Simplify this:

$\sqrt{2-\sqrt{3}}$

so that there isn't a square root of a square root.

[hide]This can be written as

$\sqrt{2-\sqrt{3}}=\sqrt{a}-\sqrt{b}$.

Squaring both sides, we get

$2-\sqrt{3}=a+b-2\sqrt{ab}$

So
$a+b=2$
$ab=\frac{3}{4}$

We see that the solution is, $(\frac{3}{2},\frac{1}{2})$ or $(\frac{1}{2},\frac{3}{2})$

But a>b, hence it is $\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.

So the answer is $\frac{\sqrt{6}-\sqrt{2}}{2}$

$\frac{\sqrt{6}-\sqrt{2}}{2}$[/hide]

well there is no general formula (try to denest $\sqrt{6+\sqrt{11}}$) ... here's why:
Consider $(\sqrt{m}+\sqrt{n})^{2}=m+n+2\sqrt{mn}$
We must always have $m+n=a$, but that's about all we can say (for the general case). Number one, we need two equation about $m+n$ in order to solve. However, $b\sqrt{c}=\sqrt{cb^{2}}=\sqrt{4mn}\implies 4mn=cb^{2}$, which doesn't help us. In fact, theoretically there are infinite solutions (this should be guessed because you have three variables $a,b,c$ and to denest you only need 2), except we are of course only looking for integer solutions, which rarely exist (outside of contest problems :wink: ).

[quote="cincodemayo5590"][b][u]Generalization[/u][/b]

Denest $\sqrt{a+b\sqrt{c}}$.

(these type of radicals are called "nested radicals")[/quote]

[hide] Continuing from me@home's equations $m+n=a$ and $4mn=cb^{2}$:

We have $m=a-n$. Substitute into the second equation to get $4n(a-n)=cb^{2}$. Simplify to get $n^{2}-an+\frac{cb^{2}}{4}=0$.
$n=\frac{a\pm\sqrt{a^{2}-cb^{2}}}{2}$.
$m=a-\frac{a\pm\sqrt{a^{2}-cb^{2}}}{2}=\frac{a\pm\sqrt{a^{2}-cb^{2}}}{2}$.

Thus, it can only be denested if $\sqrt{a^{2}-cb^{2}}\in\mathbb{Z}$. [/hide]

EDIT: $\sqrt{6+\sqrt{11}}=\frac{\sqrt{2}+\sqrt{22}}{2}$. :wink:

wow *blushes* :blush: sorry for the bad example... anyway, yes you are right with the quadratic equations...
isn't it interesting how using geometry and pythagoras to find cos15 as opposed to half angle stuff can give an alternate explanation for $\sqrt{2+\sqrt{3}}$ denested?


function
trigonometry
algebra proposed
algebra

let $ f$ and $ g$  functions from $ [a;b]$ to $ R$ such that:
$ (\forall$ $ x$ and $ y$ $ \in$  $ [a;b]$
$ g(x)\equal{}x\plus{}f(x)$
$ lf(x)\minus{}f(y)l \equal{}< 1/2lx\minus{}yl$
prove that $ g$ is $ injective$

$ remark$
$ lxl$ is the absolut value of $ x$

Obviosly $ |g(x) \minus{} g(y)|\ge \frac 12 |x \minus{} y|.$

$ |g(x) \minus{} g(y)|\equal{}|x\minus{}y\plus{}f(x)\minus{}f(y)\ge |x\minus{}y|\minus{}|f(x)\minus{}f(y)| \ge \frac 12 |x \minus{} y|.$

A result :
$ 0 < a < 1$
$ |f(x) \minus{} f(y)|\leq a|x \minus{} y|$
Then $ g(x) \equal{} f(x) \plus{} x$ is inject.
[u]Proof [/u]
[b]Solution 1[/b]
Suppose exist $ x_0,y_0$ such that $ g(x_0) \equal{} g(y_0)$ then
$ f(x_0) \plus{} x_0 \equal{} f(y_0) \plus{} y_0$ then:
$ |f(x_0) \minus{} f(y_0)| \equal{} |x_0 \minus{} y_0|$ so $ x_0 \equal{} y_0$
[b]Solution 2[/b] As Rust's solution.
Then $ a > 1$ it not true.
If $ a > 1$ we can consider the function:
$ f(x) \equal{} a\sin{x}$
$ g(x) \equal{} x \plus{} a\sin{x}$
But when $ a \equal{} 1$?

[quote="TTsphn"]A result :
$ 0 < a < 1$
$ |f(x) \minus{} f(y)|\leq a|x \minus{} y|$
Then $ g(x) \equal{} f(x) \plus{} x$ is inject.
But when $ a \equal{} 1$?[/quote]
It is true for a=1, if $ |f(x) \minus{} f(y)| < |x \minus{} y| \ \forall x\not \equal{} y.$

When $ a \equal{} 1$ if the function is continous and has deriverty  then it true .
Because $ |f(x) \minus{} f(y)| \equal{} |x \minus{} y\parallel{}f'(c)|$
so $ |f'(x)|\leq  1,\forall x\in R$
So $ g'(x)\geq 0$


integration
function
logarithms
Support
calculus
real analysis
derivative

Hello,
Please consider the following question -

Define a distribution $ T \in D'_1(-1,1)$ by
$ \langle T, \phi \rangle = \int_{0}^{1} \frac{\phi(x) - \phi(0)}{x} dx$
for any $ \phi \in D_1 (-1, 1)$.
(a) Does there exist a function $ f \in L_1 (-1,1)$, such that $ f' = T$ ?
(b) The same, but this time with $ f \in L_{\infty} (-1,1)$ ?

I have the solution of (a) - the suitable $ f \in L_1(-1,1)$ is ${ f(x) = \ln x \cdot {\mathbf{1}}_{(0,1)} (x)}$.
(The last "1" is the indicator function of (0,1) - supposed to be bolder..).

(b) - Not sure about it (the answer should be no I think, but I don't have any complete proof), so would like to get some help here.

Thanks!

- $ D_1 (-1,1)$ denotes the linear space of (ones) continuously differentiable functions $ (-1,1) \rightarrow \mathbb{C}$ with compact support in (-1,1).
- Convergence in $ D_1(-1,1)$ defined as follows: if $ \phi_n, \phi \in D_1(-1,1)$ then
$ \phi_n \rightarrow \phi$ in $ D_1 (-1,1)$ iff $ \phi_n ^{(m)} \rightarrow \phi ^{(m)}$ for $ m=0, 1$ and the support of all $ \phi_n$ are contained in a single compact subset of (-1,1).
- $ D_1 ' (-1,1)$ is the set of linear functionals $ T: D_1 (-1,1) \rightarrow \mathbb{C}$ such that
$ \langle T, \phi_n \rangle \rightarrow \langle T, \phi \rangle$ whenever $ \phi_n \rightarrow \phi$ in $ D_1 (-1,1)$.

$ <T,\phi>\equal{}<f',\phi>\equal{}\minus{}<f,\phi{'}>\equal{}\minus{}\int_{\minus{}1}^{1}f(t)\phi{'}(t)dt$.
Then using an integration by parts we find that a solution in $ L_{loc}^{1}$ is $ f(x)\equal{}\minus{}ln(x)1_{(0,1)}(x)$.
If $ g$ is another solution in $ L_{loc}^{1}$ then $ g(x)\equal{}f(x)$ for almost all $ x$ (because $ \phi{'}$ runs through the continuous functions). Thus there is no solutions in $ L^{\infty}$.

loup blanc,
Thanks for your reply!
I agree with what you said about the case when g is another solution in $ L_{loc} ^1$ etc., but I don't understand why it proves that there are no solutions in $ L_{\infty}$.
The solution, f,  we found in L_1 is not in $ L_{\infty}$, but how do we know that there are no other solutions (that are not in L_1) in $ L_{\infty}$ ? I miss something here..
Can you please explain?

--
edit: Or maybe I do understand?

Is this because the solution f is not just in $ L_1(\minus{}1,1)$, but it's in $ L_{loc} ^1 (\minus{}1,1)$ and the last contains $ L_{\infty}$... (although $ L_1(\minus{}1,1)$ does not contain $ L_{\infty}$)  ?

Hi kar,
"Is this because the solution f is not just in ..."
 is correct.
Moreover $ L^\infty(-1,1)\subset{L^}^{1}(-1,1)\subset{L^}^{1}_{loc}(-1,1)$.

Ahh, right!
(Sorry, the last thing I said previously was wrong.)

Now it's clear. You are using something like [url=http://planetmath.org/encyclopedia/TheoremForLocallyIntegrableFunctions.html]this[/url] theorem, right?
Is there a way to prove it avoiding using this theorem (or it would make one to do something like the proof of the theorem in our specific case)?
I mean maybe assume there exist such a function in L_oo and then come to some contradiction? (Like finding a test function for which it doesn't work). (Or is it not a good way to prove this?)

(Sorry for bugging you with questions.)

Thank you very much!

Let$ f\in{L}^1_{loc}$ s.t. for all smooth functions with compact support $ \phi\in{D}$: $ \int_{\minus{}1}^{1}{f(t)}\phi(t)dt\equal{}0$. Then $ f\equal{}0$ for ae $ x$.
Proof: let $ (x_0,x)\subset(\minus{}1,1)$. There exists a sequence $ \phi_n\in{D}$ s.t. $ |\phi_n(t)|\leq{1}$ and $ \phi_n(t)$ converges to $ 1_{(x_0,x)}(t)$ for ae $ t$. Thus $ \int_{x_0}^{x}{f(t)}dt\equal{}0$. Now we use the Lebesgue differentiation theorem:
$ \dfrac{d}{dx}\int_{x_0}^{x}{f(t)}dt\equal{}f(x)$ for ae $ x$.

Ok, I see...
Thanks a lot for your help!!


Have $ n \equal{} \overline{a_1a_2\ldots a_k}$, and $ f(n) \equal{} a_1^3 \plus{} a_2^3 \plus{} \cdots \plus{} a_k^3$. Then $ f(n) \leq k9^3 < 10^{k\minus{}1} \leq n$ is true for $ k \geq 5$.
It remains to do the computations for all numbers with at most $ 4$ digits - one can write a small program, or do a spreadsheet, or such, to find the final possibilities.

[quote="asdf4"]One of my math teachers proposed this problem (I don't think he knows the answer).

Let n be any positive integer. Each of the digits of n are cubed and added together to create a new n. Prove that for any n, after multiple operations, n will become a constant, and find all such constants.[/quote]

I hope that he does not know the solution, since it is wrong :

Take $ n\equal{}133$
Next is $ 1^3\plus{}3^3\plus{}3^3\equal{}55$
Next is $ 5^3\plus{}5^3\equal{}250$
Next is $ 2^3\plus{}5^3\equal{}133$  :(

And so a cycle $ (133,55,250,133,55,250...)$ and, even after infinitely many such operations, $ n$ will not become a constant.

[quote="mavropnevma"]Will anyone write a program to check all orbits (for numbers with at most $ 4$ digits) and post the results?[/quote]

I tested for all numbers in range $ [1,13000]$ and all end in one of these $ 9$ cycles :

$ (1)$
$ (55,250,133)$
$ (136,244)$
$ (153)$
$ (160,217,352)$
$ (370)$
$ (371)$
$ (407)$
$ (919,1459)$

Merci, pco.
Matter closed (so what does all this mean? another typical useless question!).

Ahh thanks!
I guess the problem is essentially solved.


trigonometry
inequalities
inequalities open

Let 4Pi > r3 > r2 > r1 > 2Pi.

Prove that

 r1 r2 Sin[r1 - r2] + r2 r3 Sin[r2 - r3] + r3 r1 Sin[r3 - r1] < 0.

The supposed inequality I posted yesterday:

Let 4Pi > r3 > r2 > r1 > 2Pi.

Prove that

r1 r2 Sin[r1 - r2] + r2 r3 Sin[r2 - r3] + r3 r1 Sin[r3 - r1] < 0

has some exceptions and therefore cannot be proved in general, but it's true "almost always," where for the present its limitations are not known.


are you going?

what subject, and (after the contest) what score?

this will be my first time, so how is the format?

I'm doing alg 2

I'm going for the second time and taking the Open Mad Hatter. The Mad Hatters are kind of interesting. The first 25 problems (multiple choice) are put up on an overhead one by one. When you're done with a problem you raise your hand, and when more than half the hands are raised the proctor takes the problem down and goes to the next one--so you are directly competing against everyone else in the room in the sense that you want to raise your hand ASAP and keep them from having enough time to do the problem. The next 25 problems (also multiple choice) are handed out on paper and you have x minutes to do them, where x is some constant that I don't remember.

(I took Algebra 2 last year, by the way. And I just found out that the teacher who leads our math club writes the Algebra 2 questions. Heh, I had no idea.)

i got a 25....:(  my teacher loves trig more, so we barely did much alg 2

ive heard a couple people say that a lot of kids who placed were in school of the contest writer.  might just be a coincedence

Uh, it definitely is a coincidence. Our math club teacher writes the algebra 2 questions, and we also happen to have John Boyle, Nathan Pinsker, and Avi Arfin in Trig/Analyt (our algebra 2 equivalent). I've heard that the first two were 1st and 2nd in the Miller Math Marathon, and I know they were 2nd and 4th in the Peninsula MathCounts chapter. I know Avi, too. Those three do not need to cheat in order to beat everyone else into the ground.

[quote="bookaholic"]we also happen to have John Boyle, Nathan Pinsker,[/quote]
and they also happen to be not from paly but from the middle school....but they still PWN.
And yea, those kids are cool, they don't cheat.

Yeah, they pound me into the ground at every math contest!!


Let $x,y,z$ be positive numbers such that $2^x=\left(\frac{5}{2}\right)^y=3^z$,and $a=2x,b=\frac{5}{2}y,c=3z$. Compare $a,b,$ and $c$.

