Support percent Stanford college geometry function HCSSiM

I'm debating to my self on whether or not we should have public school. My main ideas in support of this are

1) The government screws things up
2) Highly motivated kids can get a better education through a tutor or library
3) Kids who don't want to go to school don't need to, and taxpayer's money is not wasted on a kid who obviously does not want to learn
4) There are plenty of other educational resources for motivated children
5) Let them see the consequences of having no marketable skills when their parents kick them out and they need a job

However, what about a five year old child who does not wish to go to school, and with parents that are not supportive of education in general, but who laters wishes to get educated and cannot because s/he lacks basic skills (reading, writing, etc)? I believe the government does this because it assumes that children are too young to be making decsions such as these, and I think the entrie idea is that people can make choices like whether to get an education or not by themselves, but I don't think children can make these choices. I'm trying to balance personal responsibility versus idiots here. Collectivism is bad, but what to do about the poor kid with parents that won't let them get education that wants to?

I also need a good retort to people who say, "OMG if we don't force kiddies to go to school then our country will fall apart."
I don't think it will, but I can't think of anything convincing.

I find this argument ridiculous
1) Say what? How?
2) Where are we going to get tutors if there are no schools to school them? Who says the tutors are right?
3) Going to school is one of those things you hate, and then you get out of it and bask in your knowledge and laugh at everyone who is still experiencing it

[quote="kstan013"]I find this argument ridiculous
1) Say what? How?
[/quote]
FEMA, to name an example off the top of my head.
EDIT:
War on drugs, to name another big one.

[quote="kstan013"]
2) Where are we going to get tutors if there are no schools to school them? Who says the tutors are right?
[/quote]
A private tutor does not need a school to operate. I should have added private school, but that might have been too obvious...
[quote="kstan013"]
3) Going to school is one of those things you hate, and then you get out of it and bask in your knowledge and laugh at everyone who is still experiencing it[/quote]
Explain.

I think that the problem of lots of kids who just don't like school deciding that they're not going to go to school would be too much of a problem to just abolish public schools. However, I do think that the idea of a mass, close to identical state-funded education for all is not one which works too well. There should be more choice in ways how you (or your child) is educated, from the age of 11 or 13 at least.

Everyone having a private tutor would be way too expensive. Private schooling, I think, is much more sensible. Personally, I go to a private (well, public school in a sense but only in the sense that the meaning if [i]public school[/i] is the complete reverse here as it is elsewhere in the world, public schools being a subset of private (or independent) schools) school that is itself a bit of an oddity, and I would hate to have to be educated in a state school. However, there just isn't the money for everyone to be educated in this way (and I can only do so because my school has a huge endowment).

I agree that highly-motivated kids learn much better from a library or the internet, and I learn a lot in this way, but I would still require some teachers who could ask to explain something or discuss the merits of my work with, even if I would rather not go to all their lessons.

Saying that tax-payers money is 'wasted' on kids who don't want to learn cannot be true: by being educated regardless of what they want at age 10, they grow up to be skilled workers (in some cases anyway) such as doctors, lawyers, bankers or whatever, and repay what was sown in them.

[quote="tim1234133"]I think that the problem of lots of kids who just don't like school deciding that they're not going to go to school would be too much of a problem to just abolish public schools. However, I do think that the idea of a mass, close to identical state-funded education for all is not one which works too well. There should be more choice in ways how you (or your child) is educated, from the age of 11 or 13 at least.
[/quote]
Agree. I don't like public education because it's very inefficient, at least in the USA. If all public schools were run like Exeter, I'd put aside my quips because of conflicts with my small government ideals, and just let it be. But the current system sucks majorly, even before the ages of 11 or 13.
[quote="tim1234133"]
Everyone having a private tutor would be way too expensive. Private schooling, I think, is much more sensible. Personally, I go to a private (well, public school in a sense but only in the sense that the meaning if [i]public school[/i] is the complete reverse here as it is elsewhere in the world, public schools being a subset of private (or independent) schools) school that is itself a bit of an oddity, and I would hate to have to be educated in a state school. However, there just isn't the money for everyone to be educated in this way (and I can only do so because my school has a huge endowment).
[/quote]
Agree, although as homeschools like solafidefarms will tell you, school is not the only way to get an education. In early America, many people (most of our early presidents, included) became very smart despite the lack of formal schools.
[quote="tim1234133"]
I agree that highly-motivated kids learn much better from a library or the internet, and I learn a lot in this way, but I would still require some teachers who could ask to explain something or discuss the merits of my work with, even if I would rather not go to all their lessons.
[/quote]
Agree, though the government giving you one isn't the correct solution.
[quote="tim1234133"]
Saying that tax-payers money is 'wasted' on kids who don't want to learn cannot be true: by being educated regardless of what they want at age 10, they grow up to be skilled workers (in some cases anyway) such as doctors, lawyers, bankers or whatever, and repay what was sown in them.[/quote]
Interesting, though I think it more has to do with the fundamental idea of taxation: the goverment forces you to give them a certain about of money. It's repulsive when you've worked hard for your own house, property, wealth, etc, to have the government steal from you. Obviously reducing taxes as much as possible while leaving an efficient government in place is desireable. And having money stolen from you used on kids who won't do anything with their education just adds insult. Though I admit this a weak argument, I'd still like to defend it given the chance. However I am more interested in other defences of public school, so I will concede this point to you for now.

[quote="tokenadult"]See

http://learninfreedom.org/school_state.html

for more information.[/quote]
Would you care to weigh in with your personal opinions? I see you are well versed on this subject...

This is the worst idea I've heard. There are many kids who couldn't afford a tutor or a private school, and you're basically robbing all those kids of an education with this plan. On the other hand, just because a kid might be motivated in one subject or want to succeed doesn't necessarily mean that he's going to take the necessary steps of spending hours reading up on multiple subjects just to fuel that passion.

It seems like with your little plan, you're going to get a world where only a tiny fraction of a percent of the population is educated at all, and the rest are peasant serfs whose sole purpose of life is to serve the educated in some utopian/Randian world.

[quote="Elemennop"]This is the worst idea I've heard. There are many kids who couldn't afford a tutor or a private school, and you're basically robbing all those kids of an education with this plan.
[/quote]
Um no, we have these things called Libraries and Pell Grants. The military will pay you for stuff, also (not that I endorse the military). Even if you are the poorest kid in America, there are still places for one to learn if you look.
[quote="Elemennop"]
On the other hand, just because a kid might be motivated in one subject or want to succeed doesn't necessarily mean that he's going to take the necessary steps of spending hours reading up on multiple subjects just to fuel that passion.
[/quote]
Ok. That's his choice, and he can learn why that's a bad choice when he gets kicked to the street when he's 18 and has no marketable skills.
[quote="Elemennop"]
It seems like with your little plan, you're going to get a world where only a tiny fraction of a percent of the population is educated at all, and the rest are peasant serfs whose sole purpose of life is to serve the educated in some utopian/Randian world.[/quote]
Where would you get this from? If you want an education in this world, you can get it. If you don't, then the government doesn't have to baby you.

[quote="PenguinIntegral"][quote="Elemennop"]This is the worst idea I've heard. There are many kids who couldn't afford a tutor or a private school, and you're basically robbing all those kids of an education with this plan.
[/quote]
Um no, we have these things called Libraries and Pell Grants. The military will pay you for stuff, also (not that I endorse the military). Even if you are the poorest kid in America, there are still places for one to learn if you look.[/quote]
Why go to Stanford if you can find all their info by reading books? There's something called instructors, and they aren't free.

[quote="PenguinIntegral"]Agree, although as homeschools like solafidefarms will tell you, school is not the only way to get an education. In early America, many people (most of our early presidents, included) became very smart despite the lack of formal schools.[/quote]Because they were incredibly wealthy.  Every single American President has been rich and well connected when he took office, and the vast majority were born into that wealth.

We can't all afford private tutors.  We can't all afford private school.  What 6 year old can self-motivate themselves to learn as much as they would in school, or from a tutor?  Even today, I am able to teach myself math, but find I learn much better and enjoy it more by learning with other people and discussing- not reading a book by myself.

I may have not been able to afford anything better than the public school you seem to frown upon from your elitist perch, and I may have not been able to take the classes that someone at a private school would have been able to take, but I still managed to get a damn fine education, and it was paid for completely by the government, and I certainly wouldn't be where I am today without it.

Things "paid for completely by the government" are really paid for by taxpayers. NO form of taxation has incidence only on the rich.

The country that is the most recent example of rapid, sustained economic development without compulsory school attendance is Singapore until a decade ago. Now school attendance is compulsory in Singapore, and provision of tax-supported schooling and widespread school attendance has long been commonplace in Singapore, but there are plenty of places with different models from the local oligopolies supported by taxes ("school districts") that are the usual system in the United States. For a long time, about 10 percent of the United States population has been educated without attending government-operated schools. In former days, that percentage was much higher. I would be happy to see

a) school attendance become voluntary rather than compulsory

and

b) funding of schooling be distributed on a per-capita basis (e.g., through educational vouchers) rather than funding only certain government-operated schools.

Yes, to all the people in this thread that say "not everybody can afford a private school education" look up school vouchers. I believe that is what PenguinIntegral is getting at, and tokenadult just mentioned it. Public schooling is not the only way to organize an educational system...

Hand out vouchers and the public schools end up as dumping grounds for students who couldn't meet private school standards.

Everything goes downhill from there.

(Of course, not providing education for everybody isn't necessarily implosive to the economy.  Somebody has to pick the fruit.)

Consider other countries experiences when stating, 'this works, this doesn't'

In Germany, France, Spain and other important european countries the education is mainly taken care of by the public system. [i]Particularly in areas such as science[/i]. One year at the public university in theoretical physics, which is the best university around, costs me say 800 dollars. Of course the rest I pay during my life through taxes which I'm guessing are a lot higher than in the US. My point is, when you say, the public system is wrong or bad, I truly don't understand what you are talking about. No one, and by no one meaning no one, thinks here that you will get a better education in mathematics by going to the private university. Actually, people think that the guys that go to the private do it because they think they would fail in the public.

Ursula

[quote="PenguinIntegral"][quote="statistical"][quote="PenguinIntegral"] ...money that they earned being stolen by the government ...[/quote]
You say "stolen" as though there is some pre-existing, inexorable, God-given and of course self-evident notion of money, property ownership, etc relative to which we must conduct the discussion.   There isn't.
[/quote] There's this thing called the Constitution... [/quote]

If the Constitution is what you mean, then taxation is legal by definition, and as such cannot be a "stealing" of anything.

Ok, let me correct that. The Constitution was a really bad example.

In the same way we should have a right to free speech, we should have one to absolute control of our private property. Not because a god or philosopher tells us to, but because I believe it is essential to a true free society. Obviously being practical, we cannot just abolish taxes tomorrow, but I would like to work towards world where people fund the government because they can and because they are nice people, not at the point of a gun.

JBL: I cannot find the paragraph you refer to, could you quote it?

[quote="PenguinIntegral"][quote="eryaman"][quote="PenguinIntegral"]I would like to work towards world where people fund the government because they can and because they are nice people, not at the point of a gun.[/quote]That's funny, really funny.  You assume that human nature is essentially good.[/quote]
No, I assume people will be rational and appreciate that stable government is worth more than a necklace for the mistress. I still think that might be too much for ask for.[/quote]

Rationally, people would spend their extra $\$$50 to buy a gun or upgrade their security systems to protect their property indefinitely, rather than give it to the government to pay a policeman for a day's work.

[quote="tokenadult"]There are real-world examples of capitation payment to schools that have to compete for students, and that is more economically efficient and less expensive for whoever is paying to subsidize schools than the United States system of local public oligopolies.[/quote]

That sentence is too unspecific and rhetoric-heavy to evaluate.    What are the real-world examples, how do they contradict the assertion above about economic theory, and is "more economically efficient" the observed result of these examples or a separate claim you are making?

Are you aware of any non-public system that is not locally oligopolistic, by the way?  That is, where there is a variety of significantly different private schools that grow and shrink according to market forces, rather than an effectively captive market where so-and-so many students have to go to some school (from a small local list) regardless of quality?

