Well Christmas is more of a greed holiday. I feel bad that people die close to Christmas and life goes on.

[quote="fredbel6"]Holiday controversy need not always be about religion.
For instance, in Belgium there is a "Dynasty Day" on the fifteenth of november.  Anti-monarchists don't like this holiday.
And then there is a national Belgian holiday on the 21st of July, while the Flemish celebrate their own on the 11st of July.  Lots of Flemings don't get a day off then.

There must be many other examples of controversies , like "Columbus Day".[/quote]

I don't relaly care...but my Jewish friend celebrates Hannakuh (god my spelling sucks thats the closest i can get lol) and Christmas....got like $\$ 500$ worth of presents...

[quote="paladin8"][quote="kstan013"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress. What do you think?[/quote]

I really don't care, but imo people are overprotective of their religion and its public image.[/quote]I very much agree with that statement.

somebody's always going to feel discriminated against no matter what holiday it is. It is impossible for all humankind to reach a consensus, therefore discrimination exists.

Why should I feel discriminated against? A holiday is, after all, a holiday, and the more of them the better. We have a long holiday during durga pujas. We have a holiday on the day of each of the smaller pujas. We have christmas holidays and easter holidays and eid holidays (no halloween though). Then there is muharram and guru nanak's birthday. But we don't have jewish holidays, as there aren't many jews here.

How about a random allocation of holidays in the beginning of each year? :P

Ours is a church affiliated christian school...80% are hindus,even then we have 3 weeks for christmas and a week for easter and only 2 days for diwali and 10 days for Dussehra! :(  :(  Truly unfair...but...
[b] As long as we GET holidays at SOME point of the year..i'm not complaining! :lol: [/b]
Thankfully there are so many religions....each one's festivals SHOULD be honoured...(We get so many holidays!!)

Ours was a church affiliated christian school too.
The christmas songs are sweet, though the rest is rubbish.

a holiday is just that, each person celebrates what they want, what's everyone else complaining about?  people who get too upset over a holiday ruin what the season is about; isn't it supposed to be a time of peace, goodwill, personal reflection regardless what religion you're from?  why get mad over a bunch of holidays about peace and happiness?

Everybody says "holiday" but we all know it's christmas they're talking about.
like my "unprejudiced" school played christmas music the whole day between classes.

[quote="blahblahblah"][quote="kstan013"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress.[/quote]

Spelling aside, the very way in which you word the question is biased (e.g., you ascribe a victim's mentality to those who may be  (legitimately) troubled by how major (essentially Christian) holidays are celebrated).[/quote]

Is it only me or can no one else spot the 'spelling' mistake either?

[quote="Padmavathi"]The carols are amazing and so are the hymns...Don't forget the christmas plum-cakes they hand out each year!!
I totally agree...the rest is rubbish!
[/quote]

They didn't hand out christian plumcakes in our school. :(
And I do not (rather cannot) distinguish between carols and hymns.

[quote="bubka"][quote="blahblahblah"][quote="kstan013"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress.[/quote]

Spelling aside, the very way in which you word the question is biased (e.g., you ascribe a victim's mentality to those who may be  (legitimately) troubled by how major (essentially Christian) holidays are celebrated).[/quote]

Is it only me or can no one else spot the 'spelling' mistake either?[/quote]

You're right, I'm wrong.  My mistake.

I feel that a lot of people (my mom included) are WAY to oversensitive about the whole Christmas thing in the United States. I think that people are stupid to assume that everyone celebrates Christmas, but their stupidity doesn't really hurt anyone. Schools are now so busy being politically correct, that it ends up just annoying everyone. I would personally rather have a Christmas party than have a regular day of school, even though I am Jewish. I have a Jewish friend who goes caroling just because he likes singing and spreading holiday cheer.

it's more about many people using others to get what they want as in gifts.

yeah about holidays it has gotten a much more diverse meaning but holiday= holy day and has a christian influence

but truely I believe that some people are overprotective of their religon they really might have some cause for that but it is not as strong as they believe

also most holidays have been overy commercilized making them bigger then what they used to be

As someone who celebrates Hanukka and not Christmas, I really don't like it when we do Christmas things in school or in public places.

Now, it's not really about "holidays;" it's about religion. You see, in America at least, although it may not be so in other countries, you can't promote one particular religion in public schools; it comes from the doctrine of "Separation Between Church and State" in our laws. So, I feel uncomfortable when we're doing Christian religious things in a public school! I don't believe in those thing and they're against my beliefs, why don't we do Hanukka stuff too? You just really can't mix religion in with school or politics or anything public...

I mean wouldn't you (assuming you're Christian) feel uncomfortable if your school was promoting Jewish religious viewpoints?  :huh:

500TH POST!!!  :D  :D  :lol:  :lol:

Christmas is awesome, the movie theaters are mostly empty so no having to get there really early and we always get Chinese.

[quote="Dark_Shadow"]The only solution is to ban Christmas.

As an added bonus, it will diminish the influence of the Pope on our beloved country.[/quote]

Christmas is still a valued tradition for many people. Other people should be able to enjoy it as well. You can just not celebrate Xmas if you don't want, no one's forcing you against your will. Religion is like a decision. If you want to, you can do it. It's just like you can believe in god if you want. You can study the Bible and attend church if you would like. Yeah, you get the point.

 :lol:

I think Christmas has been commercialized enough that even though someone may mention baby Jesus here or there, the Christian doctrine isn't inherent in the celebration of Christmas so it shouldn't discriminate against many. Sure, Hannukah may not get as much press, but in reality it's not that important of a holiday, anyways (please excuse my ignorance if it is, but I'm going from what I know of the holiday and what others have told me). I'd consider Passover an 'important' holiday just for reference.

Does anyone else think it's odd that there's no poll option of "I do not feel discriminated by the holidays". Very biased, meh.

"The only solution is to ban Christmas. "

Christmas is not solely religious but is inherently cultural nowadays. Banning it would create much uproar and animosity towards whoever proposed the bill. Christmas is a time of happiness (I mean, you're off from school!) so enjoy it! It's also good for the economy.

[quote="bubka"]I was in a christian school, and it was petty of them to make sing hymns and all. But I have seen Hindu schools and heard of Muslims schools, and they are just as petty.

Religion makes men petty, god makes men petty, hence my war against religion, against god.

But holidays of course make people feel happy, so I like them. :)[/quote]

Religion makes people happy, why don't you like them?

Just felt like posting this too. :wink:
[hide][quote]Please accept with no obligation, implied or implicit, my best wishes for an environmentally conscious, socially responsible, non-addictive, gender-neutral celebration of the winter solstice holiday, practiced within the traditions of the religious persuasion of your choice, or secular practices of your choice (with respect for the religious/secular persuasions and/or traditions of others or their choice not to practice such traditions at all).*

Furthermore, I offer my (non-binding) best wishes for the onset of the generally accepted calendar year of 2006, but not without due respect for the calendars of choice of other cultures whose contributions to society have helped make America great. (Which is not to imply that America is any greater than any other country or is the only “America” in the western hemisphere.) These wishes are offered without regard to the race, creed, color, age, physical ability, choice of computer platform or sexual preference of the wishee.

Happy (if happiness is in your belief system)

Holidays (if you recognize them)

*This greeting is subject to clarification or withdrawal. It implies no promises by the wisher to actually implement any of the wishes for himself or others, is void where prohibited by law and is revocable at the sole discretion of the wisher. This wish expires within one year or until the issuance of a subsequent holiday greeting, whichever comes first. [/quote][/hide]

As long as we get to stay at home and somewhere else...someone is celebrating...thats fine with me!!

[quote="WindSlicer"]Does anyone else think it's odd that there's no poll option of "I do not feel discriminated by the holidays". Very biased, meh.[/quote]That would be "I really don't care"[quote="calcrulz"]I mean wouldn't you (assuming you're Christian) feel uncomfortable if your school was promoting Jewish religious viewpoints?[/quote]I would be more confused than anything. You have to realize how nondiverse my community is. There are no Jews in my entire school and the nearest synagogue is 20 miles away, with the next closest being God knows where. Probably in Green Bay. Even non-religious people in my community celebrate Christmas and Easter. It's pretty much a given where I live.

i am crying because of christmas it is such a bad holiday :(

im hindu, and i still kind of celebrate christmas.  hey, its a reason to have parties and make cookies and get gifts.

i really don't care that christmas gets so much attention.  its just that I think that people should have more awareness about different cultural holidays.

random, but: i was reading something interesting in the newspaper about how there are shows on christmas that are sometimes called 'kosher comedy' where people who arent really celebrating christmas come for a fun time.  they have chinese takeout because its the only thing open,

[quote="kstan013"][quote="WindSlicer"]Does anyone else think it's odd that there's no poll option of "I do not feel discriminated by the holidays". Very biased, meh.[/quote]That would be "I really don't care"[/quote]

But I do care, I have opinions on the matter and am not apathetic.

I feel that the attempts of schools to become less focused on Christmas isn't making them 'multicultural,' it is making them 'noncultural.'  I am not religious at all, but we still celebrate Christmas and Easter as a family (my mother's upbringing), and celebrate Hanukkah and others with our Jewish neighbors.

I would much rather hear Christmas songs on the radio than whatever is on the rest of the year, and am not against making Christmas tree ornaments in chemistry, instead of a metal plating lab.

I guess what it comes down to is that I would rather have culture, even if I don't agree and follow it, than plain nothingness.

Christmas is so commercialized these days that in some aspects, it seems secular. Even the atheist's association celebrates it for the "sharing and happiness" aspect.

Christmas, in America, has been rooted so much in the base of culture that it is closeminded (and impossible) to avoid it completely. Thus we get the problems mentioned above with the separation of Church and State.

Not being Christian, I view Christmas as a cultural icon and more of an American tradition than a religious holiday.

If you think about it, in all countries with one religious majority, how do religious minorities avoid discrimination (even in countries that observe separation of Church and State)?

[quote="fredbel6"][quote]Please accept with no obligation, implied or implicit, my best wishes for an environmentally conscious, socially responsible, non-addictive, gender-neutral celebration of the winter solstice holiday, practiced within the traditions of the religious persuasion of your choice, or secular practices of your choice (with respect for the religious/secular persuasions and/or traditions of others or their choice not to practice such traditions at all).*

Furthermore, I offer my (non-binding) best wishes for the onset of the generally accepted calendar year of 2006, but not without due respect for the calendars of choice of other cultures whose contributions to society have helped make America great. (Which is not to imply that America is any greater than any other country or is the only “America” in the western hemisphere.) These wishes are offered without regard to the race, creed, color, age, physical ability, choice of computer platform or sexual preference of the wishee.

Happy (if happiness is in your belief system)

Holidays (if you recognize them)

*This greeting is subject to clarification or withdrawal. It implies no promises by the wisher to actually implement any of the wishes for himself or others, is void where prohibited by law and is revocable at the sole discretion of the wisher. This wish expires within one year or until the issuance of a subsequent holiday greeting, whichever comes first. [/quote][/quote]

Rotfl, saw that on GCFL.net

I believe that people have the right to not celebrate Christmas (sorry for the split infinitive :D) but shouldn't keep others from doing so if they want to.

Wow, that grammar was pretty bad.

You could avoid that dubious construction saying 'right to forego Christmas celebration'. :)

Christmas is in Europe as Durga Puja is in India. It is primarily a Hindu festival, but celebrated over five days by people of all religions, including atheists like me. I do not object to celebration or holidaying, I only object to attaching any supernatural or divine meaning to it.

[quote=Irshad]What I really hate is that when people know you're a Christian they automatically think you celebrate Christmas. Christmas isn't even a Christian holiday. It is a pagan holiday.[/quote]

Yeah as if the pagans believed in Jesus. Just because some of the traditions have pagan roots doesn't mean it's a pagan holiday. Also I don't think the holiday celebrated by pagans on December 25th was called Christmas or had anything to do with Christ.

Yes, Christians renamed it and added that "Jesus was born on this day". When you insert pagan things in worshiping God it's called sin. Take the case of the golden calf, for example, during the times of Moses. In fact, Aaron who was organizing it said that it would be a holiday to the Lord. People indeed worshiped God, but they introduced pagan ideologies, in this case the golden statue of a calf, pagan music and dancing, etc. That was a sin. God wanted to destroy the entire nation right there and then, but Moses "persuaded" God to not do it. Easter, Christmas, icons in church, all these things are the same as what happened with the golden calf. We should not insert anything pagan in worshiping God.

[quote=Dark_Shadow]The only solution is to ban Christmas.

As an added bonus, it will diminish the influence of the Pope on our beloved country.[/quote]

Hmmm. You seem to dislike catholicism. [quote=Irshad]What I really hate is that when people know you're a Christian they automatically think you celebrate Christmas. Christmas isn't even a Christian holiday. It is a pagan holiday.[/quote]

It is meant to be a christian holiday. people turn it into "Santa's Holiday" though sadly.

[quote=bubka]Holidays make people happy without making them petty.
Religion may or  may not make people happy, but it makes them petty.

So I like holidays but denounce religion.[/quote]

It's Satan who makes men "petty". It's this funny thing nowadays, people talk about being "happy". A holiday cannot make you happy. Without a God you are empty. With God there is true happiness. I, as a person who has experienced God in a real way, and seen him work in people around me, can say that what I have just said is true.

Being an agnostic, I would prefer having the whole week off for someone's religious holiday then just one day. If I had a week whole off, I could go on vacation somewhere or drive to my country house. If I just had one day off, there really would not be that much to do.

If we were to have one holiday each year on the winter solstice exactly, and call it Jeffrey, would that solve everyone's problems?

@above that's my name...
So then people would be jealous of me :P


What is the least positive multiple of 72 that has exactly 16 positive factors?

Yes, it's $ \boxed{216}$.

Apart from guess-and-check, I'm not really sure how to solve it.


