So, this is the list of players for game 11. You are on this list if you either confirmed the PM, or signed up less than 2 days ago.

List of players:
Sly Si
samath
SysRq
jli
chesspro
ZennyK
zeb
Sunny
dts
G-UNIT
pianoforte
noneoftheabove

Normally the roles will be divided as follows:

Mafia: 4
Vigilante: 1
Angel: 2
Inspector: 1
Mason: 3
Citizen: 1

You will receive your roles in a few moments.

Sorry, there has been a roles mistake. I will re-roll all roles and send you a new PM. Please ignore the first PM if you received it somehow.

Ok, everyone should have their roles now, you can start discussing. It is day.

No one can communicate during the day other than through this thread. Only exception is that everyone may communicate with me 24/24 7/7 by PM.
Mafia, masons, angels may communicate at night.

Night phases will not take more than 24 hours. Angels/Mafia/Inspector/Vigilante choose one target per class. (so I recieve 4 targets in total)
[b]Edit:[/b] The first nigh phase may take a little more, as it's the first discussion. Every group (angels/mafia/inspector/vigilante) gives me the target for that night. [b]I recommend you to add a queue to your inspections/killings/savings, just for the case you would miss one night.[/b] You can always change it the next night. So e.g. mafia tell me they will kill person A, then person B, then person C. They don't know for further nights. But person D appears to be a valuable role next day. So the next night, they change their list, first person D, then B, then C. Or something.

Any further questions, PM me. Enjoy! :)

Peter, maybe instead of a random pick you should just make that person skip their role for the night...it's their fault.

So...like all mafia games, does anyone have a random accusation to get things started?

BTW, I will probably be gone for the rest of the day; I apologize if I don't respond until later today (or possibly tomorrow).

[quote="chesspro"]Peter, maybe instead of a random pick you should just make that person skip their role for the night...it's their fault.[/quote] Yes, I'm still thinking on that one... in game 9 we got kinda screwed because windslicer didn't inspect the first 5 turns... and that ruins the game for the others too.

I currently think on mafia/angels missing their turn if they forget, but as the inspector is alone in his role, I feel forced to enfore inspection each night. Just to keep the fun in the game. (I'll just do Math.Random(12); so it's really random)

What I also can do is allowing a queue. So e.g. angels PM me first save A, then next turn save B (unless a counter-pm comes), next turn C etc. Same for mafia and inspector. Maybe this is the best option?

Here and ready to play.

Edit: didn't realize I was copying word for word the last post...

Here and ready to play.

Oh wait I'm modding this :D

You can start discussing people, confirmation that you play has already happened ;)

For clarification, I added this little rule. It only applies to mafia / inspector / angels / vigilante.

The first night phase may take a little longer, as it's the first discussion.
Every group (angels/mafia/inspector/vigilante) gives me the target for that night.
[b]I recommend you to add a queue to your inspections/killings/savings, just for the case you would forget one night.[/b]
It is not mandatory. You can always change it the next night.

So e.g. mafia tell me they will kill person A, then person B, then person C.
They don't know for further nights. But during the day person D appears to be a valuable role.
So the night, they change their list, first person D, then B, then C. Or something.

Just the same as normal, but providing me with a backup if you forget. Please people, use this option :) it's too stupid to miss kills/saves/inspections by forgetting.

lol this is really amusing right now.  It sounds like its come down to whether or not I'm mafia or mason, hehehe.

Take your chances.  Either way, citizens will win in the end.  Though I'd like to be alive for that.

In my opinion:

Angels save samath.
Samath inspects Sunny.

As for today:

1) jli has not roleclaimed.
2) jli has not been murdered by the mafia.

So, I believe that I should lynch jli. However, I shall not do so until the rest of the town agrees with this (and I may not check this thread again until tomorrow morning; my apologies if so).

Everyone except for the suspected mafia and you agrees with this lynching.

No one has counter-claimed mason, so I think we can believe Sunny (if someone other than jli or Sunny was the real mason, then jli and Sunny would both be mafia, and if jli was the real mason he would have said so).

Edit: in other words, even if Sunny is lying, jli is mafia (unless jli is an extremely stupid mason).

That evening, jli, supreme local mafia leader, was lynched by the citizens. They found him by pure coincidence, when he fell off a chair in the grocery store (while taking his favorite cereals off the highest rack) his gun dropped out of his pocket. He got arrested, and lynched after a ton of evidence was found into his house.

And then it was night. Citizens have again lynched a mafia, and have made great progress in the battle against the organised crime. The citizens held a little party tonight, so most went to bad pretty drunk... always dangerous. ;)

Night has begun. It started quiet...

That night, the mafia were on revenge. They burglared silently into zeb's house, silently entered zeb's room and shot every single bullet they had towards his bed. Bloodhungry, they turned on the light to admire their work, but they only saw a destroyed bed. Instead, they both felt a gun in their back, carried by zeb and G-UNIT, who had been silently waiting in another room all night long. Both mafia got easily arrested.

The threatening of the mafia is gone. The city is cleared, and the villagers are coming back to town. The mission has succeeded, and the surviving citizens are heroes for life.

The game is over. The citizens have won.

Painful... just 20 minutes before the deadline, the mafia pm'ed me to change the kill... if they had left it on samath, the game was long not over yet. :)  [and I do wonder why the mafia did this? you knew you had to kill the inspector this night to have a chance right?]

I have a question to the mafia: is 24 hours enough to communicate? I think I will allow day communication next time, as you guys appeared to have failing communication... like with the lynching of pianoforte...

Huh? How did they get arrested by the angels? Im confused.

I'm really really confused.  Though I'm sure the citizens would have won in the end...how did it end?

Apparently, me and G-UNIT both forgot to change the queue... originally it was G-UNIT, zeb, G-UNIT, etc., and each night we have (for one reason or another) not changed it. So, by pure coincedence, the mafia made the decision to kill me, so now we have 5 citizens to 2 mafia. Assuming samath inspected Sunny, we would have definitely won... I guess Peter just wanted to speed things up a bit.

Well, 5 citizens vs. 2 mafia, and the citizens are now 100% sure of the mafia, since every innocent has been inspected. So that basicly terminates the game, there is no way the mafia can still win.

Had they killed samath, then it was a 4-2 with still doubt on who was the last mason. A wrong decision on who was the last mason would most likely have made the mafia win.

*sigh*

I think this game may have been stacked up against the Mafia, we're supposed to be able to claim citizen and not have to worry about the actual citizen counterclaiming...  it could have been even worse if we hadn't nailed the Vigilante the first day (then again, perhaps the Vigilante would have done our work for us).

I didn't think there was much doubt on the last mason.  The strategy and rationale, from my point of view, was this:

1. Noneoftheabove is going to get lynched.
2. I will soon be called upon to claim a role.  I can't claim citizen, inspector, or mason, as the identities of these people are established.
3. Thus, I have to claim angel.
4. If I claim angel with noneoftheabove, no one will believe me.
5. Thus, we should kill an angel, let noneoftheabove get lynched, and then I can claim to be the last angel.

Yay for the innocents!!

(AND Me, of course  :lol:  :P )


group theory
abstract algebra
superior algebra
superior algebra solved

$G$ is a group that order of each element of it Commutator group is finite. Prove that subset of all elemets of $G$ which have finite order is a subgroup og $G$.

Let $\Delta$ be the commutator group of $G$. Take $a,b \in G$ with finite order.

$(ab)(a^{-1}b^{-1}) \in \Delta$.

If $(ab)^{n}a^{-n}b^{-n}\in \Delta$, then also $(ab)(ab)^{n}a^{-n}b^{-n}(ab)^{-1}=(ab)^{n+1}a^{-n}b^{-(n+1)}a^{-1}\in \Delta$, since $\Delta$ is normal in $G$. Eventually, $[(ab)^{n+1}a^{-n}b^{-(n+1)}a^{-1}][a(b^{n+1}a^{-1})a^{-1}(b^{n+1}a^{-1})^{-1}] \in \Delta$, which means $(ab)^{n+1}a^{-(n+1)}b^{-(n+1)}\in \Delta$. This implies that, for $N$ common multiple of the orders of $a$ and $b$, we will have $(ab)^{N}\in \Delta$. Hence, by hypothesis, $ab$ will have finite order.

The fact that $a^{-1}$ also has finite order is trivial.

[quote="zhaobin"]Nice proof.
But what I can't understand is why:
1.$ab(a^{-1}b^{-1})\in \Delta$
2.$(ab)^{N}\in \Delta$
Can you tell me?sorry I am a beginer,thank you.[/quote]

1. By definition of the commutator subgroup of $G$.
2. Because we showed $(ab)^{N}a^{-N}b^{-N}\in \Delta$, but we picked $N$ such that $a^{-N}=b^{-N}=1$.

[quote="zhaobin"][quote="julien_santini"]

1. By definition of the commutator subgroup of $G$.
.[/quote]
I know 2 now.
For 1:What is the definition of the commutator subgroup of $G$?
Does it is equal to:$\{x|xg=gx,for any g \in G\}$?[/quote]

it's the subgroup generated by $xyx^{-1}y^{-1}$ for x,y in G


What is the value of $ 9342 \plus{}(\minus{}438)\times 719 \plus{} (\minus{}9340) \plus{} (\minus{}438)\times (\minus{}719)$?

The important thing to note here is that $ (\minus{}438)\times 719 \plus{} (\minus{}438)\times \minus{}719$ is 0.


If a,b,c>0 and a+b+c=6 proove that:
\[ \sum \frac{12bc(3-a)}{(b+c)^2} \leq \frac{7(a^2+b^2+c^2)}{2}  \]
               Luca V. Iliesiu


inequalities
function
integration
logarithms
induction
inequalities proposed

For $ a,b,c>0$ and $ a_k>0$ prove the following inequalities:

$ a)$ $ (a^2\plus{}b^2\plus{}c^2)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})\ge(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

$ b)$ $ (\displaystyle\sum_{k\equal{}1}^{n}{a_k}^2)(\displaystyle\sum_{k\equal{}1}^{n}\frac{1}{{a_k}^2})\ge(\displaystyle\sum_{k\equal{}1}^{n}a_k)(\displaystyle\sum_{k\equal{}1}^{n}\frac{1}{a_k})$

$ c)$ $ ({a^2}^n\plus{}{b^2}^n\plus{}{c^2}^n)(\frac{1}{{a^2}^n}\plus{}\frac{1}{{b^2}^n}\plus{}\frac{1}{{c^2}^n})\ge(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

[quote="moldovan"]For $ a,b,c > 0$ and $ a_k > 0$ prove the following inequalities:

$ a)$ $ (a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$

$ b)$ $ (\displaystyle\sum_{k \equal{} 1}^{n}{a_k}^2)(\displaystyle\sum_{k \equal{} 1}^{n}\frac {1}{{a_k}^2})\ge(\displaystyle\sum_{k \equal{} 1}^{n}a_k)(\displaystyle\sum_{k \equal{} 1}^{n}\frac {1}{a_k})$

$ c)$ $ ({a^2}^n \plus{} {b^2}^n \plus{} {c^2}^n)(\frac {1}{{a^2}^n} \plus{} \frac {1}{{b^2}^n} \plus{} \frac {1}{{c^2}^n})\ge(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]
I'll just prove the general:
$ (\sum_{k \equal{} 1}^{n}{a_k}^m)(\sum_{k \equal{} 1}^{n}{a_k}^{ \minus{} m}) \ge (\sum_{k \equal{} 1}^{n}{a_k})(\sum_{k \equal{} 1}^{n}{a_k}^{ \minus{} 1})$   $ \forall m > 1, a_i \ge 0$.

