probability calculus derivative function

Hi,
I've been in a Financial Engineering PhD program at a tier 1 US university for about a year now. The mathematics is very interesting and is a good intersection of PDE, Probability, and CS. However, whenever I see the numerical application of my work in Bloomberg, I can't help but feel that all I'm doing is some how help a fund get better at gambling on the stock market. Then when this fund gets better at gambling, the other firms have to hire better quants to improve their models. It's like a "modeling arms race". When I look back at history in the longer run, I can' explain to myself how has the advances in quantitative finance in the last 25 years made the world a better place.

Sorry if this seems rude, but it's been seriously bothering me for quite a while now. My alternative would be to pursue something in Operations Research or switch into Mechanical or Electrical Engineering. My understanding is that high level Mechanics is quite math heavy too?

John

[quote="John622"]Hi,
When I look back at history in the longer run, I can' explain to myself how has the advances in quantitative finance in the last 25 years made the world a better place.
[/quote]
Honestly, I can't explain it either.  I have always considered engineering and computer science as much more beneficial than financial mathematics to the society as a whole, here I completely agree with John622. On the other hand, it is a well-paid job with some interesting problems to solve (though, I strongly prefer to think of the problems I find interesting as opposed to those someone else finds important, that's why I chose the pure academic career), so many people good at mathematics like it. This is not to discourage anybody to go to finance (everyone is entitled to his own career choice) but just to let John622 know that he's not alone in his doubts. :)

I have been thinking about the same question lately. I'm just finishing my junior year of undergrad, and I'll have the opportunity to work for a good hedge fund after I graduate if I want (probably), so I have to decide if I want to do that or if I should be applying to graduate schools.

I can only think of one "social" justification for the work that hedge funds do. They make markets more efficient. The whole point of working for a hedge fund is trying to identify where the market is inefficient and profit from it; this will remove the inefficiency. (Personally, I think the fact that there's uncertainty in this process does NOT make it the same as gambling.) Essentially, hedge funds provide a service of improving market stability and providing liquidity.

But even if you convince yourself that hedge funds are actually successful at providing these services reasonably reliably, it's such a indirect and small contribution to society that it's hardly justification for picking quant finance over a different industry.

(I really don't think that making the investors richer is a particularly strong moral justification either, even if you are an adamant believer in the capitalist system.)

For me, the only possibly justification for working in finance is that I have some ambitious goals in life (regarding contribution to society), and I don't think that directly getting a PhD and finding a research job will put me in a position to achieve those goals. So I want to work in finance in order to give myself a platform for the stuff I want to do later on.

And as fedja said, ignoring the moral question you raised, financial math is pretty interesting. I really liked the stochastic analysis class I took. So it shouldn't be an unpleasant job.

I was talking to some other students about it and found some justifications:

1. The "trading" portion of financial math helps to make the market more efficient by removing arbitrage opportunities.
2. The risk management, derivative valuation portion helps investors to better understand the risk they are taking.  There will always be risk involved in finance, but a better understanding of what the risk is allows people to make more informed decisions.  A lack of understanding of CDOs contributed to the recent subprime meltdown.

I think the second point is the clinching argument.  That financial math is not a "crystal ball" for predicting the stock market, but rather it helps investors take more informed risks.  So it contributes information about investment.

The first point mainly applies to algorithmic trading/ quant trading types.

The macroeconomic argument could be made that quantitative financial analysis contributes to the understanding of how financial markets bahave. This in turn allows society to make better judgements about what to do with its capital, investments, and savings. It also causes greater efficiency in markets, and causes greater national/global productivity and return on investments. All of these are things that ultimately fuel progress and invention, and (arguably) improve society as a whole.

Whoops, just saw that you came to the same conclusions as I did. I don't think as Xevarion does that these are small contributions. Financial markets play a very direct and very important role in the economy as a whole and in people's lives. They effect everything from pension funds, to housing markets, to interest rates.

[quote="John622"]
2. The risk management, derivative valuation portion helps investors to better understand the risk they are taking.  There will always be risk involved in finance, but a better understanding of what the risk is allows people to make more informed decisions.  A lack of understanding of CDOs contributed to the recent subprime meltdown.
[/quote]
Alas, that's only partially true. As far as I understand it, the recent subprime meltdown was due more to the fact that people just weren't able to pay what they had to pay according to their mortgage and loan contracts than to the fact that computations and models were done badly. Of course, educating people about what they can and what they cannot afford would be a great thing, but that's not what banks, investment firms, and hedge funds have in their mind. Their task is quite opposite: to lure people into giving as much of their money to them as possible. So they have no real incentive to talk about the risks involved. It is like with credit cards: you are formally praised for your good credit rating and paying on time but what the banks would really benefit from most and what they hope for is that you put yourself in exactly such a debt that you would have to spend all your free money on just paying the interest. This way of thinking was behind such monstrous  inventions as "interest only mortgage", etc.

Basically, the investment market is assumed to be a mechanism that allows to redistribute free money so that the more successful and useful for the society competitors will get more financing and the inept and useless ones will be financially starved to death. As long as you stick to trading stocks, that more or less works. But options and other derivatives seem to be of much less importance for this particular purpose and are mainly used for large scale speculations that have a negative overall impact on the stability of the economy. And a lion share of the current hedge fund profits (and losses :P) is derived from manipulating various derivatives.

Of course, this is just my personal opinion and I am by no means an expert. It is an interesting question whether there is any correlation between the trading strategies aimed at maximizing the profit of individual investment firms (that's what you'll have to design as a quant) and the trading strategies that improve the overall performance of the market economy (that is what the society would benefit from). I do not think that the answer is simple here.

Anyway, as Xevarion said, it is an interesting job. Just be aware that, despite all possible "justifications" aimed at presenting it as "useful for the society" (which may be true or false, it doesn't matter here), it exists primarily because of the "natural" desire of "normal" people to get richer by any means (mathematical modelling seems to help so far, so they employ mathematicians, but if there were any indication that looking into cristal balls and other witchcraft tricks might be of use, you would have several professional witches in the staff as well...).

a couple of examples:

the pensions of union members may be invested in a hedge fund or investment bank.  quants are (at a certain level) responsible for the steady growth of said workers' retirement fund.

foreign companies (say, biotechnology companies) may be working with a non-us or non-euro currency.  the currency risk that they incur can be mitigated by the work done by quants, allowing the aforementioned companies the best possible chance of completing products which can raise people's standards of living.

gauss202 also raises some good points.

[quote="blahblahblah"]the pensions of union members may be invested in a hedge fund or investment bank.  quants are (at a certain level) responsible for the steady growth of said workers' retirement fund.[/quote]
Sounds nice and noble until you ask the question where this growth comes from. There is a certain part of it that is delivered by the overall development of the economy (plus adjustment for the inflation), but you do not need quants to catch that part: any decently diversified index fund would do. The performance of the quant is judged by how his ideas allow to outperform the market. Now, the only way for a quant to outperform the market is to get money from other investors by outsmarting them in trading. So, basically, quants are in charge of redistributing the wealth rather than of creating it. It is nice if the result will be taking some money from greedy millioneers and passing it to the worker's pensions but it is equally likely that it will be exactly the opposite (quants do not have much control on who invests in the fund they are working for; usually they don't even know that). So, the picture is quite complicated here. I suspect that the overall balance is exactly $ 0$.

The claim that the knowledge allows to mitigate the market is true, no doubt. But this mitigation comes as a side effect. The quants aren't paid according to how much the stock (or option) price volatility decreased over the last 5 years.

Actually, my main point is not that financial mathematics is useless per se. This would be ridiculous to claim. Any decent attempt to understand the nature and mechanisms of some process (be it stock market, weather, chemical reaction, or whatever else) will benefit the society. Being done in a purely academic setting with the main goal of acquiring and disseminating the knowledge, it is as noble and useful as the work of physicists and other natural scientists. The problem is that it is not quite what is expected of a quant working for an investment firm. Every time the knowledge is made secret and used for the sole purpose of getting richer, you may expect that the chances of using it for good and for evil causes are at least equal. You can help in the ways blahblahblah brought up but you can make an equal amount of harm just as easily.

I think perhaps we should look at the role of hedge funds as a larger system of gas particles rather than talking about the contribution to society that a SINGLE hedge fund makes.  I think the SYSTEM of investors and hedgies contribute to the market by

1) Driving prices to the correct value.  If prices are more "correct" i.e. in line with their demand, then the resources in society will be more quickly distributed to the right areas.  If there is a lag and things are out of price, then the resource adjustment won't happen as efficiently.  For example, if some how 286 computers are still worth \$1000 a piece, then all the major companies will suddenly only produce 286 that nobody needs.

2) They invent new derivatives and products that better meet the risk profile of investors.  For example, by having things like basket default swaps, people can "isolate" the credit risk of a company but not bet on the operational risk of the company.

Now, no investment bank and hedge funds are founded to serve these two principles.  What's amazing is that by pursuing their own greed, they are in fact performing those two tasks as a SYSTEM.  Every win and loss means an adjustment to the market system.  It's hard to explain what contribution one single hedge fund makes by winning 5 billion dollars on writing credit default swaps, but together, all the hedge funds synergetically function to make the market function.
The analogy is if you are a car part manufacturer, it's hard to see how this one part contributes to society outside the context of it being part of a car, but you have to realize that this car part manufacturer is part of a much larger system that contributes to society.

I'm going to take a financial economics course and see if I can learn more.  My first research survey will be on  whether the market/economy has benefited from derivatives and whether profit-seeking strategies in investment is aligned with the improvement of the whole economy.

Also we have to look under the larger theme of "do other professions ultimately contribute to society? or do they merely exists due to imperfections in human society?"
For example, professions like police, locksmith, lawyers ultimately are wasteful.  They only contribution they make is keeping other people in line with current social order.  If people can be more disciplined and adhere to social order without external pressure, then these people can be diverted to more productive jobs.

I think the alternative to free market is communism.  That didn't work out well.

[quote="John622"]If people can be more disciplined and adhere to social order without external pressure, then these people can be diverted to more productive jobs.[/quote]

That's a pretty big "if."  I don't really find the premise of statements like this realistic.  To talk about "imperfections" in human society is to presume that human society is a system with a particular goal and that it contains inefficiencies that make achieving that goal more difficult.  From the viewpoint of achieving a particular goal, the latter might be true, but the more troublesome premise is that human society has a particular goal at all (beyond survival).  (There is a related notion that human society should somehow be redesigned to remove inefficiencies which does not at all take into account how it came to be in the first place nor how difficult it would be to implement a redesign.  I also find this notion absurd.)

I don't have much to say about the original topic of this post.  I just think mathematicians need to be more careful when they start trying to be philosophers :)

I thought mathematicians are philosophers.

I mean that I've noticed students of math have a tendency to talk about the social sciences in similar terms as the terms they use to talk about the physical sciences.  Please let me know if you think I'm being rude; this is merely an observation I've made about a certain type of person.  Have you studied sociology?  Psychology?

[quote="t0rajir0u"]I mean that I've noticed students of math have a tendency to talk about the social sciences in similar terms as the terms they use to talk about the physical sciences. [/quote]
I'm certainly guilty of that too :)  but that's the way we are trained: when naming something a science, we assume that it derives its conclusions from either direct and reproducible experimental observations or from rigorous logical argument within a well-defined model and that those conclusions have predictive power. Everything else is just "not science" (which doesn't mean it is unlimately bad or completely useless; it just doesn't pass the test). Not having this clear border will result in recognizing alchemy, witchcraft, and studying UFO as "sciences" (by the way, did you study witchcraft, t0rrajir0u? :P I bet not, but still you, probably, have quite a firm opinion about its value). A good friend of mine studied psychology quite seriously, and it is a decent science but, quite suprisingly, you can talk of it in the same terms as you would use for physical sciences. You observe human behavior, you try to classify the behavioral types, you design the tests that allow you to determine which type the person belongs to, you make predictions about their behavior, you check your predictions and adjust your models according to this check; what's different here from physics? What social scientists are often guilty of is trying to find the "the ultimate and the only correct point of view" on everything (be it Freudism, class struggle, or whatever else). They see some part of the elephant and claim that that is the only relevant part and those who refuse to believe it should be condemned. The tolerance to alternatives must have its limits of course (I'll refuse to recognize living a few weeks in a museum room with lice in your hair as a form of art no matter what the professional critics might say) but I would prefer the social scientists to be a bit less self-confident, especially the philosophers. I remember fairly well my student years when we were taught Marxism as "the only scientific philosophy" and then, after the collapse of Soviet Union, were told by the same teachers with the same confident expression on their faces that "it wasn't exactly so" (I'd better not go into details now); this, plus several dozen millions of people killed to let some political, philosophical, or religious dogma prevail (and I'm not talking only about 20th century Russia here), plus clear incompetence of most philosophers talking about mathematics, make me suspect that philosophy is not exactly what I'll be willing to embrace as a decent science in the nearest future and that my tendency to try to apply to it the same criteria as to physics may be excusable after all.