[quote="statistical"][quote="PenguinIntegral"]In the same way we should have a right to free speech, we should have one to absolute control of our private property. [/quote]

Absolute or partial rights to anything only exist "at the point of a gun".  You oppose the gun when it inconveniences you (e.g., paying trivial levels of taxes) but is there a similar objection when it helps you?  It is only the point of a gun that prevents friendly people from yonder ghetto or Mexico or Albania from helping themselves to you and your property, and maintains a discrepancy of circumstances whereby such people find it economically rational to work in coal mines instead of exchanging math problems on the Internet.[/quote]
Um, no. You fail to see the difference between protecting rights and taking them away. The government is only given the power of  gun and police force and court system because no private citizen or entity could possibly protect the rights of the citizens of the USA, and protect everyone's rights. Note that it is the rational choice for people to choose government over a gun or private army for self defense. When they are doing unjust things and there is no government, there is no reason the people they are slighting should be just either. Even a private army of thousands won't protect you from a community with small arms and bombs. (See: Iraq)

Perhaps deplorably, though, people aren't yet enlightened enough to measure personal wealth in terms of the general economy (which, of course, is inexorably intertwined with the well-being of the region).

I think the long grip monarchs had on many areas of the world testifies that people do, in fact, prefer having a private army to submitting to someone else's private, even if in theory communally owned, army. (Meaning that the government's army ISN'T your army, and that if you end up on the wrong side of it, you're screwed. It DOESN'T protect everyone's rights. It couldn't possibly.)

Government wouldn't protect you from a community of small arms and bombs, either. It's just an important factor in ordering society to prevent such happenstance... but it's not the only option, at least not in its current conceptualization, which is rather outdated...

What's "just" and what are rights? You're going off the ideals of centuries-dead philosphers and politicians, taking things out of context and the like. Personally, I have no frikkin idea what function the government serves and how it could be optimalized, but I do know to be leary of the same old ideas repeated again and again until they stop meaning anything at all. I'd rather have my quality of life protected than my rights.

[quote="PenguinIntegral"][quote="statistical"][quote="PenguinIntegral"]In the same way we should have a right to free speech, we should have one to absolute control of our private property. [/quote]

Absolute or partial rights to anything only exist "at the point of a gun".  You oppose the gun when it inconveniences you (e.g., paying trivial levels of taxes) but is there a similar objection when it helps you?  It is only the point of a gun that prevents friendly people from yonder ghetto or Mexico or Albania from helping themselves to you and your property, and maintains a discrepancy of circumstances whereby such people find it economically rational to work in coal mines instead of exchanging math problems on the Internet.[/quote]
Um, no. You fail to see the difference between protecting rights and taking them away. [...][/quote]

You haven't answered a single point from the above, much less indicated any failure of vision.   There are no "rights" to protect or take away without there first being a gun (or some functionally similar monopoly of force) to put those rights in place.  A large amount of governance is and was needed to create the hierarchy of infrastructures that permits any form of economic activity at all, up to and including market-style rational choices.  That gunpoint governance includes the creation and enforcement of your (far from absolute) property rights, the viability of the US currency, the difficulty for aforementioned friendly neighbors to have an uninvited party at "your" house, and pretty much everything else that you appear to take for granted when taking rhetorical postures against things that happen "at the point of a gun".

[quote="Treething"]Does public school really hurt anybody?[/quote]

It costs money to people who realize relatively little of the benefit (such as childless taxpayers) and it dramatically shrinks the potential size of the private schooling market.    These are important effects regardless of whether or not one supports public schools.

[quote="statistical"][/quote][quote="PenguinIntegral"][quote="statistical"][quote="PenguinIntegral"]In the same way we should have a right to free speech, we should have one to absolute control of our private property. [/quote]

Absolute or partial rights to anything only exist "at the point of a gun".  You oppose the gun when it inconveniences you (e.g., paying trivial levels of taxes) but is there a similar objection when it helps you?  It is only the point of a gun that prevents friendly people from yonder ghetto or Mexico or Albania from helping themselves to you and your property, and maintains a discrepancy of circumstances whereby such people find it economically rational to work in coal mines instead of exchanging math problems on the Internet.[/quote]
Um, no. You fail to see the difference between protecting rights and taking them away. [...][/quote][quote="statistical"]

You haven't answered a single point from the above, much less indicated any failure of vision.   There are no "rights" to protect or take away without there first being a gun (or some functionally similar monopoly of force) to put those rights in place.  A large amount of governance is and was needed to create the hierarchy of infrastructures that permits any form of economic activity at all, up to and including market-style rational choices.  That gunpoint governance includes the creation and enforcement of your (far from absolute) property rights, the viability of the US currency, the difficulty for aforementioned friendly neighbors to have an uninvited party at "your" house, and pretty much everything else that you appear to take for granted when taking rhetorical postures against things that happen "at the point of a gun".[/quote]
I admit in the case of protection there must be something to deter lawbreaking, and hence the use of the gun, but I don't know a clever way to say I want to. Me sitting at home with a semiautomatic minding my own business and stopping from people taking my stuff or hurting me is different from me going out with a semiautomatic and taking stuff from rich people and giving it to poor people. Does that make more sense? It's the question of who initiates the force.

[quote="PenguinIntegral"]I admit in the case of protection there must be something to deter lawbreaking, and hence the use of the gun, but I don't know a clever way to say I want to.  [/quote]

The content has the same problem no matter how cleverly it might be phrased.

[quote] Me sitting at home with a semiautomatic minding my own business and stopping from people taking my stuff or hurting me is different from me going out with a semiautomatic and taking stuff from rich people and giving it to poor people. Does that make more sense? [/quote]

No, because you are still assuming that first there exist some meaningful notions such as "my stuff" and "taking stuff from [its legitimate owners]", and that the acceptable or unacceptable forms of gunplay occur after that point.   In fact it is the other way around.

[quote]It's the question of who initiates the force.[/quote]

A huge amount of force is initiated on behalf of you, and the legitimately-or-illegitimately-taxed rich people, and many others on a daily basis in order to create the conditions that you assume for the whole discussion, such as a functional and relatively well regulated civilization where property ownership or laws are relevant concepts.
It doesn't make sense to ignore that and then complain about one particular use of force as though it is some anomalous horrible thing.


[quote="mathnovice"]Is there any reason why anyone would use DVI when PDF is more commonly used? I have never heard of DVI before and am curious why you might use it since it does basically the same thing as PDF.[/quote]

DVI may not be the best file for distribution, but it is much preferred over PDF when writing/editing a TeX document.

DVI files are generally quicker to compile than PDF, but more importantly, there's a direct connection between TeX and DVI. That is, if you double click on a spot in a DVI file, it goes to the corresponding spot in the TeX file. And you can go the other way too.

With TeX alone, it can be excruciatingly difficult to find one particular location. Which is where DVI comes in.

If you don't clump your entire [tex]\TeX[/tex] file into one line, you should be able to find where you are pretty easily.


Hi,

I'm current a freshman in college, and have had: differential Eqn, multivariate/vector calculus, newtonian mechanics (learnt from Kleppner), E&M (learnt from purcell and Griffiths).  I did quite a few problems from Irodov problems in general physics (I did about 1/3 of the E&M and mechanics sections, as well as a few dozen in the wave section).

As far as thermodynamics is concerned, I know very little.  I have been looking through the thermodynamics chapters of books like Halliday&Resnick,  Serway, but they are all very superficial, and too high-school-ish to my taste.

OK: So I'm looking for a solid text on Thermodynamics&Statistical physics, one that is an introductory, but that gives a deeper coverage than Halliday&Resnick.  The level of depth should like the depths of Griffiths E&M, or Kleppner, so that I can use to book to tackle the thermodynamics section of Irodov, and even harder problems (for example, the ones in the Problems&Solutions for major american Universities PhD exams).  I would also like it to be concise.
Also, I would like the book to have many example problems, and quality problems.

Do you know if there is such a book?

Thanks.

I feel that Heat & Thermodynamics by Zemansky and Dittman is good.


Άλλο ένα όμορφο και εύκολο προβληματάκι που μου'ρθε στο μυαλό :)

Για να μη μπλέκουμε με παθολογίες και ασυνέχειες, ας λέμε μία συνάρτηση ομαλή όταν έχει συνεχείς παραγώγους κάθε τάξης (αυτή είναι η λεγόμενη smooth category, $ C^\infty$).

Αν θυμάμαι καλά λοιπόν, είχε μπει κάποια στιγμή στο forum η ερώτηση: αν $ f$ ομαλή και $ \lim_{x\to\infty} f(x) \equal{} 0$, μπορούμε να συμπεράνουμε ότι η $ \lim_{x\to\infty}f'(x) \equal{} 0$?

Η απάντηση σε αυτό το ερώτημα είναι εύκολη και υπάρχουν άφθονα αντιπαραδείγματα, όπως π.χ. η $ f(x) \equal{} e^{ \minus{} x}\cos\left(e^{x^2}\right)$. Ωστόσο, το ερώτημα μπορεί να γίνει κάπως πιο ενδιαφέρον ως εξής:

Αν $ f$ ομαλή, [b]μονότονη[/b] και $ \lim_{x\to\infty} f(x) \equal{} 0$, μπορούμε να συμπεράνουμε τότε ότι $ \lim_{x\to\infty} f'(x) \equal{} 0$?

Cheerio,

Durandal 1707

[quote="Demetres"]Δεν ξέρω αν αυτή την λύση έχεις υπόψη σου

[hide]
Just do it and Bump functions
[/hide]

Δεν το έχω ελέγξει αλλά πιστεύω πως δουλεύει  :)[/quote]

Βρε άσε και κάνεναν άλλον να απαντήσει :)

Μάλλον το ίδιο πράγμα έχουμε στο μυαλό μας, κάτσε όμως μια δυο μέρες να δούμε αν προκύψει κάτι διαφορετικό :)

Cheerio,

Durandal 1707

[quote="r_boris"]OXI
στο attachment υπάρχει μια στοιχειώδης κατασκευή[/quote]

Ένα τετριμμένο σχολιάκι: αν δεν κάνω λάθος, η καμπύλη του attachment δεν έχει παντού παραγώγους δεύτερης τάξης αλλά who cares, δεν ήταν εκεί το νόημα :)

H ιδέα είναι ακριβώς αυτή, να κατασκευασθεί μία άπειρη σκάλα που τα σκαλοπάτια της να παραμένουν απότομα. Εγώ αυτό είχα στο μυαλό μου και, αν δεν κάνω λάθος, σε αυτό αναφερόταν και ο Demetres.

Γίνεται ακόμα πιο στοιχειωδώς η κατασκευή, απλώς χρησιμοποιώντας για το σχήμα των σκαλοπατιών την $ y \equal{} \frac{1}{n} \tanh(n^2 x)$ (ή την $ \arctan$).

Cheerio,

Durandal 1707


function
topology
real analysis
real analysis unsolved

Let  R and S be closed disjoint subsets of a metric space (X,d).  Let R be comapact.  Show that there is an epsilon > 0 such that d(x,y) >= epsilon for all x in R and y in S.

$D(x): =\inf_{y\in S}d(x,y)$ is a continuous positive function on $R$.


It's indeed simple: In triangle $AQC$ $QB$ is the symmedian, so $\frac{BA}{BC}=\frac{QA^2}{QC^2}$ and it's fixed, so $\frac{QA}{QC}$ must also be fixed, and we're done.

This was exactly my solution. Some participants used Cartesian coordinates  :D  :D  :D , others used the trigonometric form of Ceva's theorem.

  Darij

yes this one is trivial  :)
just use some trig. it turns out that the intersection is the square root of B wrt to AC. (btw this was in australian tst 2).

