In year $ N$, the $ 300^\text{th}$ day of the year is a Tuesday. In year $ N \plus{} 1$, the $ 200^{\text{th}}$ day of the year is also a Tuesday. On what day of the week did the $ 100^\text{th}$ day of year $ N \minus{} 1$ occur?

$ \textbf{(A)}\ \text{Thursday} \qquad \textbf{(B)}\ \text{Friday} \qquad \textbf{(C)}\ \text{Saturday} \qquad \textbf{(D)}\ \text{Sunday} \qquad \textbf{(E)}\ \text{Monday}$

[hide]
First, notice that $ 100\equiv 2\pmod 7$.  Thus, if the $ 300$th day of year $ N$ was a Tuesday, the $ 200$th day of $ N$ was a Sunday.  Since $ 365\equiv 1\pmod 7$, we need to have year $ N\plus{}1$ be a leap year if day $ 200$ of $ N\plus{}1$ is a Tuesday.  Thus $ N$ is not a leap year.  The $ 100$th day of $ N$ is a Friday, so the $ 100$th day of $ N\minus{}1$ is a Thursday.  Answer is $ A$.
[/hide]

Hehehe. This came up on an old AMC, right? I distinctly recall using the TI-89's calendar feature on this one :P

[quote="apiarist1"]N has 366 days since the number of days between 300-th day of N year and 200-th day of N+1 year must be divided by 7. Now, the number of days between 100-th day of N-1 year and 300-th day of N year is 565 days. (565+2)%7=0  so since 302-d day of N year is Thursday so the 100-th day of N-1 year is Thursday.[/quote]

Doesn't N+1 have 366 days?

Alternate Solution:

$300\equiv6\mod7$ so January 6 of year $N$ was a Tuesday so January 1 of year $N$ was a Thursday.

$200\equiv4\mod7$ so January 4 of year $N+1$ was a Tuesday so January 1 of year $N+1$ is a Saturday.

$366\equiv2\mod7$. If year $N$ was a leap year, then January 1 of year $N+1$ would be on a Saturday which satisfies the conditions. Thus, year $N-1$ is not a leap year. So January 1 of the year $N-1$ is a Wednesday. Since $100\equiv2\mod7$, then the $100$th day of year $N-1$ is on a Thursday.

I just used the general pattern they gave to find that year (N-1)'s 400-365=65th day of Year N which is a Tuesday. You find 65 (mod 7)=2 so Tuesday+2 days=Thursday. Is this a valid solution?


A man has a $ 10$ m $ \times$ $ 10$ m square  garden. In the center is a $ 2$ m $ \times$ $ 2$ m square patch which he cannot use. He divides his usable space into four congruent rectangular patches. What is the number of meters in the perimeter of each rectangle?

I don't know how to draw it, but they are 6m x 4m, so the perimeter is 20m.

[quote="5849206328x"][asy]draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);
draw((6,10)--(6,4));
draw((0,6)--(6,6));
draw((4,0)--(4,6));
draw((4,4)--(10,4));[/asy]

If we let $ w$ be the width and $ l$ be the length, you get \[ w\plus{}l\equal{}10\] \[ w\plus{}2\plus{}w\equal{}10\]
Solving gets us $ w\equal{}4$ and $ l\equal{}6$, so the perimeter is $ 2(4\plus{}6)\equal{}\boxed{20}$[/quote]

You get upvoted for the awesome asymptote picture!

EXCUSE ME PLEASE TELL ME WHERE THIS COMES FROM! ASAP

Since we get $w+l=10$, and the perimeter is $2(l+w)$, we don't have to solve for them, our answer is just $2(10)=\boxed{20}$


probability blogs geometry Support calculus articles induction

I'm making this topic because some discussion about this has gotten into the probability of GOD topic, when it obviously does not belong there. This is only for scientific evidence for or against evolution, or for or against any other explanation for the diversity of life on earth. Talk about religious implications in the 'science and religion' thread.

I thought I'd also bring in the last thing Lynnelle said on that thread:

[quote="Lynnelle Ye"]Define 'religion'. Certain interpretations of the Christian religion conflict with evolution. Other interpretations don't--accounting for the theistic evolutionist majority at http://www.christianforums.com (check out the creo/evo debate forum there). I don't know as much about other religions versus science, but my point is that religion conflicts if you think it conflicts, and doesn't conflict if you don't think it conflicts. I mean, there are still a few geocentrists and even flat-earthers around, and they claim heliocentrism and a round earth conflict with their religion, which is fine because that's partially how their individual religion is defined. If you see what I mean.
That wasn't very coherent, but I hope you can get some meaning out of it...
P.S. Maybe we need a separate thread for evolution.[/quote]

According to  some (eg Aismov),  simple things like  price of  papyrus in old days have much to do with how  the theory was documneted. If you think I am not making any sense please  read  this blog:

[url=http://www.sumware.com/creation.html]How it Happened[/url]

Enjoy!

It is almost the definition of science that ANY body of ideas will come into conflict with it (or more precisely: enter the contest as one of the contenders) upon making assertions specific enough that they can be tested, compared with other theories' conclusions, etc. In other words, science is a kind of (ruthless) natural selection process within the realm of all possible sufficiently-specific theories.

The only way for religion, or any other doctrine (politics, politesse, fashion,....) to avoid conflict with science is to only make assertions nebulous enough to escape any evaluation or comparison.   Assertions that are "not even wrong", as Pauli put it.

If particular religious doctrines or their interpreters choose not to go near the arena, OK --- but once they enter the space of meaningful theories, they are subject to evaluation and comparison like all others.

Religion and evolution are completely different topics that cannot be grouped together. Religions is based on certain assumptions that cannot be argued against, and evolution is based on other assertions of mankind.

[quote="mcalderbank"]Religions is based on certain assumptions that cannot be argued against,[/quote]

Religious assumptions/assertions are unarguable only if they are too nebulous to determine; "not even wrong", in other words.  When they are more specific, such as claims that the world is several thousand years old, they become highly arguable.

[quote] and evolution is based on other assertions of mankind.[/quote]

Evolution is a mathematical phenomenon, such as faster reproducers eventually dominating a population. These theorems and mathematical phenomena are applicable to biology and many other subjects.

[quote="white_horse_king88"]Please define evolution.[/quote]
The tendency for the frequency of different alleles in a population to change over time. (Paraphrasing a definition occasionally posted on http://www.christianforums.com ; I think they got it from http://www.talkorigins.org , though I'm not sure) Common descent is also part of the theory I believe, though it's not included in the above definition.

[quote="LynnelleYe"][quote="white_horse_king88"]Please define evolution.[/quote]
The tendency for the frequency of different alleles in a population to change over time.[/quote]

It's really much broader than that.

Evolution is more or less the same thing as population dynamics (dynamics in the sense
of "dynamical systems"), which involves ideas from many areas of mathematics.
In the applications to biology one looks at the evolution of molecular, cellular, or organismic
populations, but the phenomena are generic and visible everywhere.   For example, a group of
people with 5 children per family will eventually dominate a population where everyone else has 2 children per family.

What people normally think of as evolution in Darwin's sense, is the application of such ideas
to situations where heredity, variation, and replication are all present.  But such contexts need not be biological.

[quote="boy_21"]Well, by that definition, evolution obviously occurs. Changes happen all the time.

The real question is, has Darwinian evolution (or maybe we should just call it "macroevolution" occured? Has one type of organism EVER been seen to turn into another type of organism (eg. whales into cows)?[/quote]

It's not exactly one type of organism "turning" into another.

Basically, mutations occur, and the mutants are "naturally selected", and over long period of time these mutants no longer breed with their parent population and we have speciation.

There are evidences that whales and cows are genetically related (DNA sequence, homologous structures...)

Right, so lets say we speed up the mutation rate of fruit flies and see if we get any beneficial mutations.  We could simulate millions of years of "evolution" in a few years. Can you guess what would happen?

Please, [i][b]PLEASE[/b][/i] do not mistake selection of certian alleles that allow a population to survive better in certain conditions for mutations. I've heard that done WAY to many times. (eg powdered moths)[/i]

[quote="boy_21"]Well, by that definition, evolution obviously occurs. Changes happen all the time. [/quote]

And mathematical models can be applied to quantify various aspects of the changes.

[quote] The real question is, has Darwinian evolution (or maybe we should just call it "macroevolution" occured? Has one type of organism EVER been seen to turn into another type of organism (eg. whales into cows)?[/quote]

Funny you should mention cows.  Breeding of domesticated animals, such as the hundreds of types of dogs (originally bred from wolves) and cows (there are no naturally occuring breeds) should answer your question.
The evolution in vivo (so to speak) of new strains of e.g. HIV virus is a another, current example on a much shorter timescale.  It was predicted on theoretical grounds roughly how fast this would happen.
Other examples:
Bioengineering of various bacterial strains to acquire desirable traits such as metabolization of toxins.
"Directed evolution" of fruit flies to increase lifespan by an order of magnitude.
These both happened on a lab-experiment timescale.

[quote="boy_21"]Right, so lets say we speed up the mutation rate of fruit flies and see if we get any beneficial mutations.  We could simulate millions of years of "evolution" in a few years. Can you guess what would happen?[/quote]

What is your point, if any? Is the above in answer to some statement (which one?) in this thread?

Certainly there have been a lot of experiments with flies, bacteria, and other organisms that illustrate
the evolutionary phenomenon on a short time scale.

[quote="white_horse_king88"]two major clashes in evolution. TIME [age of earth] and MEANS [God's creations are perfect and need not evolve].  This can just go on, because both sides made an assumption which cannot be proven by OBSERVATION.[/quote]
The evolutionary side of those questions is based on observation, not  mere assumption.  It is observed fact that all species, including humans, evolve (i.e. that heredity and variation are present; for humans it is debatable when and how large population differences would arise, but small ones are very apparent between families, tribes, and national groups).  The conclusions one can draw from this are proven by essentially mathematical reasoning (population dynamics of various models), not observation or assumption, and one then can compare them with further observation.

[quote="LynnelleYe"] We don't know if evolution is driven by God. It doesn't matter--it still happens. God is outside of science, so whether or not he has anything to do with it is beside the point. [/quote]

You are still making the assumption that macroevolution does happen. All that have been listed so far as "visible" evolution are cases of adaption, or microevolution, which is far from proving that macroevolution occurs. You have taken 100 small steps along the side of a canyon and claimed that you can therefore walk accross the 100-foot canyon. Each example of microevolution (adaption) is a small step, but that doesn't give any evidence to macroevoultion. Simply put, evolution requires the addition of [i]information[/i], and no one has ever proven that adaption to a new environment or resistance to certain drugs is done by the addidtion of new information. It has been proven that it is possible to create a system that will adapt, but it never been proven that the system can evolve into a completely different system.

So can you please stop listing proof of adaption, and try to show me some scientific evidance for evolution?

This is the example my teacher uses for macroevolution, the squirrels across the Grand Canyon.  Because of isolation for thousands of years, these two types of squirrels are different species from each other because they can no longer reproduce with each other.

Macro-evolution does exist. Let's say that evolution can be defined as the change of allele frequencies in a population, which is the accepted definition among the entire scientific community.

Ok.

Keep in mind that natural selection, which leads to evolution acts on the population, and if we can prove that it acts on the individuals(micro-scale) of a population, then by inference, it acts on the population(macro-scale).

Observation: Natural populations of all organisms have the potential to increase rapidly because organisms can produce far more offspring than are required merely to replace the parents. A populations consists of all the individuals of one species in a particular area.

Nevertheless, the sizes ofmost natural populations and the resources abailable to maintain them remain relatively constant over time.

Conclusion: There is competition for survival and reproduction. In each generation many individuals must die young, fail to reproduce, produce few offspring, or produce less-fit offspring that fail to survive and reproduce in their turn.

Observation: Individual members of a populatioin differ from one another in their ability to obtain resources and withstand evnironmental extremes, escape predators, and so on.

Conclusion 2. The most well-adapted, ("fittest") indivudals in one generation tend to be the ones who leave the most offspring. The environment "selects" for the better adapted traits.

At least some of the variation among individuals in traits that affect survivalor reproduction is due to genetic differences that may be passed from parent to offspring.

Conclusion 3. Over generations, differences, or unequal, reproduction among individuals with different genetic makeup changes the overall genetic composition of the population. This process is evolution.

Notice that as long as allele frequencies change, the population evolves.