I want to prove by the $ \epsilon-\delta$ definition that $ \underset{(x,y)\to (1,2)}{\mathop{\lim }}\,\frac{x^{2}}{x+2y}=\frac{1}{5}.$

Then, I started by making the following:

\begin{eqnarray*}
   \left| \frac{x^{2}}{x+2y}-\frac{1}{5} \right|&=&\left| \frac{5x^{2}-x-2y}{5(x+2y)} \right| \\
  &=&\left| \frac{5x^{2}-5x+4x-2y+4-4}{5(x+2y)} \right| \\
 & =&\left| \frac{5x(x-1)+4x-2(y-2)-4}{5(x+2y)} \right| \\
 & \le& \frac{5\left| x \right|\left| x-1 \right|+4\left| x \right|+2\left| y-2 \right|+4}{5(x+2y)},
\end{eqnarray*}

now I can suppose that $ \delta=1$ to bound the other expressions.

Another way of doing it is that given $ |x-1|<\delta$ and $ |y-2|<\delta$ and with $ \delta=1$ we get $ 0<x<2$ and $ 1<y<3,$ thus $ x+2y>2$ and $ \left| 5x^{2}-x-4 \right|<14\delta,$ hence by taking $ \delta =\min \left\{ 1,\frac{5\epsilon }{8} \right\}$ we get

$ \left| \frac{x^{2}}{x+2y}-\frac{1}{5} \right|=\left| \frac{5x^{2}-x-4-2y+4}{5(x+2y)} \right|<\frac{14\delta +2\delta }{10}=\frac{8}{5}\delta =\epsilon .$

Which one of the proceduresis the best? I like the second one.

The second one is the method I'm more used to seeing and using when proving a limit via $ \varepsilon$-$ \delta$.

Excellent, I just wanted some clarification.

Thanks for your answer.


Hi all! I am struggling with the following problem:

Let E be a finite dimensional Euclidean space and let [b]T: E -> E[/b] be a symmetric operator, which additionally satisfies the  relation [b](T(x), T(x)) = (x,x)[/b] for every vector [b]x [/b]in [b]E[/b].
Prove that there exist subspaces [b]M =< E[/b]  and [b]N =< E[/b] such that [b]E=M+N[/b] (direct sum), and [b]T(a) = a[/b] for each [b]a [/b]in [b]M [/b]and [b]T(b)= -b[/b] for each [b]b [/b]in [b]N[/b]. Check that [b]N [/b]is orthogonal complement of [b]M[/b].

Anyone got any idea???

The theorems:

A symmetric/Hermitian operator is orthogonally/unitarily diagonalizable with real eigenvalues.
An orthogonal/unitary operator has $|\lambda|=1$ for all eigenvalues $\lambda$.
Your $T$ satisfies both conditions. Diagonalize, and see what happens.

an orthogonal/unitary operator is defined as (T(x), T(y)) = (x,y). But does (T(x), T(x)) = (x,x) still mean that T is orthogonal/unitary????

Yes. You can derive the inner product from the quadratic form.


dayereye C va W yekdigar ro to P,Q ghat mikonan
dayereii be markaze P ,C ro dar S,R  va W ro dar M,N ghat mikone

agar mahale barkhorde  MS  , RN ro  K benamin :
sabet konid  PQ bar KQ amoode [/list]

nohteye $MN\cap RS=Q'$roye ghotbiye $K$ gharar darad .agar az $Q'$bar $PK$ amood konim ta $PK$ ra dar $K'$ ghat konad

$K'$ varoone $K$ khahd bood.

az tarafi $\angle PCA=\angle PCQ'=\angle PQC$ dar natige$PQ'\times PQ=PK\times PK'=r^2$($r$ shoae dayere be
markaze $P$)

pas chehar zeleiye $KK'QQ'$ mohati ast dar natige $\angle PQK=90^{\circ}$


Putnam
function
integration
logarithms
algebra
functional equation
absolute value

Find all differentiable functions $f: (0,\infty)\mapsto (0,\infty)$ for which there is a positive real number $a$ such that
\[ f'\left(\frac ax\right)=\frac x{f(x)} \]
for all $x>0.$

Let
$g(x)=f\left(\frac ax\right)f(x)$
so
$g'(x)=f\left(\frac ax\right)f'(x)-\frac{a}{x^2}f'\left(\frac ax\right)f(x)=f\left(\frac ax\right)f'(x)-\frac{a}{x}$
$g'(\frac{a}{x})=f\left(x)f'(\frac ax\right)-x=0$

so $g(x)$ is constant.

Suppose
$f\left(\frac ax\right)f(x)=c$
$f\left(\frac ax\right)=\frac{c}{f(x)}$
$f'\left(\frac ax\right)=\frac{-f'(x)c}{f(x)^2}$

So the initial condition becomes

$\frac{f'(x)}{f(x)}=\frac{a}{c}$
so
$f(x)=Bx^r$, $r=\frac{a}{c}$, $B>0$

and these give all the solutions, with the only exception that $B=1$ when $r=1$ (you get this by checking if this actually is a solution).

I got precisely this solution: $f(x) = Bx^{a/C}$, where $B > 0$ and $C$ is some constant. Then, in the end I bungled things by showing that $f(x) = x$ is the only solution that satisfies the conditions of the problem for all $a > 0$. I missed the part that I only need to find solutions that satisfy the given conditions for some $a > 0$  :(

Does anyone think I will get credit for my solution? Help!!!

Here's my solution.

We have $f''(t)=\frac{d}{dt}\frac{a/t}{f(a/t)}=\frac{d}{dt}\frac{a}{tf(a/t)}= \frac{-a}{t^2f(a/t)}-\frac{af'(a/t)}{t(f(a/t))^2}\cdot\frac{-a}{t^2}$
$f''(t)=-\frac1tf'(t)+\frac{a^2}{t^4f(a/t)}\cdot\frac{a}{\frac at f(t)}= -\frac1tf'(t)+\frac{(f'(t))^2}{tf(t)}.$

This is a second order differential equation with no singular points in the specified range, so by the uniqueness half of the existence-uniqueness theorem, $f$ is completely determined by $f$ and $f'$ at a single point. Since $f'(\sqrt{a})=\frac{a}{\sqrt{a}}$, $f$ is completely determined by $f(\sqrt{a})$. A little experimentation (how about powers of $x$?) gives us that for positive $c$, $f_c(x)=\frac{x^c}{\sqrt{ca^{c-1}}}$ works, with $f_c(\sqrt{a})=\sqrt{c}$. The previous argument shows that these are all solutions, and we are done.

This didn't take very much insight.

The only problem I see with that is that the statement of the problem only said that $f'$ exists; it didn't say anything about $f''.$

That was a proof that $f''$ exists; since $f'$ depends algebraically on something differentiable, $f'$ is differentiable.

Here is what I did in the exam and it worked ofcourse.
( infact I was really happy to see a functional equation  :D )

Define g(x) = e^f(ln x)

then

g'(x) = a /e^[g(x) + g(lna - x)]

the right hand side is symmetric with respect to x and lna -x. so if b=lna
then g'(x) = g'(b-x) so g(x) + g(b-x) = const so g'(x) = const and therefore f(x) = a x^n

-Ali

Here is my solution. It is somewhat similar to dystopia's though the technique is slightly different. Anyways, here's what I did.

$f'(\frac ax) = \frac x{f(x)}$ $... (1)$

Replace $x$ by $\frac ax$ in both sides of the equation above. We obtain
$f'(x) = \frac a{xf(\frac ax)}$ $... (2)$

Dividing $(1)$ by  $(2)$ and rearranging terms, we get
$\frac{af'(\frac ax)}{x^2f(\frac ax)} = \frac{f'(x)}{f(x)}$

Integrating both sides with respect to $x$, we get

$- \int \frac{d(f(\frac ax))}{f(\frac ax)} = \int \frac{d(f(x))}{f(x)}$

Or, $-ln|f(\frac ax)| = ln|(f(x))| + lnC_1$, where $C_1 > 0$

Since $f(x) > 0$, we can drop the absolute value sign, $| . |$. We then get
$f(\frac ax) = \frac{C_2}{f(x)}$, where  $C_2 > 0$

Differentiating both sides above with respect to $x$ and combining again with $(1)$ while noting that $x > 0$, we get
$\frac{f'(x)}{f(x)} = \frac{aC_3}x$, where $C_3 > 0$

Integrating further both sides above with respect to $x$, we finally obtain
$f(x) = Bx^{\frac aC}$ as the required solution. Here, $B > 0$ and $C$ is some real constant.

I skipped a few steps above though the original solution I wrote was quite detailed. Well, in the end I messed up by showing that $f(x) = x$ is the required solution for all $a > 0$, forgetting the fact that equation $(1)$ holds true for some $a > 0$.

Could anyone tell me how many points I might get for this?

i got a differnt answer and was wondering if it is right or not.

[hide="here is my solution"]
$f'(x)=\frac{a}{xf\left(\frac{a}{x}\right)}$

$g(x)=f(x)f\left(\frac{a}{x}\right)$

$g'(x)=f'(x)f\left(\frac{a}{x}\right)+f(x)\left(\frac{-a}{x^{2}}\right)f'\left(\frac{a}{x}\right)$

so $g'(x)=0$ and $g(x)=k$ for some constant.

$f(x)f(a/x)=k$

$f(a/x)=\frac{k}{f(x)}$

$f'(a/x)=\frac{x}{f(x)}=\frac{-kf'(x)}{f(x)^{2}}$

$xf(x)=-kf'(x)\longrightarrow\frac{-f'(x)}{f(x)}=\frac{x}{k}$

integrating both sides yields

$-\ln(f(x))=\ln\left(\frac{1}{f(x)}\right)=\frac{x^{2}}{2k}+C$ so we get

$f(x)=\frac{A}{e^{Bx^{2}}}$ where $B=\frac{1}{2k}$

i am wondering if i made a mistake or if this solution is correct.
[/hide]

Well, your answer does not survive being substituted into the original equation. It also fails to satisfy your functional equation

$f(x)f\left(\frac{a}{x}\right)=k.$

So where does it go wrong? I'd suggest looking at this step:

[quote]$f(a/x)=\frac{k}{f(x)}$

$f'(a/x)=\frac{x}{f(x)}=\frac{-kf'(x)}{f(x)^{2}}$[/quote]
The problem with that is that $f'(a/x)$ is not the same thing as $\frac{d}{dx}f\left(\frac{a}{x}\right).$

Can someone tell me how FieryHydra exactly got from step 1 to step 2.
[quote="FieryHydra"]
Dividing $(1)$ by  $(2)$ and rearranging terms, we get

[u]Step 1:[/u] $\frac{af'(\frac ax)}{x^2f(\frac ax)} = \frac{f'(x)}{f(x)}$

[u]Step 2:[/u] Integrating both sides with respect to $x$, we get

$- \int \frac{d(f(\frac ax))}{f(\frac ax)} = \int \frac{d(f(x))}{f(x)}$
[/quote]

Peace.
Faustus
 :yinyang:

Obviously, $\dfrac{d}{dx}\left[f(x)\right] = f'(x)$. So $\dfrac{f'(x)}{f(x)}\,dx = \dfrac{d(f(x))}{f(x)}$

Using the Chain Rule: $\dfrac{d}{dx}\left[f\left(\dfrac{a}{x}\right)\right]$ $= \dfrac{d}{dx}\left[\dfrac{a}{x}\right]f'\left(\dfrac{x}{a}\right)$ $= -\dfrac{a}{x^2}f'\left(\dfrac{x}{a}\right)$.

Thus, $\frac{af'(\frac{a}{x})}{x^{2}f(\frac{a}{x})}\,dx = -\dfrac{d(f(\frac{a}{x}))}{f(\frac{a}{x})}$.

Therefore, $ \frac{af'(\frac{a}{x})}{x^{2}f(\frac{a}{x})}=\frac{f'(x)}{f(x)} $ is equivalent to $ -\frac{d(f(\frac{a}{x}))}{f(\frac{a}{x})}= \frac{d(f(x))}{f(x)} $.

We did this problem in Wednesday Putnam Seminar today, and I think have a solution which is different from all the other ones posted here.  (EDIT: After reading the other solutions, I've realized that mine is essentially equivalent to many of the ones written here, just written in a different form.)

[hide="Solution"]
Rewrite the original FE as $f(x)f'\left(\frac ax\right)=x$; under the substitution $x\mapsto \frac ax$ we get $f\left(\frac ax\right)f'(x)=\frac ax$.  Now integrate both sides with respect to $x$.  The RHS is just $a\ln x + C_1$ for some $C_1\in\mathbb{R}$.  To compute the LHS, we use integration by parts.  Specifically, let $u=f\left(\frac ax\right)$ and $dv=f'(x)\, dx$.  Then $du=-\frac{a}{x^2}f'\left(\frac ax\right)\, dx$ and $v=f(x)$.  As a result, \begin{align*}\int f\left(\frac{a}{x}\right)f'(x)\, dx&=f\left(\frac ax\right)f(x)-\int f(x)f'\left(\frac ax\right)\left(-\frac{a}{x^2}\right)\,dx\\&=f\left(\frac ax\right)f(x)+\int x\cdot\frac{a}{x^2}\,dx\\&=f\left(\frac ax\right)f(x)+a\ln x+C_2.\end{align*} Thus $f(\tfrac ax)f(x)=b$ for some constant $b$. (The reason we use $b$ instead of $C$ will become apparent later.)

We're almost done with the problem now.  Remark that solving for $f\left(\frac ax\right)$ and plugging into $f\left(\frac ax\right)f'(x)=\frac ax$ gives $\frac{b f'(x)}{f(x)}=\frac ax$.  Integrating and utilizing the fact that $f>0$ for all $x$ gives \[\ln f(x)=\frac ab\ln x + C\implies f(x)=Cx^{a/b}.\] Now it just suffices to plug into the original FE for additional constraints.  Specifically, we see that \[f(x)f'\left(\frac ax\right) = Cx^{a/b}\cdot \left(\frac ax\right)^{a/b-1}=Cxa^{a/b-1}.\] Setting this equal to $x$ yields $Ca^{a/b-1}=1\implies C=a^{1-a/b}$.  As a result, $f(x)=\boxed{Cx^{a/b},\,C=a^{1-a/b}}$ works.