[b]Solution[/b]

Let $D=G \cap PQ$. Because $ADCQ$ is cyclic, we have the following.

\[\frac{AD}{CQ}=\frac{BD}{BC} \quad \textbf{and} \quad \frac{AB}{BD}=\frac{AQ}{CD}\]
Multiplying these yields the following.

\[\frac{AB}{BC}=\frac{AQ}{QC} \cdot \frac{AD}{CD} \quad (1)\]
Since $\triangle{ADP} \sim \triangle{QAP}$ and $\triangle{CDP} \sim \triangle{QCP}$, we also have the following.

\[\frac{AD}{AQ}=\frac{DP}{AP}=\frac{DP}{CP}=\frac{CD}{CQ}\]
Substitution into (1) yields that
\[\frac{AQ}{CQ}=\sqrt{\frac{AB}{BC}}\]

Let $X$ be the intersection of the bisector of $\angle{AQC}$ with $AC$. By the angle bisector theorem we have that

\[\frac{AX}{XC}=\frac{AQ}{CQ}=\sqrt{\frac{AB}{BC}}\]

Therefore the location of $X$ on segment $AC$, is dependent only on the location of $B$.

[hide="Solution"]Let $X$ be the point of intersection of $AC$ and the angle bisector of $\angle AQC$, and let $R$ be the second intersection of $PQ$ and circle $G$. It suffices to prove that $\frac{AX}{CX}$ is constant.

Noting that quadrilateral $ARCQ$ is harmonic, $\frac{AX}{CX}=\frac{AQ}{CQ}=\frac{AR}{CR}=\sqrt{\frac{AQ\cdot AR}{CQ\cdot CR}}=\sqrt{\frac{[AQR]}{[CQR]}}=\sqrt{\frac{AB}{BC}}$, which is indeed constant. We are done. $\blacksquare$ [/hide]

We see that $BQ$ is a symmedian. Let the angle-bisector intersect $AC$ at $K$. Then by Steiner's theorem $\frac{AB}{BC}=\frac{AQ^2}{AC^2}$, implying that $\frac{AQ}{QC}$ is fixed. But by the angle-bisector theroem, $\frac{AK}{KC}=\frac{AQ}{QC}$ and we have that $K$ is fixed.

[hide="Solution"]
Let the angle bisector of $\angle{AQC}$ intersect $AC$ at $X$. By the Angle Bisector Theorem, $AX/CX=AQ/CQ$. Then by Law of Sines:
\[ \frac{AQ}{CQ}=\frac{\frac{BQ\sin\angle{ABQ}}{\sin\angle{BAQ}}}{\frac{BQ\sin\angle{CBQ}}{\sin\angle{BCQ}}}=\frac{\sin\angle{BCQ}}{\sin\angle{BAQ}}=\frac{\sin\angle{QAP}}{\sin\angle{QCP}}\]
\[ =\frac{\frac{PQ\sin\angle{APB}}{AQ}}{\frac{PQ\sin\angle{CPB}}{CQ}}=\frac{CQ}{AQ}\cdot\frac{\sin\angle{APB}}{\sin\angle{CPB}}\]
\[ =\frac{CQ}{AQ}\cdot\frac{\frac{AB\sin\angle{ABP}}{AP}}{\frac{BC\sin\angle{CBP}}{CP}}=\frac{CQ}{AQ}\cdot\frac{AB}{BC}\]
so
\[ \left(\frac{AQ}{CQ}\right)^2=\frac{AB}{BC}.\]
It follows that $AX/CX=\sqrt{AB/BC}$, which is fixed, so the point $X$ is also fixed. // [/hide]

Let the second intersection of the angle bisector with the circle be $R$.
Also , Let the intersection of $AC$ and the bisector be $Y$
Then from the similarity of $\Delta ARP$ and $\Delta QAP$ , we get , $\frac{AR}{RQ}=\frac{RP}{PA}$.
Similarly ,we get  $\frac{RP}{PC}=\frac{RC}{QC}$.
So , $AQ/QC=AR/RC$ .
Now ,
$\frac{AB}{BC}=\frac{AR\sin{\angle{ARQ}}}{RC\sin{\angle{CRQ}}}$

So , $\frac{AQ}{QC}.\frac{AR}{RC}=\frac{AB}{BC}$
hence, $\frac{AY}{YC}=\frac{AQ}{QC}= (\frac{AB}{BC})^{1/2}$ , which is fixed. hence done  :)

[hide=" Solution"]
It suffices to show that $\dfrac{AQ}{QC}$ is fixed. Since $P$ lies on the $Q$-symmedian of $\triangle AQC,$ we must have $\dfrac{AQ}{QC}= \sqrt{\dfrac{AB}{BC}},$ which is fixed. $\blacksquare$ [/hide]

[hide="Solution"]
We first prove something.
[b]Fact: [/b]If $N$ is the intersection of the $A$-symmedian with $BC$, then $\frac{\sin\angle CAN}{\sin\angle BAN}=\frac{AC}{AB}$.
[b]Proof:[/b] Consider the median from $A$, $AM$, where $M$ is on $BC$. Then the Law of Sines yields that $\frac{\sin{\angle BAM}}{BM}=\frac{\sin{\angle AMB}}{AB}$ in $\Delta AMB$. Similarly, we can find that $\frac{\sin{\angle MAC}}{MC}=\frac{\sin{\angle CMA}}{AC}$ in $\Delta AMC$. Dividing the two, we have that $\frac{\sin{\angle BAM}}{\sin{\angle MAC}}=\frac{AC}{AB}$. If $N$ is the intersection of the $A$-symmedian with $BC$, we have $\frac{\sin\angle CAN}{\sin\angle BAN}=\frac{AC}{AB}$. $\Box$
By the Law of Sines in $\Delta ABQ$ and $\Delta BQC$, we find that $\frac{AB/\sin\angle AQB}{BC/\sin\angle BQC}=\frac{AQ/\sin\angle ABQ}{CQ/\sin\angle QBC}$. Since $\sin\angle ABQ=\sin\angle QBC$, our equation is $\frac{AB}{BC}\cdot\frac{\sin\angle BQC}{\sin\angle AQB}=\frac{AQ}{CQ}$. But $\frac{\sin\angle BQC}{\sin\angle AQB}=\frac{CQ}{AQ}$, so $\frac{AB}{BC}=\frac{AQ^2}{CQ^2}$. Hence $\frac{AQ}{CQ}$ is fixed, and we are done by the Angle Bisector Theorem. $\blacksquare$[/hide]

Apply Ceva(sinus) to triangle ACP and after law of sinus to tirangles APM and CPM and we are done. M is the intersection of PQ and AC.

[quote=FantasyLover]
$\sqrt{\frac{AQ\cdot AR}{CQ\cdot CR}}=\sqrt{\frac{[AQR]}{[CQR]}}=\sqrt{\frac{AB}{BC}}$, which is indeed constant. We are done. $\blacksquare$[/quote]

Why $\sqrt{\frac{[AQR]}{[CQR]}}$ gives us $\sqrt{\frac{AB}{BC}}$ ?

[quote=samirka259][quote=sjaelee]Then by Steiner's theorem $\frac{AB}{BC}=\frac{AQ^2}{AC^2}$, implying that $\frac{AQ}{QC}$ is fixed.[/quote]

What does [b]Steiner's theorem[/b] say ?[/quote]

I didn't read proofs above, but if you are asking for Steiner's theorem, it may be the next theorem(I am too lazy to read the problem and solutions :D ): for $\triangle{ABC}$ and points $X$ and $Y$ that lies on side $AC$, $BX$ and $BY$ are symmetric wrt(I am not sure for this word, may be "about" suits more) angle bissector of $\angle{ABC}$ if and only if $\frac{AX*AY}{CX*CY}=\frac{AB^2}{CB^2}$. I am pretty sure, that I used this theorem when I was solving it on the training :)

I guess this was before symmedians were well known...

Let the bisector of angle $AQC$ intersect $AC$ at $T$. By the angle-bisector theorem, $\frac{AT}{CT}=\frac{AQ}{QC}$. By the Symmedian Lemma, we have that $BQ$ is the $Q$-symmedian of $AQC$. Thus, Steiner's theorem and our work before yields that $\frac{AT}{CT}=\frac{AQ}{QC}=\sqrt{\frac{AB}{BC}}$, which is fixed.

My solution was similar to those above:

Note that $QP$ is a symmedian of $\triangle QAC$, so $\frac{AB}{CB} = \frac{QA^2}{QC^2}$, so $\frac{QA}{QC}$ is fixed, and thus the point of intersection does not depend on $G$.

$BQ$ is a symmedian of $\triangle AQB$. Then let the intersection of $PQ$ and $AC$ be $G$ and $M$ be the midpoint of $AC$.
We get $\angle AQB = \angle MQC$ and $\angle AQM = \angle BQC$.
Firstly $\frac{AB}{MC}=\frac{|AQB|}{|MQC|}=\frac{0.5AQ \cdot BQ \sin{AQB}}{0.5MQ \cdot CQ \sin {MQC}} = \frac{AQ \cdot BQ}{MQ \cdot CQ}$
In a similar way $\frac{AM}{BC} = \frac{AQ \cdot MQ}{BQ \cdot CQ}$
Note the two above quantities are constant because $A,M,B,C$ are fixed.
Multiplying the two quantities, we get $\frac{AQ^2}{CQ^2}$ and therefore $\frac{AQ}{CQ}$ is constant.
Since $\frac{AG}{GC}=\frac{AQ}{CQ}$ by angle bisector theorem on $QG$, we get $G$ is fixed.

Huh? ISL G2?!!
[quote=darij grinberg]Three distinct points $A$, $B$, and $C$ are fixed on a line in this order.  Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$.  Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$.  Suppose $\Gamma$ meets the segment $PB$ at $Q$.  Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.[/quote]

Let $T$ denote the intersection of the angle bisector of $\angle AQC$ with the fixed line $\overline{AC}$.
$  $\\
From the angle bisector theorem in $\triangle AQC$, we have that $\frac{AT}{CT} = \frac{ AQ}{CQ}$.
 Also, since $\overline{QB}$ is the $Q-$ symmedian of $\triangle AQC$, by Steiner Ratio theorem we have that : $\frac{AB}{CB}= \left(\frac{AQ}{CQ}\right)^2$.
$  $\\
Therefore we have that $$\frac{AT}{CT} = \underbrace{\sqrt{\frac{AB}{CB}}}_{\text{Fixed}}$$
So $T$ is fixed irrespective of choice of $\Gamma$.

It is well known that $QB$ is a symmedian in $\triangle AQC$. So $\frac{AQ}{QC} = \frac{AB^2}{BC^2}$. So $\frac{AQ}{QC}$ is fixed.
But by the angle bisector theorem $\frac{AL}{LC} =\frac{AQ}{QC}$ is fixed. So $L$ is fixed. Then we are done.

[hide=Solution]
[quote=ISL 2003 G2]Three distinct points $A$, $B$, and $C$ are fixed on a line in this order.  Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$.  Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$.  Suppose $\Gamma$ meets the segment $PB$ at $Q$.  Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.[/quote]
[b][color=#000]Solution:[/color][/b] Let tangent to $\odot (AQC)$ $\cap$ $AC$ $=$ $X$ $\implies$ $(X,B;A,C)$ is harmonic. Hence, $XQ,XA$ is fixed. $$\frac{AT}{TC}=\frac{AQ}{QC}=\frac{XA}{XQ}$$
Hence, $T$ doesnot depend upon choice of $\Gamma$ $\qquad \blacksquare$
[/hide]

[quote=ISL 2003 G2]Three distinct points $A$, $B$, and $C$ are fixed on a line in this order.  Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$.  Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$.  Suppose $\Gamma$ meets the segment $PB$ at $Q$.  Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.[/quote]
One line proof:
Since, $QB$ is $Q$-symmedian in $\triangle QAC\implies$ angle bisector of $\angle AQC$ meet $AC$ at the point $X$ and so, the ratio $AX/XC$ remains constant because $QA/QC$ is constant as $B$ is fixed, and thus, $X$ is fixed.                $\blacksquare$

[quote=Mewto55555]The coordinate bash is not too bad if you're careful:

[hide="Solution"]

Let $A$ be at $(-1,0)$, $B$ be at $\left(\frac{1}{a},0\right)$, $C$ be at $(1,0)$, and the center $O$ of circle $G$ be at $\left(0,-\frac{1}{k}\right)$. WLOG, $a>1$ and $k>0$. (if $B$ lies at $(0,0)$, the problem is obvious.)

