induction
modular arithmetic
number theory unsolved
number theory

Find all positive integer a such that there exists infinitely many positive odd integers n such that $ a^n \plus{} 2009^n$ is divisible by $ n$
[hide="I got "]$ a\ge 2$ because if $ a\equal{}1$ then $ n$ must be an even (well-known)[/hide]

I think that $ a \plus{} 2009 \ne 2^k (k \in N)$.

To explain tronghieu's idea we see that if $ p$ is and odd prime that divides $ a\plus{}2009$ it is easy to see that $ n\equal{}p^b$ works
(induction or lifting the exponent)
I'm still thinking the other part which seems more interesting

[quote="Rust"]For any a>0 and $ a + 2009\not = 2^k$ exist infinetely many n, suth that $ n|a^n + 2009^n$.
let $ n_0|a^{n_0} + 2009^{n_0}$ and $ p|\frac {a^{n_0} + 2009^{n_0}}{n_0},p > 2$, then for $ n_1 = n_0p$ we get
$ n_1|a^{n_0} + 2009^{n_0}|a^{n_1} + 2009^{n_1}$.[/quote]
Thank you, tronghieu, Rust and lambruscokid . But you only showed that when $ a+2009$ was not a power of 2 then $ a$ satisfied our property. I think tt is not enough, we must prove that there is no $ a$ satisfying the initial one ( now we exclude the case proved by you above!).
BTW, When $ a=1$, $ a+2009=2010\neq 2^k$. But we have no $ n$ satisfying the condition (if $ n$ does, it must be even- a well-known lemma). Can you explani more.
Thank you
D.Brown

Let $ a>1$ satisfy the conditions of the problem with $ a\plus{}2009\equal{}2^k$.  It is clear that $ gcd(a,2009)\equal{}1$.  Thus, $ gcd(n,a)\equal{}gcd(2009,n)\equal{}1$.  Let $ p$ be the smallest prime dividing $ n$.  Let $ p^r\parallel{}n$ and $ m\equal{}n/p^r$.

Then, $ 0\equiv a^n\plus{}2009^n\equiv a^m\plus{}2009^m\pmod{p}$.  It follows that $ a^m\equiv 2009^m\pmod{p}$.  Thus, since $ a,2009$ are invertible $ \mod p$, we have $ \left(\frac{a}{2009}\right)^m\equiv\minus{}1\pmod{p}$.  Thus, $ ord_p(\frac{a}{2009})|2m$ but $ \not|m$.  However, since all primes dividing $ m$ are greater than $ p$, it follows that $ ord_p(\frac{a}{2009})\equal{}2$.

Therefore, $ \left(\frac{a}{2009}\right)^2\equiv1\pmod{p}$.  So, $ p|a^2\minus{}2009^2$.  Since $ p\not|a\plus{}2009$, it follows that $ p|a\minus{}2009$.  From $ p|a^n\plus{}2009^n$ we find $ p|a,p|2009$, contradiction!


Show that for any set of five integers, we can always choose three of these integers whose sum is a multiple of 3.

[hide]Every integer is either congruent to 0 mod 3 , 1 mod 3 , or 2 mod 3.   Any set of 5 integers will fall under the category of either having 3 or more of one of the 3 mod cases or will have 2 of 2 different mod cases.

Btw, the mod cases are 0 mod 3, 1 mod 3, 2 mod 3(just to clear up any confusion since I don't know any other way to state this).

Ok, if the set of numbers has 3 or more of a single mod case then you can just pick 3 of that case to form a set whose sum is divisible by three  i.e.

3*(0 mod 3) :equiv: 0 mod 3.  3*(1 mod 3) :equiv: 3 mod 3  :equiv: 0 mod 3 or finally 3*(2 mod 3) :equiv: 6 mod 3  :equiv: 0 mod 3.  0 mod 3 is divisible by 3 so all we need to do is look at if it works under the other possibility.

The other possibility is that there are two integers under a certain mod cases for two cases and one for a third.  It is easy to prove that you can choose three of these whose sum is a mult. of three  since every set of 5 integers under this category will have at least one 0 mod 3 , one 1 mod 3, and one 2 mod 3  ( 0+1+2 mod 3 :equiv: 3 mod 3 :equiv: 0 mod 3)

[/hide]

I'm not sure that this proof is easy to understand.  I tried to make it as clear as possible but I don't think that I found the right words to do so.

[quote]I'm not sure that this proof is easy to understand. I tried to make it as clear as possible but I don't think that I found the right words to do so.[/quote]

The proof was easy enough to understand, and it was also correct. Good job! BTW, the three 'mod cases' are referred to as equivalence classes (if you need a name for them). You could also just announce that everything is in mod 3 and then assume that the ONLY elements in the set were 0, 1, and 2, and that way you don't need to have the words 'mod 3' everywhere throughout your proof.

Good job Joe. :-)

(This is a challenge question by the way, so others should still work on it before looking at Joe's solution.)

Sorry, I didn't spoiler it because it was a fair way down the page.

For joml, this is an example of what might be considered a 'clearer' proof with the same reasoning.

[hide]Let the 5 integers be given by x1 .. x5.  Consider x1 .. x5 mod 3.  If any three of x1 ... x5 are congruent mod 3, then the result follows immediately.

Otherwise, we must have that there are two pairs of mutually congruent integers and one lone integer which is not congruent to the other three (else we would have that three integers which were congruent mod 3)

Because we now have three disjoint sets, which contain elements congruent to 0,1,2 (mod 3), we can pick an element from each set whose sum will then be 0+1+2  :equiv: 0 mod 3 and hence the sum of these three numbers is divisible by 3.[/hide][/hide]

spoiler? So that other people can try it!

And there is nothing wrong with your solution blah^3, but I actually like joml's solution as it is. Words like "follows immediately" are often used (and abused) by professional mathematicians to shorten proofs - but it's a [i]very[/i] bad convention.

Especially for young proof writers, the more explicit the explaination the better - because then they can clearly see the logic in each step, and avoid making false assumptions. When it comes to the writing of proofs, shorter is not always better.

By the way, the problem can be generalized.  Let n be a positive integer.  Then for any set of 2n - 1 integers, we can always choose n of them whose sum is 0 mod n.  This result is due to Erdos and some co-authors (whom I forget now).  It's not so easy to prove.


Sea ABCD un paralelogramo y P un punto tal que
2<PDA=<ABP y 2<PAD=PCD
Demostrar que AB=BP=CP=CD.

Esta pregunta fue tomada en Argentina, no??

[quote="conejita"]Aki mi solucion: (en mi dibujo considere que AB>AD)
Tracemos un punto Q tal que ADPQ sea paralelogramo. Observamos que 2AQP=ABP y 2BQP=BAP, entonces Q es circuncentro del triangulo ABP por lo tanto QA=DP=QP=BC=DA=QB=PC.
 :lol:[/quote].

 Please , according to the wording I have made the figure , that you see in the attachment.Tell me what I have made wrong.
 Thank you - babis


[quote]When we have the sum and the product of roots, for example, a quadratic equation.
ax^2+bx+c=0
We can create a sequence,
aS_n+bS_n-1+cS_n-2=0
But why, may anyone show me a proof?[/quote]

i dont get this, can someone show, how i can make a sequence from the sum and product of the roots of a polynomial?

Let $\alpha ,\ \beta$ be the roots of the quadratic equation $ax^{2}+bx+c=0$, we have $a\alpha^{2}+b\alpha+c=0\cdots [1]$ and $a\beta^{2}+b\beta+c=0\cdots [2]$. Denote $S_{n}=\alpha^{n}+\beta^{n}\ (n=0,\ 1,\ 2,\ \cdots)$, $[1]*{\alpha }^{n-2}+[2]*{\beta }^{n-2}$ gives the result.

sorry i dont get this part , $[1]*{\alpha}^{n-2}+[2]*{\beta}^{n-2}$. wot does the [1] and [2] mean ?
the sequence it gives is $S_{n}={\alpha}^{n}+{\beta}^{n}(n=0, 1, 2,\cdots)$ if i am right, then what is this part for $[1]*{\alpha}^{n-2}+[2]*{\beta}^{n-2}$

sorry for asking i am new to this.

All right, binomial_4eva. :)

$[1]*\alpha^{n-2}$ means that $\alpha^{n-2}(a\alpha^{2}+b\alpha+c)=\alpha^{n-2}*0$.


Dados $ n\geq 4$ puntos en el plano tales que cada distancia es entera, demuestre que por lo menos $ \frac{1}{6}$ de las distancias son divisibles por $ 3$

normalmente soy my haragan para postear una solucion aqui, y menos un problema q no es corto. Pero este problema se merece! me parecio muyyy lindo y solamente por eso pongo..Esta es mi sol..[/hide]


Let n be a fixed prime number >3. A triangle is said to be admissible if the measure of each of its angles is of the form $\frac{m\cdot 180^{\circ}}{n}$ for some positive integer m.
We are given one admissible triangle. Every minute we cut one of the triangles we already have into two admissible triangles so that no two of the triangles we have after cutting are similar. After some time, it turns out that no more cuttings are possible. Prove that at this moment, the triangles we have contain all possible admissible triangles (we do not distinguish between triangles which have same sets of angles, i.e. similar triangles).

Strange...
I have following text in my hands:

Let $n$ be a fixed [b]integer number[/b], $n$ is not divisible by 2 and 3...

[quote]Let n be a fixed integer number, n is not divisible by 2 and 3...[/quote]
Yes, at first the problem started in this way, but later organizers of the competition exchanged the text of this problem to the one, that I've already posted.


greatest common divisor
number theory
least common multiple

Let $ A$ be the greatest common factor and let $ B$ be the least common multiple of 8, 12 and 24. What is the value of $ A \plus{} B$?

The GCF of 8, 12, and 24 is 4
the LCM is 24

24 + 4 = 28
voila, there's the answer!


geometry
3D geometry
parameterization
sphere
calculus
calculus computations

A wineglass is full of water and an  iron ball is dipped so that water spilled over. The iron ball rest inside and its top leveled at the rim of the wineglass. What is the maximum angle A (angle subtended by wineglass' side with its dividing vertical line) that maximizes the spill over of water?

That isn't clear to me- what shape is the glass, anyway? Even if it's a cone, there are some other size parameters you need to specify for this to make any sense. As stated, we could just make the glass arbitrarily large without changing its shape. As for the (spherical) ball- is it exactly the right size so that its top is at the surface after spillage, or is it potentially larger and we ignore the part that's sticking out?

thanks for your reply [b]JMerry[/b], The wineglass in my mind is conical in shape and the metal ball is spherical, and the spherical ball's top levels with the wineglass rim surface level . . . . what i thought was not translated well into a sentence . . . . During the exam i failed to copy the exact wordings, i'm directly relying on my faint memory. I'm sorry for the trouble . . . .

There's still a size parameter to nail down. The version that makes the most sense to me is that the volume of the glass is fixed; the other possibilities I could think of put the maximum in a degenerate case.

Sir, let's say 190 mL as the volume of the wine glass in an inverted conical shape. But i think the volume should be arbitrary. just like in any maxima/minima problems. For example what is the maximum shape of a right circular cylinder inscribed inside a sphere. the volume of the sphere is not needed to solve the proportion of the base and height of the cylinder.... My teacher's answer was 30 degrees.

It doesn't matter what the volume is. It matters that the volume is fixed; if we had that the height is fixed instead, the maximum would come as the sides approached flatness.

[b]JMerry[/b], i am sorry to pester you with this but the only thing that is constant in this problem is the height of the inverted cone (wineglass), the diameter of the spherical metal ball, the diameter of the upper lip of the cone varies. To maximize spillage, what is the angle alpha.Angle alpha is the angle formed by the vertical line with slant height of the cone.

$ 90^\circ$, at which point it's a flat plane holding an infinite amount of wine. If you sensibly rule that that isn't a wineglass, there is no maximum, but the spilled quantity increases as the sides get flatter and the ball rests lower. Actually, if the ball's size varies as well, we want to make it very large as well, for infinite spillage from our infinite glass.
This just doesn't feel like the right question.

i gave up [b]JMerry[/b], my proposed problem seems problematic in nature. Now, i think you are right after all.

[b]JMerry[/b], how about this similar problem from [b]Calculus by Granville Problem p.118 # 30.[/b]
Into a full conical wineglass of depth [i]h[/i] and generating angle [i]A[/i], there is carefully
dropped a sphere of such size as to cause the greatest overflow. Show that the
radius of the sphere is equal to (h sin A)/(sin A + cos 2A)

Ah, unlike the previous one, this is well-posed. Just for fun, here is a picture:
[asy]size(200);
path P=(0,0)--(3,5)--(-3,5)--cycle;
pair O=(0,5.3),
H=foot(O,(0,0),(3,5));
path Q=CP(O,H);
fill(Q,yellow);
clip(P);
D(P,black+linewidth(1));
D(Q,magenta+linewidth(1));
D("C",(0,0));
D("H",H,SE);
D("O",O,N);
D("D",IP(D(MA("A",(3,5),(0,0),O,1,2)--O),Q));
D("E",(0,5));[/asy]
The spillage equals to the volume of the yellow portion of the ball ;).

So, licu-an, what have you done so far?

my drawing is not that, thats why i cann't arrive at the answer. i will try your [b]well-drawn out figure[/b]. Thanks


thanks you ; arqady
your solution very nice  :roll:


1) A book has 500 pages numbered 1,2,3 and so on. How many times does the digit 1 appear in the page numbers?

Can any1 tell me some methods of solving this problem!
coz i used a lame one :blush:

This should be right.
[hide]In the 100's place the 1 only appears 100 times (100-199). In the 10's place, the 1 appears 19 times for every 100 numbers. That would mean in total it appears 19 * 5 = 95 times. In the 1's place the 1 appears 10 times every 100 numbers. So it appears 10 * 5 times total in the ones place. The total number of 1's from 1 - 500 inclusive is 50 + 95 + 100 = 245.[/hide]

[hide]
Wouldn't it be 19x4 and 10x4 instead of x5 because you already counted the 100's.
My answer:
216
[/hide]

I don't understand what you're saying. I'm saying I counted the 100's place.

