[hide]
we know it's a 3 digit number square that starts with 1, and ends with 3 or 7.  3 is more probable, so lets try 3 first.  so we have $(103 + a*10)^2 = 2xy89$, where $a$ is the middle digit.  Expand and we have $10609 + 2060a + 100a^2 = 2xy89$.  Look at the 8, only the middle term will determine it, so $6a \equiv 8 (\bmod\ 10)$, solving we have $a \equiv 3 (\bmod\ 5)$.  If $a = 3$, then the number square will be less than 20000, so $a = 8$, but then it exceeds 29999.  Do the same thing with $7$ yields the solution below :P
[/hide]

[quote="drunner2007"]The five-digit number 2xy89 is a square of an integer. Find the two digit number $xy$.[/quote]

If you use pkerichang's way with the 7, you get a= 6 so the square is 167. Then xy=78.

If a=8 your answer is larger than 29989...

[quote="pkerichang"][hide]
 Look at the 8, only the middle term will determine it, so $6a \equiv 8 (\bmod\ 10)$, solving we have $a \equiv 3 (\bmod\ 5)$.
[/hide][/quote]

Where did you get the 6a and the 8 (mod 10)?

[hide]27889 just by trial and error  :D [/hide]

The question is easy enough, and there are solutions above,  but this trick/method  may be helpful for   solving similar problems.

(How to do this without pen or paper :)

Let the perfect square be  $n$

[hide]1.  $140 < n < 180$
[hide="because"]If you know that $14^2 < 200$ and $18^2 > 300$ , actually you can find a more refined vaule by computing $\sqrt{20089}$ and $\sqrt{29989}$  and gettting $142 \le n \le 173$  but no need  to do any thing which you can't do in a few seconds without pen or paper. [/hide]
2.  And $n \equiv (25 \pm 8 )$  or $\equiv (75 \pm 8 ) \bmod 100$
[hide="because"] $n^2 \equiv 89 \bmod 100 \implies n^2 \equiv 1 \bmod 4 \ and \ n^2 \equiv 14 (\equiv 64) \bmod 25$
 $\implies$ ($n$ is odd ) and ($n \equiv \pm 8 \bmod 25$)
$\implies n \equiv (25 \pm 8)$ or $n \equiv (75 \pm 8) \bmod 100$  [/hide]
3. Only choice for n is   $n=175-8 = 167$[/hide]

Yes it takes much longer to write the solution than actually  figure out the answer. :) . Hope this is helpful.


It depends on what you are doing at school

Before I took algebra 1, I was already doing way past my grade level math.
I was thinking to take algebra 2 since I already knew a lot of algebra.

But then, I just signed up for algebra 1 since my mother suggested it.

It turns out that I had SO much to learn! It's not all just stuff like:

$ x\plus{}y\equal{}32$
$ 2x\minus{}6y\equal{}78$

That's just the beginning.
Then you learn like a [u][b]TON[/b][/u] of problem solving skills, good for stuff like future soon-to-be AMC winner. Logic is my weakness.  :maybe:

So I suggest that you take the course, even though you think that it may be easy for you.  :P

[quote="ritwik_anand"]I have taken Algebra 1 (school) so should I buy the AOPS algebra 1 book or should I just go to Algebra 2, cause I don't wan't to take a book that is pretty much review...[/quote]
I suggest Algebra 2, but you should take the  "Do you need this?" test for Algebra 1 and Algebra 2 to see if you, well, need it.

Our Intro Algebra book covers all of the algebra of a typical Algebra 1 class plus much of the algebra of a typical Algebra 2 class.  We use the book for both our Algebra 1 and Algebra 2 classes.  So, if you're asking about a class here at AoPS, I would recommend the Algebra 2 class, which uses our Intro Algebra book.

Mr. Rusczyk, is the book for Algebra 1 and 2 and 3 the same or just Algebra 1 and 2?

Thanks.

The book for Algebra 1 and 2 is Introduction to Algebra, and the book for Algebra 3 is Intermediate Algebra.


i'm getting pretty random now. exactly like the top admin competition. you can add a point and take one away. and each soda will start wiht 30. i will allow people to post as much as they want i guess
. oh and soon i'm gonna start a top chip on aops. unless someone else starts it already.

pepsi:30

sierra mist:30

coke:30

sprite:30

orange soda:31

grape soda:30

root beer:29

pepsi:30

sierra mist:30

coke:30

sprite:30

orange soda:31

grape soda:30

root beer:29

Ah, reversed.

#4pepsi:30

sierra mist:30

coke:30

sprite:30

orange soda:32

grape soda:30

root beer:28

pepsi:30

sierra mist:30

coke:30

sprite:30

orange soda:31

grape soda:30

root beer:29

This is repetitive

What, you can't minus 3!

1 root beer: 11
2 orange soda: 7
3 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 12
2 orange soda: 5
3 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 13
2 orange soda: 5
3 ginger ale: 2
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 11
2 orange soda: 5
3 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

I just found out, drop 1... so

1 root beer: 13
2 orange soda: 5
3 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 13
2 orange soda: 3
2 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 12
2 orange soda: 3
2 ginger ale: 3
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

1 root beer: 14
2 orange soda: 3
2 ginger ale: 2
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

Read the Rules, 1=2.

Hmm...people have been subtracting 2 since page 8 despite the original rule, and there was even a comment about it.

[quote="iin77"]So, uh, I see people randomly started subtracting two now. [/quote]

I'm going to subtract 2.

[color=green]1 root beer: 15[/color]
2 ginger ale: 2
[color=red]2 orange soda: 1 [/color]
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

actually, it's heal 1 hurt 2. not the other way around.

[quote="cf249"]Hmm...people have been subtracting 2 since page 8 despite the original rule, and there was even a comment about it.

[quote="iin77"]So, uh, I see people randomly started subtracting two now. [/quote]

I'm going to subtract 2.

1 root beer: 15
[color=red]2 ginger ale: PWNT[/color]
[color=green]3 orange soda: 2 [/color]
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT[/quote]

[color=green]1 root beer: 16[/color]
[color=red]2 orange soda: PWNT[/color]
3 ginger ale: PWNT
4 sierra mist: PWNT
5 sprite: PWNT
6 coke: PWNT
7 mountain dew: PWNT
8 pepsi: PWNT

Root beer wins!  (it's my favorite of the choices listed)

Sweet :) I love arbitrarily taking sides and then winning!

With respect to the +1/-1 rule vs the +1/-2 rule, I think it is sensible to use a system in which the game is forced to terminate (such as the latter system) rather than a game which is potentially endless. I mean there are plenty of pointless games that don't terminate, but people get bored of them... but hurt and heals don't end until there's a winner!

I strongly encourage anyone creating a new hurt/heal game to force it to terminate (ie, take off more points than you put back).

Dang it!!!! I wanted orange soda to win, and I missed the great finale.


geometry
calculus
integration
calculus computations

My friends and I disagree mightily over a problem as such:

"Find the area between the curves y=x^2, y=x, and y=2."
I think you have to subtract the integrals, so that you have

/int^2_0 (x^2-x)/,dx = 2/3

They disagree and say the answer is, variously, 5/6 and 1.

Can anyone solve this and verify an answer?

My answer is $ \frac{13\minus{}8\sqrt{2}}{6}$

I think there's some ambiguity here. When $ 0\leq x\leq 1$, there are two pieces: the piece above $ y\equal{}x^2$ and below $ y\equal{}x$, and the piece above $ y\equal{}x$ and below $ y\equal{}2$. Are they both considered? Similarly, when $ 1\leq x\leq \sqrt{2}$, there's also two pieces.

If all pieces are considered, then the area is
\[ \int_{\minus{}\sqrt{2}}^1 (2\minus{}x^2)\ dx \plus{}\int_1^2 (2\minus{}x)\ dx\]
Here the area is the integral of the difference between the highest curve and the lowest curve (as long as the two curves form a closed piece). Outside of $ \minus{}\sqrt{2}\leq x\leq 2$, there's no intersection, so no more closed pieces.


[quote="hatsup123"]whats the multiplication sign in latex code????[/quote]

There are two types:

\[
\times \text{ and } \cdot
\]

\times and \cdot

Examples:

\[
2 \times 2 = 5
\]
\[
1 \cdot 2 = 3
\]


linear algebra
matrix
search
linear algebra open

Hi,

I have the following matrix:

1 1/2 1/3.....1/n
1/2............1/n+1
.
.
.
.
.
1/n...............1/2n-1

Does anybody have any idea of what it's inverse should be? It's claimed that the inverse has all integer(!!) entries.

The inverse of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=163506]Hilbert matrix[/url]. Follow the links in that thread, too.

(And thanks to Kent for inserting the search keyword I used)


function
complex analysis
complex analysis unsolved

Let $ f$ be analytic on $ D\equal{}\{z<1\}$.   Suppose $ f$ is one-to-one on $ D\minus{}\{0\}$.  Prove that $ f$ is one-to-one on $ D$

If $ n$ is the largest integer so that $ f^n(0) \equal{} 0$, then $ f$ maps a neighborhood $ V$ of $ 0$ to a neighborhood of $ f(0)$ which is covered $ n\plus{}1$ times. Now, suppose there is $ w$ so that $ f(0) \equal{} f(w)$. By the above reasoning applied to $ w$, we have $ f'(w) \neq 0$. Then there is a neighborhood $ W$ of $ w$, chosen with $ V \cap W$ empty, such that $ f$ maps $ W$ topologically to a neighborhood of $ f(0)$. This is clearly a contradiction to $ f$ being injective on the punctured disc.


Let $ a,b,c > 0$ and $ a\plus{}b\plus{}c\equal{}1$ Prove that \[ \sum_{cyc}\sqrt{\frac{2b}{3\plus{}a}} \leq \sqrt{\sum_{cyc}\frac{1}{1\plus{}2\sqrt{ab}}}\]

[quote="apollo"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$ Prove that
\[ \sum_{cyc}\sqrt {\frac {2b}{3 \plus{} a}} \leq \sqrt {\sum_{cyc}\frac {1}{1 \plus{} 2\sqrt {ab}}}
\]
[/quote]
$ \sum_{cyc}\sqrt {\frac {2b}{3 \plus{} a}} \leq \sum_{cyc}b\sqrt{\frac{2}{b(3\plus{}a)}}\leq\sqrt{\sum_{cyc}\frac{2}{3\plus{}a}}$ and $ \sqrt {\sum_{cyc}\frac {1}{1 \plus{} 2\sqrt {ab}}}\geq\sqrt {\sum_{cyc}\frac {1}{1 \plus{} a\plus{}b}}.$
Hence, it remains to prove that: $ \sum_{cyc}\frac{2}{3\plus{}a}\leq\sum_{cyc}\frac {1}{1 \plus{} b\plus{}c}.$ But $ \sum_{cyc}\frac{2}{3\plus{}a}\leq\sum_{cyc}\frac {1}{1 \plus{} b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{1\plus{}a\minus{}2b\minus{}2c}{(3\plus{}a)(1\plus{}b\plus{}c)}\geq0\Leftrightarrow\sum_{cyc}\frac{a\minus{}b\minus{}(c\minus{}a)}{(3\plus{}a)(1\plus{}b\plus{}c)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)\left(\frac{1}{(3\plus{}a)(1\plus{}b\plus{}c)}\minus{}\frac{1}{(3\plus{}b)(1\plus{}a\plus{}c)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(2\minus{}c)(3\plus{}c)(1\plus{}a\plus{}b)\geq0.$


Ther are 8 boys and 6 girls in who are members of the grumpet section in the band.  From the members of the trumpet section, a quintet is formed.  IF the quintet must consist of 3 boys and 2 girls , how many possible quintets are there?


quadratics
summer program
Mathcamp
function

Let x_n be a sequence so that we have x_n=4x_(n-1)-x_(n-2) for any n>=3 with x1=2 and x2=7. Prove that we have the following thing:
for any n>=1, the sequence a_n=x_(2n+1)/2 is the sum of two consecutive squares!

thank you!

This shall be messy.  It's going to be inductive-like, several times over.  I'm just going to give the outline, not the details.  First, assume it works for two consecutive odd terms of the sequence, which means they can be written as (4a :^2: + 4a + 2) and (4b :^2: + 4b + 2).  Then find out what the next odd term of the sequence will look like.  Assume this term has the same form and solve for the resulting number whose square plus the next greater square gives you the same result.  We can use this to recursively solve the first few terms.  (Alternatively, you could just generate the sequence out to the 9th term or so and solve a bunch of quadratics like that.  Either way, we need the numbers whose squares we are considering.)  You can then make the educated guess that those numbers must also follow a similar recursion to the given recursion, which they do (although you'll need 5 terms to figure out exactly what it is, maybe.  At least 4 terms.)  So, this gives you a recursion.  However, it's not quite as nice a recursion -- I'm not sure if I know how to solve it off the top of my head.  I'll keep thinking on it.

i solved by the characteristic equation that

x_n = 1/2 ( (2+sqrt(3))^n + (2-sqrt(3))^n)

what we can do with that... i have still to figure out. But it might be useful...

Usually when the problems asks you to show something about the terms of the sequence like this, the actual formula for each term usually isn't that helpful. Although, if you really feel like it, you will probably be able to bash the way there, though it would take heapsa algebra!

One thing that might help is if we knew exactly which two squares they are.  After trying the first five numbers, here is my conjecture: an is the sum of the squares of [hide](xn+1 - xn - 1) / 2 and (xn+1 - xn + 1) / 2[/hide].  Okay, now it's up to the rest of you to prove it.