Now, it is immediately evident that $P$ lies at $(0,k)$.

The equation of line $PB$ is $y=-kax+k$.

The equation of $G$ is $x^2+\left(y+\frac{1}{k}\right)^2=\frac{1}{k^2}+1$ or $k^2x^2+(ky+1)^2=k^2+1$

So, their intersection occurs when:

$k^2x^2+(-k^2ax+k^2+1)^2=k^2+1$

$k^2(k^2a^2+1)x^2-2ak^2(k^2+1)x+k^2(k^2+1)=0$

We can divide by $k^2$ to get

$(k^2a^2+1)x^2-2a(k^2+1)x+(k^2+1)=0$ so we have

$x=\frac{a(k^2+1)-\sqrt{a^2(k^2+1)^2-(k^2+1)(k^2a^2+1)}}{(k^2a^2+1)}$ (we obviously take the lesser root here). Simplifying, we get

$x=\frac{a(k^2+1)-\sqrt{(k^2+1)(a^2-1)}}{(k^2a^2+1)}$

Then, the y-coordinate $y$ of $Q$ is just $y=-kax+k$.

We need $\frac{QA}{QC}$ to be constant, so we must show that

$\frac{(x+1)^2+y^2}{(x-1)^2+y^2}=1+\frac{4x}{(x-1)^2+y^2}$ does not depend on $k$.

As $1$ clearly does not depend on $k$, it is sufficient to show $\frac{4x}{(x-1)^2+y^2}$ does not.

For convenience, denote $k^2a^2+1=p, k^2+1=q, \sqrt{q(a^2-1)}=r, ak^2+1=s$.

Thus, $x=\frac{aq-r}{p}$.

$\frac{4x}{(x-1)^2+y^2}=\frac{4x}{(k^2a^2+1)x^2-2(ak^2+1)x+(k^2+1)}$

Subbing in $x$'s value:

$\frac{\frac{4(aq-r)}{p}}{p\frac{(aq-r)^2}{p^2}-2s\frac{aq-r}{p}+q}$

Cancelling out some $p$'s we get
$\frac{4(aq-r)}{(aq-r)^2-2s(aq-r)+pq}$

The denominator is $a^2q^2-2aqr+r^2-2saq+2rs+pq$

Subbing everything back in (except some $r$'s) we have the denominator to be:

$a^2(k^4+2k^2+1)-2a(k^2+1)r+(k^2+1)(a^2-1)-2a(k^2+1)(ak^2+1)+2r(ak^2+1)+(k^2a^2+1)(k^2+1)$

or

$-2r(a-1)+[a^2k^4+2a^2k^2+a^2-k^2+k^2a^2-1+a^2-2a^2k^4-2ak^2-2a^2k^2-2a+a^2k^4+a^2k^2+k^2+1]$

The stuff in the brackets is just:

$-2ak^2-2a+2a^2+2a^2k^2=2a(-k^2-1+a(k^2+1))=2a(k^2+1)(a-1)$

so the denominator is just $2aq(a-1)-2r(a-1)$

So all that's left to prove is that $\frac{4(aq-r)}{2(aq-r)(a-1)}=\frac{2}{a-1}$ is constant as $k$ varies, which is obviously true, as $a$ and $k$ are independent.[/hide][/quote]

This solution over-complicates things so much it's actually funny.  :-D

Storage
[quote=ISL 2003 G2]Three distinct points $A$, $B$, and $C$ are fixed on a line in this order.  Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$.  Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$.  Suppose $\Gamma$ meets the segment $PB$ at $Q$.  Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.[/quote]

Let $\ell_1$ be the angle bisector of $\angle AQC$ and $\ell_1\cap AC=T$. So, $$\frac{AQ}{QC}=\left(\frac{AB}{BC}\right)^2=\frac{AT}{TC}$$ So, $\ell_1\cap AC$ does not depend on the choice of $\Gamma$.

Let the intersection of the two objects be $R$, then note $QP$ is the $Q$-symmedian of $\triangle QAC$ so $AB/BC=(AQ/QC)^2=(AR/RC)^2$ by the angle bisector theorem, which finishes because it uniquely specifies $R$.

Let $G$ denote the intersection of the angle bisector of $\angle AQC$ and line $AC$. It is well known that $BQ$ is the $Q$-symmedian of $\triangle ACQ$. By a property of the symmedian, we know that $\frac{AB}{BC}=\left(\frac{AQ}{QC}\right)^2$. But $\left(\frac{AQ}{QC}\right)^2=\left(\frac{AG}{GC}\right)^2$ by the angle bisector theorem. This means that $\frac{AB}{BC}=\left(\frac{AG}{GC}\right)^2$. But $\frac{AB}{BC}$ is constant, so the point $G$ is uniquely determined.

Let the intersection point of the angle bisector and $AC$ be D.

Note that P is the $Q-$symmedian point of $\triangle QAC$. Thus, \[\frac{AB}{BC}=\left(\frac{QA}{QC}\right)^2\]
Thus, the ratio of $\frac{AQ}{QC}=\frac{AD}{DC}$ are purely in terms of A,B,C, therefore $D$ is always the same point determined by the initial points $A,B,C$.

Let the intersection point be $X$ and notice that $\overline{QB}$ is the $Q$-symmedian of triangle $QAC.$ We are done as $$\frac{AX}{XC}=\frac{AQ}{CQ}=\sqrt{\frac{AB}{CB}}$$ which is constant. $\square$

ISL marabot solve

Let the intersection point be $X$.

It's well-known that $\overline{PQ}=\overline{QB}$ is the $Q$-symmedian of $\triangle AQC$.

By the angle bisector theorem, $\frac{XA}{XC}=\frac{AQ}{QC}$. Now, since $\overline{QB}$ is the $Q$-symmedian of $\triangle AQC$, \[\frac{XA}{XC}=\frac{AQ}{QC}=\sqrt{\frac{BA}{BC}},\]which is constant.

We have to prove QA/QC is fixed. let PQ meet $\Gamma$ at K.
AB/BC = sin ∠BQA/sin ∠BQA . QA/QC = sin ∠KCA/sin ∠KAC . QA/QC = QA/QC . KA/KC = (QA/QC)^2.
AB/BC is fixed so (QA/QC)^2. is fixed so QA/QC is fixed.
we're Done.

Imagine forgetting some very important symmedian fact :eyes:

[img width=50]https://media.discordapp.net/attachments/925784397469331477/950541023963123802/Screen_Shot_2022-03-07_at_4.50.35_PM.png?width=1358&height=1171[/img]

We claim that $\left(\frac{AQ}{QC}\right)^2=\frac{AB}{BC},$ so it's a fixed quantity, which shows that the angle bisector intersects $AC$ at a fixed point. Note that \[\frac{AQ}{QC}=\frac{\sin\angle ACQ}{\sin\angle CAQ}\]
\[=\frac{\sin\angle PAQ}{\sin\angle PCQ}\]
\[=\frac{\frac{\sin\angle PAQ}{PQ}}{\frac{\sin\angle PCQ}{PQ}}\]
\[=\frac{\frac{\sin\angle AQP}{AP}}{\frac{\sin\angle CQP}{CP}}\]
\[=\frac{\sin\angle AQB}{\sin\angle CQB}\]
\[=\frac{\frac{AB\sin\angle BAQ}{BQ}}{\frac{BC\sin\angle BCQ}{BQ}}\]
\[=\frac{AB}{BC}\cdot\frac{QC}{AQ}\]
which proves our claim.

Let $f(X)=\frac{XA}{XC}$. Let $PQ\cap (AQC)=K.$ Then by ratio lemma,
$f(B)=f(K)\cdot f(Q)=f(Q)^2$
$f(M)=f(Q)\cdot 1=f(Q)=\sqrt{f(B)}$ where $M$ is the intersection point.
$\square$

Let $I$ be the intersection of the angle bisector with line $AC$ and let $X$ be the intersection of segment $PB$ with the circle. Then we have $\frac{AI}{IC} = \frac{AQ}{QC}$ by the Angle Bisector Theorem, and $AQ = \frac{PA \cdot AX}{PX}$ and $QC = \frac{PQ \cdot CX}{PX}$ and thus $\frac{AQ}{QC} = \frac{PA \cdot AX \cdot PC}{PX \cdot PQ \cdot CX} = \frac{PQ}{PA} \cdot \frac{AX}{PQ} \cdot \frac{PC}{CX} = \frac{AX}{CX}$. But by the Ratio Lemma we have $\frac{AX}{CX} \cdot \frac{\sin \angle AXB}{\sin \angle CXB} = \frac{AX}{CX} \cdot \frac{AQ}{QC} = \frac{AB}{BC}$. Thus, $\frac{AX}{CX} = \frac{\frac{AB}{BC}}{\frac{AQ}{QC}} = \frac{AQ}{QC}$, so $\frac{AB}{BC} = \left(\frac{AQ}{QC}\right)^2$. But this means that $\frac{AQ}{QC}$ is determined, and thus so is $\frac{AI}{IC}$ as desired.

Because $\overline{QBP}$ is the $Q$-Symmedian of $ACQ$, we have $$\frac{QA}{QC} = \sqrt{ \frac{BA}{BC} }$$ which finishes via the Angle Bisector Theorem. $\blacksquare$

[b]Remark:[/b] Because the center of $\Gamma$ doesn't lie on $\overline{ABC}$, we know $P$ isn't at infinity.

Nice problem and a good exercise in symmedians. We observe that $QP$ is a $Q-\text{symmedian}$ of $\triangle QBC$. Thus $\left(\frac{QA}{QC}\right)^2=\frac{AB}{AC} \implies \frac{QA}{QC}=\sqrt{\frac{AB}{AC}}$ so by angle bisector theorem the intersection of the bisector of $\angle AQC$ with $AC$ is a fixed point $X$ with $\frac{AX}{XC}=\sqrt{\frac{AB}{AC}}$.

Let $X$ be the desired point. Note that $QB$ is the Q-symmedian in $\triangle ACQ$. This means that the ratio $$QA/QC=\sqrt{AB/BC}$$ does not depend on $\Gamma$. By the Angle Bisector Theorem, this also means that $AX/XC$ does not depend on $\Gamma$, so the location of $X$ does not depend on $\Gamma$, QED.

Is this the easiest isl?

Let the intersection be E. It's a well known property that QP is a symmedian of AQC, so then AB/BC=AQ^2/AC^2, and by angle bisector theorem the length AE/EC=AQ/AC is constant.

@above because symmedians can be constructed by drawing the equal tangents' intersection, QB==BP.

$\color{red} \boxed{\textbf{SOLUTION}}$

Let the angle bisector of $\angle AQC$ intersect $AC$ at $K$
$QB$ is the symmedian of $\triangle AQC$
So, $$\frac{AB}{BC}={\frac{AQ^2}{QC^2}} \implies  \frac{AQ}{QC}=\sqrt{\frac{AB}{BC}} \implies \frac{AK}{KC}= \underbrace{\sqrt{\frac{AB}{BC}}}_{\text{Fixed}} \blacksquare$$

why is g1 and g2 both not hard symmedian problems

Clearly $QB$ is the $Q$-symmedian in $\triangle AQC$, so if we let $D$ be the foot of the $Q$-angle bisector in $\triangle AQC$, then
\[
\frac{AD}{DC} = \frac{AQ}{QC} = \sqrt{\frac{AB}{BC}}
\]
which is fixed, so we're done. $\square$

We consider the converse: let $K,L$ lie on $\overline{AC}$ such that $(A,C;K,L)=-1$, and consider the circle $\Gamma$ with diameter $\overline{KL}$—some Apollonius circle. Let $Q$ be a variable point on $\Gamma$ and $P$ be the intersection of the tangents to $(ACQ)$ at $A$ and $C$. Then prove that $\overline{AC} \cap \overline{QP}:=X$ is fixed.

Clearly $X$ is the foot of the $Q$-symmedian. If $M$ is the midpoint of $\overline{AC}$, then $\overline{QK}$ and $\overline{QL}$ are the $Q$-angle bisectors in $\triangle QXM$, hence $(X,M;K,L)=-1$. Since $K,L,M$ are fixed independent of $Q$, it follows that $X$ is as well.