Sorry, my mistake.  I misread the question.

I think tarquin's approach is probably the best, only I don't understand how there is 19 for ever one hundred. I only find 10...

[hide]
Take it by units, tens, and hundreds.

HUNDREDS:
Will appear only when its is 100-199, so [b]100[/b] there.

TENS:
Will appear only 10 times (10-19) every hundred, so [b]50[/b] there (one for below 100, 100, 200, 300, 400).

UNITS:
Will apear 10 times every hundred (1, 11, 21 and so on), so [b]50 [/b] times again(one for below 100, 100, 200, 300, 400).

So I get a grand total of [b]200[/b][/hide]

Yeah you're right.
 [hide]45 from my answer is 200.[/hide]

I agree.  What was with 19 in the 10's place, anyway?

Yeah, and I fell for it, too :wacko:  :no:  :rotfl:


vector
linear algebra
linear algebra unsolved

Suppose that $\{V_\alpha\}_{\alpha\in I}$ are a family of vector spaces of dimension $k$, and intersection of any two of them is a vector space of dimension $k-1$. Prove that there all of the vector spaces are contained in a vector space of dimension $k+1$, or all of them contain a vector space of dimension $k-1$.

You just play around with dimensions :).

Let $U,V$ be two of our spaces, and assume there is at least one other space $W$ such that $U\cap V\not\le W$. Then $\dim(U\cap V\cap W)=k-2$, and, since $\dim(U\cap W+V\cap W)=\dim(U\cap W)+\dim(V\cap W)-\dim(U\cap V\cap W)=k$, we have $W=U\cap W+V\cap W\le U+V$. We have thus proved that whenever $U\cap V\not\le W$ for three of our spaces, we have $U+V=U+W=V+W$. In addition, $U\cap V+U\cap W+V\cap W=U+V\ (*)$.

Now take another space $X$ in our family. $X$ cannot contain all three spaces $U\cap V,U\cap W,V\cap W$, because of $(*)$. This means that what we've shown above applies to $X$ and two of the spaces $U,V,W$, say $U,V$, and we get $X\le U+V$, as desired.


geometry
3D geometry
calculus
integration
number theory proposed
number theory

Find the smallest integer that can be represented in two ways by the sum of two cubes.  :D

[hide]It's $ >1000$ [/hide]

I guess it's $ 1729\equal{}1^3\plus{}12^3\equal{}9^3\plus{}10^3$  :D

Well it's a interesting number, eh?

Yes, that's right :D [url]http://mathworld.wolfram.com/Hardy-RamanujanNumber.html[/url]

This is a famous anecdote (Hardy, Ramanujan, Taxicabs, etc...  :) ) but the way you phrased it the answer is not, in fact, 1729. Remember the amazing coincidence:

$ 3^3\plus{}4^3\plus{}5^3\equal{}6^3$, from which it follows that $ 91 \equal{} 3^3 \plus{} 4^3 \equal{} 6^3 \plus{} (\minus{}5)^3$. You should have said, the sum of two [b]positive [/b] (integral) cubes.


Princeton college MIT Harvard CMU Duke Columbia

That is probably the most eclectic college list I've ever seen.

yeah if you happen to be a millionaire, it is.

have you looked at college life at any of these universities? cornell, for example, is an ag school. there are lots of cows. it's rural new york.

That's a pretty good list of math schools.

The crucial question in building a college application list is always, what is your "safety" school? I have considered the issue of what a safety school is from a parent's perspective.

My definition of a safety school is a school that

1) is pretty much certain to admit my child, based on its known behavior in acting on admission applications,

2) has a strong program in major(s) my child is interested in,

3) is affordable based on its known behavior in acting on financial aid applications,
and

4) is likeable to my child.

The state university in my state fits those characteristics for my oldest son. It pretty much admits people "by the numbers," and is not known to reject applicants who are successful in the accelerated secondary math program my son is now enrolled in there. Of course, we will consider and apply to other schools as well.

So, knowing your own test scores, high school grades, state of residence, etc., what's your safety school? That's the place to start in building your college list.

[quote="arp"]I'm going to be a senior this school year, and during the summer I've been trying to figure out where to apply.  I'm looking for the best schools for studying pure mathematics.  I live in PA and would somewhat prefer a school towards the east, although I will still consider and apply to some in the west.  So far I'm almost positive I'll apply to Princeton, MIT, Harvard, Harvey Mudd, University of Chicago, and CMU (because I did governor's school there and have a good chance of getting in).

However, I still have a fairly large list of other colleges I'm interested in that I would like to narrow down.  After visitng some and looking up information online, I still can't get a feel for which would be best for pure math studies and college in general.  The list right now is:
Duke
Johns Hopkins
Univeristy of Michigan
Columbia University
Cornell
Dartmouth (?? not really sure how good this is for math)
Berkeley
Stanford
Penn State Honors
Rutgers

I'd appreciate any insight into these colleges or suggestions and information about other colleges I should look into.[/quote]

First of all, I'd suggest applying to as few schools as possible.  I remember when I was applying, the application fee was usually around $\$$ 50-70.  Also, completing an application takes a lot of work.  You have to write a few essays and get teacher recommendations.  And I'd assume you have a decent amount of work from school to do at the same time.  So do fewer apps but do them well.  I'd suggest applying to a few elite schools, a few middle schools, and a few backups.  Don't just apply somewhere because someone said you'd like it there.  Do some research and find out if you really would like it there.  If there's a nontrivial chance you'll want to go there, then apply.

Out of those colleges you listed I've visited UMich, Duke, and Penn State.  They're all quite different.  Duke was the most beautiful, in my opinion, with UMich next.

Duke has a really good Putnam team and their math department puts on the Duke Math Meet for high schoolers (which is a very popular and relatively big event in the Carolinas).  So it seems that their math program should be good.  Sportswise, Duke of course has really good basketball usually.

UMich is huge in terms of number of people who go there.  Football is huge there.  If you go there, you will know about their football.  That said, they do have a honors program which is kinda isolated from the rest of campus which I find nice.  UMich has a very good math program as well.  There are basically no cars on campus for the students.  The only way you'll find a parking spot is if you live off campus at an apartment or something where they give you a spot.  The city's great for walking or using their bus system.  The campus is BIG so you'll definitely need to use a bus as some point (especially if you take any classes in the engineering department which is located on the other side of town).  One last thing to consider is that UMich probably gets the coldest weather out of all of the colleges you listed.  Since you're from PA this probably shouldn't bother you too much.  Even if it does, I was told that you don't have to be in the cold that much...only to get on the bus which picks students up right from their dorms.

I don't know much about Penn State's academics.  I'd rather live in Ann Arbor (UMich) than State College.  The dorms we stayed at for ARML weren't that great either.

Good luck!

To tokenadult, I'm not exactly sure which school will be my "safety" yet.  I know I will apply to Villanova, although it's not supposed to be really good for pure math, because my dad works there, I'll almost definitely get in, and I can go for free.  There are also some tuition exchange schools with Villanova that I can go to for free if I get in, so I'm looking into those.
         Right now I've just been trying to find the best schools that fit my interests and then see if any of those could be a safety school.  I guess this is the opposite course of action as you recommend though, so I might start trying to find safety schools first.  However, I think some of those I mentioned could be a safety because I have really high test scores, grades, etc.

joml, thanks for the information.  Yes, I'm trying to follow that advice about applying to as few colleges as possible.  In the end I want to apply to 10 or less, and preferably 8 or less.  Right now that list is just colleges that look good to me but that I want to narrow down a lot.  Anyway, your information about Duke and UMich was helpful and pretty much confirmed my ideas about the schools.  One other thing, you said to do research to find out which schools I'd like best...for all of these schools I've read about them and searched their site, but I'm having a hard time deciding if I'd really like to go there.  Besides reading biased CC forums on the schools and checking their math courses offered, how would you recommend getting a better idea of whether a school is good for me?

I don't have anything valuable to contribute, but as a Princeton student majoring in math (who loves it so far), I feel uncontrollably compelled to recommend Princeton. :D

You're just jealous because we have Andrew Wiles. :P

i would go with  Princeton, MIT, Harvard, rutgers and Umich..

i think rutgers and U mich are safety school for you.

leave berkeley or UChicago for grad studies.. (or one of harvard mit princeton )

[quote="joml88"]
I don't know much about Penn State's academics.
[/quote]

I can fill in some details.

Penn State is probably a really good safety school for you, arp, since you live in PA.  I have not yet even met a single student from PA who applied to Penn State and got rejected.  And, given the quality of colleges you are considering applying to, you will most likely get into the Schreyer Honors College, although this is somewhat more difficult.

Since Penn State accepts basically every PA resident, there is a VERY wide variety of students there.  It is an enormous campus - there are about 35000 undergraduate students at the main campus (University Park) alone!  But this means that you can also find alot of high-quality teachers, students, and classes there if you look, especially in the honors college.  I know several very intelligent students who went to Penn State Honors.  From what I've heard, there are quite a number of opportunities for undergraduate research there as well, if you are interested in that.

And the dorms really aren't that bad, in my opinion.  They're just all identical, sort of factory-like.   :D

And I agree, don't apply to too many schools.  I only applied to 4 (yes, with Penn State as my backup), and I'm completely happy with the results.

Thanks for the info 'hmmm'.  Yes I think it would be a good safety school and am definitely strongly considering it.

In case anyone has an impression otherwise, I'm really only planning on applying to 8 or less schools.  That huge list is just of ones that I'm interested, but I will surely not apply to all.

I also have some questions about math at NYU and Columbia that I could put here instead of starting a new thread.  So, I've heard that NYU is especially good for graduate school math and it looks that way from research I've done.  I suppose that their program for undergraduates is then reasonably good as well.  However, it seems like their programs are geared towards applied math.  Does anyone have any input as to the quality of a pure math education I could get at NYU?  Also, what about in comparison to Columbia?  I think I will apply to one of these schools because of their quality and appealing location.  I would be grateful for any advice.

FWIW - (personal views so take it for what it's worth -- No intention of putting any particular school down etc..)

Cal tech - Check this out too.

One place I would highly recommend you to check ou[b]t,  i[/b]s Duke. Apply to it and if they  are interested in you (you do not have to do anything special, just   use normal application process) they will try to recruit you. And they do try to attract good math students with the offer of prestigious scholarships. ( again  process is automatic .. any one who applies is also considered  for these) Duke does well in Putnam etc, and attracts many good math students who seem very happy with the choice.

On a personal note, I have two sons (and I was very involved when they were looking for colleges). One  graduated from Duke (Math/Physics dbl. major) and was very happy with his choice (His  other final  choices were MIT and Cal-tech). Other is in MIT (His other choices were Duke/Cal-tech and chose MIT).    Both kids (and their parents :) ) are very happy with their choices (and think that they will still make the same choice).

Also to put the things in perspective, honestly, any of the school you listed would not be a bad  (and in fact excellent) choice.

Best of luck.

I would be happy to share my perspective if you have specific questions (PM would be okay )

Thanks to everyone for the help I've received so far in my search.  I now have another college question which relates to this thread, so I might as well post it here.

My dad works at Villanova, and because he has for a certain amount of time I can go there for free.  Now Villanova is (according to what research I've done) not a school I would want to attend.  However, Villanova is part of The Tuition Exchange, which means I can go to another participating college on the money that would normally pay for Villanova.  The list of participating schools ( see http://www.tuitionexchange.org/partinst.cfm ) is pretty big and I do not really recognize any of the schools on it.  If you guys could look at the list a little and see if there are any schools that you know might interest me, I'd be thankful.

[quote="arp"] The list of participating schools ( see http://www.tuitionexchange.org/partinst.cfm ) is pretty big and I do not really recognize any of the schools on it.  If you guys could look at the list a little and see if there are any schools that you know might interest me, I'd be thankful.[/quote]

Let's see:  The schools that I suspect are better-than-or-equal-to Villanova for math majors would include Connecticut College, American U., GWU, Illinois Inst. of Tech., Rose-Hulman, Mt. Holyoke (all-female), Smith (all-female), Skidmore, Allegheny, Bucknell, F&M, and Lawrence.

As always, your mileage may vary at any school, and you can get a lot out of a school which doesn't rank well or have tons of resources---one of my classmates in grad school at U Michigan graduated from Villanova.

I have a friend who's going to Rose-Hulman. I think that's a fairly good school.

To add to above: I have visited Rose-Hulman too (and know friends who  graduated from there) .It has been ranked very high (Like #1 !) among the schools which do not offer PhD.

The emphasis  in  the faculty is on teaching. The place is middle of nowhere in Indiana. Similar in culture as Harvey Mudd.

Pure Engineering school though. (Have good but not the top rank  Math and Physics - some good professors who love to teach and are more accessible than bigger schools as there are no graduate students) (  but not at the top like other schools in your first list for math ). Certainly (IMO) a  serious choice, if you are interested in Engineering.


Find the inverse of A
A =
a b
c d

Show the steps using the identity matrix.

I believe you want to do it this way?
Let $A^{-1}=\left(\begin{matrix}w & x \\ y & z\\ \end{matrix}\right)$, then $AA^{-1}=I$ so $\left(\begin{matrix}a & b \\ c & d\\ \end{matrix}\right)\left(\begin{matrix}w & x \\ y & z\\ \end{matrix}\right)=\left(\begin{matrix}1 & 0 \\ 0 & 1\\ \end{matrix}\right)$ and thus $\left(\begin{matrix}aw+by & ax+bz \\ cw+dy & cx+dz \\ \end{matrix}\right)=\left(\begin{matrix}1 & 0 \\ 0 & 1\\ \end{matrix}\right)$. Since the components must be equal, now you can solve by hand the system $aw+by=1$, $ax+bz=0$, $cw+dy=0$, and $cx+dz=1$.


linear algebra
matrix
real analysis
induction
function
integration
factorial

Show that if $ a_n\equal{}a_{n\minus{}1}\minus{}\frac{2}{n}a_{n\minus{}2}$ for $ n\geq 2$ then $ \displaystyle\sum_{n\equal{}0}^{\infty}|a_n|<\infty$.