Now consider a fault in transcription or replication of DNA. This occurs one in about every 100,000 genes or so. Now these mutations code for different proteins which affect the genetic genotype and phenotype of certain characteristics of a species. Now, the environment is selective for or against phenotypes, so the advantagious phenotype will survive in the individual and the trait is carried to the next generation. Now, as more random mutations occur, they survive, and there is a large genetic drift in the entire population, cause evolution on the macro-scale. The evidence is enormous.

Your stance is basically - induction cannot be used as proof, when induction is one of the best forms of proof out there.

Induction and logic support the macroevolution of populations from observation at the micro-scale.

The squirrels Sunny refers to are the Albert and Kaibab species - they are different species because they no longer interbreed. One lives on the north ridge and the other the south ridge. This form of speciation, or isolation, is caused by geography.

You say forget about proof and show evidence? Proof is evidence.

I am not sure it is even worth it any more. If you made the effort, you would notice that the people on this forum have given you [i]several[/i] instances of macro-evolution, myself included. Without an understanding of the process or the willingness to listen and learn, this information is pointless. I retire from this argument in frustration.

[quote="boy_21"][quote="LynnelleYe"] We don't know if evolution is driven by God. It doesn't matter--it still happens. God is outside of science, so whether or not he has anything to do with it is beside the point. [/quote]

You are still making the assumption that macroevolution does happen. All that have been listed so far as "visible" evolution are cases of adaption, or microevolution, which is far from proving that macroevolution occurs. You have taken 100 small steps along the side of a canyon and claimed that you can therefore walk accross the 100-foot canyon.

So can you please stop listing proof of adaption, and try to show me some scientific evidance for evolution?[/quote]

Blah Blah.

I will use my own horrible metaphor since you choose to use one that's completely irrelevant.

Just because you haven't seen 1 billion dollars in person does not mean that 1 billion dollars does not exist. What does this example prove? Nothing more than yours'. What does your example prove? Nothing much. Settled.

I have given you fossils. I have given you evidence from our DNA. I have given you examples of us observing new species evolving--note that 'macroevolution' is defined as speciation. I could give you more if you want. Others in this thread have given you more as well. And you still say that we are 'providing no evidence'. FYI, when someone posts a link, it is usually meant to be clicked on.

P.S. Your analogy is messed up. It'd be more like walking around your block, and claiming that given enough time you could walk 20 miles. Which is true. You don't have to be able to walk those 20 miles in one step, which is what you're claiming that evolution requires.

[quote="JBL"]That's not true at all -- it would be exceedingly easy for me to write a computer program to generate a sonnet.  I would simply have it run, in lexicographic order, over all possible strings of words in the English language up to a certain length.  It would produce a lot of garbage, which an interested reader would skip over, but it would also produce poetry and, if I let it run long enough, any other written work you like.  If I had a database of english words, I could start pumping out Haikus right now.

Also, I note that you tend to respond with actual challenges only with sarcasm and it makes you a much less convincing person to listen to.  HIV yields HIV, yes, but only because we happen to know exactly where it came from.  It is demonstratably different, in fundamental ways, from pre-existing strains of HIV.  Of course, we have the difficulty of making a distinction here because HIV doesn't reproduce sexually, but HIV today is not "the same" HIV that was around thirty years ago, it's an exceedingly different creature, and, if it makes sense to talk about species of viri, it would be safe to call it a distinct species.[/quote]

YEAY! Hitchhiker's Guide to the Galaxy.

Those monkeys are going to type up Hamlet sooner or later.

boy21 - In your example of the CD from ealier, you mention that the CD is still a CD. But that's generalizing. For example, if one species of Squirrel becomes a separate species, we can say that they are organisms. Sure whatever. But, the truth is, as compared to the information stored in a CD, the information which composed the organism, DNA, has changed.

You do not need additional information for evolution. Like LynnelleYe said, it's just speciation - which comes from change in allele frequency by random mutation, or comes from isolation. It's all logic layered upon logic if you just flat out disregard scientific proof.

[quote="boy_21"] 10 systems that Behe outlines that are absolutely dependant on every major part of the system--for example the human eye. If you take away or change any major part, the system does not work (or all of a bunch of minor parts, like the rods or cones).[/quote]

That is an expected result of evolution, not contradictory to evolution.  It is amazing, and not in a good way, that Behe considers such signatures of evolution to be evidence against it.

Evolution optimizes critical systems, including specialized organs (the eye) or basic biological machinery (the Krebs cycle).  This means that needless components will usually be removed over time, leaving an "irreducible" set.   It also means that minor modifications to a system will render it less efficient, or inoperable.

[quote] Therefore the system, as long as it has been in existance, has worked exactly as it does now. [/quote]

The much likelier conclusion is that evolution reached a steady (locally optimal) state of that particular system, and since then the system has worked as it does now.  If the selective pressure changes so will the system (e.g. if humans move to a dark underground, the structure of the eye would change; note the difference in eye structure of near-surface and deep-water fish).

People here may be interested on [url=http://www.cnn.com/2005/EDUCATION/05/02/life.evolution.reut/index.html]this article[/url] about some trial that is taking place in Kansas right now about if evolution should or should not be taught in school!

[quote="boy_21"][quote]It would help to define 'new, useful information'[/quote]

Good point. What is information?

I once heard a lecture about information theory by Dr. Werner Gitt (this genius German guy) which I found absolutely fascinating. In fact, maybe we should start a new thread on information theory. Anyways, to make a really long story really short, you cannot randomly generate information. You cannot make a computer program that will write an original sensible sonnet or an ode.
[/quote]
Heck yes you can. I made a random name generator the other day. It just takes some linguistics knowledge.
[quote]
The whole point is, you need to add information for macroevolution (think whales to cows), so random mutations aren't going to do it.[/quote]
The mutations add information by changing some of the other information. Some of this is bad(cystic fibrosis) and some of this is good ([url=http://www.msnbc.msn.com/id/5278028/]more muscles[/url]), but new is really just changing a bunch of things. I can get some fossil records if you need them.

I'm new to this site and have read all these posting.

Both sides seem to be talking past each other.  This reminds me of discussions between members of two different religions.  (Perhaps it is.)  Many of the examples are clearly not on point.  (Similar to the argument that a great warrior Achilles couldn't past a turtle in a race, if the turtle had a head start and kept moving, since when the warrior reached the place the turtle was when the warrior started, the turtle would have moved a bit more. . . .)  The problem with this historical logic problem is that it only applies until the point where both are the same distance from the starting point.

Similarly the random generation of "information" doesn't occur in the stated examples.  Random data is selected by an observer, who is the "generator" of the new "information."  This is also a "problem" with the use of fossils that are sorted to "show" evolution.  Many times the location and dating of the fossil are not the critical aspect in placing a fossil in an evolution line.

The examples of micro-evolution are good and well accepted by most people on both sides of this discussion, so references to micro-evolution, such as the relative populations of dark and light moths when London went from a dark background (high coal dust deposition) to a lighter background (lower coal dust deposition) are just "noise" in this discussion.

Next, the Bible says many times that acceptance is based on faith alone, so anyone who thinks that they have a "proof" for ID is in direct conflict with the major source of inspiration for ID.  Since God says that acceptance is based on faith alone, he could easily have set up the fossil record to support a much older system.  No where is there a requirement that when God created the earth, it had to start with any form.  If the stars are really million of light years away, why wouldn't one assume that the "in transit" light was created well, in transit?

If one wants to explore faith, ask where did the material come from for the Big Bang?

If I were to find a watch on a beach, what scientific proof would be required to support the idea that there was a watchmaker somewhere?  I think none.  The simplest organism is so much more complex than the simplest living organism.

Let's get out of biology and into chemistry.  Why do all living things use the same optically active molecules?  (Just like a person's two hands are very similar, but can't be placed so the fingers are lined up when both hands face the same way, certain molecules, essential to life, are distinct.  All life uses the same orientation, but no process has ever been found that doesn't use a living force (perhaps in selecting an optically active chemical) that doesn't result in a racemic mixture (equal quantities of both the right hand molecule and left hand molecule).  First, this is relevant to evolution because it doesn't require that the material under study be alive and be studied for many, many years.  Second, even if the initial start was with God or an ID, the fact that subsequent evolution occurred would still destroy the basic idea of evolution: NO GOD.

Finally, let’s apply the second law of thermodynamics to this theory of evolution.  Everything runs down, becoming more disordered.  (If you find a way to not have this increase in disorder occur, I’d like to try it out on my room.)  The only way a room becomes more ordered and something becomes more complex is as a sub-system of a system becoming more disordered.

A tree combines the carbon dioxide it brings in through its leaves, with the water and minerals brought in through the roots, catalyzed with chlorophyll, and powered by the sun to make sugars and wood.  However, when one considers the sun that generates the energy the system is loosing order some energy is converted into waste heat, that can never be recovered.  So if everything is winding down and we don’t want to talk about the watch maker, what about asking about the watch winder?  Where did the order that we all live on come from?

This is probably enough from me.  (Some may say too much.)

As we go along, we should all remember that not many are likely to change their views, let's be reasonable in discussing this issue that is important to many people.

[quote="wizard2378"]  Second, even if the initial start was with God or an ID, the fact that subsequent evolution occurred would still destroy the basic idea of evolution: NO GOD.
[/quote]
Nowhere does evolution say there is no god. Some evolutionist also think there is no god, but evolution itself does not exclude the possibility

[quote="wizard2378"]You're right that the evolution theory itself doesn't preclude a belief in God, but this is the essence of the public and legal debate.[/quote]

Yes, I agree, the evolution debate is mostly a vieled God vs No God debate.

Also, the Second Law ONLY applies to closed systems. The Universe is a closed system, not the earth. We have the sun coming in for example. And it does not exclude local flucuations in entropy, resulting in one area actually gaining more energy.


function
algebra
domain
complex numbers
imaginary numbers

If f(x) = x^2 - 3 and g(x) = sqrt(x+2), then find f(g(x))

thanks

in f(x) the x stands not just for an x but can stand for any variable thingy...so f(g(x)) is just replacing all the x's in the f(x) equation by g(x) so we hvae

f(g(x))=(g(x))^2-3.

plug in g(x) we have x+2-3=x-1,

but you have to know that x+2>0, so $ x\geq\minus{}2$ otherwise it wont work

so its $ x\minus{}1$ with $ x\geq\minus{}2$

well imactually assuming the question said that x is in its domain because im betting you arent dealing with imaginaries here.

f(g(x)) is broken down to two simple steps
1.Given x, do g(x)
2.Take result of g(x) and treat it as x for f(x)

So in your question:

f(g(x))= [Sqrt(x+2)]^2-3=
[b]x+2[/b]-3=
[b]x-1[/b]

And as stated above, x cannot be less than -2, otherwise g(x) would be an imaginary number


[size=150]1. Suppose V = Span(v_1, ... , v_k) ⊂ Rⁿ. Prove that there are vectors w_1, ..., w_k ∈ V that are mutually orthogonal that also span V.

2. Suppose U and V are subspaces of Rⁿ. Prove that (U + V)┴ = U┴ ⋂ V┴.[/size]

Note: U┴ is the orthogonal complement of U.

1. Let $ w_1 \equal{} v_1$. Inductively define the rest. So suppose $ \{w_1, \ldots, w_m\}$ (for $ m < k$) are chosen such that for any $ l < m$ we have $ w_1, \ldots w_l$ mutually orthogonal and span the set spanned by $ v_1, \ldots, v_l$. To construct the next vector $ w_{m \plus{} 1}$ let
\[ w_{m \plus{} 1} \equal{} v_{m \plus{} 1} \minus{} \sum_{l \equal{} 1}^m\frac { < v_{m \plus{} 1} | w_l > }{\|w_l\|^2}w_l
\]
Can you fill in the details from here? If you need more help, this is called the Gram-Schmidt process. The idea is to essentially subtract off the non-orthogonal parts (that you have already taken care of) so that all you have left is something orthogonal to the previous ones (and non-zero, do you see why?).


1. Why doesn't the text wrap work for for the Shoutbox? [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=681]Example.[/url]

2. Is there a linebreak system designed for the Custom Block?

1. You mean when you have a very long word? Because CSS doesn't know how to cut that in two. You have to avoid those basically.

2. For now the Custom Block is not that customizable. I will be fully BBcodefiable in the future.


LaTeX
geometry
parallelogram
ratio
algebra
polynomial
analytic geometry

We need something challenging/interesting on here... so I'm gonna start posting some problems.  Feel free to post your solutions.  If you solve one, you can post your own questions too!