Returning to the main problem, as we vary $K$ and $L$ around $\overline{AC}$ such that $(A,C;K,L)$ holds, $X$ can achieve any point; in particular we can have $X=B$ (and $K$ lying inside segment $\overline{AC}$). Then as $Q$ varies for this fixed choice of $K,L$, $\Gamma$ can become any circle through $A$ and $C$. Hence in the original problem, the intersection of the $\angle AQC$-bisector and $\overline{AC}$ is fixed at $K$. $\blacksquare$

???

Let $D$ be the point where the angle bisector of $\angle AQC$ hits $\overline{AC}$. By construction, $QB$ is the $Q$-symmedian in $\triangle AQC$, so by the angle bisector theorem we have
$$\frac{AD}{DC} = \frac{AQ}{QC} = \sqrt{\frac{AB}{AC}},$$
and $D$ is fixed, as needed.

Let $PB$ meets the circle $\Gamma$ at $D$ and $Q$, respectively. Then note that $ADCQ$ is harmonic (from the construction of the tangents).

Let the angle bisector of $\angle AQC$ meet $AC$ at a point $X$. Then note that
\[ \left( \frac{AX}{XC} \right)^2 = \left( \frac{AQ}{QC} \right)^2 = \frac{AB}{BC} \]
where we use Angle Bisector theorem and the definition of a harmonic quadrilateral.
Thus, $X$ only depends on $A$, $B$, and $C$, as desired. $\blacksquare$


Since the deadline appears to have passed, can anyone post the problems?

Here you are:
http://www.om.edu.pl/index2.php?poziom=9&main=zadania.php
problems from 59. OM

Merry Christmas :)

I know that site... can you translate the problems into English?  :D


limit
real analysis
real analysis unsolved

Given an example of this

Let $E_{n}$ be an decreasing sequence of sets, i.e., $E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq \cdots$ such that $m^\ast(E_{n}) < \infty$ and $m^\ast(\bigcap E_{i}) < \lim m^\ast(E_{i})$.

Ok, I have solved it.

Let $W$ be a (typical) non-measurable set (constucted by choosing one element from each equivalence class where the equivalence relation is defined by $x\sim y \iff x-y$ is rational) in $[0,1]$ and list all rationals in $[0,1]$ as $\{r_{i}\}$
Then let $E_{i}= \bigcup_{k=i}^\infty (W+r_{i})$.

$\bigcap E_{i}= \emptyset$ but $\lim m^\ast(E_{i}) = m^\ast(W) > 0$

[quote="liyi"]
$\bigcap E_{i}= \emptyset$[/quote]
But how can you assert it?

[edit:OK,I know,but you must say that $W$ is the special non-measurable set...]

[quote="Soarer"]This works too, but we are talking about extended real number system :blush:
$E_{n}= (n, \infty)$[/quote]
Yes, quite right.
I forget a constraint: $m^\ast(E_{n})<\infty$.
Now I've updated the problem.


Let $ x,y,z$ be distinct non-negative real numbers . Prove that :

$ (xy\plus{}yz\plus{}zx)(\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(y\minus{}z)^2}\plus{}\frac{1}{(z\minus{}x)^2})\geq 4$

My friend solves it by AM-GM inequalities!

Replace $ (x,y,z)$ by $ (x \minus{} min(x,y,z),y \minus{} min(x,y,z),z \minus{} min(x,y,z))$, we can think that one of $ x,y,z$ is 0.

Let $ z \equal{} 0$ and $ x \minus{} y \equal{} \minus{} 1$.

It is easy to show that
\[ x(x \plus{} 1)(1 \plus{} \frac {1}{(x \plus{} 1)^2} \plus{} \frac {1}{x^2}) \geq 4.
\]

[quote="tanpham90"]Let $ x,y,z$ be distinct non-negative real numbers . Prove that :

$ (xy \plus{} yz \plus{} zx)(\frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(y \minus{} z)^2} \plus{} \frac {1}{(z \minus{} x)^2})\geq 4$[/quote]
Solution of my friend:
Suppose that $ z\equal{}min\{x;y;z\}$
We have:$ f(x,y,z)  \geq  f(x,y,0)\equal{}xy[\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{x^2}\plus{}\frac{1}{y^2}]$
$ \equal{} \frac{1}{\frac{x}{y}\plus{}\frac{y}{x}\minus{}2}\plus{}\frac{x}{y}\plus{}\frac{y}{x} \geq 4$

[quote="tanpham90"]This is the problem in pvthuan's book  :D  :!:[/quote]

 :lol: Yes, it is problem 6.27 page 248 of [size=150]The Inequalities Book [/size] by Pham Van Thuan, Le Vi.

During IMO 2007 about 50 people (IMO leaders, deputy leaders) managed to purchase copies of the book and I guess that some might appear at national olympiads this year. You will see a lot more original inequalities there I promise.

[quote="tanpham90"]Let $ x,y,z$ be distinct non-negative real numbers . Prove that :

$ (xy \plus{} yz \plus{} zx)(\frac {1}{(x \minus{} y)^2} \plus{} \frac {1}{(y \minus{} z)^2} \plus{} \frac {1}{(z \minus{} x)^2})\geq 4$[/quote]
 :(  See here http://www.mathlinks.ro/viewtopic.php?t=184883

No, see here  :D

http://www.mathlinks.ro/viewtopic.php?t=168974


[quote="abhilash_i"]can someone give me a hint on this problem
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=3295

Thanks in advance.[/quote]

First of all, how many ways are there to choose any triangle (counting degenerate ones)? The answer is ${(M+1)(N+1) \choose 3}$, since we can pick any three of the points (there are $(M+1)(N+1)$ total). However, many of these triangles are degenerate (simply a line), so we need to find a way to remove them from our total. Some of them are easy enough (vertical/horizontal), but for the rest, you'll have to find some way (since I can't think of any good one right now.  :P ).

This might not be fast enough, though (since I haven't tried it), so you may have to find an even faster method.

uhh... how can i forget this problem? We solved it during the real contest. we participated at ICPC as pre-university student just as practice and we beat all the university teams and positioned second solving 5 problems while Fudan solved 6 and got the first position. the official second position went to Dhaka University team who solved 4.  :D

Anyway during contest time i solved this problem so i quite remember the solution.

ya nebula's idea is absolutely correct. u have to count the number of degenerate triangles. you have to do that in O(MN) which is 10^6 quite resonable i guess :)


probability
calculus
integration
combinatorics unsolved
combinatorics

You first randomly generate a real number $ x$ with $ 0\le x\le 1$. Then you make a coin such that the probability of getting head is $ x$. Prove that the probability of getting $ k$ heads in $ n$ flips is $ \frac{1}{n\plus{}1}$ without using calculus.

[hide="calculus solution"]Integrate$ \binom{n}{k}\int_0^1 x^k(1\minus{}x)^{n\minus{}k}dx$by parts and use induction.[/hide]

I think somethings wrong with the problem statement. The probability is independent of $ k,x$?

[quote="nealth"]You first randomly generate a real number $ x$ with $ 0\le x\le 1$. Then you make a coin such that the probability of getting head is $ x$. Prove that the probability of getting $ k$ heads in $ n$ flips is $ \frac {1}{n \plus{} 1}$ without using calculus.

[hide="calculus solution"]Integrate$ \binom{n}{k}\int_0^1 x^k(1 \minus{} x)^{n \minus{} k}dx$by parts and use induction.[/hide][/quote]

The answer is $ \binom{n}{k}x^k(1\minus{}x)^{n\minus{}k}$

As ^ said, I can't see how this can be = (1/n+1)

You are supposed to take the average of all values of that formula when $ x$ ranges from $ 0$ to $ 1$.

He's asking the probability that you get $ k$ heads in $ n$ flips if you have no idea what $ x$ is.

There is actually a seperate [url=http://www.mathlinks.ro/index.php?f=498]Proababilty and stats Forum[/url]
PLEASE post all probability related questions therre


algebra
polynomial
integration
search
real analysis
real analysis unsolved

Let $P$ be a polynomial with degree n such that
\[ \int_0^1x^kP(x)dx=0  \]
for $k=1,2,...,n$.
Prove that:
\[ \int_0^1(P(x))^2dx=(n+1)^2(\int_0^1P(x)dx)^2  \]

It has been posted before.
Search the forum.

and guess who's posted the solution :D
here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=35895[/url]


symmetry
number theory unsolved
number theory

[quote="toan hoc"]Find all  prime numbers $x;y;z$ such that $x^{y}+y^{x}+300=z$[/quote]
Let $x\le y$. It is obviosly $x=2,$ and $y,z$ odd. If $y>3$ then $2^{y}=-1(mod 3),y^{2}=1(mod 3)$ (because y iss odd ). Therefore $y=3$ and $z=300+2^{3}+3^{2}=317$.

[quote="Rust"]
Let $x\le y$. It is obviosly $x=2,$ and $y,z$ odd. [/quote]
Why?

symmetry btwn x and y allows us to assume x leq y.
z prime, z>2, z odd
x^y + y^x odd
x^y, y^x, one odd, one even
x>2 => y>2 => x and y are odd => x^y, y^x both odd, contradiction
x=2 => x^y even => y^x odd => y odd
......


trigonometry
geometry
inequalities
inequalities unsolved

Sow that in any triangle $ABC$ we have

$\sin A \cos B+\sin B \cos C+\sin C \cos A  \leq 3\frac{\sqrt 3}{4}$

$\sin A\cos B=\frac{1}{2}\{\sin (A+B)+\sin (A-B)\}=\frac{1}{2}\{\sin C+\sin(A-B)\}\leqq\frac{1}{2}\sin C$.

Similarly we have $\sin B\cos C\leqq\frac{1}{2}\sin A$ , $\sin C\cos A\leqq\frac{1}{2}\sin B$.
Therefore L.H.S.
$=\frac{1}{2}(\sin A+\sin B+\sin C)=\frac{3}{2}\cdot\frac{\sin A+\sin B+\sin C}{3}\leqq\frac{3}{2}\cdot\sin {\frac{A+B+C}{3}}$

$=\frac{3}{2}\sin\frac{\pi}{3}$. we are done.

$\displaystyle \frac{1}{2}\{\sin C+\sin(A-B)\}\leqq\frac{1}{2}\sin C$

Why???????

Since $0<A<\pi,0<B<\pi$,we have $-\pi<A-B<\pi$,so when $A-B=\frac{\pi}{2}$, $ \sin (A-B)=1$.

And what if $C=\frac{\pi}{2},A=\frac{\pi}{4},B=\frac{\pi}{4}$? Then it surely doesn't hold. Or generaly when $|A-B|<C$

It seems, that the left hand side is equal with S/R^2 where S is the area of the triangle and R is the radius of the circumcircle. When this is correct, then the rest of the proof isn't hard at all.

wellknown, it's not correct, we have :
[tex]\frac{s}{r^2}=2sinAsinBsinC=1/2.(sin2A+sin2B+sin2C)<=sinAcosB+sinBcosC+sinCcosA [/tex]
Hence, your solution is bad!!!! But this problem is very easy!
[code]try my code[/code]

[hide="to hungkhtn"]Is this site very interesting![/hide]

This problem was created by my teacher.Here is the another solution:
Suppose $(X,Y,Z)$ a permutation of $(A,B,C)$ such that:$X\leq Y\leq Z$.
We have: $\sin X\geq \sin Y \geq \sin Z,\cos X\leq \cos Y \leq \cos Z$
Hence:
\[ LHS \leq \sin X\cos Z+\sin Y\cos Y+\sin Z\cos X
           \leq \sin (X+Z) +\sin Y\cos Y\]

\[ LHS \leq \sin Y(1+\cos Y)=\sqrt {(1-\cos ^2 Y)(1+\cos Y)^2}=\sqrt {\frac{1}{3}(3-3\cos Y)(1+\cos Y)^3}\]
From $AM-GM$, we have:
\[ LHS\leq \sqrt {\frac{1}{3}(\frac{6}{4})^4}=\frac {3\sqrt {3}}{4}\]

So  the  triangle must be  acute , right?

