algebra
polynomial
modular arithmetic
complex numbers
superior algebra
superior algebra solved

Prove that the quotient ring $C(x,y)/P(x,y)$ isn't a field, where $C(x,y)$-the ring of the polynomials in two variables, $P(x,y)$-some elemnt from $C(x,y)$. By the way,the elements of the factorazitaion are said to be equal if their difference is divisible by $P(x,y)$  (I don't know whether the word "quotient ring" is correct, because in my books it called "factor ring")

That just means that the ideal generated by $P$ is not maximal. In other words, we have to show that we can find another polynomial $Q$ with $P\not |\ Q$, s.t. the ideal generated by $P,Q$ is not the entire ring.

In particular, it suffices find $Q$ s.t. the identity $P(x,1)F(x,1)+Q(x,1)R(x,1)=1$ doesn't hold identically for any $F,R\in\mathbb C[x,y]$ (I supose $\mathbb C$ is the set of complex numbers? :?). If $P$ has degree more than $1$ in $x$, take $Q$ s.t. $Q(x,1)$ is a divisor of $P$. If $P$ ahs degree $\le 1$ in $x$ but $>1$ in $y$, do the same for $y$. If its degree is $\le 1$ in both variables, then it's easy to work out manually :). I haven't looked at this case properly.

P.S. This doesn't work if by $C(x,y)$ you didn't mean $\mathbb C[x,y]$, the ring of polynomials in two variables with [b]complex[/b] coefficients. Is this what you meant, or is your $C$ something else?

By the $C(x,y)$ i meant $\mathbb{C}[x,y]$- so wuth you solution there are no problems
I just want you to find the mistake in my solition, because it seems to me that it's almost trivial and it must be wrong
Supposing contary for every ${(f+P)\in{\mathbb{C}[x,y]}/P[x,y]}$ there must be inverse $(g+P)$ s.t.
$(f+P)(g+P)=1$, taking there $f=1-P$(which is not divisible by P apart from the trivial cases) we get
$g=1-P$, so $(1-P)^2=1$, so $P^2-2P=1$ which is possible only when $P$-constant

But the fact that the quotient ring is a field means simply this: given any $f$ which is not divisible by $P$, we can find another polynomiual $g$ s.t. $fg-1$ is divisible by $P$. Maybe I am misunderstanding what you wrote, but it seems to me like you think you have to find $g$ s.t. $fg=1$ not just $fg\equiv 1\pmod P$. Your counterewxample does not work: you simply have $f\equiv g\equiv 1\pmod P$.

thanks,grobber, for showing me my stupidness-yes, of course it needs to be $fg\equiv1(mod P)$ i forgot about equivalences
thanks


um.......this might be a weird request, but can someone start a game, like beat the mod, mafia, etc? I dont have time every day, and plus, my friends say i suck at story lines, :lol: o well........ but, can someone plz start one? o, and i join the newest game...........

There are ongoing and new games all around you in this forum. Take a look at them.

any NEW games?

i dnt relly see n e......

Try the AoPS apprentice. I THINK that's a parody of "The Apprentice" on television, but I could be wrong...

Also Empire* is still open for signups, although the next time anyone can get in is around turn 35.


(i)Each point on the side of a square is coloured one of n colours. The square has side lengths of magnitude 1, and two points that are apart by distance one are of differing colors. Find the smallest such n.
(ii) Solve the problem in (i), with the change that every single point inside the square and on the boundary are to be coloured one of n colors.
- - -
The following problem is incredibly simple and a routine counting problem and perhaps shouldn't be on this forum, but just checking that I've got the correct answer:

There are 7 lanes in a shop where one can check out, labeled from 1 to 7; but only lanes 1 to 4 are able to accept credit cards. Three shoppers A,B and C are shopping, and while A intends to use his credit card, B and C intend to pay with cash instead. How many different ways are there of the three shoppers to stand in line (Note two people can be in the same column, and if this is the case order does matter.)

For the first part of the first question, the answer is $3$:

Let $ABCD$ be the square and take $X_a\in AB,X_b\in BC,X_c\in CD,X_d\in DA$ s.t. $X_aX_b=X_cX_d=1$, and both $X_aX_b,X_cX_d$ make angles of $45^{\circ}$ with the sides of the square. Color $[AX_a],[AX_d),[CX_c],[CX_b)$ with color $1,\ [BX_b],[BX_a)$ with color $2$, and $[DX_d],[DX_c)$ with color $3$.

That shows that $3$ colors are enough. On the other hand, we need at least $3$ colors, since it's easy to inscribe an equilateral triangle of side $1$ in a unit square, and the vertices of such a triangle must have different colors.


Make one single group go from A to Z and back to B. This is 50 letters long. So, now we divide 2005 by 50 and get 5 as a remainder. Look at the sequence- A,B,C,D,E. E is 5th in a group so [b]E[/b] is the letter, I think.

When Ann first counts from A to Z, she counts 26 letters. After that, from Y to A there are 25 letters and from B to Z there are also 25 letters. So $ 25x\equal{}2005\minus{}26\equal{}1979\implies{x\equal{}79R4}.$ Since 79 is odd, Ann will be counting from B to Z. Thus, E is the 2005th letter.


watch this [url]http://www.youtube.com/watch?v=1VJwn-At244[/url] and the related videos
"Just wait", said President Bush.

Um, well, Saddam killed thousands of innocent people in cold blood, and I don't think Putin is capable of doing that....  Speaking of President Bush, on a different topic, anyone seen the videos of him grabbing the German ambassador, or cursing with Tony Blair (all at the G8)?

I think Mr. Putin said that: Moscow doesn't want the kind of violence-plagued democracy the United States has fostered in Iraq.

So why do you think that he shall be the next Saddam? Well, anyway, I believe that he will not be the next Bush who put the kind of violence-plagued democracy in Iraq.

WHEN PUTIN STRIDES FOR THE UNCONSTITUTIONAL THIRD TERM, SUPER KASPAROV IS GOING TO FLY DOWN AND UNLEASH DISCOVERD SLAP!

[quote="shobber"]I think Mr. Putin said that: Moscow doesn't want the kind of violence-plagued democracy the United States has fostered in Iraq.[/quote]
Of course, Putin doesn't want any kind of real democracy in Russia.

[quote="Buffalo"]I never know that US invaded Iraq because Saddam killed thousands of innocent people in cold blood, what I do know is, USA troops killed tens of thousands or more, in boiled blood, maybe. [/quote]
Are you familiar with the word, "non-sequitur?"

[quote="mysmartmouth"]Um, well, Saddam killed thousands of innocent people in cold blood, and I don't think Putin is capable of doing that.[/quote]
Why on earth not?

Bush is indeed a really interesting president!

http://www.youtube.com/watch?v=FOomGp8AnH8&search=bush%20G8

Well the title doesn't really describe your question.

Does he act like him, or will he end up like him?

Russia is no perfect democracy, I recently saw a documentary where they managed to interview one of his spin doctors, and I frowned at their practices (especially if you know that this was what the people were still allowed to see).

Did he kill innocents?  Well some say his war in Chechnya (the second war) was meant to be a short war to gain popularity, and some even dare to say the bombing of buildings in moscow in 1999 was connected to him somehow.

But the truth is : USA does not attack countries that can really fight back.  If Saddam really had WMD that could be deployed in 45 minutes, they wouldn't have dared.  They will never attack Russia.
Don't forget Reagan once said (unintentionally) on the radio that they were gonna bomb Moscow in ten minutes.
[url]http://politicalhumor.about.com/cs/quotethis/a/reaganquotes.htm[/url]


Hm...no real biology "problems" yet. I'll take the honors of becoming the first.

Question:

What does D.N.A stand for?

[quote="Valentin Vornicu"]At least this I know: DNA = Desoxyribo Nucleic Acid :D[/quote]Hehe, at least you were close.  :D

yeah deoxyribonucleic acid...some say it had an even longer formulaic name originally...

nucleic acids carry genetic information, which is present in every cell. They have a high molecular weight and are composed of C, H, O, N.
just a question, does deoxyribonucleic acids have $O$?

of course theres oxygen in DNA, it has lots of phosphates $\text{PO}^{3-}_{4}$ which also makes it negatively charged.

[quote="rippledance"]*laughs* my friends and I have a CHON dance... =D[/quote]
What?! lol How do you do it? I want to learn!

DNA stands for the word deoxyribonucleic acid.

DNA...... :mad: don t ask about htat non-sense thread who thinks he knows everythin and poke shis nose in all the lifeprocesses :ninja: if i find who find somebody who gave all the control to that DNA i will....beg him to exlplain me how these four nitrogen bases with that five headed sugar and phosphate are ruling everything  :rotfl:

D.N.A. stands for deoxyribonucleic acid.


Solve in N the equation:

(1+x!)(1+y!)=(x+y)!.

If a,b,c,d are pozitive real numbers then we have:

sqrt((a+b)/(c+d))+sqrt((b+c)/(a+d))+sqrt((a+c)/(b+d))>=4(a+b+c)/(a+b+c+d).

Solve 1/x*5^x+x*5^(1/x)=10.

Solve in N the equation:

(1+x!)(1+y!)=(x+y)!.

[b]Solution:[/b]

we see that if x>=2 and y>=2 then both (1+x!) and (1+y!) are odd, so their product is odd so we can never have(1+x!)(1+y!)=(x+y)! in this case! :D
WE consider that x=1. So we get that 2(1+y!)=(y+1)!. SO for y>=3 we get that 3 | (y+1)! but 3 not| 2(1+y!). So we get that y<3.
we get now the only solutions: S={(1,2);(2,1)}.

You should try numbering the problems when you post more than one in a post! :D :D

cheers! :D

If a,b,c,d are pozitive real numbers then we have:

sqrt((a+b)/(c+d))+sqrt((b+c)/(a+d))+sqrt((a+c)/(b+d))>=4(a+b+c)/(a+b+c+d).

[b]Solution:[/b]
apply the mean inequality for

sqrt((a+b)/(c+d)*1)>=2(a+b)/(a+b+c+d) and you do the same for
sqrt((b+c)/(a+d)*1) and sqrt((a+c)/(b+d)*1) and you get the ineq! :D :D
cheers! :D :D


IMO Shortlist
combinatorics solved
combinatorics

There are $2005$ students at a university. Some students join together to form several clubs( a student may belong to several clubs). Some clubs join together to form several societies (a club may belong to several societies). There are a total of $k$ societies. Suppose the following hols:

(a) Each pair of students is exactly in one club;

(b) For each stusent and for each society, the student is in exactly one club of the society;

(c) Each club has an odd number of students. In addition, a club with $2m+1$ students is in exactly $m$ societies.

Find all possible values of $k$.
[hide="Remark"]This problem is basically the same with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=258304#p258304]IMO Shortlist 2004, Combinatorics, Problem 1[/url][/hide]

Have a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=41116  :)

[quote="warut_suk"]By the way, what is IMOTC?[/quote]Read [url=http://www.mathlinks.ro/Forum/resources.php?c=78&cid=49]here[/url] about it. Continue discussion about the problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=41116 ;)


what is dimension 0? its not empty space, cause that's 3d

what is dimension 4? "space" or time?

is there such thing as 4D? if so explain it to me

But it's with a funny metric. I'd rather think of space in $n$ dimensions, for $n\ge 0$.

[quote="RobinHood3000"]4D, which comprises the known universe (I think) constitutes of a fabric called "space-time," which includes the three dimensions of space (height, width, depth) and the dimension of time.[/quote]
i doubt we have any access whatsoever to any part of the fourth dimension and we have access to time (not dimensions of it but stil...)

[quote="peter"]if we move freely in the 3d dimension and only one way in the 4th are we standing still in the 5th?[/quote]

my opinion:
if you only move one way in 4th dimension ur moving in max 4 ways or min. 1 way. That means you are not still in the 5th you are moving in1-4 directions.

1st dimension, 2nd dimenstion, 3rd dimension, 4th dimension, 5th dimesntion, 6th dimention, you guys are reallyy confusing me know.


Consider the interval [0,1] in R with the inherited metric. Consider the open cover consisting of [0, 1/10), (1/2,1], and, for each natural number n, the interval (1/(n+2), 1/n). Find the Lebesgue number of this cover.

I took the obvious finite subcovering consisting of [0, 1/10), (1/2,1] and (1/(i+2), 1/i) for i = 1, 2, ..., 9. I then considered the intersection of [0, 1/10) and (1/11, 1/9), or (1/11, 1/10) and divided the diameter of this interval by 2 to get 1/220. Does this work?

Probably, yes, if you are also able to show that this interval around any point fits into one of the covering intervals. "Probably", because I'm not sure whether the question really was about [b]the[/b] (as you wrote) or about [b]a[/b] (which would be much more agreeing with the general spirit of analysis) Lebesgue number, i.e., whether you need also to check that no larger number works.

Hmm I was wondering about that too. I don't think it's too hard to show that 1/220 is a lebesgue number, but finding the largest would seem annoying. I have seen one or two proofs of the corresponding lemma, and it's usually done by contradiction. The proof on wikipedia ([url]http://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma[/url]) takes a direct approach, but it's not all that clear how one would go about finding the lebesgue number prescribed in the proof in practice. Thanks for confirming and clarifying the problem.