Ravi's comment also explains why the squares would obey a similar recursive relation ... I didn't think to look for them in the sequence itself.

Nukular -- I don't think it works.  There isn't any nice way to manipulate the resulting expression to get rid of the :rt3:'s, as far as I can tell.

[quote="Ravi B"]My guess is that Nukular's closed form could be used to prove my conjecture in a brute-force manner.  That doesn't mean it's the most elegant solution.

This problem reminds me of some identities involving Fibonacci numbers.  For example, F2n+1 = Fn+12 + Fn2.  More generally, Fn+k = Fk Fn+1 + Fk-1 Fn.  There are several ways to prove these identities.  One pedestrian method is to prove (say the second one) by induction.  A slicker method is to use 2x2 matrices.[/quote]

I learned the second identity through combinatorics. So I guess that makes a third method.

You can also prove it with path-travelling.

hm. could you please post the proofs of these last two methods? I think I can figure out the first two...

Imagine walking on a strip like the one shown in the attachment.  There are Fn ways to reach the nth vertex from the left.  Now consider a path to the (n+k)th vertex -- if it goes through the kth vertex, there are Fk*Fn + 1 total routes, Fk to the kth vertex and Fn + 1 from there to the (n + k)th vertex.  If it doesn't go through the kth vertex, it has to go to the (k - 1)th vertex, then go directly horizontal to the (k + 1)th vertex.  Thus, there are Fk - 1*Fn total routes -- Fk - 1 to the (k - 1)th and Fn from the (k + 1)th to the (n + k)th.  Thus, we add and get the identity we wanted.

That's another nice combinatorial way to do it.... the way I had learned it was to consider fibonacci words* of length m+n-2.

I will be using the definitions that fibonacci numbers are F[n] with F[0]  = 0, F[1] = 1 and F[n] = F[n-1] + F[n-2]. **

Furthermore, i will let f[n] be the number of fibonacci words of length n.

It is well known that with these definitions, f[n] = F[n+2]. This is not too difficult to prove.

Suppose we want to prove

 Fn+k = Fk Fn+1 + Fk-1 Fn.

Then we can also prove, equivalently:

fn+k-2 = fk-2 fn-1 + fk-3 fn-2.

And that is exactly what we're going to do.

A fibonacci word of length n+k-2 can be broken up into two parts. The first part has length k-1, and the second part has length n-1. Both parts must be themselves fibonacci words. The first word ends in either 0 or 1. If it ends in 0, there are fk-2 such words, and we can pick any n-1 sized word for the second part without violating restrictions. Therefore there are fk-2 fn-1 n+k-2 sized words of this type. If the first part ends in a 1, this one must be surrounded by two 0's. Therefore, there are only fk-3 ways to pick the first part, and fn-2 ways to pick the second part, making
 fk-3 fn-2 words of this type and

fk-2 fn-1 + fk-3 fn-2 words in total. QED

* A fibonnaci word is a binary sequence with no two consecutive 1's.

** Is the actual convention F[0] = 0 or F[0] = 1 ? I always mix these up. However, the proof does not differ with the secondary definition. The only difference is that f[n] = F[n+1] instead of F[n+2].

Okay.. can someone help with my problem? I would really need a full solution! :)
thank you very much:)

Well I have been struggling over with this problem, and yet I haven't found any solution! This problem is turning into an obsesion! :)

I noticed that this recursive sequence (with different starting values) appears in Problem 8 of this year's MathCamp quiz: [url]http://www.mathcamp.org/quiz.php[/url].  So perhaps we should not discuss the problem further, at least for now?

[quote="Ravi B"]My conjecture boils down to proving x2n+1 = (xn+1 - xn)2 + 1.  That doesn't look too hard to prove, does it?  You could always use Nukular's closed form to prove it if you run out of other ideas.

P.S. Yes, Nukular, the usual convention for Fibonacci numbers is F0 = 0 and F1 = 1.  As a mnemonic, you can remember that the Fibonacci sequence starts off as the identity function.[/quote]

ahh very good thats a nice way to remember it... i think ill just write a quick function in C++ to find fibonacci's... that'll solidify it a bit more.


MIT college Stanford Columbia Princeton calculus probability

Hey I was wondering if anyone could help me compare some colleges.  I would possibly like to double major in math and engineering.  Anyway, right now, I am considering:

MIT
Stanford
Wash U
Northwestern
Illinois (urbana-champaign)
Carnegie Mellon
University of Penn
Mizzou

Especially wondering about Mizzou because I have seen little ranking them near these other schools engineering-wise, and nothing in math.

What type of engineering?  Some other schools for your list:

Berkely
Caltech
Cornell
Columbia
Johns Hopkins
Georgia Tech

JRav, I am definitely not adding more schools to my list.  Anyway, I've actually considered most of those schools at one point and even visited Johns Hopkins.

Right now I am thinking possibly electrical/computer engineering, but I'm not completely sure of that.

I don't think UPenn is good at engineering.And Northwestern seems a bad choice when considering math and engineering....

MIT
Stanford
Illinois (urbana-champaign)
Carnegie Mellon

These 4 are great especially the former two.  :D

UPenn is known for its engineering department in that it is combined with its science department (at least thats what my friend said.) It's still probably good.

galoisj, remember everything here is relative.  Penn and Northwestern aren't bad schools, but from what I've found they just don't quite measure up to MIT and Stanford in engineering.  Anyway, when I visited Northwestern their engineering program sounded very interesting at the least.   They combine the math into their freshmen engineering classes.  But I'm not sure that that is such a good idea.  But I'm still wondering if you have any more particular reasons for cutting Penn and Northwestern and for keeping Carnegie Mellon.

Anyway, in particular I was wondering where University of Missouri falls in comparison.  anywhere close?

Personally, I think that if you want to double major in math (as opposed to applied math) alongside EE, integrating math into the engineering education might not be a good thing -- the math they'd end up teaching would be very heavily geared towards the engineering contents, and might not give you as rich or though provoking or abstract an experience as you probably want from an undergraduate math degree. Of course, they probably only integrate calc and lin.alg, so you should still be fine for higher courses. But even lin.alg loses a lot of flavor when viewed in a heavily applied way...

Penn is best known for its Wharton School. If you're interested in doing engineering at an Ivy-League school, the better considerations would be Cornell, Princeton, Columbia, and perhaps Dartmouth.

Zuton force, exactly what I think too.  If the math is more integrated, then it might be more geared toward applications and less abstract/theoretical.  I remember that Northwestern integrates differential equations and linear algebra into "Engineering Analysis" (I believe calculus is supposed to be  freshman course).  Anyway, I suppose I'd rather just be required to take those subjects as math courses rather than as part of an engineering class.

Anyway, I really couldn't care if I end up at an Ivy league school.  Right now I think that I'm just going to go with the list galoisj laid out, because that is basically what I found from looking at various rankings.

Do you have any ideas of things I should look for in math/engineering curricula?  Also, what exactly is applied math and what is the difference between applied and regular math?  Somehow I think I could end up doing some of all three...

Applied math is statistics, probability, game theory, economics: math that has real-world applications.  In pure math, you study things like analysis, abstract algebra, topology, differential geometry, number theory...  Many of these things have important applications in physics and engineering but you typically study it for the theoretical applications, not the physical ones.

probability is closer to being pure math than applied math, especially when you learn that an event is really a 'measurable set' and a random variable is really a 'measurable function'.  it's been discussed here before, but calling 'applied math' math with real-world applications is somewhat dangerous because even the purest math tends to have applications somewhere.

that being said, i don't know a truly satisfactory definition of pure or applied math, other than that i (mostly) know it when i see it, and it's really not that important anyways.

I think I have to say that I'm interested in all of it.  I think I'll probably end up doing more of the theoretical, though, with possibly some applied as electives or something.  Anyway, that's not very important now.

Thanks for your help everyone!  Anyway, the last question I have that is actually somewhat important now is how does University of Missouri compare to these others?

Look up their department website.  If the department looks good to you, then it will be good for you.  If not, then it won't be.

oh come on, this is just ridiculous.  if you don't have anything of any value to post JRav, then just don't post.


Let us consider a 200-digit number that repeats itself every 10 digits.  Thus we have 9876543210...3210.  Let us pick every third digit from this number to form a new number.  Repeat until only 2 digits are left.  What is this 2-digit number?

Why would it be 98?
I don't get it.[/hide]

I just wanted to say that I figured out why it's 98.

In order to solve this, you take the third digit of the third digit of the third digit out the number created by taking the third digit out of the number.  Thus, you need to find the 81st and 162nd digit of the number.

[u]Digits[/u]

1st: 9
11th:9
21st: 9
Thus, the 81st digit is nine.  Following the same principle, we find that the 162nd digit is 8.

So, the answer is 98.

Yes it is, I've seen the problem on a test before

don't look at this till you've tried 2 solve the problem

[hide]98 because you would have to find out what exponent of three would divide 200 into two. since 81 is the closest exponent that doesn't go over 100, we would have to find the 81st number and the 162nd. seems like its 98 (unless im wrong)[/hide]


If n is a positive integer, what is the maximum possible number of zeroes at the right end of the integer expression $ 1^n\plus{}2^n\plus{}3^n\plus{}4^n$?

Is the answer [hide="answer"]$ 2$[/hide]?

I haven't tried to find a solution yet.


function
algebra
functional equation
algebra unsolved

Determine all functions $f: \mathbb R \to \mathbb R$ satisfying
\[ f(x^2 \plus{} f(y)) \equal{} y \plus{} f^2(x), \quad \forall x,y \in \mathbb R.\]

[quote="reason"]$ R$ means $ \mathbb{R^{*}}$ ?[/quote]

$ R$ means the set of real number (including 0)
And finally, I have solved it by myself  :lol:.

[quote="ll931110"]I think this functional equation is not difficult, but I haven't solved it yet  :mad:

[i]Determine all functions $ f: R \rightarrow R$ satisfying
$ f(x^2 \plus{} f(y)) \equal{} y \plus{} f^2(x)$[/i][/quote]

Let $ P(x,y)$ be the assertion $ f(x^2\plus{}f(y))\equal{}y\plus{}f(x)^2$
Let $ f(0)\equal{}a$

$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x\plus{}a^2$ and so $ f(x)$ is bijective

Let then $ u$ such that $ f(u)\equal{}0$
$ P(u,u)$ $ \implies$ $ f(u^2)\equal{}u$ and so $ f(f(u^2))\equal{}0$ But $ P(0,u^2)$ $ \implies$ $ f(f(u^2))\equal{}u^2\plus{}a^2$ so $ u^2\plus{}a^2\equal{}0$ and $ u\equal{}a\equal{}0$

$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}f(x)^2$ and so $ P(x,y)$ may be rewritten as $ Q(x,y)$ : $ f(x^2\plus{}f(y))\equal{}y\plus{}f(x^2)$

$ Q(x,f(y))$ $ \implies$ $ f(x^2\plus{}y)\equal{}f(y)\plus{}f(x^2)$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x\ge 0,\forall y$

It's then immediate to find $ f(\minus{}x)\equal{}\minus{}f(x)$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$
And since $ f(x^2)\equal{}f(x)^2$, we get that $ f(x)\ge 0$ $ \forall x\ge 0$

So solutions are solutions of Cauchy's equation and have a lower bound on $ \mathbb R^\plus{}$. So $ f(x)$ is continuous and is $ cx$

Plugging back in the original equation, we get $ c\equal{}1$ and the unique solution $ \boxed{f(x)\equal{}x}$ $ \forall x$

there is my solution:

$ P(x,y): f(x^{2} + f(y)) = y + (f(x))^{2}$

$ P(0,y)\Rightarrow f(f(y)) = y + (f(0))^{2}\Rightarrow$ $ f$ is bijective.

we have $ f(x^{2} + f(y)) = f(( - x)^{2} + f(y)) \Rightarrow (f(x))^{2} = (f( - x))^{2}$

since f is injective $ \Rightarrow$ $ f(x) = - f( - x)$ ($ f$ is odd).

we have $ f(0) = 0$ so $ f(f(y)) = y$ and $ f(x^{2}) = (f(x))^{2} > 0$

so $ f(\mathbb{R}) > 0$ $ \forall x\in\mathbb{R^{*}}$.

Let $ u = x^{2}$ and $ v = f(y)$ so $ f(u + v) = f(u) + f(v)$.

$ \Rightarrow$ $ f(x) = x$ $ \forall x\in{\mathbb{Q + }}$.

since $ f(x) > 0$ we have $ f(u + v) = f(u) + f(v)\ge f(u)$ $ \Rightarrow$ $ f$ is increasing in $ \mathbb{R + }$. so since $ f$ is odd we conclude finally that: $ f(x) = x$ $ \forall x\in\mathbb{R}$. :)


There are 1997 pieces of medicine.  Three bottles $A, B, C$ can contain at most 1997, 97, 19 pieces of medicine respectively. At first, all 1997 pieces are placed in bottle $A$, and the three bottles are closed. Each piece of medicine can be split into 100 part. When a bottle is opened, all pieces of medicine in that bottle lose a part each. A man wishes to consume all the medicine. However, he can only open each of the bottles at most once each day, consume one piece of medicine, move some pieces between the bottles, and close them.  At least how many parts will be lost by the time he finishes consuming all the medicine?

I just think every time we should move medicines to smaller bottles as long as we can .

you can put it in to a  3D  Grid table
then you can solve it easily.
sorry my english isn't very good.


