Let be $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$ . Show that :

$ x^5\plus{}(y\plus{}z)((y\plus{}z)^4\plus{}5x(2(x^2\plus{}y^2\plus{}z^2)\plus{}yz))\ge 1$

[hide="Solution"]We expand the LHS of
$ \dfrac{1}{2}[(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2] \ge 0$
to get
$ x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx \ge 0$.
Adding
$ (x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx \equal{} 1$
we get
$ 2x^2 \plus{} 2y^2 \plus{} 2z^2 \plus{} xy \plus{} yz \plus{} zx \ge 1$,
and rearranging terms we get
$ 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz \ge 1 \minus{} x(y \plus{} z)$.
Substituting $ 1 \minus{} x \equal{} y \plus{} z$ we get
$ 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz \ge 1 \minus{} x(1 \minus{} x) \equal{} 1 \minus{} x \plus{} x^2$,
and multiplying by $ 5x$ we get
$ 5x(2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz) \ge 5x \minus{} 5x^2 \plus{} 5x^3$.
Adding
$ (1 \minus{} x)^4 \equal{} 1 \minus{} 4x \plus{} 6x^2 \minus{} 4x^3 \plus{} x^4$
we get
$ (1 \minus{} x)^4 \plus{} 5x(2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz) \ge 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4$
which can be rewritten as
$ (y \plus{} z)^4 \plus{} 5x(2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz) \ge \dfrac{1 \minus{} x^5}{1 \minus{} x}$
and then as
$ (y \plus{} z)^4 \plus{} 5x(2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz) \ge \dfrac{1 \minus{} x^5}{y \plus{} z}$
which implies
$ (y \plus{} z)((y \plus{} z)^4 \plus{} 5x(2(x^2 \plus{} y^2 \plus{} z^2) \plus{} yz)) \ge 1 \minus{} x^5$
and
$ x^{5} \plus{} (y \plus{} z)((y \plus{} z)^{4} \plus{} 5x(2(x^{2} \plus{} y^{2} \plus{} z^{2}) \plus{} yz))\ge 1$
as desired.[/hide]


A paperboy delivers newpapers to 10 houses along Main Street. Wishing to save effort,
he doesn't always deliver to every house, but to avoid being fired he never misses three
consecutive houses. Compute the number of ways the paperboy could deliver papers in this
manner.

[hide]
We use a recursion.  Let $ D_n$ be the number of legal delivery sequences for $ n$ houses.  If a sequence ends with a delivery, we simply append one to $ D_{n\minus{}1}$.  If it ends in $ 1$ nondelivery, we append a nondelivery and a delivery to $ D_{n\minus{}2}$.  If it ends in $ 2$ nondeliveries, we append them and a delivery to $ D_{n\minus{}3}$.  So

$ D_n\equal{}D_{n\minus{}1}\plus{}D_{n\minus{}2}\plus{}D_{n\minus{}3}$.

Thus, since clearly $ D_1\equal{}2$, $ D_2\equal{}4$, $ D_3\equal{}7$, we have $ D_4\equal{}13$, $ D_5\equal{}24$, $ D_6\equal{}44$, $ D_7\equal{}81$, $ D_8\equal{}149$, $ D_9\equal{}274$, $ D_{10}\equal{}504$.

 [/hide]


7)Solve the equation:$ [x^3]\equal{}{x^2}\plus{}11$ ,where $ [x]$ denotes the integer part of a real number x, $ {x}\equal{}x\minus{}[x]$,   :?:

What is the $ x\equal{}x\minus{}[x]$ supposed to mean here?

Anyway,

[hide]From $ [x^3]\equal{}x^2\plus{}11$, we know that $ x$ must be an integer; we can arrange the equation to
$ [x^3]\minus{}11\equal{}x^2$, where the LHS must be an integer, so $ x^2$ and thus $ x$ must also be an integer.

 We can just write $ [x^3]\equal{}x^3$, then.  Our equation is now
$ x^3\equal{}x^2\plus{}11$
$ x^3\minus{}x^2\equal{}11$
$ x^2(x\minus{}1)\equal{}11$.  There are no integers $ x$ such that this is true, since 11 is a prime (11=1*11).  Therefore, there are no solutions for $ x$ where $ [x^3]\equal{}x^2\plus{}11$.[/hide]

$ x \in R$,
$ [x]$ is the integer part of $ x$, e.g $ [3.14] \equal{} 3$
In general $ [x] \leq x < [x] \plus{} 1$

hello, i think $ x\equal{}\sqrt{7}$ is the solution.
Sonnhard.

[quote="Obel1x"]$ x \in R$,
$ [x]$ is the integer part of $ x$, e.g $ [3.14] \equal{} 3$
In general $ x \leq [x] < x \plus{} 1$[/quote]

$ [x] \leq x < x \plus{} 1$

Arcsin: $ \lfloor x\rfloor \leq x < \lfloor x\rfloor \plus{} 1$, or $ x \minus{} 1 < \lfloor x\rfloor\leq x$.

Anyway, the problem statement is supposed to be:
\[ \lfloor x^3\rfloor \equal{} \{x^2\} \plus{} 11.\]
where $ \lfloor x\rfloor$ denotes the greatest integer less than or equal to $ x$ while $ \{x\} \equal{} x \minus{} \lfloor x\rfloor$, in other words, the fractional part of $ x$.

[hide="Solution"]
First, since $ 0\leq \{x\} < 1$ for any $ x$, we have $ 11\leq\{x^2\} \plus{} 11 < 12$. Since this equals $ \lfloor x^3\rfloor$, which has to be an integer, the only possible value for $ \lfloor x^3\rfloor$ is $ 11$, and also since $ \{x^2\}\equal{}0$, $ x$ must be in the form $ \sqrt n$ for some integer $ n$. The only value of $ n$ for which $ \lfloor n^{3/2}\rfloor \equal{} 11$ is $ n \equal{} 5$, so the solution is $ x \equal{} \sqrt 5$.

This can easily be verified: $ \lfloor \left(\sqrt 5\right)^3\rfloor \equal{} 11 \equal{} \{5\} \plus{} 11 \equal{} 0 \plus{} 11 \equal{} 11$.[/hide]

[quote="bbgun34"]What is the $ x \equal{} x \minus{} [x]$ supposed to mean here?

Anyway,

[hide]From $ [x^3] \equal{} x^2 \plus{} 11$, we know that $ x$ must be an integer; we can arrange the equation to
$ [x^3] \minus{} 11 \equal{} x^2$, where the LHS must be an integer, so $ x^2$ and thus $ x$ must also be an integer.

 We can just write $ [x^3] \equal{} x^3$, then.  Our equation is now
$ x^3 \equal{} x^2 \plus{} 11$
$ x^3 \minus{} x^2 \equal{} 11$
$ x^2(x \minus{} 1) \equal{} 11$.  There are no integers $ x$ such that this is true, since 11 is a prime (11=1*11).  Therefore, there are no solutions for $ x$ where $ [x^3] \equal{} x^2 \plus{} 11$.[/hide][/quote]

$ x$ doesn't have to be an integer. If $ x\equal{} \sqrt{42}$, $ x^2 \plus{} 11$ is an integer.

Yongyi781 is right! Good advice: when the stated relations seem odd, take a look at the latex code behind! In this case it reveals "[x^{3}]={x^{2}}+11 ". Why would the author take the trouble to surround the "x^{2}" in curled brackets? Moreover, the strange looking $ x \equal{} x \minus{} [x]$ is also revealed to be in fact "{x} = x - [x]". Now all is clear: the author does not know he has to prefix a curled bracket with a backslash in order for it to be printed correctly by latex. So we deal with $ \{x^2\}$.


linear algebra
matrix
linear algebra unsolved

The real Matrix $ A$ satisfies $ AA^* \equal{} \alpha A$ for a real scalar $ \alpha$, if and only if there exist some permutation matrix $ P$, with $ PAP^* \equal{} A_1 \oplus A_2 \oplus \cdots$, where the elements of each $ A_i$ are all zero or all positive and $ A_iA_i^* \equal{} sA_i$.

I assume $ A^*\equal{}A^T$ (?). Necessarily $ A$ is a square matrix.
$ AA^T$ is a positive symmetric matrix.
Case 1: $ \alpha\equal{}0$; then $ A\equal{}0$.
Case 2: $ \alpha\not\equal{}0$.  Then $ A$ is symmetric and $ A^2\equal{}\alpha{A}$. The eigenvalues of $ A$ are $ 0$ or $ \alpha$. Changing $ A$ into $ tA$ we may assume $ \alpha\equal{}1$. Then $ A$ is an orthogonal projector and conversely an orthogonal projector $ A$ satisfies $ AA^*\equal{}A$.
Let $ A\equal{}\begin{pmatrix}1/2&\minus{}1/2\\\minus{}1/2&1/2\end{pmatrix}$. $ A$ is an orthogonal projector and for all permutation matrices $ P$, $ PAP^*$ has always 2 positive coefficients and 2 negative coefficients.


1) Prove that Mersenne numbers ($M_q=2^q-1$ with q prime) can be written as:
\[ M_q=(8x)^2-(3qy)^2=(1+Sq)^2-(Dq)^2  \] when $q>3$. (And thus $M_q$ is prime iff $(x,y)$ is unic.)

2) Does it help for proving the primality of Mersenne numbers or for finding factors of composite Mersenne numbers ?

3) Did you already know this property and do you have a reference ?

Tony

[quote="Valentin Vornicu"]What type of numbers are $x,y,D,S$? Positive integers?  :?[/quote] Yes. All are integers greater than 0.

[quote="T.Rex"]
3) Did you already know this property and do you have a reference ?
[/quote]

Yes, it appeared in Quadrature (french math journal, see 'le coin des problèmes') :P

Pierre.

[quote="pbornsztein"] Yes, it appeared in Quadrature (french math journal, see 'le coin des problèmes') :P Pierre.[/quote] OK. I wrote this paper.  :blush:
But I guess no English-speaking people are aware of.
Now the question is: 3) Do you have a reference BEFORE this publication ?
Tony

Tony poses it ... at least I saw it here ;) :

http://www.primepuzzles.net/puzzles/puzz_298.htm

Dan


algorithm
number theory
greatest common divisor
Euclidean algorithm
number theory unsolved

Find the GCF(a, b), where a is a number composed of n 1, and b is a number composed of m 1.

I'm sorry...maybe I should have posted it at the number theory unsolved problems ;) .(just for avoiding some certain replies... :D )
Artiona

So, let p be a prime and assume p | a and p | b (so, in particular, p is not equal to a or b).  Then we have that p | c, where c is the number composed of GCD(m, n) 1's.  This follows from the Euclidean algorithm applied to these numbers, treated as if they were numbers written in unary.  (Unary: 1 = 1, 2 = 11, 3 = 111, etc.)  In other words, we subtract the smaller from the larger and drop all the resulting 0's to get 2 new numbers and then keep repeating as far as we can go.  So, I think that number should be the GCD?

well you just want the gcd of $\frac{10^n-1}{9}, \frac{10^m-1}{9}$ so this gcd is just

$\frac{gcd(10^n-1, 10^m-1)}{9} = \frac{10^{gcd(n,m)} - 1}{9}$

so yea, it is a number with gcd(m,n) 1's

[quote="JBL"]So, let p be a prime and assume p | a and p | b (so, in particular, p is not equal to a or b).quote]
I agree with the solution of zscool.
To JBL: if p is prime and p|a why should it divide b :?
Artiona

Well, Joel's solution is correct and essentially the same as mine, mine was just more compact and cited known ideas.

No, I'm considering a prime p which divides both a and b.  (In other words, it divides their GCD.)  Then I show it divides this other number.

zs -- you didn't cite any ideas, as far as I can tell, you just used them  :P

lol used the wrong word...(I rarely edit my posts for english mistakes  :blush: )

but I didn't use anything that is too tough to prove either.


Hi. I would like to start a thread to discuss former (and current) professional ATP  tennis players. I know someone started another "Tennis" thread but it sort of died down and wasn't geared toward this kind of discussion.

So, I would like to start off with a poll of the 2010 Australian Open!

Let's vote! (And discuss)

[quote="abcak"]Federer is too old, so Nadal wins.

And why are there 2 almost identical threads?[/quote]

Uh....NONONO

"Federer is too old." Nope, he's only 28. He still has sometime. Ok, he's not dominating as much now, but he's still pwning.

As for "so Nadal wins." Do any of you remember Nadal's injury that made him miss Wimbledon? And he sort of failed at ATP World Tour Finals 2009. Gee, I doubt he'll be the same again...

Currently: F. Verdasco/F. Lopez (ESP) vs. R. Stepanek/T. Berdych (CZE), 3-2 (set 1)

Come on guys! Post! We can discuss, you know.

Um...when there was US Open, the US open thread was bustling with discussion. Why not here?

First of all... why Nadal people?

Please predict the end of Roger Federer (SUI) and who will be the next #1:

Note: You know, I noticed Federer's recent falls to Davydenko and Del Potro. I also noticed Federer's almost-fall to Igor Andreev (RUS) in Australian Open Round 1

I'm looking for Djokovic or Del Potro to get to #1 after Federer because Nadal has fallen alittle and Murray needs to get some more slam pwns.

:arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow: [color=red][size=200][u][i][b]LOCK THIS[/b][/i][/u][/size][/color]

[hide]Too many smilies in your post: 10.
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[hide="Major Updates"]Round 4: (14)M. Cilic d. (5)J. Del Potro 5-7, 6-4, 7-5, 5-7, 6-3
Quarterfinals: (14)M. Cilic d. (7)A. Roddick 7-6(4), 6-3, 3-6, 2-6, 6-3 and (4)A. Murray d. (2)R. Nadal 6-3, 7-6(2), 3-0 retired knee injury[/hide]

Cilic vs. Del potro: I'm not sad that Del Po lost actually, because Cilic is one of my favorites. I was surprised though.