There's a hint:
[hide]
Put $x=\frac{a}{2},y=\frac{2}{5}b,z=\frac{c}{3}$ into $2^x=\left(\frac{5}{2}\right)^y=3^z$
[/hide]

I have a solution, but I think it's quite monotonous, so I put just another hint :D
[hide="Click to reveal hidden hint"]
After putting into equation logarithm each side and use figures about logarithms... It's easy :)
[/hide]


Let be $ a,b,c,d\in \mathbb{R}_\plus{}$ such that $ abcd\equal{}1$ . Prove that :

$ 8\plus{}(a^2\plus{}b^2)(c^2\plus{}d^2)\ge 3(a\plus{}b)(c\plus{}d)$

[quote="alex2008"]Let be $ a,b,c,d\in \mathbb{R}_ \plus{}$ such that $ abcd \equal{} 1$ . Prove that :

$ 8 \plus{} (a^2 \plus{} b^2)(c^2 \plus{} d^2)\ge 3(a \plus{} b)(c \plus{} d)$[/quote]
Put $ bc \equal{} x,ac \equal{} y$,this inequality become:
$ \frac {(x \minus{} 1)^2(x^2 \minus{} x \plus{} 1)}{x^2} \plus{} \frac {(y \minus{} 1)^2(y^2 \minus{} y \plus{} 1)}{y^2} \ge 0$, which is obvious true


calculus
integration
function
geometry
real analysis
real analysis unsolved

If continuous function f(x) on [a;b] satisfies the following conditions:
$\frac{1}{b-a}\int_{a}^{b}f(x)dx=1$
$\frac{1}{b-a}\int_{a}^{b}f(x)^2dx=1$

then prove that f(x)=1

To view all the problems of Uzbekistan Undergraduate Olympiad problems visit
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81099[/url]

in more detail, let $g(x)=f(x)^2 -2f(x)+1$. Then $g(x) \geq 0$.Since
$\int_a^b g(x) =\int_a^b f(x)^2 \, dx - 2 \int_a^b f(x) \, dx +\int_a^b f(x) \, dx = (b-a) -2 (b-a) + (b-a) = 0$.
so etiher $g(x)=0$ or
there is a $\xi \in [a,b]$ so that $\int_a^{\xi} g(x) \, dx + \int_{\xi}^{b} g(x) \, dx =0$ and $g(\xi)=0$, which mean the area on $x \in [a, \xi]$ cancel the area on $[\xi,b]$ so $g(x)<0$ for $x< \xi$ which is impossible since $g(x) \geq 0$. so the only posiblity is
$g(x)=0 \Rightarrow f(x)^2 -2f(x)+1 = 0 \Rightarrow f(x)=1$.

Obviously the question under discussion is wrong unless the term integrable is replaced by continuous.

we have to say that  $f$ is a continuous function

Oh, it was my fault. Now I edited :blush:


In interiorul unghiului ascutit $\ XAY$ se ia punctul $\ D$ si pe laturile sale punctele $\ B,C$ astfel incat sa avem
$\ <ABC=<XBD$,$\ <ACB=<YCD$.Sa se demonstreze ca centrul cercului circumscris triunghiului $\ ABC$ este pe segmentul $\ AD$.

Fara prea multe detalii:

Daca $E,F$ sunt reflexiile punctului $D$ in $AX$ si $AY$ respectiv, atunci $B,C$ sunt exact intersectiile dreptei $EF$ cu $AX$ si $AY$ respectiv. Avem $\angle DAC=\angle FAC=\angle ACB-\angle AFE$ si $\angle DAB=\angle EAB=\angle ABC-\angle AEF$. Cum $AEF$ e isoscel ($AE=AF=AD$), rezulta de aici, scazand cele doua relatii de mai sus, ca $\angle DAB-\angle DAC=\angle ABC-\angle CAB\ (*)$. Daca $O$ este centrul cercului circumscris lui $ABC$, atunci are loc $(*)$ cu $O$ in loc de $D$ (calcul simplu de unghiuri), si deci perechile de unghiuri $(\angle OAB,\angle OAC)$ si $(\angle DAB,\angle DAC)$ au aceeasi suma $\angle BAC$ si aceeasi diferenta. Concluzia rezulta de aici.

Ca sa iasa ca $O$ e pe segmentul $AD$ si nu doar pe dreapta, mai facem niste observatii simple (de exemplu, $O$ e in interiorul lui $ABC$, pentru ca $ABC$ iese ascutitunghic, dar $D$ e in afara).

Nu prea cred ca e solutie de clasa a VII-a.Eu de-abia trec a VIII-a.N-am invatat reflexiile.Oricum, problema am rezolvat-o si eu, dar in 3 zile.Si cred ca e o solutie de nivel gimnazial.

Cred c-am gasit una elementara. Am sa dau liniile generale, ca sunt sigur ca vei intelege restul: :?

Ducem perpendicularele $BI$ si $CI$ respectiv pe $AX$ si $AY$. E clar ca $AI$ este diametru al cercului circumscris triunghiului $ABC$. Pe de alta parte, $BI$ si $CI$ sunt bisectoarele unghiurilor interioare triunghiului $BCD$ deci $DI$ este si ea bisectoare ! Am notat apoi unghiul $IAC$ cu $x$ si unghiul $IAB$ cu $y$ (pentru simplicitate !),am utilizat unghiul inscris in cerc ca sa calculez unghiul $AIB$ . Obtin ca unghiul $AIB=90-y$.
In final, in triunghiul $BID$ avem unghiul $BID=90+y$(iti las calculul !) deci suma lor este de
$180$... si gata ! :roll:

Tare frumoasa! M-am chinuit vreo doua ceasuri si dupa o mica plimbare am gasit-o ! Poti sa-mi dai liniile generale ale solutiei tale ?

Off..si eu m-am chinuit cam mult la ea.Dar pana la urma am gasit o solutie.Numai ca eu cred ca m-am complicat.
  Din ipoteza rezulta ca $\ BA,CA$ sunt bisectoarele exterioare ale unghiurilor $\ B,C$ ale triunghiurilor $\ BCD$, deci $\ DA$ este bisectoarea interioara a unghiului $\ D$ al triunghiului.Sa presupunem acum ca perpendiculara pe dreapta $\ AX$ ridicata in $\ B$ intersecteaza pe $\ AD$ in $\ I$.Atunci $\ BI$ este bisectoarea interioara a unghiului
$\ B$ din triunghiul $\ BCD$, deoarece $\ m(<IBC)=90-m(<ABC)$ iar
$\ m(<IBD)=90-m(<XBD)$, cu $\ <ABC=<XBD$.Asadar, $\ I$ este punctul de intersectie al bisectoarelor triunghiului $\ BCD$, de unde avem
$\ CI\bot\ AY$.Se observa usor ca patrulaterul $\ ABIC$ este inscriptibil,
deoarece are o pereche de unghiuri opuse suplementare, iar centrul cercului sau circumscris este mijlocul ipotenuzei $\ AI$ a triunghiurilor
dreptunghice $\ ABI$ si $\ ACI$.Afirmatia care trebuie demonstrata in final rezulta acum din faptul ca acest cerc este circumscris triunghiului
$\ ABC$.
   Desi problema s-a dat la olimpiada din URSS pentru clasa a X-a, eu zic ca ar putea reprezenta o problema interesanta chiar si pentru clasa a VII-a.

mstoenescu, exact aceeasi solutie am gasit si eu aseara, dar cum nu prea mi`a mers netul, nu am reusit sa o postez


find all integers $a$ and $b$ such that there exists atleast one $k$ satisfying
 the equation
 $a^n + b^n = k *k^n$ for any given value of $n$.

I guess k has to be an integer?  Or else there's solutions in k for any (a,b,n).


[quote="ashrafmod"]Calculate all m values such that the roots of the following equation constitute an arithmetic sequence.
 $x^{4}-(3m+1) x^{2}+m^{2}= 0$[/quote]

[hide]notice that the function is even so if one of the roots is $x$, another root is $-x$.

Let the two positive roots be $a$ and $b$ so all the roots are $-b,-a, a, b$ in increasing order.  The distance between two consecutive terms is $2a$ so $b-a=2a$ or $b=3a$.  This means that the four roots are $-3a,-a, a, 3a$ in increasing order.

The equation is $(x+3a)(x+a)(x-a)(x-3a)=x^{4}-10x^{2}a^{2}+9a^{4}$.

$10a^{2}=3m+1$ and $9a^{4}=m^{2}$.  From the second equation, $m=3a^{2}$.  Plugging this back into the first, $10a^{2}=9a^{2}+1$.  so $a=\pm 1$.  Therefore, $m=\pm 3$.[/hide]


superior algebra
superior algebra unsolved

[quote="sisioyus"][quote="sinajackson"]$ \left|A\right|\;\equiv\;m\;\;(\text{mod }pm)$[/quote]
If I haven't forgotten, it should be corrected to $ m\left|A\right|\;\equiv\;m\;\;(\text{mod }pm)$[/quote]

i'm not sure of that [b]sisioyus[/b]! :wink:

[quote="sinajackson"]Let $ S$ be a finite set of order $ p^{r}m$, where $ p$ is a prime number (which may divide $ m$). Let $ A$ be the set of all subsets consisting of $ p^{r}$ elements of $ S$. Prove that: $ \left|A\right|\;\equiv\;m\;\;(\text{mod }pm)$[/quote]

$ |A| \equal{}{{p^{r}m}\choose{p^{r}}}\equal{} (p^{r}m)(p^{r}m\minus{}1)...(p^{r}m\minus{}m\plus{}1)$

We need to show $ \frac{|A|\minus{}m}{pm}\equal{}\frac{(p^{r}m)(p^{r}m\minus{}1)...(p^{r}m\minus{}m\plus{}1)\minus{}m}{pm}\in\mathbb{Z}$.

So,$ \frac{p^{r}(p^{r}m\minus{}1)...(p^{r}m\minus{}m\plus{}1)\minus{}1}{p}\in\mathbb{Z}$

Isnt that impossible? Unless maybe $ r\equal{}0$.


Support calculus MATHCOUNTS AMC AIME USA(J)MO USAMO

It might be a little confusing having calculus and classroom math together, but otherwise, I support. :yup:

we could have subforums under that for easy, medium, and hard levels

That does sound like a good idea :D

Perhaps there could also be a section for mock contests

There are some pros and cons to a competition forum.
Pros: All gathered into one place for organization
People will look at MOEMS Competitions as much as MathCounts competitions  :roll: if they're placed in the category
Probably others I can't think of...

Cons: Some work for the admins :D
Completely Useless...?
Probably others I can't think of...

Also, what will the categories be? I think 2 Middle School categories: Basics and MathCounts (where participants are Middle Schoolers and below) and 1 High School (HSB, Intermediate level, Mock AMC and AIMEs), and also one for more difficult stuff (MIMC, AoPS Kings??, Mock USAMO, and I doubt there is a Calculus competition :D )

Just something random I noticed: some moderators have not signed on (or rather, posted) in months, or years.  :huh:

EDIT: Shouldn't this be in "Questions, Suggestions, & Announcements?"

yes. i agree. The competition thread could also be a home for mock  contests, as mustafa said. and i also agree with lotrgreengrapes7926 that we should have more subcategories than just hte 3 i mentioned, because people might think moems competitions as mathcounts

EDIT: it won't take the admins more than 5 minutes to create a forum

[quote="anirudh in a PM to me"]Please help support the Competition Forum! Many have already supporting it, and it is a great thing. It is easier for the users of this site! Help by supporting it!

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=640709

Thanks a lot :)[/quote]

Please dude, stop spamming my inbox. I have nothing do do with this forum, neither I'm in any of your competitions, so there is no grounds not to call this spam. Sorry, I find this not an acceptable way to draw attention.

To be honest, I don't see what is wrong with the system as it is.  The MATHCOUNTS, MOEMS, and AMC forums are all "competition" forums, as are the National Olympiads, IMO, Putnam, IMC, etc., forums, not to mention the ubiquitous Other US Contests and Programs and Other International Contests and Programs forums.  As for mock competitions, I suspect that a forum for those would be rather self-defeating, as I think that the majority of people who participate in mock competitions are not those who are often thinking, "Gosh, I'd like to participate in a mock AMC/AIME today," and would therefore be likely to check such a forum often, but rather people who see a notice for such a mock competition in the forums they frequent, think that they would like to participate, and do so.

What exactly are you proposing, anyhow?  I don't really see the point of this.

Some people participate in a lot of competitions. It takes a lot of time to go through and check each thread, so it would be easier if all the threads were in one place.

No. The forums would be having marathons, and they would have problem-threads. It is very hard to find non-competition threads, especially in the Math Counts Forum.

[quote="SplashD"]actually the point of some competitions is to get that forum active again. if there was a competition forum, the forums would be dead.[/quote]

Not really.
Say if  you don't know how to do a problem on a competition and wants to know how to do it.
You'd post that individual question in that fourm.

[quote="anirudth"]can there be a competition forum, with only competitions. it would be easier to organize and keep track of the competitions one is in. If you are in support of it, post it here. I'll then pm an admin for creating one. i'll also be willing to moderate it.[/quote]
Why dont' we just post links to the competition in the Wiki?
And why don't one just write their competitions on a sheet of paper? If one is afraid of losing that paper, post it in one's blog or something. You can make the post private and only to yourself.

frankly, i think a competition forum would be completely useless

its like mathcounts. and amc. they already have their own forums

I got a PM from anuridh just now and I guess I m not really sure.

