Those are to be done without a calculator.

IF log 2= .3010 and log 3= .4771, the value of x when (3)^x+3=135 is approximately? (1954)

How many distinct ordered triples (x,y,z) satisfy the equations

x+2y+4z=12
xy+4yz+2xz=22
xyz=6    (1976)

For the second one:

[hide]

There is a reasonably standard little trick to use with this. Consider the polynomial (T-x)(T-2y)(T-4z). When we expand this and collect terms, it equals T:^3:-12T:^2:+44T-48. This has roots 2, 4 and 6. Now any combination of x,2y, and 4z equalling 2,4,6 will definitely satisfy the first and last equation, and doubling that second equation shows that it is also completely symmetrical, so overall (x,2y,4z) is any combination of (2,4,6), and you can list the cases if you want.

[/hide]

For the first one, I assume that is (3^x) + 3 = 135.. is there a list of possible answers, as 'approximately' could really mean anything..

In these types of problems, I think it is typical to express whatever needs to be expressed in terms of log 2 and log 3, and then do the arithmetic up to 4 (in this case) decimal places.

Fierytycoon

[quote="hello"]Those are to be done without a calculator.

IF log 2= .3010 and log 3= .4771, the value of x when (3)^x+3=135 is approximately? (1954)[/quote]

For the first one, I think it's 3^(x+3), since otherwise it would just as easily be written as 3^x=132...

I suspect this is from the free Anniversary Edition, since I doubt many people have copies of tests from '54.  As it turns out, it is.  The question is as follows:

If log 2 = .3010 and log 3 = .4771, the value of x when 3^(x+3) = 13 is approximately...
(A) 5
(B) 1.47
(C) 1.67
(D) 1.78
(E) 1.63

I'll post my answer on a separate post to distinguish question from solution.

Here's my solution in spoiler:

[hide]3^(x+3) = 135

x+3 = log3 135

x = log3 135 - 3

x = log3 135 - log3 27

At this point, we use the property of logarithms that log m - log n = log m/n:

x = log3 135/27

x = log3 5

Now, using the change of base formula (logb n = loga n / loga b)

x = log 5 / log 3

x = (log 10/2) / log 3

x = (log 10 - log 2) / log 3

x = (1 - .3010) / (.4771)

x = 1.4651 :approx: 1.47 (B)

[/hide]

Hope this helps.

How can you uncode the solutions in spoiler?

'Highlight' them.  Put your cursor over the solution, click and drag (while still clicked), just like you're highlighting text to copy.  The spoiler text will appear.

I got the first one..

Could someone please explain the second one? Sorry Triple M, I didn't find your answer to be clear.

Southpaw: we can think of x, 2y, 4z, as the roots of a cubic polynomial.  So the first equation gives the sum of the roots, the 2nd asks for the sum of the roots taken 2 at a time / 2, and the 3rd equation gives you the product of the roots / 8.

So if we multiply the 2nd equation by 2, and the 3rd by 8, the 2 equations will represent the sum of the roots taken 2 at a time, and the product of the roots respectively.  Remember that in  a cubic polynomial aT^3+bT^2+ct+d, -b/a is the sum of the roots, c/a is the sum of the roots taken 2 at  a time, -d/a is the product of the roots.

So we can make a polynomial given this information.  We'll assume the coefficient is 1.  The polynomial will then be T^3-12T^2+44T-48.  Then we find the roots of the polynomial,   which are 2, 4, and 6.  Then (x, 2y, 4z) is a combination of (2, 4, 6), but we test (plug in) to find out which pairs work.

I figured number 2 out by first noting that the factors of 6 are 1, 2, 3, and 6 so the third equation shows me that x, y, and z are 1, 2, or 3 in any order or the same integers except two are negative. So that narrows the ordered integers into 24 different choices such as (1,2,3), (1,3,2), (-1,2,3), (-1,3,2), (2,1,3), (2,3,1) and so on. Then considering the first equation( x + 2y + 4z = 12 ), one can assume that x must be an even number in order for the equation to equal 12 - because 2*y and 4*z are automatic even numbers and an odd + an even number equals an odd. This lowers the choices to (2,1,3), (2,3,1), (2,-1,-3), (2,-3,-1), (-2,-1,3), (-2,1,-3), (-2,3,-1) and (-2,-3,1). so try inputting the other numbers when x=2. the only equation that equals 12 is (2)+2(3)+4(1). Then try inputting the other integers when x is -2 and you'll see that nothing works. therefore the answer is  [hide][hide][/hide][hide][/hide][hide]1.[/hide][/hide][tab]

wow...this goes kinda far back.  that was my first post here i think.


Im not sure if this is a bug or not but on problem 3897 it shows a answer of [code]$\frac{5}{7}$[/code]

I think that it shoud show up as 5/7. Im reporting this here because I can't seem to find the button on which you can submit a report on the problem(I didn't realize the typo untill I checked my report and went to a subject and found the problem).
[url=http://www.artofproblemsolving.com/Alcumus/ViewProblem.php?problem_id=3897]Problem Link[/url]

Note: There should be a submit problem error button in the ViewProblem page.

The answer is fine. That's just how it's recorded in the database.

Thanks for the suggestion on putting the report problem link on the view problem page.


trigonometry
inequalities solved
inequalities

show that for x in (0,pi/2) we have the following ineqs:
a)sin(cos(x))<cos(sin(x))
b)sin(sin(x))<cos(cos(x))

since x > sin x
we have sin x > sin sin x and cos x < cos sin x

\pi/2 - x > sin (\pi/2-x) = cos x
we have sin (\pi/2 - x) > sin cos x and cos (\pi/2 - x) < cos cos x
i.e., cos x > sin cos x and sin x < cos cos x

combining the results above,
we get
sin cos x < cos x < cos sin x
and
sin sin x < sin x < cos cos x


[quote="Interval"]The circumference of a circular plot of land is increased by 10%. What is the best estimate of the total percentage that the area of the plot increased?

(1) 10% (3) 25%
(2) 21% (4) 31%[/quote]
The original circumference is $2\pi r$ and the area is $\pi r^{2}$ the new circumference is $(2\pi r)1.1$. so the new radius is $1.1r$ and then the area will be $\pi (1.1r)^{2}=\pi 1.21r^{2}$ which is an increase of 21% (2).

[quote="Interval"]The circumference of a circular plot of land is increased by 10%. What is the best estimate of the total percentage that the area of the plot increased?

(1) 10% (3) 25%
(2) 21% (4) 31%[/quote]

when the side is multiplied by a factor of $x$, then the the area is multiplied by a factor of $x^{2}$.

Figure out the rest based on this

[quote="Interval"]jli,

It already has be done.

Please, do not answer my questions anymore.

Interval[/quote]

i was just trying to show you the faster way.  its not very good to depend on one method all the time.  plus, you learn more this way

[quote="Interval"]jli,

It already has be done.

Please, do not answer my questions anymore.

Interval[/quote]

Improve your grammar.

jli, he/she doesn't have the real right to say that, so don't worry about it.

[quote="Interval"]jli,

It already has be done.

Please, do not answer my questions anymore.

Interval[/quote]
Yea, you can't just say to someone you can't answer there questions anymore. Thats one of the whole points coming here. Learning how to do a problem different ways. Just because one person answered it doesn't mean you're done with the problem.


ok. so im doing aops vol 1, and i couldn't really grasp their explanation of the pigeonhole principle. i understand the basics of it, (e.g. kn+1 --> n+1) but im not sure of its uses

can anybody give me a brief explanation of how it functions in various problems and how i can know when it's appropriate to use this principle?

thanks!

Well basically, we have k pigeons. and they have to go to sleep. because its sleepytime. anyway, they go to their little holes, and they must go into a hole to sleep. but, unfortunately, there are only k-1 holes! then, it is easy to see that at least one hole must contain at least 2 pidgeons.

so, as an example, let there be a road. the road is 10 miles long. 11 people are walking on the road. using the pidgeonhole principle, at least 2 of the people must be no more than one mile apart.

another example. a battlefield has area 5. the shooting range of 6 men on the battlefield is 1. then at least two men must be within shooting range of each other.

kids are at an ice cream store. they all get one scoop. there are 8 flavors and 9 kids. at least one pair of kids must have the same flavor.

and so on.

thanks. that's very interesting. good examples!

[quote="lokesh"]Here's a good basic PHP problem. Given any 5 points inside an equilateral triangle with side lengths of 2, show that there are 2 points at most 1 unit apart.[/quote]

[hide]an equilateral triangle with side length two can be divided into 4 equilateral triangles can be divided into 4 equilateral triangles of side length 1. since there are 5 points, at least 2 points must be contained within a small triangle.

since the side lengths are 1, these two points are at most 1 unit apart.[/hide]


each day there is a 10% chance of rain. what is the probability that in 4 days it will rain exactly once? express answer as a decimal.

how would you do this problem?

[hide]
Let X be the number of days it rained.
X ~ Bi(4, 0.1)
$Pr(X = 1) = \binom{4}{1}(0.1)^{1}(0.9)^{3}$
[/hide]


geometry
3D geometry
tetrahedron
Support
group theory

Evolution is like the most inspiring thing that I like to talk about...

What do you people think about our status of evolution and stuff?
For past what, like a bunch of million years, people evolved by natural seleciton and mutation.
But eversince we started making fancy medicines and developing awesome technologies, it has been slowed down in some ways since people's genes don't easily disappear or change. I bet the only evolutions that are now gonna happen will be some minor ones like hair and in our brains. I don't know how this is gonna happen, but since most of our lives(at least people in this forum) depends on our thinking and stuff, it'll grow more important.
But since evolution is happening so~ slowly, later we might even expose people to man-made mutagens to change ourselves for better. What I've always wondered is that when I saw things that had to do with aliens, I can't help but think that that could be how people will look like in few million years.

Sorry, I just like to talk.
So, what do you people think aobut evolution???

I think that by using technology we're playing against mother nature.  After all, our survival skills are decaying, and we have to go in an endless feedback cycle by getting even more technology.  However, the occasional epidemics do help us evolution-wise a little, so probably we'll be ok, but those epidemics occur less and less frequently it seems (good or bad?), so for future ones, the stakes will be higher.  Too bad nature is way better than we are at this game, but should we stop using our technology?  I think I better throw out my computer -- wait, nooooo I can't live without my computer!

I think that, as cruel as it sounds, modern medicine is actually killing evolution.  There is no survival of the fittest, because we are able to keep people with genetic disorders long enough to breed, which they wouldn't otherwise be able to done.  I think we're done evolving, and if anything, we may start devolving soon.

[quote="churchilljrhigh"]Its just a theory.[/quote]

Because it is a "theory" doesn't diminish any validity.  People hear the word theory and think that it somehow means that it may not be true.  Theories don't graduate and become laws or anything like that.  Theories and laws are equally accepted by the scientific community.

So, what should we do in the future?
Should we just let us evolve whichever way we might or should we promote the evolution for a certain beneficial way?
I heard that people these days are becoming more and more like those Neandertal or something people. You know, those ancient people with huge brains(although not really smart) big bellies, and bend shoulder and back. Sitting down and doing just about no exercises and watching TV all day in a couch is to blame, I heard. I'd like to call that as sort of de-evolving.
Anyways, back to the thingy, should we let people do whatever they want and de-evolve, or should we promote exercise and lots of brain activities just to keep people from de-evolving or should we go some kinda more radical ways? (Like exposing people to radioactive thingies to make ourselves evolve)
I always thought about people making artificial planets or stuff and creating life in there (little people) just to experiment with them and seeing which way they'll go, then applying researches from there to actual people to help us evolve. It IS possiblem isn't it?

[quote="EliasTheGreat"]Sitting down and doing just about no exercises and watching TV all day in a couch is to blame, I heard. I'd like to call that as sort of de-evolving.[/quote]

Individual actions have no effect on evolution, beyond getting oneself killed wtihout reproducing.  Sorry.

evolution...hmm finally someone brings up something interesting to talk about  :lol: .

I think that as we get more technology that we are evolving.  It is said that we use only a small percentage of our brain.  Well that percentage has been getting higher and higher.  Isn't that evolution?

