$ a)$ Prove that if the sequence $ (x_n)_{x\in\mathbb{N}}$ satisfies the relation:

$ x_{n\plus{}1}\equal{}\minus{}x_n\plus{}6x_{n\minus{}1}, \forall n\ge 1$, then there exist $ A,B \in \mathbb{R}$ such that:

$ x_n\equal{}A \cdot 2^n\plus{}B \cdot (\minus{}3)^n, \forall n \in \mathbb{N}$

$ b)$ Determine the functions $ f: (0,\infty)\to(0,\infty)$ which satisfy the relation:

$ f(f(x)\minus{}x)\equal{}6x, \forall x \in (0,\infty)$

[hide="Hint"]

denote $ g(x)\equal{}f(x)\minus{}x$

our equation become :  $ g(g(x))\plus{}g(x)\minus{}6x\equal{}0$

we define the sequence $ (x_n)$ by $ x_0\equal{}x$ and $ x_{n\plus{}1}\equal{}g(x_{n})$

so we have : $ x_{n\plus{}2}\plus{}x_{n\plus{}1}\minus{}6x_{n}\equal{}0$

and we use $ a)$ to find the solutions :)[/hide]


Prove that for all real numbers $x,y,z\in(2;4)$ we have the following inequality: \[\frac{x}{y^{2}-z}+\frac{y}{z^{2}-x}+\frac{z}{x^{2}-y}>1.\]

Note that $\frac{x}{y^{2}-z}>\frac{x}{4y-z}$. So, it suffices to prove that $\frac{x}{4y-z}+\frac{y}{4z-x}+\frac{z}{4x-y}\ge 1$. By CBS we have

\[\left(\frac{x}{4y-z}+\frac{y}{4z-x}+\frac{z}{4x-y}\right)(x(4y-z)+y(4z-x)+z(4x-y))\ge (x+y+z)^{2},\]

hence $\frac{x}{4y-z}+\frac{y}{4z-x}+\frac{z}{4x-y}\ge \frac{(x+y+z)^{2}}{3(xy+yz+zx)}\ge 1$

Dear Thales418!
You had better notice that the hypothesis gives $x,y,z\in(2;4)$, don't occur $x=y=z=4$


[quote="PowerOfPi"]The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $ \$1.43$. Some of the $ 30$ sixth graders each bought a pencil, and they paid a total of $ \$1.95$. How many more sixth graders than seventh graders bought a pencil?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$[/quote]

We know that the cost of a pencil (times 100) is integral, so the cost of a pencil is a common factor of $ 143$ and $ 195$, so the cost of a pencil is 1 or 13. Since there are only 30 sixth graders, the cost of a pencil is 13 cents, and the total is 13 seventh graders and 17 sixth graders. This gives $ 4, \boxed{D}$

[quote=jwhouston]How did you get 13 seventh graders and 17 sixth graders?[/quote]

They found the common factors of $143$ and $195$ and with that they found the cost of a pencil ($13$ cents). If you plug that in then you get $17$ sixth graders and $13$ seventh graders

[size=50]Nice 10 year bump! That's actually impressive! (CONSIDERING SOMEONE ALREADY ANSWERED THE QUESTION!!)[/size]

[quote=jwhouston]How did you get 13 seventh graders and 17 sixth graders?[/quote]

Nice $10$ year bump bud...

Yeah, but shouldn't it be 15 sixth graders and 11 seventh graders?
[size=75]I know its a decade late, but I just took a 2009 AMC 8 test as practice and wanted to check my answers. (I got 18 :D) [/size]

Well, if you prime factorize $143$, you should get $13 \times 11$. If you prime factorize $195$, you should get $13 \times 5 \times 3$. The GCF ([b][u]G[/u][/b]reatest [b][u]C[/u][/b]ommon [u][b]F[/b][/u]actor) is $13$


in quadrilateral $ABCD$, $\angle BAD \simeq \angle ADC$ and $\angle ABD \simeq \angle BCD$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

[hide]
Extend $AB$ and $DC$ to meet at point $E$.
Since $\angle EAD=\angle EDA$, $AE=ED$. (Let $s=AE$.)

Note that $\triangle EBD\sim\triangle ECB$.
We have $\frac{CB}{EB}=\frac{BD}{ED}\Longleftrightarrow \frac{6}{s-8}=\frac{10}{s}$.
Solving for $s$, we get $s=20$.
Hence $EB=12$.

Then $EC=6\cdot\frac{12}{10}=\frac{36}{5}$, so $CD=20-\frac{36}{5}=\frac{64}{5}$.

$\therefore 64+5=\boxed{69}$

[/hide]

Don't you mean $\cong$ instead of $\simeq$, agolsme?


Can i still get my school to register for the amc 12 contest? (the brochure says the late registration deadline is somewhere in late january, im still not sure because another site says that they started mailing the tests). I'm hoping on turning in the registration info by monday.   I have about ten people who are willing to take it.

http://www.unl.edu/amc/b-registration/registration.html

Looks like you're still good for regular registration.

Good luck!

thanks for answering!

I was so scared that i couldnt participate
 :)

Also, even if they already mailed the tests, you could probably still participate (your tests would just get mailed out late, but that doesn't matter as long as they arrive before the test date...)


Since people thought it was messed up, im editing it.


trigonometry
function
geometry
geometric transformation

At high tide, the average depth of water in a harbour is 22m and at low tide the average depth is 10m. The tides in the harbour complete one cycle approximately every 12h. Write a sine funcion that relates the depth of the water in the harbour, y, to the time, t in hours, after low tide.

[quote="Adunakhor"][quote="lingomaniac88"][hide="After trying out various things, I got"]$y=16+6\sin\frac{\pi}6t$[/hide][/quote]

[hide]
I think it should be $y = 16 + 6\sin \displaystyle \left( \frac{\pi}{6}t + \frac{3 \pi}{2} \right)$ since we are starting out at low tide, which is at the bottom of the sin curve.
[/hide][/quote]
[hide] Wouldn't it be $y=16+6\sin{(\frac{\pi}{6}(t+3))} \longrightarrow y=16+6\sin{(\frac{\pi}{6}t+\frac{\pi}{2})}$? [/hide]

EDIT: Never mind. :blush:

It should be

Edit: nvm. There was an error

[quote="drunner2007"]It should be $y=16+6\sin(\frac{\pi}{6}(x+\frac{3\pi}{2}))=16+6\sin(\frac{\pi x}{6}+\frac{\pi^2}{4})$[/quote]Why? It's a translation 3 units to the right or 9 units to the left. So it's $\frac{\pi}{6}(x+9)=\frac{\pi}{6}x+\frac{3\pi}{2}$.

yeah, your right. I'm kind of tired right now.

[quote="lingomaniac88"]Well, I was assuming that we could start anywhere we wanted, so I started just between high and low tide to keep it as simple as possible.  Otherwise, I would have adjusted my answer accordingly.  But it said nowhere in the problem that we had to start from a certain point, so I picked a point to start from.

Okay, I'm done talking now.[/quote]
[quote="king_23"]Write a sine funcion that relates the depth of the water in the harbour, y, to the time, t in hours, after low tide.[/quote]  :D  ;)

Didn't see that part. >.<  Then it would be
[hide]$y=16+6\sin\left(\frac{\pi}{6}t+\frac{3\pi}{2}\right)$[/hide]
Of course, if I could, I would use cosine.


Source: 360 PMC
Find the real solutions to the equation
$(x+y)^2=(x+1)(y-1)$.

let $a=x+1$ and $b=y-1$

$(a+b)^2=ab$

$a^2+ab+b^2=0$

$a=\pm b\text{cis} 120^{\circ}$

and equating imaginary and real parts, $b=0=a$ so
$(x,y)=(-1,1)$ the only real solution

[hide="solution"]
The trivial solution is (-1,1).  If we expand the whole thing out, we have:

$x^2 + y^2 + 2xy = xy + y - x - 1$

Rearanging,

$y^2 + (x-1)y + (x^2 + x + 1) = 0$

The descriminant of this quadratic is

$(x-1)^2 - 4(x^2 + x + 1) = -3x^2 - 6x - 3$

Since this has to be greater than zero for there to be any real solutions,

$-3x^2 - 6x - 3 > 0$
$-3(x^2 + 2x + 1)>0$
$-3(x+1)^2>0$
$(x+1)^2<0$

Obviously, this has no real solutions except x = -1, since any reals squared is greater than zero.  So  the only solution is (-1,1)
[/hide]

alltheman beat me to it with a nicer solution...


parameterization
inequalities
inequalities unsolved

for all positive reals $ a,b,c,k$.
do
we have the following inequality
$ \sqrt {\frac{a^2 \plus{} kbc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac{b^2 \plus{} kac}{a^2 \plus{} c^2}} \plus{} \sqrt {\frac{c^2 \plus{} kba}{b^2 \plus{} a^2}} \geq 2 \plus{} \sqrt {\frac {k}{2}}$?

This is my problem at Crux. At the time I proposed it, I haven't any solution but now I have. The answer is "negative". You can try with $ k\equal{}256$ (there is a nice inequality with this value of $ k$).


I'm pretty sure you can only create that sequence, at least from a computer standpoint, from recursion, memoization, or dp.  Btw for anyone who doesn't know if you have fraction $\frac{a}{b}$ and fraction $\frac{c}{d}$ the next term between those two terms is $\frac{a+c}{b+d}$.  I would also be very interested in a purely mathematical derivation of that sequence given the number of iterations.


quadratics
number theory
relatively prime
algebra unsolved
algebra

$a$ and $b$ are relatively prime. Prove that an equality : $ax^2+by^2 = z^2$ has infinitely much solutions (x,y) in integer numbers which are relatively prime.

Maybe an additional condition is necessary, since, for instance, the equation $2x^2+3y^2=z^2$ has no solution for $x,y,z$ relatively prime integers.
If $y$ is even, then $z$ is even as well and it follows that $2x^2$ is divisible by 4. Replace $y$ by $2y+1$ and $z$ by $2z+1$ to obtain \[ x^2+6y(y+1)+1=2z(z+1) .\] Therefore $x$ is also odd and replacing $x$ by $2x+1$ yields \[ 2x(x+1)+3y(y+1)+1=z(z+1) ,\] a contradiction.

The equation has solutions with $x,y,z$ relatively prime integers iff $a$ is a quadratic residue modulo  $b$ and $b$ is a quadratic residue modulo $a$ (which is not the case if $a=2$ and $b=3$).
I think the moderators should consider moving this topic to Number Theory Problems.

I saw this problem in book. I asked my teacher and he told me that there was a mistake in this book:

$ax^2 + by^2 = z^3$

So we've got $z^3$ instead of $z^2$. Sorry for this..


1.) Solve by elimination:
0.4x-0.5y=5.3
0.3x+0.6y=-0.9

2.) Solve the system of equations:
5x-y+z = 29
3x+2y-3z = 20
x-3y+2z = 3

3.) Solve the system of equations:
5x-2y+7z = 46
x+5y+z = 13
5y-z = 1

4.) A hardware supplier manufactures 3 kinds of clamps; types A, B, and C. Production restrictions force it to make 20 more type C clamps than the total of the other types and twice as many as type B clamps as type A clamps. The shop must produce 260 clamps per day. How many of each type are made per day?

How many type A clamps are produced?

hello, writing your first equation system in the form
$ 4x \minus{} 5y \equal{} 53$
$ 3x \plus{} 6y \equal{} \minus{} 9$
From the first equation we have $ y \equal{} \frac {4x \minus{} 53}{5}$ , inserting this in the second one we get
$ 3x \plus{} \frac {6(4x \minus{} 53)}{5} \equal{} \minus{} 9$ . This implies $ x \equal{} 7$ and inserting $ x$ in the equation above we get $ y \equal{} \minus{} 5$.
Sonnhard.

hello, sorry, this was a typo from me, i have corrected it.
Sonnhard.

[quote="leoxnlin"]TPotter26, please don't expect people to work out system after system of equations like machines. Learn the techniques.

If you really are having such a difficult time, there are plenty of online solvers.[/quote]

What about Cramer's Rule? Oh wait that's only for two variables.
For three variables, if you don't know how to do it, you can ask for help on one type of problem to learn how it's done, but don't ask us to do your homework. That isn't going to get you anywhere on a final.


[b] Is there any short way,or fastest route to determine[u]any given "x[/u] factor." "changing[b] variable[u]"
(Ex.a,c.z.e,n etc.)in any provided equation-so as to determine the possible "values[/u][/b]" thereof?Please explain.

                  Its just my beginning.I'm still on the verge of knowing thev "LATEX[b] Technique."

                                               Respectfully,
                                              A.Eddington

[quote]Its just my beginning.I'm still on the verge of knowing thev "LATEX[b] Technique."

                                               Respectfully,
                                              A.Eddington[/quote]

You can learn LaTeX here:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

and test it here:
http://www.mathlinks.ro/Forum/index.php?f=224

conceive

I think http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 is even better since it was made for the forum.


calculus
integration
trigonometry
logarithms
calculus computations

I'm drawing a blank on this one.  It's been a while I guess.  Anyways, I know how to integrate sec(x), but can't apply that method to this:

$\int\sec^3\theta\,d\theta$

Should I use a substitution?

Thanks for the help.