[quote="anirudh"]can there be a competition forum, with only competitions. it would be easier to organize and keep track of the competitions one is in. If you are in support of it, post it here. I'll then pm an admin for creating one. i'll also be willing to moderate it.[/quote]

I oppose.  If the categories in the "Competition forum" were too narrow, then we could just keep the competitions where they are now.  If they were wider, then we would have a hard time navigating the competitions.  Either way, a Competition Forum would be useless.  I also balk at the idea of you moderating the forum, considering [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=640039][this][/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=639368][this][/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=638615][this][/url].

[quote="junggi"][quote="funcia"]WE NEED 1! I am so lost with the competitions! But anyway..why not have a middle school olympiad highschool like...level ones?[/quote]
ummmm.....
the only competitions there are, are the ones in moems forum, ones in mc forum, and the mock amc/aime/usamos in the amc forum.  so....why would we need a comp forum?

[quote="anirudh"]i'll also be willing to moderate it.[/quote]
dude, are you stupid(to anirudh of course)?
or is this another part of your plan to get more post counts
ps. when you edit a post, please say what you edited. and if you don't want to do that, at least get rid of the stupid hide tag in your sig so that we know that you edited[/quote]

DUDE I M different! It is so hard but actually thanks for telling me where the comp. are I will now go and look for them! It'd be easier if they were all in the same spot!!!

 junggi you have a point anyway. But why'd anyone want more post counts? I don't get the post count thingy It confuses me!

Anyway this will be my last say in it since arguing will do no good-plus to contradict myself we'd just have a jumble of forums.

[quote="junggi"]when you edit a post, please say what you edited. and if you don't want to do that, at least get rid of the stupid hide tag in your sig so that we know that you edited[/quote]
I think hiding your post info shouldn't be allowed. It has absolutely no purpose, unless he [i]wants[/i] to annoy people.

I have seen this talked about already, actually.  I think Val is already working toward adding this, although he has more pressing issues to deal with, I'm sure.

I seriously remember bringing this up a few month ago in some thread in the Mathcoutns forum.

Are there enough competitions for a competition region forum though?

Also, can you please not spam people's inboxes with PMs informing them of this topic? Thx.

There are like 20-40 competitions together running at AoPS/ML. Look here:

Math Counts-8? competitions including mock contests
MOEMS-2 competitons currently

Other forums also have more

It might be worth having one page somewhere where there are links to all the competitions currently taking place, but a competitions forum? Seriously, who would decide one day that they felt doing a competition, go along to the forum and see if there were any of the right difficulty level? There are more than enough forums and subforums anyway.

that page would take a lot of updating to do :( It would be much easier to just create a forum with competitions and mock contests, instead of a list, and one could then keep track easier. The subforums may not be needed anyway... :|

[quote="rrusczyk"]Could this not be well-organized within the wiki?[/quote]

I think that is the best idea.  We already have pages for the Mock contests (AMC, AIME), as well as pages for tons of real contests.

EDIT: I created a page http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Competitions, but i still think it would be much easier to just click buttons on the forum, instead of having a group of persons to edit that page all the time a new one is created :(


Given that you only have an A4 paper, by folding and creasing (marking lines), and comparing lengths, show that $\text{Height : Width : :}\sqrt{2}\text{ : 1}$

What about if you had a $\sqrt{n}: 1$ (height to width, n a natural number) piece of paper, how could you verify this?

Note: No rulers or anything like that.

http://en.wikipedia.org/wiki/Square_root_of_two

[quote]The square root of two is the aspect ratio of paper sizes under ISO 216.[/quote]

I know, but the point is to show it for an A4 page, and generalise.  :D

[hide="First case"] We have rectangle $ABCD$ such that $AB = CD = \sqrt{2}, BC = DA = 1$.  Let $E$ be on $AB$ such that $AE = 1$.  Fold $DE = \sqrt{2}$.  Then fold the angle bisector of $DE$ and $CD$.  Immediately this shows that $DE = CD$ and we have the result. [/hide]


How many n-digit numbers are there,consisting only of the digits 1,2,3,4,5 in which adjacent digits always differ by 1?

@arjun:
are repetitions of digits allowed?
because If they are, then you have infinite no. of solns. (I think)
12345432123... (you can go on as long as you want to)

If not, I think it is
[hide]
25 solns.
[ 1 digit - 5
  2 digit - 8
  3 digit - 6
  4 digit - 4
  5 digit - 2 ][/hide]

Repetitions are allowed.
You have to find no of [b]n digit [/b]numbers.(ie)give an expression in n

Oh, sorry arjun.
I misunderstood the question. :oops:
will try it again.

Denote the number of $ n$ digit numbers by $ A_n$
$ A_{2n\plus{}1} \equal{} 14.3^{n\minus{}1}$
$ A_{2n} \equal{} 8.3^{n\minus{}1}$ for $ n>\equal{}1$
$ A_1\equal{}5$ and $ A_2\equal{}8$ are the initial terms of the recurrence.

I did this by recursion. I will post the complete solution later.

Yeah,Correct  :)
You could tell that $ a_1 \equal{} 5$and for n>1,it
is
$ 3^{\frac{n}{2}\minus{}1}*8$ , when n is even and
$ 3^{\frac{n\minus{}3}{2}}*14$ , when n is odd.


function
induction
trigonometry
strong induction
algebra unsolved
algebra

1. Find all function that send from nautural numbers to itself such that

 f(m+n) + f(m-n) = {f(2m) + f(2n)}/2 For all m,n are naltural numbers

 2. Find all function that send from real numbers to itself such that

 f(x+y) + f(x-y)  = 2f(x)cosy For all x,y are real numbers.

For the first one, do you mean all non-negative integers?

If so, setting $m=n$ gives us $f(0) = 0$.  Setting $n=0$ gives us $4f(n) = f(2n)$.

We will now prove by strong induction that $f(n) = f(1)n^{2}$, for all positive integers $n$.  Our base cases, $n= 0,1$ are trivial.  Now, assuming this holds for all $n\le m$,
\begin{eqnarray*}f(m+1)+f(1)(m-1)^{2}&=& \frac{f(1)4m^{2}+f(1)4}{2}\\ f(m+1)/f(1)+m^{2}-2m+1 &=& 2m^{2}+2 \\ f(m+1)/f(1) &=& m^{2}+2m+1, \end{eqnarray*}
as desired.  Thus $f(n) = kn^{2}$, for some integer $k$, and it is easy to see that any value of $k$ will work.

For the second one, first note that $f(x+\pi/2)+f(x-\pi/2) = 0$, so $f(x) =-f(x+\pi )$.  Then, setting $x=0$, $y = a$ gives us
\[f(a)+f(-a) = 2f(0)\cos a . \qquad (*)\]
Next, setting $x = \pi/2$, $y = \pi/2-a$ gives us
\begin{eqnarray*}f(\pi-a)+f(a) &=& 2f(\pi/2) \cos (\pi/2-a) \\-f(-a)+f(a) &=& 2f(\pi/2) \sin a . \qquad (**) \end{eqnarray*}

Adding $(*)$ and $(**)$ gives us $f(a) = f(0)\cos a+f(\pi/2)\sin a$.  Indeed, it is easy to see that any linear combination of $\cos a$ and $\sin a$ will work.

I liked this second problem.  It's in Engel, isn't it?

For the first one, do you mean all non-negative integers?
I mean only positive integers.

Thank you.
It's in my book.
I don't know it's source.

In that case, for any positive integers $a>b$ of the same parity, we have
\begin{eqnarray*}f(a)+f(b) &=& f\left( \frac{a+b}{2}+\frac{a-b}{2}\right)+f\left( \frac{a+b}{2}-\frac{a-b}{2}\right) \\ &=& \frac{ f(a+b)+f(a-b) }{2}\\ &=& \frac{f(2a)+f(2b)}{4}. \end{eqnarray*}
It follows that for any integer $n$, $f(2n) = 4f(n)$.  We can then induct to prove that $f(n) = f(1) \cdot n^{2}$.  For $f(1)$ this is trivial, and we have $f(2) = f(2\cdot 1) = 4f(n)$.  Then, assuming this holds for $n \le m$ ($m \ge 2$), we can induct, exactly as before.


Find the value of $ \displaystyle\sum_{x\equal{}1}^{\infty}\frac{x}{2^x}$

this simplifies to $ \frac {1}{2} \plus{} \frac{1}{4} \plus{} \frac {1}{8} \plus{} ...$

this equals:

$ \frac {\frac {1}{2}}{1 \minus{} {\frac {1}{2}}} \equal{} 1$

That's $ \sum_{x\equal{}1}^{\infty} \frac{1}{2^x}$.  The given summation is $ \frac{1}{2} \plus{} \frac{2}{4} \plus{} \frac{3}{8} \plus{} ...$.

[quote="sdbro11"][hide="Solution"]
Let $ y \equal{} \sum_{x \equal{} 1}^{\infty}\frac {x}{2^x} \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \frac {4}{16} \plus{} ...$
$ 2y \equal{} 1 \plus{} \frac {2}{2} \plus{} \frac {3}{4} \plus{} \frac {4}{8} \plus{} \frac {5}{16} \plus{} ...$
$ y \equal{} 2y \minus{} y \equal{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} ...$
So, $ y \equal{} \frac {1}{1 \minus{} \frac {1}{2}} \equal{} \boxed {2}$[/hide][/quote]

But , how do you even know that the series converges? :?:

[quote="Poincare"] But , how do you even know that the series converges? :?:[/quote]

The ratio test:

$ \lim_{x \to \infty} \frac{x\plus{}1}{2^{x\plus{}1}} \times \frac{2^{x}}{x} \equal{} \frac{1}{2}$, so the series converges.

or you could break the original summation into

$ (\frac{1}{2} \plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}...)$
$ \plus{} (\frac{1}{4} \plus{} \frac{1}{8} \plus{} \frac{1}{16} \plus{} ...)$
$ \plus{}...$

$ \equal{}$

$ 1 \plus{} \frac{1}{2} \plus{} \frac{1}{4} \plus{} ... \equal{} 2$

[hide="Alternately"] Verify the following identity:

$ \left( 1 \plus{} \frac{1}{2} \plus{} \frac{1}{4} \plus{} ... \right)^2 \equal{} 1 \plus{} \frac{2}{2} \plus{} \frac{3}{4} \plus{} \frac{4}{8} \plus{} \frac{5}{16} \plus{} ...$

The desired summation is a simple transformation away. [/hide]


$\forall m\le n, m,n$ are two integers.Prove that $n | \binom{n}{m}.(m,n)$

Suppose $m=0$?

And what about, say $\binom{6}{3}=20$?

Maybe I'm misinterpreting the problem...

For some integers $a, b$ we have
\begin{eqnarray*} ( m, n ) & = & am + bn \end{eqnarray*}
Therefore
\begin{eqnarray*} n & | & \left(\begin{array}{c} n\\ m \end{array}\right) \cdot ( m, n )\\ n & | & \left(\begin{array}{c} n\\ m \end{array}\right) \cdot \left( am + bn \right)\\ n & | & \left(\begin{array}{c} n\\ m \end{array}\right) \cdot am\\ n & | & \frac{n!}{( m - 1 ) ! \cdot ( n - m ) !} \cdot a \end{eqnarray*}
It suffice to show that $a \cdot \frac{( n - 1 ) !}{( m - 1 ) ! ( n - m ) !}$ is an integer. But this is clear since
\begin{eqnarray*} \frac{( n - 1 ) !}{( m - 1 ) ! ( n - m ) !} & = & \frac{( n - 1 ) !}{( m - 1 ) ! \cdot ( ( n - 1 ) - ( m - 1 ) ) !}\\ \frac{( n - 1 ) !}{( m - 1 ) ! ( n - m ) !} & = & \left(\begin{array}{c} n - 1\\ m - 1 \end{array}\right) \end{eqnarray*}
Seems too easy  :?

greatest common divisor,greatest common factor whatever you want
to call it.


[quote="236factorial"]I hate absolute value, and I don't get the solution in the book:

From Chinese Olympiad practice (7th grade):

We know that $a_1<a_2<a_3<....<a_n$, what is the smallest value of $|x-a_1|+|x-a_2|+|x-a_3|+....+|x-a_n|$?[/quote]

Try to sketch a graph for $|x-1|+|x-3|+|x-5|$ by dividing case and you will see that the gragh has a " quadratic-like "  shape (just that this graph consists of $4$ straight line while quadratic has only one curve line) . So you will find the minimum actually occurs when $x$ is at the median of the graph( in above case the min is $x=3$ )   .It is similar in quadratic which we have it reaches a minimum when $x$ is at the symmetry axis=median  .

Is there a more algebraic way to do this problem? I like your way, but I'm wondering if there's another possible solution to this problem.


LaTeX
advanced fields
advanced fields unsolved

a,n are intergers. then $ n|\varphi(a^{n} \minus{} 1)$ where$ \varphi(n)$ is the number of numbers that is prime to n.  (sory Latex is not available by now.)

Don't use unicode characters in $ \text{\LaTeX}$ but use e.g. \phi or \varphi.

Anyway, see http://www.mathlinks.ro/viewtopic.php?t=99900 .

thank you :)
It's a ques from Murty's <<Problems in Alge Num Theo>>.


inequalities
abstract algebra
superior algebra
superior algebra unsolved

Given two groups $ A, B$, let $ \text{hom}(A, B)$ denote the set of all group homomorphisms $ A \to B$.  Prove or disprove that

$ \left| \text{hom}(A, A) \times \text{hom}(B, B) \right| \ge \left| \text{hom}(A, B) \times \text{hom}(B, A) \right|$.

What if all homomorphisms are required to be isomorphisms?

It seems to be true for finite abelian groups, by the following inequality, which I'd never seen before: $ (\sum_{i\leq j} a_i a_j)(\sum_{i\leq j} b_i b_j) \geq (\sum_{i\leq j} a_i b_j)(\sum_{i \leq j} b_i a_j)$. (the $ a_i$ and $ b_j$ are nonnegative integers)

But surely it can't be true for all groups...?!?