Like t0rajir0u, I didn't have an intention to offend anybody here, just to express my point of view and somehow to explain the phenomenon he refered to. Of course, this is only my personal perspective, but I believe that I can qualify as a "typical student of math" :)

As to "human society having a particular goal", I believe it is the same as with an individual human being. You aren't born with a particular goal but, as you grow and become sufficiently aware of yourself and surroundings, you start setting goals beyond your physical survival. The problem seems to be that general awareness of the society is not that high at the moment. But this is starting to become philosophical, so I'll better stop.

It seems to me that the primary focus of psychology these days is offering ways to interpret the results of neuroscience, and it is in that interpretation that I would distinguish the methods of psychology as a social science from neuroscience as a "physical" science.  Behavioral psychology is very empirical, that's true, but there's a lot more to psychology than behavioralism :)

[quote="fedja"]What social scientists are often guilty of is trying to find the "the ultimate and the only correct point of view" on everything (be it Freudism, class struggle, or whatever else). [/quote]

Thank you; this was partly the point I was trying to make :oops:

Quant help in the efficient allocation of capital and other resources.
We are in a period of time where the increase in such efficiency is easily the most massive factor in the improvement of living standards of people on Earth.

We only have to look at huge failed countries like India and China to see what happens if you don't have functioning capital markets.
The socialist fascists who run China are just [b]so[/b] proud that their economy is now 4th in the world, having just over taken Britain. A country with [i]1/20th[/i] the population.
India adopted a less evil form of socialism, and murdered a vastly smaller % of their population, and given time it may get into the top level by size.

Both are growing fast, but that is the relief you get when you stop banging your head against a wall.
When your people get jobs in the global market because they earn vastly less, you have failed, not succeeded.

The biggest marginal return you get from education is learning to read, after that the marginal value only ever goes down. Same with economies.

China gets a large % of it's relative growth simply because the number of university graduates it murders per year is now only 100 times that of a civilised country.
India has reduced it's bureaucracy and now invests nearly enough in education. It's overall efficiency is now nearly as good as the more backward parts of Mexico.

Already we are seeing structural  issues in the Chinese economy driven by the fact that capital allocation is done by political connections not some combination of risk and return.

Of course the credit crunch is a dreadful mess, but western government are responding to it by various economic measures, unlike China which would murder a few thousand people as scapegoats, or India which would let them starve.

[quote="fedja"]So, basically, quants are in charge of redistributing the wealth rather than of creating it. [/quote]

I'm not convinced this is true -- suppose, say, that the involvement of quants in repackaging, securitizing, and reselling student loans creates a much larger market for student loans.  This in turn leads there to be many more student loans available, and at lower rates.  It's hard not to call this wealth creation.  And we haven't even started talking about mortgages, where the work of quants can be arguably said to have dramatically increased people's access to borrow funds to buy houses.  The last 2-3 years notwithstanding, this can also be viewed as wealth creation.

Or, even more simply, by narrowing the spread between the bid and ask, and making more trading possible, quants have created a great deal of wealth.  Trading itself is fundamentally a wealth-creating activity, no matter how hard the protectionists in the US and EU will argue otherwise.

It's not as tangibly wealth creating as making widgets, but it's creating wealth nonetheless.

Dominic Connor,

Your perception of China as a "socialist" country is clearly a sign of your anachronism on the matter.  China was very socialist about 20 years ago, it was a command economy where all production were based on what the party demanded.  Ever since Deng Xiaoping in the late 1980s, China has reformed its economy into one of the biggest free market in the world.  In fact, it is too free.  There are many loopholes and regulation deficiencies which allows the capitalists in China to greedily exploit the working class.  The Chinese market is more capitalist than the United States.  China needs more regulations now to shorten the wealth-poor disparity and add stability to its capital markets.
The "murdering" of university graduates is one of the most, interesting shall I say, that I have heard from China/India bashers so far.  Would you please elaborate who does the murdering and what your sources are?
I don't see how a discussion of quant's contribution to society has quickly led someone like yourself to delve into a tangent of rage against two rising powers: China and India.  It is perhaps advisable for your own survival in the 21st century to have a more correct perception of these two rising states.

Sincerely yours,

China still is highly socialist, you don't have to reduce the amount of socialism in a system to make it vastly less awful.

China is no an open market for capital, that isn't hurting it all that much yet, it will soon.
I am entertained by the notion that more regulation leads to more stability. I appreciate that economic history isn't part of many mats curricula, but what you are seeing is the substitution of many small shocks for a smaller number of bigger ones.

It's true that the people running China have more contempt for the working classes than Americans or Europeans but leaving aside morality that simply is not efficient.

As for state murders of graduates I commend you to Amnesty International, or indeed any news media of any kind. Again, that is not efficient, as the Nazis found when they gassed the people who might have made their atomic bomb in WWII.

Financial markets are an expression of the "freedom" in a culture. It is not coincidence that efficient markets are only to be found in democracies. I came into financial markets when my ethnic group was not only out of favour, but occasionally blowing up banks.

But most of us were nice and useful people, honest... :)
Imagine what would happen if a Tibetan applied for a senior job at a Chinese bank. OF course since he would not have been allowed to go to university that's not likely, is it ?

Again this is not a moral perspective, but an efficiency term. China is a dismally poor country because it does not manage any of its resources well, be they physical, financial or human.
That's what it means to be a rich country, doing what you do well.

[quote="rrusczyk"] It's hard not to call this wealth creation.  And we haven't even started talking about mortgages, where the work of quants can be arguably said to have dramatically increased people's access to borrow funds to buy houses.  The last 2-3 years notwithstanding, this can also be viewed as wealth creation.

Trading itself is fundamentally a wealth-creating activity, no matter how hard the protectionists in the US and EU will argue otherwise.

It's not as tangibly wealth creating as making widgets, but it's creating wealth nonetheless.[/quote]

Rather than putting forward my own arguments, I suggest that you watch
[url]http://www.youtube.com/watch?v=cy-fD78zyvI[/url] ;)

Oh boy, I'm not stepping into that one...

Or John could justify it to himself by saying that the money he will earn being a quant, he will use to better society

John, I am with you on this one :)

Quants benefit the society as much as, for example, professional poker players. Don't buy this "allocation of capital" nonsense...money shuffling isn't allocation of capital.
Venture capitalists are involved in allocation of capital, not quants....

This is from Paul Volcker interview...
http://net.educause.edu/ir/library/pdf/ff0811s.pdf

"Well I found myself sitting next to Bill Sharpe, a Nobel
Prize winner for helping to invent fi nancial engineering. So I nudged him and asked, ”What’s this all about? What does
it do for the GNP or productivity?” He whispered back to
me, ”Nothing.” Which is not the answer that I expected. I
said, “What does it do?” He said, “Well you know, it changes
around all the rents in the fi nancial system and besides, it’s a
lot of intellectual fun.”
Now I don’t know about the intellectual fun, but I’m not
sure that fi nancial engineering has produced much in the rate
of growth or productivity in the real world. We would have
just as much employment without it. But that’s what we’re
spending all our time on. It sure has created some enormous
wealth for relatively few people, including for some universities
I guess. "

[quote="Different"]Quants benefit the society as much as, for example, professional poker players. Don't buy this "allocation of capital" nonsense...money shuffling isn't allocation of capital.
Venture capitalists are involved in allocation of capital, not quants....

This is from Paul Volcker interview...
http://net.educause.edu/ir/library/pdf/ff0811s.pdf

"Well I found myself sitting next to Bill Sharpe, a Nobel
Prize winner for helping to invent financial engineering. So I nudged him and asked, ”What’s this all about? What does
it do for the GNP or productivity?” He whispered back to
me, ”Nothing.” Which is not the answer that I expected. I
said, “What does it do?” He said, “Well you know, it changes
around all the rents in the financial system and besides, it’s a
lot of intellectual fun.”
Now I don’t know about the intellectual fun, but I’m not
sure that financial engineering has produced much in the rate
of growth or productivity in the real world. We would have
just as much employment without it. But that’s what we’re
spending all our time on. It sure has created some enormous
wealth for relatively few people, including for some universities
I guess. "[/quote]

Agree 100%.


I exercised  for more than 7 consecutive days. The distances of the different routes  that I may take are 5km, 7km, or 9km. At the end of the training period, he ran a total of 47km. If I may take only one a route a day, how many times did I run the 9km route?

$ a \equal{}$ number of times the $ 5$ km route was taken.
$ b \equal{}$ number of times the $ 7$ km route was taken.
$ c\equal{}$ number of times the $ 9$ km route was taken.

$ 5a\plus{}7b\plus{}9c\equal{}47$
$ a\plus{}b\plus{}c\equal{}7$

$ c\equal{}7\minus{}(a\plus{}b)$

$ 4a\plus{}2b\equal{}16$

$ 2a\plus{}b\equal{}8$.

$ a\equal{}1, b\equal{}6, c\equal{}0$
$ a\equal{}2, b\equal{}4, c\equal{}1$
$ a\equal{}2, b\equal{}2, c\equal{}2$.

It seems that the answer could be $ 0, 1$ or $ 2$.

The answer that was given is none or 0. How come?

[quote="orangefronted"]I exercised for more than 7 consecutive days.[/quote]
And AndrewTom evaluated the case where $ a\plus{}b\plus{}c\equal{}7$.


linear algebra
matrix
Euler
number theory
greatest common divisor
least common multiple
algorithm

Yeah, matrices is probably the right idea.
You want to know what the group generated by $L=\left[\begin{array}{cc} 0&1\\1&0\end{array}\right] \hbox{ and } R=\left[\begin{array}{cc}1&0\\1&1\end{array}\right]$ is (S is the inverse of R, L is its own inverse).
Firstly every matrix in this group must have determinant 1 or -1.
Consider the matrix $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$. By repeatedly applying R (or S) you can get to $\left[\begin{array}{cc}a\%b&b\\c\%d&d\end{array}\right]$. Then apply L to get $\left[\begin{array}{cc}b&a\%b\\d&c\%d\end{array}\right]$. Repeating this is just Eulers method for finding the gcd, so you will end up with something of the form $\left[\begin{array}{cc}x&0\\y&z\end{array}\right]$, with $x=\gcd (a,b)$. But you know that the determinant is 1 or -1. If $x = z = 1$, that matrix is just $R^y$. If $x = 1 \hbox{ and }z = -1$, applying R (or S) y times gives $\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]$, which is, for example, $LRLR^{-1}LR$. So all matrices with determinant 1 or -1 (with integer entries) are in that group. So if you want to get from X to Y in the original problem, if $X^{-1}Y$ doesn't have integer entries and determinant 1 or -1, there is no solution. Otherwise do the above procedure (and list the sequence of moves backwards).

[quote="galois"]hey
need help with this one
starting from the quadruple
(a,b,c,d).
define transformations
L : (a-b,b,c-d,d)
R : (a+b,b,c+d,d)
S : (b,a,d,c)

what sequence of such operations  should be done to reach (p,q,r,s) from (a,b,c,d) or output as impossible.
given the starting and ending quadruples
i have written the transformations as matrices  and made some progress but cant finish this off

please help[/quote]

I believe number theory / LCM / euclidean algorithm might be helpful in determining impossibility. It's late so I'm going to just leave that as my gut instinct and maybe see if it's actual useful tomorrow (sorry).


trigonometry
geometry
circumcircle
inequalities
geometry unsolved

Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. If $\measuredangle MAC=15^{\circ}$, then find the maximum possible value of the angle $B$.