I think  for arbitrary triagle the inequality allway holds. ;)

But in an  obtuse  triangle....its not  true  that  [tex]X\leq Y\leq Z [/tex]   then  [tex]cosX \leq cosY \leq cosZ[/tex]  Where is  my  mistake?  Thank  you!

The inequality still true for abiraty triangle.


LaTeX
integration
calculus
derivative
linear algebra
matrix
vector

How should I type in LaTeX so that my formulas dont overlap?
[code]
\newcommand{\pder}[2]{\ensuremath{\frac{\partial #1}{\partial #2}}}

\begin{eqnarray}
	\textbf{p}=\left(\begin{array}{c}p_1\\p_2\end{array}\right)=
	\left\{\begin{array}{cc}\displaystyle \pder{\cal{L}}{q_1}&\int\\\displaystyle \pder{\cal{L}}{q_1}&4\end{array}\right.
\end{eqnarray}

[/code]

With or without \displaystyle the partial derivatives will overlap, how can this be undone.
I have seen in books where they have an entire matrix of large partial derivatives, how is this possible?
Thanks

That's what you might call a mess.
1. Never use eqnarray as that's now obsolete, replaced by align. However, there's no reason to use eqnarray here as you only have one equation so the equation environment is fine.
2. \cal is also obsolete, use \mathcal
3. Do not use arrays for vectors as they don't look good, use the pmatrix environment
4. Use the cases environment instead of an array
5. add extra spacing using \\ [ [i]vertical spacing [/i]]
6. use \dfrac in your newcommand to get displayed fractions
7. lay out your code for easy reading
[code]\newcommand{\pder}[2]{\ensuremath{\dfrac{\partial #1}{\partial #2}}}
\begin{equation}
	\textbf{p}=
  \begin{pmatrix}
		p_1 \\
		p_2
  \end{pmatrix}
   =
  \begin{cases}
		\pder{\mathcal{L}}{q_1} & \int \\[3ex]
		\pder{\mathcal{L}}{q_1} & 4
  \end{cases}
\end{equation}[/code]will look something like this (though the equation number will be better placed in a document)
\[ \begin{equation}\textbf{p}=\begin{pmatrix}p_{1}\\ p_{2}\end{pmatrix}=\begin{cases}\cfrac{\partial\mathcal{L}}{\partial q_{1}}&\int\\[3ex]\cfrac{\partial\mathcal{L}}{\partial q_{1}}& 4\end{cases}\end{equation}\]

[size=75]Edit: changed [1em] to [3ex] which makes more sense as em is a horizontal distance (width of m in the current font size) whereas ex is a vertical distance (height of x in current font).[/size]


induction
combinatorics proposed
combinatorics

Let G be a graph whose odd cycles are pairwise intersecting ( that is every two odd cycles in G have a common vertex). Prove that the chromatic numer of G is at most 5.

Lemma 1: A graph with no odd cycle can be normally colored in two colors.
Proof: If the graph is not connected, apply induction to each of its conex components.
Otherwise pick up a vertex $A$ and color each vertex $V$ according to the parity of distance from $A$ to $V$. (since there is no cycle, all paths from $A$ to $V$ have same parity of length).
Lemma 2: Any cycle with no edges in the interior can be colored in at most three colors.
Proof: obvious.

Now about the problem:
If there is no odd cycle we are done by Lemma 1.
Otherwise take a minimal odd cycle $C$. The subgraph obtained by removing $C$ has no odd cycles so can be colored in $2$
colors by Lemma 1.
Finally we color $C$ in three other colors which is possible by Lemma 2.
QED

[Edit: a flaw noticed by Myth was corrected]


Erstmal für alle Neugierigen die Teilnehmerliste der IMO-Auswahl 2006:

Alagalingam, Meru
Fintzen, Jessica
Flake, Johannes
Grinberg, Darij
Günther, Felix
Hamacher, Paul Jonas
Harrer, Daniel
Huxol, Tobias
Kraus, Nicolai
Oehme, Markus
Ohst, Matthias
Rodenhausen, Moritz
Sattler, Christian
Schönherr, Georg
Scholze, Peter
Schröter, Georg

Der Cutoff lag irgendwo zwischen 16 und 18 Punkten (d. h. mit 18 war man sicher drin), was vielleicht ein Minimalrekord der letzten Jahre ist...

Nun die Frage: Wann kommt ihr voraussichtlich in Rostock an? Mein Plan:

Karlsruhe Hbf   09.01.06        ab      08:51   3               ICE 78  InterCityExpress
Hamburg Hbf     09.01.06        an      13:35   11a/b
Hamburg Hbf     09.01.06        ab      13:44   12a/b           IC 2376         InterCity
Rostock Hbf     09.01.06        an      15:35   6
Rostock Hbf     09.01.06        ab      15:46   8               S       S-Bahn
Rostock-Lütten Klein    09.01.06        an      15:59   1
Dauer: 7:08; fährt nicht täglich, Verkehrstage

Etwas gewagt, sowohl von der Umsteigedauer her, als auch von der Aufstehzeit (für mich jedenfalls ;) ). Hoffe mal, daß alles klappt.

  Darij

Erstmal hi @all,
da ich es nun irgendwie zu den Auwahlseminaren geschafft habe, melde ich mich in diesem Forum doch mal zu Wort.
Bin ich ja erst einmal beruhigt, dass ich nicht ganz die untere Grenze markiere mit meinen 22/60 Punkten ;)
und scheinbar bin ich auch nicht der einzige mit einem seltsamen Tagesrhytmus (1 Uhr nachts über Anfahrten reden?)

jedenfalls werde ich wohl zwischen 15 und 16 Uhr in Rostock sein, höchstwahrscheinlich folgende Verbindung: (Falls Matthias Ohst irgendwann meine E-Mail liest fahren wir wohl zusammen)

Staßfurt 	09.02.06 	ab 	10:09 	1
Magdeburg Hbf 	09.02.06 	an 	10:45
Magdeburg Hbf 	09.02.06 	ab 	11:12
Wittenberge 	09.02.06 	an 	12:40
Wittenberge 	09.02.06 	ab 	12:45
Schwerin Hbf 	09.02.06 	an 	13:42
Schwerin Hbf 	09.02.06 	ab 	13:51
Rostock Hbf 	09.02.06 	an 	14:52
Rostock Hbf 	09.02.06 	ab 	15:06
Rostock-Lütten Klein 	09.02.06 	an 	15:19 	1

Bleibt also auch noch genug Zeit für die Viertelstunde Fußmarsch zur Jugendherberge aufgrund von Orientierungsproblemen eine Stunde zu benötigen...

Tobias

[quote="Tobias Huxol"]und scheinbar bin ich auch nicht der einzige mit einem seltsamen Tagesrhytmus (1 Uhr nachts über Anfahrten reden?)[/quote]
LOL...wenn das so weiter geht müssen die seminarzeiten in zukunft einen halben tag nach hinten verschoben werden, weil sonst nicht nur NoName vormittags fehlt...

herzlichen glückwunsch an alle die es geschafft haben!
vielleicht schafft ihr ja dieses jahr attack mars.

ps: ich werde vielleicht beim quaisser-seminar vorbeischauen

Den Mars machen wir natürlich platt ;)
WE MUST BUILD AN ATOMIC BLASTER!  :D

Ansonsten muss ich meine Anreise noch mit Nicolai klären, werde aber wohl mit dem NoName-Zug von Hamburg aus fahren (kommt um 17:59 an).

@Ente: weißt du denn welches das Quaisser-Seminar ist¿  :P

Den Glückwünschen kann ich mich nur anschließen :thumbup:
Neuerdings ist ja die Teilnahme [i]auf eigenes Risiko[/i], aber das wird den Spaß wohl kaum mindern  :D
Bleibt noch: Alles Gute für die Klausuren und

KICK THOSE MARTIAN BUTTS TO THE MOOOOOOOOOOOOOOOOOOOOOOOOOONN!!!!!!!!!!1111337

[quote="ZetaX"]@Ente: weißt du denn welches das Quaisser-Seminar ist¿  :P[/quote]
Ja! :D  :D  :D  :D
[quote="Christian Hirsch"]Neuerdings ist ja die Teilnahme auf eigenes Risiko, aber das wird den Spaß wohl kaum mindern   :D [/quote]
Ich hoffe, daß ihr trotzdem die warnemünde-nachtwanderung nicht ausfallen lasst.

ich möchte übrigens über alle GV-awards informiert werden (ihr wisst ja, was ich meine  ;)   )

ATTACK!!!!!!!!!!!!!!!!!!!!!!!!!

[quote="Tobias Huxol"]und scheinbar bin ich auch nicht der einzige mit einem seltsamen Tagesrhytmus (1 Uhr nachts über Anfahrten reden?)[/quote]

deshalb habe ich ja auch gemeint, daß meine aufstehzeit für den zug nach rostock etwas abenteuerlich ist ;)

[quote="Tobias Huxol"]Bleibt also auch noch genug Zeit für die Viertelstunde Fußmarsch zur Jugendherberge aufgrund von Orientierungsproblemen eine Stunde zu benötigen...[/quote]

der weg ist auch bei perfekter orientierung länger als eine viertelstunde, da haben die leute von der gartenschau gut gearbeitet... (oder sind sie vielleicht dieses jahr endlich weg?)

[quote="Werwulff"]LOL...wenn das so weiter geht müssen die seminarzeiten in zukunft einen halben tag nach hinten verschoben werden, weil sonst nicht nur NoName vormittags fehlt...[/quote]

war doch schon letztes mal nicht der fall, ich hatte in der kategorie den 2. oder 3. platz (weiß nicht ob zetax öfter als ich gefehlt hatte) - und zumindest in oberwolfach ist es ja schon tradition, daß man bis um 2 uhr nacht im institut hockt (man muß ja nicht unbedingt dabei im netz herumtrollen oder karamata-wettkämpfe veranstalten wie letztes jahr)... übrigens bin ich gespannt [hide]ob der magnet immer noch hängt[/hide] ;)

BLOW THE SAUCER OUT OF THE SKY!!!!!!  :starwars: :starwars: :starwars: :starwars:

  dg

LOL@[hide]Magnet (ich wette, dass er hängt)[/hide] :lol:

Also Platz 1 oder 2 im Seminar-Verpassen hast du sicher nicht, das machen NoName und ich unter uns aus ;)
Außerdem hast du die Aufzugfahrten und PKversuche vergessen.

Eigentlich könnten wir dieses Jahr dafür sorgen, dass die AIMO nicht mehr nach Oberwolfach darf...

NO TIME FOR TEA, BLAST THOSE MARTIANS!  :alien:  :alien:  :alien:

Hi,

Matthias Ohst hat sich mittlerweile gemeldet, wir werden wohl doch den Zug über Berlin nehmen und gegen 17:45 in Rostock sein.

mfg
Tobias

PS: wann wird eigentlich die Schule benachrichtigt? Oder muss ich mich um die Freistellung selber kümmern?

Ja die Befreiung darfst du schon selber besorgen...
Viel Spass dabei ;)


geometry
rectangle
linear algebra
linear algebra unsolved

Let $A \in {\mathbb R}^{3 \times 4}$ and  $B \in {\mathbb R}^{4 \times 3}$,
$A= \left (\begin{array}{ccccccc}1 & 6 & 2 & 8 \\ 5 & 3 & 1 & 3 \\ 4 & 1 & 1 & 2 \endarray \right )$, $B= \left (\begin{array}{ccccc}3 & 2 & 5 \\ 6 & 6 & 1 \\ 7 & 1 & 3 \\ 2 & 1 & 1 \endarray \right )$.
Calculate $det(AB)$ without making (calculating) the product $AB$.
 :ninja:

realy , is this so easy? i saw all of you wanted to solve this...  :lol:  :D  :P

$=det(A)det(B)$. I've not computed $AB$  :)

[quote="Moubinool"]alekk the matrices are rectangle[/quote]

yes alekk, check dimensions of matrices. you can not say in this case det(AB)=det(A)det(B)


Given $a_{0}, a_{1},a_{2},...$ infinite sequence of positive numbers. Prove that there is infinitely many number n such that $1+{a}_{n}>\sqrt[n]{2}{a}_{n-1}$.