My state emailed everybody the scores of the 181 individuals and 28 teams who participated at states (no names, just their rank and score) You can correspond the top lists released along with their score so I was wondering why we aren't allowed to post this yet the state released scores for everybody?

I think you can post your scores, but not the problems or any hint of the problems' topics.

That would give an indication of the difficulty of the test.  If I got a low score and still placed high, the test would be difficult.  If I got a high score and still placed low, the test would be easy.

I dont think your allowed to post scores.

[quote="rcv"]Do not discuss any problems from the MATHCOUNTS State Competition until further notice. Do not discuss your score. You may state your rank and whether or not you qualified for Nationals, but you may not solicit others to discuss any aspect of the State Competition.

[quote="Last year, MCrawford wrote:"]
Students who discuss any information including scores and difficulty level of the test will be put in the AoPS penalty box and not be allowed to post for a considerable period of time. [/quote]
...
[/quote]

From the Announcement.

The topic is only a couple of pages long, is it that hard to read it to find out what I'm asking?

[quote="ehehheehee"]Because states take tests at different times, another person from a state that has not taken the test yet could use your scores to find out the difficulty of the test.  You have taken the test, so it doesn't matter that you know your score.  But somebody who hasn't taken the test would have access this forum as well and see your scores.[/quote]

Exactly.

[size=200][color=blue]DO NOT, I REPEAT, DO NOT DISCUSS SCORES, PROBLEMS, OR DIFFICULTY OF THE 2006 STATE COMPETITION UNTILL AFTER THE COMPETITION WINDOW (should be by the end of March).[/color][/size]

[quote="Ignite168"]I know your not allowed to post scores, that's why I'm asking why my state just gave a list of everybody's score and the team scores.

Also, the difficulty of the test wouldn't really help you or hurt you. Saying "It was harder than last years" doesn't tell you what areas the test is mainly on, what kind of problems, what you need to study, etc. at all.[/quote]

Well it doesn't really matter than everybody in your state knows everybody else's score. The problem is if the information spreads to other states that haven't had their competition yet. I think that knowing whether or not the test will be easy is advantageous.

Yeah, you can talk about anything in team meetings, in your state, etc.

But:

1. Do not post it in the forums, even the communities (and private ones too!).

2. Do not talk about it in some mathcounts gathering with other states.

In my state, they wouldn't even tell anyone what the top score was, only our team's scores.  When they announced the countdown, they only talked about the tiebreakers, but nothing about scores.  I only know the scores because my mom figured them out somehow.

Meh...
The emcee told the whole crowd the score of the first place guy, basically. (He missed only X questions today!)
And this dude told me my score when I was filling out forms for nats.
But we didn't get ANY other scores, in fact the only other thing we DID get was team rankings.
We didn't even get indiv rankings...

I haven't had my state yet, and you posting your scores will help me.  We don't usually get our scores until a month after.  No one told me at the previous chapters either. Hmmm. It's a conspiracy!

[quote="Ignite168"]I know your not allowed to post scores, that's why I'm asking why my state just gave a list of everybody's score and the team scores.

Also, the difficulty of the test wouldn't really help you or hurt you. Saying "It was harder than last years" doesn't tell you what areas the test is mainly on, what kind of problems, what you need to study, etc. at all.[/quote]
As the Announcements state, and as others have repeated, [b][size=150]DO NOT DISCUSS ANY ASPECT OF THE STATE COMPETITION[/size][/b], except your own rank.

There is no point discussing this matter further.  Thread locked.


The question asked for the amount of gallons of paint, so:

$2120\div{450}=4.7\overline{1}$

So we need 5 gallons of paint.


Let $x,y \in R$ sastisfying that $x+y,x^{2}+y^{2},x^{4}+y^{4}\in Z$.Prove that $x^{3}+y^{3}\in Z$ :)

[hide="solution"]
$(x+y) \in \mathbb{Z}\Rightarrow (x+y)^{2}=x^{2}+y^{2}+2xy \in \mathbb{Z}$.
Since $x^{2}+y^{2}\in \mathbb{Z}$ also, this implies $2xy \in \mathbb{Z}$

$(x^{2}+y^{2}) \in \mathbb{Z}\Rightarrow (x^{2}+y^{2})^{2}=x^{4}+2x^{2}y^{2}+y^{4}\in \mathbb{Z}$
Since $x^{4}+y^{4}\in \mathbb{Z}$, also, this implies $2x^{2}y^{2}\in \mathbb{Z}$ also. Along with $2xy \in \mathbb{Z}$ we get that $xy \in \mathbb{Q}$.
Suppose $xy$ is not an integer. Then it can be expressed as $\frac{a}{2}$ with $a$ an odd integer. Then $2x^{2}y^{2}=\frac{a^{2}}{2}$- which is a contradiction since $2x^{2}y^{2}\in \mathbb{Z}$. Hence $xy \in \mathbb{Z}$.

Now $(x+y)^{3}=x^{3}+y^{3}+xy(x+y) \in \mathbb{Z}$ and $xy$, $x+y$ and hence $xy(x+y) \in \mathbb{Z}$. Hence $x^{3}+y^{3}\in \mathbb{Z}$.

[/hide]

[quote="lingomaniac88"][hide]$x+y\in\mathbb{Z}$, so $(x+y)^{2}\in\mathbb{Z}$.
We get $x^{2}+y^{2}+2xy\in\mathbb{Z}$.  Since $x^{2}+y^{2}\in\mathbb{Z}$, we see that $2xy\in\mathbb{Z}$, so $xy\in\mathbb{Z}$.  So $-xy\in\mathbb{Z}$, as well.  So we can say that $x^{2}-xy+y^{2}\in\mathbb{Z}$, too.
So $(x+y)(x^{2}-xy+y^{2})\in\mathbb{Z}$, meaning $x^{3}+y^{3}\in\mathbb{Z}$.
You don't even need to know that $x^{4}+y^{4}\in\mathbb{Z}$.[/hide][/quote]

um, that's not quite correct. You actually need $x^{4}+y^{4}$ to be an integer, otherwise you could end up with
[hide="this"]
\[x=\frac{3+\sqrt{7}}{2}, y=\frac{3-\sqrt{7}}{2}\]
\[x+y=3, x^{2}+y^{2}=8, x^{3}+y^{3}=\frac{45}{2}\]
[/hide]

[quote="cincodemayo5590"][b]Another Follow-Up (Why not  :P ):[/b]

After what point (if any) will $x+y, \ x^{2}+y^{2}, \ \ldots, \ x^{n}+y^{n}\in \mathbb{Z}\Longrightarrow x,y \in \mathbb{Z}$?[/quote]
[hide]Never.

[quote="Farenhajt"][quote="cincodemayo5590"][b]Follow-up:[/b]

Do $x+y \in \mathbb{Z}$ and $x^{2}+y^{2}\in \mathbb{Z}$ imply $x^{n}+y^{n}\in \mathbb{Z}$? Do they imply $x, y \in \mathbb{Z}$?[/quote]

1. Yes. Since $x^{n}+y^{n}$ can be expressed as a polynomial with integer coefficients in $x+y$ and $xy$, the conclusion follows.
[size=59]For those who wonder: Take the binomial expansion of (x+y)^n, rearrange it and then induct over n.[/size]

2. No. Take $x=1-\sqrt{2}, y=1+\sqrt{2}$. Then $x+y=2\in\mathbb{Z}$ and $x^{2}+y^{2}=6\in\mathbb{Z}$, but $x,y\notin\mathbb{Z}$[/quote]

We can use the counterexample from part 2 again: from part 1 we know that $x^{n}+y^{n}$ is an integer for all $n$, satisfying the conditions for any $n$, but $x$ and $y$ are not integers.[/hide]

[quote="randomdragoon"][quote="lingomaniac88"] we see that $2xy\in\mathbb{Z}$, so $xy\in\mathbb{Z}$.[/hide][/quote]
 :P
I nominate this for a "what's wrong with this proof" puzzle[/quote]

Sorry.  I must've thought I was working in $\mathbb{Q}$ for a second there... that's what happens when you think too fast.

[quote="cincodemayo5590"]After what point (if any) will $x+y, \ x^{2}+y^{2}, \ \ldots, \ x^{n}+y^{n}\in \mathbb{Z}\Longrightarrow x,y \in \mathbb{Z}$?[/quote]

[hide="Remark"] This is the general term of a second-order homogeneous linear recursion

$a_{n+1}= A a_{n}+B a_{n-1}$

With $A = x+y, B =-xy$.  The first two conditions imply both that the first two terms are integer and that the coefficients of the recursion are integer, so all subsequent terms are integer.  :)

A generalized version of this was the subject of an interesting AIME problem, and noting this recursion made for a very easy solution. [/hide]

[quote="cincodemayo5590"][quote="bibobeo"]I have a shorter idea.

2xy ; 2(x^2)(y^2)  :arrow: Z
~> 2xy(xy+1)  :arrow: Z
~> xy :arrow: Z

Btw can we prove that it's the same with n.

And if we have x+y ; x^2 + y^2 ; x^3 + y^3  ... Z. Can we prove that x^4 + y^4 ... Z?[/quote]

[hide]
Yes. $x^{3}+y^{3}\in \mathbb{Z}\ \text{ and }\ x+y \in \mathbb{Z}\implies x^{2}-xy+y^{2}\in \mathbb{Z}\ \stackrel{x^{2}+y^{2}\in \mathbb{Z}}{\Longrightarrow}\ xy \in \mathbb{Z}$. Then, $x^{4}+y^{4}-2(xy)^{2}= (x+y)(x^{3}+y^{3}-(x+y)(xy)) \in \mathbb{Z}\Longrightarrow \boxed{ x^{4}+y^{4}\in \mathbb{Z}}$. [/hide][/quote]

Hey, are you sure?
Just have a look at this.
x = 1/sqrt(2)      y = -1/sqrt(2)
then x + y = 0 ... Z
       x^2 + y^2 = 1 ... Z
       x^3 + y^3 = 0 ... Z
but x^4 + y^4  = 1/2  :roll:

[quote="cincodemayo5590"][b]Another Follow-Up (Why not  :P ):[/b]

After what point (if any) will $x+y, \ x^{2}+y^{2}, \ \ldots, \ x^{n}+y^{n}\in \mathbb{Z}\Longrightarrow x,y \in \mathbb{Z}$?[/quote]

the sequence can be recursively defined (see newton's sums), so if the first few are integers, then the rest of them will be integers (that is, if the first 4 sums are integers, then the rest of the sequence will be rational)

anyway, a quick example: $a_{n}=(\sqrt{2})^{n}+(-\sqrt{2})^{n}$ shows this to be false

[quote="cincodemayo5590"] $x^{3}+y^{3}\in \mathbb{Z}\ \text{ and }\ x+y \in \mathbb{Z}\implies x^{2}-xy+y^{2}\in \mathbb{Z}\$[/quote]
Of course, this is the mistaken line.  $x^{2}-xy+y^{2}= \frac{x^{3}+y^{3}}{x+y}$ need not be an integer.

[quote="t0rajir0u"][quote="cincodemayo5590"] $x^{3}+y^{3}\in \mathbb{Z}\ \text{ and }\ x+y \in \mathbb{Z}\implies x^{2}-xy+y^{2}\in \mathbb{Z}$[/quote]

Of course, this is the mistaken line.$x^{2}-xy+y^{2}= \frac{x^{3}+y^{3}}{x+y}$ need not be an integer.[/quote]

Oh. :oops:  I'll delete what I've posted.

(edit: The forum says i can't edit it!  :o  Darn. )


analytic geometry
conics
parameterization

Good morning

I need to do a piece like as barrel in a CNC (Computer Numeric Control) lathe machine.

On the piece I need to do several grooves with 15mm of depth. The barrel's radius is 6260mm. The piece length is 1000mm. The tool (disc)radius is 150mm. The grooves have to finish with 150mm from the top of each side.

So, the question is: How can I found the exact center coordinates of tool to respect the job coordinates?

I will appreciate your prompt help
Best regards

Pedro Ferrer

You're new, so you probably don't yet understand the forums, but in the Middle School Classroom Math forum, you're not going to get the appropriate feedback you're looking for.  Try to look around and get a grip on the levels of the forums so that you know best where to post about problems.  I'll move this thread.

Good morning

Ok. But can you help me?
You can't transfer my topic to the correct forum? Please, tell me where you puted my topic.

Thank you.

Best regards

Pedro Ferrer

Hi Andre

Thank for your attention

The drawing was designed in AutoCad only to show my problem. The following attachment shows what I want to do. I need to do parametric CNC programs to make the jobs an easier way. So, I already have parametric CNC programs to do grooves in a cylindrical and conic columns. CNC machines works with G code.

So, the operator that works with the machine usually knows the following parameters:

- the column length (1000mm)
- the distance of the tops to start and finish the groove (150mm both sides)
- the groove depth (15mm)
- the tool radius (150mm)
- the barrel radius

The ideia is the operator insert the values that want in the text box, then press "Compile" button and automatically the computer generates the CNC program. For that, I need to calculate the tool center coodinates.

Imagine that the point (0,0) is on the upper left side of the column, and all the coordinates are in absolute mode and not in incremental mode.

So, I need this kind of information to create myself a canned cycle, to this kind of job in an easier way many times as I want.