$1=\frac{MB}{MC}=\frac cb\cdot \frac{\sin (A-15)}{\sin 15}=$ $\frac{\sin C\sin (A-15)}{\sin B\sin 15}\Longleftrightarrow$

$\sin C\sin (A-15)=\sin B\sin 15\Longleftrightarrow$

$\cos (C-A+15)+\cos (B+15)=\sin B\sin 15\Longleftrightarrow$

$\cos (C-A+15)=2\sin 15\cdot \sin B-\cos 15\cdot \cos B.$

$|2\sin 15\cdot \sin B-\cos 15\cdot \cos B|\le 1$. I note $\tan \frac B2=t>0$. Thus

$|t^2\cos 15 +4t\sin 15 -\cos 15|\le t^2+1$. Multily with $4\sin 15$:

$|t^2+4t-1|\le 4\sin 15\cdot (t^2+1)$ a.s.o. $\left( \sin 15=\frac{\sqrt 6 -\sqrt 2}{4}\right) .$

The angle $\angle B$ is maximum when BA is tangent to the circumcircle (O) of the triangle $\triangle AMC$ at the vertex A.

nice , but why

Keeping the segment BC fixed, the circumcircle (O) of the triangle $\triangle AMC$ is also fixed, because the angle $\angle MAC = 15^\circ$ is fixed, and A is somewhere on the major arc MC of this circumcircle. The line BA (with B fixed and A on the major arc MC of fixed (O)) forming the largest possible angle with the fixed chord MC, identical with the line BC, is a tangent from B to the circumcircle (O), the one to the major arc BC. Any line though B forming a larger angle with $MC \equiv BC$ does not have a common point with the major arc MC of (O).

From the very subtle [b]Yetti[/b]'s idea $(\"max A\Longleftrightarrow \widehat {BAM}\equiv \widehat {ACM}\"),$

I note $m(\widehat {BAM})=m(\widehat {ACM})=x\Longrightarrow m(\widehat {ABC})=165^{\circ}-2x$.

Therefore, $1=\frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin x}{\sin 15^{\circ}}=\frac{\sin^2 x}{\sin 15^{\circ} \sin (165^{\circ} -2x)}\Longrightarrow$

$\sin ^2 x=\sin 15^{\circ} \sin (15^{\circ} +2x)\Longleftrightarrow 2\cos 2x=1+\cos (30^{\circ} +2x)\Longleftrightarrow$

$(4-\sqrt 3)\cos 2x+\sin 2x -2=0\ (t=\tan x)\Longleftrightarrow$

$(6-\sqrt 3)t^2-2t-(2-\sqrt 3)=0\Longleftrightarrow t=\frac{\sqrt 3}{3}\Longleftrightarrow x=30^{\circ}\Longleftrightarrow \underline {\overline {\left| B=105^{\circ}\right| }}.$

[quote="yetti"]The angle $\angle B$ is maximum when BA is tangent to the circumcircle (O) of the triangle $\triangle AMC$ at the vertex A. Then $\angle MAB = \angle MCA \equiv \angle C$ and the power of the point B to the circle (O) is $BA^2 = BM \cdot BC = 2MC^2 = \frac{BC^2}{2}$, $BC = BA \sqrt 2$. On the other hand, $BA^2 = (BO + R)(BO + R) = BO^2 - R^2$, where R is the radius of the circle (O). Since $MC = 2R \sin \widehat{MAC} = 2R \sin 15^\circ$,

$BO^2 = BA^2 + R^2 = 2MC^2 + R^2 = (8 \sin^2 15^\circ + 1) R^2 =$

$= (5 - 4 \cos 30^\circ) R^2 = (5 - 2 \sqrt 3)R^2$

From the isosceles triangle $\triangle OMC$ with the angle $\angle MOC = 2 \angle MAC = 30^\circ$, we have $\angle BCO \equiv \angle MCO = 75^\circ$. Using the sine theorem for the triangles $\triangle BAO, \triangle BCO$,

$\sin \widehat {OBA} = \frac{AO}{BO} \sin 90^\circ = \frac{1}{\sqrt{5 - 2 \sqrt 3}} > \frac{1}{\sqrt 2}$

$\sin \widehat {OBC} = \frac{CO}{BO} \sin 75^\circ = \frac{\sqrt{2 + \sqrt 3}}{2 \sqrt{5 - 2 \sqrt 3}} > \frac{1}{\sqrt 2}$

because the 1st inequality is equivalent to

$5 - 2 \sqrt 3 < 2 \Longleftrightarrow\ 3 < 3 \sqrt 3\ \Longleftrightarrow\ 9 < 27$

and the 2nd inequality is equivalent to

$4(5 - 2 \sqrt 3) < 2(2 + \sqrt 3) \Longleftrightarrow\ \ 8 < 5 \sqrt 3\ \Longleftrightarrow\ 64 < 75$

Consequently, $\angle OBA > 45^\circ, \angle OBC > 45^\circ$, the angle $\angle B = \angle OBA + \angle OBC > 90^\circ$ is obtuse and the angle ${\angle C < 90^\circ}$ is acute.  Using the sine theorem for the triangle $\triangle ABC$,

$\sqrt 2 = \frac{BC}{BA} = \frac a c = \frac{\sin \widehat A}{\sin \widehat C} = \frac{\sin(\widehat {MAB} + \widehat {MAC})}{\sin \widehat C} = \frac{\sin (\widehat C + 15^\circ)}{\sin \widehat C}$

Squaring this trigonometric equation for the acute angle $\angle C$,

$2 \sin^2 \widehat C = \sin^2 (\widehat C + 15^\circ)$

$2(1 - \cos 2\widehat C) = 1 - \cos (2\widehat C + 30^\circ)$

Substituting $\cos 60^\circ = \frac 1 2$ and using $\cos \theta = \sin(90^\circ - \theta)$,

$2(\cos 60^\circ - \cos 2\widehat C) = \sin (2\widehat C - 60^\circ)$

$4 \sin (\widehat C + 30^\circ) \sin(\widehat C - 30^\circ) = 2 \sin (\widehat C - 30^\circ) \cos (\widehat C - 30^\circ)$

Either $\sin(\widehat C - 30^\circ) = 0$, $\angle C = 30^\circ$ or

$2 \sin (\widehat C + 30^\circ) = \cos (30^\circ - \widehat C)$

However, this is not possible for an acute angle $\angle C$, because $2 \sin(\widehat C + 30^\circ) > 1$ for $0^\circ < \widehat C < 120^\circ$.  Knowing the angle $\angle C = 30^\circ$, we have $\angle AOM = 2\angle MCA \equiv 2 \angle C = 60^\circ$ and $\angle AOC = \angle AOM + \angle MOA = 90^\circ$. As a result, 3 internal angle of the quadrilateral OABC at the vertices O, A, C are equal to $90^\circ, 90^\circ, 75^\circ$, hence, the remaining angle $\angle B = 105^\circ$.[/quote]

  Here is my solution:

 The answer is in equality BA^2 = BM * BC      which results in  BC = BA * sqrt(2)

  It means that BC is a diagonal of square side BA. In other words,  if we construct on

  BC an isosceles triangle with angle 30 at vertex A'  it's side A'C will be parallel to

  BA.  This makes angle ABC = 90 + 15  :yup:

  Thank you.

  M.T.

[quote="armpist"]
  Here is my solution:

 The answer is in equality BA^2 = BM * BC      which results in  BC = BA * sqrt(2)

  It means that BC is a diagonal of square side BA. In other words,  if we construct on

  BC an isosceles triangle with angle 30 at vertex A'  it's side A'C will be parallel to

  BA.  This makes angle ABC = 90 + 15  :yup:

  Thank you.

  M.T.[/quote]

Saved for posterity.

[u]A generalization of this nice problem.[/u]

$\triangle ABC: \ M\in (BC),\ MB=MC;\ m(\widehat {MAC})=\phi$ ([u]constant[/u]) $\Longrightarrow$

$B_{max}=180^{\circ}-\phi -2C_0$, [u]where[/u] $\tan C_0=\frac{\sin \phi (\sqrt 2 +\cos \phi )}{1+\sin ^2 \phi }.$
[u]A particular cases.[/u]

$1.\ \phi =45^{\circ}\ (C_0=45^{\circ})\Longrightarrow B_{max}=45^{\circ}$.
$2.\ \phi =15^{\circ}\ (C_0=30^{\circ})\Longrightarrow B_{max}=105^{\circ}$.
$3.\ \phi =30^{\circ}\Longrightarrow \tan C_0=\frac{2\sqrt 2+\sqrt 3}{5}.$
$4.\ \phi =90^{\circ}\Longrightarrow \tan C_0=\frac{\sqrt2}{2}.$
$5.\ \phi =135^{\circ}\Longrightarrow \tan C_0 =\frac 13$ a.s.o.

When <ABC is maximum, as many peoples pointed out, AB is tangent to c(AMC) at A.
Then we have that BAM is similar to BCA, so  BA^2/BM^2 = CA^2/BA^2 = 2. If O is the center of the square on AC opposite to B wrt AC, it holds that AO = AM = sqrt(2)AC. But <MAO = 45°+15°= 60° then  AOM is equilateral from which it follows that OM = OA = OC. This means that O is the circumcenter  of c(AMC) then 60° = <AOM = 2 <ACM. And from here is easy to obtain that <ABC = 105°.


algebra
polynomial
limit
algebra unsolved

Let $P_n(x)$ be a polynomial such that :
 $P_1(x)=1+x$
 $P_2(x)=1+2x$
 $P_{2n+1}(x)=P_{2n}(x)+(n-1)xP_{2n-1}(x)$
 $P_{2n+2}(x)=P_{2n+1}(x)+(n+1)xP_{2n}(x)$
 Call $a_n$ the maximum root of $P_{n}(x)$ ($a_n \in R$)
  Prove that : $(a_n)$ is increasing and $\lim\limits_{n\to \infty}a_n =0$

I can prove $a_n$ is increasing but I can't prove $lim a_n =0$
  Can you help me ?


Let $ B_t$ a brownian motion and $ T_1 \equal{} \inf(t \geq 0: B_t \equal{} 1)$ and $ T\equal{}\inf(t \geq T_1 : B_t\equal{}0)$. How do you prove properly that $ T$ is a stopping time ?

I do not know if it is strict enough, but still:
1. $ T_1$ is stopping time: $ \{T_1 < t\} \equal{} \{\sup_{p\in(0,t)}B_p\geq1\}$
2. $ \{T < t\} \equal{} \{\inf_{p\in[T_1,t]}B_p < 0\}\in \mathbb{F}_t$.

I do not consider this very correct. Even the fact that $ T_1$ is a stopping time is a little bit more complicated. You have to use the continuity of the paths of the brownian motion, and consider for example
$ \{T_1 \leq t \}\equal{}( \omega : \inf(t \in \mathbb{Q} \cap [1;t] : |B_t|)  \equal{} 0 )$

Yes, of course you are right  :lol: We have to take suprema over the countable number of points, thus we need continuity of paths.
Finally, after editing the last post two times i think i figured out how it should be:
$ \{T<x\}\equal{}\{\inf_{t\in (T_1,x)}B_t<0\}\equal{}\bigcup_{n\equal{}1}^{\infty}\bigcup_{i\equal{}1}^{n}\left\{\{T_1\in(\frac{(i\minus{}1)x}{n},\frac{ix}{n})\}\cap \{\inf_{t\in(\frac{ix}{n},x)}B_t<0\}\right\}$


function
analytic geometry
induction
graphing lines
slope
algebra unsolved
algebra

It isn't typically algebraical problem, it's only based on a graph of some function, but I don't know in which categhory I should published it ...

Let A and B be two points with the same "y" coordinates in the orthogonal system of coordinates Oxy. Let's imagine that there's a graph of some linear broken function (a polyline) such that no point of this graph has smaller "y" coordinate than points A,B and this polyline starts in A and ends in B. Then in the same moment from both points A, B two dots start to move (not always against each other) so that they would always have the same "y" coordinate. Prove that they can meet.

It's easy to see that the meeting point should be the point with the higher "y" coordinate (or in case that there are more points with this property, any arbitrary one of them). But if we try to solve some cases of the task, we can easily lose in it... It's necessary to take it with bird's eye view that I still miss. Thanks.