Cilic vs. Roddick: Surprised and mixed feelings. Roddick is one of my favorites, but I'm happy to see cilic at his first SF

Murray vs. Nadal: Even though I like murray and not Nadal so much, I'm still sad to see nadal go. He used to actually be a great player and I feel sorry for him cuz he won the Australian last year. I also feel Federer might be out this round due to Davydenko.
But, I still say, good luck Murray!

R. Federer d. N. Davydenko: 2-6, 6-3, 6-0, 7-5
J. Tsonga d. N. Djokovic: 7-6(8-6), 6-7(5-7), 1-6, 6-3, 6-1

Federer and Davydenko: I'm so happy to see Roger get his 268th week #1 and his 23rd straight SF. GL vs Tsonga!

Tsonga and Djokovic: I have no strong feelings, because neither Tsonga nor Djokovic is my favorite player. But it was interesting to see two Aussie Finalists battling again, and I feel somewhat sorry for Djokovic, who was hoping for #2

Marin Cilic vs Andy Murray: Going to be interesting

I don't know who I'm rooting for. Both are in my top 5 players list, generally I favor Murray slightly more. Being the underdog, 21, and getting to his first major SF, I'm leaning toward Cilic there. But on the other hand, I'd like to see Murray grab a slam.

Update: Murray reaches Aussie Final, defeating Marin Cilic 3-6, 6-4, 6-4, 6-2

Cilic will probably rise to no. 10 or 11

[quote="robbiez"]From the start of the tourney, I picked a Fed-Murray final.

As for the final, it's hard for me to pick. It's difficult for a defensive player such a Murray to win the final of a GS against an offensive player like Roger, as it means the offensive player is playing very well.

However, I want Andy to win this title, so I'm gonna say Murray in 5.[/quote]

Well, Fed wins 6-3,6-4, 7-6 (11)

Well Andy was really confident this tourney...

For me Agassi rullez

Now Federer is the master ...

1. Roger Federer [b](W)[/b]
2. Novak Djokovic (QF): L Tsonga (3-->2)
3. Andy Murray: L Federer (4-->3)
4. Rafael Nadal: L Murray (2-->4)
5. Juan Martin Del Potro: L Cilic
6. Nikolay Davydenko: L Federer
7. Andy Roddick (QF) L Cilic
8. Robin Soderling (2R)
9. Jo-Wilfried Tsonga L Federer (SF) (10-->9)
10. Marin Cilic L Murray (SF!!) (14-->10)
11. Fernando Gonzalez (4R) L Roddick
12. Fernando Verdasco (4R) L Davydenko (9-->12)

[quote="Joe10112"]Whoops I accidently voted for Roddick when I meant Federer.

Oh well![/quote]

Well, the poll doesn't matter.

true.

Well, I don't watch tennis, and If i came back earlier, I would have made a Super Bowl 44 poll.

Soderling (8) and Roddick (7) switch places... not happy about that. I'm skeptical on whether Soderling should really be up there.

Juan Carlos Ferrero (22) flies to no. 16 and pushes tommy haas and gilles simon down to 19 and 18.

Cilic and Verdasco are up 1, tsonga down 2. (WOOT for Cilic) wow Robredo, Hewitt, Almagro, Ljubicic and Berdych are all falling and Isner and Querrey reach carreer highs

Even though Rodger Federer is my favorite, his time is coming. I have a strong feeling about Andy Murry this year. Nadal is just falling behind because of his recent injuries, one after the other....  :(

not a revival i think...

haven't checked tennis in LONG time

You should, tennis is a great game. I could even teach you if you want to learn  :lol:

i could do as well as you. :D probably not, but o well!

BTW, what major tennis event is going on right now?

Andy Roddick cause roger's getting old and rafa just is getting bad

Rafa is only doing bad because of his recent injuries. Lets all hope he recovers and starts kicking but again  :D

Looks like he is sort of...

Still going strong in Miami...

who wins guys?

I hope Fed...
Murray's out omg how'd that happen...

Right now Cilic is playing Baghdatis... 6-3, 4-2 (baghdatis serve)
Go CILIC!!

Well Rafa's into the quarters

But Cilic was knocked out by verdasco and federer by berdych

Oh well... so GO RODDICK

[quote="polkadotman1"]Well, roddick won in Miami

he almost looked sad when he won :lol:[/quote]

yeah i saw...but he's up to no. 7

Roddick FTW!!

[quote="polkadotman1"]he went down after winning??

he was 6th during the championship game in Miami[/quote]

6th SEED- but 8th ranked


Two players write alternatively some integer on the blackboard. The rules are the following:
a) The first player write $1$.
b) At each of the other turns, the player has to write $a+1$ or $2a$ where $a$ is any number already wrote in the blackboard and $2a\leq 1000$.
c)  One cannot write a number which has already been written, and no number is erased.
d) The player who writes $1000$ is the winner.
Determine which player has a winning strategy.

thanks very much [b]seshadri[/b] :lol:  ;)


geometry
topology
calculus
number theory
linear algebra
real analysis

Explain, and then list second, third favorites, etc.

[quote="tokenadult"]You didn't mention geometry. But then again, in most of the United States a college-level geometry course is not part of the standard curriculum for math majors, which is a pity. I like what John Stillwell writes about geometry, and of course Coxeter's books are well worth reading.[/quote]

Yes, other choices that I would have considered adding to the poll are differential geometry, algebraic geometry topology, number theory, combinatorics, and partial differential equations. However, the list would have been far too long, and these are not required in too many undergraduate curriculums.

Going back to geometry, I believe that it is a beautiful subject, but it doesn't get too much attention recently because it is not entirely useful in most major branches of mathematics. You can learn lots of planer geometry, and it will sometimes help you prove things in some other classes, but ultimately you will not need it for too many subjects.

I've been reading Maclane and Birkhoff's Algebra book which is nice so I went with that. I don't have much experience with any of the others besides calculus and linear algebra so I guess I'd consider those to be my next choices.

How did complex analysis not make it onto the list?

What I mean by "college-level geometry" is the stuff in the book [url=http://www.amazon.com/Geometry-David-Brannan/dp/0521591937/]Geometry[/url] by David A. Brannan,  Matthew F. Esplen, and Jeremy J. Gray.

My favorite topic in math so far is measure theory. So I voted for real analysis. But most of the time if someone says he is in real analysis he will not be referring to measure theory...

I also liked real analysis (in the traditional sense). I'd like to take more harmonic analysis if I have the chance.

I don't think it's really a matter of the class, though. Good classes are most of the time good because the professor is good. I think I happened to get into math (and analysis in particular) because it happened that I took intro analysis with Conway my first semester, and he was awesome. Then I had linear algebra with Gunning, who wasn't bad but he's just not as crazy or engaging or exciting as Conway. So the algebra switch got flipped off and the analysis switch got flipped on. And here I am studying more analysis as an undergraduate as some mathematicians ever will (but I've only done one intro course in algebra...).

[quote="joml88"]He keeps it real.  :coool:

So that this post doesn't get completely axed for spam...what do you have in mind when you say "college-level geometry course".  Euclidean or non?[/quote]

lol.

I haven't voted yet. I think that after taking only single cal., multi cal., and linear algebra, i'm not in the position to select a favorite from the list. However, i have been through pockets of real analysis/topology and differential equations in a book that i'm going through, and i've also taken a little bit of abstract algebra (but it was too watered down for me to consider it an thorough class).

My favorites so far would have to be Combinatorics and Number Theory.


inequalities
calculus
derivative
logarithms
algebra
binomial theorem

Show that this sequence is increasing:

\[ \Big(\frac{n}{n-1}\Big)^{n-1} < \Big(\frac{n+1}{n}\Big)^n \]

You could be lame and take the derivative of $f(x) = \left(1+\frac{1}{x}\right)^x$...

This has links to the number $e$.  It shows that your series is monotonically increasing.

why do we need to use calculus when we can use amgm

people (including me) use calculus because it gives an instant way to solve these things

hmm... would you take the derivative and show that $f'(x) > 0$ for all positive integers x?

taking the derivative you get:

$f'(x) = -\frac{1}{x} \Big(1+ \frac{1}{x} \Big)^{x-1}$

don't you? and in which case $f'(x) < 0$ ....

i think im doing something wrong

lol...you had me stumped for a while (shows how long it's been since calculus).  You can't use the power rule like that with the variable in the exponent.  You needed to use logarithmic differentiation:

\begin{eqnarray*} \frac{y'}y &=& \ln\left(1+\frac 1x\right)-(x+1)^{-1}\\ &=& \frac{(x+1)\ln\left(\frac{x+1}x\right)-1}{x+1}\\ &=& \frac{\ln\left(\frac{(x+1)^{x+1}}{ex^{x+1}}\right)}{x+1} \end{eqnarray*}

haha wow for shame... i just learned this in calc too hahaha

oops my bad

hmm thats nice, but does anyone have the AM-GM solution?

and heres the followup problem:

Show that

 \[ \Big(\frac{n+1}{n} \Big)^{n} < 3 \]

For any positive integer $n$.

I think that you can use AM-GM on $\underbrace{\left(1+\frac 1{n-1}\right), \left(1+\frac 1{n-1}\right),\ldots,\left(1+\frac 1{n-1}\right)}_{n-1}, 1$.  That gives you \[ \frac{n+1}n\ge \sqrt[n]{\left(1+\frac 1{n-1}\right)^{n-1}}. \]

Equality occurs when $1+\frac 1{n-1}=1\Rightarrow \frac 1{n-1}=0.$ but $\frac 1{n-1}$ doesn't have any zeros thus the inequality is strict.

For $\left(\frac{n+1}n\right)^n<3$ you could use the fact that the limit of $\left(\frac{n+1}n\right)^n$ as $n$ goes to infinity is $e\approx 2.718.$  But since they give 3 as a bound, I think they were looking more for something like this:

Using the binomial theorem on $\left(1+\frac 1n\right)^n$ we find it is equal to \[ \sum_{k=0}^\infty \left[\binom{n}{k}\left(\frac 1{n^k}\right)\right]=\sum_{k=0}^\infty \left[\frac{n(n-1)\cdots(n-k+1)}{k!n^k}\right].  \]

If we take the limit as $n$ goes to infinity, we have that the product $n(n-1)\cdots(n-k+1)$ approaches $n^k$.  Thus, these cancel out in our expression leaving \[ \sum_{k=0}^\infty \frac 1{k!}=\frac 1{0!}+\frac{1}{1!}+\frac 1{2!}+\frac 1{3!}+\cdots  \] which should be very familiar (remember that we already said this limit is $e$?)

Ok, so now we have that $\frac 1{0!}+\frac 1{1!}+\frac 1{2!}+\cdots<1+1+\frac 12+\frac 14+\frac 18+\cdots=3$ since $\frac 1{k!}<\frac 1{2^{k-1}}$ for $k>2.$

haha ooo nice solution

so $\sum_{k=0}^\infty \frac 1{k!}=\frac 1{0!}+\frac{1}{1!}+\frac 1{2!}+\frac 1{3!}+\cdots = e$ ?

hmmm nvr knew that

Yes, if I remember correctly,\[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}. \] It's something you might learn in the second semester of BC Calculus.

That is usually taken to be the definition of $e^x$.


[quote="NC7"]given ab+bc+ca=1 (a,b,c positive numbers) . prove that
$\frac{1}{a^{2}+ab+b^{2}}+\frac{1}{b^{2}+bc+c^{2}}+\frac{1}{c^{2}+ca+a^{2}}\geq 3$[/quote]
I think it is wrong. Try $a=b\rightarrow1^{-},c\rightarrow0^{+}.$ :wink:

[quote="NC7"][quote="NC7"]given ab+bc+ca=1 (a,b,c positive numbers) . prove that
$\frac{1}{a^{2}+ab+b^{2}}+\frac{1}{b^{2}+bc+c^{2}}+\frac{1}{c^{2}+ca+a^{2}}\geq 3$[/quote]
so can you find minimum of above expression[/quote]

When the equality holds?

[quote="NC7"]

When the equality holds?[/quote]
The equality not holds. If $a=b\rightarrow1^{-},c\rightarrow0^{+}$ then the expression $\rightarrow\frac{7}{3}.$


logarithms
trigonometry
calculus
calculus computations


J'ai l'irresistible envie de poster (anglicisme, sacrilege ;)) des problemes "pas gentil". En voici un,

[quote]Existe t-il un nombre premier $p$ tel que l'ensemble $\varphi^{-1} (2p+2)$ est nul ?[/quote]
Bien entendu $\varphi$ est la fonction totient d'Euler.

$p=37$ works.
Indeed $ \phi^{-1}(2\times 37+2)=\emptyset$ because:
i) if $ n>6$ then $ \phi(n)>\sqrt{n}.$
ii) $ \forall{n}<10^{4},\phi(n)\not\equal{}76.$ (Maple).

Moreover, $37$ is the smallest prime number that answers the question.


geometry
trigonometry
Pythagorean Theorem

In quadrilateral $ABCD$, $BC=8$, $\angle BAD=\angle ADC=135^\circ$, $AB+CD=AD$. The area of the quadrilateral is $8$. $P$ is a point (outside the quadrilateral) such that $PA=PD$ and $\angle APD=45^\circ$. Find the area of $\triangle PAD$.