I honestly don't know.  There was an interesting discussion about "categorifying" the Cauchy-Schwarz inequality back at the [url=http://golem.ph.utexas.edu/category/2007/05/the_two_cultures_of_mathematic_1.html#c009713]n-category cafe[/url], but nothing much came out of it.  An analogue of this result is true for the category of finite-dimensional representations of, say, a finite group, due to the identity

$ \left < \chi_V, \chi_W \right > \equal{} \dim \text{hom}_G(W, V)$

for the traces $ \chi$ associated to two representations.

The case of homomorphisms is completely mysterious for me, but isn't it obvious in the case of isomorphisms? If $ A$ and $ B$ aren't isomorphic, the LHS is positive and the RHS is $ 0$. If they are, choose once and for all an isomorphism $ \phi : A \to B$. Then there is an isomorphism (of sets) $ f \mapsto \phi \circ f$ from $ \text{Iso}(A,A)$ to $ \text {Iso}(A,B)$, and similarly an isomorphism $ g \mapsto \phi^{\minus{}1} \circ g$ from $ \text{Iso}(B,B)$ to $ \text{Iso}(B,A)$. Thus you actually get equality in this case.


probability
Alcumus
MATHCOUNTS
AMC
AMC 8
AMC 10
AIME

I am pretty good at easy-intermediate MC-level probability problems, but have trouble with hard ones. I usually understand the solution, but wouldn't think to approach the problem in that  way. I have done Alcumus some, but the problems seem pretty easy now. Does anyone know of a good resource for probability problems?

Thanks! :)

Alcumus has some pretty hard problems.  You probably should do that, and the easy ones will atleast reinforce you skills.

There's Mathcounts, Rocket City, AMC8, AMC10, AIME, USAMO, and etc. A lot of things you can try. Mathcounts only gives out its previous year though. But if you check the AMC web, they give you a lot of years.    :D


inequalities
integration
trigonometry
function
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that
\[ \frac{a}{1\minus{}ab}\plus{}\frac{b}{1\minus{}bc}\plus{}\frac{c}{1\minus{}ca} \le \frac{3\sqrt{3}}{2}\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that
\[ \frac {a}{1 \minus{} ab} \plus{} \frac {b}{1 \minus{} bc} \plus{} \frac {c}{1 \minus{} ca} \le \frac {3\sqrt {3}}{2}
\]
:)[/quote]

We can expand it,
use a^2+b^2+c^2=3,ac^2+ba^2+cb^2=<2+abc
and schur inequality
But is isn't nice. :(

[quote="kunny"]$ \frac {a}{1 \minus{} ab} \equal{} \int_0^{a} \frac {1}{(bx \minus{} 1)^2}\ dx$. :D[/quote]
What do you mean?  :?:   :maybe:

Can the following inequalies hold?

$ \frac{a}{1\plus{}c^2}\plus{}\frac{b}{1\plus{}a^2}\plus{}\frac{c}{1\plus{}b^2}\leq \frac{3\sqrt{3}}{4}$.

$ \frac{a}{1\plus{}\sqrt{3}c}\plus{}\frac{b}{1\plus{}\sqrt{3}a}\plus{}\frac{c}{1\plus{}\sqrt{3}b}\leq \frac{\sqrt{3}}{6}$.

[quote="kunny"]Can the following inequalies hold?

$ \frac {a}{1 \plus{} c^2} \plus{} \frac {b}{1 \plus{} a^2} \plus{} \frac {c}{1 \plus{} b^2}\leq \frac {3\sqrt {3}}{4}$.
[/quote]
It is stronger than the original and it is true. :)

For the second, let me check it later (since the first appeared in my proof) :)

hi I think , I have an approach with cauchy-shwarzz:

we have : $ LHS\equal{} \sum{\frac{a}{1\minus{}ab}} \equal{} \sum{\frac{a}{(a\minus{}b)^2\plus{}ab\plus{}c^2}}$
            so:  $ LHS \leq \sum{\frac{a}{ab\plus{}c^2}}\equal{} \sum{\frac{1}{b\plus{}\frac{c^2}{a}}}$

but  by cauchy : $ (b\plus{}\frac{c^2}{a})(b\plus{}a) \geq (b\plus{}c)^2$

so  it sufficies to prove that : $ \sum{\frac{a\plus{}b}{(b\plus{}c)^2}} \leq \frac{3\sqrt{3}}{2}$
but I don't know if the last inequality is true  :(  , plz help me to prove it   :)  or give me a couterexample

[hide="kewl way i did it"]
WLOG let $ \frac a{1\minus{}ab}\plus{}\frac b{1\minus{}bc}\plus{}\frac c{1\minus{}ca}\leq\frac b{1\minus{}ab}\plus{}\frac c{1\minus{}bc}\plus{}\frac a{1\minus{}ca}$. Then you have
$ 2\left (\frac a{1\minus{}ab}\plus{}\frac b{1\minus{}bc}\plus{}\frac c{1\minus{}ca}\right )\leq \frac {a\plus{}b}{1\minus{}ab}\plus{}\frac {b\plus{}c}{1\minus{}bc}\plus{}\frac {c\plus{}a}{1\minus{}ca}$.
Let $ a\equal{}\tan x$, $ b\equal{}\tan y$, $ c\equal{}\tan z$. So $ \tan^2x\plus{}\tan^2y\plus{}\tan^2z\equal{}1$.
Let $ f(k)\equal{}\tan^2k$, then $ f''(k)\equal{}4\sec^2k\tan^2k\plus{}2\sec^4k\geq 0$. Hence $ f(k)$ is concave up, and
$ 1\equal{}\tan^2x\plus{}\tan^2y\plus{}\tan^2z\geq 3\tan^2\left (\frac{x\plus{}y\plus{}z}3\right )$, so
$ x\plus{}y\plus{}z\leq\frac{\pi}2$. Now, we have
$ \frac{\tan x\plus{}\tan y}{1\minus{}\tan x\tan y}\plus{}\frac{\tan y\plus{}\tan z}{1\minus{}\tan y\tan z}\plus{}\frac{\tan z\plus{}\tan x}{1\minus{}\tan z\tan x}\equal{}\tan (x\plus{}y)\plus{}\tan (y\plus{}z)\plus{}\tan (z\plus{}x)\leq 3\tan\left (\frac{2(x\plus{}y\plus{}z)}3\right )\leq 3\tan\left (\frac{\pi}3\right )\equal{}3\sqrt 3$
[/hide]

With $ a\plus{}b\plus{}c\equal{}p$, $ q\equal{}ab\plus{}bc\plus{}ca$, and $ r\equal{}abc$, I am trying to maximise the function
\[ f\equal{}\frac{\sqrt{1\plus{}2q}\plus{}qr\minus{}q\sqrt{1\plus{}2q}\plus{}\sqrt{q^2(1\plus{}2q)\minus{}4q^3\plus{}2\sqrt{1\plus{}2q}(5q\minus{}2)r\minus{}27r^2}}{1\minus{}q\plus{}r\sqrt{1\plus{}2q}\minus{}r^2}\]

I don't think we can find the maximum value of your expression by hand  :D (but computer will do it very very ugly ). In my opinion it is will better if we use $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3,ac^2 \plus{} ba^2 \plus{} cb^2\le 2 \plus{} abc$ as shfdfzhjj's post. Anyway, they are not nice solutions for this one.

[quote="kunny"]

$ \frac {a}{1 \plus{} \sqrt {3}c} \plus{} \frac {b}{1 \plus{} \sqrt {3}a} \plus{} \frac {c}{1 \plus{} \sqrt {3}b}\leq \frac {\sqrt {3}}{6}$.[/quote]
THis inequality is not true, for example $ a \equal{} 1,b \equal{} c \equal{} 0$.

[quote="anas"]so  it sufficies to prove that : $ \sum{\frac {a \plus{} b}{(b \plus{} c)^2}} \leq \frac {3\sqrt {3}}{2}$
but I don't know if the last inequality is true  :(  , plz help me to prove it   :)  or give me a couterexample[/quote]
This inequality is not true, for example $ a\equal{}b\equal{}\frac{1}{\sqrt{2}},c\equal{}0$.

[quote="can_hang2007"][quote="SCHNOGGINVOGGINSCHWITZERW"]
WLOG let $ \frac a{1 \minus{} ab} \plus{} \frac b{1 \minus{} bc} \plus{} \frac c{1 \minus{} ca}\leq\frac b{1 \minus{} ab} \plus{} \frac c{1 \minus{} bc} \plus{} \frac a{1 \minus{} ca}$. Then you have[/quote]
We cannot assume this one, SCHNOGGINVOGGINSCHWITZERW.[/quote]
I thought since $ \frac a{1 \minus{} ab} \plus{} \frac b{1 \minus{} bc} \plus{} \frac c{1 \minus{} ca}$ and $ \frac b{1 \minus{} ab} \plus{} \frac c{1 \minus{} bc} \plus{} \frac a{1 \minus{} ca}$ were like symmetrical then you could assume that either was smaller.
If that's wrong, isn't there a way to make the question into $ \frac {a\plus{}b}{1\minus{}ab}\plus{}\frac {b\plus{}c}{1\minus{}bc}\plus{}\frac {c\plus{}a}{1\minus{}ca}\leq 3\sqrt 3$?

[quote="kunny"]Can the following inequalies hold?

$ \frac {a}{1 \plus{} c^2} \plus{} \frac {b}{1 \plus{} a^2} \plus{} \frac {c}{1 \plus{} b^2}\leq \frac {3\sqrt {3}}{4}$.

[/quote]

What's your proof?


abstract algebra
superior algebra
superior algebra theorems

Hello everyone,

I'm taking an advanced course on Elliptic Curves, which assumes a great degree of familiarity with abstract algebra. The TA gave us a brief review of the necessary group-theoretic concepts, but I still feel uncertain about a number of points.

My question is: Can anyone recommend a good (not too advanced) abstract algebra book? What I would like to find is detailed information on:
- homomorphisms
- modules (Z-mod-p type stuff)
- finite fields
- isomorphism among finite fields and among modules
- [b]a wealth of examples[/b]

Thank you,
K. Marinov

Look at Introduction to Abstract Algebra by Nichols.  Might be what you're looking for.  I really like Gallian's Contemporary Abstract Algebra as well.  Good luck and happy hunting.

Jon


$ a,b,c,d \succ 0 ,$prove that:

$ \frac{a}{b\plus{}2c\plus{}d}\plus{}\frac{b}{c\plus{}2d\plus{}a}\plus{}\frac{c}{d\plus{}2a\plus{}b}\plus{}\frac{d}{a\plus{}2b\plus{}c} \ge 1$

Normalize to $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Then the inequality becomes
\[ \frac {a}{1 \minus{} a \plus{} c} \plus{} \frac {c}{1 \minus{} c \plus{} a} \plus{} \frac {b}{1 \minus{} d \plus{} b} \plus{} \frac {d}{1 \minus{} b \plus{} d}\ge 1 (*)
\]
As this expression is symmetric in $ a,c$, and symmetric in $ b,d$, we may assume WLOG that $ a\ge c$, $ b\ge d$.

It suffices to show that
\[ \frac {a}{1 \minus{} x} \plus{} \frac {c}{1 \plus{} x} \ge a \plus{} c
\]
where $ x \equal{} a \minus{} c\ge 0$. But this is equivalent to
\[ a \plus{} ax \plus{} c \minus{} cx \ge a \plus{} c \minus{} ax^2 \minus{} cx^2 \Leftrightarrow (a \minus{} c)x \ge \minus{} x^2(a \plus{} c)
\]
, which is true because the LHS is nonnegative.

Similarly, this is true for $ b,d$, and hence the the LHS of $ (*)$ is at least $ a\plus{}b\plus{}c\plus{}d\equal{}1$.

another way is to use the Engel form of Cauchy

The following inequality is true too:
For all $ a_i>0$ and natural $ n>1$  ( $ a_{i\plus{}n}\equal{}a_i$ ) prove that:
\[ \sum_{i\equal{}1}^n\frac{a_i}{a_{i\plus{}1}\plus{}2a_{i\plus{}2}\plus{}...\plus{}(n\minus{}1)a_{n\minus{}i\plus{}1}}\geq\frac{2}{n\minus{}1}\]


Suppose that two polynomials $ P(x)$ and $ Q(x)$ satisfy $ P(Q(x))\equal{}Q(P(x))$ for all real $ x$. If the equation $ P(x)\equal{}Q(x)$ has no real solution, show that the equation $ P(P(x))\equal{}Q(Q(x))$ also has no real solution.

[quote="moldovan"]Suppose that two polynomials $ P(x)$ and $ Q(x)$ satisfy $ P(Q(x)) \equal{} Q(P(x))$ for all real $ x$. If the equation $ P(x) \equal{} Q(x)$ has no real solution, show that the equation $ P(P(x)) \equal{} Q(Q(x))$ also has no real solution.[/quote]

As $ P(x)\minus{}Q(x)\equal{}0$ has no solution we have WLOG, $ P(x)>Q(x)$ for all $ x$ .

So $ P(P(x))>Q(P(x))\equal{}P(Q(x))>Q(Q(x))$ for all $ x$.

So the equation $ P(P(x)) \equal{} Q(Q(x))$ has no real solutions.


category theory
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superior algebra unsolved

Hi,

as usual , please relocate if this place in appropriate

two categories A and B are said to be isomorphic if you have two covariant functors F and G from A to B and B to A respectively, such that GF is the trivial functor on A, FG is trivial on B

well how do we prove that two categories are not isomorphic?

for instance, take any ring R (not $\mathbb{Z}$), and consider the category of left R modules and the abelian groups
are they isomorphic or not?  guess not? but how to prove? my professor gave me an : cannot always define scalar multiplication on abelian group argument
but I do not understand?  Is that really enough? :|

[quote="fredbel6"]for instance, take any ring R (not $\mathbb{Z}$), and consider the category of left R modules and the abelian groups
are they isomorphic or not?  guess not? but how to prove? my professor gave me an : cannot always define scalar multiplication on abelian group argument
but I do not understand?  Is that really enough? :|[/quote]

What does he mean by scalar multiplication?