This looks slightly different than jmerry's parametrization, but I think by and large they're the same.[/hide]

Let $g(x) = f(a/x)$ then $[g(x)f(x)]' = g'(x) f(x) + f'(x) g(x) = -a/x^2 f'(a/x)f(x) + f'(x) f(a/x) = 0$ so $g(x) f(x)$ is constant. The rest of the solution is the same.

Observe:
$x\to a/x \implies f'(x)=\frac{a}{xf(a/x)} \implies f''=-\frac{a}{x^2f(a/x)}+\frac{a^2f'(a/x)}{x^3f(a/x)^2}.$
$f(a/x)\to a/(xf'(x)) \implies f''(x)=\frac{(f'(x))^2}{f(x)}-\frac{f'(x)}{x}.$
Let $g(x)=\log f(x).$ It follows $g'(x)=f'(x)/f(x),$ obviously $g'(x)$ is positive.
$g''(x)=\frac{f(x)f''(x)-f'(x)^2}{(f(x))^2}=\frac{-f(x)f'(x)/x}{(f(x))^2}=\frac{-f'(x)}{xf(x)}=\frac{-g'(x)}{x}.$
Let $\phi(x)=\log g'(x),$ so $\phi'(x)=g''(x)/g'(x)=-1/x.$
Integrating yields $\phi(x)=c-\log x,$ where $c$ is any real.
Exponentiation yields $g'(x)=k/x,$ where $k=\exp c.$
Integrate again $g(x)=d+k\log x.$
Exponentiate again $\boxed{f(x)=yx^k},$ where $y=\exp k.$
Easy to see this works.
If $k=1,$ easy to get $\boxed{f(x)=x},$ which works too.


Let $ p_n$ a sequence of positive integers such that :

if $ n \neq m$ then $ gcd(p_n,p_m)\equal{}1$  and $ p_n$ is composite for all $ n$.

Do we have $ \sum_{n \equal{}1}^{\infty} \frac{1}{p_n} < \infty$ ?

It is maybe false, but i did not succeed in finding a counterexample, could you help me?

Disclaimer: It is now 4:30 in the morning. No guarantees  :maybe:

Let $ q_n$ be the smallest prime dividing $ p_n$. Since $ p_n$ is composite we have $ p_n \geq {(q_n)}^2$. Since $ \gcd(p_n,p_m)\equal{}1$ we have $ q_n \neq q_m$ when $ m \neq n$. This means that the sequence $ \frac{1}{{(q_n)}^2}$ is a subsequence of $ \frac{1}{m^2}$ in some order. So

$ \sum_{n\equal{}0}^\infty \frac{1}{p_n} \leq \sum_{n\equal{}0}^\infty \frac{1}{{(q_n)}^2} \leq \sum_{m\equal{}0}^\infty \frac{1}{m^2} < \infty$.


geometry
incenter
blogs
analytic geometry
geometry proposed

3. The incircle of triangle $ABC$ touches $AB$ at $D$. $DE^{'}$ is a diameter of the incircle. Tangent to the incircle in point $E^{'}$ intersects $AC$ at $A^{'}$ and $BC$ at $B^{'}$. If $M$ is the midpoint of $AB$, Prove that:

$a)$ $A^{'}B,AB^{'}$ and $CM$ reach at one point.

$b)$ $M,$ the incenter of $ABC$ and the midpoint of $CD$ are on a straight line.

here's it's link in my blog:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=749422#749422

For a), $AB$//$A'B'$ -> obvious!
b), barycentric coordinate calculation!


geometry
analytic geometry
geometry open


hmm not sure if this is right but here's an attempt

[hide]
[b]Case I:[/b] Let $ 2^{i}=re^{2\pi k}$, where $ k\in \mathbb{Z}$.  Then we can find that $ r=\frac{1}{2}$ so $ \boxed{2^{i}=\frac{1}{2}e^{2\pi k}}$.

[b]Case II[/b]: Let $ 2^{i}=re^{\pi+2\pi k}$, where $ k\in \mathbb{Z}$. Then we can find that $ r=-\frac{1}{2}$ so $ \boxed{2^{i}=-\frac{1}{2}e^{\pi+2\pi k}}$.

[/hide]

I'm not sure how you're getting your results.

[hide="Solution"] $ 2^{i}= e^{i \ln 2}= \cos \ln 2+i \sin \ln 2$ [/hide]


Please ,
      say if    $AC \perp EB$

or not.

 Babis


geometry
geometric transformation
rotation
function
analytic geometry
circumcircle
trigonometry

A triangle is inscribed in a circle. The length of the sides of the triangle are 7, 8, 9 inches. Find the radius of the triangle.

Is there a special formula?

Let x^2 - sqrt(3)*x*y - 1 = 0. What is the angle of rotation from it's parent function?

How do you figure out angle of rotations?

The graph of x^2 + 6x +y^2 - 6y +15 = 0 lies in which quadrants?

How would I do this without a graphing calculator?

The point (-3,3) is rotated 30 deg. clockwise about the origin. The coordinates of the point after the rotation is?

The altitude of PQR forms 2 right triangles,PQX and RQX.
PQ=11, RX = 10, angle RQX = 48 deg.
What is PX?

[quote]the radius of the triangle. [/quote] I think you means the radius of $ R$:circumradius of $ \triangle ABC$. You can use Heron's formula to find the area of $ \triangle ABC$ and Using $ Area\equal{}\frac{abc}{4R}$ you'll get R. Or you can use law of cosines to find 1 angle, and use law of sines to find R.

[hide="3rd question"]
For this question, you simply "complete the square" and change into the standard form of a circle as follows:
$ x^2 \plus{} 6x \plus{} y^2 \minus{} 6y \plus{} 15 \equal{} 0 \implies (x^2 \plus{} 6x \plus{} 9) \plus{} (y^2 \minus{} 6y \plus{} 9) \minus{} 3 \equal{} 0 \implies (x \plus{} 3)^2 \plus{} (y \minus{} 3)^2 \equal{} 3 \implies (x \plus{} 3)^2 \plus{} (y \minus{} 3)^2 \equal{} \left(\sqrt {3}\right)^2$
This equation's graph is the circle with center $ ( \minus{} 3,3)$ and radius $ \sqrt3$, which lies only in the $ \boxed{2^{nd} \text{ quadrant}}$.
You should look at example 3 on [url=http://www.analyzemath.com/CircleEq/Tutorials.html]this website[/url] if you did not understand my method
[/hide]
[hide="4th question"]
The easiest thing to do is to change this point into [url=http://en.wikipedia.org/wiki/Polar_coordinates]polar form[/url]. Since the point $ (\minus{}3,3)$ is $ 3\sqrt2$ units away from the origin, and this point has an angular value of $ 135^{\circ}$, the point $ (\minus{}3,3)$ is also equivalent to the point $ (3\sqrt2,135^{\circ})$ in oplar form.
In order to rotate this point 30 degrees clockwise about the origin, you simply have to decrease the angle by 30 degrees, resulting in the point $ (3\sqrt2,105^{\circ})$.
Finally, we reverse the process to turn this point in polar form to its equivalent point in rectangular ((x,y) form). Since this point has a angular value of $ 105^{\circ}$, it is in the form $ (r\cos(105^{\circ}),r\sin(105^{\circ}))$, where r is the radius. The final answer is $ \boxed{(3\sqrt2 \cos(105^{\circ}),3\sqrt2 \sin(105^{\circ}))}$
[/hide]


limit
real analysis
real analysis unsolved

Find the limit :
$ lim (\sqrt{2\minus{}\sqrt{2}}.\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2}}}...\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}....\plus{}\sqrt{2}}}}})$

Call that sequence is $ a_n$ (has n product)
$ b_{n} = \sqrt {2 + \sqrt {{2}... + {\sqrt{2}}}}$
Then
$ a_n.b_n=\sqrt {2}$
but $ \lim_{n\to\infty} b_n = 2$
So $ \lim_{n\to\infty} a_n = \frac {\sqrt {2}}{2}$


inequalities
number theory
inequalities unsolved

If $a,b,c$ are positive reals number. Find all positive real numbers $x,y,z$ such that:

$x+y+z=a+b+c$

$4xyz-(a^{2}x+b^{2}y+c^{2}z)=abc$
 :wink:

I bet it is not a number theory problem. Don't know what mods do on this forum...

I can analyze it for case $x,y,z\leq \frac{a+b+c}{2}$...

Yeah, looks suspiciousally cauchy-swartzy...

[quote="Myth"]I bet it is not a number theory problem. Don't know what mods do on this forum...[/quote]
 :(
I can't even remember having seen this.
PS: you are mod here ;)

Yes, it was already posted and Myth and I gave different solutions, but it was on the inequalities forum.


Find all positive integer solutions of

$ (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$.

[hide]

We'll first write the equation in form: $ (a+1)(b+1)(c+1)=2abc$. Let's now make a substitution $ a=u+4, b=v+4, c=w+4 (-3 \le u,v,w)$.

The equation becomes: $ (u+5)(v+5)(w+5) = 2(u+4)(v+4)(w+4)$. When we expand that we get $ uvw+3uv+3vw+3wu+7u+7v+7w+3=0$.

Now note that at least one of $ u,v,w$ need to be negative because we want the LHS to equal 0. Now we just do a little case work, symmetrically we can just check the negative values for u:
1) $ u=-3$
$ 2v+2w =-18$, this is impossible, because it would require $ u \mathrm{or}v <-3$.
2) $ u=-2$
$ vw+v+w=11$
We get the following (w,v) pairs: $ (11,0), (5,1), (3,2)$.
3) $ u=-1$
$ vw+2v+2w=2$ results $ (1,0), (-1,4)$.

So the solutions $ (a,b,c)$ are: $ (2,15,4), (2,9,5), (2,7,6), (3,5,4), (3,3,8)$ and all their permutations.
[/hide]

im not sure im mad or this pro's mad :D
WLOG $ \frac{1}{a}\geq \frac{1}{b}\geq \frac{1}{c}$
then we have $ (1+\frac{1}{a})^{3}\geq (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$
equivalents to $ 1 \geq (\sqrt[3]{2}-1)a$ finally, we obtain $ a \leq 3$
similar way to the other variables


Find all points of intersection of the two graphs. One intersection occurs at x=1.

(Problem has the picture of these graphs below)

Y = x^4 + 2x^3 - 9x^2 +20 and Y = -2x^2 + 7x - 2

This is from a math book. I already know the answers. By looking at the graph I can easily tell what the answers are. Also by plugging and plotting my visual guesses from the looks of the picture of the two graphs, I can be sure of those guesses. Even though I have the answers, I want to know the "correct" method of solving polynomial interestions. I want to know of an exact and specific method to solving this problem. Unless the method I did was the way to do it? All I did was look at the two graphs, estimate where they would intersect and plugged in the the x or y coordinates to get the other and do the same for both equations. But I know this is not the "correct way" of solving this problem.

Please describe to me how to solve this problem and the general method of solving intersections of polynomials. Thank you very much for reading my post.

Intersections are where the equations equal each other.
But at x=1, these two equations don't equal each other.
1+2-9+20=14 doesn't equal -2+7-2=3.
Are you sure you typed the problem correctly?

In general, you solve the two equations (kind of like solving linear simultaneous equations.

$y=x^4+2x^3-9x^2+20$
$y=-2x^2+7x-2$.

Subtract the two equations to get something in terms of $x$ only, then solve that equation for $x$.

Then plug in those values of $x$ into the original equation to get the $y$ values.

Am sry, the second equation is -2x^2 + 8x +8. Very sorry. :(

So we have two polynomials:
$y=x^4+2x^3-9x^2+20$
$y=-2x^2+8x+8$
if they intersect somewhere they attain the same value for the same $x$ so:
$x^4+2x^3-9x^2+20=-2x^2+8x+8$
$x^4+2x^3-7x^2-8x+12=0$
We find roots among divisors of $12$ and we get :$-3,-2,1,2$

Math is one of my hardest subjects to understand, still I enjoy it a lot. If you don't mind, can you describe how you got -3, -2, 1 and 2 from the equation x^4 + 2x^3 - 7x^2 - 8x + 12 = 0? I realize that the equation factors into (x+3)(x+2)(x-1)(x-2) from those 4 numbers you had told me. Am basically asking if you may describe how you factored the equation (if there's a method / formula)? If I were to look at that equation I would have never simplified it (x+3)(x+2)(x-1)(x-2) and gotta -3, -2, 1 and 2 for the "X's". The book only goes are far as giving me a few detailed examples, but not enough for me to have a complete understanding. Please help me, and I thank you very much for helping me already with the equation.

Thank you very much dondigo for the explanation. It helped me solve the other problems. One more question though. :) This will be the last I promise. After you find the divisors, is there a definte way of checking to see if they are the roots of the equation? How did you weed out -1,-4,-6,-12,3,4,6, and 12 from the divisors of 12, and figure out that -3, -2, 1 and 2 would be the roots of the equation? Am sorry if am giving you a hard time asking so many questions. If you don't want to answer I understand and I am sorry. Thank you.

First of all. If you have some problem don't hesitate to ask. This is a Mathforum, so other people will be pleased to help you. You can ask as many question as you can. If there is something easy, that you don't understand post the topic in Getting Started, if you have more difficult problem-ask in Intermediate just like you have done it.
The answer to your question:
For example we have a polynomial:
$W(x)=x^2+2x+1$
We check whether for example $2$ is the root:
$W(2)=2^2+2\cdot2+1=4+4+1=9$
$2$ is not the root.
We check $-1$
$W(-1)=(-1)^2+2\cdot(-1)+1=0$
$W(-1)=0$, so $-1$ is the root of this polynomial

Thank you Dondigo, I understand completely now. This is my first post on MathLinks. I'll be sure to post more problems that I may have trouble with in the future. I'll be sure to browse and help others that with theirs too. :)


excuse me can anyone fill me in,
with some number theory links please..........