An 8-foot piece of tin is to be cut into three pieces in the ratio of 1:2:3. How long will the middle sized piece be if the cut is made by a pair of tin snips. Give answer in feet and inches.

The middle piece is $ \frac{2}{1\plus{}2\plus{}3} \equal{} \frac{2}{6} \equal{} \frac{1}{3}$ of the entire piece.
$ \frac{1}{3}$ of 8 is $ 2 \frac{2}{3}$ feet.
Since there are 12 inches in a foot, $ \frac{2}{3}$ of a foot is 8 inches.
That means the middle sized piece will be [b]2 feet and 8 inches[/b] long.


I can't help you much but I think I know that:

Schur inequality = 舒爾不等式

I usually use it when facing $a^{3}+b^{3}+c^{3}+3abc \geq \sum_{sym}a^{2}b$ (Which is when $r=1$), it is useful that it use with Muirhead.

Bernoulli's Inequality = 伯努利不等式

I don't know Chinese names, but I think that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7942]this thread[/url] can be useful.

1   http://www.google.com
    scholar.google.com
2   http://www.mathworld.com
3   en.wikipedia.org
    http://encyclopedia.thefreedictionary.com   国内镜像站
4   http://www.answers.com

[quote="billid"]想问问下列几个不等式的中文名是什么，请各位高手指教，哪怕知道一点相关信息或知识，最好是关于在联赛中如何运用的知识：
(I want to know the Chinese names of the following inequalities. Thanks for translating, or giving any hint or knowledge you know about them, particularly welcoming things about their application in USMO)

Majorization Inequality

Muirhead's Theorem

Schur

Maclaurin's Symmetric Mean Inequality

Power Mean Inequality

Bernoulli's Inequality[/quote]

Majorization Inequality  - 控制不等式(又或者叫优超不等式, 等..)

Muirhead's Theorem  - 米尔黑德不等式(陈计的书是这样译的)

Schur  - 舒尔不等式

Maclaurin's Symmetric Mean Inequality  - 麦克劳林(对称)不等式(我猜)

Power Mean Inequality  - 幂平均值不等式

Bernoulli's Inequality  - 贝努利不等式


trigonometry
function
LaTeX
calculus
calculus computations

So I have found the equation of the tangent lines to the graph of $ f(x)\equal{}sinx$ at $ x\equal{}0$ to be $ y\equal{}x$. Additionally I have found that the equation of the tangent lines to the graph of $ f(x)\equal{}sinx$ at $ x\equal{}pi/3$ to be $ y\equal{}\frac{1}{2}x\plus{}\frac{3*sqrt(3)\minus{}pi}{6}$. Now however I am struggling to understand what “Use the tangent line at $ x\equal{}0$ to approximate $ sin(pi/6)$" and “Use the tangent line at $ x\equal{}pi/3$ to approximate $ sin(pi/6)$" means. Any ideas?

It's a local approximation.  Use the equation $ f'(x)\approx f(a)\plus{}f'(a)(x\minus{}a)$, which locally approximates a function.  So in your first example, $ x\equal{}\frac{\pi}{6}$ and $ a\equal{}0$, so we get $ \sin\left(\frac{\pi}{6}\right)\approx 0\plus{}\frac{\pi}{6}\equal{}\frac{\pi}{6}$.  This is close to the actual value of $ \frac{1}{2}$, within about .02-.03.  By adding more terms, you can improve the approximation.

I am a little bit confused. I don't see where in $ \sin\left(\frac {\pi}{6}\right)\approx 0 \plus{} \frac {\pi}{6} \equal{} \frac {\pi}{6}$ you are solving for anything and coming up with $ 1/2$.

I don't understand your last post.  I just plugged into the formula I provided.  $ f(a)\equal{}\sin(0)\equal{}0$, and $ f'(a)(x\minus{}a)\equal{}\cos(0)\left(\frac{\pi}{6}\minus{}0\right)\equal{}\frac{\pi}{6}$.

If you want to get more accurate, you can add on the term $ \minus{}\frac{f^{(3)}(a)(x\minus{}a)^{3}}{3!}$.  This term comes from the [url=http://en.wikipedia.org/wiki/Taylor_series]Taylor series expansion[/url].  If you plug in, you get that the extra term $ \approx .0239$, so the total estimate is $ .499$, which is pretty accurate.

Telling someone who doesn't understand tangent line approximations about Taylor series makes it less likely that they will understand your explanation, not more.

A tangent line is a local approximation to a graph: near the point of tangency, we expect the function values of the graph to be very close to the values of the tangent line.  Thus, you're being asked to estimate the value of a function by the value of its tangent lines.  Since you already have the equation of the tangent lines, all you have to do is plug in the appropriate value of $ x$ in order to get your estimate for the function value.  (Your textbook doubtless has lots of material on this, including exposition and solved examples.  You would probably do well to read it and your class notes thoroughly in the future when you don't understand how to do something.)

Latex suggestions: \pi, \sin, \cos, \sqrt{} all look better than without the slash (though be careful -- \sinx doesn't mean anything, it needs to be "\sin x").  Also, \cdot makes a nicer multiplication symbol than *.

Hopefully JBL cleared that up for you a bit more.  In your example, we see that for $ \theta$ small, $ \sin\theta\approx\theta$.  This is actually a really important approximation in physics since often times, it's a lot easier to solve a differential equation (say for some simple harmonic motion) in terms of $ \theta$ than $ \sin\theta$.

For example, suppose we want to find the period of a pendulum.  If $ \theta$ is small, we can just write the equation $ \ddot{\theta}\plus{}\alpha\theta\equal{}0$, where $ \alpha$ is a constant.  This gives a pretty good approximation.


calculus
integration
function
inequalities
algebra
domain
analytic geometry

I am at the section on Jacobians (12.9) and I just need a little clarification.

How do I know if an integral is a good candidate to use this Transformation of Variables?
And more importantly: How do I know what conversion will help?

It uses the whole polar/cylindrical/spherical conversion as an example but that is relatively easy to see. (Circular boundaries, circular formulae, etc.)

Unfortunately there is no general rule. One indication is that you have the same expression in the function as in the inequalities describing the domain of integration. Then you may want to take it as a new variable. But sometimes this doesn't work. Another useful idea is to try to make the domain of integration as simple as possible (preferrably, a parallelepiped) but there are exceptions to this "rule" too. Of course, you may also try the same idea as in the one variable case: if you see a complicated expression, take it as a new variable. But this may result in a very unpleasant change in the domain of integration (not to say about ugly Jacobian). Most examples in the textbooks that are about something else than polar coordinates are artificially constructed using one of three above ideas. If you have some particular integrals in mind and wonder how to guess the right substitution for them, we can discuss that but, as to a general rule that never fails, I know of none.

I had this horrible feeling that that was going to be the answer.....oy.


vector
linear algebra
linear algebra unsolved

[quote="OwlMan"]I need some help with this one, too.

Let U, V, W be subspaces of an F-space. Using the fact that dim(W1) + dim(W2)=dim(W1+W2) + dim(W1 intersect W2), show that dim[(U+V) intersect W] + dim(U intersect V) = dim[(V+W) intersect U] + dim(V intersect W).[/quote]

$\dim{((U+V) \cap W)} + \dim{(U \cap V)} = \dim{((V+W) \cap U)} + \dim{(V \cap W)}$

Consider the left-hand-side. This is nothing else then:
\[ \dim{(U+V)} + \dim{(W)} - \dim{(U+V+W)} + \dim{(U \cap V)} \\ = \dim{(U)} + \dim{(V)} - \dim{(U \cap V)} + \dim{(W)} - \dim{(U+V+W)} + \dim{(U \cap V)} \\ = \dim{(U)} + \dim{(V)} + \dim{(W)} - \dim{(U+V+W)} \]

Do the same thing for the right-hand-side and you will get the same thing.


Here's some math links with problems and information for multiple levels:
http://mathproblems.info/
http://mathworld.wolfram.com/
http://www.math.com/  :rotfl:

$ x \equal{} sqrt{25}$
$ therefore x \equal{} 5$

Actually:

$ x\equal{}\sqrt{25}$
Thus, $ x\equal{}\pm 5$

If $ x\equal{}\sqrt{25}$, isn't x just 5 (principal root)?

Yes, I was sorta assuming that we're solving a quadratic or something, so there's two possible values of $ x$...

Haha, ok...

(I don't see a problem)  :D .

this is going entirely off topic.
but here's a link:
http://www.nerdparadise.com
not bad on the mathy stuff....


topology
real analysis
real analysis unsolved

let $ X$ be a compact metric space, $ f: X\to X$ is a isometry, ie. $ \forall x, y \in X d(x,y)\equal{}d(f(x),f(y))$. Prove that $ f$ is a one-to-one and onto (bijection )

I know how to prove $ f$ is one - to -one but I am getting stuck how to prove $ f$ is onto. Any idea will be highly appreciated. Thanks

Hint:
For each $ x$ consider the sequence $ f_n (x)\equal{}f^{n} (x)$ then  pass to the convergent subsequence. :wink:

I didn't get what leshik means. Anyway, if some point $ y\notin f(X)$ (a closed set!), then so does its neighborhood (of radius $ r$, say). Now take the maximal $ r$-net $ A$ (a set where any two points are at distance at least $ r$) of the compact. It exists and it's finite due to compactness. Its image $ f(A)$ is also an $ r$-net, moreover, $ f(A)\cup \{y\}$ is an $ r$-net, contradicting the maximality.

Hi zhoraster,
what lesha has said is the following:
consider the sequence $ f_n (x) \equal{} f((..(x)..))$ ( $ n$ times ). Because of the compactness $ X$ we can pass to the convergent subsequence $ f^{n_k} (x)$. Now from the condition of the problem we have that $ d(f^{n_k} (x), f^{n_l} (x)) \equal{} d(x, f^{|n_k \minus{} n_l|} (x))$, hence we can find the subsequence $ f^{l_k}$ of $ f_n (x)$ which is convergent to $ x$. Now choose the convergent subsequence of the sequence $ f^{l_k \minus{} 1}(x)$ (let $ a$ be its limit point ) and by continuity $ f$ we get $ f(a) \equal{} x$. That's all. :)

[quote="leshik"], hence we can find the subsequence $ f^{l_k}$ of $ f_n (x)$ which is convergent to $ x$. [/quote]

Why does subsequence converge to x?  :blush:

Because the sequence $ f^{n_k}$ is fundamental, thus $ d(f^{n_k}, f^{n_l})\to 0$ when $ n_k, n_l \to\infty$, is everything clear now?

[quote="leshik"]Hi zhoraster,
what lesha has saisd is the following:[/quote]

Oh yes, I had solved it some 10 years ago similarly :) Now too old to invent such beauties.


summer program
Mathcamp
Ross Mathematics Program
function
geometry
videos
PROMYS

Hello

I'm kind of new to the American concept of math camps (they're awesome though  :D ) and I wanted to ask a dew questions:

1. I am also currently applying to Ross, so if i get accepted into both (assuming that i ever get my application for Ross typed up, I have the proofs^^), what do I do? Do I just tell one of the: Hey I'm sorry, but I'd rather go to your competitor?

2. For the application, when it says deadline 4/25/08 what is it referring to? when it should be by them or when we send them off? (the later system is used in Germany)

3. Which one is more fun/interesting?

4. How much money should I bring for either (preferably an amount per week, since one is 8 and the other 6 weeks)

5. What amount of correct solutions gets one in? (they don't seem super hard, but i didn't look at them very intently yet... )

6. if you do get in, when do they ask you for schedule-choices?

If there are any mistakes in grammar/spelling, please tell me  :blush:

1.  Yes, this is fine.  You can do the same thing to colleges, too.

2.  Unless otherwise specified, you should interpret it as being the deadline for receiving the application.  The first case is referred to as "postmark deadline."

3.  That depends on what you think is interesting.  Clearly different things work for different people, although I've heard a lot of good things about MathCamp and Ross is quite difficult.  (The fact that I've gone to neither is your assurance that I'm being vaguely objective.)

4.  As far as I know, the cost of both programs cover room and board, so you only need spending money for outings and snacks.  Bring whatever you feel is appropriate, I guess.  (And some extra money for things you forgot to bring with you - toothpaste, etc.)

5.  Acceptance is (probably) not directly a function of how many problems you solve correctly.  I imagine your style is also important, as well as the rest of the application and how many spaces are available.  So it's hard to say.

Hello,

1. Yup, you can apply, get in, and then decide to do something else with your summer.  (Otherwise it's not really fair: you apply and don't get in and then you have nothing to do!)

2. We'd like to have applications *received* by then.  If you think you need a little bit more time, drop us a note.  You can also apply online, in which case you don't have to wait for postal mail.

3. Well, you're posting in the Mathcamp forum, so you'll get a biased answer!  You can find out more about Mathcamp from the Math Jam transcripts, which are linked to from another post in this forum.

4. It depends --- if you get in, we'll send a suggestion.

5. As far as correct solutions, that also depends on a lot of different factors.  Certainly the quiz is a major part of your admission, but other factors also come into it.

6. You decide at camp --- indeed, you can choose on the day-of and change your classes part-way through the week if you decide others are a better fit for you.

Good luck!

[quote="Smartguy"]I also have a couple questions:

1. We have to pay for the flight, correct?

2. When we are doing the quiz, can we go out of order?[/quote]

You do have to pay for the flight, although if your financial situation is quite dire we have included transportation in our financial aid in the past.