It's equivalent to:
$ \sum_{i,j \equal{} 1}^{n} {a_i}^m{a_j}^{ \minus{} m} \plus{} {a_i}^{ \minus{} m}{a_j}^m \ge \sum_{i,j \equal{} 1}^{n} {a_i}{a_j}^{ \minus{} 1} \plus{} {a_i}^{ \minus{} 1}{a_j}$

So it suffices to prove $ a^mb^{ \minus{} m} \plus{} a^{ \minus{} m}b^m \ge ab^{ \minus{} 1} \plus{} a^{ \minus{} 1}b \iff a^{2m} \plus{} b^{2m} \ge a^{m \plus{} 1}b^{m \minus{} 1} \plus{} a^{m \minus{} 1}b^{m \plus{} 1}$. And this is obvious by the rearrangement inequality. And this concludes the proof of a, b and c. :)

Edit: The last ineq can also be proved by: $ a^{2m} \plus{} b^{2m} \minus{}  a^{m \plus{} 1}b^{m \minus{} 1} \minus{} a^{m \minus{} 1}b^{m \plus{} 1} \equal{} (a^{m\plus{}1}\minus{}b^{m\plus{}1})(a^{m\minus{}1}\minus{}b^{m\minus{}1}) \ge 0$, and here is it also easier to see that it only holds when $ |m| \ge 1$.

Let $ f$ be a positive function defined on the reals.
Consider the positive:
\[ \int\int \left(\frac{1}{f^m(x)f^m(y)}\right)(f^{m\plus{}1}(x)\minus{}f^{m\plus{}1}(y))(f^{m\minus{}1}(x)\minus{}f^{m\minus{}1}(y)) dx dy\]
\[ \equal{}\int\int \left[f^{m}(x)f^{\minus{}m}(y)\plus{}f^{\minus{}m}(x)f^{m}(y)\right]\minus{}\left[f(x)f^{\minus{}1}(y)\minus{}f^{\minus{}1}(x)f(y)\right] dx dy\]
\[ \equal{}2\left(\int f^{m}(x)dx\int f^{\minus{}m}(x)dx \minus{} \int f(x)dx \int f^{\minus{}1}(x)dx\right)\geq 0\]
Now all the previous inequalities and a plethora more fall out by taking various functions $ f$.

More directly (again, $ f: \mathbb{R}\to \mathbb{R}^{\plus{}}$)
\[ \frac{\partial\left(\int f^m\int f^{\minus{}m}\right)}{\partial m}\equal{}\int \int \left(\frac{f(x)}{f(y)}\right)^m(\ln f(x)\minus{}\ln f(y))dx dy\]
[hide=" :| "]\[ \equal{}\frac12\int \int\left[\left(\frac{f(x)}{f(y)}\right)^m\minus{}\left(\frac{f(x)}{f(y)}\right)^{\minus{}m}\right]\ln\left(\frac{f(x)}{f(y)}\right)dxdy\]
[/hide]
\[ \equal{}\frac12 \int\int h\left(\frac{f(x)}{f(y)}\right)dxdy\geq 0\]
where $ h(a)\equal{}(a^m\minus{}a^{\minus{}m})\ln a\geq 0$ for real $ a$.

Those inequalities can be proved using Cauchy.

If $ x_1, ..., x_n$ are positive numbers and $ m$ and $ l$ are real numbers then, by Cauchy:
\[ \dfrac{x_1^{m \plus{} l} \plus{} ... \plus{} x_n^{m \plus{} l}}{x_1^m \plus{} ... \plus{} x_n^m}\ge\dfrac{x_1^m \plus{} ... \plus{} x_n^m}{x_1^{m \minus{} l} \plus{} ... \plus{} x_n^{m \minus{} l}}
\]
For the first one we use repeatedly
\[ \dfrac{a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\ge\dfrac{a \plus{} b \plus{} c}{1 \plus{} 1 \plus{} 1}\ge\dfrac{1 \plus{} 1 \plus{} 1}{a^{ \minus{} 1} \plus{} b^{ \minus{} 1} \plus{} c^{ \minus{} 1}}\ge\dfrac{a^{ \minus{} 1} \plus{} b^{ \minus{} 1} \plus{} c^{ \minus{} 1}}{a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2}}
\]
The general form as Mathias_DK stated, can be proved in a similar way.
:)

____
it is interesting to note the close connection between Cauchy and Rearrangement here, don't you think?

[quote="ElChapin"]
it is interesting to note the close connection between Cauchy and Rearrangement here, don't you think?[/quote]
We can actually prove the rearrangement by hölder, which is just cauchy generalised, so it doesn't suprise me much :) A lot of inequalitys are almost equivalent. E.g. is cauchy equivalent to weighted AM-QM, so Cauchy kan be proved by weighted Jensens. It is actually quite funny how all inequalities seems to be related in one way or another.

Yes, that is exactly what I meant.  :D  A chain I like a lot is the following

prove Jensen (the classical induction proof is fine)
then use Jensen to prove weighted AM-GM (in fact this can also be proved with induction)
then use AM-GM to prove Hölder
then use Hölder to prove power mean inequality
(and thanks to what Mathias_DK said) then use Hölder to prove rearrangement
 (and all the permutations [?] jajajaja)

Math is just sublime  :)


Gauss
number theory proposed
number theory

Let $(a)_n$ be the residue when $a$ is divided by $n\ (n=2,3,\cdots)$ and $a,b,c$ be positive integers.
Prove that

\[\left[\frac{a(bc)_n}{n}\right]+a\left[\frac{bc}{n}\right]=\left[\frac{b(ca)_n}{n}\right]+b\left[\frac{ca}{n}\right]=\left[\frac{c(ab)_n}{n}\right]+c\left[\frac{ab}{n}\right]\].

Note:$[\ ]$ denotes Gauss' sign.

$\left[\frac {a(bc)_n} n\right] +a\left[\frac {bc} n\right] $ $=$

  $\left[\frac{a((bc)_n+n[\frac {bc} n])} {n}\right]$ $=$ $\left[\frac {abc} n\right]$

 Similarily we have two more identities.


function
real analysis
real analysis unsolved

Let $f$ and $f_{n}, n \in \mathbf{N}$ be functions from $\mathbf{R}$ to $\mathbf{R}$. Assume that $f_{n}(x_{n}) \rightarrow f(x)$ as $n \rightarrow \infty$ and $x_{n}\rightarrow x$. Show that $f$ is continuous.

Thanks!

It should be clear that $f_{n}(x)\to f(x)$ for all $x$.
We want to prove that $f(y_{k})\to f(y)$ as $y_{k}\to y$. For each $k$ choose $n_{k}$ so that $|f_{n}(y_{k})-f(y_{k})|<1/k$ for $n\ge n_{k}$. Also, make sure that $n_{k}>n_{k-1}$. Let $x_{n}=y_{k}$, where $k$ is chosen so that $n_{k}\le n<n_{k+1}$. Since $x_{n}\to y$, we have $f_{n}(x_{n})\to f(y)$. Deduce that $f(y_{k})\to f(y)$.


What comes next in the following sequence:
1,2,3,2,1,2,3,4,2,1,2,3,4,3,...

i agree with NoSoupForYou... although i have no explanation for it, since ive seen something like this already...

That answer is not correct. How did you get it?

I was thinking along the lines of

12321
123421
1234321
12345321
123454321

See the pattern?

yea... i was thinking in the same line.. except one of the numbers on the descending order gets taken out.

[hide="Hint"]The nth term in the sequence has something to do with a certain representation of the number n.[/hide]

[quote="jmadsen"][hide="Another hint"]A very old representation.[/hide][/quote]

the hint gave it away, in my opinion.
2, 3, 4, 5, 3, 2, ...
[hide]XV, XVI, XVII, XVIII, XIX, XX,...[/hide]


Prove that the product of the two complex roots of $x^4 + x^3 - 1 = 0$ is a root of $x^6 + x^4 + x^3 - x^2 - 1 = 0$

I believe this is wrong  ;)

Look at my proof of it on here:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33610]My Proof[/url]

Where am I going wrong?

Let $a,b,c,d$ be the roots of the equation. Then by Vieta's formula we know

\[a+b+c+d=-1\]
\[abcd=-1\]
\[ab(c+d)+cd(a+b)=0\]
\[ab+(a+b)(c+d)+cd=0\]

Using the above equations we see that $c+d=-1-a-b$ and $cd=\frac{-1}{ab}$.  Substituting these into the second and third equation we obtain

\[(1) \Leftarrow -ab(1+a+b)-\frac{a+b}{ab}=0\]
\[(2) \Leftarrow ab-(a+b)(1+a+b)-\frac{1}{ab}=0\]

From (1) we obtain

\[(ab)^2+(a+b)((ab)^2+1)=0 \Leftrightarrow a+b=-\frac{(ab)^2}{(ab)^2+1}\]

Substituting for $a+b$ in (2) yields

\[ab+ \left(\frac{(ab)^2}{(ab)^2+1} \right) \left(1-\frac{(ab)^2}{(ab)^2+1} \right)-\frac{1}{ab}=0\]

Simplifying and letting $u=a+b$ we obtain

\[u+\frac{u^2}{u^2+1} \cdot \frac{1}{u^2+1}-\frac{1}{u}=0\]
\[u+\frac{u^2}{(u^2+1)^2}-\frac{1}{u}=0\]
\[u^2(u^2+1)^2+u^3-(u^2+1)^2=0\]
\[u^6+u^4+u^3-u^2-1=0\]

Edit: Wow....I kept multiplying by $u^2+1$ instead of $(u^2+1)^2$.  :blush:


1) No, try $ a\equal{}0,b\equal{}2,c\equal{}0.5$

[b]Q1:[/b] $ b(c \plus{} a) \plus{} ca \equal{} 1 \Rightarrow (a \plus{} b)(c \plus{} b) \equal{} b^2 \plus{} 1$,
similarly we have $ (b \plus{} c)(a \plus{} c) \equal{} c^2 \plus{} 1$ and $ (b \plus{} a)(c \plus{} a) \equal{} a^2 \plus{} 1$.

So $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} [(a \plus{} b)(b \plus{} c)(c \plus{} a)]^2$ is a square . I think you mean $ a,b,c$ are integers of course.

[b]Q2: [/b]
$ \begin{cases}a \equal{} \frac {1}{a \minus{} b}.............(1) \\
b \equal{} \frac {1}{1 \minus{} c}............(2) \\
c \equal{} \frac {1}{1 \minus{} d}.................(3)\end{cases}$
Q2 is true if you mean $ a,b,c$ are integers and it will give $ a \equal{} d \equal{} 1$

But I guess you made a typo, it should be $ a \equal{} \frac {1}{1 \minus{} b}$ for $ (1)$ , am I right?


just noticed that I have a loose keyboard button on my laptop. This has happened before once and I was able to press it down but this time it's resisting. Can it (the button) still be salvageable? Should I physically remove it and try to put it back in? What do you suggest? Has this happened to you before?