Rather than inventing some ad hoc method for this particular question, let me try to outline a general approach to such problems that helps more often than not :).

1) If you have a linear recursion for the terms of the series, convert it to one matrix recursion. In our particular case, put $ X_n\equal{}\left[\begin{matrix} a_n \\ a_{n\minus{}1}\end{matrix}\right]$ and note that the recursion can be rewritten as $ X_n\equal{}A(1/n)X_{n\minus{}1}$ where $ A(t)\equal{}B\plus{}tC$, $ B\equal{}\left[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}\right]$, $ C\equal{}\left[\begin{matrix} 0 & \minus{}2\\ 0 & 0\end{matrix}\right]$

2) Note that the statement of the problem is implied by (actually, equivalent to) the statement $ \sum_n \|P_n\|<\plus{}\infty$ where $ P_n\equal{}A(1/n)A(1/(n\minus{}1))\dots A(1/2)$.

3) If $ A(t)$ were numbers, we would be able to say that the norm of the product is the product of the norms and finish the story by a simple computation. Unfortunately, with matrices everything is a bit more complicated. It is true, of course, that the norm of the product does not exceed the product of the norms but, since the norm of each $ A(t)$ is at least $ 1$, it doesn't seem to help much.

4) If $ A$ were a fixed matrix not changing with $ n$, the convergence of the series $ \sum_n \|P_n\|$ would be determined by the spectral radius of $ A$ rather than by its norm. Indeed, in most cases, we would be able to write $ A\equal{}E^{\minus{}1}DE$ where $ D$ is a diagonal matrix whose entries are the eigenvalues of $ A$ and $ E$ is some invertible matrix (the transition matrix from the standard basis to the basis of eigenvectors). Then $ P_n\equal{}A^n\equal{}E^{\minus{}1}D^nE$ and the behaviour of $ D^n$ is not too hard to discern.

5) In our case $ A$ changes with $ n$ but quite slowly and in a really nice manner. So, we can also write $ A(t)\equal{}E(t)^{\minus{}1}D(t)E(t)$. $ D(t)$ is not hard to evaluate: the eigenvalues of $ A(t)$ are $ 2t\plus{}O(t^2)$ and $ 1\minus{}2t\plus{}O(t^2)$ for small $ t$ (here you have to do a direct computation), so $ \|D(t)\|\le 1\minus{}2t\plus{}O(t^2)$. If there were no $ E$'s, we would immediately get $ \|P_n\|\approx \frac 1{n^2}$ from here and that would be the end. Unfortunately, we have a couple of $ E$'s between every 2 $ D$'s.

6) It would be a big headache to compute $ E$'s explicitly (especially when the recurrence relation is really long) but, fortunately, we usually do not need to. It suffices to notice that $ E(t)$ is analytic in $ t$ near $ t\equal{}0$ (this is true whenever $ A(t)$ is analytic in $ t$ and $ A(0)$ has distinct eigenvalues). Analiticity is more than enough to conclude that $ E(s)E(t)^{\minus{}1}\equal{}I\plus{}O(|t\minus{}s|)$ as $ s,t\to 0$, so $ \|E(1/n)E(1/(n\minus{}1))^{\minus{}1}\|\le 1\plus{}O(n^{\minus{}2})$.

7) The rest should be clear.

That was pretty cool, fedja.  Here is my ad-hoc method.

We know that
\[ a_n \equal{} a_{n\minus{}1} \minus{} \frac{2}{n} a_{n\minus{}2}.
\]
for $ n \ge 2$.  Replacing $ n$ with $ n \minus{} 1$, we get
\[ a_{n\minus{}1} \equal{} a_{n\minus{}2} \minus{} \frac{2}{n\minus{}1} a_{n\minus{}3}
\]
for $ n \ge 3$.  Eliminating $ a_{n\minus{}2}$, we get
\[ a_n \equal{} \left(1 \minus{} \frac{2}{n} \right) a_{n\minus{}1} \minus{} \frac{4}{n(n\minus{}1)} a_{n\minus{}3}
\]
for $ n \ge 3$.  Multiplying both sides by $ n(n\minus{}1)$, we get
\[ n(n\minus{}1)a_n \equal{} (n\minus{}1)(n\minus{}2) a_{n\minus{}1} \minus{} 4 a_{n\minus{}3}
\]
for $ n \ge 3$.  Let $ b_n \equal{} n(n\minus{}1) a_n$.  Then the equation becomes
\[ b_n \equal{} b_{n\minus{}1} \minus{} \frac{4}{(n\minus{}3)(n\minus{}4)} b_{n\minus{}3}
\]
for $ n \ge 3$.  From here, it's pretty simple to show that $ b_n$ is bounded.  Thus $ |a_n|$ is $ O(\frac{1}{n^2})$.

Fedja, reading your post I was wondering if there is a theory behind all these stuffs?

Not that I know of any but there are many problems about the long products of matrices that arise here and there and what I did is one of the standard tricks in treating them. If you want to entertain yourself with something more difficult, here is one more question about long matrix products.

Assume that $ A_k$ is a fixed sequence of real $ 2\times 2$ matrices of determinant $ 1$. Let $ H(t)\equal{}\left[\begin{matrix}1 & t\\0 &1\end{matrix}\right]$. Let $ P_n(t)\equal{}A_nH(t)A_{n\minus{}1}H(t)\dots A_1H(t)$. Let $ B\equal{}\{t\in\mathbb R: \sup_n\|P_n(t)\|<\plus{}\infty\}$. Prove that
a) $ B$ has finite Lebesgue measure
b) $ B$ may contain any given countable set
c) $ B$ may be essentially unbounded (i.e., the measure of intersection of $ B$ with every ray may be positive)

[quote="Ravi B"]
\[ b_n \equal{} b_{n \minus{} 1} \minus{} \frac {4}{(n \minus{} 3)(n \minus{} 4)} b_{n \minus{} 3}
\]
for $ n \ge 3$.  From here, it's pretty simple to show that $ b_n$ is bounded.  .[/quote]
Hello,Can kindly show me how to prove $ b_n$ is bounded :?:

By induction, we can show that
\[ |b_n| \le |b_4| \prod_{i \equal{} 5}^n \left( 1 \plus{} \frac {4}{(i \minus{} 3)(i \minus{} 4)} \right).
\]
The product converges as $ n$ approaches infinity.  (In the text that you quoted, I should have said $ n \ge 5$, not $ n \ge 3$.)

Thank you. :)
But the $ |b_4|$ maybe replace by $ \max \{|b_1|,|b_2|,|b_3|,|b_4|\}$

Yes, thanks.

By the way, I think it may be possible to get a closed-form formula for $ a_n$ using generating functions.  The calculations look messy.

Fedja,Ravi B,Zhaobin,Alekk a lot of thanks.Absolutelyinteresting works!!! :roll:

The exact approach with generating functions:
Let $ f(x)\equal{}\sum a_nx^n$. Rewrite the recurrence as $ na_1\minus{}na_{n\minus{}1}\plus{}2a_{n\minus{}2}\equal{}0$. Multiply by $ x^{n\minus{}1}$ and sum over $ n$; we get $ f'(x)\minus{}(xf(x))'\plus{}2xf(x)\equal{}c$.
This differential equation simplifies to $ (1\minus{}x)f'(x)\minus{}(1\minus{}2x)f(x)\equal{}c$, or $ f'(x)\equal{}\frac{1\minus{}2x}{1\minus{}x}f'(x)\plus{}\frac{c}{1\minus{}x}$.

The general solution to this equation is
$ f(x)\equal{}(1\minus{}x)e^{2x}\left(a\plus{}\int \frac{ce^{\minus{}2x}}{(1\minus{}x)^2}\,dx\right)$. If $ c\neq 0$, that's not an elementary function. In the $ c\equal{}0$ case, we get the explicit form $ a_n\equal{}a\left(\frac{2^n}{n!}\minus{}\frac{2^{n\minus{}1}}{(n\minus{}1)!}\right) \equal{}\frac{(2\minus{}n)2^{n\minus{}1}a}{n!}.$

While we can express the coefficients of the $ c\neq 0$ part as finite sums, I don't see how to get a closed form. The terms without factorials in the denominator are order $ \frac1{n^2}$ as expected.

Okay, my suggestion of a closed form was an exaggeration.  But continuing your work, I think we get the following formula for $ a_n$ as a sum (for $ n \ge 3$):
\[ a_n \equal{} \frac{2\minus{}n}{n!}2^{n\minus{}2}(3 a_0 \minus{} a_1)
      \plus{} \frac{2\minus{}n}{n!}2^{n\plus{}1} (a_1 \minus{} a_0) \sum_{k\equal{}4}^n \frac{(k\minus{}4)!(k\minus{}1)}{2^k}.
\]
That's enough to give us the asymptotic behavior of $ a_n$:
\[ \lim_{n \to \infty} n^2 a_n \equal{} 2(a_0 \minus{} a_1).
\]
I hope I didn't make any errors.


does anyone have answer keys for 1997-98 nats, or the 2007 nats?

also, what was the 2007 master's round?

well in 2007 the setting is this:
the MATHCOUNTS mail service wants to place mail in some mailboxes (in a rectangular array) such that no 2 adjacent boxes are empty at the same time
*please correct if wrong*
so first it says how many ways are there to put 8 pieces of mail in an 8 by 2 array of mailboxes
is it possible for there to be less? why?
then it says how many ways are there to put 9 pieces in the same array

now we move on to a 3 by 5 array
what is the minimum # of pieses of mail possible
i think it is how many ways are there to put 8 pieces of mail
then it asks how many ways are there to put 9 pieces of mail
and now i think thats it
...

SO WOULDNT THE IDEA BE TO MAKE A CHECKERBOARD PATTERN....
SO FOR THE FIRST PROBLEM WITH THE 8X2 WITH 8 MAIL THINGS IT SHOULD BE 2....
WITH 9 MAIL THINGS THE ANSWER IS 16...
AND SO ON... RIGHT?

Please less caps :(.

Yes that is correct - master's round was a nightmare... I did 4(n-2)+4=4n right in front of the judges...they just nodded and looked at me awkwardly.   :)

To give you perspective, I don't think anybody seriously made any progress on the 3x5 case?  (Or maybe Kevin did, but of me and the two people I watched, not really).

The Masters round is always interesting to watch

No matter how well you did, it's a great accomplishment just to be in it!!

Our team predicted that Kevin Chen would win : )

The previous year, the team and our coach disagreed, and our coach was right o.O

We'll have to see this yr my last yr :(

No countdown decides all lol. Masters round = extra.

But I swear, Justin made more progress than all of you, except he doesn't have the public speaking skills.  He caught something really important that I don't really remember =P.  And of course, if kevin won countdown, there must be a little bias lol xD xD...

[quote="mathking123"]umm... srry for the caps
umm....isnt that a relatively easy masters round FOR NATS?
ehhh, also for the 3by 5, the fewest number of mail things is uhh...7?
also, does masters affect final ranking?[/quote]

It's certainly non-trivial :).  (I don't think they actually asked the fewest number of mail pieces - first, an answer isn't enough, and secondly, yeah, that would be the easy part...)

[quote="kyyuanmathcount"]I did 4(n-2)+4=4n right in front of the judges...they just nodded and looked at me awkwardly.[/quote]

Haha, I remember I only watched the Masters Round to see you beat everyone (ok, I admit it, I just wanted to look at your face), so when I was leaving, I was resisting the urge to tell you immediately after your presentation. So I think I told you later.

Yeah, I'm fairly certain that Kevin was the only one to make any progress on the 3 by 5 case.


conics
hyperbola
geometry
rectangle
analytic geometry
perpendicular bisector
similar triangles

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14938[/img]

 :(

Line $ V_1V_2$ is the major axis, perpendicular bisector $ n$ of $ V_1V_2$ is the minor axis. $ C$ is the midpoint of $ V_1V_2$ (hyperbola center), $ CV_1 \equal{} CV_2$ is the semimajor axis length.  Let the lines $ PV_1, PV_2$ cut the minor axis $ n$ at $ Q_1, Q_2.$ Circle with diameter $ Q_1Q_2$ cuts the line $ V_1V_2$ at $ R_1, R_2,$ $ CR_1^2 \equal{} CR_2^2 \equal{} CQ_1 \cdot CQ_2.$ Circle with center $ C$ and radius $ CR_1 \equal{} CR_2$ cuts $ n$ at $ Y_1, Y_2.$ $ CY_1 \equal{} CY_2$ is the semiminor axis length. The focal distance is $ CF_1 \equal{} CF_2 \equal{} V_1Y_1 \equal{} V_2Y_2.$ Complete the right $ \triangle CV_1Y_1$ to a rectangle $ CV_1DY_1.$ Circle with center $ C$ and radius $ CD$ cuts the line $ V_1V_2$ at the foci $ F_1, F_2.$

[hide="Proof"] $ N$ is foot of a normal from $ P$ to $ V_1V_2.$ $ a \equal{} CV_1 \equal{} CV_2, b \equal{} CY_1 \equal{} CY_2$ are the hyperbola semimajor / semiminor axis lengths. Taking the coordinate origin $ C$ and the major / minor axes as the coordinate axes, the hyperbola equation is

$ \frac{x^2}{a^2} \minus{} \frac{y^2}{b^2} \equal{} 1\ \Longrightarrow\ \frac{b^2}{a^2} \equal{} \frac{y^2}{x^2 \minus{} a^2} \equal{} \frac{y^2}{(x \plus{} a)(x \minus{} a)}.$

Translating this into line segment equation,

$ \frac{CR_1^2}{CV_1^2} \equal{}\frac{CY_1^2}{CV_1^2} \equal{} \frac{PN^2}{NV_1 \cdot NV_2}.$

From similar triangles $ \triangle PV_1N \sim \triangle Q_1V_1C,$ $ \triangle PV_2N \sim \triangle Q_2V_2C,$ we have

$ \frac{PN}{NV_1} \equal{} \frac{Q_1C}{CV_1},\ \ \frac{PN}{NV_2} \equal{} \frac{Q_2V_2}{CV_2}\ \Longrightarrow$

$ \frac{PN^2}{NV_1 \cdot NV_2} \equal{} \frac{Q_1C \cdot Q_2C}{CV_1^2}\ \Longrightarrow\ Q_1C \cdot Q_2C \equal{} CR_1^2 \equal{} CY_1^2.$[/hide]


Find all prime numbers of the form 100····001, where the total number of zeros between the first and last digits is even.