1. This sort of question belongs in "Calculus Computations and Tutorials." That's what the administrators set up this section for.

2. There's a reduction formula here, at least the first stages of one. Reduction formulas are derived by integration by parts.

The key to an integration by parts is to find something useful to be $dv,$ the thing that you integrate. Suffice it to say that $dv$ had better be something you know the antiderivative of, and that it won't cause too much new trouble.

Start with $dv=\sec^2\theta\,d\theta,$ so that $v=\tan\theta.$ Then you must have $u=\sec\theta.$

Do that integration by parts. There will be two more key steps left after that. One involves the use of one of the forms of the Pythagorean identity - in this case, $\tan^2\theta=\sec^2\theta-1.$ The other step is the idea that if the original integral appears on both sides of an equation, you can solve for it algebraically.

I actually got it using a substitution.  That's interesting though (integration by parts).  Thanks for the reply.

Now you've got me curious. What substitution? And what did you get for the final answer?

I was actually going to do:

$u=\tan{\theta}$

So then I have:

$\int\sqrt{1+x^2}\,dx$

Then use another substitution $x=\sinh{t}$.

I like the ways you both showed me better though.  Thanks again :)

We will use integration by parts:

$\int sec^3\theta d\theta = \int sec^2\theta * sec\theta$

$= tan\theta sec\theta - \int sec\theta tan^2\theta = tan\theta sec\theta - \int sec\theta (sec^2\theta -1)$

$=tan\theta sec\theta - \int sec^3\theta + \int sec\theta$

Rearranging,
$2\int sec^3\theta = tan\theta sec\theta + ln(sec\theta + tan\theta)$

therefore, $\int sec^3\theta= \frac{tan\theta sec\theta + ln(sec\theta + tan\theta)}2$


Galois Theory
superior algebra
superior algebra unsolved

Hello, i'm stuck with this proofs:
1- If $E/K$ and $F/K$ is algebraic extensions with $E$ and $F$ subfields from the same field then $EF/K$ is algebraic.
2- If $E/K$ is algebraic with $E$ and $F$ subfields from the same field then $EF/F$ is algebraic.
3- Show that if $E$ is finite extension of prime degree from $F$ then $E/F$ is simple, that is, $E=F(\alpha)$, $\alpha \in E$.

anyone knows good material on Finite Extensions/ Galois Theory and or Abstract Algebra book? (for begginner as you see). Algebra is really amazing but i don't understand it very well.

1.) $E=K(A)$ and $F=K(B)$ where A and B are subsets of algebraic (over K) elements. Then $EF=K(A\cup B)$ and hence it is also algebraic.
2.) Same idea: $EF=K(A\cup B)=K(B)(A)=F(A)$ and hence it is algebraic too.
3.) Let $|E: F|=p$ and let M be an intermediate field. Then $|E: M|\cdot|M: F|=|E: F|=p$ and hence one of $|M: F|$ and $|E: M|$ is 1, the other is p, so M=F or M=E. Now take any element $\alpha\in E\backslash F$ and consider $M: =F(\alpha)$.

Thanks Gockel.
Can you guys say if my proof (for 1) is correct?

Let $E$ and $F$ be finite extensions.  By adjoining each element of $F \backslash E=\{a_{1},\ldots,a_{n}\}$ at time to $E$ we get $E(a_{1},\ldots,a_{n})$ who is finite then algebraic over $K$. Since $EF$ is the smallest subfield who contains $E$ and $F$ then $EF=E(a_{1},\ldots,a_{n})$

Thanks

Nobody says that E and F are finite extensions. And even if the were, the set $E\backslash F$ is in general not finite (think at extensions over $\mathbb{Q}$...)

Ok Gockel. I was triyng to do these with algebraic, but algebraic don't implies finite, then a i could not adjoin the elements. I'm right?

It is also possible to adjoin an infinite set of elements!


A group of students consisting of 4 NCs, 5 PCs and 6 GCs. Suppose that three students are randomly selected for a committee.

(1)What is the probability that exactly two of the selected students are GCs?
(2)What is the probability that all three of the selected students belong to the same catagory?

(3)What is the probability that one student of each catagory is selected?
(4)Suppose that students are selected randomly at least a GC is selected, what is the probability that it is necessary to perform selection at least 6 times?

Note: Please help .. need it urgent !!

Is 1) $ 4/25$ I want to know if I got this probability's basics...

[quote="kratsg"]For the first question, examine the probability of picking a GC out of the 15.

6/15.

Then, there are 5 GCs left, and 14 students. So the probability of picking another GC is down to 5/14.

(6/15)*(5/14)*(4/13)[/quote]
well you don;t want to pick another GC, you only want 2 GC

(6/15)*(5/14)*(9/13)*3
6/15 is the probability one GC is picked
then 5/14 is the prob a second GC is picked
but we want EXACTLY two
so 9/13 is the probability you don't picking a remaining GC
and we multiply by 3 since one GC doesn't HAVE to chosen first Letting O stand for a non GC we could have
O Gc Gc
Gc O Gc
Gc Gc O

[quote="brconterato"]

I didn't understand the number (4)...[/quote]

i think....(correct me if im wrong) it means

assume we keep on picking randomly until a GC is selected. What is the probability that it is necessary to perform this action at least 6 times?


Can anyone provide me with the final statistics?

How many participants?
How many gold, silver, and bronze winners?

Congrats to everyone... Thanks for the prizes Mr. Rusczyk and the rest of the USAMTS team!

All of the stats are available [url=http://www.usamts.org/Leaders/U_Stats.php]here[/url].


(im not sure NT is the right forum...if not a Mod can move this. thanks!)

Let $\alpha$ be a given irrational number.
prove that the sequence $\{\{n\alpha\}\}_{n\in\mathbb{N}}$ is dense in interval $(0,1)$

note:
the sequence is reffering to the fractional part of $n\alpha$.

Sometimes on the forum, for example: http://www.mathlinks.ro/Forum/viewtopic.php?t=27475


There are 52 playing cards. We are choosing one card. If we know that the card is red, find probability that card to be of hearts .

I think The answer should be $ \frac {13}{26} \equal{} \frac {1}{2}$.

It's $ \frac12.$ The words "if we know" make this a conditional probability.


$ f(x)$ を $ x$ について微分できる関数とするとき, 次の極限値を求めよ。

$ \lim_{n\to\infty}\left(\frac{n}{k}\right)\left\{f\left(\frac{x}{n}\right)-f(0)\right\}^{k}$

ただし, $ \left(\frac{n}{k}\right)$ は相異なる $ n$ 個のものから $ k$ 個をとり出す組合わせの数を表す。とくに

$ f(x)=(x-\alpha )(x-\beta )$ のとき, 問題の極限値はどうなるか。

こっちはわけわかりませんね。さすが東工大・・・。

[hide]
\[ _nC_k \left\{ f\left(\dfrac{x}{n}\right)\minus{}f(0)\right\}^k\]
\[ \equal{}\left(\dfrac{x}{n}\right)^k\cdot_nC_k \left\{  \dfrac{f\left(\dfrac{x}{n}\right)\minus{}f(0)}{\left(\dfrac{x}{n}\right)}  \right\}^k\]
\[ \equal{}\dfrac{_nC_k}{n^k} \left\{  \dfrac{f\left(\dfrac{x}{n}\right)\minus{}f(0)}{\left(\dfrac{x}{n}\right)}  \right\}^k\cdot x^k\]

ここで，$ f(x)$は微分可能であるから，$ x$を任意に定めると，
\[ \dfrac{f\left(\dfrac{x}{n}\right)\minus{}f(0)}{\left(\dfrac{x}{n}\right)} \to f'(0)\ (n \to \infty)\]

また，$ k$を任意に定めると
\[ \dfrac{_nC_k}{n^k}\equal{}\dfrac{n}{n}\cdot\dfrac{n\minus{}1}{n}\cdots \dfrac{n\minus{}k\plus{}1}{n}\cdot \dfrac{1}{k!}\ \to \dfrac{1}{k!} \ (n \to \infty)\]

これより求める極限値は
\[ \dfrac{1}{k!}\cdot \{f'(0)\}^k\cdot x^k \equal{} \dfrac{\{f'(0)x\}^k}{k!}\]

\[ f(x)\equal{}(x\minus{}\alpha)(x\minus{}\beta)\]
のときは，$ f'(x)\equal{}2x\minus{}(\alpha\plus{}\beta)$より$ f'(0)\equal{}\minus{}(\alpha\plus{}\beta)$

このとき上の極限値は
\[ \dfrac{\{\minus{}(\alpha\plus{}\beta)x\}^k}{k!}\]
[/hide]

いやー、それにしても自信がない・・・


In a school there are 2008 students. Students are members of certain committees. A committee has at most 1004 members and every two students join a common committee.

(a) Determine the smallest possible number of committees in the school.

(b) If it is further required that the union of any two committees consists of at most 1800 students, will your answer in (a) still hold?

[hide="a"]6 commitees. To be in a commitee with every other person, every person obviously needs to be in at least 3 commitees, so their are at least 3*2008/1004 = 6 commitees. This is possible: consider the commitees split into 4 groups, each with 502 members. For every choice of 2 groups, make one commitee consist of all the people in those two groups. Then, the condition is satisfies.[/hide]


Let $[a,b]$ be a given interval of real numbers not containing integers. Prove that there exists $N>0$ such that $[Na,Nb]$ does not contain integer numbers and the length of the interval $[Na,Nb]$ exceedes $\dfrac 16$.

[b]I can solve the problem if by exceeds we mean $\geq$ rather than $>$. Perhaps someone can extend my ideas:[/b]

Firstly, if $b-a \geq 1/6$ then we are done, assume therefore that $b-a< 1/6$.

It follows that in the interval $[a,b]$ we cannot have BOTH rational numbers of the form $n/3$ and $n/2$ since otherwise our interval's length will be greater than or equal to $1/6$. So if we have no number of the form $n/3$, then the interval $[3a,3b]$ contains no integers. On the other hand, if we have no number of the form $n/2$ then we can consider the interval $[2a,2b]$, it has no integer in it. We can then continue this process until our interval is of length greater than or equal to $1/6$.

[b]edit:[/b] actually, I am right, we need exceed to mean $\geq$ rather than $>$. Because consider the interval $[1/3, 1/2]$It has length $1/6$ therefore $N=1$ is not enough.

So can $N=2$? no because then 1 is in the interval

can $N=3$? no because then 1 is also in the interval.

can $N=4$? no because then 2 is in the interval.

can $N=5$? no because then 2 is in the interval since $1/3 < 2/5 < 1/2$.

can $N=6$? no because then $2$ and $3$ are in our interval.

$N>6$ is not possible since then our interval length is greater than 1, and hence contains an integer.

$N$ needn't be an integer; choosing, for instance, $N = \frac{3}{2}$ for $[\frac{1}{3}, \frac{1}{2}]$ satisfies the condition. :D

Ok then if that's the case then you can easily add those sorts of cases in (just multiply those intervals of length exactly $1/6$ by some number a little bit bigger than 1 and you are done). I thought $N$ is an integer because otherwise the problem may be too simple. I showed that $1/6$ is best possible for integer $N$(with the extra condition that we mean $\geq$ by exceeds) if we let $N$ be any positive real number as you say would it also be best possible?

I think that this problem might be meaningless.
If $N$ is not an integer,we let $N=\frac{1}{6\sqrt{a^2+b^2}}$,obviously we have $\frac{b}{6\sqrt{a^2+b^2}}-\frac{a}{6\sqrt{a^2+b^2}}=\frac{1}{6}(\sin{\alpha}-\cos{\alpha})<\frac{1}{6}$.
 If $N$ is  an integer,when we choose $(0.1,0.9)$ ,obviously whatever $N\geq2$ is,the length of the interval is greater than $1$ and certainly contain integers.Thus it must be wrong if $N$ is not an integer.

If any real $N$ is allowed, the critical value is $\dfrac {1} {2}$, not $\dfrac {1} {6}$. Moreover, since this threshold is not reachable, we will work with open intervals.

We may assume $b> a > 0$ (otherwise multiply everything by $-1$). The example of the interval $\left(\dfrac {1} {2} - \varepsilon, 1\right)$ shows we cannot enlarge intervals of length more than $\dfrac {1} {2}$.

For any interval $(a,b)$ of length less than $\dfrac {1} {2}$, let us multiply by the largest $N$ such that $Nb$ becomes some integer $k \geq 1$, while $N(b-a) \leq 1$. Then by multiplying with $N\dfrac {k+1} {k} > N$ we should have $N\dfrac {k+1} {k}(b-a) > 1$, so $N(b-a) > \dfrac {k} {k+1} \geq \dfrac {1} {2}$, so we obtained an interval of length more than $\dfrac {1} {2}$.

In order to regain the problem on closed intervals, we can always adjust by some final multiplier close enough to $1$ so we avoid integers at the ends of the interval, while keeping the length of the interval as close as wanted to the just previous value.

Therefore I assume Valentin meant $N$ integer ...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ a group with $ 2n$ elements and $ H$ a subgroup of $ G$ with $ n$ elements. Prove that for all $ g\in G$, $ g^2\in H$.

