday's laws) and preparation and reactions of organic compounds(only   those given in ncert textbook)
 [b]mathematics[/b] was a bit challenging with even a past rmo problem being asked as an objective question.problems based on [b]modulus function [/b]are quite common.the paper demands a thorough understanding of functions(like symmetry etc.)counting problems,problems , number systems(like binary to decimal conversion,problems based on clocks are also very likely
 [b]biology was horrible [/b]and it would be better if u just leave it at peace and do well in PCM.by the way cutoffs are extremely low(I cleared the written exam despite attempting only 50% questions and even became a kvpy fellow)
JUST GO WITH A COOL MIND AND YOU WILL DEFINITELY CLEAR IT
                       BEST OF LUCK!!!!!!!

[quote="ashwinkrishnan"]By the way, when are the KVPY results likely to be out[/quote]

KVPY results have been declared. Check the website for details.


[quote="minsoens"][quote="tunganh"]Gives that $a,b,c,d$ are real numbers such that $\frac{a}{b}= \frac{b}{c}= \frac{c}{d}= \frac{d}{a}$. Find the largest possible value of:
$\frac{ab-3bc+ca}{a^{2}-b^{2}+c^{2}}$[/quote]if $\frac{a}{b}= \frac{b}{c}= \frac{c}{d}= \frac{d}{a}=k$, then is it not true that $a=b=c=d$ or $a=c=-b=-d$? because we can say $k^{4}=\left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\left(\frac{c}{d}\right)\left(\frac{d}{a}\right)=1$ and $k$ is real, so..  :maybe:[/quote]
Why it isn't true?
You say that $k^{4}=1$, but if $a=b=c=d=0 then k=0$ :)

Who can solve it?

First, notice that $a,b,c,d$ must be nonzero.. Otherwise, the value of $X$ will be undetermined..
Let $X = \frac{a}{b}= \frac{b}{c}= \frac{c}{d}= \frac{d}{a}$ and $S = \frac{ab-3bc+ca}{a^{2}-b^{2}+c^{2}}$
We have
$a = \frac{d}{X}$
$c = dX$
$b = cX = dX^{2}$
But, $a = bX = dX^{3}$. Since $d \neq 0$, then $X^{3}= \frac{1}{X}\Rightarrow X = 1$ or $X =-1$.

i) If $X = 1$, then $a=b=c=d$ and $S =-1$
ii) If $X =-1$, then $a=-b=c=-d$ and $S = 3$.

So, the maximum value of $S$ is $3$.  :)

[quote="RDeepMath91"]First, notice that $a,b,c,d$ must be nonzero.. Otherwise, the value of $X$ will be undetermined..

ii) If $X =-1$, then $a=b=-c=-d$ and $S = 3$.

[/quote]
I don't understand, can you explain it? :P

[quote="RDeepMath91"]Well, for the second one, I have fixed it..  Thanks for the correction..
About the first one, WLOG we can choose $a = 0$, but $X = \frac{d}{a}= \frac{d}{0}$ which doesn't make sense..  :wink:[/quote]
Remember the Cauchy-Schwarz inequality :D
When the inequality occurs, if the denominator is equal to zero, then the numerator is equal to zero, too :)

[quote="RDeepMath91"]Well, okay then if you consider that $a=b=c=d=0$ is possible..
But how about $S$ ?
$S$ will equal to $\frac{0}{0}$ which once again, doesn't make sense..  :([/quote]
Sorry :)
I'm just thinking of $\frac{a}{b}= \frac{b}{c}= \frac{c}{d}= \frac{d}{a}= \frac{a+b+c+d}{a+b+c+d}$
Then we must consider the case when $a+b+c+d=0$ :D

What's SMO, and why would a national olympiad have such a trivial problem? Was there a mistake in the transcribing...

[quote="me@home"]What's SMO[/quote]
Singapore Mathematics Olimpiad :D
[quote]why would a national olympiad have such a trivial problem? Was there a mistake in the transcribing...[/quote]
Yes, I was very surprise when I see that problem :)


calculus
integration
function
calculus computations

$ \int\limits_{0}^{ \infty }\frac{x^{3}}{e^{x} \minus{} 1}dx$

This shouldn't be in the high school forum.

Recall
\[ \Gamma(z)\zeta(z) \equal{} \int_0^{\infty} \frac{t^{z\minus{}1}}{e^t \minus{} 1}\] where $ \Gamma$ is the Gamma Function and $ \zeta$ is the Riemann Zeta Function.

Then $ z$ obviously is 4.  So you'd need $ \Gamma(4)\zeta(4)\equal{}6\times\frac{\pi^{4}}{90}\equal{}\frac{\pi^{4}}{15}$.

See for example [url=http://www.math.utah.edu/~milicic/zeta.pdf]this little article[/url] for further reading.


give us ur opinion why va is the most pwnage state

cuz of the pwning  schools here and smart kids, i guess?
 :oops:

1) hjoon0125 lives there

2) because longfellow ms pwns

3) the white house is there

4) the capitol is there

5) alex960531, turing, and hjoon0125 r funny

6) their going to win Nats

7) cause hjoon0125 said so

and many more

wth, capital is in virginia?

come on hjoon, you know better than that

its spelled capitol. lol

come on you know better than that  :P

or maybe im wrong?

10. It isn't the least pwnage state.

11. It isn't the second least pwnage state.

12. It isn't the third least pwnage state.

You get the idea...

