Long John Silverman has captured a treasure map from Adam McBones. Adam has buried the treasure at the point $ (x, y)$ with integer co-ordinates (not necessarily positive). He has indicated on the map the values of $ x^{2}\plus{}y$ and $ x\plus{}y^{2}$, and these numbers are distinct. Prove that Long John has to dig only in one place to find the treasure.

[quote="AndrewTom"]Long John Silverman has captured a treasure map from Adam McBones. Adam has buried the treasure at the point $ (x, y)$ with integer co-ordinates (not necessarily positive). He has indicated on the map the values of $ x^{2} \plus{} y$ and $ x \plus{} y^{2}$, and these numbers are distinct. Prove that Long John has to dig only in one place to find the treasure.[/quote]

$ x^2\plus{}y\minus{}x\minus{}y^2\equal{}(x\minus{}y)(x\plus{}y\minus{}1)$ is fixed. Let $ x\minus{}y\equal{}p , x\plus{}y\minus{}1\equal{}q$ we have $ x\equal{}\frac{p\plus{}q\plus{}1}{2} , y\equal{}\frac{q\minus{}p\plus{}1}{2}$.

As $ 4(x^2\plus{}y)\minus{}3\minus{}2pq$ is fixed we have $ p^2\plus{}q^2\plus{}4q$ is fixed $ \Rightarrow p^2q^2\plus{}q^4\plus{}4q^3$ is fixed $ \Rightarrow q^4\plus{}4q^3$ is fixed $ \Rightarrow q$ is fixed.

(Proof: Assume that $ \exists$ distinct integers $ m,n$ such that $ m^4\plus{}4m^3\equal{}n^4\plus{}4n^3$ . WLOG $ m > n$
Case 1: $ m,n \ge 0$ , impossible as $ f(x)\equal{}x^4\plus{}4x^3$ is strictly increasing on $ [0, \infty)$
Case 2: $ m>0,n\equal{}\minus{}k<0$, then the assumption $ \Leftrightarrow m^4\plus{}4m^3\equal{}k^4\minus{}4k^3$ , impossible for $ m \ge k$ (or LHS>RHS) or $ m \le k\plus{}2$(or LHS<RHS) . It's easy to see that $ m\equal{}k\plus{}1$ is impossible too.
Case 3: $ m\equal{}\minus{}k_1<0$ , $ n\equal{}\minus{}k_2<0$ . Let $ g(x)\equal{}x^4\minus{}4x^3$ then the assumption $ \Leftrightarrow g(k_1)\equal{}g(k_2)$.
As $ g(x)$ is strictly increasing in $ (3, \infty)$ , we only need to check the case where one of $ k_1,k_2$ is $ 1,2$ or $ 3$ and see they are all impossible.
So $ (m,n)\equal{}(0,\minus{}4)$  , and $ q\equal{}0 \Rightarrow x^2\plus{}y\equal{}x\plus{}y^2$ , contradiction. )

Then $ q$ is fixed $ \Rightarrow p$ is fixed also $ \Rightarrow x,y$ are fixed , done.

Assume the contrary that there exist two distinct points $ (x,y)$ and $ (a,b)$ such that $ x^2\plus{}y\equal{}a^2\plus{}b$ and $ x\plus{}y^2\equal{}a\plus{}b^2$. $ x\equal{}a$ iff $ y\equal{}b$, so $ x\ne a$ and $ y\ne b$. $ x^2\plus{}y\equal{}a^2\plus{}b\implies y\minus{}b\equal{}(a\minus{}x)(a\plus{}x)$. $ x\plus{}y^2\equal{}a\plus{}b^2\implies x\minus{}a\equal{}\minus{}y^2\plus{}b^2\equal{}\minus{}(y\minus{}b)(y\plus{}b)\equal{}\minus{}(a\minus{}x)(a\plus{}x)(y\plus{}b)$, which gives $ (a\plus{}x)(y\plus{}b)\equal{}1$. Case 1: $ a\plus{}x\equal{}1,y\plus{}b\equal{}1$. $ x^2\plus{}y\equal{}(1\minus{}x)^2\plus{}(1\minus{}y)$, so $ x\plus{}y\equal{}1$. But $ a\plus{}x\equal{}1$, so $ y\equal{}a$. Similarly, $ x\equal{}b$. So we have $ x^2\plus{}y\equal{}y^2\plus{}x$. But we are told that these numbers are distinct, so we get a contradiction. Case 2: $ a\plus{}x\equal{}\minus{}1,y\plus{}b\equal{}\minus{}1$. $ x^2\plus{}y\equal{}(\minus{}1\minus{}x)^2\plus{}(\minus{}1\minus{}y)$, so $ x\equal{}y$. Similar to case 1, we get a contradiction. Thus the point is unique. Q.E.D.

http://www.bmoc.maths.org/solutions/bmo1-2010/

Scroll down to problem 6 for official solution - but I don't know how long it will be available


Elrond is organizing a council in Rivendell. He has invited two men, two dwarves, and two elves, and plans to seat them along with himself in a circle with seven seats. He knows that dwarves and elves do not get along, so he plans to seat them so that no dwarf is sitting next to an elf. The circle has a head chair. If Elrond (himself an elf) must sit at the head of the circle, in how many ways can he seat the other six guests?

[quote=math154]It is easy to see that the men must separate the elves and the dwarves. It must look like this,

[code]
  E
 E E
M   M
 D D
[/code]

which has $ 3!2!2!\equal{}\boxed{24}$ arrangements.[/quote]

Could you explain your work a bit more?  Or maybe could someone else, since it appears he/she probably forgot about this thread.

[hide=Solution]If we call the two dwarves d_1 and d_2, then we see that there are 4 spots to choose where they sit. If we count them out, there are 6 ways to leave seating for the other elves, and there is only 1 way for the elves. So, the answer is 6 * 2! * 2! = 24.[/hide]

yeah also what about frodo baggins? isnt he the best part of the story?

[quote=zvyvyz]what about eemddme and emddmee?[/quote]

um, what?

hehe one math problem to rule them all :) (and it was quite hard for me)

[hide=sol]Elrond himself must sit at the head seat, so he has one choice. Next, for simplicity, we seat the elves and this determines where the dwarves will seat. The elves have 6 places to sit (such that none of them complicates the situation) and you can find that by counting. Remember, reflections are distinct and there are no rotations!

The dwarves must sit opposite the middle elf, and this gives them 2 ways to seat themselves.

Finally, the men have 2 ways to sit as well.

Thus, $6*2*2 = \boxed{24}$. [/hide]


calculus
derivative
calculus computations

[size=200][b]Maximum[/b][/size] Value of:
$ \frac {1}{x^x} \equal{} ?$

$ f(x)\equal{}\frac{1}{x^x}\\
f'(x)\equal{}\minus{}\frac{\ln(x)\plus{}1}{x^x}\\
\minus{}\frac{\ln(x)\plus{}1}{x^x}\equal{}0\\
\ln(x)\equal{}\minus{}1\\
x\equal{}e^{\minus{}1}$


Let me call Arqady. Do you think it's true? I am uncertain about it but I am sure that it's true for small value of $n$.

Sorry for replying this old topic of Gackiem. I think this conjecture is too hard to check, but I did find the following result

(1) The inequality is obvious for $n=2$.
(2) The solution to the case $n=3$ and $n=4$ and $n=5$ is very nice and simple.
(3) I don't try the case $n=6$, but I believe that it can be done similarly.
(4) Finally, I hope your help.

[quote="shalex"]I guess that it could be solved in such kind of way; let
$T_{m}=\sum_{1\leq i,j\leq m,i+j=m+1}\frac{1}{a_{i}a_{j}}\ (m=1,2,\ldots,n)$
Then
$\left(\sum_{i=1}^{n}\frac{1}{a_{i}}\right)^{2}\geq\sum_{m=1}^{n}T_{m}$
And it's easy to get
$T_{m}\geq m\left(\frac{1}{a_{1}^{2}a_{2}^{2}\cdots a_{m}^{2}}\right)^{1/m}\geq\frac{m^{2}}{a_{1}^{2}+a_{2}^{2}+\cdots+a_{m}^{2}}$
Then it's done. (I'm not sure if there are some mistakes in it.)[/quote]
Would anybody like to point out whether this solution is right or not? I want to know if this method works. Thanks a lot~


find all such functions $ f(R) \rightarrow R$ that
$ f(f(x)\plus{}y)\equal{}2x\plus{}f(f(y)\minus{}x)$

$ f(x)\equal{}x$ for all real number  :)

Lol ! why ?? it is really the only solution

if you want my complete solution, i will try to post it :)

Yes, could you try to post it please? :D

Thanks :)

[i][color=green]Hello everyone,

Please try to post your full solution always  :) so we can know how  :D [/color][/i]

A similar one is [url=http://www.mathlinks.ro/viewtopic.php?t=309080]here[/url] and [url=http://www.mathlinks.ro/viewtopic.php?search_id=1486438634&t=309350]here[/url]. Maybe they are helpful... :)

I think that $ f$ is injective is the key of the probleme,1h that I work to prove it but nothing for the moment,if I find something I mail the solution.. :)

the general solution is $ f(x) \equal{} x \plus{} c$ $ \forall x\in{\mathbb{R}}$

hi!!

Let $ P(x,y): f(f(x)\plus{}y)\equal{}2x\plus{}f(f(y)\minus{}x)$

$ P(x,\minus{}f(x)): f(0)\equal{}2x\plus{}f(f(\minus{}f(x))\minus{}x) \Rightarrow f$ is $ surjective$.

since $ f$ is surjective we can put $ f(a)\equal{}0$ then :

$ P(a,y): f(y)\equal{}2a\plus{}f(f(y)\minus{}a)$ we put $ t\equal{}f(y)$ then:

$ P(a,y): t\minus{}2a\equal{}f(t\minus{}a)$

if $ t\minus{}a\equal{}z$ and $ t'\minus{}a\equal{}z'$

$ \Rightarrow f(z)\equal{}f(z') \Rightarrow z\equal{}z'$ so $ f$ is injective.

$ P(0,y): f(f(0)\plus{}y)\equal{}f(f(y)) \Rightarrow$   $ f(y)\equal{}y\plus{}f(0)$

so $ f(x)\equal{}x\plus{}c$  $ \forall x\in{\mathbb{R}}$   :)

how can you tell the function is surjective?

$ f(0)\minus{}2x\equal{}f(f(\minus{}f(x))\minus{}x)$ so since $ g(x)\equal{}f(0)\minus{}2x$ is surjective then $ f$ is also surjective  :D

Hello everybody , reason is right , the general solution is $ f(x)\equal{}x\plus{}c$ with $ c\equal{}f(0)$

i will make a hint :

$ x\equal{}0 \implies f(f(0)\plus{}y\equal{}f(fy)) \implies f$ is an increasing function  because $ fof$ is increasing function , and  $ y \rightarrow y$  is also increasing function , which lead us to conclude that $ f$ is an increasing function on $ {\mathbb{R}}$ , therefor $ f$ is injective .

then $ f(f(0)\plus{}y\equal{}f(fy)) \implies f(x)\equal{}x\plus{}f(0)\equal{}x\plus{}c$ .

$ f(x)\equal{}x\plus{}c$ ;$ \forall x\in{\mathbb{R}}$

[quote="R.Maths"]Hello everybody , reason is right , the general solution is $ f(x) \equal{} x \plus{} c$ with $ c \equal{} f(0)$

i will make a hint :

$ x \equal{} 0 \implies f(f(0) \plus{} y \equal{} f(fy)) \implies f$ is an increasing function  because $ fof$ is increasing function , and  $ y \rightarrow y$  is also increasing function , which lead us to conclude that $ f$ is an increasing function on $ {\mathbb{R}}$ , therefor $ f$ is injective .

then $ f(f(0) \plus{} y \equal{} f(fy)) \implies f(x) \equal{} x \plus{} f(0) \equal{} x \plus{} c$ .

$ f(x) \equal{} x \plus{} c$ ;$ \forall x\in{\mathbb{R}}$[/quote]

Hi redwan :D

I think you have missed something :(

$ f$ is [b]monotonous[/b] $ \Leftrightarrow$ $ f$ is injective.

you have just prove that f is [b]increasing[/b]...

[quote="reason"][quote="R.Maths"]Hello everybody , reason is right , the general solution is $ f(x) \equal{} x \plus{} c$ with $ c \equal{} f(0)$

$ x \equal{} 0 \implies f(f(0) \plus{} y \equal{} f(fy)) \implies f$ is an increasing function  because $ fof$ is increasing function , and  $ y \rightarrow y$  is also increasing function , which lead us to conclude that $ f$ is an increasing function on $ {\mathbb{R}}$ , therefor $ f$ is injective .

[/quote]

$ f$ is [b]monotonous[/b] $ \Leftrightarrow$ $ f$ is injective.

you have just prove that f is [b]increasing[/b]...[/quote]

Hello Hamza  :D

i have proved it let take a see again to my solution  :wink:

[quote="R.Maths"]Hello everybody , reason is right , the general solution is $ f(x) \equal{} x \plus{} c$ with $ c \equal{} f(0)$

i will make a hint :

$ x \equal{} 0 \implies f(f(0) \plus{} y \equal{} f(fy)) \implies f$ is an increasing function  because $ fof$ is increasing function , and  $ y \rightarrow y$  is also increasing function , which lead us to conclude that $ f$ is an [color=red]increasing function[/color] on $ {\mathbb{R}}$ , therefor $ f$ is injective .

then $ f(f(0) \plus{} y \equal{} f(fy)) \implies f(x) \equal{} x \plus{} f(0) \equal{} x \plus{} c$ .