Notice that $ H$ is of index $ 2$ in $ G$, hence $ H$ is a normal subgroup of $ G$ and we can form the quotient $ G/H \cong Z_2, \plus{}$. Let $ \pi$ be the projection map of $ G$ onto $ G/H \cong Z_2$ then $ \pi(g^2) \equal{} \pi(g) \plus{} \pi(g) \equal{} 0$, so $ g^2 \in H$.


Find a recurrence relation for the number of ternary strings (0 s, 1 s and 2 s) of length n that contain none of the strings 101, 202, 102, or 201. What are the initial conditions? Use the recurrence to compute the number of such sequences of length 5.

I am trying this for a very long time. Can anyone help me with this?

I found that for the initial conditions n=1, hn= 3 and for n-2, hn=9. So, I need the recurrence relation. Help would be greatly appreciated.

Let $ a_n$ be the number of strings of length $ n$ ending in $ 1$ or $ 2$, $ b_n$ be the number of strings of of length $ n$ ending in $ 0$ which can be followed by a non-zero, $ c_n$ be the number of strings of length $ n$ ending in $ 0$ which must be followed by a zero. Then we have initial conditions $ a_1 \equal{} 2, b_1 \equal{} 1$, and
\[ a_n \equal{} 2(a_{n \minus{} 1} \plus{} b_{n \minus{} 1}), \quad b_n \equal{} b_{n \minus{} 1} \plus{} c_{n \minus{} 1}, \quad c_n \equal{} a_{n \minus{} 1}
\] Hence \[ s_n \equal{} a_n \plus{} b_n \plus{} c_n \equal{} 2[a_{n \minus{} 1} \plus{} b_{n \minus{} 1}] \plus{} [b_{n \minus{} 1} \plus{} c_{n \minus{} 1}] \plus{} [a_{n \minus{} 1}] \equal{} 3s_{n \minus{} 1} \minus{} 2c_{n \minus{} 1}.\] Thus $ c_{n} \equal{} \frac{3s_{n} \minus{} s_{n\plus{}1}}{2}$. Also, \[ c_n \equal{} a_{n \minus{} 1} \equal{} 2(a_{n \minus{} 2} \plus{} b_{n \minus{} 2}) \equal{} 2s_{n \minus{} 2} \minus{} 2c_{n \minus{} 2}\] Combining these two recursive relations yields $ c_n \equal{} \frac{3s_n \minus{} s_{n\plus{}1}}{2} \equal{} 2s_{n\minus{}2} \minus{} 2\left(\frac{3s_{n\minus{}2} \minus{} s_{n\minus{}1}}{2}\right) \Longrightarrow s_{n\plus{}1} \equal{} 3 s_n \minus{} 2s_{n\minus{}1} \plus{} 2s_{n\minus{}2}$.


Here is an easy problem, but I found it nevertheless amusing:

Let p(x) be a polynomial with integer coefficients such that

(1) p(0) = 0,
(2) 0 \leq p(1) \leq 107,
(3) p(a) = 1999 for some positive integer a,
(4) p(b) = 2001 for some positive integer b.

Find all possible values of p(1).

[i]Italy 2000, Problem 6[/i]

Ok I really hope that 1999 is a prime :) I didn't check it out but it seems to me that it is

p \in Z[X] , p(0)=0 => x|p(x), for all x => a=1 or 1999
Since b must divide 2001 and |b-a| divides 2 (|b-a| divides p(b)-p(a)) we have the following possible pairs

a=1 , b=3 or a=1999,b=2001.

First case means that p(1)=1999. Second case means that p(1999)=1999 , p(2001)=2001 so p(x)-x is divisible by x(x-1999)(x-2001).

p(x) = x(x-1999)(x-2001)q(x) + x
p(1) = 1*1998*2000*q(1)+1 and since it's positive we have q(1)>=0.

If q(1)=0 then p(1)=1
If q(1)=1 then p(1)=3 996 001
If q(1)=2 then p(1)=7 992 002
If q(1)=3 then p(1)=11 988 003>10^7. So for q(1)>=3 we do not get a valid value.

Therefore all possible values of p(1) are 1,1999,3 996 001, 7 992 002.

That's exactly what I did (and yes Andrei, 1999 is prime) :)


Show that $\mathrm{GL}_n(\mathbb{F}_q)$ has an element of oder $q^n-1$.

In the link above by grobber, it is exhibited $ A$ of order $ q^n\minus{}1$ such that $ \{0,A,A^2,\ldots A^{q^n\minus{}1}\}$ is the field $ \mathbb{F}_{q^n}$.
So here is a follow up question:
Does $ \mathrm{GL}_n(\mathbb{F}_{q})$ have an element $ A$ of order $ q^n\minus{}1$ such that $ \{0,A,A^2,\ldots A^{q^n\minus{}1}\}$ is not the field $ \mathbb{F}_{q^n}$.


algebra
polynomial
difference of squares
special factorizations

Factor completely: $ -a^{2}b^{2}+2ab^{3}-b^{4}+a^{2}c^{2}-2abc^{2}+b^{2}c^{2}$

lead me step by step through this one....for some reason im pretty bad at this higher level of factoring...anybody recommend a way that i can get better?

thanks. :) :lol:

Move the terms around to form a quadraticish polynomial around $ a$

$ a^{2}(c^{2}-b^{2})+a(2b^{3}-2bc^{2})+b^{2}c^{2}-b^{4}$

then just break each of those "coefficients" up

$ a^{2}(c+b)(c-b)-a(2b)(c+b)(c-b)+b^{2}(c+b)(c-b)$

you'll then see that all the terms contain (c+b)(c-b) so you can factor that out

$ (c+b)(c-b)(a^{2}-2ab+b^{2})$
$ (c+b)(c-b)(a-b)^{2}$

there's no standard way to do these kinda problems. you just have to recognize patterns and work from there :P

It looks pretty well organized already, in that the order of terms suggests a clear methodology.  Looking at just the coefficients, I see $ -1, 2,-1, 1,-2, 1$
which suggests 2 groups of 3.

Now I look for the term with the largest shared exponent that can be factored from each group of 3:$ -a^{2}b^{2}+2ab^{3}-b^{4}=-b^{2}\times \left(a^{2}-2ab+b^{2}\right)$ and $ a^{2}c^{2}-2abc^{2}+b^{2}c^{2}= c^{2}\times \left(a^{2}-2ab+b^{2}\right)$.

We can combine terms, as well as note the perfect square: $ \left(a^{2}-2ab+b^{2}\right) = \left(a-b\right)^{2}$ to get $ \left(c^{2}-b^{2}\right) \times \left(a-b\right)^{2}$.

Finally, noting the difference of squares, I get: $ \left(c+b\right) \times \left(c-b\right) \times \left(a-b\right)^{2}$.

Once you start breaking it down into smaller parts, you can usually recognize key expressions like perfect squares, difference of squares, etc.  The first step is probably going to be finding the right ordering of terms that lends itself to finding smaller factorizations that combine via distribution of addition over multiplication: $ a \times b+c \times b = \left(a+c\right) \times b$.  Reordering terms can be tricky, but it may help to keep track of coefficients and shared variable exponents.


What common fraction in the interval $ 0.76 <x<0.81$ has the smallest denominator?

[hide="Solution"]$\dfrac{3}{4}$ is equal to 0.75, so the next fraction would be $\boxed{\dfrac{4}{5}}$[/hide]


linear algebra
matrix
integration
calculus
inequalities
inequalities proposed

Let $ a_1,a_2,...,a_n \in \mathbb R$. Prove that:

$ (\sum^{n}_{i\equal{}1}{a_i})^2\leq \sum^{n}_{i,j\equal{}1}{\dfrac{ij}{i\plus{}j\minus{}1}a_ia_j}$

\begin{eqnarray*}\sum_{i,j}\frac {ij}{i + j - 1}a_ia_j & = & \sum_{i,j}\int_0^1ija_ia_jx^{(i - 1) + (j - 1)}dx \\
& = & \int_0^1\left(\sum_iia_ix^{i - 1}\right)^2dx \\
&\geq & \left(\int_0^1\sum_iia_ix^{i - 1}dx\right)^2 \\
& = & \left(\sum_ia_i\right)^2\end{eqnarray*}

Is there any solution that does not make use of matrix or integrals?

[quote="scheherazadex"]can someone explain to me how does the inequality step in boxedexe's solution work?

thanks a lot![/quote]

Integral form of Cauchy Schwarz inequality

[quote="me@home"]equiv to $ \sum \frac {(i \minus{} 1)(j \minus{} 1)}{j \plus{} i \minus{} 1} a_ia_j \geq 0$
which is just saying that the matrix
\[ {\bf A} \equal{} \left\{\frac {(i \minus{} 1)(j \minus{} 1)}{j \plus{} i \minus{} 1}\right\}
\]
is symmetric[/quote]

Could you (or someone else) explain this solution more clearly?  Why does the fact that 'it is a symmetric matrix' imply the inequality?

[quote="boxedexe"]\begin{eqnarray*}\sum_{i,j}\frac {ij}{i + j - 1}a_ia_j & = & \sum_{i,j}\int_0^1ija_ia_jx^{(i - 1) + (j - 1)}dx \\
& = & \int_0^1\left(\sum_iia_ix^{i - 1}\right)^2dx \\
& \geq & \left(\int_0^1\sum_iia_ix^{i - 1}dx\right)^2 \\
& = & \left(\sum_ia_i\right)^2\end{eqnarray*}[/quote]

nice :D


number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ R$ be a ring with no zero divisors.  If $ a,b \in R$ s.t. $ a^n\equal{}b^n$ and $ a^m\equal{}b^m$ where $ n,m \in \mathbb N$ are relatively prime, then prove $ a\equal{}b$.

It follows from two observations:
1. When $ R$ is a ring with no zero divisors, then
$ \forall a, b, c \in R, a \neq 0 \quad ab \equal{} ac \implies b \equal{} c$
2. If $ m, n \in \mathbb{N}$ are relatively prime, then exists $ x, y \in \mathbb{N}$ such that $ mx \minus{} ny \equal{} 1$.


Let $n$ be an integer greater than or equal to 3.  Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with a rational area.

Because any three points lattice points form a triangle with integer area, we will select $ n$ lattice points with irrational distance between any of them. We can build up this set by induction.

Base Case: $ n\equal{}3$. Choose $ (0,0),(1,1),(2,4)$.

Inductive step: Assume we have $ n$ points $ (x_1,y_1),(x_2,y_2),...,(x_n,y_n)$ with each with irrational distances from each other. Now choose $ y$ distinct from all the $ y_i$, and let $ a\equal{}\max |y\minus{}y_i|$. Now choose $ x$ sufficiently large such that $ x\minus{}\max x_i>\frac{a^2\minus{}1}{2}$ (meaning $ 2(x\minus{}\max x_i)\plus{}1>a^2$) and $ (x,y)$ is not collinear with any two points chosen already (this is possible because the first $ n$ points only determine $ \binom{n}{2}$ lines). Now I claim that the distance from $ (x,y)$ to $ (x_i,y_i)$ for any $ i$ is irrational.

$ (x\minus{}x_i)^2\plus{}(y\minus{}y_i)^2>(x\minus{}x_i)^2$.

$ (x\minus{}x_i)^2\plus{}(y\minus{}y_i)^2\le (x\minus{}x_i)^2\plus{}a^2<(x\minus{}x_i)^2\plus{}2(x\minus{}\max x_i)\plus{}1<(x\minus{}x_i)^2\plus{}2(x\minus{}x_i)\plus{}1\equal{}(x\minus{}x_i\plus{}1)^2$.

$ (x\minus{}x_i)^2\plus{}(y\minus{}y_i)^2$ lies between two squares, so it cannot be a square; therefore the distance between $ (x,y)$ and $ (x_i,y_i)$ is irrational, so this completes the induction and so we can build a set with all triangle having rational area and all distances being irrational.

Also in IMO1987/5

Let there be a square lattice such that the length of each interval is $ \sqrt{3}$. i.e. $ (x,y)\equal{}(a\sqrt{3},b\sqrt{3})$ where $ a,b$ are integers. Then choose any $ n$ non-collinear points in the lattice.
By the area formula, we get the area as $ \frac{1}{2}\sqrt{3}^{2}\cdot m$, where $ m$ is an integer. So the area is rational.
But if $ k$ is the distance between some two points, then $ k^{2}\equal{}3p^{2}\plus{}3q^{2}$, where $ p$ and $ q$ are integers. It is easy to see that $ k$ cannot be rational unless it is an integer. Suppose $ k$ is an integer: take $ mod3$, we get $ 3\mid k$, then take $ mod3$ again, we get $ 3\mid p$ and $ 3\mid q$ and we get infinite descent. Contradiction, so $ k$ must be irrational.


Let $F(x)=(x^2-17)(x^2-19)(x^2-323)$.$\forall m\in \mathbb{N}$ prove that:$F(x)\equiv 0\ (\mod\ m)$ has answer.
But$F(x)=0$ has no answer in $\mathbb{Q}$.

This can be done exactly as here :

http://www.animath.fr/cours/arithm.pdf

See p.61-62

Pierre.


Alex has four books of different sizes which he wishes to place on a bookshelf. Unfortunately, the bookshelf has a conspicuous hole at one end through which only the smallest book can fall. If he wishes for all his books to stay stationary, in how many ways can he arrange his books?

Since the smallest book can be in 3 places, the answer is 3*3!=18.