[quote="hjoon"]hjoon0125 r funny[/quote]

hjoon0125 r not funny. hjoon0125 r hilarious!!!

[quote="hjoon"]2) because longfellow ms pwns [/quote]

You go to Longfellow? Do you take math reasoning classes in George Mason HS?

Virginia is an Awesome state. Fairfax County is a great school system too.

:D because TJ is there, because I live there :D

[quote=eqjj168][b][color=darkred]Virginia Tech Football!!!!![/color][/b]
[color=orange]Currently ranked #5 in the nation!!! (AP Poll 10/8/09)[/color]
 :clap2:  :thumbup:  :clap:[/quote]

NO SUPPORT UVA

[quote=-$45!][quote="hjoon0125"]8) out of the 5 earliest presidents, 4 of them were born in va (george washington, thomas jefferson, james madison, and james monroe)

9) many [color=red]Manchester United[/color] fans

and many more[/quote]

Oh no Hjoon! Rooney got injured and they lost!![/quote]

Um... #ChelseaIsBetter

[quote=thegabster37]Again, no need to revive old threads[/quote]

[quote=gymnast19]Um... #DontRevive[/quote]

RIP

[quote=-$45!][quote="hjoon0125"]8) out of the 5 earliest presidents, 4 of them were born in va (george washington, thomas jefferson, james madison, and james monroe)

9) many [color=red]Manchester United[/color] fans

and many more[/quote]

Oh no Hjoon! Rooney got injured and they lost!![/quote]They would lose even with all their players in top form.

Lol, I'm from Virginia in California in Alabama. Is that possible?


I need to construct a sequence where a matrix of the row (2^m)-1 and kolum (m) is of the almost complete set of kombinations similar to this;

1 2 3
-------
|1 1 1| 1
|1 1 0| 2
|0 1 1| 3
|1 0 1| 4
|0 0 1| 5
|0 1 0| 6
|1 0 0| 7
-------

every kombination except the last |0 0 0| is included.

does anyone know a good, easy way to do this?

thanks

if you are trying to print this sequence on the output screen. then you can do that by using loops. I am not sure. but i will try this in C++.

[quote="Watson"]Just count in binary and convert the numbers to bit vectors, then glue them together.  Multiply the matrixes to get the numbers out:
$\left(\begin{array}{ccc} 0&0&1\\ 0&1&0\\ 0&1&1\\ 1&0&0\\ 1&0&1\\ 1&1&0\\ 1&1&1\\ \end{array}\right)\left(\begin{array}{c} 1\\ 2\\ 3\\ \end{array}\right)=\left(\begin{array}{c} 3\\ 2\\ 5\\ 1\\ 4\\ 3\\ 6\\ \end{array}\right)$

To get a proper ordering use sort or a different counting method.[/quote]
Well the order isnt important, however every kombination has to be counted in. The only information one gets is $m$

Now wheather I would do it as a binary transformation or not, I still dont see the general method.
How would I get $m=4$ for example. I kind of thought in the same lines as [b]varun[/b] however I have trouble with seeing the amount of rows at higher integers like 5,6,7....
But later I figured that by taking a general formula for the rows per (m), like $f_1(m)=m$ would be the amount of ones (every kombination) in the matrix. And continue on that path, the function for the two's would be $f_2(m)= \frac{m^2-m}{2}$ and so on: However that kind of math is to hard for me and besides, even if I would come up with the amount of rows I still have trouble in inserting the digits so that every digit is uniqe... This would be my major difficulty, I dont seem to come up with a good sequence for that function..

It all started out to be easy for me but I regret that I  have underestimated it now.

thanks for your replies


I had read lots of posts recently about a new user named sp4m. He had been posting too much spam that is way too annoying. He doesn't have any interests in posting informative, useful stuff or reasonably good questions about math or anything. All he's doing here is spamming on every board.

So, in a serious matter, he should be permanently banned from this site. sp4m does not even do anything here besides posting nonsense stuff and somewhat offensive stuff on the forums. Like, for example, he posted a gay comment in the MOEMS forum that isn't even a question. Plus, he has spammed on nine other threads. He'd better be banned forever. As soon as possible.

I would be happy and I would appreciate it if admins and mods could look this over and ban him when appropriate. Thanks.

gauss1181

P.S. sp4m and another person named omglolz had also been insulting me on the MOEMS thread that I mentioned above. Both of those users must be perma-banned. This same person also made another account called gauss sucks. Ban that person and delete all his posts too. The same person keeps making constant accounts, and he also made an account called dumbuser, as well. Ban ALL of his accounts, please. And tell him never to go on AoPS or register accounts ever again for the rest of his life.

http://www.mathlinks.ro/viewtopic.php?t=181364
There are atleast three more such users.  :mad:

really? Who?

I thought I listed them all....

[quote="gauss1181"]Ban ALL of his accounts, please. And tell him never to go on AoPS or register accounts ever again for the rest of his life.[/quote]

I doubt this will ask them to stop. An IP ban might work.

EDIT: gauss it might help if you didn't feed them.

I'm pretty sure the AoPS people are on vacation at the moment, but they will surely take action very soon.

Is it possible to email ban on phpbb? Circumventing an IP ban or email ban alone is pretty easy (proxy for former, new email or latter), but circumventing both is a hassle.