$ f(x) \equal{} x \plus{} c$ ;$ \forall x\in{\mathbb{R}}$[/quote]

$ f$ must be [b]strictly increasing[/b] ,that is the problem ;)

Yep this is it , but $ fof$ is always increasing such as $ T(y)\equal{}f(0)\plus{}y$ then it self-important to have $ f$ defined as been an increasing function

What do you think ??  :)

[quote="R.Maths"]Yep this is it , but $ fof$ is always increasing such as $ T(y) \equal{} f(0) \plus{} y$ then it self-important to have $ f$ defined as been an increasing function

What do you think ??  :)[/quote]

$ T$ is monotonous,right but $ T$ monotonous $ \Leftrightarrow$ $ foT$ is [b]monotonous[/b] :?:

no my friend , you didn't get me assumption  :)  .

it's not that what i wanted to clarify  .

it 's clear that $ fof$ is monotonous , besides it's an increasing function , thereafter $ foT$ is also increasing [ because $ fof\equal{}foT$ ], but since we know that $ T$ is increasing , then we need necessarily to have $ f$ is increasing

$ T$ and $ fof$ are always strictly increasing i think ...  :roll:

[quote="R.Maths"]no my friend , you didn't get me assumption  :)  .

it's not that what i wanted to clarify  .

it 's clear that $ fof$ is  [color=red]monotonous[/color] , besides it's an increasing function , thereafter $ foT$ is also increasing [ because $ fof \equal{} foT$ ], but since we know that $ T$ is increasing , then we need necessarily to have $ f$ is increasing

$ T$ and $ fof$ are always strictly increasing i think ...  :roll:[/quote]

dear redwan;

why $ fof$ is monotonous?  :)

Hello hamza  :)

i think you are right

$ fof$ is an increasing function , but this is not enough to conclude that $ f$ is bijective

it must be strictly increasing , ... then we can separate our solution in 2 cas :

$ 1_$ if $ f$ is strictly increasing ( the following is my proposition )

otherwise :

$ 2_$ if $ f$ is only increasing ( but i think this cas will be more complicated )

Thank you  :lol:

Public service announcement: [url=http://www.merriam-webster.com/dictionary/monotonous]monotonous[/url] is not the same word as [url=http://www.merriam-webster.com/dictionary/monotonic]monotonic[/url].


$A$, $B$ are two points on $\frac{x^2}{a^2}+\frac{25y^2}{9a^2}=1$. $F_2$ is the right focus. $|AF_2|+|BF_2|=\frac{8}{5}a$, the distance from the midpoint of $AB$ to the left directrix is $\displaystyle\frac{3}{2}$. Find the equation of the ellipse.

[quote="pkerichang"]what's a directrix of an ellipse???I know there's a directrix for parabola, I never knew there was one for an ellipse...[/quote]
all nondegenerate conics have directrix
[quote="Altheman"]the line through the focus parallel to the minor axis intersects the ellipse at 2 points, the directix is the segment formed by these 2 points[/quote]
 :?: latus rectum is not directrix...

woops, my bad, i am not that good with conics

For an ellipse: $\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$, the directrix is $x=\frac{a}{e}$ and $x=-\frac{a}{e}$.


[quote="khashi70"]Farz konid $ a,b,c>0$ , sabet konid :
$ \frac{a^{3}+abc}{b+c}+\frac{b^{3}+abc}{c+a}+\frac{c^{3}+abc}{a+b}\geq a^{2}+b^{2}+c^{2}$[/quote]
khob ino mishe injoori neveshtesh:

$ \frac{a^{3}+abc}{b+c}-a^{2}+\frac{b^{3}+abc}{c+a}-b^{2}+\frac{c^{3}+abc}{a+b}-c^{2}\geq 0$

$ \frac{a^{3}+abc-a^{2}b-a^{2}c}{b+c}+\frac{b^{3}+abc-b^{2}c-b^{2}a}{c+a}+\frac{c^{3}+abc-c^{2}a-c^{2}b}{a+b}\geq 0$

$ \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}\geq 0$

khob hala chon namosavi hamgene mishe farz kard $ a+b+c=1$ pas khahim dasht:

$ \frac{a(a-b)(a-c)}{b+c}=\frac{a(a-b)(a-c)}{1-a}$

hala age gharar bedim $ f(x)=\frac{x}{1-x}$ namosavi akharie be shekle zir dar miad:

$ f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$

ke inam chon $ f(x)=\frac{x}{1-x}$ ye tabe mohadabe doroste...

---------------------------------------------------------------
tavajoh dashte bashid ke dar akhar az ye ghazie jadid estefade kardim,ghaziehe ine:

[b]farz konid $ a,b,c \in \mathbb{R}$ va $ f: \mathbb{R}\rightarrow\mathbb{R^{+}}$ ye tabe soudi ya mohadab bashe
oonvaght namosavie zir bargharare:

$ f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$[/b]

esbate in ghaziaro mizaram be ohde [b]mehdi[/b] ke esbatesho benevise va chanta ham mesal bezare azash ( :D )

Fekr mikonam rahe hallet doroste hala ye rahe halle sade tar :
az oonjayi ke residi be in namosavi edame midam :
$ \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{c+a}+\frac{c(c-a)(c-b)}{a+b}\geq 0$ $ \Longleftrightarrow \frac{a(a^{2}-b^{2})(a^{2}-c^{2})+b(b^{2}-c^{2})(b^{2}-a^{2})+c(c^{2}-a^{2})(c^{2}-b^{2})}{(a+b)(b+c)(c+a)}\geq 0$
$ \Longleftrightarrow a(a^{2}-b^{2})(a^{2}-c^{2})+b(b^{2}-c^{2})(b^{2}-a^{2})+c(c^{2}-a^{2})(c^{2}-b^{2}) \geq 0$
hala dar namosavi Schur baraye $ a^{2},b^{2},c^{2}$ gharar midim $ r=\frac{1}{2}$ . va be sadegi namosavie bala be dast miad :)

ye zarre ghaziye motefaavete dar haghighat tooye tarife tabe fargh mikone:
farz konid $ f$ yek tabe bashad ke maghadirash namanfi hastand va rooye maghaadire mosbat tarif mishavad(tafavotesh hamine va in mosbat boodan kheili moheme) va in tabe soudi "yaa" mohaddab bashe.(yani harkodoom az sharthaye soudi ya mohaddab boodan ba dar nazar gereftane damane va bord hokm ro natije midan.)va agar $ x,y,z$ se adade mosbat bashand daarim:
$ f(x)(x-y)(x-z)+f(y)(y-x)(y-z)+f(z)(z-x)(z-y)\geq0$
esbaate ghaziye(agar az tabe tanhaa sharte soudi boodan ra dashte bashim):
ba tavajjoh be taghaarone masale farz konid ke $ x>y>z$(agar dotaye oonha barabar bashand hokm badihi khahad bood) va gharar dahid:
$ x=z+a$ va $ y=z+b$ ke $ a>b>0$ pas namosavi be in shekl dar miyad:
$ f(z+a)(a-b)a+f(z+b)(b-a)b+f(z)ab=(a-b)(f(z+a)a-f(z+b)b)+f(z)ab\geq0$ zira bana bar soudi boodane tabe darim $ f(z+a)\geq{f(z+b)}$ va $ a>b>0$ va $ f(z)\geq0$.
(agar sharte mohaddab boodan ra dashte bashim):baz farz konid$ x>y>z$.
midaanim ke $ \frac{x-y}{x-z},\frac{y-z}{x-z}<1$ pas $ \frac{x-z}{y-z},\frac{x-z}{x-y}>1$ banaa barin daarim:
$ f(x)(\frac{x-z}{y-z})+f(z)(\frac{x-z}{x-y})\geq{f(x)(\frac{y-z}{x-z})}+{f(z)(\frac{x-y}{x-z})}\geq{f(x(\frac{y-z}{x-z})+z(\frac{x-y}{x-z}))}=f(y)$                    <<< (bana bar tarife tabe mohaddab)
pas $ f(x)(\frac{x-z}{y-z})+f(z)(\frac{x-z}{x-y})\geq{f(y)}$ banabarin
$ f(x)(x-z)(x-y)+f(z)(x-z)(y-z)\geq{f(y)(x-y)(y-z)}=-f(y)(y-x)(y-z)$
ke age deghat konin hamoon namosavie morede nazare maast.pas hokm be tore kaamel saabet shod.
hala chand ta mesaal.rastesh ina ro khodam tarh kardam(bebakhshid age kasife) vali say kardam chizaye ghashangi beshan.
ba hamoon mafroozate ghaziye saabet konid:
1)$ x({x^{2}}-{y^{2}})({x^{2}}-{z^{2}})+y({y^{2}}-{x^{2}})({y^{2}}-{z^{2}})+z({z^{2}}-{x^{2}})({z^{2}}-{y^{2}})\geq0$
2)$ x(y-x)(yz-1)+y(x-y)(xz-1)+xy(1-yz)(1-xz)\geq0$
3)$ {x^{(y-x)(x-z)}}{y^{(z-y)(y-x)}}{z^{(x-z)(z-y)}\geq1}$
khob hala halle in se taa ro mizaram be ohde BaBak .BaBak bebinam chikar mikoni(:D)
( :cursing: chandin baar editesh kardam taa dorost shod :cursing: )

bebakhshid khashayar man postam neveshtanesh kheili toolani shod chon man chand baar editesh kardam.baraye hamin ham
nadidam ke to daghighan be yeki az mesaal hayi ke gozashtam be ohdeye babak residi.
dar har haal sharmande :blush:

[quote="mehdi"]2)$ x(y-x)(yz-1)+y(x-y)(xz-1)+xy(1-yz)(1-xz)\geq0$
3)$ {x^{(y-x)(x-z)}}{y^{(z-y)(y-x)}}{z^{(x-z)(z-y)}\geq1}$
khob hala halle in do taa ro mizaram be ohde BaBak .BaBak bebinam chikar mikoni(:D)
[/quote]
aval az hame soale $ 3$,ino man bedoone ghaziehe halesh kardam,ba in ghaziehe roosh fek nakardam,hala badan age toonestam ye rahe halle intoriam vasash mizaram:

$ \boxed{\tex{soale 2: }}$

biaid parantezaro baz konid mishe:

$ y^{2}za-x^{2}yz-xy+x^{2}+x^{2}yz-y^{2}zx-xy+y^{2}+x^{2}y^{2}z^{2}-x^{2}yz-y^{2}zx+xy\geq 0$

pas age sadash konim mishe:

$ x^{2}-xy+y^{2}+x^{2}y^{2}z^{2}-x^{2}yz-y^{2}zx\geq 0$

pas kafie sabet konim:

$ x^{2}+y^{2}+x^{2}x^{2}y^{2}z^{2}\geq xy+x^{2}yz+y^{2}zx\geq 0$

ke inam vazehe,kafie tooye namosavie $ a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ gharar bedid $ a=x,b=y,c=xyz$  :D

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

$ \boxed{\tex{soale 3: }}$

midoonim $ \ln 1=0$ pas age biaim az dotarafe namosavi $ \ln$ begirim kafie sabet konim:

$ \ln(x^{(y-x)(x-z)}y^{(z-y)(y-x)}z^{(x-z)(z-y)})\geq 0$

midoonim $ \log^{ab}_{c}=\log^{a}_{c}+\log^{b}_{c}$ va $ \log^{a^{b}}_{c}=b\log^{a}_{c}$

pas samte chape namosavie akhar be shekle zir dar miad:

$ \ln x^{(y-x)(x-z)}+\ln y^{(z-y)(y-x)}+\ln z^{(x-z)(z-y)})\geq 0$

pas kafie sabet konim:

$ (y-x)(x-z)\ln x+(z-y)(y-x)\ln y+(x-z)(z-y)\ln z\geq 0$

khob hala midoonim tabe $ f(x)=-\ln x$ ye tabe mohadabe pas tebghe ghazie ke goftim bara adade $ x,y,z$ darim:

$ f(x)(x-y)(x-z)+f(y)(y-z)(y-x)+f(z)(z-x)(z-y)\geq 0$

$ \Rightarrow-(x-y)(x-z)\ln x-(y-z)(y-x)\ln y-(z-x)(z-y)\ln z\geq 0$

yeni darim:

$ (y-x)(x-z)\ln x+(z-y)(y-x)\ln y+(x-z)(z-y)\ln z\geq 0$ yeni hamoon chizi ke mikhastim  :)

[quote="BaBaK Ghalebi"]rasti be joz esbate mehdi ye esbate digeam has,ziad fargh nemikone albate vali goftam shayad fahme in rahat tar bashe:

[hide]aval tavajoh dashte bashid ke $ a,b,c$ motemayezand chon age dotashoon barabar bashe namosavimoon mishe $ 0\geq 0$ ke
doroste...

pas bedoone az das dadane kolliat farz konid $ a>b>c$ dota halato dar nazar begirid:

$ \boxed{\text{halate aval}}: f(b)\geq f(c)$

khob aval az hame chon $ f$ mohaddabe darim:

$ f(a)x_{1}+f(c)x_{2}\geq f(ax_{1}+cx_{2})$ke $ x_{1}=\frac{b-c}{a-c},x_{2}=\frac{a-b}{a-c}$ pas $ x_{1}+x_{2}=1$...

hala samte raste namosavie akhar barabare ba $ f(b)$ pas darim:

$ f(a)(b-c)+f(c)(a-b)\geq f(b)(a-c)$

$ \Rightarrow f(a)\geq f(b)\frac{a-c}{b-c}-f(c)\frac{a-b}{b-c}$

$ \boxed{\Rightarrow f(a)(a-c)\geq f(b)\frac{(a-c)^{2}}{b-c}-f(c)\frac{(a-b)(a-c)}{b-c}}$

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

khob hala ma mikhastim sabet konim:

$ f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$ biaid tarafeine ino be $ a-b$ taghsim konid mishe (tavajoh dashte bashid
ke $ a-b>0$):

$ f(a)(a-c)-f(b)(b-c)+f(c)\frac{(c-a)(c-b)}{a-b}\geq 0$

khob hala midoonim $ f(c)\geq 0$ hamchenin $ c-a<0,c-b<0$ pas $ (c-a)(c-b)\geq 0$ pas $ f(c)\frac{(c-a)(c-b)}{a-b}\geq 0$

pas kafie sabet konim $ f(a)(a-c)-f(b)(b-c)\geq 0$

hala ba tavajoh be rabetei ke be das avordim kafie sabet konim:

$ [f(b)\frac{(a-c)^{2}}{b-c}-f(c)\frac{(a-b)(a-c)}{b-c}]-f(b)(b-c)\geq 0$

pas kafie sabet konim:

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

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

$ f(b)(a-b)(a-c+b-c)\geq f(c)(a-b)(a-c)$

$ f(b)(a-b)(a-c)+f(b)(b-c)\geq f(c)(a-b)(a-c)$

$ \Rightarrow [f(b)-f(c)](a-b)(a-c)+f(b)(b-c)\geq 0$

ke inam chon farz karde boodim $ f(b)\geq f(c)$ doroste...

