Find all $ f: R\to R,g: R\to R,h: R\to R$ and $ f,g,h$ is continuous on $ R$ satisfying

 $ f(x^2) \plus{} f(y^2) \equal{} [h(x) \minus{} h(y)]g(x \plus{} y)$ with $ \forall x,y\in R$

For any $ x \in \mathbb{R}$, take $ y \equal{} x$ to obtain $ 2f(x^2) \equal{} 0$. Thus $ f(x) \equal{} 0$ for all $ x \geq 0$, and in fact we're not given any information on the values of $ f(x)$ when $ x < 0$ so $ f$ can be any continuous function which is $ 0$ for non-negative reals. We are left with

$ (h(x)\minus{}h(y)) g(x\plus{}y) \equal{} 0$.

If $ g(x) \equiv 0$, then $ h$ can be any continuous function. Otherwise we have $ g(x_0) \neq 0$ for some $ x_0 \in \mathbb{R}$, and since $ g$ is continuous there is a certain $ \delta > 0$ such that $ g(x_0\plus{}t) \neq 0$ whenever $ |t| < \delta$. It follows that

$ h(x) \equal{} h(x_0\plus{}t\minus{}x)$ for all $ x \in \mathbb{R}, |t| < \delta$

so, applying this again for $ x' \equal{} x_0\plus{}t\minus{}x$ and $ t'\equal{}s$ we obtain

$ h(x) \equal{} h(x_0\plus{}t\minus{}x) \equal{} h(x_0\plus{}s\minus{}(x_0\plus{}t\minus{}x)) \equal{} h((s\minus{}t)\plus{}x)$ for all $ x$ and any $ s,t$ with $ |s|<\delta, |t|<\delta$.

Thus $ h$ is periodic $ h(x)\equal{}h(x\plus{}w)$ for a set of periods $ w$ which can be made arbitrarily small, and it's a simple exercise to show that a continuous function with arbitrarily small periods is constant. Of course, if $ h$ is constant, then the condition is satisfied regardless of $ g$.

In summary, we have:

1. $ f(x) \equal{}0$ for $ x \geq 0$, $ f(x)$ is continuous for all $ x$ but otherwise arbitrary for $ x<0$.

and either:

2a. $ g(x) \equiv 0$, $ h(x)$ any continuous function, or

2b. $ g(x)$ any continuous function, $ h(x) \equal{} \mbox{constant}$.


This will probably get moved to G&FF, since it doesn't promote "serious" discussion.

I'm a few inches short of 6' myself, so maybe 5'10"?

I'm short, I think about a bit less than 5 feet. Maybe, what, 4'11, 4'10 or 5'?

uhmmm... I believe that 5'9''  (exactly 5,71ft)... :huh:

exactly 174cm

5'3".

a little under average 8th grade girl height.

im pretty tall for a seventh grader-5 foot 9 inches

Make this a poll. 5'4" at age 14. One of the shortest kids in the high school. Must learn to stimulate and/ or artificially reproduce growth hormone. Fast

I'm kinda tall for someone who just turned 13.

5' 7''

[quote="distracted523"]5'2" I think, 8th grade boy. I'm a little short.[/quote]

wow, a little short?

4'6" 8th grade

shortness ftw!-only 12 tho so it's not that bad...

[quote="skyhog"]A little over 6' at 14 years old....  woot.
There's another freshman at my school that's 6'5", though.  It's kinda creepy.[/quote]
Aye at my school there's a freshman (I am one too) who is 6' 3 or 4, and its scary.

5'1"

a bit tall for a 7th grader.

i haven't hit my growth spurt

About 4'11"

About average for 11 year old.

5' 7"-5' 8"

The message isn't too small. I just felt like putting it here

Why are you guys all taller than me?
:(
Probably because you're all in HS or something.
:|

i feel so short :(  i'm an 8th grade girl and i'm only 4'11"...
shortness! (i get run over in the hallways a lot) :rotfl:

6'. 183 cm I guess.

I'm Asian and 16, so I'm fairly tall for my demographic.

ditto everything from Rafaelloaa. 5'1" 7th grade

6'

I'm 17 so I guess thats about average.


inequalities
search
function
inequalities proposed

Now, I am quite interested in solving inequalities by using Cauchy- Schwartz :)
Actually, I will prove the weaker inequality
The inequality is equivalent to
$ \sum_{cyc}{\frac {a^2(a\plus{}c)(a\plus{}b)}{(b \plus{} c)}}  \geq \sum_{cyc}{a^2(b\plus{}c)}$
Meanwhile, applying Cauchy- Schwartz, we get the result
$ \frac {a^2(a\plus{}c)(a\plus{}b)}{(b \plus{} c)}\plus{}\frac {b^2(b\plus{}c)(b\plus{}a)}{(a \plus{} c)}\geq 2ab(a\plus{}b)$
Likewise, we have the same inequalities. Now, the result is apparent! :)

nice my friend  :)  , but I think you have used AM-GM not cauchy  :wink:

[quote="anas"]hello mathlinkers  :)

Let $ (a,b,c)$ a triplet of positive real numbers , Prove that :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 2(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$

the weaker one is very nice :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac {2abc}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$ with the same conditions  :wink:[/quote]
See here: http://canhang2007.wordpress.com/2008/11/27/nice-inequality-anas-bentaleb/ :)

my friends,for the first inequality  i use triangle function,but find another inequality....but i think it's true.. :D
  http://www.mathlinks.ro/viewtopic.php?p=1331805#1331805

[quote="dduclam"][quote="anas"]hello mathlinkers  :)

Let $ (a,b,c)$ a triplet of positive real numbers , Prove that :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 2(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$

the weaker one is very nice :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac {2abc}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$ with the same conditions  :wink:[/quote]

The second, It's weaker then the following $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac {10abc}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 2$
See http://www.mathlinks.ro/viewtopic.php?search_id=333697881&t=195311

The fist, I think the following inequalities are also true :
1/  $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 3(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq \frac9{8}$ for all $ a,b,c$ be nonnegative real numbers.

2/ $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 10(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \leq 2$ for all $ a,b,c$ be side lengths of a triangle.[/quote]

i proofed  it ...
1/  $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 3(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq \frac9{8}$ for all $ a,b,c$ be nonnegative real numbers.

[quote="dduclam"][quote="anas"]hello mathlinkers  :)

Let $ (a,b,c)$ a triplet of positive real numbers , Prove that :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 2(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$

the weaker one is very nice :

$ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac {2abc}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 1$ with the same conditions  :wink:[/quote]

The second, It's weaker then the following $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac {10abc}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq 2$
See http://www.mathlinks.ro/viewtopic.php?search_id=333697881&t=195311

The fist, I think the following inequalities are also true :
1/  $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 3(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \geq \frac9{8}$ for all $ a,b,c$ be nonnegative real numbers.

2/ $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 10(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \leq 2$ for all $ a,b,c$ be side lengths of a triangle.[/quote]

2/ $ \sum_{cyc}{\frac {a^2}{(b \plus{} c)^2}} \plus{} \frac { 10(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{ (a \plus{} b)(a \plus{} c)(b \plus{} c)} \leq 2$ for all $ a,b,c$ be side lengths of a triangle.

http://www.mathlinks.ro/viewtopic.php?p=1331805#1331805

:oops: Why ?

$ \sum cos^4 A \leq 3/16$

Thank you very much.


inequalities
logarithms
inequalities unsolved

Clever skill :wink:
I have heard of the skill before, but I didn't think of it.

[quote="arqady"]$ \sum_{cyc}{\sqrt{3a^{2}+4}}\leq \sqrt{7}(a+b+c)\Leftrightarrow\sum_{cyc}\left(\sqrt7a-\sqrt{3a^{2}+4}-\frac{4\ln a}{\sqrt7}\right)\geq0.$
Let $ f(a)=\sqrt7a-\sqrt{3a^{2}+4}-\frac{4\ln a}{\sqrt7}.$
Then $ f'(a)=\frac{9(a-1)(a+1)}{\left(\sqrt7+\sqrt{3a^{2}+4}\right)\cdot\sqrt{21a^{2}+28}}.$
Thus, $ f(a)\geq f(1)=0.$  Done.[/quote]
How can you think of this $ -\frac{4\ln a}{\sqrt7}$? Do you have the method?

[quote="Twins"]
How can you think of this $ -\frac{4\ln a}{\sqrt7}$? Do you have the method?[/quote]
$ \sum_{cyc}{\sqrt{3a^{2}+4}}\leq \sqrt{7}(a+b+c)\Leftrightarrow\sum_{cyc}\left(\sqrt7a-\sqrt{3a^{2}+4}+\lambda\ln a\right)\geq0.$
Let $ f(a)=\sqrt7a-\sqrt{3a^{2}+4}+\lambda\ln a$ and $ f'(1)=0.$
Then $ \lambda=-\frac{4}{\sqrt7}.$ :wink:

[quote="Namdung"]It's nice trick. Thank you arqadi.

Namdung[/quote]

The trick was discussed thoroughly in my book, illustrated with various problems.

[quote="arqady"][quote="hanschinaman"]a,b,c are positive real numbers whose product is 1, prove
$ \sum_{cyc}{\sqrt{3a^{2}+4}}\leq \sqrt{7}(a+b+c)$[/quote]
$ \sum_{cyc}{\sqrt{3a^{2}+4}}\leq \sqrt{7}(a+b+c)\Leftrightarrow\sum_{cyc}\left(\sqrt7a-\sqrt{3a^{2}+4}-\frac{4\ln a}{\sqrt7}\right)\geq0.$
Let $ f(a)=\sqrt7a-\sqrt{3a^{2}+4}-\frac{4\ln a}{\sqrt7}.$
Then $ f'(a)=\frac{9(a-1)(a+1)}{\left(\sqrt7+\sqrt{3a^{2}+4}\right)\cdot\sqrt{21a^{2}+28}}.$
Thus, $ f(a)\geq f(1)=0.$  Done.[/quote]

Arqady, how do you know that when $ f'=0$, that's minimum, why it couldn't be maximum?
Sorry if it's stupid question.. :oops:

[quote="kunny"]Since $ a>0$, the sign of $ f'(a)$ is coincides with $ a-1$.[/quote]

Oh, i see it now..thx kunny

[quote="hanschinaman"]A similar problem, I wonder if it can be solved by the same trick:
a,b,c are positive real numbers whose product is 1, prove
$ 2(a^{2}+b^{2}+c^{2})+a+b+c\geq 6+ab+ac+bc$,
if we can prove that $ 2x^{2}+x-\frac{1}{x}-2-6\ln x\geq 0$ is true.[/quote]

hanschinaman, the following inequality ( Vasc ) stronger:
let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc=1.$ Prove that
$ a^{2}+b^{2}+c^{2}+6\geq \frac{3}{2}\cdot\left(a+b+c+ab+ac+bc\right).$ :wink:

[quote="hanschinaman"]It can be solved by the same trick, right? :)[/quote]
No! :D  :wink:

[quote="hanschinaman"]$ a,b,c$ are positive real numbers whose product is 1, prove
$ \sqrt {3a^{2} \plus{} 4} \plus{} \sqrt {3b^{2} \plus{} 4} \plus{} \sqrt {3c^{2} \plus{} 4}\leq \sqrt {7}(a \plus{} b \plus{} c)$[/quote]
For convenience, denote $ a \equal{} x^3, b \equal{} y^3, c \equal{} z^3$.

$ \sum{x\sqrt {\frac {3x^4 \plus{} 4y^2z^2}{7}}}\leq \sum{\frac {x}{7}(7x^2 \plus{} 4y^2 \plus{} 4z^2 \minus{} 4xy \minus{} 4xz)} \equal{} \sum{x^3}$.

[hide="Proof."]$ (7x^2 \plus{} 4y^2 \plus{} 4z^2 \minus{} 4xy \minus{} 4xz)^2 \minus{} 7(3x^4 \plus{} 4y^2z^2)$

$ \equal{} \left(\frac {y \plus{} z}{2} \minus{} x\right)^2[7(y \plus{} z \minus{} 2x)^2 \plus{} 2(9y^2 \minus{} 14yz \plus{} 9z^2)]$

$ \plus{} \left(\frac {y \minus{} z}{2}\right)^2(48x^2 \plus{} 39y^2 \plus{} 42yz \plus{} 39z^2)\geq 0$.[/hide]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3301]Generalization.[/url]

[quote="arqady"]
Let $ f(a) \equal{} \sqrt7a \minus{} \sqrt {3a^{2} \plus{} 4} \minus{} \frac {4\ln a}{\sqrt7}.$
Then $ f'(a) \equal{} \frac {9(a \minus{} 1)(a \plus{} 1)}{\left(\sqrt7 \plus{} \sqrt {3a^{2} \plus{} 4}\right)\cdot\sqrt {21a^{2} \plus{} 28}}.$
[/quote]
It's wrong! I am sorry.
$ f'(a)\equal{}\sqrt7\minus{}\frac{3a}{\sqrt{3a^2\plus{}4}}\minus{}\frac{4}{a\sqrt7}$ and $ f'(1)\equal{}0.$
$ f''(a)\geq0\Leftrightarrow(3a^2\plus{}4)^3\geq63a^4,$ which true.
Thus, $ f(a)\geq f(1)\equal{}0.$
The following inequality still can be proved by this method:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
$ \sum_{cyc}\sqrt{1\plus{}5a^2}\leq\sqrt6(a\plus{}b\plus{}c).$

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[\sqrt{1\plus{}11a^2}+\sqrt{1\plus{}11b^2}+\sqrt{1\plus{}11c^2}\leq\sqrt{12}(a\plus{}b\plus{}c)\]


[i]Let be x and y reals positive. Proof that[/i]

$x^{2}+y^{2}+1\geq y\sqrt{x^{2}+1}+x\sqrt{y^{2}+1}$

[hide] By inequality between geometric and arithmetic averages we have

$\frac{(x^{2}+1)+y^{2}}{2}\geq y\sqrt{x^{2}+1}$

$\frac{(y^{2}+1)+x^{2}}{2}\geq x\sqrt{y^{2}+1}$

Then,

$x\sqrt{y^{2}+1}+y\sqrt{x^{2}+1}\leq\frac{2(x^{2}+y^{2}+1)}{2}=x^{2}+y^{2}+1$

[b]QED[/b].[/hide]


Derive the equation of the circle circumscribing the triangle determined by the lines: $ x\plus{}y\equal{}8$; $ 2x\plus{}y\equal{}14$ and $ 3x\plus{}y\equal{}22$.
A) $ x^{2}\plus{}y^{2}\minus{}4x\plus{}6y\minus{}12\equal{}0$
B) $ x^{2}\plus{}y^{2}\plus{}4x\minus{}6y\minus{}16\equal{}0$
C) $ x^{2}\plus{}y^{2}\minus{}6x\minus{}4y\minus{}16\equal{}0$
D) $ x^{2}\plus{}y^{2}\minus{}6x\plus{}4y\minus{}12\equal{}0$
E) None of the Above

Coordinates of vertices of triangle ABC are:

A(6;2);
B(7;1);
C(8;-2);

Let the formula of circle be $ x^2\plus{}y^2\plus{}ax\plus{}by\plus{}c\equal{}0$.

$ 6^2\plus{}2^2\plus{}6a\plus{}2b\plus{}c\equal{}0$ ----->  6a+2b+c=-40
$ 7^2\plus{}1^2\plus{}7a\plus{}b\plus{}c\equal{}0$ -----> 7a+b+c=-50
$ 8^2\plus{}(\minus{}2)^2\plus{}8a\minus{}2b\plus{}c\equal{}0$ -----> 8a-2b+c=-64

We have an equation system:

$ \begin {cases} 6a\plus{}2b\plus{}c\equal{}\minus{}40\\
7a\plus{}b\plus{}c\equal{}\minus{}50\\
8a\minus{}2b\plus{}c\equal{}\minus{}64\end {cases}$

Solving we get a=-8;b=2;c=4;

Equation is $ x^2\plus{}y^2\minus{}8x\plus{}2y\plus{}4\equal{}0$

Answer: E)


i must be bored
you can suggest more

i myself chose help
you should see the titles of the threads in the HSB

ook the poll isnt working...
w/e

here is the blog post that inspired this thread
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=222372

My list:

math
lol
1337
own/pwn
beast
w00t

List:

EPIC
FAIL
cd
typo
"This message is too short..."
Lynch:
FOS:
asdf

Well lets see,

idiot
AoPS
levans
mod
rrusczyk
admin
lock
topic
/in
Latex
lol
wtf
ehh
aww
w00t
pwn
own
math
AIME
Mathcounts
number
FTW
problem
answer

cant think of any more...

Aww... you just made me lose the game!

RULE 1: 	You are playing The Game.
RULE 2: 	Whenever you think about The Game, you lose.
RULE 3: 	Loss must be announced.

The amount of time you get to forget about The Game varies from group to group. I've heard of one group that gives 30 minutes, I personally play with 30 seconds.

I lost the game (again!)

I am not good at that game

Man, I just won.


Find the sum of the terms in the sequence $ 1,4,7,\ldots,193.$

hello, with $ a_n\equal{}3n\plus{}1$ and $ n\equal{}0,1,...,64$ we get
$ \sum_ {i\equal{}0}^{64} 3i\plus{}1\equal{}6305$.
Sonnhard.

You could also use the sum of arithmetic sequence formula:
$\frac{n}{2}(a_1+a_n)$ to get $\boxed{6305}$


An infinite square grid is colored in $ n^2$ colors so that all unit squares of any $ n\times n$ square are of different colors. It is known that any infinite row contains all the colors. Show that there is an infinite column containing exactly $ n$ colors.

[quote="probability1.01"]Hmmmm... I suspect that in fact every infinite column has exactly n colors...[/quote]

Umm... how then?


Foruma gönderdiğimiz sorular için konuşuyorum telif hakkı konusunda herhangi bir problem çıkmıyor mu?Ya da çıkmayacağını nereden biliyoruz?


Show that every natural number greater than 11 can be written as the sum of two natural numbers.

There is some lack of the word [b]composite[/b]  :wink:

http://www.mathlinks.ro/Forum/viewtopic.php?t=150909


.Let ABC be an equilateral triangle and P a point in its interior . Let X Y Z be the projection of P

onto BC,CA and AB ,respectively . Prove that the sum of the areas of the triangles BXP,CYP.ACP does not

depend on P

[hide] Construct three lines which pass through P and parallel to $ BC,CA,AB$[/hide]


geometry
perimeter
MATHCOUNTS
SFFT
special factorizations

[quote="bpms"][quote="nebula42"][quote="bpms"]The perimeter of a right triangle is 60. The altitude to the hypotenuse is 12. What is the area?

Guess and check works very well for this, but I want a way without guess and check. I wrote two equations with two variables, but when I tried to solve them things got messy. Any non-messy, non-guess and check ways?[/quote]
Guess and check is definitely useful for MathCounts problems like this one  :wink: , because the perimeter and altitude are both integers, so it's highly likely that the answer will also be an integer. (Also, you can notice that it doesn't say "Express your answer as a common fraction" or "Round to the nearest hundredth"  :P )

But if you wanted to solve it (sort of) algebraically, you could let $a$ and $b$ be the legs, and then you would have the two equations \[a+b+\sqrt{a^{2}+b^{2}}=60\]  \[\frac{ab}{\sqrt{a^{2}+b^{2}}}=12\] Then you can substitute $\sqrt{a^{2}+b^{2}}=\frac{ab}{12}$ into the first equation to obtain \[a+b+\frac{ab}{12}=60\] Multiply the equation by $12$ to get \[ab+12a+12b=720\] And then from there (if you assume $a$ and $b$ are integers), you can use Simon's Favorite Factoring Trick to get \[(a+12)(b+12)=864\] and go from there, using the fact that $a+b<a+b+\sqrt{a^{2}+b^{2}}=60$ to reduce the number of possibilities.[/quote]
Nice. I had manipulated it to
$(60-x)(60-y)=1800$
where x and y where the legs. I didn't think of doing the substitution the way you did. I had tried to isolate a variable and substitute but your way is better...[/quote]
Could you show how you got to $(60-a)(60-b)=1800$? If my equation and your equation both work, then it should be pretty easy to solve for $a$ and $b$.
And how is my way better? I don't think it is...  :roll:  :P

[hide]Well, we can call the lengths of the legs $a$ and $b$.

This means the hypotenuse is $60-a-b$.

So, since the altitude to the hypotenuse is 12, then the area of the triangle is $360-6a-6b$.

Also, the area is $\frac{ab}{2}$.

So equating the two, we get $(a+12)(b+12)=864$.

I still don't see how bpms got his equation.

But I can see how nebula42 got his...lol.[/hide]

[size=75][color=darkred]Fixed a typo.
-nebula42[/color][/size]

You know that $x+y+\sqrt{x^{2}+y^{2}}=60$.
From here the first thing that came to mind was to eliminate the square root.
$\sqrt{x^{2}+y^{2}}=60-(x+y)$
$x^{2}+y^{2}=3600+x^{2}+y^{2}+2xy-120x-120y$
Canceling gives.
$3600=-2xy+120x+120y$
Dividing by -2 gives
$-1800=xy-60x-60y$
Simon's favorite factoring trick
$(x-60)(y-60)=1800$
Which can be rearranged to
$(60-x)(60-y)=1800$
.
There.
I also got the equation
$(x+12)(x-12)(y+12)(y-12)=144^{2}$ if it is of any importance or if anyone cares I can show how I got that...

[quote="cj"][hide="my solution"]

I have three equations \[a+b+c=60\] (1) \[{a^{2}+b^{2}}=c^{2}\] (2) \[ab=12c\] (3)
From (2), we have \[{a^{2}+2ab+b^{2}}=c^{2}+2ab\] (4)
or \[{(a+b)}^{2}=c^{2}+24c\] (5)
From (1), we have \[a+b=60-c\] or \[{(a+b)}^{2}={(60-c)}^{2}\] substitute this to (5), we have \[{(60-c)}^{2}=c^{2}+24c\] Solve for c and 12 c will be the result
[/hide][/quote]
That is by far the best way. I think the lesson of this is that when you can write one variable in terms of two others, you don't have to... You can make another equation with the third variable...

I would use the area of the triangle. The hyp*altitude/2 is equal to the base*height/2, so the hyp*alt=base*ht. You need the area, and to find that you need the hyp. You can use the fact that the perimeter is 60, and the pythagorean theorum to find out.

[quote="nebula42"][/quote][quote="bpms"][quote="nebula42"][quote="bpms"]The perimeter of a right triangle is 60. The altitude to the hypotenuse is 12. What is the area?

Guess and check works very well for this, but I want a way without guess and check. I wrote two equations with two variables, but when I tried to solve them things got messy. Any non-messy, non-guess and check ways?[/quote]
Guess and check is definitely useful for MathCounts problems like this one  :wink: , because the perimeter and altitude are both integers, so it's highly likely that the answer will also be an integer. (Also, you can notice that it doesn't say "Express your answer as a common fraction" or "Round to the nearest hundredth"  :P )

But if you wanted to solve it (sort of) algebraically, you could let $a$ and $b$ be the legs, and then you would have the two equations \[a+b+\sqrt{a^{2}+b^{2}}=60\]  \[\frac{ab}{\sqrt{a^{2}+b^{2}}}=12\] Then you can substitute $\sqrt{a^{2}+b^{2}}=\frac{ab}{12}$ into the first equation to obtain \[a+b+\frac{ab}{12}=60\] Multiply the equation by $12$ to get \[ab+12a+12b=720\] And then from there (if you assume $a$ and $b$ are integers), you can use Simon's Favorite Factoring Trick to get \[(a+12)(b+12)=864\] and go from there, using the fact that $a+b<a+b+\sqrt{a^{2}+b^{2}}=60$ to reduce the number of possibilities.[/quote]
Nice. I had manipulated it to
$(60-x)(60-y)=1800$
where x and y where the legs. I didn't think of doing the substitution the way you did. I had tried to isolate a variable and substitute but your way is better...[/quote][quote="nebula42"]
Could you show how you got to $(60-a)(60-b)=1800$? If my equation and your equation both work, then it should be pretty easy to solve for $a$ and $b$.
And how is my way better? I don't think it is...  :roll:  :P[/quote]
How is it pretty easy to solve for a and b since both of the solutions work?

[quote="nebula42"][hide]Well, my equation said $ab+12(a+b)=720$, and your equation said $60(a+b)-ab=1800$. If we add them together, we'll get $72(a+b)=2520$, or $a+b=35$. From there, we know $\sqrt{a^{2}+b^{2}}=60-35=25$. Then you can immediately use this to find the area (using the fact that the altitude is $12$), and if you wanted to solve for $a$ and $b$, you would use these two equations that you found: \[a+b=35\]  \[a^{2}+b^{2}=625.\] From there, it should be simple enough (there are several different well-known methods that can be used to solve this kind of system).[/hide][/quote]
I just realized
[hide]You don't need to solve the system. The problem looks for the area. The area is $\frac{ab}{2}$. $(a+b)^{2}=35^{2}$. Subtracting from the second equation gives $2ab=600$. Diving by four gives you the answer, $150$[/hide]


Ok, so there's this certain Nats problem, about circles.
Basically it says.
What is the equation of a circle whose circumference passes through $ (2,3)$, $ (2,\minus{}3)$, and $ (\minus{}3, 2)$

[hide="BIG HINT"]
The vertices form a right triangle
[/hide]

Isn't there a formula for radius knowing three points?

I just saw it like 5 seconds ago, actually.

Oh, and, doesn't it have to be a right triangle? (I'm thinking about sth else...)

Edit: I meant equation, not radius...

[hide]
Well, when it is a right triangle, then the midpoint of the hypotenuse is the center.

Radius won't help for the center. (although if you must know, the formula is
$ R \equal{} \dfrac{abc}{4A}$, where$ A$ is the area of $ \bigtriangleup ABC$.[/hide]

I don't really know a formula for three points on a circle, but I can solve one (it may be a bit long) even if it's not a right triangle.

2,3 2,-3 -3,2
[hide]
If you use the equation of a circle (x - a)^2 + (y-b)^2 = r^2 (that's how I memorize it) you have three equations with three variables yay!

plugging in x and y with the co-ordinates:

(2 - a)^2 + (3-b)^2 = r^2

(2 - a)^2 + (-3-b)^2 = r^2

(-3 - a)^2+ (2-b)^2 = r^2

You can solve it, i'm just too lazy to solve today.][/hide]

[quote="academicsrule"]Isn't there a formula for equation knowing three points?

[/quote]

there is probably one, but I don't have the general formula in my head for equation, and I'd rather not memorize that

The perpendicular bisectors of any two chords within a circle intersect at the center... Or you just look for the point equidistant from all three by using the equations $ (x \minus{} 2)^2 \plus{} (y \minus{} 3)^2 \equal{} r^2, (x \minus{} 2)^2 \plus{} (y \plus{} 3)^2 \equal{} r^2, (x \plus{} 3)^2 \plus{} (y \minus{} 2)^2 \equal{} r^2$

Obvious that $ y \equal{} 0$... and $ x \equal{} 0$

edit: err. whoops  :blush:

[hide][quote="ZhangPeijin"]Ok, so there's this certain Nats problem, about circles.
Basically it says.
What is the equation of a circle whose circumference passes through $ (2,3)$, $ (2, \minus{} 3)$, and $ ( \minus{} 3, 2)$[/quote]

I remember doing this.Btw, it only gives you the lines used to form the corners of the triangle. I got x^2 + y^2=13. So just find the center. Using distance formula, and since each point is equidistant from the center, sqrt((x-2)^2 + (y-3)^2)=sqrt((x-2)^2 + (y+3)^2

square both sides, you get (x-2)^2+(y-3)^2=(x-2)^2+(y+3)^2

after taking away (x-2)^2 from both sides you get
 (y-3)^2=(y+3)^2
y^2-6y+9=y^2+6y+9
-6y=6y
12y=0
y=0.

So now insert the 0 into a new problem.

sqrt(x+3)^2+(0-2)^2=sqrt(x-2)^2+(0-3)^2

square both sides:  (x+3)^2+(0-2)^2=(x-2)^2+(0-3)^2
x^2+6x+13=x^2-4x+13.
6x=-4x
10x=0
x=0.

So the center is (0,0)

Take any point, find distance from there to center.

sqrt(2-0)^2 + (3-0)^2
sqrt(4+9)
sqrt 13. Since that is the distance, the equation for the circle is

x^2+y^2=13.[/hide]

Yeah and Alan, I'm pretty sure it's not a right triangle.


I went to my preferences and enabled HTML, but when I write a message, in the options is still says "HTML is OFF".  How do I get the HTML to work?

Html will never work on this board because we don't allow it.


superior algebra
superior algebra unsolved

Welcome to my problem:

If $\sigma, \ \tau \ \in \ S_n$ and $\sigma\tau = \tau\sigma$, then show that $\tau$ permutes those integers which are left fixed by $\sigma$.