We find that we want to start by getting the rule down first. So there are 3 books for the hole and then there is just 3! ways left so it is 3*3*2*1=18.


well, I'm kinda new to this game and I don't really get it, you just go around acusing people?

Smiles,
lyra

What game????? Do you mean Mafia? If you do, the goal is to accuse the RIGHT ones. Once an angel saved a devil, so you have to pick and choose the right ones.

Okay, sorry I kinda posted this in the wrong place...

Smiles,
lyra :blush:

[quote="mbabbitt2003"]Once an angel saved a devil, so you have to pick and choose the right ones.[/quote]

Mwahahaha... :D  :diablo:

why did you name the topic lyra though?
and i am also new to the game, but a few days back i happened to play real mafia - everybody sleep, mafia wake up, mafia sleep, sheriff wake up, sheriff sleep, doctor wake up, doctor sleep, everybody wake up, alas, somebody was killed today, now who votes against whom? - that's the game in a nutshell (a rather big nutshell though!)

I was that angel... and dts was the only person I was able to save in the game :P .


let $ (C)$ a circle with radius $ R$. find all quadrilaterals $ ABCD$ inscriptibles on $ (C)$ such that $ AB.BC.CD.DA>\equal{}4R^2$

Are you sure its 4R^2? I think it might be 4R^4 :wink:


[quote="tranthanhnam"]a,b,c >= 0 satisfying a+b+c =9/4
prove :
               a^3 + b^3 +c^ 3 <= a^4 +b^4 +c^4 +81/256[/quote]
$3a^4 + 81/256 \ge 4 \frac{3}{4} a^3 = 3a^3$
Sum up and done. I don't really understand why $a+b+c=\frac{9}{4}$ is needed..


geometry
area of a triangle
Heron\u0027s formula

The lengths of the sides of a tirangle are 8, 10, and 14. Find the length of the altitude to the longest side

A. $ \frac{40}{7}$

B. $ \frac{45}{14}$

C. $ \frac{16\sqrt{6}}{7}$

D. $ \frac{9\sqrt{3}}{14}$

Use Herons formula to figure out area, then use area to find the length of the altitude

$ s \equal{} \frac {a \plus{} b \plus{} c}{2} \equal{} \frac {8 \plus{} 10 \plus{} 14}{2} \equal{} 16$

$ S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \equal{} \sqrt {16*8*6*2} \equal{} 16\sqrt {6}$

$ S \equal{} \frac {c*h_{c}}{2} \equal{} S$

$ h_{c} \equal{} \frac {2S}{c} \equal{} \frac {2*16\sqrt {6}}{14} \equal{} \frac {16\sqrt {6}}{7}$

So unless I made some mistakes answer should be $ \boxed {C}$

EDIT: Hopefully JRav is right and I didn't do any mistakes  :lol:

C.  Here's how I did it:
From Heron's Formula, the area is $ \sqrt{16(2)(6)(8)}\equal{}\sqrt{1536}\equal{}16\sqrt{6}$.  Then we can use the conventional formula for area of a triangle, $ A\equal{}\frac{1}{2}bh\equal{}\frac{1}{2}14h\equal{}16\sqrt{6}\Rightarrow h\equal{}\frac{16\sqrt{6}}{7}$ and we're done.


ehh this is a pretty easy trig problem that i saw somewhere a while ago and solved, but when i revisited it i couldn't seem to do it right so i'm looking for a solution to jog my memory

compute:
$cos(20)cos(40)cos(60)cos(80)$

[quote="tarquin"][hide]Using double-angle formula for sine
$\frac {\sin 40}{2 \sin 20} \cdot \frac {\sin 80}{2 \sin 40} \cdot \frac {\sin 120}{2 \sin 60} \cdot \frac {\sin 160}{2 \sin 80} =  \frac 1{16}$[/hide][/quote]
 :wow:  wow this is so neat, never thought of using double angle, i guess i must practice more trig :thumbup:

$40-20=60-40=80-60$
If the angles can form an arithmetic sequence, we can use double formula, sum to product, product to some and induced formula to find the result.


inequalities
trigonometry
geometry
ratio
circumcircle
calculus
calculus computations

Here is a problem that I asked one of my friends a few days ago:
Let $ A, B, C$ denote the angles of a triangle. It is a famous inequality that
\[ f(A, B, C) \equal{} \cos(A) \plus{} \cos(B) \plus{} \cos(C)\leq \frac {3}{2},
\] and that $ \frac {3}{2}$ is the maximum of $ f$. But, is there any way to establish the "existence" of "a" maximum for $ f$ without any calculation, say at a glance?
What can you say about the existence of a minimum for $ f$?

hello, prove the equation $ \cos(\alpha)\plus{}\cos(\beta)\plus{}\cos(\gamma)\equal{}1\plus{}\frac{r}{R}$ with $ r$ as the radius of the incircle and $ R$ the radius of the circumcircle.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, prove the equation $ \cos(\alpha) \plus{} \cos(\beta) \plus{} \cos(\gamma) \equal{} 1 \plus{} \frac {r}{R}$ with $ r$ as the radius of the incircle and $ R$ the radius of the circumcircle.
Sonnhard.[/quote]

If THAT is true, we can use R≥2r to solve the problem!

Dr Sonnhard Graubner,

And after that, one should prove that there is a triangle which the ratio of radii of its incircle and circumcircle is 1/2?


$ I$ $ could$ $ use$ $ help$ $ with$ $ this$ $ :$

$ \sqrt[3]{2\plus{}\sqrt[2]{5}} \plus{} \sqrt[3]{2\minus{}\sqrt[2]{5}} \equal{} ?$

Hint would be appreciated...

Thanks

You can try reasoning in this way:

1) Let's say $ r \equal{} \sqrt[3]{2\plus{}\sqrt{5}}\plus{}\sqrt[3]{2\minus{}\sqrt{5}}$

So,
$ r^3 \equal{} 4\minus{}3r$  (try it!),

and so the solution of r is the solution of $ r^3 \plus{} 3r \minus{}4\equal{}0$

You can try to solve it , factorizing it using Ruffini's rule.

$ (r\minus{}1)(r^2\plus{}r\plus{}4)\equal{}0$

Since the discriminant of $ r^2\plus{}r\plus{}4$ is negative, the only solution $ \in \mathbb{R}$ is $ r\equal{}1$

hello, could it be that you meant $ \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{\minus{}2 \plus{} \sqrt {5}}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, could it be that you meant $ \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{ \minus{} 2 \plus{} \sqrt {5}}$?
Sonnhard.[/quote]

No, I meant:

$ \sqrt [3]{2 \plus{} \sqrt [2]{5}} \plus{} \sqrt [3]{2 \minus{} \sqrt [2]{5}}$

[quote="Gabuntu94"]You can try reasoning in this way:

1) Let's say $ r \equal{} \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}$

So,
$ r^3 \equal{} 4 \minus{} 3r$  (try it!),

and so the solution of r is the solution of $ r^3 \plus{} 3r \minus{} 4 \equal{} 0$

You can try to solve it , factorizing it using Ruffini's rule.

$ (r \minus{} 1)(r^2 \plus{} r \plus{} 4) \equal{} 0$

Since the discriminant of $ r^2 \plus{} r \plus{} 4$ is negative, the only solution $ \in \mathbb{R}$ is $ r \equal{} 1$[/quote]

Thanks for that example of Ruffini's rule... but how does that help me solving this task... I really do not know how to solve this task I posted...

[quote="Gabuntu94"]$ r \equal{} \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}$

Since $ (a \plus{} b)^3 \equal{} a^3 \plus{} 3ab(a \plus{} b) \plus{} b^3$,

$ r^3 \equal{} \left( \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}\right)^3 \equal{}$
$ \left(\sqrt [3]{2 \plus{} \sqrt {5}}\right)^3 \plus{} \left(\sqrt [3]{2 \minus{} \sqrt {5}}\right)^3 \plus{} 3\cdot\sqrt [3]{2 \plus{} \sqrt {5}}\cdot\sqrt [3]{2 \minus{} \sqrt {5}}\cdot\left(\sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}\right) \equal{}$
$ \equal{} 2 \plus{} \sqrt {5} \plus{} 2 \minus{} \sqrt {5} \plus{} 3\cdot \sqrt [3]{4 \minus{} 5} \cdot r \equal{}$
$ \equal{} 4 \plus{} 3\sqrt [3]{ \minus{} 1}\cdot r \equal{} 4 \minus{} 3r$

It shows that $ r^3 \equal{} 4 \minus{} 3r$ (remember that $ r \equal{} \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}$ is the number you're looking for)

The only number which verify this equation is $ r \equal{} 1$...

Was it what you were looking for?

Have I understood well the question? (I'm not very good at English at school...  :roll: )[/quote]

Thank you very much about this explanation, I really overlooked part with:

$ r \equal{} \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}$

So, I was confused when you started using $ r$

Thanks once again :)

Useful lemma:
If $ a\plus{}b\plus{}c\equal{}0$, then $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$

In this problem, let $ a\equal{}\sqrt[3]{2\plus{}\sqrt{5}}\plus{}\sqrt[3]{2\minus{}\sqrt{5}}, b\equal{}\minus{}\sqrt[3]{2\plus{}\sqrt{5}}, c\equal{}\sqrt[3]{2\minus{}\sqrt{5}}$

Apply the lemma, we get
$ a^3\minus{}2\minus{}\sqrt{5}\minus{}2\plus{}\sqrt{5}\equal{}\minus{}3a$
$ a^3\plus{}3a\minus{}4\equal{}0$

Just use Horner to find the rational root, and we'll get $ a\equal{}1$
 :)

Let $ \sqrt[3]{2\plus{}\sqrt{5}}\plus{}\sqrt[3]{2\minus{}\sqrt{5}}\equal{}r\Longrightarrow (2\plus{}\sqrt{5})\plus{}(2\minus{}\sqrt{5})\minus{}r^3\equal{}\minus{}3\sqrt[3]{2\plus{}\sqrt{5}}\cdot \sqrt[3]{2\minus{}\sqrt{5}}\cdot r$

$ \Longleftrightarrow r^3\plus{}3r\minus{}4\equal{}0$


MIT college videos Yale LaTeX Harvard email

[color=blue][b]Forums/Site[/b][/color]
[url=http://physicsweb.org/]Physics Web[/url]
[url=http://www.physicsforums.com/]Physics Forum[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=332]AoPS Physics Forum![/url]
[url=http://math.ucr.edu/home/baez/]John Baez's Stuff[/url]
[url=http://www.aps.org]American Physics Society[/url]
[url=http://www.aapt.org/Contests/olympiad.cfm]USAPhO(or U.S. Physics Team) Page[/url]

[color=red][b]Formulas/Theorems[/b][/color]
[url=http://www.collegeboard.com/prod_downloads/ap/students/physics/info_equation_tables_2002.pdf]CollegeBoard Formulas & Theorems[/url]

[color=green][b]Practice Tests[/b][/color]
[url=http://www.compadre.org/psrc/evals/olympiad.cfm]Past USAPhO tests[/url]
[url=http://www.physics.utoronto.ca/~poptor/current.html]U. of Toronto Undergrad Practice Tests[/url]

[color=brown][b]Books/Lectures: [/b][/color]
[url=http://www.amazon.com/exec/obidos/tg/detail/-/0201021153/qid=1113741669/sr=8-1/ref=pd_csp_1/102-0609151-1972963?v=glance&s=books&n=507846]Feynman Lectures (Volume 1,2,3)[/url]
[url=http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures]MIT     Physics Video Lectures[/url]
[url=http://hug.phys.huji.ac.il/Phys_Machon/Colloquium/index.htm]Colloquium Lecture Notes(some in Hebrew, some in English)[/url]
[url=http://www.amazon.com/exec/obidos/tg/detail/-/0691024170/ref=pd_sim_b_1/103-3170392-1871059?%5Fencoding=UTF8&v=glance]QED: The Strange Theory of Light and Matter[/url]
[url=http://www.physicsforfree.com/]Free Textbook from Yale Prof. Frank Firk[/url]
[url=http://www.lightandmatter.com/area1.html]Light And Matter.com Books[/url]
[url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/2LDL4NCF1G4L2/ref=cm_lm_lists/103-3170392-1871059]Amazon.com  Listmania on Physics Book[/url]
[url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/21N3R7VTQFDRP/ref%3Dcm%5Flm%5Flists/102-8413543-5666523]Amazon.com Listmania on MIT Physics Books[/url]

i think there is a famous physics forum, who knows the link?

[quote="shobber"][quote="towersfreak2006"]This is the first result from looking up physics forum on Google:

http://www.physicsforums.com/

Looks famous. :)[/quote]
yep, it is famous, but it seems we cannot post in latex there.[/quote]

actually they do have it, except it looks smaller... like [url=http://physicsforums.com/showthread.php?t=71943]here[/url]

no, i dont think so. Although i do know some people who have it on pdf, but then again, printing the copyrighted stuff is illegal. ;)

[quote="shobber"]Can we find the Feyman's Lecture on line??[/quote]
you can download it using a P2P ( peer to peer) software  ( i.e : Emule )

google up HyperPhysics ... that site's got a lot of useful stuff.

anyone know when the 2005 qualifying and semifinal exams will be posted on the website?
http://www.compadre.org/psrc/evals/olympiad.cfm only has till 2004.