I believe the admins have a system where there can only be one account registered to one email.

Yeah, but the spammer probably used multiple email addresses to create his multiple accounts.

S/He would have to, its how phpBB is designed.  No more than one account per email account.


Can someone explain to me what the Chinese Remainder Theorem is?


[quote="vishalarul"][hide="A pretty good trig problem..."]Squaring both sides of the equation,
$\sec^{2}(x)+2\sec(x)\tan(x)+\tan^{2}(x) = \frac{289}{49}$
Since $\sec^{2}(x)+\tan^{2}(x) = 1$, the above equation may be simplified to:
$\sec(x)\tan(x) = \frac{120}{49}$
Taking the recipricol of both sides,
$\csc(x)\cot(x) = \frac{49}{120}$
Multiplying this by 2,
$2\csc(x)\cot(x) = \frac{49}{60}$
Adding $\csc^{2}(x)+\cot^{2}(x) = 1$ to both sides,
$\csc^{2}(x)+2\csc(x)\cot(x)+\cot^{2}(x) = \frac{109}{60}$
Taking the square root of both sides,
$\csc(x)+\cot(x) = \sqrt{\frac{109}{60}}$[/hide][/quote]
[hide]$\sec^{2}x+\tan^{2}x\ne1$  :wink: [/hide]

I'll edit my solution. Thanks for pointing that out, pianoforte.


(1)
$y=0$
$f(x) = f(x)+xf(0)$
$\boxed{f(0) = 0}$
(2)
$f(f(x+yf(x)) = f(f(x)+xf(y)) = f(x)+yf(x)$
$y=0$
$\boxed{f(f(x)) = f(x)}$
So $f(x) = x$ on its range.
(3)
Now, suppose $x$ is in the range of $f$. Then
$f(x(1+y)) = x(1+f(y))$
If $x(1+y)$ is in the range of $f$, then so is $y$ (and vice versa). Thus taking $y=x$, we have $x^{2}+x$ in the range as well. This also happens to give us $f(-1) =-1$. Thence
$f(-1-y) =-1-f(y)$
So $f$ is linear on each equivalence class of integers!
Thus $f(x) = |x|$ or $f(x) = x$ for each $x$, and $f(x+n) = f(x)+n$ for all $n$ (which really is a small restriction compared to continuity, which would force it to be $f(x) = x$ I guess).

----

Am I missing a really obvious way to show it is just $f(x) = x$ or something?  :huh:
PS I hope I didn't do anything stupid, also ;) I have no experience with this stuff...

[hide]Subbing $x = 1, y = 0$, we find $f(0) = 0$[/hide]

I haven't gleaned much else yet... :(

[quote="Xevarion"]
$f(f(x)+xf(y)) = f(x)+yf(x)$[/quote]
Can you explain how you found that  :?:

Its not clear for me? because to have that you must have before $f(f(x))=f(x)$ :!:

You are right! It is a serious oversight, I'm not sure how to correct it.  :o

[quote="Guía Rojo"]Find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that

$f\left(x+yf(x)\right)=f(x)+x\cdot f(y)$

for all $x,y\in\mathbb{R}$[/quote]
1. y=0 give f(0)=0.
2.if $f(x)=0,x\not =0$ then $f(y)=0$ for all y.
Let $f\not =0$ then $f(1)=1$ (if $f(1)\not =1$ $y=\frac{1}{1-f(1)},x=1$ give contradition f(1)=0). It give (y=1) $f(x+f(x))=x+f(x)$. If $y=-\frac{x}{f(x)}$ then $f(y)=\frac{1}{y}.$ It give $f(-1)=-1$. Therefore from $f(x+f(x))=x+f(x)$ we have $f(n)=n, \ \forall n\in Z.$
Let $y=\frac{n-x}{f(x)}$, then $f(\frac{n-x}{f(x)}=\frac{n-f(x)}{x}$.
Therefore $f(x)=x$ for all $x\in Q$.

[quote="Rust"]Let $f\not =0$ then.....[/quote]what do you mean?

We have solution $f(x)=0 \ \forall x\in R$.
$f\not =0$ mean: exist $x\in R : \ f(x)\not =0$.


If the probability it will rain today is $60\%$ and the probability it will rain tomorrow is $70\%$, what is the probability it will rain exactly one of the two days.  Assume the two events are independent.

Probability that it will rain in both days: $0.6 \cdot 0.7 = 0.42$
Probability that it won't rain in both days: $0.4 \cdot 0.3 = 0.12$

The probability you asked for is $1-(0.42+0.12)=1-0.54=0.46$, i.e. $46\%$

[hide]
both days are wet: 0.7*0.6=0.42
Neither days are wet: 0.3*0.4=0.12
if we subtract this from 1, we get the probability desired.
1-0.54=0.46
The probability is 23/50
[/hide]

[hide]the probability that it will rain today but not tomorrow is
.6 * .3 or .18.
the probability that it won't rain today but it will rain tomorrow is
.4 * .7 or .28.
Add those up and you get .46, or 46%[/hide]


If $ x > 0, y > 1$ and $ \left(\frac{y}{x}\right)xy^z \equal{} 1$, what is the value of $ z$?

simplifies to y^(z+1)=1.

anything that is raised to the 0th power, since y=\=1, is 1

z+1=0; z=-1

The denominator of the fraction (x) cancels out with the other x, and the y multiplies $ y^z$, so we have $ y^{z\plus{}1}\equal{}1$. If y is 1, then x could equal any real number, but the inequality stated that y is greater than 1, so y is positive, but not 1. Therefore, z+1 must be zero, because any real number other than 0 raised to the zeroth power is 1, so z=-1.