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

$ \boxed{\text{halate dovom}}: f(b)<f(c)$

khob alan ma bayad sabet konim:

$ f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$

amma darim:

khob chon $ f(a)\geq 0,a-b\geq 0,a-c\geq 0$ kafie sabet konim:

$ f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$

biaid ino bazesh konid mishe:

$ f(b)(b^{2}-ba-bc+ac)+f(c)(c^{2}-ca-cb+ab)\geq 0$

$ [f(b)+f(c)](b^{2}+c^{2}-bc)+[f(c)-f(b)](ab-ca)\geq 0$

ke inam doroste chon farz kardim $ f(c)>f(b),b>c$  :D[/hide][/quote]
esbaat naghese BaBak jan.yani aslan tabe soudi ma nemibinim va dar zemn dar halate dovvom ham baraye tavabe akidan nozooli saabet shode yani hamchenin halati baraye tavaabe soudi nadarim.

ye soal:manzooret az masaleye 3 hamoon akharie bood ke gozashtam?(bad az harfi ke be khashayar zadam)

ye masale minevisam ke ba schur hal mishe va besyar sadas.(albatte age man masalei tarh konam migam.in masale besyar tamize va sadas va kheili az sathe masale tarh kardane man balatare va age eshtebah nakonam kare Gabriel Dospinescu va ye nafare digas.)
agar $ a+b+c=1$ va $ a,b,c>0$ saabet konid
$ 5(a^{2}+b^{2}+c^{2})\leq6(a^{3}+b^{3}+c^{3})+1$

Dar morede soale mehdi age hamgenesh konid mishe :
$ 5(a^{2}+b^{2}+c^{2})(a+b+c)\leq 6(a^{3}+b^{3}+c^{3})+(a+b+c)^{3}$. ke inam age sade konin mishe :$ 2(a^{3}+b^{3}+c^{3}+3abc)\geq 2(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)$ .
ke namosavie Schur hastesh baraye $ a,b,c$ va $ r=1$ .

Manam ye soale dige mizaram :
farz konid $ a,b,c>0$ va dashte bashim $ a^{2}+b^{2}+c^{2}=a+b+c$ sabet konid :
$ a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\leq ab+bc+ca$

[quote="mehdi"]bache ha man be jaye oon masaleye maskhare(mesaale 1) say kardam yeki behtar tarh konam va bezaaram.faghat kasifisho be khoobi va tamizie khodetoon bebakhshid:
agar $ a,b,c>0$ va $ abc=ab+bc+ca$ saabet konid:
$ 3(\frac{c}{ab}{(a-b)^{2}}+\frac{a}{bc}{(b-c)^{2}}+\frac{b}{ca}{(c-a)^{2}})\geq abc-27$[/quote]

soale khoobi bood va oonghadram kasif kari nadasht in rahe hale man :
mishe oon farze avvaliaro injoori nevesht $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ ke inam mishe hamgen kard yani mishe farz kard :
$ a=\frac{x+y+z}{x}$ , $ b=\frac{x+y+z}{y}$ , $ c=\frac{x+y+z}{z}$ hala age gharar bedim too namosavi kasre avvale samte rast mishe :
$ 3(\frac{xy}{z(x+y+z)}.\frac{(x-y)^{2}(x+y+z)^{2}}{x^{2}y^{2}}= 3\frac{(x-y)^{2}(x+y+z)}{xyz}$ .
ke age vase kasre dovvomo sevvom ham benevisin samte rast mishe :
$ \frac{3(x+y+z)((x-y)^{2}+(y-z)^{2}+(z-x)^{2})}{xyz}$
va samte chap mishe $ \frac{(x+y+z)^{3}-27xyz}{xyz}$ .
pas bayad sabet konim :
$ 6(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx) \geq x^{3}+y^{3}+z^{3}+3x^{2}y+3x^{2}z+3y^{2}x+3y^{2}z+3z^{2}x+3z^{2}y-21xyz$
hala oon ebarate samte chap mishe $ 6(x^{3}+y^{3}+z^{3}-3xyz)=6x^{3}+6y^{3}+6z^{3}-18xyz$
pas bayad sabet konim $ 5(x^{3}+y^{3}+z^{3})+3xyz \geq 3x^{2}y+3x^{2}z+3y^{2}x+3y^{2}z+3z^{2}x+3z^{2}y$ ke inam sadast :) . ba Schur va AM-GM sabet mishe :)

[quote="mehdi"]
esbaat naghese BaBak jan.yani aslan tabe soudi ma nemibinim va dar zemn dar halate dovvom ham baraye tavabe akidan nozooli saabet shode yani hamchenin halati baraye tavaabe soudi nadarim.

ye soal:manzooret az masaleye 3 hamoon akharie bood ke gozashtam?(bad az harfi ke be khashayar zadam)[/quote]
avalan ke man faghat halati ke $ f$ mohaddab basharo esbatesho neveshtam...
asanam azin chizai ke to migi estefade nakardam :D
bebin mikhaim vase seta adade motamayeze $ a>b>c$ oon namosavi ro esbat konim dge,khob hala dota halat darim ya $ f(b)\geq f(c)$ ya inke $ f(c)>f(b)$...
man hich harfi az soudfi ya nozooli boodan nazadam,nagoftam vase har $ b,c$ ke,man faghat goftam in $ b,c$i ke ma mikhaim vasash sabet konim dota halat darano edame dastan...
------------------
are manzooram hamoon $ 3$ta soaliu bood ke gozashti ke man tamrin konam ta balke too in mabhas ghavi sham  :D

[quote="BaBaK Ghalebi"]avalan ke man faghat halati ke $ f$ mohaddab basharo esbatesho neveshtam...[/quote]
bebin BaBak jan.vaghti minevisi "esbaat" yani esbaat dar har haalat.dar zemn in harfeto ke dar morede $ b,c$ bood ghabool daram va to khoob midooni ke age eshkaal migiram male in nist ke mikham zayat konam.aslan hamchin fekri dar shane man o to nist.pas eshtebah nakon.man ahle kalkal ba hich ahadonnasi nistam.khosoosan to ke baram kheili azizi va arzesh dari.

[quote="BaBaK Ghalebi"]are manzooram hamoon $ 3$ta soaliu bood ke gozashti ke man tamrin konam ta balke too in mabhas ghavi sham  :D[/quote]

baz ham dari eshtebah mikoni.man vaghti migam to hal kon va vaghti migam "bebinam chikar mikoni" manzooram ine ke chon midoonam to be khoobi mitooni hal koni va az pasesh bar miyay va khoob tozih midi be to migam ta balke ba rahayi ke to hal mikoni va tori ke to hoselasho dari ta tozih bedi baghiyeie bache hayi ke taze mikhan in raho beran estefaadeye khoobi bebaran va chiz yaad begiran.

baba mehdi bikhial ye tikke bood endakhtam,hamin,dar sathe ye shookhi bood faghat... :D
-------------------
rajebe esbatam begam ke:
avalan man tarife esbat ro baladam (inam ti8kkas be del nagir  :D )vali khob man too hale namosavie bala azin ghazie estefade kardam:
[b]age $ f: \mathbb{R}\rightarrow\mathbb{R^{+}}$ tabei mohaddab bashe va $ a,b,c\in\mathbb{R}$ oonvagh darim:
$ f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)$[/b]
ke chanta post badam esbatesh kardam,vase inke rahe hallam kamel basheo bedoone naghs...
badesh goftam vase oonai ke alaghemandan bedoonan ke in ghazie vase tavabe soudi ham doroste ke bedoonan,toam ke lotf kardio esbatesho neveshti...
manam oon ghesmati ke vase hallam niaz boodo sabet kardam...  :)


geometry
perimeter
advanced fields
advanced fields unsolved

Prove that n^2+1 unit squares in a plane cannot cover any square of side length greater than n.
(The unit squares may be rotated.  The interior of the large square, as well as its perimeter, must be covered.)

If you solve it, would you please let me know at erickson@truman.edu.  Thank you.

Martin Erickson

here's a hint to help you do this problem: the solution is real easy.

PS. i'm not kidding.


$ f(x)\equal{}\frac{1\plus{}c x^2}{1\plus{}x^2}, c \neq 1$

Find the taylor series centered around a. I wrote down the first few terms, but can't find the general term.

$ f(x)\equal{}\frac{1\plus{}c a^2}{1\plus{}a^2}\plus{}\frac{2a^2(c\minus{}1)}{(1\plus{}a^2)^2}(x\minus{}a)\plus{}\frac{4a(c\minus{}1)(1\minus{}a^4)}{2(1\plus{}a^2)^4}(x\minus{}a)^2\plus{}...$

I can't really tell either because it looks like each new expansion introduces some new term.

[quote="Gib Z"]$ f(x) \equal{} \frac {1 \plus{} c x^2}{1 \plus{} x^2}, c \neq 1$
[/quote]
The first step what I would do is write,
$ \frac {1 \plus{} cx^2}{1 \plus{} x^2} \equal{} \frac {(1 \plus{} x^2) \plus{} (c \minus{} 1)x^2}{1 \plus{} x^2} \equal{} 1 \plus{} (c \minus{} 1) \cdot \frac {x^2}{1 \plus{} x^2}$
And so you now need to work with $ x^2/(1 \plus{} x^2)$ instead.
But $ \frac{x^2}{1\plus{}x^2} \equal{} \frac{(1\plus{}x^2) \minus{} 1}{1\plus{}x^2} \equal{} 1 \minus{} \frac{1}{1\plus{}x^2}$.

[quote="JRav"]sylow_theory, I like your signature.[/quote]
Thanks, its the general Stokes' theorem.

Yea I tried doing the series for 1/(1+x^2) but I still couldn't do that, same problem, new terms keep arising.


I want delete the topic
How to delete it......
I can only edit it......

How do I edit or delete a post?
Unless you are the board admin or forum moderator you can only edit or delete your own posts. You can edit a post (sometimes for only a limited time after it was made) by clicking the edit button for the relevant post. If someone has already replied to the post, you will find a small piece of text output below the post when you return to the topic that lists the number of times you edited it. This will only appear if no one has replied; it also will not appear if moderators or administrators edit the post (they should leave a message saying what they altered and why). Please note that normal users cannot delete a post once someone has replied.

The question is why would you want to delete a topic?


Determine all real $ x$ satisfying the equation \[ \sqrt[5]{x^3 \plus{} 2x} \equal{} \sqrt[3]{x^5\minus{}2x}.\] Odd roots for negative radicands shall be included in the discussion.

We must have $ x \geq 0$ and $ x^5 \minus{} 2x \geq 0$, so we must have $ x \in [\sqrt [4]{2}, \plus{} \infty)\cup\{0\}$

Set $ 2x \equal{} k$ and let $ f(x) \equal{} \sqrt [5]{x^3 \plus{} k}, f'(x) > 0 \Rightarrow f: 1 \minus{} 1$

Then $ f^{ \minus{} 1}(x) \equal{} \sqrt [3]{x^5 \minus{} k}$, and the equation becomes:

$ f(x) \equal{} f^{ \minus{} 1}(x) \Leftrightarrow f(x) \equal{} x \Leftrightarrow x^3 \plus{} k \equal{} x^5 \Leftrightarrow x^5 \minus{} 2x \minus{} x^3 \equal{} 0 \Leftrightarrow x(x^4 \minus{} x^2 \minus{} 2) \equal{} 0 \Leftrightarrow x(x^2 \minus{} 2)(x^2 \plus{} 1) \equal{} 0 \Leftrightarrow x \in \{0, \sqrt{2}, \minus{}\sqrt{2}\}$

But $ \minus{}\sqrt{2} \notin [\sqrt [4]{2}, \plus{} \infty)\cup\{0\}$, so the equation has the solutions $ x \equal{} 0$ and $ x \equal{} \sqrt{2}$

$ x\equal{}\pm \sqrt{2}$ are also solutions.

I've come up with this simple solution:
x=0 is a obvious root, so I only consider $ x \not\equal{} 0.$
OK. We have:
$ (x^3\plus{}2x)^3 \equal{} (x^5\minus{}2x)^5$
$ \Leftrightarrow x^3(x^2\plus{}2)^3 \equal{} x^5(x^4\minus{}2)^5$
$ \Leftrightarrow (x^2\plus{}2)^3 \equal{} x^2(x^4\minus{}2)^5$  (because $ x\not\equal{} 0) (1)$
Now, let $ y\equal{}x^2 \plus{} 2$. So $ x^4\minus{}2 \equal{} (x^4 \minus{} 4) \plus{} 2 \equal{} (x^2\plus{}2)(x^2\minus{}2) \plus{} 2\equal{}  y(y\minus{}4)\plus{}2$.
Thus, $ (1) \leftrightarrow y^3 \equal{} (y\minus{}2).[y(y\minus{}4)\plus{}2] \equal{} (y\minus{}2).[(y\minus{}2)^2 \minus{}2]^5$
All right.  Divide each side by $ y^3$:
$ (y\minus{}2).[(y\minus{}2)^2\minus{}2]^2.\displaystyle \left(\frac{y^2\minus{}4y\plus{}2}{y} \right)^3 \equal{} 1$
It's easy to see that if $ f(t) \equal{} \frac{t^2\minus{}4t\plus{}2}{t} \equal{} t \plus{} \frac{2}{t} \minus{} 4,$ then $ f(t_1)<f(t_2)$ if $ 2<t_1<t_2.$
Because $ y\equal{}x^2 \plus{} 2, x \not\equal{} 0$, $ y>2$.
Therefore, y=4 is the only nonzero root of (1).
Then, $ x\equal{}\pm \sqrt{2}$.