I think it's very unclear, man. How exactly do the points move? What do you mean by "not always against each other"? What I understand is that they may change direction however they want, so they could be moving on small curve arcs, one near $A$ and one near $B$ so that they never meet. What exactly does "they can meet" mean? Of course we can [b]make[/b] them meet if we want to, and that's what this tells me (the word "can" is, I think, a bad choice in this situation). Could you explain it again, please?

That's actually an old problem in 2-dimensional space:

* There are two guys A and B, one of them to the left of the Himalaya, one of them
to the right.
* At the beginning, both of them are at the same height H. Say, one of them starts
in (0,H), and the other starts in (100,H).
* They are in walky-talky contact with each other, and this only works if both of them
are at exactly the same height (= have the same y-coordinate).
* The Himalaya is a polygonal line between them.  No point of the Himalaya is below
(= has smaller y-coordinate) the starting height H.

The goal is that the two guys meet each other somewhere in the middle of the
Himalaya.  Allthrough their travelling, they must maintain walky-talky contact.

Prove that this is always possible.

Since the mountain is a polygonal line, I suppose it has finitely many sides, right? In this setting it's easy because we can deal with "peaks" (local maxima) one at a time, and then we can perform induction on the number of peaks. I suppose the problem should still be valid after replacing the polygonal line with an arbitrary continuous graph, and this time it should be much harder.

I think, if we replace the polygonal line with an arbitrary continuous graph then we have a chance to not be able to make any move from endpoints!
And I am not sure about straightforward induction.

I think you're right about not being able to do this when the function is a continuous function, but I really do think induction works on the initial problem.

Assume the first peak on the right is larger than the first peak on the left. Then the point on the left can get over all the peaks on the left which are smaller than the one on the right while the point on the right stays on the first "slope" of the first peak (on the right, of course). When the point on the left gets over the last peak which is smaller than the right peak, then the points are in another initial position with fewer peaks between them.

You don't take into consideration condition "No point of the Himalaya is below
(= has smaller y-coordinate) the starting height H". Do you think this condition does not matter? ;)

I think I'm somehow taking it into consideration implicitly, for if it weren't true, what I said wouldn't work: the point on the left might have to go below $H$ before reaching a peak which is as large as the first peak on the right, and then the point on the right couldn't follow as I assumed it could. This means that I did rely on that condition, I just forgot to mention it :).

The "proof" is flawed in another way: I'm not saying anything about local minima, and these also play an important role. I'll try to "save" it.

I am still unsatisfied with your arguments!  ;)  ;)
When you moved points towards each other they new "H" may be greater than some minimum between them, so you cannot apply induction proposition. Am I right?

That is precisely what I said in the last line of my post :).


Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \sqrt[4]{\frac{8a(b\plus{}c)}{(b\plus{}8c)(c\plus{}8b)}}\plus{}\sqrt[4]{\frac{8b(c\plus{}a)}{(c\plus{}8a)(a\plus{}8c)}}\plus{}\sqrt[4]{\frac{8c(a\plus{}b)}{(a\plus{}8b)(b\plus{}8a)}}\ge2\]

by am-gm,
\[ \frac{(b\plus{}8c)\plus{}(c\plus{}8b)}{2}\geq ((b\plus{}8c)(c\plus{}8b))^{\frac{1}{2}}\]
\[ \implies \frac{1}{((b\plus{}8c)(c\plus{}8b))^{\frac{1}{2}}}\geq \frac{2}{(b\plus{}8c)\plus{}(c\plus{}8b)}\equal{}\frac {2}{9(b\plus{}c)}\]
in the inequality,
\[ \sum\left (\frac{8a(b\plus{}c)}{(b\plus{}8c)(c\plus{}8b)}\right )^\frac{1}{4} \geq \sum\left (\frac{4.8a(b\plus{}c)}{81(b\plus{}c)(b\plus{}c)}\right )^\frac{1}{4}\equal{}\frac{2}{3}\sum\left ( \frac{2a}{b\plus{}c}\right )^{\frac{1}{4}}\]
by am-gm,
\[ \frac{1}{3}\sum\left ( \frac{2a}{b\plus{}c}\right )^{\frac{1}{4}}\geq \left (\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\right )^\frac{1}{12}\geq 1.\]
q.e.d

[quote="phymax"]by am-gm,
\[ \frac {(b \plus{} 8c) \plus{} (c \plus{} 8b)}{2}\geq ((b \plus{} 8c)(c \plus{} 8b))^{\frac {1}{2}}\]

\[ \implies \frac {1}{((b \plus{} 8c)(c \plus{} 8b))^{\frac {1}{2}}}\geq \frac {2}{(b \plus{} 8c) \plus{} (c \plus{} 8b)} \equal{} \frac {2}{9(b \plus{} c)}\]
in the inequality,
\[ \sum\left (\frac {8a(b \plus{} c)}{(b \plus{} 8c)(c \plus{} 8b)}\right )^\frac {1}{4} \geq \sum\left (\frac {4.8a(b \plus{} c)}{81(b \plus{} c)(b \plus{} c)}\right )^\frac {1}{4} \equal{} \frac {2}{3}\sum\left ( \frac {2a}{b \plus{} c}\right )^{\frac {1}{4}}\]
by am-gm,
\[ \frac {1}{3}\sum\left ( \frac {2a}{b \plus{} c}\right )^{\frac {1}{4}}\geq \left (\frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\right )^\frac {1}{12}\geq 1.\]
q.e.d[/quote]

Your last step is wrong .

[quote="manlio"]Nice inequality again unsolved. :([/quote]
I think this inequality is not nice, my dear friend. Anyways, a hint for you:

If $ x$ and $ y$ are nonnegative real numbers such that $ x\plus{}y>0,$ then
$ \sqrt[4]{\frac{8(x^4\plus{}y^4)}{(x^4\plus{}8y^4)(y^4\plus{}8x^4)}} \ge \frac{x\plus{}y}{x^2\plus{}xy\plus{}y^2}.$

[quote="can_hang2007"][quote="manlio"]Nice inequality again unsolved. :([/quote]
I think this inequality is not nice, my dear friend. Anyways, a hint for you:

If $ x$ and $ y$ are nonnegative real numbers such that $ x \plus{} y > 0,$ then
$ \sqrt [4]{\frac {8(x^4 \plus{} y^4)}{(x^4 \plus{} 8y^4)(y^4 \plus{} 8x^4)}} \ge \frac {x \plus{} y}{x^2 \plus{} xy \plus{} y^2}.$[/quote]

Very nice your hint, my dear friend. :)

How do you prove it? By full expansion or some nice trick ?

Thank you very much.


$ a,b,c > 0$  real number

Prove that

$ \frac {(1 \minus{} b)(1 \minus{} bc)}{b(1 \plus{} a)} \plus{} \frac {(1 \minus{} c)(1 \minus{} ca)}{c(1 \plus{} b)} \plus{} \frac {(1 \minus{} a)(1 \minus{} ab)}{a(1 \plus{} c)}\geq0$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255585[/url]


calculus
derivative
function
geometry
geometric transformation
reflection
analytic geometry

[quote="Mr. Moo"]I noticed that it's possible to find a derivative of an inverse function by taking a reciprocal of a derivative of a function(inverse of an inverse function). However, that was just an intuition from merely looking at a result, and I seem incapable of finding any valid proofs for why it is so. Does anyone have any thoughts and ideas for this discovery?[/quote]

I've seen it in Stewart's Calculus text, soon after implicit differentiation is introduced. Funky notation...

The [url=http://en.wikipedia.org/wiki/Inverse_function_theorem]Inverse Function Theorem[/url], single-variable case. Finding the derivative is just a simple calculation with the chain rule; proving that a differentiable inverse exists is what's worthy of the name "Theorem" here.

It's also easy to justify informally with a picture: taking the inverse is reflecting through the line $ y \equal{} x$.  If you reflect a curve through the line $ y \equal{} x$, the tangent lines to the reflected copy will be the reflections of the tangent lines to the original copy.  And reflecting a line through $ y \equal{} x$ changes the slope to the reciprocal of the original slope.

(Note, by the way, that the derivative of the inverse function is the reciprocal of the derivative at a different input value.)


When I view this topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180482

I get this huge stretch of blanks (the post is still there, I just have to scroll down the whole page to see it).

Internet Explorer.

Uh, actually, I do too, and I have Mozilla Firefox 3. In fact, when I scroll down, it's still not there; I can't see it at all.

As a Firefox/Mac user, I see badly done text in $ \text{\LaTeX}$, but no blankness- it's actually missing spaces between words that should be there.

There were some LaTeX errors in the post that caused the page to go wacky.  I think I fixed them.


This is my favorite trig question

Simplify
$ \cos\left(\dfrac{\pi}{7}\right).\cos\left(\dfrac{2\pi}{7}\right).\cos\left(\dfrac{3\pi}{7}\right)$

[quote="Dr Sonnhard Graubner"]hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Sonnhard.[/quote]

There is a very simple solution that does not involve complex roots, but after reading that perhaps this would be an easier question
[hide]
Proove that
$ \cos\left(\dfrac{\pi}{7}\right).\cos\left(\dfrac{2\pi}{7}\right).\cos\left(\dfrac{3\pi}{7}\right) \equal{} \dfrac{1}{8}$[/hide]

This is fairly well known:

Let $ S \equal{} \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7}$ and $ C \equal{} \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$

Now $ 8 S \times C \equal{} (2 \sin \frac{\pi}{7}\cos \frac{\pi}{7}) (2 \sin \frac{2\pi}{7} \cos \frac{2\pi}{7}) (2 \sin \frac{3\pi}{7} \cos \frac{3\pi}{7})$

= $ \sin \frac{2 \pi}{7} \sin \frac{4 \pi}{7} \sin \frac{6 \pi}{7} \equal{} \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7}(\because \sin (\pi \minus{} \theta) \equal{} \sin \theta)$

Thus we have $ 8 S \times C \equal{} S \Rightarrow C \equal{} \frac{1}{8}$


geometry rectangle MATHCOUNTS calculus integration

For the Latter question its $\sum_{k=1}^{10}k^{2}$

[hide]There are 385 (sum of the first 10 squares) squares.[/hide]

That's right, here's a hint for the first part.

[hide]two vertical lines and two horizontal lines determines a rectangle.[/hide]

[quote="Franz Joseph"]Given a $10 \times 10$ grid. How many rectangles (including squares) are there? How many squares are there?[/quote]

[hide]There are 11 vertical lines and 11 horizontal lines.  A rectangle is made by choosing 2 vertical lines and 2 horizontal lines.  So ${11\choose 2}{11\choose 2}=3025$.

I think thats right.

squares are easy.[/hide]

[hide="Rectangles"]There are 55 combinations of two vertical lines, and 55 combinations of two horizontal lines. $55^{2}=3025$[/hide]

[hide="Squares"]There are 385 squares on a $10\times 10$ grid.[/hide]

[hide="Generalization For Rectangles"]The amount of rectangles on an $n\times n$ grid is $\left(\frac{n(n+1)}{2}\right)^{2}.$[/hide]

[hide="Generalization For Squares"]The amount of squares on an $n\times n$ grid is $\sum^{n}_{k=1}k^{2}.$[/hide]

[quote="Franz Joseph"]Thank you [b]i_like_pie[/b] for your generalisations. Of course there is a further generalisation for any $m \times n$ rectangles.[/quote]
[hide="Complete Generalization For Rectangles"]The amount of rectangles on an $m\times n$ grid is $\frac{mn(m+1)(n+1)}{4}.$[/hide]

[quote="jgao"]i-like-pie, the link on your signature is dead :rotfl:[/quote]
Huh? It works. :huh:

[quote="jgao"]Sorry to bother you but it now works :oops: . By the way is the forum created by nebula42 :rotfl:[/quote]
Yes, nebula42 (Nebula) is the creator of the forum. The link is also in his signature.

Um, has everyone forgotten about rectangles with non-integer sidelengths?

[quote="b-flat"]I remember someone posted something similar to this in the Mathcounts forum a while ago - no one got a general formula.