[quote="shobber"]Can we find the Feyman's Lecture on line??[/quote]

These aren't from Feynman's Lectures, but they are great videos of lectures that Feynman gave in New Zealand: http://www.vega.org.uk/video/subseries/8.

[color=red]Gauss202's recommendation:-[/color]

Here is the Mechanics Book used for Harvard's Physics 16 course (the honors mechanics course). It has lots of interesting and difficult problems. It also introduces some advanced topics like Relativity and Lagrangian Mechanics.

[url=http://my.harvard.edu/icb/icb.do?course=fas-phys16&pageid=tk.page.phys16.dir.7f829e3adaebcfa703b7db42e0201022]Mechanics and Relativity[/url]

"rushil...........

can u plz send again the link............

plz........

Here's the link to an excellent, free-to-use book on wave phenomena:

[url=http://www.people.fas.harvard.edu/~hgeorgi/onenew.pdf]The Physics of Waves[/url]
by Howard Georgi
[i]Harvard University[/i]

Hi everyone,
I ve searching the Polish national Physics Olympiad in 1988 for so long but still havent found it.
anyone please send me the problems and solutions for Polish national Physics Olympiad 1988, I really appreciate and thank you so much :)
My email: minhduc_nguyen_1988@yahoo.com

Polish national physics olympiad archive contains all, I believe, problems from 1987/1988 olympiad (so one held in 1988) and few experimental problems from 1988/1989. The archive is available in polish at http://www.of.szc.pl/index.php?zapytanie=&rodzaj=&dzial=&numer=87&numer_zaw=&zawod=&widok=1&strona=32&szukaj=szukaj

A suggestion: I think we should continue xxreddevilzxx's work. There are a whole lot of topics about useful books, why wouldn't we gather the books into one topic, that would save everyone from searching a lot... wouldn't it?

could someone has an ebook named "Cambridge Problems in Physics and Advice on Solutions" ? I heard the book was good and the questions were challenging

Outside of the USAPHO tests which books and resources are best for preparing for the F=ma exam and the USAPHO?

It seems the link with the formulas is incomplete.

Where can we find a PDF with all theorems/formules for the IpHO or doesn't it exist?

If you looking for Problems as a supplement to [b]Feynman Lectures on Physics[/b] then go here http://www.feynmanlectures.info/

Physics Challenge Problems:

http://www.luiseduardo.com.br


Find the last three digits of $7^{9999}$.

[hide="My soln so far"]
i'm left with $143$, $343$, $543$, $743$ and $943$ after $\mod 50$ and $\mod 8$. I have no idea how to pick the good one (excluding a calculator, which points to $143$)
[/hide]

Hmmm...I really don't know how to do this one, I can get the last digit because the pattern is $7,9,3,1$ and because $7^{996}$ would have last digit of $1$ so the last digit of $7^{999}$ is $3$. Don't really know where to go from there...hmmm....
Then you can look at the pattern for the 10s digit and see that it goes $4,4,0,8,4,0,0,4,....$ so then the tens digit of $7^{999}$ would be $4$ I think, so so far we have $x43$

I really don't know where to go from there because I don't see a pattern with the hundreds digit  :(

[quote="joml88"]I can finish your solution:

[hide]
Euler phi tells us that

\[7^{\phi(125)}\equiv 1\mod 125\Rightarrow 7^{100}\equiv 1\mod 125.\]

Thus $7^{10 000}\equiv 1\mod 125.$

So 7 times our number must be congruent to 1 mod 125.  Using your numbers, we look at their product with 7 mod 7 and the only one that gives a remainder of 1 is what we are looking for (and, of course, it comes out that 143 is the only one).[/hide][/quote]

thanks.

and one more thing, how do you compute $\phi (n)$? (i remember it was a formula which ran through all $d | n$, but i forgot it)

[quote="joml88"]I can finish your solution:

[hide]
Euler phi tells us that

\[7^{\phi(125)}\equiv 1\mod 125\Rightarrow 7^{100}\equiv 1\mod 125.\]

Thus $7^{10 000}\equiv 1\mod 125.$

So 7 times our number must be congruent to 1 mod 125.  Using your numbers, we look at their product with 7 mod 7 and the only one that gives a remainder of 1 is what we are looking for (and, of course, it comes out that 143 is the only one).[/hide][/quote]

Wow...Why didn't I use mods????

If $n=p_1^{e_1}\cdot p_2^{e^2}\cdots p_n^{e_n}$ then \[\phi(n)=n\left(1-\frac 1{p_1}\right)\left(1-\frac 1{p_2}\right)\cdots\left(1-\frac 1{p_n}\right).\]


[quote="whywhywhy"]Let $ ABC$ be a acute triangle.$ A',B',C'$ are projections of $ A;B;C$ on $ BC;CA: AB$,respectively.
Circle center $ B$,radius $ BB'$ cuts $ A'C'$ at $ K;L$.Let $ O$ be circumcenter of triangle $ ABC$
Prove that $ AK;BL;CO$ are concurrent[/quote]
Could you please check your statement? sth is not right here.

I think either $ BO,AK,CL$ or $ BO,AL,BK$. :blush:


Consider that $M\subset \mathbb{N}\backslash\{0\}$ satisfy: $M\neq\varnothing$ and if $x\in M$ then $4x,[\sqrt{x}]\in M$.Prove that:$M=\mathbb{N}\backslash\{0\}$
[x] is the greatest integer $\le$ x.

For the "main idea" look here:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=TST&t=39063


I'm in the AIME Problem Series B class, but I only see this "Classes Information" Forum under the "Online School" heading while logged in.

Go to my classes, AIME Problem Solving then click the link to the class message board. If that doesn't work, then you should post something of the sort on the support page.

try to find it under online classes in the forum lists

The class message board for the AIME B has been moved to the Archive.  You should see an Archive section now.

This might sound dumb, but where is the Archive section?

After a class is over, under Online School in the forum list, there will be a folder called "Archives"
In that folder will be a subforum that will have the complete message board for your class.
It will remain there forever and you can post in it.
:)

I can only see the Classes Information and my another class under the Online School section, no Archive...

You mean it isn't a folder by itself or in Archive?
Weird... :huh:

I go to My Classes and enter my old class page and click on "message board", but the system says "The forum selected does not exist". Does this mean my class forum is deleted? By the way, my old class is AIME Problems Series B, to clarify.

That class has been moved into the Archive; you can get there by scrolling down to the Online School section of the message board on the top Forum page.

Can anyone in that class PM me a link? I did what Mr. Rusczyk said but there was no such Archive under the Online School section.

http://www.artofproblemsolving.com/Forum/index.php?f=375

No one can access it if they didn't take the class, so it's safe to post.

The forum you selected does not exist.

I'm pretty sure I took the class and have a very good attendance. But the above line was all what I got. :o

I'm currently taking another class. Will this affect the Archive?

Problem fixed. You should be able to access the link above timwu. Sorry for the inconvenience.

Thanks, it works now!  :)

um...Is there anyway to access the Archive directly?


I suppose for reals it wont be possible without additional hypothesis. f(x+y)=f(x)+f(y) is a more strict condition than this. And that's the well known cauchy's equation. And for rationals the solution is.f(x)=cx but by higher mathematics it can be shown that one can't tell anything when x & y are real there(without additional hypothesis like continuity or bound etc.).

Well, if $ P$ is supposed to be a polynomial, as stated above, you do have the continuity.

Pierre.

Sorry. I thought functional eqn. :blush:

i said this problem isnt functional eqn this problem is poly :wink: nomial!!!!

is there any solution for this problem!!!!!
u hang out??

[quote="praneeth"]Let p(x)=y,then
p(x+y)=p(x)+p(y)
=> p(x) is linear
=> p(x)=ax+b {a,b are real } satisfies the given equation.[/quote]

this is wrong... :wink:

By replacing x with zero we get p(0) = 0
Suppose n=degr(p)>1, then expand p make the necessary substitutions and obtain a polinomial of degree n^2-n+1 with the dominant quotient equal to the n-th power of the dominant quotient of p, polinomial which equals 0
Results the dominant quotient of p must be 0. Contradiction!

Then n=1. The conclusion follows

[quote="praneeth"]But this satisfies the given equation  :([/quote]

if so, we must have $ b\equal{}0$..and i didn't say the final solution is wrong, but the way you have proved it.. :wink:


problem 1

$ 4\left( {x^2  \plus{} \frac{1}{{x^2 }}} \right) \minus{} 4\left( {x \minus{} \frac{1}{x}} \right) \minus{} 27 \equal{} 0$

problem2

\[ 2\left( {\frac{{x \minus{} 2}}{{x \plus{} 3}}} \right) \minus{} 7\left( {\frac{{x \plus{} 3}}{{x \minus{} 1}}} \right) \equal{} 5;{\rm{where }}x \ne  \minus{} 3,x \ne 1.
\]

members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com

[hide="#2"]
$ 0\equal{}5(x\plus{}3)(x\minus{}1)\plus{}7(x\plus{}3)^2\minus{}2(x\minus{}2)(x\minus{}1)$
$ 0\equal{}2(5x^2\plus{}29x\plus{}22)\Longrightarrow \boxed{x\equal{}\frac{\minus{}29\pm\sqrt{401}}{10}}$

*[i]used quadratic formula[/i]
[/hide]

4) $ 4 \left ({x^2\plus{}\frac{1}{x^2}}\right)\minus{}4\left ({x\minus{}\frac{1}{x}}\right)\minus{}27\equal{}0$

$ a\equal{} x \minus{} \frac{1}{x}$
$ a^2\equal{}x^2\minus{}2\plus{}\frac{1}{x^2} \Rightarrow a^2\plus{}2\equal{}x^2\plus{}\frac{1}{x^2}$

$ 4(a^2\plus{}2)\minus{}4a\minus{}27\equal{}0$
$ 4a^2\minus{}4a\minus{}19\equal{}0  \Rightarrow  a\equal{}\frac{1}{2} \pm \sqrt{5}$

$ x\minus{}\frac{1}{x}\equal{}\frac{1}{2} \pm \sqrt{5}$

$ \Rightarrow x\minus{}\frac{1}{x}\equal{}\frac{1}{2}\plus{}\sqrt{5}$
$ x^2\minus{}\left({\frac{1}{2}\plus{}\sqrt{5}}\right)x\minus{}1\equal{}0 \Rightarrow x\equal{}\boxed{\frac{2\sqrt{5}\plus{}1 \pm \sqrt{37\plus{}4\sqrt{5}}}{4}}$
$ \Rightarrow x\minus{}\frac{1}{x}\equal{}\frac{1}{2}\minus{}\sqrt{5}$
$ x^2\minus{}\left({\frac{1}{2}\minus{}\sqrt{5}}\right)x\minus{}1\equal{}0 \Rightarrow x\equal{}\boxed{\frac{\minus{}2\sqrt{5}\plus{}1 \pm \sqrt{37\minus{}4\sqrt{5}}}{4}}$


geometry MATHCOUNTS email function \/closed

With the start of the new school year, we'll be ramping up our efforts to build the community and the school.  Specifically, we are exploring many avenues to reach more people and let them know about what we're doing.

For those of you who are willing to help out, here are some areas that you can pitch in:

1) Suggestions for websites we should try to get to link to us.

2) Suggestions for competitions or programs at which we could reach students.  We're looking for local, state, and national competitions/programs here.  Any suggestions you have are welcome, as are introductions to event hosts, or contact information.

3) Suggestions about gifted student groups, parents of gifted students groups, or homeschooling organizations we should contact.

4) Make contact yourself with competitions, websites, summer programs, etc., in which you participate to let them know about us.

5) Let us know any other ideas you have.

Thanks!

Well you could send more letters, but that would cost too much, I guess.
If we make a forum just for middle/elementary level math, then "kids" websites can link here, I guess.

how bout you just take off you shirt in math class and have "www.artofproblemsolving.com" written on it. itll get the point through.

hmm, u go do that Neal

Anyway, maybe get some famous/semi-famous mathematician to sponsor/recognize this

Or

advertise at some of the math camps...

The T-shirt might be a good idea. We can sell it at MATHCOUNTS national or USAMO/AIME/ARML (if we can. I don't know anything about these tests...)

t-shirts would be better than taking off your shirt.

Has AoPS been mentioned on the SET/Imagine magazine? I recall they did a mathematics issue recently in which there were programs, books, and maybe websites mentioned...

OOOO, i have that issue

lemme check....

that's a good idea, tho

The folks at SET have been very helpful to us; as far as we can tell, they take their role of informing their members about opportunities very seriously.  They gave us a prominent box on page 9 of the March/April 2003 Imagine (before the community even existed), and have informed SET members who elected to receive such information of our existence upon release of the community and the school.

We're big fans of SET - some parents in the San Diego area have formed a local SET group through which we've built a weekend problem solving group that's attracted many excellent students in our area (and if you're from San Diego and don't already know about this group, send me an email and I'll tell you more).

how about getting some space in local school math magazines.

i think hunter's and stuyvesant's would be happy to help....

Hmm, me and a friend are starting a math journal at our school, so we may be able to throw a reference in.

I'll put up fliers in the math classrooms.

[hide]

And maybe set my friends' startup pages to the AoPS Website.  *grins* [/hide]

Fiery - let me know if you need any information from me for your journal.