There is  avaliable a special NT forum  here in mathlinks [url]http://www.mathlinks.ro/index.php?f=6[/url].Enjoy it !

thanks friend but i think it is too assorted
what about some pdf link if anyone has??
thanks for replying

What kind of number theory?
Considering you're posting in the Pre-Olympiad forum it is sort of implied that you want the olympiad level stuff (what enndb0x linked to).

If you're looking for something introductory, AoPS has an introduction to number theory book.


As illustrated below, we can dissect every triangle $ABC$ into four pieces so that piece 1 is a triangle similar to the original triangle, while the other three pieces can be assembled into a triangle also similar to the original triangle. Determine the ratios of the sizes of the three triangles and verify that the construction works.

[asy]
import olympiad;size(350);defaultpen(linewidth(0.7)+fontsize(10));
path p=origin--(13,0)--(9,8)--cycle;
path p2=rotate(180)*p,
p3=shift(-26,0)*scale(2)*p,
p4=shift(-27,-24)*scale(3)*p,
p1=shift(-53,-24)*scale(4)*p;
pair A=(-53,-24), B=(-8,16), C=(12,-24), D=(-17,8), E=(-1,-24), F=origin, G=(-13,0), H=(-9,-8);
label("1", centroid(A,D,E));
label("2", centroid(F,G,H));
label("3", (-10,6));
label("4", (0,-15));
draw(p2^^p3^^p4);
filldraw(p1, white, black);
pair point = centroid(F,G,H);
label("$\mathbf{A}$", A, dir(point--A));
label("$\mathbf{B}$", B, dir(point--B));
label("$\mathbf{C}$", C, dir(point--C));
label("$\mathbf{D}$", D, dir(point--D));
label("$\mathbf{E}$", E, dir(point--E));
label("$\mathbf{F}$", F, dir(point--F));
label("$\mathbf{G}$", G, dir(point--G));
label("$\mathbf{H}$", H, dir(point--H));
real x=90;
draw(shift(x)*p1);
label("1", centroid(shift(x)*A,shift(x)*D,shift(x)*E));
draw(shift(130,0)*p4);
draw(shift(130,0)*shift(-27,-24)*p);
draw(shift(130,0)*shift(-1,-24)*p3);
label("2", shift(130,0)*shift(-27,-24)*centroid(F,(9,8),(13,0)));
label("3", shift(130,0)*shift(-1,-24)*(-10,6));
label("4", shift(130,0)*(0,-15));
label("Piece 2 rotated $180^\circ$", (130,10));[/asy]

This problem is very nicely.
I think that the answer is $\frac{GH}{1}=\frac{DE}{4}=\frac{BC}{5}\ \left(=\frac{HE}{2}=\frac{GE}{3}\right).$


A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring
at a distance much greater than R and allowed to fall towards the ring under the influence of the
ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same
(linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum
speed of the particle?

$ \textbf{(A)}\ \frac{1}{2}v\qquad \textbf{(B)}\ \frac{1}{\sqrt{2}}v\qquad \textbf{(C)}\ v\qquad \textbf{(D)}\ \sqrt{2}v\qquad \textbf{(E)}\ 2v$

Conservation of energy - work done by the ring on the particle = change in kinetic energy of the particle.
therefore, find the gravitational potential energy of the ring. However, this does not change when the ring is replaced, and so the velocity stays the same.
Another topic about this question can be found at: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190314[/url]

[quote="xsk13"]Another topic about this question can be found at: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190314[/url][/quote]

Thanks! Sorry, I searched for "2008 F=ma #18" and didn't find anything, so I posted this up.

Doesn't the gravitational potential energy change though? The mass of the ring change because at a radius of 2R; it is larger and will have two times as much mass as it did before. The problem is not saying that mass is remaining the same. It is saying that mass density is remaining the same.

Qualitatively, the larger radius means a greater distance between the mass and the $ 2R$ ring than the distance between the mass and the $ R$ ring, at every point along the x-axis. Thus, increase in mass and increase in distance "balances" out to produce the same net work.

Quantitatively, you can form an expression for the force, integrate it along the x-axis to find work. Since the starting distance $ x$ is much greater than $ R$, you can treat $ R \equal{} 0$ when you have something along the lines of $ \frac {something}{x \plus{} R}$. Conclusion: $ R$ doesn't matter for a sufficiently large $ x$.

Oh wow...I was considering all mass at the Center of Mass and ignored the increased distance because the center of mass didn't change.

I don't understand... Since gravitational energy can be described as $ G\frac{Mm}{r}$, if the new ring is twice as massive, wouldn't the speed then be $ \sqrt{2}$ times as much?

[quote="pascal12"]I don't understand... Since gravitational energy can be described as $ G\frac {Mm}{r}$, if the new ring is twice as massive, wouldn't the speed then be $ \sqrt {2}$ times as much?[/quote]
Gravitational Potential Energy at Centre of the ring is $ U_{centre} \equal{} \frac {GMm}{R}$.
Change in K.E in the process of the falling of the particle from $ \infty$ to the centre of the ring(where the particle has max. K.E) is equal to the change in the P.E.
So $ (K.E)_{centre} \equal{} U_{centre}$
But $ U_{centre}$ doesn't change as "M"(mass of ring) is also doubled along with the radius "R".

But if it's falling from infinity, wouldn't the radius of the ring be negligible? After all, isn't $ r$ the distance from the object to the center of mass?

By telling you that it starts from infinity, it is basically telling you that the original potential gravitational energy is zero.

When the particle gets closer to the ring, the distance is no longer negligible.


[quote="Αρχιμήδης 6"]Να λυθει στους φυσικους η εξισωση:
$ 1! \plus{} 2! \plus{} .....x! \equal{} y^2$[/quote]

Για $ x>5$ έχω $ LHS\equal{}3(mod5)$ που είναι μη τετραγωνικό κατάλοιπο $ mod 5$. Μένει να ελέγξουμε για $ x\equal{}1,2,3,4$

$ qed$

Σωστος Νικολακι..... :lol:
Αλλα να μην την γενικευσουμε....κριμα δεν ειναι....
Να λυθει η εξισωση:
$ 1! \plus{} 2! \plus{} ...x! \equal{} y^n$ (για καθε $ n$ φυσικο μεγαλυτερο του $ 2$).Νικο good luck WAGM!ΖΗΤΑΩ ΚΑΙ ΤΗΝ ΕΝΑΣΧΟΛΗΣΗ ΤΟΥ ΑΝΤΟ ΚΑΙ ΤΟΥ CRETANMAN ΓΙΑ ΤΟ ΠΡΟΒΛΗΜΑ....KAI OΛΩΝ ΤΩΝ ΑΡΙΘΜΟΘΕΩΡΗΤΙΚΩΝ ΦΥΣΙΚΑ.... :wink:

Tελικά ποια είναι η λύση σε αυτό το πρόβλημα?


This is not the appropriate forum to discuss proofs. Please keep the problems Mathcounts level. Examples of some relatively easy mathcounts problems are [url]http://www.mnspe.org/mathcounts/08_state_test/2008_sprint.pdf[/url].

This should be moved to an appropriate forum. I'd suggest Intermediate topics, since I've seen this before in there (I think). This was an ARML problem, except it was find $ \sqrt[3]{\sqrt[3]{2}\minus{}1}\equal{}\sqrt[3]{a}\minus{}\sqrt[3]{b}\plus{}\sqrt[3]{c}$, and we wanted $ a\plus{}b\plus{}c$.

[hide=Post #5 by sqing][url=aops.com/community/user/148231][b]sqing[/b][/url] · Jun 23, 2021, 1:41 PM [url=aops.com/community/p22425693](view)[/url][color=transparent]helo[/color]
[quote=Ji Chen]Prove that ${ \sqrt[3] {\sqrt [3]{2} - 1} = \sqrt [3]{\frac{1}{9}} - \sqrt [3]{\frac{2}{9}} + \sqrt [3]{\frac{4}{9}}}.$[/quote]
[url]https://artofproblemsolving.com/community/c7h2598821p22420229[/url]

-----------
[color=#5b7083][aops]x[/aops] 5[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@sqing][img]https://avatar.artofproblemsolving.com/avatar_148231.jpg[/img] [center]AoPS User[/center][/tip] why the 12-year bump?

Just use x^3+y^3=(x-y)(x^2+y^2-xy) first,notice the r.h.s has some match with this eqn simplify then use x^3-y^3=(x-y)(x^2+x-y+y^2).
Anyway as far I know these formulas are taught in class 9 in my country,I think it's taught before in some other countries.

I think some people cannot see the dates and are commenting/replying stuff randomly /shrug

@above what I told is a truth,and it doesn't matter what the date is,people learn everyday.
For your information aops is not like any social media,even 15 year old problems are also solved hear.
P.S.--I reckon you are new here so surf a little bit more.


analytic geometry
topology
complex analysis

Prove that every discrete set in $ \mathbb{C}$ is countable

See [url]http://mathworld.wolfram.com/DiscreteSet.html[/url]

There is a standard argument.
By the definition of a discrete set $ S$, for any $ x \in S$ there exists a neighborhood $ U_x$ of $ x$, such that $ U_x \cap S \equal{} \{x\}$. Then there is a ball $ B(x, \rho_x) \subseteq U_x$, and a point $ P_x(a_x, b_x) \in B(x, \rho_x)$ of rational coordinates. But $ \mathbb{Q} \times \mathbb{Q}$ is countable.

How does one prove that every open set in $ \mathbb{R}^{n}$ has atleast one point with rational coordinates

First, put the exponent $ ^n$ outside the brackets of $ \mathbb{R}$, in order to get $ \mathbb{R}^n$, and not $ \mathbb{R^n}$.

Next, any open set contains a ball, and $ \mathbb{Q}^n$ is dense in $ \mathbb{R}^n$.


if $x_{1}=3$ and $ 1+x_{n}x_{n+1}=2n^{2} (x_{n+1}-x_{n})$, then compute the limit of $x_{n}$

have fun! nice and easy! :cool:

I discover that the sequence {x_n} is 3, 5, 7, 9, 11, 13, ...

Please check your computation! It's not good!

you rewrite the sequence as:
(x_(n+1)-x_n)/(1+x_n*x_(n+1))=1/(2n^2) now you apply arctan to this equality
and we know that:

acrtan ((x_(n+1)-x_n)/(1+x_n*x_(n+1)))=arctan(x_(n+1))-arctan(x_n)
and arctan(1/(2n^2))=arctan(n/(n+1))-arctan((n-1)/n) and you sum it up and get:
arctan (X_(n+1))-arctan(x_1)=arctan(n/(n+1)) and then you get that:
x_(n+1)=(4n+1)/(-2n+1) so lim(x_n)=-2
I hope i didn't make any mistakes!

cheers!~ :D :D :cool:

oops...i made a mistake during the computation...  :(

Your answer is right, Lagrangia.


I took some time to study complex numbers today with AoPS Vol 2.  I understood all the concepts (except for roots of unity) but I failed to put them to use.  I encountered a few problems.  One of them is that I don't know how to express general numbers in the exponential form.  For example, $-64$ can be written as $64e^{i\pi}$ and $1 = e^{2ki\pi}$

The following probles are taken from the AoPS textbook.

1. Find $(1+i)^{4}(2-2i)^{3}$ - Try using polar form to complete the problem.

2.  Find the product of the $n$ $n$th roots of $1$ in terms of $n$.

3.  Find $Im(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^{6}$ (Try not to use a geometric method.)

If anyone has additional resources (webstes) for complex numbers, the help would be greatly appreciated.

Thanks for looking.

[hide="Problem 1"] \begin{eqnarray*}(1+i)^{4}(2-2i)^{3}& = & \left(\sqrt{2}\left(\cos{\frac{\pi}{4}}+i \sin{\frac{\pi}{4}}\right)\right)^{4}\left(2\sqrt{2}\left(\cos{\frac{7\pi}{4}}+i \sin{\frac{7\pi}{4}}\right)\right)^{3}\\ & = & \left(\sqrt{2}\right)^{4}\left(2\sqrt{2}\right)^{3}\left(\cos{\frac{\pi}{4}}+i \sin{\frac{\pi}{4}}\right)^{4}\left(\cos{\frac{7\pi}{4}}+i \sin{\frac{7\pi}{4}}\right)^{3}\\ & = & (4)\left(16\sqrt{2}\right) \left(\cos{\pi}+i \sin{\pi}\right)\left(\cos{\frac{21\pi}{4}}+i \sin{\frac{21\pi}{4}}\right) \\ & = & 64\sqrt{2}\left(\cos{\frac{\pi}{4}}+i \sin{\frac{\pi}{4}}\right) \\ & = & \boxed{64+64i}\end{eqnarray*} Note that DeMoivre's Theorem was used to find $\left(\cos{\frac{\pi}{4}}+i \sin{\frac{\pi}{4}}\right)^{4}= \left(\cos{\pi}+i \sin{\pi}\right)$, etc.[/hide]

[hide="Problem 2"]Do you mean the nth roots of 1?

The nth roots of 1 are of the form $\cos{\frac{2k\pi}{n}}+i \sin{\frac{2k\pi}{n}}= \mbox{cis}\frac{2k\pi}{n}$, where $k = 0,1,2,...,n$. By DeMoivre's Theorem, their product is: \begin{eqnarray*}\mbox{cis}\frac{0\pi+2\pi+4\pi+...+2n\pi}{n}& = & \mbox{cis}\frac{2\pi(1+2+3+...+n)}{n}\\ & = & \mbox{cis}\frac{2\pi n(n+1)}{2n}\\ & = & \boxed{\mbox{cis}\left(\pi(n+1)\right)}\end{eqnarray*} [/hide]

And what do you mean by Im(...)?