Added bonus: because I suck at LaTeX, you get my problems in deluxe ascii math style!

Problem #1)
[code]
Let n>1 be an integer.

Let S be the set of all complex numbers k such that

                          n
                        -----
                         \     2i
                          )   k   = 0
                         /
                        -----
                        i = 0

Let T be the set:
  n+1
{k   : k is an element of S}

Prove that T is always {-1,1}, regardless of the choice of n.[/code]

[color=darkred][Mod LaTeX addition, for those of you who prefer properly typesetted mathematics.][/color]

Let $n>1$ be an integer. Let $S$ be the set of all complex numbers $k$ such that
\[ \sum_{i=0}^{n} k^{2i} = 0  \]
Let $T$ be the set $\{ k^{n+1} : k \text{ is an element of } S \}$. Prove that $T$ is always $\{-1, 1\}$, regardless of the choice of $n$.

Elyot's backlash:
If you're gonna latex it, at least do it properly :P.  I changed the incorrect sum from 1 to n to a correct sum from 0 to n.

Actually I have nothing against latex (it's very nice), but the ascii art makes it much funnier in my opinion.  You gotta admit... that ascii sigma is PIMPIN! ;)

Additional backlash:
There is no such word as "typesetted".  The correct past tense of typeset is, in fact, typeset.

[quote="billzhao"]Here's a problem proposed by me in the Mathematical Mayhem (Crux v30 n7)

[size=150]M153[/size].
Two similar triangles $APB$ and $BQC$ are erected externally on a triangle $ABC$. If $R$ is a point such that $PBQR$ is a parallelogram, show that triangles $ARC$ and $APB$ are similar.[/quote]

[hide="Click to unhide my solution"]
I got:
- use angle chasing to prove angle ABC = angle APR = angle RQC.
Then basic similar triangle ratios gives:
- spiral similarity about B, ratio a/c, angle ABC, maps AB : CB
- spiral similarity about P, ratio a/c, angle ABC, maps AP : RP
- spiral similarity about Q, ratio a/c, angle ABC, maps RQ : CQ
Hence:
- ABC, APR, RQC are similar.
Then:
- desired result immediately follows from the ratio AC:AR:RC = AB:AP:RQ (RQ=BP, parallelogram)
[/hide]

Maybe some minor additions to the angle chasing step for special cases, but I think that's about it.  It originally seemed hard, but not so bad after you figure out what's going on.  Nice one, though.

Note: my solution also works if the triangles are erected internally, so long as P and Q are not the same point (otherwise things get degenerate, but theoretically still work)

Some teasing: the solution to my problem is about 3 or 4 lines... but hard to see.  There's very little complex number stuff involved... just polynomials, roots of unity, and a simple algebraic trick (or three).

[quote="Elyot"]
You gotta admit... that ascii sigma is PIMPIN!
[/quote]

No, no it's not "pimpin".  Stick to LaTeX.
--

Now onto the problem.

Let $x \in \mathbb{C}$, and $K(x) = \sum_{j=0}^n x^{2j}$.  Notice that $(x^2 - 1)K(x) = (x^2)^{n+1} - 1$.  (*)

Suppose $x \in S$.  Then $K(x) = 0$, and it implies $(x^{n+1})^2 = 1$, implying $x^{n+1} = -1,$ or $1$.  So $T$ can only contain $-1$ or $1$.  To show both values are possible, set $x^{n+1} = \pm 1$, where $x^2 \ne 1$.  Then (*) implies $K(x) = 0$, as required.

[quote="billzhao"]Here's a problem proposed by me in the Mathematical Mayhem (Crux v30 n7)

[size=150]M153[/size].
Two similar triangles $APB$ and $BQC$ are erected externally on a triangle $ABC$. If $R$ is a point such that $PBQR$ is a parallelogram, show that triangles $ARC$ and $APB$ are similar.[/quote]

Oooh I like it.  Nothing to post though, cuz my solution turned out to be basically the same as Elyot's.  Gotta love those roots of unity eh.   :P

Yes, good job... that's my exact solution.  Actually that polynomial has very very interesting properties.  You can prove that the set of roots of the polynomial 1 + x^2 + x^4 + ... + x^(2n) are the unique (2n+2)th roots of unity, excluding 1 and -1 (multiplying by n^2-1 puts these roots back into the polynomial).  For n=1, you get i and -i... but for n>=2, you will have 2 consecutive roots somewhere in there (if you order them by the angle in polar coordinates), one of which must be an (n+1)th root of 1 and the other which must be an (n+1)th root of -1.

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

I have another one... the topic this time is FUNCTIONS, SEQUENCES, AND RECURSION!

Part a) is easy and worth 1 point.  Part b) is hard and worth 6 points.

a) Prove that for any real constant k>0, the polynomial x^3 + k*x - 12 always has exactly one real positive root.

b) For k>0, Let f(k) be the real root of the polynomial x^3 + k*x - 12.  Consider the sequence a[1], a[2], a[3]... defined recursively by a[1]=1, a[n] = f(a[n-1]).  Prove that this sequence converges and find the limit of the sequence as n goes to infinity.

I also have one

Let $a_1, a_2, \dots, a_n$ be distinct reals, where $n \ge 2$ is an integer.  Prove the identity

$\displaystyle \sum_{cyc}^n \frac{(\prod_{i=2}^n a_i) (\sum_{i=2}^n a_i)}{\prod_{i=2}^n (a_1 - a_i)} = (-1)^{n+1} \sum_{i=1}^n a_i$

Wow.  That problem is... well... not too hard... but so ugly.  I could write up a solution... but something like that is just not fun to write up.

Do you have an elegant solution, or is this just a bunch of algebraic manipulation, ugliness, cancelling, etc?  I can prove a few lemmas that simplify it a bit... but dude... that question is a monstrosity!

Yes, I have elegant solution.

[hide="Hint"]
Lagrange Interpolation
[/hide]

[hide="Answer"]

Let $s = \sum_{i=1}^n a_i$.  Consider the unique polynomial $f(x)$ (of degree n-1) such that for $i = 1, \dots, n$, $f(a_i) = s - a_i$.
By the Lagrange interpolation formula,

$f(x) = \displaystyle \sum_{cyc}^n \frac{(\prod_{i=2}^n x -a_i) (\sum_{i=2}^n a_i)}{\prod_{i=2}^n (a_1 - a_i)}$

whence

$f(0) = (-1)^{n+1} \displaystyle \sum_{cyc}^n \frac{(\prod_{i=2}^n a_i) (\sum_{i=2}^n a_i)}{\prod_{i=2}^n (a_1 - a_i)}$ (*)

Consider $g(x) = f(x) + x - s$.  It is also a polynomial of degree $n-1$, that has $n$ zeroes (at $a_i$).  This implies that $g(x)$ is the zero function.  Thus, $f(x) = s - x$, implying $f(0) = s$.  From (*), the result follows.

[/hide]

1) CANADIAN team refers to UGLY, brute force solutions as GRUNGING.

2) CANADIAN national languages are English and FRENCH.

3) The FRENCH word for GRUNGE is LA GRANGE.

4) To have any sensible solution, your problem demands the use of LA GRANGE Interpolation.

5) Therefore, your problem is UGLY!

You just got PWNED!
 :stretcher: :stretcher:

Seriously though... most people don't use those kind of techniques in solutions... that problem would never appear on the IMO for that reason... or even on the Putnam.  I can easily make difficult problems that rely on very very deep knowledge in specialized areas of mathematics... it's much tougher to make challenging problems that can be solved with high school math.

Lagrange interpolation hardly qualifies as 'very very deep knowledge'.  It is usually very obvious when you want to use it, and this is certainly one of those cases (just look at the denominator!)

Uh... I agree with Elyot.

It's no fun for high school people to try and solve a problem that requires a formula that no high school students ever heard of.  Not only that, it's EXTREMELY easy to make up math problems that require advanced knowledge and like 2 tricks nobody would ever think of.  However, that's not math.  That's just showing off...

You did get pwned... Lagrange is ugly...  :P

And I think both of you are wrong.  Saying that you pwned someone doesn't equate to actually doing so, whatever you might like to believe.

Dude, don't blame your problems on me.  You should thank me now that you have a new trick under your belt.

[size=75]
[b][u]Anyways, the idea in the solution is not a new idea either.[/u][/b]  And the "formula" isn't really contrived:  you should know that n+1 points (say, $(x_i, y_i)_{i=0}^n$) uniquely determine an $n^{th}$ degree polynomial, and it is obvious to find such a polynomial $p(x) = \sum_{i=0}^n y_i \prod_{j=0, j\ne i}^n \frac{x-x_j}{x_i-x_j}$ satisfying the n+1 points.  Therefore it is the unique one.

Also, I could've bitched that your roots problem, part (a) was just the Descartes Rule of Signs.   [u]But I'm not a whiny bitch.[/u]

By the way, ascii was never cool... actually... maybe in 1995 but thats it.
[/size]

Another problem:

Find the largest real $k$ (in terms of m) so that $|xy+yz+zx| \ge k|x+y+z|$, for all $x,y,z \in \mathbb{C}$, each with absolute value $m$.

Solved your next problem... it's more fair I think.  Too easy for putnam though.

[hide]
- Convert x,y,z to m*cis(a),etc form.
- Multiply stuff together and common factor out m.
- Lemma: |cis(a) + cis(b) + cis(c)| = |cis(a+b) + cis(b+c) + cis(a+c)|
Proof:
- by obviousness, |cis(a) + cis(b) + cis(c)| = |cis(-a) + cis(-b) + cis(-c)|
- rotate the RHS about the origin by an angle of a+b+c.  Lemma proven.
So in fact, km <= m^2, so k<=m, so largest possible k is m.
[/hide]

Here is another.

a,b,c are reals.  Prove $\sum_{cyc} \frac{a}{bc-a^2} =0 \Rightarrow \sum_{cyc} \frac{a}{(bc-a^2)^2} =0$.  Is it true for complex numbers?

i'm sorry but what the hell is 'cyc'?
...maybe i can learn somethin here

Solution in spoiler.
[hide]
Let $P_k = \sum_{cyc} \frac{a^k}{bc-a^2}$.  Then $P_0P_1 = P_2$.
[/hide]

[quote="Elyot"]
2) CANADIAN national languages are English and FRENCH.
[/quote]
Correct, this is why I'll permit myself to put a quote in French (tough I've seen Canadians be kind of hostile to that language ...)
[quote="Elyot"]
very very deep knowledge in specialized areas of mathematics...[/quote]

To answer in French "C'est genant de dire"  that this is "very very deep" knowledge. This is basically knowledge any first year undergraduate must *at last* have :) Moreoever Lagrange interpolation formula is 18-th century mathematics. Needless to say it cannot be very very deep (something deep must be recent) or specialized (Lagrange had time to work in mathematics and physics so had no time for specialization ;))
Finally I've already seen Putnam question relying on a specific trick, just like here.

[quote="Singular"]Did someone call pizza hut?[/quote]

OMG.  You inspired me to craft a problem!  This will be an easy problem for many of you, and very hard if you're bad at geometric inequalities, but have fun with it.  I guarantee that there is a pretty, non-bash solution.

Let a "pizza hut roof" (PHR) be a hexagon ABCDEF such that AF, BE, and CD are all parallel.  (see the attached pizza hut logo... note that I used the old pizza hut logo, because all the symmetry is gone in the new, modern, fancy pizza hut logo).  We're going to assume all the nice symmetrical properties of the PHR are true (ie ABC reflected through a line perpendicular to AF maps onto FED).  Think of it like two isosceles trapezoids glued together.  I'm also going to give you two more things:
- AF < BE < CD, to preserve it's "pizza-hut-roofiness".
- B and E are inside ACDF (we want a nice, sleek roof... not a bloated, ugly one)
Hopefully I didn't forget anything.  You can ignore degenerate roofs.

[img]http://www.248am.com/images/pizzahut.gif[/img]

OK, now on to the problem.  Suppose that AD, FC, and BE are concurrent and meet at Q.  Let AE and BF meet at P, and BD and CE meet at R.

If p is a point inside the roof, let f(p) be the length of the line segment through p, parallel to AF, with both of its endpoints on the perimeter of the roof.  For example, f(Q) is just the length of BE (since Q lies on BE).

Prove f(Q) > 2*f(P)*f(R)/(f(P)+f(R))

BONUS: I'll also send a cookie to anybody who can make a nicer pizza hut problem (rule: everything before "OK, now on to the problem." must be the same.)