[quote]But eversince we started making fancy medicines and developing awesome technologies, it has been slowed down in some ways since people's genes don't easily disappear or change.[/quote]

Evolution is based on favorable mutations that appear by chance. Mutations will happen. No matter what. You could claim that technology makes some mutations that could be favorable unfavorable. So can nature. Technology can be assumed to be part of nature.

[quote]I think that by using technology we're playing against mother nature. After all, our survival skills are decaying, and we have to go in an endless feedback cycle by getting even more technology.[/quote]

What makes you think that technology is not part of evolution? Would you prefer to be an animal? Then you would have to rely on nature only.

[quote]Should we just let us evolve whichever way we might or should we promote the evolution for a certain beneficial way?
[/quote]

You cannot control evolution. It happens by random mutations. These mutations are uncontrollable. At best, you could induce mutations. But you would have to check some enormous number of mutations, which is not practically possible.

[quote]I heard that people these days are becoming more and more like those Neandertal or something people. You know, those ancient people with huge brains(although not really smart) big bellies, and bend shoulder and back. Sitting down and doing just about no exercises and watching TV all day in a couch is to blame, I heard. I'd like to call that as sort of de-evolving.
[/quote]

Unless their behavior is part of a genetic mutation, this isnt true. They would not pass on their traits to their siblings, so it couldn't matter less.

All in all, you guys seem to think about evolution as Lamarck's theory: that acquired traits are passed on. This was proved false. Darwin's theories state that random genetic mutations make an organism more favorable than another, and thus this organisms offspring are better suited to live, and their offspring, and so on until the entire population comes from this one better suited organism.

Me... I have always enjoyed evolution, but I found out recently about this new concept brought forht in the early 70's that has split up scientists between it and Evolution throuhg Group Theory: Gene Theory. It proposes that instead of evolvution through the needs of a group, it occurs instead through the needs of genes. Its very enlightening. If you would like to learn more, the proposer was Robert Dawkins and he made a book titled "The Selfish Gene"

tetrahedron: I think you're partly missing the point of what he was saying.  Evolution isn't [i]just[/i] random mutation: it's the process by which those random mutations get mixed in to the general population and either weeded out or adopted.  Bigger brains for humans aren't [i]just[/i] a random mutation -- they're a random mutation that gave those who had it an evolutionary advantage that allowed them to reproduce more and dominate the human population.  Now consider a society in which everyone, regardless of their mutations (ignoring those cases of mutations that are completely debilitating or result in death in the womb and whatnot) was able to survive and mate (much like current Western society).  In some way, it's theoretically possible that the gene pool could be "degraded" as genes which serve no advantage to humans, and in fact disadvantage us, are allowed to survive.  Now, frankly, this couldn't be happening very quickly -- industrialized societies (those with the medicines to make this really relevant) have very low birth rates and make up only a very small proportion of the human race.  Also, this situation has only been in place for a few decades, at best.  But it is "theoretically possible."

You make a very interesting point in talking about technology as an extension of evolution.  I'd love to hear other people weigh in on this.

I wouldn't worry too much about humans suddenly going extinct because of drug resistance or anything like that.  The largest portion of the human population that has ever died off from one event was the 2% that perished during the black plague.  Immune systems are amazingly robust -- especially when you take large populations into account.

It does however seem to be true that evolution is slowing down, but should this concern us?  The universe seems to filter toward higher and higher efficiency.  If we become an "augmented" species, this could simply be viewed as one more uncovering of technology and a morphing of the [i]status quo[/i].

The thing we need to worry about is some nutjob getting ahold of a big weapon.

[quote="MCrawford"]The largest portion of the human population that has ever died off from one event was the 2% that perished during the black plague.  Immune systems are amazingly robust -- especially when you take large populations into account.

The thing we need to worry about is some nutjob getting ahold of a big weapon.[/quote]

Is this really true? I think the black plague may have killed off a higher percentage of the local population that it came into contact with than that. I seem to remember hearing that it was some very high percentage.

There have also been diseases in other places that have killed off even greater percentages of the population that that. The small pox outbreaks in Native American cultures killed off huge percentages of people. And I believe AIDS in some parts of Africa will ultimately kill larger percenages as well.

I personally think that diseases (both natural and human engineered) are one of the biggest things to worry about. The problem is that diseases mutate much faster than we do.

That 2% statistic, I don't know where that's coming from. But I do know that Europe's population dropped by 1/3.
But, I've often thought about evolution in that: Because although favorable mutations can occur in populations: those with favorable mutations must reproduce more than those without. Now, that's not necessarily true. Even if some guy (or gal, but we're saying he for convenience)had a marvelous trait that helped him a lot in society, which means that he makes a lot of money and is successful. However, in this position, he is less likely to have many children and pass on his trait. It's counter-evolution!
Does this mean that evolution doesn't occur? I would doubt that: looking at trends. Contrary to what some people might think, humans are not "devolving", in fact, we're becoming smaller over time (not counting the growth spurt involved with improved nutrition), and more mentally powerful.
So there must be something else that we're missing here.
Could it be. That there is a different mechanism to evolution apart from the physical mode? That's something to consider.
Cultural standards and these things, purely in the realm of the mental, do evolve over time. That is for sure, but it shouldn't change the actual physiological capacity for humans to be better, right? Just something to munch on. Chalk it up to one of my crackpot theories.

The estimates for the black plague vary, but most recent studies suggest 20% of Europe (as opposed to the 1/3 that we read about in our textbooks) which was about 2% of the world's population.  Even if we say 3%, that's still pretty low for the most Malthusian death check in the history of the world.

Evolution: A very long time ago, there was this tiny bit of nothing, which just happened to exist. Then nothing exploded, and created the whole universe.

I'm afraid I'm quite a logical person and I cannot believe that :)

TripleM have u ever read Angels and Demons by Dan Brown?  U should it explains the Big Bang theory quite well with proof that has acctually been brought forward in todays scientific world.

The problem is that most people's conception of what the "Big Bang Theory" actually says isn't entirely accurate. The way that it's explained to young people and laymen isn't correct.

To physicists the Big Bang Theory simply states that the Universe is uniformly expanding, and that it has been explanding since it was very very small and hot. (There is extremely strong physical evidence to support this claim). That's all it says, no more - no less.

Now, how small is "very very small" and what happened before it started expanding, are NOT questions answered by the Big Bang theory; and until very recently, were questions that couldn't be answered by classical physics at all.

How you visualize the state of a very early Universe, or reconcile it with your imagination, is entirely up to you. But the [i]Big Bang Theory[/i] itself does not say that there was a period of calm followed by a sudden violent explosion. That's the part that people add themselves, and it leads to a common misrepresentation of what the Big Bang theory actually says.

Yes thats what I'm trying to say now they have the ecidence to support what happened before the big bang using a technoligy called anti matter.  correct?

Actually string theory (or some variation of it) has the best chance of explaining what happened before the current period of expansion started. But those theories are still being worked out and explored.

The problem with trying to extrapolate the laws of physics back too far is that when the Universe gets very very small and very hot - our classical laws of physics (Relativity and Quantum Mechanics) break down, and no longer apply. So we can't predict what happened in such an early universe, because we don't know what the laws of physics would be in such a situation. They may have been completely different than our current laws of physics.


How would I use seperation to find the general solution of Newton's Law of Cooling? I'm really confused...

Newtons Law of Cooling:
[img]http://i2.tinypic.com/symro4.jpg[/img]

I first did this...is this even right?
[img]http://i2.tinypic.com/symtjp.jpg[/img]

then I kinda got lost would I just subtract the ky and then take the anti-derivatives of each?

If you have an equation of the form $\frac{dy}{dx}=f(x)g(y)$, then you want to integrate
\[ \frac{dy}{g(y)}=f(x)dx  \]
In this case, write it as
\[ \frac{dy}{y-t}=kdt  \]
and work as usual.

If you start with $- dT/dt = k( T - T(o) )$ as your original equation
   where $T$ is the instantaneous temperature and $k$ is your cooling constant
  then I think the final equation comes out to be something like :

  $T(t) = T(o) + (T(i) - T(o))(e raised to (-kt) )$  [ somebody LaTeXify this for me]  :oops:
  where
  $T(t)$ is the temperature at any time t
  $T(i)$is the initial temperature
  $T(o)$ is the temperature of the surroundings
  The temperature of the body becomes equal to the surrounding temperature when
 $t = infinity$

Your problem was that in your equation
\[ \frac{dy}{dt} = k(y-t) \]
the $t$ in $dt$ is not the same as the $t$ in $(y-t)$. Instead the second $t$ is the lower temperature... so better rename the second $t$ as $T$. Now,
\[ \frac{dy}{dt} = k(y-T) \]
And now you can follow blahblahblah's method.

the temperature decreases with time. so we have.

  $-dy/dt = k(y-T)$


algebra polynomial function calculus integration

[quote="Interval"]Determine which functions below are polynomial functions.
For those that are, state the degree.  For those that are not, tell me why.

(1) F(x) = (x^2 - 5)/x^3

(2) h(x) = (sqrt{x})(sqrt({x}) - 1)

(3) F(x) = 5x^4 - pi(x^3) + (1/2)[/quote]
polynomial functions are functions with positive integral exponents for each x term. that means any x term in the denominator makes the function rational instead of polynomial. (denominator=negative exponent)

Thanks but your hint is not helpful in terms of answering the question.

well...look at which functions that have x terms in the denominators, or have square roots. the ones with x's in denominators and the ones with square roots are the non-polynomial functions.

So, x in the denominator and any function with square roots = NOT A polynomial function.

Got it.

Thanks!


algebra function domain LaTeX logarithms

can someone explain to me what the domain of an algebraic expression is?

The domain of $f(x)$ is the set of reals such that for every member of that set (let's call the element $k$), f(k) is real.

EX: The domain of $\sqrt{x}$ is the nonnegative reals, since for negative values of $x$, $f(x) = \sqrt{x}$ is imaginary.

I hope this helps...

You can google it if you want

[quote="SplashD"][quote="Treething"]want a list :p ?[/quote]

yes, i do  :help:[/quote]

There are this many: $x$ as $x\to\infty$

What is your algebraic expression?

$\frac{2}{\sqrt{x+3}}$

and

$\frac{3x-1}{x-2}, x\ne 0$

they are separate

so you just put a $y=$ in front of it and graph and see what possible values $x$ can take? ok, i got it

[quote="Treething"](what's the latex for greater than or equal to?)[/quote]
\ge $\ge$
\le $\le$

:D

[quote="Treething"]ah.

for the first one, $\sqrt{x+3}$ can't be negative so the domain is x >= -3 (what's the latex for greater than or equal to?)

for the second one, x can't equal 2 or else it'd go to never-never land, and it can't equal 0 either, so the domain is all reals excluding 0 and 2[/quote]

Actually, it's $x>-3$.  If $x=-3$, we get $\frac{2}{\sqrt{-3+3}}$, or $\frac20$.

Yes, that is correct.

$\sqrt{x+3}>0$ in this case.

Domain just means all values of the variable that are allowed in the expression.

its basically, all of the input values that give a real number as the output. to find numbers that aren't in the domain, look for 0 denominators, and negative square roots. when doing logarithims, for example log x: x cannot be negative, for any base.  :|  :wink:

[size=117]VERY SIMPLE DEFINITION:[/size]

When you have a function (lets say $y=x^{3}+2x^{2}+3$, you can choose any value of $x$ that you want.  Then, one of two things will happen to $y$:

1) You will get a value for $y$.
2) $y$ will be undefined because you chose a weird value of x.

The set of all $x$ values that will produce a $y$ value that will fit in category 1 is the [b]domain[/b].  Synonyms for domain include "abscissa."

The set of all $y$s that fit in category 1 is the [b]range[/b] of the function.  Synonyms for range include "ordinate."

[size=117]EXAMPLES:[/size]
$y=\frac{x+3}{x-5}$
The domain is ALL REAL NUMBERS EXCEPT FOR 5.  Plugging in 5 for x would cause y to be undefined.


algorithm
logarithms
function
number theory
prime numbers
prime factorization

There is no formula for large prime numbers. The easiest way to generate large prime numbers is as follows:
1) Take a random 100 digit number A.
2) Test whether it is prime.
3) If it is not, increment A; repeat step 2.
4) Print A.