As to the quiz, you can work on the problems in any order that you like; it's a little bit easier for us to grade if you write them up in order.

Thanks for the responses...

I think i am going to apply online, simply because i have no idea how long the US postal system would take to get my letter up there ^^

one more question:
what about transportation from the (Seattle, right?) airport to the university?
Is it provided and if not, how did past students get to the facilities?

Mathcamp staff will meet you at the airport and take you to campus.

[quote="e=mc2.kroll"]I have (another) question about the quiz, in particular about number 2.
If we have proven the general rule, must we prove the specific case, since it is asked before the general rule?[/quote]

Nope, you can do the general case first and then go to the specific.

[quote="Zeph"]I'm kinda a huge procrastinator, and I was wondering whether the deadline of the 25th includes Saturday itself; like, can I submit my application online on Saturday or will it be considered late? <_<[/quote]

Submitting on Saturday is definitely fine.  Good luck!

[quote="Cyrazeno"]I went to Ross last year. Of course, as you probably already know, the camp's main focus is on number theory. Each day you are given a problem set with a few terminology problems, numerical problems, and proofs. The majority of the people didn't finish a set per day though and got pretty behind, but that's understandable. At the beginning, you are placed into a group with about two other first years, a junior counselor, and your counselor. Throughout the camp, your counselors give hints to help you with problems on your sets. Each morning there is a one hour lecture and Mondays, Wednesdays, and Fridays ( I believe) you have the problem session which lasts an hour, as well.

Most everyone hung out in the lounge area till very late (3AM- 5AM sometimes all-nighters) usually doing problem sets. After the lectures and problem sessions you are basically free to do whatever you want. You can go to Market Place which is where most people ate (excellent restaurant food) and since you'll have a food card you don't have to pay. The counselors don't want you going off campus alone but... people still did occasionally, haha.

Ah, and yes... three times throughout the camp you are given a test with 20 problems total (10 numerical, 10 proofs) that you have to complete by the next morning and turn in to the professor. Most everyone pulls an all nighter on celebration nights... It isn't necessary to complete them, but... if you care about your reputation at camp, you should at least try.

Unfortunately, bringing laptops is highly discouraged and there is no wireless internet in the dorms. You can walk to the library (not far at all) to use the internet.  I brought about 60 dollars or so and was fine but then again... my parents had bought a lot of things for my dorm already. And there are the movie nights... where you watch videos having something to do with math.

For my Ross admission test, I believe I solved all of them but they weren't necessarily written *that* well. You don't have a schedule choice for Ross.

At Ross you have a tremendous amount of freedom, seriously. About 40-50 students are there... you would probably like it very much. :) Also, if you are looking to improve your proof skills, I highly recommend Ross.[/quote]

advertising for Ross?
Ross and PROMYS are pretty similar right? even in terms of their schedule?

I'm simply stating my opinion of the camp based on experience. Also, there was no need to quote my entire post since it's pretty obvious that you're talking to me. Anyways, yes, PROMYS and Ross are very similar in terms of the curriculum offered.

[quote="Cyrazeno"]
Unfortunately, bringing laptops is highly discouraged and there is no wireless internet in the dorms.

You don't have a schedule choice for Ross.

At Ross you have a tremendous amount of freedom, seriously.
[/quote]

Methinketh my definition of freedom and yours might differ? :)

Well, besides that. You are still able to walk (less than half a mile or so) to the library for internet. :) I guess that was the worst part about camp, though. Haha.

[quote="Cyrazeno"]Most everyone hung out in the lounge area till very late (3AM- 5AM sometimes all-nighters) usually doing problem sets.[/quote]

Wow... last summer at MC, my average bedtime was about 3AM. But, usually, although I worked on math a lot during the day, at night I was just talking/playing games/watching videos that late. The one night I pulled an all-nighter (besides the all-nighter on the last day of camp which basically everyone does) it was because we started a game of RISK at about 4AM. Not recommended if one wants to go to sleep soon :P . Anyway, Ross certainly sounds more serious...

Ross does sound like an awesome program, and I'm sure one knows number theory inside and out after eight (I think?) weeks. But I really love being able to choose my schedule (at least in part a result of being homeschooled), among other things, so I prefer MC freedom.

i am also apllying to ross (have all the proofs typed up since 2 moths ago, just need to get the app out... mathcamps online app is sooo much easier to complete quickly).
i was thinking of kindof alternating in between them, since one seems very rigorous and the other seems to be very diverse)
so it would be like
9-mathcamp (just wanna find out what interests me a lot)
10-ross
11-mathcamp (junior-senior summer is very important as far as i understood, thus the shorter one)
12-ross

but we'll see...
also the fact that mathcamp has standing invitation is really nice :-)

The question of which camp is more "serious" is an interesting one.  I think it's true that the average Ross student spends more time working on problems than the average Mathcamp student (although Mathcampers spend more time in classes, for example).  I don't think Ross gives more work, it's just that the variety of choices at camp tends to mean that Mathcamp students also spend time on things outside of math --- believe me, there are [i]plenty[/i] of problems to do at MC!  One thing I've enjoyed as a teacher at Mathcamp is talking to campers after the summer, who are working on doing problems leftover from their time during the summer.  I think you have to judge for yourself how important Ross' culture of working problems all night is important to you.  Will you do enough math to satisfy yourself if there is less focus on constantly doing math?  That's something you have to judge for yourself.

Alternating programs is an interesting idea and I know people who've switched between the two.  Do be wary of of strange contrasts, though.  For example, if you take a lot of number theory at Mathcamp and then go to Ross, you may find it hard to find the right niche for yourself since you'll have seen a lot of the material but not all of it.

The freedom at Mathcamp definitely extends to the amount of time you spend on problems - it truly depends on what classes you decide to take and how serious you are about them.  At Mathcamp 2008, I spent half of camp incredibly overworked, and half of camp focusing more on non-academic activities than on problem sets - a disparity that would not exist at Ross.  I personally like the ability to choose how much math I want to do any particular week.  If you want a more clearly-defined workload, you might prefer Ross.

I would tend to think that Ross before Mathcamp would be better than Mathcamp before Ross.  The set curriculum at Ross may be better than Mathcamp at providing a solid introduction to theoretical mathematics.  But I don't think anyone can beat Mathcamp for those who are ready for advanced topics.


Let $PT$ and $PB$ be two tangents to a circle $\omega$, $AB$ the diameter through $B$, and $TH$ the perpendicular from $T$ to $AB$. Then prove that $AP$ bisects $TH$.

Let $R = OA = OB = OT$ and $\alpha = \measuredangle BPO = \measuredangle OPT$ so that $\measuredangle TOP = \measuredangle POB = 90^\circ-\alpha$. Thus, $\measuredangle HOT = 2 \alpha$.
Let $W$ be the midpoint of $TH$.
We want to prove that $2 \cdot WH = TH$:
$ \, \Longleftrightarrow \, \, 2 \cdot \frac{AH}{AB}\cdot PB = TH$
$ \, \Longleftrightarrow \, \, 2 \cdot \frac{R-HO}{2R}\cdot PB = TH$
$ \, \Longleftrightarrow \, \, 2 \cdot \frac{R-R \cos \left( 2 \alpha \right)}{2R}\cdot R \cot \alpha = R \sin \left( 2 \alpha \right)$
$ \, \Longleftrightarrow \, \, \left( 1-\cos \left( 2 \alpha \right) \right) \cdot \cot \alpha = \sin \left( 2 \alpha \right)$
$ \, \Longleftrightarrow \, \, 2 \sin^{2}\alpha \cdot \cot \alpha = 2 \sin \alpha \cos \alpha$
$ \, \Longleftrightarrow \, \, \sin \alpha \cdot \frac{\cos \alpha}{\sin \alpha}= \cos \alpha$,
which is obviously true.

First:
$X$ is the intersection of $AP$ and $HT$
$Y$ the intersection of $AT$ and $BP$
$O$ is the centre of the circle

By angle chase: $\measuredangle POB=\measuredangle YAB$ $\Rightarrow$ $OP//AY$

$AO=OB$, then $BP=PY$

$AHT$ and $ABY$ are similars so: $\frac{HX}{XT}=\frac{BP}{PT}$ $\Rightarrow$ $\boxed{HX=XT}$  $\mathbb{QED}$


hi,

compute $\sum_{p=1}^{n}\arctan(\frac{1}{p^{2}+3p+3})$
 :blush: .

thanks in advance.

Use the hint  :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=111391[/url]

i haven't seen your hint Ramanujan yet but i have found the

solution  :lol: .

$\frac{1}{p^{2}+3p+3}= \frac{(p+2)-(p+1)}{1-(p+2)\times(-p-1)}$

 :P .


$ a,b,c>0$,$ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}1$  $ \implies$  $ ab\plus{}bc\plus{}ac\geq\frac{3}{4}$

[quote="cnyd"]$ a,b,c > 0$,$ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 1$  $ \implies$  $ ab \plus{} bc \plus{} ac\geq\frac {3}{4}$[/quote]
$ x\equal{}a\plus{}b;y\equal{}b\plus{}c;z\equal{}c\plus{}a$ . We got $ xyz\equal{}1$ and want to prove that

$ ab \plus{} bc \plus{} ac\equal{}\sum{\frac{1}{4}(x\plus{}y\minus{}z)(x\plus{}z\minus{}y)}\geq\frac {3}{4}$

Change it to

$ 2\sum{xy}\geq3\plus{}\sum{x^2}$

It's wrong!!! See [url=http://www.mathlinks.ro/viewtopic.php?t=50745]here[/url]

but in the  this link  $ abc \equal{} 1$

See example 2 here:
[url]http://www.math.ust.hk/excalibur/v14_n1.pdf[/url]
The sign is reversed, you don't need abc=1.

[quote="cnyd"]$ a,b,c > 0$,$ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 1$  $ \implies$  $ ab \plus{} bc \plus{} ac\geq\frac {3}{4}$[/quote]
Yes, it's wrong:  $ a \equal{} b\rightarrow\frac {1}{\sqrt [3]2}$ and $ c\rightarrow0.$

I think $ ab \plus{} bc \plus{} ac\leq \frac {3}{4}$

$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac) \equal{} 1 \plus{} abc \Longrightarrow$

$ \sqrt {3}{(ab \plus{} bc \plus{} ac)}^{\frac {3}{2}} \leq (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac) \equal{} 1 \plus{} abc \leq 1 \plus{} (\frac {ab \plus{} bc \plus{} ac}{3})^{\frac {3}{2}} \Longrightarrow$

$ ab \plus{} bc \plus{} ac\leq \frac {3}{4}$ :wink:


inequalities
integration
function
calculus
calculus computations

Suppose that $ f(x)$ be continuous ,positive function and $ f(x\plus{}1)\equal{}f(x)$ for all $ x \in R$.Let$ \alpha \in R$
Prove inequality :
       $ \int_{0}^{1}{\frac{f(x)}{f(x\plus{}\alpha)}dx}\geq 1$

we can immediately see that

$ f\left(0 \right)\equal{}f\left(1 \right)\equal{}f\left(2 \right)\equal{}f\left(3 \right)\equal{}.........$

hence  f(x) cann  be  a constant function in which case, equality occurs  in the given inequality


Let $ m$ and $ n$ be two strictly positive integers. Prove that $ \frac{GCD(m,n)}{n}\binom{n}{m}$ is an integer.

Try this.  :wink:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=98174[/url]


Are given 2 boxes with the same gas . Box 1 with volume $ 3m^{3}$ and pressure 1 atmosphere , box 2 with volume $ 2m^{3}$ and pressure 2 atmosphere . If we join the boxes with a  tube , what is the pressure in the boxes ?

What about the temperatures?

Assuming them to be same for both boxes.

$ P_1V_1 \equal{} n_1RT .... \text{ (for box 1)} \\ \\
P_2V_2 \equal{} n_2RT ....\text{ (for box 2)} \\ \\
P_f(V_1\plus{}V_2) \equal{} (n_1\plus{}n_2)RT .... \text{ (after opening the stop cock)} \\ \\
P_f \equal{}\frac{7}{5} \equal{} 1.4\text{ atmosphere}$

how can you assume the temperatures to be equal.it is not specified

[quote="aadil"]how can you assume the temperatures to be equal.[/quote]

I took it for granted that temperatures are equal. There is no way you can find the temperatures and IMO the answer would contain terms $ T_1$ and $ T_2$.

[quote="Rajiv"]
I took it for granted that temperatures are equal. There is no way you can find the temperatures and IMO the answer would contain terms $ T_1$ and $ T_2$.[/quote]

Here's how:

$ P_1V_1 \equal{} n_1RT_1 .... \text{ (for box 1)} \\
\\
P_2V_2 \equal{} n_2RT_2 ....\text{ (for box 2)} \\
\\
P_f(V_1\plus{}V_2) \equal{} (n_1\plus{}n_2)RT_f .... \text{ (after opening the stop cock)} \\
\\ \\
(n_1\plus{}n_2)R \equal{} \frac{3}{T_1}\plus{}\frac{2}{T_2} \\ \\ \\
5P_f \equal{} \left(\frac{3}{T_1}\plus{}\frac{2}{T_2}\right)T_f$

[quote="Rajiv"]Why are there no comments by anyone else? :mad:[/quote]

Because you've analyzed the problem correctly.