~ Elyot

Lemma:  Suppose ABCD is a trapezoid, with E the intersection of AC and BD.  Its left to the reader to prove that the length of the line through E parallel to AB and CD is $\frac{2|AB||CD|}{|AB|+|CD|}$.

--

On to the problem:  let $x,y,z$ =  lengths $AF, BE, CD$.

So we need to show $\frac{2xz}{x+z} > \frac{2 \frac{2xy}{x+y} \frac{2yz}{y+z} }{ \frac{2xy}{x+y} + \frac{2yz}{y+z} }$

It becomes $\frac{xz}{x+z} > \frac{2y^2xz}{xy(y+z) + yz(x+y)}$, becoming $y < \frac{2xz}{x+z}$.

Consider Q on XY, with XY parallel to AD, X on AC, Y on DF.  Then $y = BE < XY = \frac{2xz}{x+z}$ (BE is given inside ACDF) and result follows.

That's really a lot of algebra... do you realize 2ab/(a+b) is just the harmonic mean in disguise?  :D  Do you see the two-line solution now?

OK... First we replace the right hand side of your lemma with HM(AB, CD) where HM is the harmonic mean.

So we'll do the same thing as you did with x,y,z.  then:
2*f(p)*f(r)/(f(p)+f(r)) = HM(f(p),f(r)) = HM(HM(x,y),HM(y,z)) = HM(x,y,y,z) = HM(y,HM(x,z)).  By essentially the argument that you gave in the last line, we have y<HM(x,z).  Then since y<HM(y,HM(x,z))<HM(x,z), we're done.

It's basically the same thing you had... but it's much much easier to rearrange harmonic means than to do all of that ugly ugly algebra.

I've got a new one.  Maybe hard... but simple if you're clever.

Let ABCD be a convex quadrilateral.
Let P be the midpoint of AB.
Let Q be on BC such that 2*BQ = QC.
Let R be on CD such that 3*CR = RD.
Let O be on PR such that 3*PO = OR.
Let QO meet AD at S.
Find QO:OS.

7/9 is correct.

My solution is: place rocks m1, m2, m3, m4, of masses 6, 6, 3,and 1 on points A, B, C, and D.  P,Q,R, and O are the centres of gravity (using either vectors or mass-point methods) of (m1 & m2), (m2 & m3), (m3 & m4), and (m1, m2, m3, and m4).  So S is the centre of gravity of (m1 & m4), and thus the ratio is (6+1)/(6+3) = 7/9.

hey, i got a question, this was in a local contest, well anyways the whole test was pretty easy, except this one question:

A certain set S has 24 more subsets than a second set T. The number of elements in set S is:

(A)2  (B)3  (C)4  (D)5  (E)6

can someone plz tell me how one would approach this problem?

Any set of cardinality $n$ has $2^n$ subsets.  Thus we want $2^s=2^t+24$, or $2^{t}(2^{s-t}-1)=24$.  Hence $t=3$, $s=5$.

:huh:  :huh:

cardinalty?? wth is that??
whats the definition of an element? subset?

i still dont get how you did it >_<

lol... no one in my school got this question -____-

For finite sets, the cardinality of the set is just the number of elements.

Just curious,[b] blahblahblah [/b],when did you start learning abstract algebra? My teacher told me it's usually in the third year of university. I've started reading a book called Modern Algebra, the one was called "the most classical algebra textbook" , what type of textbooks do you use?
Plus, I was told that rational solutions of such equations: $y^2 = F(x) = x^3 + ax +b (a,b \in Q)$, (typical elliptic curves), forms groups. But why?

how do you know that any set of cardinality n has 2^n subsets?

sorry for all the question. i just really wanna understand what i dont understand :roll: lol

I started learning abstract algebra in Grade 12.  Nothing too significant, just the basic theory of groups (it was for one of my IB courses).  Since then, I've taken an undergraduate course on rings and fields and a graduate course in groups and rings with some representation theory thrown in.  The two algebra texts that I have the most experience with are Dummit and Foote's Abstract Algebra, which is a huge text and a good text to start learning from, and Isaacs' Algebra: A Graduate Course.

It's hard to say what the most classical algebra textbook is.  There's a brief discussion of a lot of them [url=http://www.ocf.berkeley.edu/~abhishek/chicmath.htm#i:general-abstract-algebra]here[/url], not having read most of them I can't really comment.  Personally, I'd hesitate to read anything too old; older textbooks are often very good but difficult to get much out of on a first reading.

As for your question on elliptic curves, I can't say that I know; I haven't taken a course in algebraic geometry yet.

There is $1$ set with $0$ elements, $\binom{n}{1}$ sets with $1$ element, $\binom{n}{2}$ sets with $2$ elements, etc.

Hence a set of cardinality $n$ has
\[ \sum^n_{j=0}\binom{n}{j}=(1+1)^n=2^n  \]
subsets.

[quote="Chocoboom"]how do you know that any set of cardinality n has 2^n subsets?[/quote]blahblahblah's answer makes use of the binomial expansion, which is part of the grade 12 curriculum. However, you can also think of it this way: each element of the set can either be in the subset or not in the subset, so there are 2 possibilities, and n elements. Thus $2^n$.

lol i havnt learned the gr 12 curriculum yet  :P

anyways, i think i get it now thanks everyone
AoPS rocks XD

Now, I'm sure all of you are a big fan of 3D geometry problems that are very hard to visualize but easy to solve.  This is one of those.  Once you figure out the trick, it's a simple problem.

Let a "pigeon" be a point in 3-space with integer coordinates.  Define a "flock" F of size n to be an ordered list of n pigeons [p1, p2, p3, ... pn] with the following properties:

- n is an integer and n > 3.
- No two pigeons have the same coordinates.
- For all integers x and y in {1..n}, if x-y-1 is divisible by n, then the distance between pigeons px and py is 1.
(for all you newbs out there, 0 IS divisibly by n for all n)
- For all integers w,x,y, and z in {1..n} if w-x-1, x-y-1, and y-z-1 are all divisible by n, then pigeons pw, px, py, and pz are NOT coplanar.  4 Pigeons are coplanar if there exists a plane in 3-space such that all 4 of them lie on it.

For example, [(0,0,0), (0,0,1), (0,-1,1), (1,-1,1), (1,-1,0), (1,0,0)] is a flock.

Prove:
a) If a flock is of size n, then n is divisible by 6.
b) For every integer k, there exists a flock of size 6k.


[quote="pontios"][quote="José"]Let be $P$ any point inside $\triangle{ABC}$. Prove or disprove:

a) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds)
for an acute-angled $\triangle{ABC}$.

b) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds)
for any $\triangle{ABC}$.[/quote]

When you first read this problem, you think that the part (a) is always true but the part (b) is not

Consider an isosceles acute triangle with $AB=AC$ enough larger than $BC$

Then take a point $P$ near to the base $BC$

You'll get $PB+PC<PA$

So, not true for all acute triangles => not true for all triangles[/quote]

What about equilaterals triangles???

http://www.mathlinks.ro/Forum/viewtopic.php?t=76616

[quote="pontios"][quote="pontios"]not true for all triangles[/quote]

Here, I mean that this fact could be true for some triangles, but not for all.

I don't mean that it is false for all triangles

I think that we can find a lot of triangles (not equilateral) such that this fact is true[/quote]

Ohhh, OK then.


algorithm
number theory unsolved
number theory

There is a keypad entry system on a safe door. There is no 'clear' button, but the safe will open automatically if the last n digits entered match the n digit 'password' required.
If there are m digits to choose from, what is the fewest number of button presses you would need to type in to guarantee access if the password is n digits long?
This is a well known problem, usually called the [b]shortest common superstring problem[/b]. It is NP hard. It has very pratical applications in genome mapping, and there's lots of work on generating a common superstring that is no worse than (some limit) x as long as the ideal string.
Can you prove or disprove that the minimal length sequence, is the total number of unique combinations + the length of the code - 1 ?
x(m,n) = m^n + n - 1

This is essentially asking for a proof of the existence of a de Bruijn sequence of $m$ alphabets and length $n$.  De Bruijn sequences exist for all $m, n \in \mathbb{N}$.

Indeed. I think this problem was posted a long time ago by alekk in the "Combinatorics" section. A nicer way to pose the problem would be considering the letters are arranged around a circle (in a clockwise direction) s.t. all words of $n$ letters are eventually obtained, and the last $n-1$ letters of a code are the first $n-1$ of the next code, and then we would require $m^n$ letters (I say it's nicer because $m^n$ certainly looks nicer than $m^n+n-1$ :)).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5624]Here's[/url] the thread that grobber was referring to.

The following is an algorithmic construction of a de Bruijn sequence of $m+1$ alphabets $0, 1, \ldots, m$ and length $n$.  (The sequence generated should be read cyclically, the way grobber described it.)

Consider the following array which is divided into two parts: the left part contains a single column which contains nodes $0, 1, \ldots, L - 1$ where $L = (m+1)^{n-1}$.  The right part contains $m+1$ rows each of which contains the labels $0, 1, \ldots, m$ in this order.

[b]Node --- Labels[/b]
$0 \hspace{6ex} \mbox{ } 0 \mbox{ } 1 \mbox{ } 2  \ldots m$
$1 \hspace{7ex} 0 \mbox{ } 1 \mbox{ } 2  \ldots m$
$2 \hspace{7ex} 0 \mbox{ } 1 \mbox{ } 2  \ldots m$
$\ldots$
$L-1 \hspace {2ex} \mbox{ } 0 \mbox{ } 1 \mbox{ } 2  \ldots m$

[b]Algorithm:[/b]
[b]Step 1.[/b]     Start with the node $i = 0$ and set the sequence to be generated to be empty.
[b]Step 2.[/b]     (Iterative) If the row with node $i$ is not empty, then append the rightmost label $s$ in the row with node $i$ to the sequence already generated and remove this label from the row and evaluate $i \leftarrow (i(m+1) + s) \mbox{ mod } L$.  Otherwise stop.

From Step 2, note if the last $n$ alphabets of the sequence generated are read as a number in base $L$, this is simply the node to go to in the next step.

[b]Lemma.[/b]     When Step 2 of the algorithm reaches the row with node $i$, then, if $i$ is non-zero, the row is non-empty, i.e. Step 2 always succeeds.

[b]Proof.[/b]  Suppose the row with node $i (\neq 0)$ is empty.  That means that Step 2 has reached it $m+2$ times, because there are only $m+1$ labels in the row with node $i$.  But this is impossible except when $i = 0$ since there are only $m+1$ numbers in the range from $000\ldots0$ ($n$ zeros) to $mmm \ldots m$ ($n$ $m$'s) which have the same non-zero remainder when divided by $L$.  Hence, the algorithm must terminate at the row with node $0$ when step 2 reaches this row for the $(m+2)$th time.

Now, we note that to reach the row with node $0$, we have to first empty a row with node a multiple of $(m+1)^{n-2}$ but not $L$.  Hence, if the row with node $0$ is empty, then all the rows with nodes a multiple of $(m+1)^{n-2}$ but not $L$ have been emptied.  To empty all those rows, we have to first empty all the rows which are multiples of $(m+1)^{n-3}$ but not $(m+1)^{n-2}$ and so on.  This means that when the algorithm terminates, the table is empty and the sequence generated has exactly $(m+1)^n$ alphabets.

That no subsequence with length $n$ can occur more than once follows from Step 2.  Thus the sequence generated is a de Bruijn sequence.

[b]Reference[/b]: Xie, Shenquan.  (1987).  Notes on de Bruijn Sequences.  [i]Discrete Applied Mathematics, 16[/i], pp. 157-177.

P.S. Thanks to Orlando for helping me fix the array. :)


If a,b,c are positive real numbers prove that
(a^7)+(b^7)+(c^7)>=(a+b+c)[(abc)]^2

(7,0,0) > (3,2,2)
then muirhead.
Of course it's the same to use amgm.

Also  from Chebychev  and  AM-GM  :
[tex]3(a^7+b^7+c^7)\geq(a+b+c)(a^6+b^6+c^6)\geq3(a+b+c)(a^2b^2c^2) [/tex]


greatest common divisor
algebra
polynomial
geometry

So, as a warmup to remember Visual Basic (which may be a bad thing), I'm going to make a math program to make my (and others') life easier. The CPU-intensive subroutines will probably be written in C++ for efficiency, and the GUI will be made in the aforementioned VB. So, the reason I'm posting is: I need ideas for useful tools people will need. Currently, I'm going to start with:

Prime factorisation
LCD
GCF
Solving/Factoring Polynomials, maybe graphing, although there's GraphCalc for that

So, I need ideas. Offer them up, and I'll integrate what's practical. Version 0.01 should be available pretty soon.