It's probably fairly easy to fix- if you have the tools. Find an expert.


[quote="PenguinIntegral"][quote="shadysaysurspammed"]Find the general expressions for two positive integers which are such that if their product is taken from the sum of their squares the difference is a perfect square.[/quote]
You mean
$a^2+b^2-ab=n^2$?[/quote]

Yeah

i sort of have an answer, but it's not exactly complete

[hide] we multiply both sides of the equation by a+b to get

$a^3+b^3=(a+b)n^2$

adding to both sides

$a^3+3a^2b+3ab^2+ b^3=(a+b)n^2+3a^2b+3ab^2$

$(a+b)^3=(a+b)n^2+3ab(a+b)$

dividing by a+b

$(a+b)^2=n^2+3ab$

$(a+b)^2-n^2=3ab$

$(a+b+n)(a+b-n)=3ab$

and this is the part that gets kinda freaky.  i pretended that a and b are prime, and perhaps $a+b+n=3a$ and $a+b-n=b$ but the solution to that was $a=b=n$, which does satisfy the equation.  but then i thought there must be other solutions, like what if $a+b+n=ab$ and $a+b-n=3$

$a+b-n=3$

$a+b-3=n$

substituting in this value for n in the equation a+b+n=ab

$a+b+a+b-3=ab$

$2a+2b-3=ab$

$2b-3=ab-2a$

$\frac{2b-3}{b-2}=a$

$2+\frac{1}{b-2}=a$

but then the only way a is an integer is for b=3.  and surely there are other solutions.  so maybe a and b are not necessarily prime, or something.  any ideas??
[/hide]

The quiestion asks for the general expression for the solution of the two positive integers which satisfy the given equation(like $x=3m+2n^2$(just an example))

What penguin integral wrote is not the answer.

[quote="Meghrabi"]And what about this :

[hide](a-b)²+ab=n²
 (a-b-n)*(a-b+n)+ab=0

a and/ or b = 0 , but let we assume that b=0.

(a-n)(a+n)=0

so a=n

Now can somebody tell me what i've done wrong ? :)[/hide][/quote]

Why should we assume $b=0$?

That's not correct though.

[quote="Meghrabi"]I mean by that, that you also consider the cases a=0 and a,b=0.

What is my fault ?[/quote]

Even if that is a solution,its only one of the cases,what if $a,b\neq0$?
You have to find the general expression for the solutions. :)

Some hints:

 [hide]If you consider $a=u+v$ and $b=u-v$  what you want  is $3u^2+v^2=n^2$ where you can use the standard technique aka Gausian primes...
[hide]You would get some thing like  $a=3x^2-y^2+2xy$, $b=3x^2-y^2-2xy$ so $a^2+b^2-ab = (3x^2+y^2)^2$ and you are done.
[hide]For exaple x=2,y=1 would give $15^2+7^2-15\cdot7=13^2$[/hide]
[/hide].[/hide]


The friends of a local animal shelter want to make a profit of at least $ \$2000$ by holding a raffle. If each raffle ticket is to cost $ \$3$ and the total expenses will be $ \$400$, what is the smallest number of tickets that must be sold?

Oh... I thought it meant that the profit on each of the tickets was $400... :(

No, it says in the problem that the friends what to make a profit of at least $\$2000$; expenses means costs.


How can I write a fancy I or R etc. to denote the set of all integers or reals?

Well, then set of integers is denoted by Z, not I.  But use \mathbb{}  like so:

$\mathbb{R,Z,N,C}$

Hello,

You know, something that could be nice would be to be able to define some latex commands by yourself. For example, I usually just write \R in my documents, because I have it defined as \mathbb{R}, giving $\mathbb{R}$. May be we could talk to Valentin or any other administrator to see if that is possible, this is, personalize $\LaTeX$ inside the forum, and before the message is filed inside the database, a subrutine just translate your personalized commands into the extended pure $\LaTeX$ version of them (just as the latex compiler does)

May be we could have 10K of space for such personalized commands, that is around 70MB counting all the members of the list, and I swear, 10 K is more than enough for personal commands for most of the non-ultra-expert users of $\LaTeX$ (I believe that my personal commands in my documents are less than 2K)

Just an idea in the air. Any way, I have programmed also, and I know that such ideas sound great, but at the end are a pain in the neck sometimes to implement. Best regards,

[quote="djimenez"]Any way, I have programmed also, and I know that such ideas sound great, but at the end are a pain in the neck sometimes to implement.[/quote]I think that sentence pretty much summarizes our response to your request.  We all have so many things on our "To Do" lists that this is going to be an extremely low priority, if at all.

Hello,

[quote="Valentin Vornicu"]There has been another discussion here somewhere about this. For one, not everyone uses the same definitions (I use \rr, others use \r or \R, etc.) For second, it's not easy to implement such command (although it might seem easy to the outsider). Since there is no quick solution to this, it will be, if ever, added later. I got used with writing \mathbb{R} and such :)[/quote]

As I already said, I am just putting ideas in the air, I am not saying that they have to be implemented. Nevertheless, let me reexplain my idea. I am not saying that the administrator should implement, but that in the options of each user one could have a small file (I said 10K) with Latex personalized commands. I use \R, then, in that file I, by my self, personalize

\newcommand{\R}{\mbox{$\mathbb{R}$}}

and Valentin personalize in his

\newcommand{\rr}{\mbox{$\mathbb{R}$}}

and some subrutine when I press preview or submit tranlate my \R into \mathbb{R}, but just for me, in Valentin case, it would look for \rr.

Again, just an idea in the air. As I said

[quote="David Jimenez"]Any way, I have programmed also, and I know that such ideas sound great, but at the end are a pain in the neck sometimes to implement.[/quote]

Best,[/quote]


[quote="Interval"]Ray and Jane live 150 miles apart.  Each drives toward the other's house along a straight road connecting the two, Ray at a constant rate of 30 miles per hour and Jane at a constant rate of 50 miles per hour.  If Ray and Jane leave their houses at the same time, how many miles are they from Ray's house when they meet?

I don't just want the answer.  I want to know the steps to find the answer.[/quote]

[hide]In an hour, they'll drive a combined distance of 80 miles. Then, they'll meet in $150/80 = 1 7/8$ hours. By that time, Ray would have driven $225/4$ or $56.25$ miles.[/hide]

[quote="Interval"]Ray and Jane live 150 miles apart.  Each drives toward the other's house along a straight road connecting the two, Ray at a constant rate of 30 miles per hour and Jane at a constant rate of 50 miles per hour.  If Ray and Jane leave their houses at the same time, how many miles are they from Ray's house when they meet?

I don't just want the answer.  I want to know the steps to find the answer.[/quote]
The algebraic way to think about this is this:
[hide="x being time... "]Ray's distance is $30x$ because the rate times time equals distance. Jane's distance is $50x$. The distances add up to 150, so $80x=150$ and then the time is $\frac{150}{80}$ or $\frac{15}{8}$ or $1\frac{7}{8}$. To find the answer, plug in $x=\frac{15}{8}$ to get $30x=30*\frac{15}{8}=\frac{750}{8}$ or 93.75 miles from Ray's house.[/hide]
 :P

[quote="Quevvy"][hide]$30*\frac{15}{8}=\frac{750}{8}$ or 93.75 miles from Ray's house.[/hide]
[/quote]

does it make sense that Ray would've traveled over half the distance if he's the one going slower? (check your math above)

It's just that $15 \cdot 30=450 \neq 750$!!   :)

The answer you get that way is of course coincident with the one given by buzzer11...

Oh yeah, I'm sorry! So then $30\cdot\frac{15}{8}=\frac{450}{8}$ or 56.25 miles from Ray's house.
I know it's the same as buzzer11's answer but that's the way that you should really think about it.

[hide="EXTENSION:"] What if they started at their houses and were going in the same direction (Jane drives to Ray's house and further & Ray goes from his house along the line formed by the 2 houses)? Using the same rates and stuff so that Ray is in front of Jane and they will meet up later. The question is, how many miles away from Ray's house are they? Also, how many hours later will they meet?[/hide]


Congratulations to Jonas Clarke Middle School, who for the third year showed everyone else where it's at and took the state title. Also congratulations to:

Alan Zhou, Jonas Clarke MS
Aaron Klein, Edith C. Baker MS
Jonathan Tidor, Jonas Clarke MS
David Fink, R.J. Grey MS

who will make up the 2009 national team.

Hm so apparently Alan changed his last name from Chou to Zhou. Hence his change in username.

And hopefully, Massachusetts makes top 5 in Nationals this year. :D

Yay, another Acton kid!

Congrats to Alan and the others too!

I was there! Grading, then there was an incident so I had to leave before CD and results announcement. (And I was gonna be the CD timekeeper! Wouldn't that have been fun?) How'd that go?

My timing expertise shined through quite clearly.

Although I did forget to start timing more than once...Jonathan Tidor gave me some relief however

Who were the rest of the top teams? How many Clarke kids in CD? Did Liz speak in an accent?

It was something like

1. Clarke
2. R.J. Grey
3. Diamond (which is surprising seeing that they have quite low standards there...it's all right because when they come to LHS they get cleaned up).
4. Sage
5. ?
6. ?
7. Meadowbrook (maybe they were #6)
8. Melican
9. McCall
10. ?

I'm not entirely sure of where McCall was. Notably, Roxbury Latin (they kind of have low standards too seeing that they have people named SamL coaching the team) and Chenery were not in the top 10.

Clarke had four people in countdown: Peijin came in 7th, Ivy Ren (individual) came in 9th. Hao Shen came in 11th.

And no, Liz did not speak in an accent. It was unfortunate.

Per my notes:

10. Fay
9. AMSA
8. Newton Bigelow
7. Meadowbrook
6. Winchester McCall
5. Stow Hale
4. Sage
3. Lexington Diamond
2. AB RJ Grey
1. Lexington Clarke

Given 1, 2, and 3, it doesn't look like the MML leaderboard will be changing much in the coming years.

:maybe:

What happened to Hao and Peijin?

[quote="CatalystOfNostalgia"]3. Diamond (which is surprising seeing that they have quite low standards there...it's all right because when they come to LHS they get cleaned up).[/quote]

Hey I resent that (and I'm sure the rest of Diamond agrees). Up until after my year, we had good high standards (like winning consistently and often). It's unfortunate Mosca left. Oh well. At least (to my knowledge) you guys still haven't broken 212.

We haven't broken 212, but we did tie it just last meet(with two alternates getting perfects as well).  We only need 182 to break last year's total points record too.  Since 198 is our lowest score this year, looks like we should make it :lol:  Not that we're competitive or anything.

Oh my gosh, are you guys running an underground sweatshop over there? :P

I remember that when we hit 212, Mr. Mosca mentioned that the meet marked just the 6th and 7th (or something like that) time that 200 had been broken (Clarke got exactly 200 at that meet). :lol: Gosh, Clarke, you guys gotta stop averaging 200-plus.

I assume you guys also broke our margin of victory record last year?

