How many distinct sequences of four letters can be made from the
letters in PROBLEM if each sequence must begin with L and not end with P?

For a sequence to not end with $ P$, we have $ 4$ different possibilities for the last letter.  Obviously, we only have $ 1$ for the first letter.  The second letter has $ 5$ possibilities.  The third letter has only $ 4$.

Thus, the answer is $ 1 \times 5^2 \times 4$ or $ \boxed{100}$.

Yeah, that was a double-typo which I misinterpreted.

The first letter only has one choice. Then the last letter has 5 choices. Then the second letter only has 5 choices and finally the third letter has 4 choices. Multiply to get 100.

First, we can place a L at the front of the four letter sequence: L  _ _ _ .
There are $ 6 \cdot 5 \cdot 4 = 120$ ways to do this without the restriction of the P.
With the P at the end (L _ _ P), there are $5 \cdot 4 = 20$ ways to do this.

Hence, the answer is $120 - 20 = \boxed{100}$.

[hide = Alcumus Sol]The first letter, as stated, must be L, and the fourth letter cannot be P. One way to solve this is to consider the permutations of all 7 letters where the first four satisfy these conditions, and then divide to correct for overcounting. Once we have placed the L, we have 5 options for where to place the P--- any spot except the fourth one, which will become the last one when we remove the last three letters. We can then place the remaining 5 letters without restriction, which there are $5!$ ways to do. This gives us a preliminary count of $5\times5!=600$. However, no matter what the order of the last three letters, we will still get the same sequence of the first 4 letters; for example, LPROMEB, LPROEBM, and LPROBEM and three other permutations all have the first 4 letters LPRO. Because of this, each sequence of 4 letters has been counted exactly 6 times so our answer is $\dfrac{5\times5!}{6}=\boxed{100}$.
[/hide]

[hide=solution without division]start from the end. we have L _ _ _ and the last one has five letters (R,O,B,E,M). The next one has five letters (P and R,O,B,E,M without the one chosen) and then four for the second one. Thus we have 5*5*4 = $\boxed{100}$ solutions. [/hide]


inequalities
search
geometry
geometric transformation
reflection
algebra
system of equations

From MCrawford's Livejournal :D

Solve the system of equations
(x1 - x2 + x3)^2 = x2(x4+x5-x2)
(x2 - x3 + x4)^2 = x3(x5+x1-x3)
(x3 - x4 + x5)^2 = x4(x1+x2-x4)
(x4 - x5 + x1)^2 = x5(x2+x3-x5)
(x5 - x1 + x2)^2 = x1(x3+x4-x1)
for real numbers x1, x2, x3, x4, x5.

er, I don't know if he'll appreciate something that was supposed to be for a class being spewed out onto the web...whatever

Wow, I'm surprised that turns out so nicely.  I guess that was what it was designed for, though.

[quote="churchilljrhigh"]where website is this thing mcraford journal on?[/quote]

http://www.livejournal.com

[quote="mathfanatic"][quote="Ragingg"]He has a online class Livejournal?[/quote]

No, he posted it in his personal LJ.  Churchy, stop asking invasive questions.  There is no aops class livejournal where you can steal math problems from.[/quote]

uh mathfanatic i'm not. i went to livejournal.com and searched for mcraford there wasn't any. it is not my fault people don't give the exact website.

goly gee i tried mcrawford and it still doesn't work
[url]http://www.livejournal.com/userinfo.bml?user=mcrawford[/url]

His LJ username doesn't exactly resemble MCrawford.

In any case, has anyone come up with a solution to this problem yet? (hint: [hide]consider Cauchy-Schwarz.[/hide])

SOLUTION - you've been warned.

[hide] Add up the equations. By C-S we have

5 LS :ge: (x1 + x2 + ... + x5)2

Replacing LS with RS of the equation, and moving the xk2s to the right side of the inequality, we obtain

2 (x1x3 + x1x4 + x2x4 + x2x5 + x3x5) :ge: (x1 + x2)2 + (x2 + x3)2 + (x3 + x4)2 + (x4 + x5)2 + (x5 + x1)2 + 2 (x1x3 + x1x4 + x2x4 + x2x5 + x3x5)

Therefore the sum of squares is less than or equal to 0. This implies that all squares are 0, so

x1 + x2 = x2 + x3 = x3 + x4 = x4 + x5 = x5 + x1 = 0,

so x1 = x2 = x3 = x4 = x5 = 0. [/hide]

Cauchy Schwarz is overkill.  You can do it with the trivial inequality.

Woops! Hah, of course. Just different sums of squares.

[quote="churchilljrhigh"]then what is his username?

which is the question i asked long time ago and mathfantic gets all pissed off at me when nobody seems to say where it is.[/quote]

Google search brought this up:

http://www.livejournal.com/userinfo.bml?user=agnosticessence

[quote="churchilljrhigh"]mathfantic gets all pissed off at me when nobody seems to say where it is.[/quote]

That's because I try to respect people's privacy, i.e. I don't go snooping around for people's personal livejournals.  By the way, it's mathfanatic.  There is no such thing as a fantic.

if he wanted a private journal than why did mcrawford put in livejournal where alot of people can see it.

Well, when someone puts their number in the telephone book, that doesn't mean that they want all the telemarketers in the world to call it.  Same thing here - just because he uses livejournal doesn't mean that he wants a bunch of kids to read it.  He probably uses it to communicate with his [i]friends[/i], to reflect on his life, or whatever people use them for.

P.S. I don't understand your signature.

Yes, but when someone puts their telephone number in the phone book, it means that they are making themselves available to anyone who might like to find out how to call them.  You'll note that the analogy between churchill and a telemarketer is a very bad one, on several levels.  I have no idea why you're being so defensive about readily available information.  I also don't understand why none of you simply asked Mathew.  Jeez.

The reason I didn't give out Mr. Crawford's username is because I thought it appropriate only for Mr. Crawford himself to give out such information. After all, there is the chance, however slim it is, that he would not rather have it publicised.

- but that's getting off topic...


AMC
AIME
summer program
Mathcamp
geometry
calculus
integration
\/closed

If anyone here does GHP (not sure if this is only in Georgia...), do you know if it's a good idea to take an online course on this site (I'm planning to take the AIME Problem Series A) while you're in GHP? I do have Internet access and things like that, but is it a good idea?

hmm, is it a program during the summer?

from google, it seems GHP is only about 1 month

the problem series lasts quite a long time, 12 lessons i think, so at least 3 months

you'll have time to review the transcripts and problems after GHP even if you can't do it during the program itself

Yup... it's a six week program... but it starts here only on June 11... but I think the actual series starts on the 13th. So, I'll have half the lessons there, and half the lessons back home...

Moved to Classes Information.

But yeah, I think many people have done this before. Just review the transcripts well, and do the problems and you'll be fine. :)

I guess it's a good idea...it depends on the rigour of GHP.

I know of people who went to Mathcamp and/or other summer camps, and as a result, didn't really do much in the AoPs classes.  Personally, I'd prefer to attend the actual live lectures and not have to worry about catching up reading transcripts later.  Try to attend as many of the lectures as you can.

Well, it's not exactly rigorous, you know? I mean that not many people actually study stuff there... they just go there to have fun with their friends... and you can make GHP in tons of stuff... like I made it for math. I guess it's just a program where a ton of people with various talents in different areas go to have fun (though it's supposed to teach you stuff  :D ). I think it will be a good integration, because my brain will be like in math mode in GHP, because I'll have about four hours of math everyday...

If you are a math major, then you'll be doing 4+ hours of math in the morning and then attend minors in the afternoon. If you are doing the Problem of the Day, then you might devote another half hour or so for math. That gives you about six or seven hours for fun and socializing. It turns out, however, that these six hours pass by very quickly and that you probably won't have the initiative to get online to attend classes, not to mention spending another 8-10 hours on weekly assigned problems (btw, there IS a huge library in Valdosta that has computers with [fast] internet access). Thus, to get the most out of your GHP experience (which, frankly, involves not only math but meeting a diverse group of talented folks), AoPS classes might be too time consuming. But if you do, you could print out the transcripts and read them on your spare time (and do the problems!). This is because evening hours are the BEST time to have fun in GHP and you don't want to miss the best hours of your summer!

[quote="ginasboys"]But if you do, you could print out the transcripts and read them on your spare time (and do the problems!).[/quote]

The thing is, however, you won't get the MOST out of the class if you only read the transcripts, right? And the classes are only like 1.5 hours every week (I think)... so, I [i]think[/i] that's enough time for like a week. But then again, I'm not sure about that... I'll go and check on the website again for the schedule for students there  :) But do you believe that 1.5 hours per week will be too much time to spend? Or will I have no time at all?  :?

I think it's worth it because you will only be spending half of the time in GHP while enrolled in whichever AoPS class you sign up for. You will have a LOT of free time in GHP. In fact, a lot of upcoming seniors were writing their college essays in their dorms during their free time in GHP, while others were catching up with their summer readings. So if you are an upcoming junior you'll probably have a lot more free time. Thus it comes down to this: whether you want to spend your time talking to interesting people and attending informative "seminars" or doing math problems with your friends (which is also fun, as is doing Problem of the Day). All in all, what I don't want you do is being so obsessed with math that you forget all the other, perhaps more interesting, more exciting aspects of GHP.

Oh yeah, I do have summer reading stuff to do... AP US, AP Lang, AP...  :( That's gonna be a lot.  :?


geometry
perimeter
probability
algebra
system of equations

1.A circle with radius 2 units has its center at (0,0).  A circle with a radius of 7 units with its center at (15,0).  A line tangent to both circles intersects the x-axis at (X,0) What is X expressed as a common fraction????

2.The area of two squares differ by 100 units and the perimeter differs by 10. What's the perimeter of the smaller square.

3.A guy starts with 1,2,3,4 and writes them in any order. He will find the sum of adjacent numbers till he ends up with one integer.  ex. if he starts wit 3,2,1,4 he then gets 5,3,5 then 8,8 then 16. how many possible sums are there.

4.What is the value of the following expression.
1+2-3+4+5-6+7+8-9 . . . +98-99

5.AT a random time during a 45 minute class a teacher gives a 5 minute quiz. If ur late 9 minutes to the class whats the probability that u missed the quiz.  EXpress ans. as a common fraction.

6.A 3 digit integers digits are reversed. The positive difference between the two integers are 396. what is the greatest possible value for the original integer.

7.The average of 2 2-digit positive integers is equal to the decimal number obtained by writing one of the 2-digit integers befor the decimal point and the other two digit integer after the decimal point. what is the smaller of the 2 integers.

8.Four points have courdinates A(3,3),B(5,7),C(8,7),D(12,3). A'B'C'D' are found my multiplying the abscissas of abcd. find the difference of the areas of the two quadralaterals.And tell me what abscisas means please.

thats it thank u :D  :D  :D  :D  :D  :D

o yeah u dont have to ans. all problems if u solve one post please!!!!!!

[hide="4"]
this simplifies to

$ 1\plus{}(\minus{}1)\plus{}4\plus{}(\minus{}1)\plus{}7\plus{}(\minus{}1)\plus{}10\plus{}(\minus{}1)\plus{}...97\plus{}(\minus{}1)$

$ 0\plus{}3\plus{}6\plus{}9\plus{}...96$

$ 3(1\plus{}2\plus{}3\plus{}...32)$

$ 3\cdot\frac{32(33)}{2}\equal{}\boxed{1056}$

[/hide]

[hide="6"] if its in the nine hundreds, the units digit has to be five so that the difference is in the high three hundreds. 995 - 559 is too large. 985-589 turns out to work, so the answer is $ \boxed{985}$
[/hide]

i got a few more, ill post them later.

[hide="2"] Call the side lengths of both the squares $ a$ and $ b$.

The problem gives us that "the areas of the squares differ by 100" which is algebraically represented by
\[ a^2 \minus{} b^2 \equal{} 100
\]
Next, the problem also states that the perimeters differ by ten which is algebraically represented by
\[ 4a \minus{} 4b \equal{} 4(a \minus{} b) \equal{} 10
\]
From this equation, we get that
\[ a \minus{} b \equal{} 2.5
\]
Now we go on to equation 1
\[ a^2 \minus{} b^2 \equal{} 100
\]
If we look closely, we notice that this can also be written as
\[ (a \minus{} b)(a \plus{} b) \equal{} 100
\]
But we know that
\[ a \minus{} b \equal{} 2.5
\]
So
\[ 2.5(a \plus{} b) \equal{} 100
\]
Thus,
\[ a \plus{} b \equal{} 40
\]
We solve this system of equations to get $ a \equal{} 21.25$ and $ b \equal{} 18.75$.

Obviously, the side length of the smaller square is $ 18.75$.

Now find the perimeter
\[ 18.75\cdot4 \equal{} \boxed{75}
\]
[/hide]
[hide="7"]The average of two numbers integers with a decimal has to be a decimal ending with $ .50$.

So one of the numbers is fifty. So the other one is $ \boxed{49}$.[/hide]

[hide="5"]
It had to have started in the first four minutes, so $ \boxed{\frac{4}{45}}$.
[/hide]

you didnt clarify if he was supposed to miss any of it or ALL of it.

o sory my bad whats the probability if he misses the start of the quiz

Um, then it's just $ \frac{9}{45}\equal{}\frac{1}{5}$.


Find the value of $\sin 18^\circ$ using the equation \[ z^4 + z^3 + z^2 + z + 1 = 0. \]

i believe that i posted this before, but anyway, we notice that the roots of unity, and also the roots form reciprocal pairs, so it can be factored as

$(x^2+ax+1)(x^2+bx+1)$, comparing coefficients, $a=\frac{ \sqrt{5}-1}{2}=2\cos 72^{\circ}$  then clearly, $\sin 18^{\circ}=\frac{ \sqrt{5}-1}{4}$

(note: the other value for $a$ is $b$, and it is negative, so it must be the $2\cos 144^{\circ}$)

z^2+z+1+z^-1+z^-2=0
substitue n=z+z^-1
n^2+n-1=0

then you find an approximation of the sin of 18 degrees to see which of your 4 values for z is it.


Well, most of us have seen/heard the llama song (http://www.albinoblacksheep.com/flash/llama), here's the gamma song!

http://home.iitk.ac.in/~shankars/GammaSong/

Hey, that Ravishankar was in college together with my cousin (AoPS id of my cousin - will'otherwisp or something like that)! And that Loganayagam was the weirdest guy in IIT Kanpur in her (my cousin's) time. No wonder his face was the most suitable to keep popping up in the song!

Both Ravishankar and Loganayagam are in my orkut friend list. I'll tell them about this.

By the way, had you heard the Llama song before?

[quote="akalra1"]what about the beaver song?

beaverbeaverbeaver.com[/quote]

Your link doesn't bring up anything....
I think you mean the badger song, and btw, badger song and llama song are completely different.

i meant the badger song

yea i know they r different

badgers, gamma and llamas

*sigh*

Forgot to mention: I am a fan of both Ravisankar and Loganayagam. Especially the latter: he is very similar to me according to my cousin's descriptions! see his eyes? they belong to a faraway world. he is a genius.

bubka, r u bragging?

it sounds like it no offense

sorry the last four lines came out wrong. I meant he was sloppy and nonconforming like me. The genius part was separate.

Anyway I just found out that Ravishankar is an International Physics Olympiad gold medallist in 2002. I had learnt it before but forgotten it.

thats cool

so what will we indians do next?

Here is the song.

Here's a Gamma ...
There's a Gamma ...
And another little gamma!
Real gamma!
Virtual gamma!
Gamma, Gamma ...
Look !
Gamma Gamma
Reimann Gamma
Tangent
Wick-rotate
A gamma.
Gamma Gamma
Squareroot gamma
Gamma Gamma
Look !
I was once a tensor,
I lived in a space.
But I never saw the way
the indices were raised.
I was just an A-mu
But that fixed the gauge!
And now listen little child
to this verbiage ...
Did you ever see a gamma?
Miss a gamma?
Hit a gamma?
Gamma's gamma
Curve of gamma
Gamma Gamma
Look !
Affine gamma
Dirac gamma
Chiral gamma
Basu!
Gamma ...
Gamma with psi-bar
A drama ...
Gamma, Gamma
Look !!!
Is it all in bold now?
Is it all so wierd?
Is it made of tangent bundle?
h-cross
sigma
beard
Now our song is losing spin ...
We've run out of ...
Look!
Time for me to retire now,
And become a crook.

that was very funny.....reminded me of the potter potter song(which is just a repeating song so I don't know why....)

Here is a photo if Ravishankar

[url=http://imageshack.us][img]http://img108.imageshack.us/img108/5149/8328297rb2.jpg[/img][/url]

I took it from his orkut album. How is it?

whoa he looks just like you except not really, no not really


AMC
AMC 8
AMC 12
AIME
search
AMC 10
AoPS Books

I'm going to be preparing for the AMC 10, and I'm wondering if, other then AoPS volume 1, there are other good books to use. Which of the introductory series would be the best? And any books other than aops?

Basically AoPS books to learn the concepts, and then just do lots and lots of tests: finish AMC8s, 10s,...you can also find many problems from contests like Purple Comet, etc...

Do what you're weakest on (everyone seems to be weak at geometry..)

Aops V1 and V2 are amazing.

So are the other texts.

Do AMC12 stuff too.

Hm AMC12 is kind of useless for practice given that most worthwhile problems can't be solved in 2.5 minutes. I personally think AIME is way better assuming you can do at least a few AIME problems already.

if you can solve the problems on aime from ~8 up, your pretty darn set for AMC 10.

im doing aops v1 and the AMC 10 and 12 online courses. By the time I take the ten i  hope to have finished v1 v2 and the aime online course, as well as all available practice tests.That should leave me in good stead dont you think?

Guys, please use the search button. I'm getting tired of seeing these "how can I prepare for whatever and which books should I use" thread. So, see around the forum by using nice Search button.

LOCKED

P.S. This is not just to this poster. People, new or old, need to realize that some questions are often repeated and therefore, have answers in the older threads. Do NOT PM me about this. Thanks.


Can someone solve this problem. i couldn't find the solution on the net.

Vertex E of equilateral triangle ABE  is in the interior of unit square ABCD. Let R be the region consisting of all points inside square ABCD and outside triangle ABE whose distance from AD is between 1/3 and 2/3. What is the area of R?

This problem has been posted and solved here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1051853#p1051853[/url]

For some reason, this solution isn't in the Art of Problem Solving Wiki. However, generally, you can find solutions to AMC and AIME problems there.


How many diagonals does a regular octagon have?

[asy]size(4cm,4cm);

pair A,B,C,D,E,F,G,H;
A = dir(22.5); B = dir(67.5); C = dir(112.5); D = dir(157.5); E = dir(202.5); F = dir(247.5); G=dir(292.5); H=dir(337.5);
draw(A--B--C--D--E--F--G--H--A);[/asy]

The formula for the number of diagonals in a regular, $ n$-sided polygon is $ \frac{n(n\minus{}3)}{2}$(see if you can prove it.) Plugging in 8 for $ n$ gives us $ \boxed{20}$.


geometry
3D geometry
floor function
calculus
integration
AMC

[hide]Note that they're 7 for 1, 10 for 2, 13 for 3, so you have 70 for 22. Then, you want $\frac{23^{3}-1}{22}$. The top can be factored into $(23-1)(529+23+1)$ which cancels nicely with the bottom giving you $529+1+23$ which is $\boxed{553}$.[/hide]

Can you remove your comment from the original post so this can be added into the resources section?  Thanks!

[quote="not_trig"]yep, that's what I did... at first I forgot to add $k^{3}$ and then was like, OH IT'S NOT 69[/quote]

I did the EXACT SAME THING at first!  hehe.

There is a typo in the problem statement. It should say

Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$

Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$

(The error is $k=\lfloor \sqrt[3]{n_1}\rfloor=\lfloor \sqrt[3]{n_1}\rfloor$ instead of $k=\lfloor \sqrt[3]{n_1}\rfloor=\lfloor \sqrt[3]{n_2}\rfloor.$)

[color=#f00]Fixed, thanks ~dj[/color]

[quote=not_trig][quote="Ignite168"][quote="not_trig"]yep, that's what I did... at first I forgot to add $k^{3}$ and then was like, OH IT'S NOT 69[/quote]

Treething is very useful.[/quote]

very true. he loves 69. :rotfl:[/quote]

*sigh*

Clearly we have $n_1 = k^3, n_2 = k^3+k, n_3 = k^3+2k \ldots n_{70} = k^3+69k$. Then we must have $(k+1)^3 > k^3+69k$. The smallest integer value of $k$ satisfying this is $22$, and the answer is $\frac{22^3+69 \cdot 22}{22} = \boxed{553}$.


A MATHCOUNTS competition is held in a room with many tables, and there are 4 stools at each table. Each stool has 3 legs, and each table has 4 legs. If there is a total of 368 legs on all the tables and stools in the room, how many tables are in the room?

There are $ 4 \times 3 \plus{} 4 \equal{} 16$ legs at each table.  Divide by the number of total tables and you get $ \boxed{23}$

In the question, or in 007math's solution.
Btw, 007math's solution is correct if you're wondering.....

I dont understand this problem can someone help me?

[hide="Longer Solution"]
We have $4*3+4$ legs at each table. $+4$ because of the table itself, and $4*3$ because of $4$ stools with $3$ legs on each stool.

$4*3+4=16$

$\frac{368}{16}=\boxed{23}$

Nearly the exact same thing as 007math, but just to show where the numbers come from.
[/hide]


[img]http://xs312.xs.to/xs312/07055/untitled.psd.png.xs.jpg[/img]
through point o O inside triangle lines parallel ti its sides are drawn. the lines divide the figure into 6 areas. prove S=(\sqrt s1+\sqrt s2+\sqrt s3)
where s is the area of whole triangle

All the three triangles are similar to the original triangle $ABC$ .
let the lengths of sections of side $BC$ of triangle $ABC$ be $x,y,z$ then
$\frac{S_{1}}{S}= \frac{x^{2}}{BC^{2}}$
$\frac{S_{2}}{S}= \frac{y^{2}}{BC^{2}}$
$\frac{S_{3}}{S}= \frac{z^{2}}{BC^{2}}$
taking root and adding we get
$\frac{\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}}}{\sqrt{S}}= \frac{x+y+z}{BC}= 1$


[hide="a little more involved"]
$ a^3\plus{}ax\plus{}y\equal{}0$
$ b^3\plus{}bx\plus{}y\equal{}0$
$ c^3\plus{}cx\plus{}y\equal{}0$

subtracting the first and second, then the first and third we get
$ a^3\minus{}b^3\plus{}ax\minus{}bx\equal{}0$
and we factor
$ (a\minus{}b)(x\plus{}a^2\plus{}ab\plus{}b^2)\equal{}0$
and we can divide by a-b because all the numbers are distinct. We get the same for a subtraction of the first and third.
$ x\plus{}a^2\plus{}ab\plus{}b^2\equal{}0$
$ x\plus{}a^2\plus{}ac\plus{}c^2\equal{}0$

$ x\plus{}a^2\plus{}ab\plus{}b^2\equal{}x\plus{}a^2\plus{}ac\plus{}c^2$

$ ab\plus{}b^2\equal{}ac\plus{}c^2$
$ b^2\minus{}c^2\equal{}ac\minus{}ab$
$ \minus{}b\minus{}c\equal{}a$

$ \boxed{a\plus{}b\plus{}c\equal{}0}$
[/hide]

Let $ f(t)\equal{}t^3\plus{}xt\plus{}y$. Then $ f(a)\equal{}f(b)\equal{}f(c)\equal{}0$.

So $ f(t)\equal{}(t\minus{}a)q(x)$. Now $ f(b)\equal{}(b\minus{}a)q(b)\equal{}0$ but $ b\minus{}a\ne 0$ so $ q(b)\equal{}q(c)\equal{}0$. So $ q(t)\equal{}(t\minus{}b)r(t)$ and $ r(c)\equal{}0$ similarly so $ f(t)\equal{}(t\minus{}a)(t\minus{}b)(t\minus{}c)$. Expanding, we get $ a\plus{}b\plus{}c\equal{}0$.

[quote="Altheman"]Let $ f(t) \equal{} t^3 \plus{} xt \plus{} y$. Then $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$.

So $ f(t) \equal{} (t \minus{} a)q(x)$. Now $ f(b) \equal{} (b \minus{} a)q(b) \equal{} 0$ but $ b \minus{} a\ne 0$ so $ q(b) \equal{} q(c) \equal{} 0$. So $ q(t) \equal{} (t \minus{} b)r(t)$ and $ r(c) \equal{} 0$ similarly so $ f(t) \equal{} (t \minus{} a)(t \minus{} b)(t \minus{} c)$. Expanding, we get $ a \plus{} b \plus{} c \equal{} 0$.[/quote]

No need for so much explanation, it's known that $ f(t)\equal{}g(t)(t\minus{}a)(t\minus{}b)(t\minus{}c)$, and as $ f(t)$ is of third degree, $ g(t)\equal{}1$

[quote="Bugi"][quote="Altheman"]Let $ f(t) \equal{} t^3 \plus{} xt \plus{} y$. Then $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$.

So $ f(t) \equal{} (t \minus{} a)q(x)$. Now $ f(b) \equal{} (b \minus{} a)q(b) \equal{} 0$ but $ b \minus{} a\ne 0$ so $ q(b) \equal{} q(c) \equal{} 0$. So $ q(t) \equal{} (t \minus{} b)r(t)$ and $ r(c) \equal{} 0$ similarly so $ f(t) \equal{} (t \minus{} a)(t \minus{} b)(t \minus{} c)$. Expanding, we get $ a \plus{} b \plus{} c \equal{} 0$.[/quote]

No need for so much explanation, it's known that $ f(t) \equal{} g(t)(t \minus{} a)(t \minus{} b)(t \minus{} c)$, and as $ f(t)$ is of third degree, $ g(t) \equal{} 1$[/quote]

I think that we have to be careful. If $ a\equal{}b$ or something we do not have that result.

[quote="Altheman"][quote="Bugi"][quote="Altheman"]Let $ f(t) \equal{} t^3 \plus{} xt \plus{} y$. Then $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$.

So $ f(t) \equal{} (t \minus{} a)q(x)$. Now $ f(b) \equal{} (b \minus{} a)q(b) \equal{} 0$ but $ b \minus{} a\ne 0$ so $ q(b) \equal{} q(c) \equal{} 0$. So $ q(t) \equal{} (t \minus{} b)r(t)$ and $ r(c) \equal{} 0$ similarly so $ f(t) \equal{} (t \minus{} a)(t \minus{} b)(t \minus{} c)$. Expanding, we get $ a \plus{} b \plus{} c \equal{} 0$.[/quote]

No need for so much explanation, it's known that $ f(t) \equal{} g(t)(t \minus{} a)(t \minus{} b)(t \minus{} c)$, and as $ f(t)$ is of third degree, $ g(t) \equal{} 1$[/quote]

I think that we have to be careful. If $ a \equal{} b$ or something we do not have that result.[/quote]

It says they are distinct (look at Valentin's post)

Well of course. The point is to link that hypothesis to the fact that there is a factorization. It seems to me that if you are actually going to prove something in this problem (as opposed to trivializing it by citing some theorem) you use the division algorithm to write that representation of the polynomial.

[quote="Valentin Vornicu"]Let $ a,b,c,x,y$ be five real numbers such that $ a^3 \plus{} ax \plus{} y \equal{} 0$, $ b^3 \plus{} bx \plus{} y \equal{} 0$ and $ c^3 \plus{} cx \plus{} y \equal{} 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero.

[i]Ciprus[/i][/quote]
$ (a^3\plus{}ax\plus{}y)\minus{}(b^3\plus{}bx\plus{}y) \equal{} 0 \iff (a\minus{}b)(a^2\plus{}ab\plus{}b^2\plus{}x)\equal{}0$ $ \iff a^2\plus{}ab\plus{}b^2\plus{}x\equal{}0$ (since $ a \neq b$) and likewise $ b^2\plus{}bc\plus{}c^2\plus{}x\equal{}0$ since $ b \neq c$.

Now $ (a^2\plus{}ab\plus{}b^2\plus{}x)\minus{}(b^2\plus{}bc\plus{}c^2\plus{}x)\equal{}0 \iff$ $ a^2\minus{}c^2\plus{}b(a\minus{}c) \equal{} (a\minus{}c)(a\plus{}b\plus{}c) \equal{} 0 \iff a\plus{}b\plus{}c \equal{} 0$ (since $ a \neq c$)

[quote="Bugi"][quote="Mathias_DK"][quote="Valentin Vornicu"]Let $ a,b,c,x,y$ be five real numbers such that $ a^3 \plus{} ax \plus{} y \equal{} 0$, $ b^3 \plus{} bx \plus{} y \equal{} 0$ and $ c^3 \plus{} cx \plus{} y \equal{} 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero.

[i]Ciprus[/i][/quote]
$ (a^3 \plus{} ax \plus{} y) \minus{} (b^3 \plus{} bx \plus{} y) \equal{} 0 \iff (a \minus{} b)(a^2 \plus{} ab \plus{} b^2 \plus{} x) \equal{} 0$ $ \iff a^2 \plus{} ab \plus{} b^2 \plus{} x \equal{} 0$ (since $ a \neq b$) and likewise $ b^2 \plus{} bc \plus{} c^2 \plus{} x \equal{} 0$ since $ b \neq c$.

Now $ (a^2 \plus{} ab \plus{} b^2 \plus{} x) \minus{} (b^2 \plus{} bc \plus{} c^2 \plus{} x) \equal{} 0 \iff$ $ a^2 \minus{} c^2 \plus{} b(a \minus{} c) \equal{} (a \minus{} c)(a \plus{} b \plus{} c) \equal{} 0 \iff a \plus{} b \plus{} c \equal{} 0$ (since $ a \neq c$)[/quote]

Isn't it the same as qntty's solution?  :huh:[/quote]
Yes it is.

[quote="tanuagg13"][quote="leepakhin"][hide]Since $a,b,c$ are distinct, they are the real roots of $t^3+xt+y=0$, so sum of roots equals $0$.[/hide][/quote]

Would this get a good score on an actual olympiad?[/quote]

You would have to elaborate more.

a,b,c is the roots of the one equation
k*k*k+k*x+y=0
By Vieta a+b+c=0


Prove that the sum of the reciprocals of all the primes diverges.

EDIT: try to avoid using hard-to-prove facts, like there is always a prime between $n$ and $2n$.

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

i like the first proof.

[quote="Phelpedo"]try to avoid using hard-to-prove facts[/quote]

The result is not elementary.  I doubt you could find an elementary proof.

The nicest one I know of is by a factorization of the Zeta function  :D


modular arithmetic
number theory unsolved
number theory

Find all positive interger n such that $ n|(2^n\plus{}2)$ and $ (n\minus{}1)|(2^n\plus{}1)$

We will prove that $ n\equal{}2$ is the only solution. Since $ 2^n \plus{} 1$ is odd and $ (n\minus{}1)|2^n \plus{}1$, $ n\minus{}1$ must also be odd. It is followed that $ n$ is even.
Suppose on contrary that $ n\minus{}1 > 1$, then because $ n\minus{}1$ is odd, it must have an odd prime divisor. Let $ p$ be one of them. Put $ k: \equal{}\text{ord}_p (2)$, we have: $ 2^n \equiv \minus{}1 \pmod{p}$ so $ 2^{2n} \equiv 1 \pmod{p}$. Thus $ 2n$ is divisible by $ k$ but $ n$ is not. Therefore, either $ k\equal{}2$ or $ k\equal{}2n$. But $ n$ is even so if $ k\equal{}2$ then $ n$ is divisible by $ k$, contradiction. In conclusion, $ k\equal{}2n$. This yields $ p\minus{}1 \vdots 2n$ and so $ p\minus{}1 \geq 2n$. On the other hand $ p|n\minus{}1$ so $ p\leq n\minus{}1$. So $ n\minus{}1 \geq p\minus{}1 \geq 2n$, or $ n \leq \minus{}1$, contradiction.
Thus, $ n\minus{}1\equal{}1$ and $ n\equal{}2$. QED.

[b]Notice:[/b] The condition $ n|2^n \plus{} 2$ had never been used.

[quote="Mashimaru"]We will prove that $ n \equal{} 2$ is the only solution. Since $ 2^n \plus{} 1$ is odd and $ (n \minus{} 1)|2^n \plus{} 1$, $ n \minus{} 1$ must also be odd. It is followed that $ n$ is even.
Suppose on contrary that $ n \minus{} 1 > 1$, then because $ n \minus{} 1$ is odd, it must have an odd prime divisor. Let $ p$ be one of them. Put $ k: \equal{} \text{ord}_p (2)$, we have: $ 2^n \equiv \minus{} 1 \pmod{p}$ so $ 2^{2n} \equiv 1 \pmod{p}$. Thus $ 2n$ is divisible by $ k$ but $ n$ is not. Therefore, either $ k \equal{} 2$ or $ k \equal{} 2n$. But $ n$ is even so if $ k \equal{} 2$ then $ n$ is divisible by $ k$, contradiction. In conclusion, $ k \equal{} 2n$. This yields $ p \minus{} 1 \vdots 2n$ and so $ p \minus{} 1 \geq 2n$. On the other hand $ p|n \minus{} 1$ so $ p\leq n \minus{} 1$. So $ n \minus{} 1 \geq p \minus{} 1 \geq 2n$, or $ n \leq \minus{} 1$, contradiction.
Thus, $ n \minus{} 1 \equal{} 1$ and $ n \equal{} 2$. QED.

[b]Notice:[/b] The condition $ n|2^n \plus{} 2$ had never been used.[/quote]
when u concluded that $ k \equal{} 2$ or $ k \equal{} 2n$ u have made a mistake the only thing that u can say from these facts (i mean :Thus $ 2n$ is  divisible by $ k$ but $ n$ is not) is that $ v_2(k) \equal{} v_2(n) \plus{} 1\equal{}2$.
 the only things that i could find is that  if $ q | \frac {n}{2} \implies q \equiv 1,3 \pmod{8}$.
and also if $ p |n \minus{} 1 \implies p \equiv 1 \pmod {4}$.

[quote="Mashimaru"]We will prove that $ n \equal{} 2$ is the only solution. Since $ 2^n \plus{} 1$ is odd and $ (n \minus{} 1)|2^n \plus{} 1$, $ n \minus{} 1$ must also be odd. It is followed that $ n$ is even.
Suppose on contrary that $ n \minus{} 1 > 1$, then because $ n \minus{} 1$ is odd, it must have an odd prime divisor. Let $ p$ be one of them. Put $ k: \equal{} \text{ord}_p (2)$, we have: $ 2^n \equiv \minus{} 1 \pmod{p}$ so $ 2^{2n} \equiv 1 \pmod{p}$. Thus $ 2n$ is divisible by $ k$ but $ n$ is not. Therefore, either $ k \equal{} 2$ or $ k \equal{} 2n$. But $ n$ is even so if $ k \equal{} 2$ then $ n$ is divisible by $ k$, contradiction. In conclusion, $ k \equal{} 2n$. This yields $ p \minus{} 1 \vdots 2n$ and so $ p \minus{} 1 \geq 2n$. On the other hand $ p|n \minus{} 1$ so $ p\leq n \minus{} 1$. So $ n \minus{} 1 \geq p \minus{} 1 \geq 2n$, or $ n \leq \minus{} 1$, contradiction.
Thus, $ n \minus{} 1 \equal{} 1$ and $ n \equal{} 2$. QED.

[b]Notice:[/b] The condition $ n|2^n \plus{} 2$ had never been used.[/quote]
No,$ n\equal{}6$ such that condition.
I think that only a sequence positive integer $ x_n$ s.t: $ x_1\equal{}2$ ,$ x_n\equal{}2^{x_{n\minus{}1}}\plus{}2$ such that condition problem.

nice sequence math10,
now i remember that i saw your problem somewhere in this forum,
and the person who gave the solution has used exactly the same sequence to prove that these divisibilities have infinitely many solutions,
anyway finding all solutions seems sth tough!
i found it:
http://www.mathlinks.ro/viewtopic.php?p=1632926#1632926

[quote="shoki"]nice sequence math10,
now i remember that i saw your problem somewhere in this forum,
and the person who gave the solution has used exactly the same sequence to prove that these divisibilities have infinitely many solutions,
anyway finding all solutions seems sth tough![/quote]
oh,I think that problem :prove there exist infinitely n such that condition above be very old.
But I believe we can solve this problem.
I have set out a hypothese:$ n|(2^n\plus{}2)$ equivalent $ (n\minus{}1)|(2^n\plus{}1)$
I think that it not true but I  can not be negative

[quote="math10"][quote="shoki"]nice sequence math10,
now i remember that i saw your problem somewhere in this forum,
and the person who gave the solution has used exactly the same sequence to prove that these divisibilities have infinitely many solutions,
anyway finding all solutions seems sth tough![/quote]
oh,I think that problem :prove there exist infinitely n such that condition above be very old.
But I believe we can solve this problem.
I have set out a hypothese:$ n|(2^n \plus{} 2)$ equivalent $ (n \minus{} 1)|(2^n \plus{} 1)$
I think that it not true but I  can not be negative[/quote]
u r true ,sorry i don't have a good memory!
how far did u go with this ,
could u explain?
thx in advance! :)

assuming my program ran correctly, the only $ n<200000$ that satisfy this are $ \{2,6,66\}$, which is fairly few. perhaps $ 66$ will provide more information

[quote="shallow"]assuming my program ran correctly, the only $ n < 200000$ that satisfy this are $ \{2,6,66\}$, which is fairly few. perhaps $ 66$ will provide more information[/quote]
I predict the next number such that only $ n\equal{}2^66\plus{}2$,you can trial with $ n<2^77\plus{}2$,if not there exist n then I think that solution is sequence $ x_1\equal{}2$ and $ x_{n\plus{}1}\equal{}2^{x_n}\plus{}2$

solutions for $ n|2^n \plus{} 2$ are $ 2,6,66,946,8646,180246,199606,265826,383846,...$
solutions for $ n\minus{}1|2^n \plus{} 1$ are $ 2,6,66,378,1190,1470,25806,58590,134946,137346,170586,272610,285390,420210,538734,592410,618450,680706...$

We know that $ a|b$ and  $ \frac{b}{a}$ is odd $ \implies 2^a \plus{} 1 | 2^b \plus{} 1$

$ \therefore n$ is even and $ n|2^n \plus{} 2 \implies  2^n \plus{} 1 | 2^{2^n \plus{} 2} \plus{} 1$
and $ n \minus{} 1 | 2^n \plus{} 1 \implies  2^{n\minus{}1} \plus{} 1 | 2^{2^n \plus{} 1} \plus{} 1 \implies 2^n \plus{} 2 | 2^{2^n \plus{} 2} \plus{} 2$

so $ n$ is even and $ n|2^n \plus{} 2$ and $ n \minus{} 1 | 2^n \plus{} 1 \implies m|2^m \plus{} 2$ and $ m \minus{} 1 | 2^m \plus{} 1$ where $ m \equal{} 2^n \plus{} 2$

$ \therefore$ there is a sequence of solutions $ k_i$ given by $ k_1 \equal{} 2$ and $ k_{i\plus{}1} \equal{} 2^{k_i} \plus{} 2$

i.e $ 2,6,66,2^{66}\plus{}2, ...$

It remains to determine if there are any other simultaneous solutions for these two conditions.


Find all pairs of integers $ (a,b)$ such that $ a^3\pm b^3\equal{}a^2b^2$.

$ d\equal{}\gcd(a,b)$
$ a\equal{}dm,b\equal{}dn$ where $ \gcd(m,)\equal{}1$
The equation equivalent :
$ dm^2n^2\equal{}m^3\plus{}n^3$
Because $ \gcd(m,n)\equal{}1$ so $ \gcd(m^3\plus{}n^3,m^2n^2)\equal{}1$
So $ (mn)^2\equal{}1$
$ \Rightarrow mn\equal{}1 ,mn\equal{}1$
[u]Case 1[/u] $ mn\equal{}1$ then $ (m,n,d)\equal{}(1,1,2),(\minus{}1,\minus{}1,\minus{}2)$
$ \Rightarrow (a,b)\equal{}(2,2)$
[u]Case 2[/u] $ mn\equal{}\minus{}1$ so $ m\plus{}n\equal{}0$
$ \Rightarrow (a,b)\equal{}(0,0)$


trigonometry
inequalities
inequalities unsolved

Let $ a,b,c$ be real numbers. Prove the inequality $ \frac {\cos^2a}{\sin^2(b \minus{} c)} \plus{} \frac {\cos^2b}{\sin^2(c \minus{} a)} \plus{} \frac {\cos^2c}{\sin^2(a \minus{} b)}\geq 2$


inequalities
induction
geometry
geometric transformation
reflection
inequalities proposed

[quote="hungkhtn"]If $ a,b,c,d$ are non-negative real numbers with sum $ 4$, then
\[ a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a \plus{} 23abcd \le 27.
\]
[/quote]
Let $ a \equal{} \min\{a,b,c,d\}$ and $ b \equal{} a \plus{} x,$ $ c \equal{} a \plus{} y$ and $ d \equal{} a \plus{} z.$
Then $ a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a \plus{} 23abcd \le 27\Leftrightarrow$
$ \Leftrightarrow(1824(x^2 \plus{} y^2 \plus{} z^2) \minus{} 1472(xy \plus{} yz) \minus{} 704xz)a^2 \plus{}$
$ \plus{} (176(x^3 \plus{} y^3 \plus{} z^3) \plus{} 528(x^2y \plus{} y^2z) \plus{} 1296(x^2z \plus{} y^2x \plus{} z^2x \plus{} z^2y) \minus{} 5888xyz)a \plus{}$
$ \plus{} 27(x \plus{} y \plus{} z)^4 \minus{} 256(x^3y \plus{} y^3z)\geq0.$
$ 1824(x^2 \plus{} y^2 \plus{} z^2) \minus{} 1472(xy \plus{} yz) \minus{} 704xz\geq912\sum_{cyc}(x\minus{}y)^2\geq0,$
$ 176(x^3 \plus{} y^3 \plus{} z^3) \plus{} 528(x^2y \plus{} y^2z) \plus{} 1296(x^2z \plus{} y^2x \plus{} z^2x \plus{} z^2y) \minus{} 5888xyz\geq$
$ \geq\left(3\cdot176 \plus{} 6\cdot\sqrt [6]{528^2\cdot1296^4} \minus{} 5888\right)xyz\geq0$ and
$ 27(x \plus{} y \plus{} z)^4\geq256(x^3y \plus{} y^3z \plus{} z^3x) \plus{} 473xyz(x \plus{} y \plus{} z)\geq256(x^3y \plus{} y^3z).$

sorry argady if my computer made mistake. My computer gives me the following expression:
\[ f(a,b,c,d)\equal{}27(a\plus{}b\plus{}c\plus{}d)^4\minus{}4^3(a\plus{}b\plus{}c\plus{}d)(a^3b\plus{}b^3c\plus{}c^3d\plus{}d^3a)\minus{}23\times 4^4abcd\]
I factored it and I have $ \minus{}2^{10}a^5...$
Can you check it again? :(  :wink:  maybe my calculation is wrong  :wink:

[quote="arqady"][quote="zaizai-hoang"]
\[ f(a,b,c,d) \equal{} 27(a \plus{} b \plus{} c \plus{} d)^4 \minus{} 4^3(a \plus{} b \plus{} c \plus{} d)(a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a) \minus{} 23\times 4^4abcd
\]
[/quote]
What is this? :wink:
Maybe you mean
\[ f(a,b,c,d) \equal{} 27(a \plus{} b \plus{} c \plus{} d)^4 \minus{} 4^4(a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a) \minus{} 23\times 4^4abcd
\]
$ ?$[/quote]
Yes, I am wrong    :blush: your solution is correct  :wink:

I think it is correct.

Is there any other nice proof? Notice that the equality holds for $ a \equal{} b \equal{} c \equal{} d \equal{} 1$ or $ a \equal{} 3,b \equal{} 1,c \equal{} d \equal{} 0$ or permutations. s

How could you show this case of equality, Arqady?

[quote="hungkhtn"]Notice that the equality holds for $ a \equal{} b \equal{} c \equal{} d \equal{} 1$ or $ a \equal{} 3,b \equal{} 1,c \equal{} d \equal{} 0$ or permutations. s

How could you show this case of equality, Arqady?[/quote]
Of cause!
If $ x\equal{}y\equal{}z\equal{}0$ then we obtain $ a\equal{}b\equal{}c\equal{}d$ and if $ a\equal{}z\equal{}0$ we obtain:
$ 27(x \plus{} y \plus{} z)^4\geq256(x^3y \plus{} y^3z )\Leftrightarrow(x\minus{}3y)^2(27x^2\plus{}14xy\plus{}3y^2)\geq0.$
It gives the case $ a\equal{}d\equal{}0,$ $ b\equal{}3$ and $ c\equal{}1.$

So wonderful!!! Congratulation, Arqady!

Just for curious? Did you make everything by hand?

[quote="hungkhtn"]So wonderful!!! Congratulation, Arqady!
[/quote]
Thank you! :lol:
[quote="hungkhtn"]Just for curious? Did you make everything by hand?[/quote]

With a calculator, certainly! It's not so hard, hungkhtn. :wink:

The following generalization is one of the hardest inequalities I have ever created and solved

[size=117][i]Let $ a_1,a_2,...,a_n$ be non-negative real numbers with sum $ n$, with $ n\ge 3$. Prove that[/i][/size]
\[ a_1^{n \minus{} 1}a_2 \plus{} a_2^{n \minus{} 1}a_3 \plus{} ... \plus{} a_n^{n \minus{} 1}a_1 \plus{} \left((n \minus{} 1)^{n \minus{} 1} \minus{} n\right) a_1a_2...a_n \le (n \minus{} 1)^{n \minus{} 1}.
\]
At least, for this one, no one can check this by the brute force method. Instead, my solution is also very complicated, too.

For $ n \equal{} 3$, we have: [i]If $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ then[/i]
\[ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \le 4.
\]
For $ n \equal{} 4$, we have: [i]If $ a,b,c,d\ge 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ then[/i]
\[ a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a \plus{} 23abcd \le 27.
\]
This is my inequality posted above.

I hope some one can give a solution for this one as soon as possible.

Very beautiful problem,hungkhtn! :)  What are you using method to prove it?

Thank you, I used induction (but general induction), with combination with the Entirely Mixing Variable.

Let postpone the general result for a few minute. I think the following inequality for $ 4$ variables is also very nice and stronger than the previous inequality I posted before

[i]If $ a,b,c,d$ are non-negative real numbers with sum $ 4$, then[/i]
\[ a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a \plus{} 6(abc \plus{} bcd \plus{} cda \plus{} dab) \le 27 \plus{} abcd.
\]
I hope someone can give a nice solution for this one. Notice that the equality can also hold for $ (a,b,c,d) \equal{} (1,1,1,1)$ or $ (3,1,0,0)$.

Argady's technique solved it again  :D
We have:

$ f(a,b,c,d) \equal{}$
$ 27(a \plus{} b \plus{} c \plus{} d)^4 \plus{} 4^4abcd \minus{} 6\times 4^3(a \plus{} b \plus{} c \plus{} d)(abc \plus{} bcd \plus{} cda \plus{} dab)$
$ \minus{} 4^4\times 6(a^3b \plus{} b^3c \plus{} c^3d \plus{} d^3a)$

So $ f(a,a \plus{} x,a \plus{} y,a \plus{} z)$ is equivalent to:
$ I\times a^2 \plus{} Love \times a \plus{} U \ge 0$
Which:
$ I \equal{} 672(x^2 \plus{} y^2 \plus{} z^2) \minus{} 704(xy \plus{} yz) \plus{} 64xz$
$ Love \equal{} 176(x^3 \plus{} y^3 \plus{} z^3) \plus{} 528(x^2z \plus{} z^2y \plus{} xy^2 \plus{} z^2z) \minus{} 240(x^2y \plus{} y^2z) \minus{} 992xyz$
$ U \equal{} 27(x^4 \plus{} y^4 \plus{} z^4) \plus{} 108(x^3z \plus{} z^3y \plus{} xy^3 \plus{} z^3z)$
$ \minus{} 60xyz(x \plus{} y \plus{} z) \minus{} 148(x^3y \plus{} y^3z) \plus{} 162(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)$
:D

Did you use this technique to ... conquer some girls, Hoang?   :lol:

Anyway, this is a very good solution. But please make something more precise in the last step. Actually, I expect a different solution,u

@ Zaizai. What program did you use to get the last form? Is this Maple?

yes, I will complete it but not now. Today is Christmas so I should go away my house and have a perfect night with my friend  :D
Are you really curious why I can expand them? I will answer your question. I made everything by hand. I don't know how to using Maple. I took about 7 minutes to expand upper expression. Only some minutes . Of course, I will try to find out a nicer solution, maybe I should use EMV... your method   :wink:

@anh Hùng: chúc anh giáng sinh vui vẻ  :)

That is very good, Hoang. This technique is so strong,

I think the following inequality is true, too

[i]If $ a,b,c,d\ge 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ then[/i]
\[ a\sqrt {ab} \plus{} b\sqrt {bc} \plus{} c\sqrt {cd} \plus{} d\sqrt {da}\plus{} (3\sqrt{3}\minus{}4)\sqrt {abcd}\le 3 \sqrt{3}.
\]

Moreover, I also think that the following inequality holds

[i]If $ a,b,c,d\ge 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ then [/i]
\[ a\sqrt [3]{b} \plus{} b\sqrt [3]{c} \plus{} c\sqrt [3]{d} \plus{} d\sqrt [3]{a}\le 3 \plus{} \sqrt [3]{abcd}.
\]
And the general problem is

[i]If $ a_1,a_2,...,a_n$ are non-negative real numbers with sum $ n$ then[/i]
\[ a_1\sqrt [n \minus{} 1]{a_2} \plus{} a_2\sqrt [n \minus{} 1]{a_3} \plus{} ... \plus{} a_n\sqrt [n \minus{} 1]{a_1}\le n \minus{} 1 \plus{} \sqrt [n \minus{} 1]{a_1a_2...a_n}.
\]

they are also very nice. How could you create it? Only some minute?   :)

I got this idea from one of the problem I sent to Mathematics Reflection (thus, I can not show it now) and one of the problem of VASC (case $ n\equal{}2$). Of course, the general problem is just following the particular problem.

@ hoang. Chuc em giang sinh vui ve ;)

with the notation above when solving this I used the same method as for the 3 variable one.
$ f(a,b,c,d)\le f(a\plus{}x,b\plus{}x,c\plus{}x,d\plus{}x)$ so we may assume one variable is 0 and it is very easy. (actually now i see that this is similar to the one by arqady above..)

Yes, by this brute force method, it is very easy. But I just wonder if we can make it by hand, or without a calculators.

Imagine as if you have to take IMO contest!


For a given value of k the product of the solutions of $ x^2\minus{}3kx\plus{}2k^2\minus{}1\equal{}0$ is 7. Are the roots rational, irrational, or imaginary?

I really don't understand this problem or what it's even asking. Help pls. I don't really want to use the solution book that much as I've used it a lot before and I just want to know what it's really asking (maybe a hint or 2)

[quote="BlackSwordsman"]The problem is saying that the roots of $ x^2 \minus{} 3kx \plus{} 2k^2 \minus{} 1 \equal{} 0$ have a product of $ 7$, and asking if they are rational, irrational, or imaginary.

Look at it like a quadratic equation where $ a \equal{} 1$, $ b \equal{} \minus{} 3k$, and $ c \equal{} 2k^2 \minus{} 1$

You can use the quadratic formula to get the roots, and also note that $ \frac {c}a$ is the product of the roots (7).[/quote]

So the product, or $ c/a$ is 2k^2-1 and it equals 7, so $ 2k^2\minus{}1\equal{}7$ so $ 2k^2\equal{}8$ $ k^2\equal{}4$ so it would be real!

Thanks so much!

What you said is correct, except we still need to find the roots to complete the question (using the formula)

$ x\equal{}\frac{3k \pm \sqrt{(\minus{}3k)^2 \minus{} 4 \cdot 1 \cdot (2k^2\minus{}1)}}2$

And you know that $ k^2$ = $ 4$

After some simplifications we get

$ x\equal{} \frac{3k \pm 2\sqrt{2}}2$

Because k is rational the roots are irrational, which is the answer.

[quote="BlackSwordsman"]What you said is correct, except we still need to find the roots to complete the question (using the formula)

$ x \equal{} \frac {3k \pm \sqrt {( \minus{} 3k)^2 \minus{} 4 \cdot 1 \cdot (2k^2 \minus{} 1)}}2$

And you know that $ k^2$ = $ 4$

After some simplifications we get

$ x \equal{} \frac {3k \pm 2\sqrt {2}}2$

Because k is rational the roots are irrational, which is the answer.[/quote]

Oh yeah. Oops, I should have read it more clearly

EDIT: COuld you also not replace k^2 in the original quadratic equation for 4, rather then in the middle of the formula. It doesn't really matter though, both would yield the same result


Lâu lắm rồi (từ năm 2004) Việt Nam mới có dịp mở mày mở mặt với bạn bè quốc tế như năm nay. 3 HCv, 3HCB! Chúc mừng tất cả các em, chúc cho một sự khởi đầu đầy vinh quang !

Ps. năm nay Vietnamese IMOers lên mathlinks ít quá.

Anh cũng chúc mừng :D Hùng nó hiểu ý mình quá, đang định lập 1 topic đề vào hú hét cho sướng :D Anh em vào spam máu vào đêeeeeeeeeeeeeee ! :P

Ke ke, đúng là anh em mình hiểu ý nhau thật :) Spam tí cho zui zẻ :)

Đội tuyển Việt Nam lần này tuy "phải" thi ở trong nước (anh dùng từ "phải" trong dấu nháy, vì có lẽ "phải" đi ra nước ngoài mới thấm thía hết cái "Vị" của IMO, mệt mỏi sau những chuyến bay và hồi hộp, hạnh phúc chào đón người thân chờ mình ở sân bay  :lol: ), nhưng bù lại, các em được sự "chăm sóc" rất nhiệt tình từ các tổ chức, không bị mệt do đi đường dài...

Thành tích thứ 3 lần này thực ra đã vượt qua thành tích xuất sắc nhất (ko kể năm nay) 2004 (thứ tư). Đặc biệt chúc mừng Bách, vì em đã dành được HCV (cao nhất đoàn), lại là học sinh của trường KHTN (hi, mình nói thế ko phải là thiếu thiện chí đối với các bạn ở các trường khác đâu, chỉ là chút "máu" màu cờ sắc áo tổng hợp thoai :D)

Sắp tới các em sẽ còn phải... đối mặt với rất nhiều cuộc phỏng vấn, gặp gỡ :D Lo mà nói năng cho cần thận và kiếm bạn gái sớm  :D

Ước mong sau 15 năm nữa Việt Nam sẽ .. dẫn đầu cả thế giới  :lol: Khi đó hi vọng... con anh Tuân sẽ đi thi !

[quote="N.T.TUAN"]Năm nay Vn mình ra đề dã man quá, mấy thằng giỏi tạch hết , đặc biệt là hai bài 3 và 6. Thấy bảo chỉ có một chú được 37 :D   :o . Cả ông gì người Moldoval cũng chết , năm ngoái cậu ấy được 42 cơ mà :D Nói chung sang Vn là lành ít dữ nhiều.  :D[/quote]

Tên người Moldova là Iura, rất nổi tiếng, hình như còn thi cùng với em từ 2004. Bác nói rất đúng, sang Việt nam lành ít dữ nhiều, he he, thế nên US năm nay mới xuống đến 3 bậc :D Đề năm nay cao thủ trên Mathlinks tạch hết :D

Hừm, nhưng mà anh thấy mệnh đề sau không đúng:

'' Chú giỏi ngang Peter Sarnak vì cả chú và ông ấy đều không c/m được Định Lí cuối cùng của Fermat'' :P (Thực tế nếu phấn đấu, tương lai chú có thể sẽ hơn :D )

Mình thích cái kiểu đề phải có ít nhất 1 thí sinh 42/42 ! :D

[quote="hungkhtn"] Khi đó hi vọng... con anh Tuân sẽ đi thi ![/quote]
Chú nói thế này làm anh thấy trách nhiệm làm bố lớn quá :P
[hide="Peter Sarnak"]Current editors of the Annals of Mathematics
Jean Bourgain, Institute for Advanced Study
David Gabai, Princeton University
Phillip Griffiths, Institute for Advanced Study
Nicholas M. Katz, Princeton University
[b]Peter Sarnak, Princeton University [/b]
Yakov Sinai, Princeton University
The associate editors are Mladen Bestvina, Tobias Colding, Gregory Margulis, Gopal Prasad, Oded Schramm, Bernd Sturmfels, and Kari Vilonen.[/hide]

[quote="N.T.TUAN"][quote="hungkhtn"] Khi đó hi vọng... con anh Tuân sẽ đi thi ![/quote]
Chú nói thế này làm anh thấy trách nhiệm làm bố lớn quá :P
[/quote]

Chán anh quá :D lại không hiểu ý em roài :D Em bảo là 15 năm nữa con anh đi thi, mà muốn đi thi phải học đến lớp 11, tức là 16 tuổi. Vậy 15 năm nữa con anh 16 tuổi. Tại thời điểm hiện tại anh chưa cưới vợ, mà muốn 15 năm sau đã có con 16 tuổi thì... chỉ có 2 khả năng xảy ra thôi nhỉ. Tất nhiên ... nếu anh chắc chắn con anh đích thực là... con anh thì chỉ còn 1 khả năng duy nhất  :lol:

Mot thong tin thu vi nua la hinh nhu ku Bach from KHTN cung la nguoi Nam Dinh  :lol:  :D Vui that !!!

hình như có 1 cao thủ mathlink tên là Peter Schole gì đó nữa điểm cũng khá cao :D người Germany thì phải  :D Việt Nam năm nay dù chỉ 3 HCV (thua năm anh Hùng) nhưng vẫn đứng thứ 3 :D keke năm 4 huy chương vàng thì đứng thứ 4  :D Năm sau các bác cố dành 1 HCV để đứng thứ nhất nhá  :rotfl: Có ai biết nick mấy anh trong IMO ko chia sẽ em với  :wink:

A big HURRA to the VN team! After two disappointing years, VN is back!

Ăn mừng fì mà chỉ bác Tuân và anh Hùng thế nhỉ . Dân Việt Nam đâu hết roooooooooooooooooôì :first:

:lol:  Tren Vn express co phong van Tung va thay Hoa. Ai co hung thu' thi` vao` do dat vai` cau hoi cho em ay nhe.

 :blush:

Doc truyen sau thay vui vui

http://vnexpress.net/Vietnam/The-gioi/Cuoc-song-do-day/2007/07/3B9F8502/

[quote="N.T.TUAN"]Link day : http://vnexpress.net/Vietnam/Xa-hoi/2007/07/3B9F8B5B/

Dua be kia sinh cung nam, ngay, thang voi con gai minh, vay ma con gai minh lai ko duoc nhan hoc bong, chan qua! :([/quote]---> con gái nào vậy anh :roll:

[quote="N.T.TUAN"]à, con...rơi của anh ấy mà. :D[/quote]

Kieu nay` phai di kien bac ay voi Valentin thoai :D. Scandal the nay` ma ko phai tu` chuc moi la :D Em de nghi de con anh Tuan len lam Mod thay bo no' :D

Hay đấy!  :lol: Chúc mừng đội tuyển Vn!  :wink:

[quote="DuongChinh_K41SPHN"]CHAN QUA NAM NAY SP LAI LO HEN VOI IMO.MA EM HOI THAT ANH HUNG ĐANH GIA SP HAY TN CAO HON .ANH ĐUNG THIEN VI NHA[/quote]
Tớ đỗ chuyên cả TN và SP
Tớ có hỏi mọi người ý kiến nên học TN hay SP thì ý kiến mọi người nghiêng về TN hơn,mà có lẽ TN cũng cố phần mạnh hơn SP!
Nhưng chưa biết mèo nào cắn mỉu nào đâu,SP năm nay ko có ai thi IMO chắc là đợi đến 2009,năm tụi mình đi thi IMO,chắc là 6 HCV+giải đặc biệt cá nhân+giải nhất toàn đoàn đấy!
 :wink:

[quote="IamChan"]Mấy anh à cho em hỏi cái đề thi và solution của nó mình down ở đâu anh nhỉ ! chỉ cho em được không ạ ?[/quote]
Đề IMO 2007 bạn xem ở đây nè
http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=2007

Tất cả các năm ở đây http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16

Cám ơn anh T., quên chúc mừng đội tuyển VN mình ! Tự hào quá đê :P ! :blush:


geometry
incenter
parallelogram
perimeter
geometry unsolved

Given triangle ABC and its centroid G,incenter I,try to construct its orthocenter H with only a ruler without scales.

[color=blue][b]Basic straightedge construction (1):[/b] [/color]Given 2 points A, B, the midpoint M of the segment AB, and a point P, construct a line parallel to the line AB through the point P.

Draw the lines PA, PB. Draw an arbitrary line through the midpoint M intersecting the lines PA, PB at points C, D. Let the lines AD, BC meet at a point Q. The line PQ is parallel to the line AB.

[color=blue][b]Basic straightedge construction (2):[/b] [/color]Given a parallelogram ABCD, a line k and a point P, construct a line parallel to the line k through the point P.

The parallelogram diagonals AC, BD cut each other in half at a point M. Let the line k meet the lines DA, BC at points E, F respectively. Using the basic construction (1), draw parallels to the diagonal BD through the points E, F intersecting the lines BC, DA at points G, H, respectively. EGFH is another parallelogram and its diagonals EF, GH cut each other in half at a point N. Using the basic construction (1) again, draw a parallel to the line $k \equiv EF$ through the point P.

[color=blue][b]Problem:[/b][/color] Given the triangle centroid G, we can construct the midpoints A', B', C' of the triangle sides BC, CA, AB. The incenter I of the triangle $\triangle ABC$ is the Nagel point (the concurrency point of perimeter splitters) of the medial triangle $\triangle A'B'C',$ which is centrally similar to the original triangle $\triangle ABC$ with the similarity center G and similarity coefficient $-\frac 1 2.$ Thus A'I, B'I, C'I are the perimeter splitters of the medial triangle. Let the perimeter splitter A'I meet the midline B'C' at a point T'. The point T' splits the perimeter of the medial triangle $\triangle A'B'C'$ in half, $A'B' + B'T' = A'C' + C'T'.$ Hence, it is the tangency point of the medial triangle excircle against the vertex A' with the side B'C'. This tangency point is equidistant to the tangency point of the medial triangle incircle with the side B'C' from the midpoint $A'' \equiv AA' \cap B'C'$ of the side B'C'. Therefore, T' is also the tangency point of the incircle of the triangle $\triangle AC'B'$ with the side C'B'. Since the triangles $\triangle AC'B' \sim \triangle ABC$ are centrally similar with the similarity center A and similarity coefficient $\frac 1 2,$ the line AT' meets the line BC at the tangency point T of the incircle (I) of the triangle $\triangle ABC$ with the side BC.  Now we have a line IT perpendicular to the side BC. Using the parallelogram $A'CB'C'$ and the basic construction (2), draw a line through the vertex A parallel to the line IT. This is the A-altitude of the triangle $\triangle ABC.$ Another altitude can be constructed in a similar way and the 2 altitudes meet at the triangle orthocenter H.


modular arithmetic
number theory
prime factorization
number theory unsolved

Suppose that $ m$ is a positive integer that is not square-free. Show that there exist integers $ a_1$ and $ a_2$ such that $ a_1 \not\equiv a_2 \pmod m$, but $ a_1^k \equiv a_2^k \pmod m$ for all integers $ k > 1$.

Let $ a \equal{} \text{rad } m$ be the product of the distinct prime factors of $ m$ and let $ k$ be the largest power that occurs as the exponent of a prime in the prime factorization of $ m$.  Then let $ a_1 \equal{} a^{k \minus{} 1}, a_2 \equal{} 2a^{k \minus{} 1}$.

Cool thanks, but I noticed it would also work for $ a_1 \equal{} ia^{k \minus{} 1}$ and $ a_2 \equal{} ja^{k \minus{} 1}$ where $ i \neq j \neq 0$ and $ i,j$ are integers. Is this true (in case I missed something)?

Yes.  The point is that the given condition cannot be satisfied if $ a_1$ is invertible $ \bmod m$, so why not make it as un-invertible as possible?

[hide="Formal Proof"]

Write $ m$ as $ p_1^{a_1}p_2^{a_2}...p_n^{a_n}$ by the unique factorization theorem.
Let $ k$ be the highest power of $ a_i$.
Since $ m$ is not square free, $ k > 1$ (because $ m$ is divisible by a square).

Let $ a = p_1p_2...p_n$. Furthermore, let $ a_1 = ia^{k - 1}$ and $ a_2 = ja^{k - 1}$ for some integers $ i$ and $ j$ with properties $ i - j \neq a \neq 0 \neq \text{a multiple of}\;p_i$ where $ p_i$ is the product of the prime factors in $ m$ with power $ k$. Note it is possible for it to be only one $ p_i$ depending on $ m$.

Then $ a_1 - a_2 = a^{k - 1}(i - j)$ which does not divide $ m$ (unless $ i - j = a, 0$, or $ \text{a multiple of}\;p_i$) since it is missing a factor of $ p_i$.
Hence, $ a_1 - a_2 \not\equiv 0 \pmod m \Rightarrow a_1 \not\equiv a_2 \pmod m$.

Now consider $ a_1^k$ and $ a_2^k$.

$ a_1^k = i^ka^{k(k - 1)}$ and $ a_2^k = j^ka^{k(k - 1)}$

Then $ a_1^k - a_2^k = a^{k(k - 1)}(i^k - j^k)$, which is divisible by $ m$ if you recall that $ k > 1$ so $ k(k - 1) > k$, which implies $ m | a^{k(k - 1)}$ because ${ a^{k(k - 1)}} = p_1^{k(k - 1)}p_2^{k(k - 1)}...p_n^{k(k - 1)}$ contains all of the powers of the prime factors of $ m$ (recall we assumed the highest power of any prime factor in $ m$ is $ k$).

Therefore, $ a_1^k - a_2^k \equiv 0 \pmod m \Rightarrow a_1^k \equiv a_2^k \pmod m$.

Thus, we have found integers $ a_1$ and $ a_2$ that satisfy the problem's conditions.[/hide]


2008 Nationals is over.  Contestants are returning home.  AoPS Administrators, rrusczyk, DPatrick, and levans; and AoPS MATHCOUNTS Moderators, frost13 and rcv were present for some or all of the competition.

All national contestants deserve a hearty and sincere "congratulations" for making it to the event.  No contestants need to be second-guessed for being nervous or for making dumb mistakes or for anything else.

I am certain many people want to relate their experiences at the National MATHCOUNTS.  However, other moderators have already had to lock three threads which degenerated into noise and needless criticism of others.

This thread is one more chance.  Tell of your own personal experiences.  Give the facts of the competition.  Congratulate your opponents (and yourself).  But, don't be critical of other contestants.  [If you weren't in their shoes, you don't know what they were feeling or what was in their heads.]

A Webcast of the Countdown Round is available at http://www.mathcounts.org.  If you can't simultaneously see the video and the Power Point presentation of the questions, try using a different browser.  The first 30 minutes of the WebCast is introduction of the teams, announcement of the winning teams, and announcement of the Countdown qualifiers.  The last hour of the WebCast is the Countdown Round, followed by awarding of the individual winner's medals.  [MATHCOUNTS said the Webcast would be around for about a year.  It's OK to list the winners here, since the WebCast will eventually go away.  It's OK to list your own score, your own rank, or your own team's score and rank.  By MATHCOUNTS rules, it's also OK to publically announce the names of the top 25% of competitors (ranks 1 to 57) and the top 40% of teams (ranks 1 to 24).  [I expect MATHCOUNTS will post this information very soon.]  Let's not be too repetitive, please.  Also, do not disclose any association between a contestant's AoPS screen name and a contestant's real name.]

Postings which violate the above may be deleted.  If there are excessive postings that violate the above, this thread will also be locked.

Since you said  we can discuss ranks 1-57, does MathCounts have a list that they're releasing publically? I can't find it on the website (I believe they took the one from 2007 down a few months ago).

As you may know, MATHCOUNTS just redesigned their Web Site a few weeks ago.  Some of the older material is pretty hard to find.  The official 2007 National Competition results are still available at [url]http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=298&srcid=1274[/url]

The 2008 top 12 contestants and the top 10 teams are posted in a Press Release at [url]http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=1296&srcid=344[/url]

I presume MATHCOUNTS will be posting the remainder of the public portion of the results when their tiny staff returns to the offices this week.

When will the Masters Round winners be known?

*gasp* he won Masters too? :huh:

That's awesome!!!!! :lol:


vector
LaTeX
linear algebra
matrix
advanced fields
advanced fields theorems

No equations here, just a question.

Is there a board here that I could post questions about the math part of my Physics class? (Mostly just Vector stuff, but apparently I didn't retain much from my previous math classes.)

Or maybe a recommedation of a good physics forum? (Latex would be nice but not necessary.)

Two suggestions involving AoPS:

There is a Physics forum:

Math Forum » Arts & Sciences » Physics.

And then some vector questions might be appropriate for:

College Playground » Matrix - Linear Algebra » Matrix Algebra - Computations & Tutorials.


calculus
derivative
calculus computations

If $f\left(x\right)=x^{2}e^{x}$, find a formula for the $n$th derivative of $f$.


calculus
integration
trigonometry
logarithms
calculus computations

How about this integral?
http://www.mathlinks.ro/viewtopic.php?t=152138

The integral to  be computed is: $ \int_{0}^{1} \sqrt{1\minus{}x^2} \arctan x dx$.
It can be done by changing the variable $ x \rightarrow \sin x$ and the integral becomes:
$ \int_{0}^{1} \sqrt{1\minus{}x^2} \arctan x dx \equal{} \int_{0}^{\frac{\pi}{2}} \cos x \arctan(\sin x) dx \equal{} (\sin x \arctan(\sin x) \minus{} \frac{1}{2} \ln(1\plus{}\sin ^2 x))|_{0}^{\frac{\pi}{2}}\equal{}\frac{\pi}{4}\minus{}\ln \sqrt{2}$.

if you put $ x \equal{} \sin u$, then $ dx \equal{} \cos u du$

The transformation is wrong as pointed out by Attila the Hun...

[quote]The transformation is wrong as pointed out by dina013...[/quote]
 :blush: yes, you're right... I was completely unattentive... sorry.


algebra
polynomial
superior algebra
superior algebra unsolved

if E is a splitting field of $ f(x) \in F[x]$ over F and g(x) is irreducible polynomial in F[x], show that g(x) splits completely in E[x] provided that g has a root in E.

It is well-known that the splitting field of $ f$ is the smallest normal extension containing a root of $ f$. $ E$ is a normal extension of $ F$, and it contains a root of $ g$. Thus it contains the splitting field of $ g$, so $ g$ splits in $ E$.


[quote="NoSoupForYou"]Try The Gimp. Dnas is obsessed with it. Ask him for more details.[/quote]
lol. That's an interesting way to put it. :)

Anyway, I replied to your email.

[quote="Dnas"][quote="NoSoupForYou"]Try The Gimp. Dnas is obsessed with it. Ask him for more details.[/quote]
lol. That's an interesting way to put it. :)[/quote]

:D Well, it's true.

[quote="G-UNIT"]Hi, my current photoshop that came with my computer is no longer functioning.  And I am in need of a a new photoshop that meets the following criteria:

-it must be FREE

-although long-term programs are preferred, short-terms programs or trial-programs as they are called, will work jut as well

-your sure that this program is free of ANY viruses, spyware, spam or any other harmful companions (basically meaning you've tried it)

-and of course it is descently adequate, and in your mind works well

  Along with this, a brief decription would be nice.  Any suggestions, will be greatly appreciated!

  Thanks a million![/quote]

You could download the Adobe Photoshop trial off of the Adobe main page, or you could use some other programs, such as Macromedia Fireworks or Paintshop Pro as a subsitute for Photoshop.

Get the trial for now. :) There really aren't any free programs with PS's range of capabilities that I personally know of.

Um...does Paint work? Lol, if not...give a somewhat specific list of features you'd like.

True, there aren't free programs with the full range, but, for a hobbyist who is willing to learn a new interface, the GIMP is far more than enough. And due to it's open source nature, it's developing at a ridiculously fast rate.

The only way to get Photoshop for free permanently is by pirating a CD, which is illegal. Just use GIMP. It's 100 times harder but it works very well. Think of it this way: it takes one click to do something in photoshop, but in GIMP the same thing takes around 8-10 clicks, so if you download GIMP, you better be a hardcore designer!

btw, just google "gimpshop", it's a reformatted Photoshop-like version of GIMP.

I don't really like GimpShop.  It's nice that all of of Gimp's windows are in one big window, but that's basically it.  Besides, GimpShop always crashes when I open a new document  :o  :mad: .  That's why I uninstalled it.

you could always go old school and use paint.  Paint takes a lot of effort to be able to do stuff in it but if you just play around with it you can do some pretty amazing stuff...

Try Paintshop. I think that it is free, but I do not know because I use the pro edition that is not free. It is real good; it is called the "free photoshop"

paint..old school? I can make some pretty good pictures on it( modern art mostly..) unfortunately it's on a computer that doesn't have internet so I can't show it off. is there a free animation program though?

IMO nothing competes Photoshop.
it is very unique.

x and y you are right *goggles over new paintshop*

you are right i just realised how crude paint is.. :blush:

is your avatar one of your creations funcia? If yes, with wich program? nice!

it's not mine...i found it on some forums i'll pm you the information...

ok thanks! I saw this bag today (not with the Lord V of course!)

Paint.NET is good for gentle photoediting/playing around. Free to download.

It depends, what do you need it for?

I use Paint for sprites, Bryce for environment renders, and AoI and Blender for general 3d modelling. I have access to a library of programs that I don't use, because it would involve climbing stairs.

if you're willing to learn a new language, try pov-ray, but if it's just image manipulation, i use paint.net


group theory
abstract algebra
function
complex numbers
number theory
prime numbers
superior algebra

Hello,

(first: my english is worse, and escpecially I dont know all english math vocabularys, is there in the internet a dictionary german-english with math words? I couldnt find something like that, so forgive me, I hope you'll understand me.)

My question: Its easy to show, that every infinite set has an real infinite bijektive subset (subtracting one element does not change the cardinality). I asked me, if there was an analogon for groups, and so the question is:

Has every infinite group an real isomorph subgroup? For the examples, I could find for infinite groups, it was easy to find isomorph subgroups, but a generals proof was not possible. Is that declaration true? And if, how to proof? The difficulty for me is, that I do not know, how to generate a subgroup for a general infinite group. For sets thats easy, just subtract a finite number of elements - but for groups? Thank you, and forgive me the language again.

Regards T.

It's actually false. The set of all complex numbers of the form
\[ \text{exp} \left( \frac{2\pi i a}{2^n}\right)\]
where $ a,n$ are integers, $ n\geq 0$, and $ \text{exp} (x)$ is the function $ e^x$, is a group under multiplication. But its only proper subgroups are finite groups of the $ 2^k$ th roots of 1 for some positive integer $ k$.

[quote="Tauchender"]Ok, thank you. So two new questions (in my lecture we do not analyze infinite groups as yet): Exists for the first question an not trivial answer, I mean a infinite group, which has proper infinite subgroups, but no isomorph ones.

And secondly, more general: Has every infinite group G subgroups (not equal to G or {1})? Seems so to me, cause: Idea: If not, than every cyclic subgroup <g> with elements g of G had the order 2. That would mean <1,g> is isomorph to Z_2, and so a not trivial subgroup of G. Ist that idea correct?[/quote]

Hello.

As for the first question, the direct sum of the groups $ \mathbb{Z}/p\mathbb{Z}$, where $ p$ ranges over the prime numbers, has (clearly) proper infinite subgroups but  none of its proper subgroups is isomorphic to it : a subgroup isomorphic to it must have element(s) of order $ p$ for each prime $ p$ and so it must contain each "term" of the direct sum, and so be identical with the direct sum.

Second question : let $ G$ an infinite group. Choose an element $ x$ other than 1. If $ x$ is of infinite order, the subgroup generated by $ x^{2}$ doesn't contain $ x$ (so it is not $ G$) and is not identical with {1} because $ x^{2} \not\equal{} 1$ (since $ x$ is of infinite order).
If now $ x$ is of finite order, the subgroup generated by $ x$ is different from {1} and from $ G$.

Friendly,
P.


trigonometry
geometry
trapezoid
ratio
probability
logarithms
similar triangles

The problems are here: http://www.math.sc.edu/contest/1995/exam/PDF/exam1995.pdf

I was not able to get #20 and #29. For #20 is there a better way than to trig it to death (which is certainly possible since it's a trapezoid and sins are supplementary, and you can express the area of the trapezoid using two different pairs of triangles using sin area, and using sin double angle formula in there too...but that gets messy...).

[hide="1"]$3*(2*1)=3*(5)=9+5=14$[/hide]

[hide="2"]$2^{16}-16=(2^{8}-4)(2^{8}+4)=(2^{4}-2)(2^{4}+2)(2^{8}+4)=14 \cdot 18 \cdot 260$. $13$ is the largest prime factor.[/hide]

[hide="3"] Notice that $f(1+f(x))=4f(1)=f(5)+4(4)=16$.[/hide]

[hide="4"]Let $AC=AB=5$. Thus $MC=CP=NB=BP=2$. The answer is thus $14$.[/hide]

[hide="5"] Call the respective equations $(1), (2), (3)$. Notice that $(1)-(2)+(3)=18$ which is what we wanted. [/hide]

[hide="6"] $\sin{x}+\cos{x}=1.2 \Longrightarrow \sin^{2}{x}+\cos^{2}{x}+2\sin{x}\cos{x}=1.44 \Longrightarrow \sin{2x}=.44$.
[/hide]

[hide="7"]The answer is the sum of the roots of $2x^{2}-6x+1$ which is $3$.[/hide]

[hide="8"] Let $x$ be the number that was omitted and $n$ be the number of terms. We have that $\frac{60n-x}{n-1}=70 \Longrightarrow 70=10n+x$. The largest $n$ we can have is $6$.[/hide]

[hide="9"]The sum of each row of the $5 \times 5$ square is $\frac{1+2+3+...+25}{5}=\frac{26(25)}{10}=65$. Thus, $A=22$.[/hide]

[hide="10"]Since we just want the units digit, we look at $7^{89}$. Since $89 \equiv 1 \mod{4}$, the answer is $7^{1}=7$.[/hide]

[hide="11"]$x^{2}-x+1=\frac{x^{3}+1}{x+1}=0 \Longrightarrow x^{3}=-1$.[/hide]

[hide="12"]Let the two poles be $AB$ and $CD$, where $BC$ is on the "ground." Let $P$ be the intersection of the chords and let $Q$ be on the ground (where $PQ=24$). $\triangle{PCQ}\sim \triangle{ACB}$ and $\triangle{PBQ}\sim \triangle{BDC}$. Let $BC=90$. Thus, using similar triangles, $BQ=32$. Thus $32/24=90/DC$ and $DC=60$.[/hide]

[hide="13"]Trial and error gives us the answer of $1.0045$.[/hide]

[hide="14"]Notice that $a_{0}+a_{1}+a_{2}+...+a_{21}=2187$ (after letting $x=1$). And $a_{0}+a_{1}+a_{2}=\binom{7}{0}+\binom{7}{1}+\binom{7}{2}=29$. Thus the answer is $2158$.[/hide]

[hide="15"]Using demoivres, the answer is $\sin{180}+i^{60}\cos{180}=-1$. $i$ is affected since we're not in the form $\cos{x}+i\sin{x}$.
[/hide]
[hide="16"]The answer is a geometric series with first term $1$ and common ratio $\sqrt{2}/2$. The answer is thus $2+\sqrt{2}$.[/hide]

[hide="17"]The probability that Atlanta wins the series is the probability that Cleveland does not win the series. The probability that Cleveland wins the series is $1/4$. Thus the probability that Atlanta wins is $3/4$.[/hide]

[hide="18"] We must have $x679y$ divisible by $72$ and thus by $9$ and $8$. Thus $79y$ must be divisible by $8$ and thus $y=2$. Then we must have $x=3$.
[/hide]
[hide="19"]Trial and error shows that $x=4$ works in which there is there are two gallons of wine in each jug.[/hide]

[hide="21"]Notice that $\log{5}=\log{(10/2)}=\log{10}-\log{2}=1-.301=.699$. The number of digits of $5^{80}$ therefore is $[80 \cdot .699+1]=56$.[/hide]

[hide="22"]Notice that the largest number of $S_{k}$ is $1+2+3+4+...+k$. It follows that the largest number of $S_{100}=\frac{101(100)}{2}=5050$.[/hide]

[hide="23"]We can do this problem by proving a counterexample for each set (looking at the elements $\mod{3}$). Let $S_{k}$ be a set with $k$ elements.

$S_{3}=1,0,1$
$S_{4}=1,0,0,1$

We get to $S_{5}$ and see that there are no counterexamples. Thus a set of $5$ is the minimum.[/hide]

[hide="24"]The number of even divisors is the total number of divisors minus the number of odd divisors. Which is $(3)(5)(2)-(3)(2)=24$.[/hide]

[hide="25"]The answer is $\binom{7}{2}(4!-4 \cdot 3!+6 \cdot 2!-4 \cdot 1!-1)=924$.[/hide]

[hide="26"]The fastest way I think is just to fill out the square. $a=b=1$ and the answer is $2$.[/hide]

[hide="27"]We must find the smallest integer $n$ such that $(1-.20)^{n}<.4 \Longrightarrow n=5$.[/hide]

[hide="28"]A picture can be drawn. The line segments are: $P_{1}P_{4}, P_{2}P_{4}, P_{2}P_{3}, P_{3}P_{4}, P_{4}P_{5}, P_{3}P_{5}$. There are two lines going through $P_{5}$.[/hide]

[hide="30"]We have:

$(1)$ $x+y+z+t+w=1$
$(2)$ $x+y+2z+2t+2w=1$
$(3)$ $x+2y+3z+4t+5w=2$
$(4)$ $x+y+3z+2t+w=3$

Thus:

$(4)-(1)=2z+t=2$
$(4)-(2)=z-w=2$
$(2)-(1)=z+t+w=0$

From these, we get that $z=2, w=0, t=-2$.

After plugging these into $(1)$ and $(3)$, we get:

$x+y=1$
$x+2y=4$

From these, we have $y=3$ and $x=-2$. Thus, $(a)$ is the answer.[/hide]

[hide="29"]
We note that the expressions given can be rewritten as \begin{eqnarray*}\text{I.}&& \frac{x^{3}-1}{x-1}\\ \text{II.}&& \frac{x^{3}+1}{x+1}\\ \text{III.}&& \frac{x^{5}-1}{x-1}\\ \text{IV.}&& \frac{x^{5}+1}{x+1}\end{eqnarray*} We see that [b](I)[/b] is true if $x^{3}=1$ and $x\ne 1$.  So for [b](I)[/b] to divide our original expression, plugging $x^{3}=1$ into $x^{81}-x^{10}-x+1$ would have to reduce to 0.  So plugging in we get \begin{eqnarray*}x^{81}-x^{10}-x+1 &=& (x^{3})^{27}-x\cdot(x^{3})^{3}-x+1 \\ &=& 1-x-x+1\\ &=&-2x+2 \end{eqnarray*} Thus [b](I)[/b] does not work.

Plugging in for [b](II), (III),[/b] and [b](IV)[/b] all reduce to 0.  Thus, (E) is the answer.[/hide]

Please stop using WLOG.  Unless you've changed the meaning of it to 'With Loss of Generality' you aren't using the term correctly.

You should try another method than trial and error on #19.

For #24, try changing the wording of the question to "how many positive multiples of 4 divide $720=2^{4}\cdot 3^{2}\cdot 5$?"

[quote="joml88"]Please stop using WLOG.  Unless you've changed the meaning of it to 'With Loss of Generality' you aren't using the term correctly.
[/quote]

 :rotfl:

and also, here's another way to do #24:

We know that the number of divisors of $2^{4}\cdot 3^{2}\cdot 5$ is $(4+1)(2+1)(1+1)=30$. However, for the even factors, we can't have $2^{0}$, therefore there's one less case for the first factor. In other words, the number of even divisors would be $(4+1-1)(2+1)(1+1)=24$.

Now for your question, we want it so that we restrict the case by taking out $2^{0}$ and $2^{1}$. So the number of divisors now would be $(4+1-1-1)(2+1)(1+1)=18$.


Hi,I am a new member of this famous forum,so I want to seek acquaintance with you,all members here,I think we will be good friends,soon.

My name is Chu Yến Ly,Username Trịnh Lưu Ly,I'm from Việt Nam,where the IMO 2007 was held.(I am sorry,my English is not good,I will try my best... :blush: ).I live in Hà Nội city,in glass 10TN4,Việt Đức school.I love math,and I think you also love math,hi hi...

Come here and post some information about you for me,thanks very much! :lol:

[quote="hunter34"]Ok, I am hunter34. I am from Dallas, Texas. I am currently taking Calculus A, Geometry, English, Physics and Computer Programming. I am Japanese.

If you want more go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15747&start=340]here[/url][/quote]

Thank you,thank you very much,It's my pleasure to meet you!
How old are you?What grade are you in?
I am learning Japanese and I think It's very interesting,but..very difficult! :lol:

I didn't skip a grade either, despite scoring 96 and 97 on my past two math finals. I could easily have taken calculus during high school if my stupid public school wasn't so ghetto and under-funded  :(

[quote="hunter34"]I am 13 years old and in 8th grade.[/quote]

Uh,you are a good boy!(Your English is very good,but my English is...so bad :blush: )
hi hi
In my country,A person of 13 is in 7th grade.Is the knowledge difficult?

[quote="hunter34"]No, I just finished 7th and now starting 8th.[/quote]

Is Calculus a language?I 've never heard about it? :P

Do you know him?I like him very much.His film is very interesting,i think! :lol:

[quote="hunter34"]http://en.wikipedia.org/wiki/Calculus[/quote]

Oh I know!Thank you.I know what is "calculus".It is a new word in English.I think I must improve my English soon. :blush:
Now I must out,See tou later,my friends! :lol:

[quote="hunter34"]Just saying I never skipped a grade either, but my school was "Oh look a prodigy, better challenge him"[/quote]

Seems like a fairly rare instance  :dry:

Hi Trinh! Welcome to AoPS  :)

------------

[b]@ above:[/b]

I know, seriously, public schools are supposed to be like:

[quote="Public School"]Oh look, a prodigy!

Let's give the kid more work to keep 'em shut up, and if s/he's still done before everyone else, then make 'em help everyone else.[/quote]

Hi, I'm from Upstate NY, Taking Calculus and trying to get through AoPS volume one s quick as possible,

12 yrs old, 7th grade. Skipped and got held back(I like math competition!)

I am back now.Wow,there are many friends here.Hic,I am very happy.Please introduce more and more

( If there are mistakes in my sentences,I hope you will sympathize with me.In a long time,I didn't use English,so now It's getting and getting worse :maybe:  :oops: )

It's nice to know you  :)

I'm Vietnamese girl too and I'm from 11CT, LHP, HCM city

I bet if the OP was a guy there would have been like two posts >____>

[quote="The Zuton Force"]I know, seriously, public schools are supposed to be like:

[quote="Public School"]Oh look, a prodigy!

Let's give the kid more work to keep 'em shut up, and if s/he's still done before everyone else, then make 'em help everyone else.[/quote][/quote]Agreed.  My public school forced me to sit through 3rd graders learning how to multiply 4 by 7 even after taking the morning off to do Trig.  Fortunately, I dropped out of that school.

[quote="Nerd_of_the_Ages"]Wow, 8th grade and taking Calculus. Lol, im like the only person here that hasnt skipped a grade. I just started my first day of 7th grade today, am 11 years old (soon to become 12), and am taking prealgebra (WAYYY too easy. do we honestly need and entire chapter teaching you how to add and subtract?)[/quote]

Ewww.... I had to take Prealgebra in 7th grade too... I learned exactly NOTHING from the class. My 6th grade teacher informed me that I could take a test to skip, but that was like the day before school ended, so I never got to do it. You know what? Compared to everyone else here, I seem to suck at math, but I got to state last year...  :huh:

[quote="aanjohri"]Hi! I just started 6th grade and I live Rockville, Maryland although I am originally from Northern India.

I was too young to go to Mathcounts or Math Team, but I am going to start this year. Currrently, my favorite courses are computer science and Algebra.[/quote]

You are a Indian!So great!!Hi,I like Indian,they're so hard-working,And I think you are too :blush: It's nice to meet you,too.Help me learning math,ok? :lol:

I'm an idian too, but not hard-working at all :wink:
I can't believe I'm only taking geometry...I hate school math

But that not say you are bad at math,ok?  Try your best,this forum is very good for everyone,so don't be nervous! :lol:

[quote="shaggy75"]hi trinh and welcome to aops! i am from california. i am 2 lazy 2 write down anything else[/quote]

Thank you very much! :lol: .You are from California?Oh!A big city.I know your city because I travelled to it last year.What a beautiful city!!It makes me feel that the world is very big!

[quote="Trịnh Lưu Ly"]It makes me feel that the world is very big![/quote]

When you get a map of your village/town/city, scale it to your country, and then scale it to the world, you'll really get that sensation.

If you are learning pre-algebra through algebra, go to:
http://www.artofproblemsolving.com/Forum/index.php?f=343
http://www.artofproblemsolving.com/Forum/index.php?f=300
http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=11

If you are learning Algebra II through precalculus, go to:
http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php
http://www.artofproblemsolving.com/Forum/index.php?f=132
http://www.artofproblemsolving.com/Forum/index.php?f=149

If those are too easy for you, try going to the intermediate forum, the pre-olympiad forum, and the olympiad forums. If you want to be challenged at the college level, go here:
http://www.artofproblemsolving.com/Forum/index.php?f=218

If you want to work a few tests, go here:
http://www.artofproblemsolving.com/Forum/resources.php

[quote="hunter34"]Ok, I am hunter34. I am from Dallas, Texas. I am currently taking Calculus A, Geometry, English, Physics and Computer Programming. I am Japanese.

If you want more go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15747&start=340]here[/url][/quote]

wow...  u make me feel so bad. :(

im in 9th grade and still in Algebra II

welcome to the forum!

[quote="frostmourne"]
wow... u make me feel so bad.

im in 9th grade and still in Algebra II

welcome to the forum!
[/quote]

I sorree.

[quote="Nerd_of_the_Ages"]Wow, 8th grade and taking Calculus. Lol, im like the only person here that hasnt skipped a grade. I just started my first day of 7th grade today, am 11 years old (soon to become 12), and am taking prealgebra (WAYYY too easy. do we honestly need and entire chapter teaching you how to add and subtract?)[/quote]
no, you're not.
I even wish i could take geometry in 8th grade. (my school only goes up to algebra one)

[quote="Lawrence Wu"]my name is my username. i'm in 7th grade and taking algebra, and it's STILL TOO EASY, seeing as it's the only thing I have a 100 in.[/quote]

Whee... exact same here, except I don't have a 100. I'm good at math, but test taking is a different story (I make SO many careless mistakes... forget to put negative sign, add instead of subtract, stuff like that  :mad: )

Forgot to say, I thought a lot of you were in highschool, but turns out you're in middle school also (I thought aanjohri was a sophomore or something).

[quote="mz94"]i am in the seventh grade, i have the go to 8th grade honnors, i am there and it is the easierest. i have a 100 perceent so far[/quote]

lol good job.
*copies*
i am in the tenth grade, i have the go to 11th grade honnors, i am there and it is the easierest. i have a 95 perceent so far

i am in the seventh grade, i have the go to 12th grade honnors, i am there and it is the easierest. i have a 97 perceent so far

lol, i think i have like an 85 right now. theres a difference between math and test taking though  :ninja:  Really easy though. The teacher keeps on going like "this is review, so you should know this stuff". If it's review, why not go on to harder topics?

i am in the seventh grade, i have the go to more than one grade honnors, i am there and it is the easierest. i have a 95 perceent so far


geometry
geometric transformation
rotation
rectangle
analytic geometry
calculus
integration

We're doing solids of revolution in class right now, and all of the problems seem to involve rotating the curve around either the x or y axis or a line parallel to one of them. I've been trying to generalize that to rotation about a line y = mx.

Let's say the curve in question is y = x^2, and we want to rotate about y = 2x.

The basic idea I've been working with is breaking up the y = 2x axis into intervals and finding the heights of the approximating rectangles by using the formula for the distance from a point to a line. I've also considered trying to make y = x^2 into something of the form y' = f(x'), where (x', y') are the coordinates of P(x,y) when the x-axis is rotated onto y = 2x.

So far I've just succeeded in confusing myself with a profusion of variables and transformations. There's a better way, right? (or, relatedly, would I even be able to do something along the lines of the method I've been attempting?)

(I apologize if this is the wrong forum...I get the impression that calculus stuff doesn't belong in the HS forums)

[url=http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem]Pappus's centroid theorem[/url] helps a lot here. For one thing, you can find the coordinates of centroid in the usual way, and then calculate its distance to the axis. The distance from $ (x_0,y_0)$ to the line $ y\equal{}mx$ is $ \frac{|y_0\minus{}mx_0|}{\sqrt{1\plus{}m^2}}$ if I remember correctly.

Pappus's theorem can also be used to derive a general formula for this kind of volumes. Suppose that the region ($ g(x)\le  y\le f(x)$, $ a\le x\le b$) lies to the right of the line $ y\equal{}mx$. A thin vertical slice with position $ x$ and thickness $ dx$ has area $ (f(x)\minus{}g(x))dx$, and its centroid is at $ (x,(f(x)\plus{}g(x))/2)$. Therefore, the volume obtained by revolving this slice is $ (f(x)\minus{}g(x))\frac{mx\minus{}(f(x)\plus{}g(x))/2}{\sqrt{1\plus{}m^2}}dx$. The total volume is $ \frac{1}{2\sqrt{1\plus{}m^2}}\int_a^b (f(x)\minus{}g(x))(2mx\minus{}f(x)\minus{}g(x))dx$.

[url=http://www.mathlinks.ro/viewtopic.php?t=109728&search_id=2050951989]You can see the solution here.[/url]


At Hamburger Heaven, it costs $ \$4.25$ for a burger and a drink, $ \$4.00$ for a burger and fries, and $ \$7.00$ for two burgers and a drink. How many dollars would it cost for one burger? Express your answer to the nearest cent.

We don't need the burger and fries info. We have the two equations:

$ b\plus{}d\equal{}4.25$
$ 2b\plus{}d\equal{}7$

Now we can subtract the first equation from the second and get:

$ \boxed{b\equal{}2.75}$


inequalities
function
inequalities unsolved

Hi! As I know Jensen's inequality for convex functions states that:

$ \frac{f(x) \plus{} f(y) \plus{} f(z)}{3} \geq f(\frac{x\plus{}y\plus{}z}{3})$

for all convex functions...

And Nesbitt's inequality states that:

$ \frac{a}{b\plus{}c} \plus{} \frac{b}{c\plus{}a} \plus{} \frac{c}{a\plus{}b} \geq \frac{3}{2}$

Then my question is how to apply Jensen's Inequality to solve Nesbitt's inequality....

Thanks a lot!  :lol:

[quote="shaaam"]Hi! As I know Jensen's inequality for convex functions states that:

$ \frac {f(x) \plus{} f(y) \plus{} f(z)}{3} \geq f(\frac {x \plus{} y \plus{} z}{3})$

for all convex functions...

And Nesbitt's inequality states that:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3}{2}$

Then my question is how to apply Jensen's Inequality to solve Nesbitt's inequality....

Thanks a lot!  :lol:[/quote]
Of cause!
Let $ a \plus{} b \plus{} c \equal{} 1.$ Hence, we need to prove that $ \sum_{cyc}f(a)\geq\frac {3}{2},$ where $ f(x) \equal{} \frac {x}{1 \minus{} x},$
which is a convex function on $ (0,1).$

[quote="arqady"][quote="shaaam"]Hi! As I know Jensen's inequality for convex functions states that:

$ \frac {f(x) \plus{} f(y) \plus{} f(z)}{3} \geq f(\frac {x \plus{} y \plus{} z}{3})$

for all convex functions...

And Nesbitt's inequality states that:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3}{2}$

Then my question is how to apply Jensen's Inequality to solve Nesbitt's inequality....

Thanks a lot!  :lol:[/quote]
Of cause!
Let $ a \plus{} b \plus{} c \equal{} 1.$ Hence, we need to prove that $ \sum_{cyc}f(a)\geq\frac {3}{2},$ where $ f(x) \equal{} \frac {x}{1 \minus{} x},$
which is convex function on $ (0,1).$[/quote]

Arqady, can you please tell me, why do you suppose a + b + c=1? i can choose arbitrary positive numbers that may not have 1 as a sum. Thanks!  :)

Here is a way of using Jensen:

First normalize to $ a \plus{} b \plus{} c \equal{} 1$. Then note that $ f(x) \equal{} \frac {x}{1 \minus{} x}$ implies $ f''(x) \equal{} \frac {2}{(1 \minus{} x)^3} > 0$ for $ x\in (0,1)$. Thus $ f$ is convex in $ (0,1)$ and from Jensen's inequality we conclude that
\[ \frac {f(a) \plus{} f(b) \plus{} f(c)}{3}\ge f\left(\frac {a \plus{} b \plus{} c}{3}\right) \equal{} f\left(\frac 13\right) \equal{} \frac 12
\]
Oops new replies already posted! :oops:

[quote="dgreenb801"]Weighted Jensen's is more powerful than normal Jensen's. It says
$ xf(m) \plus{} yf(n) \plus{} zf(p) \ge (x \plus{} y \plus{} z) \cdot f(\frac {xm \plus{} yn \plus{} zp}{x \plus{} y \plus{} z})$
So for the function you gave, we have
$ a \cdot f(b \plus{} c) \plus{} b \cdot f(a \plus{} c) \plus{} c \cdot f(a \plus{} b)$, where $ f(x) \equal{} 1/x$.
Using weighted Jensen's, this is $ \ge (a \plus{} b \plus{} c) \cdot f(\frac {2(ab \plus{} bc \plus{} ac)}{a \plus{} b \plus{} c}) \equal{} \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ac)}$
So the problem simplifies down to $ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ac)$.
This is clearly true, as it is equivalent to $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ac$, or $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \ge 0$, which is true because all squares are positive.[/quote]

Thank you for taking time and writing this..  :)

[quote="dgreenb801"]We have proved it for a+b+c=1. But if the inequality is true for a,b, and c, then it is true for an, bn, and cn (you can see this by plugging them in). But an+bn+cn=(a+b+c)n=n, so you can let them add up to any n.[/quote]

So you mean , plugging $ a, b ,c$ and plugging $ an, bn, cn$ will give the same result?

Yes. For example, if you plug in an, bn, and cn into $ \frac {a}{b \plus{} c}$ you get $ \frac {an}{bn \plus{} cn} \equal{} \frac {a}{b \plus{} c}$, which is the same fraction. For example, let's say you had the numbers 2,3, and 6. You could instead use the numbers 2/11, 3/11, and 6/11, and when they are all multiplied by 11, it must also be true. That is why they only had to prove it if a+b+c=1.


number theory
greatest common divisor
algebra
polynomial
number theory solved

this one is going to be difficult
prove that the sum is an integer
(1/4i)  \sum (i^k) tan (k \pi /4n) .
sum ranges over all k from 1 to 4n satisfying gcd (k,n)= 1
my solution spans two pages.i need sumthing shorter
cheers {and ya ,i is good ol  \sqrt (-1) }

Hey, did ne1 put any thought into this? I think I can show this for an even n, but I haven't thought about odd n's up until now (I doubt it's a very nice solution):

Since n is even, we have (k,n)=1 iff (k,4n)=1.

Let z=cis(Pi/4n). It's well known that cos (kPi/4n)=(z^2k+1)/2z and sin(kPi/4n)=(z^2k-1)/2iz. After all the calculations we get our sum (S) =   (1/4)*\sum i^k*(1-z^2k)/(1+z^2k). If we write (1-z^2k)/(1+z^2k) as -1+2/(1+z^2k) we get S= 1/2  \sum i^k/(1+z^2k), because  it's easy to see that \sum i^k = 0.

Now let a be cis(2Pi/4n) = z^2. It's obvious that a^2n=-1.
S= (1/2) \sum i^k/(1+a^k), the sum being taken for 1<=k<4n, (k,n)=1=(k,2n)=(k,4n). We now group the terms of this sum 2 by 2, the k'th term with the 2n+k'th term, so we find
S = 1/2* \sum (i^k/(1+a^k) + i^(2n+k)/(1-a^k)), for k<2n, (k,n)=1. We calculate this, taking into consideration the fact that 4|2n, so i^(2n+k)=i^k, so we find
S = \sum i^k/(1-b^k), where k<2n, (k,n)=1, and b=a^2=cis(2Pi/2n). This means that b^2n=1, b^n=-1 and b^(n/2)=i. It shouldn't be too hard now since b^k covers every root of the cyclotomic polynomial of order 2n, which has integer coefficients, and i^k can be written as (b^k)^(n/2). I have the rest (the complete soln), and the rest isn't even that long, but this post is already too long.

How about the case n odd?


abstract algebra
group theory
inequalities
combinatorics unsolved
combinatorics

Let $ G$ be a finite abelian group,and $ A,B$ are two subsets of $ G$,such that $ A\cap\{ \minus{} B\} \equal{} \{0\}$,prove:
\[ |A \plus{} B| \ge |A| \plus{} |B| \plus{} 1\]

[Moderator edit: This is wrong; see below for a correction.]

[hide="A thought"]Would the problem simple imply that we have to prove that $ \exists a\in A,b\in B$ so that $ a+b\notin A\cup B$. This is true if $ A\ne \{0\}$ and $ B\ne \{0\}$.[/hide]

- Note that if $ |A| \equal{} 1$ or $ |B| \equal{} 1$, the result doesn't hold. The previous counterexample is just a technicality, I know... However, the claim is also false if we take $ G \equal{} (\mathbb{Z}_{6}, \plus{} )$, $ A \equal{} \{0,3,4\}$, and $ B \equal{} \{0,1\}$. Are you sure that you didn't mean [i]two subgroups[/i] instead of [i]two subsets[/i]?

- It is pretty easy to give a counterexample where we observe the worst imaginable scenario... A case where the inequality that you give allows us to ascertain the existence of a group $ G$ and subsets $ A$ and $ B$ such that $ A \plus{} B$ has a cardinality greater than that of $ G$. Go figure! [hide]In order to corroborate this asseveration of mine, just take $ G \equal{} (\mathbb{Z}_{4}, \plus{} )$, $ A \equal{} \{0,2\}$ and $ B \equal{} \{0,3\}$.[/hide]

I am sorry....

The inequality must be:
\[ |A\plus{}B|\ge |A|\plus{}|B|\minus{}1\]

I can't edit my post now,can somebody help me.Sorry again.

That's easy... Obviously, the zero of $ G$ belongs to both $ A$ and $ B$. Therefore, $ A \subseteq A \plus{} B$ and the same is true for $ B$. Since $ A \cap [B\setminus\{0\}] \equal{} \emptyset$, and $ A \plus{} B \supseteq A \cup [B\setminus\{0\}]$, we conclude that $ |A \plus{} B| \geq |A \cup [B\setminus\{0\}]| \equal{} |A| \plus{} |B\setminus\{0\}| \equal{} |A| \plus{} |B| \minus{} 1$.

Thank you for your reply.But Please notice that the condition is $ A\cap \{\minus{}B\}\equal{}\{0\}$,not $ A\cap B\equal{}\{0\}$

where $ \{\minus{}B\}\equal{}\{\minus{}x|x\in B\}$

[Moderator edit: discussion continued at http://www.mathlinks.ro/viewtopic.php?t=308539 ]