[hide=" :D$_{1}$"]

\begin{eqnarray*}(1+i)^{4}(2-2i)^{3}& = & \left(2^{3}\right)\left(1+i\right)\left(1+i\right)^{3}\left(1-i\right)^{3}\\ & = & \left(2^{3}\right)\left(1+i\right)\left(2\right)^{3}\\ & = & 64+64i\end{eqnarray*}
[/hide]
[hide=" :D$_{2}$"] \[\text{The n n-th roots of 1 satisfy the equation: }x^{n}-1=0\]  \[\text{Let this equation be F(x,n). We factor it into n roots: }\prod_{i=1}^{n}(x-r_{i}0\]  \[\text{The constant term of F(x,n) is the only term without a power of x. Therefore it is the product of the roots: }\]  \[\prod_{i=1}^{n}(x-r_{i}) \ \ \text{mod x}\equiv \prod_{i=1}^{n}(-r_{i})=(-1)^{n}\left(\prod_{i=1}^{n}r_{i}\right) \text{(which we know equals }-1 \text{)}\] $\therefore \prod_{i=1}^{n}r_{i}=\frac{(-1)}{(-1)^{n}}\implies \boxed{\prod_{i=1}^{n}r_{i}=(-1)^{n+1}}$
Notice that: $\mbox{cis}\left(\pi(n+1)\right)=\mbox{cis}(n\pi)\mbox{cis}(\pi)=(-1)^{n+1}$
[/hide]
[hide=" :D$_{3}$"]
\begin{eqnarray*}(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^{6}& = & \left(\mbox{cis}(30^\circ)\mbox{cis}(18^\circ)+\frac{\mbox{cis}(30^\circ)}{\mbox{cis}(18^\circ)}\right)^{6}\\ & = & \left(\mbox{cis}(18^\circ)+\frac{1}{\mbox{cis}(18^\circ)}\right)^{6}\\ & = & \left(\cos(18^\circ)+i\sin(18^\circ)+\cos(-18^\circ)+i\sin(-18^\circ)\right)^{6}\\ & = & (2\cos(18^\circ)^{6}\end{eqnarray*}
And that is precisely the answer, so the imaginary part ($\text{Im(...)}$) is 0. The geometric method uses vectors, but it is really the same as the algebra :)
[/hide]

[quote="t0rajir0u"][quote="xsquaredster"]And what do you mean by Im(...)?[/quote]

It refers to the imaginary part.  That is, given $a+bi \in \mathbb{C}$ with $a, b \in \mathbb{R}$ we define

$\Re(a+bi) = a$
$\Im(a+bi) = bi$[/quote]
Isn't $\Im(a+bi) = b$? :maybe:

Hmm.  I'm not so sure.  It doesn't make too much of a difference, since it's only ever used in isolation  :P


inequalities
trigonometry
geometry
circumcircle
IMO Shortlist

Let $ ABCD$ be a convex quadrilateral, and let $ R_A, R_B, R_C, R_D$ denote the circumradii of the triangles  $ DAB, ABC, BCD, CDA,$ respectively. Prove that $ R_A \plus{} R_C > R_B \plus{} R_D$ if and only if  $ \angle A \plus{} \angle C > \angle B \plus{} \angle D.$

$ R_A = \frac {AD}{2sin\angle ABD}$
$ R_B = \frac {BC}{2sin\angle CAB}$
$ R_C = \frac {BC}{2sin\angle BDC}$
$ R_D = \frac {AD}{2sin\angle DCA}$

WLOG $ \angle DAB$ + $ \angle BCD$ $ > =$ $ \angle CDA$ + $ \angle ABC$
The sum of all angles is $ 360$ and that implies $ S =$ $ \angle DAB$ + $ \angle BCD$ $ > =$ $ 180$

1)If $ S = 180$ then A,B,C and D lie on the same circle, so $ R_A + R_C = R_B + R_D$

2)If $ S > 180$ then $ \angle DCA > \angle ABD$ and $ \angle CAB > \angle BDC$ which implies
$ R_A > R_D$ and $ R_C > R_B$
Sum these two inequalities and you get the result

@Bugi: $\angle DCA > \angle ABD$ does not imply $\sin \angle DCA > \sin \angle ABD$ or the opposite inequality, since some of these angles may be obtuse.

[hide]
By capital letters $A, B, C, D$, I mean those angles of the quadrilateral $ABCD$ with the specified point as vertex. Also, let $R(XYZ)$ denote the circumradius of triangle $XYZ$.

Clearly the if and only if statements are equivalent. Assume then that $A + C > B + D$ First we dispose of most quadrilaterals with one method, and then we look for those for which this method is invalid and prove the problem for them separately.

Assume WLOG $\sin C < \sin A$ so that $R_C > R_A$ by the law of sines. Extend $AC$ to meet the circumcircle of $ABD$ at point $C'$. We have $R_A + R_C > R_A + R(BC'D)$ since $R(BC'D) = R_A$, and $R_A + R(BC'D) = R(ABC') + R(ADC')$. Next, if $BC' > BC$ and $CD' > CD$, then we would have $R(ABC') = \frac{BC'}{\sin A} > \frac{BC}{\sin A} > R(ABC)$, and similarly $R(ADC') > R(ADC)$, so we would be done.

Hence it remains to consider $ABCD$ for which $BC' > BC$ and $CD' > CD$ are not both true. Clearly at least one of $\angle ACB$, $\angle ACD$ is acute, so at most one of these inequalities is false. WLOG, assume $BC' < BC$. Since the other inequality $CD' > CD$ holds we can show as before that $R_A > R_D$. We are left with proving $R_C > R_B$.

From $BC' < BC$ it follows that $\angle BCA$ is obtuse. Next, we show that $\angle BDC < \angle BAC$ and both are acute. The acuteness follows from the obtusenesses of $\angle BCA$ and $C$. The inequality is obvious from the fact that the ray $DC$ meets the circumcircle of $ABD$ at a point on minor arc $BC'$. Hence $\sin \angle BDC < \sin \angle BAC$ so $\frac{BC}{\sin \angle BDC} > \frac{BC}{\sin \angle BAC}$ and $R_C > R_B$.
[/hide]


inequalities
Cauchy Inequality
inequalities proposed

[b]By: Pham Dinh Khanh -10A2t-khtn[/b]
Let $ x,y,z > 0$ Prove that
$ \frac {x^2}{\sqrt {x^2 \plus{} xy \plus{} y^2}} \plus{} \frac {y^2}{\sqrt {y^2 \plus{} yz \plus{} z^2}} \plus{} \frac {z^2}{\sqrt {z^2 \plus{} zx \plus{} x^2}} \geq \frac {x\plus{} y \plus{} z}{\sqrt {3}}$

This problem can be written as follows
$ \sum_{cyc}\sqrt{x^2\plus{}xy\plus{}y^2} \ge \sum_{cyc}\frac{y(x\plus{}y)}{\sqrt{x^2\plus{}xy\plus{}y^2}} \plus{}\frac{x\plus{}y\plus{}z}{\sqrt{3}}$
After that, using Cauchy inequality, we derive
$ x^2\plus{}y^2\plus{}xy \ge \frac{3}{4} (x\plus{}y)^2$
We are done!

Could you show me how do you use Cauchy?

Just rewrite $ x^2\plus{}y^2\plus{}xy \ge \frac{3}{4} (x\plus{}y)^2$:

$ 4(x^2\plus{}y^2\plus{}xy) \ge 3(x\plus{}y)^2 \equal{}  3x^2\plus{}6xy\plus{}3y^2$
$ x^2 \plus{} y^2 \minus{} 2xy \ge 0$
$ (x\minus{}y)^2 \ge 0$.

I'm so sorry. This $ x^2 \plus{} y^2 \plus{} xy \ge \frac {3}{4}(x \plus{} y)^2$ is obvious, but I thought about something else. Now I understand the whole proof, but previously I didn't, but I don't know why...

Sorry for creating confusion  :blush:

[quote="dragon1"][b]By: Pham Dinh Khanh -10A2t-khtn[/b]
Let $ x,y,z > 0$ Prove that
$ \frac {x^2}{\sqrt {x^2 \plus{} xy \plus{} y^2}} \plus{} \frac {y^2}{\sqrt {y^2 \plus{} yz \plus{} z^2}} \plus{} \frac {z^2}{\sqrt {z^2 \plus{} zx \plus{} x^2}} \geq \frac {x \plus{} y \plus{} z}{\sqrt {3}}$[/quote]
It's weaker than the well-known:
$ \frac{a}{\sqrt{a\plus{}b}} \plus{} \frac{b}{\sqrt{b\plus{}c}} \plus{} \frac{c}{\sqrt{c\plus{}a}} \geq\ \frac{\sqrt{a} \plus{} \sqrt{b} \plus{} \sqrt{c}}{\sqrt{2}}$
With this ineq, I remember that hxy09 and me have found a nice proof for it  :lol:

Guys,

Unfortunately this solution is wrong, because the inequality is reversed: $ x^2 \plus{} xy \plus{} y^2 \geq \frac34(x \plus{} y)^2 \iff \frac 1{\sqrt {x^2 \plus{} xy \plus{} y^2}} \leq \frac 2{\sqrt3(x \plus{} y)}$.

Also unfortunately, I couldn't come up with a solution too. But I'll keep trying!

EDIT: I was referring to the post above nguoivn's. By the way, I tried to use the Fermat point of a triangle with sides $ \sqrt{x^2\plus{}xy\plus{}y^2}$, $ \sqrt{y^2\plus{}yz\plus{}z^2}$ and $ \sqrt{z^2\plus{}zx\plus{}x^2}$ (so $ x\plus{}y\plus{}z$ is a natural minimum), but I still haven't found anything nice about the LHS.

[quote="cyshine"]Guys,

Unfortunately this solution is wrong, because the inequality is reversed: $ x^2 \plus{} xy \plus{} y^2 \geq \frac34(x \plus{} y)^2 \iff \frac 1{\sqrt {x^2 \plus{} xy \plus{} y^2}} \leq \frac 2{\sqrt3(x \plus{} y)}$.

[/quote]
No! nguoivn means that $ x^2 \plus{} xy \plus{} y^2\leq\frac {3(x^2 \plus{} y^2)}{2}.$ :wink:
[quote="nguoivn"]
It's weaker than the well-known:
$ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}} \geq\ \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt {2}}$
With this ineq, I remember that hxy09 and me have found a nice proof for it  :lol:[/quote]
What is your proof for this one, nguoivn? Thank you!
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071

Here is our poof (hxy09 and me)  :)
We can write it into: $ \sum\ \frac {a^2}{\sqrt {a^2 \plus{} b^2}} \geq\ \frac {a \plus{} b \plus{} c}{\sqrt {2}}$
After square and use Cauchy Schwarts and Am-Gm, we only need to prove:
$ \sum\ \frac {a^4 \plus{} b^4}{a^2 \plus{} b^2} \plus{} \frac {2(ab \plus{} bc \plus{} ca)^2}{a^2 \plus{} b^2 \plus{} c^2} \geq\ (a \plus{} b \plus{} c)^2$
And the last step is a nice hint of hxy09 :lol:
[quote="hxy09"][b]AND[/b] it is [b]EQUIVALENT TO[/b]
$ \sum \frac {(a^2 \plus{} b^2 \minus{} ca \minus{} cb)^2(a \minus{} b)^2}{a^2 \plus{} b^2} \ge 0$
Which is obviously true.:) [/b][/quote]

I have checked  it. It's very interesting proof.    My congratulations!


calculus
derivative
function
trigonometry
algebra unsolved
algebra

prove that for all $x\in \mathbb{R}$ function $f(x)=\sum_{i=1}^{\infty}\frac 1{2^{i}}.\cos (3^{i}x)$ is continious and isn't derivative.

${f(x)=\sum_{i=1}^{\infty }\frac{\cos \left(3^{i}x\right)}{2^{i}}}$

${f'(x)=\frac{\partial \left(\sum_{i=1}^{\infty }\frac{\cos \left(3^{i}x\right)}{2^{i}}\right)}{\partial x}}$
${f'(x)=\sum_{i=1}^{\infty }\frac{\partial \frac{\cos \left(3^{i}x\right)}{2^{i}}}{\partial x}}$
${f'(x)=-\sum_{i=1}^{\infty }\left(\frac{3}{2}\right)^{i}\sin \left(3^{i}x\right)}$

but  $1\geq \sin \left(3^{i}x\right)\geq$

And

${\sum_{i=1}^{\infty }\left(\frac{3}{2}\right)^{i}}$

Diverge

i dont think if this is a proof.... but, the derivative diverges ...


VOne side of a triangle has length 8 and a second side has length 5. Which of the following could be the area of the triangle?

I 24
II 20
III 5

[hide]
Assume the triangle is right and the given sides of the legs, then $A = \frac{ab}{2}= \frac{5 \cdot 8}{2}= \frac{40}{2}= 20$
So $II$
[/hide]

I forgot to mention that there are 2 correct answers.  What is the 2nd 1?  and why is it.

[hide="Furthered"]
Then from the area formula of any triangle $A = \frac{a b \sin \theta}{2}$.
Again let $a=5$ , $b=8$ and then let $A=24$.
Rearranging $\theta = \sin^{-1}\left( \frac{24}{20}\right)$, this then eliminates this as a solution because the $\sin^{-1}$ function is not defined at numbers greater than one.
So the two correct ones must be $II \ , \ III$.
[/hide]

What if you didn't know there are 2 answers?

and it was either

5
20
or both.  How would you know its 5.

[hide="Furthering the Previously Furthered"]
Using the same method, just let $A=5$
Then $\theta = \sin^{-1}\left( \frac{5}{20}\right)$, since $\frac{5}{20}= \frac{1}{4}< 1$ where the $\sin^{-1}$ is defined.
The $sin^{-1}$ being defined there implies that there exists such a triangle.
[/hide]

What was the first formula that was used in the beginning of the thread? I've never seen it before, A=ab/2

[quote="rofler"].. You have never seen the forumla for the area of a right trangle, but you know your trig?  :rotfl:

Sry, just had to say.[/quote]
He actually never demonstrated 'knowing his trig'.
That was another user...