Combining Boyle's law and Avogadro's law:

\[ \frac{P_1V_1}{n_1} \equal{} k \equal{}\frac{P_2V_2}{n_2} \\ \\ P_f(V_1 \plus{} V_2) \equal{} n_1k \plus{} n_2k \\ \\ \implies P_f(V_1 \plus{} V_2) \equal{} P_1V_1 \plus{} P_2V_2 \\ \\ \therefore P_f \equal{}\frac{P_1V_1 \plus{} P_2V_2}{V_1 \plus{} V_2} \equal{}\frac{(1\mbox{ atm })(3\mbox{ m}^3) \plus{} (2\mbox{ atm })(2\mbox{ m}^3)}{3\mbox{ m}^3 \plus{} 2\mbox{ m}^3} \\ \\ \equal{} \boxed{\frac{7}{5}\mbox{ atm }}\]

[quote="Dr. No"]$ \boxed{\frac {7}{5}\mbox{ atm }}$
[/quote]

Thank you! :lol:


The planes containing the faces of a cube divide the space into several regions. How many regions?

[hide] it would be 6*2 + 8 + 1 (if you include the cube) which is 21, i think[/hide]

[hide]27. If we draw the 1 of the planes, there are 2 regions. Drawing the plane parallel to that plane we get 3 regions. Drawing the plane that intersects those 2 planes we get 6 regions, and then drawing the plane parallel to that we get 9 regions. Then drawing the final 2 planes that intersect it, we add 18 more, which makes 27.[/hide]

[hide] Yeah, I got 27 too. The planes are just like the dividing planes in a 3x3x3 cube which divide it into unit cubes. And of course 3^3=27.[/hide]

[hide]Basically, you have 2 planes in each dimension, which divides it into 3, making 3^3=27.[/hide]

gosh, Im soo stupid! I forgot about the regions containing the edges :blush:  :blush:  :blush:

[hide]If you guys have read the other problems I've posted, I've said that these have been from the USME (Utah State Math Exam).  It was multiple choice, and the first answer I got was 24.  Luckily, when I checked it, I got 27.[/hide]


function
abstract algebra
integration
trigonometry
limit
calculus
complex numbers

Suppose that $a_n$ is a decreasing sequence that tends towards 0 and that $f(x)=\sum_{n\geq 1}{ a_n sin(nx)}$ is continuous. Does it follow that $a_n=o(\frac{1}{n})$?

I don't have time today to write the whole proof, but the answer will be yes. A sufficient condition for the Fourier coefficients of $f$ to be $o\left(\frac1n\right)$ is that $f$ be absolutely continuous. So we concentrate on showing that $|f'(x)|$ is integrable.

Of course this would be false if you took out the word "decreasing."

This would require showing that the series is the Fourier series of its sum, which I don't know how to do. Let me ask a simpler question: is it true that under the assumptions of the problem, there exists $\gamma>0$ such that $\liminf_{n\to\infty} n^\gamma a_n=0$?. If somebody can show that, I can do the rest. Of course, if the series is the Fourier series of a continuous function, we have the Parceval identity, which immediately implies $a_n=o(n^{-1/2})$ but, as I said, I do not know how to show that it is a Fourier series  :( .

My original answer was too hasty, as shown by fedja's objections. It may still be a possible direction to look in, but there are obstacles.

Seems I finally found a neat proof. Recall that the Fejer kernel is $K_n(x)=\sum_{k=-n}^n\left(1-\frac{|k|}{n}\right)e^{ikx}$. It is a standard exercise written in many textbooks on Fourier series to show that $K_n\ge 0$, $\int_{-\pi}^\pi K_n(x)\,dx =2\pi$ and, for every $\delta>0$, we have $\int_{(-\pi,\pi)\setminus(-\delta,\delta)}K_n(x)\,dx\to 0$ as $n\to\infty$. Now let $G_n(x)=\Im (e^{inx}K_n(x))=\sum_{k=-n}^n\left(1-\frac{|k|}{n}\right)\sin(n+k)x$. Note that $G_n(x)=O(|x|)$ as $x\to 0$. Since the partial sums $f_m(x)=\sum_{k=1}^m a_k \sin kx$ of our series are uniformly  bounded by $\frac C{|x|}$ on $(-\pi,\pi)$, we can use the dominated convergence theorem to conclude that, for every fixed $n$, $\int_{-\pi}^\pi G_n(x)f(x)\,dx=\lim_{m\to\infty}\int_{-\pi}^\pi G_n(x)f_m(x)=\pi\sum_{k=-n}^n \left(1-\frac{|k|}{n}\right)a_{n+k}\ge c na_n$ with some absolute constant $c>0$. On the other hand, it follows from the above properties of $K_n$ and from continuity of $f$ at $0$ that $\int_{-\pi}^\pi G_n(x)f(x)\,dx\to 0$ as $n\to\infty$. Thus, we, indeed, must have $a_n=o(1/n)$.

$K_n$ is real-valued, so the way you defined $G_n$ you should have $G_n(x)=\sin(nx)K_n(x)$.
if that's the case, it's not necessarily true that $\int G_n(x)f(x)\to 0$.

[quote="cimabue"]$K_n$ is real-valued, so the way you defined $G_n$ you should have $G_n(x)=\sin(nx)K_n(x)$.[/quote]
That's true; I wrote the imaginary part of the product just to make the representation of $G_n$ as a sum of sines obvious.

[quote="cimabue"] if that's the case, it's not necessarily true that $\int G_n(x)f(x)\to 0$.[/quote]
Why? :? The integral over $(-\pi,\pi)\setminus(\delta,\delta)$ tends to $0$ because $f$ is bounded and the integral of $|G_n|\le K_n$ tends to $0$, while the integral over $(\-\delta,\delta)$ is arbitrarily small because $f$ is small on $(-\delta,\delta)$ (it is continuous at $0$ and $f(0)=0$) and the integral of $|G_n|$ over $(-\delta,\delta)$ does not exceed $2\pi$.

sorry, my bad. i failed to notice that $f(0)=0$.


geometry
incenter
number theory
greatest common divisor
cyclic quadrilateral
geometry proposed

Suppose that $ ABCD$ is a cyclic quadrilateral and $ CD\equal{}DA$. Points $ E$ and $ F$ belong to the segments $ AB$ and $ BC$ respectively, and $ \angle ADC\equal{}2\angle EDF$. Segments $ DK$ and $ DM$ are height and median of triangle $ DEF$, respectively. $ L$ is the point symmetric to $ K$ with respect to $ M$. Prove that the lines $ DM$ and $ BL$ are parallel.

$ BD$ is the bisector of angle $ ABC$ and $ \angle EDF\equal{}\frac{1}{2}\angle ADC\equal{}90^o\minus{}\frac{1}{2}\angle EBF$ therefore $ D$ is the B-excenter of triangle $ EBF$.
Thus $ L$ is the tangency of incenter $ (I)$ of triangle $ EBF$ and $ EF$. Let $ J\equal{}BD\cap EF$
$ \frac{r_b}{r}\equal{}\frac{DK}{IL}\equal{}\frac{KJ}{LJ}\equal{}\frac{ML}{LJ}\plus{}\frac{MJ}{LJ}\equal{}1\plus{}2\frac{MJ}{LJ}$
$ \frac{r_b}{r}\equal{}\frac{DJ}{IJ}\equal{}\frac{DB}{IB}$ then it easy to show that $ \frac{MJ}{LJ}\equal{}\frac{DJ}{BJ}$
$ \Rightarrow BL//DM$

Let $ G \in AC$. $ CG=EA \Rightarrow \triangle EAD = \triangle GCD$, $ BEDG$ is cyclic, $ DF$ is a bisector of $ \angle EDF$.
$ DF \cap (BEDG)=M$,
$ BM \cap DE=N$,
$ ME \cap BD=W$,
$ FE \cap WD=S$,
$ I$-midpoint of $ WD$.
$ DM$ is a diameter of $ (BEDG) \Rightarrow ME$ and $ DB$ are altitudes of $ \triangle MND$ and $ BWEN$ is cyclic $ \Rightarrow \angle NDM=\angle NBE=\angle NWE$.
$ \angle EWD=\angle DFG=\angle DFE \Rightarrow FWFD$ is cyclic $ \Rightarrow \angle EWF=180^0 - \angle NDM$.
Then $ \angle NWE+\angle EWF=180^0$ and $ N, W, F$ lie on the same line and $ NF$ is an altitude.

We shall prove:
For an arbitrary quadriateral $ XYZT$ such that $ \angle Y=\angle T=90^0$, if $ X'$ and $ Z'$ are the projections of $ X$ and $ Z$ on $ YT$, then $ YX'=TZ'$.
$ \triangle X'XY \sim \triangle Z'YZ$
$ \triangle XX'T \sim \triangle TZ'Z$
Then,
$ \frac {XX'}{YX'}= \frac {YZ'}{ZZ'}$ and $ \frac {XX'}{TX'}= \frac {TZ'}{ZZ'} \Rightarrow \frac{TX'}{YX'}= \frac{YZ'}{TZ'} \Rightarrow YX'=TZ'$

Then $ WL \perp FE$ and $ \triangle SWL \sim \triangle SIM \Rightarrow \frac{SW}{SI}=\frac{SL}{SM}$.
$ \angle FIE=2 \angle FDE$
$ \angle FBE=\angle FBW+\angle EBW=\angle EMD+\angle FND=180^0-2\angle FDE$.
Then, $ BEIF$ is cyclic. We know that $ FWED$ is cyclic, so
$ BS \cdot SI=FS \cdot SE=WS\cdot SD \Rightarrow \frac{SW}{SI}=\frac{BS}{SD} \Rightarrow\frac{SL}{SM}=\frac{BS}{SD}\Rightarrow \triangle SBL\sim\triangle SDM \Rightarrow BL \parallel MD$.

[quote="stergiu"][quote="livetolove212"]................................

 then it easy to show that $ \frac {MJ}{LJ} \equal{} \frac {DJ}{BJ}$
$ \Rightarrow BL//DM$[/quote]

 Can someone explain how we arrive to the first relation ?

 Babis[/quote]
$ 2\frac{MJ}{LJ}\plus{}1\equal{}\frac{DB}{IB}\equal{}\frac{EB\plus{}BF\plus{}EF}{EB\plus{}BF\minus{}EF}\equal{}\frac{2EF}{EB\plus{}BF\minus{}EF}\plus{}1$
$ \equal{}2\frac{1}{\frac{EB\plus{}BF\minus{}EF}{EF}}\plus{}1\equal{}2\frac{1}{\frac{EB\plus{}BF}{EF}\minus{}1}\plus{}1$
$ \equal{}2\frac{1}{\frac{BF}{\frac{BF.EF}{EB\plus{}BF}}\minus{}1}\plus{}1\equal{}2\frac{1}{\frac{BF}{FJ}\minus{}1}\plus{}1$
$ \equal{}2\frac{1}{\frac{BD}{DJ}\minus{}1}\plus{}1\equal{}2\frac{1}{\frac{BJ}{DJ}}\plus{}1\equal{}2\frac{DJ}{BJ}\plus{}1$

it's easy to prove that $D$ is the excenter of triangle $BEF$
hence the result is trivialized by ex-Boone theorem.


calculus derivative function integration

I'm going to be doing a science fair project this summer. (I just finished 8th grade.) In a different thread, somebody suggested learning some calculus and modeling the cooling of an object. I think you would use Newton's Law of Cooling for this, but I'm not sure. I learned how to find the derivative of a function at a given point, and how to differentiate a function. I think I also need to learn how to integrate a function, in order to solve a 1st order differentiation equation or something like that. What else do I need to know? Any suggestions? :)

[quote="DPopov"]Hmm....that's actually about all you need to know. Integration of a differential equation is all Newton's Law of Cooling is.[/quote]

Really? :) Okay. But it's pretty confusing :wacko:  :what?: Are you supposed to use definite or indefinite integral for differential equations?

I think all you really have to worry about is  how to use a graph (There are certain advantages of being a physics student -  You just draw a graph   and  if you tried to get a  graph paper which has 'logs'  - the  diff eq part will take care of itself)

Basically Newton's law   simply is.  that rate of cooling  varies directly as the difference of the temperatures between the object and the  environment.  (Unless the environment happens to be your hand, in that case,  it may seem that the object is hotter than it should be)

So  just small ammount of math  will get you  that the temperature is  something like  $e^{-kt}$ where t is time and k is some constant which you can calculate by looking at the graph...

have fun, make sure that you don't [b]touch[/b] the hot object :)

Okay so I finished the project. What should I call it? "Calculus and Pies"?? :lol: (I made a pie for the project. Surprisingly, there wasn't much calculus. If you want to see my work, PM me please.) Thanks. :)


Let ABC is a  triangle,prove that:
$ \frac{r_a}{r_b} \plus{} \frac{r_b}{r_a}\geq \frac{A}{B} \plus{} \frac{B}{A}\geq \frac{a}{b} \plus{} \frac{b}{a}.$
BQ

[quote="xzlbq"]Let ABC is a  triangle,prove that:
$ \frac {r_a}{r_b} \plus{} \frac {r_b}{r_a}\geq \frac {A}{B} \plus{} \frac {B}{A}\geq \frac {a}{b} \plus{} \frac {b}{a}.$
BQ[/quote]
$ r_a, r_b, r_c$,         radii of excircles, respectively.


What method of factoring quadratic equations do you use?

Say I have $7x^2+9x+2$ and want it factored. What method of factoring would you use?

Theres this method where you use $2$ factors of $ac$, $F_1$ and $F_2$, which $F_1 + F_2 = b$. You use the $2$ factors to replace b and use factor by grouping to factor the rest of the equation. Why is the initial multiplication of $a$ and $c$ required?