I believe my old school got like a 50 recently...It's scary that all the math fact stuff I knew when I got second at states in 06, I learned from IMLEM...(This is why I did not know what an x-intercept was at nats...Well that and it never came up in practice, or in AoPS 1, which I did between states and nats)

dude meeee toooo (on the mathcounts thing). except i did know what an x-intercept was? look how that turned out for us...

I believe there were like 5 combo problems on the last two pages and I got all of them wrong.

Haha that happened to me too...Which explains why we tied on sprint?

[quote="CatalystOfNostalgia"]Winning 4 meets by a total of 54 points is just inexcusable. Especially when all the point values are doubled for some seemingly arbitrary reason.[/quote]

We're only up by 54, but it's not like we had much room to improve on that with the highest team total after 4 meets ever!!

As for points being doubles arbitrarily, I finally got a reason for that.  Back in the day problem #1 in each category was worth 1, #2 was worth 2pts and #3 was worth 3.  Later they decided to make all questions be of equal value but wanted to keep the totals the same as to not change typical team scores or make any records obsolete.

[quote="frost13"]I don't know what your margin of victory record was, but last year we were first with 1006 (first 1000+ ever) and Diamond was 2nd with 884.  Scores are all higher this year, so we won't win by as much this year.  Only up like 54 so far.

We've broken 200 in 6 of last 8 meets(198 the other two).  200 is always the goal.  It's really not a sweatshop, just that the only way to even make our team now is to really know your stuff!![/quote]

It seems that within the next year you guys are going to have doubled whatever number of 200s had already been accomplished.

I remember when I was in 7th grade we had like 970 points, which I think far exceeded the previous record, and when I was in 8th grade we won by around 70 over Winchester. Our margin of victory for the last meet that year wasn't larger than 10, so we were winning by around 70 after four meets that year. (lol I don't have the exact numbers with me.) [i]Those[/i] were the days...

As for the scoring, that explains the previous meet record (I think by Marblehead; as I recall it was set sometime in the 80s) of 205 before we got 212.


$x,y,z$ are positive real numbers. Let $S=x+y+z$. Prove the following inequality
$\sum{\frac{S^{2}+z^{2}-2xy}{\sqrt{(S+z)^{2}-4xy}}}\geq \frac{4\sqrt{3}}{3}S$.
  :wink:

$x,y,z$ are positive real numbers, such that $x+y+z=1$.
Prove the following inequality

$\sum\frac{1+z^{2}-2xy}{\sqrt{{(1+z)}^{2}-4xy}}\geq \frac{4\sqrt{3}}{3}$.

What do you think about this inequality?
 :wink:


equation
algebra
number theory
infinitely many solutions
IMO Shortlist
Vieta Jumping
Pell equations

Is there a positive integer $m$ such that the equation \[ {1\over a}+{1\over b}+{1\over c}+{1\over abc}={m\over a+b+c}  \] has infinitely many solutions in positive integers $a,b,c$?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We take $m=12$. What follows is a standard argument: given a solution $(a,b,c)$ with $a\le b\le c$, we show that we can find a solution $(b,c,a')$ with $b\le c\le a'$.

Start with $(a_0,b_0,c_0)=(1,1,1)$ (this is a solution of the equation). Let $P(n)$ be the property $a_n|b_nc_n+1, b_n|c_n+a_n, c_n|a_nb_n+1$ and let $Q(n)$ be the property $a_n\le b_n\le c_n$. Furthermore, let $R(n)$ be $(a_n,b_n)=(b_n,c_n)=(c_n,a_n)=1$. It’s clear that $P(0),Q(0),R(0)$ hold.

Let’s now take $(a_1,b_1,c_1)=(b_0,c_0,\frac{b_0c_0+1}{a_0})$. $Q(1)$ holds because we already know $b_0\le c_0$ from $Q(0)$, and $b_1\le c_1$ because otherwise we would have $c_0a_0>b_0c_0+1$, which is false. $R(1)$ is equally easy to check. Let’s now prove $P(1)$: $c_1|a_1b_1+1$ because of the definition of $c_1$, $a_1|b_1c_1+1\iff a_0b_0|b_0c_0^2+c_0+a_0$, which is true because $a_0|b_0c_0+1,b_0|c_0+a_0,(a_0,b_0)=1$, and $b_1|c_1+a_1\iff c_0a_0|a_0b_0+b_0c_0+1$, which is true because $a_0|b_0c_0+1,c_0|a_0b_0+1,(c_0,a_0)=1$. We have thus shown that $P(1),Q(1),R(1)$ hold, and, in particular, we can define $(a_2,b_2,c_2)=(b_1,c_1,\frac{b_1c_1+1}{a_1})$, and so on. We use the exact same arguments to prove that $P(n),Q(n),R(n)$ are true, so we can define $(a_{n+1},b_{n+1},c_{n+1})=(b_n,c_n,\frac{b_nc_n+1}{a_n})$.

On the other hand, a simple computation reveals that if $(a_n,b_n,c_n)$ is a solution, then so is $(a_{n+1},b_{n+1},c_{n+1})$, so we have found infinitely many solutions (that’s because the inequalities in $Q(n)$ become strict for $n\ge 2$, due to $R(n)$).

Consider the sequence x[0] = x[1] = 1 and, for k>1, x[k+1] = 3.x[k]-x[k-1].
For k>1, we have x[k+1]>x[k].
As x[k+1]^2 -3.x[k+1].x[k] + x[k]^2 + 1 = 0, it follows, that the given relation is verified with m = 12 for
a = x[k], b = x[k+1], c = x[k] + x[k+1].

I think $ m\equal{}12$ is the only solution.

Assume $ a\ge b\ge c$, then

\[ a^2\plus{}a(b\plus{}c\plus{}\dfrac{1\minus{}(m\minus{}1)bc}{b\plus{}c})\plus{}bc\plus{}1\equal{}0\]

We have $ a|bc\plus{}1$. Suppose $ a\plus{}b\plus{}c$ has the smallest sum, and let $ a_1$ be the other solution to this equation.

We have $ aa_1\equal{}bc\plus{}1$, but $ a_1\ge a\ge b$. So either $ a^2\equal{}bc\plus{}1$ or $ a\equal{}b\equal{}c\equal{}1$. The former is impossible. Therefore the only possible $ m$ is $ 12$. We can use root flipping to generate the other solutions. $ (a,b,c)$ to $ (b,c, (bc\plus{}1)/a)$.

Finally, I [i]think[/i] root flipping basically guarantees $ (bc\plus{}1)/a$ to be an integer from the quadratic equation itself.

HEY! we need a solution with m=12 to start root flipping. In the case (1,1,1), m=9, and the root flipping gives nothing...

[quote="nealth"]I think $ m=12$ is the only solution.[/quote]
According to the official solutions [url=http://anhngq.files.wordpress.com/2010/07/imo-2002-shortlist.pdf]here[/url], "using a little more theory on quadratic forms, it can be shown that if the equation is soluble for a given value of $m$ then there are infinitely many solutions for that value of $m$." They also construct solutions for infinitely many $m$, but don't provide any details on how to actually use quadratic forms for fixed $m$ when there exists a solution.

In particular, does anyone see how to construct infinitely many solutions for $m = 15$ ($(a,b,c) = (1,4,5)$ works)? The root flipping argument also generates $(1,4,1)$, $(21,4,5)$, and $(21,4,17)$, but nothing else from there...

Note that we can expand to get $(a+b+c)(ab+bc+ca+1)=mabc$, so rearranging gives a quadratic in $a$: $(b+c)a^2+((b+c)^2+(1-m)bc+1)a+(b+c)(bc+1)=0$. Using a Vieta Jumping argument, if $(a,b,c)$ is a solution then so is $\left(\dfrac{bc+1}{a},b,c\right)$. So we take $a=b=c=1$, so then $m=12$. We define the sequence $(a_n)_{n\ge 1}$ by $a_1=a_2=a_3=1$ and $a_n=\dfrac{a_{n-1}a_{n-2}+1}{a_{n-3}}$ for all $n\ge 4$. It is easy to show that $a_n>a_{n-1}$ for $n\ge 4$, so if we can show the sequence only contains integers, then note that $(a_n,a_{n+1},a_{n+2})$ gives us a different solution to the equation for $m=12$ for all positive integers $n$. We will induct to show that $a_n=4a_{n-2}-a_{n-4}$ for all $n\ge 5$, from which the result will follow. Note that $a_4=2$ and $a_5=3$, so the base case holds. Now, assume $a_k=4a_{k-2}-a_{k-4}$ for some $k\ge 5$, and we will show that $a_{k+1}=4a_{k-1}-a_{k-3}$:

\begin{align*} a_k&=4a_{k-2}-a_{k-4}\\
\implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-a_{k-1}a_{k-4}\\
\implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-(a_{k-2}a_{k-3}+1)\\
\implies a_{k-1}a_k+1&=a_{k-2}(4a_{k-1}-a_{k-3})\\
\implies a_{k+1}a_{k-2}&=a_{k-2}(4a_{k-1}-a_{k-3})\\
\implies a_{k+1}&=4a_{k-1}-a_{k-3},
\end{align*}

as desired.

[quote=Systematicworker]My solution is quite different from others; therefore I wanna share it with Aops users. :D
Granted, we take $m=12$ and choose $a=b+c$. After some algebra, it remains to show that there are infinitely many positive integers $b,c$ such that $b^2$ $+$ $c^2$ $+$ $1$ $=$ $3bc$. As this problem is very famous and quite easy to deal with it, we can even find that the solution set of aforementioned equation is increasing, giving us solution set of $a,b,c$ infinite. :)[/quote]

How famous? Never heard of the equation $(b-c)^2 = bc-1$ before.

[quote=benstein]In fact, for the equation to have integer solutions, $5c^2-4$ would have to be a square, which occurs infinitely often by Pell's equations and the initial solution $(1,1)$.[/quote]

Can you explain more about why it is connected with Pell’s function?

The answer is yes with $m=18$.
[color=#00f]Lemma. [/color]The equation $3c^2-11k^2=4$ has infinitely many solutions in positive integer.
[color=#00f]Proof.[/color] Let $c=4u+22v$ and $y=2u+12v$ the equation become
$$u^2-33v^2=1$$
Notice that $(u,v)=(23,4)$ is a solution. Therefore by Pell's equation we obtain infinitely many solution of it. Hence there are infinitely many solutions of $3c^2-11k^2=4$.

Now letting $m=18$. We show that a solution exists for infinitely many values of $c$.
After clearing denominator, the equation transform into
$$18abc=(a+b+c)(ab+bc+ca+1)$$
Subsitiute $x=ab$ and $y=a+b$. Notice that for any pair of integer $x,y$, there exists positive integers satisfying $x=ab$ and $y=a+b$ if and only if $y^2-4x=k^2$ for some $k\in\mathbb N$. Now treating $c$ as a constant we have
\begin{align*}
18xc&=(y+c)(x+cy+1)\\
x(17c-y)&=(y+c)(cy+1)
\end{align*}
We now make the crucial step: there exists a solution which satisfies $c=2y$ for infinitely many $y$. Indeed, substituting $c=2y$ we have
\begin{align*}
33xy&=3y(2y^2+1)\\
x&=\frac{2y^2+1}{11}
\end{align*}
Hence
\begin{align*}
k^2&=y^2-4x\\
&=\frac{3c^2-4}{11}\\
3c^2-11k^2&=4
\end{align*}
From the lemma there exists infinitely many solutions, each of which induces a pair of $(a,b,c)$. This completes the proof.