Uber-Furthered:

[hide]
The generic triangle ABC has area $\frac{1}{2}ab\sin{C}$ for two sides and their included angle.

Thus all possible areas of this triangle have area $\frac{1}{2}(5)(8)\sin{x}$ where x is the angle between the sides of length 5 and 8. We don't really care about the length of the third side, but clearly the triangle exists for all values of x degrees, 0<x<180.

The range of $\sin{x}$ on 0<x<180 degrees is (0,1]. Thus the area function $A=\frac{1}{2}(5)(8)\sin{x}=20\sin{x}$ takes values (0,20] on that domain.

So the area can be any real number on (0,20] and only choices II and III fall in that range.
[/hide]


algebra
polynomial
induction
calculus
derivative
superior algebra
superior algebra unsolved

First define f(x) = (x^p)-x. It is obvious that f(x) = x(x-1)...(x-p+1) = 0 (mod p) for all y in Z_p.
Let g(x) =x(x-1)...(x-(p^n)-1) be a monic polynomial of degree p^n such that for every y in Z_(p^n), f(y) = 0 (mod p).
For every prime and n >=2 (except for the case p=2 and n=2), find an explicit monic polynomial g(x) of degree strictly less than p^n such that for every y in Z_(p^n), g(y) = 0 (mod p^n).

I'm not sure why you gave an expression for $g(x)$ and then ask for an expression for $g(x)$, but if I understand your question correctly, this problem is very easy by induction as long as the formal derivative of $f(x)$ is not $0 \bmod p$.


calculus
integration
trigonometry
calculus computations

I think here is what Kent wrote :D
[hide="HINT"]$e^xsin x=\mathrm{Im}\left(e^{(1+i)x}
\right)$[/hide]

In Japan,ordinary high school student don't study the integral in using Euler's formula.
This problem is solved by those who are 17 or 18.(Pre college). That's why Integral by parts will do or be sufficient.
But I have another solution more basically in only one minute. :D

Even with the complex exponential, there will be one integration by parts; it's just that without that simplifying notation, the organization of the several necessary integrations by parts gets fairly messy. Do-able, certainly, but long-winded.

Here is my solution.

$(xe^x\sin x)'=e^x\sin x+xe^x\sin x+xe^x\cos x$
$(xe^x\cos x)'=e^x\cos x+xe^x\cos x-xe^x\sin x$

Subtracting of both sides we obtain $(xe^x\sin x)'-(xe^x\cos x)'=2xe^x\sin x-(e^x\cos x)'$.

Therefore $\int  xe^x\sin xdx=\frac{1}{2}e^x\{x{(\sin x-\cos x)+\cos x\}+Const.}$

As you see, Integral by parts is not necesary except special case.


geometry
trigonometry
circumcircle
ratio
projective geometry
geometry proposed

Assume WLOG that $ \angle {B} \leq \angle{C}$ (in this case $ F$ will be on the segment $ AE$, as in the statement). Denote by $ Y$, $ Z$ the tangency points of the incircle with the sides $ CA$, and $ AB$. Let $ T$ be the intersection point of the lines $ YZ$ and $ BC$. Since $ DZEY$ is an harmonic quadrilateral (because of the concurrency of the tangects at $ Y$, $ Z$ on the line $ DE$), the line $ TD$ is tangent to the incircle at $ D$, and thus $ \angle{TED} \equal{} \angle{TDE}$. But $ \angle{TED} \equal{} \angle{TFE}$, and so the lines $ TD$ and $ CF$ are parallel. Now, since the quadruple $ (B, E, C, T)$ is harmonic, the pencil $ D(B, E, C, T)$ is harmonic, and by intersecting it with the line $ CF$, we conclude that $ F$ is the midpoint of $ CG$ (according to the parallelism of $ DT$ and $ CF$).

If we denote $ K\in\omega(I)\cap BD$,$ L\in\omega(I)\cap CD,N\in\omega(I)\cap AB,M\in\omega(I)\cap AC$,then problem follows by the fact that the following fours are harmonic:$ D(DMEN)\cap l,D(DNKE)\cap l,D(DMLE)\cap l$,where $ l\equal{}CF$.

Let the incircle interescts $ BD$ and $ CD$ at $ K,L$ respectively

Then $ \frac{GF}{CF}\equal{}\frac{DG \times EK}{DC \times EL}\equal{}\frac{DL \times EK}{DK \times EL}$

I think I have proved that $ DL \times EK\equal{}DK \times EL$,

but the proof is long and complicated.

Does anyone have a nice proof for this?

See here: [url]http://www.mathlinks.ro/viewtopic.php?t=2619[/url] (Read especially the 5th post).
For some extensions, see also here: [url]http://www.mathlinks.ro/viewtopic.php?p=887161#887161[/url].

Let tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\parallel CF$. Line parallel to $ DK$ through $ B$ intersect line $ AE$ in $ M$ Let $ L$ be a 'point' respective to a set of lines parallel to $ DK$.
Then $ (GFCL)=(DFEM)=(KCEB)=-1$ So $ GF=FC$

:oops:
I have made one unnecessary turn.
Let tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\parallel CF$. Let $ L$ be a 'point' respective to a set of lines parallel to DK.
Then $ (GFCL)\equal{}(KCEB)\equal{}\minus{}1$. So $ GF\equal{}FC$.

I will use [b]pohoata[/b]'s notation.

Since $ T$ lies on polar of $ A$, $ A$ lies on polar of $ T$. But $ TE$ is tangent to given circle so $ AE$ is polar of $ T$ and we already have that $ D$ is on both $ AE$ and incircle, so $ TD$ is tangent to incircle.

Now it is easy to conclude that $ CF$ and $ TD$ are parallel and by Menelaus theorem:

$ \frac{GF}{FC}\equal{}\frac{EB}{EC}\frac{DG}{BD}\equal{}\frac{BZ}{ZY}\frac{CT}{TB}\equal{}\frac{BZ}{TB}\frac{CT}{CY}$ $ \equal{}\frac{\sin \angle BTZ}{\sin \angle AZY}\frac{\sin \angle AYZ}{\sin \angle CTY}\equal{}1$ which implies $ CF\equal{}FG$.

See from the post #2 that $DT\parallel CF$ and $T$ is the harmonic conjugate of $E$ w.r.t. $BC$, then apply Menelaos to $\triangle BCG$ and the transversal $EFD$, tol get $\frac{CE}{BE}\cdot \frac{BD}{DG}\cdot \frac{GF}{FC}=1$, but $\frac{BD}{DG}=\frac{BT}{CT}$ from parallel lines and $\frac{BT}{CT}=\frac{BE}{CE}$ from the harmonic division and remains $\frac{GF}{FC}=1$.

Best regards,
sunken rock

I will use pohoatza's notation. In addition, let $BD$ intersect the incircle again at $Q$. Let $I$ be the incenter.

We have $\angle EQD = \angle CEF = \angle CFE = \angle DFG$, thus, $GQFE$ is cyclic.

Furthermore, note that $C$ is the circumcenter of $\triangle EFY$, so $\angle EFY = 180 - \frac {C} {2} $, and $\angle DFY = \frac {C} {2}$. Furthermore, $\angle FDY = 90 - \frac {C} {2}$, so $\angle DYF = 90$. Thus, $\triangle DYE \sim \triangle IYC$.

Now I will calculate $GF$. We have $\frac {GF} {QE} = \frac {DF} {DQ}$ from $GQFE$ cyclic, so $GF = QE \cdot \frac {DF} {DQ}$. Now using $DF = \frac {DY} {\sin {\frac {C} {2}}}$, we get $GF = \frac {QE} {DQ} \cdot \frac {DY} {\sin {\frac {C} {2}}}$.

$ZE$ is the symmedian of $QEDZ$, so $\frac {QE} {DQ} =  \frac {ZE} {2ZD}$. But we also have $\frac {ZE} {ZD} = \frac {AE} {AZ} = \frac {AE} {AY} = \frac {YE} {DY}$. Thus, we can multiply out to get $GF = \frac {YE} {2 \cdot \sin {\frac {C} {2}}} = CE$, as desired.

I found a solution without using harmonic properties, though admittedly quite a bit more computational.

[hide="hidden for length"]Let $ A' $ be the foot of the altitude from $ A $ to $ BC $. Firstly, using the usual notation, \[ 4s(s-c)-2ab=(a+b+c)(a+b-c)-2ab=(a+b)^2-c^2-2ab=a^2+b^2-c^2, \] so \[ \frac{2s(s-c)}{ab}-1=\frac{4s(s-c)-2ab}{2ab}=\frac{a^2+b^2-c^2}{2ab}=\cos C. \] We also have $ \cos C=\frac{CA'}{b}=\frac{s-c-A'E}{b} $, so \[ \frac{2s(s-c)}{ab}-1=\frac{s-c-A'E}{b}\implies a\cdot A'E=sa-ac-2s(s-c)+ab=(s-a)(c-b). \]

For brevity, let $ \angle AEC=\theta $ and $ A=[ABC] $. Then, from $ \triangle AA'E $, we have

\[ \tan\theta=\frac{a\cdot AA'}{a\cdot A'E}=\frac{2A}{(s-a)(c-b)}. \]

Now let $ C' $ be the projection of $ C $ onto $ AE $, and let $ E' $ be the point on the incircle diametrically opposite $ E $. Then, since $ \angle DE'E=\angle CEC' $ and $ \angle CC'E=\angle DE'E=90^\circ $, so $ \triangle CC'E\sim\triangle EE'D $. Then $ \frac{DE}{CC'}=\frac{EE'}{CE}=\frac{2r}{s-c} $. Also, $ \frac{CC'}{C'E}=\tan\theta $, so multiplying these two ratios, we find that

\[ \frac{DE}{C'E}=\frac{2r\tan\theta}{s-c}=\frac{2A\tan\theta}{s(s-c)}=\frac{4A^2}{(s)(s-a)(s-c)(c-b)}=\frac{4(s-b)}{c-b}. \]

Since $ FE=2\cdot C'E $, then $ \frac{DE}{FE}=\left(\frac{1}{2}\right)\left(\frac{4(s-b)}{c-b}\right)=\frac{2(s-b)}{c-b} $. Therefore, $ \frac{DF}{FE}=\frac{DE}{FE}-1=\frac{2(s-b)-(c-b)}{c-b}=\frac{a}{c-b} $.

Therefore, $ (BC)(FE)=(DF)(c-b)=(DF)((s-b)-(s-c))=(DF)(BE-CE) $, and $ DF\cdot CE+FE\cdot BC=BE\cdot DF $.

Rearranging,

\[ 1+\frac{FE\cdot BC}{DF\cdot CE}=\frac{BE}{CE}. \]

Now, consider $ \triangle BDE $ and the line $ GFC $ passing through it. By Menelaus, $ \frac{DF\cdot CE\cdot BG}{FE\cdot BC\cdot GD}=1 $, so $ \frac{FE\cdot BC}{DF\cdot CE}+1=\frac{BG}{GD}+1=\frac{BD}{GD} $, and therefore,

\[ \frac{BE}{CE}=\frac{BD}{GD}. \]

A mass point argument then finishes off the problem with $ \triangle CDB $ and cevians $ DE $ and $ CG $.[/hide]

Let the ray $DN$ intersect the circle $\omega$ with center in point $C$ and has radius $CE$, the second time in point $P$. \\ 1)$\angle CNP=\angle DNA=\angle DEN=\frac{\angle FCN}{2} $ so $\angle FNP=\angle FNC+\angle CNP=90^{\circ}-\frac{\angle FCN}{2}+\frac{\angle FCN}{2}= 90^{\circ}$
because $EC=FC=NC$. And since  point $C$ is the center of $\omega$, we must have points $F,C,P$ collinear. \\ 2)$\frac{2FC}{FD}=\frac{FP}{FD}=\frac{\sin\angle FDP}{\sin \angle FPD}=\frac{\sin\angle FDP}{\sin \angle DEN}=\frac{EN}{DN}$\\ 3) Let the incircle touch $AB$ and $AC$ at points $Z,N$ respectively. Let  point $M$ be the middle of $ZE$. Then since $BD$ is the symmedian of $\angle ZDE$, we have $\angle ZDM=\angle GDF=90^{\circ}-\angle MDE-\angle \frac{ABC}{2} $.Also we find that $\angle EZD=\angle DNA+\angle ENC=90^{\circ}-\angle \frac{FCE}{2}$.\\ 4)$\angle ZMD =180^{\circ}-\angle ZDM-\angle DZM=\angle \frac{ABC}{2}+\angle \frac{FCE}{2}+\angle MDE$.\\$
\angle DZA=\angle ZED =\angle ZMD -\angle MDE$. \\ 5)
$\angle DGF= \angle FCE+\angle GBC=\angle FCE+\angle ABC-\angle  ZBD=\angle FCE+\angle ABC-\angle AZD+\angle ZDB=\angle FCE+\angle ABC-\angle ZMD +\angle MDE+\angle ZDB=\angle \frac{ABC}{2}+\angle \frac{FCE}{2}+\angle ZDB . $. So we found that $\angle DGF=\angle ZMD$ and $\angle ZDM=\angle GDF$ . Which gives us that $\triangle GDF\sim\triangle MDZ$.\\ 6)So $\frac{2GF}{DF}=\frac{2ZM}{ZD}=\frac {ZE}{ZD}=\frac{EN}{DN}=\frac{2FC}{FD}$, because $AE$ is the symmedian of $\angle ZEN$.\\ From (6) we find $GF=FC$ and we are done.

Projective geometry yields the following quick solution.
Let the incircle touch $AB$ and $AC$ at $F$ and $G$ respectively. Also, let $BC \cap GF \equiv X$ and $AE \cap GF \equiv Y$.
By the Symmedian Lemma, $DFEG$ is harmonic. This implies that $XD=XE$ from tangency, thus $XD$ is parallel to $CF$. Therefore we can write
\[ -1 = (X, Y; G, F) = (X, E; C, B) = (\infty, F; C, G) \]
and the conclusion follows.

[hide=Solution][quote=China TST 2008 P1]Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.[/quote]
[b][color=#000]Solution:[/color][/b] Let $(I)$ be tangent to $AB,AC$ at $I_B,I_C$. Let the tangent at $D$ to $(I)$ intersect $BC$ at $X$. Since, $DI_BEI_C$ is a Harmonic Quadrilateral, hence, $X$ $\in$ $I_BI_C$. From the Statement, $\angle CEF$ $=$ $\angle XDE$ $=$ $\angle CFE$ $\implies$ $XD$ $||$ $CF$.
\begin{align*} -1= (I_C, BI_C \cap (I) ; E, I_B) \overset{I_C}{=} (C, B ; E, X) \overset{D}{=} (C, G; F, \infty_{CF}) \end{align*}
This implies, $CF=FG$ $\qquad \blacksquare$
[/hide]

Hmm easy even for a P1
[quote=Fang-jh]Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.[/quote]
Let the incircle touch $AC,AB$ at $G,H$ respectively.Let $T=GH\cap BC$.Then since $GHDE$ is harmonic hence $TD$ is tangent to the incircle.Hence $TD=TE\implies TD\parallel CF$.Now \[-1=(T,E,B,C)\overset{D}{=}(\infty_{CF},F,G,C) \implies CF=FG\square\]


Prove that a line through M to bisect any side of the quadrilateral is perpendicular to the opposite side.(M is not hte centre of the circle.)

[quote="Arne"]This is the Brahmagupta theorem, you can search for it.[/quote]
Are you sure I think I can do it without it.
[hide="Solution"]
As the diagonals of the quadrilateral are perpendicular to each other then the quadrialteral is a rhombus or a square.
It is a square in this case because in a cylic quadrilateral the sum of the opposite sides is $180^\circ$ So, then a perpendicular to one side will be parallel to the adjacent side and hence it would be perpendicular to side opposite the bisected side.
[/hide]

[quote="Arne"][quote="riddler"]
Are you sure I think I can do it without it.[/quote]

Read my post carefully -
I said that it IS the Brahmagupta theorem, not that you have to use it...

[quote="riddler"]As the diagonals of the quadrilateral are perpendicular to each other then the quadrialteral is a rhombus or a square.
[/quote]

And that is false.[/quote]
As for the $1$st thing sorry :blush:  but as for the $2$nd thing I fail to understand why it wouldnt be a rhombus or a square.

[quote="riddler"]...I fail to understand why it wouldnt be a rhombus or a square.[/quote]

If you think it's true, then you have to prove it. Otherwise it remains a conjecture.

[quote="Arne"]Take the points (1, 0), (-3,0), (0, 5) and (0, -2). Do these points form a quadrilateral? Are the diagonals perpendicular? Is the quadrilateral a rhombus?[/quote]
 :oops:  :oops:  :oops:  :oops:

SO this question is the brhmagupta theorem....it's in our 10th grade math school text book as a question...thnx for tellin me 'll search :lol:


1. An isosceles trapezoid has lengths of 5,5,5 and 8. What is the sum of the lengths of its diagonals?

I got an answer of [hide]2 :sqrt: 65[/hide]. I believe that this is a cyclic quadrilateral with can then be used with the theorem: ab+cd=ef, (Ptolemy's?) Can anyone shed light on this?

Your answer is right, it is $2 \sqrt{65}$. Actually, what you used is

[b](a)[/b] the fact that every isosceles trapezoid is cyclic;
[b](b)[/b] the Ptolemy theorem for cyclic quadrilaterals;
[b](c)[/b] the fact that the diagonals of an isosceles trapezoid are equal in length.

But note that with the usual notations (ABCD cyclic quadrilateral, AB = a, BC = b, CD = c, DA = d, AC = e, BD = f), the Ptolemy theorem is ac + bd = ef, and not ab + cd = ef as you wrote.

  Darij

Oh, I do apologize for writing the wrong thing for Ptolemy. I did mean that the a and the b were supposed to be opposite sides, same for the c and d. Thank you for verifying the answer though.


Let $ c_{n}$ be the number of permutations $ \sigma$ of the set $ \{1, 2, \ldots, n\}$ which have the property that there do not exist $ i < j < k$ such that $ \sigma(i) < \sigma(k) < \sigma(j)$.  Thus, for example, $ \sigma = 32541$ is not such a permutation because $ 2 = \sigma(2) < 4 = \sigma(4) < 5 = \sigma(3)$, but $ \pi = 34521$ is.  Compute $ c_{n}$.  (These are called $ 132$-avoiding permutations.)

Similarly, let $ d_{n}$ be the number of permutations $ \sigma$ of $ \{1, 2, \ldots, n\}$ which have the property that there do not exist $ i < j < k$ such that $ \sigma(i) < \sigma(j) < \sigma(k)$.  Thus, $ \sigma = 32541$ is such a permutation but $ \pi = 34521$ is not because $ 3 = \sigma(1) < 4=\sigma(2) < 5 = \sigma(3)$.  Compute $ d_{n}$.  (These are called $ 123$-avoiding permutations.)

Maybe it helps:
[url]http://www.combinatorics.org/Volume_13/PDF/v13i1r51.pdf[/url]


Un tetraedru $ ABCD$ este inscris intr-o sfera de raza $ R$ si are baza $ ABC$ un triunghi echilateral.Se stie ca sfera de raza minima care contine $ ABC$ si care admite pe $ D$ ca punct interior are raza egala cu $ m\equal{}\frac{R}{2}$ sa se arate ca V =$ m(6m^2\minus{}DA^2\minus{}BD^2\minus{}CD^2)$

Ceva nu-i in regula cu enuntul! Sfera cea mai mica care contine atat triunghiul ABC cat si punctul D,
contine tetraedrul ABCD. Insa cea mai mica sfera care contine tetraedrul ABCD este sfera circumscrisa!

$ \mbox{In tetraedrul }A_1A_2A_3A_4, \mbox{ notand cu } a_{ij} \equal{} |A_iA_j| \mbox{ si cu } \\
t \equal{} \frac {a_{12}a_{34} \plus{} a_{13}a_{24} \plus{} a_{14}a_{23}}{2}\mbox{; avem: } \\
6.R.V \equal{} \sqrt {t(t \minus{} a_{12}a_{34})(t \minus{} a_{13}a_{24})(t \minus{} a_{14}a_{23})}.\ \ (1) \\
\mbox{In cazul in care triunghiul }A_1A_2A_3 \mbox{ este echilateral, avem: }\\
a_{12} \equal{} a_{13} \equal{} a_{23} \equal{} m \mbox{, iar relatia }(1) \mbox{ devine: } \\
24.R.V \equal{} m^2.\sqrt {(a_{14} \plus{} a_{24} \plus{} a_{34})(a_{24} \plus{} a_{34} \minus{} a_{14})(a_{14} \plus{} a_{34} \minus{} a_{24})(a_{14} \plus{} a_{24} \minus{} a_{34})} \\
\mbox{sau: }
24.R.V \equal{} m^2.\sqrt {2(a_{14}^2a_{24}^2 \plus{} a_{14}^2a_{34}^2 \plus{} a_{24}^2a_{34}^2) \minus{} (a_{14}^4 \plus{} a_{24}^4 \plus{} a_{34}^4)}.$

Problema apare sub aceasta forma la ONM 1991.Intr-adevar ,la rezolvari, se specifica faptul ca o astfel de situatie e practic imposibila,insa exista o rezolvare (destul de neclara).

[quote="mihai miculita"] Insa cea mai mica sfera care contine tetraedrul ABCD este sfera circumscrisa!

[/quote]
Nu neapărat. De exemplu, cel mai mic cerc care conţine un triunghi obtuzunghic nu e cercul circumscris.

[quote="Adriana N."]Intr-adevar ,la rezolvari, se specifica faptul ca o astfel de situatie e practic imposibila,insa exista o rezolvare (destul de neclara).[/quote]
Daca te referi la culegerea de la editura Bîrchi, afirmatia "sfera de raza minima in general nu exista" cred ca e gresita. Solutia data acolo pare OK.


I faxed off my solutions in time, and it says that the results are available, but I don't have a score. Why not? How can I see what I got? I've gone to the "My Usamts" but it shows the - of it not being graded yet. :(

Please send an email with your full name, username, and USAMTS Id to usamts@usamts.org; if you didn't put your username or USAMTS Id on your paper, then we probably have the paper but it's in our stack of papers we can't map to id numbers.


994 cages each contain 2 chickens. Each day we rearrange the chickens so that the same pair of chickens are never together twice. What is the maximum number of days we can do this?

do you take guesses?

there are 994 cages, and 994 pairs of chiken. to count the maximum number of days, we need to move only two chickens a day. we also only need to move one chicken from each cage and fix the other chicken.

so on the first day, we move the chicken in cage 1 to 2 and 2 to 1. on day two, we move the chicken in box 2 to 3, and 3 to 2. keeping this up, on the 993rd day, the first chicken will be at the 994th cage, and every other chicken displaced one cage down.

then we count the chicken that is now in the 1 box, moving it up to the 993th box, we have 992 days to count.

then we count the third chicken, which is in the first box, we have 991 days.

doing this on to the 994th chicken, we should have 993+992+...+2+1 days, or $d=\frac{993(1+993)}{2}$.

EDIT: oops. i think the answer should be $d=993+991+989+987+...+5+3+1=\frac{497(1+993)}{2}=247,009.$

EDIT: actually i'm not sure any more.

please tell me if i reasoned incorrectly.

[quote="arsemesomething"]do you take guesses?

[hide="be warned"]there are 994 cages, and 994 pairs of chiken. to count the maximum number of days, we need to move only two chickens a day. we also only need to move one chicken from each cage and fix the other chicken.

so on the first day, we move the chicken in cage 1 to 2 and 2 to 1. on day two, we move the chicken in box 2 to 3, and 3 to 2. keeping this up, on the 993rd day, the first chicken will be at the 994th cage, and every other chicken displaced one cage down.

then we count the chicken that is now in the 1 box, moving it up to the 993th box, we have 992 days to count.

then we count the third chicken, which is in the first box, we have 991 days.

doing this on to the 994th chicken, we should have 993+992+...+2+1 days, or $d=\frac{993(1+993)}{2}$.[/hide]

please tell me if i reasoned incorrectly.[/quote]

Well considering you only have 1988 chickens, if you consider the possible pairs the first chicken could have and there are always two chicken in a cage, the maximum possible number of days is only 1987.. :maybe: maybe I have the question wrong
But if your method did work, then I think you'd have to prove that by moving the first chicken you can achieve the maximum..so moving chickens might not be the best idea  :|


From 2001 National Sprint Round, # 25:

The lengths of the perpendiculars drawn to the sides of a regular hexagon from an interior point are 2, 4, 5, 9, 10, and 12 centimeters. What is the number of centimeters in the length of a side of this hexagon? Express your answer as a common in simplest radical form.

This problem has been posted before, I suggest you go search for it (you'll probably find a better explanation than mine), but if you're too lazy (like me), you can read my solution:

[hide]Use the area of a triangle 6 times ($\frac{bh}{2}$), so you get the area of the hexagon to be $\frac{(2+4+5+9+10+12)s}{2}=21s$, where s is the length of a side. Then you find it using equilateral triangles (since it is a REGULAR hexagon), $\frac{s^2\sqrt{3}}{4}\cdot6=\frac{3s^2\sqrt{3}}{2}$. Then you do
$21s=\frac{3s^2\sqrt{3}}{2}$
$s=\frac{14\sqrt{3}}{3}$
[/hide]


Sea $T$ un conjunto de $2007$ puntos en el plano donde no hay 3 puntos colineales. Sea $P$ un punto cualquiera de $T$. Pruebe que el numero de triangulos que contienen a $P$ en su interior y cuyos vertices estan en $T$ es par.

[color=darkblue]
Sea $\Delta_{P}$ el numero de triángulos que contienen a $P$ en su interior. Ahora veamos cada grupo de cuatro puntos cualesquiera de $T$ distintos de $P$. Si un triángulo tiene en su interior el punto $P$ este triangulo se encontrara en un total de 2003 conjuntos. Asi sumados de cada conjunto obtendriamos en total $2003 \Delta_{P}$.

Ahora por otro lado en cada conjunto de cuatro puntos tenemos que estos formaran o bien un cuadrilatero convexo, o bien un cuadrilatero no convexo que lo podemos ver como un triángulo sub dividido en tres triangulitos internos, es facil ver que cada conjunto de cuatro puntos tiene o bien 0 triángulos que contienen a $P$ en su interion o 2 triángulos que contienen a $P$ en su interior. Si sumamos el total de tráingulos  en cada conjunto de cuatro puntos obtenemos un numero par, digamos $2k$.

Tenemos por las dos formas de contar que $2k = 2003 \Delta_{P}$.

De donde se concluye que $\Delta_{P}$ es par.[/color]

 :ninja:   :pilot:


Let $ \triangle ABC$ and $ \triangle DEF$  be inversely congruent.Show that midpoint of $ AD,BE,CF$ are collinear

Very nice fact and Lemma 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=564145#564145]http://www.mathlinks.ro/Forum/viewtopic.php?t=82897 post #7[/url].

  Darij


what is the greatest area for a square that can fit in a rhombus with side 20

express you answer to the nearest whole number

[hide]Area (square) =< Area (rhombus)

say an angle of the rhombus is x, the area is 400*sin(x)

if x=90, rhombus is at a max,

Area(square) =< 400

equality does occur since if Area (rhombus)=400, x=90, the rhombus is the square... $\boxed{400}$[/hide]

oops...sorry

i forgot to mention that one of the angles of the rhoumbus is 50 degrees

 :blush:


0.000100010001.... repeating
is written in base sqrt(2).

What is this number in base 10?

of course drunner is correct, but the problem statement is confusing because i don't see how you can have digits for base $\sqrt{2}$

Since 1 is less than sqrt(2), using the digit 1 seems legit, but yes weird.