[hide="solution"]The only prime of this form is $11$.

Numbers of this form can be expressed as $10^{2k+1}+1$ for some nonnegative integer $k$. Examine this expression in mod 11. $(-1)^{2k+1}+1 \mod{11}$. Because $2k+1$ is odd, $(-1)^{2k+1}=-1$, and the entire expression is congruent to 0 mod 11. Therefore the number is divisible by 11 and is not prime, with the exception of the only prime divisible by 11.[/hide]

[hide]Use the divisibility rule for 11.
$100\cdots 001=10^{2k+1}+1\equiv (-1)^{2k+1}+1\equiv 0 \pmod{11}$, for all integer k. Thus all numbers of said form are divisible by 11, so the only prime is $\boxed{11}$.[/hide]


These are just some simple math and logic problems see if you can figure them out! :wacko:

1.If farmer Brown has 45 :rt3: balles of hay and farmer ted has 20 :rt2: balles of hay how many balles of hay do they have together?(express your answer in the simplest radical form)

2.If Jhon wants to dig a hole 3 feet deep, and 3 feet wide he will need to go to the hardware store and by a shovel and a yard stick then he finds out that he will need to use water on the side of the hole to keep the dirt from falling in. Now how many cubic feet of dirt are in the hole when he is done?

3. If a cow ways 75kg on 3 legs how much does it way on all 4?

Answer :downto:
[hide]1.$\sqrt 1$ duh if you stack hay balles its just one big one!
2.NONE it's a hole!
3.Still 75 it doesn't way ant more on one more leg!
If you got all of these right, good job you can think logicly! :first:  :cool:  And if you missed all of them ... then you need to think more before you answer! :blush:  :blush:   :rotfl: [/hide]

pretty good problems except for number one

[quote="tarquin"][hide="1"]$\sqrt 1$[/hide]
That's simplest radical right?[/quote]

Well I guess so I have to change it . Thanks for the tip :D

[hide="Number 3"]
75 plus the weight of one leg, assuming "standing on three legs" means with one leg chopped off.
[/hide]

wouldn't $\sqrt{1}$ be 1, or does it have to be a radical?

I wouldn't call this simple.....

and for the best, change the forum of this topic..

[hide]
1.  :sqrt: 1
2.  none, it's a hole
3.  75Kg


If I remember right this problem was posted in this forum  ( not in this face, ) and was solved/


What is your favorite round (or round that you think helps you the most)?

That's better with BB code

[color=blue]Moderator kyyuanmathcount:  Alright, added Official Countdown.[/color]

target, because you have 3 minutes per problem :w00tb:
And also stupid mistakes are rarer.

eh...not for me, without target, I probably would have gotten first in chapter and state (obviously, I didn't)

I also like the AMC tests because if I mess up, chances are, it won't be one of the multiple choices

Countdown is my favorite :D. Seriously that should be one of the options. Out of the two up there though, I definitely prefer sprint. In target the problems are much harder. Also it's easier to make stupid mistakes. For example. Will edit after discussion is allowed :lol:
EDIT: One number something on the target I didn't see the word percent until a minute before. I had to work my but off for the next minute to solve it  :blush: .

Sprint.

I'm more fast than thorough and my calculator only has the basics, I know some people who have programs that can calculate a whole problem for them and that have nifty functions that makes stuff much easier.

[quote="ehehheehee"]I will put "official countdown" up there if there is a way to edit the poll...[/quote]  :idea: Ask a mod to do it.

Countdown is my worst round, I make the stupidest mistakes including ones where I have the answer written down, but I get nervous and say it wrong or something like that.

Target is my favorite. I don't usually make many mistakes on target, while I make tons on Sprint.

[quote="ehehheehee"]a mod did it...[/quote]
Aww, i would've voted for it.
But i voted before he added the poll -_-

Countdown definitely.  If VA's countdown was official, I would have made nationals!!!!!!  :(

My favorite is target because I don't usually make mistakes, while I make lots on sprint.

Target. It's less stressful, less stupid mistakes, and since you can use calculators I can play calculator games between each pair of problems  :lol:
Each problem is also weighted more...so yeah.

I like Target better... more time on each problem.

Sprint completely

I always get high scores in the sprint, the lowest spring score I got in the two years I've done it in the seventh and eight grade was 27.  :lol:

The target I get around 7s and 8s, so that is my least favorite.  :mad:

Countdown I also like, but this year I got owned in countdown but I usually own other people.  :(

[quote="Xantos C. Guin"]Target:
I own on targets.
At nationals I got 6. My teammates got 1, 2, and 3.

If only the target round was used for tiebreakers. Then I might have made countdown at nationals last year.[/quote]

What state are you from again?
What place did you get at nats.

My favorite round by far is the Team Round!   :rotfl:  I don't see how you forgot it!    :rotfl:

Oops--I voted Sprint for some reason. That's the round where I do the worst, because I make too many stupid mistakes. Target is my favorite round.

[quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Xantos C. Guin"]State: GA
Place at 2005 nationals: 16 (Tied #12)
I got a 18/6 for a score of 30.[/quote]

You made it to nationals this year right?[/quote]

He's in high school this year I believe.[/quote]

Yes, I should chang my sig now.

EDIT: sig changed.

Target -- I'm AWESOME at it in practice. I don't miss any (exaggeration, but you get the jist of it)

But it's SCREWED ME OVER the last two years at state. Both years it's saved me at chapter. %!@%@#.

But I love countdown and it's the ultimate key to success so....

[quote="ehehheehee"]I have enough speed, but I have stupid mistakes...so I picture running into a wall at high speeds[/quote]

Me too.
Last year at the GA countdown round (All 10 participants at once) I buzzed in on about 30 problems and got 15 right. Since there is no guessing penalty, I got   :winner_first:

Target- more time for each problem and easier to make up ground on opponents, and it's easier to find and fix mistakes.

I  :love: Target (the round, not the store) because I got a perfect on it, but only a 20 on Sprint.

Sprint is still winning even with the renewed support for target =)


Find all the polynomial $P$ which has roots on $R$  :

$P(x) P(-x) =P(x^{2}-1)$

It posted before.
Idea: if x is root P(x), then -x and $x^{2}-1$ are roots.

Set $g(x)=x^{2}-1$ and call a number periodic under $g$, if $g^{n}(x)=g(g(\cdots(x)\cdots))=x$ for some $n\geq 1$.
First I'll show, that $-1,0$ and $\frac{1\pm \sqrt{5}}{2}$ are the only under $g$ periodic real numbers.
Set $\lambda=\frac{1+\sqrt{5}}{2}>0$ and $\mu=\frac{1-\sqrt{5}}{2}\in ]-1,0[$.
Then $g(-1)=0,g(0)=-1,g(\lambda)=\lambda$ and $g(\mu)=\mu$. So $-1,0,\lambda$ and $\mu$ are periodic under $g$.
Using the factorizations
$g(x)=(x-1)(x+1)$,  $g(x)-x=(x-\mu )(x-\lambda )$,  $g^{2}(x)-x=x(x+1)(x-\mu )(x-\lambda )$
and $g(x)=x^{2}-1\geq-1$ and $g(x)=\mu+(x-\mu)(x+\mu)$
it's easy to check:
(1) $g(x)>x$ for all $x>\lambda$
(2) $g(x)<x$ for all $0<x<\lambda$
(3) $g(]-1,\mu [)=]\mu ,0[$ and $g(]\mu ,0[)=]-1,\mu [$
(4) $g^{2}(x)>x$ for all $\mu <x<0$

Now if there is an under $g$ periodic number $x\in\mathbb{R}\setminus \{-1,0,\mu ,\lambda \}$ with $g^{n}(x)=x,n\geq 1$. Then $g^{m}(x)=g(g^{m-1}(x))=(g^{m-1}(x))^{2}-1\geq-1$ for all $m\geq 1$ and $x=g^{n}(x)\geq-1$.
If $x>\lambda$, then using (1) we get $x<g(x)<g^{2}(x)<\cdots$. Contradiction!
So $x<\lambda$ and if $g^{m}(x)>0$ for all $m\geq 0$, then (2) gives $x>g(x)>g^{2}(x)>\cdots$. Contradiction!
So $g^{m}(x)\in [-1,0]$ for some $m$, and then using (3) $x=g^{kn}(x)\in [-1,0]$ for $k$ large enough.
If $x\in ]\mu ,0[$ then using (3) and (4) we get $g^{2m}(x)\in ]\mu ,0[$ for all $m\geq 0$ and $x<g^{2}(x)<g^{4}(x)<\cdots$
and if $x\in ]-1,\mu [$ then analog we have $g^{2m+1}(x)\in ]\mu ,0[$ for all $m\geq 0$ and $g(x)<g^{3}(x)<\cdots$.
Thus $x$ can't be periodic under $g$ and $-1,0$ and $\frac{1\pm \sqrt{5}}{2}$ are the only under $g$ periodic real numbers.

Now set $P_{1}(x)=x(x+1)$, $P_{2}(x)=x-\lambda$ and $P_{3}(x)=x-\mu$. Then using $\lambda^{2}=\lambda+1$ and $\mu^{2}=\mu+1$ we get
$P_{1}(x)P_{1}(-x)=-x^{2}(x+1)(-x+1)=x^{2}(x^{2}-1)=P_{1}(x^{2}-1)$ and
$P_{2}(x)P_{2}(-x)=(x-\lambda)(-x+\lambda)=-(x^{2}-\lambda^{2})=-(x^{2}-1-\lambda)=-P_{2}(x^{2}-1)$
and analog $P_{3}(x)P_{3}(-x)=-P_{3}(x^{2}-1)$.
So the polynomials $P_{1},P_{2}$ and $P_{3}$ satisfy
(*) $P(x)P(-x)=\pm P(x^{2}-1)=\pm P(g(x))$
and the product of solutions of (*) is again a solution of (*).
Now assume there is a solution $P$ of (*) with only real roots other then $P_{1}^\alpha P_{2}^\beta P_{3}^\gamma$, $\alpha,\beta,\gamma\geq 0$
and take $P$ with minimal degree.
If $P=c$ constant, then using (*) we get $c^{2}=c$ and $c\in \{0,1\}$. But $0$ has nonreal roots and $P\not =1$ was assumed.
So $P$ is not constant and has some root $x$, which is real.
Then using (*) we see that $g^{m}(x)$ is a root of $P$ for all $m\geq 0$.
But $P$ has only finitely many roots. Therefore $g^{m}(x)=g^{n}(x)$ for some $0\leq m<n$.
But then $y=g^{m}(x)=g^{n-m}(g^{m}(x))$ is an under $g$ periodic root of $P$ and using above we must
have $y\in\{-1,0,\lambda ,\mu \}$. But $g(0)=-1$ and $g(-1)=0$, so $P(0)=0\Leftrightarrow P(-1)=0$.
And so $P_{i}|P$ for some $i=1,2$ or $3$, say $P=P_{i}Q$, and since $P$ and $P_{i}$ solve (*), so must $Q$
and it has also only real roots. But $Q$ then has smaller degree than $P$.
So it must have the form $P_{1}^\alpha P_{2}^\beta P_{3}^\gamma$ and so does $P$.

But now $P=P_{1}^\alpha P_{2}^\beta P_{3}^\gamma$ satifies $P(x)P(-x)=(-1)^{\beta+\gamma }P(x^{2}-1)$.
So the solutions of the original problem are the polynomials
$x^\alpha (x+1)^\alpha (x-\frac{1+\sqrt{5}}{2})^\beta (x-\frac{1-\sqrt{5}}{2})^\gamma$ with $\alpha,\beta,\gamma\geq 0$ and $\beta,\gamma$ both odd or even.


Rectangle ABCD is folded in half to form rectangle AEFD,which is then folded in half to form rectangle AGHD. The perimeter of rectangle AGHD is 23 cm and the perimeter of ABCD is 47 cm. What is the area of rectagle ABCD?

[hide]I remember this from 2005 state. Just let the sides of the small one be x and y and you get systems of equations. I think the answer was 120.[/hide]

[hide]this is more of an intuitive method, but that is usually how I do most of the questions. It lose 24 from two folds, so 4x-x=24 or 3x=24 so the length of the smaller square is 8/2 or 4. From this you can see that the width is 7.5. So quadruple the length to get back to the original shape 4*4=16. To find the area, do length times width.

16*7.5=120 as xpmath said.[/hide]


This is the last puzzle in the list provided by the Snark site.

Replace each letter by a digit (same letter, same digit; distict letter, distinct digit) such that the equality holds

SEND+MORE=MONEY

[quote="probability1.01"]Does the Snark mailing list, by any chance, cost money? :D[/quote]
No, it does not. It is a group of people who like a lot Puzzles and Brainteasers. You can easely subscribe, but it is totally handled in Spanish (It was originally part of the Culturan and Scientific Mailing list of the Universidad de Buenos Aires), but you may give it a try if you fairly understand Spanish!