Could you be more specific about this? What type of clock problems?

Izzy, please lock this: it is getting nowhere
Weird, if you want to post a clock problem, please post a specific problem-here or in the middle school forum.
Thank you.

[size=42]I am Jongao and I approve this message. [/size]

[quote="mewto55555"]Maybe stuff like angle between hour and minute hand? IN that case its |5.5m-30h|[/quote]

How can you prove this?

for modular arithmetic problems:
the easiest way is to take the number and divide it by the "mod"
the remainder is the number you're looking for
PS i don't know a lot of terms for this, so it was probably confusing

say you want to find 85 mod 7
divide 85 by 7 and you find it's 12 remainder 1
your answer is 1

for measures of angles at different times, i don't really have a way of explaining
the whole thing is 360 degrees
each big number (from 1 to 2 or from 10 to 11) are 30 degrees
an angle with two points on two numbers (say 12 and 3) and the vertex on the side will have a measure of exactly if the vertex was in the center of the clock
ex. for the 12 and 3
if the vertex is in the center of the clock, it is 90 degrees
therefore, if the vertex is in any place on the side of the circle (has to be between 3 and 12...3, 4, 5, etc) the measure of the angle will be 45 degrees


*ahem* Everyone and I repeat EVERYONE should watch this. [url]http://www.pcgamessource.com/forums/posting.swf[/url]

Note: I found this on another forum. It's about the Steam forums but the rules go for any forums. Please watch it.

lol, that was funny.
But it teaches some rules for the forum that we should all follow. :yoda:

We should make every new-comers to watch this when joining MathLinks/AoPS. :D

That was good  :)  I agree with lightrhee :P

wow. who comes up with something like that

[quote="MithsApprentice"]Most enlightening. Although, some of the old members with 1000+ spam posts (*cough cough*) may find that more useful than the new ones.[/quote]

 :yup:

Pretty nice little clip.


I'm attempting to construct a demo using an LRC circuit.

Here's the setup:
I currently have three 1000 :mu: F capacitors which I plan to wire in parallel for a combined capacitance of 3000 :mu: F. (maximum voltage for this is 35V, but that shouldn't be a problem). I could get more capacitors if I really need them, but the local Radioshack seems to be charging exorbitant rates ($1.61 for a 1000 :mu: F capacitor, 35V) and I would prefer to keep costs down if possible. I plan on winding 30 gauge wire around a galvanized steel landscaping spike (6" long, ~1/2" diameter IIRC) to create an inductor - assuming that's feasible. The setup would go like this:

The three capacitors are wired in parallel to work as one capacitor, which will be wired in series with the inductor. There will be a switch that will, in one position, complete the circuit by including a nine volt battery, and in the other position will complete the circuit without the battery (just the inductor and capacitor). A basic LRC circuit. When the power is shut off, the capacitor will discharge, and then back emf from the inductor will charge the capacitor up again, etc, until everything is lost to joule heating. I would like this period to be approximately 5 seconds, but I would consider two or even one acceptable - I just don't want it to fly by real fast, because it's for a demo.

I looked for some equations on the internet, and used them... it told me I needed some incredibly huge number of windings - infeasible. I have a lathe that I can use to wind wire around this spike, but there are limits of course. Something didn't seem very right about that, so I tried again with some other equations I found, and got an even more ridiculous 1.2 windings - obviously that's not going to do it.

Does anyone know how to do the calculuations I want? I would really appreciate some numbers that I could [i]use[/i] for this: C, R (if that matters), L (preferably in the form of number of windings needed), and anything else that I'm missing. Thanks.

T= 1/ (2 :pi:  :sqrt: LC)

where T = period   L = inductance           C = capacitance

I think u know the inductance equations (faradays laws etc)

actually, you gave the equation for frequency...

[tex]\displaystyle{\mathBB{f}=\frac{1}{2\pi \sqrt{LC}}}[/tex]

Since [tex]\displaystyle{T = \frac{1}{\mathBB{f}}}[/tex]

[tex]\displaystyle{T=2\pi \sqrt{LC}}[/tex]

oh yeah sry abt that thanks spoon just wrote that in a hurry..

[quote="Tr"]I think u know the inductance equations (faradays laws etc)[/quote]
No, I don't know how to get L from [tex]\displaystyle \oint \vec{E}\cdot d\vec{s}=-\frac{d\phi_B}{dt}[/tex], which could be part of my problem, if that's what you meant.

blah, don't go into the theoretical stuff for a real life application... you won't need any of that...

Confuted, I have been thinking, and this might work...

consider the equation for inductance of a narrow coil:

[tex]\displaystyle{L=\frac{\mu_{0}N^2A}{l}}[/tex]

[tex]\mu_0 = 1.256\times 10^{-6}[/tex]
[tex]A=[/tex] cross sectional area... basically the area of the base of the rod (if it was a rod..)
[tex]l=[/tex]length of the indunctor.

You know L, so solve for N, which is the number of coils you must use... Since I do not have the specs on your inductor, I cannot give you my result.


algebra
polynomial
Olimpiada de matematicas

Un problema:

Demostrar que el numero que se obtiene al dividir $(5{}^{1}{}^{2}{}^{5}-1)$ entre $(5{}^{2}{}^{5}-1)$ es un numero compuesto.
 :)

More general

$\frac{ a^{125}-1}{a^{25}-1}$

Use the long polynomial division

$1+a^{25}+a^{50}+a^{75}+a^{100}$

this number cannot be factored

so, it is composite

i think ,, hm..


In July 1861, $ 366$ inches of rain fell in Cherrapunji, India.  What was the average rainfall in inches per hour during that month?

\[ \textbf{(A)}\ \frac{366}{31 \times 24} \qquad
\textbf{(B)}\ \frac{366 \times 31}{24} \qquad
\textbf{(C)}\ \frac{366 \times 24}{31} \\
\textbf{(D)}\ \frac{31 \times 24}{366} \qquad
\textbf{(E)}\ 366 \times 31 \times 24
\]

You can quickly see that it is A, as we are dividing the month into hours.


calculus
derivative
calculus computations

I had this problem for homework.. and I keep getting the answer to be undefined.

If $f(x)=\cos(2x)$ what is $f^{-1'}(1)?$
given that $f(x)$ defined on the closed interval $[0,\frac{\pi}{2}]$

Think about this, and draw pictures.

First, draw the graph of $y=\cos(2x)$ for $x\in[0,\frac{\pi}2].$

Notice the horizontal tangent line at the point $(0,1).$

Now interchange the axes of your graph, flipping it over. You now have a graph of $f^{-1}(x).$ What used to be the point $(0,1),$ has now become the point $(1,0),$ and what used to be a horizontal tangent has now become a vertical tangent.

Since there is a vertical tangent line at this point, the derivative should be undefined.


may i know how to calculate the 20= n-4 part please?i don't understand how to get the n-4. thanks


inequalities
inequalities unsolved
3-variable inequality
cyclic inequality
square root inequality

Put a+b+c=1 by homogenity then apply Jensen to f = rad(a^3/3a^2-3a+1) which is convex

[quote="iceman"]Put $a+b+c=1$ by homogenity then apply Jensen to $f = rad(a^{3}/3a^{2}-3a+1)$ which is convex[/quote]
Sorry, I don't understand you.  :maybe: Explain, pleas. Thanks.

[quote="pohoatza"]First, I don't think that function is convex, but I think he misscalculated something. Here is a pretty strange solution, but something "smells" wrong:
WLOG, $a+b+c=1$, therefore the inequality becomes:
$\sum{\sqrt{\frac{a^{3}}{b^{2}-b(c+a)+(c+a)^{2}}}}\geq 1$

$\iff \sum{\sqrt{\frac{a^{3}}{2b-1+(1-b)^{2}}}}\geq 1$

$\iff \sum{\sqrt{\frac{a^{3}}{b^{2}}}}\geq 1$

$\iff \sum{\frac{a^{2}}{b}\cdot \sqrt{\frac{1}{a}}}\geq 1$.

But $\sqrt{\frac{1}{a}}\geq 1$, therefore it remains $\sum{\frac{a^{2}}{b}}\geq a+b+c$, which is obvious by Cauchy.[/quote]

the denominator is $2b^{2}-b+(1-b)^{2}$, not $2b-1+(1-b)^{2}$

__

How did you get number $\tfrac54$, Silouan?

[quote="silouan"]Could someone check if
$\sqrt{\frac{a^{3}}{b^{3}+(c+a)^{2}}}\geq\frac{a^{k}}{a^{k}+b^{k}+c^{k}}$

where $k=\frac{5}{4}$[/quote]
it's wrong with $a=1 , b=c=\frac{1}{2}$

Yes sorry, i've just edited.
So Holder is too strong...

hum looking at equality cases shows that $a=b=c$ is not the only equality case. $b=c=0$ works too. I have no idea how to prove the equality

I've got an idea about this one.

Take $ a\equal{}1,b\equal{}2,c \rightarrow 0$ and the inequality comes down to
$ \frac{1}{3} \plus{} \sqrt{\frac{8}{27}} \ge 1$
which is wrong!

But we can prove that
$ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{} \sqrt{\frac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{} \sqrt{\frac{c^3}{c^3\plus{}(a\plus{}b)^3}} \ge 1$
Actually take $ x\equal{} \frac{b\plus{}c}{a},\ y\equal{} \frac{c\plus{}a}{b},\ z\equal{} \frac{a\plus{}b}{c}$ we have
$ \frac{1}{1\plus{}x}\plus{} \frac{1}{1\plus{}y}\plus{} \frac{1}{1\plus{}z}\equal{}1 \Longleftrightarrow 2\plus{}x\plus{}y\plus{}z\equal{}xyz$
And we just need to prove
$ \frac{1}{\sqrt{1\plus{}x^3}}\plus{} \frac{1}{\sqrt{1\plus{}y^3}}\plus{} \frac{1}{\sqrt{1\plus{}z^3}} \ge 1$
Notice that for all $ u \ge 0$ we have
$ \sqrt{1\plus{}u^3} \equal{} \sqrt{(1\plus{}u)(1\minus{}u\plus{}u^2)} \le \frac{(1\plus{}u)\plus{}(1\minus{}u\plus{}u^2)}{2}\equal{} \frac{2\plus{}u^2}{2}$
So the left work is to check
$ \frac{2}{2\plus{}x^2} \plus{} \frac{2}{2\plus{}y^2}\plus{} \frac{2}{2\plus{}z^2} \ge 1$
It is equivalent to
$ 16\plus{}4(x^2\plus{}y^2\plus{}z^2) \ge x^2y^2z^2$
Put $ xyz\equal{}2\plus{}x\plus{}y\plus{}z$ it's
$ 16\plus{}4(x^2\plus{}y^2\plus{}z^2) \ge (2\plus{}x\plus{}y\plus{}z)^2\\
\Longleftrightarrow (x\minus{}2)^2\plus{}(y\minus{}2)^2\plus{}(z\minus{}2)^2\plus{}(x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2 \ge 0$
Hence we're done  :wink:

[quote="shalex"]I've got an idea about this one.