I think you would just have to count out all of the different non-integral side length rectangles, unless the problem meant to ask for only those of integral sides.[/quote]
I'm almost positive that I've seen the formula somewhere before for non-integral sides as well. I'm pretty sure that there is one, and that it isn't very complicated.

[quote="Treething"]non-integral? wouldn't you replace m with the floor of m and n with the floor of n? unless I'm missing something...[/quote]
I think it's the rectangles whose sides aren't parallel to the axes.


trigonometry
inequalities
geometry theorems
geometry

Prove the formula

$\cot\omega=\cot A+\cot B+\cot C$,

where $\omega$ is the Brocard angle of a triangle $ABC$.

[i]Remark.[/i] The [i]Brocard angle[/i] of triangle $ABC$ is defined as follows:

There exists one and only one point $\Omega$ in the plane of triangle $ABC$ such that $\measuredangle\Omega BC=\measuredangle\Omega CA=\measuredangle\Omega AB$. This point $\Omega$ is called the [i]first Brocard point[/i] of triangle $ABC$, and the respective angle $\measuredangle\Omega BC=\measuredangle\Omega CA=\measuredangle\Omega AB$ is called the [i]Brocard angle[/i] of triangle $ABC$.

i belive mihai would like to explain this :)

Unfortunately, Mihai isn't here,  :D so here goes. In triangles BTC and ATC (T is Brocard's point) we have sine thm so: $\frac{TC}{\sin\omega}=\frac{a}{\sin\left(A+B\right)}$ and $\frac{TC}{\sin\left(A-\omega\right)}=\frac{b}{\sin\left(B+C\right)}$. If we divide them we get $\frac{\sin\left(A-\omega\right)}{\sin\omega}=\frac{a}{b}\cdot\frac{\sin\left(B+C\right)}{\sin\left(A+B\right)}$,  but $\frac{a}{b}=\frac{\sin A}{\sin B}$ and $\sin\left(A+B\right)=\sin C$, so this can be written as $\frac{\sin\left(A-\omega\right)}{\sin A\cdot\sin\omega}=\frac{\sin\left(B+C\right)}{\sin B\cdot\sin C}$. If you look closely, you'll see that the expression on the left is $\cot\omega-\cot A$ and the one on the right is $\cot B+\cot C$, so these two are equal. We got our conclusion.

thanks grobber,
and this way we can easily prove $\omega\leq\frac{\pi}{6}$.

Hey People!!!My name is Johann and it is my first message for the MathLinks. my English, it isn't very good :(

This is a problem of IMO in Sigtuna,Sweden:

"Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30 degrees."

The solution is obvious with the Brocard point! The anterior messages contains all necessary theorems. P is the Brocard poin of the triangle ABC.
If X=P then aren't nothing to prove.
If X<>P then X is in the interior of one of the triangles PAB,PBC,PCA (wlog PAB).Then the problem is obvious (connect the point X to point A and verify the angle).

We can prove it thanks to this following useful formula
$\cot A=\frac{b^{2}+c^{2}-a^{2}}{4S}$.
  Let M be the Brocard point.
  We have $\cot\omega=\frac{MA^{2}+a^{2}-MB^{2}}{4S\left(MAB\right)}=...=\frac{a^{2}+b^{2}+c^{2}}{4S\left(ABC\right)}=\cot A+\cot B+\cot C$.

That problem Johann was talking about can also be solved by using the Erdos-Mordel inequality: $PA+PB+PC\geq 2(PD+PE+PF)$, where $D,E,F$ are the projections of $P$ on the sides of $\triangle ABC$.


vector
linear algebra
linear algebra unsolved

All spaces here are finite-dim real vector spaces.

If $E'\buildrel{u}\over{\rightarrow}E\buildrel{v}\over{\rightarrow}E''$ is exact, prove: $Hom(F,E')\buildrel{w_{1}}\over{\rightarrow}Hom(F,E)\buildrel{w_{2}}\over{\rightarrow}Hom(F,E'')$ is exact.

where $w_{1},w_{2}$ are defined as follows:

for any $f \in Hom(F,E'), g\in Hom(F,E), w_{1}(f)=u \circ f, w_{2}(g)=v \circ g$

[to show $Ker(w_{2}) \supset Im(w_{1})$ is easy, I have trouble to show

$Ker(w_{2}) \subset Im(w_{1})$. please help, thanks.]

[quote="owk"]$g\in\ker w_{2}\iff v\circ g=0\iff\mathrm{im}\,g\subseteq\ker v=\mathrm{im}\,u$, but $E'\to\mathrm{im}\,u$ has a section. owk[/quote]

so how to show if $g\in\ker w_{2}$, then we can find a $f \in Hom(F,E')$, such that $g=u \circ f$? how to construct this $f$?

Call the section $s$. Then (by definition of a section) $u\circ s=\mathrm{id}_{\mathrm{im}\,u}$, so $f=s\circ g$ does it.

The core of this is the following lemma: Any surjective linear mapping $f\colon V\to W$ has a section, i.e., there is a linear mapping $s\colon W\to V$ such that $f\circ s=\mathrm{id}_{W}$. Proof: Choose a basis $\{w_{i}\}$ of $W$; for each $i$ choose a preimage $v_{i}\in V$ (i.e., $f(v_{i})=w_{i}$) and extend linearly, i.e. $s\Bigl(\sum\lambda_{i}w_{i}\Bigr)=\sum\lambda_{i}v_{i}$.

owk


haha I actually logged in to ask exactly the same question.  :D

According to a message I just received from David Linker, NYSML will take place this year on Saturday, April 8, 2006.  It will be held at [url=http://www.nyackschools.com/high/high.html]Nyack High School[/url] in Rockland County.

Who's going?

I guess NYC will never host.
Bronx Science is getting completely stomped this year...  :(  :ninja: (ninja hood=paper bag)
How do you apply to assistant coach? I might not make NYSML but I'm pretty sure I'll be free by ARML.

The reason NYC doesn't host is because it's difficult to find places to house people in NY -- also, upstaters get worried about sending their kids to the city.  If you'd like to be an assistant coach, just send an e-mail to Mr. Linker.  This is the second year in a row they've scheduled NYSML the week after my spring break -- I am greatly disappointed.

well, they travel on Saturday morning, so that might mean you could make it, i'm trying to figure out if i should go.... hope practice is during spring break though :)

the timing is pretty bad...
i hear that 3 of the 5 current stuy captains won't be going because it conflicts with MIT campus preview weekend.

alan, I guess that gives the other teams a chance this year. :P  j/k, I'm sure the team will be just as strong as last year's.
My spring break also ends the week before NYSML. I hope I can make the training session too. Crossing my fingers.

Do we know when the practice session is, yet?

"The NYSML training session will be held at Stuyvesant High School at 3:00 pm on a date to be determined."
*hopes it does happen during spring break*
so is there a chance anyone is actually going (as an assistant coach)? i am gonna try to!

wow...still have two more months to go....

hey, wensi!  its been a while...
your reluctance to install AIM will be a problem for the relay practices ;)

Yea, that was a bit annoying last year.
Everyone should get Gaim. You can install a plug-in that lets you use $\LaTeX$.
I'll be glad to help again with relays this year. In fact, let's start right now :D ... as soon as I find some problems.

Hi! Alan! havn't seen u for a long long long time!!

Aim..... is a problem....I'll see...

[quote="Bictor717"]
How do you apply to assistant coach? I might not make NYSML but I'm pretty sure I'll be free by ARML.[/quote]

Just e-mail Linker and let him know you are interested in doing so.  I did it last year, and I coached for NYSML and ARML.  I had an absolute blast doing so as well.

Oh, and I'm definitely going to be around for this year's NYSML and ARML.  Going to college in the city still has its benefits.

Thus, I'll definitely be seeing a lot of you again come the spring, and perhaps I may be working with you.  Cheers!

As for the contest itself, what is posted is accurate.  All I have is what I have as a coach--nothing about teams, the practice session, and so on.  I'm actually conducive to having it at any time...

As for practice, I am definitely considering using IRC over AIM...since I've gotten to the point of using the former constantly.  Thing is, I'm absoultely clueless as how to set it up (it being a room with guardian), but I can try to give it a shot.

And I hope to see a lot of returning members as coaches.  It's quite different from the other side of the looking glass, but it's still just as fun (if not more).
--KC

no that's not it, it's my first time competing and so far everyone has only given a vague description of the selection process and i just wanted to know if such a "cutoff score" actually existed and how high it could POSSIBLY be.[/u][/b]

Well, I have my magic 8-ball at school and I'll put its perdictive powers to work later this week...now were did I put my tarot cards?

oh ...maybe i've been unclear about what i mean, what i'm asking is how significant is your NYSML score for team assignments. sorry if you thought i was inquiring about something else.

I think the point others are making is that there is no hard-and-fast formula and so no really appropriate answer to your question.  Having a good NYSML score bodes well for your being invited to ARML.  Having a not-so-good NYSML score bodes less well, but there are lots of other factors that are taken into consideration.  In my experience, most people who go to NYSML also get invited to ARML.  (That might be different this year because there were more NYSML teams than usual -- is NYC taking all 6 teams to ARML?)

oh really i thought only 5 but whatever, thanks for explaining it to me!

We're taking all six teams?  Are there going to be more buses than just one Manhattan and one Queens bus?

Well, we have 59 person busses.  6*15=90 kids.  Yes, if someone's absent, we're screwed...unless I have some stuy kiddies in super reserve the day of.  The busses accomidate 118 people.  That leaves 28 spots for coaches and what not.  Hopefully, a coach will drive and take a few others.  You figure that we'll take me, oana, tim, joe, richard, patrick, scott, tom, a few alumni, linda berman, oleg, david linker, deana a., and others that I'm too tired to remember to list.  It should (hopefully work)

Oleg has driven in the past and carted a few Russians with him :-)

Sadly, that might be illegal.  I'll have to check.

Does Deeps refer to your better half?

Wow.  I thought you said before we only had 5 teams.  I suppose, then, that we can pull another ARML 2002 again with "New York City S,"...or not?

On the note of the buses, I remember last year there was a lot of commotion on the buses, since I recall Linker freaking out about the buses, especially for the Manhattan students.  At least we can hope that ARML 2005 won't repeat itself.

[quote="JBL"]n my experience, most people who go to NYSML also get invited to ARML.[/quote]

That holds mostly true--though in the past, some of those invited to NYSML were rejected for ARML because of people who couldn't compete at the former contest (this year, Tony and Katherine come to mind.)  With the expansion to a sixth team, I don't think this should be too much of a problem.  For those of you who have never gone to ARML, please look forward to it--I've always found it the most amazing experience I've had on math team when I was still competitive.

I don't recall Oleg's coming solo being illegal though...

~KC

P.S. About how many practice sessions should be expected?  As I've just finished up with my term as of about next Friday or so, I'll be available to come in.

How would the ARML-SAT situation be fixed? They are both on June 2nd.

Every year there is an alternate SAT held at Stuyvesant sometime afterwards.

Indeed.

Here is [url=http://www.arml.com/SAT%20Alternate%20Test%20Day%20Policy%202006-07.pdf]some reading[/url] about it.  Helps a little more with the conflict.

For the specifics, the exam is usually held the Tuesday after the contest.

Apologies for the double post.

According to the NYC Math Team site, the team rosters are up, but I can't find them anywhere...

~KC


calculus
derivative
analytic geometry
trigonometry
real analysis
real analysis unsolved

Find the minimum possible value of $ x^2 \plus{} y^2$ given that x and y are
real numbers satisfying
$ xy(x^2  \minus{}  y^2) \equal{} x^2 \plus{} y^2$ and x 6= 0.

I'm a bit stuck, I haven't studied partial derivatives and this problem can be done without them.

i know all i need to do is minimise $ xy(x^2  \minus{}  y^2)$

just not sure how, help please

One approach is to use polar coordinates: $ x\equal{}r\cos t$, $ y\equal{}r\sin t$. Since $ x^2\plus{}y^2\equal{}r^2$ and $ xy(x^2\minus{}y^2)\equal{}r^4\sin t\cos t\cos 2t\equal{}\frac{1}{4}r^4\sin 4t$, the problem reduces to the following: find the smallest positive $ r$ for which the equation $ \frac{1}{4}r^4\sin 4t\equal{}r^2$ has a solution. Since $ \sin 4t\le 1$, the smallest $ r$ that works is $ r\equal{}2$.