[quote="AndrewTom"]You have two watering cans. One holds five litres and the other three. How can you measure exactly one litre?[/quote]
[hide]We are solving a diophantine equation $ 5x\minus{}3y\equal{}1$. Taking the equation in mod 3 gives $ 2x\equiv 1\bmod{3}$ or $ x\equiv 2\bmod{3}$. Therefore $ x\equal{}2\plus{}3j$ for all integers $ j$ that are greater or equal to 0. Substituting this back into the original equation results in $ 10\plus{}15j\minus{}3y\equal{}1$ or $ y\equal{}3\plus{}5j$. Thus we have infinite possibilities to measure exactly one liter. The first is when $ j\equal{}0$ or measuring the 5 liters twice and the 3 liters three times and subtracting. [/hide]

Thanks, I meant to post this in the 9-13 forum, where I thought I'd get a response like: fill the three litre can, pour it into the five litre can; fill the three litre can again, use it to fill the five litre can, leaving one litre in the three litre can.


geometry
geometric transformation
rotation
reflection
group theory
superior algebra
superior algebra solved

Does anyone know a [b]nice[/b] (purely algebraic) way to prove that \[ Z\left(\mathcal{D}_{2n - 1}\right) = \{e\},\ Z\left(\mathcal{D}_{2n}\right) = \left\{e,a^n\right\}?  \] I hope the notation I am using is "clear". I guess it's fairly standard. Ask if there is any problem :)

Well, we usually denote the dihedral group with 2(2n-1) elements as $D_{4n-2}$, and so $D_{2n-1}$ would make no sense in our notation... But I've seen both here on the site.

Anyway, isn't this geometrically trivial? The only nontrivial rotation which commutes with all reflections is the rotation over $\pi$, thus for an even number of points, that element is in (your $a^n$) and for an odd number of points, that element is not in.

Purely grouptheoretical: the simplest way would look to me to translate the geometrical idea to group theory... I see no direct other way. So then we would have to show that for all $a^m$ we have $a^m b a^m b=e$, and by commutativity (center) that means that $a^{2m}=e$, which is only possible for trivial rotation and $\pi$.


A class collects $ \$50$ to buy flowers for a classmate who is in the hospital. Roses cost $ \$3$ each, and carnations cost $ \$2$ each. No other flowers are to be used. How many different bouquets could be purchased for exactly $ \$50$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 7 \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 17$

[hide="Solution"]
Let r be the number of roses and c be the number of carnations.
3r+2c = 50

One bouquet will contain 25 carnations and 0 roses.

For every 3 carnations we take away, we can add two roses.

So our possibilities are (c,r) = (25, 0); (22, 2); (19, 4); ... (1, 16)

We can count how many possibilities there are based on the number of roses:
The series 0, 2, ... 16 has the same length as 0, 1, ... 8, and thus as 1, 2, ... 9, hence there are 9 possible bouquets, and the answer is [b](C)[/b].[/hide]

[hide="a bad way"]just find the number of solutions to $ 3x \plus{} 2y \equal{} 50$ and they are,

(16,1)
(14,4)
(12,7)
(10,10)
(8,13)
(6,16)
(4,19)
(2,22)
(0,25)

so the answer is $ 9$ $ \fbox{\mbox{C}}$[/hide]


search algebra polynomial calculus integration function domain

Are there solutions in integers to $ x^3\plus{}y^4 \equal{} 19^{19}$?

so my next question is what motivated mod 13. I mean I miserably plodded through mods one after another till 9 and gave up. Is there a way to narrow down the search for an appropriate modulus?

[b]Lemma:[/b]  There are $ \frac {p \minus{} 1}{\gcd(k, p \minus{} 1)}$ nonzero $ k^{th}$ powers $ \bmod p$.

Proof:  the polynomial $ x^{p \minus{} 1} \minus{} 1 \bmod p$ has as roots all the nonzero residues $ \bmod p$.  Hence the polynomial $ x^{\frac {p \minus{} 1}{\gcd(k, p \minus{} 1)} } \minus{} 1$ has as its roots all the $ k^{th}$ powers.

Hence there are $ 4$ third powers and $ 3$ fourth powers, which narrows down the possible values of the LHS $ \bmod 13$.  So if a problem involves second and third powers, you might want $ \bmod 7$ or $ \bmod 9$ - that kind of thing.  The above result is harder to generalize for composite moduli (you run into problems counting the $ k^{th}$ powers not relatively prime to the modulus) but, for example, the only squares $ \bmod 8$ are $ 0, 1, 4$, which can come in handy.

There are no integral solution.

Proof:
Let  $ f(x)\equal{}\sqrt[4]{19^{19} \minus{} x^3}$. Then, $ f(x)$ is defined for all $ x\in(\minus{}\infty,\, \sqrt[3]{19^{19}}]$.  It is easy to see that $ f(x)$ is decreasing throughout its domain. So it takes the minimum value at the end point, i.e. at the point $ x_0\equal{}\sqrt[3]{19^{19}}$, and $ f(x_0)\equal{}0$. Accordingly, the minimum value of the function is $ 0$ while at all the interior points of its domain, it remains positive.
That also means that the only root of $ f(x)$ is at $ x_0$. But $ x_0$ is not an integer, and at the integer just less than $ x_0$, $ f(x)>0$. Thus, $ f(x)$ has no integral solution and hence we conclude that the given equation has no integral solution.

"$ f(x)$ has no integer zero" is a very different statement from "$ f(x)$ takes no integer values."  Your argument works without modification if we replace $ 19^{19}$ by any other constant, such as $ 5^3 \plus{} 6^4$, where an integer solution clearly exists.

Ooops, terrible me!! yeah you are right.. I acknowledge my mistake.... I got a bit carried away :oops:


hola  amigos  aqui les va uno
x-1=1,x-2=2,x-3=3,x-4=4
ademas 4x-n=x-n-1 +x-n-2 +x-n-3 +x-n-4  n>=5
píden calcular el  limite de x-n

animense a resolverlo[/quote][/list]

amigos disculpen no se nota bien (  el problema  es de sucesiones )
espero que les guste
 no recuerdo la fuente

Bueno, .. una re-edicion de la pregunta seria bueno:

Sea una secuencia  $x_n$, tal que:
$x_1 = 1$, $x_2=2$, $x_3=3$ y $x_4=4$.
Ademas, $4x_n = x_{n-1} + x_{n-2} + x_{n-3} + x_{n-4}$, para todo $5 \leq n$.

Hallar  $\lim x_n$.

Carlos Bravo
Lima- Peru

He colgado esta pregunta a la seccion advanced en Algebra ....

este es el [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55135]Link[/url].

Y he visto que ya una chica de New Zealand  ha presentado su solucion,
como tambien, se ha recibido un link de parte de DARIJ de algun post anterior relacionado al tema...

espero que la gente interesada lo revise ...

Carlos Bravo
LIMA-PERU

Xq me parece el prob familiar.....
a si..ya me pregunto :rotfl:
chrv !!

:?  bueno aca les va una:::
Sea P(x) un polinomio con coeficientes enteros y k un natural, tales que k no divide a P(i); para todo i = 1, 2, 3, 4, 5, ..., k-1, k
Probar que P(x) no posee ninguna raiz entrea


I can't get into the classroom when there's no class going on. Is it supposed to be this way?

I'm pretty sure its suppose to be like that, but why would you want to go in when there is no class?

Only WOOTers can get in 24/7.  Others have designated open classroom times (aside from class time).  These are outlined in the forums of each individual class.


Would it be possible to add a feature so that you could see what you voted for in polls?  I don't remember what I voted for in some of the older polls and was wondering...

Not that its a big deal or anything, but just a suggestion if Valentin has any spare time.

hmm and possibly present an option where multiple choices can be selected? some questions require you to choose between multiple similar answers.


Hi everyone i love math and i really wanna do good in class. Can some one teach me how i get the percent?
What i need to know-
What is 21% of 53

How do i find that out? Any kinda problem liek that. THANK YOU.

Well, percent is over 100.

So take $\frac{21}{100} * 53$ or whatever number in place of 53. Remember to put percent over 100 and multiply by number you want that percent of.

remember: of means multiply

so 21% multiplied by 53 = .21 * 53

to find a percent, say 21, move divide it by 100 or put a decimal in front of the 2

123%=1.23, 1234%=12.34, 12345%=123.45 and so on

hope i didnt confuse you.

i think i will also move this

[hide]23% of 50 means .23 8 50 which is 11.5[/hide]

21% = .21 of = * .21 * 53= 11.13
( I put this in LaTeX but i wouldnt work so I just took the latex off..)

Alight guys, thank you very much. I am very glad about the amout of answers i got and i really appreciate it. THANK YOU!

Any of you casio owners please help me out here

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=328204#p328204


Let $ a,b,c > 0$ and $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \equal{} 10$. Prove that $ \sqrt {\frac {b}{a}} \plus{} \sqrt {\frac {c}{b}} \plus{} \sqrt {\frac {a}{c}} \le 10$

your problem is totally wrong, at first $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$ can not be equal as $ 1$ because by AM-GM
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3$ and this one  $ \sqrt{\frac{b}{a}}\plus{}\sqrt{\frac{c}{b}}\plus{}\sqrt{\frac{a}{c}}\le 1$ is also wrong, because by AM-GM $ \sqrt{\frac{b}{a}}\plus{}\sqrt{\frac{c}{b}}\plus{}\sqrt{\frac{a}{c}}\geqslant 3$

obviously true

$ \frac{a}{b} \geq\ \sqrt {\frac{a}{b}} \leftrightarrow\ \frac{b}{c} \geq\ \sqrt {\frac{b}{c}}\leftrightarrow\ \frac{c}{a} \geq\ \sqrt {\frac{c}{a}}$

adding them them result follows

your proof solves only the case when a>=b>=c>=a, or the trivial equality case a=b=c.

Proof:
\[ \sum x^6 \geq xyz\sum x^3\] by Muirhead, so let x^6=a/b, etc cyclically to yield
\[ \sum\frac{a}{b}\geq\sum\sqrt{\frac{a}{b}}\]

i am extremely sorry for my stupid post :(  :o
i was sleepy and just got this foolish idea :blush:


calculus
derivative
function
algebra
domain
calculus computations

Let f(x, y) be a function such that d(df/dx)/dy = d(df/dy)/dx  holds for every (x, y) in the domain of f.

Show that if x = s + t and y = s - t, then the function f satisfies the partial differential equation

d^2 f/dx^2 = k (d^2 f/ds^2 + 2*d^2 f/dsdt + d^2 f/dt^2), where k is a constant, and find the value of the constant k.

Note: the d represent partial derivative.

$ f_s\equal{}f_xx_s\plus{}f_yy_s\equal{}f_x\plus{}f_y$
$ f_t\equal{}f_xx_t\plus{}f_yy_t\equal{}f_x\minus{}f_y$
$ f_{ss}\equal{}f_{xx}x_s\plus{}f_{xy}y_s\plus{}f_{yx}x_s\plus{}f_{yy}y_s\equal{}f_{xx}\plus{}2f_{xy}\plus{}f_{yy}$
$ f_{tt}\equal{}f_{xx}x_t\plus{}f_{xy}y_t\minus{}f_{yx}x_t\minus{}f_{yy}y_t\equal{}f_{xx}\minus{}2f_{xy}\plus{}f_{yy}$
$ f_{st}\equal{}f_{xx}x_t\plus{}f_{xy}y_t\plus{}f_{yx}x_t\plus{}f_{yy}y_t\equal{}f_{xx}\minus{}f_{yy}$

$ f_{ss}\plus{}2f_{ts}\plus{}f_{tt})\equal{}$
$ (1\plus{}2\plus{}1)f_{xx}\plus{}(2\plus{}0\minus{}2)f_{xy}\plus{}(1\minus{}2\plus{}1)f_{yy}\equal{}4f_{xx}$

so $ k\equal{}\frac{1}{4}$


AMC
AIME
function
limit
quadratics
linear algebra

The function $f$ defined by $\displaystyle f(x)= \frac{ax+b}{cx+d}$. where $a,b,c$ and $d$ are nonzero real numbers, has the properties $f(19)=19, f(97)=97$ and $f(f(x))=x$ for all values except $\displaystyle \frac{-d}{c}$. Find the unique number that is not in the range of $f$.

First, we use the fact that $f(f(x))=x$ for all x in the domain. Substituting the function definition, we have $\displaystyle{\frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d}=x}$, which reduces to $\displaystyle{\frac{(a^2+bc)x+b(a+d)}{c(a+d)x+(bc+d^2)}=\frac{ex+f}{gx+h}=x}$. In order for this fraction to reduce to $x$, we must have $f=g=0$ and $e=h\not=0$. From $c(a+d)=b(a+d)=0$, we get $a=-d$ or $b=c=0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a=-d$, so $f(x)=\frac{ax+b}{cx-a}$.

The only value that is not in the range of this function is $\displaystyle\lim_{x\to -d/c}f(x)=\frac{a}{c}$. To find $\frac{a}{c}$, we use the two values of the function given to us. We get $2(97)a+b=97^2c$ and $2(19)a+b=19^2c$. Subtracting the second equation from the first will eliminate $b$, and this results in $2(97-19)a=(97^2-19^2)c$, so $\displaystyle{\frac{a}{c}=\frac{(97-19)(97+19)}{2(97-19)}=58}$.

You need to first understand that the problem is basically asking for the value of -d/c since
f(f(x))=x, and it asks for the number that is not in the range.