To factor your example, we get $(7x+2)(x+1)$. I don't really use any method for factoring; I just look at it and see what it can be split into. Your example was fairly simple to factor, because there are very few ways to consider.

I think your method can only be used when 2 of the coefficients are 1, like in the example. In this case the 2 factors would be 7 and 2, which add up to 9. I believe the intitial multiplication of ac in your method is required because it is equal to the product of the coefficients of the factored form. For example, $8x^2+26x+15=(2x+5)(4x+3)$. $ac=120$, which equals $2\cdot5\cdot4\cdot3$.

My approach to factoring is that if the polynomial is easy to factor, then I factor it. If it's not, then I use the quadratic formula. If factoring is required, then that should be simple after you find the x values.

Hope this helps. :)

Thanks for the advice.

I did some thinking on why the method taught to me works, and I think it is because by multiplying finding factors of ac to replace b, you have 2 numbers which are either multiples or factors of a and c. Therefore it is very simple to factor it out by grouping.

The coefficients in factored form must equal ac because the 2 leading coefficients of each binomial multiply together to form a and the trailing coefficients' product is c.


Ok I really don't get this!  [geogebra]168649eb1051012ef3a4d0ab8c96b071624aac58[/geogebra]
I am supposed to find the area of the leave like things.  I tried separating into quarter squares and I found out that the area of half of the leavy things is pi-1. There are 8 halves so I multiplied by 8 and got 8pi-8. The answer is 8pi-16. Where did I go wrong and what is an easier way to do this? BTW the side length of the square is 4,[/img]

Though I can't see the picture, maybe you can try quarter circle - area of triangle?

We see that the angle between the intersection points of adjacent circles and the midpoint of a circle is 90 degrees. We can find the area of one of the leaves by finding the area of two quarters, or one half, of a circle and subtracting 4. Here is a picture that might clarify some things:
[geogebra]fd1f18af9d390f18b661c3df304aa9277940ec28[/geogebra]
The side length of the square is 2. To find the area, as I said, of the leaf we find the area of the square plus the leaf (the area of two quarters/half of a circle) and subtract the square so we have the leaf left over. The area of half the circle is $ 2\pi$ and the area of the square is 4. The area of one leaf is then $ 2\pi\minus{}4$. Multiply by four to find the area of all four:

$ 4(2\pi\minus{}4)\equal{}\boxed{8\pi\minus{}16}$

I hope you understand!

The area of one of the triangles is 2 not 1.
Thus, the area of half the leaves is pi-2.
Multiplying this by 8 we get 8pi-16.

EDIT beaten

Ok ytrewq I understand that mistake I made, but Power Of Pi, I quite don't get yet. I don't understand the part of adding the square and the leaf. I do get subtracting the square, but I don't understand how you get 2pi for adding the square and the leaf. Thanks!

I'll say it this way:

When you find the area of half the circle, you are counting the leaf twice (once with each quarter of the circle). You can find from that $ square\plus{}leaf\equal{}2(\frac{1}{4}circle)$.


See in the book:

  <<  Μπάμπης Στεργiου και Νίκος Σκομπρής  : Αλγεβρικές Ανισότητες , Εκδόσεις Σαββάλας 2004 ,  Problem 4.48 >>

 There ,  you will find a solution , based on the fact that this inequality is symmetric.

   [u]Babis[/u]

[quote="tµtµ"]Rewrite it $\sum a(a-b)(a-c)$ : it is Schur[/quote]

OK, but this is $a,b,c \in \mathbb{R}^+$

Here is $a,b,c \in \mathbb{R}$

I thought it was a typo because it is false:

take $b=c=0$, we would have $a^3 \geq 0$, $\forall x \in \mathbb{R}$, for instance

This inequality is true for positive numbers

OK, then let's prove Schur:
$a,b,c,r \in R^+$, prove:
\[ \sum a^r(a-b)(a-c)\geq 0 \]

Easy. :)

Since, the ineq. is symmetrical, we can assume, WLOG, that : $a \geq b \geq c \geq 0$.
Thus, the inequality is true since : $a^r(a-b)(a-c) \geq b^r(a-b)(b-c) \text{ and } c^r(c-a)(c-b) \geq 0$.

[quote="varun"]what is schur? :?:[/quote]
View the posts above yours.

mathmanman: Thanks for the proof.

what mathmanman meant to say?
was it that all the three terms are $\ge 0$

Then its OK.
And thanks for thenew inequality.

In my book the following inequality holds: \[ a^{t}(a-b)(a-c)+b^{t}(b-c)(b-a)+c^{t}(c-a)(c-b)\geq 0, \]
 where $a,b, c>0$ and $r\in\mathbb{R}.$ ;)

Well yes...that is Schur and its proven in the posts before

Look carefully: $r\in\mathbb{R}.$ I have seen Schur only for $r>0.$ :)

it also can be proven like this:
like it has been said a b c are positive then divide all by abc, and WLOG $a \geq b \geq c$ then:

$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}+3 \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$

and take $(\frac{a}{b},\frac{b}{c},1,1,1,\frac{c}{a})$ and $(\frac{a}{c},1,1,1,\frac{c}{b},\frac{b}{a})$

So by some kind of arrangement:

$\frac{a}{b}\frac{a}{c}+\frac{b}{c}\frac{b}{a}+\frac{c}{a}\frac{c}{b}+3 \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$
like we wanted to see

Cesar could you post the solution with $r\in R$. Thank you very much

Well, that's still easy :)
If $r \geq 0$, read my proof above.
If $r < 0$, then we can assume WLOG that : $0 < a \leq b \leq c$.
And, the rest is exactly the same thing as in my previous post.

[quote=xitim]$a,b,c \in \mathbb{R}$

Prove that $a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ac(a+c)$[/quote]
 [url=https://artofproblemsolving.com/community/c893771h1869509p12676711]All Soviet Union MO 1975[/url]


geometry
calculus
integration
function
symmetry
real analysis
Euler

Suppose that $ S$ is a 2-dimensional surface in $ \mathbb R^3$ whose boundary is the red brocken line shown below:
[asy]size(150);
D((-1.5,-1.5)--MP("z",(1.5,1.5),NE),black);
D(MP("y",(0,7),N)--(0,0)--MP("x",(7,0),E),black);
D(MP("1",(1,1),W)--(1,7)--(-1,5)--MP("-1",(-1,-1),W)--(4,-1)--(6,1)--cycle ,red+linewidth(1));
MP("b",(0,6),W);
MP("a",(5,0),S);[/asy]
Show that the total area of $ S$ is at least $ 2(a\plus{}b\minus{}2)$.

I'd like to present this as an application of differential forms to estimating areas of minimal surfaces but I'm hesitating whether just to make a lecture about it or to give it to the students as an exercise with many hints (a la ENS exams). So, I decided to post it here to see if AoPSers can solve it easily. The suggested idea is to find a 2-form $ \omega$ such that $ \|\omega\|\le 1$ at all points inside the convex hull of the red contour (you may assume without loss of generality that $ S$ is contained in that convex hull), $ d\omega\equal{}0$, and the integral of $ \omega$ over $ S$ (which does not depend on $ S$ because of the second condition) is at least $ 2(a\plus{}b\minus{}2)$.

if  I understand everything well,  then your answer is strange (incorrect). Let's set $ b \to 0$ then answer must tend to $ 2a$.

It is just a simple lower bound, not the sharp estimate! Of course, it is of interest only when both $ a$ and $ b$ are large. I don't really know if the exact value of the minimum in this problem can be expressed in any reasonable terms.

[quote="fedja"]I don't really know if the exact value of the minimum in this problem can be expressed in any reasonable terms.[/quote]
I agree with you.
This is what I'm thinking for sharp estimate.
this surface can be expressed as $ S \equal{} (x,u(x,z),z)$  where $ x*z \in [0,a]*[ \minus{} 1,1] \equal{} K$
I can assume that $ u(x,z)$ is smooth function of two variables. Since, if it's not smooth , then this can be approximated sufficiently well with smooth function.
One can note by symmetry that $ u(x,z) \equal{} u(x, \minus{} z)$ and $ u'_{x}(x,z) < 0$. Also note that $ u(0,z) \equal{} b$ and $ u(a,z ) \equal{} 0$ and
$ u(x,1) \equal{} u(x, \minus{} 1) \equal{} 0$
I assume  $ ds$ is  measure on our surface.
We need to find area of our surface $ S$ which is equal to $ \int_{S}ds$
 But $ ds \equal{} \sqrt {\det{\Gamma(w)}} dw$  where    $ dw \equal{} dzdx$ is Lebesgue measure on $ K$ and $ \Gamma(w)$ is Gramian matrix. Thus we obtain:
$ \int_{S}ds \equal{} \int_{K}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx \equal{} \int_{0}^{a}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx$
Now we need to minimise  such functional as $ I[u] \equal{} \int_{K}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx \equal{} \int_{0}^{a}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx$
After some calculation we will get euler equation (if I have no mistakes in my calculations):
$ u_{xx} \plus{} u_{zz} \plus{} u_{z}^{2}u_{xx} \plus{} u_{x}^{2}u_{zz} \minus{} 2u_{x}u_{z}u_{xz} \equal{} 0$
If I remember well, this is  wellknown elliptic equation in mathematical physic.
Today, I don't know how to find any solution of such equations.
I believe that such equations are well investigated in math physic's.
I note that, all of this also works for any 2-dimensional manifolds with boundaries   and you always get this elliptic equations.
But if you need to find just a simple lower bound , what fedja has written, then fedja's hint is  one of the possible way for the solution.

[quote="Extremal"]
If I remember well, this is  wellknown elliptic equation in mathematical physic.
Today, I don't know how to find any solution of such equations.
I believe that such equations are well investigated in math physic's.
I note that, all of this also works for any 2-dimensional manifolds with boundaries   and you always get this elliptic equations.[/quote]
Oh, yes, the minimal surface equation itself is well known. Still, there is absolutely no hope to find an explicit solution for all possible boundary curves (actually, the only "explicitly" solved case with a broken line as the boundary I know of is a skewed quadrilateral), so I do not think this one is an exception. What has been studied extensively is existence, uniqueness, and possible singularities of solutions and it is a completely separate story.

So, Extremal, can you prove this bound (using the hint or by some other method)?

I have found other bounds, which is not  so sharp as your.
I was estimating this integral   by using simplest inequalitys like this $ \sqrt {2(a \plus{} b)} \geq\sqrt {a} \plus{} \sqrt {b}$ :
\[ S \equal{} \int_{0}^{a}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx \plus{} \lim_{c \to 0} \int_{ \minus{} c}^{0}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx
\]
I believe that it's useful to think ab0ut such estimation since :
\[ S \equal{} \int_{0}^{a}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx \plus{} \lim_{c \to 0} \int_{ \minus{} c}^{0}\int_{ \minus{} 1}^{1}\sqrt {1 \plus{} u_{x}^{2} \plus{} u_{z}^{2}}dzdx \geq 2a \plus{} 2b \minus{} C
\]
where we need to show that   $ C \approx 4$.
We can assume  that $ |u_{x}| \leq \frac {\sqrt {b^{2} \minus{} c^{2}}}{c}$ where $ |AO| \equal{} b$
also  we can think about such estimation as:
on $ [ \minus{} c,0]$  $ |u_{x}| \geq \frac{\sqrt {b^{2} \minus{} c^{2}}}{a \plus{} c}$ (if $ a > b$) and similary for the interval $ [0,a]$

[asy]size(150);
D((-1.5,-1.5)--MP("z",(1.5,1.5),NE),black);
D(MP("y",(0,7),N)--(0,0)--(-4,0)--MP("x",(7,0),E),black);

D(MP("",(-2,6),N)--(0,0),black);

D(MP("",(-2,8),N)--(-2,0),black);
D(MP("O",(1,1),W)--(-1,7)--(-3,5)--MP("-1",(-1,-1),W)--(4,-1)--(6,1)--cycle ,red+linewidth(1));
MP("A",(-3,6),W);
MP("a",(5,0),S);
MP("-c",(-2.9,0),N);[/asy]


function
limit
number theory
prime numbers
number theory proposed

Let $d(x)$ be the function, that maps the number of prime factors of the natural number $x$.
Let $S_{1}=d(1)+...+d(n)$ with $n\in N$ and $n>1$
Let $p_{1},p_{2},...,p_{k}$ all the prime numbers less than $n$, and $S_{2}=\frac{1}{p_{1}}+...+\frac{1}{p_{k}}$.
Show that: $\lim_{n\rightarrow\propto}({S_{2}-\frac{S_{1}}{n})=0}$

[u][b][i]Proposed by Sef[/i][/b][/u]

I have a (rather large) solution (if there's no mistake somewhere). I will gladly post it here (it's already in the greek forum) if everyone else is too bored to post their own :lol:

Don't name topics in Number Theory with Number Theory, it's a bit redundant ;)


geometry
geometric transformation
reflection
circumcircle
incenter
homothety
cyclic quadrilateral

Denote K be the midpoint of arc CD of (O). Easy to see that K is the excenter of triangle PAB. O is the circumcenter of triangle ABK, so it is also the midpoint of segment HK where H is the incenter of triangle ABP. Thus P,H,O,K all lie on the bisector of angle APB. This implies $ OP\equiv OK\perp CD$

Chasing angles, it's very easy to see that APBO is cyclic.Now, consider triangles POC and POD.
We have that PO is a common side and OC=OD (radii).So, if we prove that <POC=<POD, the problem follows, because with effect, PC=PD => PDC is isosceles and PO is bisector of <CPD.Then, PO is perpendicular to CD.But to prove this angle equality, it's only necessary chase angles using the fact that APBO is cyclic.