I will appreciate your prompt help
Best regards

Pedro Ferrer


logarithms
geometry
rectangle
SFFT
calculus
calculus computations

I need help on this problem.

Find the constant k such that the equation

dy/dx = (ln x)(ln y) + ln( kxy) becomes seperable.

I tried using the identity ln( kxy) = ln(k) + ln(x) + ln(y) but I can't seem to figure out where to go from there.

For what value of $\ln k$ does the right hand side factor as a product of two "linear" terms? Complete the rectangle.

"Complete the rectangle" also goes by the AoPS nickname of "Simon's favorite factoring trick." Just remember to think of $\ln x$ and $\ln y$ as your variables for the purpose of the factorization.

Sorry but I am not sure how to "complete the rectangle" or even what you mean by that.

SFFT is the factorization $ab+a+b+1 = (a+1)(b+1)$ and other variants. Then follow Mr. Merryfield's hint and you should be set.


calculus
integration
algebra proposed
algebra

Show that for any nonzero integer $ a$ one can find a nonzero integer $ b$ such that the equation $ ax^2\minus{}(a^2 \plus{} b)x \plus{} b \equal{} 0$ has integral roots.

[quote="tdl"]Show that for any nonzero integer $ a$ one can find a nonzero integer $ b$ such that the equation $ ax^2 \minus{} (a^2 \plus{} b)x \plus{} b \equal{} 0$ has integral roots.[/quote]

That's wrong : $ a\equal{}2$ implies equation $ 2x^2 \minus{}(b\plus{}4)x\plus{}b\equal{}0$, so $ b$ even and $ b\equal{}2c$ and so $ x^2\minus{}(c\plus{}2)x\plus{}c\equal{}0$ which can only have integers roots if $ c\equal{}0$, which would imply $ b\equal{}0$, which is forbidden since you are looking for nonzero $ b$.

if $ a\neq 2$
$ b \equal{} 2a(2 \minus{} a)$ gives $ ax^2 \minus{} (b \plus{} a^2)x \plus{} b \equal{} a(x \minus{} 2)(x \minus{} 2 \plus{} a)$

if $ a \equal{} 2$ if $ 2x^2 \minus{} (4 \plus{} b)x \plus{} b \equal{} 2(x \minus{} i)(x \minus{} j)$ for some $ i,j\in N$ then $ 1 \equal{} (1 \minus{} i)(j \minus{} 1)$ so $ b \equal{} 2ij \equal{} 0$


complex analysis
complex analysis unsolved

Show that X(t, 0, 1 , infinity) = t and deduce that four distinct points z[size=75]i[/size] lie on the circle if and only if X(z1,z2,z3,z4) is real.

Please use different titles for different topics, so your posts can be distinguished (You can still edit a new title/description in for the next day). In particular, "Complex Analysis" as a title is so generic as to be useless, especially in this forum. I thought I was going to have to delete some threads for multiposting there.


combinatorial geometry
superior algebra
superior algebra unsolved

Recall that a [i]convex polytope[/i] $\mathcal{P}$ in $\mathbb{R}^n$ is defined as the convex hull of a finite set of points in $\mathbb{R}^n$ (or, equivalently, as a bounded intersection of finitely many half-hyperplanes). A [i]face[/i] of the polytope $\mathcal{P}$ is the set of all points in $\mathcal{P}$ that maximize some linear function.

For any sets $A,B \subset \mathbb{R}^n$, define the [i]Minkowski sum[/i] $A + B := \{a+b \mid a \in A, b \in B \}$.
It is obvious that if $A$ and $B$ are convex polytopes, then so is $A + B$.

[b]Problem. [/b] (a) Let $A$ and $B$ be convex polytopes in $\mathbb{R}^n$. Show that every face of $A+B$ is the Minkowski sum of unique faces of $A$ and $B$.
(b) Strengthen (a) as follows: show that if $A_1 \subset A$ and $B_1 \subset B$ are convex subsets such that $A_1 + B_1$ is a face of $A+B$, then $A_1$ and $B_1$ must be faces of $A$ and $B$, respectively.

First of all, it's clear that every vertex of $C=A+B$ is the unique sum of a vertex from each of $A$ and $B$.

Let $A_1+B_1=C_1$ be a representation of a face $C_1$ of $C$ as a sum of convex subsets of $A,B$ (the notation is the same as in (b) above). If $c_1,\ldots,c_n$ are the vertices of $C_1$, then let $a_i+b_i=c_i$ be the unique representations of the $c_i$'s as sums of vertices of $A,B$ (note that the $a_i$'s need not be distinct, and the same holds for the $b_i$'s). $A_1$ must contain $\{a_i\}$, and $B_1$ must contain $\{b_i\}$. Let $A_1'\supseteq A_1$ be the face of $A$ spanned by $\{a_i\}$, and define $B_1'$ similarly.

Take any $i,j$. We have $a_i+b_i=c_i,a_j+b_j=c_j$. Let $f$ be a linear functional which reaches its minimum $m$ on $C$ in $C_1$. We have $\frac{f(a_i+b_j)+f(a_j+b_i)}2=f\left(\frac{a_i+b_i+a_j+b_j}2\right)=\frac{f(c_i)+f(c_j)}2=m$, so $f(a_i+b_j)=m$, meaning that $a_i+b_j\in C_1$. This shows that $\mbox{conv}(a_i)+\mbox{conv}(b_i)\subseteq C_1$, and since the reverse inclusion is obvious, we have $\mbox{conv}(a_i)+\mbox{conv}(b_i)=C_1\ (*)$.

Notice that $f$ takes the same value in all the $a_i$'s, and the same value in all the $b_i$'s, so it takes a constant value on $A_1'$, and also a constant value (which may be different from the other one, of course) on $B_1'$. This means that $A_1'+B_1'\subseteq C_1$, so, since $A_1'\supseteq\mbox{conv}(a_i),B_1'\supseteq\mbox{conv}(b_i),\ (*)$ gives us $A_1'+B_1'=C_1$.

Suppose we can find a point $u\in A_1'\setminus\mbox{conv}(a_i)$. We can then find a linear functional $g$ for which $g(u)<\min_{\mbox{conv}(a_i)}g$, and if $g$ reaches its minimum on $\mbox{conv}(b_i)$ in $v$, then $u+v$ cannot belong to $\mbox{conv}(a_i)+\mbox{conv}(b_i)=C_1=A_1'+B_1'$, which is a contradiction. This shows that $A_1'=\mbox{conv}(a_i)$, and the same type of argument gives $B_1'=\mbox{conv}(b_i)$. Now, since $A_1\supseteq\mbox{conv}(a_i)=A_1',B_1\supseteq\mbox{conv}(b_i)=B_1'$, and the same type of argument as a few lines above shows that $A_1,B_1$ cannot contain points outside $A_1',B_1'$ respectively, we get $A_1=A_1',B_1=B_1'$, and we're done.

I hope it's correct.


function
calculus
derivative
quadratics
trigonometry
algebra
algebra unsolved

[i]I have put the problem in the College section too.In this section the problem can be solved ,as well more elementary.MLinkres in Romany have similar problems in their school marerial.[/i]

Find the functions  $f , g$ with $f'(x) = f(x) +g(x) , g'(x) = g(x) - f(x), f(0) = 0 , g(0) = 1$

        for all  $x\in IR$ and  $f , g$ are not constant functions.

    [u]Babis[/u]

 I suspect $f(x) =sinx e^x , g(x) = cos x e^x$. Please try with elementry way, not with the general form of a  linear differential equation.

Finally I solved it. The function

  $H(x) = (f(x)e^{-x} - sinx)^2 + (g(x)e^{-x} - cosx)^2$

 has derivative 0 ect , so H(x) = 0.We used the fact that $f' - f = g , g' - g = -f$
 I still try to find them , without to guess them.

 [u]Babis[/u]

From $f'=g+f$ $\implies f''=g'+f'=2g=2f'-2f$ So $f''-2f'+2f=0$ .The roots for the quadratic equation $m^2-2m+2=0$ are $m=1\pm i$ , which mean the general solution for $f$ is

$f(x)=e^x(A\cos x+B\sin x)$ . Then using $f(0)=0$ and $f'(0)=g(0)+f(0)=1$ to get $A=0,B=1$ .

Hence $f(x)=e^x\sin x$

We do the same thing for $g$ to get $g''-2g'+2g=0$ $\implies g=e^x(C\cos x+D\sin x)$. Since $g(0)=0 and g'(0)=1$ , we get $C=1,D=0$ so $g(x)=e^x\cos x$

Edit : Opps , i didnt see that you want an elementary way ....  :roll:

Thank you.

   Really this was the simlest way - I mean your solution - based on homogenous linear differential equations theory. You helped me to refresh my poor  steps in Dif. Equations( I was taught them in 1978 !!!).
 If you find something more elementary, please remember me !

   [u]Babis[/u]


1) Determine all $n\in\mathbb{N}$, such that $\{(a,b)\in[2,\infty)^2\cap\mathbb{N}^2|(a^a)^n=b^b\}$ is non-empty.
2) Solve $(a^a)^5=b^b$ in $\mathbb{N}^2$.

This problem is from Belarus 2000 (or 2002), if I remember correctly.

What is N^2...

Bomb,  I think prime factorization would help.

[quote="bomb"]What is N^2...[/quote]
$\mathbb{N}^2=\{(x,y)|x,y\in\mathbb{N}\}$.

There is a non-trivial solution for the second part: $a = 2^8$ and $b = 2^{10}$.


How does one determine the sorting algorithm used (insertion, selection, merge, quick, heap) based on a chart of Size of Data and Time to Sort? Specifically:

Data in sorted order:
[u]Size of Data Set and Time to Sort (microsec)[/u]
100,000 and 10
200,000 and 40

Data in random order:
[u]Size of Data Set and Time to Sort (microsec)[/u]
300,000 and 84
600,000 and 336

I looked around a lot for an explanation, but I couldn't find one... thanks in advance!

You have an algorithm that takes time quadratic in the size of the data set and performs better on sorted data than unsorted data.  Which of those algorithms have both of these properties?

I know quadratic is probably insertion or selection... And I [i]think[/i] insertion works better for sorted data?

Whoops; what I wanted to say is, it takes time quadratic in the size of the data set regardless of whether it was already sorted or not.  Insertion, selection, and bubble sort are all quadratic in the average case; which one is also quadratic in the best case?

Well insertion is O(n) in the best case. Selection is O(n^2) in both cases. So the answer would be selection?
(I'm not sure what a bubble sort is...)

Well bubble sort is also quadratic in the worst case :maybe:.

Isn't bubble sort O(n) in the best case?

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

I don't think anyone uses bubble sort anymore anyways...

That's irrelevant to this question.  Of the five algorithms you listed, one satisfies both requirements.

True, I'm just saying that bubble sort isn't the answer.

Well bubble sort was the first sorting algorithm I learned [i]ever[/i]. But apparently they thought us a different variation... sorry for confusing you :).

(It went like this:[code]    for (int i = 1; i < n; i ++)
        for (int j = 0; j < n - i; j ++)
            if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);
[/code])


triangle ABC is isosceles triangle with AB=AC, AD is perpendicular to BC, DE is perpendicular to AC . F is midpoint of DE.      AF and BE are drawn  they intersect in point G. then prove that AG is perpendicular to BE. (assume triangle ABC is an acute angled triangle and  don't use co-ordinate  geometry for this.)

Here if  you use co-ordinate geometry problem becomes very simple.

Let $H$ be the midpoint of $CE$. Then $DH$ is a midline in triangle $BCE$, so $DH||BE$.
Now, observe that $F$ is the orthocenter of triangle $ADH$. Indeed, $DF\perp AC$ and $HF\perp AD$ since $HF||CD$. It follows that $AF\perp DH$, hence $AF\perp BE$, as desired.

This problem is a particular case of the problem proposed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18084 . In order to make the relation clear, I rewrite the initial problem with shifted notations.

[color=blue]Let ABC be an isosceles triangle with CA = CB, and let D be the orthogonal projection of its apex C on its base AB. Let E be the orthogonal projection of the point D on the side BC. Let F be the midpoint of the segment DE. Then prove that $CF\perp AE$.[/color]

Now, since the triangle ABC is isosceles with CA = CB, the orthogonal projection D of its apex C on its base AB is simply the midpoint of this base AB. Thus, $\frac{AD}{DB}=1$. On the other hand, since F is the midpoint of the segment DE, we have $\frac{EF}{FD}=1$. Thus, $\frac{EF}{FD}=\frac{AD}{DB}$, and hence this problem is a particular case of the one discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18084 .

  darij

Precisely the same construction ($F' \in \overline{BE}$ such that  $\overline{BD} \parallel \overline{FF'}$) manages in other problem as well.


topology
function
real analysis
real analysis unsolved

A map $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is called a $C^r$ diffeomorphism, if $f$ is a homeomorphism and $f$ and its inverse $f^{-1}$ are r-times continuously differentiable, i.e. they are both $C^r$ maps.

How can one prove the following:

If $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a $C^1$ diffeomorphism and a $C^r$ map, then $f$ is a $C^r$ diffeomorphism.

Can you please help me?