[quote="José"][quote="sponge008"]
LCD
GCF[/quote]

What's that??

An idea: If you write a number n, you're shown all divisors of n.[/quote]

Good idea, as well as the sum, product, and number of divisors.

LCD=least common denominator/multiple
GCF=greatest common factor.

How about... solving triangles (given 2 sides and an angle, or something like that) or solving linear equations.

Also, you should see your local calculator.. it has a lot of things that people find useful :)

Edit--I'm talking about stuff like TI-83 graphing calculators.
Edit II--you should also check out Mathematica.

[quote="solafidefarms"]But if you don't want to pay for Mathematica, try Maxima.

Aren't most of these things already out there, or am I just spoiled by my OS? For instance, a good graphing program, K Interactive Geometry, and Python for raw computation does everything I need.[/quote]

Target users are:
1) Computer noobs or seminoobs who don't know how to do most of this stuff and would be more able to use this than other programs (such as the aforementioned Python, or in my case, C++)

2) Lazy bastards like me who don't feel like doing each case individually.  :D


I'm experiencing trouble with my paragraphs in LaTeX. I use the book class, but after linebreaks (\\) I don't get any indents.

It might be different with the book class, but I doubt it: Make sure you have a blank line before the new paragraph starts (\\ does a line break, NOT a paragraph break), and if you REALLY REALLY REALLY need to put indents in [i]artificially[/i], use \indent.

But isn't standard to get an indent after a linebreak in a paragraph?

A line break simply puts you on the next line of the [i]same[/i] paragraph. It doesn't start the next paragraph, and so it doesn't indent. If you want to go to the next paragraph, leave a blank line before you start the paragraph.


Remember [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1472359248&t=148061]Stretch[/url]?  It's back.

[b]Rules:[/b]
You may post whenever you like, as long as you were not the last poster.
When you post, you must add one letter to the current word and anagram it.
The resulting word must be a real English word by the official game dictionary found   [url=http://www.scrabulous.com/twl_dictionary.php]here[/url].
Please post all previous words as well as your new word [b]in bold[/b].
Use your word in a sentence to assure you know what it means.
You may not post a word with the same basic root as the previous (LOBSTERS may not follow LOBSTER).
If you believe that no word can follow the previous word, you may type the word [b]FORFEIT[/b] to end the round.

[b]Scores:[/b]
When you end a round, scores are added as follows:
Whoever posted the last word receives N-5 points, where N is the number of letters in the word.
You receive 3 points as well, just for ending the round.
If you end a round when there was another possible word which a player points out, you lose 3 points instead of gaining 3.
There is also a 3 point penalty for posting a word that does not exist in the official dictionary.  Check it before posting.
Please post the scoreboard after ending a round.

[b]Example:[/b]
SIT
SUIT
QUITS
SQUINT
INQUEST
ANTIQUES
QUAINTEST
QUANTITIES
EQUIDISTANT

---

[b]PIG[/b]

Sentence: "A pig is an animal."

[b]PIG
GRIP[/b]
"The man had a strong grip."

[b]PIG
GRIP
GRIPE[/b]

"His tone of voice GRIPES me"

Scores:

miyomiyo: -2
cf249:-1
BOGTRO: 0

lol...

To clear things up: nobody receives points until the round ends.  There are no scores currently.  Also, only the most recently added word should be bolded.  Sorry if that was unclear.

PIG
GRIP
GRIPE
The [b]PINGER[/b] continued to ping obnoxiously.

(I think that's a word.)

PIG
GRIP
GRIP
PINGER
I had fun [b]REAPING[/b] the benefits.

Problems:
-The dictionary we are using does not recognize words that are greater than 8 letters.
-Is there a penalty for saying a word that does not exist?
-Soup?

PIG
GRIP
GRIP
PINGER
REAPING
I'm [b]REPAYING[/b] Sam the fourteen fifteenths of my soul that I owe him.

Solutions:
- I replaced with this [url=http://www.scrabulous.com/twl_dictionary.php]better dictionary[/url].
- 3 point penalty.  I'll add that into the rules.
- Absolutely.

PIG
GRIP
GRIP
PINGER
REAPING
REPAYING
[b]PREPAYING?[/b] (does that count?)
I am [b]prepaying[/b] 4/5 of the lease on the car

Sure, though REPLAYING was more obvious.

PIG
GRIP
GRIP
PINGER
REAPING
REPAYING
PREPAYING
[b]REAPPLYING[/b]

There's no better way to not get sunburned than reapplying sunscreen every fifteen minutes.
(As a challenge, I'm using 15 in every sentence I write.)

TEA
TEAL
[b]STALE[/b]

"The bread was stale."

TEA
TEAL
STALE
[b]STEALS[/b]

miyomiyo:7

TEA
TEAL
STALE
STEALS
[b]ATLASES[/b]

i looked up the location of Kazakhstan on many ATLASES.

miyomiyo has 5. read the rules

TEA
TEAL
STALE
STEALS
ATLASES
[b]LACTASES[/b]

Milk has many lactases (I think).

ABY
BABY
TABBY
[b]BATBOY[/b]

the guy stunk at hitting, but he was so overpaid they decided to make him a batboy and force him to do all the work in the dugout

[b]FORFEIT[/b]

New Word:
[b]HIS[/b]
The boy said the fallen quarter was his.

Last round, both yongyi and james were given 3 points, when forfeiting and "flagons" should both earn just 2 points.  Previous to that, 33286 earned a point and somehow I missed that he was already on the list, and put him on again.  These should be combined into one score.  The scores have been adjusted accordingly.
[hide="Score:"]
1. mathforlife: 16
2. BOGTRO: 14
3. james4l: 13
3. 33286: 13
5. mz94: 12
6. Flame: 9
6. miyomiyo: 9
8. woser456: 7
9. Yongyi781: 6
10. Quickster94: 5
10. Brut3Forc3: 5
10. i_like_pie: 5
13. lingomaniac88: 4
14. distracted523: 3
15. infinity4ever: 2
15. cf249: 2
15. krakola45: 2
18. pacman2812: -6[/hide]

HIS
[b]THIS[/b]

I could have formed an inappropriate word, but instead I decided to use this.

[hide="Score:"]
1. mathforlife: 16
2. BOGTRO: 14
3. james4l: 13
3. 33286: 13
5. mz94: 12
6. Flame: 9
6. miyomiyo: 9
8. woser456: 7
9. Yongyi781: 6
10. Quickster94: 5
10. Brut3Forc3: 5
10. i_like_pie: 5
13. lingomaniac88: 4
14. distracted523: 3
15. infinity4ever: 2
15. cf249: 2
15. krakola45: 2
18. pacman2812: -6[/hide]

HIS
THIS
[b]SHIFT[/b]

The boy shifted his position on the couch.

HIS
THIS
SHIFT
[b]FIFTHS[/b]

I will divide the pie into fifths, then eat 20 percent.

[b]Forfeit.[/b]

New Word:
[b]DAY[/b]

An apple a [b]day[/b] keeps the doctor away.

[hide="Scores"]1. mathforlife: 17
2. BOGTRO: 14
3. james4l: 13
3. 33286: 13
5. mz94: 12
6. Flame: 9
6. miyomiyo: 9
8. woser456: 7
9. Yongyi781: 6
10. Quickster94: 5
10. Brut3Forc3: 5
10. i_like_pie: 5
13. cf249: 4
13. lingomaniac88: 4
15. distracted523: 3
16. infinity4ever: 2
16. krakola45: 2
18. pacman2812: -6[/hide]

[b][i][u]DRAY.[/u][/i][/b]
Medieval people used a dray to haul objects.

must be hard for you scorekeepers  [hide=":jump:"][hide=":jump:"][hide=":jump:"]o.O Those smileys stop idk[/hide][/hide][/hide]

I dispute 1=2's sentence... but w/e.

DAY
DRAY
HARDY
[b]HARDLY[/b]

The poor family [i]hardly[/i] had any food.

DAY
DRAY
HARDY
HARDLY
[b]HALYARD[/b]

you use a halyard on a ship to lower the flag

DAY
DRAY
HARDY
HARDLY
HALYARD
[b]CHARLADY[/b]

Charlady is the British equivalent of "charwoman", which is a woman who cleans in a large house or office.

[b]FORFEIT[/b]

New Word:
SAW

I saw all the ARML teams in their bright shirts.

[hide="Score"]
1. mathforlife: 19
2. BOGTRO: 14
3. james4l: 13
3. 33286: 13
5. mz94: 12
6. Flame: 9
6. miyomiyo: 9
8. woser456: 7
9. Yongyi781: 6
10. Quickster94: 5
10. Brut3Forc3: 5
10. i_like_pie: 5
13. cf249: 4
13. lingomaniac88: 4
15. panjia123: 3
15. distracted523: 3
17. infinity4ever: 2
17. krakola45: 2
19. pacman2812: -6
[/hide]

SAW
[b]PAWS[/b]

The dog used its paws to change the channel to a nature documentary

SAW
PAWS
[b]WARPS.[/b]
Warps are science fictional thingys to get from one place to another, usually.

SAW
PAWS
WARPS
[b]SPRAWL[/b]

My cats sprawl on the floor and sleep all day.


geometry
3D geometry
tetrahedron
function
geometry solved

I assume $P$ can also be on the border of $ABCD$, because otherwise the maximum can't be reached (in case the maximum is really what we want; are you sure the problem didn't ask for the minimum?).

I'll use the fact that the distance between two points in (or on) the regular tetrahedron with side $\ell$ is at most $\ell$. Through $P$ draw a plane parallel to $(BCD)$, cutting $AB,AC,AD$ in $B',C',D'$ respectively. In the plane $(B'C'D')$ draw a parallel through $P$ to $C'D'$, cutting $B'C',B'D'$ in $C'',D''$ respectively.

We now have $PA\le AB'\ (1)$, $PB\le PB'+BB'\le B'C''+BB'\ (2)$, $PC+PD\le DD'+D'D''+D''C''+C''C'+C'C\ (3)$. If you draw the picture properly you'll see that the sum of the three RHS's is exactly $3\ell$ ($3$, in our case, since $\ell=1$). The equality is reached only when $P$ is one of the vertices of the tetrahedron.

Really P can be on the boundary.
Nice solution.

[quote="hardsoul"]Given a regular  tetrahedron   ABCD   with edge length 1 and a point P inside it.
What is the maximum value of   |PA|+|PB|+|PC|+|PD|.[/quote]

I am just seeing that this is a corollary of the well-known fact that the maximum of a convex function inside a convex polyhedron is reached in a vertex of the polyhedron. This time, the convex function is |PA| + |PB| + |PC| + |PD| (the distance function is convex!). Actually, Grobber's solution was just an elementary formulation of this.

[b]EDIT:[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=19485 for details.

  Darij

[color=darkred][size=150]Excuse me .
i wanted to know why the function |PA| + |PB| + |PC| + |PD| is convex .
thanks for your help.[/size][/color]

[quote=hardsoul]Given a regular  tetrahedron   $ABCD$   with edge length $1$ and a point $P$ inside it.
What is the maximum value of   $\left|PA\right|+\left|PB\right|+\left|PC\right|+\left|PD\right|$.[/quote]

We claim that the maximum value is $3$. Equality occurs at any one of the vertices. Let $f(P) = AP + BP + CP + DP$. Suppose $P$ is the point for which $f$ attains it maximum value.

[b]Claim 1:[/b] $P$ lies on one of the face of the tetrahedron

[i]Proof.[/i] Suppose not then consider the ellipsoid $\mathcal{C}_1, \mathcal{C}_2$ with foci $(A,B) ; (C,D)$. with sum of distance from foci $AP + BP$ and $CP +DP$ respectively. Note that there exists another ellipsoid $\mathcal{C}_3$ with foci $(C,D)$ and sum of focal distance slightly larger than that of $\mathcal{C}_2$ such that $\mathcal{C}_3$ meets $\mathcal{C}_1$ inside the tetrahedron. This leads to a contradiction and we get the desired result. $\square$

[b]Claim 2:[/b] $P$ lies on the sides of equilateral triangle $\triangle ABC$ if $P$ maximizes $g(P) = AP + BP + CP$

[i]Proof.[/i] Similar to proof of [b]Claim 1[/b]. $\square$

By our [b]Claim 1[/b] and [b]Claim 2[/b], we have $P$ lies on the edges of the tetrahedron let’s say $AB$. Note that $AP + BP + CP + DP = 1 + CP + DP \leq 3$. $\blacksquare$


Let $ f(x,y) $ be a periodic function satisfying the condititon $ f(x,y) = f ((2x+2y),(2y-2x)) \forall x,y \in \mathbb{R} $. Now define a function $ g(x) = f ( 2^X , 0 ) $ . Then prove that $g$ is periodic.