I never thought to use polar coordinates to resolve the problem.

thanks a lot for the insight mlok.


There exist non-negative integers $ a$ and $ b$ such that for any $ x$ satisfying $ a < x < b$, it is true that $ \sqrt{x} > x$. Find the ordered pair $ (a,b)$.

The only way $ \sqrt{x} > x$ is when $ 0 < x < 1$. (i.e. $ \sqrt{\frac14} \equal{} \frac12$)  Thus, the answer is $ (0,1)$.

Note:

$ \sqrt{x}>x$

Dividing by $ \sqrt{x}$ where $ \sqrt{x}\neq0$

$ \sqrt{x}<1$

Thus as ernie said 0<x<1

If you draw a graph it becomes obvious.


A prime p has demical digits \[p_{n}p_{n-1}...p_{0}\]
 with \[p_{n}> 1 \]
. Show that
the poynomial \[p_{n}x^{n}+p_{n-1}x^{n-1}+...+p_{1}x+p_{0}\]
 cannot  be represented  as a product  of two  nonconstant
polynomials with integer coefficients.

WHO CAN WRITE WHERE CAN I GET FUNCTIONAL  EQUATIONS BOOK. THANKSSSS


topology
advanced fields
advanced fields theorems

Hello people, It is easy to see geometrically the generators for the fundamental group of a torus.  However, I was trying to see them for the fundamental group of the Klein Bottle, but I cannot imagine which ones are.    Maybe, one can be the one which goes inside the bottle and then goes outside looking for the base point, and the other generator maybe is the one which start in some point  and passes under the handle and goes for the base point to make a loop.   I have no idea if this idea is correct , maybe it is totally wrong, because I cannot prove that the relations in the presentation of the Klein Bottle hold with these generators.

I appreciate you can help me out.

Thank you!

Let me answer myself.    We can see the Klein Bottle as a square with the edges identified as everyone knows.  Then, just look at the edges of the square via the identification.

The answer was obvious  :blush:

What you've realized is in a sense a special case of a phenomenon described by the Siefert-Van Kampen theorem. Basically the idea is that if you have a polygon with sides glued together to make a topological space, then the fundamental group is generated by the nontrivial loops which go around the identified edges.


I took the AMC 12A and we got our scores back on Friday. Our teacher finally got us the results, and it said that I got a 130.5. Here's the catch: It's ABSOLUTELY IMPOSSIBLE to get a 130.5 on ANY AMC! Something really weird is going on here...

Have you seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76023]this[/url] thread yet?  I'm guessing you got 21 right, 1 wrong, and 3 blank?

Thanks a lot, joml88. So that's why all of our predicted scores were an arbitrary amount greater than the "real" scores.

I'll tell my teacher to keep checking his e-mail...  :roll:

And yes, that is what I got...1+4+9 is not 13...


Find the smallest integer $k$ such that there exist two sequences ${a_{i}},{b_{i}}(i=1, \cdots , k)$ such that
(a) For $i=1, \cdots, k, a_{i}, b_{i}\in{1, 1996, 1996^{2}, \cdots}$.
(b) For $i=1, \cdots, k, a_{i}\neq b_{i}$.
(c) For $i=1, \cdots, k-1, a_{i}\leq a_{i+1}$ and $b_{i}\leq b_{i+1}$.
(d) $\sum_{i=1}^{k}a_{i}= \sum_{i=1}^{k}b_{i}$.

[quote="Myth"]Denote 1996 as $m$. The answer is $m+1$: \[m\cdot 1+1\cdot m^{2}= (m+1) \cdot m.\] So we are to prove $k\neq 1,2,...,m$. For $k\in\{1,2,...,m-1\}$ it is evident since both sums will represent distinct numbers in base $m$. For $k=m$ it is also quite clear.[/quote]

Is it possible for to please explain the question itself? I can't deduce the 2 sequences from the question.


probability
function
logarithms
arithmetic sequence
number theory unsolved
number theory

Here is one of the most beautiful problem I have ever seen. I don't have any idea. I found it in Mathematics Magazine:
   Are there arbitrarily long sequences of positive integers no two of which have the same number of divisors?

I think you forget something, unless the problem is trivial with the sequence (2 n )

Pierre.

I'm completely stupid today. The correct problem is the following:
  Are there arbitrarily long sequences of consecutive positive integers no two of which have the same number of prime divisors?

A sketch of the proof.

Lemma. Let $ A$ be an arithmetic progression with positive integer terms, in which the first term and  the difference are coprime. Let $ k$ be a positive integer. Than the density of elements of $ A$, which are divisible by some $ k$-th power does not exceed some constant $ c(k)$, which does not depend on $ A$, and $ c(k)$ tends to 0 when $ k$ tends to infinity.

I omit the proof, it is rather standard.

Now fix a positive integer $ m$, which is a number of desired consecutive integers. Fix primes
$ p_{1}<p_{2}<\dots<p_{m}$ such that $ m<p_{1}$ and $ m\cdot c(p_{1}-1)<1$. By Chinese Remainders Theorem find a remainder $ C$ modulo $ M: =\prod_{i=1}^{k}p_{i}^{p_{i}}$ such that $ C+i$ gives remainder $ p_{i}^{p_{i}-1}$ modulo $ p_{i}^{p_{i}}$ for any $ i=1,2,\dots,m$. Then for any positive integer $ d$, consider $ m$ consecutive numbers $ C+dM+1,\,C+dM+2,\dots,C+dM+m$. Note that the number of divisors of $ C+dM+i$ is divisible by $ p_{i}$. We try to find such $ d$, that this number is not divisible by other $ p_{j},\,j\ne i$. It suffices to find such $ d$ that $ (C+dM+i)/(p_{i}^{p_{i}-1})$ is not divisible by any $ (p_{1}-1)$-th power. By lemma, this happens with positive probability.

At fact, I am sure that the probability that $ \tau(n)=\tau(n+d)$ is zero for any fixed $ d$ (here $ \tau(n)$ is the number of divisiors of $ n$). The problem immediately follows from this. But I do not see, how this may be proved.

Very beautiful solution. The fact that the counting function of the number of solutions of $ \tau(n)=\tau(n+1)$ is between $ \frac{x}{(\ln(\ln x))^{3}}$ and $ \frac{x}{\sqrt{\ln(\ln x)}}$ suggests that it should not be easy to prove your last assertion. These bounds appear in an article of Adolf Hildebrand in Pacific Journal of Mathematics 129, No 2/ 1987.

I would be astonished, if his methods do not work for the equation $ \tau(n)=\tau(n+d)$.

Actually, all you need to prove Fedor Petrov's assertion is a few standard probability techniques and a bit of common sense ;) Here is a brief sketch of the proof.

1) Since there are very few numbers that are divisible by squares of big primes, it suffices to show that for every $ P>0$ and $ C>0$, the probability of the event $ |\eta(n)\minus{}\eta(n\plus{}d)|<C$ ($ 1\le n\le N$) tends to $ 0$ as $ N\to\infty$ where $ \eta(n)$ is the number of the prime divisors of $ n$ exceeding $ P$.

2) It suffices to show that $ \frac1{\sqrt {S_N}}(\eta(n)\minus{}\eta(n\plus{}d))$ converges to the normal distribution in probability where $ S_N\equal{}\sum_{P< n\le N}\frac 2p$.

3) For every prime $ p\in(P,N]$, consider the random variable $ \zeta_p(n)$ that is $ 1$ if $ p|n$ and $ 0$ otherwise.
Then $ \eta(n)\minus{}\eta(n\plus{}d)\equal{}\sum_p[\zeta_p(n)\minus{}\zeta_p(n\plus{}d)]$.

4) Note now that for every $ \delta>0$, the last sum can be restricted to $ p<N^\delta$ with introducing an error at most $ 1/\delta$ that will disappear after division by $ S_N$.

5) For the restricted sum we can easily see that its moments of order $ m< 1/(3\delta)$ are pretty much the same as if the variables on the right hand side were independent.

6) The rest should be clear.

Thanks to fedja!

It is quite similar to famous Turan's proof that almost any number from 1 to n has approximately log(log(n)) prime divisors, which is written in Alon and Spencer's
"Probabilistic method".


In triangle ABC, AB=20, BC=24, and the measure of angle B is 60 degrees. What is the area of triangle ABC?

Apply the formula

$ [ABC] \equal{} \frac {ab \sin C}{2}$

[hide="Solution"]

$ \frac {(20)(24)}{2} \times \frac {\sqrt {3}}{2} \equal{} 120 \sqrt {3}$
[/hide]

[hide="Another way"]
Apply Law of Cosine to get the third side. Then use Heron's or whatever method that you prefer. [/hide]

there may be another way to do this(for those who don't know the Law of Cosine)

let X be the point so that BX=20  since B=60°  ,AX=BX=AB

now the height on BC is the same as the height on BX and we know that the heigth od BX is

20*sqrt(3)/2

so the area is

20*sqrt(3)/2*24/2

120*sqrt(3)


IF the outbox used for sending messages, then what is the sentbox used for?  :oops: :fool:

[quote="D.P.L"]IF the outbox used for sending messages, then what is the sentbox used for? :oops:[/quote]
This is definetly not a pre-olympiad question. You should have posted it in the support forum. As for your question, I'm not sure.

[quote="pianoforte"]I believe that the outbox is for sent messages that the receiver has not read, while the sent box is for all the old sent messages that have been read.[/quote]
Yes, i think thats right. I was thinking along the same lines. :)

can some mod move this or delete this :huh:


In triangle $ ABC$ point $ D$  lies on segment $ AC$ and point $ E$ on segment $ BC$ and  quadrilateral $ ADEB$ is cyclic. Lines $ DE$ and $ AB$ intersect at $ F$ and $ CF$ , $ BD$ at $ M$. Prove that   $ M$ is a middle of $ CF$ if and only if $ MC^2 \equal{} MB * MD$

In triangle $ BCF$ we have that cevians $ BM,CA,FE$ concur at point $ D$. By Ceva's theorem, this happens if and only if $ \frac{BE}{EC}\times\frac{CM}{MF}\times\frac{FA}{AB}\equal{}1$.

Since we know $ M$ lies on line $ CF$, we can say that $ M$ is the midpoint of $ CF$ if and only if $ \frac{CM}{MF}\equal{}1$. Then, $ M$ is the midpoint of $ CF$ if and only if $ \frac{BE}{EC}\times\frac{FA}{AB}\equal{}1 \rightarrow \frac{BE}{EC} \equal{} \frac{AB}{FA} \rightarrow AE \parallel CF$.

So, $ M$ is the midpoint of $ CF$ if and only if $ \angle DBA \equal{} \angle DEA \equal{} \angle DFC$, which happens if and only if triangles $ MDF$ and $ MFB$ are similar, with $ \frac{MD}{MF} \equal{} \frac{MF}{MB} \rightarrow MF^2 \equal{} MD \times MB$. But also, $ M$ is the midpoint of $ CF$ if and only if $ MF\equal{}MC$, so this happens if and only if $ MC^2 \equal{} MD \times MB$. The proof is complete.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $D, E, F$ be collinear points on the side lines $BC, CA, AB$ of a triangle $\triangle ABC$. Let $K, L, M$ be the midpoints of the segments $AD, BE, CF$ and $(K), (L), (M)$ circles on the diameters $AD, BE, CF$. Show that:

[color=blue][b](1)[/b][/color] The circles $(K), (L), (M)$ belong to the same pencil (this is simple, but necessary for (2)).
[color=blue][b](2)[/b][/color] The circumcenter of the triangle formed by the lines $AD, BE, CF$ is on the radical axis of this pencil (this is beautiful).

This is a problem from the 1st Round of the 5th MathLinks Contest. It's actually an old result of a rather famous mathematician [EDIT: this mathematician is Möbius], but I won't give any hints for now. Please don't post anything about the problem until April 24, 2005.

  Darij

Ha-ha  :lol: Hoda net - hodi s bubei.  :lol:  :lol:  :lol:

Here is another one: if you don't know the answer, say "Pushkin", it works!

This problem is not hard at all, but the solution is quite long, yet straightforward...

[quote="pestich"]Yetti ,

    Are you trying to prove something here by taking the motto under my
    signature (the first one)  too seriously?