First you place x into the equation f(f(x)) and you should wind up with x.
Substitute this in and you get an ugly equation:
(a((ax+b)/(cx+d))+b)/(c((ax+b)/(cx+d))+d)=x
Now you expand this and
you get an equation where (a+d) is factor, and thus a=-d.
Place 97 and 19 into the equation, and you get
19(c+d)*19=19a+b
97(c+d)*97=97a+b
Subtract the two equations from one another and substitute a=-d
Doing algebra, you get -d/c=58.

thanks for writing solutions
i'm sorry for grammatical mistakes :)

Alternatively, note that $f$ is a Möbius transformation, inversions are their own inverse function, and there exists a unique circle centered on the real line that passes through two given points on the real axis.

[quote="Mildorf"]Alternatively, note that $f$ is a Möbius transformation, inversions are their own inverse function.[/quote]

Could you say that in English?!!! :P  :lol:  :blush:

This function must be its own inverse. f(x) only has one vertical asymptote (at x=-d/c), so this must also be the one horizontal asymptote (y=-d/c). Since there is no quadratic term, the intersection of the two asymptotes must be the midpoint of the segment connecting the two given points (because they're both on y=x), so the answer is (19+97)/2=58.

...so this argument seems to work, but I'm not sure how to make it rigorous.

Möbius functions, also known as linear fractional transformations, are functions of the form

$x\mapsto \frac{ax+b}{cx+d}.$

The thing to remember - the computational hook - is that such such functions compose exactly as matrices compose. We associate to the general function above the matrix

$M=\begin{bmatrix} a & b \\ c & d\end{bmatrix}.$

If $f$ has associated to it the matrix $M$ and $g$ has associated to it the matrix $N$, then $f\circ g$ has assiociated to it the matrix $MN.$

The correspondence between functions and matrices is not one-to-one in that many matrices correspond to the same function, but it still works nicely.

The algebra at the beginning of fuzzy-logic's post is best understood as writing out the condition on the matrix $M$ that $M^2=I.$

Here's another case. Consider $f(x)=\frac1{1-x}.$ Can we figure out what $f$ composed with itself 2005 times is? The matrix formalism associates with this transformation the matrix

$M=\begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix}.$

It's not that hard to just calculate the powers of $M.$ We notice fairly quickly that $M^3=-I$ which would make everything periodic of period 6. Actually, the iterates of $f$ are periodic of period 3, since $-I$ represents the same function as $I.$

(Or, if you know about eigenvalues, you can use those.)

Meh, I don't think there's any hurt in copy pasting my solution from algebra class.[hide]Since $f(f(x))=x$ for all $x$ except $-\frac{d}{c}$, then $f$ must be its own inverse. Thus, the only value not in the range of $f$ is the value not in the domain of $f$.

$f(f(x))=\frac{a\left(\frac{ax+b}{cx+d}\right)+b}{c\left(\frac{ax+b}{cx+d}\right)+d}=x$,
$a(ax+b)+b(cx+d)=x(c)(ax+b)+x(d)(cx+d)$,
$(ac+dc)x^2+(bc+d^2-bc-a^2)x-ab-bd=0=0x^2+0x+0$

Setting the coefficients equal, we find that $a=-d$. Hence, \[ f(19)=\frac{-19d+b}{19c+d}=19,  \] \[ -19d+b=361c+19d,  \] \[ f(97)=\frac{-97d+b}{97c+d}=97,  \] \[ -97d+b=9409c+97d.  \] Subtracting the second equation from the fourth, \[ -156d=9048c,  \] \[ -\frac{d}{c}=58.  \]
Thus, the range of $f$ is $f(x)\neq58$ and the desired answer is $\boxed{058}$.[/hide]

[hide="Cheat"]First, the unique number not in the range of $f$ can be found by solving the equation $y=\frac{ax+b}{cx+d}$ for $x$:\[x=\frac{b-dy}{cy-a}\] Therefore, the value we are looking for is $\frac{a}{c}$. Since $f(f(x))=x$ holds for all $x$, we can substitute any value in for $x$ to find the answer. The easiest value to substitute in is $0$, which yields \[f(f(0))=0\]\[f(b/d)=0\]\[\frac{\frac{ab+bd}{d}}{\frac{bc+d^2}{d}}\] For this entire fraction to be equal to $0$, the numerator must be $0$. Thus, \[\frac{ab+bd}{d}=0\]\[b(a+d)=0\]\[a+d=0\]\[a=-d\] because the problem specifies the constants and coefficients as nonzero real numbers. Turning to the other information in the problem, we can set up two equations: \[19a+b=(19^2)c+19d\]\[97a+b=(97^2)c+97d\] Subtracting the first equation from the second, we find: \[78a-78d=(97^2-19^2)c\]\[78(a-d)=(116)(78)(c)\]\[a-d=116c\] Plugging in $-a$ for $d$, we get: \[2a=116c\]\[\frac{a}{c}=\boxed{58}\].[/hide]

[hide=Slightly flawed solution(?) but still gets the correct answer]Consider the function $h(x)=f(x)-x$. From the information given in the problem, the equation $h(x)=0$ has two solutions, one at $x=97$ and the other at $x=19$.

We now take a look at $f(x)-x$ itself. With some simple algebra, we see that this is equivalent to $\frac{-cx^2+(a-d)x+b}{cx+d}$. In order for this to equal $0$, the numerator must equal $0$, so the numerator is simply the quadratic with roots $19$ and $97$. Solving then gives us $c=-1$, $b=19*97$, and $a-d=-116$.

Substituting these values into $f(x)$ gives us $f(x)=\frac{(d-116)x+19*97}{-x+d}$. We now take a look at the second condition of the problem, which is $f(f(x))=x$ for all values other than $\frac{-d}{c}$. Plugging in $f(0)$ gives us $\frac{19*97}{d}$, so $f(\frac{19*97}{d})=0$, implying that the numerator is $0$. Plugging that in gives us $19*97-\frac{116*19*97}{d}+19*97=0$, and solving gives us $d=58$ and $a=-58$.

We now know that $f(x)=\frac{-58x+19*97}{-x+58}$. It is clear that the only value not in the range of $f(x)$ is its limit. To compute this, we only have to consider $\frac{a}{c}$, as $b$ and $d$ will become negligibly small as $x$ gets larger. Therefore, our answer is $\frac{-58}{-1}=\boxed{58}$.[/hide]

Sorry for the bump :(

Extend $f$ to be from $\mathbb P^1 \to \mathbb P^1$. Since $f$ is an involution fixing $19, 97$, it takes infinity to $\frac{19+97}{2} = 58$.

[hide=Super cancerous solution]From $f(f(x))=x$, it is obvious that $\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\frac{b}{d}$, $f(\frac{b}{d})=0$ which means $a(\frac{b}{d})+b=0$ so $a=-d$. Using $f(19)=19$, we have that $b=361c+38d$; on $f(97)=97$ we obtain $b=9409c+194d$. Solving for $d$ in terms of $c$ leads us to $d=-58c$, so the answer is $\boxed{058}$.

When I did this problem, I just realized that the value that wouldn't be in the range would have to be the arithmetic mean of the two points such that $f(n)=n$ on the graph, done.

[s]to be fair, I wasn't able to really prove it too rigorously, so I prayed to god this would work[/s]

p1: We first use the $f(f(x)) = x$ condition. It gives $\frac{a \cdot \frac{ax+b}{cx+d} + b}{c \cdot \frac{ax+b}{cx+d} + d} = \frac{x(a^2+bc)+(ab+db)}{x(ac+dc)+(bc+d^2)} = x$, or $x(a^2+bc)+(ab+db)=x^2(ac+dc)+x(bc+d^2)$. This means $ac+dc=0, ab+db=0, a^2+bc = bc+d^2$. Simplifying these, we have $c(a+d)=0, b(a+d)=0, a^2 = d^2$. It is given that $a,b,c,d$ are nonzero, so we must have $a+d=0 \implies a = -d$. Our function is hence $\frac{-dx+b}{cx+d}$. It's given that $f(19) = 19, f(97) = 97$, so $\frac{-19d+b}{19c+d} = 19, \frac{-97d+b}{97c+d}=97$. We can expand and solve to get $b = 1843p, c = -p, d = 58p$ for some $p$. Now our function is $\frac{-58px+1843p}{-px+58p} = \frac{-58x+1843}{-x+58}$. The horizontal asymptote is at $\boxed{58}$.


linear algebra
matrix
vector
search
linear algebra unsolved

Is it always true that rank(A) = rank (BA) where B is the transpose matrix of A.

Hi,

We have to be more precise on the field.

For instance, if A is on the real field then we have rank(A)= rank(BA). It's easy, we just have to show that ker A = ker BA, in fact

if X is a column vector, then AX=0 implies BAX=0.
For the converse, we have  BAX=0 implies 0=transpose(X)BAX= ||AX||^2, so AX=0. Where we denote ||Y||=\sum (y_i)^2

But it's not true if the field is not R. For instance, let K be a field such that (-1) is a square.

Of course, the complex field is convenient, and it's classic to show that the finite Fp field of p elements (with p prime) is convenient if 4 divde p-1.

Let a such that (-1)=a^2 and   let A=(1,a)(a,-1), we have

rank(A)=1
B=transpose(A)=A
and BA=A^2 = (0,0)(0,0)
so rank(BA)=0

I haven't search about other fields.

rafik

The complex version of this is that $ A^*A$ has the same rank as $ A$.

This is about inner product structure; if $ Ax\neq 0$, $ 0\neq\langle Ax,Ax\rangle\equal{}\langle x,A^*Ax\rangle$ and $ A^*Ax\neq 0$. If $ Ax\equal{}0$, then of course $ A^*Ax\equal{}0$. This means that $ A$ and $ A^*A$ have the same null space, and thus the same rank as well.


Solve this trigonometric equation for $ x\in R$:

$ \sin^{10}{x}\plus{}\cos^{10}{x} \equal{} \frac{29}{64}$

i have done the solution using graph
x is approx equal to -0.3927-8.236.10^(-17)i

hello, applying the addition formulas we get
$ 1\plus{}12\cos(4x)\plus{}\cos(8x)\equal{}0$
setting $ 4x\equal{}t$ we have
$ 1\plus{}\cos(t)\plus{}\cos(2t)\equal{}0$
with $ \cos(2t)\equal{}2\cos^2(t)\minus{}1$ we get
$ \cos(t)\plus{}2\cos^2(t)\equal{}0$
Sonnhard.

[quote="thanhnam2902"]Solve this trigonometric equation for $ x\in R$:

$ \sin^{10}{x} \plus{} \cos^{10}{x} \equal{} \frac {29}{64}$[/quote]

$ u^5\plus{}v^5\equal{}(u\plus{}v)^5\minus{}5uv(u\plus{}v)^3\plus{}5(uv)^2(u\plus{}v)$ $ \forall x,y$

So, considering $ u\equal{}\sin^2(x)$ and $ v\equal{}\cos^2(x)$, and so $ u\plus{}v\equal{}1$, we get :

$ \frac{29}{64}\equal{}1\minus{}5uv\plus{}5(uv)^2$
$ \iff$ $ (uv)^2\minus{}uv\plus{}\frac{7}{64}\equal{}0$
$ \iff$ $ uv\equal{}\in\{\frac 18,\frac 78\}$
$ \iff$ $ \sin^2(2x)\equal{}4uv\in\{\frac 12,\frac 72\}$
$ \iff$ $ \sin^2(2x)\equal{}\frac 12$

Hence the solutions : $ x\equal{}(2k\plus{}1)\frac{\pi}{8}$


number theory
relatively prime
number theory proposed

Find all the pairs of prime numbers $ (p,q)$ such that $ pq|5^p\plus{}5^q.$

Let $ p\le q$. If $ p \equal{} q$, then $ p^2|25^p\to p \equal{} q \equal{} 5$.
If $ p \equal{} 2$, then $ q > 2,q|5^{q \minus{} 1} \plus{} 5\to q|6\to q \equal{} 3$ or q=5.
If $ p \equal{} 3$, then $ 3|5^{q \minus{} p} \plus{} 1\to q \minus{} p \minus{} odd$ contradition with $ q > p$.
If $ p \equal{} 5$, then $ q|5^{q \minus{} 1} \plus{} 625\to q|626\to q \equal{} 313.$
Let $ 5 < p < q$, then $ pq|5^{p \minus{} 1} \plus{} 5^{q \minus{} 1}$.
Let exist minimal $ n \equal{} n(p)$, suth that $ p|5^n \plus{} 1$, then for all odd m $ p|5^{nm} \plus{} 1$ and $ pq|5^p \plus{} 5^q\to \frac {q \minus{} 1}{n(p)} \minus{} odd$.
By anology $ \frac {p \minus{} 1}{n(q)}$ is integer and odd.
If $ p \equal{} 3\mod 4$, then for exist $ n(p)$ must be $ (\frac 5p) \equal{} \minus{} 1\to p \equal{} 3,7\mod 20$. But if $ p \equal{} 3\mod 4$, then $ n(p)$ odd.
For example $ n(7) \equal{} 3$. If $ n(p)$ odd $ \frac {q \minus{} 1}{n(p)}$ - even. Therefore p,q must be $ 1\mod 4$.
If $ ord_2(q \minus{} 1) \equal{} k$, then $ ord_2(n(p)) \equal{} k\to ord_2(p \minus{} 1)\ge ord_2(n(p)) \plus{} 1\to ord_2(p \minus{} 1) > ord_2(q \minus{} 1)$. By anology $ ord_2(q \minus{} 1) > ord_2(p \minus{} 1)$. Therefore thereare not solutions $ p > 5$.
All solutions ($ p\le q$) are $ (2,3),(2,5),(5,5),(5,313)$.