[quote=benstein]By Pell theory, any equation of the form $px^2-y^2=q$ has infinitely many solutions iff it has $1$. In this case, the equation is $5c^2-y^2=4$, which has the initial solution $(c,y)=(1,1)$.[/quote]

I see. Thank for your explaining.

Good problem. It takes some time to get an equation where you can use vieta jumping.

The answer is yes, and the example is $m=12$.
First the equation is the same as: $$ \frac{(a+b+c)(ab+bc+ca+1)}{abc}=m$$
So $c \mid (a+b)(ab+1)$ after seeing that the degree of $c$ is $1$ (This was my motivation). We can guess that $c=a+b$, so if $m=12, c=a+b$ we have:
$$\frac{2(ab +(a+b)^2+1}{ab}=12 \implies \frac{a^2+b^2+1}{ab}=3$$
Now we need to prove that exists infinites pairs such that: $a^2 + b^2+1-3ab=0$, the trival solution is $(1,1)$ and if $(a,b)$ is a solution with $a\leq b$ implies that $(b,3b-a)$ is also a solution but with largest maximum number so there are infinites pairs. $\blacksquare$

We claim that $m=12$ works. Let $c=a+b$, then we will find infinitely many solutions to:
$$\frac1a+\frac1b+\frac1{a+b}+\frac1{ab(a+b)}=\frac6{a+b}$$
$$\Leftrightarrow a^2-3ab+b^2+1=0.$$
We start with the initial solution $(a,b)=(1,1)$. Using the method of Vieta jumping, suppose a solution $(a,b)$ exists with $a+b$ maximal. WLOG $a\le b$, then $(3b-a,b)$ is another solution but $3b-a+b\ge a+b$, a contradiction.
So infinitely many solutions must exist.

[i]Motivation.[/i] Observe the degree of each variable after multiplying both sides by $abc(a+b+c)$. Once you know that the degree is $2$, or equivalently we have a quadratic equation wrt each variable individually, Vieta Jumping idea hits instantly. So, if we have at least one triple $(a, b, c)$, we can construct the "larger" one. For one solution, just take $(a,b,c)=(1,1,1)$ for which $m=12$.

[i]Solution.[/i] $m=12$ works.
Proof. WLOG assume that $\min\{a, b, c\} = a$. Rewrite the equation as $(b+c)a^2 + (b^2 - 9bc + c^2 - 1)a + (b+c)(bc+1) = 0$. Hence, $a$ is a solution to the quadratic equation $(b+c)x^2 + (b^2 - 9bc + c^2 - 1)x + (b+c)(bc+1) = 0$. Let $a'$ be the second root. by Vieta, $a'a=bc+1 \implies a' = \frac{bc+1}{a} \ge \frac{a^2 + 1}{a} > a$. So we can construct the triple $(a', b, c)$ from $(a, b, c)$, having larger minimum, so we can construct an infinite number of new triples and we are done.

We'll prove that for $m = 12$, there are infinitely many triples $(a, b, c)$ such that $a = b + c$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc} = \frac{12}{a + b + c}$.

Note that the given equality is equivalent to $a^2 + a(b + c - \frac{11bc - 1}{b + c}) + (bc + 1) = 0$. Substituting $a = b + c$, we get $b^2 - b \cdot 3c + (c^2 + 1) = 0$. Thus we only need to prove that the above equation has infinitely many solutions. Note that $b = c = 1$ is a solution of above equation. WLOG, assume $b \le c$. Then root of the quadratic equation $x^2 - x \cdot 3c + (c^2 + 1) = 0$ is $b$ and let $b_1$ be the other root. Then $b + b_1 = 3c$ and $bb_1 = (c^2 + 1)$, hence $b_1$ is integer and $b_1 = \frac{c^2 + 1}{b} > b$. Hence swapping $(b, c) \to (b_1, c)$, we see that there are infinitely many solutions. $\blacksquare$


MATHCOUNTS
AMC
USA(J)MO
USAMO
summer program
MathPath

What was your rank at states?
No posting of scores.

lol that was a prediction. States for me is in 7 days.

my goal=2nd place states... if possible, 1st  :)

unfair, can you make it like less than 35?

cause i got 34th

ill describe what happened when we are allowed to talk.

I am so damn angry!

now i can't see leet, preet (rhyming skills), dan, richard, kstan013, etc. at nats

i'm seriously angry

-jorian

If your state is like indiana, 5,123,485 people come with anyways. Yes I pulled that number out of nowhere.

Edit: Ubemaya, I like you, I really do, but your "OMG I SUCK" posts are

A) Annoying
B) Spamtastic.

I hope to et third or second, as I can't beat leet, and will probably be beat by at least one other person.

[quote="frost13"]Little bit of an exaggeration there Fanatic, but Indiana does bring more like 18-20 extra people  :lol: The last couple years there were times when I could barely get into my room(I was roommates with Mr Fischer) it was so full of Hoosiers!!!!  Definitely the state with the most supporters, although this year I bet Texas has more!![/quote]
What's a hoosier?

a person who lives in or is from Indiana.

[quote="jka386"]well, I'm not bragging or anything, but I got from a 26th to 2nd. I was like... whoa....[/quote]

I got from 25th to second.

before countdown: 2nd
after (unofficial)countdown 1st.

[quote="kyyuanmathcount"]Didn't everybody get 3-0'ed by Mathew? I have not yet won a countdown match at states and I have not yet lost a countdown match at chapters. I do countdown with the same people from chapters to states. How does that work?[/quote]

Which state?

Here are the people who got to Nat's from my state:

1. Dan Mendelsohn 8th
2. Matthew Babbitt(I was expecting fourth..... :huh: ) 7th
3. Felix Sun 8th
4. Schuyler Smith 8th

Team: Home Educators Enrichment Group, Dr. O'Keeffe.

[quote="luke(mathnerd)"]hate to take the "we're all winners" approach, but seriously, how can anyone be disapointed with state rank?  By making it to state alone youve beat tons of other people out from your state.  I won my chapter and got 17th in ky state-- i would have loved a scholarship but its still amazing that i even had a chance to go to state.  Don't be disappointed with 4th, 5th, 20th, or even dead last.  By being at states you are the cream of the crop, by even participating in mathcounts you are the cream of the crop.  i cant count the number of kids i know who are intimated by the word math alone, can you imagine what they would say if they saw some of the problems we work for mathcounts.  Be proud, cause we rock!

ok well thats my BLAAH BLAH BLAH for the day :)[/quote]

The first thing that comes to my mind after reading that was:

I suck at math.  I only got 4th place and only got 3rd written.  I wish I was better at math.    :( :rotfl:  :rotfl:

[quote="LawOfSigns"]I don't think so.  I was there and I only think 2 people from fairview got top 20.  I know someone from fairview was like 18-19th or something.  I got 39th...made too many stupid mistakes and didn't have enough past tests (I heard one school from my chapter's gone every year since 1985!).  Had I even kept up with people I beat at chapter, I would have ranked in the teens.[/quote]
And you don't even notice me. The other fairview kid that got 6th. Thanks. JK. You beat eli.

whats your name luke?
you can pm it to me if you feel uncomfortable about releasing your name on the forum
i think you probably know my name

[quote="kyyuanmathcount"]Didn't everybody get 3-0'ed by Mathew?  I have not yet won a countdown match at states and I have not yet lost a countdown match at chapters.  I do countdown with the same people from chapters to states.  How does that work?[/quote]

Kevin Wu got 3-1, everyone else did...

Yeah, Allen, I think you mean, you never won, not not yet and never lost.

I don't know, it just happens like that.

I mean, the only people I lost to were people who were below me before written, and yeah, just like you.

Was anyone here 4th before countdown and 5th after (and the countdown is official in his/her state)

All states are done now, right? Does that mean we can discuss problems?

We can't until mathcounts.org releases the state test on its website.

Check the stickies at the top of the forum page.

[quote="Fanatic"]The posts that are stickied (always up there even if they're never bumped)[/quote]

You mean the announcements?

Yeah i guess they're called announcements here.

Every other forum I've been to has called them stickies.

Whooops.

Top 4 in FL:
1st: Liam Bressler
2nd: Anthony Grebe
3rd: Preet Patel
4th: Andrew Harn

Sorry if I misspelled anyone's names.

See you all at nats.


modular arithmetic
number theory unsolved
number theory

How many $ x\in N , 0\leqslant x \leqslant376$  is solution $ 2x^2\plus{}3x\plus{}4\equal{}0(mod 377)$

$ 377 \equal{} 13 \cdot 29$
Solve mod 13 and 29 and use the CRT (Chinese Remainder Theorem).

Notice that $ (2,377) \equal{} 1$, so we can multiply by 2.
Complete the square: $ (2x \plus{} 3)^2 \equiv 1 \pmod {377}$
Again, because $ (2,377) \equal{} 1$, we can replace $ 2x \plus{} 3$ by $ t$. The problem is reduced to solving $ t^2 \equiv 1 \pmod {13,29}$. There are 2 solutions for each prime, thus 4 solutions in total between $ 0$ and $ 376$ (the substitution doesn't change it because it is surjective and $ 0, \cdots ,376$ is a complete residue system)


For $ a,b,c$ positive reals prove that

$ \frac{\sum ab}{\sum a^2}\plus{}2\sum \frac{a}{b\plus{}c} \geq 4$

[quote="arqady"][quote="manlio"]For $ a,b,c$ positive reals prove that

$ \frac {\sum ab}{\sum a^2} \plus{} 2\sum \frac {a}{b \plus{} c} \geq 4$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$ Hence, your inequality is equivalent to a linear inequality of $ w^3.$
Id est, it remains to check only two cases:
1) $ b \equal{} c$ ;
2) $ c\rightarrow0^ \plus{} .$[/quote]

What does it mean to have a linear inequality?  :oops:

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \frac {\sum ab}{\sum a^2} \plus{} 2\sum \frac {a}{b \plus{} c} \geq 4$[/quote]
$ \frac {\sum ab}{\sum a^2} \plus{} 2\sum \frac {a}{b \plus{} c} \geq \frac {\sum ab}{\sum a^2}\plus{}2\frac{(\sum{a})^2}{2\sum{ab}}$
$ \equal{}\frac{\sum{ab}}{\sum{a^2}}\plus{}\frac{\sum{a^2}}{\sum{ab}}\plus{}2 \geq 4$


calculus derivative integration function

If    $ \displaystyle\int_{t_i}^{t_f}{F}(t) dt \equal{} \vec{p_{f}}\minus{}\vec{p_{i}}$   implies   $ {F}(t)\equal{}\frac{dp}{dt}$

and   $ \displaystyle\int_{x_i}^{x_f}{F}(x) dx \equal{} \minus{}(U_f \minus{}U_i)$   implies   $ F\equal{} \minus{}\frac{dU}{dx}$

Does the Kinteic Energry relation

$ \displaystyle\int_{x_i}^{x_f}{F}(x) dx \equal{} (KE_f \minus{}KE_i)$ imply $ F\equal{}\frac{d(KE)}{dx}$

This has been bothering me for some time...