     Maj.  Pestich
____________________________________________________
A wise man will investigate what a fool takes for granted.[/quote]

Why would I do such a thing ? It is your motto, not mine. I do not even think that it is true. But I will not disturb you with the details (because the details would disturb you).

Yetti.

Continuation of this problem is here: http://www.mathlinks.ro/Forum/viewtopic.php?p=216159#p216159
Topic closed.

There is another topic where two solutions were published.


I need help, or rather just a nudge in the right direction. A question on my homework is asking me to "conjecture a formula for [summation from k=1 to n of 1/(k(k+1)] from the value of small integers n and to prove it's right using induction."

I realize this is a basic telescoping sum of 1/k-1/k+1 that sums to 1, but I don't understand what they mean by a formula for the sum? Isn't the summation notation a formula? 1/k-1/k+1 doesn't seem like a formula for the sum itself to me. I know I'm overthinking this and probably missing something basic. Thanks for helping, though.

By "a formula", they probably mean a expression with no running summations.  You have the right idea.

Well you basically have the key to the solution

You telescope out and collect the terms, and show that the summation is always equivalent to
[hide="expression"]
$ 1\minus{}\frac{1}{n\plus{}1}\equal{}\frac{n}{n\plus{}1}$
[/hide]
You can show that there are N pairs of cancelling terms with induction if you want.

[quote="jpps1"]I realize this is a basic telescoping sum of 1/k-1/k+1 that sums to 1[/quote]

Careful; this is for the infinite case.  This question is about the finite case.  It's the difference between

$ \sum_{k \equal{} 1}^{\infty}\frac{1}{2^{k}}\equal{} 1$

and

$ \sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{} 1\minus{}\frac{1}{2^{n}}$

(and, in fact, this summation can be proven in a similar way).  The summation up to a finite number $ n$ is known as a [i]partial sum[/i], and studying the sequence of partial sums is one way to compute the infinite sum.  (The second equation is a formula for the partial sum.  Do you see what is meant now?)


We consider an alphabet with $ n$ letters. We denote by $ M$ the set of words with the following property: when two letters of the word are the same, all the letters between them are different. For example, when the alphabet is $ \left\{a,b,c\right\}$, the word $ abbacb$ is not in $ M$ but the word $ bbacbac$ is in $ M$.

[b](a)[/b] What is the maximal length of a word in $ M$ ?
[b](b)[/b] How many words of maximal lengths are in $ M$ ?

It is clear that a word in $ M$ can contain at most three copies of each letter. On the other hand, there exists a word in $ M$ of the form [i]aaabbbccc...[/i] with exactly three copies of each letter. Therefore, the maximum length is $ 3n$.

The number of words of maximal length is $ n! 2^{n\minus{}1}.$


Can some of you guys give me some nice geometry theorems and concepts that are very useful in the AIME and USAMO?

[url=http://mathworld.wolfram.com/CevasTheorem.html]Ceva's Theorem[/url]
[url=http://mathworld.wolfram.com/MenelausTheorem.html]Menelaus' Theorem[/url]
[url=http://mathworld.wolfram.com/StewartsTheorem.html]Stewarts' Theorem[/url]
:D


Three 3-digit factors are to made up of the nine digits,
1,...,9, each one used once only.  Let "d" stand for a
yet-to-be-determined digit, and "X" for "times."

Determine the 3 factors without using a calculator/computer, and without figuring on paper and pen(cil).  Also, show
their product (by your choice of method).

ddd X ddd X ddd = product

*Edit*

Explanation:

The leading digits should be 7,8, and 9.

As in a product of two whole numbers, a product of three factors would be as close together, but the final digits with a particular leading digit would be such that they would give a larger, rather than smaller number.

76_ X 85_ X 94_ = product

763 X 852 X 941 = 611,721,516

I don't know if this is correct.


linear algebra
matrix
function
real analysis
real analysis unsolved

Consider the set of solutions of the matrix differential equation $ dA/ (dt)=A^2$ where $A$ is a complex matrix function. What are the possible values of $ A(0)$ when $ A(t)$ is a solution of this equation?

perhaps I say stupidities but it seems that $A(0)$ can take
any arbitrary value.

The function $M -> M^2$ is $C^1$ so one can apply
Cauchy-Lipschitz theorem that claim that there exists a
(unique) solution to the differential equation:
$\frac{d}{dt}(A(t))=A^2(t)$ and $A(0)=M$
Where do I make mistakes ?

Ps: for the case where $M$ is diagonalisable it is easy to
find a solution. One can take $f:R->C$ such that $f_i'(t)=f^2(t)$
and $f_i(0)={\lambda}_i$ : $A(t)=P.diag(f_1(t),..,f_n(t)).P^{-1}$ works.


If I wrote cot 3x as cos 3x/sin 3x and expanded all that stuff with the identities for cos 4x and sin 4x and whatnot and bashed away long enough, I'd probably get it, but I'm hoping there's an elegant algebraic or possibly geometric solution.

hmmm
i think the identities are the way to go...

[quote]i think the identities are the way to go...[/quote]

Huh? What identities? The ones for the expansions of sin/cos(kx)? So you're agreeing with MysticTerminator, ja?

actually, im thinking sum to product

just use it once

and i believe ur done

yeah you're right. sigh i'm now going to seriously bang myself on the head...excuse me...

lol
i shoulda labled this one "easy trig #3"
cuz really, its only one step


geometry
trigonometry
AMC
AIME
angle bisector
Pythagorean Theorem

If in a triangle $ABC$ we have that: $AB=9$ and $\frac{BC}{CA}=\frac{40}{41}$, what's the greatest value $[ABC]$?

[hide="possible way to do the problem"]
you know that the difference between bc and ac is less than nine, so we keep writing equivalent factions to 40/41 and wait until the difference is 9. at that point we have 360/369, but that can't be right, so we use 320/328. so bc is 320 and ac is 328, then use heron's.  i don't think that works though, because it never said sides had to be integers.
[/hide][/hide]

[hide="My solution"]
Let $h$ be the altitude of $C$ in $\triangle ABC$. Then $S = [ABC] = \frac{1}{2}AB\cdot h$

Let $CD, CE$ be the internal and external angle bisector of $\angle ACB$ respectively $(D,E \in AB)$

Then $\frac{DB}{DA}= \frac{EB}{EA}= \frac{CB}{CA}= \frac{40}{41}\ \ (1)$

$\frac{DB}{DA}= \frac{40}{41}\Rightarrow  \frac{DB}{AB}= \frac{40}{81}\Rightarrow $

$DB = \frac{40}{9}$ and $DA = \frac{41}{9}$

The locus of the point $C$ such that $(1)$ holds is the circle with diameter $DE$.

So, the maximum altitude $h$ is the radius of this circle.

$DE$ is the harmonic mean of $DA$ and $DB : \boxed{\frac{2}{DE}= \frac{1}{DA}+\frac{1}{DB}}\Rightarrow $

$h_{max}= \frac{DE}{2}= \frac{DA \cdot DB}{DA+DB}\Rightarrow $

$S_{max}= \frac{1}{2}AB\cdot \frac{DA \cdot DB}{DA+DB}= \frac{DA \cdot DB}{2}= \frac{1}{2}\cdot \frac{40}{9}\cdot \frac{41}{9}= \frac{820}{81}$[/hide]

[hide="trig solution"]
Let $AC = 41x$ and let $BC = 40x$ then we have $1/2\sin{\theta}\cdot 40\cdot 41x^{2}= [ABC]$ which is obviously maximal when $\sin{\theta}= 1$.

So then we can use the pythagorean theorem to solve for $x$ and plug back in to find an area of $\frac{66420}{3281}$ ?
[/hide]

this is an old AIME problem :wink:

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1992&p=431380

Now that I saw the correct answer to the problem I noticed that I used an incorrect formula for the harmonic mean

It sould be $\frac{2}{DE}= \frac{1}{EA}+\frac{1}{EB}$

From $\frac{EB}{EA}= \frac{40}{41}$ we get $EB=40 \cdot AB$ and $EA = 41 \cdot AB$

Then $h_{max}= \frac{DE}{2}= \frac{EA \cdot EB}{EA+EB}= \frac{40 \cdot 41}{40+41}AB = \frac{40 \cdot 41}{9}$

and $S_{max}= \frac{1}{2}\cdot 9 \cdot \frac{40 \cdot 41}{9}= 820$


can any one answer these 2 questions

1.As temperature increases how does the frequency of tuning fork change?
2.What happens to frequency when one of the prongs is cut??

That's a nice question. In fact, I don't even know why the frequency of a tuning fork is constant.

I do know that if you cut of one of the prongs, the tuning fork won't vibrate, the vibrations will be damped out by whatever is in contact with the handle.


Given that $ \overline{MN}\parallel\overline{AB}$, how many units long is $ \overline{BN}$?

[asy]pair A,B,M,N,C;
M = 1.2*dir(255); N = dir(285);
A = 3*M; B = 3*N;
draw(M--N--C--A--B--N);
label("A",A,W);label("M",M,W);
label("3",C--M,W);label("5",M--A,W);
label("2.4",C--N,E);label("N",N,E);label("B",B,E);[/asy]

Since these triangles are similar by Angle-Angle Similarity, we can set up a proportion and solve.

$ \frac {2.4}{3} \equal{} \frac {x}{5}$

$ x \equal{} \frac {12}{3}$

$ x \equal{} \fbox{4}$


[quote="vaibhav2903"]S.T.cosec2x + cosec4x + cosec8x + ........+ cosec2^2009x =cotx - cot2^2009x[/quote]

$ cosec2x \plus{} ...... \plus{} cosec2^{2009}x \plus{} cot2^{2009x} \equal{} cotx$
consider ,$ cosec{2^2009}x \plus{} cot2^{2009x}$

$ 1/sin2^{2009x} \plus{} cos2^{2009}x/sin2^{2009x}$

$ 1 \plus{} cos2^{2009}x/sin2^{2009}x$

which  is  equal  to  $ cot2^{2008}x$

add $ cot2^{2008}x$ with $ cosec2^{2008}x$

u wud get $ cot2^{2007}x$ and so on

at last you ll get cosec 2x + cot2x

which is equal to cotx

Q.E.D


geometry
circumcircle
analytic geometry
geometric transformation
geometry solved

For any convex polygon, no 4 vertices concyclic, define a triangle thick if its vertices are that of the polygon and the circumcircle of the triangle contains all the points of the polygon in its interior. Define a triangle thin if its vertices are that of the polygon and the circumcircle of the triangle contains no points except those three on its circumcircle. Proof number of thick triangles is the same as number of thin triangles.

I think I can remember Pierre solving this one or a very similar one r4ecently. Am I wrong? :?

Oh, this was the hardest problem on the Ukrainian MO last year.
There were two proposed solutions. First required considering some figures in ${\mathbb R}^3$. Second used coordinates.

When I proposed it to Tsimerman, he gave the idea of moving one point. One can prove, that during this process the numbers of thin and thick triangles don't change. But how we can use this?

hmm how can you prove the moving one point result; and how do you use coordinates?

[quote="kueh"]hmm how can you prove the moving one point result;[/quote]
It's quite easy. Just watch what's happening when the moving point crosses one of the circumcircles and consider few variants.

[quote="kueh"]and how do you use coordinates?[/quote]
To say the truth, I don't remember  :blush:
But I have one solution by Alexey Teplinskiy, in Ukrainian. If you give me some time, I'll translate it and post here.

So... Let's go! :)
BTW, on our olympiad the 2004-gon was given.

Let's prove that there are 2002 triangles of each kind. We'll consider only thick triangles -- for thin triangles the proof is the similar.

The existence of 2002 different thick triangles can be proved constructively. On the first step we consider some side AB of the original 2004-gon, and the one vertex X is added to A,B, for which <AXB is minimal. So, ABX is thick.

By next step we consider side of ABX which is diagonal of 2004-gon (i.e. AX). Then we can find such vertex Y, which lies in another half-plane of AX than BX, and which makes <AYX minimal (among all vertexes in that half-plane). So, AXY is also thick.