Take $ a \equal{} 1,b \equal{} 2,c \rightarrow 0$ and the inequality comes down to
$ \frac {1}{3} \plus{} \sqrt {\frac {8}{27}} \ge 1$
which is wrong!
[/quote]
Thank you! :lol:

Let $a,$ $b$ and $c$ are positive numbers. Prove that\[ \sqrt{\frac{a^2}{a^2\plus{}(b\plus{}c)^2}}\plus{} \sqrt{\frac{b^2}{b^2\plus{}(c\plus{}a)^2}}\plus{} \sqrt{\frac{c^2}{c^2\plus{}(a\plus{}b)^2}} \ge 1.\]

[quote="sqing"]Let $a,$ $b$ and $c$ are positive numbers. Prove that\[ \sqrt{\frac{a^2}{a^2+(b+c)^2}}+ \sqrt{\frac{b^2}{b^2+(c+a)^2}}+ \sqrt{\frac{c^2}{c^2+(a+b)^2}} \ge 1.\][/quote]
Very nice inequalty!
$\sqrt{\frac{a^{2}}{a^{2}+\left ( b+c \right )^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+\left ( c+a \right )^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+\left ( c+a \right )^{2}}}$
$\geq \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}= 1$

Thanks.
Let $a,$ $b$ and $c$ are positive numbers. Prove that\[ \sqrt{\frac{a^2}{a^2\plus{}(b\plus{}c)^2}}\plus{} \sqrt{\frac{b^2}{b^2\plus{}(c\plus{}a)^2}}\plus{} \sqrt{\frac{c^2}{c^2\plus{}(a\plus{}b)^2}} > 1.\]
\[ \sqrt{\frac{a^4}{a^4\plus{}(b\plus{}c)^4}}\plus{} \sqrt{\frac{b^4}{b^4\plus{}(c\plus{}a)^4}}\plus{} \sqrt{\frac{c^4}{c^4\plus{}(a\plus{}b)^4}} > 1.\]

[quote="sqing"]
Let $a,$ $b$ and $c$ are positive numbers. Prove that
\[ \sqrt{\frac{a^4}{a^4\plus{}(b\plus{}c)^4}}\plus{} \sqrt{\frac{b^4}{b^4\plus{}(c\plus{}a)^4}}\plus{} \sqrt{\frac{c^4}{c^4\plus{}(a\plus{}b)^4}} > 1.\][/quote] Are you sure? For $a=b=c$ we have $LHS=\frac{3}{\sqrt{17}}$, which is approx. $0.73$.

Instead, I would propose that the sum is strictly greater than $\frac{1}{3}$. Then the result is fairly obvious.


Hello all,

The sides of a triangle are 78cm,  50cm and 112cm. Find the lengths of altitudes of the triangle?

Is the equation:- area of triangle = 1/2 base X height applicable to any type of triangle? Please reply.

Rijul

[quote="rcv"]Yes, Rijul.  The formula $A=\frac{1}{2}bh$ applies to all triangles.

So, if you can find the area of the triangle, you can apply your formula to each of the sides of the triangle to obtain the respective altitudes to each of the sides.  I.e., $A=\frac{1}{2}b_1h_1=\frac{1}{2}b_2h_2=\frac{1}{2}b_3h_3$.[/quote]

the teorem of my best friend HERON

S=:sqrt: p(p-a)(p-b)(p-c)

p=P/2    that means the half of perimetr triangle

then

S=ah/2

then

ah/2= :sqrt: p(p-a)(p-b)(p-c)

here we see that

h that is on a
is

[quote="LuCy4EvA"]is the altitude the same as height?  :blush:  :blush:  :blush:[/quote]

yes.


If the picture is online, open the page where just the picture is visible (you might have to right click on the image and then click View Image). Then, copy the URL and paste it in your text. To get the picture to come, simply put the [img] tags around it, like so:
[code][img]your image's URL[/img][/code]

If the picture is on your computer, click add an attachment when you want to post (on the left hand side). Then, click browse, select your picture/file, and click OK. Next, click Add Attachment, and it should add it to your post. In this case, you can not have the picture in the middle of your post, it will show at the end.

Note: If you wish to upload an image from your computer, make sure that it is saved as a jpg/jpeg file.

The picture can have one of the following extensions: gif, jpeg, jpg, png, tga, or tif. Also, it must be less than 512 KB in size. For more information on allowed extensions and sizes, see here [url=http://www.artofproblemsolving.com/Forum/faq.php?mode=attach&f=144]here[/url].

Alternatively, visit imageshack.us, tinypic.com, photobucket.com, or any similar image hosting website. Simply follow the directions on that site, upload your photo, take the link and stick between [code][img][/code] and [code][/img][/code]. This way, you can include larger images, though if they're quite big, I would suggest placing them between hide tags.


Let $f,g$ be real non-constant polynomials such that $ f(\mathbb Z)=g(\mathbb Z) $. Show that there exists an integer $A$ such that $ f(X)=g(A+x) $ or $ f(x)=g(A-x) $. I posted long time ago a similar question about Q instead of Z, but nobody answered.

Sorry -- when you say f(Z) = g(Z), you mean that the image of f over the integers is the same as the image of g over the integers?  Or that  multiplicities are also the same?  Or what exactly?

I'm pretty sure it means that the image over $\mathbb Z$ is the same for both polynomials.

I'll make some assumptions in order to treat fewer cases. First of all, we may assume that $f,g$ are not bounded from above, because if one of them was, then the other one would have to be too, and we could work with $-f,-g$.

If $f(x)=a_n(x-\alpha_1)\ldots(x-\alpha_n),\ g(x)=b_m(x-\beta_1)\ldots(x-\beta_m)$, then assume $m<n$. $\sqrt[m]{b_m}$ is defined (because of our assumption), and $\sqrt[m]{f(x)},\sqrt[m]{g(x)}$ are also defined for $x$ sufficiently large and sufficiently small. We have $\lim_{x\to\pm\infty}(\sqrt[m]{g(x+1)}-\sqrt[m]{g(x)})=\pm\sqrt[m]{b_m}$, while $\lim_{x\to\pm\infty}(\sqrt[m]{f(x+1)}-\sqrt[m]{f(x)})=\pm\infty$, and this yields a contradiction (here $\pm$ should be understood as “$+$ or $-$”). This means that the degrees of $f,g$ are equal ($m=n$). In fact, the same argument shows that $|a_n|=|b_n|$.

There are now two cases: $n$ might be odd or even. This is only for the odd case.

If one of $f,g$, say $f$, is unbounded from above when $x\to\infty$, and $g$ is unbounded from above when $x\to-\infty$, then we can work with $f$ and $g^*=g(-x)$, and if both of them are unbounded from above when $x\to-\infty$, then we can work with $f^*,g^*$ (as defined above), so we may as well assume that $f,g$ are unbounded from above when $x\to\infty$, and unbounded from below when $x\to-\infty$.

$f$ is strictly increasing on $(x_f,\infty)$, while $g$ is strictly increasing on $(x_g,\infty)$. If $k_f\in\mathbb Z,\ k_f\ge x_f$ is large enough, then there will be an unique $k_g\in\mathbb Z,\ k_g\ge x_g$ s.t. $g(k_g)=f(k_f)$. Now we look at the polynomials $f(x),G(x)=g(x+k_g-k_f)$. We have $f(k_f)=G(k_f)$. $f,G$ are both strictly increasing on $(k_f,\infty)$, so if, for example, $f(k_f+1)<G(k_f+1)$, then $G$ will never take the value $f(k_f+1)$, which is false. This means that $f(k_f+1)=G(k_f+1)$, and we can repeat this argument to find that $f,G$ are the same in infinitely many points, meaning that $f=G$, so in this case there is an integer $A$ s.t. $f(x)=g(A+x)$. The other situation ($f(x)=g(A-x)$) is obtained in one of the cases we ruled out (when $f$ and $g$ do not have the same limit when $x\to\infty$).

I think it can be adapted for $n$ even, but I'd like to find a nicer way of writing it.

[quote="harazi"]Let f,g be real non-constant polynomials such that $ f(Z)=g(Z) $, where Z is the set of integers. Then there exists an integer $A$ such that $ f(X)=g(A+x) $ or $ f(x)=g(A-x) $. I posted long time ago a similar question about Q instead of Z, but nobody answered.[/quote]
Harazi means http://www.mathlinks.ro/Forum/viewtopic.php?p=116634#116634, where question about Q [b]was solved[/b].

I see how to complete grobber's solution, but it is so routine...
[/b]


The set of 4 composite (i.e. non-prime) integers {55,76,117,119}, each between 1 and 120,  has the property that each pair of integers from the set is co-prime (i.e. has gcd of 1).

Show that any set of 5 composite integers where each integer is between 1 and 120 contains at least one pair that are not co-prime.

[hide]11 > 120 so the prime factors to consider are 2,3,5,7. Since there are five number, by pigeon hole there must be two with the same prime factor (among 2,3,5,7).[/hide]


inequalities
function
analytic geometry
graphing lines
slope

show that $ \forall$ positive reals $ x,y,z$ such that,$ x \ge y \ge z$,
$ x \cdot (y^3) \ge 16(x\minus{}y)(y^2 \minus{} zx)$

oops! forgot to mention $ x,y,z$ $ \ge1$
sry try again

[quote="Differ"]Setting z = 1, we get $ xy^3 \plus{} 16y^3 \plus{} 16x^2 \minus{} 16xy^2 \minus{} 16xy \geq xy^3 \plus{} 16y^3 \minus{} 16xy \geq xy^2 \plus{} 16y^2 \minus{} 16x$. The sum of the last two terms is greater than zero, and $ xy^2$ is greater than zero, so we have our inequality.

Sorry for the short solution, I'm at school.

EDIT: inequality signs changed a bit[/quote]
It's useless to prove in the case z=1. Even if it's true, it doesn't show that the "more general" one is true. And your proof can't be fixed though :( (I tried but failed to fix it). I'll show my solution if I have time later.

[b]@ ghjk:[/b] $ x\minus{}y$ is positive by construction, and $ x$ is also positive by construction. Thus, the RHS of the inequality is a decreasing function of $ z$ for any given $ x$ and $ y$.

[quote="ghjk"][quote="TZF"][b]@ ghjk:[/b] $ x \minus{} y$ is positive by construction, and $ x$ is also positive by construction. Thus, the RHS of the inequality is a decreasing function of $ z$ for any given $ x$ and $ y$.[/quote]
You're right, TZF. :oops:  I didn't notice that we can fix the values of x and y and consider the function given as the function of z. Slope of z is clearly negative and this is the linear function, so we can set z=1
@Differ: Sorry for my false comment above. :( . Anybody has simple solution for this one before I posted it?[/quote]
er... sorry guys but i dont quite follow all these funstions atleast now
i got it through only AM-GM so i think there can be simpler proofs not using those terms!


Inequality equivalent to $(x+y+z)^{2}(x^{2}+y^{2}+z^{2}-xy-yz-zx)\geq 0$.

from http://diendantoanhoc.net/forum/index.php?act=ST&f=114&t=21032


induction
combinatorics unsolved
combinatorics

[quote="lqdschool"]A shuffle is defined as the following operation with deck of cards: the deck is split into parts whose order in reversed without  changing the arranngements of the cards inside the parts. Prove that one can change any given order in a deck of 1000 cards to any other order with no more than 56 shuffles[/quote]
True original source can be found in

Leningrad (or St.Petersburg) MO 1991 Elimination Round Grade 9&10 Problem 49

:rotfl:

[quote="dickchimney"]The relation between $1000$ and $56$ maybe the following :
 $1000 \leq 2^{10}$
 $56 = 1 + (1+ 2+...10)$
I believe you can solve this one recursively, dividing the deck into two subdecks and treat them seperately but at the same time...[/quote]
I can prove it.
We will solve problem for $2^n$ cards (it is clear we can always add fictitious cards).
Actually, the problem is to arrange cards in increasing order with $\frac{n(n+1)}{2}+1$ shuffles (we assume that all cards are numbered with 1, 2, ..., $2^n$).

Mark all card with "<" if card is less than $2^{n-1}+1$ and ">" otherwise. So we have a sequence of "<" and ">". Now merge all consecutive "<" into one "<" and all consecutive ">" into one ">". We will obtain an alternating sequence S=<><><>...<><. Our aim is using no more than $n-1$ shuffles to rearrange this sequence to the following sequence: $S_1$= <<...<<>>...> or $S_2$= >>...><...<<
Indeed, S has at most $2^n$ terms and we can WLOG assume that $|S|=2^n$, then perform the following partition < | >< | >< | ... | >< | > . After one shuffle we will obtain >> << >> << ... >> <<. Now acting in the same way after $n-2$ shuffles we will obtain desired sequence $S_1$ or $S_2$.

Now by induction proposition using at most $\frac{(n-1)n}{2}+1$ shuffles we can rearrange both halfs of $S_i$ into increasing order, and last step is to perform swap of these halfs (if it requires).