In what base b>0, does 1000 = 110?

(Note for bases b less than your answer, 1000<110.)

[quote="mathgeniuse^ln(x)"]Um, how can a number equal to another number in any base?  Am I missing something here?[/quote]

uh, $10_2=2_{10}$. A number in base 2 is equal to another number in base 10.

Is that what you want?

we have $b^3 = b^2 + b$, so rewrite it to be $b(b^2 - b - 1) = 0$, since base 0 and base negative is absurd, I think the answer is $\frac{1+\sqrt{5}}{2}$.

Good job.  BTW, yes both numbers were in the same base, b.  Weird things happen for non-integral bases. :D
Note that the solution for b is the Golden Ratio :)

I did specify b>0, Pekerichang, but you shouldn't have mentioned the 'absurdity' of negative bases :lol:

A few more challenges!
Using only 0 and 1 (and no negative signs), in base negative 2:

1.  Count from 1 to 10
2.  Count from -1 to -10
     (Note: in negative bases, all integers can be written WITHOUT the negative symbol!)
3.  Convert the base ten numbers 100 and -100 to base negative 2.
4.  Write one-third in base negative two.
5.  Write negative one-third in base negative two.

Here's a blurb from Keith Devlin, a British author, about bases in general.  http://www.maa.org/devlin/devlinfeb.html

BTW: Keith presents an additional 'base' problem at the end of his article.  No solution given and I can't solve it.  Anyone?

I'll do #1:

[hide]1, 110, 111, 100, 101, 11010, 11011, 11000, 11001, 11110.[/hide]

Wouldn't it be fun to always use negative bases? They're more versatile since you no longer need negative signs.

The notion of "base" is being stretched very badly here.  Any reasonable definition of the word base requires that there be [i]unique[/i] representations in that base.  If you want to talk about base-like representations, go ahead, but these are not bases in any conventional sense of the term.

Yes, 'useful' bases probably would require the 'extra' property of unique representation.  But who would have appreciated the need for the extra property without playing around with the base-like b's?  Ergo, 'weird bases' challenge us to think more carefully about exactly what a base is (or should be).

As far as I can tell, representations in negative integer bases are unique.  In addition, the Devlin article mentions that some 1950's UMC-1 computers were developed and sold that were based on base negative 2.  Actually, I've liked them since a high school friend of mine did a neat class presentation on them.

BTW, any takers on 1/3 or -1/3?  Both answers are pretty nice.

With two's complement you can do any integer without negative signs (and i think you can do fractions?).

My guess is that programmers quickly realized that twos complement was a much more practical way to "avoid the negative sign" than base negative 2.

I believe one of the USACO online competition program is to convert a number from base 10 to base negative 2


We throw a dice.  What is the probability of seem each number after $ n$ rolling?

I think this is what you're asking:  "We throw a dice.  What is the probability of seeing a given number after $ n$ rolls?

[hide]The probability of seeing a given number on the first roll is $ \frac{1}{6}$, and the probability of not seeing it is $ \frac{5}{6}$.  This probability remains the same for each subsequent roll (if the number hasn't already been rolled).  The answer to your question (I think) is $ 1\minus{}(\frac{5}{6})^n$.[/hide]

I do not mean that $ cf249$. I want to  see  all of 1,2,3,4,5,6 after rolling dice $ n$ times. Of course $ n>6$.

OK, then, I think this is the same as the first question under [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186425]this topic.[/url]

It's a different question as I see it; we want to know the probability that it works within the first $ n$ steps, which is essentially the cumulative distribution.

It will be complicated.

The probability of not seeing the number $ 1$ is $ \left(\frac56\right)^n$. The probability of not seeing $ 1$ or $ 2$ is $ \left(\frac46\right)^n$, and so on. Build an inclusion/exclusion sum; the probability of getting everything is $ \sum_{S\subset \{1,\dots,6\}} (\minus{}1)^{|S|}\left(\frac{6\minus{}|S|}{6}\right)^n$
$ \equal{} 1\minus{}6\left(\frac56\right)^n\plus{}15\left(\frac46\right)^n\minus{}20\left(\frac36\right)^n\plus{}15 \left(\frac26\right)^n\minus{}6\left(\frac16\right)^n\plus{}\left(\frac06\right)^n$

This works for all $ n$, as long as we interpret the last term as $ 1$ when $ n$ is zero.


geometry
circumcircle
trigonometry
angle bisector
perpendicular bisector
geometry unsolved

Let $\Gamma$ be a circle with radius $r$.  Let $A$ and $B$ be distinct points on $\Gamma$ such that $AB < \sqrt{3}r$.  Let the circle with centre $B$ and radius $AB$ meet $\Gamma$ again at $C$.  Let $P$ be the point inside $\Gamma$ such that triangle $ABP$ is equilateral.  Finally, let the line $CP$ meet $\Gamma$ again at $Q$.

Prove that $PQ = r$.

Let $\triangle A_0BP_0$ be an equilateral triangle inscribed in the circle $\Gamma$ with the center O. Let the point A move on the minor arc $A_0B$. The equilateral triangles $\triangle ABP \sim \triangle A_0BP_0$ remain similar, the vertex B is fixed and the vertex A moves on the circle $\Gamma$. It follows that the vertex P also moves on a circle $\Gamma'$. When $A \longrightarrow A_0,\ P \longrightarrow P_0$, when $A \longrightarrow B,\ P \longrightarrow B$ and when $AB \longrightarrow r,\ P \longrightarrow O$. Therefore, the locus circle $\Gamma'$ is the circumcircle of the isosceles triangle $\triangle BOP_0$ with the angle $\angle BOP_0 = 120^\circ$. It follows that the circles $\Gamma' \cong \Gamma$ are congruent, the center O' of $\Gamma'$ lies on $\Gamma$, and the angle $\angle BOO' = 60^\circ$. Hence, the lines $OO' \perp BP_0$ are perpendicular and the line OO' meets the circle $\Gamma$ again at $A_0$ and $OA_0 = r$.

Since the angle $\angle BAP = 60^\circ$, the line AP passes through $P_0$. The angles $\angle CAP_0 = \angle CBP_0$ spanning the same arc $CP_0$ of $\Gamma$ are equal. On the other hand, $\angle CAP_0 \equiv \angle CAP = \frac{\angle CBP}{2}$, because $\angle CBP$ is the central angle of the major arc CP of the circle (B) with radius AB and the angle $\angle CAP$ spans this arc. Consequently, $BP_0$ bisects the angle $\angle CBP$. Since the triangle $\triangle CBP$ is isosceles with CB = PB, the angle bisector $BP_0 \perp CP$ is perpendicular to its base CP. As a result, the lines $OA_0 \equiv OO' \parallel CP$ are parallel, both being perpendicular to $BP_0$. But then form the congruence of the circles $\Gamma' \cong \Gamma$ with their centers separated by $OO' = r$, we get $PQ = OA_0 = r$.

This problem is simply and nicely. [hide="Here is a short proof."][color=darkred]Denote $2x=m(\widehat {CBP})$ and calculate the values of the all angles from this figure. Suppose w.l.o.g. $CB=BA=AP=PB=1$. Thus, $m(\widehat {BPC})=m(\widehat {BCP})=90^{\circ}-x$, $m(\widehat {APQ})=30^{\circ}+x$, $m(\widehat {BAC})=m(\widehat {BCA})=60^{\circ}-x$, $m(\widehat {ACQ})=m(\widehat {PBQ})=\boxed {m(\widehat {ABQ})=30^{\circ}}$, $BQ\perp AP$, $AQ=PQ$ $\Longrightarrow$
$m(\widehat {CAP})=x$, $m(\widehat {AQB})=m(\widehat {BQP})=60^{\circ}-x$.
Therefore, $1=AB=2r\sin \widehat {BCA}=2r\sin (60^{\circ}-x)\Longrightarrow \boxed {r=\frac{1}{2\cos (30^{\circ}+x)}}\ .$
$\frac{PQ}{\sin \widehat {PBQ}}=\frac{PB}{\sin \widehat {BQP}}\Longrightarrow \frac{PQ}{\sin 30^{\circ}}=\frac{1}{sin (60^{\circ}-x)}\Longrightarrow \boxed {PQ=\frac{1}{2\cos (30^{\circ}+x)}}\ .$
At last, we draw the conclusion $\boxed {\ PQ=r\ }\ .$[/color][/hide]

Since $ABP$ is equilateral, $P$ lies on the circle with centre $B$ and radius $AB$.
By the Inscribed Angle Theorem, $\angle ABQ=\angle ACP=\frac{\angle ABP}{2}=30^{\circ}$, which means $BQ$ is the perpendicular bisector of $AP$.
By the Sine's Law, $PQ=AQ=2r\sin\angle ABQ=2r\sin30^{\circ}=r$, QED.


Find all integers  satisfying $y(x-y)(x+2y)\neq0$ and $(x-y)^{4}+4y^{2}(x+2y)^{2}=z^{2}$.

You should be able to get all solutions by writing $z=r^{2}+s^{2}$, $r^{2}-s^{2}= (x-y)^{2}$, and $rs = y(x+2y)$ for some positive integers $r\ne s$.


Let $ u_1,u_2,...,u_n(n>2)$ be a sequence of positive real numbers such that

1)$ \frac{1004}{k}\equal{}u_1\ge u_2\ge ....\ge u_n$ for some positive integer $ k$
2) $ u_1\plus{}u_2\plus{}...\plus{}u_n\equal{}2008$
Prove that it is possible to select $ k$ elements from the set $ (u_n)$ such that in this collection of $ k$ numbers, the smallest one is at least half of the largest

Not really hard,if I got correctly.
assume the contrary at first,afterwards note that the sum $ 1004(1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\dots)$ is less than $ 2008$.
the last step is to divide all these $ n$ numbers into the groups,each with $ k\minus{}1$ elements,(the last one may contain less number of elements).
I can elaborate it,if necessary.


superior algebra
superior algebra unsolved

Show that :
$ V$ is simple left $ R$-modul  if and only if exist $ M$ maximal left ideal of $ R$ and $ V\simeq R\backslash M$

Which direction are you having trouble with?

Construct a nonzero map $ R\to V$.  Why is it surjective?  If $ V\cong R/I$, why must $ I$ be maximal?


Let $BC$ meet perpendicular line from $A$ on $BC$ at $D$ in triangle $ABC$. Let $E$ and $F$ be incenters of triangles $ABD$ and $ACD$ respectively. Let $EF$ meet $AB$ and $AC$ at $K$ and $L$ respectively. Prove that $AK=AL$ iff $AB=AC$ or $\angle BAC = 90$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=78401 .

  darij


So I got inspired from Poland Inequality problem so I decided to write one up. It follows from exactly same method as the original problem so it's probably very easy but I think proofs shouldn't be at Intermediate so I'm posting here.  :D

For real numbers $ x,y,z,w$, prove that:

\[ (x\plus{}y\plus{}z\plus{}w)^{2!} \plus{} 2! w^{\frac{3!}{2!}} \leq 2! \left(x^{2!} \plus{} y^{2!} \plus{} z^{2!} \plus{} w^{2!}\right) \plus{} \frac{4!}{3!} \left((x\plus{}1)(y\plus{}1)(z\plus{}1) \minus{} (x\plus{}y\plus{}z\plus{}xyz\plus{}1!) \plus{} \frac{w^{\frac{3!}{2!}}}{2!}\right)\]

What about $ x\equal{}y\equal{}z\equal{}0$, $ w \rightarrow \infty$?

Then it reduces to $ w^2 \plus{} 2w^3 \le 2w^2 \plus{} \frac{2w^3}{3}$, and $ w^3$ clearly grows faster...

(also, what is up with the factorials?)

Yeah obviously I haven't realized that condition (this is my FIRST one) so ignore that...

And there shouldn't be $ \frac{2w^3}{3}$ on RHS. $ 4! \equal{} 4*3*2*1$ and $ 3! \equal{} 3*2*1$ so $ 4!/3! \equal{} 4$ and you should end up with $ 2w^3$ on RHS as well (which cancels LHS's one).

Factorials are nothing. I just added them at the last minute to make problem more fancy.  :rotfl:

[quote="Silverfalcon"]Factorials are nothing. I just added them at the last minute to make problem more fancy.  :rotfl:[/quote]
Yeah, it looks quite silly with all the factorials  :D

[hide="As for the problem,"]after simple expanding it reduces to $ (x\plus{}y\plus{}z\minus{}w)^2\geq 0$.[/hide]

:D haha.

[hide]
My intended solution was (by Cauchy)

$ ((x \plus{} y \plus{} z) \plus{} w)^2 \leq 2((x \plus{} y \plus{} z)^2 \plus{} w^2)$ [/hide]


pigeonhole principle
algebra open
algebra

guys try to solve

If n is odd and a_1; a_2;.......... ; a_n are the numbers 1;
2;...............; n arranged in some order, prove that the product
(a_1 - 1)(a_2 - 2)...........(a_n - n) must always be even. :?:

Assume that all $ a_i\minus{}i$ are odd numbers. We have $ \sum(a_i\minus{}i)\equal{}0$. But $ \sum(a_i\minus{}i)$ is odd (it follows from n-odd). Contradiction.

use the pigeonhole principle.

in the numbers 1,..., n, if n is odd, there will be (n+1)/2 odd numbers and (n-1)/2 even numbers.

among the 2n numbers a1, a2, ...., an and 1, 2, 3, ..., n, there are n+1 odd numbers. but there are only n places in which to fill them. therefore, one of the factors will contain two odd numbers, whose difference is even.

therefore, the product is even.


[hide="Click for solution"]
The distance from the center of the circle to the intersection point of the tangents is $ \frac{3}{8}$, so $ CD\equal{}\frac{3}{8}\plus{}\frac{3}{16}\equal{}\frac{9}{16}\equal{}x\plus{}\frac{1}{2} \implies x\equal{}\frac{1}{16}$ or $ \boxed{\textbf{(E)}}$.
[/hide]