Fie $ \triangle ABC ,AD,BE,CF$ inaltzimile sale, $ A',B'C'$ mijloacele seg $ (BC),(CA),(AB)$ si $ D',E',F'$ mijloacele inalztimilor $ AD,BE,CF$.Demonstratzi ca $ A'D',B'E',C'F'$ sunt concurente in punctul $ K$(punctul lui Lemoine)

[hide]  dak poate cineva sa dea o solutzie folosind numere complexe i-as fi recunoscator:)[/hide]

[b]Parca am mai amintit pe undeva, tot pe acest forum, faptul ca este cunoscuta urmatoarea teorema mult mai generala:
[i]Daca AD, BE si CF sunt trei ceviene concurente si punctele D', E' si F' sunt mijloacele segmentelor [AD], [BF] si [CF], iar A', B' si C' sunt mijloacele laturilor
[BC], [CA] si respectiv [AB], atunci dreptele A'D', B'E' si C'F' sunt concurente in centrul conicei tangente la dreptele suport ale laturilor triunghiului ABC[/i].[/b]

Am o demonstratie frumoasa cu vectori si una echivalenta folosind coordonate baricentrice.., dar mai astept (poate vine o solutie din partea unui elev)


yes kunny the inequality can be written as you have done  but I want to know if that new look will contribute to solve the inequality.

[img]8109[/img]
I checked. It's flipped. Equality at $x=y=z=0$.

[quote="vietnam2007"]I wanted to write the RHS
I m so sorry :blush:.I've edited my post[/quote]
 :)

So the first inequality is right after the sign is flipped but the second "equivalent" one by kunny is wrong, try the triplets $(1.31,1.56,1.23)$ and then $(2,10,1)$

[quote="vietnam2007"]the inequality you have to prove is :

$3\frac{a^{2}+b^{2}+c^{2}-3}{a^{2}+b^{2}+c^{2}+3}\geq \frac{1}{2}[\frac{2a^{2}+2b^{2}-3}{2a^{2}+2b^{2}+3}+\frac{2b^{2}+2c^{2}-3}{2b^{2}+2c^{2}+3}+\frac{2c^{2}+2a^{2}-3}{2c^{2}+2a^{2}+3}+\frac{2a^{2}-3}{2a^{2}+3}+\frac{2b^{2}-3}{2b^{2}+3}+\frac{2c^{2}-3}{2c^{2}+3}]$

I wanted to edit it  but i can 't[/quote]

Okay this one makes sense! :)


geometry
area of a triangle
Heron\u0027s formula
geometry solved

doesnt a  triangle  have fixed  area  if  the  length  of  its  sides  are  fixed??

  anyway  just  use  herons  formula  and  its done.

It's a very easy problem, but here is a detailed solution for the sake of completeness:

[color=blue][b]Problem.[/b] Let a, b, c be the sidelengths and S the area of a triangle ABC. Denote $x=a+\frac{b}{2}$, $y=b+\frac{c}{2}$ and $z=c+\frac{a}{2}$. Prove that there exists a triangle with sidelengths x, y, z, and the area of this triangle is $\geq\frac94 S$.[/color]

[i]Solution.[/i] Since a, b, c are the sidelengths of a triangle, we have b + c > a, c + a > b and a + b > c. Thus, $\left(b+\frac{c}{2}\right)+\left(c+\frac{a}{2}\right)=\left(b+c\right)+\frac{c+a}{2}>a+\frac{b}{2}$, or, equivalently, y + z > x. Similarly, z + x > y and x + y > z. Hence, there exists a triangle with sidelengths x, y, z. Remains to prove that the area of this triangle is $\geq\frac94 S$.

The numbers b + c - a and c + a - b are positive (since b + c > a and c + a > b), and thus we can apply AM-GM to the three numbers b + c - a, b + c - a and c + a - b to get

$\left(b+c-a\right)+\left(b+c-a\right)+\left(c+a-b\right)\geq 3\sqrt[3]{\left(b+c-a\right)\left(b+c-a\right)\left(c+a-b\right)}$.

Hence,

$y+z-x=\left(b+\frac{c}{2}\right)+\left(c+\frac{a}{2}\right)-\left(a+\frac{b}{2}\right)=\frac12\cdot\left(\left(b+c-a\right)+\left(b+c-a\right)+\left(c+a-b\right)\right)$
$\geq \frac12\cdot 3\sqrt[3]{\left(b+c-a\right)\left(b+c-a\right)\left(c+a-b\right)}=\frac32\sqrt[3]{\left(b+c-a\right)^2\left(c+a-b\right)}$.

Similarly,

$z+x-y\geq\frac32\sqrt[3]{\left(c+a-b\right)^2\left(a+b-c\right)}$;
$x+y-z\geq\frac32\sqrt[3]{\left(a+b-c\right)^2\left(b+c-a\right)}$.

Finally,

$x+y+z=\left(a+\frac{b}{2}\right)+\left(b+\frac{c}{2}\right)+\left(c+\frac{a}{2}\right)=\frac32\left(a+b+c\right)$.