[quote="santosh_blr"]How do u define periodicity of functions of MULTIPLE varaibles ??  :?:[/quote]

I ask it too...


I had a question about subsequences and de limit inferrior and superrior

1. I have a bounded convergent sequence xn

Is it then always correct to say that lim inferrior=limit superrior=Limit of xn?
Is there a theorem that concludes this?

2. I have a bounded non-convergent sequence yn

I want to make a subsequence for the limit superrior, How do I do that without knowing yn? (I only know it is bounded and non-convergent)

1 is correct, you should use the fact that if a sequence is convergent then every its subsequences are also convergent to the same limit.


Let $ABCD$ be a cyclic quadrilateral, let $M$ be the midpoint of $AB$, let $N$ be the intersection of $AC$ and $BD$, and let $O$ be the point where the line passing through $M$ and $N$ intersects $CD$.

Prove that $\angle DON=\angle CON=90^{\circ}$.

edited to correct problem statement.

Sorry but I still don't think the problem's statement is correct. Please look at the picture I added.

Indeed.  Maybe I transcribed the problem wrong; I'll check when I get home.  Thanks.


calculus
integration
limit
LaTeX
trigonometry
complex analysis
complex analysis unsolved

1. What is the best way of getting $ Res(f,0)$
   while $ f(z) \equal{} \frac {z}{sinz(1 \minus{} cosz)}$
    I know it cat can be done with $ \displaystyle\lim_{z\to0}\frac {d}{dz}(z^2f(z))$
    but - is there a better way?

2. what is $ \int_{|z| \equal{} 2} zsin(\frac {z \plus{} 1}{z \minus{} 1})$
    (not the answer of course but some direction).

Thanks

Hey Nick, suppose $ f(z)$ is analytic for $ |z| > R_0$, then for $ R > R_0$:

$ \displaystyle\mathop\oint\limits_{|z| \equal{} R} f(z)dz \equal{} 2\pi i\mathop\textbf{Res}\limits_{z \equal{} 0}\left[\frac {1}{z^2}f\left(\frac {1}{z}\right)\right]$

In case you're interested in checking it numerically via Mathematica:

[code]
In[21]:=
g[z_] := z*Sin[(z + 1)/(z - 1)]
NIntegrate[g[z]*2*I*Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}]
N[4*Pi*I*(Cos[1] - Sin[1])]

Out[22]=
-7.771561172376096*^-16 - 3.7845972565184955*I

Out[23]=
0. - 3.7845972369939314*I
[/code]

I would calculate the residue like this (and I think everyone else would too). It takes a long time to type in latex, but it's quick on paper.

$ \frac {\sin z}{z} \equal{} 1 \minus{} \frac {z^2}{6} \plus{} O(z^4)$ and $ 1 \minus{} \cos z \equal{} \frac {z^2}{2} \plus{} O(z^4)$ so their product is $ \frac {z^2}{2}(1 \plus{} O(z^2))$. Using $ \frac {1}{1 \plus{} z} \equal{} 1 \plus{} O(z)$ we get

$ f(z) \equal{} \frac {2}{z^2} \plus{} O(1)$ so the residue is zero.

[quote="Kalle"] (and I think everyone else would too)[/quote]
$ f(z)\equal{}f(\minus{}z)$ hence residue at point $ 0$ is zero.  :blush:


number theory
least common multiple
modular arithmetic
number theory unsolved

Given $ a,b,n\in\mathbb{N}^*$ satisfy $ \gcd(a,n) \equal{} \gcd(b,n) \equal{} 1$. Prove that:
$ \mathrm{ord}_nab \equal{} \mathrm{lcm}(\mathrm{ord}_na;\mathrm{ord}_nb)$

That's wrong, you just have $ \mathrm{ord}_n(ab) | \mathrm{lcm}(\mathrm{ord}_n(a),\mathrm{ord}_n(b))$. Otherwise already $ a\equiv b\equiv \minus{}1 \mod n$ is a counterexample.

Oh, I see, so sorry.

The problem first is: If $ \gcd(\mathrm{ord}_na;\mathrm{ord}_nb)\equal{}1$ then $ \mathrm{ord}_nab\equal{}\mathrm{ord}_na.\mathrm{ord}_nb$.

Then if we remove the condition $ \gcd(\mathrm{ord}_na;\mathrm{ord}_nb)\equal{}1$, could you please tell me how can we express $ \mathrm{ord}_nab$ in terms of $ \mathrm{ord}_na$ and $ \mathrm{ord}_nb$

I think that $ ord_n a$ is the least number such that

$ a^{ord_n a}\equiv 1$ $ mod$ $ n$

Right.

Mashimaru: In general, you cannot express $ \mathrm{ord}_n(ab)$ solely with $ \mathrm{ord}_n(a)$ and $ \mathrm{ord}_n(b)$. It depends on $ a,b$ itself. The most general thing is the one I mentioned: $ \mathrm{ord}_n(ab)$ divides $ \mathrm{lcm}(\mathrm{ord}_n(a),\mathrm{ord}_n(b))$.

Greatly thanks to you, ZetaX.

We can put $ \mathrm{lcm}(\mathrm{ord}_na;\mathrm{ord}_nb)\equal{}k_a\mathrm{ord}_na\equal{}k_b\mathrm{ord}_nb$ then:
$ (ab)^{\mathrm{lcm}(\mathrm{ord}_na;\mathrm{ord}_nb)}\equal{}(a^{\mathrm{ord}_na})^{k_a}.(b^{\mathrm{ord}_nb})^{k_b}\equiv 1\pmod n$ so $ \mathrm{ord}_nab|\mathrm{lcm}(\mathrm{ord}_na;\mathrm{ord}_nb)$


floor function
ratio
geometric sequence
algebra proposed
algebra

Prove that among the elements of the sequence $\left\{ \left\lfloor n\sqrt{2003} \right\rfloor \right\}_{n\geq 1}$ one can find a geometric progression having any number of terms, and having the ratio bigger than $k$, where $k$ can  be any positive integer.

[i]Radu Gologan[/i]

Fix two positive integers $k,m>1$. Now take some $n$ for which $\{n\sqrt{2003}\}<\frac 1{k^m}$. Now, for each $i\in\overline{1,m-1}$, we have $\lfloor k^{i-1}n\sqrt{2003}\rfloor\le\frac{k^in\sqrt{2003}}k<\lfloor k^{i-1}n\sqrt{2003}\rfloor+\frac 1k$, which, after multiplication with $k$, gives $k\lfloor k^{i-1}n\sqrt{2003}\rfloor\le k^in\sqrt{2003}<k\lfloor k^{i-1}n\sqrt{2003}\rfloor+1\Rightarrow\lfloor k^in\sqrt{2003}\rfloor=k\lfloor k^{i-1}n\sqrt{2003}\rfloor$.

This means that the numbers $\lfloor k^in\sqrt{2003}\rfloor,\ i\in\overline{0,m-1}$ are $m$ numbers in geometric progression with ratio $k$.


Let a>0 and the sequence xn=a^1/n. Calculate the limit of
(n^3)(x_n-x_(n+1)-x_(n+2)+x_(n+3)).

One idea wich conduct to 4lna, an asymptotic expansion of x(n) when n
tends to +\infty
x(n) = 1 +lna/n + (lna)^2/2n^2 + (lna)^3/(6n^3) + o(1/n^3)
do the same for x(n+1), x(n+2), x(n+3) and get 4.lna wich probably the coefficient of 1/n^3 in the asymptotic expansion of
x(n)-x(n+1)-x(n+2)-x(n+3)


Ok, as asked by some members, we will try to make a guide in order to defend yourselves in spanish there in Mexico:

[b]Basic Supervivence[/b]

Hi My name is (your name here)= Hola, me llamo (your name here).

I am from (your country here)= Yo soy de (your country here)

what time is it?= Que hora es?

Whats your name?= Como te llamas?

How did you do in the test?= Como te fue en el examen?

Where is the bathroom?= Donde queda el baño?

I want some tequilla please!= Quiero Tequila por favor!

Could you help me please, i am lost= Me puede ayudar, estoy perdido

Where is the dinner?= Donde es la comida?

I am in room (your room number)= Estoy en la habitacion (room number)

I love Mexico= Amo Mexico

I am a Mathlinker, my nickname there is (your nickname)= Yo soy un  Mathlinker, mi nombre de pila es (nickname).

Where is (name of hotel) Hotel?= Donde queda el hotel (nombre del hotel)?

We need a wake up servie at 7 am= Necesitamos servicio de despertador a las 7 (siete!) am

I cant find my sit for the exam= No puedo encontrar mi puesto para el examen

How much is it?= Cuanto vale eso?

Please give me a (name of product you wanna buy)= Por favor me vende un (name of product)

[b]More advanced stuff[/b]

Hi girl, you are very good looking= Hola nena, eres muy linda

Can you give me your phone number?=Me puedes dar tu numero?

So we can party tonight?= para que salgamos esta noche?

I will be going with 600 friends that are staying with me at the hotel!= Voy con 600 amigos que estan conmigo en el hotel :D

I hope it helps, wanna know something else? let me know! :D

[quote="Pascual2005"]what time is it?= Que hora es?

I hope it helps, wanna know something else? let me know! :D[/quote]This is no good as long as we don't know how to count in spanish :P :D

[quote="Valentin Vornicu"]This is no good as long as we don't know how to count in spanish :P :D[/quote]
Actually, "Voy con 600 amigos que estan conmigo en el hotel" would be even more amusing, because I'm sure that anybody who's learning from the above list wouldn't be able to say 600 in Spanish, and I certainly can't imagine saying "Voy con [i]six hundred[/i] amigos que estan conmigo en el hotel". :D

1=uno
2=dos
3=tres
4=cuatro
5=cinco
6=seis
7=siete
8=ocho
9=nueve
10=diez
11=once
12=doce
13=trece
14=catorce
15=quince
16=dieciseis
17=diecisiete
18=dieciocho
19=diecinueve
20=vainte
30=treinta
40=cuarenta
50=cincuenta
60=sesenta
70=setenta
80=ochenta
90=noventa
100=cien
600= seicientos

How do you say 21? Vainte e uno? (just in case someone decides to play some blackjack :D :D).

The only sentence I know in Spanish is: "Ola chica, como estas, tu es muy querida..."
"querida" may be substituted by "caliente"... :D

[quote="mahbub"]Well, what will be spanish of:

1) I don't know spanish much.

2) Can you speak english?

3) Thank You

4) Good morning

5) Good night

6) Bye[/quote]
Well,
2)¿Puede usted hablar inglés?
3)Gracias.
4)Buenos días.
5)Buenas noches.
6)Adios.
   (There's several for this one) Some are more formal than others.

***And remember, if you are talking to someone older than you, someone you don't know, or someone who respect needs to be shown to, use the usted form and not tu. ***

Hope this helps!!!  :)

1) is "no se mucho espanol"

2) might also be said "habla usted ingles?"

[quote="matty"]Well...I know this in Spanish:

lindos ojos !!

is that right??
 :D[/quote]
If you wanted to say "Cute Eyes" yes, it is right.
By the way, if someone wants to know his/her guide name, just tell me, maybe i already meet him/her.
See you in Mérida

[quote="mahbub"]i am noot in any imo. so what do u mean by guide? will the host country give a guide with each country? if so then who is for our country?[/quote]yes, every country has a guide, to help that team familiarize with the hosting country's way of things :)

[quote="Dave Vather"][quote="matty"]Well...I know this in Spanish:

lindos ojos !!

is that right??
 :D[/quote]
If you wanted to say "Cute Eyes" yes, it is right.
[/quote]

yup!
this is what i meant!
 :D

[quote="Valentin Vornicu"][quote="Dave Vather"][quote="mahbub"]AHA! that's good!

well how long they will stay with us daily?[/quote]
I heard that they must wait you patiently to you arrive, and must be whit you everytime, Go to bed after the last one of your country and be awake before the first one.[/quote]That's a tough life :) Especially when we all know some teams don't go to bed to early :D :heli:[/quote]
Its a little prize considerating that you are going to the IMO free. Also most of the guides are math students that make lots of partys, so they are used to going to bed late. (But don´t tell your guide that i told you this)

[quote="kueh"]How do I say: Did you score 21 today?  ;)[/quote]I think you need to learn how to spell that in Chinese :D :D

To Kueh: "Lograste 21 puntos hoy?" look for how to spell 21 "veinta ei ouno"

To Valentin: Yes, some teams dont even get to bed!