I think we could use the fact that \[ \tau=\sigma\tau\sigma^{-1} = (\sigma(a_1) \ldots \sigma(a_k)) \ \text{if} \ \tau=(a_1 \ldots a_k)  \]
Thanks a lot for your help.

Mark.

Setting $\tau = \sigma$ gives an commuting counterexample...

So we shall just precise $\sigma \not= \tau$

Then take $\tau = \sigma^{-1}$  ;)
I don't think it's that easy to handle...


abstract algebra
superior algebra
superior algebra unsolved

let $ M$ and $ N$ be finite genrator $ R$ module,
can we deduse that $ HOM(M,N)$ is finitly generator too?

[quote="Third Edition"]It is not true in general. But it is true if, e.g., R is Noetherian.[/quote]

please give me a counterexample,if possible?

[quote="Third Edition"][quote="hamidreza"]please give me a counterexample,if possible?[/quote]What did you try out so far? Certainly you need to construct a counterexample over a non-Noetherian ring. Let's see what you have done. ;)[/quote]

I can not construct a counterexample.
please give me a counterexample,if possible?(with complete details)
Thanks  for your attention.

Lemma: If $ R$ is a ring with unit element and $ I$ is a
left ideal of $ R$  and $ M$ is an $ R$-module then
$ Hom_R(R/I,M) \cong_R Ann_M(I)$
Proof: Let $ g: Hom_R(R/I,M)\to Ann_M(I)$ such that $ g(f)\equal{}f(1\plus{}I)$ it is easy to verify g is a $ R$ isomorphisim.

Now let $ K\equal{}R[x_1,x_2,...]$ where $ R$ is the field of real numbers,
we put $ I\equal{}Kx_1\plus{}Kx_2\plus{}Kx_3\plus{}...$ and $ J\equal{}I^2$ so
By above lemma we have $ Hom_K(K/I,K/j) \cong_K Ann_K/J(I)\equal{}I$ but $ K/I$
and $ K/J$ are cyclic and $ I$ is not finitly generated. :)

 "Forgive him who wrongs you; join him who cuts you off; do well to him who does evil to you; and speak the truth even if it be against yourself."
                                                                                Prophet of Islam


calculus
integration
limit
function
geometry
geometric transformation
real analysis

It's True that:

${\int_{\alpha }^{\beta }f(x) \, dx=\int_{\alpha+\epsilon }^{\beta+\epsilon }f(x-\epsilon ) \, dx}$

So, we can have

${\int_{0}^{\infty }f(x) \, dx=\int_{\epsilon }^{\infty }f(x-\epsilon ) \, dx=\lim_{\epsilon \to \infty }\, \int_{\epsilon }^{\infty }f(x-\epsilon ) \, dx}$

but it is too much strange for me.

It's like a Paradox. a false one.

because it look like that the integral is zero. but this is false.
but i cant understant the behavior of the function when $\epsilon$ goes to infinity.

this is not a problem, it's just a annoying thing.

Part of the confusion, I think, is that the integral $\int_{\epsilon}^{\infty}f(x-\epsilon) \, dx$ is itself a limit; in other words, it is the limit

$\lim_{\delta \to \infty}\int_{\epsilon}^{\delta}f(x-\epsilon) \, dx$

So to take the lower bound to infinity as well is problematic (it's similar to finding $\lim_{(x, y) \to (a, b)}f(x, y)$ generally - if the limit is different when approaching $(a, b)$ along different "paths" then the limit does not exist).  The expression needs to be defined more tightly.

Welcome to the world of convergence theorems and classic counterexamples. What I mean by a convergence theorem is a theorem that states that if $f_{n}$ tends to $f$ (in some fashion) and if certain other conditions are met, then $\int f_{n}$ tends to $\int f$.

For instance, if $f_{n}$ tends to $f$ uniformly on $[a,b],$ then $\int_{a}^{b}f_{n}$ tends to $\int_{a}^{b}f.$

Two other famous theorems - I'll save their statements for later - are the Dominated Convergence Theorem and the Monotone Convergence Theorem.

But we need those certain other conditions.

Let's make your example concrete. Make $f$ a function whose graph is a nice easy "bump". Let $f$ be continuous and let it be zero outside of the set $[0,1].$ On $(0,1),$ it is positive, and just for convenience (multiplying by some constant if necessary), assume that

$\int_{-\infty}^{\infty}f(x)\,dx=\int_{0}^{1}f(x)\,dx=1.$

Now define $f_{b}(x)=f(x-b).$ This is a translation of the original function by $b,$ and $f_{b}$ is a simple bump, but on the interval $[b,b+1].$

It is true that $\int_{-\infty}^{\infty}f_{b}(x)\,dx= \int_{-\infty}^{\infty}f(x)\,dx=1.$

That's the observation about translation invariance that you made.

But for each $x,\ \lim_{b\to\infty}f_{b}(x)=0.$ (Wait at one place while the bump moves - sooner or later it has moved past and the value is back to zero.) Hence,

$\lim_{b\to\infty}\int_{-\infty}^{\infty}f_{b}(x)\,dx\ne \int_{\infty}^{\infty}\lim_{b\to\infty}f_{b}(x)\,dx$

After all, the left hand side is $1$ and the right hand side is $0.$

What are the Dominated Convergence Theorem and the Monotone Convergence Theorem? They are theorems that start from the hypothesis that $f_{n}(x)\to f(x)$ pointwise (actually, the condition is "pointwise almost everywhere") on some set $E.$ Then, we can conclude that $\int_{E}f_{n}\to\int_{E}f$ provided we also have:

1. In the case of the MCT, we must have that $0\le f_{1}\le f_{2}\le f_{3}\le\cdots,$ or

2. In the case of the DCT, we have one fixed function $g$ such that $|f_{n}(x)|\le g(x)$ for all $x$ and $\int_{E}g(x)\,dx$ is finite.

In the example above, the sequence of functions $f_{b}$ is certainly not monotone - for any fixed $x,$ we have $f_{b}(x)$ first increasing and then eventually decreasing. And the only "dominating function" $g$ we could build would be essentially a positive constant on $[0,\infty)$ - and that constant has an infinite integral. Hence neither the MCT nor he DCT apply, and indeed, the conclusion of such theorems fails for this example.

woow. Kent Every make excelent solutions.

$\int_{\epsilon }^{\infty }f(x-\epsilon ) \, dx$

If we make a substituition of $f=g'$ we get:

$g(\infty)-g(0)$  (I did not write the limit to become more simple.

but this equation is independent of $\epsilon$

we get the same result if
$\int_{\epsilon }^{2\epsilon }f(x-\epsilon ) \, dx = g(\epsilon)-g(0)$

So the interval $(\epsilon,2.\epsilon)$ has more than 1 value.

This is a false paradox because the integral seens to be

$\int_{\infty }^{\infty }f_{\infty }(x) \, dx = 0$
but this is false

but a better approach is difficult and long


We have 5 balls, such that any 4 of them can be divided into 2 groups (each containing 2 balls) with equal weight. Prove that all the balls have equal weight.

5 can be changed to 2009, or 23, or any odd number. :D

see here: http://www.mathlinks.ro/viewtopic.php?t=224724


what's wrong in this code? :S

[code]
\noindent
\begin{equation*}
\begin{align*}

\displaystyle \tan 12^\circ \cdot \tan 24^\circ \cdot \tan 48^\circ \cdot \tan 96^\circ &= \frac{\sin 12^\circ}{\cos 12^\circ}\cdot \frac{\sin 24^\circ}{\cos 24^\circ}\cdot \frac{\sin 48^\circ}{\cos 48^\circ}\cdot \frac{\sin 96^\circ}{\cos 96^\circ}\\
\\
& = \displaystyle \frac{(\sin 48^\circ)(\sin 12^\circ) \cdot (\sin 96^\circ)(\sin 24^\circ)}{(\cos 48^\circ)(\cos 12^\circ) \cdot (\cos 96^\circ)(\cos 24^\circ)}\\
\\
& = \displaystyle \frac{(\cos (48 - 12) - \cos (48 + 12))(\cos (96 - 24) - \cos (96+24))}{(\cos (48 - 12) + \cos (48 + 12))(\cos (96 - 24) + \cos (96+24))}\\
\\
& = \displaystyle \frac{(\cos 36^\circ - \cos 60^\circ)(\cos 72^\circ - \cos 120^\circ)}{(\cos 36^\circ + \cos 60^\circ)(\cos 72^\circ + \cos 120^\circ)}\\
\\
& = \displaystyle \frac{\left(\frac{1+\sqrt 5}{4} - \frac{1}{2}\right) \left(\frac{\sqrt 5 - 1}{4} + \frac{1}{2}\right)}{\left(\frac{1+\sqrt 5}{4} + \frac{1}{2}\right) \left(\frac{\sqrt 5 - 1}{4} - \frac{1}{2}\right)}\\
\\
& = \displaystyle \frac{\left(\frac{\sqrt 5 - 1}{4}\right) \left(\frac{\sqrt 5 + 1}{4}\right)}{\left(\frac{\sqrt 5 + 3}{4}\right) \left(\frac{\sqrt 5 - 3}{4}\right)}\\
\\
& = \displaystyle \frac{\left(\frac{\sqrt 5 - 1}{4}\right) \left(\frac{\sqrt 5 + 1}{4}\right)}{\left(\frac{\sqrt 5 + 3}{4}\right) \left(\frac{\sqrt 5 - 3}{4}\right)}\\
\\
& = \displaystyle \frac{(\sqrt 5 - 1)(\sqrt 5 + 1)}{(\sqrt 5 + 3)(\sqrt 5 - 3)} =  \frac{5-1}{5-9} = \frac{4}{-4} = -1

\end{align*}
\end{equation*}
[/code]

thanks

2 things:

1) You do not need \begin/\end{equation*}
2) You cannot skip lines (or use \par) between \begin{align*} and \end{align*}

Once you correct these two things, the code works without any error messages.

Also, as the align environment automatically uses display mode you don't need any of the \displaystyle commands. Similarly \noindent can be removed as it has no effect on align.


I know how to use it, but does anyone know why it works?

http://mathworld.wolfram.com/CrossProduct.html

The Shoelace Theorem is basically the sum of a whole bunch of cross products that correspond mathematically to the area of the polygon.


Solve the equation $ x^x \plus{}y^y \equal{}\overline{xy}\plus{}3$  where   $ \overline{xy}\equal{}10x\plus{}y$

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:

are $ x$ and $ y$ supposed to be digits?

If they are digits, we can test for every $ 0,1,2,..,9$. After all, we've just 10 cases.

[hide="My solution"]
$ \overline{xy}\plus{}3\leq 99\plus{}3\leq 102 \implies x^x\plus{}y^y\leq 102 \implies x,y\leq 3$.

Furthermore, $ 0^0$ is undefined so neither digit can be 0.

Case $ x\equal{}1$: $ 1\plus{}y^y\equal{}13\plus{}y\implies y^y\minus{}y\equal{}12\implies y\notin \mathbb N$.

Case $ x\equal{}2$: $ 4\plus{}y^y\equal{}23\plus{}y\implies y^y\minus{}y\equal{}19\implies y\notin\mathbb N$.

Case $ x\equal{}3$: $ 27\plus{}y^y\equal{}33\plus{}y\implies y^y\minus{}y\equal{}6\implies y\notin\mathbb N$.

So, there are no solutions in $ \mathbb N$.[/hide]


Prove that there exists an integer $n$, $n\geq 2002$, and $n$ distinct positive integers $a_1,a_2,\ldots,a_n$ such that the number $N= a_1^2a_2^2\cdots a_n^2 - 4(a_1^2+a_2^2+\cdots + a_n^2) $ is a perfect square.

Obviously you mean n>m for arbitrary m right?

Bomb

[quote="orl"]Prove that there exists an integer $ n$, $ n\geq 2002$, and $ n$ distinct positive integers $ a_1,a_2,\ldots,a_n$ such that the number $ N \equal{} a_1^2a_2^2\cdots a_n^2 \minus{} 4(a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2)$ is a perfect square.[/quote]
Does anyone have solution?

Let $m \geq 3$ be a positive integer. We first claim that if there exists some positive integer $S$ and distinct integers $a_1, a_2, \ldots, a_m$ such that
\[ a_1 a_2 \cdots a_m = a_1^2 + a_2^2 + \cdots + a_m^2 + S, \]
then there exists some positive integer $a_{m+1}$ distinct from each of $a_1, a_2, \ldots, a_m$ such that
\[ a_1 a_2 \cdots a_m a_{m+1} = a_1^2 + a_2^2 + \cdots + a_m^2 + a_{m+1}^2 + S-1. \]
Indeed, if we take $a_{m+1} = a_1 a_2 \cdots a_m - 1 > \max(a_1, a_2, \ldots, a_m)$, we get

\begin{align*}
a_1 a_2 \cdots a_m a_{m+1}
&= (a_1 a_2 \cdots a_m)^2 - a_1 a_2 \cdots a_m
= (a_1 a_2 \cdots a_m - 1)^2 + a_1 a_2 \cdots a_m - 1 \\
&= (a_1 a_2 \cdots a_m - 1)^2 + (a_1^2 + a_2^2 + \cdots + a_m^2 + S) - 1
= a_1^2 + a_2^2 + \cdots + a_m^2 + a_{m+1}^2 + S-1.
\end{align*}

We now observe that we can apply the claim repeatedly to find some $a_{m+1}, a_{m+2}, \ldots, a_{m+S-1}$ such that
\[ a_1 a_2 \cdots a_{m+S-1} = a_1^2 + a_2^2 + \cdots + a_{m+S-1}^2 + 1. \]

Now define $a_i = i$ for $1 \leq i \leq 2002$ and define
\[ S = a_1 a_2 \cdots a_{2002} - (a_1^2 + a_2^2 + \cdots + a_{2002}^2) = 2002! - (1^2 + 2^2 + \cdots + 2002^2) > 0. \]
Let $n=2002+S-1 \geq 2002$. By the above claim, we can find $a_{2003}, a_{2004}, \ldots, a_n$ with
\[ a_1 a_2 \cdots a_n = a_1^2 + a_2^2 + \cdots + a_{m+S-1}^2 + 1. \]
It follows that
\[ a_1^2 a_2^2 \cdots a_n^2 - 4(a_1^2 + a_2^2 + \cdots + a_n^2) = a_1^2 a_2^2 \cdots a_n^2 - 4a_1 a_2 \cdots a_n + 4 = (a_1 a_2 \cdots a_n - 2)^2 \]
is a perfect square, so our construction is complete.

We will find for $n = 3a^2 - 14 \ (a\ge 5)$, $a_1,...,a_n$ such that $N = (\prod_{i=1}^n a_i - 2a)^2$. Equating this with the original expression for $N$ we end up with the equation
$$\sum_{i=1}^n a_i^2 - 2a\prod_{i=1}^n x_i + 4a^2 = 0. \qquad \qquad (*)$$
Construct an infinite sequence of positive integers as follows: let $x_1 = ... = x_{n-2} = 1, x_{n-1} = 4, x_n =  a$, and
$$ \forall m \in \mathbb N, x_{n+m} = 2a\prod_{j=1}^{n-1} x_{j+m} - x_m.$$ Induction shows that $(x_{n+m})_{m\in \mathbb N}$ is strictly increasing and every $n$ consecutive terms of $(x_i)_{i\in \mathbb N}$ is a solution to $(*)$ (ignoring "pairwise distinct" constraint). However as $x_{n+1},...,x_{2n}$ are pairwise distinct, we can take $a_i = x_{n+i}$ to satisfy the original problem.


Solve the equation:
$ \sin 2x \plus{} \cos 5y \equal{} 2$ if $ x \equal{} E [0, \frac {\pi}{2} ]$ and $ y \equal{} E[0, \frac {\pi}{2}]$

Sorry, I can't find the latex code for the symbol that looks like E. HOpe you understand what I mean...

Uh.
[hide]$ \sin{2x} \equal{} 1 \Rightarrow x \equal{} \frac {\pi}{4}, \cos{5y} \equal{} 1 \Rightarrow y \equal{} 0$.[/hide]

Solve the equation:
$ \sin 2x \plus{} \cos 5y \equal{} 2$ if $ x \in [0, \frac {\pi}{2} ]$ and $ y \in[0, \frac {\pi}{2}]$

[hide="hinto"]Well i think davidkyo's answer is correct but just for kicks :

$ \sin{2x}+\cos{5y}=2\cdot\sin{(\frac{2x+5y}{2}+45})\cdot\sin({\frac{2x-5y}{2}-45})$

So we get that
$ \sin{(\frac{2x+5y}{2}+45})\cdot\sin({\frac{2x-5y}{2}-45})=1$[/hide]

For $ y$, my answer sheet says $ \frac{2\pi}{5}$.

[quote="t0rajir0u"][quote="sonny"][hide="hinto"]Well i think davidkyo's answer is correct but just for kicks :

$ \sin{2x} + \cos{5y} = 2\cdot\sin{(\frac {2x + 5y}{2} + 45})\cdot\sin({\frac {2x - 5y}{2} - 45})$

So we get that
$ \sin{(\frac {2x + 5y}{2} + 45})\cdot\sin({\frac {2x - 5y}{2} - 45}) = 1$[/hide][/quote]

Not really useful.  Either way, you will have to use the fact that $ \sin, \cos$ have a maximum of $ 1$; using your method, you'll have to solve a system of linear equations, so you actually end up doing two extra calculations that cancel each other out (as opposed to davidyko's answer, which is straightforward).

[quote="jeez123"]For $ y$, my answer sheet says $ \frac {2\pi}{5}$.[/quote]

Both values of $ y$ work (plug them in).[/quote]

Ya my way is definitely much lenghtier and harder to work with. But since jeez123 posted the question again after davidkyo posted a straigthfordward solution I just thought of another way of solving the problem. I am sorry if my way didn't help jeez123.


MATHCOUNTS AMC USA(J)MO USAMO geometry ratio

Create a list of top things intellecutalish at competitions set by people from MO:

For example:

Highest MC Individual: 3rd (Mario DiDonna, 2002)

Add on!

Well why don't we start a competition. Let's have a game shall we, our half of the town Palo Alto that goes to Gunn High School versus the whole state of Missouri. Population is around 25000, for reference, no noteworthy private schools in the vicinity. Also I will use my veto to disqualify mathcounts achievements, because they clearly take much more luck than skill.
USAMO winners:
Rishi Gupta
Inna Zakharevich
USAMO honorable mention:
Rishi Gupta
Richard Ho

Yeah one MO person did IMO too, another one made red MOP.

You're not being consistent, you didn't name anyone. Right now the score is:
IMO: Gunn to Missouri 2-0
Winner: Gunn to Missouri 2-0
HM: Gunn to Missouri 2-0

[quote="pianoforte"]KS...

IMO: Gunn to Kansas 2-1 (C.Bosley)
Winner: Gunn to Kansas 2-1? (C.Bosley?)
Honorable Mention: Gunn to Kansas 2-3 (Jeffrey Amos, Nicholas Ma, C.Bosley)

Highest MC Individual: MO to KS 3rd-1st (C.Bosley)

MOSP: KS had 4 (C.Bosley, D.Zhang, Jeffrey Amos, Nicholas Ma)

(and dude, Mario Didonna is effin tall, perhaps tallest math person ever?)[/quote]
Darn you. Okay well
real MOP:
Taylor Han (lol)
the 3 listed before
Darn I cannot remember the other one, which I am pretty sure existed.

So MOP: KS v. Gunn, 4-4 unless you counted red, which would be cheating, then it would be 4-6.

red MOP:
Kevin Yang
John Boyle (at the time he was going to Paly, but he lives in the Gunn area)

Oh man I think Missouri got pwned, still 0 in every category :O.
Darnit I need to like, get so lucky on USAMO that I get HM, so Gunn can be tied with Kansas. That would be pretty funny, getting HM then going to MOP and being like hi I fail at math.

Are we counting multiple times as one person or multiple?

Fine, 3 real moppers, 1 red mopper. (3-4, 4-6 KS to Gunn)

Hmmm good question...well until Missouri finds a person and needs to rely on their multiple times making MOP, then we'll leave the scoreboard the way it is.

[quote="Mewto55555"]
I will appreciate the validity of that statement when you have a silver+ medal at IMO.
[/quote]

Do unofficial silvers count?

[quote="Temperal"]nebraska has had several thousand moppers throughout its history.[/quote]

And that is really quite believable due to:

40 or so USAMO tests
50 MOSP positions POST-EXPANSION (i.e. 8 years)
and 25 MOSP positions PRE-EXPANSION (i.e. 30 years)

total: 750+400=1150.

several thousand implies over 2000.

FAIL temperal again.

[quote="ZzZzZzZzZzZz"]Matt Ince: 2003 MOP, 2004 IMO silver medalist with a 31.
Nate Ince: 2003, 2004 MOP

WHAT NOW?!  :agent:[/quote]
Okay so the scoreboard is now:
(Gunn-Kansas-Missouri)
IMO:
2-1-1
Winner:
2-1-1
Honorable Mention:
2-3-1
real MOP:
4-3-1
fake MOP:
2-1-1

Man mr. zzzzzzzzzzzzzz (however many there are, what a stupid username), I do not see your motivation for saying "WHAT NOW", when an area with just 20000 people is pwning your entire state by far in every category. Your neighboring state is also completely pwning your entire state.

Man do you guys want to switch Gunn to San Jose? San Jose has gasp 500000 people, and about like 5 times the figures up there.

@hunter: Wth why did you actually change your name to Emily?  And no comment about the enhancements.

Does my being a mod count as an achievement?

I AM A GIRL! DO YOU THINK THEY COULD PUT ME IN FOR CGMO?

guys guys, we should turn you all into girls and then

usa

can like pwn cgmo

edit: david yang is sufficiently young that i think if we start giving him estrogen treatments, the results might be quite successful.

just like that guy in china who uh...ate too much kfc

my (then) 92 year old grandpa ate kfc in china. who are you talking about gnaw?

[quote="serialk11r"][quote="Mewto55555"][quote="serialk11r"]Hmmm Mewto you probably aren't aware of this but in the near future it is a possibility that CGMO will be referred to unofficially as white MOP. I think I will start to propagate that now. So you're a girl?[/quote]

Uh I'm afraid I cannot comment on that. You know, top secret and whatnot...[/quote]
Oh I completely understand. I suppose I won't be disclosing any of your private medical information, especially not ones regarding certain operations and hormone treatment.[/quote]

Erm no operations were needed....hey how do you know my secret medical info?

[quote="Mewto55555"][quote="pythag011"][quote="ZzZzZzZzZzZz"]Matt Ince: 2003 MOP, 2004 IMO silver medalist with a 31.
Nate Ince: 2003, 2004 MOP

WHAT NOW?!  :agent:[/quote]

2004 was an extremely trivial imo, so that 31 is fail.[/quote]

I will appreciate the validity of that statement when you have a silver+ medal at IMO.

tytia

PS George Yu made red MOP in 07.[/quote]

IT SHOULD BE NOTED THAT I APPRECIATE THE VALIDITY OF MR. PYTHAG'S STATEMENT.

Victor Wang is Missouri's first IMO gold medalist!  I think, at least, that this is the case.

Yay Missouri!  Yay Victor Wang!

Anyways, what did Max Schindler get on I[s]M[/s]LO?

The same ranking as what James Tao got at IMO.


Let $f: \mathbb{N}\to \mathbb{N}$ and
\[f(n) = \left\{ \begin{gathered}k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, n = 2k \hfill \\ 6k+4\,\,\,\,\,\,\,n = 2k+1 \hfill \\ \end{gathered}\right. \]
Prove
\[\forall n \in \mathbb{N},\exists m \in \mathbb{N},\underbrace{f(f( \cdots f(n) \cdots )) = 1.}_{m{\text{ times}}}\]

we can rewrite your problem like

${f(n)=\begin{cases}n/2 & n \bmod 2=0 \\ 3 n+1 & n \bmod 2=1 \end{cases}}$

since this is collatz conjecture. and since it is still a conjecture, no one here will be able to proove that.

since in more than 70 years no one prooved it.

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

[quote]since this is collatz conjecture. and since it is still a conjecture, no one here will be able to proove that.[/quote] why so may be in the futur :huh:


inequalities
triangle inequality
geometry
perpendicular bisector
geometry proposed

Find the largest value of n, such that there exists a polygon with n sides, 2 adjacent sides of length 1, and all his diagonals have an integer length.

Let $ V_1, V_2, V_3 \cdots V_n$ be the vertices of the polygon.

WLOG $ V_1V_2 \equal{} V_2V_3 \equal{} 1$ then by triangle inequality on $ V_1V_2V_3$, $ V_1V_3 < V_1V_2 \plus{} V_2V_3$

If $ n\ge 4$ then $ V_1V_3 \equal{} 1$ then $ V_1V_2V_3$ is an equilateral triangle.

Suppose that $ n \ge 6$ then $ V_1V_4$ and $ V_2V_4$ are diagonals then

triangle inequality on triangle $ V_1V_2V_4$ implies $ V_1V_4 \equal{} V_2V_4$

Similarly $ V_1V_5 \equal{} V_2V_5$ however $ V_3$, $ V_4$ and $ V_5$ are on the perpendicular bisector of $ V_1V_2$ Contradiction, therefore $ n \le 5$

and similarly with $ V_2V_5 \equal{} V_3V_5$ to get the example for $ n \equal{} 5$ and concave polygon

if you add the convex condition $ n \equal{} 4$ is the maximum.


11. A line with slope 3 intersects a line with slope 5 at the point (10, 15). What is
the distance between the x-intercepts of these two lines?

$ (A) 2 \qquad (B) 5 \qquad (C) 7 \qquad (D) 12 \qquad (E) 20$.

We have a slope and a point for each line.
[hide]The first line is $ y\minus{}15\equal{}3*(x\minus{}10)$, or in slope-intercept form, $ y\equal{}3x\minus{}15$. We have our first intercept at -15. Repeat for the second line. $ y\minus{}15\equal{}5*(x\minus{}10)$, or $ y\equal{}5x\minus{}35$. Our second intercept is at -35. The actual difference between them is [b]20, [size=150]E[/size][/b][/hide]


$a,b,c>0$ , prove that

$2a^2+8b^2+5c^2\geq4ab+4ac+4bc$

[hide]
well , I use factoring method to get $(a-2b)^2+(2c-a)^2+(2b-c)^2\geq 0$ . Any other method without factorising ?
[/hide]

[quote="shyong"]$2a^2+8b^2+5c^2\geq4ab+4ac+4bc$[/quote]
$4b^2+a^2\ge4ab$
$4b^2+c^2\ge4bc$
$4c^2+a^2\ge4ac$

I think it's pretty straightforward? The ineq is even strict...

[quote="Peter VDD"][quote="arqady"]Prove that $2a^2+8b^2+2\frac{2}{3}c^2\geq4ab+4ac+4bc.$ :)[/quote]a=b=c=1 --> 11+1/3 < 12, so that is wrong...[/quote]

No, it holds for $a=b=c=1.$  I think you misunderstood that $2\frac 23$  means $2+\frac 23$ and not $\frac 43.$

[quote="shyong"]$a,b,c>0$ , prove that $2a^2+8b^2+5c^2\geq4ab+4ac+4bc$[/quote]
If $a,b,c>0$ I think we can even say $2a^2+8b^2+5c^2 > 4ab+4ac+4bc$ (strictly)

[quote="arqady"][quote="arqady"]Prove that $2a^2+8b^2+2\frac{2}{3}c^2\geq4ab+4ac+4bc.$ :)[/quote]
Is wrong: $a=4,b=1,c=3$
The next  is true:
Prove that $2a^2+8b^2+4\frac{2}{3}c^2\geq4ab+4ac+4bc.$[/quote]

It is equivalent to

$2(a-b-c)^2+6(b-\frac{2c}{3})^2\geq 0$ . The equality holds when $a=5k,b=2k,c=3k$ for all integer $k$
 :)

And I think that the inequality i give doesnt even need to be positive  :D

[quote="arqady"][quote="shyong"]
$2(a-b-c)^2+6(b-\frac{2c}{3})^2\geq 0$ . The equality holds when $a=b=c=0$
 :)
[/quote]
The equality holds when $a=5t,b=2t,c=3t,t\in\mathbb R.$ :D[/quote]

 :D  I editted just after you post  :P


geometry
advanced fields
advanced fields unsolved

Not really.. They are two cylindrical surfaces having the $(Oz,(Oy$ lines as their axes (respectively).


function
calculus
derivative
calculus open

The function
\[f(x)=\displaystyle\sum_{n=0}^{+\infty}\frac{sin 3^nx}{2^n}
\]
is continuous everywhere but has derivative nowhere.
Find the extreme values of  $f(x)$  on the closed interval $ [0,1] $

i) Inf(f(x))=f(0)=0. The idea of the proof is as follows:
Let An(x) the general term of the series; for n=1,2,...
f(x)>=A0(x)+...+An(x)-2^(-n) ; the right hand is positive except for small values of x; for these values the sinus are quasi linear and
A0(x)+...+An(x)~((3/2)^(n+1)-1)*2*x; then the right hand is positive except if x<Bn+o(Bn),  where Bn=1/(3^(n+1)-2^(n+1)); limit(Bn)=0 therefore f(x)>0 for all x>0.
ii) f(Pi/4)=sqrt(2) (easy). Numeric calculations with Maple "show" that
sup(f(x))=sqrt(2) and is obtained only in Pi/4.
The series An is uniformly convergent towards f; thus sup(f) is the limit
of sup(A0(x)+...+An(x)). This last bound is obtained in  a single point Xn; it remains to be proven that limit(Xn)=Pi/4.
   I don't know how to solve this question..


The day before yesterday, I was 17 years old. The next year I will turn 20. How is this possible?

Thats a easy one ,
let the others try it out .But its a good one for school students to engross them in dates :)

[quote="Shogia"][hide="solution"]Today is the 1st of january and your birthday is on December 31st[/hide][/quote]Yep, that is right!


function
real analysis
real analysis unsolved

Suppose that f, f' is continuous on [a, b], and f'' exists on (a, b). If f(a) = 0, f(b) = 1 and f(c) > 1 where a < c < b. Prove that there is at least one value d between a and b for which f''(d) < 0.
How would I do this?  Thanks in advance.

You probably have learned or proved a theorem stating that a continuous function takes its maximum on a closed and bounded interval; if you haven't proved it it's not very hard and is a good exercise with the sequential criterion for limits and suprema of sets. Anyhow, let d equal the maximum value of f(x) on (a,b), and you should be able to prove that f'(d) = 0 and f''(d) < 0.

Unfortunately, that does not work: consider the function $ f(x) \equal{} \frac {16 \minus{} (3x \minus{} 2)^4}{15}$ on $ [0,1]$. It attains its maximum at $ d \equal{} 2/3$, but $ f''(2/3) \equal{} 0$.

Use the mean value theorem a lot:
[hide]First, there exists $ u\in (a,c)$ such that
\[ f(c) \minus{} f(a) \equal{} (c \minus{} a)f^\prime (u)
\]
and $ v\in (c,b)$ such that
\[ f(b) \minus{} f(c) \equal{} (b \minus{} c)f^\prime (v)
\]
Since $ c \minus{} a,b \minus{} c$ positive, $ f(c) \minus{} f(a)$ positive, and $ f(b) \minus{} f(c)$ negative, we have $ f^\prime (u)$ positive and $ f^\prime (v)$ negative. Also, $ a < u < c < v < b$. In particular, $ u < v$.

Use Mean Value Theorem one more time to get the existence of $ d\in (u,v)$ such that
\[ f^\prime (v) \minus{} f^\prime (u) \equal{} (v \minus{} u)f^{\prime \prime }(d)
\]
and this makes $ f^{\prime \prime }(d)$ negative.[/hide]

It's also possible to use the idea of [b]hadjimurad[/b]: Let $ g(x)\equal{}f(x)\plus{}\epsilon (x\minus{}a)(x\minus{}b)$, where $ 0<\epsilon<\frac{f(c)\minus{}1}{(c\minus{}a)(b\minus{}c)}$. The function $ g$ satisfies the same assumptions as $ f$. It attains its maximum on $ [a,b]$ at a point $ d\in (a,b)$, where $ g''(d)\le 0$. But $ g''(x)\equal{}f''(x)\plus{}2\epsilon$, hence $ f''(d)\le \minus{}2\epsilon<0$.

This also gives a (sharp) estimate $ f''(d)\le \minus{}2\frac{f(c)\minus{}1}{(c\minus{}a)(b\minus{}c)}$. When $ c$ is close to an endpoint, even a modest excess of $ f(c)$ over $ 1$ forces a huge negative value of $ f''$.


What do you mean by $ \minus{}\frac {1}{2}$ hours?

I play to whenever my parents call me to stop  :D which is normally about $ 45$ minutes to $ 60$ minutes

Define "play".  Does staying in the lobby and chatting count as playing?

If not, about 3 minutes (1 or 2 unrated games).

uhh... i come on once a month... but 30 minutes? if you dont count waiting for a game in the lobby...

almost never??
lol but when i do (usually in the morning when no one is on), about 5 minutes...

I don't play every day but when I do I usually play for at least an hour.. It's really addicting.

About 1 and 1/2 hours a day, from 7 to 8:30 pm. ET
I used to do like 1/2 a hour as a guest, but I got bored.

[quote="isabella2296"]Maybe 4-6 hours a day?[/quote]
That's a lot! Are you addicted?

Hm, when I did Mathcounts, I used to go on FTW for like 20ish minutes per day for some days, but not as much anymore since I'm now in high school.

[quote="Mewto55555"]I pop in every now and then to appease my adoring fans  :wink:[/quote]

You mean you talk to yourself?

And yes, I'm addicted. :D

[quote="isabella2296"][quote="Mewto55555"]I pop in every now and then to appease my adoring fans  :wink:[/quote]

You mean you talk to yourself?
[/quote]

Wow you could be the next JBL!

Last year: 1 hour
Now: Almost never (if I do, 5 minutes)

nvm wut i said earlier..

after i posted it i started going on more..

now-- 3 hrs??  :lol:

I go on for about 3-4 hours, but mostly as a guest since I keep getting flashies, not being able to answer, huge lag, etc.

I usually go on for about 30 minutes-1 hr. 30 min. and sometimes 2 hours.

[quote="Dillinger"]Eggylv.999 Do you play maplestory? Your avatar looks like a monster at Easter event.[/quote]

Wrong place Dilinger, but I see that you are new to the forums. You can always send a PM to eggy about this.

Depends. Used to be on 12+ hours, but cut down to about 8-9. (most is afk though)


Let $ ABCD$ be a convex quadrilateral. Square $ AB_1A_2B$ is constructed such that the two vertices $ A_2,B_1$ is located outside $ ABCD$. Similarly, we construct squares $ BC_1B_2C$, $ CD_1C_2D$, $ DA_1D_2A$. Let $ K$ be the intersection of $ AA_2$ and $ BB_1$, $ L$ be the intersection of $ BB_2$ and $ CC_1$, $ M$ be the intersection of $ CC_2$ and $ DD_1$, and $ N$ be the intersection of $ DD_2$ and $ AA_1$. Prove that $ KM$ is perpendicular to $ LN$.

This is a very well know theorem of Van Aubel. The segments quoted are not only perpendicular, but also of the same lenght.

In the page of Dick Klingens, http://www.pandd.demon.nl, clicking in the link Meetkunde (Geometry in Dutch) it is possible see the proof and an Cabri Applet of this theorem of Van Aubel (BTW, van Obel is incorrect spelling of the name).
Also in the british journal The mathematical Gazette, vol.90, number 517, march 2006, pp.2-12, one can find very nice extensions of this result, by John R. Silvester.


algorithm
algebra
polynomial
Rational Root Theorem
advanced fields
advanced fields unsolved

Is there a pure algebraic algorithm decomposit polynomial? (one variable polynomial Z[x])
There is an "analytic one", finding all the roots and try each "possible" combination.

Try every possible pair of factors of the constant coefficient and/or the top coefficient as the constant coefficients and/or top coefficients of the two factors together with every possible pair of degrees, then solve systems of equations for the other coefficients.

Rational root theorem is essentially this but gives it as an easy-to-implement strategy for finding individual rational roots.

Thanks for answering but I didn't mean to find a rational root but a factor of that polynomial.
Your solution seem to be an infinit process at first sight. for example, when you want to try $ a_0\equal{}c_1c_2$ then eq $ a_1\equal{}xc_1\plus{}yc_2$ already has infinitly many solutions, if there is one.

The slicker way to do this is to generalize the idea behind Eisenstein's criterion:  factor the polynomial modulo small primes (which is a finite computation) and put together the resulting information via CRT.  This is assuming you want to find the irreducible factors over $ \mathbb{Z}[x]$.  To be more precise, for any fixed polynomial there exists $ n$ such that information about the factorization of the polynomial modulo the first $ n$ primes uniquely determines its factorization over $ \mathbb{Z}$.

[b]Edit:[/b]  Apparently finite fields are in fact the way to go, but there are [url=http://en.wikipedia.org/wiki/Berlekamp%27s_algorithm]better[/url] [url=http://en.wikipedia.org/wiki/Cantor%E2%80%93Zassenhaus_algorithm]standard[/url] algorithms.

Thanks.
Is there any prescribed boundary for the prime one has to try? I mean if one consistently get non-trivial factors with each prime. Or, can it be proved that for sufficiently large prime, factors of poly $ f$ on $ F_p$ "coinside" with factors of $ f$ on Z[x]? I heard that there exist irreducible poly that can factor on "many" $ F_p$.

[quote="mathiq"]I heard that there exist irreducible poly that can factor on "many" $ F_p$.[/quote]
This is true; for example, the polynomial $ x^4 \plus{} 16$ is irreducible over $ \mathbb{Z}$ but factors modulo every prime.  The point is that knowing the factorization modulo several primes uniquely determines the possible, say, constant terms of each irreducible factor, since the constant factors must still multiply to a fixed constant (the constant factor of the original polynomial).  I believe - although I haven't thought about it particularly hard - that the non-constant terms should be similarly restrained.


factorial
limit
integration
calculus
algebra
partial fractions

Prove that, for any real number z and integer n:

$ \sum_{k \equal{} 0}^{n} \binom{n}{k} ( \minus{} 1)^k \frac {1}{z \plus{} k \plus{} 1} \equal{} \frac {n!}{(z \plus{} 1)(z \plus{} 2)...(z \plus{} n \plus{} 1)}$

[hide="Hint"] What is $ \lim_{z \to \minus{}k\minus{}1} \frac{n! (z  \plus{} k \plus{} 1)}{(z \plus{} 1)...(z \plus{} n \plus{} 1)}$? [/hide]

hmm, how would you solve it that way?

The two ways I solved it by were:

[hide="1"]
Perform partial fractions on the RHS, and note that for the $ \frac{1}{z\plus{}k\plus{}1}$ term the coefficient would be $ (\minus{}1)^k \binom{n}{k}$
[/hide]
[hide="2"]
Noting that the LHS is $ \int^1_0 t^{z\minus{}1} (1\minus{}t)^n dt$ and then using integration by parts.
[/hide]

It's just a nice way to do partial fractions.


modular arithmetic
quadratics
number theory
superior algebra
superior algebra unsolved

sir
  In number theory, if one of the primitive root of prime number is  known, then it is possible to find all other primitive root using given root of that prime number.

 If 3 is the primitive root of mod(43) then it is possible to find all the primitive root of mod (43) using 3?

   clear my doubt...........

Stop posting completely different problems into your older threads! That was already the second time.

For the problem: what are the generators of $ \mathbb Z/n$ (with addition)¿

If $ g$ is a primitive root, then it's easy to show that $ g^{\dfrac{p\minus{}1}{2}} \equiv \minus{}1 \pmod p$, so it's easy to deduce that all the primitive roots are of the form $ g^{2k\plus{}1}$, with $ k$ a positive integer, and there are exactly $ \dfrac{p\minus{}1}{2}$ distinct primitive roots modulo $ p$.

Would you please explain me, ZetaX, which part isn't true.
It's clear that $ x^2 \equiv 1 \pmod p \longrightarrow x \equiv 1$ or $ \minus{}1 \pmod p$. Then, by the definition of $ g$, $ g^{\dfrac{p\minus{}1}{2}}$ must be $ \minus{}1 \pmod p$.
Look that if we change $ g$ by $ x\equal{}g^{2k\plus{}1}$, we'll have $ x^{\dfrac{p\minus{}1}{2}} \equiv (\minus{}1)^{2k\plus{}1} \equiv \minus{}1$, which implies that $ g^{2k\plus{}1}$ must be a primitive root.

$ a^{\frac{p\minus{}1}{2}} \equiv \minus{}1 \mod p$ doesn't imply that $ a$ is a primitive root, but only that it is a quadratic nonresidue.

Only Fermat prime (prime of the form $ 2^{2^n}\plus{}1$ satisfy all quadratic nonresidues are primitive mod $ p$ .

Please, see this:
$ a$ is a primitive root modulo $ p$ if the $ h$ for which $ a^h \equiv 1 \pmod p$ is satisfied is a multiple of $ p \minus{} 1$. Then $ a^{\dfrac{p \minus{} 1}{2}}$ is $ 1$ or $ \minus{} 1$ modulo $ p$, but it cannot be $ 1$, because would contradict $ a$ as a primitive root modulo $ p$; so it's $ \minus{} 1 \pmod p$.
I'm getting in trouble, please clarify me!
ZetaX, what I though was that all non-quadratic residues modulo $ p$ were too primitive roots.

A very trivial example ,let consider $ p\equal{}4k\plus{}3$ and $ a\equal{}p\minus{}1$ ,it is not hard to see that
$ (p\minus{}1)^{\frac{p\minus{}1}{2}}\equiv \minus{}1 (\mod p)$ ,although it can't be a primitive root mod p because
$ (p\minus{}1)^2\equiv 1 (\mod p)$
As I has said that ,all n has this property are Fermat's prime.

Yes your proof works fine for the quadratic nonresidues, and in fact the nonresidues are, as you say, exactly thos of type $ a^{2k \plus{} 1}$. So your proof is fine to that point.
But primitive roots are those that generate the full multiplicative group (or equivalently $ 1,a,a^2,...$ are all nonzero residues $ \mod p$), whereas quadratic nonresidues are only those that are not squares $ \mod p$. For example, $ 3$ is a primitive root $ \mod 7$, but $ 3^3 \equiv \minus{} 1 \mod 7$ isn't. So it's just the very last conclusion (nonresidue -> prim. root) that is wrong; in fact, it's the other way around (prim. root -> nonresidue).


The arithmetic mean of 7, 2, $ x$ and 10 is 9. What is the value of $ x$?

If the mean of 7,2,x, and 10 is 9, then $ \frac{7\plus{}2\plus{}x\plus{}10}{4}\equal{}9$, so $ x\plus{}19\equal{}36$, so $ x\equal{}\boxed{17}$.
The way I do it (harder to explain) is:
7-9=-2
2-9=-7
10-9=1
and -2, -7, and 1 add to -8, so x is 9-(-8)=17

Explaining aleph0's method slightly more, if you take the differences between each number and the average, they must all add up to 0. the difference between 7 and 9 is -2, between 2 and 9 is -7, between 10 and 9 is 1, so the difference between x and 9 should be -(-7-2+1) which is 8. 8+9 is 17 so x is 17.


Dear Members,
                     How would you convert the speed of 186,300 miles in metric whose constant velocity run in 3 parsec?

                                               Sincerly,
                                             Mikhail S.

Maybe this will help $1 \ \text{Parsec}=3.08568025 \cdot 10^{16}\text{meters}$.


im studying for my next exam and i can not find an answer of the cell membrane architecture is made of, im confuse between lipid and proteins building blocks or lipid and protein with glycolipids and cholesterol?, or it is an oil film of lipid only?

I believe the fluid mosaic model is generally accepted that the structure of the membrane. It consists of a phospholipid bilayer with hydrophilic heads and hydrophobic tails. This bilayer has embedded proteins (some glycoproteins, some transport proteins, some ion channels, etc.) and also cholesterol helps to maintain the integrity of the membrane. There are also cell-to-cell interactions between extracellular matrices.


What is the ordered pair of real numbers $ (x, y)$ which satisfies the equation $ |x\plus{} y\minus{}7|\plus{} |4x \minus{} y\plus{} 12|\equal{} 0$?

Since absolute values must be $ \geq 0$, we have $ x \plus{} y \minus{} 7 \equal{} 0 \iff \boxed{x\plus{}y\equal{}7}$ and $ 4x\minus{}y\plus{}12\equal{}0 \iff \boxed{4x\minus{}y\equal{}\minus{}12}$. Adding the equations gives $ 5x\equal{}\minus{}5 \iff x\equal{}\minus{}1$, and plugging this back in gives $ y\equal{}8$, so the solution is $ \boxed{(\minus{}1,8)}$.

but the answer says (-18)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

btw, my location is the best out of all of yours

[color=#FF0000][size=89]no, it's not[/size][/color]


function
number theory unsolved
number theory

we say two positive integer are amicable if $\phi (n)=\phi (m)=m+n$ find all a such that a and a+1 are amicable.

What is this strange function? Is it the sum of the divisors?

[url=http://www.kalva.demon.co.uk/putnam/psoln/psol7612.html]Here[/url]'s a proof of the fact that if $\sigma(a)=2a+1$ then $a$ is $k^2$, for odd $k$.

You can also see there why when $\sigma(n)$ is odd, $n=2^tu^2$ for some odd $u$. $a+1$ must have this form, since $\sigma(a+1)=2a+1$, so $a=k^2,a+1=2^tu^2$, with odd $k,u$. Since $k^2+1$ can't be a multiple of $4$, it means that $2^t\in\{1,2\}$. If it's $1$, we get $u^2-k^2=1$, and this can never happen with positive integers $k,u$. When $2^t=2$, we get $k^2-2u^2=-1$, so we're looking for solutions among the solutions of this negative Pell equation. I don't really know how to continue right now..

it might help:
you can have all the solutions of the eq $x^2-2y^2=-1$ working with in $\mathbb{Z}[\sqrt{2}]$ with the norm $N(u+\sqrt{2}v)=u^2-2v^2$  (using the multiplicity
of the norm ie $N(z\cdot w)=N(z)N(w)$)
in fact, if we have a solution of $x^2-2y^2=-1$, $x+y\sqrt{2}$  has  norm -1, since $1+\sqrt{2}$ has norm $-1$, $(x+\sqrt{2}y)(1+\sqrt{2})$ has norm 1 that is $(x+2y,x+y)$ is a solution of the pell eq $u^2-2v^2=1$ which are well known. (also $(x-2y,y-x)$ is a sol.).


geometry
3D geometry
prism
sphere
tetrahedron
symmetry
perpendicular bisector

A regular polyhedron is a polyhedron that is convex and all of its faces are regular polygons. We call a regular polhedron a "[i]Choombam[/i]" iff none of its faces are triangles.
a) prove that each choombam can be inscribed in a sphere.
b) Prove that faces of each choombam are polygons of at most 3 kinds. (i.e. there is a set $\{m,n,q\}$ that each face of a choombam is $n$-gon or $m$-gon or $q$-gon.)
c) Prove that there is only one choombam that its faces are pentagon and hexagon. (Soccer ball)
[img]http://aycu08.webshots.com/image/5367/2001362702285797426_rs.jpg[/img]
d) For $n>3$, a prism that its faces are 2 regular $n$-gons and $n$ squares, is a choombam. Prove that except these choombams there are finitely many choombams.

(a)

For a face $\mathcal F$ of our polyhedron, let $\ell(\mathcal F)$ be the perpendicularto the plane of $\mathcal F$ passing through the center of $\mathcal F$. In order to prove that the polyhedron can be inscribed in a sphere, it suffices to prove that for every vertex $A$ of the polyhedron, all $\ell(\mathcal F)$ as $\mathcal F$ ranges through the faces containing $A$ intersect.

The angles of a face with $n$ sides are $\frac{n-2}n\pi\ge\frac\pi 2$, and the angles at $A$ of the faces containing $A$ must sum up to less than $2\pi$. This means that we can never have at least four faces meeting in $A$, or, in other words, for every vertex $A$ of the polyhedron, exactly three faces meet in $A$. If the three edges passing through $A$ are $AA_{1},AA_{2},AA_{3}$ and the faces containing $A$ are $\mathcal F_{1},\mathcal F_{2},\mathcal F_{3}$, contained in the planes $(AA_{2}A_{3}),(AA_{3}A_{1}),(AA_{1}A_{2})$ respectively, then $\ell_{\mathcal F_{i}}$ intersect simply because a tetrahedron (in this case $AA_{1}A_{2}A_{3}$) can be inscribed in a sphere.

(b)

Start placing the faces of our polyhedron as follows: place the first one arbitrarily, then one which is adjacent to the first one (“adjacent” meaning that they share a side), and then, from the third step onwards, each new face must have a vertex in common with two faces which have already been placed. Suppose in the current step we are placing down the face $\mathcal F$, adjacent to two previously posted faces $\mathcal F_{1},\mathcal F_{2}$ (which share a side). If $\mathcal F_{1},\mathcal F_{2}$ are the first two faces, then $\mathcal F_{1},\mathcal F_{2},\mathcal F$ are of at most three types. If, on the other hand, $\mathcal F_{1},\mathcal F_{2}$ are not the first two faces, then the other face which is adjacent to both $\mathcal F_{1},\mathcal F_{2}$ has already been placed, and, by symmetry with respect to the perpendicular bisector plane of the common side of $\mathcal F_{1},\mathcal F_{2}$, it must be congruent to $\mathcal F$. This shows that from the fourth step onwards all faces which appear must be of a type which has appeared before, and proves the assertion in (b).

(c)

It suffices to prove that in some point of the polyhedron two hexagons and one pentagon meet, because then, by using a procedure of placing down the faces one by one as described above, there is only one polyhedron we can obtain (in other words, a choombam is uniquely determined by the three faces which meet at any one of its vertices).

The three faces meeting at a point cannot all be hexagons, because then the angles would add up to $2\pi$, and they must add up to less than $2\pi$. If they are all pentagons, then the polyhedron is the regular dodecahedron. It is also impossible that they be two pentagons and a hexagon. In fact, in general, the neighboring faces of a face with an odd number of sides in a choombam must all be congruent, as can easily be seen. This leaves just one possibility: the three faces meeting at a vertex are two hexagons and a pentagon, and this is precisely what we wanted to prove.

(d)

 Again, we make use of the observation made above that a choombam is uniquely determined by the triple of plane angles meeting at any one of its points. This means that there is an injection from the set of choombams into the set of triples of the form $\left(\frac{n-2}n,\frac{m-2}m,\frac{p-2}p\right)$ with $m,n,p\ge 4$ and such that $\frac{n-2}n+\frac{m-2}m+\frac{p-2}p<2$.

If such a triple contains two angles $\frac12$, then the corresponding choombam is a prism, as described in Omid’s post. Since there are only finitely many such triples with only one $\frac{1}{2}$, we’re done.


how do you draw a triangle on asymptote with specific side lengths say 2, 3, and 4 and label them?

I would draw one side, and then find the intersection of two circles with centers at the endpoints of the segment and radii $ 3$ and $ 4$, thus finding the third point of the triangle.
[code][asy]size(200);
pair A=(0,0), B=(2,0), C=IP(circle(A,4),circle(B,3));
D(MP("$A$",D(A),SW)--MP("$B$",D(B),SSE)--MP("$C$",D(C),NNE)--A);
MP("$2$",midpoint(A--B),S);
MP("$3$",midpoint(B--C),ESE);
MP("$4$",midpoint(A--C),NW);[/asy][/code]
[asy]size(200);
pair A=(0,0), B=(2,0), C=IP(circle(A,4),circle(B,3));
D(MP("$A$",D(A),SW)--MP("$B$",D(B),SSE)--MP("$C$",D(C),NNE)--A);
MP("$2$",midpoint(A--B),S);
MP("$3$",midpoint(B--C),ESE);
MP("$4$",midpoint(A--C),NW);[/asy]

this does not seem to be working.  i copied and pasted the code in texnic center, like this:

[code]\begin{figure}[ht]
\begin{asy}
size(200);
pair A=(0,0), B=(2,0), C=IP(circle(A,4),circle(B,3));
D(MP("$A$",D(A),SW)--MP("$B$",D(B),SSE)--MP("$C$",D(C),NNE)--A);
MP("$2$",midpoint(A--B),S);
MP("$3$",midpoint(B--C),ESE);
MP("$4$",midpoint(A--C),NW);
\end{asy}
\end{figure}
[/code]
plz help

I_like_pie's code uses cse5. Do you have it installed? If you have, did you add the import cse5; command in the beginning of your code? AoPS uses two non-standard packages written by users on its site: olympiad.asy and cse5.asy. Many codes posted here use at least one of them and import command is not necessary when posting on the Forum but necessary when running the code at home.

ok thanks but how do i get olympiad.asy and cse5.asy?

Read 2 my stickies in this forum. The first one (about cse5) has a link to the page with the latest version in the very first post; the second one (about the olympiad package) has the file attached.

ok now where do i put the two .asy files once i have them?

In the same directory where asy.exe and all other packages are. Asymptote will find them there automatically. Alternatively, you can put them wherever you want and add the lines
import settings;
dir=[i]path to your folder[/i];
to the config.asy file.

ok one more question-how do i write square root symbols like $ \sqrt{6}\minus{}\sqrt{2}$ in asymptote

can these be lengths for segments

i need to use them for a label

Asymptote allows (almost) all $ \text{\LaTeX}$ constructions in the labels. So, just write
[code]
label("$\sqrt 6-\sqrt 2$",A,S);
[/code]
or whatever your point and direction are.
The MP command in cse5 adds dollar signs to your string automatically, so you do not need to write them:
MP("\sqrt 6-\sqrt2",A,S);
has essentially the same effect.
MC is quite similar to MP but allows you to mark curves rather than points.
[asy]size(100);
pair A=(0,0), B=(2,5), C=(-2,3);
MC(66, "\sqrt 3-\sqrt 2", 8, D(D(A)--D(B)),W);[/asy]

ok when i do my second diagram, it gives me this warning:

Package asymptote Warning: file xxxxxxxxxx_2.pdf does not exist on input line 82.

what could i have done wrong

That's quite hard (read [i]impossible[/i]) to tell without seeing your code. I surmise that you are running Asymptote from Latex and you made some error in the second diagram that was so bad that no output file was created but this is only my wild guess.


Prove there is an integer $ k$ for which $ k^3 \minus{} 36 k^2 \plus{} 51 k \minus{} 97$ is a multiple of $ 3^{2008.}$

Anyway... There is a standard procedure to solve this. We make $ k \equal{} 3x \plus{} 1$ and get (except if I have miscalculated) $ 3^{2005}|P(x) \equal{} x^3 \minus{} 11x^2 \minus{} 2x \minus{} 9$. Then we show this: if $ 3^n|P(x)$ and $ 3|x$, there is $ y \equal{} x \plus{} 3^na$ such that $ 3^{n \plus{} 1}|P(y)$. The case $ n \equal{} 1$ is obvious.

Yes, thats the way Hensels lemma is done.
In fact, I think that's one of the most stupid problems AIMO ever had...


calculus
integration
function
algebra
domain
calculus computations

My MV teacher mentioned a few days ago that, in order to integrate $e^{x^2}$ and $e^{-x^2}$, we'd have to use polar integrals.  Now that we've learned polar integrals, I'm still not sure how to go about integrating these functions, if it's even possible.

Say I want to find the value of $\int_{-1}^1 e^{x^2} \, dx$.  My teacher suggested that I multiply this integral by $\int_{-1}^1 e^{y^2} \, dy$, giving the double integral $\int_{-1}^1 \int_{-1}^1 e^{x^2 + y^2} \, dy \, dx$.  This is convenient, because $r^2 = x^2 + y^2$.

But I'm kind of stuck here.  I can substitute $r^2$ into the integral, but I'm not sure what to do with the $dy \, dx$ or the integrals' bounds.  I could change the bounds to be a circle, as in $\int_0^{2\pi} \int_0^1 r e^{r^2} \, dr \, d\theta$, but I don't know how to translate the result of this double integral, the volume under some surface, into my original integral.

But maybe $e^{x^2}$ isn't solvable without numerically integrating...I got that impression from my brief perusal of old AoPS posts.

Still, we should be able to evaluate $\int_0^\infty e^{-x^2} \, dx$, right?  Doesn't that (or some similar function) converge to some number involving pi?  Needless to say, I'm not sure how to integrate that one, either.

Anyway, I'm curious: What's going on here?

-Starwiz

As you've found, you can, in closed form, find the integral over a circular disk:

$\int\int_{x^2+y^2\le 1}e^{x^2+y^2}\,dx\,dy=\int_0^1\int_0^{2\pi}e^{r^2}r\,d\theta\,dr.$

You've also found that that is most emphatically [b]not[/b] the same as the integral over a square:

$\int_{-1}^1\int_{-1}^1e^{x^2+y^2}\,dx\,dy=\left(\int_{-1}^1e^{x^2}\,dx\right)^2.$

The latter cannot be evaluated in closed form (that is, in a finite number of steps involving the familiar list of functions).

If you give up on closed form by accepting an infinite number of steps, you can do some nice things with power series:

$\int_{-1}^1e^{x^2}\,dx=2\int_0^1e^{x^2}\,dx=2\int_0^1\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}\,dx$

$=2\sum_{k=0}^{\infty}\frac1{k!}\int_0^1x^{2k}\,dx=2\sum_{k=0}^{\infty} \frac1{k!(2k+1)}.$

That sum converges very rapidly. Using a spreadsheet, I found the value $2.9253034918144$ using just 14 or 15 terms. That's accurate to as many digits as are shown.

As for the integral from zero to infinity, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63767]here.[/url]

[quote="starwiz"]Still, we should be able to evaluate $\int_0^\infty e^{-x^2} \, dx$, right?  Doesn't that (or some similar function) converge to some number involving pi?  Needless to say, I'm not sure how to integrate that one, either.
[/quote]
This one you can do by squaring and converting to a polar integral. The integration domain $\{(x,y)|x, \,y \geq 0\}$ becomes $\{(r,\theta) | r \geq 0, \, 0 \leq \theta \leq \pi/2\}$. I'll leave the rest to you, since you have worked out most of it already.

That's exactly what I wanted to know.

Thanks a lot, guys!


A computer is provided with a program which generates pairs $ (x,y)$ with $ x,y \in (0,1)$   and then it finds the real numbers
 $ a \equal{} x \plus{} ym$ and $ b \equal{} y \plus{} xm$ ( where $ m$ is a positive integer). The process is keeping on until  all pairs $ (a,b)$ with integers $ a , b$ are determined. The computer needs $ 5$ sec to determine each of such a pair. Find $ m$ , if we know that the computer needed $ 595$ sec to determine all the possible integer pairs $ (a,b)$.

$ \forall x,y \in (0,1): f(x,y)\equal{}(x\plus{}ym,y\plus{}xm)\equal{}(a,b)$. if m is fixed, so $ \forall a,b \in N, f(x,y)\equal{}(a,b)\implies x\equal{}\frac{bm\minus{}a}{m^2\minus{}1}, y\equal{}\frac{am\minus{}b}{m^2\minus{}1}$(*), so (x,y) is unique.

if 0 < x,y < 1 then 0 < a,b < (m+1).

if x,y>0, by relation (*) we have $ b>a/m$ and $ a>b/m$. in conclusion, if m is fixed, we can choose all the value (a,b) with $ 1 \le a \le m$ and $ 1 \le b \le m$ except (m,1) and (1,m).
so, we have $ m^2\minus{}2$ possibilities. so, $ m^2\minus{}2\equal{}\frac{595}{5} \implies m\equal{}11$

bye
paolo[/hide]


Can you cover the chessboard 10x10 with bars 4x1?

hmm intuitively im pretty sure the answer is no... but surprisingly this is kinda hard to figure out?

im pretty sure u use a coloring argument... and i think it might revolve around the fact that each block u put down, the colors of the two ends are different (so one end will be black and one end will be white)

[hide]No, I cannot.
Any rectangle made by 4x1 tiles always has 1 dimension divisible by 4.  By expanding the 1 side, 4xN rectangles can be made.  Notice that to change the 4 to a number not divisble by 4, 1x4s have to be added to the 4xN, but it only forms a rectangle if N is divisible by 4.  If two rectangles are added, they only form a new rectangle when they share at least 1 dimension.  THe rectangles are merged along the dimensions with the same length; therefore, that dimension is the same in the sum.  If that dimension is a multiple of 4, the sum will also have a dimension that's a multiple 4.  If it isn't, the dimensions getting added are both multiples of 4, so the sum will again have a dimension that's a multiple of 4.  SInce 10 isn't divisible by 4, a 10x10 square can't be covered by 4x1 squares.[/hide]  I don't think that's a complete proof.
Is it true that any NxM rectangle cannot be covered by Zx1 tiles unless either N or M is divisible by Z?  If so, is there a similar rule for ZxY tiles?

Or a 4 color argument could work.  It ends up being 24,25,25,26 of each of the four colors if you do it so that every 4x1 will cover 4 different colors like
a b c d a b c d a b
b c d a b c d a b c
c d a b c d a b c d
d a b c d a b c d a
etc
So no.


inequalities
calculus
calculus computations


[quote="Ignite168"]Do you think The Bible could be read as a piece of literature?[/quote]
why not?

Many people say you learn many new things each time you read the Bible, so you can never read it too much. In fact reading it from a new perspective each time probably helps with that. So why not?

Also, many of the lessons taught by the Bible are given in the form of parables (especially by Jesus in the New Testament but not only). Those are some of the most enduring tales of all time.

There is also much subtle use of language in the Old Testament (I'm afraid I don't know so much about the New) so if you can read it in Hebrew, there is certainly much interest on that front as well.

I have heard tha Bible is a great piece of English literature. Many say that if you have a dictrionary in one hand and the Bible in the other, you can learn fine English soon.

As for the stories, those in the New Testament are plain stupid, but those in the Old Testament  are pretty interesting. They have suspense, crime, romance, love, war - most of the elements of modern story telling. It will be quite enjoyable.

[quote="Ignite168"]Do you think The Bible could be read as a piece of literature?[/quote]

Sure, it [i]could[/i] be, but I'm not sure you would actually want to:

[quote="Genesis 5:1-14"] 1 This is the written account of Adam's line.
      When God created man, he made him in the likeness of God. 2 He created them male and female and blessed them. And when they were created, he called them "man. [a] "

 3 When Adam had lived 130 years, he had a son in his own likeness, in his own image; and he named him Seth. 4 After Seth was born, Adam lived 800 years and had other sons and daughters. 5 Altogether, Adam lived 930 years, and then he died.

 6 When Seth had lived 105 years, he became the father [b] of Enosh. 7 And after he became the father of Enosh, Seth lived 807 years and had other sons and daughters. 8 Altogether, Seth lived 912 years, and then he died.

 9 When Enosh had lived 90 years, he became the father of Kenan. 10 And after he became the father of Kenan, Enosh lived 815 years and had other sons and daughters. 11 Altogether, Enosh lived 905 years, and then he died.

 12 When Kenan had lived 70 years, he became the father of Mahalalel. 13 And after he became the father of Mahalalel, Kenan lived 840 years and had other sons and daughters. 14 Altogether, Kenan lived 910 years, and then he died. [/quote]

That comes from the New International version, available (along with many others, in many languages) at http://www.biblegateway.com

Also, David Plotz of Slate magazine has been blogging the bible.  It begins here:
http://www.slate.com/id/2141050/

[quote="spider_boy"][quote="Ignite168"]Do you think The Bible could be read as a piece of literature?[/quote]
Also I advice you to read "Koran".I can 100% guarantee you will not regret.[/quote]

I really don't have enough time to read anything that isn't required by school unless it is a short article in a magazine or a novel.

http://www.amazon.com/gp/product/0195283481/103-9296734-0109416?%5Fencoding=UTF8&v=glance&n=283155

That is the bible I ordered.


group theory
abstract algebra
algorithm
superior algebra
superior algebra unsolved

$ S_n$ can be generated by a set $ S$ of $ n \minus{} 1$ transpositions.
if $ S'$ runs over all subsets of $ S$ can $ S'$ generate all subgroups of $ S_n$ ?

can this be generalized to "subsets of the largest generator set generate all the subgroups ?"

What about subgroups with no transpositions (like $ A_n$)?

Steve

[quote="spx2"]$ S_n$ can be generated by a set $ S$ of $ n \minus{} 1$ transpositions.
if $ S'$ runs over all subsets of $ S$ can $ S'$ generate all subgroups of $ S_n$ ?[/quote]

Perhaps I misunderstand your question.  Obviously no non-identity subgroup of odd order can be generated by transpositions.

The subgroups generated by subsets S' are called parabolic subgroups (assuming you chose S correctly).  They look like direct products of copies of Sn.

you're both right sorry.
I'm still searching for some kind of algorithm to generate all subgroups.
Guess I'll read a bit more from [i]G. Butler - Fundamental algorithms for permutation groups[/i]


If $ a,b,c \ge 0$ and $ a\plus{}b\plus{}c\equal{}2$
Prove that $ 9abc \plus{} 8 \ge 8(ab\plus{}bc\plus{}ca)$

[hide="Solution"]We are to prove that $ 8\ge 8ab\plus{}8bc\plus{}8ca\minus{}9abc$.

Note that since $ (8\minus{}9a)(8\minus{}9b)(8\minus{}9c)\equal{}81(8ab\plus{}8bc\plus{}8ca\minus{}9abc)\minus{}640$, we have $ 8ab\plus{}8bc\plus{}8ca\minus{}9abc\equal{}\frac{640\plus{}(8\minus{}9a)(8\minus{}9b)(8\minus{}9c)}{81}$.

Hence, it suffices to prove that $ 8\ge \frac{640\plus{}(8\minus{}9a)(8\minus{}9b)(8\minus{}9c)}{81}$, which is true by applying AM-GM on $ (8\minus{}9a)(8\minus{}9b)(8\minus{}9c)$.[/hide]

[quote="FantasyLover"]
Hence, it suffices to prove that $ 8\ge \frac {640 \plus{} (8 \minus{} 9a)(8 \minus{} 9b)(8 \minus{} 9c)}{81}$, which is true by applying AM-GM on $ (8 \minus{} 9a)(8 \minus{} 9b)(8 \minus{} 9c)$[/quote]

AM GM only works for positive numbers and $ (8\minus{}9a)$ isn't necessarily positive

[hide="solution"]
sub in $ a\plus{}b\plus{}c\equal{}2$ to give
$ 9abc \plus{} (a\plus{}b\plus{}c)^3 \ge 4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
Expanding gives
$ a^3\plus{}b^3\plus{}c^3 \plus{} 3abc \ge \sum a^2(b\plus{}c)$

$ \sum a(a\minus{}b)(a\minus{}c) \ge 0$ which is Schur's inequality

[/hide]

[hide="My Solution"]
Let c be the min of a,b,c.
We have to show
$ 8 \ge ab(8 \minus{} 9c) \plus{} 8c(a \plus{} b) \equal{} ab(8 \minus{} 9c) \plus{} 8c(2 \minus{} c)$
Since $ c \le \frac {2}{3}, 8 \minus{} 9c \ge 0$. Also $ ab \le (\frac {a \plus{} b}{2})^2 \equal{} (\frac {2 \minus{} c}{2})^2$ by AM-GM
So we have to show $ 8 \ge (8 \minus{} 9c)(\frac {2 \minus{} c}{2})^2 \plus{} 8c(2 \minus{} c) \iff c(3c\minus{}2)^2 \ge 0$, which is obvious.[/hide]

[hide="The way I would do it."]Substitute $ 8\equal{}(a\plus{}b\plus{}c)^3$ and $ 8(ab\plus{}bc\plus{}ac)\equal{}4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)$

Now our inequality is equivalent to:

$ 9abc\plus{}(a\plus{}b\plus{}c)^3 \ge 4(ab\plus{}bc\plus{}ac)(a\plus{}b\plus{}c)$

$ 15abc \plus{} \sum a^3 \plus{} 3\sum(a^2b\plus{}b^2a) \ge 12abc \plus{} 4\sum(a^2b\plus{}b^2a)$

$ \sum a^3 \plus{} 3abc \ge \sum (a^2b\plus{}b^2a)$ which is true by Schur's.[/hide]

[quote="ocha"][quote="KrazyFK"][hide="The way I would do it."]Substitute $ 8 \equal{} (a \plus{} b \plus{} c)^3$ and $ 8(ab \plus{} bc \plus{} ac) \equal{} 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac)$

Now our inequality is equivalent to:

$ 9abc \plus{} (a \plus{} b \plus{} c)^3 \ge 4(ab \plus{} bc \plus{} ac)(a \plus{} b \plus{} c)$

$ 15abc \plus{} \sum a^3 \plus{} 3\sum(a^2b \plus{} b^2a) \ge 12abc \plus{} 4\sum(a^2b \plus{} b^2a)$

$ \sum a^3 \plus{} 3abc \ge \sum (a^2b \plus{} b^2a)$ which is true by Schur's.[/hide][/quote]

perhaps it would be a good idea to check what other people have written before you post an almost identical solution. :wink:[/quote]

Whoops, sorry.

divide both sides by abc

get:

9+8/abc > or = to 8/a + 8/b +8/c

multiply both sides by a+b+c = 2

and use cauchy on the rhs

to get 18 + 16/abc > or = to 54

abc < or = to 8/27 which is true by am-gm

Ecstatic Potter, You cannot use cauchy on RHS because that may reduce the value of the expression even though it could be larger than LHS.

oh ya ur right varun thx...what if i multiplied both sides by -1 to switch the signs, used cauchy, and then reversed the signs again? lol iono im terrible at inequalities

Ecstatic for a second I forgot who you were! I coudn't!
remember cauchy deals with positive real numbers because we square things!

lol ya i got a better solution now. purely am-gm. =)

Can you post it? I don't think you can have a solution purely by AM-GM because equality holds at (1,1,0) and (1/3,1/3,1/3) and AM-GM usually only works when the only equality case is when all the variables are equal.

I think s/he means the following inequality that is purely solved by Cauchy inequality

$ \prod_{cyc}(a \plus{} b \minus{} c) \le abc$

I created this problem using that inequality  :)

My solution:
$ \prod_{cyc}{\left( \frac a 2 \plus{} \frac b 2 \minus{} \frac c 2 \right)} \le \frac {abc} 8$ , using the inequality [b]great math[/b] mentioned.

$ \implies 8(1\minus{}a)(1\minus{}b)(1\minus{}c) \le abc$ , since $ \frac a 2 \plus{} \frac b 2 \plus{} \frac c 2 \equal{}1$

Expand and rearrange the terms. Voila! We get the required inequality.

Proof of the inequality used:
$ (a\minus{}b\plus{}c)(a\plus{}b\minus{}c)  \equal{} a^2  \minus{} (b\minus{}c)^2 \le a^2$

similarly $ (b\minus{}c\plus{}a)(b\plus{}c\minus{}a) \le b^2$ and $ (c\minus{}a\plus{}b)(c\plus{}a\minus{}b) \le c^2$

Multiplying the three, we get the required inequality.


inequalities
calculus
derivative
function
inequalities unsolved

Prove that if $a,b,c,d>0$ and $abcd=1$ then:

$\frac{1}{1+a+b+c}+\frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b} \le \frac{1}{3+a}+ \frac{1}{3+b}+ \frac{1}{3+c}+ \frac{1}{3+d}$

It's a natural result from Bungari Mo 2000.