Now, by Heron's formula, the area S of triangle ABC equals

$S=\frac14\sqrt{\left(a+b+c\right)\cdot\left(b+c-a\right)\cdot\left(c+a-b\right)\cdot\left(a+b-c\right)}$,

while the area of the triangle with the sidelengths x, y, z is

$\frac14\sqrt{\left(x+y+z\right)\cdot\left(y+z-x\right)\cdot\left(z+x-y\right)\cdot\left(x+y-z\right)}$
$\geq\frac14\sqrt{\frac32\left(a+b+c\right)\cdot\frac32\sqrt[3]{\left(b+c-a\right)^2\left(c+a-b\right)}\cdot\frac32\sqrt[3]{\left(c+a-b\right)^2\left(a+b-c\right)}\cdot\frac32\sqrt[3]{\left(a+b-c\right)^2\left(b+c-a\right)}}$
$=\frac94\cdot\frac14\sqrt{\left(a+b+c\right)\cdot\left(b+c-a\right)\cdot\left(c+a-b\right)\cdot\left(a+b-c\right)}=\frac94 S$,

and the problem is solved.

  Darij

[quote=Fermat -Euler]Let a, b, c be the sidelengths and S the area of a triangle ABC. Denote $x=a+\frac{b}{2}$, $y=b+\frac{c}{2}$ and $z=c+\frac{a}{2}$. Prove that there exists a triangle with sidelengths x, y, z, and the area of this triangle is $\geq\frac94 S$.[/quote]
[Hide=solution]
Here's a sketch of my solution
Proceed in a similar way as @above to prove that $x,y,z$ form sides of triangle.
Now, let $T$ denote the area of this triangle.
We know by heron's formula that $$16S^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)$$
Now,
$$16T^2=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)$$
$$256T^2=3(a+b+c)(a+3b-c)(b+3c-a)(c+3a-b)$$
We need to prove
$T\geq \frac{9}{4}S\Leftrightarrow 256T^2\geq 27*16S^2$
Or, we need to prove
$(a+3b-c)(b+3c-a)(c+3a-b)\geq 27(a+b-c)(b+c-a)(c+a-b)$ where $a,b,c$ are sides of triangles.
By ravi substitution($a=p+q,b=q+r,c=r+x$), this is equivalent to proving
$(2p+q)(2q+r)(2r+p)\geq 27pqr$
(Which itself is an RMO problem LOL)
$2p+q\geq\sqrt[3]{p^2q}$ by AM-GM
and similarly, $2q+r\geq \sqrt[3]{q^2r}$
$2r+p\geq \sqrt[3]{r^2p}$
Multiplying all, we get the desired inequality
[b]Q.E.D.[/b]
[/Hide]
[Hide=comment]
This problem required nothing but patience(and AM-GM). Even without ravi substitution, one could easily prove the inequality by expanding.

simply herons formula. i dont know how this made it to INMO :wacko:


calculus
integration
real analysis
real analysis unsolved

It doesn't appear that the indefinite integral has a nice form.  You mean reals $a$, right?


function
floor function
algebra unsolved
algebra

So, you have two functions f, g from N = {1,2,...} to N satisfying:
1. for all x, y in N you have f(x) is different from g(y);
2. for all z in N either there is x such that f(x)=z, or there is x such that g(x) = z;
(in other words, 1 and 2 together say that Range(f) and Range(g) are non-intersecting and cover the whole N as their union, where i mean Range(f) = {f(x) | x in N} )
3. for all x in N, g(x) = f(f(x))+1.
Proove the existance and uniqueness of f and g and find explicit formulas for them.
Actually, this was given at the final round of our national informatics olympiad this year, just asking for calulating f and g for 1 <= x <= 10^15. But the solution i read claimed for having found some explicit formulas.

If you've seen stuff like this before, then guessing the functions shouldn't be that hard. I haven't checked the computations, but I think that if you put $\tau=\frac{\sqrt 5+1}2$, then $f(n)=\lfloor\tau n\rfloor,g(n)=\lfloor\tau^2n\rfloor$

This is problem 3, IMO 1978, Bucharest. I've been there...;)

grobber, i'm amazed :)
that's exactly what the author of the solution claims :))
but he didn't give any ideas for proof, just noticing that writing the program is thus extremely easy and effective. I suppose it isn't that hard to proove 1 and 2, and 3 too, but i wonder can it be done in an elegant way ?

I think its like Beatty sequence problem. you probably get some conditions on $\alpha$ and $\beta$, given  that $g(x) = f(f(x)) + 1$. As for uniqueness and existance...


What do we get when we treat an alkyl halide R-X with alc. $ KNO_2$?

an alkyl nitrite or a nitro alkane??  :maybe:  :maybe:

in terms of substitution, where's the electron pair on nitrite?

or maybe, the product is actually an alkene...
nitrite is a base after all...(weak)

In general, one obtains a mixture of alkyl nitrites and nitroalkanes. However, with other solvents (specially Et2O, DMF, and DMSO) the yield of nitro alkane is significantly better.


Could anyone help me this? Thx

Suppose that f : (a; b) -> R is a differentiable function
and that there is a constant M > 0 such that lf'(x)l < and = M for all x in (a; b).

Show that limx->a f(x) and limx->b f(x) both exist.