I am no specialist in this, but i think when you have a $C^r$ function that has a jacobian determinant nonzero in some point p, there is a neighbourhood of p on which it works as a $C^r$ diffeomorphism (this is a theorem)

when you know that it is a $C^1$ diffeomorphism, it gives you jacobian determinant nonzero everywhere and I think this should help

Putting $f(u)=v$ we know that $matr(df(u))=matr(df^{-1}(v))$ so if the functions in the rows of
 $matr(df(u))$ are
$C^{r-1}$ then the functions in the rows of $matr(df^{-1}(v))$ are $C^{r-1}$ and vice versa.


inequalities
inequalities proposed
algebra

Can this be useful?

$\displaystyle (a+b+c)^5-a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$

Maybe,It's very straight.
But in my solution I didn't use it.
 :)

So what 's yours ? give some hints about your solution .....

A tighter inequality is this one $\frac{a^2+b^2+c^2}{a+b+c} \geq \sqrt{3}R$, which is the Garfulkel 's inequality.

I 'm sorry for my bad memory. It should be $\frac{a^2+b^2+c^2}{a+b+c}\leq \sqrt{3}R$.

Oh,along time and no one post the solution for this problem
Here is then hint ,very neccessary:
$sinAsin3A+sinBsin3B+sinCsin3C \le 0$
Please check it!

Then, the following better inequality holds

$\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \ge \sqrt{3}R$.

I start with the ineq:

$\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$

Which is equal with:

$sinAsin3A+sinBsin3B+sinCsin3C \le  0$

And easy to prove $(\frac{a^5+b^5+c^5}{a^4+b^4+c^4})^2 \ge \frac{a^4+b^4+c^4}{a^2+b^2+c^2}$
Because $(a^5+b^5+c^5)(a^5+b^5+c^5)(a^2+b^2+c^2) \ge (a^4+b^4+c^4)^3$ (Holder)

Vasc,Please proof $\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \ge \sqrt{3}R$
I think it's tronger.

It is not very hard.
We get the desired result by multiplying the inequalities

$(a^4+b^4+c^4)(a^2+b^2+c^2) \ge (a^3+b^3+c^3)^2$
and
$a^4+b^4+c^4 \ge 3R^2(a^2+b^2+c^2)$.  ;)

Can  anybody  please  give a  hint  to  that  Jack  Garfunkel's inequality?  And also  does  anybody  knows  in  what  magazine (issue-year)  was  published?

Thanks

[quote="hungkhtn"]I start with the ineq:

$\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$

Which is equal with:

$sinAsin3A+sinBsin3B+sinCsin3C \le  0$[/quote]
Yesterday I found in my collection the inequality
$\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$.

The proof is the following. Since
$R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,
using the notations $a^2=\frac{y+z}{2}$ etc., the inequality becomes as follows
$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$ ;)

Also, I nearby found the following complementary inequalities for any triangle:

(a) $(a+b+c)R \ge \sqrt{ab^3+bc^3+ca^3}$;

(b) $(a+b+c)R \ge \sqrt{a^2b^2+b^2c^2+c^2a^2}$.

Why do you use this substitution?

$\displaystyle a^2=\frac{y+z}{2} $

In my opinion $a^2,b^2,c^2$ are not sides of a triangle.

$a^2+b^2 \geq \frac{(a+b)^2}{2} \geq \frac{c^2}{2}$

instead of

$a^2+b^2 \geq c^2$

Where am I wrong, Vasc?

manlio,
\[ a^2+b^2-c^2 = \frac{y+z+z+x-x-y}{2} = z \]
and so on, so they are sidelengths of a triangle.

Manlio, the triangle is acute.  ;)

[quote="hungkhtn"]Show that in acute triangle ABC we have:
$\frac{a^5+b^5+c^5}{a^4+b^4+c^4} \ge \sqrt{3}R$[/quote]

generalization :
Let $ k,p\in N,k-p\geq 1 $ , then  we :  $ \frac{\sum a^{k}}{\sum a^{p}}\geq (\frac{a+b+c}{3})^{k-p}\geq (\sqrt{3}R)^{k-p} $

[quote=Vasc]Also, I nearby found the following complementary inequalities for any triangle:
(b) $(a+b+c)R \ge \sqrt{a^2b^2+b^2c^2+c^2a^2}$.[/quote]
We have
\[LHS^2 - RHS^2 = \left[12Rr+(p^2+5r^2-16Rr)\right] (4R^2+4Rr+3r^2-p^2)+4r(4R^2+3Rr+2r^2)(R-2r) \geqslant 0.\]

[quote=sqing]Show that in acute triangle $ABC$ , prove or disprove:

$\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt{3}R$ $(n\in R^+,n\ge 2)$

$\frac{a^{n+2}+b^{n+2}+c^{n+2}}{a^n+b^n+c^n} \ge 3R^2$ $(n\in R^+,n\ge 2)$[/quote]

The first one holds for $\boxed{n\ge 3}$  :!: . And they are both trivial because of the very nice inequality in [b]non-obtuse-angled[/b] triangles which has already been mentioned, that is:
\[\boxed{\frac {a^4 + b^4 + c^4}{a^2 + b^2 + c^2} \ge 3R^2}\ .\]
This was also proven by other means [url=https://artofproblemsolving.com/community/c6h361604p1980265][b]here[/b][/url].

Now, $n\mapsto \frac {a^{n+1} + b^{n+1} + c^{n+1}}{a^n + b^n + c^n}$ is decreasing and this is an immediate consequence of CS. And the base case for $n = 3$ follows again from CS together with the above-mentioned inequality.

Similarly, $n\mapsto \frac {a^{n+2} + b^{n+2} + c^{n+2}}{a^n + b^n + c^n}$ is also a decreasing function on $n$ because $f(n)\ge f(n - 1)$ reduces to:
\[
\begin{array}{lc}
     & \displaystyle \sum \left(b^{n-1}c^{n+2} + c^{n-1}b^{n+2}\right) \ge \sum \left(b^nc^{n+1} + c^nb^{n+1}\right) \\\\
\iff & \displaystyle \sum b^nc^n\left(\frac {c^2}b + \frac {b^2}c\right) \ge \sum b^nc^n\left(b + c\right),
\end{array}\]
which is obvious. As the base case coincides with the previous "boxed" inequality, there is nothing else to be done ;).

#14Feb 14, 2005, 9:05 pm
hungkhtn wrote:
I start with the ineq:

$\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$

Which is equal with:

$sinAsin3A+sinBsin3B+sinCsin3C \le  0$
Yesterday I found in my collection the inequality
$\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$.

The proof is the following. Since
$R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,
using the notations $a^2=\frac{y+z}{2}$ etc., the inequality becomes as follows
$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$ ;)
right??????


The interior angles of a convex polygon form an arithmetic progression with a common difference of $4^\circ$. Determine the number of sides of the polygon if its largest interior angle is $172^\circ.$

[hide]The external angles forms an arithmetic progression where $d = 4^\circ$ and $a_1 = 8^\circ$. Since the sum of the external angles of a convex polygon is equal to $360^\circ$, we can determine $a_n$:
$S = \frac{(a_1 + a_n)d}{2}$
$360 = \frac{(8 + a_n)4}{2}$
$a_n = 172^\circ$

$a_n = a_1 + (n - 1)d$
$172 = 8 + (n - 1)4$
$n = 42 \Rightarrow \boxed{\mbox{42 sides}}$[/hide]


In triangle ABC, L,M,N is the midpoint of side BC, CA, AB, respectively. P is another point that isn't on the circumcircle of ABC. X, Y, Z are the foot of perpendicular from P to BC, CA, AB, respectively. Line MN and YZ, NL and ZX, LM and XY meet at D, E, F, respectively. Prove that:
(1) The three lines XD, YE, ZF meet have a common point S.
(2) S lies on the circumcircle of $ \triangle LMN$ and $ \triangle XYZ$.

Actually, the first part of this one is not very difficult. Applying Ceva of Menulaus, we can solve it (and it is not nessery for X,Y,Z to be the foot). However, the second part is rather difficult that till now it remains unsolved without lots of caculation.
Thanks!

Combine Theorems 1.1, 1.6, 1.7, in [url=http://www.cip.ifi.lmu.de/~grinberg/GenFeuerPDF.zip]"Generalization of the Feuerbach point"[/url] on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url]. (Your point S is called L in this note.)

  Darij


Let $ a;b;c;d$ be real numbers satisfying
\[ \Longleftrightarrow\begin{cases}
0<a\le b \le c\le d\\
\frac{1}{a}+\frac{2}{b}+\frac{d}{c}\ge 3\\
\frac{2}{b}+\frac{d}{c}\ge 2
\end{cases}
\]
Prove that
\[ a^4+b^4+c^4-d^4\le 17
\]

No one likes it. Must I post my solution?
@nguoivn: Can you prove it by $ AMGM$ inequality?. I am sure you are able to do because I did it well.


Prove that in any triangle we have:

[tex]a^{2}+b^{2}+c^{2} \geq 4sqrt{3}S[/tex]

Thread title changed, promoted to next level.

This would look a little betterBottema wrote:Prove that in any triangle we have: [tex]a^{2}+b^{2}+c^{2} \geq 4\sqrt{3}S[/tex], where [tex]S[/tex] is the area of the triangle.

Hint:(spoiler)

[hide]use [tex]c^2=a^2+b^2-2ab\cos{C}[/tex] and [tex]S=1/2ab\sin{C}[/tex], we can calculate that

[tex]a^{2}+b^{2}+c^{2} - 4\sqrt{3}S = 2(a-b)^2 + 4ab(1-\cos(\pi/3-C))\geq0[/tex][/hide]

Just got Arthur Engel's book, found out that it's a 1961 IMO problem (E21 on p.170). Amazingly, the book provided [b]11 proofs[/b]!


For a set with one hundred elements, how many subsets are there whose cardinality is a multiple of 3?

Use generating functions, then look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165573

Basically a simpler version of this problem.


I was just curious:

How many posts do you need to get 1 star (under the username), 2 stars, 3 stars, etc.?

Well, there's a few halves too

1/2 Star (P versus NP): 20
1 Star (Hodge Conjecture): 50 posts
2 Stars (Poincare Conjecture): 100 posts
2 1/2 Stars (Riemann Hypothesis): 250 posts
3 Stars (Yang Mills Theory): 500 posts
4 Stars (Navier-Stokes Equations): 1000 posts
5 Stars (Birch & Swinnerton Dyer): 2500 posts

Admins all have 6 stars, don't let that confuse you...

EDIT: Oops, someone beat me to it.


a printer is used to print out the squares of the first 15 positive integers. how many digits are printed by the printer?

no calc allowed

Correct: [hide]1-3 (3 numbers) squared is one digit

4-9 (6 numbers) squared is two digits

10-15 (6 numbers) squared is three digits, so 1*3+2*6+3*6=33, 33 digits.[/hide]


inequalities
logarithms
inequalities proposed

Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
Can you prove it without the lema $ a^xb \plus{} b^xc \plus{} c^xa$ ???

2) $ \frac {ab}{b^{3} \plus{} 1} \plus{} \frac {bc}{c^{3} \plus{} 1} \plus{} \frac {ca}{a^{3} \plus{} 1}\leq\frac {3}{2}.$

3) $ \frac {ab^2}{b^{3} \plus{} 1} \plus{} \frac {bc^2}{c^{3} \plus{} 1} \plus{} \frac {ca^2}{a^{3} \plus{} 1}\leq\frac {3}{2}.$

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
 2) $ \frac {ab}{b^{3} \plus{} 1} \plus{} \frac {bc}{c^{3} \plus{} 1} \plus{} \frac {ca}{a^{3} \plus{} 1}\leq\frac {3}{2}.$[/quote]
$ \frac {ab}{b^{4} \plus{} 1} \plus{} \frac {bc}{c^{4} \plus{} 1} \plus{} \frac {ca}{a^{4} \plus{} 1}\leq\frac {3}{2}\,$  holds,  but

$ \frac {ab}{b^{5} \plus{} 1} \plus{} \frac {bc}{c^{5} \plus{} 1} \plus{} \frac {ca}{a^{5} \plus{} 1}\leq\frac {3}{2}\,$  doesn't.

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
3) $ \frac {ab^2}{b^{3} \plus{} 1} \plus{} \frac {bc^2}{c^{3} \plus{} 1} \plus{} \frac {ca^2}{a^{3} \plus{} 1}\leq\frac {3}{2}.$[/quote]
Analogous inequalities hold up to

 $ \frac {ab^2}{b^{15} \plus{} 1} \plus{} \frac {bc^2}{c^{15} \plus{} 1} \plus{} \frac {ca^2}{a^{15} \plus{} 1}\leq\frac {3}{2},\,$ but

 $ \frac {ab^2}{b^{16} \plus{} 1} \plus{} \frac {bc^2}{c^{16} \plus{} 1} \plus{} \frac {ca^2}{a^{16} \plus{} 1}\leq\frac {3}{2}\,$ is not true.

[quote="Gibbenergy"]I am interested in the proof. Do you have it ? I think it is hard for me at least. :)[/quote]
Sorry, I have no proofs to them in traditional sense.

[quote="Guest"][quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
 2) $ \frac {ab}{b^{3} \plus{} 1} \plus{} \frac {bc}{c^{3} \plus{} 1} \plus{} \frac {ca}{a^{3} \plus{} 1}\leq\frac {3}{2}.$[/quote]
$ \frac {ab}{b^{4} \plus{} 1} \plus{} \frac {bc}{c^{4} \plus{} 1} \plus{} \frac {ca}{a^{4} \plus{} 1}\leq\frac {3}{2}\,$  holds[/quote]
Yes you are right! We can check it by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=522084]BW[/url] and computer.