There are very efficient methods for primality testing that you can use at step 2.
If you have Maple, the following code will solve the problem:

A:=rand(10^100)();
while not isprime(A) do A:=A+1: end do:
A;

BTW. If you want to use these prime numbers for cryptographic applications you should use a good random number generator.

Take the product of the first a lot of primes, and add 1 to it, and you will have a prime.

$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13+1$ is not prime.  (It's $59 \cdot 509$.)

bpms, that is only a proof that there are infinite primes, but it doesn't generate new primes (it works because you assume that those are the only primes).

Right.  You can turn the proof into an algorithm for generating primes: Every prime factor of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13+1$ must be greater than 13.  But prime factorization is a slow way to generate primes.

Just curious, but why do you need such long prime numbers?

Umm...there are quite a lot of primes in that range. I believe a very inaccurate formula is $\frac{x}{\ln x-1}$. Which gives about $2.88441834\times10^{148}$. :wink:

Why do you need such a list?

No function will generate all the primes, unforetunately.

However, if you have a program such as mathematica, you can figure it out fairly simply, but that would require programming knowledge.

[quote="funcia"]I need a list of prime numbers with 100-150 decimal digits and don't know where to find them!!!! Is there a formula for generating such primes without programming knowledge?!
I hope this is in the right forum!

Funcia[/quote]

This should probably in Round Table.
For cryptology purposes maybe?  :ninja:

What do prime numbers have to do with cryptology?

You are crazy if you didn't realise it was cryptology....OK not really but yes thanks for the advice  :lol:
I have some programming knowledge but my dad is a programmer so yeah i could maybe get through it..
Unfortunately I can't get more than a(i think) something billion with programmes...what bit? 32? 64? some mult. like that!
 Thanks! I am going to read more about primes!

BTW I am completely crazed about cryptology and primes so that might explain this random question....i am fine with it being moved since i was hesitant of putting it here  :)

[quote="mustafa"]But what does prime numbers have to do with cryptology? :huh:[/quote]

http://en.wikipedia.org/wiki/Prime_number#Applications

Wikipedia is your friend.

[quote="funcia"]Unfortunately I can't get more than a(i think) something billion with programmes...what bit? 32? 64? some mult. like that![/quote]

Depending on what language you're using, you may have a built-in arbitrary-precision integer class that you can use.

If you're using C/C++, or a language with a FFI you might want to give GMP a try :)

yeah i am thanks! but just so you guys know these numbers are BIG like 900-bit numbers  :lol:
i am though just recovering from a break through with prime units digits..
EDIT: I will not tell you what i discovered but I will show you:
47277214610743530253622307197304822463291469530209
7116459852171130520711256363590397527
That's a pretty prime.... :love:

Yeah, this probably does not belong here, most likely round table, but maybe not.

LOL, my favorite number is 7907 (999th prime).

[quote]You're likely to find some on the prime pages. http://primes.utm.edu/[/quote]

To be more specific: http://primes.utm.edu/lists/small/small2.html

Selfwritten prime tester using what I said above.
Afterwards I put it in Mathemnatica just to be sure  :D

The prime number generating program I wrote just uses the sieving algorithm... but then again I've only used it for 4-5 digit numbers.

You could probably find a list of such on the internet.  Try google.

Taking primes for cryptography from a public list is a very bad idea.

but there is so many of them when you get higher  :(
i now have a way to generate them "my they only change by like 5 digits in an hour since there is so many in the 200 digit range...." :oops:
*goes back to drawing board*
well at least java has a way....


Let $a$ and $b$ be two primes having at least two digits, such that $a > b$.

Show that \[240|\left(a^4-b^4\right)\] and show that 240 is the greatest positive integer having this property.

For $a=13,b=11$ we have $a^4 - b^4 = 240 \cdot 2\cdot 29$
For $a=17,b=13$, we have $a^4 - b^4 = 240 \cdot 229$ and $226$ is not divisible by $29$. This proves that $240$ cannot be improved.

In another hand :
Since $a,b>5$ we deduce from little Fermat that $a^4-b^4 = 0$ mod[5] and $a^4-b^4 = (a^2)^2 - (b^2)^2 = 0$ mod[3].
Moreover, $a^4-b^4 = (a^2-b^2)(a^2+b^2)$ and $a^2 = b^2 = 1$ mod[8] since $a,b$ are odd.  Thus $a^4 - b^4 = 0$ mod[16].
Therefore $a^4 - b^4 = 0$ mod[$16 \cdot 3 \cdot 5 $], as desired

Pierre.


Fie M,N,Q proiectiile lui P pe laturile BC,AC si respectiv AB.( nu stiu in latex cu indicii )
Din lema lui Carnot obtinem ca $ BM^{2}\plus{}NC^{2}\plus{}AQ^{2}\equal{} MC^{2}\plus{}AN^{2}\plus{}BQ^{2}$ (1)
Fie $ BM \equal{} x$, $ NC \equal{} y$ si $ AQ \equal{} z$ si a latura triunghiului
Atunci $ MC \equal{} a\minus{}x$ $ AN \equal{} a\minus{}y$ si $ BQ \equal{} a\minus{}z$ si inlocuind in relatia (1) obtinem ca $ 2(x\plus{}y\plus{}z) \equal{} 3a$ c.c.t.d
La a doua nu am avut timp sa ma gandesc

PS:ACESTA TREBUIA SA FIE UN RASPUNS LA ACEST POST DA AM DAT DIN GRESEALA 'POST NEW TOPIC '
http://www.mathlinks.ro/Forum/viewtopic.php?t=167058

La a 2 problema se aplica prima. Acesta a fost motivul pentru care am grupat cele 2 probleme impreuna.


let x,y,z,t>0 be real numbers such that x+y=z+t=2
proove that
| \sqrt (xy)- \sqrt (zt)| \leq | \sqrt x+ \sqrt y- \sqrt z- \sqrt t|
i like very much this inequality... i don't think that it is to hard

Kuba

sooooory for this mistake
your right
it was
| \sqrt (xy)- \sqrt (zt)|<=2| \sqrt x+ \sqrt y- \sqrt z- \sqrt t|
sorry again  for my mistake
 :D
nice solution Diogene

my solution is:
 \sqrt x+ \sqrt y=  \sqrt 2 \sqrt (1+ \sqrt xy)
 \sqrt z+ \sqrt t=  \sqrt 2 \sqrt (1+ \sqrt zt)
then
i denote with p= \sqrt (xy) and q= \sqrt (zt)
now we have:
|p-q|<=2 \sqrt 2| \sqrt (1+p)- \sqrt (1+q)|
now i took 2 cases when p>=q and p<q
for case 1
2 \sqrt 2 \sqrt (1+q)-q<=2 \sqrt 2 \sqrt (1+p)-p
and we have that f(x)=2 \sqrt 2 \sqrt (1+x)-x and
f'(x)>=0 for x<=1 (we know that p,q<=1 from AM>=GM) so f / for 0<=x<=1
that means that f(p)>=f(q) (because p>=q)
q.e.d.
same for the second case
_____________________


modular arithmetic
number theory
relatively prime
partition
algebra
IMO Shortlist

a) Show that the set $ \mathbb{Q}^{ + }$ of all positive rationals can be partitioned into three disjoint subsets. $ A,B,C$ satisfying the following conditions:
\[ BA = B; \& B^2 = C; \& BC = A;
\]
where $ HK$ stands for the set $ \{hk: h \in H, k \in K\}$ for any two subsets $ H, K$ of $ \mathbb{Q}^{ + }$ and $ H^2$ stands for $ HH.$

b) Show that all positive rational cubes are in $ A$ for such a partition of $ \mathbb{Q}^{ + }.$

c) Find such a partition $ \mathbb{Q}^{ + } = A \cup B \cup C$ with the property that for no positive integer $ n \leq 34,$ both $ n$ and $ n + 1$ are in $ A,$ that is,
\[ \text{min} \{n \in \mathbb{N}: n \in A, n + 1 \in A \} > 34.
\]

Part a) Take $p$ be a prime of the form $3k+1$ then the following mapping is a homomorphism :
\[ f: \mathbb Z_p^*\rightarrow\{1,\omega,\omega^2\} \]
where $\omega$ is the primitive $3$rd root of unity and the function is defined by
\[ f(g^i)=\omega^i \]
where $g$ is a primitive root in $\mathbb Z_p^*$.
Now we can write every rational $q$ in the form $p^kq'$ where $q'$ does not have any $p$ (here $k$ can be negative too)
So we define our three sets to be
\[ A=\{p^kq'|(q',p)=1,f(q')=1\} \]
\[ B=\{p^kq'|(q',p)=1,f(q')=\omega\} \]
\[ C=\{p^kq'|(q',p)=1,f(q')=\omega^2\} \]
Note that $q'$ is also a member of $\mathbb Z_p^*$ so $f(q')$ is well defined.
It's easy to check by our homomorphism that the three sets have the desired property.

Part b) We show that $A^3=B^3=C^3=A$
\[ BC=A\rightarrow B(B^2)=A\rightarrow B^3=A \]
\[ BC=A\rightarrow (BA)C=A\rightarrow A(BC)=A\rightarrow A^2=A\rightarrow A^3=A^2=A \]
\[ BC=A\rightarrow BBC=BA=B\rightarrow C^2=B\rightarrow C^3=BC=A \]

Part c) We modify our construction for part (a) and the result will become obvious.
First take $p=13$ now define $A,B,C$ to be
\[ A=\{p^kq'|(q',p)=1,\omega^kf(q')=1\} \]
\[ B=\{p^kq'|(q',p)=1,\omega^kf(q')=\omega\} \]
\[ C=\{p^kq'|(q',p)=1,\omega^kf(q')=\omega^2\} \]
First of all mention that cubes in $\mathbb Z_{13}^*$ are $1,5,8,12$.
Suppose that $n,n+1\in A$ then both cannot be relatively prime to $13$ otherwise they should be of the form $13k+\{1,5,8,12\}$ which is impossible.
So at least one of them is divisible by $13$. Also because $n\leq 34$ then none can be divisible by $13^2$.
So we show that if $m=13k$ and $m\leq 35$ then $m$ is not in $A$.
For it ($13k$) to be in $A$ then $k$ should be in $C$ but $k$ can be only $1,2$ and if we chose $g$ (the primitive root in $Z_p^*$ to construct $f$) to be $2$ then $1\in A$ and $2\in B$ so no one is in $C$ and we are done.

Solution by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=3930]seshadri[/url]

it is easier to back-work our way in this case. note that if such a partition is indeed made then one must also have the following:
$ CC \equal{} B$;
$ AC \equal{} C$;
$ AA \equal{} A$;
 this is seen as follows; for instance, suppose $ a\in A,c\in C$. then $ ac \equal{} ab_1b_2$ (since $ BB \equal{} C)$, $ \equal{} (ab_1)b_2 \equal{} bb_2\in C$. conversely, since $ 1\in A$(since $ 1\notin B,C$), $ C\subset AC$ and so we have the second identity. likewise, $ a_1a_2 \equal{} a_1bc \equal{} (a_1b)c \equal{} b'c\in A$ and so on.
this suggests skind of modulo $ 3$ addition i.e. if we denote $ A: \equal{} 0,B: \equal{} 1,C: \equal{} 2$ then the multiplication above translates as addition modulo $ 3$. hence we have the following idea:let
$ A: \equal{} \{r\in \mathbb{Q}^ \plus{} |r \equal{} \prod_i p_i^{e_i},\sum_i e_i\equiv 0\pmod 3\}$,
$ B: \equal{} \{r\in \mathbb{Q}^ \plus{} |r \equal{} \prod_i p_i^{e_i},\sum_i e_i\equiv 1\pmod 3\}$,
$ C: \equal{} \{r\in \mathbb{Q}^ \plus{} |r \equal{} \prod_i p_i^{e_i},\sum_i e_i\equiv 2\pmod 3\}$.
everything works as desired.