If one of the numbers is $ 5$ we get $ 5p \,|\, 5^p \plus{} 5^5$ so either $ p \equal{} 5$ or $ p \,|\, 626$. So there are the solutions $ \{2,5\}, \{5, 5\}, \{313, 5\}$. If $ p, q \neq 5$ and one of the numbers is $ 2$, we get $ p \,|\, 30$ and only $ p \equal{} 3$ works, so there is also the solution $ \{3, 2\}$. So providing $ p, q \not\in \{5, 2\}$, we always get these solutions.
Otherwise, we have that $ p$ also divides $ 5^{q\minus{}1} \plus{} 1$, hence the order of $ 5$ modulo $ p$ divides $ 2(q\minus{}1)$ but does not divide $ q \minus{} 1$. As this order divides $ p \minus{} 1$, it must be that $ \nu_2 (q \minus{} 1) < \nu_2 (p \minus{} 1)$. The same reasoning applies in the reverse case too, however, so we have no more solutions.

Well... I imagined a problem this simple would have a much less trivial solution, but apparently its quite simple:

First we try $ (5,5)$ and notice it works. Then we set $ p\equal{}5$, $ q\neq 5$ and end up with $ 5q|5^{5}\plus{}5^{q}$. Dividing, we get $ q|5^{4}\plus{}5^{q\minus{}1}$. Placing it in $ mod$ notation we end up with $ 625\plus{}5^{q\minus{}1}\equiv 0(mod q)$. By Fermat's Little Theorem, $ 626\equiv 0(mod q)$, which gives the solutions $ (5,313)$ and $ (5,2)$.

We now work with $ p,q$ relatively prime to $ 5$:
$ 5^{p}\plus{}5^{q}\equiv 0(mod pq)\Rightarrow 5\plus{}5^{q}\equiv 0 (mod p)\Rightarrow 5^{q\minus{}1}\equiv \minus{}1 (mod p)$. For our next argument we will exclude the case $ p\equal{}2$ (we exclude it since, in that case, $ \minus{}1\equiv 1 (mod p)$.

We have $ 5^{q\minus{}1}\equiv \minus{}1(mod p)\Rightarrow 5^{2(q\minus{}1)}\equiv 1(mod p)$. Similarly, we also have $ 5^{p\minus{}1}\equiv \minus{}1(mod q)$. Let $ 2^{k}\parallel{}ord_{p}5$. Then $ 2^{k}\parallel{}ord_{p}5|p\minus{}1$. Also, $ 2^{k}|2(q\minus{}1)$, but $ 2^{k}$ does not divide $ q\minus{}1\Rightarrow 2^{k\minus{}1}\parallel{}q\minus{}1$. Therefore, we have that the maximum power of $ 2$ that divides $ p\minus{}1$ is larger than the maximum power of $ 2$ that divides $ q\minus{}1$.

Doing the same process with $ 5^{p\minus{}1}\equiv \minus{}1(mod q)$, we get that the maximum power of $ 2$ that divides $ q\minus{}1$ is larger than the maximum power of $ 2$ that divides $ p\minus{}1$, which contradicts our previous statement.

This implies our only option is $ p\equal{}2$ (or $ q\equal{}2$). In this case, $ 2q|25\plus{}5^{q}\Rightarrow 25\plus{}5^{q}\equiv 0(mod 2q)$ and, by Fermat's Little Theorem, $ 30\equiv 0(mod 2q)$. The only primes that divide $ 30$ are $ 2$ and $ 3$, and we can easily verify the last congruence is not met when $ q\equal{}2$, but is when $ q\equal{}3$.

Then the only possible pairs are $ (2,3); (5,5); (5,2); (5,313)$.

[quote=bobthesmartypants][hide="solution"]If $p$ or $q$ is $2$, then WLOG $p=2$. Clearly $p=q=2$ does not work, so assume $q$ is odd. Now $2q\mid 25+5^q\implies 5^q+25\equiv 5+25\equiv 0\pmod{q}\implies q\mid 30$ so this case has solutions $$\boxed{(p,q)=(2,3), (2,5), (3,2), (5,2)}$$

Otherwise, if $p=q=5$ then it works: $$\boxed{(p,q)=(5,5)}$$
If $p=5$ and $q\ne 5$ then $5q\mid 5^5+5^q\implies 5^5+5^q\equiv 5^5+5\equiv 0\pmod{q}\implies q\mid 5^4+1=3\times 313$ so $q=2\text{ or }313$: $$\boxed{(p,q)=(5,2), (5,313), (2,5), (313,5)}$$

If $p,q\ne 5$ then $5^p+5^q\equiv 5^p+5\equiv 0\pmod{q}\implies 5^{p-1}\equiv -1\pmod{q}$. Then $\text{ord}_q(5)\mid 2(p-1)$. In addition, since $\text{ord}_q(5)\nmid p-1$ because $q>2$, then $v_2(\text{ord}_q(5)) = v_2(2(p-1))=1+v_2(p-1)$. Similarly, $v_2(\text{ord}_p(5)) = 1+v_2(q-1)$.

If $p=q$ then $p^2 \mid 2\cdot 5^p\implies p=5$ so if $p\ne 5$ then we may assume WLOG $p > q$. Then $pq\mid 5^p+5^q = 5^q(5^{p-q}+1)\implies 5^{p-q}\equiv -1\pmod{pq}$. So $5^{p-q}\equiv -1\pmod{q}$ so $\text{ord}_q(5)\mid 2(p-q)$ and since $\text{ord}_q(5)\nmid p-q$ therefore $v_2(\text{ord}_q(5)) = v_2(2(p-q))=1+v_2(p-q)$. This means $v_2(\text{ord}_q(5))=v_2(\text{ord}_p(5))\implies v_2(q-1)=v_2(p-1) = v_2(p-q)$.

If we let $p-1=k_12^e$ and $q-1=k_22^e$ for $k_1, k_2$ odd, then $p-q=(k_1-k_2)2^e$; however, since $2\mid k_1-k_2$ then $v_2(p-q) > v_2(p-1)$ contradiction. Thus there are no more solutions. [/hide][/quote]

Can you explain, why we get $v_2(ord_q(5))=v_2(2(p-1))$ ?

[quote=EthanWYX2009]I will post this in English if I know how do we say "阶" in English.
------
[/quote]

order.

Suppose one of $p$ or $q$ is 2 or 5. We get (what we claim to be) our only solutions:
\[\boxed{(2,3),(3,2),(2,5),(5,2),(5,5),(5,313),(313,5)}\]

We want to show a contradiction if we assume otherwise. Our condition is equivalent to
\begin{align*}
&5^q \equiv -5^p \implies 5^{q-1} \equiv -1 \implies 5^{2(q-1)} \equiv 1 \quad \pmod{p} \\
&5^p \equiv -5^q \implies 5^{p-1} \equiv -1 \implies 5^{2(p-1)} \equiv 1 \quad \pmod{q} \\ \\
&\operatorname{ord}_p5 \mid \gcd(p-1,2(q-1)) \quad \pmod p \\
&\operatorname{ord}_q5 \mid \gcd(2(p-1),q-1) \quad \pmod q \\
\end{align*}

Note $\operatorname{ord}_p5$ must divide $2(q-1)$ but not $q-1$, and symmetrically for $\operatorname{ord}_q5$, so $v_2$ tells us
\begin{align*}
v_2(\operatorname{ord}_p5) = v_2(2(q-1)) \leq v_2(p-1) \implies v_2(p-1)-v_2(q-1) \ge 1 \\
v_2(\operatorname{ord}_q5) = v_2(2(p-1)) \leq v_2(q-1) \implies v_2(p-1)-v_2(q-1) \le -1 \\
\end{align*}

which is a contradiction. $\blacksquare$

The answers are $\{p,q\} = \boxed{\{2,3\}, \{2,5\}, \{5,5\}, \{5,313\}}$, which can easily be checked to work. Now, we prove they are the only ones.

If either of $p$ or $q$ equals $5$, WLOG say $q$, we have

\[5p \mid 5^5+5^p \implies p \mid 5^4+5^{p-1}\]
\[\implies 5^{p-1}+625 \equiv 626 \equiv 0 \pmod{p}.\]

Hence, $p=2,313$, but also note that this applies if and only if $p=5$, which also happens to work.

If neither of $p$ or $q$ equals $5$, but WLOG $q=2$. Then, we have

\[2p \mid 5^p+25 \implies 5^p+25 \equiv 0 \pmod{p},\]

if $p$ is odd. This simplifies to $p \mid 30$, so the only new case is $p=3$.

If none of the above conditions holds true, we must have

\[5^p+5^q \equiv 0 \pmod{p,q}\]

by CRT. Considering modulo $p$ first, we get

\[5^p+5^q \equiv 0 \pmod{p} \implies 5^{q-1} \equiv -1 \pmod{p}.\]

This means that the order of $5$ modulo $p$ divides $2(q-1)$ but does not divide $q-1$. As the order divides $p-1$, it must be that $\nu_2(q-1) < \nu_2(p-1)$. However, the same reasoning applies in modulo $q$, giving $\nu_2(q-1) > \nu_2(p-1)$, a contradiction.

Hence, we have shown that no other cases exist, so we are done.

Solved with [b]dolphinday[/b]. Very very nice problem. Ok we claim that the answers are $(p,q)=(2,3),(2,5),(3,5)$ and $(5,313)$ which can be checked to work (the last with some work).

Now, we deal with showing that these are the only solutions. First, note that when $q=2$ we note that
\[2p \mid 25(5^{p-2}+1) \implies 5^{2p-4} \equiv 1 \pmod{p}\]
By Fermat's Little Theorem, we also have $5^{2p-2} \equiv 1 \pmod{p}$. Thus, $25\equiv 1 \pmod{p}$ which implies that $p=3$. When $q=3$ we do a similar argument as before and show that there are no solutions. When $q=5$ we repeat the exact same process and obtain the possible solutions $(5,5),(5,13)$ and $(5,313)$ of these $(5,13)$ can be checked to not work so we are left with only the desired solutions.

We assume that $(p,q)$ is a pair of solutions such that $q\geq 7$. WLOG, assume that $p\geq q$. If $p=q$ then we have
\[p^2 \mid 2 \cdot 5^p \implies p=q=5\]
Thus, assume that $p>q$. Now, we note that,
\[pq \mid 5^q(5^{p-q}+1)\]
So, (since $p,q\neq 5$) this implies that,
\[pq \mid 5^{p-q}+1\]
Thus, $5^{p-q}\equiv -1 \pmod{p}$. This implies that $5^{2p-2q} \equiv 1 \pmod{p}$. We know that by Fermat's Little Theorem, $5^{2p-2} \equiv 1 \pmod{p}$. Thus,
\[5^{2q-2} \equiv 1 \pmod{p}\]
We also know that by Fermat's Little Theorem, $5^{2q-2} \equiv 1 \pmod{q}$. Since $p$ and $q$ are relatively prime, this implies that
\[5^{2q-2} \equiv 1 \pmod{pq}\]
With an entirely similar argument, we obtain that
\[5^{2p-2} \equiv 1 \pmod{pq}\]
This means, $5^{q-1} \equiv \pm 1 \pmod{pq}$ and $5^{p-1} \equiv \pm 1 \pmod{pq}$. But then,
\[0 \equiv 5^p+5^q \equiv \pm 5 \pm 5 \pmod{pq}\]
Thus, it is quite clear that one of $p,q$ must be $5$ which is a clear contradiction. Thus, there exists no solution in this case and we have found all solutions as required.


[quote="sinankaral53"]f(1)=1  f:R--R  find all functions for all x and y such that

f(x+y)=3^y.f(x)+2^x.f(y)[/quote]

$ f(x+y)=3^{y}f(x)+2^{x}f(y)$
$ f(y+x)=3^{x}f(y)+2^{y}f(x)$

So $ 3^{y}f(x)+2^{x}f(y)=3^{x}f(y)+2^{y}f(x)$

So $ \frac{f(x)}{3^{x}-2^{x}}=\frac{f(y)}{3^{y}-2^{y}}=a$ for $ x\neq 0$

So $ f(x)=a(3^{x}-2^{x})$ even for $ x=0$ (look at initial equation)
Then $ f(1)=1$ $ \implies$ $ a=1$ and $ f(x)=3^{x}-2^{x}$

And it is easy to check that these necessary conditions are enough.


I will not be on this forum for a while... to study. Good luck for any AP takers. I guess this is my last post until after the B test.

[quote="joml88"]Good luck!

I'm gonna try to go through all of the Princeton Review again today.  I don't want to screw up the Physics B exam...[/quote]

I just did last night; I think I got it down pretty good, despite the teacher not teaching Thermodynamics and Optics at all. As well as teaching Waves (SHM) and Nuclear Physics this last week in 3 days (together).

Anyway, good luck to everyone! Should be a piece of cake compared to my world test ;)

Good luch everyone!!
Be careful and patient in the exams, and forget it after it's over! :D


function
real analysis
real analysis unsolved

Determine all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[ f(ax+(1-a)y)+f(ay+(1-a)x)=f(x)+f(y) \] holds for every $x,y\in\mathbb{R}$. $\displaystyle a\in(0,\frac{1}{2})$.

I just realized this is easier than i thought  :oops:

I think that one can prove that the sequence $u_0=x$,$v_0=y$, $u_{n+1}=au_n+(1-a)v_n$, $v_{n+1}=av_n+(1-a)u_n$ converge towards $(\frac{x+y}{2}, \frac{x+y}{2})$. Hence $f(x+y)=\frac{f(x)+f(y)}{2}$ because of the continuity of $f$, and thus $f$ is linear.


What wil i do to learn this so -called latex technique
helpful to one in expressing ideas,mathematically? Is there any
code require? If so,how will i use it? Or,do i've to download this
technique?

                                                     Respectfully,
                                                     A.Eddington

http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

You don't need to download anything to use LaTeX in this forum. However, you can download it and create documents on your own computer as well.

LaTeX is widely used by mathematicians, so you may be able to find a guide in your native language (I guess English is not your first :) ).

You can practice and ask about LaTeX here: http://www.artofproblemsolving.com/Forum/index.php?f=123


How do you graph curves with Asymptote?
For example, how would you graph $ y\equal{}x^2$?