What?! I can't find out the way you get this answer. When $ y \equal{} 4$ (or $ x \equal{} \pm \sqrt{2}$), we have $ LHS \equal{} 1.$  :huh:

[quote="iTAQ"]What?! I can't find out the way you get this answer. When $ y \equal{} 4$ (or $ x \equal{} \pm \sqrt {2}$), we have $ LHS \equal{} 1.$  :huh:[/quote]

Sorry, you are right, i did a mistake while trying to solve the equation and another one while checking if $ \sqrt {2}$ is a solution of the equation :blush:  :blush:

I deleted my post above and I corrected my solution...

Now $ \sqrt {2}$ is a solution of the equation, but I can't understand why $ \minus{} \sqrt {2}$ is a solution too. As I know from the high-school, if we have $ x^k \equal{} y < 0, k: odd, x \in R$, then we cannot write $ x \equal{} \sqrt [k]{y}$, but we write $ x \equal{} \minus{} \sqrt [k]{ \minus{} y}$, because under the root we cannot have a negative number

here my solution:

obviously $ \minus{}\sqrt{2}$ , $ 0$ and $ \minus{}\sqrt{2}$ are solutions to the equation:
By assuming $ x \neq 0$ we may substitute $ u\equal{}x^2$ and obtain following:

$ (u\plus{}2)^3 \equal{} u(u^2\minus{}2)^5$

I will now show that the only positive solution of the last equation is 2.

[b]case 1:[/b] $ 0<u<2$

$ (u\plus{}1)(u\minus{}2) < 0$
$ u^2\minus{}u\minus{}2 < 0$
$ u^6<(u\plus{}2)^3$
and
$ u^2\minus{}2<u$
$ (u^2\minus{}2)^5 < u^5$
$ u(u^2\minus{}2)^5 < u^6$

Therefore for case 1 the inequality $ u(u^2\minus{}2)^5 < u^6<(u\plus{}2)^3$ holds and we have no solution.

[b]case 2:[/b] $ u \equal{} 2$

the equation is true

[b]case 3:[/b]$ 2 < u$

 $ (u\plus{}1)(u\minus{}2) > 0$
 $ u^2\minus{}u\minus{}2 >0$
 $ u^2 > u\plus{}2$
 $ u^6 > (u\plus{}2)^3$
and
$ u^2\minus{}2 > u$
$ (u^2\minus{}2)^5 > u^5$
$ u(u^2\minus{}2)^5> u^6$

Therefore for case 2 the inequality $ (u\plus{}2)^3 < u^6<u(u^2\minus{}2)^5$ holds and we have no solution.

As you can see, $ u\equal{}2$ is the only solution. By solving the equation x^2 = u = 2 we can verify that  $ \minus{}\sqrt{2}$ , $ 0$ and $ \minus{}\sqrt{2}$ are the only solutions. $ \Box$

Wow, incredible solution. However, it's a bit unnatural.
One more thing, the roots are $ 0, \pm \sqrt{2}.$

To NickNafplio:
Only $ \sqrt[k]{y}$ is nonsense if k is even and y<0. Is that right?

no. The problem also included negative numbers when k is odd. Look at the first post.


MATHCOUNTS
search
blogs
number theory
modular arithmetic

[quote="lightspeed"]Sprint round #24:
Reverse the two digits of my age, divide by three, add 20, and the result is my age.  How many years old am I?

How do you do this one quickly? :lol:[/quote]

Don't you have to know the result to do this? Either that or i'm totally clueless... (there's a lot of two digit numbers that can be reversed and divided by three)

let his age be 10x+y
Then (10y+x+60)/3=10x+y
from this we get: 29x-7y=60, and fooling around with this for a little gives x=4,y=8 , he is 48 years old

84/3 +20=48 so it works.

[quote="frost13"][hide]if age is two digits, write it as 10x  + y where x is 10's digit and y is units digit.

Now make this equation : $10x+y = \frac{10y+x}{3} +20$

if you simplify, you get $29x - 60 =7y$

Start plugging in values for x and look for 29x to be a multiple of 7 more than 60.

Don't bother with 1 or 2 since multiplying by 29 won't get you over 60.  At 4 you get

116 - 60 = 56 =7y so y=8

So 48 is the age

[/hide][/quote]

Oh. I get it. Stupidi me. I've done questions like this before...

EDIT: Never mind. Did something stupid...

this involves both algebra and logic, but it isn't too hard, once you figure it out

[hide="solution"]
x=tens digit
y=ones digit
10x+y=(10y+x)/3+20  (multiply each side by 3)
30x+3y=10y+x +60  (simplify)
29x=7y+60

now, just try each integer possibility for x:
x=1
7y+60=29 (wrong)
x=2
7y+60=58 (wrong)
x=3
7y+60=87
7y=27 (has to be an integer)
x=4
7y+60=116
7y=56
y=8

your age is

[b]48[/b] answer
[/hide]

[quote="Hexahedron"][quote="DarkKnight"][quote="236factorial"]I saw this problem too, and I did it by guess and check. I finally got 48 after a bit of time (but I was done with the rest of the problems). I knew that some numbers were impossible to be the answer (like 11, 21, 31, etc). My other classmates all got stumped!  :lol:[/quote]

Hmm? Why is 21 impossible?
21 reversed is 12
12/3=4
4+20=24?[/quote]

21 does not equal 24  :D[/quote]

Whoops... Lol. Sry wasn't thinking.

Never mind, never mind, I don't even know how I got it now.  :(

[quote="rcv"]Lightspeed:  In another thread, you are asking about the use of modular arithmetic.  This is an example where modular arithmetic can get you the answer without any guess-and-check.

[hide="Borrowing heavily upon frost13"]

If age is two digits, write it as $10x  + y$ where $x$ is tens digit and $y$ is units digit.

\[\begin{eqnarray*}
10x+\phantom{1}y &=& \frac{10y+x}{3} +20 \\
30x+3y&=&10y+x+60\\
29x &=&7y + 60
\end{eqnarray}\]

Now, we take both sides, modulo 7.  (The remainder, when divided by 7:

\begin{eqnarray*}
29x &=&7y + 60 \\
(28+1)x &=&7y + (56+4) \\
28x+x &=& 7y + 56 + 4 \\
0 + x &\equiv& \phantom{y}0 + \phantom{0}0 + 4 \hbox{\qquad\qquad} (\hbox{modulo }7)\\
x &=& 4\hbox{ or }11\hbox{ or }18\hbox{ or }\ldots
\end{eqnarray*}

The only "digit" possible is $x=4$.  Then solve equation (1):

\begin{eqnarray*}
29x &=& 7y + 60 \\
29\cdot 4 &=& 7y+60 \\
116 &=& 7y+60 \\
56 &=& 7y\\
8 &=& y
\end{eqnarray*}

So, the required age is $48$.  Always check your work, especially when using modular arithmetic.[/hide][/quote]

Wow thats a good explanation.  Thanks!

I learned how to do problems like this the monday after chapter... but i still did okay even though i skiped the problem. I hate problems that needs to find a number after you reverse the digits and do stuff, and it just so happens that there's a lot on mathcount

I remember this problem from this years' chapter. :D

[hide]10x+y is the age

(10y+x)/3+20=10x+y

10y+x+60=30x+3y

7y+60=29x

I just guesed and checked and got 48[/hide]

There aren't a high percentage, but there are a good number. sky9073, LynnelleYe, Lucy, Ladyknight, KBabe, etc...

Search the forums for a topic like your question and you will get a interesting topic, I think in F&G.

Is there a way I can just type it in somewhere if I want to search for a topic, or do I have to look for it on my own?  Thanks

I have no idea why.... girls get.... um.... distracted sometimes.... like in our school, even the girls in mathteam gets really crazy.... i'm kinda crazy too, but i started early. In CHina, you have to do like lots of homework and stuff.... so i know the basics pretty well

There is a button under the Forum tab beside AoPs Blogs that says search.

[hide]48. You must make up an algebraic equation first and then you have your answer by trying to solve your equation. You cannot fully solve it, but you just need to do at most, five guess and checks or use mod and you get your answer.[/hide]

It took me 5 minutes to guess and check, but I got it right... :blush:

[quote="rcv"]Lightspeed:  In another thread, you are asking about the use of modular arithmetic.  This is an example where modular arithmetic can get you the answer without any guess-and-check.

[hide="Borrowing heavily upon frost13"]

If age is two digits, write it as $10x  + y$ where $x$ is tens digit and $y$ is units digit.

\[\begin{eqnarray*}
10x+\phantom{1}y &=& \frac{10y+x}{3} +20 \\
30x+3y&=&10y+x+60\\
29x &=&7y + 60
\end{eqnarray}\]

Now, we take both sides, modulo 7.  (The remainder, when divided by 7:

\begin{eqnarray*}
29x &=&7y + 60 \\
(28+1)x &=&7y + (56+4) \\
28x+x &=& 7y + 56 + 4 \\
0 + x &\equiv& \phantom{y}0 + \phantom{0}0 + 4 \hbox{\qquad\qquad} (\hbox{modulo }7)\\
x &=& 4\hbox{ or }11\hbox{ or }18\hbox{ or }\ldots
\end{eqnarray*}

The only "digit" possible is $x=4$.  Then solve equation (1):

\begin{eqnarray*}
29x &=& 7y + 60 \\
29\cdot 4 &=& 7y+60 \\
116 &=& 7y+60 \\
56 &=& 7y\\
8 &=& y
\end{eqnarray*}

So, the required age is $48$.  Always check your work, especially when using modular arithmetic.[/hide][/quote]

How do you know when to use modular arithmetic?

I hated that problem!  I did it by brute force (after the competition was over, while driving home) and I got it after about 5 minutes.  Thanks for the formula.

[quote="lightspeed"]Lucy, you're one of the few girls I've seen at these forums.  How come there aren't many girls?[/quote]

i just remembered that at our chapter contest, out of the top 10, only two people were girls...

In our chapter, the top 16 went into the countdown (Unofficial), and the only two girls that made it were in our school...

same here, I got fourth, and she got third.... in her case, countdown made her fall down a place, it got me up 2 places  :lol:

I loved the chapter countdown (I kind of dominated, minus the stupid error from time to time)

It is really noticable, in my school's math team, there are only 2-3 girls (all three come about two thirds of the time), and at MA states, there was only one girl in the countdown...I ought to stop ending my posts with triple periods. There.

um... i got as far as i could then after reading frost13's answer it kind of made me feel dumb.

[hide="my pathetic answer that included guessing and cheking"]
so i did:
10x+y=(10y+x)/3 +20
times three on both sides
30x+3y=(10y+x)+60
  -x -3y    -3y -x
-------------------------------
29x=7y+60
so used a ton of guess and checking and got that
x=4 and y=8

so thats my pathetic answer. haha um.. ya

48 :D [/hide]

[quote="mathlover"]um... i got as far as i could then after reading frost13's answer it kind of made me feel dumb. [/quote]
Reading this thread, and talking to my own team, almost everybody who solved the problem solved it with guess and check.  This problem was one of the hardest at my Chapter.  Less than 10% of the mathletes solved it correctly.

frost13 is a well-known as an outstanding coach.  He was just trying to offer a "better" solution than most students found on their own, under the time pressures.  No need to feel dumb.

[quote="rcv"][quote="lightspeed"]Frost is a coach?  I didn't know that! :blush:[/quote]
Sorry, frost13.  :ninja:  Your cover is blown.  :ninja:[/quote]

Lol very funny.  After reading your post, I saw one of his posts, revealing his true identity...
:D:D:D:D

I never knew my coaching identity was such a mystery :D   Sorry if I passed myself off as a kid.  I chack in here 20 times a day, but usually only post when people need help :)

RCV is too kind in his compliments of my coaching skills.  I have been lucky to have so many bright students on my teams.  I have posted a few times about "my kids" or my past experiences at Nationals(the last 4 years).  So sorry if I misled you guys, or if my insights make any of you feel less bright :(   Heck, sometimes RCVs solutions make me feel that way, but in the end it just mean I've learned something new!!!!

The only one of my students who spends much time here is dwx314 who is working hard to try for Nats as a 7th grader.

Good luck to all of you.

Coach Frost13 :lol:

I missed this questoin on the chapter (and spent time on it, too!) and we just learned how to do this in Algebra class, say, 10 days after chapter!  :(  [color=darkblue][size=75]<snip snip.  Watch the language, even if abbreviated, please.>[/size][/color]

Just watch the language fanatic, please. We like math to be civilized even if you're <snip> about the evil chapter questions... :)

[hide="my simple way of  "modular arithmetics""] After we get 29x=60+7y, we write:  29x-60  :equiv: 0 mod 7.
i.e. x-4 :equiv: 0 mod 7. That gives x=4 and y = 8. [/hide]


I'm a new immigrant to the US. My school doesn't offer the AMC 12 test and I really don't know how to participate.

Can I have the AMC in another school or in a college?
Do I have to pay any fees for that?
And what more should I do to participate?

Thank you for any help.

Please see the answer to Frequently Asked Question #4 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Also see the list of college and University adminstration sites and contact the mathematics department of a nearby college or university if a close site is not already listed.

http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

I'm sorry but I'm not sure understand this:

"If there is no site listed for your area yet, and there is a College or University nearby, call the Mathematics Department and ask for the MAA (Mathematical Association of America) Representative, and explain to them that you are looking for a site to take the AMC 10B/AMC 12B."

So I have to call the Mathematics Department of the university nearby, and ask them for the MAA?

Does anybody know if there are any colleges or universities in Canada (preferably near Mississauga (or at least in the GTA)) that are taking the AMC's because my school does not want to participate...  :maybe:  :maybe:

[quote="Kent Merryfield"]Since you live in San Jose, I can't help you. For Southern California, ask me.

I would suggest contacting Ted Alper at EPGY.[/quote]

Excuse me,
what is EPGY?

[quote="humanbeing"]what is EPGY?[/quote]
[url=http://epgy.stanford.edu/]EPGY[/url].

He has an AoPS user name: TMA. (I don't think he checks that account very often.)

Well, the school has to pay to get the AMC tests, so they usually ask you to donate some money or something. It shouldn't be too much, the fee is not much to worry about.

Can anyone help me, please?
I'm worrying about the deadline of registration.

The AMC offices are closed for the holidays.  We will be back at work on Wednesday, January 2.  At that time, you can contact the AMC office (use the contact information at our website and printed on everything we produce) and we will point you to higher education sites in the San Jose area who will be offering an AMC administration.   In the meantime, I suggest again that you follow the advice in our Frequently Asked Question #4.

Registration continues until February 20 for the B-dates, so you still have plenty of time to get enrolled for hte contests in one way or another.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="humanbeing"][quote="serialk11r"]San Jose? There's a lot of high schools around that offer it. Mine, for instance, does. I dunno if you have to pay a little fee or something.[/quote]

Can you tell me your school name or a list of high schools in San Jose participating in the AMC?

And what should I do to contact them? Do I have to go there and ask the Math Department or I can send an email?
( I'm new to the country so I have no idea what to do :maybe:  )[/quote]
In the area, the high schools I can name off the top of my head are...Lynbrook, Saratoga, Gunn, Paly (palo alto high), Fremont, Bellarmine, are all in Santa Clara County. The ones that offer AMC...hmmm Gunn, Paly, Saratoga, Lynbrook, dunno about the others.


complex analysis
complex analysis unsolved

Find Dual space of Hardy space when $ 1<p<\infty$

http://www.mathlinks.ro/viewtopic.php?p=1371257#1371257
Locked!

[color=red][And I'm also locking that one, because of [url=http://www.mathlinks.ro/viewtopic.php?t=250064]triple posting.[/url] -mathmanman][/color]


Today a father's age is five times his son's age. Exactly three years ago, the sum of their ages was 30. How old is the son today?

Since the sum of the ages $ 3$ yrs ago is $ 30$, the sum of the current ages is $ 36$.

Set up an equation.

$ s \plus{} 5s \equal{} 36$

$ 6s \equal{} 36$

$ s \equal{} 6$

So the son is $ \boxed{6}$ years old today.

Let the son's age be x and the father's current age to be 5x.
Thus, 3 years ago, the son was x-3 years old and the father was 5x-3 years old.
Thus, 6x-6=30.
Since you are solving for the son's current age, or x, you solve the equation.
6x-6=30
Add 6, so 6x=36.
Divide by 6 and get [b]x=6.[/b]


Math is the path for which to travel,

it's like uncovering dirt out of gravel,

Time will tell what mysteries you uncover,

it's the whimsical wonder of the world,

Whenever I have time,

a ring in my head chimes,

and can you guess what it chimes?

MATH MATH MATH!

time for math. magnificent old math  :)

today i had a really cold bath
which is not entirely an untrodden path
the dover strait separates calais and dover
and i don't like semen on my bed cover

no ... hopeless


Let $ x$, $ y$ and $ z$ be positive real numbers. Prove that

\[ \prod{(y^4\plus{}y^2z^6\plus{}1)}\ge x^3y^3z^3(x\plus{}y\plus{}z)^3\]

:)

hint

[hide]

there is some useful information for solving this problem in this tread

http://www.artofproblemsolving.com/viewtopic.php?t=249721

[/hide]

Well my truly intentions when posting this problem was to know if there is a natural way of proving the inequality

\[ y^4\plus{}y^2z^6\plus{}1\ge yz(y^2\plus{}yz^3\plus{}1)\] for positive reals $ y$ and $ z$


When a USAMTS problem asks us to find something that is self-evident once shown, can we just write the answer and be done?

For example, if a question asks us to place plus or minus signs between the digits 123456789 so that the result is 100, can we write:
123-4-5-6-7+8-9=100?

[quote="rrusczyk"][quote="juggle7"]
For example, if a question asks us to place plus or minus signs between the digits 123456789 so that the result is 100, can we write:
123-4-5-6-7+8-9=100?[/quote]

What you have there is a 'complete solution' to the question you have stated.[/quote]
But would you get full credit for the answer?  I think I can guess at the problem he's having trouble with (though I won't say which), and it seems like he's asking whether you'd get full credit for a problem by giving an answer that is obviously true once found (like the one he gave above), but does not tell you how he got the answer.  For example, if a problem asked "find a solution to [tex]\frac{dy}{dx} = x + y[/tex]" and you gave an answer of [tex] y = k \: e^x - x - 1[/tex], with no work, would you get full credit, since it is obvious once you have the solution that it is a solution (by plugging in), but you have not shown how you solved it?

[quote="zanttrang"]But would you get full credit for the answer?  I think I can guess at the problem he's having trouble with (though I won't say which), and it seems like he's asking whether you'd get full credit for a problem by giving an answer that is obviously true once found (like the one he gave above), but does not tell you how he got the answer.  For example, if a problem asked "find a solution to [tex]\frac{dy}{dx} = x + y[/tex]" and you gave an answer of [tex] y = k \: e^x - x - 1[/tex], with no work, would you get full credit, since it is obvious once you have the solution that it is a solution (by plugging in), but you have not shown how you solved it?[/quote]

That is an inherent difficulty with show-your-work contests such as the USAMTS. Usually, we ask for some kind of proof or verification with the problems, because what we really want to grade is how well you understand and manipulate the mathematics of the problem. (A proof with algebra errors is good understanding with poor manipulation. A computer-generated solution is good manipulation with no evidence of understanding.) Even when we forget to add some token "with proof" clause, we want to see a line of reasoning.

That leaves people wondering, "Which part of my reasoning do I show?" In some problems, such as 5/2/16 from the last round, it is sufficient to show enough work to verify that your answer is correct. In other problems, such as 1/2/16 from the last round, verifying an answer is trivial, so we want you to show that you mastered a technique for creating such answers, or that you know the natural bounds on such answers. And in a few rare problems, the only way to find the correct answer is to have done all the mathematics, so we give full credit for the solution alone (but the work is necessary to earn a commendation).

Some people write a stream-of-consciousness commentary on how they found the solution. Such commentaries are a pain to read, but they do cover almost all the bases. Unfortunately, if a person found a solution by a little trial and error, then giving the full story of finding the solution won't be enough for full credit. We want to see mathematical insight. If you find a solution by luck, try to find another solution. Try to find all the solutions in any problem, just for completeness.

I recommend that if you used any interesting mathematics, some calculation or insight that you are proud to have discovered, in finding your solution, go describe that.

By the way, it is nice to get feedback like this on the wording of the problems. It tells us what is important in how we phrase the problems, so that we don't make the same mistakes again.  Any comments on the wording of the problems from Round 2? Those we can talk about freely now, though we won't have the grading rubric for all of them created until next week.

Erin Schram
USAMTS grader

I felt that the wording on problems 1/2/16, 3/2/16, 4/2/16 and 5/2/16 was solid and allowed for little misinterpretation.  Despite the necessary revision to 2/2/16 which stipulated that c and d were integers and the few other quibbles people brought up with the problem, I had no problem intrepreting it, and the merit of the problem outweighs any slight issues with its wording.

[quote="Erin J. Schram"]
Any comments on the wording of the problems from Round 2? Those we can talk about freely now, though we won't have the grading rubric for all of them created until next week.

Erin Schram
USAMTS grader[/quote]
I just have one minor problem with 4/2/16 from the last round (the quadrilateral problem); the phrase "rigorously verify" left me a bit nervous when I submitted my answers, as I didn't quite understand what that phrase meant.  I used mathematics rather than brute-force arrangement to find all the answers, but I didn't know whether that was "rigorously" verified.  I guess now, from reading others' solutions, it meant you were encouraging people to prove mathematically (as I did) that it was true, and not just to give all the arrangements and say "here they are, and there are 94 of them."  Still, "with proof" would have been less ambiguous (at least to me) than "rigorously verify".

[quote="zanttrang"]I just have one minor problem with 4/2/16 from the last round (the quadrilateral problem); the phrase "rigorously verify" left me a bit nervous when I submitted my answers, as I didn't quite understand what that phrase meant.  I used mathematics rather than brute-force arrangement to find all the answers, but I didn't know whether that was "rigorously" verified.  I guess now, from reading others' solutions, it meant you were encouraging people to prove mathematically (as I did) that it was true, and not just to give all the arrangements and say "here they are, and there are 94 of them."  Still, "with proof" would have been less ambiguous (at least to me) than "rigorously verify".[/quote]

I am the person who wrote the sentence, "Rigorously verify that no circle was skipped or counted more than once," for Problem 3/2/15 last year, so I know exactly what the similar sentence in Problem 4/2/16 was for. I had graded other counting problems on the USAMTS. Often, a solution was pretty much: "I constructed all the cases I could and found these, for a total of 44." My reply was often, "You missed 12 of them. 3 points." As zanttrang surmised, I wanted to force the people working on Problem 3/2/15 and Problem 4/2/16 to justify their counting or construction technique rigorously so that they would notice the gaps in their method before submitting the solutions. And it gave me a proof, rather than an unsorted collection, to grade.

So "Rigorously verify" was too extreme. Would "Verify" without the adverb "Rigorously" convey the idea better?

If you check the USAMTS of farther past years, say Year 13, you will notice that we almost never used the phrase, "with proof". It was always given in the instructions to show relevant work and proofs. For example the current USAMTS web site says, "Remember to fully justify all your answers - don't just send numerical answers." However, a good number of students for Year 14 Round 1 ignored that instruction, that I decided it should be written into the problems themselves. Since I was typesetting the USAMTS for years 14 and 15, I had more influence than usual (or perhaps simply more ego to put my ideas into effect), so the problems became infested with "Justify your answer" and "with proof". However, throwing in a "with proof" clause serves as a reminder but does not precisely say what should be proven. I wonder whether more detail or less detail is better.

Erin Schram
USAMTS grader

P.S. All problems for this year's Round 2 are graded as of Monday. Unfortunately, shipping the solutions and the scoring sheets back to the AoPS to enter into their database will take time.

[quote="zfzf21"]If a problem asks you to find an answer or several answers, do you have to prove that your answer is unique and/or find all possible answers? If you cannot show that the answer(s) is (are) unique because a slight bit of trial and error was involved, then can you still get full credit?[/quote]
If the problem does not ask you to give all the solutions, then you do not have to prove that you found all the solutions. Sometimes such a proof is way more difficult than a USAMTS solution should be. If you have found such a proof anyway, then include it in your solution -- it could earn you a commendation. If the problem asks you to find [i]all[/i] solutions, then you should prove that you found all solutions, because without such a proof you would not know when to stop looking for more solutions. If the problem asks you to find the unique solution, that is usually a hint to help you find the solution. You do not have to prove uniqueness unless asked to.

If you found a solution by fiddling with numbers until something worked, then merely giving the solution you found won't earn five points.  What you should do in that case is look at your method of fiddling with numbers, and see if you can describe it as a form of mathematical reasoning. Something intuitive like, "I tried this number, and the result looked too small, so I tried these numbers and the result looked too big, so I tried these numbers in between and one worked," can be formalized as, "Case 1, shown here, provides a lower bound on the solution because any smaller numbers would give a smaller result. Case 2, shown here, provides an upper bound on the solution because any larger numbers would give a larger result. Between these bounds there are only five possibilities: a, b, c, d, and e.  We do not have to examine cases d and e, because property X clearly shows that they would not work. Examining the other possibilities, we see that case c works."

Or you might look at your solutions discovered by luck and say, "Hey, all these numbers are multiples of 5. I bet I can come up with a proof that the answer has to be a multiple of 5." Such a proof would enhance your solution.

Erin Schram
USAMTS grader


Does anybody know how long the trainig camp for IMO in Turkey is?

i think two weeks in summer. two weeks in winter. and one week in spring.

Thank you xeroxia. :roll:

My professor has told me that in in turky the IMO-preparations takes a half year.

two  weeks in summer, two weeks in winter ,and for the team,2+2 weeks in spring.


abstract algebra
linear algebra
group theory
superior algebra
superior algebra theorems

I read in Hoffman & Kunze's "Linear algebra" that this strange to say that the field $\mathbb{Q}$ has characteristic $0$.
Where does this terminology come from ? I guess this has to do with the fact that the homomorphism $\mathbb{Z} \rightarrow \mathbb{Q}$ given by $n \mapsto n.1$ has kernel $0$, right ? So I do not understand why Hoffman & Kunze would be surprised by such a terminology ...

If consistency (with fields such as $\mathbb{F}_p$) was our first priority, we'd say that the rationals have characteristic $\infty$. Somehow, this is reversed to zero in all tradition.

The characteristic of a ring $R$ with identity is the non-negative generator of the kernel of the unique ring-hom. $\mathbb{Z} \to R$.  So, $\mathbb{Q}$ has characteristic $0$.

We've an analogue situation in group theory: The order of $g \in G$ is the non-negative generator of the kernel of the canonical group-hom $\mathbb{Z} \to G, z \mapsto g^z$. So, the characteristic of a ring is just the order of the $1$ in the additive group.

[b]But[/b] it's a common to say "infinite order" instead of "order 0". It's only justified by the formula $|< g > | = \text{ord}(g)$, which otherwise would not hold in general. But this convention is artificial and hides the algebraic commonness of every order ...

In contrast to this vice, it's not common to say "infinite characteristic" instead of "characteristic 0". This also would by just motivated by an infinite prime subring. But this seems not to be as important as infinite orders ...  :rotfl:


search
logarithms
number theory open
number theory

Does anyone know a generalization or similar theory to the Pell's equation theory when higher powers are considered?? (I'm thinking about the equation $x^{k}-Dy^{k}=1$). For example: what about equation \[x^{3}-2y^{3}=1\] Does it have any solution in integers x,y>0 ? If yes, then how many ?

for $k=2$ the solution comes from properties of $Q(\sqrt{D})$
for arbitrary k maybe one should try to search in $Q(\sqrt[k]{D})$
for unit elements and repeat the discussion as for $k=2$
i am not aware if this is possible or not
is it ?

For all fixed $k\geqslant3$ and $D$ number of solutions is finite. Moreover, there's method that gives upper bound for solutions. But it's extremely difficult, it uses linear forms of logarithms of algebraic numbers. About existence of solutions I don't know.

The equation $X^{3}-Dy^{3}=1$ can only have  one nontrivial solutions for all $D$ except $(1,0)$.
For $x^{3}-2y^{3}=1$ does not exist solutions except (1,0) (proven Euler).

[quote="juna"]The equation $X^{3}-Dy^{3}=1$ can only have  one nontrivial solutions for all $D$ except $(1,0)$. [/quote]
Do you know how to prove this??

This demonstration is very long that give it here.
Please, see Delone "Theory irrationality third degree"
At this monographs is considered also the equation for $k=4$.


vector
trigonometry
rotation
analytic geometry
algebra
system of equations

What is the minimum force you must apply to pull a 35-lb wagon up a frictionless ramp, inclined at an angle of 15 degrees?

Free body diagrams!

EDIT: Apparently we measure force in pounds now.  Hokay.

Thank you, but could you tell me what I did wrong in my solution?

The wagon is exerting a force straight down of 35 lbs, and the force needed to keep it there is the proportion of that force going in the direction of the ramp. Thus we get a right triangle, and we can say $ \sin{15} \equal{} \frac{x}{35}$, and $ x \equal{} 9.1 \text{ lbs}$.

I think you are right, but I don't really understand this problem... (haven't taken any physics). Sorry if my solution's confusing :(

[quote="buzzer11"]Thank you, but could you tell me what I did wrong in my solution?[/quote]
Inclined plane problems like this are typically solved in a rotated coordinate frame. With enough practice one can usually write the equilibrium equations in this rotated frame by inspection of the appropriate free body diagram. However, since I suck at generating graphics to include with my posts here I'll have to set-up the solution algebraically.

\[ \left[\begin{array}{cc}F_t \\ F_n\end{array}\right] +\left[\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]\left[\begin{array}{cc} 0 \\ -mg\end{array}\right] = \left[\begin{array}{c} 0\\0\end{array}\right]\]

Solving this system of equations you get:

\begin{eqnarray*}F_t &=& mg\sin\theta \\ F_n &=& mg\cos\theta\end{eqnarray*}

$ F_t$ is the tangential component of force required to keep the wagon from rolling down the inclined plane -- the only force you need be concerned with; it's the answer to your problem. $ F_n$ is the normal (contact) force that the inclined plane exerts on the wagon.

\[ \boxed{F_t = 35\mbox{ lbs }\sin 15^{\circ} = 9.06\mbox{ lbs.}}\]

Which is equivalent to my answer? (I rounded up.) That's not what kevmo got though...

Ignore kevmo. That post was based on mistaken ideas of how to convert units, and we don't need to convert units anyway.

Total force along the ramp due to gravity, $ F$ = force applied up the ramp to pull it up.

See diagram...


function
inequalities
quadratics
calculus
derivative
analytic geometry
graphing lines

Given that $ y\equal{}\frac{x^{2} \plus{} x \plus{} 1}{(x\minus{}1)^{2}}$, prove that $ y\ge\frac{1}{4}$ for all $ x\neq 1$.

hello, your inequality is equivalent to
$ \frac{x^2\plus{}x\plus{}1}{(x\minus{}1)^2}\geq\frac{1}{4}$
$ 4x^2\plus{}4x\plus{}4\geq x^2\minus{}2x\plus{}1$
$ 3x^2\plus{}6x\plus{}3\geq0$
$ 3(x^2\plus{}2x\plus{}1)\geq0$
$ 3(x\plus{}1)^2\geq0$
and this is true.
Sonnhard.

Vielen Dank für Ihre Antwort, Sonnhard.

AndrewTom.

We can also do this as follows:

Cross-multiplying, we get

$ (y\minus{}1)x^2 \minus{}(2y\plus{}1)x \plus{}(y\minus{}1) \equal{} 0$

This is a quadratic in $ x$, where $ x$ can take all real values, apart from $ 1$, so the discrimnant must be non-negative.

This gives

$ 12y \ge 3$

$ \Rightarrow y \ge \frac {1}{4}$.

Are there other ways of doing this?

For simpler calculation, put $ u: =x-1\iff x=u+1, u\neq 0$ to get

\begin{eqnarray*}{x^2+x+1\over(x-1)^2}&=& {(u+1)^2+u+1+1\over u^2\\ &=& {u^2+3u+3\over u^2}\\ &=& 1+{3(u+1)\over u^2}\\ &=& 1+{3\left(\left({u\over 2}+1\right)^2-\left({u\over 2\right)^2\right)\over u^2\\ &=& 1+3\left({1\over 2}+{1\over u}\right)^2-{3\over 4}\\ &=&{1\over 4}+3\left({1\over 2}+{1\over u}\right)^2}}}\end{eqnarray*}

and the conclusion follows. Equality for $ u=-2\iff x=-1$

hello, we can also prove this showing that $ f(x)$ has at the point $ x\equal{}\minus{}1$ the global minimum $ y\equal{}\frac{1}{4}$
Sonnhard.

Thanks, Farenhajt and Sonnhard.

I've differentiated, using the quotient rule, set the derivative to zero to get $ x \equal{} \minus{} 1$ and checked that this gives a minimum by looking at the slopes on either side of the point where $ x \equal{} \minus{} 1$.

$ y\equal{}\frac{x^{2} \plus{} x \plus{} 1}{x\minus{}1)^{2}}$

$ \Rightarrow\, \frac{dy}{dx} \equal{} \frac{(x\minus{}1)^{2}(2x\plus{}1)\minus{}(x^{2}\plus{}x\plus{}1)\times 2(x\minus{}1)}{(x\minus{}1)^{4}} \equal{} 0$

$ \Rightarrow \, (x\minus{}1)((2x^{2} \minus{}x\minus{}1)\minus{}(2x^{2} \plus{} 2x \plus{}2)) \equal{} 0$

$ \Rightarrow \, x \equal{} \minus{}!$, since $ x\neq 1$.

$ y'(0) > 0$ and $ y' (\minus{}2) < 0$ $ \Rightarrow \, (x, y) \equal{} (\minus{}1, \frac{1}{4})$ is a local minimum.

This is the only stationary point ($ y$ approaches $ 1$ as $ x$ tends to $ \pm \infty$), so $ (\minus{}1, \frac{1}{4})$ is the global minimum.

Are there any other ways of doing this?

Does anyone have another way of looking at this?

I'm still curious about this and am interested in finding out if there are other ways of looking at this problem.


binomial theorem/ selections/arrangements/w/repetition w/o/repetition

I always get stuck when I try to problems concerning these topics

Our Introduction to Counting & Probability text covers those topics (and a lot more).  You can check it out [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3]here[/url].


I was wondering whether anyone here read any of Ayn Rand's books like Fountainhead and Atlas Shrugged? To me it seems that she always speaks in favor of capitalism but promotes ideas that would not be too popular in the U.S and around the world ( I am refering to egoism). What do you think of her philosophy?

All I can say is that I read the Fountainhead, which for me, is quite a considerable achievement. I'm not too sure of my grasp of objectivism, so I won't comment until I actually understand what I am saying. Otherwise, I would have to say that her books are marvelous.

If objectivism is viewing the world non-subjectively, then maybe I kinda understand it. Otherwise, I'm in the dark.  :lol:

I have read both The Fountainhead and Atlas Shrugged, and they were interesting, though I think they could be condensed quite a bit.

From what I understand, objectivism means that people should be rewarded for their abilities, and nothing else. In addition, you should always act in your own self interest.
The highest virtue that one can attain is to act in your own interest and to achieve excellence. In this way both you and society benefit from your productivity.

I guess fundamentally, I agree with the philosophy. I plan on acting for my own self interest throughout my lifetime (note that this will inevitably also consist of helping others and being compassionate.) However, her unrestricted capitalism is a extreme in my opinion. I believe that all people should be provided for in the most basic ways.

I read anthem..... and it was pretty lousy.. fountainhead looked intimidating..

There's usually a small-but-noticeable Rand subculture at
engineering schools, and the Internet brought a lot of her
followers out of the woodwork.  Not for nothing is the standard
term of derision for her followers, "Randroids".

Her works holds a certain psychological appeal for two types
of people especially (the intersection of which is, again,
found fairly often among engineering, CS and some
science types):
1. those who like certitude and clarity and tend toward
rigidity or binarisms in their thinking (if you
know of engineer types who overuse the word "reality" and
"either A or not-A" dichotomies, those are prime
though not conclusive indicators of Rand-amenability).
2. those who identify with the heroes in Rand's novels, feeling
that they are noble spirits fettered by the mediocrities
surrounding them such as schoolmates, family members,
and countrymen.

The more recent venture capitalism culture that developed
at engineering schools in the last few decades is a further point
helping Rand's popularity, due to her endorsement of
self-enrichment and entrepreneurship.

As a novelist, she was limited.  "We the Living" was best
of the ones I read.

Her critical and cultural essays and interviews are quite good.
In the radio interviews she comes across better than when
pictured on TV or in magazines, as she looked like a cartoon
eccentric, dressed all in black with a huge dollar sign (in place
of a cross or other religious symbol) on a chain around her neck.

Sorry to barge in like this, but fleeting_guest, there seems to be some problem with your email and/or private messages. I've sent you a couple of messages thourgh those methods, can you confirm receiving / not receiving them?
Also please try to check out [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23560]this topic[/url] in the support forum.

i read fountainhead, and loved it!! All i have to say is that she is wecome to express her opinion and philosophy.

I've read Anthem, The Fountainhead, and We the Living. I can definately appreciate her philosophy, yet am a tad too superstitious to follow it to the fullest. The whole "reality" deal would apply to me, say, 90% of the time, but I totally respect her thinking form.

I recently read Anthem for school. I liked everything except for the last few pages...in my opinion, her views of an ideal man are a bit extreme. It was very interesting to compare the British version with the American edition though.

[quote="jeffrey"]
From what I understand, objectivism means that people should be rewarded for their abilities, and nothing else.

In addition, you should always act in your own self interest.
The highest virtue that one can attain is to act in your own interest and to achieve excellence. In this way both you and society benefit from your productivity.
[/quote]

the first "stanza" of this response is a bit off; the idea of "rewarding" someone is not part of the philosophy, that I can tell, after all, it kinda contradicts with the rest (how can you pursue your own stuff while rewarding someone else?)

anyhow.  Personally, I've read the Fountainhead and plan to read Atlas Shrugged as soon as I can get the book and 30 billion hours in which to read it.  I've also read "Old School" which is something of an attack on Rand and [size=200]SPOILS THE ENDING OF THE FOUNTAINHEAD! DO NOT READ IT UNTIL YOU'VE READ THE FOUNTAINHEAD[/size] I was sitting there wish they had a wikipeida style spoiler notice, or a AOPS style (click to reveral hidden content)

anyhow.  The Fountainhead was good.  A very abnormal novel.  I like some aspects of her philosophy quite a bit, but I feel like the book itslef was somewhat spoiled by all my previous exposure to Ayn Rand in the context of popular culture and other books and stuff.

Anyhow, one thing I really worry about with the book is this:
The book reads almost entirely as a vigorous assertion of Rand's ideas of what is good and what is bad, &c.  Approval or Disapproval creates an extremely powerful emotional response in most people.  Simply by asserting that something is good (not in the form of opinion, but as an objective assertion) has, from my observation, created a HUGE slant in most persons (mine included) perspectives.  I tend to think I'm pretty self aware on most things, but I really felt a bit like I was being brainwashed after I read The Fountainhead.  I mean, I agree with Rand on some things, for example, I think Roark's a great guy and all, but I find my thought falls, unconciously, into Rand's categories for viewing the world.  i.e. I'll start thinking about someone as either a second-hander and therefor vastly inferior, or simply try to identify things in the terms of objectivism.

Also, I am disappointed that we get NOTHING of Roark's past/youth.  We get everyone elses, all the people who are essentially failures from her POV, but not his.  So how is a person like Roark created?  My meta-answer to this hypothetical question is that a person like Roark is not a product of situations and can not be determined by background or external forces.

[quote="MCrawford"]I think Rand has a few good things to say.  I agree with her that individualism is a more direct path toward value, but I disagree with some of her conclusions.  Some of her followers however elevate her to a cult status.  Objectivism sometimes feels more like a religion than a philosophy and that's fairly ironic considering she was an avowed atheist.[/quote]
Same here. I would be interested in seeing what conclusions you disagree with and for what reasons. Even if I do not agree with some articles in her collections of essays, I find them very hard to refute totally.


trigonometry
LaTeX
inequalities
inequalities unsolved

LHS = sum of cos(b+c)/cos(b-c)... peters off...

[quote="ARTI"]I know that it might not be difficult but I need the solution now! ;)
If A+B+C=90 prove that
(sinA)/[sin(A+2B)] + (sinB)/[sin(B+2C)] + (sinC)/[sin(C+2A)] >= 3/2[/quote]

First, I rewrite your problem with LaTeX:

[i]If A, B, C are three positive angles such that A + B + C = 90°, then prove

$\frac{\sin A}{\sin\left(A+2B\right)}+\frac{\sin B}{\sin\left(B+2C\right)}+\frac{\sin C}{\sin\left(C+2A\right)} \geq \frac32$.[/i]

Now, I note that A + B + C = 90° implies A = 90° - (B + C), so that cos A = sin (B + C). On the other hand,

A + 2B = (A + B + C) + (B - C) = 90° + (B - C)

leads to sin (A + 2B) = cos (B - C). Thus,

$\cos A\sin\left(A+2B\right) = \sin\left(B+C\right)\cos\left(B-C\right)$
$= \frac12\cdot\left(\left(\sin\left(B+C\right)\cos\left(B-C\right)+\cos\left(B+C\right)\sin\left(B-C\right)\right)+\left(\sin\left(B+C\right)\cos\left(B-C\right)-\cos\left(B+C\right)\sin\left(B-C\right)\right)\right)$
$= \frac12\cdot\left(\sin\left(\left(B+C\right)+\left(B-C\right)\right)+\sin\left(\left(B+C\right)-\left(B-C\right)\right)\right) = \frac12\cdot\left(\sin\left(2B\right)+\sin\left(2C\right)\right)$
$= \frac12\cdot\left(2\sin B\cos B + 2\sin C\cos C\right) = \sin B\cos B + \sin C\cos C$.

Thus,

$\frac{\sin A}{\sin\left(A+2B\right)}=\frac{\sin A\cos A}{\cos A\sin\left(A+2B\right)}=\frac{\sin A\cos A}{\sin B\cos B + \sin C\cos C}$.

Similarly,

$\frac{\sin B}{\sin\left(B+2C\right)}=\frac{\sin B\cos B}{\sin C\cos C + \sin A\cos A}$;
$\frac{\sin C}{\sin\left(C+2A\right)}=\frac{\sin C\cos C}{\sin A\cos A + \sin B\cos B}$.

Now, since the angles A, B, C are positive and we have A + B + C = 90°, the angles A, B, C are acute; thus, the numbers sin A cos A, sin B cos B and sin C cos C are all positive. Hence, we can apply the Nesbitt inequality to these numbers and get

$\frac{\sin A\cos A}{\sin B\cos B + \sin C\cos C}+\frac{\sin B\cos B}{\sin C\cos C + \sin A\cos A}+\frac{\sin C\cos C}{\sin A\cos A + \sin B\cos B}$
$\geq \frac32$.

In other words,

$\frac{\sin A}{\sin\left(A+2B\right)}+\frac{\sin B}{\sin\left(B+2C\right)}+\frac{\sin C}{\sin\left(C+2A\right)} \geq \frac32$.

This solves the problem.

  Darij

[b]Adding a little extension to darij's inequality[/b]
From
$\displaystyle \frac{\sin A\cos A}{\sin B\cos B + \sin C\cos C}+\frac{\sin B\cos B}{\sin C\cos C + \sin A\cos A}+\frac{\sin C\cos C}{\sin A\cos A + \sin B\cos B}\geq\frac32$
we can get
$\displaystyle \frac{\sin (2A)}{\sin (2B) + \sin (2C)}+\frac{\sin (2B)}{\sin (2C) + \sin (2A)}+\frac{\sin (2C)}{\sin (2A) + \sin (2B)}\geq\frac32$

Let $a=\sin (2A)$, $b=\sin (2B)$ and $c=\sin (2C)$. Then the above inequality is equivalent to
$\displaystyle \frac{a}{b + c}+\frac{b}{c + a}+\frac{c}{a + b}\geq\frac32$ with condition $2A+2B+2C=\pi$, where $0\leq a,b,c\leq 1$


geometry
circumcircle
algebra
polynomial
linear algebra
matrix
rectangle

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14020]IMO Problem 1[/url]

Let $ABC$ be an acute-angled triangle with ${AB\ne AC} $. The circle with diameter $BC$ intersects the sides$AB$ and  $AC$ at $M$ and $N$, respectively. Denote by $O$ the midpoint of $BC$. The bisectors of the angles $BAC$ and $MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the line segment $BC$.

[i]By Valentin Vornicu and Dinu Serbanescu, Romania[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14021]IMO Problem 2[/url]

 Find all polynomials $P(x)$ with real coefficients which satisfy the equality

$\begin{matrix*}
P(a-b)+P(b-c)+P(c-a)=2P(a+b+c)
\end{matrix*}$

for all triples $a,b,c$ of real numbers such that ${ab+bc+ca=0}$.

[i]By Hojoo Lee, Korea[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14023]IMO Problem 3[/url]

Determine all $m\times n$ rectangles that can be covered with "hooks" made up of $6$ unit squares, as in the figure. See the hook below.

Rotations and reflections of hooks are allowed. The rectangle must be covered without gaps and overlaps. No part of a hook may cover area outside the rectangle.

[i]By Jan Wil, Estonia[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14091]IMO Problem 4[/url]

 Let ${n\ge 3}$ be an integer and ${t_1,t_2,\dots,t_n}$ positive real numbers such that

$\begin{matrix*}
n^2+1>(t_1+t_2+\cdots +t_n)\left(\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n}\right).
\end{matrix*}$

Show that $t_i,t_j,t_k$ are the side lengths of a triangle for all $i,j,k$ with ${1\le i<j<k\le n}$.

[i]By Hojoo Lee, Korea[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14092]IMO Problem 5[/url]

In a convex quadrilateral $ABCD$ the diagonal $BD$ does not bisect the angles $ABC$ and $CDA$. The point $P$ lies inside $ABCD$ and satisfies

$\begin{matrix*}
\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.
\end{matrix*}$

Prove that $ABCD$ is a cyclic quadrilateral if and only if ${AP=CP}$.

[i]Proposed by Poland. [/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14093]IMO Problem 6[/url]

We call a positive integer [i]alternative[/i] if its decimal digits are alternatively odd and even. Find all positive integers $n$ such that $n$ has an alternative multiple.

Proposed by Mohsen Jamali and Armin Morabbi, Iran.


Not sure if this is Basic or Intermediate.. should just be some geometry, so here it goes

There is a semi-circle that is inscribed in another semicircle. The "apex" (not sure of the word) of the inscribed semicirrcle divides the other semicircle's diameter into equal portions. What is the ratio of the smaller circle's radius to the larger circle's radius.

I'm not too interested in the ansswer as in the way you get it.

here's a picture that might help you visualize it, note that although the figures may not appear to be semicircles, they are. [img]http://home.1asphost.com/WindSlicer/semiinsemi.JPG[/img]

This problem is actually quite simple.

[hide]You have a smaller semicircle inscribed in a larger semicircle. Since the apex of the smaller semicircle splits the base of the larger semicircle in half, it must also be the center of the large semicircle. Call this point $O$. Now, call the center of the smaller semicircle $C$. Finally, the base of the smaller semicircle touches the larger semicircle at two points. Call one of those points $D$.

We draw right triangle $\triangle OCD$. Notice that the hypotenuse of the triangle is the radius of the larger semicircle and the two legs are radii of the smaller semicircle. This means the triangle is a 45-45-90 and the ratio between leg to hypotenuse, or in this case, the ratio of the smaller radius to the larger radius is $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.
[/hide]

Oh, it actually wasn't that hard!

Thanks for the help

[hide="solution"]

from the center, O, of the bigger semicircle, we draw two line segments to each end corner of the smaller semicircle, and call the end corners D and F.

OD and OF are both radii of the bigger semicircle, while they form the angle DOF, which is a right triangle, because it forms a triangle with DF, the diameter of the smaller semicircle.

so if we call OF or OD $x$, then DF becomes $x \sqrt{2}$.

since we want the ratio of the radii of two semicircles, we simply obtain the radii of the smaller semicircle by dividing DF by two, thus getting $\frac{x \sqrt{2}}{2}$.

since OF is the radius of the bigger semicircle, the ratio is
$\frac{\frac{x \sqrt{2}}{2}}{1} = \frac{x \sqrt{2}}{2}$
[/hide]

that should have been in getting started... but i can never measure difficulty... :lol:

yeah it should have been, don't know why I didn't see it was that simple *shrugs* ah well

it's okay. You will learn to pick out easier way to solve as you learn. just like  me. :D


Let $S$ be a set of integers such that [list][*] there exist $a, b \in S$ with $\gcd(a, b)=\gcd(a-2,b-2)=1$, [*] if $x,y\in S$, then $x^2 -y\in S$.[/list] Prove that $S=\mathbb{Z}$.

We begin with a quite standard
[i]Lemma[/i]
If $ a_1,\ldots,a_n$ are integers such that $ \gcd(a_1,\ldots,a_n) = 1$ then any $ m\in\mathbb{Z}$ can be expressed as a linear combination of $ a_1,\ldots,a_n$, i.e. there exist $ \epsilon_1,\ldots,\epsilon_n\in\mathbb{Z}$ such that $ m = \epsilon_1a_1 + \ldots + \epsilon_na_n$.

[i]Proof of Lemma:[/i]
We will use induction. The base case is $ n = 2$. This one is really well-known, but for completeness, here's its proof:
Let $ \gcd(a,b) = 1$ and $ m\in\mathbb{Z}$. The numbers $ 0,a,2a,\ldots,(b - 1)a$ are all different $ \pmod b$. This means one of them, say $ a\epsilon_1$ is $ 1\pmod b$. Then $ 1 - ax = b\epsilon_2$ for some $ \epsilon_2$, hence $ a\epsilon_1 + b\epsilon_2 = 1$. Then $ m = amx + bm\epsilon_2$.
Now the induction step. Assume we're done for $ n - 1\ge2$. Let $ d = \gcd(a_1,\ldots,a_{n - 1})$. Since $ \cgd(d,a_n) = 1$ then any integer $ m$ is representable as $ m = d\epsilon + a_n\epsilon_n$. Consider now the numbers $ b_i = \dfrac{a_i}d$, for $ i = \overline{1,n - 1}$. Then $ \gcd(b_1,\ldots,b_{n - 1}) = 1$ and according to the Induction Hypothesis, $ 1 = \alpha_1b_1 + \ldots + \alpha_{n - 1}b_{n - 1}$, for some $ \alpha_1,\ldots,\alpha_{n - 1}\in\mathbb{Z}$. Then $ d = \alpha_1a_1 + \ldots + \alpha_{n - 1}a_{n - 1}$, and therefore $ m = \epsilon d + a_n\epsilon_n = \epsilon_1a_1 + \ldots + \epsilon_na_n$, where $ \epsilon_i = \epsilon\alpha_i$, for $ i = \overline{1,n - 1}$.
___
Now back to the problem.
Take 3 elements $ a,b,c$ of $ S$ (it's trivial that $ S$ has at least 3 elements).
Then $ b^2 - a\in S$, and $ c^2 - (b^2 - a) = a + (c^2 - b^2)\in S$. We prove by induction on $ k$ that $ a + k(c^2 - b^2)\in S$.
Base case $ k = 1$ is done. Assume the statement true for $ k$. Then $ b^2 - [a + k(c^2 - b^2)]\in S$, and consequently $ c^2 - b^2 + a + k(c^2 - b^2)\in S$.
Changing $ b,c$ with places we obtain the following result: $ a + n(c^2 - b^2)\in S$, for all $ n\in\mathbb{Z}$. Immediately follows that $ a + n_1c_1 + \ldots + n_tc_t$ are in $ S$ for all $ n_i\in\mathbb{Z}$, where $ c_i$ are of the form $ x^2 - y^2$ with $ x,y\in S$. We are left to show that there is a set of numbers of the form $ c_i = x^2 - y^2$ with $ x,y\in S$ such that $ \gcd(c_1,\ldots,c_?) = 1$ (after that we just apply our [i]Lemma[/i]).
Let $ M = \{x^2 - y^2 > 0|x,y\in S\} = \{m_0,m_1,\ldots\}$. Let $ b_i = \gcd(m_0,\ldots,m_i)$. It is clear that $ b_i$ is decreasing. If $ b_i > 1$ for all $ i$, then from some point on it's constant, thus there exists a prime number $ p$ dividing all positive differences $ x^2 - y^2$, for $ x,y\in S$. We show that this is impossible.
Suppose $ p|x^2 - y^2$, for all $ x,y\in S$. Take $ a,b\in S$ such that $ \gcd(a,b) = \gcd(a - 2,b - 2) = 1$. WlOG, $ p\not|a$. Since $ p|a^2 - b^2$ we get $ p\not|b$. Also $ p|(a^2 - a)^2 - b^2$ and we have $ p|a^3(a - 2)$, thus $ p|a - 2$. Similarly $ p|b - 2$, hence $ \gcd(a - 2,b - 2)\neq1$, contradiction.


algebra
polynomial
quadratics
group theory
abstract algebra
Galois Theory
superior algebra

Let $ F \subseteq E$ be a finite field extension such that

- each polynomial in $ F[X]$ of odd degree has a root in $ F$,
- each quadratic polynomial in $ E[X]$ splits over $ E$.

Show that $ E$ is algebraically closed, using Galois theory and the Sylow theorems.

With a bit of knowledge about formally real fields:
The first condition says that $ F$ is formally real. So it's algebraic closure has degree $ 2$ over it, so it's $ E$.

I don't think I have the knowledge you are mentioning :)

just copy the standard proof for $ \mathbb{R} \subseteq \mathbb{C}$ using sylow theorems, exactly the two properties are needed there ... and characteristic 0. *g*

Hm...last proof failed, here's another try. Still assuming the field is perfect, maybe I don't need that assumption though...

Lemma 1: $ [K: F]$ is a power of $ 2$ for every Galois extension $ K$.
Proof: Let $ G$ be the Galois group of the extension. Consider the fixed field $ K$ of the $ 2$-Sylow subgroup of $ G$. This has odd degree over $ F$. Suppose that the lemma is false and $ K \neq F$. Pick an element $ \beta$ in $ K$ that is not in $ F$. By the tower law, $ [F(\beta): F]$ is odd. Thus the irreducible polynomial for $ \beta$ is odd, but this is impossible since all polynomials of odd degree have a root. Thus $ [K: F]$ must be a power of $ 2$, as was desired.

Now induct on the degree of any Galois extension $ K/F$ with Galois group $ G$, which must be a $ 2$-group. The inductive hypothesis is that if $ [K: F] \equal{} d$ and $ d > 2$, then $ K$ does not exist. Consider the center $ Z$ of $ G$ and consider its fixed field $ L$. By Galois' fundamental theorem $ L/F$ is also Galois (since $ Z$ is normal). Thus by the inductive hypothesis, the fixed field should have degree $ 2$. This means that the center has degree $ 2$, but you can show that if $ G/Z$ is cyclic then $ G$ is Abelian. This implies that $ G/Z$ is trivial, which contradicts the fact that it must have size $ 2$.

Thus all Galois extensions of $ F$ have degree $ 2$. Since $ E$ is the splitting field for all polynomials of degree $ 2$, $ E$ necessarily contains all of these extensions. Thus in particular $ E$ contains the splitting field of the irreducible polynomial of every $ \alpha \in F$.  and since $ E$ is necessarily the splitting field for all of these extensions, it is also the algebraic closure of $ F$. Here we use the fact that anything algebraic over $ E$ is necessarily algebraic over $ F$ since $ E$ is algebraic over $ F$.


I searched for this problem but all the hits come up with placing real numbers on the chessboard which is a (significantly) harder problem.

An infinite chessboard is filled with positive integers such that each entry is the arithmetic mean of the entries directly above, below, and on either side of it. Prove that all the entries must be equal.  (Paraphrased, don't have the book on me at the moment.)

Here's my solution, I don't think I phrased my restriction exactly as I would have liked...
[hide]
Since the chessboard contains only positive integers, by the Well Ordering Principle, there exists some least element $ c$ on the board. Assume that all the entries are not equal, ie. one of $ x_1,x_2,x_3,x_4>c$, where $ c\equal{}\frac {x_1\plus{}x_2\plus{}x_3\plus{}x_4}{4}$. WLOG let $ x_4>c$ (and thus $ x_1,x_2,x_3\equal{}c$ due to our assumption of minimality).
Rearranging yields: $ 4c\minus{}(x_1\plus{}x_2\plus{}x_3)\equal{}x_4$ which implies $ c\equal{}x_4$, a contradiction.
Therefore all the entries on the board are equal.

Question: Is it enough to show that we derive a contradiction from having only one element be larger than $ c$?
[/hide]

Yes.  If the entries on the chessboard are not all equal, then $ c$ must appear somewhere next to an element greater than $ c$.

[quote="max_tm"]WLOG let $ x_4 > c$ (and thus $ x_1,x_2,x_3 \equal{} c$ due to our assumption of minimality).

[/quote]

can you explain this step?  i don't get what an "assumption of minimality" is, or where it was made.

By the well ordering principle, every non-empty subset of the natural numbers contains a least element, which I called $ c$. Since I specified that one of $ x_1,x_2,x_3,x_4>c$ then the other three must equal $ c$ because $ c$ is the smallest element in the set. Whether or not this imposes too many restrictions is what I'm still worried about.

I misread your solution slightly.  You can't actually conclude that $ x_1 \equal{} x_2 \equal{} x_3 \equal{} c$, nor do you need to.  As long as [i]at least[/i] one of $ x_1, x_2, x_3, x_4$ is greater than $ c$ (which you can conclude from your assumption), $ \frac{x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4}{4} > c$, which provides the desired contradiction.

Yes I realized that as I was coming out of the shower just now  :P, any number of them could be greater than $ c$ and the nature of the argument would not change. Thanks for the input!


algebra
polynomial
irrational number
Rational Root Theorem
algebra unsolved

Let $ P(x) \in Z[x]$, $ deg P \ge 2$.
Prove or disprove:
There exists an irrational number $ x$ such that $ P(x)$ is rational

Let $ P(x)\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_nx^n\in Z[x]$, Let $ P_0(x)\equal{}x^n\plus{}a_{n\minus{}1}x^n\plus{}a_{n\minus{}2}a_nx^{n\minus{}2}\plus{}...\plus{}a_0a_n^{n\minus{}1}$.
If (1) $ P(x)\equal{}r$, then (2)$ P_0(a_nx)\equal{}ra_n^{n\minus{}1}.$ Therefore we can prove it only for $ P_0(x)$.
If $ P_0(x)\equal{}m\in Z$ had not root in $ Z$, then it had not root in $ Q$.
Let $ S_N\equal{}\#\{ P_0(x), x\in Z, |P(x)|\le N\}.$ Obviosly $ \S_N\equal{}O(N^{1/n})$ and had 0 density in $ Z$. Therefore exist infinetely many $ m\in Z$
suth that exist root $ P_0(x)\equal{}m,x\not \in Z$. It mean $ x$ irrational.

[quote="Rust"]Let $ P(x) \equal{} a_0 \plus{} a_1x \plus{} ... \plus{} a_nx^n\in Z[x]$, Let $ P_0(x) \equal{} x^n \plus{} a_{n \minus{} 1}x^n \plus{} a_{n \minus{} 2}a_nx^{n \minus{} 2} \plus{} ... \plus{} a_0a_n^{n \minus{} 1}$.
If (1) $ P(x) \equal{} r$, then (2)$ P_0(a_nx) \equal{} ra_n^{n \minus{} 1}.$ Therefore we can prove it only for $ P_0(x)$.
If $ P_0(x) \equal{} m\in Z$ had not root in $ Z$, then it had not root in $ Q$.
Let $ S_N \equal{} \#\{ P_0(x), x\in Z, |P(x)|\le N\}.$ Obviosly $ \S_N \equal{} O(N^{1/n})$ and had 0 density in $ Z$. Therefore exist infinetely many $ m\in Z$
suth that exist root $ P_0(x) \equal{} m,x\not \in Z$. It mean $ x$ irrational.[/quote]

Could you explain more details? At least, thanks for your hint. I've just had my own proof./.

Here's a nice approach I learned from the Quattomaster 6000 (I think) a while ago: Shift the polynomial down by its constant term and call this new polynomial $ Q(x)$. Now consider the sequence of primes $ p_i$ and in particular the values of $ x$ such that $ Q(x) \equal{} p_i$; call these $ x_i$. Now note that the polynomial $ Q(x) \minus{} p_i$ has root $ x_i$, so by the rational root theorem we have $ x_i \equal{} \pm \left\{\frac {p_i}{k},\frac {1}{k}\right\}$, where $ k|a$, the leading coefficient; both give an immediate contradiction by considering sufficiently large $ p_i$.


a,b,c>0 a 2 +b 2 +c 2 =1
Prove that:
a+b+c+1/abc \geq 4 \sqrt 3

Apply the AM-GM inequality and obtain that: (a+b+c)+1/(3abc
)+1/(3abc)+1/(3abc)\geq 4 * 4 \sqrt[ (a+b+c)/(27a 3 b 3 c 3 )]. But now, also by the AM-GM inequality: a+b+c=(a+b+c)(a 2 +b 2 +c 2 ) 4 \geq 243a 3 b 3 c 3 , and finally we get that a+b+c+1/abc \geq 4*4 \sqrt 9=4 \sqrt 3.

Here's my soln (it's a bit longer that pcalin's):
add the 3 ineqs of the type a+1/9abc >= 2/(3*(bc)^1/2). We have to prove that
 \sum 2/(3*(bc)^1/2)+2/3abc >= 4*3^1/2. By AM-GM applied to a^2, b^2, c^2 we get 2/3abc >= 2*3^1/2, so it's enough to prove that  \sum 2/(3*(bc)^1/2) >=2*3^1/2 or, if we divide it by 2 and multiply it by 3,  \sum 1/((bc)^1/2) >= 3*3^1/2. This is easy, we use (\sum 1/((bc)^1/2))* (\sum (bc)^1/2) >= 9 and \sum (bc)^1/2 <= 3^1/2 iff (\sum (bc)^1/2)^2 <= 3 but (\sum (bc)^1/2)^2 <= 3* \sum bc <= 3*  \sum a^2=3, so what we are trying to prove is true.


I suppose that x_1n means (x_1)^n. Let's proceed. It's trivial to prove that x_k=1-1/(k+1)!. Thus (sum (x_k)^n)^1/n=((1-1/2!)^n+(1-1/3!)^n+...+(1-1/(n+1)!)^n)^1/n, which obviously tends to 1, since it is greater than 1-1/(n+1)! and smaller than (n*(1-1/(n+1)!)^n)^1/n, and these sequences tend to 1.


quadratics
MATHCOUNTS
algebra
polynomial
geometry
3D geometry
search

Say it was...
$ ax^3\plus{}bx^2\plus{}cx\plus{}d$
Instead of a quadratic which is
$ ax^2\plus{}bx\plus{}c$
Would you solve it in a similar way?

From
http://www.math.vanderbilt.edu/~schectex/courses/cubic/
Solves for one solution, then you can factor and use quadratic formula...
If $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$, then:
[img]http://www.math.vanderbilt.edu/~schectex/courses/cubic/cubic.gif[/img]

Much more practically, you could use Rational Roots Theorem or something...

this is not a mathcounts level question and the best way is probably to use the Rational Root theorem, and if there is a rational root, then use synthetic division to get a quadratic, which you can solve however way you want.

Of course, for $ ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$, you start by making it a monic polynomial (coefficient of the term with greatest degree is 1), and using a few clever substitutions bring it down to a quadratic.  It's a long story though, and many weird things happened a long time ago when the formula was first discovered.

[quote="ZhangPeijin"][quote="alanchou"]this is not a mathcounts level question and the best way is probably to use the Rational Root theorem, and if there is a rational root, then use synthetic division to get a quadratic, which you can solve however way you want.

Of course, for $ ax^{3} \plus{} bx^{2} \plus{} cx \plus{} d$, you start by making it a monic polynomial (coefficient of the term with greatest degree is 1), and using a few clever substitutions bring it down to a quadratic.  It's a long story though, and many weird things happened a long time ago when the formula was first discovered.[/quote]
So you need to try to make it into a quadratic?
 :huh:
I'm still confused.
 :mad:[/quote]

You can reduce the equation to the form $ x^3\plus{}px\plus{}q$ (try to prove this yourself). Then set $ u\plus{}v\equal{}x$ and obtain
\[ u^3\plus{}v^3\plus{}(u\plus{}v)(p\plus{}3uv)\plus{}q\equal{}0
\]and we can require $ 3uv\plus{}p\equal{}0\implies u^3v^3\equal{}\minus{}\frac{p^3}{27}$ which reduces the first equation to $ u^3\plus{}v^3\plus{}q\equal{}0\implies u^3\plus{}v^3 \equal{} \minus{}q$ and $ u^3,v^3$ are the solutions of the quadratic $ \xi^2\minus{}q{27}\xi\minus{}\frac{p^3}{27}\equal{}0$ (Vieta). Calculate $ u^3$ and $ v^3,$ take the cube root respectively and add up. Note that in general this formula is next to useless, as it can involve complex numbers which mess calculations up. In nearly any high school problem guessing one root and factoring it out will work, too.

(The formula above gives only one value $ x;$ the other two have the form $ \omega u\plus{}\omega^2v$ and $ \omega^2 u \plus{} \omega v$ with $ \omega^2\plus{}\omega\plus{}1\equal{}0.$)

No reasonable contest will force you to solve a cubic in which it takes more than 30 seconds to guess a rational root. If you ever find yourself with a cubic with either a ridiculous amount of rational roots, or no rational roots, you should probably find some other way to solve the problem.

There are "tricks" to solving polynomial equations directly, though, such as Descartes' Rule of Signs and the Intermediate Value Theorem. The Intermediate Value Theorem is rather intuitive. Search for them.

" You can reduce the equation to the form  (try to prove this yourself). Then set  and obtain
and we can require  which reduces the first equation to  and  are the solutions of the quadratic  (Vieta). Calculate  and  take the cube root respectively and add up. Note that in general this formula is next to useless, as it can involve complex numbers which mess calculations up. In nearly any high school problem guessing one root and factoring it out will work, too.

(The formula above gives only one value  the other two have the form  and  with ) "
The above method is also called as Cardan's Method
 :wink:  :) [/quote]

The problem here is that while the quadratic formula is simple and can be used easily by hand, the cubic formula is [b]extremely[/b] unwieldy and not at all practical to use.  You will never use it.  Its existence is of purely theoretical interest.  (Interestingly enough, while a "quartic formula" exists (which is even worse), a corresponding formula does not exist for polynomials of degree five or higher.)

If you ever need to actually find the roots of a cubic, it will usually have integer coefficients and a rational (or better yet, integer) root.  If such a root exists, use the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]rational root theorem[/url] along with [url=http://www.purplemath.com/modules/synthdiv.htm]synthetic division[/url] to test for them.  If you find a rational root $ r$, then you can divide out $ x \minus{} r$ from your cubic.  The resulting quadratic can be solved with the quadratic formula.

Cardano's method (which is the analogue of completing the square) is as impractical as the cubic formula.


There is a stack of oranges, 7 oranges tall, with the oranges so arranged that each orange rests atop 4 other oranges. The base is in a square. A borer starts at the top of the stack and bores into the first orange. He then randomly selects a orange below to bore into. He continues until the bottom. What is the probability that he comes out on the bottom from a orange that is exactly 1 orange in from the edge?

Hint:[hide][hide][hide]Make a 3D Pascals Triangle[/hide]
[/hide][/hide]
Original, by Billy.[/hide]

I get $\frac{152}{633}$. Don't know if I'm right though.

i think the N is right (N for ur last name). Daiwei is the L, i'm kentucky, and patricia is CBR

i get the answer of $21/{64}$ tell me if it's right