Let $x_n\in ]a,b[$ be a convergent sequence to $a$ . You can write : ${|f(x_{n+m} - f(x_n}| = |(x_{n+m} - x_n)f'(t)| \leq M|x_{n+m} - x_n|$
And now it's easy :$x_n\ is\ convergent \Longrightarrow x_n\ is\ a\ Cauchy's\ sequence \Longrightarrow f(x_n)\ is\ a\ Cauchy's\ sequence$$\ \Longrightarrow f(x_n)\ is\ convergent$
:cool:


1. Find all polynomials W with integer coefficients, satisfying following condition : for every positive integer n, W(n) divides 2^n - 1.

2. let n>=2 be an integer. Given a polynomial p(x)=x^n + a_(n-1)x^(n-1)+............+a_1 x +1, with positive integer coefficients satisfying a_k=a_(n-k) for every 0<k<n. Prove that there exists infinitely many pairs a, b of positive integers, for which a divides p(b) and b divides p(a).

Thank you!

이상훈이구나 ㅋㅋ

1. $ (a,b)\equal{}1 \Rightarrow (W(a),W(b))\equal{}1$. If $ W\not\equiv 1$, There exist $ N$ such that $ |W(x)|>1$ for all $ x>N$.
It is easy to prove that $ f(a)|f(a\plus{}f(a))$, therefore $ (a,W(a))\equal{}(a,a\plus{}W(a))>1$ for all $ a>n$.
But we can choose $ a$ to coprime with constant term of $ W(x)$... so $ x|W(x)$, $ x|2^{x}\minus{}1$. It is obviously contradiction!

2. see this : [url]http://www.mathlinks.ro/viewtopic.php?t=187267[/url]

Thank you very much Lee Soo Hong!! *^^*  고마워 ㅋ
아 근데 통신강좌 1호 2번 문제에 대해 힌트좀 주면 안될까 ㅋ 못풀었음 ㅠ

그거 거의 걍 노가다밖에 방법이 없는것같애.... 난 막 넓이 최대인 삼각형 잡고 별짓 다했음


Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$ respectively.
Show that $pq=R^{2}$, where $R$ is the circumradius of $\Delta ABC$.

See the attached diagram.

Since $PA=PB$, $P$ must lie on the bisector of $AB$. Similarly, $Q$ lies on the bisector of $AC$. Those two bisectors, of course, intersect in circumcenter $O$. By the condition of touching, we have $PB\perp BC$ and $QC\perp BC$. Let $M$ be midpoint of $BC$ and $\alpha,\beta,\gamma$ the angles of $\triangle ABC$

Now $\angle PBO=\angle BOM=\alpha$ and $\angle POB=\gamma$, hence $\angle BPO=\beta$. Therefore

${p\over\sin\gamma}={R\over\sin\beta}\qquad (1)$.

Similarly, we find $\angle OCQ=\alpha,\angle COQ=\beta, \angle CQO=\gamma$, hence

${q\over\sin\beta}={R\over\sin\gamma}\qquad(2)$

Multiplying $(1)$ and $(2)$ we get $pq=R^{2}$

Note: Sine Law could be avoided (plain similarity works as well), but this way it's easier to follow.


function
algebra
rational function
partial fractions
complex analysis

Problem says:

Find Laurent expansion of

$ \frac{1}{(z^2\minus{}1)(z^2\minus{}4)^2(z\minus{}3)}$

in

$ 1<|z|<2$
$ 2<|z|<3$
$ 3<|z|$

Problem is that my calculations are too tedious and too big. Is there a faster and more efficient way than brute-forcing it all the way?

Without seeing just what level of complexity you're not satisfied with, I can't say. I would do it all by starting with partial fractions.

Honestly, [b]jmerry[/b], when I saw the partial decomposition of it, I was horrified with it:

$ \frac {1}{200 ( \minus{} 3 \plus{} z)} \minus{} \frac {1}{48 ( \minus{} 2 \plus{} z)^2} \plus{} \frac {5}{288 ( \minus{} 2 \plus{} z)} \minus{} \frac {1}{36 ( \minus{} 1 \plus{} z)} \plus{} \frac {1}{72 (1 \plus{} z)} \minus{} \frac {1}{240 (2 \plus{} z)^2} \minus{} \frac {61}{7200 (2 \plus{} z)}$

When it comes to Laurent series, I've never worked with anything more complicated than i.e. $ \frac {1}{z(z \plus{} 1)}$, so this was pretty repulsive.

Yes, this is ugly. I guess the textbook author was trying to make the point that the Laurent series of the same rational function can be completely different in different annuli and just got carried away with this great idea. Unless it is a compulsory homework assignment that is going to be graded, I would just take the point and skip the problem. Otherwise, what you've done (partial fractions) is the  "approach of choice". Just check that you didn't make any stupid error in the arithmetic and forget the entire thing: there seems to be nothing of real value in it. Even if some clever trick exists, it is not what the subject is about... :).

It is in fact a homework problem (worth 0.5p  :lol:  ). But I really doubt it is worth that much hard work...  :lol:

If that's the point, I prefer [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=681920#681920]this example[/url].


geometry
geometric transformation
rotation

On an old-fashioned bicylce the front wheel has a radius of $ 2.5$ feet and the back wheel has a radius of $ 4$ inches.  If there is no slippage, how many revolutions will the back wheel make while the front wheel makes $ 100$ revolutions?