To Igor: Dont worry Igor, Latin girls are hot! :)  Ask Orl! :D

[quote="Igor"]Dave Vather: Did you meet the french guide  :P  ?[/quote]
Yes, In fact, you have one of the better guides of the contest. He is one of the few guides that know guides and he speaks french very well, since he lived in France. I don´t know his name, everybody here know him as "El french"

As far as i know there are no bad words in spanish?
is there any.

[quote="Vahe Musoyan"]Dave Vather:
Do you know if Armenia guide is a girl?  :P
And what's her(his) name?[/quote]
It seems to be not yet defined. You should ask niño, maybe he knows.

By the way, there are lots of bad words in spanish

[quote="Dave Vather"]By the way, there are lots of bad words in spanish[/quote]I would really appreciate if you guys don't start posting them here!  ;)

Who knows anything about the Albanian guide.


to play: Label the letters A-Z with points 1-26, respectively. Then think of a word that begins with A in less than one minute. If you fail, you have to use the letter A. Write your word down. Repeat with the rest of the letters. Add up your total points, and that's your score.

[hide="mine"]

aerodynamics
bullion
californium
dynamite
euphemism
francium
granite
hallway
igneous
jocular
kiss
lonely
mullet
novice
overestimate
pollution
quotation
rainbow
sodium
tainted
underestimate
vociferous
wasteful
xylophone
yellow
zebra[/hide]

My score: 2357

So in the first minute, you think of a word that begins with the letter a, and in the second minute you think of a word that begins with the letter b, so on. Then calculate your score.

you add the point value of your letters, like the word "am" will get 1+13=14 points. So will the letter n.

Okay, lemme try:

[hide="An attempt"]
Assassinations
Bosses
Cluttering
Dizziness
Extraction
Fuzzy
Grosses
Happiness
Isosceles
Jazzy
Kisses
Lozenges
Mosses
Nastiness
Organizations
Preppily
Questioningly
Razorblades
Seizures
Testosterone
Unanimous
Verifyingly
Wastefully
Xylophones
Yellow (Y is a hard one)
Zyzzyva[/hide]

I think it's better with 10 sec.  That's how I did it.  I don't feel like adding mine up.

I can write a program easily to calculate scores

If I ever feel like it

[hide="blargh"]
argumentative
beryllium
condominium
dysprosium
ectoplasm
fluorine
gadolinium
heterozygous
indexing
juxtaposition
kindling
luminous
mendelevium
neodymium
osmosis
praseodymium
quadrillionth
rubidium
strontium
typewriters
uranium
vanadium
wryly
xylophone
ytterbium
zyzzyva (yeah I stole this one)
[/hide]
score: 3335

I realized after I was past [hide]c[/hide] that I missed [hide]chlorofluorocarbon[/hide].

miyomiyo: you got 3258

[hide]antidisestablishmentarianism
bystander
cretaceous
disillusionment
epidemiologist
fraternity
geophysicist
hydrothermal
integumentary
janissary
kaleidoscope
lithification
mesosphere
neanderthal
ornithologist
pseudoscience
quintessential
reexamination
syzygy
technological
usurper
viviparious
wee-wee
xylophone
ytterbium
zulu

beh[/hide]


geometry
AMC
USA(J)MO
USAMO
Putnam
probability
IMO Shortlist

He has already won 3 gold medals in IMO 2000,2001,2002. If he won a gold medal in IMO 2003, then he will have 4 gold medals. He will be the second IMO contestant after Reid Barton to have 4 gold medals.

Yes he did. He also had won a bronze medal before his four gold ones.

You mean  he won 5 medals at the IMO 1 bronze and 4 gold medals?

so basically reid is not a record man anylonger :D

Yes he is .. He also got a gold medal 2 times in the IOI and as far as I know no one so far has won 2 different international olympiads in the same year :D He won the compsci one too in 2001 with a score of 580 out of 600.

and he also got another gold medal in some other olympiad, biology or chemistry - that's a rumor that i coudn't infirm or accept by now

IOI(1 gold medal), IMO(4 gold medals) and (chemistry or biology) he must be a genius! :D ( i'm talking about reid barton)

well Christian Reiher in maths won more medals than Reid Barton, so in maths C.R. is the record holder!

:D interesting debate

I got to meet both of them very briefly at the IMO 2002 (Reid was in the army of USA leaders )

one more point in favor of Reid : Reid was 18 when he won the last gold medal at the IMO, and won the highest score ever in the IOI, while Christian is older, 19 or even more

I can't imagine him staying so long in high school because he has study problems, because I heard he even did some research, so it's just a trick to keep him in business

Nikolai Dourov is also very strong, his average IMO score is higher than Reid Barton.

Nikolai Dourov:
IMO gold medals 1996,1997,1998
IOI gold medals 1996
IOI silver medals 1995,1997,1998
IMO 1996 7 7 7 7 2 7
IMO 1997 7 7 7 4 7 7
IMO 1998 7 7 7 7 7 4
average IMO score = 38.33

Reid Barton
IMO 1998 4 7 7 7 0 7
IMO 1999 7 7 7 7 2 4
IMO 2000 7 7 7 7 7 4
IMO 2001 7 7 7 7 7 7
average IMO score = 36.75

Reid has submitted this year one of the most beautiful and hardest problems ever to be found on the IMO ShortList - it's a shame that it didn't got elected - we will have to wait up untill next year to check it out!

Mihai Manea also scored 3 gold medals and next year proposed a problem at the IMO 2002: problem no. 4.

What problem is that? Algebra, Geometry, Combinatorics or Number Theory?

Talking about the numbers of medals and scores , it seems that every country has his genius! :D

hey wait,but isnt ciprian the only one to have three perfect scores(if i'm not mistaken).but i really luv to know which of these math guys were brilliant at physics!!.that wud be the sign of real genius ,in my view :D

ciprian manolescu was a genius at almost everything he put his mind into - unfortunately in Romania you CANNOT participate to more than ONE international olympiad ... damn :(

well I am from ROmania and I didn't know that... anyway.. the people who came out with this "rull" should take the example of other contries! some of them won in the same year gold at  IMO and IOI, perhaps we have fellows that can do that too, but they are stopped by this "rull" ..  :(

There are many people who are good in both math and physics. For example, Gregory Price, Timothy Abbott, Daniel Gulotta, Ari Turner, Steven J. Byrnes. All of them are from USA, and they have good results in USAMO and USAPhO. Ari Turner is an IPhO gold medallist, and he is also a Putnam Fellow(top 5). Gregory Price is good in Math, Physics and Infomatics, he represented USA in IPhO and IOI and performed very well in USAMO, being a winner of USAMO 2001 and 2002 (top 12).

Since fredbel6 accused Germany and C.Reiher of cheating and no-one expressed a different opinion, let me do it. I do not know Reiher nor R.Barton personally and I have not met them. I just know of them. I believe that both of them are brilliant and I would not like to compare them. How could I?

However, I think that the above accusation for a student like Reiher and a country as Germany is very bad and above all unfair.

I cannot accept that it is so easy to fool the IMO committees and so many math leaders and have students compete at the Olympiad without being eligible. Especially, if these students have attracted enough publicity, like Reiher, with their achievements. If there was something that violates the rules of the IMO, the other countries would react. Moreover, if he was eligible because they "kept" him in high school so that he breaks Barton's record as fredbel6 suggested, how were they so sure that he would get a gold medal in the IMO 2003? And without accusations of this type...? It is like a conspiracy and it does not make any sense to me.

I checked the website

http://www.bundeswettbewerb-mathematik.de/imo/

and it says that C.Reiher attends KLASS 13 and that this is his fifth olympiad. If you check the previous years at that site, you will see that it is not at all rare for the German team to have members that attend that klass; i.e. to be so "old". In 2002, he attended KLASS 11 and so on. I must say that I am not familiar with the German system though.

Also, I do not know how old he is and so I do not base my judgement on rumors. Even if he is 19 (is this true 100%?) the eligible age is 20, as long as you do attend high-school because of your country's system.

Also, even if he did some research, this does not imply anything suspicious either. It is not at all rare for students of that level to do some kind of research. You may find many examples, especially here in the USA.

I hope you all agree!

How come a 13-grader could be regarded as cheating if he's under 20? Hong Kong Team has many F.7-ers. (equivalent to grade 13) Without knowing the education systems one just shouldn't accuse another of that.

edit: removed (silly mistake, didn't see this had multiple pages)

[quote="darij grinberg"]Concerning the German education system, it is not seldom to find 19 year olds at school; we have 13 grades and there is usually no reason for anybody to jump over a grade unless he is very bored and very good as I was when I jumped over the 2 grade. When I jumped over the 6 grade I was not so good any more and I indeed have missed some important stuff so that my grades in biology, chemistry, sports now are not really pleasant :-) And moreover, I have lost two olympiad years - damn!

  dg[/quote]
are you still in high-school? :)

Of course I am - in the 11 grade, and I still have 3 years of math olympics before me. I have actually a chance to get into the IMO this year - if I will belong to the 6 best of the 16 training participants.

  Greetings, Darij

[quote="Anonymous"][...][/quote]

Sorry - I just had forgotten to login.      Darij

[quote="Achilleas Sinefakopoulos"]Moreover, if he was eligible because they "kept" him in high school so that he breaks Barton's record as fredbel6 suggested, how were they so sure that he would get a gold medal in the IMO 2003? And without accusations of this type...? It is like a conspiracy and it does not make any sense to me.[/quote]

 s it illegal to keep them in high school btw :?

[quote="Valentin Vornicu"]that's nice, so I will expect you on the contestants list on MathLinks contest? :D (in case I put more geometry questions :D:D).[/quote]

Well, I'll try... thanks anyway, btw, what do we get as prizes? an upgrade of the user level?

  Darij

well I gaved diplomas and in the first edition an insignia, and in the second some money (not much though - some 100 dollars) to the winners.

It's called a surprise prize you know :)

But pretty much you earn by learning from the contest :P.

Aha, clear, it was really a joke question...

  dg

PS. I have finally decided to take part in the contest after I have solved one problem...

Well, I don't really understand why do people speak so mcuh about Reiher who won his first gold medal in the 10th grade and they forget about so many Russians who score higher gold medals in the 9th grade?
It's a lack of luck that Russia has just 11 grades, or the Russians could smash everybody!

well scoring 4 golds is a great achievement, even if you start in 10th grade.

I see that you are from Moldovia, iura. There is a Moldovian student, Boreico which has scored last year (8th grade!) a silver medal with 28p. He participated at the Romanian NMO and went first on the NMO at 9th grade, and also 1st at the Romanian TST (unofficially). He will probably score 4 gold medals as well at the IMO. It remains to be seen how many full scores will he achieve (I belive he's the only one by now capable of making Ciprian Manolescu's Record :D). Do you know him?

Yes, it's me.
Anyway, it's early to speak about 4 gold medals or even 3 perfect scores!
I will do my best, I hope. Time will show what is this "best"

Hi,

I doubt your statement that Russia would have many more IMO record if they had 12 grades at school. I agree that it increases their opportunities for sure but not too much.

Contestans usually cannot participate too often in IMO since there is a lot of competition among the participating students. Imagine a huge country with many people, what are the chances, that a student of 9th (or even younger) defeats somebody from 10th or 11th grade ?

I want to provide an example. Imagine a student from Sankt-Peter(s)burg (formerly Leningrad) which have a superb math tradition. There students have to be the best at 3-4 rounds, then the Russian NMO follows. Finally they still have to be among the TOP6 at the Russian TST(s). It is possible but it requires a unique performance, doesn't it ?

Anyway I wish you good luck at IMO !!!

Maybe you are right, but imagine the the Russian team at 2002 IMO(6 gold medals) would participate at 2003 IMO (that would probably happen if there were 12 grades). Don't you think they would smash China?

Of course they would smash China, but they wouldn't have smashed Bulgaria (which distroyed China). China had the worst team in years last year ... so I guess we'll see an incredible come back this year ;)

anyway, as I see it Moldovia is approaching her first presence in top 10 countries, with her best 3 students better than all Romanian students this year :(

there is another side with this 12 grade thing. Think how many of the Russians which went to IMO in 9th grade would go in if 12th grades were allowed ... same number as USA 9th graders, Romania 9th graders, China 9th graders ... at most 1 (USA had an exception in 1999 when they had two 9th graders with gold medals: Reid Barton and Gabriel Caroll -> but that was a big exception). So again the probability of scoring a 4-times-gold medal at the IMO is smaller ...

[quote="iura"]Maybe you are right, but imagine the the Russian team at 2002 IMO(6 gold medals) would participate at 2003 IMO (that would probably happen if there were 12 grades). [/quote]

I agree with Valentin that introducing a further grade, e.g. a 12th grade, does not increase chances too much. And your argument is tricky iura as you suppose that right now the new grade would have been introduced. So if the team would have participated in IMO with students mostly in grade 12 you cannot suppose anymore they also would have been at IMO with mostly the same students being in grade 11 back then. It is much more probable that in this year the elder guys would have made up the team, isn't it ?

Just a little correction Valentin..
Moldova, not Moldovia   :)

upss :blush: I ment the English name of Moldova which is [url=http://www.nationmaster.com/encyclopedia/Moldovia]Moldovia[/url]

one thing you forgot is that Reid Barton also had the absolute #1 spot  on the 2001  IMO while Christian Reiher never had an absolute #1 score. that's why Reid Barton is my favourite among those two :)


Three people's first names are Josie, Nicki and Sara. Their last
names are, not necessarily in this order, Croll, Liu and Menkin.
Their favorite colors, not necessarily in this order, are red,
yellow and blue. The number of letters in a person's favorite
color is the same as the number of letters in either their first
name or their last name. What is the first name and the last
name of the person whose favorite color is blue?

blue has 4 letters.

so it's neither Josie or Nicki, so Sara.

and yellow is Menkin, Red is Liu.

so the last name of person who likes blue is Croll.

Answer : Sara Croll


function
trigonometry
combinatorics unsolved
combinatorics

Let $ A(x) \equal{} \frac{1}{\cos x} \equal{} \sum_{n\ge0}a_nx^n$.  Recall that

$ \cos x \equal{} \sum_{n\ge0}(\minus{}1)^n\frac{x^{2n}}{(2n)!} \equal{} 1 \minus{} \frac{x^2}{2!} \plus{} \frac{x^4}{4!} \minus{} \frac{x^6}{6!} \plus{} \ldots$

Compute $ a_0$ through $ a_4$.  (Hint: $ A(x) \cos x \equal{} 1$)

We have $ \cos x \sum_{n \ge 0} a_n x^n \equal{} (a_0\plus{}a_1 x \plus{}a_2 x^2\plus{}a_3x^3\plus{}a_4 x^4\plus{}...) \left(1\minus{}\frac{x^2}{2!} \plus{} \frac{x^4}{4!} \minus{} ....\right) \equal{} 1$

Comparing Coefficients on both sides

$ a_0 \equal{} 1; a_1 \equal{} 0$

$ a_2 \minus{} \frac{a_0}{2!} \equal{} 0 \Rightarrow a_2 \equal{} \frac{1}{2}$

$ a_3 \minus{} \frac{a_1}{2!} \equal{} 0 \Rightarrow a_3 \equal{} 0$

$ a_4 \minus{} \frac{a_2}{2!}\plus{}\frac{a_0}{4!} \equal{} 0 \Rightarrow a_4 \equal{} \frac{5}{24}$

And you can check that these are correct here: http://en.wikipedia.org/wiki/Trigonometric_function

http://en.wikipedia.org/wiki/Euler_number
http://mathworld.wolfram.com/SecantNumber.html
http://mathworld.wolfram.com/EulerNumber.html

  darij


function
trigonometry
real analysis
real analysis unsolved

f:R-->R and f(x) is continuous function such that [tex] f(x)f(-x)=cos^2x[/tex]
Find f(x)?

i don't think there is just one continuous function which satisfies that equation. e.g. you might choose $f(x)$ for positive $x$ quite arbitrary(but continuous) and then choose $f(-x)$ s.t. the equation is satisfied. this always gives as a solution(maybe not in some particular cases, but in general...e.g. you have to take care of the positive $x$ where $\cos^2 x=0$; if $f(x)=0$ there, you have to take some $f(-x)$ s.t. $f$ stays continuous at $-x$...).

:( may be...my english is very very bad
my problem is :
Consider a continuous function f(x):R-->R satisfies that equation:[tex]f(x)f(-x)=cos^2x[/tex]
Find f(x)
I think[tex] f(x)=cosx\sqrt{\frac{g(x)}{g(-x)}}[/tex]
with g(x)>0 and g(x) is a continuous function ;but I haven't proved :(  :D

This is a really minor issue, but I feel like saying it. The word "continuous" in English has two [i]u[/i]'s: [u]contin[b]u[/b]ous[/u]. Around here it sometimes seems that the one-[i]u[/i] misspelling occurs more frequently than the correct spelling.

(/pedantry)

[quote="Peter Scholze"]your conjecture can't be true, a counterexample is $f(x)=1+\sin x$ (the square root would always be nonnegative, therefore the sign of $f(x)$ would be determined just by $\cos x$ if you're statement was true; however the above function is always nonnegative -> contradiction).[/quote]
thank you
I had a mistake
I thought [tex] 1+sinx=cosx\sqrt{\frac{1+sinx}{1-sinx}}=f(x)[/tex](choose g(x)=1+sinx)
But[tex] f(x) =cosx\sqrt{\frac{1+sinx}{1-sinx}}[/tex] isn't continuous function :blush:
Oh;can you post your solution? :D
I haven't solved this problem


trigonometry
inequalities
function
inequalities unsolved

Let $A, B, C$ be real numbers in the interval $\left(0,\frac{\pi}{2}\right)$. Let \begin{align*} X &= \frac{\sin A\sin (A-B)\sin (A-C)}{\sin (B+C)} \\ Y &= \frac{\sin B\sin(B-C)\sin (B-A)}{\sin (C+A)} \\ Z &= \frac{\sin C\sin (C-A)\sin (C-B)}{\sin (A+B)} . \end{align*} Prove that $X+Y+Z \geq 0$.

[quote="hardsoul"]Hope that the following solution is correct.
Suppose that C>=B>=A,then sinC>= sinB>= sinA,sinAsin(A-B)sin(A-C)/sin(B+C)>=0.
The inequality reduces to show that
sinCsin(C-B)sin(C-A)/sin(A+B) >=sinBsin(C-B)sin(B-A)/sin(A+C).
Note that sinC>= sinB,sin(C-B)>=0.So  it would sufficient to show that
sin(C-A)sin(C+A)>=sin(B-A)sin(B+A)is equivalent to cos2A>=cos2B<=> sin 2 A>=sin 2 B.[/quote]

Nice!  :D Well, I didn't find any mistake except the last part, which should be $\cos{(2A)} \geq \cos{(2B)} \Leftrightarrow \sin^2{A} \leq \sin^2{B}$, which is true since $A \leq B$.

Or, you could have finished the last part with by noting that $\sin{(C-A)} \geq \sin{(B-A)}$, and $\sin{(C+A)} \geq \sin{(B+A)}$.  I think it's right... anyways, cool solution.

Does anyone notice that this inequality is a lot like Schur's except with trig?

as me ,I use 'Shur :lol: '.thank you problem very easy

Denote $a=\sin A, b=\sin B, c=\sin C$

Then it is equivalent to,

$\sum \sin A\sin (A-B)\sin (A-C)\sin (A+B)\sin (A+C)\\ =\sum \sin A(\sin ^2A-\sin ^2B)(\sin ^2A-\sin ^2C)\\ =\sum a(a^2-b^2)(a^2-c^2)\\ \geq 0$

The last inequalitiy refers to Schur's.

Assume WLOG that $ A \ge B \ge C $.  Note that $ \sin $ is an odd, nonnegative, monotonically increasing function in the interval $ \left(0, \frac{\pi}{2}\right) $.  Moreover, since $ \frac{\sin{A}}{\sin{(B + C)}} + \frac{\sin{C}}{\sin{(A + B)}} \ge \frac{\sin{B}}{\sin{(C + A)}} $ we have that Theorem $ 3c $ in http://web.mit.edu/~darij/www/VornicuS.pdf immediately solves the problem.


You are given 3 equations:
$x-y-2z=-3$
$x+3y=30$
$y+4z=10$
What is $xyz$?

[quote="davidlizeng"]i brute forced it
[hide]
x=6, y=8, z=1/2
so
24[/hide][/quote]
That's right but what equations did you use?

i used the three given ones....
[hide]
x+3y=30
y+4z=10
x-y-2z=-3

so first solve for x in terms of y
you get x=30-3y from the first equation
then solve for y in terms of z
from the second one you get y=10-4z
plug this one into the x=30-3y to get
x=12z
plug this and y=10-4z into x-y-2z=-3
and solve for z which is 1/2
x would be 6 and y =8
[/hide]
i'm sure there's a better solution

[hide]Multiply the last equation by $-2$ so you can cross out the z, then using that to the second. So $2x-2y-4z=-6$
                                               +  $y+4z=10$ Solving for this you get $2x-y=4$. Then multiply the middle equation by -2 so we have $2x-y=4$
              +$-2x-6y=-60$.
Solving this gets you y=8. Putting y into the middle and the last you can get x=8 and z=1/2. So $xyz=24$.[/hide]

[quote="mathgeniuse^ln(x)"]He probably used all the equations.[/quote]
No, I meant how did he get there with the given equations.

[quote="akech"]It's easy to set up an augmented matrix and then row reduce it, perhaps this is not applicable here (but took three steps to find z by elimination method.[/quote]

dude, this is mathcounts...

[hide="Ok, here is the full solution"]Subtract the second equation from the first:
$-4y-2z=-33$
Multiply this by $2$ and add this to the third equation:
$-7y=-56$
$y=8$
From the second and third equations,
$x=6$
$z=\frac{1}{2}$
So $xyz=24$[/hide]


[hide]Well $ a^{2^{m}}\plus{}2^{2^{m}}|a^{2^{n}}\minus{}2^{2^{n}}$ if $ m<n$.  Therefore, because $ a^{2^{n}}$ is odd, the desired conclusion follows[/hide]


topology
real analysis
advanced fields
advanced fields unsolved

This is a problem from Hatcher.

Let $X$ be the subspace of $\mathbb{R}^{2}$ consisting of the horizontal segment $[0,1]\times\{ 0 \}$ together with all the vertical segments $\{r\}\times [0,1-r]$ for $r$ a rational number in $[0,1]$. Show that $X$ deformation retracts to any point in the segment $[0,1]\times\{0\}$, but not to any other point.

Suppose that $\{f_{t}\colon 0\le t\le 1\}$ is a homotopy between identity $f_{0}$ and the constant map $f_{1}$ into a point $(r,q)$, $q>0$. Since $f_{t}(x,y)$ is uniformly continuous in both $t$ and $(x,y)$, there is $\delta>0$ such that $|f_{t}(x,y)-(r,q)|<q$ whenever $|(x,y)-(p,q)|<\delta$. Thus, the image of any point $(x,y)$ under $f_{t}$ must stay in the open upper half-plane, which leads to a contradiction.

Why must $f_{t}(x,y)$ be uniformly continuous in $(x,y)$?

Touché. Attempt #2: Choose a sequence $(x_{n},y_{n})$ converging to $(r,q)$. For each $n$ choose $t_{n}\in [0,1]$ so that $f_{t_{n}}(x_{n},y_{n})$ lies on the real axis. Passing to a subsequence if necessary, we may assume that the sequence $t_{n}$ has a limit $T\in [0,1]$. By continuity $f_{t_{n}}(x_{n},y_{n})\to f_{T}(r,q)=(r,q)$, hence $q=0$.


Equilateral triangles $ \triangle ABX$ and $ \triangle ACY$ are drawn on sides $ \overline_{AB}$ and $ \overline_{AC}$ of a $ \triangle ABC$.
Prove that $ CX = BY$

since AX=AB, AC=AY and <CAX=60+<BAC=<BAY, thus triangles AXC and ABY are congruent and CX=BY.

Thanx greentreeroad
Sry I couldnt get such an easy problem  :blush: