Who do think is going to win?
I say MENS: Roddick, Federer, or Safin if he plays well.
        WOMENS: Henin-hardene or sharapova

MENS: Federer or Hewitt
WOMAN:H-H or Williams, if she comes back.

Roddick has to prove that he is out of the major-less slump.... or whether he could be the player he was when he used to be No.1 in the world.

Nadal has to prove that he is better than what people think he is [b] outside [/b] of clay.

Safin has to prove that he can control himself.

As for the woman's side:

Sharapova needs to prove that she can beat H-H or Serena if she were to face them before finals, or even before semi.

Personally, I dont think Davenport deserves to be No. 1, and she would have to prove that she is, by winning the Wimbledon.

But my picks still stands as it is.

MEN: Federer... I don't even think I need another name here, but we'll see. Next would be Roddick or Hewitt.
WOMEN: Henin-Hardenne or Sharapova

Federer is just unstoppable on grass and, frankly, he's due for a Grand Slam now :P. I'd like to see Roddick make it to the semis at least since grass is one of his better surfaces, and Hewitt is just fast so he's always a possibility imo. Anyway, Federer just took out Safin in a tournament, so I'd say Safin has relatively little chance unless some fluke knocks Federer out before he plays Safin. I doubt either of the two other players I mentioned could beat Federer either, so yeah, we'll see...

On the women's side, I'd go for Sharapova but she's really young, so who knows. Henin-Hardenne has done pretty well coming back, but I don't know how she'll fair on grass. Then again, I don't pay attention to the women as much so I wouldn't be one to call it.

When does it start/when does the coverage of it start?

[quote="WindSlicer"]One does it start/when does the coverage of it start?[/quote]

June 20, so about a week from now... yay! :D

Awesome, can't wait. Just started playing tennis, and when I watch them play I improve a lot, like after the French Open I played tons better.

On a side note, I think it's funny how sometimes we type stuff depending on how sometimes it sounds in our head.. i.e. how I put "one" instead of "when". *shrugs* heh

Hmm...someone with more tennis experience could probably shed more light on this, but in my opinion Sharapova is not experienced enough to be reliably consistent in major tournaments like Wimbledon.

On the other side, Federer is probably my pick for most likely to win. A few of my friends got his autograph when they were playing in Houston. :-) He seems like a nice fellow actually.

sharapova is young but she is more than fit and perfect for W.
and she super hot

[quote="plokoon51"]hopefully, federer will be able to tackle the grass surface better than he did at the French open. personally, i want roddick to win cause he hasn't won a major in a while, but federer will probably win it all.[/quote]

Federer's game is way better on grass than on clay, or at least compared to his competitors

[quote="cadge_nottosh"]An absurdly optimistic prediction:

Henman will win![/quote]

God save the queen!

ANDY RODDICK! err, that's how you spell it right? but dude rafael-something is really good too for a beginner. but he's not always a great player.

Rafael Nadal is only good on clay, he stinks on grass!

wow,serena wiiliams lost to Craybas, who is no. 85 in the world.  Did any of you guys see thr shot Federer madeb when he was playing Kiefer?  It was in the last game of the match, Kiefer was at the net, and Federer hit an amazing crosscourt backhand.  After the match, he said that the shot was the best shot of his life.

[quote="GoBraves"][quote="math92"]If they played i think Roddick would win in 5 sets, but hewitt is 6-1 agains roddick[/quote]

Which is exactly why Lleyton would win..because he's better. Roddick is overrated a bit too much.[/quote]

He's overrated, yeah, but not that much. There's just a lot of hype about him because he's American. I think he deserves his #4 spot in the world right now. A match between Roddick and Hewitt might actually be interesting, instead of both being owned by Federer.

[quote="juicybooty911"][quote="plokoon51"][quote="manuel"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane!
btw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]

 :o thats awespme!!!

too bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.[/quote]

Safin has 2[/quote]

Yeah he has One U.S open and an Australian, but he's made the finals tons of times. I wonder why he isnt doing well in U.S Open anymore, when he won he played amazingly.

[quote="GoBraves"][/quote][quote="juicybooty911"][quote="plokoon51"][quote="manuel"]The only one that can beat federer is marat safin!!! he's the guy with most ability on tour! but he is insane!
btw, i hit yesterday with a guy that beat andy 2 times in boy's 16 !!![/quote]

 :o thats awespme!!!

too bad safin isn't very dedicated to practicing tennis. he has won one grand slam tournament already and could (should) do better in wimbledon.[/quote]

Safin has 2[/quote][quote="GoBraves"]

Yeah he has One U.S open and an Australian, but he's made the finals tons of times. I wonder why he isnt doing well in U.S Open anymore, when he won he played amazingly.[/quote]

oops...sorry :blush: i was talking about this year. i forgot about the US Open win over Sampras (how could i forget!!!)

Safin is inconsistent and seems to be bad on grass (at least he seriously complained about it after last year's Wimbledon.)

safin can be the best in the world when he is on fire and he can be the worst when he is angry.

yeah... but i thik he will win like a couple of  grand slams more...

[quote="math92"]maybe...i dunno...possibly one more.
but for Nadal i see 4 more.[/quote]

I dont see anymore for Nadal...

well there is this really good 15 year old named Donald Young who might have 4 slams.

i think nadal can become fairly competitive on grass in a few years time. dont forget, he's only 19. however, i doubt anyone's gonna beat federer anytime soon.


8*sin(20)*sin(70)*sin(10)*sin(50) = ?
i got it down to
8*sin(20)*cos(20)*sin(10)*sin(50)

dont know what to do from there.

Just use a bunch of co-function and double angle identities.

[hide]
\begin{align*}
\ 8\sin(20)\sin(70)\sin(10)\sin(50) &=8\sin(20)\cos(20)\sin(10)\cos(40) \\
\ &=4\sin(40)\cos(40)\sin(10) \\
\ &=2\sin(80)\cos(80) \\
\ &= \sin(160) \\
\ &= \sin(20).
\end{align*}
[/hide]

this does not go in intermediate, more like HSB.


Has the jenks team been chosen yet?
If so, who is on the team this year?

Hi Basm? Is dat how u spell ur name? yeah.. This is the team:
Jonathan-Me/7th
Cheeho -8th
Eric-8th (me bro)
Callie-7th
Arnav-8th
Mitchell-8th
Caleb-8th
Shawn-8th

BTW: Regionals already over and I won!!!

Yo. State are me, carphead, aznkid, mrk, and shortstack

What about your "brother" jonathanchou711?

same person as aznkid. duh. i don't have multi tho.

Whoa dude! I've never met you! Dang...

I get it now!

[size=200]Basem!!! [/size]

thank you!!!
and kill arnav for starting the "[b]basem[/b]ent" joke
please

Basement? It's okay. I already beat him up. Despite his size, I beat him up with disses... :) jk jk jk. I'm only 4 ft 10 though.... :(  :|


I am trying to write a conjecture for an isosceles trapezoid without using an altitude. I hope you can help me.

do you mean construct?

otherwise, i have not idea what your question is

draw a circle, draw a line that intersects the circle 2 times (and does not go through the center of the circle), connect these 2 intersections and the center of the circle, this is an isosceles triangle because 2 sides are congruent because they are radii of a circle

He wants an isosceles trapezoid, not triangle.  I think you could do it similarly though.  Draw a secant that doesn't go thru the center.  Draw the radii to the intersection points.  Now draw another secant that intersects both radii that is parallel to the first secant.  Then you have an isosceles trapezoid.

Thanks, how would you write the proof for this?


I found these a bit tough. It would be great if you could provide the steps long with the answer. Thanks in advance!  :P

1. A swimming pool can be filled either using a pipe, a hose, or both. Using the pipe alone takes 12 hours. Using both it takes 60/7 hours. How long does it take using the hose alone?

2. Wayne drove 3 hours on a freeway at 55 mi/h and then drove 10 miles in the city at 35 mi/h. What was his average speed?

3. Donna drove half the distance of a trip at 40 mi/h. At what speed would she have to drive for the rest of the distance so that the aberage speed for the entire trip would be 45 mi/h?

4. At what time after 4:00 will the minute hand and the hour hand of a clock first be in the same position?

I solved a bunch of these types of questions, and for some reason, I kept getting the incorrect answers for these.

[quote="apriori"]1. A swimming pool can be filled either using a pipe, a hose, or both. Using the pipe alone takes 12 hours. Using both it takes 60/7 hours. How long does it take using the hose alone?[/quote]

[hide="Solution"]Okay, so in one hour, using the pipe would fill up $ \frac1{12}$th of the pool.  Using both, it would fill up $ \frac {7}{60}$ths of the pool.  So to find how long it would take to use the hose, we take the reciprocal of $ \frac {7}{60} \minus{} \frac1{12} \equal{} \frac {2}{60} \implies \frac {1}{30} \implies \boxed{30 \text{ hours}}$[/hide]

[quote="apriori"]2. Wayne drove 3 hours on a freeway at 55 mi/h and then drove 10 miles in the city at 35 mi/h. What was his average speed?[/quote]

[hide="Solution"]Wayne drove $ 3 \times 55$ or $ 165$ miles on the highway.  Thus, we must do a weighted average:

$ \frac{165 \text{ miles} \times 55 \frac{\text{miles}}{\text{hour}} \plus{} 10 \text{ miles} \times 35 \frac{\text{miles}}{\text{hour}}}{175 \text{ miles}} \implies \frac{(9075 \plus{} 350) \frac{\text{miles}^2}{\text{hour}}}{175 \text{ miles}} \implies \frac{9425 \frac{\text{miles}^2}{\text{hour}}}{175 \text{ miles}} \implies \boxed{53 \frac67 \text{ miles}/\text{hour}}$[/hide]

[quote="apriori"]I found these a bit tough. It would be great if you could provide the steps long with the answer. Thanks in advance!  :P

I solved a bunch of these types of questions, and for some reason, I kept getting the incorrect answers for these.[/quote]

apriori, you see what I highlighted?

You should post at least some of your work on all of these problems, along with
your final answer attempt, so we can see your thinking, and then we not doing
your work.

[hide="3"]
$ \frac {d}{2} \equal{} 40t$
$ t \equal{} \frac {d}{80}$
Let $ r$ be the rate for the second half of the distance.
Then, $ t \equal{} \frac {d}{2r}$.

Then, we have:
average $ r \equal{} \frac {total distance}{total time} \equal{} 45$

average $ r \equal{} \frac {d}{\frac {d}{80} \plus{} \frac {d}{2r}} \equal{} \frac {1}{\frac {1}{80} \plus{} \frac {1}{2r}} \equal{} 45$
Solve for $ r$.
[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=548908835&t=172357]Hint for Number 4.[/url]


This question comes from The Art and Craft of Problem Solveing.

Find the minimum value of $xy+yz+xz$, given that $x$, $y$, $z$, are real and $x^2+y^2+z^2=1$.

oi... this is pretty easy with lagrange multipliers, which we just learned to use...
but that's calc, so i'm guessing you want a more basic approach?

[hide]$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=1+2(xy+yz+zx)\geq 0\Rightarrow xy+yz+zx\geq -\frac{1}{2}$

Equality is attained when $x=\frac{\sqrt{2}}{2},\ y=-\frac{\sqrt{2}}{2},\ z=0$. Therefore minimum is $-\frac{1}{2}$.[/hide]

[quote="maokid7"]$2x^2+2y^2+2z^2-2(xy+xz+yz)$
and since $x^2+y^2+z^2=1$ we have
$2-2(xy+xz+yz)=2(1-xy-yz-xz)$

sorry but i do not see how this helps. :([/quote]

Sorry, I was mistaken about the Maximum :blush:  value!

o i see now. i feel so stupid now. i tried AM-GM and other inequality tactics on $(x+y+z)^2$. but never thought about the trivial inequality.  :oops:  :blush:


I was wondering what cyclic and symmetric sums were, and possibly some of their common uses.

Cheers,
The Idiot without a village

Basically cyclic sums are when we have some formula where relabelling the variables doesn't change anything. For example ab+bc+ca, or a:^2:+b:^2:+c:^2:+2ab+2bc+2ca+abc.

As to what you'd use them for.. apart from just simplfying how much you need to write (for the first thing above you could just write cyc  :Sigma: ab (not sure if this is official language or anything but I think its usable).. they just often pop up all over the place because they are 'nice'.  For example most inequalities use cyclic sums. Can't really think of any special use of them though..

Cyclic sums are of outmost importance in some part of Algebra, the Galois theory for instance.

Just to give a small example :

Let's consider an equation  :

123 * x^56 + 456 * x^12 + 789 * x^5 + 123456789 = 0

There are 56 roots x_i (not necessary all different) to this horrible equation.
I have NO idea of its roots but by Viete relations I can state that :

sum x_i = 456/123
sum x_i * x_j (i < j) = 0
sum x_i * x_j * x_k (i < j < k) = 0
....

prod x_i = 123456789/123.

What about symmetric sums?  What are they?

I'm pretty sure they're just the same thing.

There's a slight difference. For example, assume we have the variables a, b, and c. Then the symmetric sum of a^2*b is a^2*b+a^2*c+b^2*a+b^2*c+c^2*a+c^2*b. But the cyclic sum of a^2*b is a^2*b+b^2*c+c^2*a. Sorry, but I can't think of a way to explain it better. I hope this is sufficient.

Are there any specific links or theorems you may provide me with so that I may view exactly where these cyclic/symmetric sums occur?

Also are there any links that contains elementary mathematics topics not coverred in the High School Curricula throughout the nation.

Thank you, I appreciate your aid on this matter,
The Idiot without a village

Is it safe to assume that this has nothing to do with cyclic quadrilaterals?

Why yes Pierre [if that is indeed your real name]... I do believe that it is safe to assume that this has nothing to do with cyclic quadrilaterals... see ya tommorow in class..

Cyclic quads taste yummy...  but what really turn's me on is modelling photosynthetic biochemistry via PDEs.

Peace from the Idiot...

I also like it when people rub my tummy.

[quote=achen29]Just 15 years late but; Newton's Theorem or something like that about symmetric sums[/quote]

If you had a question, you could start a new topic. Just necroing a 15 year old post with no major contribution other than "Hey just wanted say Newton's Theorem" is bad, especially since the OP's question already got answered long long LONG time ago. I remember a bunch of these guys when we were in high school. I'm sure they all graduated college by now....


If you take [url=http://www.artofproblemsolving.com/Forum/index.php?f=271]this[/url] link, and "close" the Vietnam tab, India, Iran, Pakistan, Tajikistan become hidden as well.

Just something small I'd like to mention.

how come India, Iran, Pakistan, & Tajikistan are part of Vietnam now? :P


search
Ross Mathematics Program
inequalities
IMO Shortlist
graph theory
\/closed

Hi,  :)

I would like to know your opinion. Do you think we should mention the source to all problems we are posting ? One of the advantages of not mentioning the sources: Nobody will be seduced to the search the web for the problem, e.g. kalva and others.
So if we want to mention the sources. Do you think we should edit the old postings, e.g. from the unsolved section or even the solved ones (this would mean to unlock all the problems first) ? It also should be nice if others will tell us about the source where they have seen the problem as well. If the source is unknown, it should be indicated as unknown. Otherwise we would assume that you forgot to mention the source. Another advantage is that we can remember a problem easier though we might have forgotten about the problem's statement. Moreover we get to know about the preferences, meaning which olympiads they like, of the mathlinks users.

searching the web for problems doesn't seem like a suitable idea for those pursuing math (olympiad :D) knowledge.

so my guess is that we should try to post the sources when we can. :) for quicker indexing and searching of the problems.

So my idea is the following one: Please indicate the source of the problem below the problem statement in Italic letters. If you want to prevent others from looking up the problem somewhere, e.g. Kalva. In this case you might post the source when the problem is solved. Therefore we don't try to lock/remove the problems quickly so that you have time to add the source.

Suggestions for citing sources are given at: http://www.mathlinks.ro/phpBB/viewtopic.php?t=1053

Honestly, I don't know why anybody would want to prevent somebody else from looking up a problem on the web.  After all, it does seem like "cheating", so I don't think anybody who really wants to do the problem seriously would consider the idea.  It's much simpler to state the source when posting the problem.

I think, more importantly, since most questions are not our own, it's only correct that we acknowledge where we got the problem from (so if you forgot, just say so).  And of course, it is convenient for future reference.

Orlando, regarding your idea of including the year, stage, class, problem no., I think a simple way to get around the issue of people omitting certain information would be to put a letter in front of each number e.g. 2003 S4 C13 P3.

Many thanks for your suggestion Valiowk. I hope everybody will keep to the rule. Otherwise I will always ask for the source. Maybe we should publish a document on the site where you can read about how your posting should look like, meaning some general summarized rule. Every new member should be informed about the document when subscribing to mathlinks.

well I have made some annoucements earlier concerning that. I am planning to gather them up ... thanks for the sugestions, and keep up the good work orl :)

In my opinion it is a good idea to indicate the source of the problem only after its solution is posted.
A question. I have in my notebook a lot of problems of which I don't remember the source so shall I write  "source unknown" or "source forgotten" ?

Why is it so important to know the source of a problem  :?  Problems exist by them-selves, don't they ?

Each week hundreds of new problems emerge on the site. already some of them are repeatting themselves. It is a good ideea to be able to search for specific problems to see if they are posted. :)

[quote="Valentin Vornicu"]Each week hundreds of new problems emerge on the site.[/quote]
Sure they do. :D
Lets not exaggerate. This forum has yet to grow to reach 100 new topics a week.

Maybe you could even add some dropdown/text inputbox in the "post a reply" window?

I am not sure what you mean mindspa. maybe you can ellaborate a little bit ? :)

I think it is very important to indicate the source. There are three kinds of sources:

(1) self-created  - you created the problem by yourself and you are not aware that it has already been proposed anywhere else
(2) "competition" - tell the source in the way valiowk suggested it
(3)  unknown       - in the case you don't know the source

Maybe it is possible for (2) and (3) to say where you have found the problem, meaning which journal, book etc. There are several problem which are used in competitions again, sometimes known by the guys proposing problems and sometimes not. Anyway I feel it is an appreciation for a certain problem to emerge in many sources.

Please have a look at the following postings:

(a)  http://www.mathlinks.ro/phpBB/viewtopic.php?t=1985
(b)  http://www.mathlinks.ro/phpBB/viewtopic.php?t=1028

To provide an example, problem (b) was used in many sources and I even did not remember every source back then. So I try to mention all sources I am right now aware of:

(I)   "early" Chinese mathematical olympiad
(II)  "early" AMM volume which covered Chinese math olympiad problems from (I)
(III) Honsberger, Ross; Mathematical morsels; MAA, 1978
(IV)  Richard K. Guy, The strong law of small numbers, American Mathematical Monthly, 95: 697-712, 1988
(V)   E. A. Morozova, I. S. Petrakov, V.A. Skvortov; International Mathematical Olympiads; Moscow
      (comment: the problem in this source is taken from an early IMO shortlist)
(VI)  Paul Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999
(VII) I. Tomescu, R. A. Melter; Problems in Combinatorics and Graph Theory; John Wiley & Sons, 1985

Another reason might be the Valentin's idea to browse the forum easier. Maybe it is possible to establish a menu for each problem where the user can write down the source of a problem. Other users may edit the menu. But they can only add different sources where they run across the problem.

Please also indicate the soure below the problem statement though the subject (of your created) might be something like "country" + "year".

I also think that mentioning problem sources is important. Maybe even mention similar problems or generalizations etc (with sources). And to clarify my earlier post I was thinking about adding some extra fields to the "Post a reply" page, and of course connecting these fields with the search engine. Some plausible fields would be a drop down box which would have a list on the majority of olympiads, another one with the year, another one with problem number etc etc, those last could be entered manually though. I would also like to stress the importance, according to me of couse -- I'm sure many don't agree, in naming problem entries unambigously, giving them some distinct name that enlightens the problems nature or statement. Having 59 entries named inequality xx doesn't describe the problem that well, but as I said before, this are only my thoughts.

Technically, the search engine already searches the entire post, so it would not make that much of a difference to link the separate parts to the search engine.  The drop down box is a good idea though; of course, there should always be the option labelled "others".

I was mostly refering to that the search output could create tabulated information about the source of the problems.


I'm wondering if there's a simpler solution to this problem than what I have.

The lengths of the three medians of a given triangle are 9, 12, 15. Find the length of the side which the median of length 15 is drawn to.

[hide="outline of my solution"]I let the lengths of the three sides be $2x,2y,2z$, then wrote three equations involving $x,y,z$ using Stewart's Theorem on all three medians, I ended up with pretty symmetric system of equations, and solved for the side I wanted.[/hide][/i]

i think that is pretty much what you do...just to note though...your systems should be very simple...linear if you do m=x^2, n=y^2, p=z^2


geometry circumcircle calculus integration perimeter function inequalities

ARO (= All-Russian Olympiad):

[b]problems:[/b] http://www.mccme.ru/olympiads/vmo/31/zadachi.htm
[b]results:[/b] http://www.mccme.ru/olympiads/vmo/31/protocol.htm

The proposers of the problems are in the post below.

Here are the links of the problems. Thanks to [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=270]Fedor Petrov[/url] for providing the translation. :)

[hide="9th grade problem"]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35296]9.1[/url] Given parrallelogram $ABCD\ (AB<BC)$. Prove that the circumcircles of triangles $APQ$, where points $P$ and $Q$ are chosen on sides $BC$ and $CD$ respectively such that $CP=CQ$, have a common point, different from $A$.

[i] proposed by T. Emelyanova[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35325]9.2[/url] Lesha put numbers from 1 to $22^{2}$ into cells of $22\times 22$ board. Can Oleg always choose two cells, adjacent by the side or by vertex, the sum of numbers in which is divisible by 4?

[i] proposed by O. Podlipsky[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35324]9.3[/url] Given three  reals $a_{1},\,a_{2},\,a_{3}>1,\,S=a_{1}+a_{2}+a_{3}$. Provided ${a_{i}^{2}\over a_{i}-1}>S$ for every  $i=1,\,2,\,3$ prove that $\frac{1}{a_{1}+a_{2}}+\frac{1}{a_{2}+a_{3}}+\frac{1}{a_{3}+a_{1}}>1.$

[i] proposed by S. Berlov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35323]9.4[/url] Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?

[i] proposed by M. Garber[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35322]9.5[/url] Ten mutually distinct non-zero reals are given such that for any two, either their sum or their product is rational. Prove that squares of all these numbers are rational.

[i] proposed by O. Podlipsky[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35321]9.6[/url] Find the number of subsets  $A\subset M=\{2^{0},\,2^{1},\,2^{2},\dots,2^{2005}\}$ such that equation $x^{2}-S(A)x+S(B)=0$ has integral roots, where $S(M)$ is the sum of all elements of $M$, and $B=M\setminus A$ ($A$ and $B$ are not empty).

[i] proposed by Nazar-Bashi, I. Bogdanov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]9.7[/url]  = [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]10.6[/url]

[i] proposed by A. Akopyan[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35320]9.8[/url] 100 people from 50 countries, two from each countries, stay on a circle. Prove that one may partition them onto 2 groups in such way that neither no two countrymen, nor three consecutive people  on a circle, are in the same group.

[i] proposed by S. Berlov[/i]
[/hide]

[hide="10th grade problem"]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35319]10.1[/url] Find the least positive integer, which may not be represented as ${2^{a}-2^{b}\over 2^{c}-2^{d}}$, where $a,\,b,\,c,\,d$ are positive integers.

[i] proposed by V. Senderov[/i]

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35304]10.2[/url] Positive numbers are written in cells of a board $2\times n$ such that the sum of numbers in each of $n$ columns equals 1. Prove that   it is possible to select one number in every  column such that the sum of selected numbers in each of two rows does not exceed ${n+1\over 4}$.

[i] proposed by E. Kulikov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]10.3[/url] Given 2005 distinct numbers $a_{1},\,a_{2},\dots,a_{2005}$. By one question, we may take three different indices $1\le i<j<k\le 2005$ and find out the set of numbers $\{a_{i},\,a_{j},\,a_{k}\}$ (unordered, of course).  Find the minimal number of questions, which are necessary to find out all numbers $a_{i}$.

[i] proposed by I. Bogdanov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35317]10.4[/url] $w_{B}$ and $w_{C}$ are excircles of a triangle $ABC$. The circle $w_{B}'$ is symmetric to $w_{B}$ with respect to the midpoint of $AC$, the circle $w_{C}'$ is symmetric to $w_{C}$ with respect to the midpoint of $AB$. Prove that the radical axis of $w_{B}'$ and $w_{C}'$ halves the perimeter of $ABC$.

[i] proposed by P. Kozhevnikov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35306]10.5[/url] Some 16 cells are chosen on a chessboard. Find the minimal number of pairs of cells, which are in the same row or column.

[i] proposed by E. Kulikov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35298]10.6[/url] Given acute triangle $ABC$, $AA'$ and $BB'$ are its altitudes.  A point $D$ is chosen on an arc $ACB$ of a circumcircle of $ABC$. Lines $AA'$ and $BD$ meet in $P$,  lines $BB'$ and $AD$ meet in $Q$.  Prove that the line $A'B'$ contains a midpoint of $PQ$

[i] proposed by A. Akopyan[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35316]10.7[/url] Positive integers $x>1$ and $y$  satisfy an equation $2x^{2}-1=y^{15}$. Prove that 5 divides $x$.

[i] proposed by V. Senderov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35315]10.8[/url] A white plane is partitioned onto cells (in a usual way). A finite number of cells are coloured black. Each black cell has an even (0, 2 or 4) adjacent (by the side) white cells. Prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjacent cells.

[i] proposed by A. Glebov, D. von der Flaas[/i]

 [/hide]

[hide="11th grade problem"]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35314]11.1[/url] Find the maximal possible  finite number of roots of the equation  $|x-a_{1}|+\dots+|x-a_{50}|=|x-b_{1}|+\dots+|x-b_{50}|$, where $a_{1},\,a_{2},\,\dots,a_{50},\,b_{1},\dots,\,b_{50}$ are distinct reals.