No, recall from the first equation you wrote, that force is the derivative of momentum.  Now watch:
$ \frac{d}{dx}(KE)\equal{}\frac{d}{dx}\left(\frac{mv^{2}}{2}\right)\equal{}\frac{dm}{dx}\frac{v^{2}}{2}\plus{}\frac{m}{2}\frac{dv}{dx}\equal{}0\plus{}mv\equal{}mv\equal{}p$

Alternatively,
$ KE\equal{}\frac{mv^{2}}{2}\equal{}\frac{p^{2}}{2m}\Rightarrow\frac{d}{dx}\left(\frac{p^{2}}{2m}\right)\equal{}mv\equal{}p$

So again, this yields momentum.

Actually, for conservative forces
      $ \vec F = - \nabla W_p \;\;\;$, where $ W_p$  means potential energy.

   If  $ \vec F$ is the resultant (net force) of conservative forces, $ \vert F * dP = dW_k$,  ($ \;W_k$ stands for kinetic energy) , and in the unidimensional case we have yours.

 Brief, work or energy is the spatial version for momentum in time.

JRav I'm a little confused by your arguments

[quote]$ \frac{d}{dx}(KE)\equal{}\frac{d}{dx}\left(\frac{mv^{2}}{2}\right)\equal{}\frac{dm}{dx}\frac{v^{2}}{2}\plus{}\frac{m}{2}\frac{dv}{dx}\equal{}0\plus{}mv\equal{}mv\equal{}p$[/quote]

First I, understand that you are applying the product rule but I dont understand how you got the second term in the equation that comes after the second equal sign (shouldn't it be $ mv\frac{dv}{dx}$).

Second if your equation is correct can you explain how you got the last step where one of the quantities becomes zero and the other becomes mv.

Lastly I'm also confused by the alternative derivation.  Particularly how the spatial deriative of kinetic energy as a function of momentum gives momentum. Maybe an elucidation of the previous derivation would help me understand this step

[/img]

There is a small mistake here:

Assuming m constant

$ \frac {d}{dx}(KE) \equal{} \frac {d}{dx}\left(\frac {mv^{2}}{2}\right) \equal{} mv\frac {dv}{dx} \equal{} m\frac {dx}{dt}\frac {dv}{dx} \equal{} m\frac {dv}{dt} \equal{} ma \equal{} F$   (resultant).

Ahh you fixed it before I could reply...

[quote]Lastly I'm also confused by the alternative derivation. Particularly how the spatial deriative of kinetic energy as a function of momentum gives momentum. Maybe an elucidation of the previous derivation would help me understand this step[/quote]

I didn't do anything different in my calculations except re-express kinetic energy.  Particle physicists frequently describe kinetic energy as $ \frac{p^{2}}{2m}$ instead of the traditional way because this way makes it easier to measure.  You can check easily that this is the same thing as before by simply noting that $ p^{2}\equal{}m^{2}v^{2}$.  Then just divide out the $ m$ and we get the traditional expression.

JRav I understood that $ \frac {1}{2}mv^2 \equal{} \frac {p^2}{2m}$ I was just wondering how the derivative of the last term gives momentum. But since Immanuel confirmed my first conjecture I guess the issue is unnecessary.

So from Immanuel's post, the derivative (in respect to x) of Kinetic Energy is in fact Force?

[quote="JRav"]No, recall from the first equation you wrote, that force is the derivative of momentum.  Now watch:
$ \frac {d}{dx}(KE) \equal{} \frac {d}{dx}\left(\frac {mv^{2}}{2}\right) \equal{} \frac {dm}{dx}\frac {v^{2}}{2} \plus{} \frac {m}{2}\frac {dv}{dx} \ldots$[/quote]

There is a basic differentiation error here.

\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{mv^2}{2}\right) \equal{} \frac{1}{2}\left(\frac{\mathrm{d}m}{\mathrm{d}x}v^2 \plus{}m\frac{\mathrm{d}v^2}{\mathrm{d}x}\right) \\ \\ \equal{}\frac{1}{2}\left(\frac{\mathrm{d}m}{\mathrm{d}x}v^2 \plus{}2mv\frac{\mathrm{d}v}{\mathrm{d}x}\right) \\ \\ \equal{} \frac{\mathrm{d}m}{\mathrm{d}x}\frac{v^2}{2}  \plus{}mv\frac{\mathrm{d}v}{\mathrm{d}x} \]

[quote="emjay285"]If    $ \displaystyle\int_{t_i}^{t_f}{F}(t) dt = \vec{p_{f}} - \vec{p_{i}}$   implies   $ {F}(t) = \frac {dp}{dt}$

and   $ \displaystyle\int_{x_i}^{x_f}{F}(x) dx = - (U_f - U_i)$   implies   $ F = - \frac {dU}{dx}$

Does the Kinteic Energry relation

$ \displaystyle\int_{x_i}^{x_f}{F}(x) dx = (KE_f - KE_i)$ imply $ F = \frac {d(KE)}{dx}$

This has been bothering me for some time...[/quote]

I realize Immanuel has already answered this in part, but consider this:
\[ KE =\frac{p^2}{2m}\Rightarrow \frac{\mathrm{d}(KE)}{\mathrm{d}x}= \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{p^2}{2m}\right)\\ \\ =\frac12\frac{(m\cdot 2p\frac{\mathrm{d}p}{\math{d}x} -p^2\frac{\mathrm{d}m}{\mathrm{d}x})}{m^2}\]

In most cases the mass is considered a point particle where $ \displaystyle \frac{\mathrm{d}m}{\mathrm{d}x} = 0$, therefore the above reduces to:
\[ \frac{\mathrm{d}(KE)}{\mathrm{d}x} = \frac12\frac{(m\cdot 2p\frac{\mathrm{d}p}{\math{d}x})}{m^2} = \frac{p}{m}\frac{\mathrm{d}p}{\mathrm{d}x}\\ \\ = \frac{\mathrm{d}x}{\mathrm{d}t}\frac{\mathrm{d}p}{\mathrm{d}x}\\ \\ = \frac{\mathrm{d}p}{\mathrm{d}t}\]

\[ \boxed{\therefore \frac{\mathrm{d}(KE)}{\mathrm{d}x} = \frac{\mathrm{d}p}{\mathrm{d}t}}\]


inequalities
AMC
USA(J)MO
USAMO
search
function
inequalities theorems

[quote="Arne"]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \]
[/quote]

[quote="Cezar Lupu"]Nice and easy. In fact, this is too easy! ;) Rewrite the inequality in the form:

        $ \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\geq 6$. This one follows immediately from AM-GM. ;)[/quote]

No Cezar Lupu, you are really wrong. This is very hard. Nobody at the USA IMO training camp 1994 solved this, and the USA IMO team 1994 got 6 x 42. You probably copied the solution from some book or so?

:mad: You don't tell me that I copied the solution from a book,cause this is not true.I'm not a cheater :mad: .It is very easy and I maintain my opinion. What's so hard to observe that $\frac{1+abc}{a+ab}=\frac{1+a+ab+abc}{a+ab}-1=\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}-1$ :P

Well, I guess I'd better not say anything anymore :D but it's hard to believe you solved it so quickly, especially if you look at some of your previous posts.

Man, why don't you see that this inequality is too easy? :?  There are thousands of inequalities much more interresting than this one. ;)

[quote="harazi"]Cezar, you know, much of the work of some people can be seen by looking at some of their posts. I don't want to be rude, but when you look at the level of inequalities you posted, I might think Arne is right. Anyway, call it simple or trivial or whatsoever, but I think this inequality is one of the best things I have ever seen. It is a beauty, even if for some on the forum the difficulty or beauty of a problem consists in the length of the proof. And since I'm really tired to see how one of the best forums of the site is going down, I promis not to enter anymore on the inequalities forum. Anyway, I see now too many experts proving what other people have already proved years ago and proving how good they are. So be it, it was a nice job, but now it's nothing. Bye![/quote]

Don't leave.....
Myth has left, I do hope other experts won't leave...  :(  :(

I have seen too many experts on this part of the forum and I have never been an expert. I really hope one day Myth will forgive me and come back. Anyway, this was my last post on this part of the site.

You departure will be a great loss to mathlinks. :(

What happened between Harazi and Myth??? Why hasn't Myth been around for so long, like more than half a year???

Bomb

:(  :(  :(  :(
please harazi ,dont leave ,You are one of the best guys here and i learned a lot from you,
Arne is there any thing i can do for good :(  :( ?

Try to limit spam on the inequalities forum (and ML in general), don't make useless posts, etcetera. I don't think you can do much, except for that.

I understand that in the days before happened many things.(Harazi left :( )(I hope he will return and we try to make some rules to continue better. But I think that many times we get out of the limits. Don't forget this is a $MATHEMATICAL$ FORUM not a forum for very much discussion. So I believe that we should left the mathematic  symbols to speak ;)  :)

Silouan is right here. Harazi has left the forum. I hope one day he will forgive me and return on this forum. :(

 So, please for God sake stop $SPAMMING$  :mad:

Forgive me for making another non-mathematical post, but... maybe the moderators could make some clear rules to state what is allowed to post or not, what is considered spam and what is not?
just a suggestion.

[and I think chances for harazi coming back are greater if something really happens about it...]

[quote="Arne"]No, nothing happened between Harazi and Myth, it's just that they became pissed off because some people posting in the inequalities forum just keep posting nonsense or spamming: that's why they left. To be honest, I am getting pissed off too because of what's happening here.[/quote]Yes, I will soon start to be more personal with the people doing these spammings ... for now the thread is closed.

Well, I don't delete many posts in this forum (although I'm tempted to delete everything on this page) but as a moderator, this is what annoys me:

1) Problems saying "This is just Cauchy" or "Use Chebyshev then Muirhead".  Write up most of a solution, if you have time.  If you don't have time, just outline the solution, and come back to it later.

2) Posts which attempt to provide a partial solution are welcome.  Usually these should provide non-trivial steps, especially if it's implied that the inequality is difficult, but there are so many users at different levels that this is difficult to enforce.

3) I don't have a problem with people saying 'this is really easy', but 1) + 3) instantly raises my blood pressure.  Don't do it.

4) Inequalities that are obviously false (and not just a sign reversed) - that is, that aren't true as either $\geq$ or $\leq$.  Do your homework, people.  This is much worse if it's coupled with a "Source: me" - typos when copying from a book or the internet are natural.

5) Impatience when waiting for a problem to be solved.  Sometimes you have to wait a week.  Sometimes two months.  Sometimes a couple of years (see the USAMO '00 solution by Ravi B, there :) )

6) Please try as hard as you can to find the correct source for problems.  I know that most of you are not willing to use the search function, but some of us are, and make it easy on us.

7)  Excessive non-mathematical discussion is bad.  Many discussions would be better off over PM.  Use that function sometimes, please.


[quote="mathgeniuse^ln(x)"][quote="Quevvy"]If the interior angle of a regular polygon is $(\frac{1140}{7})^{\circ}$, how many sides does it have?

What is the formula for a question like this?[/quote]

[hide]Well, the formula is for an n side polygon is 180-360/n.

Now you figure it out.[/hide][/quote]
[hide]
$180-\frac{360}{n}=\frac{1140}{7}$

$\frac{360}{n}= \frac{120}{7}= \frac{360}{21}$

$n=21$[/hide]

Thanks for the help.

Just a question about $\LaTeX$, can I make the parentheses bigger? They look weird like this: $(\frac{1140}{7})^{\circ}$

[code](\frac{1140}{7})^{\circ}[/code]

[quote="Quevvy"]Thanks for the help.

Just a question about $LaTeX$, can I make the parentheses bigger? They look weird like this: $(\frac{1140}{7})^{\circ}$

[code](\frac{1140}{7})^{\circ}[/code][/quote]

$\left(\frac{1140}{7}\right)^{\circ}$

[code]\left(\frac{1140}{7}\right)^{\circ}[/code]

You can also do this:
$\LaTeX$

[code]\LaTeX[/code]


calculus
integration
trigonometry
limit
calculus computations

(1) Calculate $ \int \frac {x^3}{x \minus{} 4}dx$

(2) Evaluate $ \int_0^{\frac {\pi}{6}} \frac {1}{\cos x}dx$

(3) Find $ \lim_{n\to\infty} \frac {1}{n^2}\sum_{k \equal{} 1}^n k\sin \frac {k\pi}{2n}$

(1) Long division. Or first let $ u\equal{}x\minus{}4,x\equal{}u\plus{}4,$ multiply it out, and then do the division.

(2) The antiderivative of the secant is one of those things it's best to just have memorized.

(3) Clearly a Riemann sum. The question is one of detail - what to let $ x$ equal. If we try $ x\equal{}\frac{k\pi}{2n}$ then we need to make it

$ \lim_{n\to\infty}\frac4{\pi^2}\sum_{k\equal{}1}^{n}\left(\frac{k\pi}{2n}\right)\sin
\left(\frac{k\pi}{2n}\right)\cdot\left(\frac{\pi}{2n}\right)\equal{}
\frac4{\pi^2}\int_0^{\frac{\pi}2}x\sin x\,dx$

It shouldn't take too long to finish things from here.

[quote="skywalkerJ.L."](2)$ \int \sec{x}dx \equal{} \ln|\sec{x} \plus{} \tan{x}| \plus{} c$[/quote]

That's correct, but Japanese highschool students don't study the integral formula.

I myself don't remember it. :oops:

$ \int\sec{x}dx\equal{}\ln|\sec{x}\plus{}\tan{x}|\plus{}c$

Let $ u\equal{}\sec{x}\plus{}\tan{x}$, then $ du\equal{}(\sec{x}\tan{x}\plus{}\sec^2{x})dx$
$ \int\sec{x}dx$
$ \equal{}\int\frac{\sec{x}(\sec{x}\plus{}\tan{x})}{\sec{x}\plus{}\tan{x}}dx$
$ \equal{}\int\frac{(\sec^2{x}\plus{}\sec{x}\tan{x})}{\sec{x}\plus{}\tan{x}}dx$
$ \equal{}\int\frac{1}{u}du$
$ \equal{}\ln|\sec{x}\plus{}\tan{x}|\plus{}C$

I think perhaps the proof can help to memorize it.

$ \int \frac{dx}{\cos x} \equal{} \int \frac{\cos x }{1\minus{}sin^2{x}} dx \equal{} \left\|\begin{array}{c} t\equal{}\sin x  \end{array}\right\| \equal{} \ldots$


function
conics
parabola
calculus
derivative
complex numbers
complex analysis

[b]If [i]f[/i] is a differentiable function on an open disk [i]D[/i] such that its range lies on the parabola {[i]u[/i] + [i]iv[/i] : [i]u, v[/i] are real numbers and [i]v[/i] = [i]u[/i]^2}, then show that [i]f[/i] must be constant.[/b]

 :idea: Here are my thoughts so far:
If [i]f[/i] has the "y-values" [i]u[/i] +[i]iu[/i]^2, then for some [i]z[/i] in the complex numbers, $f(a + ib) = u + iu^2$
Since [i]f[/i] is differentiable, $f'(a + ib) = (u + iu^2)' = 1 + 2iu$.

So, if I could show $1+2iu = 0$, perhaps I could find a solution :!: .

There is the open mapping theorem:

Suppose $G$ is open and $f: G\mapsto \mathbb{C}$ is analytic and nonconstant. Then for all $z_0\in G$ and all $\delta>0$ such that $B(z_0;\delta)\subset G,$ then $\exists\,\epsilon>0$ such that $f(B(z_0;\delta))\supset B(f(z_0);\epsilon).$

Then you can focus less on the exact shape of that parabola and more on the fact that the parabola has no interior.

using the cauchy-riemann equations with partial derivatives, you can show it as well. if you work it through you can see that the derivative of [i]u[/i] is zero, so [i]u[/i] is constant. then if [i]u[/i] is constant, [i]v[/i] must be constant, and hence, the function [i]f[/i] must also be constant.


integration
analytic geometry
trigonometry
calculus
calculus computations

Evaluate $ \int \int \int _{E} x^2 dV$ , where $ E$ is bounded by the xz-plane and the hemispheres $ y\equal{}\sqrt{9\minus{}x^2\minus{}z^2}$ and $ y\equal{}\sqrt{16\minus{}x^2\minus{}z^2}$

$ \iiint_{E}x^2dV\equal{}\iiint_{E}x^2dxdydz$. Changing into spherecial polar coordinate system, we get
$ \iiint_{E}x^2dV\equal{}\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{R}(r\sin\theta\cos\varphi)^2(r^2\sin\theta{dr}d\theta{d\varphi})$.
 After calculation, the value is $ \frac{2R^5}{15}\pi$.
 $ R\equal{}3$ : $ \frac{162\pi}{5}$, $ R\equal{}4$ : $ \frac{2048\pi}{5}$.


I have a set $ A$ of the numbers $ 1$ through $ 10$.  For each number, I want to match it with a another number from set $ B$, also of numbers $ 1$ through $ 10$.  Each element of $ B$ may only be used once.  A match between sets $ A$ and $ B$ may not occur if $ A \equal{} B$ (i.e. you cannot pair $ 1$ with $ 1$, or $ 4$ with $ 4$) and it also may not occur if $ B \equal{} A \plus{} 1$ (i.e. $ A\equal{}4$, $ B \equal{}5$ is illegal, $ A \equal{} 10$, $ B \equal{} 1$ is illegal).

Given these restrictions, how many distinct orders are there in which I can pick from set B? (e.g. 3,4,5,6,7,8,9,10,1,2 is an order which satisfies the conditions)

(Sorry this problem is kind of hard to explain)

It's the derangement problem -- a search for that word on the forum brings up many, many hits.


2. Show that, for every positive integer n,
121^n − 25^n + 1900^n − (−4)^n
is divisible by 2000.

Some hints in answering this question please?

Induction is definitely not the way to go.

Start by considering the prime factorisation of 2000, and then you can kill it with modular arithmetic.

Would you have to work modulo 16 and modulo 125?

Really too easy for an Olympiad.

Its trivial that $ 1900^n$ and $ 25^n$ are multiples of $ 125$. So it's enough to proof that $ 125^n \minus{} ( \minus{}4)^n$ is a multiple of 125...

Similarly its enough to show that $ 121^n \minus{} 25^n$ is a multiple of $ 16$...

I mean, you can simplify more your solution!


function
integration
calculus
real analysis
real analysis theorems

[b]Theorem[/b]. Let $f(x)$ be a density function of the random variable $X$.  If the function given by $y=u(x)$ is differentiable and is either increasing or decreasing for all $x$ in which $f(x)\neq0$, then for these values of $x$, $y=u(x)$ can be uniquely solved for $x$ to give $x=w(y)$ and the density function for $Y=u(X)$ is given by $g(y) = f[w(y)]\cdot |w'(y)|$ for $u'(x) \neq0$.

The proof begins with the strictly increasing case:
$P(a<Y<b) = P[w(a)<X<w(b)]$
$=\int_{w(a)}^{w(b)}f(x)dx \qquad \qquad \qquad (1)$
$=\int_{a}^{b}f[w(y)]\cdot w'(y)dy \qquad \quad (2)$

My question is what action is taken to get from $(1)$ to $(2)$?
I understand the change of variable and the change of the limits of the integral, but why does the integrand change to what it is in $(2)$?

How do you know the change of variables theorem? That is exactly the textbook statement of a change of variables theorem for me. In the language of Leibniz or whatever we have $x = w(y)$, so that $dx = w'(y) dy$.


factorial
number theory unsolved
number theory

Let $ P\equal{}(1!)(2!)(3!)...(2007!)(2008!)$ be the product of the factorials of the first $ 2008$ positive integers. Decide whether it is possible to cancel one of the factorials in the expression of $ P$ in such a way that the resulting number is a perfect square.

One?  There are at least two primes between $ 1004$ and $ 2008$.

Yes, but those primes are getting multiplied in more than once, and we can cancel them all simultaneously by picking $ 2008$. :P

Sorry; let's try this again.  If $ p$ is a prime between $ 670$ and $ 1004$ then it appears an odd number of times, so computations probably reveal that there are at least two of these.

We have $ 2008!2007! \equal{} (2007!)^2\cdot 2008$, $ 2006!2005! \equal{} (2005!)^2\cdot 2006$ and so on. Hence

$ 1!2!\ldots 2008! \equal{} 2\cdot 4\cdot \ldots \cdot 2008 \cdot (1!3!\ldots 2007!)^2 \equal{} 2^{1004}1004!(1!3!\ldots 2007!)^2.$

Now it's clear that deleting $ 1004!$ the remaining product will be a perfect square.

This exercise is widely known. I've seen a number of times the version with 100 instead of 2008...


Will points be taken off if I just say, "So I evaluated this huge expression on [insert program here], and I got [answer]"? Of course, I would have to have done some work to get that huge expression, but... will that earn a 5?

on #4 in round 1, i just put like, i typed (giant equation) into my calculator and used the code (some code) and i got the answer 6. i ended up getting a 5 on it.

...
wow.
I mean if I knew that I don't have to write 3 pages of stuff (also omitting 8 pages of work) for this round's problem 5...

HINT: Don't mention any of the round's problems. :wink:  :wink:  :wink:


algebra
system of equations
algebra unsolved

Sorry about the formatting, but I'm new to this:

I'm working on a problem, and I have it down to the following form. I need to solve for X.

AX^(Z) - BX^(2Z) + X = C

A, B, C, and Z are all constants.
Thanks in advance for any help you can give me
-Simon

Where did you get the problem from?  Most "arbitrary questions" don't have closed-form solutions.

Sorry, I may have been trying to over-simplify my problem. This might not even be the sort of thing that you deal with on this forum. But here is my problem. I'm a graduate student working with mass spectrometers. I've been reading a research paper on how to correct the data that I get from the machine, but I can't seem to solve the equations that are given in the paper.

The following equations are from the paper I was reading:

(R45) = 2(R15) + (R17)
(R46) = 2(R15)(R17) + (R18) + (R15)^2
(R17) = A(R18)^B

I need to solve for (R15) and (R18). (R45) and (R46) are what I measure directly using the mass spectrometer, and A and B are constants.
The paper states that this system of equations can be solved numerically. I've been trying to work it out on paper all day, and I'm stumped. The form that I originally posted was what I managed to get it down to at one point, but in hindsight it may be wrong, so just ignore it.
I'd appreciate any help that you could give me.
Thanks,
-Simon

[quote="sjthuss"]The paper states that this system of equations can be solved numerically. [/quote]

What exactly does "numerically" mean?  Perhaps the paper only meant that solutions could be calculated numerically (Newton's method, etc.), not that they could be expressed in closed form.


Hello,

1. your code is incompatible with the last version of Asymptote;
2. your code is not optimized at all.

Update Asymptote and your code to version 1.86

I don't mind the code length. So how would I update if I made this on the forums? Thanks.


This really is very difficult.  Put a^1/4=x and so on. Then x^3+y^3+z^3=3 and by a problem of Vasile Cartoaje it follows that sum (xy)^4<=3, which means sum ab<=3. So, 4 sum sqrt(a+b)(a+c)<=4 sum sqrt(3+a^2)<= sum a^2+21, since it reduces to sum (2-sqrt(a^2+3))^2>=0.


probability
function
integration
calculus
real analysis
inequalities
advanced fields

Definition: Let $(\Omega,\Sigma,\mu)$ be a probability space. i.e. a measure space with $\mu(\Omega)=1$. Suppose $\Sigma_{1}$ is a sub-$\sigma$-algebra of $\Sigma$. Then for every $\mu$-integrable function$f$, there is a function $g$ that is $\Sigma_{1}$-measurable and $\mu|_{\Sigma_{1}}$-integrable such that $\int f \chi_{E} d\mu = \int g \chi_{E} d\mu$ for all $E \in \Sigma_{1}$. g is called a conditional expectation of f with respect to $\Sigma_{1}$.

Question: Show that if $f \in L^{p}(\Omega,\Sigma,\mu)$,$1 \leq p \leq \infty$, then $g \in L^{p}(\Omega,\Sigma,\mu)$ and $||g||_{p} \leq ||f||_{p}$.

The case $p=1$ is clear from the definition.
For $1<p<\infty$, the hint given was to first assume that g is nonnegative and show that $||\phi||_{p} \leq ||f||_{p}$ for any $\Sigma_{1}$-simple function $\phi$ where $0 \leq \phi \leq g$. Also note that if $c \in \mathbb{R}$, then $cg$ is a conditional expectation for $cf$. I am having difficulty with this case as well as for $p=\infty$.

Note that the existence of the conditional expectation $g$ is directly due to Radon-Nikodym theorem. And if you have a close look at the proof of this theorem (not the one using the Riesz representation theorem, but the one using the Hahn decomposition and the $sup$ of certain integrals) you will have your result immediately (by construction of $g$).

You can find it for instance in Foundations of Modern Analysis, th. 2.12.14 but I guess you can find that anywhere on the web. I think the Rudin uses the Riesz representation theorem (since it's proved right at the beginning).

I had a closer look, this is not immediate.

In the following, all the inequalities are meant almost everywhere in respect to the measure $\mu$ restricted to $\Sigma_{1}$. We can assume $f \geq 0$ (because we can work with $f^{+}$ and $f^{-}$ separately) and $g \geq 0$.

For every measurable set $E \in \Sigma_{1}$ ($\mu (E_{1}) < \infty$) and $c \geq 0$, $c1_{E} \leq g \Rightarrow c1_{E} \leq f1_{E}$ (otherwise we would have a measurable set $F \in \sigma_{1}$, $\infty > \mu (F)>0$ and  $c1_{F} > f1_{F}$ and then $\int_{F} {g} d\mu > c\mu (F) > \int_{F} {f}$, which is absurd). Whence $c^{p}1_{E} \leq f^{p}1_{E}$ for every $p \geq 1$, and for any (positive) integrable simple function $h$ we will then have $h<g \Rightarrow \int_{E} {h^p} < \int_{E} {f^{p}}$ . But since $g^{p}$ will be written as an increasing limit of a sequence of such functions, $\int_{E} {g^{p}} \leq \int_{E} {f^{p}}$.

For $p=\infty$ it is trivial.


function
trigonometry
geometry
geometric transformation
analytic geometry

Find the equations for a sine/cosine function with a peak at (-pi/6 , 7) and the valley following the next peak at (4pi/3 , 1). Write both a sine and cosine function.

Thanks for your time!

[hide="Hints"]Well, you can determine the amplitude and the translation in the $y$-direction very easily from the $y$-coordinates of those points.

For the period of the function, note that the distance in the $x$-direction between a peak and the following valley will be $\frac{1}{2}$ of the period. Look at any sine or cosine graph and this should be evident.

Finally, for the phase shift, simply graph the function that you have created thus far without the phase shift and determine it from that information. I'm sure there's a rigorous, non-graphical way to do this, but I think that this will be easier, particularly for the cosine graph (since if you graph it without a phase shift, it will be at a peak when $x=0$.[/hide]

For some reason this isn't working for me. I don't have a graphing calc on me so ya. Any more help??

thanks

Very well. Since you don't want hints, you want a solution...

[hide="Solution"]
The amplitude is the maximum value, $7$, minus the minimum value, $1$, all over $2$. So the amplitude is $3$. Using this amplitude, we can see that the vertical translation is $4$ (so that would make the function range from $1$ to $7$).

The period would be twice the difference in the x-coordinates of a valley and the previous peak, in this case $2(\frac{4\pi}{3}-\frac{-\pi}{6}) = 3\pi$. Using the formula $\frac{2\pi}{a}= p$ where $p$ is the period of the function and $a$ is the coefficient on $x$, we see that $a = \frac{2}{3}$.

This gives us the nearly-complete functions $f(x) = 3\sin \frac{4x}{3}+4$ and $f(x) = 3\cos \frac{4x}{3}+4$. Looking at the latter function, it has a peak at $(0, 7)$. We need to shift that peak over $\frac{\pi}{6}$ units to the left in order to have a peak at the desired point. Thus, the phase shift is $\frac{\pi}{6}$ times $a$, so our completed cosine function is:

$f(x) = 3\cos (\frac{2x}{3}+\frac{\pi}{9})+4$.

Looking at the sine graph, we see that our incomplete function has  a valley at $(\frac{-3\pi}{4}, 1)$, which we wish to shift $\frac{4\pi}{3}-\frac{-3\pi}{4}= \frac{25\pi}{12}$ units to the right. Multiplying this by $a$ to give us the final part of the function, we see that our desired function is:

$f(x) = 3\sin (\frac{2x}{3}-\frac{25\pi}{18})+4$[/hide]


geometry
rhombus
inradius
geometry proposed

A tangent $l$ to the circle inscribed in a rhombus meets its sides $AB$ and $BC$ at points $E$ and $F$ respectively.
Prove that the product $AE\cdot CF$ is independent of the choice of $l$.

Let the tangent $ l$ of the rhombus incircle (I) meet CD, DA at G, H and let (I) touch AB at T. $ \triangle AEH \sim \triangle BEF \sim \triangle CGF$ are similar, having parallel sides. Denote AE = h, EH = a, HA = e sides, s semiperimeter, r the inradius and $ r_a, r_e, r_h$ the A-, E-, H-exradii of $ \triangle AEH.$ (I) is the H-excircle of $ \triangle AEH,$ B-excircle of $ \triangle BEF,$ and G-excircle of $ \triangle CGF.$ Thus

${ \frac {AE}{AT} = \frac {h}{s - e},\ \ \ \frac {CB}{CF} = 1 + \frac {BF}{CF} = 1 + \frac {r_e}{r_{a}}} = 1 + \frac {s - a}{s - e} = \frac {h}{s - e}$

As a result, $ \frac {AE}{AT} = \frac {CB}{CF},$ $ AE \cdot CF = AT \cdot CB.$ But AT, CB are rhombus segments independent of the tangent $ l.$


calculus
integration
real analysis
real analysis unsolved

It's not- that's a meaningless pile of symbols. An integral sign doesn't mean anything without the $ d(\cdot)$, and that's especially important with more than one possible variable. We can guess what you probably meant in this case, but not in general.

Define $ I_n\equal{}\int_0^{1}\frac{x^n}{(1\plus{}x)^{n\plus{}2}}dx$, prove that $ I_n\equal{}\frac{1}{(n\plus{}1)2^{n\plus{}1}}$ by using I. by P.

[color=darkblue]For $ n\in N$, denote $ I_n \equal{} \int_0^1\frac {x^n}{(x \plus{} 1)^{n \plus{} 2}}\ \mathrm {dx}$. Observe that $ \left(\frac {x}{x \plus{} 1}\right)' \equal{} \frac {1}{(x \plus{} 1)^2}$.

Therefore, $ I_n \equal{} \int_0^1\left(\frac {x}{x \plus{} 1}\right)^n\cdot\left(\frac {x}{x \plus{} 1}\right)'\ \mathrm {dx} \equal{}$ $ \left\|\frac {1}{n \plus{} 1}\cdot \left(\frac {x}{x \plus{} 1}\right)^{n \plus{} 1}\right\|_0^1$ $ \Longrightarrow$

for any $ n\in N$, $ \boxed {I_n \equal{} \frac {1}{(n \plus{} 1)2^{n \plus{} 1}}}\ .$

I"ll propose two more difficult similar problems, where $ \{m,n\}\subset N\ :$

$ 1\blacktriangleright$ $ I_{m,n} \equal{} \int_0^1\frac {x^m}{(x \plus{} 1)^n}\ \mathrm {dx}\ .$

$ 2\blacktriangleright$ $ I_{m,n} \equal{} \int_0^1\frac {x^m}{(x^2 \plus{} 1)^n}\ \mathrm {dx}\ .$

[/color]


Who has one? Anyone care to share?

BlogSpot, MySpace, etc. don't count  :P

[b]Edit:[/b] [i]To address the poster below --[/i] Yes, I do mean "own website" -- specifically, you're own website about you.

Personal website?
As in your own website?

What if you're the creator of Google? Does that count?

well...
it's not about me.. nor does it have that much updates..
i'm gonna post a blog pretty soon....
http://www.freewebs.com/kimmystar94/
i'm putting the chat back up soon, too...
hope those spammers are gone...

Meh?
Kimmy has a website?

[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=799]Mine[/url]


Prove that if $ f$ is a continuous and injective function on an interval, then $ f$ is strictly monotone.

Let $ f$ be a continuous and injective function on an interval $ I$. Consider two points $ a$ and $ b$ of $ I$ with $ a<b$. Assume $ f(a)<f(b)$. Let $ c$ be any real number such that $ a<c<b$. If $ f(c)<f(a)$, by the Intermediate Value Theorem, there must be a point $ d$ on $ I$ such that $ f(a)\equal{}f(d)$. So $ f(c)>f(a)$. Similarly we can prove that $ f(c)<f(b)$. Hence $ f(a)<f(c)<f(b)$. If $ c>b$, $ f(c)$ must be greater than $ f(b)$ (this follows from our previous argument). If $ c<a$, $ f(c)<f(a)$. The case $ f(a)>f(b)$ is done by the same way. Hence $ f$ is strictly monotone.