Also, for graphing lines, would you use the draw command? or is there a special graphing command?

Thanks!

[asy]import graph;
size(200,0);

real f(real x) {return x^2;}
pair F(real x) {return (x,f(x));}

xaxis("$x$",EndArrow);
yaxis("$y$",EndArrow);

draw(graph(f,-3,3,operator ..));

label("$y=x^2$",F(2),2.25*E);[/asy]

Thanks!

I have some more questions  :)
For tick marks, what do the two numbers following the f mean?
Ticks(f, 8, 0.5)

Can you specify the tick marks so they display numbers like $ \sqrt{2}$ instead of 1.4142...

Can you graph trig functions in degrees?

Thanks again!

[quote="tornado.adv4"]For tick marks, what do the two numbers following the f mean?
Ticks(f, 8, 0.5)

Can you specify the tick marks so they display numbers like $ \sqrt {2}$ instead of 1.4142...[/quote]

[asy]import graph;
size(200,0);

Label g;
g.p=fontsize(5);
xaxis(-3,3,Ticks(g,10,2));
yaxis(0,9,Ticks(g,15,2));

real f(real x) {return x^2;}
pair F(real x) {return (x,f(x));}

xaxis("$x$",EndArrow);
yaxis("$y$",EndArrow);

draw(graph(f,-3,3,operator ..));

dot(Label("$(\sqrt{2},2)$"),(sqrt(2),2),SE);
draw((0,2)--(sqrt(2),2)--(sqrt(2),0),dashed);
label("$\sqrt{2}$",(sqrt(2),0),S);
label("$y=x^2$",F(2),2.25*E);[/asy]

[quote="tornado.adv4"]Can you graph trig functions in degrees?[/quote]
http://www.mathlinks.ro/viewtopic.php?t=263417


algebra
polynomial
vector
geometry
geometric transformation
trigonometry

When a convex polygon's internal angles are all the same, we call that  the convex polygon is equiangular.
Try to prove that $p>2$ and $p$is a prime if and only if the equiangular polygon which all the sides are rational numbers IS a regular polygon.

My English is very poor, so it must be many mistakes in my artical.
I hope you can tell me about it and I will correct it!

[b]Lemma[/b]
[color=blue]Let $a_{1},a_{2},...,a_{n}$ be positive real numbers and $\varepsilon$ is $n^{th}$  root of unity. $a_{1},a_{2},...,a_{n}$ are sides of equiangular polygon (in counterclockwise order) iff
\[a_{1}+a_{2}\varepsilon...+a_{n}\varepsilon^{n-1}=0. \]
[/color]
For proof of this Lemma consider the polygon's sides as vector and with translation in $\mathbb{C}$ it follows easily.

For the proof of the problem:
Suppose that $p$ is a prime and let $a_{1},a_{2},...,a_{p}$ be the sides of the equiangular polygon in this order. By the Lemma it follows that $\varepsilon=\cos\frac{2\pi}{p}+i\sin\frac{2\pi}{p}$ is a root of the polynomial
\[a_{1}+a_{2}x...+a_{p}x^{p-1}\in\mathbb{Q}[x] \]
But on the other hand $\varepsilon$ is a root of $1+x+...+x^{p-1}.$ This means that $a_{1}=...=a_{p}$ or $f(x)=1+x+...+x^{p-1}$ is reducible. But the second case is impossible(for example consider $f(x+1)$ and Eisenstein criterion).
Now suppose that $p=mn$ is not prime and every equiangular polygon with rational sides is regular. This means that on one side $1+\varepsilon^{n}+\varepsilon^{2n}+...+\varepsilon^{n(m-1)}=0$ and on the other side $1+\varepsilon+...+\varepsilon^{p-1}=0.$ Adding up these equalities we deduce that $\varepsilon$ is a root of polynomial with degree $p-1$ and some coefficients equals to 1 and other to 2. By the Lemma it follows that there exists equiangular polygon with some sides 1 and other 2 which is not reguilar.

it is our solution.
Can anyone use another way to solve it?

http://reflections.awesomemath.org/2006_1.html

Second article is about equiangular polygons may help you.


geometry
similar triangles
calculus
calculus computations

I definitely think I screwed up, but here's my work:

The right triangle VOC has height |VO| = 25 ft, and base |OB| = 14/2 = 7 ft. If h(t) is the height of the water at time t seconds then I can sketch the triangle and label it.

Using similar triangles the relationship 25-h(t)/s(t) is contrived and hence

s(t) = 3/8 (25 - h(t))

The top surface of the water has area (2*s(t))^2 = 4*s(t)^2 and hence the volume of water in the tank is

V(t) = 1/3*196*25-(1/3)*4*s(t)^2*(25 - h(t))

Hence

V'(t) = (4/3)s(t)^2*h'(t)+(8/3)s(t)*s'(t)*(25-h(t))

V'(t) = 3 cubic feet per second and h(t) = 4 feet when you want to find h'(t), thus you can solve for h'(t).

h'(t)= -5.297 cubic feet per second


S, T, U are three distinct points on line P can ST=TU?

Have I misunderstood the question, isuki? T is the mid-point of SU?

T is the midpoint... so the answer is yes right? hahaha!!! i just want to clear things up... ^^


algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

Let $ F$ be a field of characteristic not $ 2$, and let $ K$ be an extension of $ F$ with $ [K: F]\equal{}2$. Show that $ K\equal{}F(\sqrt{a})$ for some $ a\in F$; that is, show that $ K\equal{}F(\alpha)$ with $ \alpha^{2}\equal{}a\in F$. Moreover, show that $ K$ is Galois over $ F$.

pick some $ b\in K\backslash F$. then $ F(b) \equal{} K$, and the minimal polynomial of $ b$ has the form $ f \equal{} x^{2}\plus{}px\plus{}q$ for some $ p,q\in F$. since $ b$ is a root, the other root is $ \minus{}p\minus{}b\in K$. thus $ K$ is the splitting field of $ f$, in particular normal. observe that $ f$ is irreducible and it's derivative is $ \neq 0$ since the characteristic of $ F$ is not $ 2$. thus $ f$ is separable. this shows that $ K$ is galois. finally, the equation $ (b\plus{}p/2)^{2}\equal{} (p/2)^{2}\minus{}q$ shows that $ F(\sqrt{(p/2)^{2}\minus{}q}) \equal{} K$.

Consider $ l\in K\minus{}F$. We have that $ \{1,l\}$ are independent, so they form a base of $ K$ over $ F$. Obvious, $ K \equal{} F(l)$, so the extension is simple. Let $ m_{l}$ be the min. poly. of $ l$ in $ F[X]$. Because $ [K: F] \equal{} 2$, $ deg m_{l}\equal{} 2$ so $ l^{2}\plus{}al\plus{}b \equal{} 0$. Again, it is clear that $ K \equal{} F(l) \equal{} F(l\minus{}\frac{a}{2})$ (because $ 2\neq 0$). Then, $ \alpha \equal{} l\minus{}\frac{a}{2}$ has the property $ \alpha^{2}\equal{}\frac{a^{2}}{4}\minus{}b\in F$ and $ K \equal{} F(\alpha)$.
To prove that it is a Galois extension, we have to prove that it is finite(it is simple, so QED), normal and separable.
Let $ x\in K$.
If $ x\in F$, the min. poly. has deg 1, so it is splitted.
If $ x\in K\minus{}F$, we will show that de min. poly. has deg 2. Because it has one root in $ K$, it will split in $ L$(from Bezout).
$ x \equal{} m\plus{}n\alpha$, with $ m,n\in F$, $ n\neq 0$. We must find $ u,v\in F$ such that $ x^{2}\plus{}ux\plus{}v \equal{} 0$. Calculating, we get that $ 2mn\alpha\plus{}un\alpha \equal{} 0$ and $ m^{2}\plus{}n^{2}\alpha^{2}\plus{}um\plus{}v \equal{} 0$. Because $ n\neq 0$ we get $ u$, and because $ \alpha^{2}\in F$ we get $ v$.
So, the extension is normal.
To prove that it is separable, look at $ D(m)$ where $ D(m)$ is the formal derivative of the minimal polynom $ m$. Because $ 2\neq 0$, $ D(m)\neq 0$ for any $ m$, so the extension is separable.
So, it is Galois.

[Edit]: Sorry -oo-, we posted almost simultaneously and didn't see your post.


For $a,b,c$ positive reals and $k$ integer prove that

$\frac{a^k}{b}+\frac{b^k}{c}+\frac{c^k}{a} \geq 3(\frac{a^k+b^k+c^k}{3})^{\frac{k-1}{k}}$

Can be remove $3$ by integer $k>0$ in to have a general inequalities.


function
algebra
polynomial
algebra unsolved

Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all reals $ x,y$,

$ f(x \minus{} f(y)) \equal{} f(x \plus{} y^{2009}) \plus{} f(f(y) \plus{} y^{2009}) \plus{} 1$

Let $ n\equal{}2009$ and $ a\equal{}f(0)$.

(1) $ P(f(y),y): f(f(y)\plus{}y^n)\equal{}\frac{a\minus{}1}2$
(2) $ P(x\plus{}f(y),y): f(x)\equal{}f(x\plus{}f(y)\plus{}y^n)\plus{}\frac{a\plus{}1}2$.
Let $ y\equal{}0$ above,
(3) $ f(x)\equal{}f(x\plus{}a)\plus{}\frac{a\plus{}1}2$
Compare (2) and (3)
(4) $ f(x)\equal{}f(x\plus{}f(y)\plus{}y^n\minus{}a)$.

For any $ y$, we see that $ f(y)\plus{}y^n\minus{}a$ is a period of $ f(x)$. Obviously there exists nonzero period. Otherwise $ f(y)\equal{}a\minus{}y^n$ and it is easy to check this is not a solution. Let $ p\neq0$ be a period. Then $ f(y\plus{}mp)\plus{}(y\plus{}mp)^n\minus{}a\equal{}f(y)\plus{}(y\plus{}mp)^n\minus{}a$ is period of $ f$ for integer $ m$. Since the difference of periods is period, $ (y\plus{}2p)^n\minus{}(y\plus{}p)^n$ and $ (y\plus{}p)^n\minus{}y^n$ are periods. Again, we have period $ g(y)\equal{}(y\plus{}2p)^n\plus{}y^n\minus{}2(y\plus{}p)^n$. Note that $ g(y)$ is a polynomial of $ y$ with odd degree $ n\minus{}2$. Hence $ g(y)$ can be any real number, which means $ f$ can have any period. So $ f$ is a constant. It is easy to see $ f(x)\equal{}\minus{}1$.


[quote="kunny"]Find the pair of integer $(a,b,c)$ such that $a^3+2b^3+4c^3=2abc$.[/quote]

[hide] If im not wrong the only solution is $(0,0,0)$ .

Suppose $a,b,c\neq 0$ then from the equation $a$ must be even . Plugging $a=2a'$ and you will get

$4a'^3+b^3+2c^3=2a'bc$ . Now we see that the new equation is similar to the original equation which mean we can keep reducing the equation by halfing $a,b,c$ without end . So this implies $a,b,c$ can only be $0$ .[/hide]

yeah but if im not mistaken you have to get exactly the same equation to apply infinite descent  :?

But he can repeat it twice and he will got the same equation...

well can i see the complete solution? i dont get it. because i dont think with the new equation you can affirm parity :?


function
Putnam
algebra
floor function
number theory proposed
number theory

If x and y are positive irrational numbers such that [tex]\displaystyle {\frac{1}{x} + \frac{1}{y} = 1}[/tex], prove that [x], [2x], [3x],... and [y], [2y], [3y],... (where [] is the Floor Function) together contain all positive integers without repetition.

You are right this is a very well-known problem, and these sequences are knwon as Beatty's sequences.
Note that the converse is also true, that is :
Let $S(x) = \{[nx]/n =1,2,3,.... \}$.
If $S(x)$ and $S(y)$ form a partition of $N^*$ then $x,y$ are irrational numbers such that $ \frac {1} {x} + \frac {1} {y} = 1$.

Moreover, there is no $x,y,z$ such that $S(x),S(y), S(z)$ form a partition of $N^*$. This is problem B-6 of Putnam 1995.

Pierre.

From what I remember, the solution is a straightforward application of the properties of the integer part (I mean I've never seen or given a cool solution).

If for some $n$, $n<ix<n+1$, $n<jy<n+1$, then $\frac{i+j}{n+1}<\frac1x+\frac1y<\frac{i+j}n$ and $n<i+j<n+1$.
If for some $n$, $ix<n$, $(i+1)x>n+1$, $jy<n$, $(j+1)y>n+1$, then $\frac{i+j}n<\frac1x+\frac1y<\frac{i+j+2}{n+1}$ and $n-1<i+j<n$.


Let $a_{0}=b_{0}=1$ and $a_{n+1}=a_{n}+b_{n},b_{n+1}=3a_{n}+b_{n}$. Prove that exist unique $c\in\mathbb{R}$ such that $n|ca_{n}-b_{n}|<2\forall n\geq 0$.

Solving the system equation of reccurence gives
\[a_{n}=\frac{1}{2\sqrt{3}}\{(1+\sqrt{3})^{n+1}-(1-\sqrt{3})^{n+1}\}\]

\[b_{n}=\frac{1}{2}\{(1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1}\}\]
This problem smells the flavour of Pell equation. :lol:

And therefore...? :D I know a other solution, I will post it later. Post concrete your solution, please! :P

[quote="kunny"]The answer is $c=\frac{1}{\sqrt{3}}?$ :wink:[/quote]
No, $c=\sqrt{3}$.  Post concrete your solution, please! :P

Oops! The answer is $c=\sqrt{3}.$ :blush:  :wink:

We are to prove that $\lim_{n\infty}\frac{b_{n}}{a_{n}}=c.$