I'll get in touch with Hunter & Stuy about theirs if I can...

[quote]Fiery - let me know if you need any information from me for your journal. [/quote]

I thought I'd just write up a paragraph about the site, throw it into a box, and then place it somewhere on the journal...unless you have some official description you would like me to include instead.

What you suggest sounds fine to me - thanks!

How about advertising at some random fairly small scale competitions, like Math Field Day in Flint, MI?

You could also contact the math department at my school.  There are a lot of kids that have the [i]potential[/i] to be great, but the school system hasn't taught us problem solving for contests really.  They love to win competitions, and the math program is pretty good, so they might be interested in doing some stuff with AoPS, especially in the younger grades.

Sounds great - can you give me contact information for both the event and your school (you can pm them to me).

Thanks.

how bouts i tell me mathcounts teacher to tell the coordinators at state and chapter mathcounts to plaster it on the screen before the countdown round. or you could make pencils or somat with http://www.artofproblemsolving.com written on it, and have them at the tables to take the test with. my ideas may be lame, but im trying

I sent an email to my school's math competition mailing list telling them about this site, but I'm not sure how many of them signed up.

[quote="darkquantum"]If you want, i can help u .
Im close with both math departments 8-)[/quote]

For Stuy, talk to the AP, Danny Jaye
For Hunter, talk to Mr. Weinstein
and for added bonus, give Mr. David Hankin an email, too

Syntax - ideas aren't lame.  We'll probably be doing pens/pencils/Frisbees or something at some point in the future.  You're in OK, right?  Is Tom Carlisle still the state coordinator?  He and I did the 'Cool Down Round' at National MATHCOUNTS in 1994; was a blast.

darkq - do you have email addresses for them?  If so, can you pm them to me?

Thanks all for spreading the word.

[quote="rrusczyk"]Syntax - ideas aren't lame.  We'll probably be doing pens/pencils/Frisbees or something at some point in the future.  You're in OK, right?  Is Tom Carlisle still the state coordinator?  He and I did the 'Cool Down Round' at National MATHCOUNTS in 1994; was a blast.
[/quote]

a quick check on the mathcounts site ([url=http://mathcounts.org/queries/coordinator_search.taf?_function=list&_UserReference=175D69C89A2D8EA63F55145C&_start=1://url]here)[/url] and i get that hes the coordinator for Northeast Chapter MATHCOUNTS. that basically covers a small town called Bartlesville. im in the East Central Chapter where Perry Brown is the coordinator. but im sure that my coach could call him up or somat, or i could. whatever. i dunno, just give me the goodly details

(located in Texas) I'll do whatever I can if you need me to. Also, I have an idea for the Instructors which I will PM to Mr. Rusczyk sometime in the near future when i finish organizing my thoughts and have it in some presentable form (it could help a lot with fundraising). I think it's something that would have to pass through the Admin before released out to the suggestions board and what not. Also, we should start Xanga/Live Journal/Dead Journal Webrings for AoPS members. I know of people in at least 12 states that check my Xanga at least occasionally. I think the key here would be mass advertisement student by student. It's unlike we'd reach too many kids just through commercial ad's and it would be too slow if we tried doing school by school. With hundreds (and soon, hopefully, to be thousands)  of kids in an online network spread out across the United States helping us reach schools, it might work a lot faster.

Another idea is, if Mr. Rusczyk could work up a flyer that we could print and hand out at Math competitions we could do that. Also, (just for the convience of the members) I'd like to be able to purchase like a name tag with our AoPS membership names on them that we could put on when going to State/National meets so that we could recognize fellow AoPS'ers

[quote="MithsApprentice"]Also, we should start Xanga/Live Journal/Dead Journal Webrings for AoPS members. [/quote]

I started the Mathematics community on Livejournal a couple of years ago.  If you visit you'll notice that the name of the community is The Art of Problem Solving and it links to this site.  Feel free to create other good links to this site through other community spaces.

Math Club is officially starting at my school, so I'll try and advertise there. Does anyone else's school have a math club? If so, we could hand out fliers or something, and get people to come here (and/or buy the books)

um...Tare...most of the schools that we AoPS'ers go to have a math club

OK then...correction: Out of middle-schoolers here, does anyone have a math club?

I know most high schools have math clubs. I just sometimes forget you're all high-schoolers...

its not odd for middle schools to have math clubs either. All the middle schools in our district have them.

I successfully managed to advertise "artofproblemsolving.com," so there should be at least a couple, if not 40 to 50 new members this evening by 7:30. But please don't get mad at me if no one joins :)

...yet it got my mom really mom, but that's another story...

[quote="Tare"]

...yet it got my mom really mom, but that's another story...[/quote]

*confused*

[quote="Tare"]See, my mom got "worried sick" when she thought I was going to not go to Math Club, so now my dinner... :cry:[/quote]

man...lol...my mom wouldnt care...

With regards to contests, I would look more into GPML (Great Plains Math League), although I noticed AoPS did have a hand in the prizes for the top overall student at state. Also, locally, we have a number of "excellence in mathematics" contests in St. Louis hosted by MEGSL, the Math Educators of Greater St. Louis. These contests have almost become a state competition recently.

If I reveal that I am part of this community, I feel quite certain that my classmates will find me quite geeky (even more so than as it is), even though my school is a moderately selective one (possibly best within 100 miles unless IMSA is within 100 miles).  Does anyone else have this problem?  BTW, sorry if I misinterpreted this thread's title; I admit to being too lazy to read previous messages thoroughly.


limit
ratio
logarithms
integration
inequalities
calculus
real analysis

$ \underset{n\rightarrow\infty}{\lim}\frac{n!}{n^{n}}\equal{}0$

-

Use that $ n! \sim  \sqrt{2\pi n}(\frac{n}{e})^n$.

[quote="Symbian"]According to the hierarchy of sequences, since $ n^n$ grows faster than $ n!$ it follows that the limit is 0.[/quote]  This is a just a restatement of the problem.  And Kirill's solution is huge overkill -- simple approximations of $ n!$ coming from AM-GM would suffice, but even they are much stronger than you need:
\[ \frac {n!}{n^n} \equal{} \prod_{i \equal{} 1}^n \frac {i}{n} \leq \frac {1}{n}
\]
since every term in the product after the first is at most 1.

[quote="JBL"]
\[ \frac {n!}{n^n} \equal{} \prod_{i \equal{} 1}^n \frac {i}{n} \leq \frac {1}{n}
\]
since every term in the product after the first is at most 1.[/quote]
O, thank you. Nice.

Alternatively, apply the ratio test to the convergence of $ \sum \frac{n!}{n^n}.$

You get a limiting ratio of $ e^{\minus{}1},$ which is smaller than $ 1,$ so the series converges, so the terms tend to zero.

What you have actually shown is this: $ \frac{n!}{n^n}$ equals $ e^{\minus{}n}$ multiplied by something whose rate of growth or decay is slower than geometric.

More formally, you could say that for any real number $ a$ and $ b$ such that $ a<e^{\minus{}1}<b,$ we have $ a^n<\frac{n!}{n^n}<b^n$ for sufficiently large $ n.$

You can consider this result to be a very weak version of Stirling.

===========================================

Here's another very weak version of Stirling:

$ \ln((n\minus{}1)!)\equal{}\sum_{k\equal{}1}^{n\minus{}1}\ln k<\int_1^n\ln x\,dx\equal{}n\ln n\minus{}n\plus{}1$

Exponentiating, we get that

$ (n\minus{}1)!<\frac{n^n}{e^{n\minus{}1}},$ which becomes $ n!<(ne)\cdot\frac{n^n}{e^n}.$

That would suffice to answer the original question.

To see the inequality between the sum and the integral, just draw a picture. It's a case of a left-hand endpoint Riemann sum being less than the integral of an increasing function.

I see it quite simply. The numerator is a product of $ n$ increasing integers. The denominator is the product of $ n$ integers, all of which are greater than every factor in $ n!$ with the exception of $ n$ itself. Or, just write it out and you'll see that the result is perfectly natural. \[ \lim_{n\to\infty} \frac{1\cdot 2\cdot 3\cdots n}{n\cdot n\cdot n\cdots n}\equal{}0\]

$ n! \leq \left(\frac{n\plus{}1}{2}\right)^n$

$ \frac{n!}{n^n} \leq \left(\frac{n\plus{}1}{2n}\right)^n$

$ \frac{1}{n^n} \leq \frac{n!}{n^n} \leq \frac{1}{2^n} \left(1 \plus{} \frac{1}{n}\right)^n$

$ 0 \leq \lim_{n \to \infty} \frac{n!}{n^n} \leq 0 \cdot e$

Right, this is the elementary method with AM-GM that I referred to.  Actually, we can make it "prettier" by saying
\[ n! \equal{} n \cdot (n \minus{} 1)! \leq n \cdot \left(\frac {n}{2}\right)^{n \minus{} 1} \equal{} \frac {n^n}{2^{n \minus{} 1}},
\]
where the second step comes (just as Carcul's first step) from AM-GM: the geometric mean of the first $ k$ positive integers is $ (k!)^{1/k}$ and the arithmetic mean is $ \frac {k \plus{} 1}{2}$.  (In this "prettier" solution, we apply AM-GM to the first $ n \minus{} 1$ positive integers after pulling out a factor of $ n$, instead of applying it to $ 1, \ldots, n$ as Carcul did.)


Geico commercials are pretty funny, with the "celebrity" translating what the "non-celebrity" is saying.  Sometimes it gets pretty lame though...

However, the commercials I HATE are the car commercials.  People must be really desperate to get their cars sold, because stinkin' commercials for Chevy and SUVs and a bunch of other cars come on like every 10 seconds.   :mad:

I hate almost every commercial on TV.  Off the top of my head, I can't think of any that I like.  Maybe if there is a commercial that plays a song I like, but otherwise, no.

And with regards to your signature, US has the most medals so reverse the order.   Sure China has more golds, but look at the events they're winning in: Weightlifting, Shooting, and Diving.  And most of the track and field events have yet to take place.

So easy a caveman could do it. I also like the one where they use the honey bottle. I especially like the quote. "O look they have just put a logo over my head." Here, I'll add a youtube link.
[youtube]vyF1mrFO9XM[/youtube]

[quote="JRav"]And with regards to your signature, US has the most medals so reverse the order.   Sure China has more golds, but look at the events they're winning in: Weightlifting, Shooting, and Diving.  And most of the track and field events have yet to take place.[/quote]

golds matter more than the number.
think of giving medals point values. gold=5 silver=3 bronze =1

I always thought of the medal values as Gold=3 Silver=2 Bronze=1

My score would make it US 102, China 114. However their are still alot of events to go. Most of which China has almost no chance in. Event Gymnastics for instance should be a pretty nice even split between China and the US. However Track and Field the US well guarantee have more medals than China. Factor in the rest of the sports and it will be pretty even.

Lets get back on topic now  :wink: .

Shoulda gone to freeeeeeeeeeeeecreditreport.com...

lol

for some reason i get really, REALLY pissed off at this one cinnamon toast crunch commercial. its the one where the kids with the cereal make fun of that amazon guide/ppl protector person. they even say that he sees really dangerous things, like snakes and alligators, and saves ppl from them, but them they make fun of him with the whole "he just doesnt see it" thing! dunno why, it makes me really pissed off.

i like the one where this guy accidently hits these 2 old timers car and then this happens:

guy: im sorry plz forgive me.
old lady: forgive you i got your forgiveness right here!
(slams car door into guy and starts beating him with her purse
old guy: HIT HIM IN THE HEAD
old lady hits him in the head
guy: OWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW

What is that a commercial for?  I have never seen it before.


limit
real analysis
real analysis unsolved

solution of charlotte and nicolas  high school students

all terms are positive, set $a$ for  liminf $x_{n}$

$b$ for limsup $x_{n}$

$a\leq b$

$a\geq 3\sqrt{a}$

$b\leq 3 \sqrt{b}$

a=b=9

In the Moubinool's proof you have to show that the cases $a=0$ or $b=\infty$ are not possible.
That's right because $\min (9,x_{0},x_{1},x_{2}) \leq x_{n}\leq \max (9,x_{0},x_{1},x_{2})$
 :cool:


Let $ p,q,r$ be distinct primes.
Show that $ \frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r}$ cannot be an integer.

[quote="Anavel_Gato"]Let $ p,q,r$ be distinct primes.
Show that $ \frac {1}{p} \plus{} \frac {1}{q} \plus{} \frac {1}{r}$ cannot be an integer.[/quote]

[hide="answer"]
\[ S \equal{} \frac 1p \plus{} \frac 1q \plus{} \frac 1r \equal{} \frac {pq \plus{} qr \plus{} pr}{pqr}
\]
Clearly: no one of them is even.

So:
\[ 1 > \frac 13 \plus{} \frac 15 \plus{} \frac 17 \ge \frac 1p \plus{} \frac 1q \plus{} \frac 1r > 0\]
so $ S$ can not be an integer.

[/hide]

You can also generalize it to n distinct primes.