Anyway, I would go for universality properties like initial objects, direct sums, products and such. I think, but I am not sure, that these should be preserved by isomorphisms.

Just the multiplication with an element in the ring.

Can you elaborate?  It really seems weird.

[quote="amfulger"]Anyway, I would go for universality properties like initial objects, direct sums, products and such. I think, but I am not sure, that these should be preserved by isomorphisms.[/quote]

I would not try these for the problem of determining whether the category of left $R$-modules and the categroy of abelian groups are isomorphic.
Isomorphic categories, just as about everywhere else in algebra, have the same "properties".
I just descibed some "properties". It happens that for every ring $R$ with $1$, the category of left $R$-modules has initial and terminal object (the trivial $0$ module), direct sums (the direct sum of modules) and direct products (the direct product of modules) so these would not work to distinguish $\mathbb Z$ from the rest.

[quote="fredbel6"]Take any ring R (not $\mathbb{Z}$), and consider the category of left R modules and the abelian groups
are they isomorphic or not?[/quote]

Is this what you want to prove? Or do you just want an example when the category of left $R$-modules and the categeory of abelian groups are not isomorphic.

If you just want a counter-example, take $R$ to be any commutative field $k$.
Denote $\mathcal K$ the categeory of left $k$-modules. It is the same as the category of $k$-vector spaces.
Denote $\mathcal Ab$ the category of abelian groups.

Assume there is an isomorphism $\mathcal F: \mathcal K\to\mathcal Ab$. It is easy to prove that $\mathcal F(0)=0$. $0$ is regarded here as the $0$ $k$-vector space and $0$ abelian group respectively.
Every object in $\mathcal K$ is free, but what I need is that for every nonzero object $V$ in $\mathcal K$, $Hom(V,W)\neq 0$ for every nonzero object $W$ of $\mathcal K$. But then $Hom(\mathcal F(V),G)\neq 0$ for every nonzero abelian group $G$.
Take $V$ such that $\mathcal F(V)=\mathbb Z/2\mathbb Z$. Then $Hom(F(V),\mathbb Z)=0$ and we have reached a contradiction.

Nice, did you assume that the vector spaces over the field were all of finite dimension somewhere?

I was thinking yesterday of a product argument to prove non isomorphicness of categories
for instance in rings i can make categorical products for every couple of rings, in fields i cannot...but is my last statement really correct, can i really find two fields that have no categorical product??

[quote="fredbel6"]Nice, did you assume that the vector spaces over the field were all of finite dimension somewhere?[/quote]
No, I didn't.

[quote="fredbel6"]
I was thinking yesterday of a product argument to prove non isomorphicness of categories
for instance in rings i can make categorical products for every couple of rings, in fields i cannot...but is my last statement really correct, can i really find two fields that have no categorical product??[/quote]

Let $p$ and $q$ be two distinct primes. Assume there exists a product $F=\mathbb F_p*\mathbb F_q$ in the category of fields. First of all, this means that there exist $\pi_p: F\to \mathbb F_p$ and $\pi_q: F\to\mathbb F_q$ field morphisms. They are either $0$ morphisms or isomorphisms.
If you don't allow the existence of $0$ field morphisms, the problem ends here because you get $\mathbb F_p\simeq F\simeq\mathbb F_q$.

I allow them.
We have some cases:

1. If $\pi_p$ and $\pi_q$ are isomorphisms, then, just like before, we get the contradiction $\mathbb F_p\simeq\mathbb F_q$.

2. If $\pi_p$ is iso and $\pi_q=0$, then $F\simeq F_p$. Let $f_p: \mathbb F_q\to\mathbb F_p$ be the null field homomorphism and let $f_q=1_{\mathbb F_q}$ be the identity map of $\mathbb F_q$. By the universality property of the product, there exists $f: \mathbb F_q\to\mathbb F_p$ such that $\pi_pf=f_p$ and $\pi_qf=f_q$. Since $f$ is a homomorphism between fields of different charcteristics, $f=0$. $0=\pi_q\circ 0=f_q$ is a contradiction.

3. If $\pi_p$ is null and $\pi_q$ is iso, then the method in (2) applies with obvious changes.

4. If $\pi_p$ and $\pi_q$ are null then take again $\mathbb F_q$ and the maps in (2). This time we don't know $F$, but it does not matter because $f_q=0\circ f$ is a contradiction.

[quote="amfulger"]
It is easy to prove that $\mathcal F(0)=0$[/quote]

I'm sorry I have a vague idea how but it would be very interesting if you gave me your idea

About the finite fields, we always demand in ring morphisms that 1 is mapped onto one, so that was indeed quite obvious

Now another question, more generally, are there any two fields that have a categorical product?  Guess not?

Like $\mathbb{Q}$ with itself?

About your question, you've helped me so much, I don't think I will be able to help you with that
However didn't you kinda prove that with$R$ field and $\mathbb{Z}$ for the other, then the latter category is just the category of abelian groups?

The categorical product of $\mathbb{Q}$ and $\mathbb{Q}$ is $\mathbb{Q}$, if we don't allow zero morphisms.

[quote="fredbel6"][quote="amfulger"]
It is easy to prove that $\mathcal F(0)=0$[/quote]

I'm sorry I have a vague idea how but it would be very interesting if you gave me your idea

[/quote]

$0$ is the only object in both categories for which there exists exactly one morphism from and to any other object.

[quote="amfulger"]Are there nonisomorphic rings $R$ and $S$ such that the categories $\mathcal R$ and $\mathcal S$ of left $R$-modules respectively $S$-modules are isomorphic?[/quote]Yes. One canonical example is when $R$ is commutative. Then the respective categories of left modules over $R$ and $M_n^n(R)$ ($n \times n$ matrices over $R$) are equivalent.

[quote="dryga"]Yes. One canonical example is when $R$ is commutative. Then the respective categories of left modules over $R$ and $M_n^n(R)$ ($n \times n$ matrices over $R$) are equivalent.[/quote]

I've heard about this before, but since I've never bothered to check it, I am not sure that this equivalence of categories can be turned into an isomorphism.


Let $DEF$ be a pedal triangle of $P$ wrt $ABC$. ($D\in BC, E\in CA, F\in AB$)
$U=BC \cap EF, V=CA\cap FD, W=AB \cap DE$.
Prove that the circumcircles of triangle $VWD$, $WUE$, $UVF$ concur at a point.

Anyone can solve this? This is a nice problem (or even a nice theorem).

Oh I think it is Miquel theorem( for that pedal triangle ), only check it :)

Oh, I had not been thought like that.  :oops:


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Hi, I have recently been covering perturbation solutions to ODEs, and am generally able to follow it but I haven't been able to do this question:

[quote]Find the first two terms of the perturbation solution for the nonlinear initial value problem

$ \ddot{x} \plus{} \varepsilon x\dot{x} \plus{} x \equal{} 0$

subject to $ x(0) \equal{} 0$ and $ \dot{x}(0) \equal{} 1$

for $ |\varepsilon | \ll 1$ by posing an expansion of the form

$ x(t) \equal{} x_0(t) \plus{} \varepsilon x_1(t) \plus{} ...$

Hint: You may use the identity $ cos(t)sin(t) \equal{} \frac {1}{2}sin(2t)$[/quote]

I can't see where the hint comes into it. I did what I have been doing for previous questions where we can solve x_0 which in turns allows us to get x_1 etc but that doesn't work with this question. I'm guessing that's because it's nonlinear. So, can anyone help?

Thanks, and best wishes

I think it works ... why don't you show your workings?  Also, \sin and \cos are built-in LaTeX commands, just like most other common functions.

Well, here's what I did:

$ x \equal{} x_0 \plus{} \varepsilon x_1 \plus{} ... \\ \\ \rightarrow \dot{x} \equal{} \dot{x_0} \plus{} \varepsilon\dot{x_1} \plus{} ... \\ \\ \rightarrow \ddot{x} \equal{} \ddot{x_0} \plus{} \varepsilon\ddot{x_1} \plus{} ...$

I then subbed these into my ODE, and grouped them in terms of the power of $ \varepsilon$ to get:

$ (\ddot{x_0} \plus{} x_0) \plus{} \varepsilon (\ddot{x_1} \plus{} x_0\dot{x_0} \plus{} x_1) \plus{} ... \equal{} 0$

(neglecting all terms where the power of epsilon is 2 or greater, as is implied in the question (I think!!!!))

Setting $ \varepsilon \equal{} 0$ implies that $ \ddot{x_0} \plus{} x_0 \equal{} 0$. From here, I don't know where to go, or where the hint comes into it. Normally when I get to this point, I have something like $ \ddot{x_0} \plus{} p \equal{} 0, p \in \mathbb{R}$ so I can find $ \ddot{x_0}$ explicitly in terms of number, but here I can't do that.

What am I doing wrong, or what am I not seeing? Thanks a lot for your help!

[quote="rorylando45"]$ \ddot{x_0} \plus{} x_0 \equal{} 0$[/quote]  This is a completely standard second order linear differential equation with constant coefficients ... have you not met it before?  (Its general solutions can easily be written down in terms of functions you're definitely familiar with.)

The hint will come in after you've solved this equation and you move on to solving for $ x_1$.  (When solving the equation I've quoted, don't forget to use your initial values.)

Christ, how did I miss that! Thanks. Here's what I've now done:

Solved $ x_0$ as you suggested by inspecting it's characteristic polynomial then applying the initial conditions to get $ x_0 \equal{} \sin{t}$ (there is no particular solution as $ x_0$ is homogeneous by default).

Inserting this into my second polynomial (the one containing terms multiplied by $ \varepsilon$) is where the hint comes into it. I ended up having to solve $ \ddot{x_1} \plus{} x_1 \equal{} \minus{}\frac{1}{2}\sin{2t}$ to get $ x_1 \equal{} \frac{1}{6}\sin{2t} \minus{} \frac{1}{3}\sin{t}$ after applying the initial conditions.

Substituting this all back into my perturbation expansion yields:

$ x(t) \equal{} \sin{t} \plus{} \varepsilon (\frac{1}{6}\sin{2t} \minus{} \frac{1}{3}sin{t}) \plus{} O({\varepsilon}^2)$

Does this all seem correct, and does it seem that I have an alright understanding of the subject matter?

Also, is this an acceptable enough approximation, or should I have gone as far as the $ {\varepsilon}^2$ term?

[quote="rorylando45"]Does this all seem correct, and does it seem that I have an alright understanding of the subject matter?

Also, is this an acceptable enough approximation, or should I have gone as far as the $ {\varepsilon}^2$ term?[/quote]  That looks right to me.  One of the nice things about differential equations is that you can always check your work -- just plug it back in and see that all the terms of degree 0 and 1 in $ \varepsilon$ cancel out.  You seem to have a perfectly fine grasp on how to tackle problems of this sort, though you might benefit from doing more examples (especially trickier ones, if you have access to them).

The meaning of "acceptable enough approximation" depends on context -- here you were asked to give the first two terms, and you did.  It's quite possible to give more, though the arithmetic gets less and less enjoyable.  For many purposes, my understanding is that linear approximations are really what you want because they're "good enough" estimates and very easy to work with, while going to higher-order approximations improves your estimates slightly but decreases ease of manipulation substantially -- but this is also extremely context-dependent.

Okay, many thanks for your help and comments. I've got my first year final tomorrow!


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11/15일에 개최된 경시대회로, 대학교 1-2학년 수준의 수학적 배경지식만을 요구하는 문제입니다.
A competition held in 11/15, which only presumes mathematical background of at most sophomore level.

[hide="Korean"][b]제 27회 전국 대학생 수학 경시대회 / 제 1차 문제[/b]

[b]1-1.[/b] 꼭지점의 좌표가 $ (0, 1, 0), (1, 2, 1), (1, 3, 3), (3,2,1)$ 인 사면체의 부피를 구하여라.

[b]1-2.[/b] $ 0 < a, b, c, d < 1$ 일 때, 다음 부등식이 성립함을 보여라
\begin{eqnarray*} \sqrt {(1 - ab)(1 - cd)} & \geq & \sqrt {ac(1 - b)(1 - d)} + \sqrt {bd(1 - a)(1 - c)} \\
& & + \sqrt {(1 - a)(1 - b)(1 - c)(1 - d)} \end{eqnarray*}
[b]1-3.[/b] 다음 적분을 계산하여라.
\[ \int_{0}^{\infty} \frac {e^{ - x} - e^{ - 2x}}{x} \, dx
\]
[b]1-4.[/b] $ \alpha > 1$ 일 때, 다음 적분값이 존재함을 보여라.
\[ \int_{1}^{\infty} \cos(x^{\alpha}) \, dx
\]
[b]1-5.[/b] 다음 값을 구하여라.
\[ \sin \frac {\pi}{10} \sin \frac {2\pi}{10} \cdots \sin \frac {9\pi}{10}
\]
[b]제 27회 전국 대학생 수학 경시대회 / 제 2차 문제[/b]

[b]2-1.[/b] 모든 성분이 양수인 $ 3 \times 3$ 행렬 $ A$의 역행렬 $ A^{ - 1}$이 존재한다. $ A^{ - 1}$의 성분 중에서 양수인 성분이 6개일 때, $ A^{ - 1}$의 성분 중에서 음수인 성분은 3개임을 보여라.