THis can't be true. THe LHS is always smaller than 1 and the RHS is always greater than 1. Anyway, I hope soon we will have in mathlinks contest a very nice related problem... :D

Sorry Every one!MY problem in my opion in's that.I has a mistake in typing.I've Edit it .
Now easy to check two hand side is less than 1

I hope it's correct.
I guess the stronger inequality ($abcd=1$ is replaced by $abcd \ge 1$) still holds.
Let \[
f\left( {a,b,c,d} \right) = \sum {\frac{1}
{{3 + a}}}  - \sum {\frac{1}
{{1 + a + b + c}}}
\]
then prove that if $t=\frac{a+b}{2}$, $f(a,b,c,d) \ge f(t,t,c,d)$.
\[
\begin{gathered}
  f(a,b,c,d) - f(t,t,c,d) \hfill \\
   = \frac{1}
{{3 + a}} + \frac{1}
{{3 + b}} - \frac{2}
{{3 + t}} + \frac{1}
{{1 + b + c + d}} + \frac{1}
{{1 + c + d + a}} - \frac{2}
{{1 + 2t + d}} \hfill \\
\end{gathered}
\]
We will show $\frac{1}{{3 + a}} + \frac{1}{{3 + b}} - \frac{2}{{3 + t}} \ge 0$ and the other is just the same.
\[
\begin{gathered}
  \frac{1}
{{3 + a}} + \frac{1}
{{3 + b}} - \frac{2}
{{3 + t}} \hfill \\
   = \frac{{\left( {6 + 2t} \right)(3 + t) - 2(3 + a)(3 + b)}}
{{(3 + a)(3 + b)(3 + t)}} \hfill \\
   = \frac{{2\left[ {(3 + t)^2  - (3 + a)(3 + b)} \right]}}
{{(3 + a)(3 + b)(3 + t)}} \geqslant 0 \hfill \\
\end{gathered}
\]
So we have $f(a,b,c,d) \ge f(y,y,y,y)$ where $y=\frac{a+b+c+d}{4}$, thus, it remains to prove that
\[
\frac{1}
{{3 + y}} - \frac{1}
{{1 + 3y}} = \frac{{2(y - 1)}}
{{(3 + y)(1 + 3y)}} \geqslant 0
\]
that's obvious because \[
y = \frac{{a + b + c + d}}
{4} \geqslant \sqrt[4]{{abcd}} \geqslant 1
\]

[quote="siuhochung"][quote="hungkhtn"]Prove that if $a,b,c,d>0$ and $abcd=1$ then:

$\frac{1}{1+a+b+c}+\frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b} \le \frac{1}{3+a}+ \frac{1}{3+b}+ \frac{1}{3+c}+ \frac{1}{3+d}$

[/quote]
I guess the stronger inequality ($abcd=1$ is replaced by $abcd \ge 1$) still holds.
Let \[
f\left( {a,b,c,d} \right) = \sum {\frac{1}
{{3 + a}}}  - \sum {\frac{1}
{{1 + a + b + c}}}
\]
then prove that if $t=\frac{a+b}{2}$, $f(a,b,c,d) \ge f(t,t,c,d)$.
\[
\begin{gathered}
  f(a,b,c,d) - f(t,t,c,d) \hfill \\
   = \frac{1}
{{3 + a}} + \frac{1}
{{3 + b}} - \frac{2}
{{3 + t}} + \frac{1}
{{1 + b + c + d}} + \frac{1}
{{1 + c + d + a}} - \frac{2}
{{1 + 2t + d}} \hfill \\
\end{gathered}
\]
We will show $\frac{1}{{3 + a}} + \frac{1}{{3 + b}} - \frac{2}{{3 + t}} \ge 0$ and the other is just the same.
\[
\begin{gathered}
  \frac{1}
{{3 + a}} + \frac{1}
{{3 + b}} - \frac{2}
{{3 + t}} \hfill \\
   = \frac{{\left( {6 + 2t} \right)(3 + t) - 2(3 + a)(3 + b)}}
{{(3 + a)(3 + b)(3 + t)}} \hfill \\
   = \frac{{2\left[ {(3 + t)^2  - (3 + a)(3 + b)} \right]}}
{{(3 + a)(3 + b)(3 + t)}} \geqslant 0 \hfill \\
\end{gathered}
\]
So we have $f(a,b,c,d) \ge f(y,y,y,y)$ where $y=\frac{a+b+c+d}{4}$, thus, it remains to prove that
\[
\frac{1}
{{3 + y}} - \frac{1}
{{1 + 3y}} = \frac{{2(y - 1)}}
{{(3 + y)(1 + 3y)}} \geqslant 0
\]
that's obvious because \[
y = \frac{{a + b + c + d}}
{4} \geqslant \sqrt[4]{{abcd}} \geqslant 1
\][/quote]

Nice!!!

[quote="spdf"]Siuhochung,Ithink you used mixxing variable in wrong way.The condition is abcd=1,then you can't replace (a,b,c,d) by ((a+b)/2;(a+b)/2;c;d)[/quote]

I tried to prove a stronger inequality by replacing $abcd=1$ by $abcd \ge 1$, in this case, my way of mixing variables still holds.

Sorry,siuhochung,I read your solution too quickly :blush: .Your idea is very nice.But there are some mistake.
[tex]F(a;b;c;d)-F(t;t;c;d)\geq\ 0[/tex] <=>
[tex]\frac{t^2-ab}{(3+a)(3+b)(3+t)}\geq\frac{t^2-ab}{(1+c+d+a)(1+c+d+b)(1+c+d+t)}[/tex] <=>
[tex] c+d\geq\ 2[/tex](1)
So,let me complete your solution.WLOG,assume that [tex]a\geq\ b\geq\ c\geq\ d[/tex].Then,from [tex]abcd\geq\ 1[/tex],we have [tex] a+b\geq\ a+c\geq\ 1[/tex]
,use (1),we can see that the minimum of F(a;b;c;d) is attained in (a;m;m;m).
Now,we have to prove:[tex]\frac{3}{3+x}+\frac{1}{3+y}\geq\frac{3}{1+2x+y}+\frac{1}{1+3x}[/tex] with [tex]x^3y\geq\ 1;y\geq\ x[/tex].The rest is easy

you are right!  :blush:
Sorry for my ignorance..

I have made all computations to have

$24 +77x+33y \geq 71x^2 +39xy+10x^2y+8xy^2+6x^3$

with as you said $y \geq 1$ and $x^3y \geq 1$

Is it easy ? :?

manlio, can you check your expansion? I get
\[
3x^2  + 5x^2 y + 22x^2  + 4xy^2  + 6xy \geqslant 25x + 3y + 12
\]
I don't have a nice way to solve it, but anyway,
First,
\[
\frac{9}
{5}xy^2  \geqslant \frac{9}
{5}y^{\frac{5}
{3}}  + \frac{6}
{5} \geqslant 3y(xy^{\frac{1}
{3}}  \geqslant 1)
\]
by weighted amgm, so we are going to prove
\[
\begin{gathered}
  3x^3  + 5x^2 y + 22x^2  + \frac{{11}}
{5}xy^2  + 6xy - 25x \geqslant \frac{{66}}
{5} \hfill \\
   \Leftrightarrow 15x^3  + 25x^2 y + 110x^2  + 11xy^2  + 30xy - 125x \geqslant 66 \hfill \\
\end{gathered}
\]
Now, WLOG assume $x^3y=1$ (because if $x^3y \ge 1$, take y' so that $x^3y'=1$, and it's obvious that the value of L.H.S. decreases)
Then by substitution, we want to prove,
\[
15x^3  + 25\frac{1}
{x} + 110x^2  + 11\frac{1}
{{x^5 }} + 30\frac{1}
{{x^2 }} - 125x \geqslant 66
\]
Differentiation of the L.H.S. gives
\[
5\left( {9x^2  + 44x - 25 - \frac{5}
{{x^2 }} - \frac{{12}}
{{x^3 }} - \frac{{11}}
{{x^6 }}} \right)
\]
\[
\begin{gathered}
  0 \geqslant 5\left( {9x^2  + 44x - 25 - \frac{5}
{{x^2 }} - \frac{{12}}
{{x^3 }} - \frac{{11}}
{{x^6 }}} \right) \hfill \\
   \Leftrightarrow 25 + \frac{5}
{{x^2 }} + \frac{{12}}
{{x^3 }} + \frac{{11}}
{{x^6 }} \geqslant 9x^2  + 44x \hfill \\
\end{gathered}
\]
As $x \le 1$, L.H.S. $\ge 53 \ge$ R.H.S.
Therefore, \[
15x^3  + 25\frac{1}
{x} + 110x^2  + 11\frac{1}
{{x^5 }} + 30\frac{1}
{{x^2 }} - 125x
\]
is a decreasing function with $0 \le x \le 1$ and at x=1, L.H.S. = R.H.S., which proves the inequality.

Thank you very much.

OK, I will check it! ;)

It seem so difficult to check,Isn't it!
I think use expanding is useful to proof,but I don't like that.
I hope Vasc will have a nice solution to help me solve it. :)

As the complete solution is not posted yet, maybe i post it now after combining all the posts here.  (hungkhtn, do you have a solution to this inequality,\[
3x^2  + 5x^2 y + 22x^2  + 4xy^2  + 6xy \geqslant 25x + 3y + 12
\] with $y \ge x$ and $x^3 y \ge 1$?)

assume $c \ge a \ge d \ge b$, then we have $ca \ge cd \ge cb$, therefore, we have  $ca \ge cd \ge 1$ and then $c+a \ge c+d \ge 2$.
Now, mixing variables $f(a,b,c,d)-f(t,t,c,d)$, where $t=\frac{a+b}{2}$, is equal to
$(t^2-ab) \frac{(1+c+d+t)(1+c+d+a)(1+c+d+b)-(3+t)(3+a)(3+b)}{(1+c+d+t)(1+c+d+a)(1+c+d+b)(3+t)(3+a)(3+b)}$. Now it's obviously that $t^2 \ge ab$, so we only need $(1+c+d+t)(1+c+d+a)(1+c+d+b)-(3+t)(3+a)(3+b) \ge 0 $. Certainly if $c+d \ge 2$, this inequality is true.
By our order of variables, we have $c+d \ge 2$, so $f(a,b,c,d) \ge f(t,t,c,d)$.
Now we repeat the process again, as  $c+ \frac{d+t}{2} \ge 2$(By weighted amgm and $c \ge 1$), therefore $f(t,t,c,d) \ge f(y,t,c,y)$  where $y=\frac{c+\frac{d+t}{2}}{2}$.
Repeat this process again and again we will finally arrive at $f(a,b,c,d) \ge f(c,z,z,z)$ , where $z=\frac{a+b+d}{3}$.

Now,we have to prove:[tex]\frac{3}{3+x}+\frac{1}{3+y}\geq\frac{3}{1+2x+y}+\frac{1}{1+3x}[/tex] with [tex]x^3y\geq\ 1;y\geq\ x[/tex].

Expanding this inequality,
\[
3x^2  + 5x^2 y + 22x^2  + 4xy^2  + 6xy \geqslant 25x + 3y + 12
\]
I don't have a nice way to solve it, but anyway,
First,
\[
\frac{9}
{5}xy^2  \geqslant \frac{9}
{5}y^{\frac{5}
{3}}  + \frac{6}
{5} \geqslant 3y(xy^{\frac{1}
{3}}  \geqslant 1)
\]
by weighted amgm, so we are going to prove
\[
\begin{gathered}
  3x^3  + 5x^2 y + 22x^2  + \frac{{11}}
{5}xy^2  + 6xy - 25x \geqslant \frac{{66}}
{5} \hfill \\
   \Leftrightarrow 15x^3  + 25x^2 y + 110x^2  + 11xy^2  + 30xy - 125x \geqslant 66 \hfill \\
\end{gathered}
\]
Now, WLOG assume $x^3y=1$ (because if $x^3y \ge 1$, take y' so that $x^3y'=1$, and it's obvious that the value of L.H.S. decreases)
Then by substitution, we want to prove,
\[
15x^3  + 25\frac{1}
{x} + 110x^2  + 11\frac{1}
{{x^5 }} + 30\frac{1}
{{x^2 }} - 125x \geqslant 66
\]
Differentiation of the L.H.S. gives
\[
5\left( {9x^2  + 44x - 25 - \frac{5}
{{x^2 }} - \frac{{12}}
{{x^3 }} - \frac{{11}}
{{x^6 }}} \right)
\]
\[
\begin{gathered}
  0 \geqslant 5\left( {9x^2  + 44x - 25 - \frac{5}
{{x^2 }} - \frac{{12}}
{{x^3 }} - \frac{{11}}
{{x^6 }}} \right) \hfill \\
   \Leftrightarrow 25 + \frac{5}
{{x^2 }} + \frac{{12}}
{{x^3 }} + \frac{{11}}
{{x^6 }} \geqslant 9x^2  + 44x \hfill \\
\end{gathered}
\]
As $x \le 1$, L.H.S. $\ge 53 \ge$ R.H.S.
Therefore, \[
15x^3  + 25\frac{1}
{x} + 110x^2  + 11\frac{1}
{{x^5 }} + 30\frac{1}
{{x^2 }} - 125x
\]
is a decreasing function with $0 \le x \le 1$ and at x=1, L.H.S. = R.H.S., which proves the inequality.

do we have the general inequality?
If $x_1,..,x_n$ are positive reals such that their product equals 1, then
\[
\sum {\frac{1}
{{n - 1 + x_i }}}  \geqslant \sum {\frac{1}
{{1 + S - x_i }}}
\]
where $S=x_1+...+x_n$.
It looks very very difficult!

But I thinks You Solution is really needed to check again,Siuhochung. :(

Do we can take $y'$?Because Two Handsides are all decreasing.

[quote="siuhochung"]
\[
\begin{gathered}
   \Leftrightarrow 15x^3  + 25x^2 y + 110x^2  + 11xy^2  + 30xy - 125x \geqslant 66
\end{gathered}
\]
Now, WLOG assume $x^3y=1$ (because if $x^3y \ge 1$, take y' so that $x^3y'=1$, and it's obvious that the value of L.H.S. decreases)
[/quote]

Why two hand-sides decrease?
We fix x and changes y only..


Let a,b,c be the sides of a triangle. Prove that

a^4+b^4+c^4  + (ab+bc+ca)(a 2 +b 2 +c 2 ) \geq 4(a 2 b 2 +b 2 c 2 +c 2 a 2 )

A very nice inequality. First, we observe that it can be written:
 sum a^4+abc(a+b+c)+sum a^3(b+c)>=4 sum a^2*b^2. From Schur inequality we deduce that sum a^4+abc(a+b+c)>=sum a^3(b+c). So it is enough to prove that sum a^3(b+c)>=2 sum a^2b^2, which is true since sum a^3(b+c)= sum ab(a^2+b^2)>=2 sum a^2b^2.