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a + b + c = 3$ Prove

2) $ \frac {ab}{b^{3} + 1} + \frac {bc}{c^{3} + 1} + \frac {ca}{a^{3} + 1}\leq\frac {3}{2}.$

3) $ \frac {ab^2}{b^{3} + 1} + \frac {bc^2}{c^{3} + 1} + \frac {ca^2}{a^{3} + 1}\leq\frac {3}{2}.$[/quote]

Very powerful inequalities !

Here is my solution for the third inequality.

I proved the following lemma.
If $a,b,c$ are positive reals such that $a+b+c=3$ then
\[\boxed{\frac32 \geqslant \sum_{cyc}\frac{ab^3}{b^3+1}}\]
We have
\[\sum_{cyc} \frac{ab^3}{b^3+1}=\sum_{cyc}\frac{a(b^3+1)-a}{b^3+1}=\sum_{cyc}a-\sum_{cyc}\frac{a}{b^3+1} \leqslant 3-\frac32 =\frac32\]

\[ \sum_{cyc}\frac{a}{b^3+1} \geqslant \frac32 \]

This is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=72996]Zhaobin & Arqady's inequality[/url]
Thus we can apply your second inequality and my lemma for the third
we have
\[ \frac32 \cdot \frac 32 \geqslant \sum_{cyc}\frac{ab}{b^3+1} \cdot \sum_{cyc}\frac{ab^3}{b^3+1} \geqslant \left( \sum_{cyc} \frac{ab^2}{b^3+1}\right)^2\]
Q.E.D

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove

2) $ \frac {ab}{b^{3} \plus{} 1} \plus{} \frac {bc}{c^{3} \plus{} 1} \plus{} \frac {ca}{a^{3} \plus{} 1}\leq\frac {3}{2}.$

[/quote]

Second inequality is so tough. Here is my traditional C-S solution for it.
\[ \sum_{cyc} \frac{xy}{y^3+1}=\sum_{cyc}\frac{xy+xy^4-xy^4}{y^3+1}=\sum_{cyc}\frac{xy(1+y^3)-xy^4}{y^3+1}\]
Thus we need to prove
\[ \sum_{cyc} \frac{xy^4}{y^3+1} \geqslant \sum_{cyc}xy -\frac32\]

Using C-S we have
\[\sum_{cyc} \frac{xy^4}{y^3+1} \cdot \sum_{cyc}x(y^3+1) \geqslant \left( \sum_{cyc} xy^2\right)^2\]

It remains to prove that
\[ \left( \sum_{cyc}xy^2\right)^2 \geqslant \sum_{cyc}x(y^3+1) \cdot \left( \sum_{cyc}xy-\frac32\right)\]

The last inequality is 6 degrees and can be proved by SDM method.

[quote="My_dad"]

The last inequality is 6 degrees and can be proved by SDM method.[/quote]
What is it? :maybe: Is there a link?

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
Can you prove it without the lema $ a^xb \plus{} b^xc \plus{} c^xa$ ???
[/quote]

I don't know how you manage to prove this monster. Can you show me your proof, if it is possible ? I think this one and the one by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=465014&hilit=tourish]Tourish[/url]'s are among the most difficult inequalities I have ever encountered.

My ugly proof is as following:\[ \left( \sum_{cyc}\frac{x}{y^5+1}\right) \cdot \left[ \sum_{cyc} \frac{x(y^4-y^3+y^2-y+1)}{(y+1)^3}\right] \geqslant \left[ \sum_{cyc} \frac{x}{(y+1)^2}\right]^2\]
Thus it remains to prove
\[ \frac23 \cdot \left[ \sum_{cyc} \frac{x}{(y+1)^2}\right]^2  \geqslant \sum_{cyc} \frac{(y^4-y^3+y^2-y+1)x}{(y+1)^3}\]

Notice that \[ \frac{y^4-y^3+y^2-y+1}{(y+1)^3} = y-4 +\frac{10y^2+10y+5}{(y+1)^3}\]
hence we need to prove

\[ \frac23 \cdot \left[ \sum_{cyc} \frac{x}{(y+1)^2}\right]^2  \geqslant \sum_{cyc}xy -12 +5\sum_{cyc}\frac{x(2y^2+2y+1)}{(y+1)^3}\]

The last inequality is true but I have no nice solution.

Looking forward to see ingenious approach

From curiosity what is the lemma that you talking about Gibbenergy and how can be proved?

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove

3) $ \frac {ab^2}{b^{3} \plus{} 1} \plus{} \frac {bc^2}{c^{3} \plus{} 1} \plus{} \frac {ca^2}{a^{3} \plus{} 1}\leq\frac {3}{2}.$[/quote]

I have a very simple solution for $3$,
We have
\[\sum_{cyc} \frac{ab^2}{b^3+1} \leqslant \sum_{cyc} \frac{ab^2}{2b^{\frac32}}=\frac12 \sum_{cyc}ab^{\frac12}\]
Thus, it remains to prove $\sum_{cyc}ab^{\frac12} \leqslant 3$, with condition $a+b+c=3$.

This problem is too well-known.

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
Can you prove it without the lema $ a^xb \plus{} b^xc \plus{} c^xa$ ???
[/quote]

I am curious.

For all positive reals $x,y,z$ such that $x+y+z=3$. What is the best constant $k$ such that
\[ \sum_{cyc} xy^k \leqslant 3\]

[quote="arqady"]$\log_{\frac{3}{2}}2$ :maybe:[/quote]
How do you prove this result ? Obviously, Gibbenergy cannot use that lemma to the original problem.

[quote="My_dad"][quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove

2) $ \frac {ab}{b^{3} \plus{} 1} \plus{} \frac {bc}{c^{3} \plus{} 1} \plus{} \frac {ca}{a^{3} \plus{} 1}\leq\frac {3}{2}.$

[/quote]

Second inequality is so tough. Here is my traditional C-S solution for it.
\[ \sum_{cyc} \frac{xy}{y^3+1}=\sum_{cyc}\frac{xy+xy^4-xy^4}{y^3+1}=\sum_{cyc}\frac{xy(1+y^3)-xy^4}{y^3+1}\]
[/quote]
Second inequality was proved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=543819&p=3141615&hilit=gibbenergy#p3141615]here[/url].  Magnificent solution by Quykhtn-qa1.

[quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
Can you prove it without the lema $ a^xb \plus{} b^xc \plus{} c^xa$ ???

$[/quote]

The first inequality was discussed in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=441639&p=2489367&hilit=gibbenergy#p2489367]here[/url]. As I suspected, the lemma did not work.

[quote=My_dad][quote="Gibbenergy"]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
Can you prove it without the lema $ a^xb \plus{} b^xc \plus{} c^xa$ ???
[/quote]

$\frac23 \cdot \left[ \sum_{cyc} \frac{x}{(y+1)^2}\right]^2  \geqslant \sum_{cyc}xy -12 +5\sum_{cyc}\frac{x(2y^2+2y+1)}{(y+1)^3}$

[/quote]

Is there any simple solution for it ?

[quote=Gibbenergy]Let $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove
1) $ \frac {a}{b^{5} \plus{} 1} \plus{} \frac {b}{c^{5} \plus{} 1} \plus{} \frac {c}{a^{5} \plus{} 1}\geq\frac {3}{2}.$
[/quote]

Prove that for all $ a,b,c \geq0$ and $ a \plus{} b \plus{} c \equal{} 3$
$ \frac {a}{b^{4} \plus{} 1} \plus{} \frac {b}{c^{4} \plus{} 1} \plus{} \frac {c}{a^{4} \plus{} 1}\geq\frac {3}{2}.$


I am going to Mathpath this summer, it will be my second year. I am just wondering if anyone else from AoPS is coming. If so post here.

Uhhh... there's another thread 3 pages long.  Maybe you want to use that.

Sorry I looked for one and I completely looked over it. Can someone delete this thread?

Edit: Changed my mind don't delete this.

The other thread is 3 pages long.  It was created in 2003, but was recently revived.  In searching that thread, there are *no* users acknowledging their attendance at MathPath.

In response to JMI's query, my daughter will be attending MathPath.  Yesterday, she received an e-mail containing the "pre-camp briefing".  Is anyone else attending?


[code]NAGIBNI

Railway network infrastructure in one country consists of a certain numbers of individual railway segments.

Each segment connects two different towns. We know the lengths of all segments, and the network is such that between any two towns there is a unique path connecting them.

Recently, the management of the railway company decided to replace their old trains with ultramodern and fast trains imported from Japan.
A route of a train consists of a sequence of two or more towns, with all neighboring towns on the route connected by a segment. Length of the route is defined as the sum of the lengths of all the segments included. Because trains are very fast it is not allowed that two different routes pass through the same town.

Write a program that will determine the new routes in the network under the following restrictions: Exactly one route has to pass through each town, and sum of the lengths of all the routes should be as large as possible.

Input

Input will contain several cases. First line of each test case contains an integer N, 1 ≤ N ≤ 2000, the number of towns. Towns are designated with numbers from 1 to N.

Following N-1 lines contain the descriptions of the segments. Each line contains three integers A, B and C representing a segment between town A and town B of length C, 1 ≤ C ≤ 1,000,000.

Note: there will always exist a solution for the given test data.

Output

For each test case, your output should contain the maximal sum from the problem statement.

Sample Input

5
1 2 10
1 3 10
1 4 10
4 5 1
6
1 2 1
1 3 3
1 4 7
2 5 4
2 6 6
9
1 2 1
2 4 2
3 4 1
4 5 5
6 5 1
5 7 2
7 8 1
8 9 2

Sample Output

21
20
11
[/code]

Looks like a reasonably typical tree problem.
Root it at any vertex.
For each node, work out two values: 1) Maximum value for this subtree given that there is no route coming *down* to this node, 2) Maximum value for this subtree given that there *is* a route coming down to this node (from its parent). Each of them is easy to do recursively.

i think i got it.

for a node, get two information from its children, max1 and max2. (max1 is the first maximum)

Now, it will return max1.

And it will check max1+max2 with answer. for single child it will be max1 to be checked.

I think i am correct!


Orice subgrup finit al grupului unităţilor unui domeniu de integritate este ciclic.
1) Sigur este valabil si daca caracteristica e 0?
2) Ar putea cineva schita o demonstratie a acestei afirmatii?
3) Cum $Z_p$ este corp finit, deci inel integru finit, putem zice ca $(Z_p^*,\cdot)$ e ciclic? Stiu ca e ciclic, dar poate fi considerat o consecinta a afirmatiei de mai sus?

In programa aia sunt multe chestii care nu au demonstratie elementara. Eu am vazut o demostrati care foloseste nusch module de tip finit pe inele principale  :rotfl:  Daca nu vrei sa inveti jumate din materia de facultate pana la nationala invat-o asa. E adevarat pentru orice fel de inel.

La intrebarea 3) e o demonstratie elementara a faptului ca orice corp finit e ciclic. Arati ca exista un elem de ordin $|K|$. Demonstratia foloseste faptul ca $X^{p^\alpha}-1$ are cel mult $X^{p^\alpha}$ radacini si ca $|K|=\sum_{d\mid |K| } \phi(d)$.


Find all positive integer $ (x,y)$ such that:

$ \sqrt{x}+\sqrt{y}= \sqrt{1992}$.

[hide]The obvious solutions are $ (0,1992)$ and $ (1992,0)$. Since $ \sqrt{1992}=2\sqrt{498}$, another solution is $ (498,498)$.[/hide]

Oh shoot. :blush: Then just the other one? Though I doubt there's only one solution. Trying to find more now...

[hide="Hint"] Try squaring it and explore certain aspects of $ \sqrt{xy}$[/hide]

[hide="Answer"] $ (x,y) = (498,498)$.[/hide]


[quote="AnhHungThamLang"][i]Let[/i] $\large a,b,c\ge0$ [i].prove that[/i]
...............
$\large(1+\sqrt{\frac{a}{a+b}})(1+\sqrt{\frac{b}{b+c}})(1+\sqrt{\frac{c}{c+a}})\ge \frac{10+7\sqrt{2}}{4}$
[b]Undefined Mixing Variables[/b]  :wink:
[b]@pvthuan:[/b]  Thầy thấy bài toán này thế nào ạ!  :rotfl: Thấy hãy nhắn lại cho em qua nick [b]silenthero_k09[/b] nhé. :D

[/quote]

Please don't use another language in your post  :mad:

see here http://www.mathlinks.ro/Forum/viewtopic.php?t=138774


[quote="winsome"]When the parking garage opened in the morning, there were 100 cars in lot a and 20 cars in lot b. After the attendants moved 20 cars from lot a into lot b, what was the ratio of the number of cars in lot a to the number of cars in lot b?[/quote]

Explanation:

20 cars were moved from Lot A to Lot B, so Lot A then had 80 cars while Lot B had 40 cars. Thus the ratio is $ \boxed{2: 1}$.


I have noticed that when taking contest counting/probability is one of my weakest areas.  What are some good resources directed toward these types of questions?  I have read through AOPS vol.1, and am planning to do so again, I am currently working through the second book.

Here:

[url]http://home.wlu.edu/~rushm/probabilityhelp.htm[/url]

[url]http://www.nzmaths.co.nz/Statistics/Probability/CountingOnProbability.htm[/url]

[url]http://www.ltcconline.net/greenl/courses/102/Probability/Probability.htm[/url]

Or just google it.


The 2004 Team Scramble will occur on Thursday, October 21st, 2004.  This contest consists of 100 problems which a school's entire team (as many students as are interested) collaborates to answer questions from across the mathematical spectrums of topics and difficulties.  As with our other contests, all participating schools receive detailed solutions for all problems, and so there will probably not be enormous discussion of the problems from the test (but please ask questions if you have them).

It doesn't seem like there are many ppl on this site doing these tests but I m sure there are some.  To those who are doing it good luck tomorrow on the Team Scramble.  It should be fun!

We got a 30.... yeah it was hard.  We could have used a few more people.  We had about 8 ppl.

This test tends to be fairly difficult, so scores of 44 or 30 are not bad, and might even earn awards.  We have not received many official results yet, but there is a report of an 89, which I would guess is the highest we'll hear of.  Of course, the important thing is not the score, but rather the experience.  We've received a number of e-mails from coaches regarding how much fun their teams had, and how much fun the coaches had watching the madness.  How many people formed your teams?  How did you organize your team's efforts?

holy...

an 89?? that is amazing. Although I must say that I'm pretty skeptical of that...unless their team had LOTS of people...or if it was TJ...or i dunno, maybe we just suck.

we had a team of size 10, 4 did the first 30, 4 did the second 30, 2 people did the last 40.

My group was so lame.  We probably got like 15, or so.  Who are you people that are making these tests?  Did you people also make an 100 problem test for an individual back about a month ago?  I did horribly on that (a 21 as I recall), what did others get?

Last year my team got a 12 so don't feel too bad...we even missed the problem that goes something like this:

If a cow weighs 1000 lbs standing on 4 legs how much does it weigh standing on 3.

Obviously the answer is 1000 but someone on my team put 750... :?   O well...we had the better ppl do the harder ones and found out that didn't work too well.

Our strategy this year was a little better.  We had each person work a different set of problems.  The problems are usually grouped by category so we switched a little.  For example, one of my teammates got the trig section so he let me do those and he did mine.  We didn't collaborate very much however.  We just did our problems and put down the answers near the end.  I m not sure how collaboration would have affected us.

Yeah they did make the other test with 100 problems.  The website is under tom_clymers signature.  There are results there so you can see the top ppl.

Damn, my team only got 25 (we would have gotten higher but the scribe miswrote two my my solutions which would have otherwise been correct!).

do your teams utilize scribes? The major problem is of course, miswriting esp. when there's also a board with all of the problem numbers.

We hope to have the official results posted by Monday.  We did receive the paper with the 89, and I cannot imagine that our double-checking will reduce it by much.  The scores do fall off fairly quickly from there, however.  I'll leave details until we post the official results.

As far as strategies go, we do encourage the use of a scribe (someone in charge of the official answer key) as Simfish described, but we believe you should watch while they write the answer to be sure they get what you said.  We also encourage splitting up the work, as joml88 was describing.  If you don't have many people working on the test, you can't double-check much... team size is very helpful on this test.  We think that's a good thing, because it encourages teams to reach out to the rest of the school.  Even if you can only pull in ten algebra I students, they can do all the arithmetic and some of the algebra problems, which can free up your most experienced members for other work.

We do write the other contest that themonster referred to, as joml88 noted.  That contest was the 2004 Fall Startup Event, and had a high score of 82.  Detailed results are available at our website.


[quote="beta"][quote="MysticTerminator"]then everybody who scored higher than the index would qualify for the USAMO, as well as all that business with floors for tiny people, etc etc

why?[/quote]

I think his point is that we'll have a floor value of like 0... then they'll select top 80ish 9th and 10th grade AIME qualifiers by index.[/quote]

Well if the index is less than 150, chances are the AMC was really hard too, so I think it woulnd't be SO bad... though yeah the floor is probably not sufficient and they'll pick the 80. That would be pretty killer if the highest score on the AIME was like 2.

[quote="probability1.01"]On a related note, what happens if the floor is -16, and the AIME cutoff is 62.5? Seriously, what do you mean?[/quote]

That's an interesting situation. I'd probably have a tough time qualifying, I mean, my mad guessing skillz are lacking.

Oh yeah what if less than 160 people take the AMC12?

[quote="MysticTerminator"][quote="probability1.01"]On a related note, what happens if the floor is -16, and the AIME cutoff is 62.5? Seriously, what do you mean?[/quote]

I dunno, but it would really suck if you decided not to leave everything blank and got < 62.5[/quote]

We had someone in my grade who had that happen to her as a freshman.

Oh yeah, and what would happen if the USAMO cutoff was a 16 AIME?

[quote="lowest_reynolds"][quote="MysticTerminator"]btw wouldn't it suck if you were just walking down the street and then someone shot you, I'd be pretty pissed[/quote]

That...or you'd be dead. :rotfl:[/quote]

Not necessarily.  People get shot all the time and live through it.

Anyway...everyone knows the USAMO index will be at least 300 this year :)

Lol, depends where the shot is.  Sigh, what has this world come to.  :roll:

And what happened to this thread...heh.

well what's up with all this "what ifs", they are not going to happen and if they happen there's nothing we can do lol.

[quote="chess64"]Everyone on AoPS should leave all the questions blank so the cutoff will be really low...[/quote]

Go ahead...we'll follow you  :roll:

[quote="lowest_reynolds"]And what happened to this thread...heh.[/quote]

Well, where was the thread to begin with?...

[quote="h_s_potter2002"]I know this is almost impossible, but what would happen if the USAMO qualifying score was less than 150?[/quote]

It hasn't really strayed much from the first post which is quoted above.

[quote="joml88"][quote="chess64"]Everyone on AoPS should leave all the questions blank so the cutoff will be really low...[/quote]

Go ahead...we'll follow you  :roll:
[/quote]

OK, everyone do it on the B test! :D

(After the B test, chess64 logs on to AoPS, only to find that he was the only one to leave them all blank after all :rotfl:)

What if none of the MC answers were right on the AMC? So if you guessed you automatically got it wrong. What now? And what if all the AIME answers were *gasp* radicals? *runs away*

Then all the people who didnt take answer any questions make USAMO.... and my world shatters to pieces.


Find all solutions in positive integers $ x,y,z$ and $ t$ with $ x \le y \le z\le t$ of the equation:
$ \frac {1}x\plus{}\frac{1}y\plus{}\frac{1}z\plus{}\frac{1}t\equal{}1$

we have $ \frac{1}{x}\geq\frac{1}{y}\geq\frac{1}{z}\geq\frac{1}{t}$
then: $ 1\equal{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}\frac{1}{t}\leq \frac{4}{x}$
$ \Leftrightarrow 1<x\leq4$
if $ x\equal{}4$ we have $ \frac{3}{4}\equal{}\frac{1}{t}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\leq\frac{4}{y}$
then $ 1<y<5$ because $ x\geq4$ if $ y\equal{}4$ we have $ \frac{1}{z}\plus{}\frac{1}{t}\equal{}\frac{1}{2} \Leftrightarrow (z\minus{}2)(t\minus{}2)\equal{}4$
then $ z\equal{}t\equal{}4$ or $ z\equal{}3$ and $ t\equal{}6$ wich is not true because $ z\geq y\equal{}4$
after we see the second case $ x\equal{}4$ ; $ y\equal{}3$ and $ x\equal{}4$ ; $ y\equal{}2$ after $ x\equal{}3$ ; $ y\equal{}3$ and $ x\equal{}3$ ; $ y\equal{}2$ after $ x\equal{}2$ ; $ y\equal{}3$ and $ x\equal{}2$ ; $ y\equal{}2$
use this for simplicating the job $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{n} \Leftrightarrow (a\minus{}n)(b\minus{}n)\equal{}n^2$.
P.S: sorry for my bad english!


inequalities
logarithms
inequalities unsolved

Please help me  :(
How can I prove Bernoulli inequality by Jensen inequality?
Thanks.

Let $x>-1\wedge \alpha>1$.

\[ (1+x)^{\alpha} > 1+\alpha x,  \]
\[ \alpha\ln(1+x)>\ln(1+\alpha x).  \]

It's Jensen for $f(x)=-\ln x$.

($0=\left(1-\frac{1}{\alpha}\right)\ln 1$).


algebra
polynomial
geometry
3D geometry
ratio
trigonometry
number theory

[hide="Results"]Big Schools

Level 1
1st Place Individual - Sean
2nd Place Individual  - ?
3rd Place Individual  - Sage
1st Place Team - AC Flora
2nd Place Team - ?
3rd Place Team - ?

Level 2
1st Place Individual  - John B.
2nd Place Individual - Peter, Harry, Jim, Brian (tied, in order of "tiebreakers")
1st Place Team - Spring Valley
2nd Place Team - AC Flora
3rd Place Team - Hoggard

Overall
1st Place Team - AC Flora
2nd Place Team - ?
3rd Place Team - ?

I have a really bad memory, so someone fill in the blanks.
[/hide]

It was a much better contest than last year, but I thought level 1 was too hard (a 12 won it in small schools) and level 2 was way too easy.  But I liked the problems, and it was cool.

So I'm gonna post some level 1 problems some time soon.  Can someone post the hard level 2 problems?

Well, here are the hardest level 2 problems. As was said, it was pretty easy. They should have known last year that they needed to make it harder, after getting a 23, 24, and 25. At least this year they specifically said they needed to make it harder.

[hide]13. The equation $\sqrt{x+5}-\sqrt{x-2}+1=0$ has how many real roots and how many complex roots?

14. When the polynomial $5x^{4}+4x^{3}+3x^{2}+Mx+N$ is divided by $x^{2}+1$, the remainder is $0$. If $M$ and $N$ are real numbers, what is $M-N$?

16. Suppose that $sin{x}+cos{x}=a$. Write $sin^{4}{x}+cos^{4}{x}$ in terms of $a$. [i]These were multiple choice, so the only way I found to do this is plugging in values to the possibilities to see which works...[/i]

20. If a 4x4x4 cube is made from 32 white unit cubes and 32 black unit cubes, what is the largest possible percentage of black surface area?

21. Circle $C_{1}$ with center $P_{1}$ and circle $C_{2}$ with center $P_{2}$ intersect in points $Q$ and $R$ such that angle $QP_{1}R=30^{o}$ and angle $QP_{2}R=60^{o}$. What is the ratio of the area of circle $C_{1}$ to the area of circle $C_{2}$?

24. In triangle ABC, the point D lies on BC, and AD is the bisector of angle $BAC$. If $|AB|=c$, $|AC|=b$, and angle $CAD=w$, then what is $|AD|$, in terms of $b$, $c$, and $w$?[/hide]

Hmm... complaints about the test: Problems 12-19, with the exception of 15, were all blah classroom problems. The last page was exclusively geometry (with not a single geometry problem occurring before that last page). There was a total of one number theory problem, and that was "what is the last digit of 517^{1003}?" Also, I tend to think poorly of any test that contains more than one messy trig problem (I mean, basic law of cosines with 120 degree angles for example is fine, but nothing with 75 and 15 degree angles).

i tried getting my school to go to ccu, but unsuccessfully.
and assuming i may be able to find more than one other person in my school to go next year, does anyone happen to want to, say, post all the level II problems?  thanks^_^


I am the ruler of shovels
I have a twin
I am thin as a knife
I have a wife

Who am I?

[quote="kstan013"]Mary-Kate Olson :rotfl:[/quote]
Lol. Could be, but not quite the answer i was looking for

[quote="anyone"][hide="guess"]
king of spades?[/hide][/quote]
Good guess

[hide="Answer, I think."]The king of spades.[/hide]


Sorry, but I haven't been able to find this on the wiki...

How do you create a large brace for systems of equations? Thanks!

Well actually, I dunno if you want something like

[code]\begin{tabular}{rcl|}
$$ & = & $$ \\
$$ & = & $$ \\ \hline
\end{tabular}[/code]

Ohhh, that's easy

$ x_{k}\equal{}\left\{\begin{array}{cr}0 &\text{if }x_{k\minus{}1}\equal{} 0,\\ \left\{\frac{p_{k}}{x_{k\minus{}1}}\right\}&\text{if }x_{k\minus{}1}\ne0,\end{array}\right.$

[color=red]x_k=\left\{\begin{array}{cr}0&\text{if }x_{k-1}=0,\\
\left\{\frac{p_k}{x_{k-1}}\right\}&\text{if }x_{k-1}\ne0,\end{array}\right.[/color]

Does that make sense?

The cases environment is simpler
$ x_{k}\equal{}\begin{cases}0 &\text{if }x_{k\minus{}1}\equal{}0,\\ \left\{\frac{p_{k}}{x_{k\minus{}1}}\right\}&\text{if }x_{k\minus{}1}\ne 0\end{cases}$

[code]x_k=
\begin{cases}
0 & \text{if } x_{k-1}=0,\\
\left\{\frac{p_k}{x_{k-1}}\right\} & \text{if } x_{k-1} \ne 0
\end{cases}[/code]

Thanks for your help! And what package is the cases environment included in?

It's in amsmath which you should use in every document along with amssymb (which means you can write $ x\in\mathbb{R}$ for example).


Hi, my calculator gives the answer as 2012, which may serve as a hint.


[hide="hint"]$A^{x}=B$ and $B^{x}=A$

multiplying them seems like a good idea....[/hide]

[quote="now a ranger"]Suppose $A$ and $B$ are positive real numbers for which $\log_{A}B = \log_{B}A$. If neither $A$ nor $B$ is $1$ and $A\not=B$, find the value of $AB$.[/quote]
[hide]Let $\log_{a}b=x$
$x=\frac{1}{x}$
$x^{2}=1$
$x= \pm 1$
x clearly can't be 1 (they told us that $a \neq b$)
So x=-1
meaning that $A^{-1}=B$
so AB=1.[/hide]

But what he doesn't get is why $\log_{b}a=\frac{1}{\log_{a}b}$.

[hide="Proof."]$\log_{b}a=x\\b^{x}=a\\b=\sqrt[x]{a}\\b=(a)^{\frac{1}{x}}\\\rightarrow\boxed{\log_{a}b=\frac{1}{x}}$. Q.E.D.[/hide] :D


I love sudoku.

I love PBN and cross-sum as well.

Sudoku and other similar games can also be found through [url=http://www.puzzle.jp/index-e.html]Puzzle Japan[/url].  I am personally a fan of Edel.

Edel is PBN!
I knew in Korean it's 네모네모로직, but I did not know it was also called Edel.

Yeah.  I've seen it.  In fact, the math teacher I work for has gotten her eighth grade students practically addicted to those puzzles.

As for me, I love them, but I have done them in another incarnation--in the bimonthly issues of "Dell Math Puzzles and Logic Problems."  I can say I am pretty fast at these things.

[quote="bleumoose"]ah, yes, i found that article on MSN and read about it. downloaded and solved my first sudoku from DailySudoku. It's incredibly fun, but a warning- I had a "hard" one and I finished it during an errand trip. So the "easy" and "mediums" might be boring.[/quote]

I like to solve ones published in [b]The Times[/b] online newspaper. My another favorite among Japanese puzzles is PBN for more than a year already.

[quote="themonster"]Have any of you ever played [url=http://www.sudoku.com/]this game[/url]?  It's pretty fun and simple: You have a 9 X 9 grid separated into 9 3 X 3 smaller squares.  There are already integers between 1 and 9 in some cells.  Your objective is to fill in all the cells w/ integers between 1 and 9 such that every row, column, and 3 X 3 box has each integer between 1 and 9 in it exactly once.  It's pretty fun and involves lots of logic.  Try it![/quote]We've been playing this on the board already in the NEW [url=http://www.mathlinks.ro/Forum/index.php?f=360]Math & Strategy Games[/url] forum :) It's a ton of fun!

[quote="math92"]I LOVE SUDOKU!!!!!!!!!!!![/quote] i love them too in a matter afact i just did one!!!!!

[quote="math92"]i'm doing one right now!!![/quote]post it here :P

The Los Angeles Times will add them starting Monday- they're definitely spreading.

[quote="Geehoon"][quote="math92"]6**
*5*
**6[/quote]

wait... 6th box has two 6s...[/quote]
i must have typed it wrong :(
i solved it correctly though(without the six in the wrong place)
I don't know where the paper with the Sudoku on it is :(  :(


A prjectile of intial speed V m/s is fired at an angle of elevation alpha from the origin O towards a target T, which is moving away from O along the x-axis.

At the instant the projectile is fired, the target T is d meteres from O and it is moving away at a constant speed of u m/s. It can be easily derived that if the projectile hits the target when fired at a n angle of elevation alpha, that

u = V cos(alpha) - gd/(2Vsin(alpha))

i) Show that if u > V/(root 3), the target cannot be hit by the projectile no matter what angle of elevation the projectile is fired.

ii) Suppsoe u < V (root 3), Show that the target can be hit when it is at precisely two distances from O.

Different approaches for doing part i) would be appreciated, but part 2 is givine me problems. Thanks for any help.

ahh sorry guys, for part ii) it is actually V/root (3), not V root (3).  anyone got any ideas on how to do it? any solutions??

=D


calculus
integration
abstract algebra
advanced fields
advanced fields unsolved

Let $ X$ be a scheme of finite type over a field $ k$(not necessarily algebraically closed).
(a)Show that the following three conditions are equivalent(in which case we say that $ X$ is geometrically irreducible).

(i)$ X \times _k \bar{k}$is irreducible.where $ \bar{k}$ denotes the algebraic closure of $ k$(By abuse of notation,we write $ X \times _k \bar{k}$ to denote $ X \times _{Spec k} Spec{ \bar{k}}$).
(ii)$ X \times _k k_s$is irreducible.where $ k_s$ denotes the separable closure of $ k$.
(iii)$ X \times _k K$ is irreducible for every extension field $ K$ of $ k$.

(b)Show that the following three conditions are equivalent(in which case we say that $ X$ is geometrically reduced).

(i)$ X \times _k \bar{k}$is reduced.
(ii)$ X \times _k k_p$is reduced,where $ k_p$ denotes the perfect closure of $ k$.
(iii)$ X \times _k K$ is reduced for every extension field $ K$ of $ k$.

I can't translate the problem locally,and I can't tackle the problem when $ X$ is affine.Thank you for help.

Some hints for the nontrivial direction(s):

1. If $ X,Y$ are reduced schemes of finite type over $ k$ algebraically closed, then $ X \times_k Y$ is reduced: Reduce to $ X,Y$ affine, write a global section of $ \mathrm{Spec} (A \otimes_k B)$ as $ \sum a_{ij} f_i \otimes g_j$, $ (f_i),(g_i)$ $ k$-bases of $ A,B$. Restricting this to every closed point of $ X$ and using $ \bigcap_{\mathfrak{m} \in \mathrm{Spec}_\mathrm{max} A} \mathfrak{m} \equal{} (0)$ (reduced) gives $ \sum a_{ij}f_i(x) \equal{} 0$ for all $ j$, so $ a_{ij} \equal{} 0$. [Similar for integral]
2. If $ f \in A \otimes_k K$ is nilpotent, this holds already in a finitely generated extension of $ k$. Then apply 1. and that every $ k$-module is flat. [Similar for integral]
3. See http://en.wikipedia.org/wiki/Radicial_morphism

Sorry,Can somebody explain it,I can't understand it, :(

if you are still interested in this, I could elaborate it. but what is exactly unclear? have you followed myan's solution (sketch)?

bump...

If you could elaborate a bit more that would be great.


search
number theory proposed
number theory

Find $ a;b;c \in N$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3abc$

Hint: [hide="click text"] Do a Google search on "Markov numbers"[/hide]

take a look at:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159235[/url]
post #8


Which of the following numbers is a perfect square?
\\
A. $ 98! \cdot 99!$
\\
B. $ 98! \cdot 100!$
\\
C. $ 99! \cdot 100!$
\\
D. $ 99! \cdot 101!$
\\
E $ 100! \cdot 101!$

let's first look at option a

$ 98!\cdot99!$

we know that every number up to $ 98$ is repeated, and therefore makes a perfect square
however, we would still have a factor of $ 99$ left, and since $ 99$ is not a perfect square, this expression is not a perfect square

for option b, we use the same principle to see that the factors remaining are $ 99$ and $ 100$
since the product of these numbers is not a perfect square, the expression is not a perfect square

for option c, we can see that every factor up to $ 99$ is multiplied twice
the factor left is $ 100$, which is a perfect square, making the whole expression a perfect square

for options d and e, we would have a factor of 101 left over
since 101 is prime and it doesn't have a repeat, both d and e are not perfect squares

so our answer is $ \boxed{c}$


I'm trying to install LaTeX on my new computer, a vista. I have installed MikTex just fine and am having trouble downloading the TeXnic Center installer. I go to the SourceForge download page, the download options screen comes up and I click on "save", but then the options screen just disappears and won't let me choose a place to save the file to. Clicking on "run" doesn't seem to put the file anywhere either. Is this a problem with the site, my computer, or just the operator? Any help is appreciated. Thanks.

Is Internet Explorer blocking the download? Click on the yellow bar at the top to allow the download to proceed.

I do that, the screen refreshes, and then the menu comes up asking if I want to run, save, or cancel. I click on save, but it doesn't go into "my computer" and ask where I want to save it to. When I click on "run" it downloads as normal but then it's not saved anywhere, which doesn't do me any good.

It sounds like a permissions issue. Are you able to download other files from other places? 'Run' does save it to a temporary directory and you may be able to find it using Vista's search.

What happens if you click on this link to TexnicCenter http://ovh.dl.sourceforge.net/sourceforge/texniccenter/TXCSetup_1Beta7_01.exe
Does that allow you to download?

Do you have or are you able to download Firefox? If so, can you download from Sourceforge using Firefox?


Prove the following inequality $ (3a\plus{}2b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\le \frac{45}{2}$
where $ a,b,c\in [1;2]$.When does equality occur?

This problem is in an ongoing contest in Vietnam.


i use aops because it is :coool: er

first i was introduced to this site as http://www.mathlinks.ro though

[quote="robinhe"][quote="mathnerd314"]I find that it takes less effort to type "www.mathlinks.ro" than "www.artofproblemsolving.com". And I have less chance of making a mistake.[/quote]

Just put it as a favorite, like me.  :P[/quote]

Thats exactly what I did. :)


group theory
abstract algebra
linear algebra
superior algebra
superior algebra unsolved

How many Sylow $ p$-subgroups are there in the group $ GL_n(F_p)$?

A typical $ p$-Sylow subgroup is the matrices of the form $ I\plus{}N$, where $ N$ is a strictly upper triangular nilpotent matrix. All other choices are of course similar.

Now the structure: we have an eigenspace $ V_1$ for which $ Nv\equal{}0$ whenever $ N$ is any of these matrices and $ v\in V_1$, a space $ V_2$ for which $ N^2v\equal{}0$, and so on.
The $ p$-Sylow subgroups are in 1-1 correspondence with chains of subspaces $ V_0\subset V_1\subset V_2\subset\cdots \subset V_n\equal{}F_p^n$ where each $ V_k$ has dimension $ k$.

Those chains of subspaces go by the name of [url=http://en.wikipedia.org/wiki/Flag_(linear_algebra)]complete flags[/url], and it's not too bad to count them directly.  One can also embed the Sylow $ p$-subgroup $ S$ given by jmerry into the [url=http://en.wikipedia.org/wiki/Borel_subgroup]Borel subgroup[/url] $ B$ of upper triangular matrices and argue that the normalizer of $ B$ is $ B$ (as well as arguing that $ S$ is normal in $ B$).


ratio
geometry
3D geometry
calculus
integration

[quote="grizzland"]It takes you 1 gallon of paint to paint the inside of a box. The next day you paint a box that has the same ratios of length to width and length to height, but it takes you 2 gallons to paint the box. You conclude that the total inside area of the larger box is twice that of the smaller box. You conclude that the volume of the larger box equals_______  times that of the smaller box.

 You have two boxes of the same shape that hold flour. The larger box holds twice as much flour as the smaller one. You conclude that it would take you______  times as much paint to paint the larger box than it would take you to paint the smaller box.[/quote]

For the first one:

[hide]Ratio of Areas is $1: 2$ thus ratio of sides is $\sqrt{1}: \sqrt{2}$ $\rightarrow$ $1: \sqrt{2}$.  The ratio of volumes is the cube of the sides which is $1^{3}: \sqrt{2}^{3}$

Thus the second box is $\sqrt{2}^{3}=2^{3/2}=\boxed{2\sqrt{2}}$ times as large as the first in terms of volume.[/hide]

For the second:

[hide]Same as above with volumes.  $\text{Small}: \text{Big}\rightarrow 1: 2$ in terms of volume ratios.  Thus the side ratios are $1^{1/3}: 2^{1/3}\rightarrow 1: 2^{1/3}$ and the ratio of the areas would be $1^{2}: (2^{1/3})^{2}\rightarrow 1: 2^{2/3}$.

Thus the second box would require $\boxed{2^{2/3}}$ times as much paint as the first.[/hide]

Basically you have to look at it as:
Side to side is $a: b$
Area to area is $a^{2}: b^{2}$
Volume to volume is $a^{3}: b^{3}$

The first blank is $2\sqrt{2}$, since the an amount of paint covers a certain surface area, two gallons implies twice as much surface area as the one gallon; that then implies the product of two sides is twice as big.  So, the sides increase in proportion by $\sqrt{2}$, $\sqrt{2}\sqrt{2}= 2$. Then the volume increases accordingly:
$xyz=V \Rightarrow x\sqrt{2}\left(y\sqrt{2}\right)\left(z\sqrt{2}\right)= 2V\sqrt{2}$ .

The second blank is $\sqrt[3]{4}$.
To get the increase per side:
$2^{\frac{1}{3}}$ .
Then per surface area:
$\left(2^{\frac{1}{3}}\right)^{2}= \sqrt [3]{2^{2}}= \sqrt [3]{4}$ .

This may be redundant, to what was just posted. :P

I guess to prove the concepts behind scaling that grizzland was confused about requires calculus but it is really easy to formalize then. (just consider integration over large dimensions)


trigonometry
geometry
geometric transformation
reflection
LaTeX
angle bisector

In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.

Siuhochung, you forgot the additional condition $\angle A\ge\angle B$.
Let M be the mid-point on AB and $\theta$ be the degree of the four equal angles,
then \[1=\frac{AM}{MB}=\frac{[AMC]}{[MBC]}=\frac{AC\cdot CM \sin3\theta}{BC\cdot CM \sin\theta}=\frac{AC}{BC}\frac{\sin3\theta}{\sin\theta}=\frac{\sin(90^\circ -3\theta)}{\sin(90^\circ -\theta)}\frac{\sin3\theta}{\sin\theta}=\frac{\sin6\theta}{\sin2\theta}\]
$(\frac{AC}{BC}=\frac{\sin(90^\circ -3\theta)}{\sin(90^\circ -\theta)}$ is obtained by sine rule in $\triangle ABC. )$
So $6\theta=2\theta$ or $6\theta+2\theta=180^\circ$
Rejecting the first one, we have $\theta=22.5^\circ$

[quote="shobber"]Is there a geometrical way to find the answer?[/quote]
mecrazywong's solution:
Since O and H are isogonal conjugates, O is on the median, thus O is the midpoint of AB, then angle C is 90 and the answer comes out readily.

I got mine now:

Let $M$ be the midpoint, $CD$ is the bisector of $\angle{C}$, $CH \perp AB$. Let $BH=HD=y$ and $MD=x$, therefore $AM=x+2y$.  Use the angle bisector theorem:
\[\frac{2x+y}{x}=\frac{AC}{DC}=\frac{AC}{BC}=\frac{x+y}{y}\].
Hence: $\frac{x}{y}=\sqrt{2}$.

Then use the angle bisector theorem again in $\triangle{CMH}$: $\frac{x}{y}=\frac{CM}{CH}$. And because $\angle{CHM}=90^o$, so we have: $2CH^2=CH^2+(x+y)^2$, which gives: $CH=x+y$. Then $CM=\sqrt{2} CH=2y+\sqrt{2}y=BM=AM$. So $\angle{C}=90^o$.

[quote="leepakhin"]Would it be too stupid for me to ask what are 'isogonal conjugates'?[/quote]

You may read darij's post on isogonal conjugates.
http://www.mathlinks.ro/Forum/viewtopic.php?t=18472

In brief, if you reflect three concurrent cevians with respect to their angle bisectors, the reflected cevians still concur and the concurring point is called the isogonal conjugate of the origianl point.

I know that this does not help , but

  a) the problem is from Prasolov' s << Plane geometry >> , where there is a simple solution.
  b) this problem was given again in the Greek Math Competition 2004 - 2005. I have one more elementary , beautiful  and simple solution.
  I' ll  ask to a friend to write this solution in Latex for the ... mathlinkers.
 (excuse me if I have done something wrong. It ' is the first time I write here ...)

Sorry ,
  the comments of my last post concern the topic
 << http://www.mathlinks.ro/Forum/viewtopic.php?p=322058#p322058 >>

  I kindly ask from the moderators to cancel these two posts from here.

Thank you

It is kownthat $CH$ and $CO$ are isogonal conjugates. From this we get that $O$ lies on the median from $C$. Hence, the triangle is rightangled or issoscelles. If it is issoscelles the hipothesys connot hold. Hence it is rightangled. Siince, in the right-angled triangle the median is a half of the side it bisects, we get that each of the four angles from $C$ is equal to $B$. Hence $B=\frac \pi 8$


Express the following as a mixed number:

\[ \minus{}7\minus{}(\minus{}7)^2\plus{}(\minus{}7)(\minus{}7)\plus{}7(\minus{}\frac{1}{7})^2\]

to solve this, we simply evaluate the expression

$ \minus{}7\minus{}(\minus{}7)^2\plus{}(\minus{}7)(\minus{}7)\plus{}7(\minus{}\frac{1}{7})^2$

$ \minus{}7\minus{}49\plus{}49\plus{}7(\frac{1}{49})$

$ \minus{}7\plus{}\frac{1}{7}$

$ \boxed{\minus{}6\frac{6}{7}}$


Suppose that $a_n$ is a sequence of non-zero real numbers that goes to 0 at infinity. Prove that for any real number $x$ one can find integer sequences $k_n, m_n$ such that $x$ is the sum of the infinite series of general term $a_n k_n$ and  also product of all $m_n a_n$.

It seems we should just use the approach of Riemann when he shows that a conditional convergent series can be arranged to do anything.

We just go along and pick the smallest integer $k_N$ such that the partial sum of $k_na_n$ up to $N$ is greater than $x$. Then for the $N+1$ we pick the largest integer so that the partial sum is less than $x$. It is not hard to see we must get within $a_n$ each time, so it converges.

Then for the $m_n$ one it is similar; we want to pick the one that makes teh product just greater/just smaller each time. Now we are sure we're multiplying by a number $1 + a_n \ge a_nm_n \ge 1 - a_n$. It seems it converges as well but for the moment it is not so obvious to me the reason. Maybe we take logs? I shouldn't be so lazy, I bet it is not really too hard.


Given triangle $ ABC$. $ <C\equal{}90$. $ M$ is a midpoint of $ AB$. $ CH$ ia an altitude.Point $ P$ is internal point of triangle $ ABC$. We know that $ AP\equal{}AC$. Prove that  $ PM$ is bisector of $ <BPH$ if and only if $ <A\equal{}60$.

[quote="matex (L)(L)(L)"]Given triangle $ ABC$. $ < C \equal{} 90$. $ M$ is a midpoint of $ AB$. $ CH$ ia an altitude.Point $ P$ is internal point of triangle $ ABC$. We know that $ AP \equal{} AC$. Prove that  $ PM$ is bisector of $ < BPH$ if and only if $ < A \equal{} 60$.[/quote]

It's easy to achieve that if $ \angle A\equal{}60^o$ then that $ PM$ is bisector of $ \angle BPH$.
So i will prove only this:
If $ PM$ is bisector of $ \angle BPH$ then $ \angle A\equal{}60$
We have :
 $ \frac{AH}{AP}\equal{}\frac{AH}{AC}\equal{}\frac{AC}{AB}\equal{}\frac{AP}{AB}$ and $ \angle HAP\equal{}\angle PAB$
therefore $ \triangle AHP \sim  \triangle APB$
then we have:
$ \angle APH\equal{} \angle ABP$
$ \angle APM\equal{}\angle APH\plus{}\angle HPM\equal{}\angle MBP\plus{}\angle BPM\equal{}\angle AMP$
therefore $ AC\equal{}AP\equal{}AM\equal{}\frac{1}{2}AB$
Therefore $ \angle A\equal{}60^o$


geometry
geometric transformation
conics
parabola
analytic geometry
geometry unsolved

We have a circle C and an exterior line L. Find the geometric locus of the centers of the circles which are tangent both to the circle C and the line L.

If r is the radii of C(O) and R the radii of the variable circle. Apply a translation that maps L into L'  in the perpendicular direction of L and magnitude r, then the center of C equidist form L' and O, the locus is a parabola with foci O and director L'.

Sorry if i am silly . but could you explain this to me ?  :oops:  tx in advance

[quote="Kenshin24"]Apply a translation that maps L into L'  in the perpendicular direction of L and magnitude r, then the center of C equidist form L' and O.[/quote]

I have an ugly approach using coordinates , i'm interested in understanding what you did without coordinates.


its 'HMMT'

ill be going

are you going with your school?

its havard mit november tournament so its hmnt

i am going with lexington


quadratics
number theory proposed
number theory

Let n be a positive intenger such that $2+2 \sqrt{28n^{2}+1}$ is an intenger.Show that $2+2 \sqrt{28n^{2}+1}$ is the square of an intenger.

actually the solution can be simplified :)
$2+2\sqrt{28n^{2}+1}=4m$ implies $7n^{2}=m(m-1)$ we have two cases
1) $m=k^{2}$ then we are done because $2+2\sqrt{28n^{2}+1}=(2k)^{2}$
2) $m=7k^{2}$ but then $m-1=r^{2}$ which contradicts the fact that $-1$ is not a quadratic residue $(\mod 7)$
So we are finished...

Thanks.I did like Eagle, but I haven't conclude the second case, thanks

See also here
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=43334[/url] :wink:

set $2+2\sqrt{28n^2+1}=m$ where $m \in \mathbb{Z^{+}}$\\

we get $4(28n^2+1)=(m-2)^2$, now clealry $m$ has to be even\\

set $m=2k$ for $k \in \mathbb{Z}$\\

so we get $4(28n^2+1)=4(k-1)^2 \implies 28n^2=k(k-2)$ , now again k has to be even so we set $k=2j$ for $j \in \mathbb{Z}$\\

hence we get $7n^2=j(j-1)$ set $n=xy$ for some integers $x,y$ s.t. $\gcd(x,y)=1$, now $\gcd(j,j-1)=1$ we get:\\
[list]\\

[*][color=#00f]Case 1:-[/color] $j=7x^2 , j-1=y^2 \implies 7x^2-y^2=1$ but reducing this $\pmod 7$ we get no solution\\

[*][color=#00f]Case 2:-[/color] $j-1=7x^2, j=y^2 \implies 7x^2+1=y^2$ which works for every $(x,y) \in
\mathbb{Z}$ we choose.
[/list]\\

hence we get $k=2y^2$ and $m=4y^2=(2y)^2$ which is square of an integer. $\blacksquare$


limit
integration
trigonometry
calculus
real analysis
real analysis unsolved

Calculate $ \lim_{n\leftarrow \infty}\int_{0}^{\pi}\frac{\sin x}{1\plus{}\cos ^2 nx}dx$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146991]Two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]solutions[/url] in more generality. The only detail is to compute the integral of $ \sin$ on $ [0,\pi]$ ($ 2$) and the average value of $ \frac1{1\plus{}\cos^2 x}$ ($ \frac1{\sqrt{2}}$).

ok, so well known I thought it must have been posted before. BTW this or something very similar is an IMC 1994 problem. The exact one is taken from a very old schweitzer competition.


:huh:

Is there a number between 1 and 1 thats not 1?
Irrational numbers count.

No.  You're thinking too hard.

The reals satisfy a property known as [i]trichotomy[/i], which says that any number must either be positive, zero, or negative.

Suppose there is such a number; call it $ \alpha$.  What is $ 1 \minus{} \alpha$?  It can't be zero because $ 1 \neq \alpha$.  It can't be positive because then it wouldn't be less than $ 1$.  And it can't be negative because it wouldn't be greater than $ 1$.

Well, is [i]i[/i] greater, less, or equal to zero?

$ 1\minus{}i$ is a complex number. Not a real number.

The complex numbers don't satisfy trichotomy no matter how you redefine positive and negative, so it doesn't make sense to talk about whether a complex number is greater than or less than another.

So, if you graph 1 on a complex plane, wouldn't it lie on the line 1+[i]n[/i][b][i]i[/i][/b], and wouldn't it thus mean that 1+[i]n[/i][b][i]i[/i][/b] is equal to 1 in the real axis, but yet not be equal to 1?

On the complex plane, $ 1$ is a point.  The set of points $ 1 \plus{} ni, n \in \mathbb{R}$ is a line.  What do you mean "equal to $ 1$ in the real axis?"  If you mean that its projection onto the real axis is a point, you'd be correct.  That has nothing to do with the definition of "greater than" or "less than."

What's interesting, is that $ i$ can be graphed on a polar system and $ i$ will approximately be on $ \left(1, \frac{\pi}{2}\right)$

[quote="7h3.D3m0n.117"]What's interesting, is that $ i$ can be graphed on a polar system and $ i$ will approximately be on $ \left(1, \frac {\pi}{2}\right)$[/quote]
If we plot a point like $ \sqrt {i}$ in polar coordinates, is it correct to read it as $ \left(1, \frac {\pi}{4}\right)$ or as $ \left(\frac {\sqrt {2}}{2},\frac {\sqrt {2}}{2}\right)$ from knowing $ \sqrt {i} \equal{} e^{i\frac {\pi}{4}} \equal{} \frac {\sqrt {2}}{2} \plus{} i\frac {\sqrt {2}}{2}$?

Oh it was $ \frac{\pi}{4}$? Wow I really need to work on my memory.

[quote="7h3.D3m0n.117"]Oh it was $ \frac {\pi}{4}$? Wow I really need to work on my memory.[/quote]

No, he said $ \sqrt{i} \equal{} 1\ \text{cis}\ \frac{\pi}{4}$. Indeed the coordinates of $ i$ are $ (r,\theta) \equal{} (1, \pi/2)$.

As for how to write it, I guess, just make sure there is no ambiguity in the notation. How the coordinate should be read should follow from context.

The complex numbers don't have trivial comparisons of magnitude like the real numbers do.

Also, $ i$ in polar coordinates is $ \left(\frac{\pi}{2},1\right)$; you were correct, Henry. Patterns_34 was talking about $ \sqrt{i}$.

I apologize for confusing everyone by putting in $ \sqrt {i}$.  :oops:
Anyways, thank you for the clarification.

[quote="Temperal"]The complex numbers don't have trivial comparisons of magnitude like the real numbers do.[/quote]

The complex numbers can be partially ordered by magnitude; the issue, as I have already said, is that trichotomy is violated.  I'm not sure what you meant by "trivial."