[hide="solution"]
The front wheel has a radius of \[ 2.4\text{ feet}\left(\frac{12\text{ inches}}{\text{foot}}\right)\equal{}30\text{ inches}.\] We want to know how many rotations the back wheel can make in $ 100$ of the front wheel's rotations. This is basically $ \frac{(\text{the circumference of the front wheel})\times100}{\text{the circumference of the back wheel}}.$
When the front wheel makes $ 100$ rotations, the back wheel makes \[ \frac{2\pi\times30\times100}{2\pi\times4}\equal{}25\times30\equal{}\boxed{750}\] rotations.[/hide]

Oops.  Fail.

I misread the question thinking it was all in inches and i put the answer as 125/2.  I have to multiply by twelve and therefore, the answer is 750.

:(


I have read, in an excerpt from a book by two Chinese military administrators, the phrase

[quote]the financial attack by George Soros on East Asia[/quote]
What is this? I was unable to find anything about it on the [url=http://en.wikipedia.org/wiki/George_Soros]Wikipedia page[/url] about him, even though there is a "criticism of financial activities."

There are pretty much two possible explanations.

1) Westerners don't really know or care what Soros did, but he did something.

2) Soros was doing what Westerners would consider normal financial activities. However, these Chinese authors were pissed that a Westerner made so much money by activities in Asian markets and therefore referred to his activities as "an attack."

I'd be delighted if someone would clear this up for me. Are my suspicions about the Western media well-founded? Are my suspicions about Chinese military officials well-founded? Both?

[quote="Xevarion"]I was unable to find anything about it on the [url=http://en.wikipedia.org/wiki/George_Soros]Wikipedia page[/url] about him[/quote]

Hmm?

[quote="Wikipedia"]In 1997, during the Asian financial crisis, then Malaysian Prime Minister Mahathir bin Mohamad accused Soros of using the wealth under his control to punish ASEAN for welcoming Myanmar as a member. Later, he called Soros a moron. [4] Thai nationals have called Soros "an economic war criminal" who "sucks the blood from the people". [5][/quote]

The two links refer to

http://www.asiaweek.com/asiaweek/97/1003/cs1.html
http://www.asiaweek.com/asiaweek/magazine/nations/0,8782,98437,00.html

By the way, why do you use Wikipedia as your sole representative of the Western media?

Thanks, apparently I can't read.

Wikipedia is representative of the knowledge and opinions of its editors. Its editors comprise a large cross-section of Western web users. Therefore it is a good indicator, though not infallible (as other discussions here have revealed).

And I don't only use Wikipedia. I did some Googling, but apparently I was too lazy.... :|

Okay I found an [url=http://papers.ssrn.com/sol3/papers.cfm?abstract_id=58650]academic article[/url] about it. However, it was not written by Asians.

I don't understand money markets well enough yet to really comprehend the financial aspects of what happened. What I'm more interested in is the impact of East/West feelings toward each other on interpretations of the incident.

Clearly despite that 1998 article, some Chinese authorities (and probably other Asians, but the paper in which I saw it mentioned was Chinese) still think Soros was (at least in part) responsible. So what I'm trying to investigate is which is the side more tainted by irrelevant political bias?

Can someone provide any scholarly work that concludes that Soros was (at least in part) responsible? The Chinese paper I have is on a different topic and only mentioned it in passing.

Edit: I also found an [url=http://www.pbs.org/wgbh/pages/frontline/shows/crash/interviews/soros.html]interview[/url] with Soros, but the interviewer doesn't directly ask him about his role in the crisis. I have seen plenty of news articles but they're all media bs. I am getting the impression that the Malaysian and Thai governments wanted to shift blame away from themselves, but I am still suspicious of the bias in the sources I've seen so far.... Maybe I should give up and stick to math, where political tensions rarely cause any problems. :P

[quote]In 1997, during the Asian financial crisis, then Malaysian Prime Minister Mahathir bin Mohamad accused Soros of using the wealth under his control to punish ASEAN for welcoming Myanmar as a member. Later, he called Soros a moron. [4] Thai nationals have called Soros "an economic war criminal" who "sucks the blood from the people". [5][/quote]

Until the current Iranian president, Malaysian PM Mahathir was probably the most significant head of state who routinely made public anti-Semitic pronouncements.  According to Mahathir, not only do "the Jews control the world", but they also invented communism, socialism, human rights and democracy in order to unfairly prevent the persecution of Jews.   Soros, who is (1) famous,  (2) Jewish,  (3) a billionaire currency speculator,  is a natural target for morons such as Mahathir who are obsessed with Jewish financiers.
See, for example,

http://www.usatoday.com/news/world/2003-10-16-malaysia-summit_x.htm

[quote="kahsi"]Maybe what he did to us swedes in 1992: (he and some friends) came with alot of speculations which made the swedish krona (our currency) fall alot. And ofcourse he made alot of money on the "deal". And he has done it before, breaking small countries currency. He is evil, indeed  :yup:[/quote]

One can argue over whether Soros' work has value or whether he should become rich for his activities.  However, if the government of a nation such as Sweden misprices its currency, thus creating opportunities for currency speculators, it is in a sense defrauding its citizens and in the short or long term making them pay for the bad fiscal policy.   Currency crashes are just one manifestation of the problem, but the citizens can pay dearly without any Soros to accelerate the process.