Bye,
Secco.

Let $ r$ and $ s$ meet the line $ CD$ at $ E$ and $ F$, respectively. Angle chasing easily shows that $ A\hat{E}D\equal{}B\hat{F}C$. Note that triangles $ ABC$ and $ ADE$ are similar, and also are the triangles $ ABD$ and $ CBF$. This yields

$ \frac{DE}{BC}\equal{}\frac{AD}{AB}\Longrightarrow DE\equal{}\frac{AD.BC}{AB}$

$ \frac{CF}{AD}\equal{}\frac{BC}{AB}\Longrightarrow CF\equal{}\frac{AD.BC}{AB}$

That is, $ DE\equal{}CF$, so $ r$ and $ s$ are symmetrical relative to the perpendicular bisector of $ CD$. The conclusion follows.

Let the symmetrical of AB wrt AP and BP intersect the circle O at B’ and A’ respectively. By construction, BB’, AA’ and CD are perpendicular to OK, where K is the midpoint of the arc CD, therefore parallel and the sketch is symmetrical w.r.t. OK, therefore P will lie on it.

Best regards,
sunken rock

[quote="cyshine"]Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \angle CAD$ and $ \angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.[/quote]
Since this is my first time applying homothety in a formal(and terribly foolish) manner, I am reviving this thread, and anyways, this is also Switz MO 2011. :)

Let the angle bisectors of $\angle CAD$ and $\angle CBD$ meet at $E.$ We obviously have that $2\angle EBC=\angle DCB=\angle DAB=2\angle EAC,$ leading to the fact that $E$ lies on $\odot(ABCD).$
Now since $AE, BE$ are the external angle bisectors of $\angle PAB$ and $\angle PBA,$ it is obvious that $E$ is the $P$- excentre of $\triangle PAB.$ Let $PA, PB$ meet the circumcircle of $ABCD$ again at $F,G$, respectively.
Now, we construct the midpoints of $%Error. "bighat" is a bad command.
{BC}, %Error. "bighat" is a bad command.
{AG}$ and $%Error. "bighat" is a bad command.
{AB},$ name them as $J,K,L$ respectively. If $XX$ denotes the tangent to the circumcircle of $ABCD$ drawn at a point $X$ on it, we let $N=JJ\cap LL, Q=JJ\cap KK$ and $M=LL\cap KK.$ Obviously $OJ\perp JJ$ and $OJ\perp BF$ lead to $QN\parallel PB.$ Similarly we have $QM\parallel PA$ and $MN\parallel AB.$
Also, $O$ is the $Q$- excentre of $\triangle QMN.$ This time, homothety about centre $MA\cap NB\cap PQ$ that maps $\triangle PAB$ to $\triangle QMN,$ must also map their respective $P$ excentre to the $Q$-excentre. So $E,O,X$ are all collinear where $X$ is the centre of homothety. Also $P,Q,X$ are collinear, and $OQ$ bisects $\angle MQN,$ and $OP$ bisects $\angle APB,$ so that $O,P,X;$ and $O,Q,X$ are collinear. So $P,O,Q,X$ all lie in the same line.
With this in hand, $\angle EOD=2\angle EAD=\angle CAD=\angle CBD=2\angle EBC=\angle EOC,$ so that $E$ is the midpoint of $%Error. "bighat" is a bad command.
{CD}.$
So $OE\perp CD$ and hence $PO\perp CD.\Box$

Sorry for such a foolish proof. :lol:

Let $M$ be the midpoint of arc $\widehat{CD}$, and let $r, s$ cut $(O)$ for a second time at $B', A'$, respectively. Note that $AM$ is the bisector of $\angle CAD.$ Because reflection preserves angles, $AM$ bisects $\angle BAB'$ as well. Hence, $M$ is the midpoint of $\widehat{BB'}$, and it follows that $BB' \parallel CD.$ Similarly, $AA' \parallel CD$, and it follows that $\overline{AA'}, \overline{BB'}, \overline{CD}$ share a common perpendicular bisector $\tau.$ Then symmetry in $\tau$ implies that $P \equiv AB' \cap A'B \in \tau$, so $OP \equiv \tau \perp CD$ as desired. $\square$

[b] Solution: [/b]
Let the bisectors of $\angle CAD $ and $\angle CBD $ intersect at $E $. It is obvious that  $E $ lies on  $\odot (ABCD) $. Join $PE $. Let $B'$ and $A'$ be the reflection of $B $ and $A $ in $AE $ and $BE $ respectively. Let $H = AA'\cap BB'$. Since, $H $ is the orthocenter of  $\Delta ABE $, so, $EH\perp AB $.

Now, $\measuredangle PBE = \measuredangle PB'E $ (by trivial angle chasing). This gives $\measuredangle PEB = \measuredangle PB'B = \measuredangle AEH$. Thus, $EH $ and $EP $ are isogonals w.r.t  $\angle AEB $. Hence, $EP $ passes through the center of  $\odot (AEB) $, i.e., $O $. Thus, $EP\perp CD $.

Here's my approach:
 Let $T$ and $L$ be the second intersections of $s$ and $r$ with the circumcircle of $ABCD$ ,respectively.

$\frac{\overarc{TD}}{2} = \angle TAD = \angle CAD = \frac{\overarc{BC}}{2}$
$\implies \overarc{TD} = \overarc{BC} \implies TB\parallel CD $   $(*)$

Similarly we get $\overarc{AT} = \overarc{LB}$
Hence, $ALBT$ is an isosceles trapezoid. $\implies PT=PB$. And also we have $OT=OB$, so $OTPB$ is a kite. $\implies PO\perp BT$
From $(*)$ we conclude $CD\perp PO$

Meh
Nice problem

Hint 1:
[hide]We just need $P$ to be in the perpendicular bissector of $CD$.[/hide]

Hint 2:
[hide]Take the intersections of $EA$ and $EB$ with the circle (call them $F$ and $G$, respectively).[/hide]

Hint 3:
[hide]Angle chase to show that $AC\cap DG$ belongs to the perpendicular bissector of $CD$.[/hide]

Again angle chase to prove $BF||CD||AG$, thus $P$ is in the perpendicular bissector of $AG,BF$, hence it is in the $CD$.

Very nice problem. An angle chase yields that $PAOB$ is cyclic, thus $\overline{PO}$ bbisects $\angle APB$. On the other hand, the intersection point $I$ between $\overline{PO}$ and the two given bisectors is the $P$-excenter of triangle $PAB$; thus, it is also the arc midpoint of $\widehat{CD}$, and the result follows.

Let $A\neq E=r\cap (ABCD)$ and $B\neq F=s\cap (ABCD)$, then $\measuredangle EAD=\measuredangle BAC$ and $\measuredangle FBC=\measuredangle DBA$, so $AF\parallel CD\parallel BE$ and $P=AE\cap BF$. Clearly $OP\perp AF$, so $OP\perp CD$.


let x,y,z are reals.Prove that:
3(x 2 -x+1)(y2 -y+1)(z 2 -z+1) \geq (xyz) 2 +(xyz)+1.

I think it was already discussed. By the way, as far as I remember, this problem was created by Mircea Lascu and Vasile Cartoaje.


[quote="CDP100"]Given is triangle ANC with AN = AC and $ \angle CAN \equal{} 90^o$. Points M and B lie on AN. If $ \angle ACM \equal{} \angle MCB \equal{} \angle BCN \equal{} 15^o$. Prove that BN = AM + BM/2.[/quote]
[hide="Solution"]
Let $ CA \equal{} a$.  Then, we have that
\[ AM \equal{} a\tan \angle ACM \equal{} a\tan 15 \equal{} \frac {a\sin 15}{\cos 15} \equal{} \frac {\frac {a(\sqrt {6} \minus{} \sqrt {2})}{4}}{\frac {\sqrt {6} \plus{} \sqrt {2}}{4}} \equal{} \frac {a(\sqrt {3} \minus{} 1)}{\sqrt {3} \plus{} 1} \equal{} a(2 \minus{} \sqrt {3})
\]
Notice that $ BA \equal{} a\tan 30 \equal{} \frac {a}{\sqrt {3}}$.  Finally, we notice that $ BN \equal{} AN \minus{} AB \equal{} a \minus{} \frac {a}{\sqrt {3}} \equal{} \frac {a(\sqrt {3} \minus{} 1)}{3}$.  Hence,
\[ AM \plus{} \frac {BM}{2} \equal{} AM \plus{} \frac {AB \minus{} AM}{2} \equal{} \frac {AB \plus{} AM}{2} \equal{} \frac {a(2 \minus{} \sqrt {3} \plus{} \frac {1}{\sqrt {3}})}{2} \equal{} \frac {a(\frac {2\sqrt {3} \minus{} 2}{3})}{2} \equal{} \frac {a(\sqrt {3} \minus{} 1)}{3} \equal{} BN
\]
as desired. [/hide]

to: cdp100

It's obvious that you wanted something in connection with the triangle MCN and CB internal bisector of the angle MCN, then its external one!

However, some simple direct calculations show that the required equality is equivalent with: 3.BN = AN +AM ( 1 ). Therefore we shall draw S, the symmetrical of M w.r.t. A and we have only to prove 2BN = BS ( 2 ); using the sine teorem in the triangles NBC and BSC, (2) implies after calculations
2.sin(15).sin(75)=sin(45).sin(45)  ( 3 ), substitute sin(75) by cos(15) and (3) becomes obvious.

Best regards,
sunken rock

A blind man sees better the details


We consider the proposition $ p(n)$: $ n^2\plus{}1$ divides $ n!$, for positive integers $ n$. Prove that there are infinite values of $ n$ for which $ p(n)$ is true, and infinite values of $ n$ for which $ p(n)$ is false.

Yea this one looks like a classic. I don't understand why its there but ...

1. Let $ p \equal{} 1\mod 4$ prime, then exist $ n_0 < p/2$, suth that $ p|n_0^2 \plus{} 1$. Obviosly $ p\not |n_0!$ or $ n_0^2 \plus{} 1\not |n_0!.$
2. Let primes $ p>q\ge 17$ and $ p\equal{}1\mod 4, q\equal{}1\mod 4$. Exist $ pq/2<n_0<pq$, suth that $ pq|n_0^2\plus{}1$. Obviosly $ n_0/2<\frac{n_0^2\plus{}1}{pq}<n_0$. Therefore $ p^4q^4|n_0!$ and $ \frac{n_0^2\plus{}1}{pq}|n_0!$. It give $ n_0^2\plus{}1|n_0!$.

1.The fact that there are infinitely many values of $ n$ for which $ p(n)$ is false can be easily proven by taking $ n\equal{}p_{1}p_{2}...p_{m}$, where $ p_{i}$ are all odd primes smaller or the same as $ n$. Then $ n^{2}\plus{}1\equiv 2 (mod 4)$. Hence $ n^{2}\plus{}1\equal{}2l$, where $ l$ is the product of some odd primes greater than $ n$. Since $ l$ does not divide $ n!$, also $ n^{2}\plus{}1$ does not divide $ n!$.

2. I don`t understand the existence of $ n_{0}$ and $ pq$ in Rust`s proof. Can anyone explain it to me? Thank you.

Sorry, now I see I have made a huge mistake in 1. Obviously it is not so easy as it seemed.

1. Let $ p\equal{}1\mod 4, \ a\equal{}(\frac{p\minus{}1}{2})!\mod p$. Then by Wilson's theorem $ p|a^2\plus{}1$. Therefore $ p|a^2\plus{}1$, but $ p\not |a!$, because $ a<p$ and p- prime. It give $ p(a)$ is false for any $ p\equal{}1\mod 4$.
2. Let $ p,q\equal{}1\mod 4$ let $ p|a^2\plus{}1,q|b^2\plus{}1,p,q\ge 5$ and let $ n_0\equal{}caq\plus{}dbp\mod pq$, were $ c,d$ choose by condition $ cq\equal{}1\mod p, \ bp\equal{}1\mod q$. If $ n_0<\frac{pq}{2}$, then we can change $ n_0\to pq\minus{}n_0$ and it give $ pq/2 <n_0<pq, \ pq|n_0^2\plus{}1.$
Because $ pq\ge n_0\plus{}1$ we get $ g\equal{}\frac{n_0^2\plus{}1}{pq}<n_0$ and $ g|n_0!$.
Because $ p,q\ge 5$ $ ord_p(\frac{n_0!}{g})\ge [\frac{n_0}{p} \minus{}1]\ge [\frac{q}{2}\minus{}1]\ge 1$
and $ ord_q(\frac{n_0!}{g})\ge 1$. Therefore $ pqg\equal{}n_0^2\plus{}1|n_0!$ for any primes $ p,q\equal{}1\mod 4$.

Where in PEN is it? I can't find using search.

See [url]http://www.mathlinks.ro/Forum/index.php?f=456[/url] and [url]http://projectpen.wordpress.com/[/url].


Let k be the largest integer so that $2^k < n$.  Then choose $2^j$ for j = 0 ... k, and n.  We have chosen at most $\lfloor log_2{n} \rfloor + 1$ numbers.