[hide="solution"]
We have $ \frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r}\equal{}\frac{pq\plus{}pr\plus{}qr}{pqr}$, so for this to be an integer, $ pq\plus{}pr\plus{}qr\equal{}pqrn$ for some integer $ n$.  Taking this mod $ pq$ we get \[ r(p\plus{}q)\equiv 0\pmod{pq}\Rightarrow p\plus{}q\equiv 0\pmod{pq}\] Since $ p$ and $ q$ are distinct primes, $ 0<p\plus{}q<pq$ and therefore $ p\plus{}q$ can't be $ 0$ in mod $ pq$, so we have a contradiction.
[/hide]


[quote="GacKiem"]If $a,b,c \ge 0$ and $abc=1$ then
\[\frac{1}{a^{2}-a+1}+\frac{1}{b^{2}-b+1}+\frac{1}{c^{2}-c+1}\le 3. \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87574
It's stronger! :wink:


given a circle with center $O$, and diameter$AB$,points $C,D$ are chosen  on the semicircle of diameter AB such that
$AC<AD$,$D$ distinct  from$B$.let $E$ be any point on the other semicircle of diameter $AB$ ,$E$ distinct from $A,B$.let$CE$ intersect  $AD$ at$I$, and let  $IO$ meet $BE$ at$K$. prove that angle $CDK=90^{o}$

We will prove a slightly different but equivalent result, as follows:
AB is the diameter of a circle centered at O. C and D (AC<AD) are points on the circle and also lying on the same side of AB. C' is a point on the circle lying on the opposite side to C wrt AB, such that $\angle ADC=\angle BDC'$. Let E be an arbitrary point on the circle lying on the same side of C' wrt AB, $CE\cap AD=I$. Then the lines IO, DC', BE are concurrent.

Note that O lies on CC'. It is easy to see that triangles OBC' and OAC are congruent. Let I' be the corresponding point of I in triangle OBC'. In other words, with the same barycentric/trillinear coordinates. Now it is obvious that IO, DC', BE concur at the isogonal conjugate of I'.

Another solution:

[color=blue][b]Problem.[/b] Let k be a circle with center O, and let AB be a diameter of this circle. Let C, D, E be three points on this circle k. Let $CE\cap AD=I$ and $IO\cap BE=K$. Prove that < CDK = 90°.[/color]

[i]Solution.[/i] Let R be the point diametrically opposite to the point C on the circle k. Then, the segment CR is a diameter of the circle k and thus passes through its center O. Of course, the diameter AB of the circle k must also pass through its center O. Now, applying the Pascal theorem to the cyclic hexagon CRDABE, we see that the points $CR\cap AB$, $RD\cap BE$ and $DA\cap EC$ lie on one line. Since $CR\cap AB=O$ and $DA\cap EC=I$, we can thus conclude that the points O, $RD\cap BE$ and I lie on one line. In other words, the point $RD\cap BE$ lies on the line IO. Thus, the point $RD\cap BE$ coincides with the point $IO\cap BE=K$. Hence, the point K lies on the line RD, so that < CDK = < CDR. But since the segment CR is a diameter of the circle k, we have < CDR = 90°. Thus, < CDK = 90°. Problem solved.

  Darij


inequalities
geometry
inequalities proposed

Prove that in any triangle we have the following inequality:

$\displaystyle h_{a} ( r_{b}+ r_{c} ) + h_{b} ( r_{a}+ r_{c} ) + h_{c} ( r_{a}+ r_{b} ) \geq \frac{2}{3} \left( h_{a}+h_{b}+h_{c} \right)^{2}.$

cheers! :D :D

[quote="Lagrangia"]Prove that in any triangle we have the following inequality:

$ \displaystyle h_{a} ( r_{b} \plus{} r_{c} ) \plus{} h_{b} ( r_{a} \plus{} r_{c} ) \plus{} h_{c} ( r_{a} \plus{} r_{b} ) \geq \frac {2}{3} \left( h_{a} \plus{} h_{b} \plus{} h_{c} \right)^{2}.$[/quote]

Where $ h_a$, $ h_b$, $ h_c$ are the altitudes and $ r_a$, $ r_b$, $ r_c$ are the exradii of triangle ABC. (Please say this explicitely!)

The solution is rather easy:

[hide="Solution"]

First, after the AM-HM inequality,

$ \frac {r_b \plus{} r_c}{2}\geq \frac {2}{\frac {1}{r_b} \plus{} \frac {1}{r_c}}$.

But if S is the area of triangle ABC and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$ is the semiperimeter, we have $ r_b \equal{} \frac {S}{s \minus{} b}$, so that $ \frac {1}{r_b} \equal{} \frac {s \minus{} b}{S}$, and similarly $ \frac {1}{r_c} \equal{} \frac {s \minus{} c}{S}$. Thus,

$ \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {s \minus{} b}{S} \plus{} \frac {s \minus{} c}{S} \equal{} \frac {2s \minus{} b \minus{} c}{S} \equal{} \frac {\left(a \plus{} b \plus{} c\right) \minus{} b \minus{} c}{S} \equal{} \frac {a}{S}$.

But after a well-known area formula for triangles, $ S \equal{} \frac12 ah_a$; thus, $ \frac {a}{S} \equal{} \frac {2}{h_a}$. Thus, we have

$ \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {a}{S} \equal{} \frac {2}{h_a}$,

and consequently,

$ \frac {r_b \plus{} r_c}{2}\geq \frac {2}{\frac {1}{r_b} \plus{} \frac {1}{r_c}} \equal{} \frac {2}{\left(\frac {2}{h_a}\right)} \equal{} h_a$,

so that $ r_b \plus{} r_c \geq 2h_a$. Similarly, $ r_a \plus{} r_c \geq 2h_b$ and $ r_a \plus{} r_b \geq 2h_c$.

Using these inequalities, we have

$ h_a\left(r_b \plus{} r_c\right) \plus{} h_b\left(r_a \plus{} r_c\right) \plus{} h_c\left(r_a \plus{} r_b\right) \geq h_a\cdot 2h_a \plus{} h_b\cdot 2h_b \plus{} h_c\cdot 2h_c$
$ \equal{} 2\cdot\left(h_a^2 \plus{} h_b^2 \plus{} h_c^2\right)$.

On the other hand, for any three real numbers x, y, z, we have $ x^2 \plus{} y^2 \plus{} z^2\geq \frac13\left(x \plus{} y \plus{} z\right)^2$ (since $ x^2 \plus{} y^2 \plus{} z^2 \minus{} \frac13\left(x \plus{} y \plus{} z\right)^2 \equal{} \frac13 \left(\left(y \minus{} z\right)^2 \plus{} \left(z \minus{} x\right)^2 \plus{} \left(x \minus{} y\right)^2\right)$). Applying this to $ x \equal{} h_a$, $ y \equal{} h_b$, $ z \equal{} h_c$, we see that $ h_a^2 \plus{} h_b^2 \plus{} h_c^2 \geq \frac13\left(h_a \plus{} h_b \plus{} h_c\right)^2$, and thus

$ h_a\left(r_b \plus{} r_c\right) \plus{} h_b\left(r_a \plus{} r_c\right) \plus{} h_c\left(r_a \plus{} r_b\right) \geq 2\cdot\left(h_a^2 \plus{} h_b^2 \plus{} h_c^2\right)$
$ \geq 2\cdot\frac13\left(h_a \plus{} h_b \plus{} h_c\right)^2 \equal{} \frac23\cdot\left(h_a \plus{} h_b \plus{} h_c\right)^2$,

what proves Lagrangia's inequality.[/hide]

  Darij

Here is another one :
$\Sigma h_a(r_b+r_c)=\Sigma\frac{2A}{a}(\frac{A}{s-b}+\frac{A}{s-c})=2s^2=$$\frac{2}{3}3s^2=\frac{2}{3}3\Sigma s(s-a)\geq\frac{2}{3}3\Sigma h_a^2\geq\frac{2}{3}(\Sigma h_a)^2 $

We  used  the well  know  ineq  [tex] h_a^2\leq s(s-a) [/tex]
A  is  area , s is  semiperimeter....

Hmm.. a one line solution.. congrats Nickolas!

And with this post I would like to wish everyone a Happy New Year! A New Year full of joy, happyness and everything you ever wanted!

A Happy New Year everyone!

 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D


A two-digit number is divided by the sum of its digits. What is the greatest remainder obtained?

[i]Solutions Appreciated[/i] :wink:
[i]No Calculators[/i] :wink:

[hide]To get a big remainder, we need a big divisor. To get a big divisor, we need a big number. By testing 90-99, the answer is $98 \equiv13\pmod{17}$[/hide]

i'm not very good with mods, so could you please elaborate?

EDIT: it is now easier to follow that ht elatex is fixed, but could you please show your steps?
EDIT2: sorry-i didn't see your post after that :blush:

[quote="anirudh"]i'm not very good with mods, so could you please elaborate?[/quote]
I'm not that good either... In $a\equiv b\pmod{c},$ (note the use of \equiv, not =)  $a$ and $b$ have the same remainder when they are divided by $c.$

Don't worry too much about mod.  Look at this solution.
[hide]This is a solution from http://www.mathlesson.com :
"The largest remainder we can hope for is 17 on account of the fact that the sum of the digits in any two digit number is not bigger than 18. But the only way this could occur is if our number is 99. But 99 = 5×18 + 9 so we don’t get a remainder of 17.

Next, try 16. The only numbers that can give this are those whose sum is larger than 16, i.e. 17 or 18. So the candidates are 89,98,99. But 89 = 5×17 + 4, 98 = 5×17 + 13, 99 = 5×18 + 9. Therefore no remainder of 16 will ever happen.

Next try 15. You immediately find 79 = 4×16 + 15 [b]so therefore 15 is the highest remainder possible.[/b]"[/hide]

Yes. I already read that solution. It is from the MOEMS answer guide. However, I wanted a faster way to do so. That might keep on going, and I might not be able to solve it within the given 6 minute time frame.


geometry
incenter
vector
trigonometry
circumcircle
geometry proposed

$\triangle{ABC}$ is an equilateral triangle. $D$ is on side $BC$. Let $I_{1}$ and $O_{1}$ be the incentre and circumcentre of $\triangle{ADB}$, $I_{2}$ and $O_{2}$ be the incentre and circumcentre of $\triangle{ADC}$ respectively. Prove that: \[I_{1}O_{1}^{2}+I_{2}O_{2}^{2}= I_{1}I_{2}^{2}\]

my method is meaningless,
but :D ,it comes using vectors and dr. trig

It surely is a very nice question.
Here is my solution:

$O_{2}=$ perpendicular bissector $AC$  $\cap$ perpendicular bissector $AD$ = angle bissector $\widehat{ABD}$  $\cap$ perpendicular bissector $AD$

So it follows that $O_{2}$ lies on the circumcircle of $\Delta ABD$.
Analogously we show that $O_{1}$ lies on the circumcircle of $\Delta ACD$. This gives us that the radii of those two circles are equal.

Call $L$ the intersection point of line $O_{1}O_{2}$ and $AD$.
Call $K$ intersection point of $BO_{2}$ and $AC$.
Clearly, $O_{2}KLA$ is a cyclic quadrilateral.
So $\angle I_{1}O_{2}O_{1}= \angle CAD = \angle I_{2}O_{1}D$.
Now $\Delta I_{2}O_{1}D \cong \Delta I_{1}O_{2}O_{1}$
So \[\left|I_{1}O_{1}\right| = \left|I_{2}D\right|\] Analogously, after calling $CO_{1}\cap AB = M$ and noticing that $LO_{1}MA$ is cyclic, we show that $\Delta O_{2}O_{1}I_{2}\cong DO_{2}I_{1}$, so it follows that \[\left|I_{2}O_{2}\right| = \left|I_{1}D\right|\] It's trivial that $I_{2}DI_{1}= \frac{\pi}{2}$
So \[I_{1}O_{1}^{2}+I_{2}O_{2}^{2}=I_{2}D^{2}+I_{1}D^{2}= I_{1}I_{2}^{2}\] $\mathbb{QED}$

[quote="Jan"]Shobber, do you know if there exists a solution using Euler's formula ?[/quote]
I am sure that there is, but I haven't found it...... :(


abstract algebra
superior algebra
superior algebra unsolved

Is there a finite abelian group, of order at least two, so that its right regular representation is irreducible?

Such a group would have exactly one class of conjugated elements since the regular representation is a sum of alle the irreducible ones and the number of irrducible representations equals the number of conjugation classes. The only group that has only one conjugation class is the trivial group which has 1 element.

Hence: No, such a group doesn't exist.


I know how to get the answer by plugging in each possibility for b, but I cannot seem to derive the answer naturally.  Any insight would be helpful.

[b]23)[/b]  For what value of $ b$ is the line $ y\equal{}10x$ tangent to the curve $ y\equal{}e^{bx}$ at some point in the $ xy$-plane?

It's multiple choice on the test, but I want to know how to do it without the answer choices :].

A curve $ y \equal{} f(x)$ is tangent to a curve $ y \equal{} g(x)$ at a point $ x \equal{} a$ if and only if we have both $ f(a) \equal{} g(a)$ and $ f'(a) \equal{} g'(a)$.  For any choice of $ b$ there are at most two possible values for $ a$.  Alternately, determine the equation for the tangent line to $ y \equal{} e^{bx}$ at $ x \equal{} a$ and set the constant term to zero.