[b]
The inscribed circle of ABC is tangent at  (AB), (BC) and (CA)
respectively in I, J and K.
Prove that : BC/IJ + CA/JK + AB/KI > ou = 6.
Thank you very much.
 :) [/b]

You can look at http://www.mathlinks.ro/Forum/viewtopic.php?p=19560 .

  Darij


function
calculus
derivative
probability and stats

Assuming $ x, t$ and a are related as in $ a\equal{}2x/t^2$, find the value for %σ_t if   x = 6.0 ± 0.2m and a =8.0 ± 0.2 m/s^2.

 If α = 0.30±0.01, and t=0.40±0.02, what is the value of σ for z=exp(-αt)? (exp is the exponential function, and σ stands for standard deviation)

I'm not really sure how to do these problems.

I'm not even sure what you mean (i.e., I understand neither what exactly is given, nor what exactly is asked) :( One reasonable interpretation is that $ x\equal{}a\pm \alpha$ means that $ x$ is approximately normally distributed with mean value $ a$ and the standard deviation $ \alpha$.

Assuming that it is the case, I would proceed as follows:

If $ z\equal{}f(x,y)$, $ x\equal{}a\pm\alpha$, $ y\equal{}b\pm\beta$ and $ \alpha,\beta$ are small and $ x,y$ are independent, I would use the linear approximation $ f(x,y)\approx f(a,b)\plus{}f_x(a,b)(x\minus{}a)\plus{}f_y(a,b)(x\minus{}b)$ and say that $ f(x,y)$ is approximately normally distributed with mean $ f(a,b)$ and standard deviation $ \sqrt{f_x(a,b)^2\alpha^2\plus{}f_y(a,b)^2\beta^2}$. Of course, all the caveats are hidden in the word "approximately" :)

Assuming that it is not the case, I would go to the instructor and ask him to state the problem in a way I could understand without making wild guesses :P.

For the first problem, you're trying to find the standard deviation of t given 2 of the values that satisfy the equation.

For the second problem, you just plug in the values, but I'm not sure how you get the answer.

I understand that much. But what is the meaning of $ 8.0\pm 0.2$? :? Do you have the answers? We can compare what they are with what would follow from my interpretation of what is written and see what's going on this way.

No, I don't have the answers. 8+-0.02 is an uncertainty. I guess you're supposed to solve for t but I'm not sure how the algebra would go with a number like       8+-0.02.

Do you have [b]any[/b] example of a problem like that together with an answer (from your textbook, notes, whatever)? It would be helpful if you could post it. That is the only way to figure out what was meant by "uncertainty" and other things you haven't defined (it seems to me that you don't understand that yourself :P). You see, as it is now, the problem allows multiple interpretations. I described what you should do if my guess about how to interpret the data is correct, but it is just my "educated guess", not certainty :(.

ey, this might help

http://web.mit.edu/fluids-modules/www/exper_techniques/2.Propagation_of_Uncertaint.pdf

Yes, of course, it is relevant but, as you can see from there, there are at least two different meanings of "uncertainty" with two different laws governing their "propagation". As I said, I'm inclined to believe that we are talking about "random uncertainties", in which case the general computation rule is as I posted before (multiply by partial derivatives, add with squares and take the square root) but I'm not 100% certain that it is what was meant. You can compute this way and see what happens. Most likely, you'll be fine but when in doubt about the meaning of the conditions of the problem, you'd better ask your teacher for clarification.


geometry
rectangle
geometric transformation
rotation
arithmetic sequence

There has to be faster ways to do these...

2005 Sprint #13
In how many patterns can a 2 by 5 rectangle be tiled with five 2 by 1 tiles?
(rotations of the same figure=different figures)
The answer:[hide]8[/hide]I had to list almost every possibility. Is this necessary?

#22:
The sum of four positive integers that form an arithmetic sequence is 46. Of all such possible sequences, what is the greatest possible third term?
The answer is[hide]15[/hide]Considering it's a sprint question, I would like to know how do do it without making lists.

#25:
What is the area of the region bounded by the three lines with equations 2x+y=8, 2x-5y=20, and x+y=10?
Answer:[hide]42[/hide]Is there a fast way to do it other than drawing or finding intersection points, then using shoelace?

Thanks!

[quote="deliky6328"]#22
[hide]If a is the first term, the sequence is a, a+x, a+2x, a+3x.
4a + 6x = 46 and a is positive, so if we want x to be as great as possible (for a+2x), a must be the smallest possible positive integer, 1. Therefore
4 + 6x = 46,
6x = 42
x = 7.

The greatest possible value of a + 2x is 1 + 14 = 15.

Strange solution maybe, but it worked  :o [/hide][/quote]
Thanks!

Oh, and, to illusiat, that's pretty much what I did, but then it's still possible to miss something especially if you have a different scenario with larger numbers, right?

So... right now 1 down, 2 more to go.

#13

ya listing is the best. there is only 8 ne way

[quote="academicsrule"]What I meant is that is there any faster way.
But I guess not. So, two down, one left.[/quote]

u make it sound like 8 is so bad.

U WILL EVENTUALLY HAVE TO BE READY TO DO 30 MAX!

No, it's not bad, but what if I get to a problem when there's, say, 82? What then?
Anyway, I already accepted what you guys said, so it's all cool.

For 25, not that I know of for sprint. Mathgeek had a recent post about this.

Oooooook.

Alright then. Three down... kind of. Thanks! :)

#13 has a very nice solution using Fibonnaci.
Think of it this way:
Suppose you want to tile a 2 x n rectangle with 2 x 1 rectangles
If you place the first rectangle up-down from the top left corner, that leaves a 2 x n-1 to tile.
If you place it sideways from the top left corner, then you must put another rectangle directly under it, leaving you to tile a 2 x n-2 rectangle.
So the number of tilings of a 2 x n rectangle = #(2 x n-1) + #(2 x n-2).


Anyone here go to any of the Math Olympiad classes at Northwest Chinese School?

[quote="janieluvsmusic"]I'll probably go in the fall.[/quote]
Yay!
I'm probably not going this year, since Mr. Li is not teaching the highschool course D:

[quote="cowpi"]I'm going too. Are you, PigsOnTheWing?[/quote]
No, cuz Interlake (my school) just created a math team, and there would be no  point.

Is NWCS going to have a team for high school?

[quote="janieluvsmusic"]I know, but the person who is teaching the high school course is someone who writes problems for the AIME.[/quote]
Yea, Chris. We actually got him to help Interlake out.

Do you think your team has a shot at winning?

I know Kaiying. I think I saw your team though. Our team accidentally went into your room (thanks to Eddie).
Ok, Eddie and Aiven did not do ANYTHING! It was so annoying.
And then Feynman couldn't find the bathroom so he went outside on a wall, and the event coordinator happened to walk by. That's definitely not all Feynman did...

[quote="janieluvsmusic"]Oh ok. I wasn't really paying attention to everyone, so I don't remember. Sorry about that. All I remeber is Kaiying and this short guy talking to Feynman. Well, I was the only girl, so it would be easy to tell me apart from them.
What I found amazing was that Feynman got 2 questions in our college bowl against Johnathan Hung's team (at least I think it was him).[/quote]
DOnt' worry about it, I wasn't paying attention either xP.
Feynman can be smart sometimes... SOMEtimes....

[quote="janieluvsmusic"]But I give him credit for getting some against Jonathan.
Didn't you guys get like 3rd in team at states?[/quote]
4th

[quote="janieluvsmusic"]I was 1st! (thanks to Sam Keller not being there)[/quote]I noticed, congrats <3

sam keller is my friends brother (hana keller)

You do? What class are you in? I used to go, but then I started taking classes here, so I quit. :)

[quote="Doink"]yes, i currently go.

but then again, i'm like way younger than other people posting onthis thread.[/quote]

Lol I just realized all the other people are from like 2007...

I used to go.   It was fun.

[quote="rsngdmns"]wow. it's creepy reading old posts and noticing people talking about you. -___-[/quote]

True.

[quote="hwl0304"]so do i, but i am in HO1 and in 5th grade[/quote]
Such nubness
Much wow
Get betr

$\tiny\text{just joking}$

hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha

what's NWCS

no offense, but I kind of think that the MO4 was much easier than the MO3 in the school. (MO1 is the easiest class)

This year I somehow got 2nd in my MO4 class.


Medians $AD$ and $CE$ of triangle $ABC$ intersect at $M$. The midpoint of $AE$ is $N$. Let the area of triangle $MNE$ be $k$ times the area of triangle $ABC$. Then $k$ equals:
(A) $\frac{1}{6}$
(B) $\frac{1}{8}$
(C) $\frac{1}{9}$
(D) $\frac{1}{12}$
(E) $\frac{1}{16}$

EDIT: On second thought:
[hide]
Say ABC is an equilateral triangle, setting each side equal to $6$, then the area of the triangle would be $9(\sqrt{3})$ also, you know the height which is also equal to the medians, and the medians would be split in a $2:3$ ratio so thus $MNE$ would have area of $\frac{3(\sqrt{3})}{4}$ thus the area of the smaller triangle $MNE$ is $1/12$ the area of triangle $ABC$.[/hide]

Use my [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31174]Area lemma.[/url]

[quote="agolsme"]isn't is sufficient to realize
[hide]
$NE = \frac{AB}{4}$ and $EM = \frac{CE}{3}$ thus it is $\frac{1}{12}$ of the whole thing when you multiply those two together. [/hide]

I am asking because I am not sure if the ratio of the median transfers to the ratio of the altitude.[/quote]
Yep, the answer is $\frac{1}{12}$. And if you read my post, you will know that it is not necerssary to transfer medians and segments into altitude.

[quote="riddler"]Answer is $1/12$ but I used a different method.
Used medians are divided in the ratio $2:1$[/quote]

That is exactly how I did it. And by picking sides for the triangle and calling it equilateral, it was cake :lol:

Area of $ABC$=$2.$area$ACE$
Area of $ACE$=$\frac{CE.AE.\sin{AEC}}{2}$, so area $ABC$ is $CE.AE.\sin{AEC}$

Area of $MNE$=$\frac{1}{2}.\frac{AE}{2}.\frac{CE}{3}.\sin{AEC}$ = $\frac{1}{12}$ area $ABC$


What do you guys think were the most significant events of 2006? (On on a global, math, or personal scale)

I think it's weird that the 3,000 American death coincided with Dec. 31st. (Maybe because of reporting)

I'd say this was:

Ice shelf breaks free in Canadian Arctic
The Ayles ice shelf, which is the size of more than 11,000 football fields and is located just south of the North Pole, has broken free in the Canadian Arctic and scientists say that global warming has played a "major" role in the break.

[quote="hillarryous"]I think it's weird that the 3,000 American death coincided with Dec. 31st. (Maybe because of reporting)[/quote]

The death occured on December 28th, but it wasn't official until December 31st.

@Scrambled: Nice avatar.

[quote="Piggypi314"][quote="Scrambled"]Democrats win Congress majority, (hopefully) marking a shift towards a democrat as president (which of course, will have major international impacts).[/quote]

A woman president or a black president?  Although, even though this is so improbable it's impossible, I think it'd be funny if both Bush and Cheney died and Pelosi became President.  No prizes for guessing what would happen to America, but it'd be funny.[/quote]

People dying is not funny at all.

[quote="bubka"][quote="Piggypi314"][quote="bubka"]Probably the overthrow of Nepal's monarchy and the rise of the Maoists. Though they have set up a democracy and things are looking comfortable now, this is probably more of a worry to Bush than Osama bin Laden.[/quote]

It seems that Bush is a bigger villain than Bin Laden, anyway.[/quote]

Not necessarily. Bush maybe more evil than bin Laden, but bin Laden is a bigger enemy of communism than Bush.[/quote]

Nah.  http://youtube.com/watch?v=wNGTF1sVQpE

[quote="kstan013"][quote="bubka"]Probably the overthrow of Nepal's monarchy and the rise of the Maoists.[/quote]Oh yeah. That was [i]all over[/i] the news. *cough* sarcasm *cough*[/quote]
 :huh:
Yes, without any coughs... it was all over the news?