To prove $\lfloor log_2 {n} \rfloor \le 2\lfloor \sqrt{n} \rfloor$; we prove $log_2 n \le 2\sqrt{n} + 2$, equiv. to $n \le 4*4^{\sqrt{n}}$.  Wrt n, the deriv. of the RS is $\frac{4 \log(4) (\sqrt{n} + 1) }{ \sqrt{n}}$, of LS: $1$.  Then its clear the RS increases faster than the LS; also, RS > LS for small cases.  It is enough.

How do you (e.g.) get the difference $5$ by this¿
(the bound is sharper than you may think)

[quote=ZetaX]Prove:
From the set $\{1,2,...,n\}$, one can choose a subset with at most $2 \left\lfloor \sqrt n \right\rfloor +1$ elements such that the set of the pairwise differences from this subset is $\{1,2,...,n-1\}$.
($\left\lfloor x \right\rfloor$ means the greatest integer $\leq x$)[/quote]

Choose $1,2,\ldots,\lfloor\sqrt{n}\rfloor-1$, all multiples of $\lfloor\sqrt{n}\rfloor$ in $\{1,2,...,n\}$ and $n$.


Show that b_n=19*8^(n)+17 is composite for any positive integer n

[hide]
Considering cases

$ n \equal{} 4k$:
$ 19 \cdot 8^{4k}\plus{}17 \equiv 1 \cdot 64^{2k} \plus{} 2 \equiv 1^{2k} \plus{} 2 \equiv 0 \pmod 3$

$ n \equal{} 4k\plus{}1$:
$ 19 \cdot 8^{4k\plus{}1}\plus{}17 \equiv 6 \cdot 8 \cdot {8^4}^k \plus{} 4 \equiv 9 \cdot 1^k \plus{} 4 \equiv 0 \pmod{13}$

$ n \equal{} 4k\plus{}2$:
$ 19 \cdot 8^{4k\plus{}2}\plus{}17 \equiv 1 \cdot 64 \cdot 64^{2k} \plus{} 2 \equiv 1^{2k} \plus{} 2 \equiv 0 \pmod 3$

$ n \equal{} 4k\plus{}3$:
$ 19 \cdot 8^{4k\plus{}3}\plus{}17 \equiv 4 \cdot 8^3 \cdot {8^4}^k \plus{} 2 \equiv 4\cdot 2 \cdot 1^k \plus{} 2 \equiv 0 \pmod 5$[/hide]


analytic geometry
algebra solved
algebra

Find all rationals x,y such that  x^2+y^2 =2

The idea for this one is to write x=p/q and y=m/n with p,q,m,n in Z and (p,q)=1 and (m,n)=1 and the equation becomes:

p^2n^2+m^2q^2=2q^2n^2

see what happens! :)

We can use the method of intersection to solve this kind of equations: We have to find all rational points in the circle (C) x^2+y^2 = 2. Denote the set of all those points by S. Note that A(1, 1) belongs to S. Now, take an rational number k and consider the line y = k(x-1) + 1. This line meets circle (C) again at the point B with coordinates (k^2-2k-1)/(k^2+1), (-k^2-2k+1)/(k^2+1). Of course B belongs to S. So  x = (k^2-2k-1)/(k^2+1), y= (-k^2-2k+1)/(k^2+1) is rational solution of x^2+y^2=2. It easy to show that all rational solution can be received by this way.

Namdung


Voir le titre du message ! :)

Comme c'est un exercice proposé par les français (il me semble ... :D  !!!)

Est-il possible de voir une résolution  ou plusieurs  résolutions ... soyons fous ....de celui-ci et surtout..... de quel contexte est-il sorti ?

Pour les solutions, voir le fil spécifiquement dédié à cet exo dans le forum IMO Spain 2008.

Pour l'origine, la version initiale vient d'un décompte de chemins de longueur fixée d'un sommet à un sommet opposé le long des arêtes d'un hypercube de dimension $ k$.
Comme l'hypercube n'avait aucune chance de passer aux OIM, on a changé l'habillage en parlant d'interrupteurs.

Pierre.


inequalities
function
calculus
derivative
integration
calculus computations

If f is twice differentiable on [-a,a] and $ M_0  \equal{} \mathop {\sup }\limits_{x \in I} |f(x)|$, $ M_1  \equal{} \mathop {\sup }\limits_{x \in I} |f'(x)|$, $ M_2  \equal{} \mathop {\sup }\limits_{x \in I} |f''(x)|$ then

$ |f'(x)| \le \frac{{M_0 }}{a} \plus{} \frac{{x^2  \plus{} a^2 }}{{2a}}M_2$

Consider the function $ y \equal{} f'(x)$. It has derivative bounded by $ M_2$ and its integral over any interval $ [\alpha,\beta]$ does not exceed $ 2M_0$. Given these constraints, how can we make $ M: \equal{} f'(x)$ large? By making $ f'$ drop down linearly on both sides of $ x$, as in $ f'(t) \equal{} \max(M \minus{} M_2|t \minus{} x|,0)$. Therefore,
\[ \int_{ \minus{} a}^a \max(M \minus{} M_2|t \minus{} x|,0)\,dt\le 2M_0
\]
This can be integrated and solved for $ M$, but the computation splits into cases depending on the signs of $ M \minus{} M_2|a \minus{} x|$ and $ M \minus{} M_2|a \plus{} x|$. Being lazy, we drop $ \max$ from the integral (this only decreases it) and arrive at
\[ \int_{ \minus{} a}^a (M \minus{} M_2|t \minus{} x|)\,dt\le 2M_0
\]
The integral easily evaluates to $ 2Ma\minus{}M_2(a^2\plus{}x^2)$, hence $ M\le \frac{M_0}{a}\plus{}\frac{M_2(a^2\plus{}x^2)}{2a}$ as required.

However, our laziness did not go unpunished: the estimate is not sharp. Indeed, it does not approach $ 0$ as $ M_0\to 0$ - as it should.

Could you explain the 'making f' drop down linearly on both sides of x' part? I don't understand what's going on there.

Fix $ x\in [\minus{}a,a]$. We may assume $ f'(x)\ge 0$. For all $ t\in [\minus{}a,a]$ we have $ |f'(t)\minus{}f'(x)|\le M_2|t\minus{}x|$ (say, by Mean Value Theorem applied to $ f'$). This implies $ f'(t)\ge f'(x)\minus{}M_2|t\minus{}x|$. To shorten notation, let $ g(t)\equal{}f'(x)\minus{}M_2|t\minus{}x|$. Notice that $ g(t)\ge 0$ when $ |t\minus{}x|\le f'(x)/M_2$. Let $ [\alpha,\beta]$ be intersection of $ [\minus{}a,a]$ with the interval $ \{t\colon |t\minus{}x|\le f'(x)/M_2\}$.

Here is the key point: since $ f'(t)\ge g(t)\ge 0$ on $ [\alpha,\beta]$, it follows that
\[ \int_{\alpha}^{\beta}g(t)dt\le \int_{\alpha}^{\beta}f'(t)dt\equal{} f(\beta)\minus{}f(\alpha)\le 2M_0\]
This is essentially the same as the first displayed formula in my previous post.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int \left( \cot 2x\cot 3x\minus{}\tan 2x\tan 7x\right) \tan 5xdx$
 :ddr:

$ \cot 5x\equal{}\frac{\cot 2x\cot 3x\minus{}1}{\cot 2x\plus{}\cot 3x}$

$ \cot 2x\cot 3x\tan 5x\equal{}\cot 2x\plus{}\cot 3x\plus{}\tan 5x$

$ \tan 5x\equal{}\frac{\tan 7x\minus{}\tan 2x}{1\plus{}\tan 2x\tan 7x}$

$ \tan 2x\tan 7x\tan 5x\equal{}\tan 7x\minus{}\tan 2x\minus{}\tan 5x$

I think this should suffice


I was doing research, and I found that in 1917, Jeanette Rankin was the first woman in US house of reps.  I know that women did not get the right to vote until 1920.  Is the first "fact" for real?  just needed some clarification for a research paper.

The right to run for office is independent of the right to vote, as far as I can tell.

If you need clarification for a research paper, what you should do is go to a library and acquire an actual historical text or biography, which you can then cite appropriately.  That is, you should do actual research.  "Asking a bunch of people on the internet" does not count for this.


. Decartes' Rule of Signs says that for a polynomial P(x), the number of positive roots to the equation is equal to the number of sign changes in the coefficients of the polynomial, or is less than that number by a multiple of 2. It say IS LESS THAN THAT NUMBER BY A MULITPLE OF 2. What does that mean? any examples?

any help would be awesome!

'Less than a multiple of 2' You have to consider that polynomial equations can have [u]complex roots[/u]. If the coefficients of the polynomial are real numbers, the complex roots come in conjugate pairs, that's why you have to subtract by 2

Short quadratic example

Suppose we have $ x^2\minus{}x\plus{}1\equal{}0$. Now use Descartes Rule. It's telling you two sign changes so there are 2 positive roots??? No, that's impossible, this equation has no real roots! The roots are complex roots, so we have 2-2=0 positive roots.

Understood?


Substituting $ x \equal{} 5^{25}$, we have $ \frac { 5^{125} \minus{} 1 }{ 5^{25} \minus{} 1 } \equal{} \frac {x^5 \minus{} 1}{x \minus{} 1}$, and obviously $ x \minus{} 1|x^5 \minus{} 1$.

EDIT: Wow, fail...

I misread the problem as 'Prove that the given fraction is integral'...

er...........
@geniusbliss
If  in $ \frac {a}{b}$  if b divides a ....$ \frac {a}{b}$ is an integer
So you cant claim its composite or prime ,can you

OK you have edited it now.

[quote="geniusbliss"][hide="@Dr.Sonnhard Graubner"]am sorry if this is a spam but i think if that product you have got in the left doesnt have a beautiful way to get it then it is a total waste of posting the problem for the idea of the problem is lost. why be so comptational??[/hide][/quote]

[quote="Poincare"][quote="Dr Sonnhard Graubner"]hello, we have
$ {\frac {5^{125} - 1}{5^{25} - 1} = 4032808198751\cdot 767186663625251\cdot 24687045214139234043375683501\cdot28707251\cdot3597751}$
Sonnhard.[/quote]

Just saying, but Mr. Graubner, unless the subject is trig, your posts are amazingly unhelpful.[/quote]

From Bictor717's link, this is an IMO-level problem, OP should know better this problem should, at least, belong to Pre-Olympiad. I will PM this user.

Now, AoPSers should refrain from posting anymore similar posts like above. If you think his post is a 'low quality post', please use the post rating accordingly. If you really need to say something, PM that user. Let's help keep the posts in AoPS 'top quality' and reduce SPAM.

[quote="Mewto55555"]Hey, it may not be a reasonable way to approach the problem, but it proves that it is true (assuming the numbers are correct). I see no reason to rate a correct post spam.[/quote]That's your opinion and it counts if you want to give him a good post rating. The reason why I suggest using this feature is because it serves as a [i]collective[/i] rating. It may not be efficient but, assuming that it does what it's supposed to do, it's as good as it gets to reduce posts like those that I have quoted above. Note that I have not said anything about Sonnhard's post.

And I should stop discussing about this because it is getting off-topic.


function
trigonometry
analytic geometry
conics
ellipse
parameterization
calculus

Hi guys. This is my first post. I would just like to see how the graph of this function would look like. I know this is a helix but Im unsure how it would look like.

The equation is $x = \sin t$  $y = t$  $z = \cos t$.

The best way to find out is to examine this on your own "manually" or by say MATLAB.

Lets say $t\geq 0$ (it will continue in the same way in the $-t$-direction, so lets simplify)
then the helix starts at $(0,0,1)$ and turns clockwise (as seen in the direction of the$y$-axis) around the $y$-axis while simultaneously the $y$-coordinate increase (the pitch is $\lambda$ in $y=\lambda t$). The projection onto the $xz$-plane will always remain the unit circle in that plane. This is becase you chose $a=b=1$ in $(x=a\cos{t},y=\lambda t, z=b\sin{t})$ otherwide it would have been an ellipse $\frac{x^{2}}{a^{2}}+\frac{z^{2}}{b^{2}}=1$. Changing the order of the parameters like $(x=a\cos{t},y=b\sin{t}, z=\lambda t)$ yields an helix around the $z$-axis

You can find more at for example

[url]http://mathworld.wolfram.com/Helix.html[/url]
[url]http://en.wikipedia.org/wiki/Helix[/url]

As a general principle, graphing anything three-dimensional is hard, since the paper is two-dimensional. You have to lose information- this is traditionally done by projecting onto some plane and "viewing from an angle".
To do this by hand, draw your axes, mark them, and graph a few key points.


complex numbers
superior algebra
superior algebra solved

A is in M2(R) s.t. detA=-1. Show that det(A^2+I)>=1.

I seem o have thought there is something wrong with this one too because I wrote a question mark beside it. Maybe I considered it too easy, I don't remember.

Hi grobber can you check my proof

A \in M2(R)

A=
[x  y]
[z   t]

det A = xt- yz = -1  (1)

det(A2 +I) = det((A+iI)(A-iI))= det(A+iI).det(A-iI)
= |det(A+iI)|2 (2)
because det(A-iI) is the complex conjugate of det(A+iI)

A+iI =
[x+i     y ]
[z       t+i]

so

det(A2 +I)
= | (xt -yz-1) + i(x+t)|2
=(-1-1)2 + (x+t)2   \geq 4 >1

What was the question mark ?

I think it's correct. What confused me is the fact that you can get a so much better result: >=4 instead of >1 ! I don't like weak ineqs, you know... :D They can be pretty confusing.