[quote="ABRORBEK"]have you old and new method of inequality:?:[/quote]

"Old and new methods" by Vasile Cirtoaje ? the inequality book ? I have it , but why do you ask :D ? But I have the book. I dont think there is an electronic version


linear algebra
matrix
vector
linear algebra unsolved

I have a matrix $ F\in \mathbb{C}^{m\times n}$ with $ m\geq n$ such that $ \parallel{}F x\parallel{}^2\equal{}c\parallel{}x\parallel{}^2$ where $ x\in \mathbb{C}^n$ and $ c$ is a complex scalar. Is this true always that $ F^H F\equal{} cI$? where $ I$ is $ n\times n$ identity matrix. Is this true for the norms?

bhupala

You say that $ c\in \mathbb{C},$ but note that since norms are nonnegative real numbers, we must actually have that $ c$ is a nonnegative real number. Furthermore, if $ c\equal{}0,$ then the whole setup is trivial, so WLOG assume that $ c>0.$

Now replace $ F$ throughout by $ \frac1{\sqrt{c}}\,F.$ This reduces the problem to the following:
\[ \|Fx\|^2\equal{}\|x\|^2\,\forall\,x\in\mathbb{C}^n\ \iff\ F^*F\equal{}I.\]
(I'm using $ F^*$ to mean the same thing you meant by $ F^H.$ It's the notation I'm used to.)

The forward direction is the greatest challenge. To prepare for it, prove this:

Lemma: for $ x,y\in\mathbb{C}^n,$ $ x^*y\equal{}0$ iff $ \|x\plus{}y\|^2\equal{}\|x\|^2\plus{}\|y\|^2$ and $ \|x\plus{}iy\|^2\equal{}\|x\|^2\plus{}\|y\|^2.$

Prove this by expanding out
$ \|x\plus{}y\|^2\equal{}(x\plus{}y)^*(x\plus{}y)\equal{}x^*x\plus{}x^*y\plus{}y^*x\plus{}y^*y\equal{}\|x\|^2\plus{}\|y\|^2\plus{}2\,\text{Re}\,(x^*y)$

$ \|x\plus{}iy\|^2\equal{}(x^*\minus{}iy^*)(x\plus{}iy)\equal{}x^*x\plus{}ix^*y\minus{}iy^*x\plus{}y^*y\equal{}\|x\|^2\plus{}\|y\|^2\minus{}2i\,\text{Im}\,(x^*y)$

Now for the main argument.

Suppose we have $ \|Fx\|^2\equal{}\|x\|^2\,\forall\,x\in\mathbb{C}^n.$ Suppose $ x^*y\equal{}0.$ Then

$ \|Fx\plus{}Fy\|^2\equal{}\|F(x\plus{}y)\|^2\equal{}\|x\plus{}y\|^2\equal{}\|x\|^2\plus{}\|y\|^2\equal{}\|Fx\|^2\plus{}\|Fy\|^2$ and

$ \|Fx\plus{}iFy\|^2\equal{}\|F(x\plus{}iy)\|^2\equal{}\|x\plus{}iy\|^2\equal{}\|x\|^2\plus{}\|y\|^2\equal{}\|Fx\|^2\plus{}\|Fy\|^2$

So we conclude that $ x^*y\equal{}0$ implies $ x^*F^*Fy\equal{}(Fx)^*Fy\equal{}0.$

Choose $ x$ to be the $ j$th standard basis vector of $ \mathbb{C}^n$ and $ y$ to be the $ k$th standard basis vector. Then $ x^*F^*Fy$ is the $ (j,k)$th element of the $ n\times n$ matrix $ F^*F.$ But the the argument above, this element is zero, meaning all of the off-diagonal entries of $ F^*F$ are zero. At the same time, if we just let $ x$ equal the $ j$th standard basis vector, then $ x^*F^*Fx$ is the $ (j,j)$the element of the matrix. This equals $ \|Fx\|^2\equal{}\|x\|^2\equal{}1,$ so the on-diagonal elements of the matrix are $ 1.$

We have shown that $ F^*F\equal{}I.$ (Meaning that the columns of $ F$ form an orthonormal set.)

The converse is much easier. If we have that $ F^*F\equal{}I,$ then $ \|Fx\|^2\equal{}(Fx)^*(Fx)\equal{}x^*F^*Fx\equal{}x^*Ix\equal{}x^*x\equal{}\|x\|^2.$


let a denot odd number,and b an even number such that  a 2 +b 2 =p (p is a prime number) prove that the equation  x 2 -py 2 =a has integer solution.

Please check komal and see if it is from a current contest...

[b]Source:[/b]Kmal January 2004 Contest problem A.336

I'm not sure if the contest is over yet for this month...


[quote="hollandman"]Prove that if $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$, then

$ (1 \minus{} \frac {1}{1 \plus{} x^2})(1 \minus{} \frac {1}{1 \plus{} y^2})(1 \minus{} \frac {1}{1 \plus{} z^2}) > \frac {1}{2}$[/quote]

Let $ a \equal{} \frac {1}{x}, b \equal{} \frac {1}{y}, c \equal{} \frac {1}{z}$, we have $ a \plus{} b \plus{} c \equal{} 1$ and
$ (1 \minus{} \frac {1}{1 \plus{} x^2})(1 \minus{} \frac {1}{1 \plus{} y^2})(1 \minus{} \frac {1}{1 \plus{} z^2}) > \frac {1}{2}$
$ \Leftrightarrow 2 > (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)$.
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} a^2b^2c^2 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 1$
$ < (ab \plus{} bc \plus{} ca) \plus{} ab \plus{} bc \plus{} ca \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 1 \equal{} 2$
($ \because 0 < a,b,c < 1$)

[quote="hollandman"][quote="NextPeace"]($ \because 0 < a,b,c < 1$)[/quote]

Really?  :wink:[/quote]

Yes. $ a\plus{}b\plus{}c\equal{}1$.


Would anybody be kind enough to post the 2006 MOP list of participants?

Don't ask why, I'm just a little curious.

Zachary Abel, Noah Arbesfeld, Gye Hyun Baek, David Benjamin
Yakov Berchenko-Kogan, John Berman, Sergei Bernstein
Noah Blach, Timothy Black, Jason Bland, Zarathustra Brady
Paul Christiano, Joseph Chu, Karl Cobbe, Peter Diao
Carlos Dominguez, Miles Edwards, Edward Gan
Michael Gottlieb, Jeremy Hahn, Yi Han, Jason Hoch
Shui Hu, Alan Huang, William Johnson, Seungsoo Kim
Shaunak Kishore, Taehyeon Ko, Meir Lakhovsky, Eric Larson
Tedrick Leung, Particia Li, Tony Liu, Jeffrey Manning
Delong Meng, Kevin Modzelewski, Daniel Poore, David Rolnick
Alex Ryan, Adrian Sanborn, Jason Steinberg, Jacob Steinhardt
Yi Sun, Christian Tessier-Lavigne, Nicholas Triantafillou
Arnav Tripathy, Andy Tsao, Yixin Wang, Yichen Wu, Alex Zhai
Bohua Zhan, Tong Zhan, Mark Zhang

I think I left somebody out who wasn't on the original list... I can't think of who.

Hmm. . . . Shui Hu and Paul Christiano were left out  :(  :D

There; fixed

[quote="MysticTerminator"]Greg Brockman actually did not exist! (got silver at IChO)[/quote]

Oh yeah, good point... I was just kind of blindly copying the list.

[quote="Sly Si"]Yeah, I'm kind of surprised AMC hasn't posted it yet; it's usually up soon after the phone calls go out.[/quote][url]http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2006-ma/06mosppart.html[/url]

At the moment, I believe that there are actually some errors on the list posted on the AMC site (i.e., some people who are currently on that list weren't at MOP this year).  The one jb05 currently has posted is correct.

two of those people goes to my school
and one from harker than used to go to my school

[quote="rnwang2"][quote="Boy Soprano II"][quote="13375P34K43V312"]how many of those are aopsers?[/quote]
Practically everybody, I think, although many aren't really active.[/quote]

Do you happen to know Alex Zhai's username?[/quote]

probability1.01


Prove that $ \forall n$: \[ \overline{6\dots 67}^2\minus{}1\equal{}\overline{4\dots 48\dots 8}\]
if there are $ n$ '4's and $ n$ '8's on the right, whereas there are $ (n\minus{}1)$ '6's and only 1 '7' on the left side.


The batting averages of David Ortiz and Manny Ramirez in this year's ALCS were, respectively, .292 and .409 (Ortiz, for instance, made a hit in 29.2 percent of his times at bat). Suppose that Ortiz and Ramirez each come to bat 4 times in the first game of the World Series (tomorrow). Assuming that whether they get hits or not is determined solely by their ALCS batting averages and random chance, what is the probability that neither either Ortiz or Ramirez doesn't get a single hit? (A calculator is useful for this problem.)

Would this be the answer?

[hide]$ (\frac{708}{1000}\plus{}\frac{501}{1000})\times\frac{1}{2}\equal{}\boxed{\frac{1209}{2000}}$[/hide]

wouldnt you rather do...

[hide]
Ortiz can hit 29.2 percent of the time, 1000-292, to get 708
so Ortix cannot hit 70.8% of the time

do the same with Ramirez
so Mr. R cannot hit 59.1% of the time

708/1000 is Ortiz's % of not hitting the ball in a fraction form
because ortiz is going to swing 4 times and not hit, do (708/1000)^4

same for Mr. R
(591/1000)^4

and multiply those two together.

DonkeyKong's solution seems reasonable too....[/hide]

[quote="Chinaboy"]huh? neither either?[/quote]
Huh???
neither, either???
I don't get it..

[quote="DonkeyKong"]what is the probability that [b]neither either [/b]Ortiz or Ramirez doesn't get a single hit?
Would this be the answer?

[hide]$ (\frac {708}{1000} \plus{} \frac {501}{1000})\times\frac {1}{2} \equal{} \boxed{\frac {1209}{2000}}$[/hide][/quote]

The wording... is confusing.

Oh dear - I wrote that, and I'm sorry about the typo.  I meant to write just "either."  Sorry!

variable

If it's either, would be asnwer be right??

You did it all in one step, which I don't think can be done.

Try to break this into 3 cases. Find the probability for each. And then what do you do? (Add or multiply, DK?)

The answer is on Vermont Forum...(Solution for POD 23)

@Chinaboy: I realized my mistake...

[quote="danb80"]DK:  You can link to the page on the other forum by just copying the url (e.g. http://www.art.....), paste it in this forum as I have done below.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172314[/quote]

Okay. Thanks for the tip. :lol:


Let $ S$ be the set of all sequences of non-negative integers.If $ a_1,a_2,......$ be the terms of any sequence of $ S$ then $ a_1\geq a_2\geq ......$ and there exists a positive number $ N$ such that $ a_n \equal{} 0$ for all $ n\geq N$

Define the dual of the sequence $ (a_1,a_2,......)$ belonging to $ S$ to be the sequence $ (b_1,b_2,....)$, where for $ m\geq 1$, $ b_m$ is the number of $ a_n$'s which are greater than or equal to $ m$

Show that:

$ (i)$the dual of a sequence in $ S$ belongs to $ S$
$ (ii)$ the dual of the dual of the sequence is the original sequence itself
$ (iii)$the duals of distinct sequences in $ S$ are distinct.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273902 .

Where'd the problem come from?

$ (i)$ is obvious
$ (ii)$ is a Ferrer's diagram
$ (iii)$ is also a Ferrer's diagram

@Differ:Can you elaborate how u used Ferrer's diagram

My bad, it's a Ferrers diagram, not a Ferrer's diagram. See http://en.wikipedia.org/wiki/Ferrers_diagram#Ferrers_diagram

Each sequence is represented by a string of dots. Taking the dual of any sequence is equivalent to flipping the Ferrer's diagram over the 45 degree line, and if you flip a diagram twice, you get the original.

Ferrers diagrams: not just good for snake oil :wink:


modular arithmetic
function
number theory proposed
number theory

Let $p$ and $q$ be distinct odd prime numbers. Prove that \[\sum_{x_{1}+x_{2}+\cdot\cdot\cdot+x_{q}\equiv q\pmod{p},1\leq x_{i}\leq p-1}(\frac{x_{1}\cdot x_{2}\cdot\cdot\cdot x_{q}}{p})\equiv 1\pmod{q}.\]

Sketch: For a summand having
\[x_{1}= x_{2}= \cdots = x_{q}\;,\]
the common value of the $x_{i}$ must be 1. That summand contributes 1 to the sum. The function on the right side is symmetric in all the $x_{i}$, which means that all the other summands fall into equivalence classes, such that any two summands in the same equivalence class make the same contribution, and the number of summands in any equivalence class is a multinomial coefficent
\[\frac{q!}{a_{1}!\cdots a_{t}!}\]
where the sum of the $a_{i}$ is $q$ and at least one of the $a_{i}$ satisfies $0<a_{i}<q$. Such a multinomial coefficient is divisible by $q$ if $q$ is prime.

The condition $1\le x_{i}\le p-1$ may be replaced by $1\le x_{i}\le k$ for any $1\le k <p$. And the Legendre symbol may be replaced by any symmetric function $F$ if we change the conclusion to
\[\text{left side }\equiv F(1,1,\ldots,1)\pmod q\]

I hope I haven't blundered. :)