[b]2-2.[/b] 폐구간 $ [0, 1]$에서 정의된 함수열 $ \phi_n$이 다음을 만족시킨다.
(a) $ \phi_0 (t) \equiv 0$
(b) $ \textstyle \phi_n (t) = 1 + 2 \sin\frac {t}{2} - \frac {1}{2} \int_{0}^{t} \phi_{n - 1}(s) \, ds, \quad n = 1, 2, \cdots$
폐구간 $ [0, 1]$에서 함수열 $ \phi_n$이 수렴함을 보이고 $ \lim_{n\to\infty} \phi_n = \phi$ 을 구하여라.

[b]2-3.[/b] 모든 $ n = 0, 1, 2, 3, \cdots$ 에 대해서 $ a_n$ 은 $ 0$ 또는 $ 1$ 이고, $ P$ 와 $ Q$ 는 다항식으로서 $ P$ 의 차수가 $ Q$ 의 차수보다 낮으며 $ |x| < 1$ 인 영역에서 다음이 성립한다.
\[ \sum_{n = 0}^{\infty} a_n x^n = \frac {P(x)}{Q(x)}.
\]
이때, 적당한 자연수 $ p$가 존재하여 모든 $ n \geq 0$ 에 대하여 $ a_{n + p} = a_{n}$ 이 성립함을 보여라.

[b]2-4.[/b] $ p_i, q_i > 0$ 이고 $ \sum_{i = 1}^{n} p_i = \sum_{i = 1}^{n} q_i = 1$ 일 때, 다음 부등식이 성립함을 보여라.
\[ \frac {1}{2} \sum_{i = 1}^{n} (p_i - q_i)^2 \leq \sum_{i = 1}^{n} p_i \ln \left( \frac {p_i}{q_i} \right)
\]
[b]2-5.[/b] 수열 $ \{ x_n \}$ 이 다음 조건
\[ x_m > 0, \quad x_{m + m} \leq x_m + x_n \quad m, n = 1, 2, 3, \cdots
\]
을 만족할 때, 수열 $ \{ x_n / n \}$ 이 수렴함을 보여라.
[/hide]

[hide="English"][b]27th Math Competition for University Students / 1st Exam[/b]

[b]1-1.[/b] Calculate the volume of tetrahedrom with vertices $ (0, 1, 0), (1, 2, 1), (1, 3, 3), (3,2,1)$.

[b]1-2.[/b] Show that the following inequality holds for $ 0 < a, b, c, d < 1$:
\begin{eqnarray*} \sqrt {(1 - ab)(1 - cd)} & \geq & \sqrt {ac(1 - b)(1 - d)} + \sqrt {bd(1 - a)(1 - c)} \\
& & + \sqrt {(1 - a)(1 - b)(1 - c)(1 - d)} \end{eqnarray*}
[b]1-3.[/b] Evaluate the following integral:
\[ \int_{0}^{\infty} \frac {e^{ - x} - e^{ - 2x}}{x} \, dx
\]
[b]1-4.[/b] Show that the following integral exists for $ \alpha > 1$.
\[ \int_{1}^{\infty} \cos(x^{\alpha}) \, dx
\]
[b]1-5.[/b] Evaluate this:
\[ \sin \frac {\pi}{10} \sin \frac {2\pi}{10} \cdots \sin \frac {9\pi}{10}
\]

[b]27th Math Competition for University Students / 2nd Exam[/b]

[b]2-1.[/b] Supposse $ A$ is a $ 3 \times 3$ matrix of positive entries which has the inverse $ A^{ - 1}$. If 6 entries of $ A^{ - 1}$ is known to be positive, show that 3 entries of $ A^{ - 1}$ is negative.

[b]2-2.[/b] A sequence $ \phi_n$ of functions on $ [0, 1]$ satisfies the following conditions:
(a) $ \phi_0 (t) \equiv 0$
(b) $ \textstyle \phi_n (t) = 1 + 2 \sin\frac {t}{2} - \frac {1}{2} \int_{0}^{t} \phi_{n - 1}(s) \, ds, \quad n = 1, 2, \cdots$
Show that $ \phi_n$ converges on $ [0, 1]$, and find the limit $ \lim_{n\to\infty} \phi_n = \phi$.

[b]2-3.[/b] Suppose $ a_n$ is either $ 0$ or $ 1$ for all $ n = 0, 1, 2, 3, \cdots$, $ P$ and $ Q$ are polynomials with degree of $ P$ less then that of $ Q$, and the following equality holds for $ |x| < 1$.
\[ \sum_{n = 0}^{\infty} a_n x^n = \frac {P(x)}{Q(x)}.
\]
Show that there exists a positive integer $ p$ such that $ a_{n + p} = a_{n}$ for all $ n \geq 0$.

[b]2-4.[/b] Given $ p_i, q_i > 0$ with $ \sum_{i = 1}^{n} p_i = \sum_{i = 1}^{n} q_i = 1$. prove the following inequality.
\[ \frac {1}{2} \sum_{i = 1}^{n} (p_i - q_i)^2 \leq \sum_{i = 1}^{n} p_i \ln \left( \frac {p_i}{q_i} \right)
\]
[b]2-5.[/b] Suppose $ \{ x_n \}$ satisfies
\[ x_m > 0, \quad x_{m + m} \leq x_m + x_n. \quad m, n = 1, 2, 3, \cdots
\]
Show that $ \{ x_n / n \}$ converges.
[/hide]

For 2nd exam; 2-4. I have tried to solve the problem, but I couldn't it.

Maybe we could use the inequality $ 1\minus{}\frac{1}{x}<\ln x<x\minus{}1\ (x>0)$, but failed.

The problem seems to associate with Informatic Theory, say, Shannon Inequality.

Could you give us a hint?

for first exam 1-1:
\[\begin{array}{l}
A = (0,1,0),B = (1,2,1),C = (1,3,3),D = (3,2,1)\\
\overrightarrow {AB}  = (1,1,1)\\
\overrightarrow {AC}  = (1,2,3)\\
\overrightarrow {AD}  = (3,1,1)\\
 \Rightarrow V = \frac{1}{6}\left| {\overrightarrow {AB}  \cdot (\overrightarrow {AC}  \times \overrightarrow {AD} )} \right| = \frac{1}{6}\left| {\left| {\begin{array}{*{20}{c}}
1&1&1\\
1&2&3\\
3&1&1
\end{array}} \right|} \right| = \frac{{\sqrt {10} }}{2}
\end{array}\]


What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster board if every cut must be made parallel to a side of the poster board? No taping or gluing is allowed.

[hide]
Since the dimensions of the notecards are even, 25*27 is equivalent to 24*26. And, since there is a way to cut notecards from 24*26 without any left over, the answer is $ \boxed{26}$.[/hide]


A function f is even if f(x) = f(-x). Give two even functions, one non-constant polynomial and one non-polynomial.

YOU DO IT! :P

[rant]Seriously. You are a smart little boy. You can think of a non-constant polynomial, and a non-polynomial. You have the definition of even functions. There are infinitely many even functions. I just generated millions in my head.

And why are these in pre-olympiad?[/rant]


What is the distance, in units, between the points $ (\minus{}3, \minus{}4)$ and
$ (4, \minus{}5)$? Express your answer in simplest radical form.

By distance formula, we have $ \sqrt{(\minus{}3\minus{}4)^2\plus{}(\minus{}4\minus{}(\minus{}5))^2}\equal{}\sqrt{49\plus{}1}\equal{}\sqrt{50}\equal{}5\sqrt{2}$.


function
trigonometry
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[quote="countdownmath"]Show that the function $f: R \to R$

$f(x) = x^{2}\cos\frac{1}{x}, (x\neq 0)$ and $f(x) = 0, (x=0)$

is differentiable $\forall x\in R$ but the derivative $f'$ is not continuous at $0$.[/quote]
obviusly : there not exist limit in point $0$.
begin solve: supppose that there exist limit in point 0. it mean $forall{\epsilon}>0$ exist $\delta >0$:... simpl


Let a perfect product be a number $n$ more than the product of three consecutive positive integers with $n$ being the middle of the three.  Find the smallest positive integer $k$ for which there is no perfect product that can be written as the sum of $k$ consecutive positive integers.

for $n \ge 2$, $n^{3}$ is a perfect product, and these are the only perfect products.

$(n-1)(n)(n+1)+n = n(n^{2}-1)+n = n^{3}-n+n = n^{3}$.  $n-1$ has to be a positive integer, so $n \ge 2$.

somebody else can finish (or i will, later)

[hide="Continuing from previous post"]
$k = 4$

It's not less than that since $3^{3}= 27 = 13+14 = 8+9+10$. However, $4$ consecutive positive integers sum to a number equivalent to 2 mod 4, which is a residue no cube can have.[/hide]

[quote="miyomiyo"][quote="MellowMelon"] However, $4$ consecutive positive integers sum to a number equivalent to 2 mod 4, which is a residue no cube can have.[/hide][/quote]
Explain.[/quote]
First, $x+(x+1)+(x+2)+(x+3) = 4x+4+\boxed{2}$.

Then, cubing possible residues in modulo $4$:
$0^{3}\equiv 0 \bmod 4$
$1^{3}\equiv 1 \bmod 4$
$2^{3}\equiv 0 \bmod 4$
$3^{3}\equiv 3 \bmod 4$.

Since $2 \bmod 4$ isn't a possible residue of a cube, a cube cannot be written as the sum of four consecutive numbers.

Follow-up question:

Describe all integers $k$ that work.

[hide="Well..."] When $k$ is odd, a sum of $k$ consecutive positive integers with middle term $a$ can be written in the form $ak$, so let $a = k^{2}$.

When $k = 2m$ is even, a sum of $2m$ consecutive positive integers with middle terms $a, a+1$ can be written in the form $m(2a+1)$.  This obviously cannot be a cube if the greatest power of $2$ that divides $m$ is not a cube.  In other words,

$k \neq 2^{3r+1}o$

Where $o$ is odd.  If $k$ is of this form, then simply let $a = \frac{o^{2}-1}{2}$.

In other words, $k$ satisfies the constraints of the original problem if and only if it is of the form

$k = 2^{3r+2}o, k = 2^{3r+3}o, r \ge 0$ [/hide]


calculus
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[hide="Hint"]Use the identity $ \frac1{\sin^2x}\equal{}\csc^2x$ and then integrate the above integral with integration by parts...[/hide]

[quote="Carcul"]
Now try $ I' \equal{} \int_{\pi/4}^{\pi/2}(x \csc x)^2\,dx$[/quote]
It's the same, we apply integration by parts twice.

That's right. Now find the last integral.

It's on the link I gave, the solution involves Catalan's Constant  :wink:


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find all functions with reel numbers wich satisfy
$f(\frac{x+y}{1+xy})=\frac{f(x)f(y)}{|1+xy|}$. :)

Setting $y=1$ gives us $f(1) = f\left( \frac{x+1}{1+x}\right) = \frac{f(1)f(x)}{|x+1|}$, so either $f(x) = |x+1|$ (except possibly when $x=-1$), or $f(1) = 0$.

Setting $y=-1$ gives us $f(-1) = f\left( \frac{x-1}{1-x}\right) = \frac{f(-1)f(x)}{|x-1|}$, so either $f(x) = |x-1|$ (except possibly when $x=1$) or $f(-1) = 0$.

We have now found two solutions, $f(x) = |x+1|$ for $x\neq-1$, and $f(x) = |x-1|$, for $x \neq 1$.  Indeed, we have just seen that these functions work for $x =-1, 1$ in both cases.  It is easy to see that both of these solutions meet the problem's requirements.  We will now look for other solutions.

Letting $y=0$ gives us $f(x) = f(x)f(0)$, which means that either $f(x)= 0$, or $y=1$.  The former satisfies the problem's conditions; we will look for more solutions.

Setting $y=-x$ gives us $f(0) = \frac{f(x)f(-x)}{|1-x^{2}|}$, so for all $x$, \[f(x)f(-x) = |1-x^{2}|. \qquad (*) \]

We note that if we let $y = \frac{-2x}{x^{2}+1}$, then we have $\frac{x+y}{1+xy}=-x$.  Thus $f(-x) = \frac{f(x)f\left( \frac{-2x}{x^{2}+1}\right)}{ \left| 1-\frac{2x^{2}}{x^{2}+1}\right|}$, or (for $x \neq \pm 1$)
\[\frac{f(-x)}{f(x)}= \frac{ f \left( \frac{-2x}{x^{2}+1}\right)}{ \left| \frac{x^{2}-1}{x^{2}+1}\right| }. \]
Since $\left| \frac{x^{2}-1}{x^{2}+1}\right|$ is an even function, we may write
\[1 = \frac{f(-x)}{f(x)}\cdot \frac{f(x)}{f(-x)}= \frac{ f \left( \frac{-2x}{x^{2}+1}\right) }{ f \left( \frac{2x}{x^{2}+1}\right) }. \]
Note that $\frac{2x}{x^{2}+1}$ takes on exactly the values in the interval $(-1,1)$, when $x \neq 1$.  Indeed, it takes the values $-1$ and $1$ when $x =-1, 1$, so it must take on all values in between, but by AM-GM, $2|x| \le x^{2}+1$, with equality only when $|x| = 1$.  Thus by $(*)$, we have $f(x) = \sqrt{1-x^{2}}$, for $x \in (-1,1)$.

This means that for any $x \neq \pm 1$, we must have
\[\frac{f(-x)}{f(x)}= \frac{ \sqrt{ 1-\frac{4x^{2}}{(x^{2}+1)^{2}}}}{ \left| \frac{x^{2}-1}{x^{2}+1}\right| }= \frac{ \sqrt{ \frac{(x^{2}-1)^{2}}{(x^{2}+1)^{2}}}}{ \left| \frac{ x^{2}-1}{x^{2}+1}\right| }= 1 . \]

Since $f(1) = f(-1) = 0$, it follows that for any $x$, $f(x) = \sqrt{ \left| 1-x^{2}\right| }$.  It is not difficult to see that htis function works as well.

Thus the functional equation has four solutions :  $f(x) = |1-x|, |1+x|, 0 , \sqrt{ \left| 1-x^{2}\right| }$.


Is there a formula or general method for distributing identical objects into identical groups? An example might be candy into piles.

So far, I have come up with a strategical way of listing them. For example, if 3 candies were to be distributed into 2 piles, the possibilities are (3,0) or (2,1), but anything else will be the same as those two. But I want a way to find the number of ways to distribute $ n$ identical things into $ m$ groups, where I don't know $ n$ or $ m$.

Thanks!

I know about that, but isn't that for identical balls into [b]distinguishable[/b] urns? I'm looking for both to be identical.

EDIT: AoPS Wiki says [url=http://alt1.artofproblemsolving.com/Wiki/index.php/Distinguishability#Indistinguishable_to_indistinguishable]this[/url], which is just casework.

Look at this link: http://www.artofproblemsolving.com/Wiki/index.php/Distinguishability

You want indistinguishable to indistinguishable.

EDIT: meh...you found the link yourself...

Casework isn't at all the best way.  The pentagonal number theorem implies a [url=http://en.wikipedia.org/wiki/Partition_%28number_theory%29#Generating_function]pretty efficient recurrence[/url] for computing the partition function.

I looked at partitions, but that has "groups of any size". What is the pentagonal number theorem?

[quote="PowerOfPi"]I looked at partitions, but that has "groups of any size".[/quote]
Oh, sorry.  You want partitions into a fixed number of groups; that turns out to be "easier," depending on your point of view.  The number $ p_m(n)$ of partitions of $ n$ identical objects into $ m$ groups turns out to be equal to the number of partitions of $ n$ identical objects into groups of size at most $ m$, and this has generating function

$ \sum_{n \ge 0} p_m(n) x^n \equal{} \frac{1}{(1 \minus{} x)...(1 \minus{} x^m)}$.

For fixed $ m$ this gives you a closed formula.  There is also a straightforward recurrence [url=http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29#Intermediate_function]here[/url] which should be faster than casework.


integration
trigonometry
calculus
function
calculus computations

R is the region that lies between the curve $y=\frac{1}{x^{2}+4x+5}$ and the x-axis from x = -3 to x = -1.

Find:(1) the volume of the solid generated by revolving R around the y-axis.

(2)the volume of the solid generated by revolving R around the x-axis.

[quote="Zeppelin"]R is the region that lies between the curve $y=\frac{1}{x^{2}+4x+5}$ and the x-axis from x = -3 to x = -1.

Find:(1) the volume of the solid generated by revolving R around the y-axis.

(2) the volume of the solid generated by revolving R around the x-axis.[/quote]

(1) The volume of the solid generated by rotating the curve $y=f(x)$ between $x=a$ and $x=b$ about the y-axis is given by

\[V=2\pi\int_{a}^{b}xf(x) dx , \qquad \mbox{ for }0\leq a\leq b ,\]

but is given by

\[V=2\pi\int_{a}^{b}(-x)f(x) dx=-2\pi\int_{a}^{b}xf(x) dx , \qquad \mbox{ for }a\leq b\leq 0 ,\]

since the radius of each cyldrical shell is negative in that case. In particular, the volume of the solid generated by rotating the region bound by $y=\frac{1}{x^{2}+4x+5}$ and the x-axis from $x =-3$ to $x =-1$ about the y-axis is

\[V=-2\pi\int_{a}^{b}xf(x)dx=-2\pi\int_{-3}^{-1}\frac{x dx}{x^{2}+4x+5}=-2\pi\int_{-3}^{-1}\frac{xdx }{(x+2)^{2}+1}\]

now let $x+2 = \tan\theta \Rightarrow dx=\sec^{2}\theta d\theta$ so that $x= \tan\theta-2$ and so that $-3\leq x\leq-1\Rightarrow-1\leq x+2\leq 1 \Rightarrow-1\leq \tan\theta\leq 1\Rightarrow-\frac{\pi}{4}\leq\theta\leq \frac{\pi}{4}$ and the integral becomes

\[V=-2\pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\left(\tan\theta-2\right)\sec^{2}\theta d\theta }{\tan^{2}\theta+1}= 2\pi\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}}\left(2-\tan\theta \right)d\theta\]
\[= 4\pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}d\theta-2\pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan\theta d\theta = 4\pi\left[\theta\right]_{\theta =-\frac{\pi}{4}}^{\frac{\pi}{4}}-0 = 4\pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = 2\pi^{2}\]

where the fact that $\tan\theta$ is an odd function has been used to determine that $\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}}\tan\theta d\theta=0$ and the trig identity $\tan^{2}\theta+1 = \sec^{2}\theta$ has been observed.

(2) The volume of the solid generated by rotating the curve $y=f(x)$ between $x=a$ and $x=b$ about the x-axis is given by

\[V=\pi\int_{a}^{b}[f(x)]^{2}dx, \qquad \mbox{ for }a\leq b .\]

In particular, the volume of the solid generated by rotating the region bound by $y=\frac{1}{x^{2}+4x+5}$ and the x-axis from $x =-3$ to $x =-1$ about the x-axis is

\[V=\pi\int_{a}^{b}[f(x)]^{2}dx = \pi\int_{-3}^{-1}\frac{dx}{\left( x^{2}+4x+5\right)^{2}}= \pi\int_{-3}^{-1}\frac{dx}{\left[ (x+2)^{2}+1\right]^{2}}\]

now (as in part (1)) let $x+2 = \tan\theta \Rightarrow dx=\sec^{2}\theta d\theta$ so that $-3\leq x\leq-1\Rightarrow-\frac{\pi}{4}\leq\theta\leq \frac{\pi}{4}$ and the integral becomes

\[V=\pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\sec^{2}\theta d\theta}{\left( \tan^{2}\theta+1\right)^{2}}= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\sec^{2}\theta d\theta}{\left( \sec^{2}\theta\right)^{2}}= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{d\theta}{\sec^{2}\theta}\]
\[= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos^{2}\theta d\theta = \frac{\pi}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1+\cos 2\theta\right) d\theta = \frac{\pi}{2}\left[\theta+\frac{1}{2}\sin 2\theta\right]_{\theta =-\frac{\pi}{4}}^{\frac{\pi}{4}}\]
\[= \frac{\pi}{2}\left[\left(\frac{\pi}{4}+\frac{1}{2}\sin \frac{\pi}{2}\right)-\left(-\frac{\pi}{4}+\frac{1}{2}\sin \left(-\frac{\pi}{2}\right)\right) \right] = \frac{\pi}{2}\left(\frac{\pi}{2}+1\right) \]


Solve in $ \mathbb{Z^\plus{}}$ the equation $ x^2\plus{}y^2\equal{}1999xy$.

write the equation as $ x^2\minus{}1999yx\plus{}y^2\equal{}0$. the discriminant is $ (1999^2\minus{}2^2)y^2\equal{}(1997)(2001)y^2$, which has to be a perfect square...

Other way :
$ t=\frac{x}{y}$ then
$ t^2+1=1999t$
$ \t^2-1999t+1=0$
This equation has not rational solution so has no pair $ (x,y)\in N$ satisfy.
More than the equation $ x^2+y^2=kxy$ where $ k\in Z$ has positive integer solution if and only if $ k=2$


Find all functions f : N -> N such that

I :  f(m) = 1 <=> m = 1

II : if d = gcd(m,n), f(mn) = f(m)f(n)/f(d)

III : For every m in*  N , we have f^2000(m) = m.

(f^2000(m) is f applied to m 2000 times)

-----

a side note:

is the belongs to symbol accessible on this messsage board?

I claim [hide]that no such functions exist.

Proof:

III implies injectivity, since if f(a) = f(b), a = f2000(a) = f1999(f(a)) = f1999(f(b)) = f2000(b) = b.

However, by II,

f(4) = f(2*2) = f(2)*f(2)/f(gcd(2, 2)) = f(2)*f(2)/f(2) = f(2).

Contradiction.  Thus there is no such function f.[/hide]

There is no belongs symbol in the emoticons ... I'm sure you could find one somewhere and copy it in or something.


[quote="aspiringchemlete"]Approximate $ \sqrt[3]{67}$.

Hint: Show that $ (1+x)^{\frac{1}{3}}\approx 1+\frac{1}{3}\, x$, when $ |x|$ is small.[/quote]

[hide="Proof hint"]Use extended binomial formula:

$ (1+x)^{1/3}= 1+\frac{x}{3}+\binom{1/3}{2}x^{2}+...$

Just take into consideration the first two terms. This isn't actually a proof (how do you prove an approximation?), but I think this is what you wanted.[/hide]

[quote="vishalarul"]This isn't actually a proof (how do you prove an approximation?)[/quote]

Well, there are well-known methods to prove that a series approximation approximates to some particular given degree of accuracy.


Let $M=\{1,2,\dots,49\}$ be the set of the first $49$ positive integers. Determine the maximum integer $k$ such that the set $M$ has a subset of $k$ elements such that there is no $6$ consecutive integers in such subset. For this value of $k$, find the number of subsets of $M$ with $k$ elements with the given property.

consider the partition {1, ..., 6}, {7, ..., 12}, ..., {43, ..., 48}, {49}. at most 5 elements can be chosen from each of the first 8 parts, and 1 from the last part, for a total of at most $8\cdot 5 +1 = 41$ elements. the set which is the union of {1, ..., 5}, {7, ..., 11}, ..., {43, ..., 47}, and {49} shows that 41 can be achieved.

number the first 8 parts of the partition listed above 1 through 8. the argument above shows that a maximal subset is gotten by including 49 and removing exactly one element from each of the parts 1 through 8. for each $i$, $i = 1, ..., 8$, let $a_i$ denote the number of the element of the $i$-th part, counting from the left of that part, which is removed; so, for example, if we remove 11 from part 2, we denote that by $a_2 = 5$, because 11 is the 5th number counting from 7. note that, because we include 49, $a_8 \geq 2$. in addition, note that we must have $a_i \geq a_{i+1}$ for every $i = 1, ..., 7$, or else we're left with a run of at least 6 consecutive integers. conversely, if we have $a_8 \geq 2$ and $a_i \geq a_{i+1}$ for every $i = 1, ..., 7$, the resulting set is a valid set.

so our problem becomes to find the number of non-decreasing sequences $a_8 \leq a_7 \leq a_6 \leq ... \leq a_1$ with $2 \leq a_i \leq 6$ for every $i$. if $a_8 = k$ ($2 \leq k \leq 6$), then choosing the remaining 7 $a_i$ is the same as putting 7 objects in $6-k+1$ boxes marked $k, ..., 6$. this can be done in $\binom{7+6-k+1-1}{7} = \binom{13-k}{7}$ ways. the answer is then $\sum_{k=2}^{6} \binom{13-k}{7}$

edit: ...which actually is the same as $\binom{12}{8}$. we could have just gotten that by covering all the $k$'s in one case... by placing 8 objects in 5 boxes marked $2, ..., 6$. anyway, another case of the identity $\binom{n}{n}+\binom{n+1}{n}+...+\binom{m}{n} = \binom{m+1}{n+1}$.


modular arithmetic
number theory unsolved
number theory

Prove that there exist infinitely set which contains $ 2000$ consecutive integers such that each number has a divisor which can be represented as $ a^2000$ with $ a\in Z_\plus{} ,a>1$

Use the Chinese Remainder Theorem: Let $ p_k$ denote the $ k$th prime. Then there exists a positive integer $ x$ such that
\[ x\equiv 0\pmod{2^{2000}},x\equiv \minus{}1\pmod{3^{2000}},x\equiv \minus{}2\pmod{5^{2000}},\cdots\]
\[ x\equiv \minus{}1999\pmod{p_{2000}^{2000}}\]
In fact, there are infinitely many such $ x$, since if $ x$ solves this congruence, so does $ x\plus{}\prod_{k\equal{}1}^{2000} p_k^{2000}$. For each such $ x$, the set $ \{x,x\plus{}1,x\plus{}2,\cdots ,x\plus{}1999\}$ works.


[quote="tytus"]Imagine we have a rope. We are also given some (max. 100000) points (range 1-1000000000). Now we are ask to choose n of them, but the distance between each other has to be maximal.

So for example we have:
Points: 1,2,4,8,9 and we have to choose 3 of them. So the answer is 1,4,9.

But when the number of points is bigger it's hard to find the answer. I see that DP (Dynamic Programming) is needed here, but I can't see how to solve it...

Can anyone give me a hint, how to solve it??

Please help...[/quote]

I think your statement of the problem is a bit ambiguous.

What does it mean that "the distance between each other has to be maximal"?  Does that mean that the minimum distance must be maximal?

I mean - we want to choose points on rope, such that the minimum distance between any two of them is as large as possible. The question is: How to find the largest minimum distance?

[quote="MCrawford"]This sounds more like a computing problem.  You're looking for an efficient algorithm to a problem whose generalization takes an enormous amount of information into account.  I don't think think this is the best forum for your question, though I'm not sure what forum would be.[/quote]

Probably in 'Other problem solving topics', perhaps.

As for the problem: heres a way to approach it, though I'm not sure what efficiency you are looking for.

Two parts: 1) choosing the minimum gap, 2) seeing if picking the numbers is possible with that minimum gap.

As an example of how it would work, for the numbers 1, 4, 6, 10, 13, 20, 25 where you have to pick 4 of them:

If we fix the minimum allowable gap at 1, we can choose 1, 4, 6, 10
Gap 2: 1, 4, 6, 10
Gap 3: 1, 4, 10, 13
Gap 4: 1, 6, 13, 20
Gap 5: 1, 6, 13, 20
Gap 6: 1, 10, 20, impossible.

So the answer here would be a gap of 5 (numbers 1, 6, 13, 20).

This just works by continuously picking the next smallest number that keeps that gap between the numbers.

You could then improve that by using binary searches for both parts (for choosing the gap, if a particular gap doesn't work you are in the lower half, if it does work you are in the upper half, and the same thing for finding the next smallest number in the list.)

There may be a faster way though..

[quote="TripleM"][quote="MCrawford"]This sounds more like a computing problem.  You're looking for an efficient algorithm to a problem whose generalization takes an enormous amount of information into account.  I don't think think this is the best forum for your question, though I'm not sure what forum would be.[/quote]

Probably in 'Other problem solving topics', perhaps.
[/quote]

Ha!  Some days I'm a little too tired to understand the forum I help operate.  Duh.


A particle of mass m is moving in a circular path of radius R. The kinetic energy of the particle varies as K.E.=k$ s^2$, where k is a constant and s is the distance covered. Find the force acting on it.

[hide="hint"]
when  a particle moves in a curve the accelaration can be divided into two perpendicular components
$ a_{t}=\frac_dv}{dt}$
$ a_{n}=\frac{v^{2}}{r}$ where $ r$ is the radius of curvature
what is given to u is
$ \frac{mv^{2}}{2}=ks^{2}$
just differentitae both sides to get $ \frac{dv}{dt}$
also u can get $ v$ from the above expression
$ r$ is given so now just add vectorally the two perpendicular components[/hide]

[quote="pardesi"][hide="hint"]
when  a particle moves in a curve the accelaration can be divided into two perpendicular components
$ a_{t} = \frac_dv}{dt}$
$ a_{n} = \frac {v^{2}}{r}$ where $ r$ is the radius of curvature
what is given to u is
$ \frac {mv^{2}}{2} = ks^{2}$
just differentitae both sides to get $ \frac {dv}{dt}$
also u can get $ v$ from the above expression
$ r$ is given so now just add vectorally the two perpendicular components[/hide][/quote]

thanks pardesi
I got it.


hmm I'm trying to use the USAMTS template, but I'm getting an error.

It says Bibtex is:
"I found no \citation commands while reading solutions.aux"
same for \bibdata and \bibstyle

It also says
"Couldn't find input index file solutions.idx"
The usamts.tex and fancyhdr files are both in the same folder, so I'm not sure what's wrong...

(I just installed Miktex and Texniccenter, I was able to compile a normal file and preview PDF fine)

Try deleting all the auxiliary files and compiling again - occasionally the compiler does weird things to the auxiliary files and then spits out errors.


number theory
relatively prime
number theory proposed

[quote="manjil"]Prove that the equation $ x^2 \plus{} y^2 \equal{} z^{2009} \plus{} z$ has infinitely many relatively prime integer solutions.[/quote]

Let $ a,b$ relatively prime integers. Then : $ (a(a^2\plus{}b^2)^{1004}\plus{}b)^2\plus{}(b(a^2\plus{}b^2)^{1004}\minus{}a)^2$ $ \equal{}(a^2\plus{}b^2)^{2009}\plus{}(a^2\plus{}b^2)$


[quote="Iver"]Four circles are given. Every circle is tangent in appearance to two from remaining. To show that points of contact lie on one district.[/quote]

Let $ r_A,r_B,r_C,r_D$ denote the radii of the four circles, and $ A,B,C,D$ the centers respectively. Then the equalities $ AB\equal{}r_A\plus{}r_B,BC\equal{}r_B\plus{}r_C, CD\equal{}r_C\plus{}r_D, DA\equal{}r_D\plus{}r_A$ imply that the quadrilateral $ ABCD$ can be circumscribed around a circle. The points mentioned in the problem lie on this circle.

Second Solution: Number the circles $ 1,2,3,4$ and let $ A,B,C,D$ denote their centers respectively and furthermore let $ P,Q,R,S$ denote the points of tangency of $ 1,2;\ 2,3;\ 3,4;\ 4,1$ respectively. Then we have $ \angle APS\equal{}\angle ASP\equal{}w$, $ \angle BPQ\equal{}\angle BQP\equal{}x$, $ \angle CQR\equal{}\angle CRQ\equal{}y$, $ \angle DRS\equal{}\angle DSR\equal{}z$, which imply

$ \angle SPQ\plus{}\angle SRQ\equal{}(\pi\minus{}w\minus{}x)\plus{}(\pi\minus{}y\minus{}z)\equal{}2\pi\minus{}(w\plus{}x\plus{}y\plus{}z)$

$ \equal{}\frac 12(\pi\minus{}2w\plus{}\pi\minus{}2x\plus{}\pi\minus{}2y\plus{}\pi\minus{}2z)$

$ \equal{}\frac 12(\angle SAP\plus{}\angle PBQ\plus{}\angle QCR\plus{}\angle RDS)\equal{}\pi$

Hence $ PQRS$ is cyclic.


algebra
polynomial
linear algebra
linear algebra unsolved

Let $f(x)=(x-x_{0})\prod_{k=0}^{n}(x-x_{k})$. Since $A[f]=0$, it follows that $A[e_{n+2}]$ is the opposite of the coefficient of $x^{n+1}$ in $f$. Hence $A[e_{n+2}]=x_{0}+S_{1}$.

The same approach should work for the second part, with some clever choice of a polynomial of degree $(n+3)$.  :maybe:


geometry
geometric transformation
reflection
MATHCOUNTS

Results for Comprehensive (as these are the only ones I remember:)

[hide="Top 10"]

Edited to reflect Jeremy's post:

1. Jeremy Hahn
2. Vivek Bhattacharya
3. Mikhail Lavrov
4. William Schleiper
5. Daniel Vitek
6. Yasha B-K
7. Damien Jiang
8. Lizi Chen
9. Varun Puvanesarajah
10. ?????

[/hide]

I'm especially mad because the TOP 6 MADE IT OUT OF 72 AND THE REST OF THE CONTESTS HAD THEIR TOP 10% MAKE IT!!

EDIT: yeah, I just looked at the contest website, and I'm like, OH DARN

Congrats to everyone. Wow, you couldn't remember who won? That is odd.

Sorry. But there's always someone good who doesn't make it from that site, so it shouldn't be too bad. And please stop saying you lost after every contest.

The policy for the entire state is for the top 7% to progress, with a minimum of 6. This is true for every comprehensive site.

Well, look at it this way - you lost to a bunch of really good people (most were on the A team last year).

Congratulations to everyone!

Actually, I won, Vivek was second, and Misha was third.

Varun Puvanesarajah came in 9th place

I don't remember the 10th place spot  :(

I am guessing this regional will be the most competitive of the year, as several people who used to go to Elon seem to have migrated

OK, I edited my first post.

Elon will just be Arnav and John Pardon... only they're important :wink:

10th place was Haoyu Chen (he's from Enloe, that's how I know)

Here's a list of the top 10 with scores (if they're wrong, post or PM me and I'll update, just going on my shaky memory)

[b]1. Jeremy Hahn - 119
2. Vivek Bhattacharya - 114
3. Mikhail Lavrov - 106
4. William Schlieper - 103
5. Daniel Vitek - 101
6. Yakov Berchenko-Kogan - 86[/b]
7. Damien Jiang - 76 (edit)
8. Lizi Chen - 74?
9. Varun Puvanesarajah - 72?
10. Haoyu Chen - 62

(Bold = qualified for State)

 - Daniel

Btw Damien, not sure if you remember me, but we had lunch with you last year at UNCG since your mom knew my best friend's mom. And you placed really high for a freshman! Congrats!

...those Wake Techies can't do much right. >.<
I'm from Ligon, and I did the Geometry competition. They gave me a 15 on my test at first, then I had to get my ponsor to pester them into checking my test. Turns out they marked an 80 as a 0, called it a "very human mistake" =\, but gave me my 95, which ranks me third.
Ligon could've swept 1st in all courses -- we got 1st in AlgII, [i]now[/i] we have 1st in Geo once they corrected my score, and would've gotten 1st in AlgI if only our best [AlgI] kid hadn't gone off to California the day before... [incase ya'll went to MathCounts, he's that kid who was answering all the trivia questions before the countdown round @ regionals].
So this year, there are two 1st Place plaques and two 3rd place medals... which is just weird.

Sorry for the ranting! =D Not that anyone read that...

And last year, UNCG graders messed up my best friend's score and gave her 5th instead of 4th. =\

Ligon has fallen far if Calvin Deng is the best there...

Oh well, I had my share of embarrassing moments at Ligon.  Take for example my 7th grade year at Wake Tech.  I got 2nd place in geometry, and not 5 minutes later as our team exited the room I dropped my trophy and it broke.  I finally had it fixed about a year ago.

Oh by the way Damien - if you saw my MOP post, I was talking about blue/black MOP.  You'll probably go to red MOP, or if not come very close (as I did last year)

Oh, nice job.

I guess he's not the best, but he will be when he grows up some... he's just 10 years old.

Oh, and the [unedited >.< with mistakes, if mine wasn't the only one] results are here!:
[url]http://www.waketech.edu/~mdpearce/contest/contest2/2007%20Results.doc[/url]

a. Calvin is annoying
b. he only beat me in the countdown round because I couldn't find the buzzer
c. Claire fluked
d. I didn't mean that (hopefully she doesn't have an AoPS account)
e. Damien got unlucky. He would've sweeped if he didn't only get 1 of the not multiple choice questions right.
f. That competition is annoying
g. Damien stop saying that I spam
h. Cheese


quadratics
linear algebra
matrix
integration
function
algebra
polynomial

$N^* = N -(0) $

Let $f:R\rightarrow R$ such that $ \forall k \in N^*$ , $f(k)>0$ , and
$ \forall (k,j) \in (N^*)^2, (f(k+j))^2 = (f(k))^2 + (f(j))^2 $
What is the signature of the quadratic form defined on $R^n$  by $Q(x) = \sum_{k,j=1}^{n} \frac{1}{f(k+j)}x_k.x_j $

the signature of $Q(x)$ is the same to the signature of
\[R(x) = \sum_{i,j=1} \frac{x_i x_j}{\sqrt{i+j}}\]
$R(x)$ seems to be positive definite... :P

[quote="Moubinool"]You are wright, give a proof for the signature of $R(x)$[/quote]

eum? the sum of positive items is positive??

I finally realized how to do this- I looked at it many months ago but couldn't do it then. This solution is even a generalization:

We write the matrix $M$ ($m_{ij}=(r_i+r_j)^{-a},\,a>0,\,r_i>0$) as a Gramian. Consider the inner product $\langle f,g\rangle=\int_0^\infty \frac{x^{a-1}}{\Gamma(a)}f(x)g(x)\, dx$ for bounded functions $f,g$ from $\mathbb{R}^+$ to $\mathbb{R}$ which decay faster than any polynomial at $\infty.$ Let $f_r(x)=e^{-rx}.$ We have $\langle f_r,f_s\rangle=\int_0^\infty\frac{x^{a-1}}{\Gamma(a)}e^{-(r+s)x}\, dx=\frac1{\Gamma(a)}\int_0^\infty\left(\frac t{r+s}\right)^{a-1} e^{-t}\frac{dt}{r+s}$
$=\frac1{(r+s)^a}\cdot\frac1{\Gamma(a)}\cdot\int_0^\infty t^{a-1}e^{-t}\, dt=\frac1{(r+s)^a}.$

Finally, $m_{ij}=\langle f_{r_i},f_{r_j}\rangle$. The question asked is the special case $a=\frac12, r_i=i$.


vector
floor function
combinatorics proposed
combinatorics
Baltic Way
2005
#7

A rectangular array has $ n$ rows and $ 6$ columns, where $ n \geq 2$. In each cell there is written either $ 0$ or $ 1$. All rows in the array are different from each other. For each two rows $ (x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})$ and $ (y_{1},y_{2},y_{3},y_{4},y_{5},y_{6})$, the row $ (x_{1}y_{1},x_{2}y_{2},x_{3}y_{3},x_{4}y_{4},x_{5}y_{5},x_{6}y_{6})$ can be found in the array as well. Prove that there is a column in which at least half of the entries are zeros.

If we think of a row as being the indicator vector for a subset of a set $X$ of $6$ elements, the problem can be stated as, given a family $\mathcal{F}$ of at least $2$ subsets of a set $X$ of $6$ elements, closed to intersection, then there exists some element $x\in X$ such that $x$ is missing from at least half the subsets in $\mathcal{F}$. This rewording has the advantage to offer a natural interpretation to the stuff on the result of that operation on rows.

Here's a sketch of a solution, which I wrote on my phone during a field trip in October.  I remember commenting that I was melting.

[quote]So uh assume for contradiction that each column has >1/2 ones. So on average there are more ones and zeros and for n>3 that means there will either be a row with exactly four ones or a row with exactly five ones. (Case of 111111 is annoying though).

If there's a row 111110 then last column has more than half zeros since anything that ends with 1 has an analogous row ending with 0.

If there's a row 111100 then let A be the set of those ending with 01 and B with 10. Note that rows ending with 00 are a "subset" of those ending with 11 like before.
So if B>A then last column works; otherwise penultimate column works. [/quote]

To elaborate on the parenthetical note in the first paragraph: note that if 111111 is a row, and all other rows have at most three ones, then there are at most $6+3(n-1) = 3n+3$ ones.  This is pretty bad, because we need at least $6\left\lfloor \tfrac{n+1}{2} \right\rfloor$ ones, and so we deduce that all other rows have three $1$'s, and furthermore, $n$ must be odd.  But then by multiplying two distinct rows with three ones, we get a row with strictly fewer than three ones, and that's a contradiction.


How come there isn't a Tennesse Forum? can we add one?

What you are interested in would be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21296]here[/url].