One could even argue that currency speculators might spur progress in how governments handle their currency. I mean, the simple forms of arbitrage disappeared pretty quickly (like different exchange rates for currencies in different places), once people started profiting from them.

But this neglects (a) the damage that the currency crash does to the innocent citizens of the suffering nation(s) [so someone like kahsi might argue that the long term benefits are not worth it, and the fact that the government made a mistake to leave its currency vulnerable is not much of an excuse] and (b) the question of how much influence the speculation actually has on the currency crash (like statistical said, would it happen anyway?).

Unfortunately it starts to get complicated. I think the paper I linked to about the Asian currency crisis describes some modern methods you might use to analyze the situation, but that's no general resolution to the question.

Anyway I think I have the answer to my question now (currency speculation exacerbates currency problems that are in fact the government's fault + it's hard to tell how much influence the speculation has + nobody wants to accept the blame when someone goes wrong).


Is there a way to numerate equations in \begin{eqnarray} NOT from $1$ but from any given higher number? If I want, for example, to continue the numeration where I left off before some block of text?

Found out myself - include \setcounter{equation}{n} inside the array, where n is the desired starting number.

Example: \begin{eqnarray}\setcounter{equation}{7}x &=& 1\\ y &=& 2\\ z &=& 3\end{eqnarray} gives \begin{eqnarray}\setcounter{equation}{7}x &=& 1\\ y &=& 2\\ z &=& 3\end{eqnarray}


I am attracted to tin whistle long ago due to its simpleness and beautiful sound quality. And I'd like to have one and learn to play it. But because here in Ecuador there is no tin whistles to sell so it could be cool if you are able to bring a tin whistle to the IMO for me  :)  . I will either pay you cash or bring you something of similar value from Ecuador. Please contact me if you can (and want) to do this favor for me so we can talk for additional info. Thank you.

No Problem! Tin Whistle is on the way, just as soon as my exams are over I'll get one! :D  What is your name btw, I realise you're from ecuador?

I'm Galin.... Hi


A swimmer finds himself in a circular pool. His coach is on the border of the pool and cannot get into the pool. He's very angry and wants to catch the swimmer when this one will get out of the pool. The speed of the coach on the land is four times the speed of the swimmer in the water.
What strategy must the swimmer apply to get out of the pool without being caught by the coach?

This was a problem on a blue MOP test this year!

Indeed? :)  Actually it's an old problem (I know it from one of our MOSP tests - about 10years ago).

From the center, he can swim out one quarter of the radius, keepin the coach on the other side of the pool.

Now, if he swims straight out from it, he has to cover $\frac{3r}{4}$ while the coach has to cover $\pi r$

So by the time the swimmer gets to the edge, the coach has moved 3r, a tiny bit short of the $\pi r$ needed. Run for it!

OK, but as the boy swims out one quarter of a radius, the coach will also move $r$ on the edge of the pool, so he won't have to cover $\pi r$, but $\pi r-r$ !  :)

Okay, the swimmer should swim out to 2/9r away from the center of the pool.  The coach now can only travel 8/9 angular units for every 1 unit the swimmer travels by staying at that same distance from the center.  So, the swimmer can manever himself until the center of the pool is between himself and the coach.  Then he swims for the side.  By the time that he makes it to the side (swims 7/9r), the coach will have traveled 28/9r which is short of the pi*r that he needs.

Actually, this stragety will work with the swimmer swimming out a distance of [i]x[/i] for all [i]x[/i] such that \[ 1-\frac{\pi}{4} < x < \frac{1}{4}  \]

Actually, we can obtain a general formula for this situation.  Let v be the
ratio of speeds of the coach to the swimmer.  Let x be the distance the
swimmer swims out.  Then the above strategy will work iff \[ 1-\frac{\pi}{v}<x<\frac{1}{v}  \] So, this implies that all values of x are invaid iff \[ v\ge\pi+1  \] (I suspect that this stategy isn't ever worse than any other, so  I don't think any strategy will work when $v \ge \pi+1$.)

Sorry, i was not clear with my solution. He can move out one quater of the radius [i]keeping the coach on the opposite side of the centre[/i]. The rest is the same.

Heres a question i solved, twas fun! Given ANY shaped pool, what is the largest n such that if the coaches speed is n times the swimmer, the swimer can get out.

Peek if you must, its only a hint:

[hide]Choose the biggest circle enclosed by the pool...

[hide]Hey geeze thats gotta touch at least three sides![/hide][/hide]

[quote="seamusoboyle"]He can move out one quater of the radius [i]keeping the coach on the opposite side of the centre[/i][/quote]

Incorrect.  As the swimmer approaches being 1/4r from the center, he will have to devot more and more of his speed to keeping his coach on the other side, since his angular velocity will be approaching that of his coach.  He swimmer forever in circles whose radii approaches 1/4r.

I ment he could get as close as he wants, which for pratical purposes is good enough. Since we have a gap at the end, we can neglect any difference.

Right, and this fits the situation I put in an earlier post in which I said that there was an interval of distances from the center to which the swimmer could swim, and then circle till he was on the opposite the coach (or, similarily, he could keep the coach across from him while he swam out to that distance).

That interval's lower bound is created by the swimmer's need to be able to beat the coach when he (the swimmer) makes a beeline for the side of the pool.
That interval's upper bound is created by the need to have a greater angular speed than the coach so that the swimmer may properly position himself.