We have two thick triangles. And we have quadrilateral ABXY, which also has a side, which is diagonal of 2004-gon. So we can repeat our step with respect to this side etc. On the n-th step we get convex (n+2)-gon triangulated to n thick triangles. On the 2002-th step we get 2004-gon, wich is divided to 2002 thick triangles.

Absence of other thick triangles we can prove wuth help of following statements.
1) Each side of the original 2004-gon is side of exactly one thick triangle.
2) Each diagonal of original 2004-gon which is a side of some thick triangle is the side of exactly 2 thick triangle, which lies in the different half-planes of this diagonal.

From second statement: if we have some thick triangle, we can build the whole triangulation of 2004-gon. By the first statement we have, that triangle built on first step is included by this triangulation. And, finally, from second we have that triangulations are the same, so, each thick triangle is included to our 2002.

If there are any questions about my translation (sorry for poor English) -- you're welcome.
If there are any questions about the solution -- try to think a bit more on it ;)

It is very very close to ARO 1998.
At least, part of your solutions about 2002 thick triangles repeats that problem.
I wrote solution on forum recently. You can find it here.

a triangulation! ahh a light bulb suddenly went on in my head (= thanks Feuerbach.

It is related to this problem :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=delaunay&t=263

Pierre.


trigonometry
Trigonometric Equations
algebra
exponential equations

[quote="gagarin"]$ 2^{\sqrt [12]{sinx}} \plus{} 2^{\sqrt [4]{sinx}} \equal{} 2.2^{\sqrt [6]{sinx}}$[/quote]

Denote $ t \equal{} \sqrt[12]{\sin x}$. Now we can write the equation as:

$ 2^{t} \plus{} 2^{t^3} \equal{} 2^{t^2 \plus{} 1}$
$ 2^{t^3}(2^{t \minus{} t^3} \plus{} 1) \equal{} 2^{t^2 \plus{} 1}$

It is quite clear(using modulo arguments) that for both $ t \minus{} t^3 > 0$ and $ t \minus{} t^3 < 0$, thre exists no solutions.

So consider $ t \equal{} t^3$. But this implies $ 2^{t^3 \plus{} 1} \equal{} 2^{t^2 \plus{} 1} \Rightarrow t^3 \equal{} t^2 \Rightarrow t^3 \equal{} t^2 \equal{} t \Rightarrow t \equal{} 1,0$

Thus for $ x \in \left[0,\frac {\pi}2\right], x \equal{} 0,\frac {\pi}2$ are the only solutions.

EDIT: :oops: sorry modulo arguments dont work. t can be any real :(

Let $ t \equal{} \sqrt [12]{sinx}$, with $ t \ge 0$.

we have $ 2^t \plus{} 2^{t^3} \equal{} 2.2^{t^2}$

$ \Longleftrightarrow 2^t \plus{} 2^{t^3} \equal{} 2^{t^2 \plus{} 1}$

$ \Longleftrightarrow 2^{t \minus{} t^2 \minus{} 1} \plus{} 2^{t^3 \minus{} t^2 \minus{} 1} \equal{} 1$

But $ 2^{t \minus{} t^2 \minus{} 1} \plus{} 2^{t^3 \minus{} t^2 \minus{} 1} \ge 2\sqrt {2^{t^3 \minus{} 2t^2 \plus{} t \minus{} 2}}$

Let $ y \equal{} t^3 \minus{} 2t^2 \plus{} t \minus{} 2$

We have $ y' \equal{} 3t^2 \minus{} 4t \plus{} 1$ $ \Longrightarrow y \ge y(0) \equal{} y(1) \equal{} \minus{} 2$

$ \Longrightarrow 2^{t \minus{} t^2 \minus{} 1} \plus{} 2^{t^3 \minus{} t^2 \minus{} 1} \ge 1$

So equation $ \Longleftrightarrow$ $ t \equal{} 0$ or $ t \equal{} 1$

$ \Longrightarrow sinx \equal{} 0$ or $ sinx \equal{} 1$.

OK!

Hi [color=darkblue]isomorphism[/color]! Why $ x\in [0;\frac{\pi}{2}]$? I think it's not true.

Roots of this equation are $ \boxed{x \equal{} k\pi, k \in Z}$ or $ \boxed{x \equal{} \frac {\pi}{2} \plus{} k2\pi, k \in Z}$.

[quote="thanhnam2902"]Hi [color=darkblue]isomorphism[/color]! Why $ x\in [0;\frac {\pi}{2}]$? I think it's not true.[/quote]

No, I was just lazy to write the general form  :lol:


Hi! I would appreciate if someone started a BTM. Please do this?


analytic geometry
trigonometry
quadratics
geometry
circumcircle
algebra
quadratic formula

Point $ A$ lies at $ (0, 4)$ and point $ B$ lies at $ (3, 8)$.  Find the $ x$-coordinate of the point $ X$ on the $ x$-axis maximizing $ \angle AXB$.

[hide="Algebraic Solution"]For acute angles, maximizing the angle is the same as maximizing $ t = \tan \angle AXB$.

$ \angle OXA + \angle AXB = \angle OAB$. Taking the tangent of both sides,

$ \frac {\frac {4}{x} + t}{1 - \frac {4}{x}t} = \frac {8}{x - 3}$, where $ t$ is the tangent of $ \angle AXB$.
Solving for $ t$, we get $ t = \frac {4(x + 3)}{x^2 - 3x + 32} = \frac {4x_'}{x_'^2 - 9x_' + 50} = \frac {4}{x_' - 9 + \frac {50}{x_'}}$, where $ x_' = x + 3$

$ t$ is maximized when the denominator is minimized, which by AM-GM occurs when $ x_' = \frac {50}{x_'} \implies x_' = 5 \sqrt {2} \implies \boxed{x = 5 \sqrt {2} - 3}$[/hide]
If I recall correctly, there was a nice geometric solution as well.

[hide="Sort of geometric, but also algebraic"]
[note: the circle is not tangent to the y-axis as it may appear below]. Edit: Thanks Altheman.
[asy]size(200);
defaultpen(fontsize(10));
real r = 5 * 2^0.5 - 3;
pair A=(0,4), B=(3,8), C=(r,0), D=circumcenter(A,B,C);
draw(circumcircle(A,B,C));
draw((0,10)--(0,0)--(10,0));
dot(A); dot(B); dot(C); dot(D);
draw(A--C--B,dashed+linewidth(0.65));
/* draw(A--(3,0)--B,dashed); */
draw(A--(8,0)--B,dashed);
draw(anglemark(B,C,A,16));
draw(anglemark(B,(8,0),A,16));
label("\(A(0,4)\)",A,W);
label("\(B(3,8)\)",B,NW);
label("\((x,r)\)",D,SSE);
label("\(X\)",C,S);
label("\(Y\)",(8,0),S);[/asy]

Consider the circles that pass through $ (0,4),\,(3,8)$ and which are tangent to the x-axis. Then the point $ X$ on the x-axis that maximizes the angle $ AXB$ is the point of tangency between the circle and the x-axis.

In case this isn't obvious, suppose their existed a different point $ Y$ that resulted in a larger angle $ AYB$. Let $ \overline{AY}$ intersect the circle at $ Y'$; then $ \angle AYB < \angle AY'B = \angle AXB$, contradiction.

Let the radius of the circle be $ r$, and the x-coordinate of $ X$ be $ x$. Then the center is at $ (x,r)$, and we can use the distance formula twice:
\begin{align*} \sqrt {(x - 0)^2 + (r - 4)^2} = r & \Longrightarrow x^2 + 16 - 8r = 0 \\
\sqrt {(x - 3)^2 + (r - 8^2} = r & \Longrightarrow x^2 - 6x - 16r + 73 = 0 \end{align*}
We can eliminate $ r$ by multiplying top equation by two and subtracting, giving $ x^2 - 6x - 41 = 0$. The quadratic formula yields $ x = \frac { - 6 \pm \sqrt {200}}{2}$, and discarding the negative root, $ x = \boxed{5\sqrt {2} - 3}$.
[/hide]

Very interesting problem!
I did it the same way as Zuton Force, taking the tan of the angle.

It is well-known that $ X$ is such that the circumcircle of $ \triangle AXB$ is tangent to the $ x$ axis at $ X$. Let $ a$ be the $ x$ coordinate of the center, and $ r$ be the radius of the circle. Then the equation of the circle is
\[ (x \minus{} a)^2 \plus{} (y \minus{} r)^2 \equal{} r^2
\]
Plug in the two points and you get a system for $ (a,r)$ that your calculator can solve very quickly. Of the two solutions, it is clear that $ a \equal{} 5\sqrt {2} \minus{} 3$ gives the smallest angle.

Another comment: There is not one unique circle with this property. So you cannot say "consider the circle through those two points touching the x-axis" because there are two of them. The rest of the solutions are more or less correct though. Also I'm not saying that my solution is rigirous, I'm just commenting on a common oversight (one that I've made myself).


Several sets of prime numbers, such as $ \{ 7, 83, 421, 659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?

$ \textbf{(A)}\ 193\qquad\textbf{(B)}\ 207\qquad\textbf{(C)}\ 225\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 447$

[quote="chess64"]Several sets of prime numbers, such as $\{ 7, 83, 421, 659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?

\[ \textbf {(A) } 193\qquad\textbf {(B) } 207\qquad\textbf {(C) } 225\qquad\textbf {(D) } 252\qquad\textbf {(E) } 447  \][/quote]
[hide]
i got C)225

so obviously 2,4,5,6,8 cannot be the ones digits, so the smallest sum (i think) is if they are tens digits

2_, 4_, 5_, 6_, 8_

since we need one one digit number that is prime, 2 or 5 works

so the ones digit shouldnt matter because they add up to the same anyways and since ten 2s is smaller than ten 5s the single digit prime is the 5

so randomly placing i got 29+47+61+83+5=225[/hide]

[hide="Answer"]Obviously, the even digits must be in the tens place except for $2$. Therefore, we need three two-digit numbers, and we can place the remaining numbers in the units places so that we have only prime numbers. Therefore, our answer is $40+60+80+1+2+3+5+7+9=207\Rightarrow \boxed{B}$. (The primes can be 2, 3, 5, 41, 67, 89.)[/hide]

For reference, this was 2002 AMC 12A Problem 17.
[hide=my sol]
we start by using 2,3,5, and 7 as one digit primes
this leaves us with 1,4,6,8, and 9 as possible options
we can pair up 4 and 1 to get 41, 8 and 9 to 89, and combine the 6 with 7 (which we already had) to get 67
this gives us a sum of 41+89+67+2+3+5=207
since 4,6,8 have to be in tens place or higher, it's not hard to see that this is the minimum
[/hide]


In a river with a steady current, it takes Superman 6 minutes to swim a certain distance upstream, but it takes him only 3 minutes to swim back. How many minutes would it take a Superman action figure to float this same distance downstream?

[hide]

This doesn't sound like High School Basics, more like MathCounts. Anyways...

$ 6x-6y=d$ is for for when superman goes upstream. When x=superman speed, y=flow, and d=total distance.
$ 3x+3y=d$ is downstream.

Now lets multiply the downstream by 2 and subtract with upstream:

$ 2(3x+3y=d)$
$ 6x+6y=2d$

Now, subtract...

$ \;\;6x+6y=2d$
$ (-)6x-6y=d$

To get $ 12y=d$

Thus giving us an answer of 12 minutes flowing downstream.

[u][b]Answer to the Superman Problem:[/b][/u] $ \boxed{12\;\mathfrak{Minutes}}$

[/hide]

[hide]First you set up the equations $ 6(x-y)=1$ and $ 3(x+y)=1$, where $ x$ is how fast Superman swims and $ y$ is the rate of the river's currents. This gives you $ 6x-6y=1$ and $ 3x+3y=1$. By solving the system of equations, you find that $ x$ equals $ \frac{1}{4}$ and $ y$ equals $ \frac{1}{12}$. Next, you get the equation $ t(\frac{1}{12})=1$, where $ t$ is time. If you solve for $ t$, you find that $ t=\boxed{12}$, which is your answer.[/hide]


Prove that for every odd prime p there exist infinitely many positve integers n such that p|n^2^n+1

I think condition must be. If exist n, s.t. $ p|m^{2^m}\plus{}1$, then exist infinetely many n, s.t. $ p|n^{2^n}\plus{}1$.