[i] proposed by I. Rubanov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]11.2[/url] = [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35318]10.3[/url]

[i] proposed by I. Bogdanov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35313]11.3[/url] Let $A',\,B',\,C'$ be points, in which excircles touch corresponding sides of triangle $ABC$. Circumcircles of triangles $A'B'C,\,AB'C',\,A'BC'$ intersect a circumcircle of $ABC$ in points  $C_{1}\ne C,\,A_{1}\ne A,\,B_{1}\ne B$ respectively. Prove that a triangle $A_{1}B_{1}C_{1}$ is similar to a triangle, formed by points, in which incircle of $ABC$ touches its sides.

[i] proposed by L. Emelyanov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35312]11.4[/url] Integers $x>2,\,y>1,\,z>0$ satisfy an equation $x^{y}+1=z^{2}$. Let $p$ be a number of different prime divisors of $x$, $q$ be a number of different prime divisors of $y$. Prove that $p\ge q+2$.

[i] proposed by V. Senderov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35311]11.5[/url] Do there exist a bounded function $f: R\to R$ such that $f(1)>0$ and $f(x)$ satisfies an inequality $f^{2}(x+y)\ge f^{2}(x)+2f(xy)+f^{2}(y)$?

[i] proposed by Nazar-Bashi[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35310]11.6[/url] Do there exist 12 rectangular parallelepipeds  $P_{1},\,P_{2},\dots,P_{12}$ with edges parallel to coordinate axes $OX,\,OY,\,OZ$ such that $P_{i}$ and $P_{j}$ have a common point iff $i\ne j\pm 1$ modulo 12?

[i] proposed by A. Akopyan[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35309]11.7[/url] A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

[i] proposed by A. Zaslavsky, M. Isaev, D. Tsvetov[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35308]11.8[/url] 100 people from 25 countries, four from each countries, stay on a circle. Prove that one may partition them onto 4 groups in such way that neither no two countrymans, nor two neighbours will be  in the same group.

[i] proposed by S. Berlov[/i]

 [/hide]

Enjoy. I am in haste, so the translation is even worse then usually.
I do not know how define normal capital R for reals. Probelms of 9 klass are in Russian, ifa anybody wants, he may translate them, else please wait a little.

\q{9.1.} Given parrallelogram $ABCD\ (AB<BC)$.
Prove that the circumcircles of triangles
$APQ$, where points $P$ and $Q$ are chosen on sides
$BC$ and $CD$ respectively such that
$CP=CQ$, have a common point, different from
$A$.

\q{9.2.} Lesha put numbers from 1 to $22^2$
into cells of $22\times 22$ board.
Can Oleg always choose two cells,
adjacent by the side or by vertex, the
sum of numbers in which is divisible by 4?

\q{9.3.} Given three  reals $a_1,\,a_2,\,a_3>1,\,S=a_1+a_2+a_3$
Provided ${a_i^2\over a_i-1}>S$ for every
$i=1,\,2,\,3$ prove that
$\frac{1}{a_1+a_2}+\frac{1}{a_2+a_3}+\frac{1}{a_3+a_1}>1.$

\q{9.4.} Given 365 cards, in which distinct numbers are written.
We may ask for any three cards, the order of numbers
written in them. Is it always possible to find out
the order of all 365 cards by 2000 such questions?

\smallskip
\q{9.5.} Ten mutually distinct non-zero
reals are given such that for any two, either
their sum or their product is rational.
Prove that squares of all these numbers
are equal.

\q{9.6.} Find the number of subsets
$A\subset M=\{2^0,\,2^1,\,2^2,\dots,2^{2005}\}$
such that equation $x^2-S(A)x+S(B)=0$ has integral root,
where $S(M)$ is the sum of all elements of $M$,
and $B=M\setminus A$ ($A$ and $B$ are not empty).

\q{9.7.} see 10.6.

\q{9.8.} 100 people from 50 countries,
two from each countries, stay on a circle.
Prove that one may partition them
onto 2 groups in such way that neither no two
countrymans, nor thre consecutive people
on a circle are in the same group.

\q{10.1.} Find the least positive integer,
which may not be represented as ${2^a-2^b\over 2^c-2^d}$,
where $a,\,b,\,c,\,d$ are positive integers.

\q{10.2.} Positive numbers are written in cells
of a board $2\times n$ such that the sum of numbers
in each of $n$ columns equals 1. Prove that
it is possible to select one number in every
column such that the sum of selected numbers in each of
two rows does not exceed ${n+1\over 4}$.

\q{10.3.} Given 2005 distinct numbers $a_1,\,a_2,\dots,a_{2005}$.
By one question, we may take three different
indecies $1\le i<j<k\le 2005$ and find out the set of numbers
$\{a_i,\,a_j,\,a_k\}$ (unordered, of course).
Find the minimal number
of questions, which are necessary to find
out all numbers $a_i$.

\q{10.4.} $w_B$ and $w_C$ are excircles of a
triangle $ABC$. The circle $w_B'$ is symmetric to $w_B$
with respect to the midpoint of $AC$,
the circle $w_C'$ is symmetric to $w_C$
with respect to the midpoint of $AB$.
Prove that the radical axis of $w_B'$ and $w_C'$
halves the perimeter of $ABC$.

\q{10.5.} Some 16 cells are chosen on a chessboard.
Find the minimal number of pairs of cells, which are
in the same row or column.

\q{10.6.}
Given acute triangle $ABC$,
$AA'$ and $BB'$ are its altitudes.
A point $D$ is chosen on an arc
$ACB$ of a circumcircle of $ABC$.
Lines $AA'$ and $BD$ meet in $P$,
lines $BB'$ and $AD$ meet in $Q$.
Prove that the line $A'B'$ contains
a midpoint of $PQ$.

\q{10.7.} Positive integers $x>1$ and $y$
satisfy an equation $2x^2-1=y^{15}$.
Prove that 5 divides $x$.

\q{10.8.} A white plane is partioned onto cells
(in a usual way). A finite number of cells
are coloured black. Each black cell
has an even (0, 2 or 4) adjacent (by the side)
white cells. Prove that one may colour each white cell
in green or red such that every black cell
will have equal number of red
and green adjacent cells.

\q{11.1.} Find the maximal possible number
of roots of an equation
$|x-a_1|+\dots+|x-a_{50}|=|x-b_1|+\dots+|x-b_50|$,
where $a_1,\,a_2,\,\dots,a_50,\,b_1,\dots,\,b_50$
are distinct reals.

\q{11.2.} see 10.3

\q{11.3.} Let $A',\,B',\,C'$ be points,
in which excircles touch corresponding
sides of triangle $ABC$. Circumcircles of triangles
$A'B'C,\,AB'C',\,A'BC'$ intersect
a circumcircle of $ABC$ in points
$C_1\ne C,\,A_1\ne A,\,B_1\ne B$ respectively.
Prove that a triangle $A_1B_1C_1$ is similar
to a triangle, formed by points, in which
incircle of $ABC$ touches its sides.

\q{11.4.} Integers $x>2,\,y>1,\,z>0$
satisfy an equation $x^y+1=z^2$.
Let $p$ be a number of different prime divisors of
$x$, $q$ be a number of different prime divisors of
$y$. Prove that $p\ge q+2$.

\q{11.5.} Do there exist a
bounded function $f:R\to R$ such that $f(1)>0$
and $f(x)$ satisfies an inequality
$f^2(x+y)\ge f^2(x)+2f(xy)+f^2(y)$?

\q{11.6.} Do there exist 12 rectangular parallelepipeds
$P_1,\,P_2,\dots,P_{12}$ with edges parallel to
coordinate aces $OX,\,OY,\,OZ$ such that $P_i$
and $P_j$ have a common point iff $i\ne j\pm 1$
modulo 12?

\q{11.7.} A quadrilateral $ABCD$ without
parallel sides is circumscribed arounda circle
with centre $O$. Prove that $O$ is a poiint
of intersection of middle lines of quadrilateral
$ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$)
iff $OA\cdot OC=OB\cdot OD$.

\q{11.8.} 100 people from 25 countries,
four from each countries, stay on a circle.
Prove that one may partition them
onto 4 groups in such way that neither no two
countrymans, nor two neighbours will be
in the same group.

[quote="Myth"]Unfortunately, despite of I-st diploma, Kirill wasn't included into the russian IMO team (because of Winter TSTs).[/quote]I am really sorry to hear that :(

But anyway it's great to be the first in the Russian MO!!! :first: :first: Congrats!

[quote="Myth"]My students got the following results:

11th grade
Belousov Kirill - I-st Diploma  :first:  :first:  :first:

10th grade
Smotrov Dmitrii - II-nd Diploma  :D :D :D

Also one students from my school, Dmitrii Yarushin, took II-nd Diploma in the 9th grade.

I congratulate them!!!

Unfortunately, despite of I-st diploma, Kirill wasn't included into the russian IMO team (because of Winter TSTs).[/quote]

Congrats to teacher Myth! You are great teacher!!!

Did any bulgarians participated?
Congratulations,Myth,I wish I had such a teacher like you!

did Chinese students participate??
What's their results?

Bulgaria

11 klass

Kralev Rosen	65774761	43	1diploma
Dimitrov Ivan	07777770	42	2diploma
Lishkov Alexandr	31774070	29	3diploma
Dimitrov Veselin	01077000	15             no diploma

10 klass

Baronov Emil	77607770	41	2diploma
Simeonov Dimitar	76077711	36	2diploma

China (I apologize for transcription of names from Russian)

10 klass

Lyu Chziyu                77777747   53	1diploma
Van Czenang	55777710	 39	2diploma
Czan Tau	                71177770	 37	2diploma
Ing Cze		57107710	 28	3diploma

9 klass

Ing Li                        77707770   42            2diploma
Vang Yuli                 70707670   34            3diploma

[quote="Myth"]My students got the following results:...
I congratulate them!!!
[/quote]
Congratulations, Mikhail! I wish you that your students do even better the next year!  :)

[quote="Fedor Petrov"]Bulgaria

11 klass

Kralev Rosen	65774761	43	1diploma
Dimitrov Ivan	07777770	42	2diploma
Lishkov Alexandr	31774070	29	3diploma
Dimitrov Veselin	01077000	15             no diploma

1diploma means first place in Russia?


Should we create a new thread to link together all the mock tests?

Mathfanatics thread hasn't been updated in a while.

If you're planning to provide difficulty ratings for the tests, I can provide input, as I've taken quite a bit of them.

Not really, since they can just be added to the AoPSWiki.  There's already a section on mock amc stuff.


linear algebra
matrix
Gauss
algebra
system of equations
linear algebra unsolved

Suppose your system of equations is $ Ax \equal{} b$, where $ A$ is an $ m\times n$ matrix. Obviously $ n\leq m$ and $ n \equal{} \text{rank}(A)$. Without loss of generality, assume that the first $ n$ rows of $ A$ are linearly independent and $ A \equal{} \begin{bmatrix}C \\
D\end{bmatrix}, b \equal{} \begin{bmatrix}b_1 \\
b_2\end{bmatrix}$, where $ C$ is an invertible $ n\times n$ submatrix of $ A$, and $ b_1\in \mathbb{Z}^n$. Since $ Cx \equal{} b$ has a unique solution, which is clearly in $ \mathbb{Q}^n$, the hypothetical solution to your original system should also be in $ \mathbb{Q}^n$.

[quote="Kirill"]After transforming $ E$ you will have equations system with rational coefficients.[/quote]
Yes. But you haven't pointed out the way this fact is related to the solutions of the given system.

[quote="The Captain"]
But you haven't pointed out the way this fact is related to the solutions of the given system.[/quote]
I think, it's easy, and there is no reason to extend this idea to full solution.

[quote="Kirill"][quote="Invariant"]Let $ E$ be linear equations system with all integer coefficients. Let us suppose that $ E$ has only one solution $ (x_{1}, ..., x_{n})$ (all real numbers). Prove that all $ x_{i}$ are rational.[/quote]
Use Gauss Method for solving this problem. After transforming $ E$ you will have equations system with rational coefficients.[/quote]

According to your idea - For me it is not so easy. It seems it is interesting problem for begginers and I'f like someone to post a complete solution, because this question suffices it.

Thank you

[b]Problem.[/b] Suppose $ Ax\equal{}b$ is a consistent system of equations, where $ A, b$ have rational entries only. Then there exists a solution to the aforementioned system whose components are all rational numbers.

[b]Solution.[/b] (Note that the set of rational numbers is closed under addition and multiplication.)
Apply the Gauss-Jordan elimination method to the augmented matrix $ [A|b]$. It is clear that the resulting matrix has only rational entries, therefore you can assign arbitrary rational values to the free variables, if they exist of course, and get some rational values for the pivot variables. $ \blacksquare$

PS. Note how the original problem follows from this one.


[hide]Well you know that the # of digits between 1-9 is 9. Then 10-99 is 180. And 100-999 is 2700. 1000-1999 is 4000. 2000-2006 is 28. Adding all this up gets 6917.[/hide]

[hide="solution"]

9  one digit numbers
90 2 digit numbers
900 3 digit numbers
1007 4 digit numbers

9+90*2+900*3+1007*4=6917[/hide]

[quote="xxxxx"]Johnny starts writing down numbers starting with 1, then 2, then 3, ... He gets tired and stops. If the last number he wrote was 2006, how many digits did peter write in all?[/quote]
Technically, you cannot figure out how many Peter has written because we are given information about Johnny!


Please post here any problems you may have and we shall try to fix them as soon as possible.

The source field problem is solved.

No, it's not supposed to do that... strangely. I'll look into it.

The automatic log in seems to work fine for me..

[quote="jeffrey"]since the site merger i'm no longer able to log in automatically. I select the log me in automatically box, but i still have to log in manually every time.[/quote]

Strange... it works for me.

The only thing that I can think of is to clear your cookies.  It works for me and since noone else has complained that leads me to believe that it is just you.

[quote="jeffrey"]can anyone give me insight into this problem? its not that inconvenient, but it would still be great if i could fix it.[/quote] This is usually a problem associated with phpBB's management of cookies. Deleting the AoPS/MathLinks cookie is the simplest way to solve the problem. Then log in checking the box with the automatical login and you are done.

this isnt so much of a problem as it is an inconvenience- every time i log in, i i always check new stuff on the site by looking at "see posts since last visit"- the problem is, it will say "seach found 164 matches" (or something to that effect) when in reality, there have only been about 10-20 posts since i last logged in. never had this problem previously, anything you can do about it? its not a big huge problem, but you know, if you could fix it, that would be nice too.

One thing you can do for now is hit 'Mark all forums read' (at the top of the forum on the index page) when you're done with a visit.

Thanks- I tried that, didn't work... Hmm... very odd. But its a very small prob/inconvenience, really. I wouldn't worry too much about it.

[quote="rcv"]The old AoPS site did not require login to read the "public" forums.  However, there are some forums on the new, combined site that *do* require login.  For example, under "Advanced --> Algebra", forums #37 and #38 require login to view the posts.  [I don't know how to get an exhaustive list.  I've just noticed these two.]

Is this an intentional change or is this an oversight?[/quote]

Looking in to it. . .

Here's an interesting one.  [I'm guessing it's probably because some forums are not openly readable.]  It's amusing, anyway.

- I just log on to AoPS, after having been off the computer for hours.

- I happen across this post:  [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=101506&highlight=#101506[/url]

- I'm interested in reading the referenced article, which points to [url]http://www.mathlinks.ro/viewtopic.php?t=14167[/url].

- I am taken to the Mathlinks login screen.

- I login to Mathlinks.

- I am taken to the "Page does not exist" page.

- After clicking "back" twice, I get to see article 14167.  Yeah!

By the way, when I use more modern browsers, the security icons light up like Christmas trees, as http://www.artofproblemsolving.com and http://www.mathlinks.ro try to share cookies between each other.  I presume you folks know it isn't Kosher (with modern security recommendations) to share cookies across domains.  I'm wondering if some of the cookie problems and different behaviors folks are seeing might be due to differences in security settings.  Cheers!

We've solved (I think) the links in posts problem.  As you'll see in the post above, the links to topics now both point to whatever server you're coming from.  I'll have to look into more detail for the cookies issue.  Please let me know if you see instances where this little fix didn't work out.

[quote="rcv"]The old AoPS site did not require login to read the "public" forums.  However, there are some forums on the new, combined site that *do* require login.  For example, under "Advanced --> Algebra", forums #37 and #38 require login to view the posts.  [I don't know how to get an exhaustive list.  I've just noticed these two.]

Is this an intentional change or is this an oversight?[/quote]It is an intentional change, or rather a remain from the old mathlinks site. The forums that required users to be logged in are the solved problems forums. We will remove this soon.

I'm having troubles on the forums. First of all, numbers in Latex don't show up. There's just a white square with a red X in it followed by the latex in normal text. Also, on the top where Forum, Math Jams, and My Classes should be are squares with red X's. Same thing goes with the side headings like Enroll and Resources. I have HTML and everything else enabled so I don't know what the problem is.

I see no one else complaining about this, so I assume it must be a particular thing.
If you can answer to these questions maybe we can help: what url do you use and does not work ? and what browser (name and version) & os ?

URL?  Isn't that like http://www.yahoo.com?  I just type in artofproblemsolving.com
My browser is Internet Explorer but I'm not sure which version, and my operating system is Microsoft 98.

[quote="NightFlarer"]My browser is Internet Explorer but I'm not sure which version, and my operating system is Microsoft 98.[/quote]

Click the "Help" menu and click "About".  It should tell you which version you are using.

A comment: If you login on mathlinks and look through all the forums and stuff, then login on artofproblemsolving, it will still say that all the stuff you looked at is unread.

well apparently there's still a very very small flaw in the new site: http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=4

it doesn't disturb me, but sometimes these 'peeks' into your source code can lead to hackers easily finding security holes in it, which can lead to eum... unwished effects :D (not by me, all I now know is that there must be more forums out there than you actually display)

could easily be fixed by first checking if there's any user with that user_id before going on (cuz now the user_id apparently gets lost somehow if there is no such user :?)

I'm still seeing two distinct, weird things.  I suspect they are both related to cookie handling.

Twice, in the last 7 - 10 days, I have gotten an error that reports a cookie was too long for the Apache Web Server.  [I think the error occured when logging on, and I think it said the Apache limit was 8KB.]  The solution was to delete the cookies.  I note your recent instruction for users to delete their cookies.  Perhaps this issue will never occur again.  Or perhaps my cookies will grow gradually, until I hit this limit again.

The other issue occurs only when I'm using MS XP SP 1, with IE 6.0.28.  I go to the Logon screen, and enter my Userid and Password.  I am then taken to the Forums page, without any errors, but I do not appear to be logged on.  [I don't have a link to view postings since last visit, I cannot post, and the Logon button is still active.]  So, I go through the logon process a second time.  The second time, the logon really succeeds, but it reports that my last logon was ~1 minute ago.  I deleted all IE cookies yesterday, but this problem bit me again this morning.  [I have never observed this behavior when using IE 5, Netscape, or Mozilla.]

I hope that first error is dead now - let me know if it happens again.

Will think about that other one - I have no immediate idea what's causing it.

could be the append_sid() being forgotten in a page redirection that was used.

can you post exactly what happens and how we can reproduce it?

when i was browsing the forums(not logged in) things appeared in yellow saying there was a new message.
Is this supposed to happen?

[quote="Valentin Vornicu"]What things appeared in yellow? A window? A pop up? ... How many times did it happen to you?

Are you sure you weren't logged in?[/quote]
Um it says there are new messsages.
and the login icon was on the side

[quote="Valentin Vornicu"]Do you use a pop-up blocker? Maybe that's what's causing your problem. Everytime you receive a private message a small pop-up appears when you loggin announcing you of that event. If you have a pop-up blocker, most probably this will not be shown, but instead you will see a warning message, usually colored in yellow, to attract attention. That's all ;)[/quote]
i dont have a popup blocker


didn't this come up in mathcounts a while ago?

go [url=http://www.rocketcitymath.org/]here[/url], click on archives, and go to apollo round 2. They have answers.

so basically you are looking for the sum of the harmonic series.
it divergest to infinity.
there are many different ways to prove it.

or look here: [url=http://mathworld.wolfram.com/HarmonicSeries.html]mathworld.wolfram.com/HarmonicSeries.html[/url]


function
induction
algebra proposed
algebra

Find all continues functions $ f: \mathbb R\to\mathbb R$ such that for all $ x,y,z\in\mathbb R$,
\[f(x\plus{}f(y\plus{}f(z)))\equal{}f(x)\plus{}f(f(y))\plus{}f(f(f(z))).\]

[quote="hana1122"]Find all continues functions $ f: \mathbb R\to\mathbb R$ such that for all $ x,y,z\in\mathbb R$,
$ f(x \plus{} f(y \plus{} f(z))) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(f(z)))$[/quote]

Let $ P(x,y,z)$ be the assertion $ f(x \plus{} f(y \plus{} f(z))) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(f(z)))$

$ P(0,0,0)$ $ \implies$ $ f(0) \plus{} f(f(0)) \equal{} 0$ and so either $ f(0) \equal{} f(f(0)) \equal{} 0$, either $ f(0)$ and $ f(f(0))$ have opposite signs and $ \exists u$ between them such that $ f(u) \equal{} 0$

So $ \exists u$ such that $ f(u) \equal{} 0$

$ P(x,y,u)$ $ \implies$ $ f(x \plus{} f(y)) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(0))$ and so $ g(x \plus{} f(y)) \equal{} g(x) \plus{} g(f(y))$ where $ g(x) \equal{} f(x) \plus{} f(f(0)) \equal{} f(x) \minus{} f(0)$. So :

$ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ $ \forall x\in\mathbb R,\forall y\in f(\mathbb R)$ and also (using $ x \minus{} y$ instead of $ x$) :
$ g(x \minus{} y) \equal{} g(x) \minus{} g(y)$ $ \forall x\in\mathbb R,\forall y\in f(\mathbb R)$

If $ f(x) \equal{} c$ constant, then $ P(x,y,z)$ $ \implies$ $ c \equal{} 3c$ and so $ f(x) \equal{} 0$ $ \forall x$, which indeed is a solution.

If $ f(x)$ is not constant, then $ \exists b\ne0$ and $ y$ such that $ g(y) \equal{} b\ne 0$ and an immediate induction gives :
$ g(x \plus{} ny) \equal{} g(x) \plus{} nb$ $ \forall x,\forall n\in\mathbb N$
$ g(x \minus{} ny) \equal{} g(x) \minus{} nb$ $ \forall x,\forall n\in\mathbb N$

and so, since $ g(x)$ is continuous,  $ g(\mathbb R) \equal{} \mathbb R$ and so $ f(\mathbb R) \equal{} \mathbb R$ and so :

$ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ $ \forall x,y$

And so $ g(x) \equal{} ax$ (continuous solutions of Cauchy) and $ f(x) \equal{} ax \plus{} b$

Plugging this back in the original equation, we get $ b(a \plus{} 2) \equal{} 0$ hence the solutions :

$ \boxed{f(x) \equal{} ax}$ and $ \boxed{f(x) \equal{} b \minus{} 2x}$


trigonometry
algebra
system of equations

If $\sin a+\cos a=.2$, what is $\sin a-\cos a$?

[hide]$\sin a-\cos a=\frac{-\cos 2a}{\sin a+\cos a}$
[/hide]

If sin a+ cos a=.2, then (sin a)^2+(cos a)^2+2sinacosa=0.04, so 2sina*cosa=-0.96 so -2sinacosa=0.96

so then sina-cosa=x, (sina)^2+(cosa)^2-2sinacosa=x^2, so we have x^2=1+0.96, so x^2=1.96, so $x=1.4$.

[hide="In other words"] $(\sin a+\cos a)^{2}+(\sin a-\cos a)^{2}= 2$ [/hide]
[hide="Why?  Because..."] $\sin^{2}a+\cos^{2}a+2 \sin a \cos a+\sin^{2}a+\cos^{2}a-2 \sin a \cos a = 2 (\sin^{2}a+\cos^{2}a) = 2$ [/hide]
[hide="But also because..."] $\sin a+\cos a = \sqrt{2}\cos (a-\frac{\pi}{4})$
$\sin a-\cos a = \sqrt{2}\cos (a-\frac{3\pi}{4}) =-\sqrt{2}\sin (a-\frac{\pi}{4})$ [/hide]

This can be transformed into an interesting system of equations problem: let $\sin x=a$ and $\cos x=b$, then we have that:

$a+b=\frac{1}{5}$
$a^{2}+b^{2}=1$

Determine $t=a-b$.

I guess this might seem contrived, but it usually helps to square when you have $a+b$ and $a-b$.

$t^{2}+\frac{1}{25}=(a+b)^{2}+(a-b)^{2}=a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}=2(a^{2}+b^{2})=2$.

From this we determine that $|t|=\frac{7}{5}$.


probability
search
floor function
limit
number theory
prime numbers
number theory proposed

Let $ n$ be a number in the form $ 0.abcd\cdots$ for example $ \frac{\pi}{10}$
It's true that this number can be written in the form $ 0.(p_1)(p_2)(p_3)\cdots$? ($ p_i$ is a prime number) for the same example above:

$ 0.(314159)(2)(6535897)(93238462643)(3832795028841971693)\cdots$

each number in () is a prime.

the answer is probably a yes, since each digit in pi is random (it really is?), and the probability of forming a finite prime of any length is non-zero.

But the real question is the following.
let $ p_1$ the biggest prime number possible to form by this "sequence", no matter how further you search, you can't form another prime number. (althought each digit of pi is random, will exist a prime numbers in this exact form? will it grow infinitely? i assume it will not, and this biggest prime in this form exist)
So you stop there and start searching for $ p_2$ and etc.

The questions I have are:
a) there is this biggest $ p_1$ prime? If no, ends. if yes, there is $ p_2$? etc.?

I made some search in this, but the density of prime numbers found start to get really thin, so the purpose of the question.

thanks

By the conjectured normality of $ \pi$, you would be able to find arbitrary long primes in it's decimal expansion.
More exactly, there are arbitrary long primes in any such "prime decomposition" of $ \pi$: if there would be a maximal prime of length $ n$ in one such sequence, assuming normality, we can find $ 2n$ consecutive $ 0$'s (or alternatively and easier to find: just even digits; or ones divisible by $ 3$ or $ 5$ or $ 7$) in $ \pi$. Then any way to write $ \pi$ as such a sequence of primes must get over this subsequence, which can only be done if there is a prime of length at least $ n\plus{}1$ (note that no prime can be completely in that $ 2n$ digits). Contradiction.

I think there is a little bit confusion on this.
The question I proposed can be formuled as "How many primes there are in a specific form"
The answer is not just that.
How many primes start with $ n$ specific digits?
If it were like that you could prove that there is infinitely many fermat numbers, because it is isomorphic to this problem, isn't it?

Correct if I am wrong, but I don't belive in your "proof", there are infinitely many primes, but the existence or not of this maximum is not yet prooven or disprooved

I don't see how Fermat numbers should come into the game. And by the prime number theorem, infinitely many primes start with $ n$ specific digits (note that we can have as many digits at the end as we like).
Well, normality is definitely not proven, but it it's necessary for randomness (a random number is normal almost surely). So is this the only doubt of yours¿

Consider the number $ 0.1111111111111...$ in binary.
There exist this prime decomposition.
Suppose there is a maximum $ p_1$ of length $ n$.
To overcome the series of $ m$ $ 1$'s next, there must exist a primer number greater than $ n\plus{}1$. And there is.
So there is no maximum.
And this prime numers are only fermat numbers, So there is infinitely many fermat numbers?

I was talking about $ \pi$ as an example where you don't have maximal primes for sure (as you have in any number that contains arbitrary long sequences of $ 0$. And you should reread my proof as we don't need to overcome $ 1$'s but $ 0$'s (a big difference).

So, Since there is no maximum prime. We can say that:
$ \lfloor 10^n \pi \rfloor$ is prime for infinitely many $ n$ and
$ \lim_{x \to \infty} \lfloor 10^n \pi \rfloor$ is a prime number? (is infinity,but you can think as a infinity prime, since there is no prime upper bond)

It was that shown any "subdivision" of $ \pi$ into primes must contain arbitrary long primes. This doesn't show (and is in some sense unrelated to) the possible length of a prime at the beginning, which is probably infinite, but somewhat more difficult.
And I don't see any usefull concept to define an "infinity prime" in this meaning (that limit is somewhat strange as even if the first statement is true, still not all these numbers are prime (in fact, only a very small fraction of them is)).

So for any normal number $ x$: $ S \equal{} \left\{\left\lfloor x 10^n\right\rfloor : n\in \mathbb{Z}\land \left\lfloor x 10^n\right\rfloor \in \mathbb{P}\right\}$, is a incontable set.

This is somewhat similar to mill's constant. (but is not truel for all $ n$, but for some (and not so related at all))

I saw my stupid dumb mistake now.
Sorry.


ratio geometry circumcircle vector symmetry search

Could anyone show that the medians intersect at a point known as the centroid and that this centroid splits the medians into a $2: 1$ ratio?

[quote="fireforce"]Medians?? I though it was the Perpendicular Bisectors?[/quote]
The perpendicular bisectors are concurrent at the circumcenter.

Perpendicular bisectors intersect at the circumcenter, not the centroid.

There's a really short proof using areal co-ordinates:

Let $AD$ and $BE$ be two medians. The co-ordinates of $A$ and $B$ are $(1, 0, 0)$ and $(0, 1, 0)$.  $D$ and $E$ have co-ordinates $(0, \frac{1}{2}, \frac{1}{2})$ and $(\frac{1}{2}, 0, \frac{1}{2})$. From this we can see the point they intersect at $G$ must have co-ordinates are $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$. $\overrightarrow{AG} = \frac{2}{3} \overrightarrow{AD}$. $\blacklozenge$

[hide]
In $\triangle ABC$, let $D$, $E$ and $F$ be the midpoints of $AB$, $BC$ and $CA$. Let $G$ be the centroid.

Note that $DE\parallel AC$ and $DE=\frac{1}{2} AC$ (Midline theorem).
$\triangle DEG\sim \triangle CAG$, so $DG: GC=EG: GA=DE: CA=1: 2$.
[/hide]

See i learnt how to prove it using vectors, but i'm not a big fan of vectors so i was just wondering if there was a better proof.

Let $\triangle ABC$ have midpoints $D$, $E$, and $F$ on $AB$, $BC$, and $CA$, respectively.

--
Lemma:
$DC$, $EA$, and $FB$ are concurrent
by ceva, we have that
$\frac{AD}{DB}*\frac{BE}{EC}*\frac{CF}{FA}=1$ using the midpoints, so they are concurrent. Say this point is $G$
--

Assign a mass of $1$ to point $A$ or $w_A=1$, then to balance along the midpoints, $w_B=w_C=1$, then using mass points again,
$w_A+w_B=w_D=2$, symmetrically, we have $w_E=w_F=2$, then

$w_A*AG=w_E*EG$
$\frac{AG}{EG}=2$
the other results follow by symmetry

P.S. if you don't know mass points search "mass points"
this question has been asked a million times

----

another solution

assuming that they are concurrent, apply menelaus:

$\frac{AD}{DB}*\frac{BC}{CE}*\frac{EG}{GA}=1$
$1*2*\frac{EG}{GA}=1$
$\frac{AG}{GE}=2$ as desired

----

another...

--
Lemma:
If $A$, $D$,  and $B$ are collinear, and $P$ is a point not on that line, then $\frac{[ADP]}{[BDP]}=\frac{AD}{DB}$ (where $[ABC]$ is the area of $\triangle ABC$)

consider the height from $P$ to $AB$, this is the height for both triangles, so dividing the area formula $[ADP]=\frac{1}{2}*AD*h$, we get the result
--

then using the midpoints, $[AGF]=[CGF]$ and $[AFB]=[CFB]$, subtract those to get

$[AGB]=[CGB]$ (1)

then we have that $[CGE]=[BGE]$ and $[AGD]=[BGD]$, so put that into (1) to get:

$2[CGE]=2[AGD]$

using transivity, we have that all the "little triangles" are equal in area

then $\frac{[ABG]}{[EBG]}=\frac{AG}{GE}$

$[ABG]=2[EBG]$ by the little triangles all having the same area, so

$\frac{AG}{GE}=2$


Find all positive integers $ x, y ,z $ s.t.

$ x^ {y^z} . y^ {z^x} . z^ {x^y} =\ 5 xyz $

It's from a book!!!! And the answer satisfies!!!

Rushil, are you sure, all of the variables are distinct :? ?

Actually you'll have to make some cases since we require integer solutins!!!

[quote="nat mc"]Maybe log it?

$\log{x^{y^z}\cdot{y}^{z^x}\cdot{z}^{x^y}}=\log{5xyz}$

$\log{x^{y^z}}+\log{y^{z^x}}+\log{z^{x^y}}=\log5+\log{x}+\log{y}+\log{z}$

$y^z\log{x}+z^x\log{y}+x^y\log{z}=\log5+\log{x}+\log{y}+\log{z}$

Does that help any?[/quote]

that's exactly what I did nat mc, however, I was kind of lost after that, and didn't know exactly what to do :?

[quote="peeta"]it is obvious that WLOG $x=5k$[/quote]
What do you mean. How does that help any?

Has nobody got any solution. Should I post the one which I have??

[quote="Hamster1800"]Here's how I got my solution.

Starting with the original equation:

$x^{y^{z}}y^{z^{x}}z^{x^{y}} = 5xyz$

divide both sides by $xyz$ to get:

$x^{y^z-1}y^{z^x-1}z^{x^y-1}=5$

since 5 is prime, we have to have 2 factors of 1, and 1 factor of 5. If $z=1$, we know that the last two factors will always be 1. So, we have to find $x$ and $y$ such that $x^{y-1} = 5$ Since we want integer solutions, we have to make the exponent 1 and the base 5. So:

$x=5$
$y=2$

So we have the solution (5,2,1). Since the exponents always go in the same order, the positions of the numbers don't matter, so all the solutions (that i found) are:

(5,2,1)
(2,1,5)
(1,5,2)[/quote]
very nice, and those are the only integer solutions


ratio
geometry
angle bisector
geometry proposed

(O) is a circle which touches side AB,BC,CD,DA of the quadrilateral ABCD. The diagonal BD meets (O) at E and F. H is the projection of O on AC. Prove that $ \angle BHE\equal{}\angle CHF$

Should be: Prove that $ \angle BHE \equal{} \angle DHF.$

The diagonals $ AC, BD$ meet at $ X.$ $ AB, BC, CD, DA$ touch $ (O)$ at $ P, Q, R, S.$ $ OH$ cuts $ (O)$ at $ M, N$ on the same sides as $ B, D.$ $ BD, QR, SP, MN$ meet at the pole $ K$ of $ AC$ and $ OK \equal{} \frac {r^2}{OH}$, where $ r$ is radius of $ (O)$ $ \Longrightarrow$ the cross ratio $ G(K, H, M, N) \equal{} \minus{} 1$ is harmonic. $ QR, SP$ meet $ AC$ at $ U, V.$ Since $ MQ, NR$ and $ MP, NS$ meet on the polar $ AC$ of $ K,$ the cross ratios  $ (K, X, E, F) \equal{} (K, H, M, N) \equal{} \minus{} 1$ and $ (K, X, B, D) \equal{} (K, V, P, S) \equal{} (K, H, M, N) \equal{} \minus{} 1$ are also harmonic. Since the angle $ \angle KHX$ is right, $ HX \equiv AC$ is the common angle bisector of the angles $ \angle BHD, \angle EHF.$

Dear yetti,
Can you detail your solution?

Could you first confirm that $ \angle BHE \equal{} \angle DHF$ is the correct proposition ? You typed $ \angle BHE \equal{} \angle CHF$ twice, they are not equal and I am guessing.

What is the [quote="mr.danh"]projection of $ O$ on $ AC$[/quote]?????


Let $\Sigma$ and $\Sigma '$ be cuboids with $\Sigma$ lying inside $\Sigma '$. Suppose $\Sigma$ has dimensions $a \times b \times c$ and $\Sigma '$ has dimensions $a' \times b' \times c'$. Must we have $a + b + c \leq a' + b' + c'$?

i can remember the answer... but not know how to prove it


geometry
perimeter
ratio
trigonometry
analytic geometry
similar triangles

Delroy’s sailboat has two sails that are similar triangles. The larger sail has
sides of 10 feet, 24 feet, and 26 feet. If the shortest side of the smaller sail
measures 6 feet, what is the perimeter of the smaller sail?
(1) 15 ft     (3) 60 ft
(2) 36 ft     (4) 100 ft

[hide]choice 2: 36 is the answer. Use similar triangles. [/hide]

How would you set up similar triangles to solve this question?

[hide]Shortest and longest sides correspond in similar triangles, so the ratio of the smaller sail to the bigger one is $3: 5$. Perimeters also have the same ratio, so $p=\frac{3}{5}\cdot(10+24+26)=36\Longrightarrow(2)$.[/hide]

i_like_pie said:

"Perimeters also have the same ratio."

Do you mean all perimeters or just questions dealing with similar triangles?

Also, does this site answer trigonometry questions and geometric proof questions using the famous high school statement vs reason chart?

[quote="i_like_pie"]In similar triangles, everything has the same ratio: corresponding sides, heights, medians, perimeters, etc.

If the ratio of corresponding sides is $3: 5$, then the ratio of the perimeters will also be $3: 5$.[/quote]

not EVERYTHING

the ratio of the area is the ratio of the sides (or the perimeter or altitudes or whatever) squared.

so if the ratio of perimeters is $\frac{3}{5}$, the ratio of the areas would be $\frac{9}{25}$

(also, if the figure was 3-D, then the ratio of volumes would be the ratio of sides cubed)

[quote="Interval"]
Also, does this site answer trigonometry questions and geometric proof questions using the famous high school statement vs reason chart?[/quote]

Please understand that this site is a [u]forum[/u], [b]NOT[/b] a tutoring site; so it does not usually act as a collective. Thus, the usage of the "statement vs. reason chart" will vary depending on who is responding; but in most cases, the chart is unnecessarily tedious to write out, as everything which can be put onto the chart can also be put into plain, intelligible English sentences. I do not like the chart as I see it as unwieldy, but also since it has given me many a wrist cramp in freshman year geometry :P .

[quote="Interval"]I think the "statement vs reason charts" does not help high school students.  What it creates is more confusion and fear in the lives of students who are so afraid of math and math teachers trying to intimidate rather than teach numbers clearly.[/quote]

Well for some it is easier to organize their thoughts in that manner; for me, a paragraph proof works just as well, usually even better, than that organizational plan.

Can coordinate geometry or analytical geometry be used to solve geometric proofs?

If so, do you have an example?


trigonometry conics parabola calculus derivative function vector

Find the maximum and minimum value of

$ \sin^2\theta\plus{}\sin\theta\plus{}\cos\theta$.

[quote="Temperal"][hide="Maximum"]Write as $ x^2 \plus{} x \plus{} \sqrt {1 \minus{} x^2}$ with $ \minus{} 1\le x \le 1$. Differentiating gives the answer easily: Solve $ \minus{} \frac {x}{\sqrt {1 \minus{} x^2}} \plus{} 2x \plus{} 1 \equal{} 0$ and check that the value is attainable.[/hide]

[hide="Minimum"]Write as $ x^2 \plus{} x \minus{} \sqrt {1 \minus{} x^2}$ with $ \minus{} 1\le x \le 1$. Use the same technique as above and check that the value is attainable.[/hide][/quote]

Other than Differentiating Method, anybody has any idea?

[quote="vinskman"]Find the maximum and minimum value of

$ \sin^2\theta \plus{} \sin\theta \plus{} \cos\theta$.[/quote]

We are to find the condition such that the parabola $ x\plus{}\left(y\plus{}\frac{1}{2}\right)^2\equal{}k\plus{}\frac{1}{4}$ has intersection point with the unit circle $ x^2\plus{}y^2\equal{}1$.

[quote="kunny"][quote="vinskman"]Find the maximum and minimum value of

$ \sin^2\theta \plus{} \sin\theta \plus{} \cos\theta$.[/quote]

We are to find the condition such that the parabola $ x \plus{} \left(y \plus{} \frac {1}{2}\right)^2 \equal{} k \plus{} \frac {1}{4}$ has intersection point with the unit circle $ x^2 \plus{} y^2 \equal{} 1$.[/quote]

would you please post the details ?

 :o  :)

If $ \sin ^ 2 \theta \plus{}\sin \theta \plus{}\cos \theta \Longleftrightarrow \left(\sin \theta \plus{}\frac{1}{2}\right)^2\plus{}\cos \theta \minus{}\frac{1}{4}$ takes some value $ k$,

the parabpla $ x\plus{}\left(y\plus{}\frac{1}{2}\right)^2\equal{}k\plus{}\frac{1}{4}$ has more than one intersection points with the unit circle $ x^2\plus{}y^2\equal{}1$.

[quote="kunny"]If $ \sin ^ 2 \theta \plus{} \sin \theta \plus{} \cos \theta \Longleftrightarrow \left(\sin \theta \plus{} \frac {1}{2}\right)^2 \plus{} \cos \theta \minus{} \frac {1}{4}$ takes some value $ k$,

the parabpla $ x \plus{} \left(y \plus{} \frac {1}{2}\right)^2 \equal{} k \plus{} \frac {1}{4}$ has more than one intersection points with the unit circle $ x^2 \plus{} y^2 \equal{} 1$.[/quote]

May be I stupid, why are their points of intersection related to the max. & min. values of $ \sin ^ 2 \theta \plus{} \sin \theta \plus{} \cos \theta$?

[quote="kunny"]Ｉｆ $ k$ could take value of $ \frac {3}{4} \plus{} \frac {\sqrt {3}}{2}$, can't it be possible to exist the values of $ x,\ y$ in $ x^2 \plus{} y^2 \equal{} 1$?

The answer is Yes. For $ x \equal{} \frac {\sqrt {3}}{2},\ y \equal{} \frac {1}{2}$, we can make a confirmation the existence of the point in unit circle, that is to say,
the parabola and the unit circle have more than one points of the intersection.[/quote]

Just you mean that under the constraints $ x^2\plus{}y^2\equal{}1$ ,the the two values of k by counting the points of intersections is justified as max. & min. values for $ \sin^2\theta\plus{}\sin\theta\plus{}\cos\theta$?

Is it?

How would you find the maximum and minimum vales of $ x\plus{}\sqrt{1\minus{}x^2}$ without calculus or trig?

[quote="kunny"]Yes. :)[/quote]

Then , as your examples to raise the value of $ k \equal{}\frac{3}{4}\plus{}\frac{\sqrt 3}{2}$
just by your obseveration nad not by solving equations?

[quote="kunny"]How would you find the maximum and minimum vales of $ x \plus{} \sqrt {1 \minus{} x^2}$ without calculus or trig?[/quote]

Sorry , I have tried , but in vain.

[quote="vinskman"][quote="kunny"]$ x \plus{} \sqrt {1 \minus{} x^2}$ would take some value of $ k$

$ \Longleftrightarrow$ there exists some real value $ x\ ( \minus{} 1\leq x\leq 1)$ such that $ x \plus{} \sqrt {1 \minus{} x^2} \equal{} k$

$ \Longleftrightarrow$ there exists some real value $ x\ ( \minus{} 1\leq x\leq 1)$ such that $ \sqrt {1 \minus{} x^2} \equal{} \minus{} x \plus{} k$

$ \Longleftrightarrow$ the graph of $ y \equal{} \sqrt {1 \minus{} x^2}$ has more than one points of intersection with the line $ y \equal{} \minus{} x \plus{} k$.

We are to maximize and minimize the $ y$ intercept $ k$.[/quote]

Just you say,

firstly , we find out all the points of intersection between the unit circle and parabola.
secondly , we test all such points in the ways you mentioned y-intercept k to jusify the points for max.and min. value for the trigo. expressions, is it?
 :D[/quote]

For which problem?

[quote="kunny"]If $ \sin ^ 2 \theta \plus{} \sin \theta \plus{} \cos \theta \Longleftrightarrow \left(\sin \theta \plus{} \frac {1}{2}\right)^2 \plus{} \cos \theta \minus{} \frac {1}{4}$ takes some value $ k$,

the parabpla $ x \plus{} \left(y \plus{} \frac {1}{2}\right)^2 \equal{} k \plus{} \frac {1}{4}$ has more than one intersection points with the unit circle $ x^2 \plus{} y^2 \equal{} 1$.[/quote]

Let's consider that if k becomes a variable then the vertex of parabola will shift along positive /negative x-direction.When the parabola just touches the unit circle, just may count the max./min.  value of k. Is it?

[quote="kunny"]That may be true.[/quote]

[quote="Temperal"][hide="Maximum"]Write as $ x^2 \plus{} x \plus{} \sqrt {1 \minus{} x^2}$ with $ \minus{} 1\le x \le 1$. Differentiating gives the answer easily: Solve $ \minus{} \frac {x}{\sqrt {1 \minus{} x^2}} \plus{} 2x \plus{} 1 \equal{} 0$ and check that the value is attainable.[/hide]

Minimum]Write as $ x^2 \plus{} x \minus{} \sqrt {1 \minus{} x^2}$ with $ \minus{} 1\le x \le 1$. Use the same technique as above and check that the value is attainable.[/quote]

By the way,
By using differentiating method , can you solve the derivative equation:
$ \minus{} \frac {x}{\sqrt {1 \minus{} x^2}} \plus{} 2x \plus{} 1 \equal{} 0$

[quote="kunny"]The calculation of the differentiation is wrong.[/quote]

why not? :)

$ f(x)\equal{}x^2\plus{}x\minus{}\sqrt{1\minus{}x^2}$, we have $ f'(x)\equal{}2x\plus{}1\plus{}\frac{x}{\sqrt{1\minus{}x^2}}$

[quote="kunny"]$ f(x) \equal{} x^2 \plus{} x \minus{} \sqrt {1 \minus{} x^2}$, we have $ f'(x) \equal{} 2x \plus{} 1 \plus{} \frac {x}{\sqrt {1 \minus{} x^2}}$[/quote]

I think he is typo ,in fact $ f(x) \equal{} x^2 \plus{} x \plus{}\sqrt {1 \minus{} x^2}, \sin\theta\equal{}x$,

$ \implies f'(x) \equal{} 2x \plus{} 1 \minus{} \frac {x}{\sqrt {1 \minus{} x^2}}$

if it is so , how to solve it when $ f'(x)\equal{}0$

We can use slightly more straightforward math if we use Lagrange Multipliers instead:
[hide]$ x \equal{} \sin \theta, \; \; y \equal{} \cos \theta$. We want to optimize $ f \equal{} x^2 \plus{} x \plus{} y$ subject to $ g \equal{} x^2 \plus{} y^2 \minus{} 1 \equal{} 0$.
$ (2x \plus{} 1,1) \equal{} \lambda (2x,2y)$
$ \implies \frac {2x \plus{} 1}{x} \equal{} \frac {1}{y} \iff 2 \equal{} \frac {x \minus{} y}{xy}$
$ \implies 4 \equal{} \frac {x^2 \plus{} y^2 \minus{} 2xy}{x^2 y^2} \equal{} \frac {1 \minus{} 2xy}{x^2 y^2}$
$ \implies 4 (xy)^2 \plus{} 2(xy) \minus{} 1 \equal{} 0$
$ \implies xy \equal{} \frac { \minus{} 1 \pm \sqrt {5}}{4}$

$ (x \plus{} y)^2 \equal{} 1 \plus{} 2xy \equal{} \frac {1 \pm \sqrt {5}}{2} \implies x \plus{} y \equal{} \sqrt {\frac {1 \pm \sqrt {5}}{2}}$

Clearly we cannot take the squareroot of a negative, therefore we change the $ \pm$'s into $ \plus{}$

We thus have that $ x,y$ are the roots of the equation:

$ t^2 \minus{} \sqrt {\frac {1 \plus{} \sqrt {5}}{2}} \plus{} \frac { \minus{} 1 \plus{} \sqrt {5}}{4}$[/hide]

[quote="TZF"]We can use slightly more straightforward math if we use Lagrange Multipliers instead:
[hide]$ x \equal{} \sin \theta, \; \; y \equal{} \cos \theta$. We want to optimize $ f \equal{} x^2 \plus{} x \plus{} y$ subject to $ g \equal{} x^2 \plus{} y^2 \minus{} 1 \equal{} 0$.
$ (2x \plus{} 1,1) \equal{} \lambda (2x,2y)$
$ \implies \frac {2x \plus{} 1}{x} \equal{} \frac {1}{y} \iff 2 \equal{} \frac {x \minus{} y}{xy}$
$ \implies 4 \equal{} \frac {x^2 \plus{} y^2 \plus{} 2xy}{x^2 y^2} \equal{} \frac {1 \plus{} 2xy}{x^2 y^2}$
$ \implies 4 (xy)^2 \plus{} 2(xy) \minus{} 1 \equal{} 0$
$ \implies xy \equal{} \frac { \minus{} 1 \pm \sqrt {5}}{4}$

$ (x \plus{} y)^2 \equal{} 1 \plus{} 2xy \equal{} \frac {1 \pm \sqrt {5}}{2} \implies x \plus{} y \equal{} \sqrt {\frac {1 \pm \sqrt {5}}{2}}$

Clearly we cannot take the squareroot of a negative, therefore we change the $ \pm$'s into $ \plus{}$

We thus have that $ x,y$ are the roots of the equation:

$ t^2 \minus{} \sqrt {\frac {1 \plus{} \sqrt {5}}{2}} \plus{} \frac { \minus{} 1 \plus{} \sqrt {5}}{4}$[/hide][/quote]

$ \implies \frac {2x \plus{} 1}{x} \equal{} \frac {1}{y} \iff 2 \equal{} \frac {x \minus{} y}{xy}$
$ \implies 4 \equal{} \boxed{\frac {x^2 \plus{} y^2 \plus{} 2xy}{x^2 y^2} }\equal{}\boxed{ \frac {1 \plus{} 2xy}{x^2 y^2}}$

Is it typo? :lol:

Right, it should be $ \minus{}2xy$ in those; I copied it off my paper incorrectly! But I copied the remaining steps correctly so nothing else changes  :)

[quote="TZF"]Right, it should be $ \minus{} 2xy$ in those; I copied it off my paper incorrectly! But I copied the remaining steps correctly so nothing else changes  :)[/quote]

sorry ,I am unfamiliar with applying for Largrange Multiplier in your proof, even though after reading "http://en.wikipedia.org/wiki/Lagrange_multipliers"

Could you please introduce it definitely?

Suppose you wanted to minimize $ f \equal{} 3x\plus{}4y$, given that $ y\equal{}x^2$. What would you do, visually?

You would plot the curve $ y\equal{}x^2$ and all lines of the form $ 3x\plus{}4y\equal{}k$ (we call these "level curves"), and find the line with the least possible $ k$ that still intersects $ y\equal{}x^2$. In doing this, you would notice that at the point of optimization, the line $ 3x\plus{}4y\equal{}k$ is tangent to the curve $ y\equal{}x^2$. So, for this two-variable case, we can take a shortcut and find the place where $ 3x\plus{}4y\equal{}k$ is tangent to $ y\equal{}x^2$.

We can extend this notion into multiple dimensions. Suppose we want to optimize a function $ f(x \ldots)$ given a constraint $ g(x \ldots)\equal{}0$. Optimization will occur if $ f$ and $ g$ are tangent, which happens where $ f$ and $ g$ have a [i]gradient[/i] (normal vector) in the same direction. Since vectors in the same direction are scalar multiples, we say that:

$ \nabla f \equal{} \lambda \nabla g$

where $ \lambda$ is the scale factor and $ \nabla f$ denotes the gradient of $ f$. This is the essence of the method of lagrange multipliers.

In this case, we have $ f \equal{} x^2\plus{}x\plus{}y$, which we want to optimize subject to the constraint $ g\equal{}x^2\plus{}y^2\minus{}1 \equal{} 0$

$ \nabla f \equal{} \left( \frac{\partial}{\partial x} f , \; \frac{\partial}{\partial y} f  \right) \equal{} (2x\plus{}1, \; 1)$

$ \nabla g \equal{} \left( \frac{\partial}{\partial x} g , \; \frac{\partial}{\partial y} g  \right) \equal{} (2x, \; 2y)$

Thus, there exists a scalar $ \lambda$ such that:

$ (2x\plus{}1 , \; 1) \equal{} \lambda (2x, \; 2y)$

We can write this as a system:

$ 2x\plus{}1 \equal{} \lambda 2x \implies \lambda \equal{} \frac{2x\plus{}1}{2x}$
$ 1 \equal{} \lambda 2y \implies \lambda \equal{} \frac{1}{2y}$

Equating the two expressions for $ \lambda$, we find that $ \frac {2x \plus{} 1}{2x} \equal{} \frac {1}{2y}$

[quote="TZF"]Suppose you wanted to minimize $ f \equal{} 3x \plus{} 4y$, given that $ y \equal{} x^2$. What would you do, visually?

You would plot the curve $ y \equal{} x^2$ and all lines of the form $ 3x \plus{} 4y \equal{} k$ (we call these "level curves"), and find the line with the least possible $ k$ that still intersects $ y \equal{} x^2$. In doing this, you would notice that at the point of optimization, the line $ 3x \plus{} 4y \equal{} k$ is tangent to the curve $ y \equal{} x^2$. So, for this two-variable case, we can take a shortcut and find the place where $ 3x \plus{} 4y \equal{} k$ is tangent to $ y \equal{} x^2$.

We can extend this notion into multiple dimensions. Suppose we want to optimize a function $ f(x \ldots)$ given a constraint $ g(x \ldots) \equal{} 0$. Optimization will occur if $ f$ and $ g$ are tangent, which happens where $ f$ and $ g$ have a [i]gradient[/i] (normal vector) in the same direction. Since vectors in the same direction are scalar multiples, we say that:

$ \nabla f \equal{} \lambda \nabla g$

where $ \lambda$ is the scale factor and $ \nabla f$ denotes the gradient of $ f$. This is the essence of the method of lagrange multipliers.

In this case, we have $ f \equal{} x^2 \plus{} x \plus{} y$, which we want to optimize subject to the constraint $ g \equal{} x^2 \plus{} y^2 \minus{} 1 \equal{} 0$

$ \nabla f \equal{} \left( \frac {\partial}{\partial x} f , \; \frac {\partial}{\partial y} f \right) \equal{} (2x \plus{} 1, \; 1)$

$ \nabla g \equal{} \left( \frac {\partial}{\partial x} g , \; \frac {\partial}{\partial y} g \right) \equal{} (2x, \; 2y)$

Thus, there exists a scalar $ \lambda$ such that:

$ (2x \plus{} 1 , \; 1) \equal{} \lambda (2x, \; 2y)$

We can write this as a system:

$ 2x \plus{} 1 \equal{} \lambda 2x \implies \lambda \equal{} \frac {2x \plus{} 1}{2x}$
$ 1 \equal{} \lambda 2y \implies \lambda \equal{} \frac {1}{2y}$

Equating the two expressions for $ \lambda$, we find that $ \frac {2x \plus{} 1}{2x} \equal{} \frac {1}{2y}$[/quote]

Hello, TZF
Thank you very much!!
You are nice and great...
 :P  :clap:

[quote="kunny"]Here is the solution for Japanese high school students, because they don't study Lagrage's Multipliers.

$ f(\theta) \equal{} \sin ^2 \theta \plus{} \sin \theta \plus{} \cos \theta\ \Longrightarrow f'(\theta) \equal{} 2\sin \theta \cos \theta \plus{} \cos \theta \minus{} \sin \theta$.
Set $ \cos \theta \minus{} \sin \theta \equal{} t\Longleftrightarrow \cos \theta \plus{} \sin \theta \equal{} \pm\sqrt {2 \minus{} t^2} \equal{} \pm\sqrt {1 \minus{} t}\ \because t^2 \minus{} t \minus{} 1 \equal{} 0$, we have $ f'(\theta) \equal{} \minus{} (t^2 \minus{} t \minus{} 1)$ for $ \minus{} \sqrt {2}\leq t\leq \sqrt {2}$. Let $ \alpha \equal{} \frac {1 \minus{} \sqrt {5}}{2},\ \beta \equal{} \frac {1 \plus{} \sqrt {5}}{2}$, $ f(\theta)\equiv g(t)$ has a local minimum $ g(\alpha)$ and a local maximum $ g(\beta)$.   Since $ \sin \theta \equal{} \frac {\pm\sqrt {1 \minus{} t} \minus{} t}{2}$, remark that $ \alpha ^ 2 \minus{} \alpha \minus{} 1 \equal{} 0$, we have $ g(\alpha)$ is either

$ \left(\frac {\sqrt {1 \minus{} \alpha} \minus{} \alpha}{2}\right)^2\pm\sqrt {1 \minus{} \alpha} \equal{} \frac {1}{2}(1 \minus{} \alpha \sqrt {1 \minus{} \alpha}\pm \sqrt {1 \minus{} \alpha})$ or

$ \left(\frac { \minus{} \sqrt {1 \minus{} \alpha} \minus{} \alpha}{2}\right)^2\pm\sqrt {1 \minus{} \alpha} \equal{} \frac {1}{2}(1 \plus{} \alpha \sqrt {1 \minus{} \alpha}\pm \sqrt {1 \minus{} \alpha})$

we also get the similar result for $ g(\beta)$.

Finally there wasn't a solution without calculus. :wink:[/quote]

Thanks for your instructions and sorry to waste your much time to find the details.
 :blush:


In how many ways can a president and a 2-person committee be chosen from a group of 8 people (where the order in which we choose the 2 people doesn't matter)?

Firstly, the president can be chosen in eight ways. The two person committee, can be chosen from the 7 people, but the order doesnt matter.
Committee=7*6/2=21

8*21=168

(i think that's right...)

ANSWER:168


Factor completely:

[size=150]1. (a-b)^3+(b-c)^3+(c-a)^3
2. (x+1)(y+1)(xy+1)+xy[/size]

Maybe these are relatively easy for intermediate, but anyway, have fun! :)

[quote="Singular"]1.

$\sum (a-b)^3 = \sum a^3 - b^3 + 3ab(a-b) = 3[ a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 ] = 3(b-a)(c-b)(a-c)$[/quote]

i got $-3(b-a)(c-b)(a-c)$... and i plugged in values in a calculator and it worked  ;) you might wanna check your expansion of $(a-b)^3$ i think

1. Let $a-b=x,b-c=y,c-a=$,....

or you can use the eqality $A^3+B^3+C^3-3ABC=$....

2. Regard $y$ as constant number and arrange with respect to $x$.

[quote="riddler"]shouldnt it be $-xy$ in the $2nd$ one? :?[/quote]
No, it's not -xy.

One well known trick which make  things easier:

[hide]One can easily check that if  $a=b$ LHS is zero .
[hide]Because  it becomes  $0^3 +(a-c)^3 +(c-a)^3 $ [/hide]

so $(a-b)$ is a factor, similarly $(b-c)$ and $(c-a)$ are also   factors;
so LHS = :

$(a-b)^3+(b-c)^3+(c-a)^3  = K(a-b)(b-c)(c-a)$
where K is a constant.   (Because you have 3rd degree on both sides)
put   a=0, b=1, c=-1  (or any simple value to plug in) and you get K=3) [/hide]

Another easy way  if you happen to remember, and it is nice identity to remember:
[hide]$x^3+y^3+z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx) $
[hide]if $x+y+z=0 $ then $x^3+y^3+z^3 = 3xyz $
and here x+y+z  becomes 0 if x=a-b, y=b-c and z=c-a :) [/hide][/hide]

[quote="DPopov"]You can also use the fact that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ to prove AM-GM for three variables. Can anyone do it?[/quote]
I'm not sure if this is right, but...
[hide]Let  A=a^(1/3), B=b^(1/3), C=c^(1/3).

Then, (a+b+c)/3 - (abc)^(1/3)
=(A^3+B^3+C^3-3ABC)/3
=(A+B+C)(A^2+B^2+C^2-AB-BC-CA)/3
=(A+B+C)((A-B)^2+(B-C)^2+(C-A)^2)/6

Given that a,b,c are positive numbers, A,B,C are also positive. (--> A+B+C is positive)
So, the expression above is positive.
Therefore, (a+b+c)/3 >= (abc)^(1/3) (for a,b,c >=0)
[/hide]
Sorry, it looks really messy without using LaTeX :(
I definately should learn LaTeX...

You are right, frt.

Good luck, for Tex. Here is my first Tex.[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16677$highlight=[/url]

If you make a mistake, someone would correct like this kindly! :)  First I didn't know how to use edit,but it's the very

Practice makes perfect!

Kunihiko


inequalities
algebra
polynomial
function
limit
quadratics
real analysis

Your inequalities prove nothing, Gohan, since you cannot have equality. Now I remember, indeed I put it in Old and New Inequalities. But I'm not going to post the solutions here, since there are too many formulae and I hate Latex.

I have a sketch of a solution, but it's horrendously messy.

Let $s_1=x+y+z=1$, $s_2=xy+yz+zx=a$ and $s_3=xyz=\frac{1}{32}$.

Then I get $P=2(s_2-1)^2-\frac{15}{16}$.

We can determine the range of values of $s_2$ subject to the condition that $x,y$ and $z$ all be real by finding the range of values of $a$ for which the polynomial discriminant of $x^3-x^2+ax-\frac{1}{32}$ is less than or equal to zero.

The discriminant will be $\left(\frac{3a+1}{9}\right)^3+\left(\frac{1}{32}+\frac{2}{27}-\frac{a}{3}\right)^2$, so the solution could be completed by finding where this is positive and negative.  I just couldn't be bothered to go through all that mess...

Silouan, do you have a nicer way to finish off the proof once you have the form of $P$?  Do you have a nice way to find what values of $s_2$ are valid?

I think you should use Horner in a point.

My solution but  in Greek because I have not time to write it to English.

[hide="A hint"]Assume that $x+y+z=1,xy+yz+zx=p$, then $xyz=\frac{1}{32}$. Called $D$ is set of values of $p$.
1) What is $D$?
2) Write $P=f(p)$.
3) Use real analysis for $f(p)$.[/hide]

I did something similar (maybe a little bit easier):

[hide="A hint"] W.l.o.g. $x+y+z=4$, then $xyz=2$. W.l.o.g. $x \ge y,z$ ,
0) By AM-GM applied to $y$ and $z$ we have $x \le 2$, so $4/3 \le x \le 2$
1) Express $y^{4}+z^{4}$ as a function of $z+y = 4-x$ and $zy = 2/x$.
2) Write $P=f(x)$.
3) Use real analysis to show $f(x) \le f(2)$.[/hide]

silouan, I have just solved this problem completely. You can have the solution by sea mail now. Greece is far from Japan. :lol:

[quote="silouan"]Let $x$, $y$, $z$ be positive reals satisfying $\left(x+y+z\right)^{3}=32xyz$
Find the minimum and the maximum of $P=\frac{x^{4}+y^{4}+z^{4}}{\left(x+y+z\right)^{4}}$[/quote]

Here is my solution:

Let $\frac{x}{x+y+z}=a,\ \frac{y}{x+y+z}=b,\ \frac{z}{x+y+z}=c\ (a>0,\ b>0,\ c>0)$, the given condition gives $abc=\frac{1}{32}$ and $a+b+c=1$.

Set $ab+bc+ca=k$, since $a,\ b,\ c$ are the positive roots of the cubic equation with respect to $t$, $t^{3}-t^{2}+kt-\frac{1}{32}=0\ \cdots [*]$.By $t>0$,

Rewrite this as $k=-t^{2}+t+\frac{1}{32t}: =f(t)$. Now we are to find the condition of $t$ for which $[*]$ has more than two positive roots including double root,

that is to say, the condition such that two graphs of the curve $y=f(t)$ and the line $y=k$ have more than two intersection points at $t>0$.

$f'(t)=-2t+1-\frac{1}{32t^{2}}=\frac{(1-4t)(16t^{2}-4t-1)}{32t^{2}}$, so $f'(t)=0\ (t>0)\Longleftrightarrow t=\frac{1}{4},\ \frac{1+\sqrt{5}}{8}$.

Let $t_{0}=\frac{1+\sqrt{5}}{8}$,the behabiour of $f(t)$ is $\lim_{t\rightarrow+0}f(t)=+\infty$, which is decreasing at $0<t<\frac{1}{4},\ t_{0}<t$ and

increasing at $\frac{1}{4}<t<t_{0}$ and $\lim_{t\rightarrow \infty}f(t)=-\infty,\ \lim_{t\rightarrow \infty}\{f(t)-(-t^{2}+t)\}=0$, yielding the local minimum

$f\left(\frac{1}{4}\right)=\frac{5}{16}$, since $t_{0}$ satisfies $16t_{0}^{2}-4t_{0}-1=0\Longleftrightarrow \frac{1}{32t_{0}}=t_{0}-2t_{0}^{2}$,

we have the local maximum $f(t_{0})=-t_{0}^{2}+t_{0}+\frac{1}{32t_{0}}=2t_{0}-3t_{0}^{2}$

$=2t_{0}-\frac{3}{16}(4t_{0}+1)=\frac{1}{16}(20t_{0}-3)=\frac{5\sqrt{5}-1}{32}\because 16t_{0}^{2}-4t_{0}-1=0$.