I find it hard to answer this question :

On a (sub)national level :

- the racist shooting of three random innocent people in Antwerp  [url=http://en.wikipedia.org/wiki/Hans_Van_Themsche]link[/url]

- the Frenchspeaking public channel falsely declaring the breakup of Belgium : [url=http://en.wikipedia.org/wiki/Flemish_Secession_hoax]link[/url]

On an international level :

- North Korea testing missiles and nuclear weapons
- the Democratic Republic of Congo, a former Belgian colony, electing its first president after more than forty years of independence that have been plagued by dictatorship and civil war.

[quote="lingomaniac88"][quote="Scrambled"]Democrats win Congress majority, (hopefully) marking a shift towards a democrat as president (which of course, will have major international impacts).[/quote]Definitely true.  I'd like to see what the democrats will do about the situation in the Middle East now that they're in the hot seat.[/quote]
Maybe the same thing Spain and Italy did as soon as they had a new prime minister? :roll:

[quote="kstan013"][quote="kstan013"][quote="bubka"]Probably the overthrow of Nepal's monarchy and the rise of the Maoists.[/quote]Oh yeah. That was [i]all over[/i] the news. *cough* sarcasm *cough*[/quote]I had a Geography Bee today and there was a question about protests with the Maoists in Asia. Weird. But the questions were hard and the events were insignificant![/quote]
Which events were insignificant?

[quote="lingomaniac88"][quote="Scrambled"]Democrats win Congress majority, (hopefully) marking a shift towards a democrat as president (which of course, will have major international impacts).[/quote]Definitely true.  I'd like to see what the democrats will do about the situation in the Middle East now that they're in the hot seat.[/quote]

Aaaah, don't worry about that now.  Let's stay home and laugh about how all the Republicans, namely Bush, are being forced to suck up to all the Democrats, namely one in whose state Bush wanted nuclear wastes dumped.  Enjoy the moment(s)!

[quote="Klebian"]I would personally think that the most significant events of 2006 wouldn't be solely from the US.[/quote]

oh pish posh...  :lol:

it's because most of the events you know of that are of significance to you are from the US since you live there.

I'm sure more significant events happened that no one here has ever heard of.  :lol:

I haven't heard of any others since I live in the US. lol .

[quote="hsiljak"][quote="now a ranger"]

I'm sure more significant events happened that no one here has ever heard of.  :lol:

I haven't heard of any others since I live in the US. lol .[/quote]

You should thank CNN, NBC, CBS for that... :lol:  :lol:  :lol: I think that your "mainstream media" don't inform you about some important things that [b]do[/b] concern your country... So it doesn't surprise me at all that they don't inform you about the important events happening somewhere across the sea! No offence...[/quote]
What are your most significant events hsiljak?

I don't think a democratic election in Congo will change very much; 30 african countries have democratic systems in place, but internecine war and oppression still rages on throughout most of africa, and most of the democracies are riddled with corruption; for example, I believe the residing president of Zimbabwe assassinated his opposing candidate to take the leadership of his country. Might be a small step, but probably not the most significant event in 06.

[quote="fredbel6"][quote="hsiljak"][quote="now a ranger"]

What are your most significant events hsiljak?[/quote][/quote][/quote]

Hmm... I haven't really thought about it... But here are some of the events  that made an impact:

Death of Slobodan Milosevic  :mad:
Death of Saddam Hussain (On sacred Eid-ul-adha!)  :mad:
The escalation of shia'i-sunni hostilities in Iraq  :mad:
North Korea and their nuke-tests  :mad:

Those were "the serious" events, but here are some things because of which I'll always remember 2006 (concerning international affairs):

Topalov-Kramnik FIDE Chess Championship match  :mad:
Chavez in UN saying "Devil is here... I can smell the sulphur" :D
Grisha Perelman and his Fields Medal  :huh:


[quote="4everwise"]A square ABCD has line segments drawn from vertex B to the midpoints N and M of sides AD and DC respectively. Find the ratio of the perimeter of quadrilateral BMDN to the perimeter of square ABCD.[/quote]

[hide]
l = length

The length of BN is $\sqrt{l^2/4 + l^2}$ according to Pythagoras.
BM is the same.
We now have $2(\sqrt{l^2/4 + l^2}) = \sqrt{l^2 + 4l^2} = \sqrt{5l^2}$
Add on DN and DM to get $\sqrt{5l^2} + l$

The ratio is $4 : \sqrt{5} + 1$
[/hide]

ccy was wrong

so just plug in numbers or variables

$\displaystyle 4l : \sqrt{5l^2} + l $

simplifies to $\displaystyle 1l : \sqrt{5} + 1 $

after divisions

[quote="jimli"]4l:2:sqrt:3+3[/quote]
Your notation is very difficult to decipher.

Might I suggest you click the LaTeX links in Valentin's signature, when he answered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224282#p224282]your question on making fractions look good[/url].

shouldn't the perimeter of bndm be 3l, because $ 2 \sqrt \frac 1/2 l^2+l^2$ is simplified to $ 2 \sqrt 1.5l^2$, and then to $2*1.5l$

how do you make the square root sign extend over the entire 1/2l^2+l^2?


AMC
AIME
quadratics
algebra
polynomial
system of equations

Uh, the book says its from AIME 1994 but I can't find it there so I'll post it here.

For any sequence of real numbers $ A \equal{} (a_1 , a_2, a_3, . . . )$, define $ \Delta A$
to be the sequence $ (a_2 \minus{} a_1, a_3 \minus{} a_2, a_4 \minus{} a_3, . . .)$ whose nth term is $ a_{n \plus{} 1} \minus{} a_n$. Suppose
that all of the terms of the sequence $ \Delta (\Delta A)$ are $ 1$, and that $ a_{19} \equal{} a_{94} \equal{} 0$. Find $ a_1$.

I am not getting an integer answer...so I suspect its not really from an AIME.

[hide]Well, if the second degree finite differences are the same, then it must be a quadratic.  Given that they are all 1, the leading coefficient must be $ \frac {1}{2}$.  Also given that two roots of this quadratic are 19 and 94, the equation must be $ a_n \equal{} \frac {1}{2}(n \minus{} 19)(n \minus{} 94)$.  For 1 this gives 837 as an answer.[/hide]

But you're right.  It was not under 1994 AIME, even though the problem statement seems like 1994 would be the year...

Wait, what? Where did the second degree come from? Perhaps I am misinterpreting what you are saying. What are the second degree finite differences? However, I realized that I read the question wrong, and I have gotten the same answer as you.

[hide="My Solution"]

If all the terms of $ \Delta (\Delta A) \equal{} 1$, then $ \Delta A$ must be a set of real numbers differing by one. Thus, we can write $ \Delta A \equal{}{y, y \plus{} 1, y \plus{} 2, y \plus{} 3...}$

Now, $ A \equal{} {a_1, a_2, a_3...}$

$ a_2 \minus{} a_1 \equal{} y \implies a_2 \equal{} y \plus{} a_1$
$ a_3 \minus{} a_2 \equal{} y \plus{} 1 \implies a_3 \equal{} a_2 \plus{} y \plus{} 1 \implies a_3 \equal{} 2y \plus{} a_1 \plus{} 2$

It can be easily proven that $ a_n \equal{} (n\minus{}1)y \plus{} a_1 \plus{} \dfrac{(n\minus{}1)(n\minus{}2)}{2}$

Thus, $ a_{19} \equal{} 18y \plus{} a_1 \plus{} 153 \equal{} 0$
And, $ a_{94} \equal{} 93y \plus{} a_1 \plus{} 4278 \equal{} 0$

Solving the system of equations, we find $ a_1 \equal{} 837$

[/hide]

Well, finite differences are just what the problem asks.  You take the second number in a sequence and subtract the first, and this is the first number in the new sequence.  You then take the third number and subtract the second, and this is the second number in the new sequence.

The reason it is second-degree, or second-order, is because it is taking the differences of the differences.  There is a theorem that says that if the n'th order finite differences are constant, then the original sequence is an $ n$ degree polynomial, and that if the constant at the end, is $ k$, then the leading coefficient of the polynomial is $ \frac{k}{n!}$.

If you're not familiar with finite differences, you should look them up.  They're quite interesting.  Here's a [url=http://en.wikipedia.org/wiki/Finite_differences]link.[/url]


Correct me if I'm wrong but 10 years ago no one in the math contest community know about exerter, and now its producing some serious talent.

I'm familiar with Prep schools and the type of students that go there, highly anti-correlated with interest in math contests. Just wondering if there is a story there.

Story goes like this: Zuming Feng went to Exeter.

I would love to hear about Exeter's math club from someone who actually goes there, i.e. how/when they practice. There's no doubt that Exeter has a great math program, but also remember that a lot of their people (Tiankai Liu, Tony Zhang, Oleg Goldberg, Sherry Gong) were already good when they got there. In fact, I think Zuming recruited them from the IMO.

Congratulations goes out to Exeter for winning ARML, pretty impressive.

I doubt that all of PEA's Olympiad Math instruction is through the math team's practices- just look at their math course MAT600.  Here's a link to their math curriculum:

http://www.exeter.edu/documents/COI/MATH07.pdf

Hm I suppose I could answer this. The math courses are far more accelerated than any other school's, and the teachers are good and make sure the students learn. On top the that, the math club meets twice a week, and thats the group that does all the math competitions. Zuming only "recruits" 1 person every two years, or so he says (I don't know who the last one was, but it was Sherry before that, and Tony even before that), but he "highly recommends" many of the good math students in the country. There are also a lot of foreign students who are really good at math.

how hard is it to get accepted? i got accepted for 08-09, but something's come up and i'm not sure if i can go this year. is there a high chance i wont get accepted for 09-10?

If you're REALLY desperate to have your questions answered, you could probably try private messaging [b]zman[/b].

I just applied to Exeter, the competition's tough. Strong school.

[quote="Phelpedo"]Story goes like this: Zuming Feng went to Exeter.[/quote]

I don't think that Zuming Feng went to Exeter; he only teaches there.


algebra
polynomial
geometric sequence
algebra unsolved

Let the polynomial $P(x)=16x^{4}-ax^{3}+(2a+17)x^{2}-ax+16$, where $a \in R$, if the roots of $P(x)$ are in a geometric progression. Find $a$.


I  would  like  to  study  pdfscreen.

Is  there  any  material   for it ?

There   are   so   mang   material  for  beamer,  but I  have difficulty in  finding  something  to  learn   pdfscreen

Thanks  a  lot!


Let  $ x_1<x_2$ roots of the equation $ x^2\minus{}(2m^2\plus{}1)X\plus{}m^4\equal{}0$, where m is a real parameter.Show that in the interval  $ (x_1,x_2)$ it is at most one perfect square.

$ x_1\equal{}\frac{2m^2\plus{}1\minus{}\sqrt{4m^4\plus{}4m^2\plus{}1\minus{}4m^4}}{2}\equal{}\frac{2m^2\plus{}1\minus{}\sqrt{4m^2\plus{}1}}{2}$

$ x_2\equal{}\frac{2m^2\plus{}1\plus{}\sqrt{4m^2\plus{}1}}{2}$

$ 2m^2\plus{}1\minus{}\sqrt{4m^2\plus{}1}\le 2n^2\le 2m^2\plus{}1\plus{}\sqrt{4m^2\plus{}1}$

Let $ \sqrt{4m^2\plus{}1}\equal{}t$, then $ m^2\equal{}\frac{t^2\minus{}1}{4}$

$ 4m^2\plus{}2\minus{}2t\le 4n^2\le 4m^2\plus{}2\plus{}2t$

$ t^2\plus{}1\minus{}2t\le 4n^2\le t^2\plus{}2t\plus{}1$

$ (t\minus{}1)^2\le 4n^2\le (t\plus{}1)^2$

$ (\frac{t\minus{}1}{2})^2\le n^2\le (\frac{t\plus{}1}{2})^2$
The bounds differ by 1, so the statement follows.


In the decimal representation of the number
$ (3 \plus{} \sqrt{5})^{2003}$
what is the digit immediately to the right of the decimal point?

[hide="solution"] Consider the number $ (3\plus{}\sqrt{5})^{2003}\plus{}(3\minus{}\sqrt{5})^{2003}$. This is clearly an integer as the irrational terms cancel each other out. The number we want is $ (3\minus{}\sqrt{5})^{2003}$ less than this. As $ (3\minus{}\sqrt{5})^{2003}<\frac{1}{10}$, our answer is $ 9$.[/hide]


A woman buys a soda. She pays $\$20.00$ and receives $\$20.05$ in change. No mistake was made.

The dollar sign in both cases probably rules out that possibility. Does the woman buy ONLY a soda?

[quote="t0rajir0u"][quote="Phelpedo"]The dollar sign in both cases probably rules out that possibility.[/quote]

Not if part of the transaction was conducted in a different currency, and during the transaction, exchange rates [i]changed.[/i]   :ninja:[/quote]

I guess that's a possibility, albeit very unlikely.

did she get a giftcard?
was the soda for a negative amount of dollars?

[quote="t0rajir0u"][hide="Does it have something to do with..."] Exchange rates?[/hide][/quote]
Correct. In fact many countries use the $\$$ symbol, especially Canada where the U.S. $\$$ is also accepted at many places.

This reminds me of a time when my friend showed me a deck of cards that had the price tag of $\$40$ ... however, the price was in Taiwan dollars.

Oh yeah! exchange rates...
So is it that she gives him 20 of his currency and he gives her back 20.05 of her currency?

[quote="probability1.01"]She probably just poured the soda out, sold the drink back, and recycled the can for five cents. All this was done instantaneously.[/quote]

No. In that case, the can of soda would have cost five cents more (to account for the deposit). The purpose of the 5 cent deposit is so that people who don't recycle will lose their 5 cents.


probability
function
integration
calculus
limit
real analysis
expected value

The exercise can be calssified both to the probability theory as well as to the real analysis but in my opinion it is more analysis than probability. The task is as follows:

Proof the theorem:
If $g$ is a continous function on interval $[0;\infty)$ and exists $E[|g(X)|]$ where $X$ is a random variable and $X\geq 0$ then
$E[|g(X)|]=g(0)+\int_{0}^\infty g'(x)[1-F_{x}(x)]dx$
where $F_{x}$ is a distribution function of random variable $X$.

I'll be very very grateful for for help as this is the only one missing exercise that I cannot do.
Thank you in advence for help.

Use partial integration on
$E(g(X)) = \int_{0}^\infty g(x)f_{X}(x) \, dx$ with $F_{X}(x)-1$ as primitive to $f_{X}$.

I got stuck in one place: I did as you advised me and here is the beggining of the solution:

$E(g(X))=\int_{0}^\infty g(x)f_{x}(x)dx=\int_{0}^\infty [1-F_{x}(x)]g'(x)dx-\int_{0}^\infty [1-F_{x}(x)]g(x)dx$

To get the thesis I have to show that $-\int_{0}^\infty [1-F_{x}(x)]g(x)dx=g(0)$.
Could you help me how to do it?

And two questions to the solution:
1. At the begginig you wrote that $E(g(X))=\int_{0}^\infty g(x)f_{x}(x)dx$. Do we know it from any theory about function of random variables?
2. Under the expected value there is absolut value of $|g(X)|$  (I wrote $E[|gX|]$). Will this absolute value change anything?

And thank you Kalle for hint

Ok. The general formula is
$E(g(X)) = \int_{-\infty}^\infty g(x)f_{X}(x) \, dx$
It is well known. Since $X \ge 0$ it is ok to only integrate over $(0,\infty)$.

I didn't see the absolute value sign and interpreted the question as $g$ a non-negative function. You will have to modify the argument a little for the absolute value sign.

The correct partial integration is
$E(g(x)) = \int_{0}^\infty [1-F_{X}(x)]g'(x) \, dx+\left[-[1-F_{X}(x)]g(x)\right]_{0}^\infty =$
$= \int_{0}^\infty [1-F_{X}(x)]g'(x) \, dx+g(0)+\lim_{x\to\infty}-[1-F_{X}(x)]g(x)$

We must show that $\lim_{x\to\infty}[1-F_{X}(x)]g(x) = 0$. Actually, I can't think of how to do this. If $g$ is an increasing function it is clear, since, for a fixed $x$,
$[1-F_{X}(x)]g(x) = g(x)\int_{x}^\infty f_{X}(t) \, dt \le \int_{x}^\infty g(t)f_{X}(t) dt$
and as $x\to \infty$ the rightmost integral will go to $0$, since by assumption the expected value exists.

However, I don't know how to do it for any $g$. Maybe someone with a firmer ground in integration theory can do it for us?


[color=purple] .Let arbitrary acute triangle ABC in plane.M in triangle.Prove that :
  $\frac{S_{DEF}}{S_{ABC}}\le 1-\frac{1}{3}(1+\frac{r}{R})^{2}$[/color]

I think you mean $DEF$ is pedal triangle of an arbitrary point but it is pity that it is worng because $S_{pedal}\le \frac{1}{4}S_{ABC}$ and exist the equality it means $\frac{1}{4}S$ is maximum value of $S_{pedal}$ you can't have any better inequality than inequality of max.

This inequality is true  and weaker than the inequality $\frac{S_{DEF}}{S_{ABC}}\leq \frac{1}{4}$.

That's right treegoner I don't think it is so funny here. I think evarist should check his problem before post it on mathlinks to make everybody fun !


I don't know much about the AMC series of contests.  I've read that 360 or so students get invited to the USAMO based on their AMC 12 index (amc12 score + 10x aime score) and 140 or are invited based on highest amc10 indexes with aime score at least as high as the lowest of all those who qualified first.

What was the cutoff in amc 12 index this year for the first USAMO pick?

What was the cutoff in amc 10 index this year for the second USAMO pick?

What was the floor AIME score for the second pick?

I know I can make AIME, and I'll know enough math for the AMC 12 in 9th grade so I'm wondering what kind of score I need to qualify for the USAMO.  What is good preperation for the AIME?  I've got "1st steps for math olympians" for the AMC, and also the intro. books for # theory, counting/prob, and geometry.

I'd really appreciate some answers.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291[/url] for all of the qualifying info.

Thanks, I had no clue whether I needed a 170 or a 220 to qualify for the USAMO.


LaTeX
linear algebra
matrix
geometric sequence

Solve the system of recurrence (i.e., find the explicit formulas in terms of $ a_0,b_0$)

$ a_{n\plus{}1}\equal{}\frac{1}{3}a_n\plus{}\frac{2}{3}b_n$

$ b_{n\plus{}1}\equal{}\frac{2}{3}a_n\plus{}\frac{1}{3}b_n$

$ \left[\begin{array}{c}a_{n + 1} & b_{n + 1}\end{array}\right] = \frac13 \left[\begin{array}{cc}1 & 2 \\
2 & 1\end{array}\right]\left[\begin{array}{c}a_n & b_n\end{array}\right] < = > X_{n + 1} = \frac13 A X_n$

and so $ X_n = \frac {1}{3^n} A^n X_0$. Each column of A add up to 3 and the trace is 2, therefore its eigenvalues are 3 and -1, with eigenvectors [1 1] and [1 -1]. So,

$ A^n = \left[\begin{array}{cc}1 & 1 \\
1 & - 1\end{array}\right] \left[\begin{array}{cc}3^n & 0 \\
0 & ( - 1)^n\end{array}\right] \left[\begin{array}{cc} \frac12 & \frac12 \\
\frac12 & - \frac12 \end{array}\right]$

$ = \frac12 \left[\begin{array}{cc} 3^n + ( - 1)^n & 3^n + ( - 1)^{n + 1} \\
3^n + ( - 1)^{n + 1} & 3^n + ( - 1)^n \end{array}\right]$

and

$ \left[\begin{array}{c}a_n & b_n\end{array}\right] = \frac {1}{2 \cdot 3^n}\left[\begin{array}{cc} 3^n + ( - 1)^n & 3^n + ( - 1)^{n + 1} \\
3^n + ( - 1)^{n + 1} & 3^n + ( - 1)^n \end{array}\right] \left[\begin{array}{c}a_0 & b_0\end{array}\right]$

$ <=> \left[\begin{array}{c}a_n & b_n\end{array}\right] = \frac12 \left[\begin{array}{cc} 1 + (-1/3)^n & 1 - (-1/3)^n \\ 1 - (-1/3)^n & 1 + (-1/3)^n \end{array}\right] \left[\begin{array}{c}a_0 & b_0\end{array}\right]$.

Equivalently, we can do the following:

From the second equation, we have $ a_n \equal{} \frac {3}{2} b_{n \plus{} 1} \minus{} \frac {1}{2}b_n$.  Plugging this in to the first equation for $ a_n$ and $ a_{n \plus{} 1}$ gives $ \frac {3}{2} b_{n \plus{} 2} \minus{} \frac {1}{2}b_{n \plus{} 1} \equal{} \left(\frac {1}{2}b_{n \plus{} 1} \minus{} \frac {1}{6} b_n\right) \plus{} \frac {2}{3}b_n$.  Rearranging this equation, we have $ 3b_{n \plus{} 2} \minus{} 2b_{n \plus{} 1} \minus{} b_n \equal{} 0$.  This equation has characteristic equation $ 3x^2 \minus{} 2x \minus{} 1 \equal{} 0$, with roots $ 1, \minus{} \frac {1}{3}$ (do these look familiar?), and so $ b_n \equal{} A \plus{} B\cdot \left( \minus{} \frac {1}{3}\right)^n$ for all $ n$.  Now use initial values to find $ A, B$ and plug into the second equation to get $ a_n$.

Incidentally, this is a very special set of equations, and we actually can express the answer in another way by saying that $ (a_n, b_n)$ is "as close as you can get" to $ (\frac{a_0 \plus{} b_0}{2}, \frac{a_0 \plus{} b_0}{2})$ using fractions with denominator $ 3^n$.  (Compute a few terms and you'll be able to state this more precisely.)

Carcul, a LaTeX suggestion: you can use \Leftrightarrow or \Longleftrightarrow in place of <=> for an equivalence symbol.

The process [b]Carcul[/b] used is called "diagonalization", and is also sometimes called [url=http://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix]eigendecomposition[/url]. You'll find it in any introductory linear algebra textbook. This is the most mathematically illuminating approach.

[b]Edit:[/b] [i]JBL wrote the same thing while I was in the bathroom, it seems![/i]

Alternately, we can decouple the system. Begin with:
$ 3 a_{n \plus{} 1} \minus{} a_n \equal{} 2 b_n$
$ 3 b_{n \plus{} 1} \minus{} b_n \equal{} 2 a_n$

From the first equation, we get:
$ 9 a_{n \plus{} 2} \minus{} 3a_{n \plus{} 1} \equal{} 2 (3b_{n \plus{} 1})$
$ \minus{} 3 a_{n \plus{} 1} \plus{} a_n \equal{} \minus{} 2 b_n$

Which we can add to get:
$ 9 a_{n \plus{} 2} \minus{} 6a_{n \plus{} 1} \plus{} a_n \equal{} 2 (3b_{n \plus{} 1} \minus{} b_n)$
$ \iff 9 a_{n \plus{} 2} \minus{} 6a_{n \plus{} 1} \plus{} a_n \equal{} 4 a_n$
$ \iff 3 a_{n \plus{} 2} \minus{} 2a_{n \plus{} 1} \minus{} a_n \equal{} 0$

We can do the same with the second equation, and we get the following system of decoupled homogeneous second-order recurrence equations:
$ 3 a_{n \plus{} 2} \minus{} 2a_{n \plus{} 1} \minus{} a_n \equal{} 0$
$ 3 b_{n \plus{} 2} \minus{} 2b_{n \plus{} 1} \minus{} b_n \equal{} 0$

Solving the two equations independently gives:
$ a_n \equal{} \alpha_1 \left( \minus{} \frac {1}{3} \right)^n \plus{} \alpha_2$
$ b_n \equal{} \beta_1 \left( \minus{} \frac {1}{3} \right)^n \plus{} \beta_2$

Or we can use the particular properties of the system:

Add up the equations to get

$ a_{n+1}+b_{n+1}=a_n+b_n$

Therefore $ a_n+b_n$ is constant, and equal to $ a_0+b_0$

Subtract the second equation from the first one to get

$ a_{n+1}-b_{n+1}=-{1\over 3}(a_n-b_n)$

Therefore $ a_n-b_n$ is a geometric sequence, hence $ a_n-b_n=\left(-{1\over 3}\right)^n(a_0-b_0)$

From

$ \left\{\begin{eqnarray*}a_n+b_n &=& a_0+b_0\\a_n-b_n &=& \left(-{1\over 3}\right)^n(a_0-b_0)\end{eqnarray*}$

we simply obtain

$ a_n={a_0\over 2}\left(1+\left(-{1\over 3}\right)^n\right)+{b_0\over 2}\left(1-\left(-{1\over 3}\right)^n\right)$

$ b_n={a_0\over 2}\left(1-\left(-{1\over 3}\right)^n\right)+{b_0\over 2}\left(1+\left(-{1\over 3}\right)^n\right)$


Let $ABC$ be a triangle and let $D$ be a point on $AB$ such that $4AD = AB$. The half-line $l$ is drawn on the same side of $AB$ as $C$, starting from $D$ and making an angle of $\theta$ with $DA$ where $\theta = \angle ACB$. If the circumcircle of $ABC$ meets the half-line $l$ at $P$, show that $PB = 2PD$.

I'm not understanding your question, can you please post some sort of diagram? Sorry :oops:

$\widehat{ACB}\equiv\widehat{APB}\equiv\widehat{ADP}$ $\Longrightarrow$ $\triangle ADP\sim\triangle APB$ $\Longrightarrow$ $\frac{AD}{AP}=\frac{DP}{PB}=\frac{PA}{BA}$ $\Longrightarrow$ $\frac{\frac{c}{4}}{AP}=\frac{DP}{PB}=\frac{PA}{c}$ $\Longrightarrow$ $PA=\frac{c}{2}$ and $PB=2\cdot PD\ .$

[b]Remark.[/b] $\lambda\cdot AD=AB$ $\Longrightarrow$ $PB=\sqrt{\lambda}\cdot PD\ .$

This may help you. Buy : The Mathematical Olympiad Handbook by Tony Gardiner (former IMO coach of UK). This book contains all BMO problems and solutions from the first BMO to the 32nd one in 1996.

It is a great book (although it doesn't so much have solutions, as walkthroughs as to how to solve it, which are a lot more helpful I think).

Tony Gardiner is the greatest. Geometry makes so much more sense after his lectures. :)

[quote]Tony Gardiner is the greatest. Geometry makes so much more sense after his lectures.[/quote]

Very very true!

But I don't like 'The Mathematical Olympiad Handbook' that much. Most of the questions pre-1990 (roughly) are very similar to school type maths and not very interesting, and it has no BMO round 2 questions/answers. I think a new version would be good (or someone should write one) since recent bmo questions are much nicer.

Yeah, a new handbook with BMO2s as well would be great. His "some elementary mathematics" chapter was very useful for me this time last year in getting from IMOK level up to BMO (which actually requires a little knowledge of stuff).

Hmmm, maybe I'll start a PDF, on the off chance that it becomes useful. I need to do some writing now I'm no longer doing any essay subjects at school...


Let $a, b, c$ three non-negative real numbers such that $abc=1$.
   Show that
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}+\frac{4(ab+bc+ca)}{(a+b)(b+c)(c+a)}\geq(ab+bc+ca). \]

cute inequality!
after we make the substitution $a=\frac1x,b=\frac1y,c=\frac1z$
the inequality reduces to:
$\sum_{cyc}\left(\frac{x^{2}}{y+z}\right)+\frac{4(x+y+z)}{(x+y)(y+z)(z+x)}\geq x+y+z$
but $\sum_{cyc}\left(\frac{x^{2}}{y+z}\right)=\frac{(x+y+z)(x^{3}+y^{3}+z^{3}+1)}{(x+y)(y+z)(z+x)}$
so all we need to prove is
$x^{3}+y^{3}+z^{3}+5\geq \sum_{sym}x^{2}y+2$
which is just Schur's inequality  :)
it was not that cruel after all. :lol:


I need some help with the following problems:

1.) A farmer went to a state fair. On Monday he bought 12 sheep and 4 goats for 60 dollars. On Tuesday he bought some sheep for 24 dollars and on Wednesday he bought some goats for 30 dollars. He bought 3 more sheep on Tuesday than goats on Wednesday. The cost of the sheep and a goat was the same on all three days. If $s and $g are the prices of a sheep and a goat respectively, compute the ordered pair (s, g).

2.) Quadrilateral TRIO is inscribed in a circle and has angle O = 120 degrees, TO = IO, TR + IR = 7, and TR/IR = 6. Compute the area of TRIO

Thanks

[quote="Mario_KaRt"]... Quadrilateral TRIO is inscribed in a circle and has angle O = 120 degrees, TO = IO, TR + IR = 7, and TR/IR = 6. Compute the area of TRIO...[/quote]
Since the sum and the ratio of the sides $TR, IR$ are known, these sides can be calculated:

$TR = 6 \cdot IR$, [color=white].[/color] $TR + IR = 6 \cdot IR + IR = 7 \cdot IR = 7$, [color=white].[/color] $IR = 1$, [color=white].[/color] $TR = 7 - IR = 6$

The sum of the opposite angles of a cyclic quardilateral is $180^o$:

$\measuredangle O + \measuredangle R = 180^o$, [color=white].[/color] $\measuredangle R = 180^o - 120^o = 60^o$

Since the sides $IR = 1, TR = 6$ and the angle $\measuredangle R = 60^o$ are known, the diagonal $IT$ can be calculated by the cosine theorem for the triangle $\triangle ITR$:

$IT^2 = IR^2 + TR^2 - 2 \cdot IR \cdot TR \cos{\widehat R} = 1 + 36 - 12 \cdot \frac 1 2 = 31$, [color=white].[/color] $IT = \sqrt{31}$

Since the triangle $\triangle ITO$ is eqilateral with $IO = TO$ and the angle $\measuredangle = 120^o$ is known, the remaining angles $\measuredangle ITO = \measuredangle TIO$ can be calculated:

$\measuredangle ITO = \measuredangle TIO = \frac{180^o - \measuredangle O}{2} = \frac{60^o}{2} = 30^o$

Now it is possible to calculate the sides $IO = TO$ using the sine theorem for the triangle $\triangle ITO$

$\frac{IO}{IT} = \frac{\sin 30^o}{\sin 120^o} = \frac{\frac 1 2}{\frac{\sqrt 3}{2}} = \frac{\sqrt 3}{3}$

$IO = TO = IT \cdot \frac{\sqrt 3}{3} = \frac{\sqrt{93}}{3}$

Now the areas of the triangles $\triangle ITR, \triangle ITO$ can be calculated:

$|\triangle ITR| = \frac 1 2\ IR \cdot TR\ \sin{\widehat R} = \frac{1 \cdot 6}{2} \sin{60^o} = \frac{3\sqrt 3}{2}$

$|\triangle ITO| = \frac 1 2\ IO \cdot TO\ \sin{\widehat O} = \frac{93}{18} \sin{120^o} = \frac{31 \sqrt 3}{12}$

The quadrilateral area is the sum of the areas of these 2 triangles:

$|TRIO| = |\triangle ITR| + |\triangle ITO| = \frac{3\sqrt 3}{2} + \frac{31 \sqrt 3}{12} = \frac{(18 + 31) \sqrt 3}{12} = \frac{49 \sqrt 3}{12}$

Just created three equations:
Price of sheep=s
Price of goats=g
# of sheep bought on Tuesday=t

12s + 4g=60
ts=24
(t-3)g=30

Start with one equation, solve for a variable, then substitute that into another equation, solve, and then substitute into the last equation. Then you should find the number value of a variable. (You may need to use quadratics, and then get two answers. The real answer is the root that is a natural number.) Then solve for the others. You should get  the ordered pair (3,6).


In how many ways can 5 books be arranged on a shelf if 2 of the books must remain together?

[hide]take the two books away, there are 3 books, 6 ways to arrange them. there are 4 ways to arrange the two books that have to stay together, but their order can be changed,  so 2*4*6=48. [/hide]

Hmm... I think you got it wrong, 236factorial. This is what I got:
[hide]

There are 4 ways to rearrang the books so that the 2 books stay together.
You have to multiply by 3!, because for the first, 3 books, second, 2 and last, 1. But, there are still the 2 books that have to be together, then you have to multiply by 2, because it could be AB or BA so the answer is $\framebox{48}$[/hide]

[quote="G-UNIT"]Pascal is correct, assuming the two books have already been picked.[/quote]

I know...  :roll:
I have fixed the solution.  :roll:


Given matrices:
B_1 =
[2 2 1]
[2 1 1]
[0 1 0]

and B_2:
[1 1 0]
[1 0 1]
[0 1 -1]

find an invertible matrix D such that A_1 =D^(-1) A_2 D

There's a problem with what you wrote. You give specific matrices $ B_1$ and $ B_2,$ then suddenly switch the notation to ask if we can find $ D$ such that $ A_1\equal{}D^{\minus{}1}A_2D.$ What are $ A_1$ and $ A_2$ supposed to be? Should those really have been $ B_1$ and $ B_2?$

Let's assume the latter: assume that for the given $ B_1$ and $ B_2,$ we seek $ D$ such that $ B_1\equal{}D^{\minus{}1}B_2D.$ That is, we would like to show that $ B_1$ and $ B_2$ are similar matrices.

If that's the question, then it has a quick, short answer: $ B_1$ and $ B_2$ are not similar, meaning that one cannot find any such matrix $ D.$ How can it be answered so quickly? If two matrices are similar, then they share a number of properties, including that similar matrices must have the same trace and the same determinant. The trace can be computed at a glance: $ \operatorname{trace}B_1\equal{}3$ and $ \operatorname{trace}B_2\equal{}0.$ So we know from that that these two matrices are not similar.


$ p$ is an odd prime.
show that all devisors of $ x^{p-1}+x^{p-1}+\cdots+x+1$ are congruent to either 0 or 1 modulo $ p$.

See my [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461]latest entry[/url] for a generalization.

[hide="Solution"]Suppose a prime $ q$ divides $ x^{p-1}+x^{p-2}+\cdots+1$.  Then $ q$ also divides $ (x-1)(x^{p-1}+x^{p-2}+\cdots+1)=x^{p}-1$.  Therefore $ x^{p}\equiv 1\pmod{q}$, so the order of $ x\pmod{q}$ divides $ p$.  So the order of $ x\pmod{q}$ (I don't know how to put orders in LaTeX  :( ) is either 1 or $ p$.  Now we do some casework:

Case 1: The order of $ x\pmod{q}$ is 1.  Then $ x\equiv 1\pmod{q}$.  Now go back to the original expression.  We have $ x^{p-1}+x^{p-2}+\cdots+1\equiv p\pmod{q}$, but also $ x^{p-1}+x^{p-2}+\cdots+1\equiv 0\pmod{q}$, so $ q|p$.  Therefore $ p=q$, so that $ q\equiv 0\pmod{p}$.

Case 2: The order of $ x\pmod{q}$ is $ p$.  By Fermat $ x^{q-1}\equiv 1\pmod{q}$.  Therefore $ p|q-1$, so $ q\equiv 1\pmod{p}$.

Therefore all prime divisors of $ x^{p-1}+x^{p-2}+\cdots+1$ are 0 or 1 $ \pmod{p}$.  Since every divisor of $ x^{p-1}+x^{p-2}+\cdots+1$ is a product of its prime divisors, every divisor is 0 or 1 $ \pmod{p}$, and we're done.[/hide]


[color=purple]Let ABCis the arbitrary triangle in plane has three medians $m_{a},m_{b},m_{c}$.Prove that
  $\frac{m_{a}}{m_{b}}+\frac{m_{b}}{m_{c}}+\frac{m_{c}}{m_{a}}\ge \frac{27(a^{2}+b^{2}+c^{2})}{4(m_{a}+m_{b}+m_{c})^{2}}$
  Sorry if it was posted in this forum.Have fun ! :D [/color]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=802436#802436]the link[/url] with the last post of me!


We painted a few sides of a cube, and then we cut it up to smaller but equally sized cubes. We got 45 smaller cubes with no paint on them. How many sides of the original cube did we paint?

$45 = 3^2 \cdot 5$

thus it is a 3 by 3 by 5 rectangular prism.

It started as a 5x5x5 cube. You painted 4 sides, leaving 2 ends.

This problem is from 1983 Australian Mathematics Competition and this problem was discussed on Crux/Mathematical Mayhem, March 2005.


function algebra domain calculus derivative limit integration

Let f be a function with domain the set of all real numbers and having the following proerties:

(i) f(x+y)=f(x)f(y) for all real numbers x and y.

(ii) lim h->0 (f(h)-1)/h = k, where k is a nonzero real number.

a) Use these proerties and a definition of the derivative to show that f'(x) exists for all real numbers x.

b) Let f^(n) denote the nth derivative of f. Write an expression for f^(n) of x (((( f^(n)(x) )))) in terms of f(x).

c) Given that f(1)=2, use the Mean Value Theorem to show that there exists a number c such the 0<c<3 and f'(c)=7/3

... if u could help me out that would be fantastic =)

[quote="Michi"]Let f be a function with domain the set of all real numbers and having the following proerties:

(i) f(x+y)=f(x)f(y) for all real numbers x and y.

(ii) lim h->0 (f(h)-1)/h = k, where k is a nonzero real number.

[/quote]
We have $ f(0+0)=f(0)f(0)$.  This gives $ f(0)=0$ or $ 1$.  Taking $ f(0)=0$ implies $ f(x+0)=f(x)f(0)=0$ i.e., $ f(x)=0$ for all x.  This contradicts the second condition.  Hence we take $ f(0)=1$, so we can write $ \lim_{h\to0}{\frac {f(h)-1}{h}=k}$ as $ \lim_{h\to0}{\frac {f(h)-f(0)}{h}=k}$ which implies $ f'(0)=k$.
Now, $ f'(x)=\lim_{h\to0}{\frac {f(x+h)-f(x)}{h}=\lim_{h\to0}{\frac {f(x)f(h)-f(x)}{h}=\lim_{h\to0}{f(x)\frac {f(h)-1}{h}=kf(x)}}}$
Its easy to observe that $ f(x)=\ e^{kx}$
nth derivative of f(x) is given by $ \ f^{n}(x)=k^{n}f(x)$

If $ f(1)=2$, we have $ 2=\ e^{k}$ i.e., $ k=ln2$.  This gives $ f(x)=\ 2^{x}$
Using Mean Value theorem for $ f(x)$ on $ [0, 3]$, we have $ \frac {f(3)-f(0)}{3}=f'(c)$ for some $ c$ in $ (0, 3)$.  So, $ \frac {8-1}{3}=\frac {7}{3}=f'(c)$.  Hence proved.

Since $ f'(x) \equal{} kf(x)$, Shouldn't $ f^{n}(x) \equal{} kf^{n\minus{}1}(x)$?

thanks u guys for trying to help me out =) also, how do u get things to come up in like picture form for limits and integrals and stuff? *noob*

The pretty math is $ \text{\LaTeX}$- see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]here[/url].


function
number theory unsolved
number theory

How do you prove that $\sum_{d|n}d(n)^{3}=(\sum_{d|n}d(n))^{2}$ ? Where $d(n)$ is the number of positive divisors of $n$. I think it's well-known, but I didn't find anything on the forum.

No typo's?
If $n$ has $d(n)$ divisors then $\sum_{d|n}c= d(n)\cdot c$ where $c$ is a constant. In the $LHS$ our constant is $d^{3}(n)$ and the result is $d^{4}(n)$ while inrthe $RHS$ side the constant is $d(n)$ and the result is $(d^{2}(n))^{2}=d^{4}(n)$.
late edit:
[quote="N.T.TUAN"]Put $\text{LHS}=F(n)$ and $\text{RHS}=G(n)$, then $G,F$ are multiplicative. We need only prove that $F(p^{k})=G(p^{k})$, here $p$ is a prime.[/quote]
with this observaton things become clear because $F(p^{k})=\sum_{i=1}^{k+1}i^{3}$ and $G(p^{k})=(\sum_{i=1}^{k+1}i)^{2}$and obviously $\sum n^{3}=(\sum n)^{2}$

[quote="ZetaX"]Correct version:

[quote="Stewie"]How do you prove that $\sum_{k|n}d(k)^{3}=(\sum_{k|n}d(k))^{2}$ ? Where $d(k)$ is the number of positive divisors of $k$. I think it's well-known, but I didn't find anything on the forum.[/quote][/quote]
Yes, I'm an idiot  :lol:.

So it's still open, because the last two posts refer to the wrong version.


Given that $ \minus{} 4\le x\le \minus{} 2$ and $ 2\le y\le 4$, what is the largest possible value of $ (x \plus{} y)/x$?

$ \textbf{(A)}\ \minus{}\!1\qquad
\textbf{(B)}\ \minus{}\!\frac {1}{2}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ \frac {1}{2}\qquad
\textbf{(E)}\ 1$

[quote="#H34N1"]Given that $-4\le x\le-2$ and $2\ge y\ge 4$, what is the largest possible value of $(x+y)/x$?

A. -1
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1[/quote]
[hide]with x negative we need the numerator negative so if
(2-4)/-4=1/2 is the largest value
but
(2-4)/-2=1 but since x=-4 in the numerator I am not sure if I am allowed with substituting values of inequalities  :huh: [/hide]

[quote="#H34N1"]Given that $-4\le x\le-2$ and $2\ge y\ge 4$, what is the largest possible value of $(x+y)/x$?

A. -1
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1[/quote]

[hide]the denominator is always negative so we want the negative so we want the top to be negative too.  the smallest negative number possible on the top is $-4+2=-2$ so $\frac{-2}{-4}=\boxed{\frac{1}{2}}$[/hide]

[hide="Perhaps easier"]Easier to work with $\frac{x+y}{x}=1+\frac{y}{x}$. $\frac{y}{x}$ is always negative, so largest $x$ and smallest $y$ yields the answer.[/hide]

[quote="Altheman"][quote="#H34N1"]Given that $-4\le x\le-2$ and $2\ge y\ge 4$, what is the largest possible value of $(x+y)/x$?

A. -1
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1[/quote]

uhh...hello? the inequality for y does not work[/quote]
Yeah, I noticed that because if you had that, then $y$ can be anything...


analytic geometry
function
algebra
domain
LaTeX
geometry
3D geometry

To input a square root type:[code]\sqrt{x}[/code]where x is the value inside the radicand

To input $ \pi$ type:[code]\pi[/code]

A coordinate pair should be entered as: (x,y)[code](4.2,5.2)[/code]

A fraction should be entered as x/y[code]1/2[/code]

A mixed fraction should be entered as X y/z[code]5 1/3[/code]

Coefficients to variables should be written without the multiplication sign[code]3x[/code]

Multipliers square roots should be written without the multiplication sign. Example  $ 3\sqrt{5}$[code]3\sqrt{5}[/code]

The problem in which you had to find the mean of a bunch of numbers with floor functions asked for the answer as a common fraction (3/10), and I gave the answer 0.3, which was accepted.

Is there a way around this?

[quote="CatalystOfNostalgia"]The problem in which you had to find the mean of a bunch of numbers with floor functions asked for the answer as a common fraction (3/10), and I gave the answer 0.3, which was accepted.[/quote]

I think the game design is accepting it in any form that is equivalent at the moment, because they are both equal.

That doesn't make a whole lot of sense. If it accepts anything equal, then, hypothetically, if a question asked "What is $ \sqrt50$ in simplest radical form?" someone could type "\sqrt50" and get it right.

Yes, but have you ever seen one of these probs?

I think only certain problems accept equivalent answers, especially ones that don't specify whether your answer should be expressed as a fraction or a decimal. Like in questions where you have to express your answer as a mixed number, if you put it as a common fraction you'll get it wrong.

One of the questions the answer is sunday, and i amswered sundya and got right :D

wow
i'm surprised that happened
i typo those q's a lot and i always get them wrong

Maybe you [i]think[/i] you answered sundya but you really put sunday?

well it was a cd against you and we both saw it as sundya...

Oh, right!! Yeah, that was strange...I'd forgotten that. Probably a glitch.

Maybe it was just because "Sundya" is closer to "Sunday" than any other "dya" of the week.

I'd suggest not to :spam: on this thread. :wink:

also, it says to input pi as \pi and everyone does it as plainly pi, and get a problem right
likewise, you can just type sqrt x and get it right
 :o  :o  :o

For quadrants should we put numbers or roman numerals?

Some of them marked it wrong with Roman Numerals while others marked it correctly.

I want to know the way that is used more often...so that I minimize chances of getting it wrong as it cost me a game today...

I don't think there is a definite way to find out. There is a pool of questions and it'd be hard for mods to check the answers. Just stick to RN, it doesn't really matter if you lose a bit of rating; its ultimate purpose is teaching math.

[quote="thdanh90"]I don't really think so ernie. I bet it is
[code]x\times 10^y[/code][/quote]

nah: probably x/{randomstuffies}10/{boomboomfirepower}y

Currently, you cannot add problems to ftw

^What he said. You can submit to the moderators, try it out.

I don't think you should.  It would just take valuable time away from AoPS, since there would be a lot of trolls, unfortunately.

EDIT: 1700th post!

what...
I fail to see your logic in this. You're saying that one person asking to add a problem to FTW! would result in lots of trolls taking time away from AoPS?


Given square $ABCD$ inscribed $(O),M\in\text{ small arc }AD$ prove that
$MA-MD=\sqrt{2}(MB-MC)$

Are you sure about the problem?
This is the result I get:

Say WLOG that the sidelength of the square is $1$.

By Ptolemy: \[AM+MC = \sqrt{2}MB\]  \[MD+MB = \sqrt{2}MC\] So \[AM-MD = (\sqrt{2}+1)(MB-MC)\] ?

Ok you are right Jan, I have a typo!
It really is $MA-MD=(\sqrt{2}-1)(MB-MC)$ there is a general case for any quadrilateral $ABCD,M\in\text{ small arc }CD$ prove that
$(\tan\frac{(OD,OA)}{4}+\tan\frac{(OA,OC)}{4})MA+(\tan\frac{(OA,OD)}{4}+\tan\frac{(OD,OB)}{4})MD =(\tan\frac{(OD,OA)}{4}+\tan\frac{(OA,OC)}{4})MB+(\tan\frac{(OB,OD)}{4}+\tan\frac{(OD,OA)}{4})MC$
here we use $(OX,OY)$ to show angle with direction
if $ABCD$ is quare we have that problem!


The shaded region shown consists of 11 unit squares and rests along the $ x$-axis and the $ y$-axis. The shaded region is rotated about the $ x$-axis to form a solid. In cubic units, what is the volume of the resulting solid? Express your answer in simplest form in terms of $ \pi$.

[asy]fill((0,0)--(0,5)--(1,5)--(1,2)--(4,2)--(4,0)--cycle, gray(.8));
draw((0,0)--(0,5)--(1,5)--(1,2)--(4,2)--(4,0)--cycle);
draw((0,1)--(4,1));
draw((0,2)--(1,2));
draw((0,3)--(1,3));
draw((0,4)--(1,4));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
draw((3,0)--(3,2));

draw((4,0)--(5,0));
label("x", (5,0), E);

draw((0,5)--(0,6));
label("y", (0,6), N);[/asy]

When rotated, the vertical strip of 5 blocks will be a cylinder of radius 5 and height 1, so it's volume is $ 25\pi$.  The rectangle will have radius 2 and height 3, so it's volume is $ 12\pi$.  Together, their volume is $ 37\pi$.

I disagree. That would equate the rectangle 3-D version to a cylinder, which it is not. Rather, I say find the volume of the entire base rectangular cylinder, which is a cylinder of height 2 and of radius 4. This has a volume of $32\pi$. Now the volume of the vertical cylinder-a cylinder of radius 1 and height 3-is simply $3\pi$. Adding these up, we get $\boxed{35\pi}$, which I believe is the correct answer as opposed to $37\pi$.

@above, it is rotated about the $x$ axis, not the $y$ axis.