[hide]Extend $AB$ and $CD$ until they interesect at $E$. We then have $AE = DE = \frac{x}{\sqrt 2}$. Let $AB = a_1$, $CD = a_2$, and $AD = x$.

We obtain two equations using the Pythagorean Theorem and the area of the triangle

\[ (\frac{x}{\sqrt 2} + a_1 )^2 + (\frac x{\sqrt 2} + a_2)^2 = 64  \]
\[ \frac 12(\frac x{\sqrt 2} + a_1)(\frac x{\sqrt 2} +a_2) = 8  \]

Expanding the first equation, remembering that $a_1 + a_2 = x$ yields

\[ x^2 + 2x  \frac{a_1 + a_2}{\sqrt 2}  + a_1^2 + a_2^2 = (1+\sqrt 2)x^2 + a_1^2 + a_1^2= 64  \]

Expanding the second equation,

\[ \frac{x^2}{2} + \frac{x(a_1+a_2)}{\sqrt 2} + a_1a_2 = 16 \Rightarrow (1 + \sqrt 2)x^2 + 2a_1a_2 = 32  \]

Adding the two results,

\[ (2 + 2\sqrt 2)x^2 + (a_1+a_2)^2 = 96 \Rightarrow x = \frac{4 \sqrt 6}{\sqrt{3 + sqrt 2}}  \]

Then using a bit of trig we find the area of $\triangle PAD$ to be (assuming I did not mess up the algebra) $\frac{9 \sqrt 2 - 6}{14}$

edit: screwed up algebra $x=\frac{4\sqrt6}{\sqrt 3 + 2\sqrt 2}$so the area should be $24\sqrt 2 - 24$[/hide]

Sorry, that's not correct :(  (The answer turns out to be a nice number.)

By the way, while you can solve this problem using trig, there's also a pure geometric solution.

I think the area of PAD is 8. Is that right, my friend? :?

No, that's not correct either...

[hide="Hint for geometric solution"]
Make a regular octagon ;) [/hide]

:(

Well, I'll post the correct answer now.
[hide][hide="Click one more time"][hide="The answer is"]
$12$
[/hide][/hide][/hide]

Try to find a solution :)

I'm so sorry about my mistake, the area should be 12 instead of 8. :blush:  This is my solution, please follow the link. Thanks.
 :) [url=http://i45.photobucket.com/albums/f68/me-Toma/aaa.gif]Click to reveal hidden content[/url]


So this is along the lines of the question about the Spanish placement test from a few weeks ago. I have the option of taking prelim exams upon entering my grad school (normally taken after the first year of classes). Do you think that it is a bad idea to study and with some luck pass upon entering?

I guess the main dilemma is: should the prelims be considered as something to "get through" and the sooner you do it the better, or would that year of the class that I skip be beneficial regardless of the level I enter (since you can't review something too many times)?

Or is this pointless, because it probably isn't the case that you would pass a prelim without the proficiency required, so if you pass, the class would have been mostly a waste of time?

If you have already had a class at the appropriate level, take the relevant prelim now. There's no penalty if you don't pass, and success means you have a lighter load in your first year. You can always move on to a different class in the same field once you've cleared the requirement matching the introductory course.

This is how I got credit for my Berkeley graduate courses I took as an undergrad- passing algebra and analysis prelims on entry.

I would agree about taking the prelims now.  If you somehow pass the prelim and later find that you don't have the necessary set of skills, presumably you can always go back and take the class.

In other words, it sounds like you don't have a lot to lose.


If $a,b,c,d,e,f>0$ then we have:

$ \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+e} + \frac{d}{e+f} + \frac{e}{f+a} + \frac{f}{a+b} \geq3 $.

cheers! :D :D :cool:

Use Cauchy and reduce it to:
  sum a^2+2(ad+be+cf)>= ab+bc+cd+de+ef+fa+ac+bd+ce+df+ae+bf (*). But with a little care you find that sum a^2+2(ad+be+cf)=(a+d)^2+(b+e)^2+(c+f)^2>=sum (a+d)(b+e) and the last quantity is the monster in the RHS of (*)


algebra
polynomial
function
vector
superior algebra
superior algebra solved

hi, how can i prove that the ring of polynomial mappings map(n,K) is isomorphic to K[X1,...Xn]/I, where I is the ideal generated by the elements Xi^q - Xi, 1<=i<=n. K is a finite field. i have proven that map(n,K) consists of q^n^n elements and K is a finite field with q elements.

Let $\mathcal I(K^n)=\{f\in K[X_1,\ldots, X_n]|f(x)=0\ \forall x\in K^n\}$.
Then $map(n,K)=K[X_1,\ldots,X_n]/\mathcal I(K^n)$ i.e. two polynomials yield the same polynomial function if and only if they differ by an element of $\mathcal I(K^n)$.
Obviously $I\subset \mathcal I(V)$, so there is a canonical surjection $K[X_1,\ldots,X_n]/I\to K[X_1,\ldots,X_n]/\mathcal I(K^n)$.
By interpolation, for all $(y_1,\ldots,y_{q^n})\in K^{q^n}$ there exists $f\in K[X_1,\ldots,X_n]$ such that $f(k_i)=y_i$ for all $i=\overline{1,q^n}$, where $k_1,\ldots, k_{q^n}$ denote all the elements of $K^n$. As grimster said (or wanted to say), this implies ${|K[X_1,\ldots,X_n]/\mathcal I(K^n)}|=|map(n,K)|=q^{q^n}$.
To finish the problem it would suffice to prove $|K[X_1,\ldots,X_n]/I|=q^{q^n}$. Rewrite $K[X_1,\ldots,X_n]/I$ as $K[x_1,\ldots,x_n]$ with $x_i^q=x_i$.
$K[x_1,\ldots,x_n]$ is a $K$-vector space with a basis given by $x_1^{a_1}\cdot\ldots\cdot x_n^{a_n}$ with $0\leq a_i\leq q-1$ for all $i=\overline{1,n}$. Since there are $q^n$ elements in this basis and $q$ elements in $K$, we finally get $|K[x_1,\ldots,x_n]|=q^{q^n}$.


Find all positive integers $ a$ and $ b$ such that $ \frac{a^{4}\plus{}a^{3}\plus{}1}{a^{2}b^{2}\plus{}ab^{2}\plus{}1}$ is an integer.

$ \frac {a^{4} \plus{} a^{3} \plus{} 1}{a^{2}b^{2} \plus{} ab^{2} \plus{} 1}\equal{}k\ge 1$
gives $ (a^2\plus{}a)(a^2\minus{}kb^2)\equal{}k\minus{}1\ge 0$
if $ k\equal{}1$ then $ a\equal{}b$
if $ k>1$ then $ k> a^2>a^2>kb^2$ gives $ b^2\equal{}0$

then $ a\equal{}b$ or $ b\equal{}0$

Exactly as solution that I wrote on exam paper  :)

Can you explain the second line please? How did you get the factored form (a*a+a)(a*a-k*b*b) ?

Standard transformation Notice that second line is equivalent to first one.

Already been posted:
http://www.mathlinks.ro/viewtopic.php?t=204778

just same as: if $m+n+1|2(m^2+n^2)-1$,then $m=n$

[quote=Paragdey12]$(a^2b^2+ab^2+1)\le (a^4+a^3+1)$
[/quote]
how did you get that inequality?


1)

A student is trying to pass a competency exam. Each time she takes the exam she has a 20% chance of passing, and she is allowed a maximum of three attempts.

What is the expected number of times she will take the exam?
What is the expected number of times she will fail the exam?

2)

Tom and Alice work [b][i]independently[/i][/b] in an attempt to solve a certain problem, i.e. whether one of them solves it [b][i]does not[/i][/b] affect the chances that the other will solve it. The probability that Tom solves it is 0.2 and the probability that Alice solves it is 0.55.

a) What is the probability that the problem will be solved?
b) If the problem is solved, what is the probability that Tom solves it?

[hide="number 1?"] Expected value is the reciprocal of the probability.  So in this case, it would be $ \frac{1}{.2}\equal{}5$.  But there is a restriction, so she will take 3 tests?

Then, the number of times an event occurs it the probability times the trial numbers.  In 3 tests, she has a $ \frac{4}{5}$ failure rate, so she will fail $ 3\cdot \frac{4}{5}\equal{}\frac{12}{5}\equal{}2.2$ times.[/hide]

[hide=" number 2"]  The Problem will not be solved if both do not solve the problem, which is $ .2\cdot.55\equal{}.11$. .  Using properties of complements, the probability it will be solved is $ .89$.

The next one is conditional probability.  $ P(A|B)\equal{}\frac{P(A \cap B)}{P(B)}$.  In this case, A= Bob solved, B=problem solved.  Thus, this is $ \frac{.2}{.89}\equal{} \frac{20}{89}$.[/hide]

[quote]The Problem will not be solved if both do not solve the problem[/quote]

No, for #2 thats not true, thats why independent was highlighted. A few other people have given me similar answers to yours, and they would be correct, except the problem is an independent probability problem....

The answer (from another forum):

[quote]I'll use the following definitions:
S - the problem is solved
T- The problem is solved by Tom
A- The problem is solved by Alice

a)
P(S) = P(T or A) = P(T) + P(A) - P(T and A) {By set theory, and definition of probability}
P(S) = .2 + .55 - (.2*.55) {By the definition of independent events}
= .75 - .11 = .64.

b)
P(T|S) = P(T and S)/P(S) (By Definition of conditional probability)
= P(T)/P(S) (Because if T happens, S is automatic-- T is a subset of S, so their intersection is all of T)
= .2/.64 = .3125[/quote]

Number 1 is similar in that your answers would be correct, but its independent so they are not. Expected value is not the actual expected number of tests....

Thanks for the help though, don't know if my response helps you solve it...


Hey, this is my first time posting in this forum. I am taking a class on linear algebra and so far I've found it rather interesting.
I have a question that says:

Consider an $ n$ x $ m$ matrix $ A$ with more rows than columns ($ n > m$). Show that there is a vector $ \vec{b}$ in $ \mathbb{R}^{n}$ such that the system $ A\vec{x}\equal{}\vec{b}$ is inconsistent.

I can understand that there is such a way to make the system inconsistent, but I have no idea how to show it generally. The coefficient matrix $ A$ has more equations than variables, but I haven't made much more progress.

Any help is appreciated. Thanks

The system is consistent if and only if the rank of A is equal to the rank of the augmented matrix. In the case of your A, there will always be zero rows at bottom when you put A in reduced row echelon form. If the corresponding rows in the augmented matrix are non zero then the system is inconsistent.

Another way is to show that a linear transformation whose range is a subspace of a finite dimensional space is not onto. But confusing matrices with linear transformations is frowned upon nowadays. :roll:

Thanks for your help Southstar. After writing it out and going through it very slowly, I believe I realize what is happening in the matrix.

I appreciate it


geometry
geometric transformation
homothety
geometry solved

Consider the triangle ABC . P , Q and R are points on the sides AB , AC and BC respectively. A' , B' and C' are points on the lines PQ , PR and QR respectively such that ABllA'B' , ACllA'C' and BCllB'C' . Prove that  AB/A'B' = SABC/SPQR     (SABC means the area of the figure ABC)

In the thread

  http://artofproblemsolving.com/Forum/viewtopic.php?t=6139

Amfulger considered the same triangles as you, but the triangles he called XYZ, ABC and DEF correspond to the triangles ABC, PQR and A'B'C' in your problem. What he called k is the factor of the homothety mapping the triangle ABC to the triangle A'B'C' (in your notation); thus, k = AB / A'B'. Now, Amfulger proved that [PQR] = k [ABC], where [$P_1 P_2 ... P_n$] means the area of a polygon $P_1 P_2 ... P_n$. Thus, we have [PQR] / [ABC] = k, and therefore, [ABC] / [PQR] = 1 / k = 1 / (AB / A'B') = A'B' / AB, as you claimed.

  Darij


[quote="shfdfzhjj"][quote="tronghieu"]Problem : Let $ a,b,c\ge 0$, $ a \plus{} b \plus{} c \equal{} 1$.Prove that
1)$ \frac {a}{{\sqrt {\frac {1}{b} \minus{} 1} }} \plus{} \frac {b}{{\sqrt {\frac {1}{c} \minus{} 1} }} \plus{} \frac {c}{{\sqrt {\frac {1}{a} \minus{} 1} }} \le \frac {{3\sqrt 3 }}{4}\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}$
2)$ \frac {a}{{\sqrt [3]{{\frac {1}{b} \minus{} 1}}}} \plus{} \frac {b}{{\sqrt [3]{{\frac {1}{c} \minus{} 1}}}} \plus{} \frac {c}{{\sqrt [3]{{\frac {1}{a} \minus{} 1}}}} \le \frac {3}{{2\sqrt [3]{2}}}\sqrt [3]{{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$

 :oops:[/quote]

Any other proof? :maybe:[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=216138 :)
By the way, The Cauchy Schwarz still works here. :)


The inequality is equivalent to :sqrt: (x 2 +y 2)/2=<(x+y)/2+(x 2-y 2  ) 2 /4
If x=y the result is obvious,else squaring both sides we obtain (x-y) 2 /4=<(x+y)(x 2 -y 2 ) 2 /4+((x 2-y 2  ) 2 /4 ) 2
dividing both sides by (x-y) 2 /4 we obtain (x+y) 2 +(nonnegative term)>=1, but this is obvious because x+y>=1


How many ordered triples of integers (a,b,c) are there such that a^2 + b^2 + c^2 = 101?

[quote="lordWings"]WLOG, $ a^{2}\geq b^{2}\geq c^{2}\geq 0$.[/quote]

The problem is asking for ordered triples... so wouldn't you have to consider all possible permutations or something?

True, but he can use this step as a start up, and then go from there.


vector
real analysis
real analysis unsolved

Can you show me why $x\in \overline{U+V}$? It's seems difficult...

[quote="RIP"]Consider the vector x=(0,1,0,1/2,0,1/3,0,1/4,...).
Show that $x\notin U+V$, but $x\in\overline{U+V}$[/quote]

I think $x\notin\overline{U+V}$, we can take $x=(0,1,0,\frac{1}{2\sqrt{2}},...,0,\frac{1}{k\sqrt{k}},...)$, but i'm not sure...


if ceramic tiles are 203mm x 203mm , how many tiles would it take to cover an area 1meters x 2 meters?

&

If a single roll of wallpaper will cover 3 square meters, how many single rolls will it take to cover a wall 2 meters x 5 meters?

Well, the area of each tile is: $(203)^2 mm^2$; and the area the region is: $2 m^2$. Now, first we have to match up the dimentions, so lets convert $mm^2$ to $m^2$.

Thus,
$(203)^2 mm^2 \cdot \left ( \frac{10^{-3}m}{1mm} \right )^2 = (203)^2 \cdot 10^{-6}m^2$

Now, we divide $2m^2$ by this number to see how many of these tiles you need to fill up this region. So,

$\frac{2m^2}{(203)^2 \cdot 10^{-6}m^2} \approx 48.533$

Thus, you need at least $49$ tiles to fill up this region.  :)

I would have to disagree
[hide]I would say it's 50, you are assuming the ceramic tile can be broken up into pieces and moved. You have to look at the side lengths of the ceramic tiles and the sidelengths of what you need to cover.[/hide]


function
real analysis
real analysis unsolved

If there is $ l$ for any real value of $ x$ statisfying $ f(2x)\equal{}2^l.f(x)$
a. If $ l>1$, show that the function is differentiable at $ x\equal{}0$.
b. If $ l<1$, show that the function is not differentiable at $ x\equal{}0$
c. If $ l\equal{}1$, show that the function is differentiable at $ x\equal{}0$ if and only if $ f$ linear.
Thanks in advance.

Dear Qruni, again, 2 out of 3 your statements are false without additional assumptions. Despite it is not hard to guess what correction is required this time, you should understand that people are getting more and more irritated with your apparent inability to state a single analysis problem correctly from the first time :(. I do not know whether it is a language problem, your severe lack of basic analytic knowledge, or something else, but if you continue posting manifestly wrong problems, the most likely outcome is that your posts (at least, in the Analysis section) will be just ignored. So, please, put more effort into ensuring that you don't miss or forget anything when stating your questions :).

Sorry, $ f$ is a continuous function. THanks in advance.


At Olympic High School, $\frac25$ of the freshmen and $\frac45$ of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?

$\text{(A)}$ There are five times as many sophomores as freshmen.
$\text{(B)}$ There are twice as many sophomores as freshmen.
$\text{(C)}$ There are as many freshmen as sophomores.
$\text{(D)}$ There are twice as many freshmen as sophomores.
$\text{(E)}$ There are five times as many freshmen as sophomores.

[hide]to get the total number of freshman and sophomores, flip it over.
2/5 = 5/2
4/5 = 5/4
5/2 is 2 times 5/4
so [b]D[/b][/hide]

[hide]Call the number of Freshmen F and the number of Sophomores S. From the problem, we can discern that 2F/5=4S/5. Cross-multiply and you get 10F=20S, or F=2S, making answer D true.[/hide]

OR

[hide]Use common sense. ;) [/hide]

Common sense always has a solution.  There is a reason for common sense, so just don't say common sense.

[hide]Use common sense to get [b]D[/b][/hide]
OR

[hide]Plug in. Say 4(Wow, this is a tiny school :| ) took it. That means there are 5 sophomores and 10 freshmen. Twice as many freshmen is [hide="..."][b]D[/b][/hide][/hide]

[hide]
uhhh $\frac{\frac 45}{\frac 25}=2$
so twice as many[/hide]

[hide]well, assuming 100 of each took it
250 freshmen
125 sophomores
D[/hide]

Let x=number of freshman
and y=number of sophomores
(2/5)y = (4/5)x
y = (4/5)/(2/5)x
y = 2x
Therefore, the answer is D.

:D I see the mistake that I made now...... its the number of people in the entire grade, not only taking the test......

Is there really any reason to post 10+ solutions on a question like this?

Increase your post count... :D don't listen to me though I'll get in trouble with the mods or something.

This is a MathCounts forum so the problems are roughly equivalent to AMC10, but keep in mind that there is a separate forum for the AMC problems.

[quote="mathgeniuse^ln(x)"][hide]I used my head to solve the question

D[/hide][/quote]

I do that a lot.

Also, I encourage having some AMC 10, 8 and even 12 problems, solving tougher problems is one great strategy for Nationals, sometimes they throw that stuff in.

[hide]
let f = total freshmen and s = total sophomores
$\frac{2f}{5}=\frac{4s}{5}$

$f = 2s$

therefore there are twice as many freshmen as sophomores
[/hide]

Amazing, 7 people managed to say the exact same thing as xxm0rg07hxx in a different way. Heck, let's make it 8.
[hide="Completely original solution"]
Okay, so we say there are x freshmen, and y sophomores. So, $\frac{2x}{5}=\frac{4y}{5}$, so our answer is D.
[/hide]

Guys stop reviving. This was from 8 years ago.

[hide=Solution]We have $\frac{2}{5}F=\frac{4}{5}S \implies F=2S$, where $F,S$ are the freshman and sophomores respectively. This makes our answer $\boxed{\text{(D)}}$.[/hide]

Guys stop reviving. This was from 7 years ago.

[quote=Alex-131]Guys stop reviving. Most of this was from over 15 years ago.[/quote] FTFY

[hide = Solution]Let the number of freshman students be $x$ and the number of sophomore students by $y$. By the second sentence, we have that $\frac{2}{5}x = \frac{4}{5}y \Rightarrow x = 2y$. This means that there are twice as many freshmen as sophomore. So, $\text{\bf (D)}$ is the correct answer.[/hide]


I don't know how to use the symbols when I post something. For example I want to write a*a and I know that the formula for it is "a^a". Can you help me, please? I just wanna post an inequality.

The forum is equipped to use LaTeX -- if you already know LaTeX, just put dollar signs around the formulas and it will work.  If you don't, you can read about it [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url].  None of the stuff about downloads, documents, etc. is relevant if all you want to do is post in the forum.

We also have a LaTeX subforum somewhere that you can go to test stuff out.


[quote="Ulanbek_Kyzylorda KTL"]Find natural numbers a,b,c: a!+2009b!=c![/quote]

We have $ a<c$ and $ b<c$

1) $ a<b$
We get then $ 1+2009\frac{b!}{a!}=\frac{c!}{a!}$ which is impossible since $ a+1$ divides two of the three numbers.

2) $ a>b$
We get then $ 2009=(b+1)(b+2)...(a-1)a(\frac{c!}{a!}-1)$. So $ b+1$ divides $ 2009$ and, since $ 2009=7\cdot\7\cdot 41$, we get $ b\in\{6,48,286,2008\}$, so $ b+2$ does not divide $ 2009$ and so $ a=b+1$ and we just have four tests to do :
$ 7!+2009\cdot 6!$ $ =4\cdot 9!$ $ \implies$ not a solution
$ 49!+2009\cdot 48!$ $ =42\cdot 49!$ $ \implies$ not a solution
$ 287!+2009\cdot 286!$ $ =8\cdot 287!$ $ \implies$ not a solution
$ 2009!+2009\cdot 2008!$ $ =2\cdot 2009!$ $ \implies$ not a solution

3) $ a=b$
$ 2010a!=c!$ and so $ 2010=(a+1)(a+2)...c$
But $ 2010=2\cdot 3\cdot 5\cdot 67$ cant be split in a product of more than one consecutive numbers.

Hence the unique solution $ \boxed{2009!+2009\cdot 2009!=2010!}$


geometry
inequalities proposed
inequalities

Let a,b,c,S be 3 sides and area of a triangle.Prove that
$ \frac{a^2(b\plus{}c\minus{}a)}{b\plus{}c}\plus{}\frac{b^2(c\plus{}a\minus{}b)}{c\plus{}a}\plus{}\frac{c^2(a\plus{}b\minus{}c)}{a\plus{}b}$≥$ 8\sqrt{3}S$

[quote="SUPERMAN2"]Let a,b,c,S be 3 sides and area of a triangle.Prove that
$ \frac {a^2(b \plus{} c \minus{} a)}{b \plus{} c} \plus{} \frac {b^2(c \plus{} a \minus{} b)}{c \plus{} a} \plus{} \frac {c^2(a \plus{} b \minus{} c)}{a \plus{} b}$≥$ 8\sqrt {3}S$[/quote]
MUST BE:
$ \frac{a^2(b\plus{}c\minus{}a)}{b\plus{}c}\plus{}\frac{b^2(c\plus{}a\minus{}b)}{c\plus{}a}\plus{}\frac{c^2(a\plus{}b\minus{}c)}{a\plus{}b)}\geq\ 2\sqrt{3}S$  :)
PUT $ b\plus{}c\minus{}a\equal{}2x;c\plus{}a\minus{}b\equal{}2y;a\plus{}b\minus{}c\equal{}2z$ and use$ (mn\plus{}np\plus{}pm)^2\geq\ 3mnp(m\plus{}n\plus{}p)$; $ S\equal{}\sqrt{xyz(x\plus{}y\plus{}z)}$
SEE HERE:
http://www.mathlinks.ro/viewtopic.php?p=1414504#1414504  :oops:


[size=150]Let x,y,z are  positive real numbers,prove that:

$ \frac{z}{x}\plus{}\frac{z}{y}\plus{}\frac{x\plus{}y}{z}\plus{}\frac{64}{3}\frac{xyz}{(y\plus{}z)(z\plus{}x)(x\plus{}y)} \geq \frac{20}{3}$[/size]

[quote="xzlbq"][size=150]Let x,y,z are  positive real numbers,prove that:

$ \frac {z}{x} \plus{} \frac {z}{y} \plus{} \frac {x \plus{} y}{z} \plus{} \frac {64}{3}\frac {xyz}{(y \plus{} z)(z \plus{} x)(x \plus{} y)} \geq \frac {20}{3}$[/size][/quote]

let $ x \equal{} tz, y \equal{} uz$ and $ p \equal{} t \plus{} u, q \equal{} tu$, that we have $ 0 < q \le \frac {p^2}{4}$ and ineq becomes

$ f(q) \equal{} (3{p}^{2} \minus{} 20p \plus{} 64)\left(q \plus{} \frac {p(3{p}^{2} \minus{} 14p \minus{} 20)}{2(3{p}^{2} \minus{} 20p \plus{} 64)}\right)^{2}$ $ \plus{} \frac {{p}^{2}( \minus{} 9{p}^{2} \plus{} 84p \plus{} 92)(p \minus{} 2)^{2}}{4(3{p}^{2} \minus{} 20p \plus{} 64)} \ge 0$,

easy to prove that $ 3{p}^{2} \minus{} 20p \plus{} 64 > 0$, so

(1), if $ \minus{} 9{p}^{2} \plus{} 84p \plus{} 92 \ge 0$, then ineq clear holds;

(2), if $ \minus{} 9{p}^{2} \plus{} 84p \plus{} 92 < 0$, we have $ 3{p}^{2} \minus{} 14p \minus{} 20 > 28p \plus{} \frac {92}{3} \minus{} 14p \minus{} 20 \equal{} 14p \plus{} \frac {32}{3} > 0$, so we get $ f(q) > f(0)$, but $ f(0) \equal{} 3p^2(p \plus{} 1) > 0$, so ineq holds.

done.


Determine all positvie integers that are expressible in the form $a^{2}+b^{2}+c^{2}+c$ where a,b,c are integers

All are expressible that way:

$a^{2}+b^{2}+c^{2}+c=n \ \iff \ (2a)^{2}+(2b)^{2}+(2c+1)^{2}=4n+1$
Thus everything we have to prove is: every integer $m \equiv 1 \mod 4$ is the sum of two even and one odd square.
By the 3-squares theorem, $m= x^{2}+y^{2}+z^{2}$ for some integers $x,y,z$. but looking $\mod 4$ tells us that two of these numbers have to be even, the other odd, so our requirement is automatically fulfilled.


What is the sum of all integer solutions to |n|<|n-3|<9?

[hide="Answer"]Obviously, no integers $\ge 3$ will work, and no integers $\le -6$. Upon checking, we see that $2$ does not work either, so we are left with $1+(-1)+(-2)+(-3)+(-4)+(-5)=\boxed{-14}$.[/hide]

so here is the way you go about doing it[hide]yeah all you have to do is answer the problem systematically as long as you get the outer and inner range you can basically figure out the answer by checking each one within that range as long as you do this process efficiently[/hide]


How many factors of $2^{95}$ are there which are greater than 1 Million?

[hide]
$2^{95}$ has 96 factors.

Since $2^{20}>1000000$ but $2^{19}<1000000$, there are 20 factors smaller than a million.

It follows that there are $\boxed{76}$ factors that are greater than a million. [/hide]

Nice hint and solution.

I got the hint before the solution was posted, but I wasn't sure. Thanks  :lol:

[hide]$2^{2}0=1,048,576$, so the factors greater a million are $2^{2}0\cdots 2^{9}5$, which there are $95-20+1=76$ of.[/hide]

[quote="jli"]im seeing lots of solutions that just say $2^{20}>1000,000$.  Just because this is true, doesnt meant that $2^{19}$ doesnt work.[/quote]
[hide]Well $2^{20}$ is barely above one million because it's $1024^{2}$ and 1024 is barely above 1000. $2^{20}$ would have to be over 2 mil. Anyways, $1024^{2}=(1000+24)^{2}=1000000+48000+576=1048576$[/hide]


Find all m,n natural numbers for this           m^n=n^m+1


geometry
complex numbers
geometry proposed

The inscribed circle of ABC touches side BC at D. A' is an arbitrary pt on BC. Let A'M be the other tangent to the incircle (M is the pt of tangency). AM meets BC at D'. Prove that (DB/DC)*(D'B/D'C)=(A'B/A'C)^2.

I only have a soln involving proj geom (duality and stuff), but I didn't look for other solns. Maybe it can be solved in a simpler way?

Can you post the projective geometry solution ?

Yeah, but in a few days. I don't have the time right now (it might be kind of long), and besides, it wouldn't be nice of me to post a solution so quickly. I'd like to let people think abt this one  :D

Here's a hint for now: consider the duality with respect to the incircle.

I don't know anything about projective geometry ...
I heard of duality but I don't exactly know what it is :(

I don't know too much either, but I want to learn more (even if it's not useful for olympiads anymore, since I'm in the 12'th grade, which means that I'll never get geom at olympiads again  :( ).

Here's the idea: for a point A consider the locus of the points P s.t. if a line through A intersects a certain circle (O) at B and C then PB/PC=AB/AC, which means that  A, B, P, C are harmonicly conjugated (I have no idea if it's the same in English, but this is what it's called in Romanian). It can be proved (complex numbers I think) that this locus, for a certain pt A, is a line, which is called the polar of A with respect to the circle (O). A duality with respect to a circle is a transformation of the pts and lines of the plane (including improper elements, meaning the infinity pt and line) s.t. every pt corresponds to its own polar with respect to that circle and each line corresponds to its pole.

Another useful and well known result is that, by applying duality to 3 collinear pts we get 3 concurrent lines and vice-versa.

For a course on Geometry Projective

8 page 46-64

http://www.unl.edu/amc/a-activities/a4-for-students/index.html

A packet on Euclidean geometry

 :D


ya know all the different types of waves like micro wave, visible light, and ultraviolet and whatnot..

why do they have different names? like do they have property differences or anything..

cuz if they dont.. then we could do ultraviolet, visible, and infrared.. only 3 types.. since we could actually see visible light, which makes it significant..
so why are there different names?

The names are there to indicate a range of frequency, because they have varying properties and (consequently) applications.

waves waves waves...i barely know anything about them except for the ultraviolet thingamabboby

Doubtful; chances are you know more about visible light than you think.

to show u the difference

they all have different physical properties..such as,, they are made differently and they are used differently...

go look up on dictionaries

they are very useful

The wave lengths or the frequinces determines the energy. It looks like a sine function.

microwaves  (not the appliance the wave) cook.
radio waves are used to transmit signals
x-rays are used to c ur bones and locate electrons
gamma-rays are results of nuclear fusion (i think)
infared waves are heat.
UV rays are emitted after nuclear fusion (i think)

And visible light, of course, is the part of the spectrum we see.

[quote]
they all have different physical properties..such as,, they are made differently and they are used differently... [/quote]

I disagree. All are made of photons. take energy away from one and it becomes a different wave on the Em spectrum.


A polynomial $ P(x)$ of degree $ n$ is given. If $ P(k) \equal{} \frac{1}{k}$ where $ k\equal{}1, 2, ..., n\plus{}1$, Find $ P(n\plus{}2)$.

The polynomial $ xP(x)\minus{}1$ has roots $ 1,2,\cdots n\plus{}1$.
So $ (x\minus{}1)(x\minus{}2)\cdots (x\minus{}n\minus{}1)$ divides $ xP(x)\minus{}1$.

Since $ xP(x)\minus{}1$ is a polynomial of degree $ n\plus{}1$,
$ xP(x)\minus{}1\equal{}(x\minus{}1)(x\minus{}2)\cdots (x\minus{}n\minus{}1)c$, where $ c$ is a constant.

Putting $ x\equal{}0\implies \minus{}1\equal{}(n\plus{}1)!(\minus{}1)^{n\plus{}1}c\implies c\equal{}\frac{1}{(n\plus{}1)!(\minus{}1)^n}$

Putting $ x\equal{}n\plus{}2\implies (n\plus{}2)P(n\plus{}2)\minus{}1\equal{}(n\plus{}1)!\frac{1}{(n\plus{}1)!(\minus{}1)^n}\equal{}\frac{1}{(\minus{}1)^n}\equal{}(\minus{}1)^n$

$ \implies P(n\plus{}2)\equal{}\frac{(\minus{}1)^n\plus{}1}{n\plus{}2}$


We know that in triangle ABC,we have the ineq $ a^2\plus{}b^2\plus{}c^2$≥$ (4\sqrt3)S\plus{}(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$ where a,b,c are sides and S is the area.But i have just found out a stronger ineq in triangles:
In acute-angled triangle,we have:
$ (a^2\plus{}b^2\plus{}c^2\plus{}(4\sqrt3)S)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})$≥$ 18$
My solution is very long(about 5 pages),I use a very strong ineq in algebra.
I hope someone will find another way to solve this.

[quote="great math"]Use the your given inequality and we set
$ a \plus{} b \minus{} c \equal{} x....$
It becomes
$ LHS \ge 4\left(xy \plus{} yz \plus{} zx \right)\left(\sum_{cyc}\frac {1}{(x \plus{} y)^2}\right) \ge 18$
The second inequality is very famous, it was a remarkable problem from Iran 98.[/quote]
I don't think your solution is right.so you mean that you let a=y+z,b=z+x,c=x+y.
So you want to prove that $ (x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(z\plus{}x)^2\plus{}2\sqrt(3xyz(x\plus{}y\plus{}z))$≥3(xy+yz+zx) to
apply IRAN 96 ineq,but $ (x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(z\plus{}x)^2\plus{}2\sqrt(3xyz(x\plus{}y\plus{}z))$≥3(xy+yz+zx) is wrong if x=0.01,y=z=9(LHS-RHS=-68.30898166)
I have another way to express this geometric ineq:
Prove that x,y,z>0,then
$ ((x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(z\plus{}x)^2\plus{}4\sqrt(3xyz(x\plus{}y\plus{}z)))(\frac{1}{(x\plus{}y)^2}\plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(z\plus{}x)^2})$≥18
I hope to see beautiful solution for this problem.

[quote="SUPERMAN2"]Prove that x,y,z>0,then
$ ((x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (z \plus{} x)^2 \plus{} 4\sqrt(3xyz(x \plus{} y \plus{} z)))(\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2})$≥18
I hope to see beautiful solution for this problem.[/quote]

I think it's not trues. Try $ a\equal{}b$; $ c\equal{}0$  :wink:

[quote="nguoivn"][quote="SUPERMAN2"]Prove that x,y,z>0,then
$ ((x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (z \plus{} x)^2 \plus{} 4\sqrt(3xyz(x \plus{} y \plus{} z)))(\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2})$≥18
I hope to see beautiful solution for this problem.[/quote]

I think it's not trues. Try $ a \equal{} b$; $ c \equal{} 0$  :wink:[/quote]
You forgot that a,b,c are 3 sides of an triangle so c cannot be zero.

nguoivn means that your inequality is wrong if $ a \to b,c \to 0$. :)

Infact, with $ z \to 0$ then we have
\[ LHS \to \left[ {x^2  \plus{} y^2  \plus{} \left( {x \plus{} y} \right)^2 } \right]\left[ {\frac{1}{{x^2 }} \plus{} \frac{1}{{y^2 }} \plus{} \frac{1}{{\left( {x \plus{} y} \right)^2 }}} \right]
\]
with $ x \to y$ we get
\[ \left[ {x^2  \plus{} y^2  \plus{} \left( {x \plus{} y} \right)^2 } \right]\left[ {\frac{1}{{x^2 }} \plus{} \frac{1}{{y^2 }} \plus{} \frac{1}{{\left( {x \plus{} y} \right)^2 }}} \right] \to 6 \cdot \left( {2 \plus{} \frac{1}{4}} \right) \equal{} \frac{{54}}{4} < 18
\]
Moreover, a note for you that in inequality when we say $ c \equal{} 0$ then it also means $ c \to 0$. That is a basic knownledge that we need to know. :)

[quote="Honey_S"]nguoivn means that your inequality is wrong if $ a \to b,c \to 0$. :)

Infact, with $ z \to 0$ then we have
\[ LHS \to \left[ {x^2 \plus{} y^2 \plus{} \left( {x \plus{} y} \right)^2 } \right]\left[ {\frac {1}{{x^2 }} \plus{} \frac {1}{{y^2 }} \plus{} \frac {1}{{\left( {x \plus{} y} \right)^2 }}} \right]
\]
with $ x \to y$ we get
\[ \left[ {x^2 \plus{} y^2 \plus{} \left( {x \plus{} y} \right)^2 } \right]\left[ {\frac {1}{{x^2 }} \plus{} \frac {1}{{y^2 }} \plus{} \frac {1}{{\left( {x \plus{} y} \right)^2 }}} \right] \to 6 \cdot \left( {2 \plus{} \frac {1}{4}} \right) \equal{} \frac {{54}}{4} < 18
\]
Moreover, a note for you that in inequality when we say $ c \equal{} 0$ then it also means $ c \to 0$. That is a basic knownledge that we need to know. :)[/quote]
It is complicated.But this ineq is true.If you want to prove it's wrong,then you have to point out a,b,c such that the inequality is wrong.I think that you cannot let $ z \to 0$ to show that this ineq is wrong,just point out an example.(remember that ABC is acute-angled triangle so there are some other details,this ineq is not true for all triangle,but it is true for acute-angled triangles)

Try $ a\equal{}b\equal{}1$ and $ c\equal{}0,01$.

[quote="Honey_S"]Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0,01$.[/quote]
We can rewrite the ineq by let $ (4\sqrt3)S\equal{}\sqrt (3(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b))$
$ (a^2\plus{}b^2\plus{}c^2\plus{}\sqrt(3(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)))(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})$≥18
When $ a \equal{} b \equal{} 1$ and $ c \equal{} 0,01$,LHS-RHS=20333.47531>0=>My ineq is true.

[quote="Honey_S"]With $ a \equal{} b \equal{} 1$ and $ c \equal{} 0,01$
\[ \left( {a \plus{} b} \right)^2 \plus{} \left( {b \plus{} c} \right)^2 \plus{} \left( {c \plus{} a} \right)^2 \plus{} 4\sqrt {3abc\left( {a \plus{} b \plus{} c} \right)} \equal{} 4 \plus{} 2 \cdot 1,0201 \plus{} 4\sqrt {3 \cdot 0,01 \cdot \left( {2 \plus{} 0,01} \right)} \equal{} 7,022442333
\]

\[ \frac {1}{{\left( {a \plus{} b} \right)^2 }} \plus{} \frac {1}{{\left( {b \plus{} c} \right)^2 }} \plus{} \frac {1}{{\left( {c \plus{} a} \right)^2 }} \equal{} \frac {1}{4} \plus{} \frac {2}{{1,0201}} \equal{} 2,210592099
\]

\[ \left[ {\left( {a \plus{} b} \right)^2 \plus{} \left( {b \plus{} c} \right)^2 \plus{} \left( {c \plus{} a} \right)^2 \plus{} 4\sqrt {3abc\left( {a \plus{} b \plus{} c} \right)} } \right]\left[ {\frac {1}{{\left( {a \plus{} b} \right)^2 }} \plus{} \frac {1}{{\left( {b \plus{} c} \right)^2 }} \plus{} \frac {1}{{\left( {c \plus{} a} \right)^2 }}} \right] \equal{} 7,022442333 \cdot 2,210592099 \equal{} 15,52326405 < 18
\]
[/quote]
You don't understand my problem.My problem is:
"If a,b,c are 3 sides of an acute-angled triangle,then $ (a^2\plus{}b^2\plus{}c^2\plus{}(4\sqrt3)S)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})$≥18"
Or
"If a+b,b+c,c+a are 3 sides of an acute-angled triangle,then $ \left[ {\left( {a \plus{} b} \right)^2 \plus{} \left( {b \plus{} c} \right)^2 \plus{} \left( {c \plus{} a} \right)^2 \plus{} 4\sqrt {3abc\left( {a \plus{} b \plus{} c} \right)} } \right]\left[ {\frac{1}{{\left( {a \plus{} b} \right)^2 }} \plus{} \frac{1}{{\left( {b \plus{} c} \right)^2 }} \plus{} \frac{1}{{\left( {c \plus{} a} \right)^2 }}} \right]$≥18"
You let a=b=1,c=0.01=>a+b=2,a+c=b+c=1.01=>a+b,b+c,c+a are not 3 sides of an acute-angled triangle(because $ (1.01)^2\plus{}(1.01)^2\minus{}2^2\equal{}\frac{\minus{}9799}{10201}<0$=>this triangle is not acute-angled)
PS:It is not necessary to prove this ineq is wrong.I have a solution for this and there is no mistake.

Superman2,can you post your solution or give me some hints?I am really interested in this problem.I try it but it is too hard for me.

[quote="SUPERMAN2"]
In acute-angled triangle,we have:
$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} (4\sqrt3)S)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})$≥$ 18$
[/quote]

The maximal $ k,$ for which the inequality
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 4k\sqrt3S)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)\geq9(1 \plus{} k)\]
holds for all acute-angle triangle is $ k \equal{} \frac {9 \plus{} 5\sqrt3}{6}.$ :wink:

[quote="arqady"]
The maximal $ k,$ for which the inequality
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 4k\sqrt3S)\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)\geq9(1 \plus{} k)\]
holds for all acute-angle triangle is $ k \equal{} \frac {9 \plus{} 5\sqrt3}{6}.$ :wink:[/quote]
Very nice problem!How can you find that?My solution's method  for k=1 cannot be apllied in your problem.Click [hide="here"]http://www.mathlinks.ro/Forum/viewtopic.php?t=267399[/hide]  to see another nice ineq that looks like this .

[quote="SUPERMAN2"]We know that in triangle ABC,we have the ineq $ a^2 + b^2 + c^2$≥$ (4\sqrt3)S + (a - b)^2 + (b - c)^2 + (c - a)^2$ where a,b,c are sides and S is the area.But i have just found out a stronger ineq in triangles:
In acute-angled triangle,we have:
$ (a^2 + b^2 + c^2 + (4\sqrt3)S)(\frac {1}{a^2} + \frac {1}{b^2} + \frac {1}{c^2})$≥$ 18$
My solution is very long(about 5 pages),I use a very strong ineq in algebra.
I hope someone will find another way to solve this.[/quote]

${ \Leftrightarrow \frac{h_a}{a}+\frac{h_b}{b}+\frac{h_c}{c} \ge \frac{18} {2\sqrt{3}+ (\frac{a}{h_a}+\frac{b}{h_b}+\frac{c}{h_c}}} )$

$ \frac{h_a}{a}=\frac{1}{ctg(\beta)+ctg(\gamma)}$ , $ \dots$

Let : $ x = ctg(\alpha)$ , $ y = ctg(\beta)$ , $ z = ctg(\gamma)$ , $ xy+yz+zx = 1$

$ \Leftrightarrow \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge \frac{9}{\sqrt{3}+x+y+z}$

$ p = x+y+z$ , $ r =xyz$ , $ q = xy+yz+zx=1$

$ \Leftrightarrow  \boxed{\frac{p^2+1}{p-r} \ge \frac{9}{\sqrt{3}+p} , q=1}$ (*)

$ p = ctg(\alpha) + ctg(\beta) + ctg(\gamma) \ge \sqrt{3}$

1. Acute-angled triangle :

$ x,y,z \ge 0$ , $ p \ge \sqrt{3}$

By Shur : $ 9r \ge 4p - p^3$ and (*) $ \Leftrightarrow (\sqrt{3}p-1)(p-\sqrt{3}) \ge 0$ which is true .

2. Condition for all triangles :

Let $ \alpha \ge \beta \ge \gamma$ => $ y=ctg(\beta) > 0$ and $ z=ctg(\gamma) > 0$

$ t= tg(\frac{\alpha}{2})$ , $ x = ctg(\alpha) = \frac{1-t^2}{2t}$ , ($ t \ge \frac{\sqrt{3}}{3}$)

$ yz \le \frac{(p-x)^2}{4}$ and $ y+z \ge 2t$

So : $ r = xyz \ge \frac{1}{4}x(p-x)^2$ and $ p=x+y+z \ge 2t+\frac{1-t^2}{2t}=p_0$

(*) $ \Leftrightarrow f(p)=\frac{9}{4}x(p-x)^2-9p+(p^2+1)(p+\sqrt{3}) \ge 0$ , ( $ p =x+y+z \ge x$)

$ f(p)$ - increase for $ p \ge \sqrt{3}$

So we only need to prove $ f(p_0) \ge 0$ $ \Leftrightarrow \frac{9}{2}t(1-t^2)-9(\frac{3}{2}t+\frac{1}{2t})+(\frac{9}{4}t^2+\frac{1}{4t^2}+\frac{5}{2})(\sqrt{3}+\frac{3}{2}t+\frac{1}{2t}) \ge 0$

- holdes for : $ \frac{\sqrt{3}}{3} \le t \le \frac{2\sqrt{3}+\sqrt{15}}{3}$

So : $ (a^2 + b^2 + c^2 + (4\sqrt3)S)(\frac {1}{a^2} + \frac {1}{b^2} + \frac {1}{c^2})\ge 18$ -  holdes for all $ \triangle$ with angles : $ \boxed{ \alpha \le 2*arctan( \frac{ 2\sqrt{3}+\sqrt{15} }{3} ) }$

Your solution is very nice.
Here is my solution using special point in triangle:
First,we notice the very famous ineq for $ x,y,z > 0$
$ \sum_{cyc}\frac {1}{x^2 \plus{} xy \plus{} y^2} \geq \frac {9}{(x \plus{} y \plus{} z)^2}$
We consider $ \triangle ABC$($ A,B,C$≤$ 120^{o}$).Let $ M$ be the point inside the triangle for which $ \angle AMB \equal{} \angle BMC \equal{} \angle CMA \equal{} 120^{o}$.
Let $ MA \equal{} x,MB \equal{} y,MC \equal{} z$.Apply the above ineq,we have:
$ \sum_{cyc}\frac {1}{x^2 \plus{} xy \plus{} y^2} \geq \frac {9}{(x \plus{} y \plus{} z)^2}$
We find that $ x^2 \plus{} xy \plus{} y^2 \equal{} a^2,y^2 \plus{} yz \plus{} z^2 \equal{} b^2,z^2 \plus{} zx \plus{} x^2 \equal{} c^2$ and $ (x \plus{} y \plus{} z)^2 \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}S}{2}$
The ineq is proved!

Nice problem and nice solution,Superman2.Thanks a lot!

[quote="Sergic Primazon"]
$ \Leftrightarrow \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\ge \frac {9}{\sqrt {3} \plus{} x \plus{} y \plus{} z}$

[/quote]
It's an old problem. See my proof here:  :)
http://www.mathlinks.ro/viewtopic.php?t=265841

So I think the 2 following problems are equivalent.Is that right?:)
Problem 1:$ \sum_{cyc}\frac {1}{a^2 + ab + b^} \geq \frac {9}{(a + b + c)^2}$
Problem 2:$ \sum_{cyc}\frac {1}{a + b} \geq \frac {9}{a + b + c + \sqrt {3(ab + bc + ca)}}$

Math is fun!:)


calculus
integration
calculus computations


function
floor function
algebra proposed
algebra

Find all funcions $ f: \mathbb{R}\to\mathbb{R}$ such that for all reals $ x$ and $ y$ the following equality holds:
\[f(x)f(yf(x)-1)=x^{2}f(y)-f(x).\]

Let $ *$ denotes the initial equation.
Plugging $ x=0$ into $ *$, we get $ f(0)=0$,
 then plugging $ y=0$ we get $ f(-1)=-1$.
Plugging $ x=-1$ into $ *$ we get $ f(-y-1)=-f(y)-1$.
Using that, the inital equation becomes $ -f(x)f(-yf(x))=x^{2}f(y)$, denote it by $ **$,
and plugging here $ y=-1$ we get $ f(x)f(f(x))=x^{2}$.  $ (1)$

Let's suppose $ f(x_{1})=f(x_{2})$. Then also $ f(yf(x_{1}))=f(yf(x_{2}))$, so $ f(x_{1})=f(x_{2})\rightarrow x_{1}^{2}=x_{2}^{2}$. $ (2)$

Look at $ (1)$. It's obvious that $ f(x)<0\rightarrow f(f(x))<0$ and $ f(x)>0\rightarrow f(f(x))>0$, meaning that $ t$ and $ f(t)$ have the same sign $ \forall t\in Im(f)$.
Then also $ f$ is injective $ \forall t\in Im(f)$ because of $ (1)$. $ (3)$

Now take $ t\in Im(f)$, $ t\not=0$. We know that $ f(t)f(f(t))=t^{2}$. Because of $ (3)$ we know that $ f$ is nonincreasing or nondecreasing for such $ t$'s. So suppose $ f(t)>t \rightarrow f(f(t))\geq f(t)>t$. By multiplying it, we get f$ (t)f(f(t))>t^{2}.$ Contradiction.. We'll get the same for $ f(t)<t$. So $ f(t)=t\longleftrightarrow f(f(x))=f(x) \forall x \in \mathbb{R}$.
Using that in $ (1)$, we get $ [f(x)]^{2}=x^{2}$.

Should be easy to prove that there isn't $ a$ such that $ f(a)=-a$, but i am in a hurry now so i'll finish it later..

[quote="goc"][quote="behemont"] Because of $ (3)$ we know that $ f$ is nonincreasing or nondecreasing for such $ t$[/quote]
why should it be monotone? :huh:  there are plenty of nonmonotone bijections(and therefore injections) in the world...
monotone->bijective and not the other way arround
edit:
one more thing
[quote="behemont"]we get $ f(-1)=-1$[/quote]
that's not quite true either. you let a solution slip by here. from that condition you can get that $ f(x)=0$ for all $ x$, or $ f(-1)=-1$...[/quote]

ok i skipped that trivial solution, but why do you say it's not monotone? :(

[quote="delegat"]Find all funcions $ f: \mathbb{R}\to\mathbb{R}$ such that for all reals $ x$ and $ y$ the following equality holds:

$ f(x)f(yf(x)-1)=x^{2}f(y)-f(x)$[/quote]

1) $ x=0$ implies $ f(0)=0$

2) $ y=0$ implies $ f(x)f(-1)=-f(x)$ and either $ f(x)=0$ $ \forall x$, either $ f(-1)=-1$

Assume now $ f(x)$ is not the null function.
3) $ x=-1$ implies $ f(-y-1)=-f(y)-1$ and $ f(-\frac{1}{2})=-\frac{1}{2}$

4) $ x=-\frac{1}{2}$ and $ y=-2$ implies $ f(-2)=-2$

5) $ x=-1$ and $ y=-2$ implies then $ f(1)=1$

6) $ x=1$ implies $ f(y-1)=f(y)-1$ and so $ f(x+1)=f(x)+1$ $ \forall x$

7) $ y=-1$ implies $ f(x)f(-f(x)-1)=-x^{2}-f(x)$ and, with point 3, $ f(x)f(f(x))=x^{2}$

8) Replace $ x$ by $ x+1$ in 7 above, and, using 6 : $ f(x+1)f(f(x+1))=(x+1)^{2}$ implies $ (f(x)+1)(f(f(x))+1)=x^{2}+2x+1$, so $ f(x)f(f(x))+f(x)+f(f(x))+1=x^{2}+2x+1$ and so $ f(f(x))+f(x)=2x$

9) $ f(f(x))f(x)=x^{2}$ and $ f(f(x))+f(x)=2x$ implies $ f(f(x))$ and $ f(x)$ are the two roots of $ Y^{2}-2xY+x^{2}=0$ and then $ f(x)=x$

So the only two solutions could be $ f(x)=0$ and $ f(x)=x$ and it is easy to check that these two solutions are OK.

[quote="pco"][quote="behemont"] ..., but why do you say it's not monotone? :([/quote]

It exist non monotone function which are injective. For example $ f(x)=2\lfloor x\rfloor-x+1$. In this example, we have :
$ f(t)$ and $ t$ has the same sign
$ f(x)$ is a bijective function.
$ f(x)$ is not monotone.[/quote]

oh, it was constantly in my mind that f is continuous... :oops:


calculus
derivative
trigonometry
function
algebra
polynomial
calculus computations

Here's a challenge for all of you just learning Calculus for the first time (if you're an American AP student, this is "BC" material.)

Define $f(x)=\frac{1-\cos x}{x^2}$ for $x\ne 0$ and $f(x)=\frac12$ for $x=0.$

Calculate $f^{[20]}(0).$  (The twentieth derivative at zero.)

Hi Kent, I am new in Calculus and I am glad that you gave a problem for beginners, thanks a lot.It will be very good if you post more problems for beginners.  :)
Now about the problem.I have an idea: let $u(x)=1-\cos{x}, v(x)=x^{-2}$ for $x\ne{0}$.So we have to find $(uv)^{[20]}$ at $0$. But there is a well-known Leibnits formula :
[i] Suppose we have   functions u and v.If u and v have all derivatives from 1 to n, when $y^{(n)}=(uv)^{n}=\sum_{i=0}^{n}{C_{n}^{i}u^{(n-i)}v^{(i)}}$. [/i]

In this problem n=20 and we can calculate twentieth derivative using the formula above. I haven't finished all calculations, but I want to know is your solution without Leibnits formula  and is it easy?If it is, when please post it, I think it will be very usefull for me.        :)

P.S. For people who don't know Leibnits formula: it can be proved by induction.

I think, Kent meant a completely different approach ;)
Try again!

Don't you think Leibnitz formula is too laborious? You have to find 40 derivatives, 20 derivatives for each $u(x)$ and $v(x)$, before you get $f^{(20)}$. And then you have to plug in for $x=0$ before getting an answer.

Thx billzhao for the short answer, but I am sure you are no beginner.  :D  :P

[quote="3X.lich"]Don't you think Leibnitz formula is too laborious? You have to find 40 derivatives, 20 derivatives for each $u(x)$ and $v(x)$, before you get $f^{[20]}.$ And then you have to plug in for $x=0$ before getting an answer.
[/quote]
Except that you can't just "plug in" $x=0$ because the formulas you get will be undefined at zero. You would have to take a truly horrific limit as $x\to0$ to get the value. I would change the words in 3X.lich's quote from "too laborius" to "not humanly possible with pencil and paper only."

But it's not a difficult problem at all. Billzhao's approach is THE way to do it. Come to think of it, how do you know this function has 20 derivatives at zero? The power series makes it obvious that it has arbitrarily many derivatives - 0 is what is known as a "removable singularity."

I use this problem in teaching calculus as part of my efforts to persuade students that Taylor polynomials, Taylor series, and power series are not some exotic extra subject arbitrarily attached to calculus, they are an inseperable part of the fabric of the subject. I also teach a course on the history of mathematics; if you study what Newton and Leibniz actually did (as well as some of their immediate predecessors like Wallis and Gregory), you see that the series were there from the beginning.

Oh yeah, how can I miss a removable singularity at $x=0$? :blush:
Thx Kent, and hey, you can change my wording because it is inhumanly possible by pencil-and-paper only than what I have assumed  :D


I need some help with this calculus homework problem: A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

The "ladder over the fence" problem is dual to, hence essentially identical as a problem to the "pipe around the corner" problem.
In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118401]this[/url] recent topic, jmerry gave [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]this link[/url] to where it's been done before.


[quote="zmli"]This situation happens all the time.  You are a math wiz and takes far more advanced math and problem solving classes outside of the school. Your school operates a math club, but the problems they are working on were the ones you solved 2 or 3 years ago.

When this happens, do you still think it is a good idea to join the school math club?[/quote]

I think it's still a good idea because you can help other kids that are having trouble and in competitions they might be on ur team so you won't regret teaching them.  :P

Well, if you don't join the club, I don't think you can do Mathcounts.

I agree completely with DarkNight.

I learned all of Algebra I in 7th grade (with tutoring) and moved on to Geometry.  To get the Gifted Algebra I Credit, I need to take it at my school, and at the same time, I take Geometry at the highschool.

I compete in Mu Alpha Theta and I am the only kid from my middle school that goes to the geometry competitions.  My teacher already lets me go without having to come to any of our practices.  Yet I still come to the practices to help the Algebra I Team.  It is good practice for me, and I get to pass my knowledge on to other students.

I don't even have a choice... my school doesn't even have a math club! I took Algebra I in 6th grade, Geometry in 7th, and now, I'm in Algebra II. I just do everything myself, and then make the school look good with the perfect score. I don't know how far my team will get...

Seungjun

Our school doesnt have a math club, but if it did, I belive it would be excellent practice for anything really.

Just my $0.02
^_^

[quote="zmli"]This situation happens all the time.  You are a math wiz and takes far more advanced math and problem solving classes outside of the school. Your school operates a math club, but the problems they are working on were the ones you solved 2 or 3 years ago.

When this happens, do you still think it is a good idea to join the school math club?[/quote]

yes, i think any type of math club is helpful as long as its within your math skill level. However, if the problems are 2 or 3 years overdue for u, i suggest u still go. You never know what you might've missed. My mathcounts team used to meet every Monday and although we rarely got much done, we still learn quite a bit.

even if your math club is below your level, you could still do the problems and review yourself. Your friend s who aren't so knowledgeable may need help and you have a chance to enter more math competitions. Personally, I woudl ask the coach if she could find some harder problems and maybe get some from the other math teachers in the school

For our school, the Math Club really is only Mathcounts, and attendance is one of the biggest factors in making the team and/or being an individual. Either way, joining the weak Math team lets you help others and helping others is actually a really good refresher for practice, (something similar happens to me in homeroom) in fact for me, helping others keeps me fresh on the topics that you did years ago, because it keeps you from getting rusty, say in Algebra when you're in Geometry, and so forth...


trigonometry
calculus
inequalities proposed
inequalities

Let $0\leq a\leq 1$  and  $n$ a natural number $n\geq 2$

Find the maximum of  $a^{n-1}(1-a)+a(1-a)^{n-1}$

[quote="Nerevar"]my solution, perhaps wrong.

let b = 1-a b>0

S = ba^(n-1) + ab^(n-1) = ab(a^(n-2) + b^(n-2))

fix ab=k

since a+b=1

a,b = 1/2 + 1/2(root(4k+1)), 1/2 - 1/2(root(4k+1))

=> S = (k/(2^(n-1)))(1 + (4k+1)*((n-2)C2) + (4k+1)^2*((n-2)C4)...)

[all negative coefficients of k dissapear, as well as the roots]

increasing k increases S, therefore S is maximal when k is maximal,

since k= ab <= ((a+b)^2)/4 = 1/4, we have S is maximal when ab=1/4 that is, a=1/2, b=1/2

in this case, S = 1/(2^n-1), so this is the maximum.[/quote]

your result only holds for $2\leq n\leq 4$ , for $n=5$ if you choose $a=0.1$ you will see the value is $0.059058$ but $\frac{1}{2^5-1}\approx0.0323$ ....

to libra_gold:
    It's very nice to use derivations,but what's your further idea?

I want to ask you a question ,silouan.
Do you have a good solution for this problem?

[quote="Lovasz"]Surely, trigonometric substitution worked.
[hide="Click to reveal hidden content"]Let $a=\sin^2x$.[/hide][/quote]

even by using this substitution , we obtain

$\frac{1}{4}\sin^22x(\sin^{2n-4}x+\cos^{2n-4}x)$

and I dont have any idea to continue .... :huh:

It's a special case of topic [b]Nice[/b] of me. The orginal problem is that

Find the maximum of $a^k(b+c)+b^k(c+a)+c^k(a+b)$ with $a,b,c\ge 0$ and $a+b+c=1$.

If $k\ge 2$ the the maximum is attained when $min(a,b,c)=0$. In this case, we need to find the maximum of $P=a^kb+b^ka$ if $k \ge 2$.

We can't find a detailed value of $\max P$, because it depends on the root of an equation which we can't find exactly what its root. But in some small case of $k$, the problem becomes easy. With $k$ real, $k \ge 2$, we can't find the detailed value. Also for a natural $n=k \ge 3$, we can't find its maximum.

Oh very nice hunghtn you find the trick to cope with this difficult problem .I had same solution but I hadn't realised that it was a special case of your problem .
I had also exactly the maximum for $n\leq 7$ .For n>7 I used mathematica since n=11 .Then I bored to find solutions of third degree equations ;)

to silouan,
what is the "methematica"?
I also have the excactly max when $n\leq 8$
but when $n=7,8$ I should solve a equation with degree 3 and 4.


x,Y are positive real numbers

S is the least of X, 1/Y ,Y +1/X

Find the maximum value of S

If $ S \geq\sqrt{2} $ then $ x \geq\sqrt{2} $, therefore $ \frac{1}{y} \geq\sqrt{2} $
so $ y+\frac{1}{x} \leq\sqrt{2} \leq\ S $

So $ S \leq\sqrt{2} $ and maximum of S is $\sqrt{2}$

Source: All-Soviet-Union Olympiad 1972 (Chelyabinsk !!!)


function
limit
integration
probability
probability and stats

John is self employed and his monthly income varies.  His monthly income is normally distributed with a mean of $ \$3900$ and a standard deviation of $ \$320.$ He is trying to determine the amount B that he can budget for his total monthly expenses.  He needs to be pretty certain that his mean monthly income will be greater than B.  What amount should he budget if he wants only a 0.05 chance that his mean monthly income for the next year is less than B?

His annual income is normally distributed with mean $ 12\mu$ and standard deviation $ \sigma\sqrt {12}$.  Thus we are solving
\[ \frac {1}{20} \equal{} \frac {1}{2}\left(1 \plus{} erf \left(\frac {12B \minus{} 12\mu}{\sigma \sqrt {24}}\right)\right)
\]

\[ \Rightarrow B \equal{} \mu \plus{} \frac{\sigma}{\sqrt{6}} erf^{ \minus{} 1}\left( \minus{} \frac {9}{10}\right)\approx 3748.05
\]

Would you please either explain the formula you used (the variables) or cite a a site with it?

Sure.  The first fact we use is that the sum of two independent normally distributed variables has mean $ \mu_1 \plus{} \mu_2$ and standard deviation $ \sqrt {\sigma_1^2 \plus{} \sigma_2^2}$.  By applying this fact twelve times to each of the monthly income distributions, this tells us that the annual income (the sum of the twelve months) is normally distributed with mean $ 12\mu$ and standard deviation $ \sqrt {12\sigma^2} \equal{} \sigma\sqrt {12}$ where $ \mu \equal{} 3900$ and $ \sigma \equal{} 320$.

The next idea we use is the [i] cumulative distribution function[/i] (CDF).  The CDF of a random variable $ X$ is a function $ \Psi(x)$ such that $ \text{Pr}(X\leq x) \equal{} \Psi(x)$.  In other words, $ \Psi(x)$ measures how likely a random variable is to appear 'to the left' of the point $ x$.  Note that any CDF must be nondecreasing, right-continuous, and must satisfy $ \lim_{x\to \minus{} \infty}\Psi(x) \equal{} 0$ and $ \lim_{x\to\infty}\Psi(x) \equal{} 1$ (in fact, subject to some reasonable assumptions you can show that any function satisfying these requirements is a CDF of some random variable!).

Now the CDF of a normally distributed random variable with mean $ \mu_0$ and standard deviation $ \sigma_0$ is
\[ \Psi(x) \equal{} \frac {1}{2}\left(1 \plus{} \text{erf}\left(\frac {x \minus{} \mu_0}{\sigma_0\sqrt {2}}\right)\right)
\]
where $ \text{erf}(x) \equal{} \frac {2}{\sqrt {\pi}}\int_0^x{e^{ \minus{} t^2}dt}$ is the so-called [i] error function [/i].  Our random variable has mean $ 12\mu$ and standard deviation $ \sigma\sqrt {12}$ so the CDF in our case is
\[ \Psi(x) \equal{} \frac {1}{2}\left(1 \plus{} \text{erf}\left(\frac {x \minus{} 12\mu}{\sigma\sqrt {24}}\right)\right)
\]
Now we are looking for a monthly budget $ B$ such that there is a $ \frac {1}{20}$ probability that the average monthly income does not fall below $ B$.  Note this does [i] not [/i] mean that the income mustn't fall below $ B$ each month, it only means that over the length of the year the income should be no less than $ 12B$.  In other words, we are looking for a $ B$ such that $ \text{Pr}(X\leq 12B) \equal{} \frac {1}{20}$.  In terms of the CDF, this is equivalent to
\[ \frac {1}{20} \equal{} \Psi(12B) \equal{} \frac {1}{2}\left(1 \plus{} \text{erf}\left(\frac {12B \minus{} 12\mu}{\sigma\sqrt {24}}\right)\right)
\]
Now we just solve this using the [i] inverse error function [/i] $ \text{erf}^{ \minus{} 1}$ (this is the function that satisfies $ \text{erf}^{ \minus{} 1}(\text{erf}(x)) \equal{} x$) to find,
\[ B \equal{} \mu \plus{} \frac {\sigma}{\sqrt {6}}\text{erf}^{ \minus{} 1}\left( \minus{} \frac {9}{10}\right)
\]
For a numerical result, plug in $ \mu\equal{}3900$ and $ \sigma\equal{}320$, then you'll need a table of values of the inverse error function, or a calculator that can compute it.
Some helpful references on the Normal distribution, PDF's, CDF's, and the error function:
http://en.wikipedia.org/wiki/Normal_distribution
http://en.wikipedia.org/wiki/Probability_density_function
http://en.wikipedia.org/wiki/Cumulative_distribution_function
http://en.wikipedia.org/wiki/Error_function


i know that this is a weird question to ask on a math forum, but how many christian brothers and sisters do i have out here on the internet, i was just wondering? i believe that christ was born of a virgin, led a perfect life, healed the diseased and blind, raised the dead, died on a cross for the sins of the world, and rose again three days later. another question to everyone in general. what do you all believe if you dont believe in god. i dont understand how someone could look at the earth and believed it all happened on accident or something!

There's a religion topic here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33938
There are also somewhat older and longer topics that you can revive if you want.
Anyway, I am an agnostic. I have seen no scientific evidence either for or against the existence of a god/gods. Until I do, I'm not going to just assume either possibility.

[quote="ash101189"] i dont understand how someone could look at the earth and believed it all happened on accident or something![/quote]

Given the fact that you have adopted those beliefs,
it is impossible for you to understand.

Many would claim that they don't understand how someone
could look at the earth and claim that it was created by a being
whose existence is not proven by physical evidence.

As a fellow Christian, I agree with your statement.
However, I know that I cannot truly understand the position
of a person who was not raised in the manner in which I was
raised or has come to that conclusion, because neither of those
is within my scope of experience.

I definately believe in God, and the truths in the Bible are probably as correct as possible for a document so old and possibly altered so many times. I suggest all agnostics give praying a try for like a week, if nothing happens it never hurt, and if you feel a connection, that's always good.  :)

ash, u kno my views on this, as does just about everyone else by now, i think. i am a Christian, and agree with all that ash said as well as what efuzzy said. :)

The Bible is a mix of Myth, History, Legend, Allegory, Poetry.

Too many people confuse it all as a solid, written history.

I too believe in God.

[quote="EFuzzy"]I suggest all agnostics give praying a try for like a week, if nothing happens it never hurt, and if you feel a connection, that's always good.  :)[/quote]
As an agnostic I would say that what you ask is somehow like if I would ask you to practice wicca for a week or so, or ask you to pray to Ba'al, or to Jupiter... if nothing happens, it never hurts!!

First than all, pray to what?, the agnostic do not believe in the existence of a god, hence, why would we waist our time and energy doing such a worthless thing for us. I honestly prefer the meditation.

Guys let's not argue again in here. If you still are in the mood of arguing about this, please do it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33938]in the other topic[/url] :D, cuz this one has just died :)


let x(q) denonate the amount of odd coefficients in (1+x)^q. if q1<<q2<q3<...<qn</sub,
prove that x(q1)+x(q2)+...+x(qn)>=x(q1)

um, i may be wrong, but isnt x(3+5) also equal to 4?

No.  The binomial terms $8 \choose k$ are 1,8,28,56,70,56,28,8,1.

This is DEFINITELY not pre-olympiad.

[url]http://www.kalva.demon.co.uk/imo/isoln/isoln853.html[/url]


trigonometry
inequalities
inequalities unsolved

let $ ABC$ a triangle , find minimum of

$ \tan^2\bigg(\frac{A}{2}\bigg)\plus{}\tan^2\bigg (\frac{B}{2}\bigg)\plus{}\tan^2\bigg(\frac{C}{2}\bigg)$

I will prove that $ \tan^2\bigg(\frac{A}{2}\bigg)\plus{}\tan^2\bigg (\frac{B}{2}\bigg)\plus{}\tan^2\bigg(\frac{C}{2}\bigg)\geq 1$.
If we set $ a\equal{}y\plus{}z$, $ b\equal{}x\plus{}z$, and $ c\equal{}x\plus{}y$, the inequality is equivalent to $ \frac{r^2}{x^2}\plus{}\frac{r^2}{y^2}\plus{}\frac{r^2}{z^2}\geq 1$.
Since $ r\equal{}\sqrt{\frac{xyz}{x\plus{}y\plus{}z}}$ we have that $ \frac{xyz}{x\plus{}y\plus{}z}\left(\frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2}\right)\geq 1\implies \frac{yz}{x}\plus{}\frac{xz}{y}\plus{}\frac{xy}{z}\geq x\plus{}y\plus{}z$ which is true by cauchy: $ \left(\frac{xz}{y}\plus{}\frac{xy}{z}\plus{}\frac{yz}{x}\right)\left(\frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{xz}{y}\right)\geq (x\plus{}y\plus{}z)^2$

thank's

$ \tan^2\bigg(\frac {A}{2}\bigg) \plus{} \tan^2\bigg (\frac {B}{2}\bigg) \plus{} \tan^2\bigg(\frac {C}{2}\bigg)\geq 1$

http://www.mathlinks.ro/Forum/viewtopic.php?t=214054

we have

$ \tan\bigg(\frac {A}{2}\bigg) \plus{} \tan\bigg(\frac {B}{2}\bigg) \plus{} \tan\bigg(\frac {C}{2}\bigg) \geq \sqrt {3}$

$ \bigg(\tan\bigg(\frac {A}{2}\bigg) \plus{} \tan\bigg(\frac {B}{2}\bigg) \plus{} \tan\bigg(\frac {C}{2}\bigg)\bigg)^2 \geq 3$

$ \tan^2\bigg(\frac {A}{2}\bigg) \plus{} \tan^2\bigg (\frac {B}{2}\bigg) \plus{} \tan^2\bigg(\frac {C}{2}\bigg)$ $ \plus{}$ $ 2\bigg(\tan\bigg(\frac {A}{2}\bigg)\tan\bigg(\frac {B}{2}\bigg) \plus{} \tan\bigg(\frac {B}{2}\bigg)\tan\bigg(\frac {C}{2}\bigg) \plus{} \tan\bigg(\frac {C}{2}\bigg)\tan\bigg(\frac {A}{2}\bigg)\bigg) \geq 3$

we now that in any trinagle

$ \tan\bigg(\frac {A}{2}\bigg)\tan\bigg(\frac {B}{2}\bigg) \plus{} \tan\bigg(\frac {B}{2}\bigg)\tan\bigg(\frac {C}{2}\bigg) \plus{} \tan\bigg(\frac {C}{2}\bigg)\tan\bigg(\frac {A}{2}\bigg) \equal{} 1$

so

$ \tan^2\bigg(\frac {A}{2}\bigg) \plus{} \tan^2\bigg (\frac {B}{2}\bigg) \plus{} \tan^2\bigg(\frac {C}{2}\bigg)\geq 1$

or this would help

$ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$


group theory
abstract algebra
geometry
geometric transformation
function
superior algebra
superior algebra unsolved

Prove there are no simple groups of order $ 180\equal{}2^2 3^2 5$.

Classic Sylow techniques get you that the number of Sylow 5-subgroups are 1, 6, or 36. If there is only one, then we are done. I have two arguments for why there must be a normal subgroup, if there are 36.

So supposing there are 6, why must there be a normal subgroup? We can discard the case when there are 1 or 4 Sylow 3-subgroups (I have an argument for this as well), so we can WLOG assume there are 10. Counting elements and the standard tricks don't seem to work.

Let $ G$ be the group of order $ 180$ under investigation. Let $ N$ be the normalizer of one of the $ 5$-Sylow subgroups. If there are six of them, then $ N$ is a subgroup of index $ 6$. $ G$ acts by left-translation on the cosets of $ N$, which provides a homomorphism $ \phi: G \to S_6$. The kernel of $ \phi$ must be trivial (otherwise it's a normal subgroup of $ G$), so $ \phi$ is in fact 1:1. Letting $ G' \equal{}\phi(G)$ be the image of $ G$ under $ \phi$, we see that $ G'$ is isomorphic to $ G$, and is a subgroup of $ S_6$.

Consider $ K\equal{}G' \cap A_6$. This is a normal subgroup of $ G'$ (since $ A_6$ is normal in $ S_6$), so it must either be $ G'$ or $ 1$. The latter is impossible since the index of $ K$ in $ G'$ is $ 2$, at most. So $ K\equal{}G'$, namely $ G' \subset A_6$. But $ |G'| \equal{} 180$ and $ |A_6| \equal{} 360$ so $ G'$ has index $ 2$ in $ A_6$, which can't happen since $ A_6$ is simple.

Ha! This is [i]precisely[/i] the argument I came up with today and got on the computer to come post.  :lol:

Great - so I didn't spoil the fun for you  :)

Also, can anyone show the case where $ |Syl_5(G)|\equal{}36$ in an elementary way? I relied on a slick trick using the transfer, but that requires some heavy machinery in how the transfer operates which is a function where it isn't even obvious that it is well-defined (see Isaacs or Gorenstein).

If all you have is a massive counting argument, you could post that, but I've worked through most of it and it isn't pretty. I'm wondering if there are nice tricks like the $ |Syl_5(G)|\equal{}6$ case above that only relies on Sylow theory as the hardest part.

I am using randomgraph's argument here (which I've never seen before), so if I've misunderstood why it works I am sorry.

Suppose there are 36 $ Syl_5$-groups and that $ G$ is simple. By the embedding argument above if there is more than one $ Syl_3$ group then there cannot be 4, so there are 10. They cannot all intersect trivially by counting. So there are two, $ H$ and $ K$, that intersect in a subgroup of order 3. $ N_G(H \cap K)$ contains $ HK$ which has 27 elements. So the index of $ N_G(H \cap K)$ is 1,2,4 or 5 and (if assuming $ G$ is simple) $ G$ embeds in $ S_1$, $ S_2$, $ S_4$ or $ S_5$ all contradictions because $ 5! < 180$

Is it correct? If not, there is a giant counting argument here: http://www.mathreference.com/grp-fin,loword.html

That's a perfectly good argument, and much more efficient than the one in the link you provided (I guess the guy who wrote that argument missed the fact that the subgroup of order $ 3$ cannot have too many conjugates). The only thing that may not be fully transparent in your argument is why $ HK$ cannot have order $ 18$: the reason I can see is that $ HK$ has two $ 3$-Sylow subgroups of order $ 9$, which contradicts the fact that in a group of order $ 18$, any group of order $ 9$ is normal.

I used the formula $ |HK| \equal{} \frac{|H\parallel{}K|}{|H \cap K|} \equal{} 27$

But this formula is not true, in general. We don't know that $ H$ normalizes $ K$. In fact, $ HK$ can be of order greater than $ 27$ - I didn't notice you said it actually equals $ 27$. Your argument remains valid either way.

No, this formula holds for all $ H,K\le G$.
$ HK$ is disjoint union of $ K$-cosets of the form $ hK, h\in H$, let the set of those be $ HK/K$.
Then the map $ H/H\cap K\to HK/K, h(H\cap K)\mapsto hK$ for $ h\in H$ is a well-defined bijection,
which gives $ |HK|\equal{}|HK/K|\cdot |K|\equal{}|H/H\cap K|\cdot |K|\equal{}{|H|\cdot |K|\over |H\cap K|}$.

We're talking about different things. When I wrote $ HK$ I meant the smallest group containing $ H$ and $ K$, that is the group generated by the products $ hk$ with $ h \in H$ and $ k \in K$. You're taking it to mean simply the set of such $ hk$.

You are correct that Kalle's proof works fine if we interpret $ HK$ as a set, and it's also true that my pontification about $ HK$ having $ 18$ elements is silly in any interpretation.

Note that if $ H \cap K$ is normal in both $ H$ and $ K$, as in the situation above, then $ H \cap K$ is being normalized by $ HK$ under my interpretation as well, which is a stronger statement ($ HK$ could be quite large).

I think it is more common to use $ HK$ for the subgroup rather than the subset, just like $ IJ$ denotes the ideal generated by $ I$ and $ J$. But I could be wrong about this.

I might as well post the transfer method just to be complete. Suppose $ |Syl_5(G)|\equal{}36$. Then for any $ P\in Syl_5(G)$ interpret this as $ [G: N_G(P)]\equal{}36$, or $ |N_G(P)|\equal{}5$. But $ P<C_G(P)<N_G(P)$, so $ C_G(P)\equal{}N_G(P)$ so the transfer $ V: G\to P$ is onto. This means that $ P\cong G/ker(V)$, so the kernel of the transfer is not trivial or all of $ G$ and it is a normal subgroup.

Note that this method will work whenever $ |G|\equal{}mp$, i.e. any product of primes with one of them being to the power 1.


Euler
number theory unsolved
number theory

Let p be prime number .prove that with each $ n$ $ \in$ $ N^{\plus{}}$ we have
    $ \large\varphi$($ p^{n}$-1) $ \equiv$ 0($ mod$ $ n$).where  $ \large\varphi$ is Euler function.

For any a>1 it is true, that
$ m|\phi(a^{n}\minus{}1)$, were \[ m\equal{}(a\minus{}1)\prod_{p|n}p^{k(p)(k(p)\minus{}1)/2},\ k(p)\equal{}ord_{p}(n).\]

This one is really obvious, since $ n$ is the order of $ p$ mod $ p^{n}\minus{}1$, so it divides $ \phi(p^{n}\minus{}1)$.


algebra
polynomial
induction
greatest common divisor
algebra unsolved

Let $ p(x)\equal{}a_n x^n\plus{}...\plus{}a_1 x\plus{}a_0$ and $ q(x)\equal{}b_m x^m\plus{}...\plus{}b_1 x\plus{}b_0$ be two polynomials with integer coefficients. Suppose that all the coefficients of $ p(x) q(x)$ are even, but not all of them are divisible by $ 4$. Show that one of $ p(x)$ and $ q(x)$ has all even coefficients and the other has at least one odd coefficient.

We show first that at least one polynomial has all coefficients even by induction on $ m \plus{} n$.  The base case $ m \plus{} n \equal{} 0$ is trivially true, since $ p(x) \equal{} a_0$ and $ q(x) \equal{} b_0$, and $ p(x)q(x)$ is a constant polynomial that is an even integer iff at least one of $ a_0$, $ b_0$ is even.  Assume now that the result is true for all sums $ m \plus{} n < s$, and consider polynomials $ p(x)$ and $ q(x)$ such that $ m \plus{} n \equal{} s$.  Clearly $ a_nb_m$ is even.  Hence at least one of them (wlog $ a_n$) is even.  Take $ p'(x)\equal{}p(x)\minus{}a_nx^n$, and since $ p'(x)q(x)\equal{}p(x)q(x)\minus{}a_nx^nq(x)$ is the result of subtracting from $ p(x)q(x)$ a polynomial with all coefficients even, then $ p'(x)q(x)$ has all coefficients even, and degree less than $ s$.  The hypothesis of induction is then true, and either $ p'(x)$ has all coefficients even, and hence so does $ p(x)$ since we add a term with even coefficient, or $ q(x)$ has all coefficients even.

Now, if all coefficients of BOTH polynomials are even, each coefficient of the product is the sum of products of two even numbers, hence the sum of multiples of $ 4$, hence a multiple of $ 4$.  Contradiction, hence one of the polynomials has at least one odd coefficient.

Given a polynomial, $ f$, with integer coefficients, define $ c(f)$ to be the greatest common divisor of the coefficients.

It is well known that $ c(fg)\equal{}c(f)c(g)$.

In this case $ 2^1\parallel{}c(fg)\equal{}c(f)c(g)$... and the problem is trivial.


Find the domain of the function $ f(x) \equal{} 1 \plus{} \sqrt{ \frac{x^2}{(x\plus{}2)(3\minus{}x)} }$

Could you explain please...

Everything in the square root must be non-negative.  The denominator can't be 0, so it is not defined at $ x\equal{}3$.  It's also not defined at $ x\equal{}\minus{}2$.  If $ x$ is greater than 3 or less than -2, then the denominator is negative, which is not valid.  So the domain is $ (\minus{}2,3)$.

Well, we don't want the value under the square root to be negative, and we don't want the denominator of the fraction under the radical to have a denominator of zero. In order for $ f(x)$ not to have zero as the denominator of the fraction under the radical, $ x$ can't be equal to $ 3$ or $ \minus{}2$. If $ x$ is real then $ x^2$ must be positive. So that means the denominator of the fraction under the radical can't be negative. This will make the value under the square root negative, which will make the value of $ f(x)$ imaginary. $ x$ can't be greater than three, and it can't be less than negative two.

So $ x$ must satisfy the following inequality;

$ \minus{}2 < x < 3$