Therefore the range of $k$ for which $k$ should be taken is $\frac{5}{16}\leq k\leq \frac{5\sqrt{5}-1}{32}$.

Then $P=a^{4}+b^{4}+c^{4}=(a^{2}+b^{2}+c^{2})^{2}-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$

$=\{(a+b+c)^2-2(ab+bc+ca)\}^2-2\{(ab+bc+ca)^2-2abc(a+b+c)\}$

$=(1-2k)^{2}-2\left(k^{2}-\frac{1}{16}\right)=2k^{2}-4k+\frac{9}{8}$

$=2(k-1)^{2}-\frac{7}{8}: =g(k)$ where $\frac{5}{16}\leq k\leq \frac{5\sqrt{5}-1}{32}$.

For this interval, the quadratic function $y=g(k)$ is decreasing and concave down, so $g\left(\frac{5\sqrt{5}-1}{32}\right)\leq P\leq g\left(\frac{5}{16}\right)$.

Consequently the desired answer is

minimum value         $\frac{383-165\sqrt{5}}{256}$ for $a+b+c=1,\ ab+bc+ca=\frac{5\sqrt{5}-1}{32},\ abc=\frac{1}{32}$

maximum value        $\frac{9}{128}$ for $a+b+c=1,\ ab+bc+ca=\frac{5}{16},\ abc=\frac{1}{32}$

kunny

Without losing of generality, we let $ x + y + z = 4 $ and then $ abc = 2 . $
Let $ f(x) = x^4 , $and then $$ f'(x) = 4x^3 ,  f''(x) = 12x^2 > 0, $$
which means $ f $ is convex on the interval $ ( 0 , + \infty ) . $
Using [b]EV[/b] method, we can let $ y = z $, then there's nothing left.


Can be solved without using trigonometry?

Let $P$ a point within $\triangle{ABC}$ equilateral, such that $\overline{PA}= 5$, $\overline{PB}= 7$ and $\overline{PC}= 8$. Find the measure of $\overline{AB}$

 :)

If $x$ is the length of the side of the triangle then $x$ satisfies $3(5^{4}+7^{4}+8^{4}+x^{4}) = (5^{2}+7^{2}+8^{2}+x^{2})^{2}$. This is equivalent to $x^{4}-138x^{2}+1161 = 0$. Thus $x = 9$ or $x= 129$. $x=129$ is not admissible. Therefore $x=9$.

Thank you

I put $x=\sqrt{129}$ in geometry program and it works.

Can you tell me why $x$ satisfies that equation.

 :)


This a cyclic acetal, and the carbon "bearing" the two oxygens have one hydrogen, so the aqueous acid product must have two -OH groups and an aldehyde group, -CHO.

[hide="Answer"]$ [HO \minus{} (CH_2)_3]_2 \minus{} CH \minus{} CHO$[/hide]

The mechanism is essentialy:

1) Protonation of one oxygen;

2) Attack of one water molecule on the carbon bearing the two oxygens;

3) Deprotonation of oxygen "from" water, and protonation of the other oxygen;

4) Departure of HO-... assisted by one of "water oxygen" 's lone pair, followed by deprotonation of the "water oxygen".


Λυστε τις διοφαντικες στους φυσικους αριθμους
$ 1) 2^{x \minus{} 1} \plus{} 1 \equal{} y^{z \plus{} 1}$
$ 2)2^{x \plus{} 1} \minus{} 1 \equal{} y^{z \plus{} 1}$

Αν θελει να τις προσπαθησει καποιος ας το κανει....Αξιζουν τον κοπο....
Αλλιως αυριο θα ποσταρω τις λυσεις τους για να μην μενουν αλυτα προβληματα στο forum....


[u][i]Problem 1[/i][/u] [i]Let a,b,c be positive real numbers such that[/i] $ abc \equal{} 1$
[i]   Prove that[/i]$ \sqrt{\frac{a}{b\plus{}3}} \plus{} \sqrt{\frac{b}{c\plus{}3}}\plus{}\sqrt{\frac{c}{a\plus{}3}}\geq \frac{3}{2}$

[u][i] Problem 2
 [/i][/u]
 [i] Let a,b,c be positive real numbers such that[/i] $ a\plus{}b\plus{}c \equal{} \frac{1}{a}\plus{} \frac{1}{b}\plus{} \frac{1}{c}$
  [i]and[/i] $ a \leq b \leq c$
[i] Prove that
 [/i]
               $ ab^2c^3 \geq 1$

Problem 2

Replacing $ x\equal{}bc$ and so on, the condition becomes $ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\plus{}z$. Also, $ a\le b\le c$ implies $ x\ge y\ge z$. Now we need to prove $ x^2 y\ge 1$.

Since $ (x\plus{}y\plus{}z)^2 \ge 3(xy\plus{}yz\plus{}zx) \equal{} 3(x\plus{}y\plus{}z)$, we have $ x\plus{}y\plus{}z\ge 3$ and $ xy\plus{}yz\plus{}zx \ge 3$. Hence $ x\ge 1$, $ xy\ge 1$ and the conclusion follows.

[quote="nkht-tk14"][u][i]Problem 1[/i][/u] [i]Let a,b,c be positive real numbers such that[/i] $ abc \equal{} 1$
[i]   Prove that[/i]$ \sqrt {\frac {a}{b \plus{} 3}} \plus{} \sqrt {\frac {b}{c \plus{} 3}} \plus{} \sqrt {\frac {c}{a \plus{} 3}}\geq \frac {3}{2}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142431

I think i can give a nicer solution :D.
$ abc \equal{} 1\rightarrow S \equal{} a \plus{} b \plus{} c\geq3\rightarrow \sum\sqrt {\frac {a}{b \plus{} 3}}\geq\sum\sqrt {\frac {a}{a \plus{} 2b \plus{} c}} \equal{} \sum\sqrt {\frac {a}{S \plus{} a}}.$
Let $ f(x) \equal{} \sqrt {\frac {x}{S \plus{} x}}$, and $ f$ is convex, then from Jensen we have that $ \sum\sqrt {\frac {a}{S \plus{} a}}\geq3f(\frac {S}{3}) \equal{} \frac {3}{2}$.
The equality occurs when $ a \equal{} b \equal{} c \equal{} 1$.

Unfortunately, your $ f$ is concave for positive $ x$.


Let $ a,b,c,x,y,z$ are real numbers. If $ |x| \plus{} |z|\geq |y|$, prove the inequality

$ x^2(a \minus{} b)(a \minus{} c) \plus{} y^2(b \minus{} c)(b \minus{} a) \plus{} z^2(c \minus{} a)(c \minus{} b)\geq 0$

You need not only $ \left|x\right| \plus{} \left|z\right|\geq\left|y\right|$, but also $ \left|y\right| \plus{} \left|z\right|\geq\left|x\right|$ and $ \left|x\right| \plus{} \left|y\right|\geq\left|z\right|$. In other words, you need $ \left|x\right|$, $ \left|y\right|$ and $ \left|z\right|$ to be sides of a triangle.

This is one of the cases of the Vornicu-Schur inequality; in my note [url=http://www.cip.ifi.lmu.de/~grinberg/VornicuS.pdf]The Vornicu-Schur inequality and its variations[/url], it is Theorem 1 [b]h)[/b]. It is not due to Vornicu and neither it is to Schur, but it was not invented by me either - if you are looking for the source, here is an equivalent version:

[quote="[url=http://www.mathlinks.ro/viewtopic.php?p=117296#117296]http://www.mathlinks.ro/viewtopic.php?t=4357 post #5[/url]"][b]Kvant Problem M1166, solved in Kvant 11/1989, page 26:[/b]

Prove the inequality $ a^2pq \plus{} b^2qr \plus{} c^2rp \leq 0$, where a, b, c are the sidelengths of a triangle, and p, q, r are three real numbers satisfying p + q + r = 0.[/quote]

I forgot the author, but you can look up in Kvant.

  Darij

I am sorry. This inequality is true if $ a\geq b\geq c$ or $ a\leq b\leq c$. So, I think it is stronger than Vornicu-schur because it is true for all real numbers a,b,c,x,y,z. And, can you prove it? or any idea? :)

[quote="manlio"]Let $ a,b,c,x,y,z$ are real numbers, and $ |z| \plus{} |x|\geq |y|$ etc, prove the inequality

$ x^2(a \minus{} b)(a \minus{} c) \plus{} y^2(b \minus{} c)(b \minus{} a) \plus{} z^2(c \minus{} a)(c \minus{} b)\geq 0$[/quote][quote="thegod277"]This inequality is true if $ b \equal{} {\rm mid}\{a,b,c\}$. I think it is stronger. And, can you prove it? [/quote]$ x^2(a \minus{} b)(a \minus{} c) \plus{} y^2(b \minus{} c)(b \minus{} a) \plus{} z^2(c \minus{} a)(c \minus{} b)$

$ \equal{} \left[|x|a \minus{} \left(|z| \plus{} |x|\right)b \plus{} |z|c\right]^2 \plus{} \left[\left(|z| \plus{} |x|\right)^2 \minus{} y^2\right](a \minus{} b)(b \minus{} c)\geq 0$.
[quote="darij grinberg"]if you are looking for the source, here is[/quote]A. Oppenheim and Roy O. Davies, Inequalities of Schur's Type,  [url=http://www.m-a.org.uk/resources/periodicals/the_mathematical_gazette/]The Mathematical Gazette[/url], Vol. 48 (1964), pp. 25-27.


linear algebra
matrix
linear algebra unsolved

Let $\zeta$ be a primitive $n$-th root of unity.

Prove that all eigenvalues of the matrix $\left(\begin{matrix} 1 & 1 & 1 & \cdots & 1\\ 1 & \zeta & \zeta^2 & \cdots & \zeta^{n-1}\\ 1 & \zeta^2 & \zeta^4 & \cdots & \zeta^{2(n-1)}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \zeta^{n-1} & \zeta^{2(n-1)} & \cdots & \zeta^{(n-1)^2} \end{matrix}\right)$ have modulus $\sqrt{n}$.

[hide="Edit - ignore this"]Terminology issue: I've seen this matrix given several names, but but it is clearly [b]not[/b] a Vandermonde matrix.[/hide]
Call the matrix $F$. (F is for Fourier - this is the matrix on which the discrete Fourier transform is based.)

Prove that $F^*F=nI.$ That is, $F$ is a constant multiple of a unitary matrix. This will lead to Peter's result.

sorry for terminology... but it is a vandermonde matrix i think, though not a general one.

Sorry for asking, but if that is not a Vandermonde... then what`s a Vandermonde?

Wait a minute - now I realize that it is a Vandermonde matrix, albeit a very specific and special one.


limit
logarithms
algebra unsolved
algebra

Solve: $ x.\ln(x\plus{}1)\minus{}(x\plus{}1).\ lnx\equal{}0$

hello, let $ f(x)=x\ln(x+1)-(x+1)\ln(x)$,then we have $ \lim_{x\to 0+}f(x)=+\infty$, $ \lim_{x\to \infty}f(x)=+\infty$, further we get $ f^'(x)=\ln\left(\frac{x+1}{x}\right)-\frac{2x+1}{x(x+1)}<0 \forall x>0$, so it exist an uniquely determined intersection point with the $ x$ axis.
Sonnhard.


function
number theory theorems
number theory

Hello.. Does anyone know if there is a formula for $ \displaystyle \sum_{i\equal{}1}^{n} \phi(i)$?

(here $ \phi$ is the number of positive integers coprime with n and less than n)

You can describe it asymptotically. Should be $ \sim \frac{3}{\pi^2} n^2$.

Will you please prove your asymptotic formula?

See here:

http://en.wikipedia.org/wiki/Proofs_involving_the_totient_function#Average_order_of_the_totient


let $ m \in N$ and $ m>4$ and it isn't a prime  number.
prove that  $ m$ divides $ (m\minus{}1)!$

if $ m\equal{}a^2$ and $ m>4$
then $ 2a<m$
if $ m$ is not a square and not prime then $ m\equal{}ab$ with $ a$ different than $ b$ and $ a,b<m$


Hi,

This doesn't have to do anything with [i]100 Math Problems[/i]. It just something that I got from geometry worksheet that I get answer but not quite sure.

Given: BD is perpendicular to plane M, AB :cong: BC

Prove:  :tri: ABD = :tri: CBD

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1. BD :perp: plane M, AB :cong: BC | 1. Given
2. BD :perp: BC, BD :perp: AB | 2. If line is :perp: to plane, the line is also :perp: to other lines in same plane
3. :ang: DBC :cong: DBA | 3. Def. :perp: lines
4. DB = DB |4. Reflexive Prop.
5. :tri: ABD = :tri: CBD |5. SAS

Hope this is right.

Looks good to me.

[hide]If line is perpendicular to a plane, it is perpendicular to all lines in the plane that pass through its foot[/hide]

Proof is basic, simple, and sound. Getting started stuff...
BTW Rep123max, nice avatar, I love the show [i]Family Guy[/i] :lol: