There was also : http://www.mathlinks.ro/Forum/viewtopic.php?t=86201 .

So, what are all $k \in \mathbb{N}$ for which this inequality holds? (with $a^k+b^k+c^k = 3$)

[quote="mathmanman"]
So, what are all $k \in \mathbb{N}$ for which this inequality holds? (with $a^k+b^k+c^k = 3$)[/quote]
For $k=6$ this is wrong: $a=b=1.1$ and $c=1.$ ;)
Remain case $k=5.$ :)

Here is my proof for the case $ k \equal{} 5.$ Ugly but proof.
hungkhtn, you are right! SOS works here. For the case $ k \equal{} 4$ it's easy enough.
We need to prove that $ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq\frac {3}{2}$ for positive $ a,$ $ b$ and $ c$
such that $ a^5 \plus{} b^5 \plus{} c^5 \equal{} 3.$
$ \sum_{cyc}\frac {a^2}{b \plus{} c}\geq\frac {3}{2}\Leftrightarrow\sum_{cyc}\left(\frac {a^2}{b \plus{} c} \minus{} \frac {a}{2}\right)\geq\frac {3 \minus{} a \minus{} b \minus{} c}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a(2a \minus{} b \minus{} c)}{b \plus{} c}\geq\frac {81(a^5 \plus{} b^5 \plus{} c^5) \minus{} (a \plus{} b \plus{} c)^5}{81 \plus{} 27(a \plus{} b \plus{} c) \plus{} 9(a \plus{} b \plus{} c)^2 \plus{} 3(a \plus{} b \plus{} c)^3 \plus{} (a \plus{} b \plus{} c)^4}.$
$ \sum_{cyc}\frac {a(2a \minus{} b \minus{} c)}{b \plus{} c} \equal{} \sum_{cyc}\left(\frac {a(a \minus{} b)}{b \plus{} c} \minus{} \frac {a(c \minus{} a)}{b \plus{} c}\right) \equal{}$
$ \equal{} \sum_{cyc}\left(\frac {a(a \minus{} b)}{b \plus{} c} \minus{} \frac {b(a \minus{} b)}{a \plus{} c}\right) \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(a \plus{} b \plus{} c)}{(b \plus{} c)(a \plus{} c)}.$
$ 81(a^5 \plus{} b^5 \plus{} c^5) \minus{} (a \plus{} b \plus{} c)^5 \equal{}$
$ \equal{} 5\sum_{cyc}(16a^5 \minus{} a^4b \minus{} a^4c \minus{} 2a^3b^2 \minus{} 2a^3c^2 \minus{} 4a^3bc \minus{} 6a^2b^2c) \equal{}$
$ \equal{} 5\sum_{cyc}(a \minus{} b)^2((5a^2 \plus{} 8ab \plus{} 5b^2)c \plus{} 8a^3 \plus{} 10a^2b \plus{} 10ab^2 \plus{} 8b^3).$
Since, $ a^4 \plus{} b^4 \plus{} c^4\leq3,$ $ a^3 \plus{} b^3 \plus{} c^3\leq3,$ $ a^2 \plus{} b^2 \plus{} c^2\leq3$ and $ a \plus{} b \plus{} c\leq3$ then
it's remains to prove that
$ 2(a \plus{} b \plus{} c)^5 \plus{} 3(a \plus{} b \plus{} c)^3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 9(a \plus{} b \plus{} c)^2(a^3 \plus{} b^3 \plus{} c^3) \plus{}$
$ \plus{} 27(a \plus{} b \plus{} c)(a^4 \plus{} b^4 \plus{} c^4)\geq$
$ \geq5(c^2 \plus{} (a \plus{} b)c \plus{} ab)((5a^2 \plus{} 8ab \plus{} 5b^2)c \plus{} 8a^3 \plus{} 10a^2b \plus{} 10ab^2 \plus{} 8b^3),$
which is equivalent to the following inequality:
$ 41c^5 \plus{} 64(a \plus{} b)c^4 \plus{} (16a^2 \plus{} 36ab \plus{} 16b^2)c^3 \minus{} (24a^3 \plus{} 37a^2b \plus{} 37ab^2 \plus{} 24b^3)c^2 \plus{}$
$ \plus{} (24a^4 \minus{} 39a^3b \minus{} 62a^2b^2 \minus{} 39ab^3 \plus{} 24b^4)c \plus{}$
$ \plus{} 41a^5 \plus{} 24a^4b \minus{} 9a^3b^2 \minus{} 9a^2b^3 \plus{} 24ab^4 \plus{} 41b^5\geq0.$
But we see that
$ 41c^5 \plus{} 64(a \plus{} b)c^4 \plus{} (16a^2 \plus{} 36ab \plus{} 16b^2)c^3 \minus{} (24a^3 \plus{} 37a^2b \plus{} 37ab^2 \plus{} 24b^3)c^2 \plus{}$
$ \plus{} (24a^4 \minus{} 39a^3b \minus{} 62a^2b^2 \minus{} 39ab^3 \plus{} 24b^4)c \plus{}$
$ \plus{} 41a^5 \plus{} 24a^4b \minus{} 9a^3b^2 \minus{} 9a^2b^3 \plus{} 24ab^4 \plus{} 41b^5\geq$
$ \geq41c^5 \plus{} 64(a \plus{} b)c^4 \plus{} 16(a \plus{} b)^2c^3 \minus{} 24(a \plus{} b)^3c^2 \minus{} \frac {23}{4}(a \plus{} b)^4c \plus{} \frac {7}{2}(a \plus{} b)^5.$
Let $ \frac {c}{a \plus{} b} \equal{} t.$ Then it remains to check out that
$ 164t^5 \plus{} 256t^4 \plus{} 64t^3 \minus{} 96t^2 \minus{} 23t \plus{} 14\geq0,$ which is true.

[quote="can_hang2007"][quote="arqady"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b} \ge \frac {3}{2}\sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}.
\]
[/quote]
We have a complete solution using Cauchy Schwarz Inequality. :)[/quote]
Could you post your solution with Cauchy Schwarz Inequality ?

[quote="can_hang2007"][quote="arqady"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b} \ge \frac {3}{2}\sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}.
\]
[/quote]
We have a complete solution using Cauchy Schwarz Inequality. :)[/quote]
My hint is to write the left hand side as the following
\[ \sum \frac {a^2}{b \plus{} c} \equal{} \sum \frac {(b \plus{} c \minus{} a)^2}{b \plus{} c}
\]
The second step is to use Cauchy Schwarz Inequality
\[ \sum \frac {(b \plus{} c \minus{} a)^2}{b \plus{} c} \ge \frac {(\sum (b \plus{} c \minus{} a)^2)^2}{\sum (b \plus{} c)(b \plus{} c \minus{} a)^2}
\]
Finally, we only need to prove the stronger but easier :)
I think the following is really interesting, isn't it?
The original expression
\[ \sum \frac {a^2}{b \plus{} c}
\]
We may use the Cauchy Schwarz
\[ \sum \frac {a^2}{b \plus{} c} \ge \frac {(\sum a^2)^2}{\sum a^2(b \plus{} c)}
\]
But the inequality
\[ \frac {(\sum a^2)^2}{\sum a^2(b \plus{} c)} \ge \frac {3}{2}\sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}
\]
is not true indeed.
But if we write it as
\[ \sum \frac {(b \plus{} c \minus{} a)^2}{b \plus{} c}
\]
Many things has changed!!
:)

With a,b,c donot < 0 .I see the $ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{a\plus{}c}\plus{}\frac{c^2}{b\plus{}a}\geq\frac{3}{2}\sqrt[4]{\frac{a^4\plus{}b^4\plus{}c^4}{3}}$ easy the
$ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{a\plus{}c}\plus{}\frac{c^2}{b\plus{}a}\geq\frac{3}{2}\sqrt[5]{\frac{a^5\plus{}b^5\plus{}c^5}{3}}$
canhang_2007 did that by p,q,r ?
Thank arqady very much.

[quote=Jettofaiyafukushireikan]How about this?
$\frac{a^3}{a^2 + b^2} + \frac{b^3}{b^2 + c^2} + \frac{c^3}{c^2 + a^2} \ge \frac{3}{2}$
when $a^2+b^2+c^2=3$[/quote]
We need to prove that $\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)\geq\frac{\sqrt{3(a^2+b^2+c^2)}-a-b-c}{2}$ or
$\sum_{cyc}\frac{a^3-ab^2}{a^2+b^2}\geq\frac{\sum\limits_{cyc}(a-b)^2}{\sqrt{3(a^2+b^2+c^2)}+a+b+c},$ for which it remains to prove that
$\sum_{cyc}\frac{a^3-ab^2}{a^2+b^2}\geq\frac{\sum\limits_{cyc}(a-b)^2}{2(a+b+c)},$ which is
$\sum_{cyc}(3a^5b^3+a^5c^3-2a^4b^4+a^5b^2c+a^5c^2b+a^4b^3c-a^4c^3b-6a^4b^2c^2+2a^3b^3c^2)\geq0$ and since by AM-GM
$\sum_{cyc}a^5b^3\geq\sum_{cyc}a^4b^2c^2$ and $\sum_{cyc}(a^5b^3+a^5c^3)\geq2\sum_{cyc}a^4b^4$, it remains to prove that
$\sum_{cyc}(a^4b+a^4c+a^3b^2-a^3c^2-4a^3bc+2a^2b^2c)\geq0$, which is obvious by $uvw$'s technique.

[quote=arqady] it remains to prove that
$\sum_{cyc}(a^4b+a^4c+a^3b^2-a^3c^2-4a^3bc+2a^2b^2c)\geq0$, which is obvious by $uvw$'s technique.[/quote]
I am afraid it is cyclic not symmetric...
Do you have the detail? :(

how to do uwv with cyclic not symmetric ineq?

[b]Jettofaiyafukushireikan[/b] I mean the following.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $\sum_{cyc}(a^4b+a^4c+a^3b^2-a^3c^2-4a^3bc+2a^2b^2c)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^4b+a^4c-4a^3bc+2a^2b^2c)\geq\sum_{cyc}(a^3c^2-a^3b^2)\Leftrightarrow$
$\Leftrightarrow81u^3v^2-81uv^4-9u^2w^3+15v^2w^3-4(9u^2-6v^2)w^3+6v^2w^3\geq(ab+ac+bc)(a-b)(b-c)(c-a)\Leftrightarrow$
$\Leftrightarrow27u^3v^2-27uv^4-15u^2w^3+15v^2w^3\geq v^2(a-b)(b-c)(c-a)$.
Easy to show that $27u^3v^2-27uv^4-15u^2w^3+15v^2w^3\geq0$.
Hence, we can assume that $(a-b)(b-c)(c-a)\geq0$
 and after squaring of the both sides we'll get a quadratic inequality of $w^3$ and the rest is smooth.


Care este inegalitatea voastra preferata? :D

E foarte greu sa ma decid. Fiecare inegalitate o aplici la un moment dat in cariera si nu prea ai cum sa alegi... :D

Care este inegalitatea lui Karamata?Si Bergstrom?

Inegalitatea lui Bergstrom este intr-o oarecare masura echivalenta cu Inegalitatea lui Cauchy, dar am vazut inegalitati care nu se pot rezolva cu Cauchy, dar ies usor cu Bergstrom.Du-te pe[b] Advanced Section [/b]>[b]Inequalities[/b]>[b]Inequalities Theoerems and Formulas[/b].Le poti gasi acolo pe amandoua.Pentru Karamata trebuie sa stii si Jensen.(du-te pe [b]Karamata Inequality[/b]).Pentru Bergstrom du-te pe [b]Inequality sent by stancioiu sorin[/b], autor Kunny. ;)

[quote="Cezar Lupu"]Pai va zic eu care este cea mai frumoasa si cea mai tare inegalitate si totodata preferata mea: \[ x\in\mathbb{R}  \] rezulta \[ x^{2}\geq 0.  \]

Din asta se obtin toate inegalitatile posibile si imposibile. :)[/quote]Oare de ce eram sigur ca o sa o bagi pe asta Cezare?  :D

[quote="stancioiu sorin"]Inegalitatea lui Bergstrom este intr-o oarecare masura echivalenta cu Inegalitatea lui Cauchy, dar am vazut inegalitati care nu se pot rezolva cu Cauchy, dar ies usor cu Bergstrom.[/quote]

M-am uitat si eu acolo. Se pare ca inegalitatea asta a lui Bergstrom este de-a dreptul echivalenta cu celebra inegalitate Cauchy/Buniakowski/Schwarz. Asadar, inegalitatile care nu se pot rezolva cu Cauchy, dar ies usor cu Bergstrom, pur si simplu nu exista.

Apropo, de unde ai auzit tu ca inegalitatea aia ar purta numele de Bergstrom?

Vorbesc chiar serios.Am vazut denumirea asta intr-o carte scrisa de Dan St. Marinescu si Dorin Lint.Ei asa o numeau.Daca o ai, iti spun si pagina. :?

Eu am vazut la "Inegalitati idei si metode" M.O.Drambe(cred)

Da, asa e.Am si eu cartea si am vazut ca scrie de inegalitatea asta.

Eu zic ca cea mai tare inegalitate e...."TITU" :)) ( pt Cezar) :P

"TITU" e pur cauchy,doar un caz particular..(Daca vbim de "TITU" de care stiu eu :D )

pooi...asta era si faza cu "pt Cezar" :) stiu ca e Coşi :))

[quote="Cezar Lupu"][quote="csabafarkas"]Am gasit...nu am stiut ca inegalitatea acela poarta numele "Karamata" ....[/quote]

Va rog frumos nu mai spuneti inegalitea lui Karamata, pentru ca ea a fost descoperita inainte de o tripleta de aur a
matematicii secolului trecut si anume: [b] Hardy-Littlewood-Polya. [/b][/quote]

Nu am stiut asta numai am vazut la cartea lui M.O.Drambe..inegalitatea lui Karamata...deci acolo e greseala....Bine ai dreptate :blush:

[quote="Cezar Lupu"]Pai ca sa vezi faza. Intr-un tratat urias a celor trei granzi aceasta inegalitate este o consecinta amarata, iar Karamata publica un an mai tarziu sub numele lui aceasta frumoasa teorema. Ca sa vezi chestie. Asa si cu presupusa inegalitate a lui Cauchy-Buniakovski-Schwarz, in ordine cronologica: "Engel form"  sau "Lemma Titu". Stiu ca toti vrem sa inventam ceva original, dar totusi......... :)[/quote]

Ai dreptate!Stiu ca tu esti regele inegalitatii in studenti..... ;)


induction
number theory proposed
number theory

$ p_{1},p_{2}$ are prime and $ p_{n}$ is the most large prime factor of $ p_{n\minus{}1}\plus{}p_{n\minus{}2}\plus{}2000$
show that this sequence has upper bounded.

PS. Is this right sentence...? sorry for poor english...

$ q_n=max\{p_n,p_{n+1}\}$
In case , either $ p_n$ or $ p_{n+1}$ equals $ 2$ then $ p_{n+2} \leq p_{n+1}+p_{n}+2000=q_{n}+2002$
In other case, none of $ p_n$ or $ p_{n+1}$ equal to 2 then both ones are odd, so $ p_{n+1}+p_n+2000$ is even
${ \Rightarrow p_{n+2} \leq \frac{1}{2}(p_n+p_{n+1}+2000} \leq q_n+1000$
In conclusion, we have $ q_{n+1} \leq q_n+2002$
Existence $ t$ is large enough satisfying $ q_1 \leq 2003!t+1$.We prove that $ q_n \leq 2003!t+1$ for all $ n$
Using induction , if we have this statement hold for some n
$ \Rightarrow q_{n+1} \leq q_n+2002 \leq 2003!t+2003$
If $ q_{n+1} > 2003!t+1$ then let $ A=q_{n+1}-2003!k \Rightarrow 1<A \leq 2003 \Rightarrow A | 2003! \Rightarrow m|q_{n+1}$,which is impossible because $ q_{n+1}$ is prime
Hence, the sequence is bounded ,$ p_n\leq q_{n} \leq 2003!t+1$ for all $ n$

Wow GREAT!

Thank you for wonderful solution :lol:


We call a number a descending number if each digit is strictly smaller than the digit that comes before it.  For example, 863 is a descending number.  How many 3-digit descending numbers are there?

Given any three distinct digits, we can form one descending number, so our answer is $ \dbinom{10}3\equal{}\boxed{120}$.

Of course, you never really answered his question :P

$ \binom{n}{r}$ denotes the number of ways $ r$ things can be chosen from $ n$ things.  Also, $ \binom{n}{r}\equal{}\frac{n!}{r!(n\minus{}r)!}$.  So $ \binom{10}{3}\equal{}\frac{10!}{3!7!}\equal{}\frac{10 \times 9 \times 8}{6}\equal{}120$.

Also, $ \binom{n}{r} \equal{} \binom{n}{n\minus{}r}$.

At first I used casework and found a pattern that it would be the sum of the first 8 triangular numbers. Is there a combinatorical identity relating to this?

Sorry that I'm reviving this, but I think I have a legit question. xP

What would happen if it was a 7 digit number would the answer be C(10,7)?

For every 3 different digits, there is one corresponding descending number, which is just the digits in descending order. So the answer is the number of combinations of three different digits, which is  $\binom{10}{3} = \boxed{120}$

[hide=Sol]So, we have to find the number of 3-digit descending numbers. Well, once we choose $3$ distinct digits, we can form a descending number.
Hence, the number of 3-digit descending numbers there are is $\binom{10}{3} = \boxed{120}$ because there are $10$ total numerals, and we have to choose $3$. [/hide]

[quote=GreenLightningPuma][quote=sp33dcub3r]what does the 2 numbers in the parentheses mean?[/quote]

It means choosing. $5\choose2$ is 5 things, and choosing 2 of them.[/quote]

you can just do [code]$\binom{a}{b}$[/code]
$\binom{a}{b}$
looks better


If $a$,$b$,and $c$ ,$\in\mathbf{R^{*}+}$ prove that :
$\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}\geq 1$

[hide]By Cauchy in Engel form,

$\sum_{\text{cyc}}\frac{a}{b+2c+d}\ge\frac{(a+b+c+d)^{2}}{2(ab+ac+bc+ad+bd+cd)+2(ac+bd)}\ge 1$

The last inequality is the equivalent of $a^{2}+b^{2}+c^{2}+d^{2}\ge 2ac+2bd$

which can be easily proven by AM-GM[/hide]

By Cauchy in Engel form, we have
$\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}= \frac{a^{2}}{ab+2ac+ad}+\frac{b^{2}}{bc+2bd+ab}+\frac{c^{2}}{cd+2ac+bc}+\frac{d^{2}}{ad+2bd+cd}\ge \frac{(a+b+c+d)^{2}}{2ab+4ac+2ad+2bc+4bd+2cd}\ge 1$
$\Leftrightarrow (a-c)^{2}+(b-d)^{2}\ge 0$


Is it just me or is the 2003 AIME II a lot easier than usual? Just wondering because I took it for practice and scored dramatically higher than usual, though I'm not saying my score.

Umm...I didn't notice that it was really easy during practice. I had scored a 9, and 10 was my maximum, with my average around 8.6ish.

That could be an unusual burst of brain power though. And it usually depends on what the problems are on. If it's, say, primarily geometry and thats your thing, it'll be easier for you, while others who absolutely hate geometry will do worse.


[quote="fjwxcsl"]Let $ (x_1,x_2,\ldots,x_n)\in{R^n_{ + }},$ prove that:
  $ (\sum_{i = 1}^n{x_i})^n - n^n\prod_{i = 1}^n{x_i}\ge{\sum_{i = 1}^n{x_i^n} - n\prod_{i = 1}^n{x_i}.}$[/quote]
$ (\sum_{i = 1}^n{x_i})^r \ge{\sum_{i = 1}^n{x_i^r} + (n^r - n)(\prod_{i = 1}^n{x_i})^{\frac {r}{n}}.}$
$ r\geq \frac {n}{n - 1},r\in R,n\geq 2$
[hide="文献"](陈计,王振,一个分析不等式的证明,宁波大学学报,1992.2
张小明老师用最值压缩定理也证明过(不等式研究通讯))[/hide]

Just subtract $ \sum_{i\equal{}1}^{n}{x_{i}^{n}}$ from $ (\sum_{i\equal{}1}^{n}{x_{i}})^{n}$ and do AMGM on the remaining variables.


For all $ a,b,c$ be positive real numbers, show that
\[ \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}\plus{}\frac{10abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge\frac{17}4\]

[quote="dduclam"]For all $ a,b,c$ be positive real numbers, show that
\[ \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}} \plus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge\frac {17}4\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, our inequality is equivalent to $ f(v^2)\geq0,$ where
$ f(v^2) \equal{} (12u^2 \minus{} 17uw)v^2 \plus{} 9w^4 \minus{} 4uw^3.$
But $ f$ is a linear function.
Hence, $ f$ gets a minimal value, when $ v^2$ gets an extremal value, which happens when, two numbers from $ \{a,b,c\}$ are equal.
Id est, it remains only to check the case $ b \equal{} c \equal{} 1.$
For $ a\equal{}t^3$ it gives $ (t\minus{}1)^2(4t^7\plus{}8t^6\minus{}5t^5\minus{}2t^4\plus{}t^3\minus{}10t^2\minus{}t\plus{}8)\geq0,$ which is indeed true.

$ \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}} > 3$.so we need to prove that $ \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge \frac {5}{4}$.
the minimum value of $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$ is $ 8abc$. so $ \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge \frac {10abc}{8abc} \equal{} \frac {5}{4}$

Oh god!! aadil!!! u seem to be making this mistake too often!!
[b]if u substitue a lesser value in the denominator then the value of the fraction increases[/b]!!! so here u cant substitute the denominator by 8abc!!!

oh..sorry.but i have a doubt.if $ \frac{10abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge\frac{5}{4}$ then $ 8abc\ge (a\plus{}b)(b\plus{}c)(c\plus{}a)$ which is not true by am gm

It is not necessary that $ \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge\frac {5}{4}$. Even if $ \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} < \frac {5}{4}$, the other term $ \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$ can be large enough for the inequality to hold. Basically, the inequality is tougher than you think.


AoPSwiki algebra function domain blogs email articles

If you go to the special pages in the AoPSWiki, there is an "upload files" button.  It's untested as of yet, but that's the way you can upload pictures for use in the wiki.  For now, go to a math page on wikipedia to see the diagram syntax.  I or someone else will add the syntax to the AoPSWiki tutorial.

Note: You can even upload files that scroll through pictures.  This allows you for instance to use one file to gradually walk through a diagram construction.

[quote="mysmartmouth"]What is the deal with copyrights on posted images?  Can we use images as long as we cite the source, or can we only use original images?[/quote]

Do not use copyright images.  That's a no-no regardless of wiki or any other use, unless the holder of the copyright gives permission.  In our case, that permission would require relinquishing the copyright altogether because they would have to accept the wiki documentation policy.

We'll need to use either public domain images (and there are many), or create our own (which will become much easier in time).

Public domain images?  According to Wikipedia, "A creative work is said to be in the public domain if there are no laws which restrict its use by the public at large."  Most images are copyrighted.  Those that are not are licensed by licenses AoPS does not accept.  There are "public domain" images on sites like Flickr, but those won't help us much...perhaps I'm confused.  No matter, I have Geometer's Sketchpad now, and I will try to add some images of my own creation.

EDIT:  Also according to Wikipedia: "According to U.S. law, at least, an author's original works are covered by copyright, even without a formal notice incorporated into the work.....Technically, any Internet posting (such as blogs or emails) could be considered copyrighted material unless explicitly stated otherwise."

Yes, restrictions are very high, but there's nothing we can do about that.

In any case, I made a template to put in an article if an image is needed.  Just put {{image}} wherever an image is needed but you don't have one.  You can put multiple {{image}}'s in an article in case there are multiple spots where an image is needed. ([url=http://www.artofproblemsolving.com/Wiki/index.php/Template:Image]this[/url] is what it looks like)

Additionally, I also made a template for when you make an article where problems are needed but you haven't added any yet.  Just put {{problems}} there.  ([url=http://www.artofproblemsolving.com/Wiki/index.php/Template:Problems]this[/url] is what it looks like)

Well, the posting images thing definitely works ([url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_USAMO_by_probability1.01_dropped_problems#Problem_2[/url])


number theory
relatively prime
superior algebra
superior algebra unsolved

Hello, I need help to prove :

"Let $ K$ be a field generated over $ F$ by two elements $ \alpha,\beta$ of relatively prime degrees $ m,n$ respectively. Prove that $ [K: F]\equal{}mn$."

thx

Three steps:
First, show that $ m| [K: F]$ and $ n| [K: F]$.
Second, show that $ [K: F]\le mn$. These first two steps need no assumptions on $ m$ and $ n$.
Third, combine the previous two and the fact that $ m$ and $ n$ are relatively prime to get $ mn\le [K: F]\le mn$.


geometry
incenter
vector
calculus
trigonometry
geometry unsolved

Let $ABC$ be a triangle. If $M \in (AB)$ and $N \in (AC)$ so that $\frac{BM}{MA}=x$ and $\frac{CN}{NA}=y$. Prove that :

$M, I, N$ are on the same line $\iff bx+cy=a$, where $I$ is the incenter and $a,b,c$ the sides of our triangle.

We're going going to use position vectors.
So that:
$\overrightarrow{r_{M}}(x+1)=\overrightarrow{r_{B}}+x\overrightarrow{r_{A}}$
$\overrightarrow{r_{N}}(y+1)=\overrightarrow{r_{C}}+y\overrightarrow{r_{A}}$
The position vector of the incircle center is well known:
$\overrightarrow{r_{I}}=\frac{a\overrightarrow{r_{A}}+b\overrightarrow{r_{B}}+c\overrightarrow{r_{C}}}{a+b+c}$
Now:
$\overrightarrow{MI}=\overrightarrow{r_{A}}(\frac{a}{a+b+c}-\frac{x}{x+1})+\overrightarrow{r_{B}}(\frac{b}{a+b+c}-\frac{1}{x+1})+\overrightarrow{r_{C}}(\frac{c}{a+b+c})$
In the same way:
$\overrightarrow{IN}=\overrightarrow{r_{A}}(\frac{y}{y+1}-\frac{a}{a+b+c})+\overrightarrow{r_{B}}(\frac{-b}{a+b+c})+\overrightarrow{r_{C}}(\frac{1}{y+1}-\frac{c}{a+b+c})$
$M-I-N$ on the same line $\Longleftrightarrow \overrightarrow{MI}=K\cdot\ \overrightarrow{IN}$
$\overrightarrow{MI}=K\cdot\ \overrightarrow{IN}\Longleftrightarrow\ \frac{\frac{a}{a+b+c}-\frac{x}{x+1}}{\frac{y}{y+1}-\frac{a}{a+b+c}}=\frac{\frac{b}{a+b+c}-\frac{1}{x+1}}{\frac{-b}{a+b+c}}$
After a long calculus we obtain
$\overrightarrow{MI}=K\cdot\ \overrightarrow{IN}\Longleftrightarrow\ bx+cy=a$, and, with this, the problem is solved...

[quote="OHO"]$M, I, N$ are on the same line.
$\iff S(AMN)=S(AMI)+S(ANI)$
$\iff \frac{1}{2}AM\cdot AN\sin A=\frac{1}{2}AM\cdot r+\frac{1}{2}AN\cdot r$
$\iff \frac{\sin A}{r}=\frac{1}{AM}+\frac{1}{AN}$
$\iff \frac{a+b+c}{bc}=\frac{1}{AM}+\frac{1}{AN}$
$\iff a+b+c=b(\frac{AM+MB}{AM})+c(\frac{AN+NB}{AN})$
$\iff a=bx+cy$[/quote]

nice, this was my solution too.


我选了计算机。这里有没有计算机高手啊，可以指点一下不？
我之前是基本上没有编程基础的，不晓得进了大学能不能学好。
对了，我进的是NUS(National University of Singapore).
这里有没有同志？ :D  :roll:

hei,must you take an ENTRANCE test? or gao kao is enough :D

Who will go to NTU? :P

cuong can read chinese?! :o This stuns me.

:D 嘻嘻，我不用参加高考拉．我是ｓｍ２的学生，不知道你有没有听说过．[/i]


Vieta
geometry
geometric transformation
reflection
circumcircle
rotation

The roots of x^3+a*x^2+b*x+c form an equilateral triangle in complex plane. Which of the following must hold?
A a^2 = 3b
B a^2+b^2 =c
C c^2= ab
D a^3 = 8c
E b^2= c^2+z, for some z.

[hide="Hint"]
[b]Lemma 1[/b]

A triangle with vertices $ x, y,z \in \mathbb{C}$ is equilateral iff $ x^2 \plus{} y^2 \plus{} z^2 \equal{} xy \plus{} yz \plus{} zx$
[/hide]

[hide="Solution"]
$ x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} yz \plus{} zx) \equal{} 0$

Therefore $ (x \plus{} y \plus{} z)^2 \minus{} 3(xy \plus{} yz \plus{} zx) \equal{} 0$

Using Vieta's, $ x \plus{} y \plus{} z \equal{} \minus{} a$ and $ xy \plus{} yz \plus{} zx \equal{} b$

Therefore $ ( \minus{} a)^2 \minus{} 3b \equal{} 0$

Hence $ a^2 \equal{} 3b \Rightarrow \boxed{A}$
[/hide]

After coming up with this solution, I realized that I am assuming that $ a,b,c$ are real, although this is not necessarily true.

[hide]Suppose these vertices are $ A,B,C$.  Every cubic has at least one real root.  WLOG let this be $ A$.  The other two roots can be expressed in the form $ a\pm bi$.  Thus they are reflections of each other through the real axis.

Let $ O$ be the centroid of $ \triangle ABC$.  It is at $ k\plus{}0i$.  Let $ R\equal{}AO\equal{}BO\equal{}CO$, since $ O$ is also the circumcenter of $ \triangle ABC$.  We can express the points $ A,B,C$ by the equation $ (x\minus{}k)^3\equal{}R^3$.  Expanding yields $ x^3\minus{}3kx^2\plus{}3k^2x\minus{}(k^3\plus{}R^3)\equal{}0$.  The only choice that this equation always satisfies is $ \boxed{\text{A}}$, where both sides simplify to $ 9k^2$.

It turns out that when $ R$ is positive, the triangle points to the right, and when $ R$ is negative, the triangle points to the left.[/hide]

This is a very nice problem ! I think it is surprising that the fact if we have an equilateral triangle is independent of $ c$ is quite interesting. In fact, this is just saying that we can rotate a triangle and preserve equilateral-ness.


Prove that $ x^3\plus{}y^3\plus{}2\equal{}z^3$ has infinitely many solutions in natural numbers

For any $ a>0$, the triple $ (x,y,z)$ with $ x\equal{}6a^2$, $ y\equal{}6a^3\minus{}1$, $ z\equal{}6a^3\plus{}1$, satisfies the equation.

Intuitively we should be able to find $ x,y$ such that one of them is "very close" to $ z$ and the other can compensate for whatever is left. So let $ y$ be the "very close" one. First try $ y \equal{} z \minus{} 1$ so
\[ x^3 \equal{} 3z^2 \plus{} 3z \minus{} 1,\]
which doesn't help in any obvious way, so try $ y \equal{} z \minus{} 2$ to get
\[ x^3 \equal{} 6(z \minus{} 1)^2.\]
Then $ x$ should be a quadratic term with leading coefficient $ 6$ and $ z \minus{} 1$ should be a cubic term with leading coefficient $ 6$ to make leading coefficients and degrees equal. So we can let $ x \equal{} 6a^2$ and $ z \minus{} 1 \equal{} 6a^3$.

Exactly :)
Besides, what's your solution?


Let $ a$, $ b$, $ c$ be positive real numbers. Prove that

$ \frac {2\left(a^3 \plus{} b^3 \plus{} c^3\right)}{abc} \plus{} \frac {9\left(a \plus{} b \plus{} c\right)^2}{a^2 \plus{} b^2 \plus{} c^2} \geq 33$

Hey I saw in G.M an inequality almost the same proposed by Gil which can be solved using Schur. :huh:

you mean you do it by brute force and then you use schur???

there's a much better way to do this.

Ceasar could you post your solution with expanding and using Shur?Thanks

You don't need brute force to come up with Soarer's factorization from post #4. In fact, equivalently transform your inequality:

$\frac{2\left(a^3+b^3+c^3\right)}{abc}+\frac{9\left(a+b+c\right)^2}{a^2+b^2+c^2}\geq 33$
$\Longleftrightarrow\ \ \ \ \ \left(\frac{2\left(a^3+b^3+c^3\right)}{abc}-6\right)+\left(\frac{9\left(a+b+c\right)^2}{a^2+b^2+c^2}-27\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ 2\frac{a^3+b^3+c^3-3abc}{abc}+9\frac{\left(a+b+c\right)^2-3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\geq 0$.

Now, it is well-known that

$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)$ and
$\left(a+b+c\right)^2-3\left(a^2+b^2+c^2\right)=-2\left(a^2+b^2+c^2-bc-ca-ab\right)$,

so the inequality above becomes

$2\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)}{abc}+9\frac{-2\left(a^2+b^2+c^2-bc-ca-ab\right)}{a^2+b^2+c^2}$
$\geq 0$.

Now, according to the well-known inequality $a^2+b^2+c^2\geq bc+ca+ab$, we have $a^2+b^2+c^2-bc-ca-ab\geq 0$, so that we can divide this inequality by $a^2+b^2+c^2-bc-ca-ab$ to obtain

$2\frac{a+b+c}{abc}+9\frac{-2}{a^2+b^2+c^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ 2\frac{a+b+c}{abc}-\frac{2\cdot 9}{a^2+b^2+c^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \frac{a+b+c}{abc}\geq \frac{9}{a^2+b^2+c^2}$
$\Longleftrightarrow\ \ \ \ \ \left(a+b+c\right)\left(a^2+b^2+c^2\right)\geq 9abc$.

But this is evident, since AM-GM yields $a+b+c\geq 3\sqrt[3]{abc}$ and $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}$, so that $\left(a+b+c\right)\left(a^2+b^2+c^2\right)\geq 3\sqrt[3]{abc}\cdot 3\sqrt[3]{a^2b^2c^2}=9abc$.

Proof complete.

  darij

Here is my solution for this topic
Suppose that $ c \equal{} min\{a;b;c\}$
We need prove that:
$ \frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \plus{} \frac {9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\geq33$
or $ \frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \minus{} 6 \plus{} \frac {9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \minus{} 27\geq0$
$ \Longleftrightarrow$ $ \frac {2(a \plus{} b \plus{} c)[(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)}{abc} \minus{} \frac {18(a \minus{} b)^2 \plus{} 18(a \minus{} c)(b \minus{} c)}{a^2 \plus{} b^2 \plus{} c^2} \geq 0$
or $ [\frac {2(a \plus{} b \plus{} c)}{abc} \minus{} \frac {18}{a^2 \plus{} b^2 \plus{} c^2}][(a \minus{} b)^2 \plus{} (a \minus{} c)(b \minus{} c)] \geq 0$
By AM-GM,easy to prove that $ \frac {2(a \plus{} b \plus{} c)}{abc} \minus{} \frac {18}{a^2 \plus{} b^2 \plus{} c^2} \geq 0$
Done!

[quote="Detective"]Let $ a, b, c$ be positive real numbers. Prove that

$ \frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \plus{} \frac {9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \geq 33.$[/quote]$ \frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \plus{} \frac {9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \minus{} 33$

$ \equal{}\left(\frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab} \minus{} 3\right)\left[2 \minus{} \frac {18abc}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}\right]\geq0.$

This problem is introduced as Exercise 1.20 on page 32 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.wl.cn/5800797]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

I just started learning SOS, try to do it:
\[ \frac{{2\left( {a^3  \plus{} b^3  \plus{} c^3 } \right)}}{{abc}} \plus{} \frac{{9\left( {a \plus{} b \plus{} c} \right)^2 }}{{a^2  \plus{} b^2  \plus{} c^2 }} \ge 33\]
\[ \Leftrightarrow \frac{{2\left( {a^3  \plus{} b^3  \plus{} c^3 } \right)}}{{abc}} \minus{} 6 \plus{} \frac{{9\left( {a \plus{} b \plus{} c} \right)^2 }}{{a^2  \plus{} b^2  \plus{} c^2 }} \minus{} 27 \ge 0\]
\[ \Leftrightarrow \frac{{2\left( {a^3  \plus{} b^3  \plus{} c^3  \minus{} 3abc} \right)}}{{abc}} \plus{} \frac{{9\left( {\left( {a \plus{} b \plus{} c} \right)^2  \minus{} 3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)} \right)}}{{a^2  \plus{} b^2  \plus{} c^2 }} \ge 0\]
\[ \Leftrightarrow \frac{{\left( {a \plus{} b \plus{} c} \right)\sum {\left( {a \minus{} b} \right)^2 } }}{{abc}} \minus{} \frac{{9\sum {\left( {a \minus{} b} \right)^2 } }}{{a^2  \plus{} b^2  \plus{} c^2 }} \ge 0\]
\[ \Leftrightarrow \left( {\frac{{a \plus{} b \plus{} c}}{{abc}} \minus{} \frac{9}{{a^2  \plus{} b^2  \plus{} c^2 }}} \right)\sum {\left( {a \minus{} b} \right)^2 }  \ge 0\]
\[ \Leftrightarrow \frac{{\left( {a \plus{} b \plus{} c} \right)\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \minus{} 9abc}}{{abc\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}}\sum {\left( {a \minus{} b} \right)^2 }  \ge 0\]
\[ \Leftrightarrow \frac{{\sum {\left( {a \plus{} b \plus{} 3c} \right)\left( {a \minus{} b} \right)^2 } }}{{2abc\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}}\sum {\left( {a \minus{} b} \right)^2 }  \ge 0.\]


advanced fields
advanced fields unsolved

Can anyone help me with the following:

Let $ N$ be a simple point process on $ E$ such that $ N(A)$ is Poisson distributed for every $ A \in B$. Prove that $ N$ is a Poisson process.

Thanks in advance

I suppose the problem is to show that the numbers of points of the process in disjoint sets are independent.
The main idea for this is to show the following:

If $ X_{1},X_{2},X$ are random variables with Poisson distributions such that $ X_{1}+X_{2}=X$,
then $ X_{1}$ and $ X_{2}$ are independent.


analytic geometry
geometry
angle bisector

Prove that all three angle bisectors of a triangle pass through a single point.

It's obvious but I don't know how to prove it. :(
Haven't learned proofs yet.

i know it starts with the angle bisector theorem, but i don't know how to do it from there.

[hide="Two hints"]1. (Angle bisector theorem.) If an angle is given, then a point belongs to its bisector if and only if it's at the equal distance from both sides of the angle.

2. Prove that the intersection of two bisectors must belong to the third bisector.[/hide]

[quote="lotrgreengrapes7926"][quote="Farenhajt"]1. (Angle bisector theorem.) If an angle is given, then a point belongs to its bisector if and only if it's at the equal distance from both sides of the angle.[/quote]

I believe the Angle Bisector Theorem Captain Sugar was refering to is:
[hide] If triangle $ABC$ has an angle bisector $AD$ ($D\in BC$), then $\frac{AB}{BD}=\frac{AC}{CD}$. [/hide][/quote]

Nope, THAT one has no use in the given problem. The clue is in the "basic" angle bisector theorem.

[quote="lotrgreengrapes7926]I believe the Angle Bisector Theorem Captain Sugar was refering to is:
 If triangle $ABC$ has an angle bisector $AD$ ($D\in BC$), then $\frac{AB}{BD}=\frac{AC}{CD}$.[/quote]

Using this Angle Bisector Theorem, we can make use of Ceva's Theorem.  :)  So either one you choose you can still solve the problem.

[quote="daermon"]I'm not sure what your hints are reffering too, but the proof I am familiar with is [hide]call the triangle $ABC$.  an angle bisector is equidistant from the two sides that make up the angle it bisects.  so, the angle bisector of $A$ is equidistant from $b$ and $c$ and the angle bisector of $B$ is equidistant from $a$ and $c$.  call the intersection of these two lines $x$.  x is equidistant from $b$ and $c$, and is equidistant from $a$ and $c$.  by transitivity, $b$ is equidistant from $a$, and this intersection point is on the angle bisector of $C$ and our proof is complete[/hide][/quote]

correct :D  this is the one that i know

So, using The Big Angle Bisector Theorem AND Ceva's Theorem for a frigging simple problem as this? Killing a fly with a cannon...

Using a strong result to prove a lemma used in the proof of the result...  :|

My thoughts exactly.

Complementary edit by PenguinIntegral:
[size=150][b]Circular logic is NOT good.[/b][/size]

[quote="vishalarul"]One method is to use coordinate proof. They do not use circular logic.

Let the vertices be $(0,0)$, $(0,a)$, and $(b,c)$. Using the 'more complicated' version of Angle-Bisector Theorem, find the equations of the lines, and prove that they intersect at the same point.

This is a little hard, but it is a good proof.[/quote]

This actually seems more like a "constructed" problem, conceived to illustrate the use of "Big" Angle Bisector Theorem. But the most natural and simple proof of the initial problem is like 6th grade stuff or so. Confer [b]daermon[/b]'s post.


It was working before, but I've been sitting here for the past hour and a half trying to get this to work.
I'm following everything it says on the 'Basics' part of the LateX, but it's just not working. When I hit ctrl+f7 for the compiling part of the process, some tab opens up, but nothing happens and then when I hit f5 nothing happens at all, when it should be showing me the output! (And I went onto my dad's computer, redownloaded everything and it's still not working! So it's not a computer problem.. I don't know what the deal is - maybe I'm missing something obvious or they have a new version or something??? I just don't know..)

I am supposed to have a document turned in tonight, in less than an hour or two, and this is annoying me.

Please help!! Thanks

-Muy Frustrated

Try using the menu commands for compiling and viewing.
Then if it doesn't work there must be something wrong with your document or computer.

Yeah, I did that too. And it also did not work on my dad's computer, so I'm not really sure what the deal is -_-"

Wich distribution do you use and wich editor?

It looks as if it can't find the latex files. Look in Build -> Define Output Profiles (or press Alt-F7) and check that all the paths are correct. You can use the Wizard (bottom left button) to help.

[quote="WindSlicer"]I am supposed to have a document turned in tonight, in less than an hour or two, and this is annoying me.

Please help!! Thanks

-Muy Frustrated[/quote]You are having problems with Texnic, you might want to consider [url=http://www.winedt.com]WinEdt[/url], as it has a trial version of 30 days with no restrictions whatsoever, especially if you say that you have to turn in just one document and quickly.

Ok, the WinEdit is working pretty well - I like the " (PDF) Texify" feature, it's very easy to use.

I just have to nitpick through my latex with the spacing and then I'll be done. Hopefully they'll accept it a bit late.

Thanks for everyone's help.

EDIT-

Yes! I am done.

Thanks again.


Hey, Darij is the only one member of German team, who solved Problem number 3 perfectly. He got 7 points, while others got 0 or 1.

[quote="Fermat -Euler"]Darij it's the best[/quote]

Now he has 28 points, but two problems was not yet coordinated, but he told me he did not do these two too well.. but as Arne said, maybe 28 would be sufficient to gold..

I wish it to him.

[quote="Arne"]If he gets one more point, he can be almost sure of gold.

If not, there still is a chance :)[/quote]

He said me, that he is not sure if he will receive in these two examples 3 or 4 points, so...

He does deserve it.

http://www.esnips.com/doc/8b39134a-eb4c-4a18-a28c-04b97e95be9f/Maria.JPG

photo of Darij with his gold

[quote="rem"]Who is the guy on the left part of the photo(in a white t-shirt)? He looks like a Math gangster :gleam:[/quote]

Also a gold winner, forum member ondrob.

yes this one is a good mathematician ( I think so ...  :blush: ) and he is from Slovakia who got last years silver medalist  also I know that he had  participated in IMO 2004 but don't know the results of him.

davron latipov
khujand, tajikistan

He got two silver medals (2004, 2005) and gold this year.


Let $f$ be the function defined by $f(x)=ax^2-\sqrt2$ for some positive $a$. If $f(f(\sqrt{2}))=-\sqrt2$,then $a=$

$\text{(A)}\ \frac{2-\sqrt2}{2} \qquad \text{(B)}\ \frac12 \qquad \text{(C)}\ 2-\sqrt2 \qquad \text{(D)}\ \frac{\sqrt2}{2} \qquad \text{(E)}\ \frac{2+\sqrt2}{2}$

[hide]Because $f(f(\sqrt{2}))=a(2a-\sqrt2)^2-\sqrt2=-\sqrt2$, $a(2a-\sqrt2)^2=0$. Therefore, $2A=\sqrt2$, and the answer is $a=\frac{\sqrt2}{2}$.

Cool, I found a good answer to a function problem. Lol. :P [/hide]

[hide]$f(\sqrt{2})=2a-\sqrt2$

$f(2a-\sqrt2)=a(2a-\sqrt2)^2-\sqrt2=-\sqrt2 \implies a(2a-\sqrt2)^2=0 \implies 2a-\sqrt2=0 \implies a=\frac{\sqrt2}{2} \implies \boxed{D}$ [/hide]


Let $ X$ and $ Y$ be nonempty sets and let $ h: X\times Y\rightarrow \Re$ have bounded range in $ \Re$.

Establish that $ \sup_{x,y} h(x,y)\equal{}\sup_{x}\sup_{y} h(x,y)\equal{}\sup_{y}\sup_{x} h(x,y)$

Can someone please help.Thank you very much.  :P

Do anyone have any idea? :?:
Please help. :!:

[hide="hint"]$ \sup_{x,y}h(x,y)\minus{}\epsilon< h(z,t)\leq \sup_x\sup_yh(x,y)$ for some $ (z,t)$ :D [/hide]


Hi!
I use MikTEX 2.8 with TEXnic Center.....
When I had Windows Wista x32 (Russian version). I had no problems with russian language.. I only typed
\usepackage[babel]{russian}
and everything was OK....

Now I changed my Windows to Vista x64(English) and problems started... Firstly russian letters were displayed as herogliphes in TEXnic Center.. Then I changed something in Language controls and this problem disappeared... But when I compile it, no letters appear in the pdf document... Sometimes it writes in Log that no hyppentation patterns were found, sometimes not....

Any suggetions????
Maybe there is another editor like Texnc center?? I tried to download another... But it was completely different... I cannot find how to compile etc...

Thanks for advice....

Please give us a minimal example of a document you are having trouble with. It must be complete (so can be compiled) but short.

TexnicCenter cannot do unicode so may have trouble with some characters. The new version being tested may work better. Also TeXworks is supplied with MiKTeX 2.8 so you could try that as an alternative editor.

[code]
\documentclass[A4paper,11pt]{article}

\usepackage{amsthm, amssymb, amsmath, amsfonts}
\usepackage[russian]{babel}
\linespread{1.2}
\pagestyle{empty}
\setlength{\oddsidemargin}{-0.2in}
\setlength{\evensidemargin}{0in}
\setlength{\textwidth}{7in}
\setlength{\textheight}{9.80in}
\setlength{\topmargin}{-0.4in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parskip}{1pt}
\setlength{\labelsep}{1pt}
\setlength{\parindent}{15pt}
\setlength{\smallskipamount}{3ex}
\setlength{\smallskipamount}{1ex}
\def\nn{{\mathbb{N}}}
\def\rr{{\mathbb{R}}}
\def\zz{{\mathbb{Z}}}
\def\cc{{\mathbb{C}}}
\def\qq{{\mathbb{Q}}}
\righthyphenmin=30
\def\1{\^{\i}}
\def\2{{\u a}}
\def\3{{\c s}}
\def\4{{\^ a}}
\def\5{{\c t}}
\def\6{{\^I}}
\def\7{{\c S}}
\def\8{{\c T}}

\begin{document}

\textbf{Неравенство Шура}
Дана функция $f:\mathbb R\rightarrow \mathbb R_{+}$ и $a,b,c\ge 0$. Тогда:

\centerline{$f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\geq0$}

\end{document}
[/code]

and the log file

[code]*************************************
* Local config file bblopts.cfg used
*
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\00miktex\bblopts.cfg"
File: bblopts.cfg 2006/07/31 v1.0 MiKTeX 'babel' configuration
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\generic\babel\russianb.ldf"
Language: russianb 2008/03/21 v1.1r Russian support from the babel system

("C:\Program Files (x86)\MiKTeX 2.8\tex\generic\babel\babel.def"
File: babel.def 2008/07/06 v3.8l Babel common definitions
\babel@savecnt=\count99
\U@D=\dimen112
)

Package babel Warning: No hyphenation patterns were loaded for
(babel)                the language `Russian'
(babel)                I will use the patterns loaded for \language=0 instead.

\l@russian = a dialect from \language0
[/code]

1. The problem with the current version of TexnicCenter is that when you type in Russian it will show the "hieroglyphics" because the editor can't show Cyrillic letters. If you can cope with this, then use
\usepackage[russian]{babel}
\usepackage[utf8]{inputenc}
and that will tell LaTeX that the "hieroglyphics" are really Russian and you will get Russian in your PDF

If you want proper Cyrillic letters when you type then use Texworks which is included with MiKTeX. You must still use
\usepackage[russian]{babel}
\usepackage[utf8]{inputenc}
there as well

2. The log contain warnings rather than errors so are not serious. They tell you that you haven't told LaTeX how you want to break up a long Russian word at the end of the line. You can ignore this or follow the instructions at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=newlang]Using a new language with Babel[/url]. The only bit you need is:
[quote]On a MiKTeX distribution v2.0 or later, the whole procedure can be done via the GUI. To select the new language, do:
Start-> Programs-> MiKTeX 2-> MiKTeX Options, and select the Languages tab. Select your language from the list, press the Apply button, and then the OK button. Then select the General tab and press the Update Now button[/quote]
3. There are some problems in your file. On the first line, [i]A4paper[/i] should be [i]a4paper[/i] (that's one of the log warnings).
There are other things you may have copied from older tex files which are obsolete now:
a) paper size, margins etc should be done using the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]geometry package[/url]. It's much simpler and makes sure everything works
b) do not use \def when making new definitions, use \newcommand because it checks for clashes etc. You can get problems with \def unless you really know what you are doing. So, for example,
\def\nn{{\mathbb{N}}} (why so many brackets?)
should read
\newcommand{\nn}{\mathbb{N}}

Hm... the problem is that it doesn't want to create a pdf file.....

You'll have to give more detail than that

what details do you need to help me???
I have MikTEX 2.8 and I've just downloaded the TEXnic Center 2 Alpha....

The document above compiled fine for me as the attached shows, so:
What document did you use?
The same as the one posted above?
Did you add the inputenc line?
What did the log say?
Did the file compile at all?
Were you using LaTeX=>PDF profile?
etc etc

Have you tried it in the old version of Texnicenter or in Texworks? Remember the new version is Alpha so may have its own problems.

Yep... the smae document
Yes... I added....
Log file
[code]This is pdfTeX, Version 3.1415926-1.40.10 (MiKTeX 2.8) (preloaded format=pdflatex 2009.11.26)  26 NOV 2009 23:18
entering extended mode
**C:/Users/Scotch/Documents/sample.tex
(C:/Users/Scotch/Documents/sample.tex
LaTeX2e <2009/09/24>
Babel <v3.8l> and hyphenation patterns for english, dumylang, nohyphenation, german, ngerman, german-x-2009-06-19, ngerm
an-x-2009-06-19, french, russian, loaded.
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\base\article.cls"
Document Class: article 2007/10/19 v1.4h Standard LaTeX document class
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\base\size11.clo"
File: size11.clo 2007/10/19 v1.4h Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\classes\amsthm.sty"
Package: amsthm 2004/08/06 v2.20
\thm@style=\toks14
\thm@bodyfont=\toks15
\thm@headfont=\toks16
\thm@notefont=\toks17
\thm@headpunct=\toks18
\thm@preskip=\skip43
\thm@postskip=\skip44
\thm@headsep=\skip45
\dth@everypar=\toks19
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\amsfonts\amssymb.sty"
Package: amssymb 2009/06/22 v3.00

("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\amsfonts\amsfonts.sty"
Package: amsfonts 2009/06/22 v3.00 Basic AMSFonts support
\@emptytoks=\toks20
\symAMSa=\mathgroup4
\symAMSb=\mathgroup5
LaTeX Font Info:    Overwriting math alphabet `\mathfrak' in version `bold'
(Font)                  U/euf/m/n --> U/euf/b/n on input line 96.
))
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\math\amsmath.sty"
Package: amsmath 2000/07/18 v2.13 AMS math features
\@mathmargin=\skip46

For additional information on amsmath, use the `?' option.
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\math\amstext.sty"
Package: amstext 2000/06/29 v2.01

("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\math\amsgen.sty"
File: amsgen.sty 1999/11/30 v2.0
\@emptytoks=\toks21
\ex@=\dimen103
))
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\math\amsbsy.sty"
Package: amsbsy 1999/11/29 v1.2d
\pmbraise@=\dimen104
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\ams\math\amsopn.sty"
Package: amsopn 1999/12/14 v2.01 operator names
)
\inf@bad=\count87
LaTeX Info: Redefining \frac on input line 211.
\uproot@=\count88
\leftroot@=\count89
LaTeX Info: Redefining \overline on input line 307.
\classnum@=\count90
\DOTSCASE@=\count91
LaTeX Info: Redefining \ldots on input line 379.
LaTeX Info: Redefining \dots on input line 382.
LaTeX Info: Redefining \cdots on input line 467.
\Mathstrutbox@=\box26
\strutbox@=\box27
\big@size=\dimen105
LaTeX Font Info:    Redeclaring font encoding OML on input line 567.
LaTeX Font Info:    Redeclaring font encoding OMS on input line 568.
\macc@depth=\count92
\c@MaxMatrixCols=\count93
\dotsspace@=\muskip10
\c@parentequation=\count94
\dspbrk@lvl=\count95
\tag@help=\toks22
\row@=\count96
\column@=\count97
\maxfields@=\count98
\andhelp@=\toks23
\eqnshift@=\dimen106
\alignsep@=\dimen107
\tagshift@=\dimen108
\tagwidth@=\dimen109
\totwidth@=\dimen110
\lineht@=\dimen111
\@envbody=\toks24
\multlinegap=\skip47
\multlinetaggap=\skip48
\mathdisplay@stack=\toks25
LaTeX Info: Redefining \[ on input line 2666.
LaTeX Info: Redefining \] on input line 2667.
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\generic\babel\babel.sty"
Package: babel 2008/07/06 v3.8l The Babel package

*************************************
* Local config file bblopts.cfg used
*
("C:\Program Files (x86)\MiKTeX 2.8\tex\latex\00miktex\bblopts.cfg"
File: bblopts.cfg 2006/07/31 v1.0 MiKTeX 'babel' configuration
)
("C:\Program Files (x86)\MiKTeX 2.8\tex\generic\babel\russianb.ldf"
Language: russianb 2008/03/21 v1.1r Russian support from the babel system

("C:\Program Files (x86)\MiKTeX 2.8\tex\generic\babel\babel.def"
File: babel.def 2008/07/06 v3.8l Babel common definitions
\babel@savecnt=\count99
\U@D=\dimen112
)[/code]
Yes, it's compiled...
what do you mean Were you using LaTeX=>PDF profile?
Yes I tried the older versionsof Texnic Center....
I don't know how to compile files in Texworks.... I tried to read a manual, but it to hard to understand...

You seem to have cut off the log file in the middle so it doesn't show any errors or warnings as that's later on in the file. So you haven't said why the compilation doesn't produce a pdf.

The LaTeX=>PDF profile is the one indicated in the dropdown box in the toolbar and is the standard one. You'll also find LaTeX=>DVI there, though you don't want that.

Texworks is easy. To compile choose pdflatex from the dropdown box and press the green arrow next to it and the result will then appear.

Ohh.. great!!!
Thanks [b]stevem[/b]..
I tried in TEXwirks as you said (i couldn't compile first cause it was pdftex).... But while compiling it asked me to install some packges (cyrillic and hi)... Now everything works great in TEXnic Center toO!!!!!!!!!!!


One card is drawn at random from a standard deck of $ 52$ cards (with the four suits-spades, hearts, diamonds, and clubs). What is the probability that the card drawn is a heart or an ace? Express your answer as a decimal to the nearest hundredth.

There are 13 hearts and 4 aces but one of those cards is a heart and an ace so we are counting it twice so the answer is:
$ \frac{13\plus{}4\minus{}1}{52} \equal{} \frac{16}{52} \equal{} \frac{4}{13} \equal{} \boxed{.31}$


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral such that $ AD\plus{}BC\equal{}AB$. Prove that the bisectors of the angles $ \hat{ADC}$ and $ \hat{BCD}$ meet on the line $ AB$.

Let $ M$ be the point on $ AB$ such that $ AM\equal{}AD$ and $ BM\equal{}BC$. Then both bisectrises go through another intersection point of circumcircle of triangle $ DMC$ and line $ AB$ :)
Bye

Let $ E$ be the intersection of the two bisectors. $ DE \cap AB \equal{} F,CE \cap AB \equal{} G$ and let $ M$ be a point on $ AB$, which $ AM \equal{} AD$. $ \angle ADC \equal{} \alpha,\angle DCB \equal{} \beta$.
$ \angle DMA \equal{} \frac {180^{\circ} \minus{} \angle DAB}{2} \equal{} \frac{\beta}{2} \equal{} \angle ACG \Longleftrightarrow DCMG$ is cyclic,
with a similar way $ DCMF$ is cyclic, and according to $ G,F,M$ are collinear, we should have $ G \equal{} F \equal{} E$
and it is proved that $ E \in AB$


search
superior algebra
superior algebra unsolved

If $ R$ is a ring, and has the property that $ x^3\equal{}x,\forall x\in R$, then show $ R$ is commutative.

I did it when the property for idempotent elements, and I couldn't for the above question.

Also, how we could prove it for $ x^n\equal{}x$, for some $ n\in \mathbb Z$.

I think it will be harder mission.

 :)

thank u dude   :)

But , I would like to see the proof of Jacobson result :

[quote="coquitao"]
- Let $ \mathrm{R}$ be a ring with center $ \mathbf{Z}$ such that for a fixed $ n > 1$, $ x^{n} \minus{} x \in \mathbf{Z}$ for all $ x \in \mathrm{R}$. Then $ \mathrm{R}$ is a commutative ring.
[/quote]

I couldn't have the book that's coquitao pointed for it .

 :)


Let $ a$ be a real number such that the polynomials $ x^3 \plus{} (a \minus{} 1)x^2 \minus{} ax \plus{} 1$ and $ x^2 \plus{} ax \plus{} 1$  have a common root.  What is the value of $ a$?

[hide]
Let $ x^3 \plus{} (a\minus{}1)x^2 \plus{}ax \plus{}1$ have roots $ r_1,r_2,r_3$

Let $ x^2 \plus{} ax \plus{} 1$ have roots $ r_1,r_4$

Using Vieta's formula;
$ r_1r_2r_3\equal{}\minus{}1\qquad(1)$
$ r_1r_2\plus{}r_2r_3\plus{}r_3r_1 \equal{} a\qquad(2)$
$ r_1\plus{}r_2\plus{}r_3\equal{}1\minus{}a\qquad(3)$

$ r_1\plus{}r_4\equal{}\minus{}a\qquad(4)$
$ r_1r_4\equal{}1\qquad(5)$

$ (1),(5)\Rightarrow r_4\equal{}\minus{}r_2r_3$
$ (3)(4)\Rightarrow r_2r_3 \equal{} 1\plus{}r_4$
$ (2)(5)\Rightarrow r_1(r_2\plus{}r_3\plus{}1)\equal{}0$

Combining there we get $ r_2r_3\equal{}\minus{}2$ and Therefore $ r_1 \equal{}\minus{} \frac{1}{2} \Rightarrow r_4\equal{}\minus{}2$

$ \left(x\plus{}\frac{1}{2}\right)(x\plus{}2) \equal{} x^2 \plus{} \frac{5}{2}x\plus{}1$

So $ a\equal{}\frac{5}{2}$
[/hide]

[hide]Rewrite the first equation as $ x^3\minus{}x^2\plus{}1\equal{}a(x\minus{}x^2)$. $ x\equal{}0$ or $ x\equal{}1$ can't be roots for any a so $ a\equal{}\frac {x^3\minus{}x^2\plus{}1}{x(1\minus{}x)}$. From the second equation $ a\equal{}\minus{}\frac{x^2\plus{}1}{x}$.

Solving $ \frac {x^3\minus{}x^2\plus{}1}{x(1\minus{}x)}\equal{}\minus{}\frac{x^2\plus{}1}{x}$ we get $ x\equal{}2$, and $ a\equal{}\minus{}\frac{5}{2}$.[/hide]

[hide="Hint"]Two polynomials have a common root $ R$ iff they share a common factor $ (x \minus{} R)$.[/hide]

[hide="Solution"]Let's find such a common root. Let those two polynomials be $ y$ and $ z$, since I'm in a hurry.

$ GCD(y,z) \equal{} GCD(y \minus{} xz \plus{} z,z) \equal{} GCD(2 \minus{} x,z)$. The only possible common root, therefore, is that of $ 2 \minus{} x \equal{} 0 \implies x \equal{} 2$.

We thus have $ x^2 \plus{} a x \plus{} 1 \equal{} (x \minus{} 2)(x \minus{} r)$, and it is clear that $ r \equal{} 1/2$, so $ z \equal{} (x \minus{} 2)(x \minus{} 1/2)$, and $ a \equal{} \minus{}(2 \plus{} 1/2) \equal{} \minus{}5/2$[/hide]


the following problem may be incorrect because i am writing it from memory.

you have a string and you make a mark every seventh and every thirteenth of the string.  Then you divide it into 20 equal pieces.  Prove that every piece except for the first and last piece will have only one marking.

DO NOT promote this because this is in my book JUST after we learn adding fractions.

edited

Your problem is poorly stated.  I could easily divide such a string into 20 pieces, 19 of which would have no markings on them.  [i]:edit -- the problem has been restated so that this isn't a problem any more:[/i]  You really should work on making sure that your problem statements are clear, well-written, and correct.

This is too similar to a problem on the application for HCSSiM, with the numbers changed.


A real number $ r\equal{}0.d_{1}d_{2}d_{3}...$ is called repetitive if its decimal expansion contains arbitrarily long blocks that are the same; that is, for every $ k$ there exists distinct $ m$ and $ n$ such that $ d_{m}\equal{}d_{n}$, $ d_{m\plus{}1}\equal{}d_{n\plus{}1}$, ..., $ d_{m\plus{}k}\equal{}d_{n\plus{}k}$. Prove that all numbers are repetitive.

[quote="Pythagorean12"]A real number $ r \equal{} 0.d_{1}d_{2}d_{3}...$ is called repetitive if its decimal expansion contains arbitrarily long blocks that are the same; that is, for every $ k$ there exists distinct $ m$ and $ n$ such that $ d_{m} \equal{} d_{n}$, $ d_{m \plus{} 1} \equal{} d_{n \plus{} 1}$, ..., $ d_{m \plus{} k} \equal{} d_{n \plus{} k}$. Prove that all numbers are repetitive.[/quote]

Let for every k let $ D_n\equal{}(d_{n},d_{n\plus{}1}, ... , d_{n\plus{}k})$. Clearly there are only $ 10^{k\plus{}1}$ different values of $ D_n$, where $ n \in N$. So by pigeon-hole principle , in $ \{D_1, D_2, ... ,D_{10^{k\plus{}1}\plus{}1} \}$ there exists distinct $ i,j$ so that $ (d_{i}, d_{i\plus{}1}, ... ,d_{i\plus{}k})\equal{}(d_j,d_{j\plus{}1}, ... , d_{j\plus{}k})$ , done.


Let a,b,c be positive real numbers.
Prove or disprove that:
$(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a})^{3}\ge 9(\frac{a^{4}}{b}+\frac{b^{4}}{c}+\frac{c^{4}}{a})$


A stack of 2000 cards is labeled with the integers from 1 to 2000, with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card in the stack is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack.This process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table.It is foudn that, readign from left to right, the labels on the cars are now in ascending order: 1,2,3,...,1999,2000. In the original stack of cards how many cards were above the cards labeled 1999?

Use the search function, and if you have an issue with an old problem, just revive the old thread.


Given: $ x > 0, y > 0, x > y$ and $ z\not \equal{} 0$. The inequality which is not always correct is:

$ \textbf{(A)}\ x \plus{} z > y \plus{} z \qquad\textbf{(B)}\ x \minus{} z > y \minus{} z \qquad\textbf{(C)}\ xz > yz$
$ \textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$

[hide="Click for solution"]
$ \textbf{(C)}$ is not always correct, because if $ z$ is negative then $ xz<yz$.
[/hide]


MIT
college
geometry
geometric transformation
reflection

I'm a high school sophomore, and I've always been good academically here, but I'm not sure how I rank nationally. What exactly does it take to stand out in the sea of applicants at a college like MIT? Also, how much do non-math-related things like debate matter?

Academically, you should be in the top tier GPA/SAT/AP/Whatever wise. While these won't make or break your application, there is no better way to show that you can handle the work and you're good enough for the school.

To stand out, it's more than a single national ranking that will set you apart. The admissions offices look hard for things that are unique, and the application process is designed to bring that out. Do something you are passionate about and do it well. Develop a love for it and show that you will become a leader in it.

The things that matter are the things you are really interested in and devote yourself to. If that's debate, great; show it in your application. On a side note, MIT isn't the only school out there :wink:. There are other [url=http://www.stanford.edu]awesome schools[/url] that might be a better fit, so don't set yourself on MIT too soon.

That's my best 30-second breakdown of the college admissions process :P.

A few more comments:

Rather than picking the $ n$ "best" applicants (whatever that means) in their pool, many selective colleges strive instead to pick a [b]spectrum[/b] of applicants that they think best reflects the qualities of the students they want to have.  Thus every year they will admit what admissions officers call "well-balanced" applicants - students who handle many activities and interests well - and also admit what admissions officers (perhaps tongue-in-cheek) call "well-lopsided" applicants - students who excel at one or two activities and interests.  There will be the music kids, the sports kids, the film kids.  Diversity is important to a lot of colleges, so your job (to the extent that you want to be admitted to a selective college only) is to find a niche.  What do you have to offer the collegiate environment?

So much for what colleges consider.  What is important for you to consider is [b]fit[/b] (as paladin8 has already mentioned).  Don't ask yourself whether you're good enough for a college, or whether the college wants you; ask yourself whether you want the college.  Different college environments suit different people.  Before you've settled on a college like MIT, ask yourself what you want in a college as far as campus size, class sizes, proximity to urban centers, housing, extracurriculars, Greek life....

ulness you have something like imo medalist on your application or something equally ridiculous, there's nothing yo ucan really do other than just do well on sats or whatever and get lucky


superior algebra
superior algebra unsolved

[quote="soccer022483"]Prove that the set of all rational numbers of the form $3^m6^n$, where $m$ and $n$ are integers, is a group under multiplication.[/quote]

1. $3^m6^n$ has closure, the product of any two elements of $3^m6^n$ is itself a member of $3^m6^n$. (The demonstration is trivial).

2. Inverses exist, e.g. $3^{-m}6^{-n}$ is itself a member of the relation.

3. Elements of $3^m6^n$ are associative.

4. An identity exists; namely, $3^06^0$.

Any relation for which these four properties hold is a group by definition.

[quote="mchis68"]You're in the wrong place here. We don't do anybody's homeworks.[/quote]

Then what does "Unsolved and Proposed Problems" mean? For some odd reason I thought it meant people post "Unsolved and Proposed Problems" for a response. What was I thinking...

By the way, you say "We" don't do anybody's homework. That implies your speaking for everyone when, in fact, I have gotten plenty of answers on this forum before from people.

I totally agree that MathLinks is not there to produce solutions for homeworks!
We (yes I want to speak/write for all!) are doing this because of the fun of it or because we want to help somebody (and get some help ourself indeed). But we don't want to make your school or university problems just because you want an easier life! You can ask questions or give problems, but if you just want a solution to copy and hand in, you are wrong here: it's some work for us to write it down and lesser work just for you! And Dr. No already gave a solution, you have only to work it out!

[quote="soccer022483"]Yes I know the properties of a group. The post was "Prove..." Can you please prove those properties work?[/quote]

What do you consider a proof that something is a group then?

Being antagonistic when you've posted something on the wrong forum isn't going to help you out much.  There are places that exist for classroom help in abstract algebra, and alas, this isn't one of those places.

[quote="soccer022483"]Yes I know the properties of a group. The post was "Prove..." Can you please prove those properties work?[/quote]I did prove that they work.

Proving closure is so simple I refuse to waste the time writing the Tex.

Inverses exist because the negative integers are integers too.

Of course it's associative under multiplication; its elements are integers.

The identity element exists because I explicitly gave you what it was.

If you know the properties of a group, why ask the flippin' question?


inequalities
function
complex numbers
algebra unsolved
algebra

Hi, everybody! Please explain to me the reason for this problem
Problem 5. Prove that
$ \sqrt \frac{7}{2} \leq |z\plus{}1|\plus{}|z^2\minus{}z\plus{}1| \leq \sqrt\frac{7}{6}$
for all complex numbers with $ |z|\equal{}1$
The solution is setting t=|z+1| and then we have the next step is
$ |z^2\minus{}z\plus{}1|\equal{}\sqrt{7\minus{}2t^2}$
and my trouble is I really do not understand the next step means but actually, I also got the different result $ |z^2\minus{}z\plus{}1|\equal{}\sqrt{3\minus{}t^2}. : o$.
Thanks in advance.

For one thing, the inequality states $ \sqrt {\frac {7}{2}}\leq \sqrt {\frac {7}{6}}$, which should already rouse suspicion. ;)

If $ |z| \equal{} 1$ it is true that
\[ \sqrt {3}\leq |z \plus{} 1| \plus{} \left\vert z^2 \minus{} z \plus{} 1\right\vert\ \leq \frac {13}{4}
\]
though showing it is somewhat tedious, so I'm assuming you've just got a typo in the problem statement somewhere.  And I've no idea in what sense it might be true that $ \left \vert z^2 \minus{} z \plus{} 1\right \vert \equal{} \sqrt {7 \minus{} 2|z \plus{} 1|^2}$ either; nor $ \left \vert z^2 \minus{} z \plus{} 1\right \vert \equal{} \sqrt {3 \minus{} |z \plus{} 1|^2}$, for that matter.  Substituting $ z \equal{} \minus{} 1$, for example, would yield either $ 1 \equal{} \sqrt {7}$ or $ 1 \equal{} \sqrt {3}$.

Thanks for response!
This problem I took from Complex number book of Titu Andresscu.

I will post the full problem and full solution for you to examine:
Problem:
Prove that:
$ \sqrt\frac{7}{2} \leq |1 \plus{} z| \plus{} |1 \minus{} z \plus{} z^2|\leq \sqrt\frac{7}{2}$
for all complex numbers with $ |z|\equal{}1$
Solution:
Let $ t \equal{} |1 \plus{} z|$ belongs to the interval     $ [0, 2]$. We have
$ t^2  \equal{} 2 \plus{} 2Re(z)$, so  $ Re(z) \equal{}\frac{t^2\minus{}2}{2}$
Then $ |1 \minus{} z \plus{} z^2| \equal{}\sqrt{|7 \minus{} 2t^2|}$.
It suffices to find the extreme values of the function
$ f : [0, 2] \rightarrow R, f (t) \equal{} t \plus{}\sqrt{|7 \minus{} 2t^2|}.$
We obtain
$ f\left(\frac{7}{2}\right) \leq t \plus{}\sqrt{|7 \minus{}2t^2|} \leq f\left(\frac{7}{6}\right)$
I can make sure that there is no mistake in typing. I need your help because I have just studied complex number so I do not have many skills in solving this interesting set. :)

First, the problem should be
$ \sqrt \frac {7}{2} \leq \left|z \plus{} 1\right| \plus{} \left|z^2 \minus{} z \plus{} 1\right| \leq 3\sqrt\frac {7}{6}$
(at least this is what my version of Andreescu's Complex Numbers says, and it is downloaded off the Springerlinks site from our university, so it should be the final version).

But if I am not mistaken, $ \left|z^2 \minus{} z \plus{} 1\right| \equal{} \sqrt {7 \minus{} 2t^2}$ is wrong indeed, and you are almost right: $ \left|z^2 \minus{} z \plus{} 1\right| \equal{} 3 \minus{} t^2$ (the right hand side needs not be square-rooted).

I don't have time to solve this problem (correctly) now, but just read on. Not all problems in this book are that stupid (actually, few of them are).

  darij

The idea of the proof is correct, except that $ \sqrt {\left \vert 7 \minus{} 2t^2\right\vert}$ should have been $ \left \vert 3 \minus{} t^2 \right \vert$, as darij grinberg noted.  In that case we are looking for the minimum and maximum of $ t \plus{} \left \vert 3 \minus{} t^2\right \vert$ on $ t\in [0,2]$ which yields the correct bounds $ \sqrt {3}$ and $ \frac {13}{4}$.

Thanks for your responses, I appreciate those. Your results are the same as my different result that I mentioned. Actually, I felt very confused because this book is written by a famous person so I do not know whether I was right or wrong. Anyway, the your attention to the problem is highly appreciated.! :)


#15

Given that $ log_32 \equal{} 0.631$, find the smallest integer $ a$ such that $ 3^a > 2^{102}$. My answer is $ 65$. Am I correct?

Do you mean the inequality to be $ 3^{a}> 2^{102}$?

[quote="pascal12"]Do you mean the inequality to be $ 3^{a} > 2^{102}$?[/quote]

yes, i already fixed it.

[quote="mr. math"]Yes, that is correct:

3 ^ (0.631x) = 2.
Raise both sides to the 102nd power and you have 3^(0.631 x 102) = 2^102.

3^64.362 = 2^102

The smallest possible number that will satisfy your inequality is therefore 65.[/quote]

i used logarithm in my solution. your solution however is simpler and more elegant.

Yes, the way I would do it with logs is to take the $ \log_3$ of each side.

$ 3^a>2^{102}$

$ \log_3{3^a} > \log_3{2^{102}}$

$ a> 102 \cdot \log_3{2}$

A quick multiplication reveals that $ 102\cdot \log_3{2} \approx 64.3$ so the lowest integer $ a$ which is greater than this is $ 65$.


geometry
incenter
circumcircle
advanced fields
advanced fields unsolved

Given simplex $A_{1}A_{2}..A_{n+1}\subset R^{n}$ denote $F_{i}$ is hyperplane that is through $n$ point $\{A_{1},A_{2},..,A_{n+1}\}/{A_{i}}$
$1)$ Prove that exist a point $I$ that $d(I,F_{i})=d(I,F_{j})$ for all $i\not= j$ ($d$ denote distance):Incenter of simplex.
$2)$ Prove $I$ satify $1)$ then $\sum_{i=1}^{n+1}S_{i}\vec{IA_{i}}=\vec{0}$ with $S_{i}=V[X_{i}], X_{i}$ is the simplex that contruct on $\{A_{1},A_{2},..,A_{n+1}\}/{A_{i}}$ and it $\subset F_{i}$ $V$ denote volume.
$3)$ Prove that exist a point $O$ that $d(O,A_{i})=d(O,A_{j})$ for all $i\not= j$:Circumcenter of simplex.

Hi guys, I think they are quite basic, can anybody help me?  :D


The third and fourth terms of a geometric sequence are 12 and 16, respectively. What is the first term of the sequence?

$ 16/12 \equal{} 12/x \rightarrow x \equal{} 9$

$ 12/9 \equal{} 9 / y \rightarrow y \equal{} \boxed{6.75}$

The common ratio $ r$ between the numbers in the sequence is $ \frac{16}{12}\equal{}\frac{4}{3}$. So the first term is $ 12\times{(\frac{3}{4})^{2}}\equal{}\boxed{\frac{27}{4}}$.


AMC
AIME
USA(J)MO
USAMO
blogs
MATHCOUNTS
AMC 10

I apparnelty got a <official csores not yet opened>. 100.5 on amc 12.

Well it'd be sad if you ddin't make USAMO, no offense or anything. You are like OH's best guy when it comes to math xD

[quote="tjhance"][quote="#H34N1"]You are like OH's best guy when it comes to math xD[/quote]

There are many counterexamples to that.[/quote]

Such as?

[quote="tjhance"][quote="#H34N1"][quote="tjhance"]A lot of people.[/quote]

lol names?[/quote]

tong[/quote]

you beast tong

Er... right

well in any case its good to see more qualifiers than last year, (9 as opposed to 5) thought not many of the people seem to be the same.

[quote="tjhance"][quote="tenniskidperson3"]Yeah! Only 9th grader to make USAMO!  I got a 9 on AIME <unofficially> and a 138 on AMC 10B.[/quote]

Yay good job. Now go make Red MOP  :D[/quote]

I need to make Red MOP next year.

[quote="#H34N1"]OK Teacher opened and showed me all results.

132 on the AMC 10 (I don't get it!?)
100.5 on the AMC 12.
7 on the AIME.

Index: 202.[/quote]

ok would you stop lying??  Its one thing to actually get that and be mad but its another to lie about it.

Also, I am sure everyone realizes you are lying so why dont we just stop now? (check your blog, everyone even posts in your blog that you lie, but of course, you delete them so no one else can see)

Anyways, I dont even think the AMC has mailed out the AIME scores yet :rotfl:  :rotfl:

Also, CHECK THE AMC WEBSITE FOR OHIO STATS, NO 8TH GRADER GOT A 132 ON THE AMC 10A OR 10B!  WANT ME TO POST THE LINK SO YOU WILL BE EMBARRASED EVEN MORE?

And I cant wait until the AIME stats come out.  THERE WILL BE NO SCORE OF 7 for 8th graders.

But of course, you purposely put your index as 202 to make it sound like you almost made USAMO to make your self look smart.

so yea stop bragging and definently stop lying.

Theres no way youll make MOP (or even USAMO :rotfl: ) if you just lie about everything and kid yourself into thinking your all good and stuff (like IM GOING TO GET 1st at STATE AT MATHCOUNTS-LOL how did that work out not getting into the CD ROUND let alone first)

And your just embarring yourself.
Check this link out where you said you got a 9 on AIME as a 7th grader.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138972
Like people actually believe you.

Dude official results haven't even been mailed out for aime ii

i said that was what I think I got based on checking answers on the unl site.

ok i admit i lied about last year. i'll stop doing it.

[quote="tenniskidperson3"][quote="tjhance"][quote="tenniskidperson3"]Yeah! Only 9th grader to make USAMO!  I got a 9 on AIME <unofficially> and a 138 on AMC 10B.[/quote]

Yay good job. Now go make Red MOP  :D[/quote]

I was planning on it. :lol: I figure that if I make it this year, I will be able to do WOOT and that will help me make it next year, too.[/quote]

Lol, after my Red MOP I took WOOT and was so excited when contest time came around.  That year I got 4 or 5 on the AIME and didn't get to go to USAMO.   :P

But this year I got 130.5 AMC12/11 AIME, so I'm hoping for Black or at least Blue MOP.  :D

(junior this year, btw.)

Dude black MOP would be awesome

But I personally will be excited just to make Blue.

because MOP was awesome and I want to go again

hopefully I will not fail usamo


let $ x\equal{}\frac ab,y\equal{}\frac bc,z\equal{}\frac ca$ hence $ xyz\equal{}1$ now we have to prove that:

a)$ x^2\plus{}y^2\plus{}z^2\geq\frac 1x\plus{}\frac 1y\plus{}\frac 1z$

b)$ x^4\plus{}y^4\plus{}z^4\geq\frac 1x\plus{}\frac 1y\plus{}\frac 1z$

now note that because of $ xyz\equal{}1$ a)  can be written as:

a)$ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx\iff (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2\geq 0$

now for b) note that we have to prove that:

$ x^4\plus{}y^4\plus{}z^4\geq xy\plus{}yz\plus{}zx$

but note that by AM-GM we have:

$ x^4\plus{}y^4\plus{}1\plus{}1\geq 4xy$

hence:

$ x^4\plus{}y^4\plus{}z^4\geq 2(xy\plus{}yz\plus{}zx)\minus{}3$

so it remains to show that:

$ 2(xy\plus{}yz\plus{}zx)\minus{}3\geq xy\plus{}yz\plus{}zx\iff xy\plus{}yz\plus{}zx\geq 3$

which is clearly true because of AM-GM:

$ xy\plus{}yz\plus{}zx\geq 3\sqrt[3]{x^2y^2z^2}\equal{}3$

I want another proof for the  second inequlity please

ok,here you go,first of all note that for positive numbers $ x,y,z$ we have:

$ \frac{x^2}{y^2}\plus{}\frac{y^2}{z^2}\plus{}\frac{z^2}{x^2}\geq\frac xy\plus{}\frac yz\plus{}\frac zx$

now let $ x\equal{}a^2,y\equal{}b^2,z\equal{}c^2$ so we have:

$ \frac{a^4}{b^4}\plus{}\frac{b^4}{c^4}\plus{}\frac{c^4}{a^4}\geq\frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}$

so for b) it's sufficient to show that:

$ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\geq\frac ba\plus{}\frac cb\plus{}\frac ac$

which is exactly part a) :)

Wandering trick . Thank you a lot Babak Ghalebi  :)

[quote="matt276eagles"][quote="boxedexe"]Also, notice that a) is an immediate consequence of the Cauchy Schwarz inequality.[/quote]

Could you explain?  I'm probably just being stupid, but I don't see it.[/quote]
[hide]By Cauchy,
$ \left( \frac{b^2}{c^2}\plus{}\frac{c^2}{a^2} \plus{} \frac{a^2}{b^2} \right) \left( \frac{c^2}{a^2} \plus{} \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2} \right) \ge \left(\frac{b}{a} \plus{} \frac{c}{b} \plus{} \frac{a}{c} \right)^2$.[/hide]

Somebody can prove that $ \frac{x^2}{y^2}\plus{}\frac{y^2}{z^2}\plus{}\frac{z^2}{x^2}\geq\frac xy\plus{}\frac yz\plus{}\frac zx$

it is more difficult than the first one

*210*, here you go.

By Chebyshev's inequality,
\[ \frac {x^2}{y^2} + \frac {y^2}{z^2} + \frac {z^2}{x^2}\ge\frac {1}{3}\left(\left|\frac {x}{y}\right| + \left|\frac {y}{z}\right| + \left|\frac {z}{x}\right|\right)^2.
\]
By AM-GM,
\[ \frac {\left|\frac {x}{y}\right| + \left|\frac {y}{z}\right| + \left|\frac {z}{x}\right|}{3}\ge1,
\]
so
\[ \[\frac {x^2}{y^2} + \frac {y^2}{z^2} + \frac {z^2}{x^2}\ge\left|\frac {x}{y}\right| + \left|\frac {y}{z}\right| + \left|\frac {z}{x}\right|\ge\frac {x}{y} + \frac {y}{z} + \frac {z}{x}.
\]

[quote="*210*"]Somebody can prove that $ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq\frac xy \plus{} \frac yz \plus{} \frac zx$

it is more difficult than the first one[/quote]
[hide="standard AM-GM bash"]$ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq\frac xy \plus{} \frac yz \plus{} \frac zx \iff \sum_{cyc}x^4z^2 \ge \sum_{cyc} x^3yz^2$.
By AM-GM, $ f(x,y,z) \equal{} \frac {2x^4z^2 \plus{} y^4x^2 \plus{} z^4y^2}{6} \ge x^3yz^2$, and similarly
$ f(y,z,x) \ge y^3zx^2$ and
$ f(z,x,y) \ge z^3xy^2$.
So $ \sum_{cyc}x^4z^2 \equal{}  f(x,y,z) \plus{} f(y,z,x) \plus{} f(z,x,y) \ge \sum_{cyc} x^3yz^2$. [/hide]

[quote="*210*"]Somebody can prove that $ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq\frac xy \plus{} \frac yz \plus{} \frac zx$

it is more difficult than the first one[/quote]

well I have two solutions for this one:

$ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq \frac xy \plus{} \frac yz \plus{} \frac zx$

$ \iff\frac {x^2 \minus{} xy}{y^2} \plus{} \frac {y^2 \minus{} yz}{z^2} \plus{} \frac {z^2 \minus{} zx}{x^2}\geq 0$

now note that $ x^2 \minus{} xy \equal{} (x \minus{} y)^2 \minus{} y^2 \plus{} xy$ so we can rewrite the above inequality as:

$ \underbrace{\frac {(x \minus{} y)^2}{y^2} \plus{} \frac {(y \minus{} z)^2}{z^2} \plus{} \frac {(z \minus{} x)^2}{x^2}}_{\geq 0} \plus{} \frac xy \plus{} \frac yz \plus{} \frac zx\geq 3$

so it's sufficient to show that:

$ \frac xy \plus{} \frac yz \plus{} \frac zx\geq 3$

which is clear by AM-GM...
_______________________________________________________
the other solution is a directly consequence of cauchy:

$ (1 \plus{} 1 \plus{} 1)\left(\frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\right)\geq\left(\frac xy \plus{} \frac yz \plus{} \frac zx\right)^2$

so it's sufficient to show that:

$ \frac xy \plus{} \frac yz \plus{} \frac zx\geq 3$

which is again,clear by AM-GM


Hello,

$ i)$ Evaluate :  \[ \lfloor \sqrt{n^2\plus{}n\plus{}1} \rfloor , \;\, \forall n \in \mathbb{N}\]

$ ii)$ Prove that if $ p, q \in \mathbb{N}^{*}$ are prime between them ( $ p\wedge q \equal{} 1$ ), then :
\[ \Large\sum_{k\equal{}1}^{q\minus{}1}\lfloor k\frac{p}{q}\rfloor \equal{} \frac{(p\minus{}1)(q\minus{}1)}{2}\]

[hide="(i)"]
Squaring, we can form the inequality $ n^2<n^2\plus{}n\plus{}1<(n\plus{}1)^2$. Therefore, $ n<\sqrt{n^2\plus{}n\plus{}1}<n\plus{}1$. Thus, $ \lfloor \sqrt{n^2\plus{}n\plus{}1}\rfloor\equal{}n$ for every $ n \in \mathbb N$.[/hide]

The second is very classic and has a very beautiful proof, which I'll suggest with [hide="this huge hint"]: how many lattice points are in the interior of the triangle with vertices $ (0, 0), (q, 0), (q, p)$? [/hide]

I was writing this proof of (ii) while t0rajir0u was posting.

\begin{eqnarray*}\sum_{k = 1}^{q - 1}\left\lfloor{kp\over q}\right\rfloor & = & \sum_{k = 1}^{q - 1}\left({kp\over q} - \left\{{kp\over q}\right\}\right) \\
& = & {q(q - 1)\over 2}\cdot{p\over q} - \sum_{k = 1}^{q - 1}\left\{{kp\over q}\right\}\end{eqnarray*}

Since $ p,q$ are coprime, numbers $ kp$ for $ k = \overline{1,q - 1}$ give the full set of non-zero residues modulo $ q$, hence the sum in the last row is $ {1\over q} + {2\over q} + \dots + {q - 1\over q} = {1\over q}\cdot{q(q - 1)\over 2} = {q - 1\over 2}$, so the initial sum is

$ S = {p(q - 1)\over 2} - {q - 1\over 2} = {(p - 1)(q - 1)\over 2}$


If McDonalds is doing a give away in which there are ten pictures on a card. One picture is there twice, and one is an X. What is the probability of winning the game if you win when both of the picture that is there twice is before the X from left to right?

[quote="bpms"][quote="SplashD"]I think it was on a precious math contest
[b]
[size=59]Just Kidding, you are right.[/size][/b][/quote][/quote]

why didi you close that :?:

did someone edit my post? :maybe:

where's the just kidding, you are right?

The answer was what you posted is what I meant...
Anirudh put what I said in you post and unwhitened it for some strange reason...
EDIT: Is anyone going for a solution?

[quote="bpms"][quote="SplashD"]I think it was on a precious math contest

[hide]anyway, i think the answer is $\frac13$, is that right?[/hide][/quote]
Notice that I didn't say express as a common fraction... The answer is 0 because it is impossible to win anything in a McDonalds give away...
[size=59][color=white]Just Kidding, you are right.[/color][/size][/quote]

that's not true, i remember winning a box of chicken nuggets from some giveaway at mcdonalds

I got [hide]1/3, because the X has to be after the two pictures and there is a 1/3 chance of that.[/hide]


AMC
USA(J)MO
USAMO
calculus
integration
abstract algebra
function

1) (warm-up) Prove that the equation $x^{n}+y^{n}= z^{n}$ has no solution in integers x,y,z for integral $n=3,4,5$.

2) (non-obvious)Prove that for projective algebraic varieties, Hodge cycles are rational linear combinations of algebraic cycles.

3) (hard) Let P denote the set of problems known to be solvable by exponential compouter time, but whose solutions can be checked in less than exponential time. Prove that such problems are not solvable in less than exponential time.

4) (warm-up)Let us call an object "simply connected" if we can stretch a string around it and shrink it down to a point by moving it slowly, without tearing it and without allowing it to leave the surface. Determine, with proof, whether a 4-dimensional hypersphere is "simply connected."

5) (non-obvious)Prove that all interesting solutions to ζ(s) = 0, where    ζ(s) $= 1+1/2^{s}+1/3^{s}+1/4^{s}+...$ lie on a vertical straight line.

6) (hard) When the whole number solutions of algebraic expressions of an abelian variety, prove that the size of the group of rational points is related to the behavior of an associated zeta function ζ(s) near the point s=1. In particular, show that if ζ(1) is equal to 0, then there are an infinite number of rational points (solutions), and conversely, if ζ(1) is not equal to 0, then there is only a finite number of such points.

[quote="jclarkemathL314159"]What is up with #6???

Fermat's Last Theorem?[/quote]

yeah it is.

omg aren't those the millenium problems? and wasn't poincare and fermat's already solved...

Sheesh just because they're hard you have to accuse them of being millenium problems?

And yes that is Fermat's it's perfectly solvable.

[hide]

All you have to do is prove (a) a solution to Fermat's has elliptic curve that is semistable but not modular. (b) all semistable elliptic curves with rational coefficient are modular. [/hide]

EDIT: this shall be dedicated to Gabriel Dospinescu.

Isn't one of those statements the Taniyama-Shimura conjecture? Which in itself was the only thing Wiles proved, and took 7 years of work?

very intresting forum. gives you something to think about :)


Jason has the same number of red and of blue marbles. He puts them in two jars so that the ratios of the number of red marbles to blue marbles in jar I is 2:5 and in jar II is 9:5. If there are 84 marbles in jar I, how many are there in jar II ?

[hide] In jar I, we have the ratio $ \frac{2x}{5x}$, such that $ 2x\plus{}5x\equal{}84 \implies x\equal{}12$. Therefore, there are $ 24$ red marbles and $ 60$ blue marbles in jar I. The difference between the blue and red marbles in jar I must be the same as the difference between the red and blue marbles in jar II. Thus, we have the ratio $ \frac{9y}{5y}$, such that $ 9y\minus{}5y\equal{}36 \implies y\equal{}9$. Thus, the nukmber of marbles in jar II is $ 9y\plus{}5y\equal{}14(9)\equal{}\boxed{126}$.[/hide]


I don't think the 4 given equations are consistent -- they appear to over-determine the system.  So, we don't need the fourth equation (or at least, we will get a different answer depending which 3 we use), but it also raised the possibility that there were meant to be 4 variables there.

you can make a polynomial $x^3-x^2-\frac{x}{2}-\frac{1}{6}$ that has roots $a$, $b$, and $c$ that satisfy the first equation, but like joel said this isn't consistent with the last equation, because it's clear that we can't make it so that $a^4+b^4+c^4=4$.


Prove that complex number $z$ satisfies $|z|-Re\,z\le\frac{1}{2}$ if and only if there exist complex numbers $u,v$ such that $z=uv$ and $|u-\overline{v}|\le1.$
(V. Radchenko)


geometry
3D geometry
combinatorics unsolved
combinatorics

The edges of a cube are labeled from $1$ to $12$. Show that there must exist at least eight triples $(i, j, k)$ with $1 \leq i < j < k \leq 12$ so that the edges $i, j, k$ are consecutive edges of a path. Also show that there exists labeling in which we cannot find nine such triples.

For the first part:

Each vertex has three edges eminating from it.  Mark the edge of those three which has the second largest label.  This creates a total of 8 markings, though some edges may have been marked twice.  Still, at least 4 edges have been marked.  If suffices to show that any marked edge appears as the middle edge in at least two distinct paths.

Suppose the edge from vertex $x$ to vertex $y$ is the middle label of the three edges eminating from $y$.  If $w$ is a vertex other than $y$ adjacent to $x$, we construct a path as follows.

-If $wx$ has a smaller label than $xy$, go from $w$ to $x$, from $x$ to $y$, then along whichever edge containing $y$ has a larger label than $xy$.
-If $wx$ has a larger label than $xy$, go along whichever edge containing $y$ has a smaller label than $xy$, then along $xy$, then along $wx$.

Since each marked edge has two choices for $w$, each marked edge is part of two triples.

For part 2:
Label the four edges on top of the cube 1, 3, 2, 4 (starting at any edge and proceeding clockwise)
Label the four edges on bottom 5,7,6,8 (ditto)
Label the four edges connecting the two levels 9,10,11,12 (any order you want).

editted to fix my labelling for part 2


[quote="hien"]Không biết các bạn có biết thông tin kỳ thi quốc gia năm 2007 sẽ diễn ra sớm không? Chúng ta thử dự đoán đề thi năm nay sẽ ra như thế nào nhé[/quote]
 ban muon co' thong tin thi` vao` day xem cung dc
   http://diendantoanhoc.net/forum/index.php?showtopic=19933

Theo thông tin mới nhất thì kì thi quốc gia năm nay sẽ diễn ra sớm hơn . Cụ thể là vào 22 tết âm lịch .Chúc các bạn sẽ làm bài tốt để ăn tết cho nó ngon  :)  :lol:


topology
advanced fields
advanced fields unsolved

$ 1/$ Prove that if $ G$ is connected topological group then the fundamental group of $ G$ is abelian.

$ 2/$ $ \pi_1 SU(n)$ is trivial forall $ n>0$ and $ \pi_1 SO(n)\cong \pi_1 SO(3)$ for all $ n\ge 3$.
 :)

If you take two loops based at the identity $ f(t)$ and $ g(t)$ ($ f(0)\equal{}g(0)\equal{}f(1)\equal{}g(1)\equal{}e$), we have two operations to combine the loops: pointwise product and concatenation. How are these related (up to homotopy)?

(Hint- we can write the concatenation as a product with at least one of the terms equal to $ e$ everywhere)


inequalities
logarithms
calculus
calculus computations

prove that $\sum_{k=1}^{n}\frac{e^{-k}}k \leq \frac 1{e-1}$

[hide] Try using $\ln{x}+1 \le x$ with $x = \frac{1}{e-1}$ and expanding the Taylor series on the LHS.[/hide]

[quote="JBL"]Extremal, $a(n) \sym b(n)$ and $b(n) \leq c(n)$ does not imply that $a(n) \leq c(n)$.

For an easy solution, just use $\frac{1}{k}\leq 1$.  (This is a very bad approximation.)[/quote]
Thanks . I will be right if $n\to+oo$

For an easy solution, just use$\frac{1}{k}\leq 1$. (This is a very bad approximation.)
please can you do it ?? :)

You can use the "quote" button (or even quotation marks) to make it clear that the first line of that was quoting me.

$\sum_{k =1}^{n}\frac{e^{-k}}{k}\leq \sum_{k = 1}^{n}e^{-k}\leq \sum_{k = 1}^\infty e^{-k}= \frac{e^{-1}}{1-e^{-1}}= \frac{1}{e-1}$.

The exact value of the sum is $-\ln(1-\frac1e)=\ln(1+\frac1{e-1})$; the result follows from the general fact $\ln(1+x)\le x$.


If $ a,b$ and $ k \in N$ and it's given that $ a$ is a prime and :
$ a^2 \plus{} 2ak \plus{} b^2 \equal{} n^2$
is it compulsorily b=k?

[hide]$ a^2 \plus{} 2ak \plus{} b^2 \equal{} n^2$ <=> $ (a \plus{} k)^2 \plus{} b^2 \equal{} n^2 \plus{} k^2$[/hide]
i'm almost sure that this isn't compulsorily but just almost  , not absolutely :huh:

It is not true. I will find all solution of $ (a,b,k)$
Let $ n \equal{} a \plus{} b \plus{} m$ where $ m\in N$
Then $ a^2 \plus{} b^2 \plus{} 2ak \equal{} (a \plus{} b \plus{} m)^2$
$ \Longleftrightarrow 2ak \equal{} 2ab \plus{} 2m(a \plus{} b) \plus{} m^2$
So $ 2|m$
Let $ m \equal{} 2p$ so
$ \Longleftrightarrow 2ak \equal{} 2ab \plus{} 4p(a \plus{} b) \plus{} 4p^2$
$ \Longleftrightarrow k \equal{} \frac {ab \plus{} 2p(a \plus{} b) \plus{} 2p^2}{a}$
So $ a|2p^2 \plus{} 2pb$
[u]Case 1[/u] $ a \equal{} 2$
So $ k \equal{} b \plus{} p(b \plus{} 2) \plus{} p^2$
So if chose $ p > 0$ then it is not true.
[u]Case 2[/u] $ a > 2$ $ a|p(p \plus{} b)$ and a is a prime so $ a|p$ or $ a|b \plus{} p$
[color=red]Case 2.1[/color] $ a|p$
So $ k \equal{} ha$
replace to relative we have : $ k \equal{} b \plus{} 2h(a \plus{} b) \plus{} 2h^2a$
[color=red]Case 2.2[/color] $ a|p \plus{} b$
$ k \equal{} b \plus{} \frac {2p(a \plus{} b) \plus{} 2p^2}{a}$ where $ b,p$ satisfy $ a|b \plus{} p$
If we chose $ p > 0$ then your statment not true.

yes , i see .
i've thought that is wrong anyway , but it's the best to ask if you aren't absolutely sure about sth
now how can i delete the topic , it's useless all the same :maybe:


[quote="Twins"]We have $ xyz=x+y+z+2$, $ x,y,z \in R^{+}$, prove that:
$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{3}{2}\leq \frac{1+x}{1+y}+\frac{1+y}{1+z}+\frac{1+z}{1+x}$[/quote]
Let $ x=\frac{b+c}{a}$ and $ y=\frac{a+c}{b},$ where $ a,$ $ b$ and $ c$ are positive numbers.
Then $ z=\frac{a+b}{c}$ and $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{3}{2}\leq \frac{1+x}{1+y}+\frac{1+y}{1+z}+\frac{1+z}{1+x}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{a}{c}-\frac{a}{b+c}-\frac{1}{2}\right)\geq0.$ Let $ c=\min\{a,b,c\}.$
Hence, $ \sum_{cyc}\left(\frac{a}{c}-\frac{a}{b+c}-\frac{1}{2}\right)\geq0\Leftrightarrow M(a-b)^{2}+N(c-a)(c-b)\geq0,$ where
$ M=\frac{1}{ab}-\frac{1}{(a+c)(b+c)}\geq0$ and $ N=\frac{1}{bc}-\frac{2c+a+b}{2(a+b)(a+c)(b+c)}=$
$ =\frac{1}{bc}-\frac{1}{2(a+b)(b+c)}-\frac{1}{2(a+b)(a+c)}\geq0.$ :)

$ x=\frac{1}{a}-1$,$ y=\frac{1}{b}-1$,$ z=\frac{1}{c}-1$
$ \Rightarrow a+b+c=1$
We only need to prove $ \sum{\frac{a}{1-a}}+\frac{3}{2}\leq \frac{b}{a}+\frac{c}{b}+\frac{a}{c}$
That is $ \frac{3}{2}\leq \sum{\frac{ab}{c(b+c)}}$
It's esay to prove
$ \sum{\frac{ab}{c(b+c)}}=\sum{(ab)^{2}}{abc(b+c)}\geq \frac{(ab+bc+ca)^{2}}{2abc(a+b+c)}\geq \frac{3}{2}$


Without guess and check, is it possible to do this problem?

Using only digits 0-9 makes this a true equation

[ ] + [ ] [ ] [ ] - [ ] [ ] - [ ] [ ] + [ ] [ ] = 100

Can the same digit be used multiple times?

I don't think so, but there is a fair amount of logic involved in reaching the solution.

Most quick ways to solve mathcounts problems rely heavily upon logic/formulas.


a) Let $ x_1,x_2,..,x_n$ are positive real number such that $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$
 Prove that :
         $ \frac {1}{x_1^2} \plus{} \frac {1}{x_2^2} \plus{} ... \plus{} \frac {1}{x_n^2} \plus{} \frac {a}{x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_{n \minus{} 1}x_n \plus{} x_nx_1} \geq n(n^2 \plus{} a)$ ,$ \forall a \in \mathbb{R}^{ \plus{} }$
 b)  Let $ x_1,x_2,..,x_n$ are  real number such that $ x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \equal{} 1$
 Prove that :  $ \|x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_{n \minus{} 1}x_n\|\leq cos( \frac {\pi}{n \plus{} 1})$
 ......... :play_ball: ....... :play_ball:

[color=red]For $ a$) that it is correct with $ n \equal{} 1,2,3,4$ , With $ n \geq 5$ It isn't sure I hope you try  it and for answer about it thanks ![/color]
Add $ a_1$) Let $ x_1,x_2,..,x_n$ are positive real number such that $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$
Prove that :
$ \frac {1}{x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2} \plus{} \sum_{1 \leq i < j \leq n}\frac {1}{x_ix_j} \geq \frac {n(n^3 \minus{} n^2 \plus{} 2)}{2}$


The polynomial $ x^3 \plus{} ax^2 \plus{} bx \plus{} 192 \equal{} 0$ has three real roots in geometric sequence. Determine the value of b/a.

let the three roots be $ \frac{x}{k}, x, xk$

$ \minus{}a\equal{}\frac{x}{k}\plus{}x\plus{}xk$
$ b\equal{}\frac{x^2}{k}\plus{}x^2\plus{}x^2k$
$ \minus{}192\equal{}x^3$

$ \frac{b}{a}\equal{}\minus{}\frac{x^2(\frac{1}{k}\plus{}1\plus{}k)}{x(\frac{1}{k}\plus{}1\plus{}k)}\equal{}\minus{}x$

we have $ x\equal{}\minus{}\sqrt[3]{192}$ from the third equation

answer : $ \sqrt[3]{192}$

vieta's theorem (or w/e you call it)

in equation $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$, let the three roots be $ r,s,t$

then $ r\plus{}s\plus{}t\equal{}\minus{}\frac{b}{a}$
$ rs\plus{}st\plus{}tr\equal{}\frac{c}{a}$
$ rst\equal{}\minus{}\frac{d}{a}$

[url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas]Vieta's Formulas[/url] (well, actually the guy's name is Viète) and [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton's Sums[/url] are often useful in these sorts of problems (usually manipulation of roots), as well as the fact that $ P(r) \equal{} 0$ iff $ r$ is a root of $ P(x)$.


calculus
derivative
function
logarithms
limit
real analysis
real analysis unsolved

I dont know is this the right place to post this question or not,so sorry anyway:

I was solving a problem and I came with this question:

the problem first gave a function $f(y)=x^{y}_{1}+x_{2}^{y}+...x_{s}^{y}$ (in this function $x_{1},...,x_{s}$ are constant numbers) and asked some questions about it...

in my solution I had to show that this function is increasing...

so I tried to show that $f'(y)\geq 0$,somehow I managed to show this (because $x_{1}...x_{s}=1$ and etc.) but the solution
isnt complete yet because I have to show one more thing:

[b]does $f'(y)$ exists?I mean is $f(y)$ derivative able?[/b]

but Im stock in this question could ant body help me?  :maybe:

If $x_{1},\dots,x_{s}>0$, then the derivative exists and is equal to $x_{1}^{y}\ln x_{1}+\dots+x_{s}^{y}\ln x_{s}$. To prove this, write $x_{i}^{y}$ as $e^{y\ln x_{i}}$.

yeah I computed that to,but is this enough?
could you please explain more?
I mean is this statement correct:[b]every function $f$ that we can compute its`s derivative,then it`s derivative exists?[/b]  :maybe:
because as far as I know we call a function $f$ derivative able on interval $(a,b)$ iff the limit:
$\lim_{x\to x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ exists for every $x_{0}\in (a,b)$
so is what you said enough to show this?or there is another theorem that I dont know?  :huh:

In this case the function is derivable as it is a sum of derivable functions.
For your general question I think that the answer may vary according to what you consider derivating a function. One may consider that derivating a function in a point is calculating that limit which you stated and if it exists the function is derivable in the $x_{0}$. You apply the known derivation formulaes for [u]known[/u] simple functions, as in the general functions are combinations of different types of elementary functions. Otherwise, I suppose computing that limit is in order.

so this statement is true:
[b]if $f(y)=f_{1}(y)+f_{2}(y)+...+f_{n}(y)$ where $f_{i}(y)$s are derivable functions.then $f(y)$ is derivable...[/b]  :maybe:
 :?:

wow thanx,I didnt know this theorem...
so my problm is completely solved,thanks to you guys :lol:
thanx alot...

In fact, all of the standard rules for combining derivatives- addition, constant mutiplication, the product rule, the quotient rule, the chain rule- are also theorems about such functions being differentiable.


[quote="galois01"]Solve the system of the equations:
$ 5752x+4248y=25752$
$ 4248x+5752y=24248$[/quote]
this looks like it can be real confusing without a calculator but its not. i did it entirely(almost) in my head. just use what we learned in like 2nd grade: expanded form

[hide]add both sides both together
and get
$ 10000x+10000y=50000$
so
$ x+y=5$
and
$ x=5-y$
substitute it into any equation, i chose the bottom, its not as dificult as it looks.
$ 4248(5)-4248y+5752y=20000+4284$
writing it like this helps a lot. playing the arthmatic game helps. $ 52-48=4$ and $ 5700-4200=1500$ so
$ 4248(5)+1504y=20000+4284$
then subtract $ 4248(5)$ from both sides and get $ 1504y=20000-4248(4)$. first do $ 20000-4000(4)$, then you do $ 4000-200(4)$, then $ 3200$.....you get the point. you end up with $ 3008=1504y$ so $ y=2$. substitute it into the equation $ x=5-y$. it saves you a WHOLE bunch of time doing this than substituting it into the 1st or 2nd like your teachers tell you. then just simply do $ 5-2$(hope we all know what that is) and get the ordered pair $ (3,2)$[/hide]

[quote="vishalarul"][quote="archimedes1"]abacadaea, why not subtract the two equations, also?  See what happens.[/quote]

Yeah, that makes it a lot easier, and you don't have to deal with big numbers (well, at the first you do, but you don't later.)

[hide]Add the two equations to get:
$ 10000x+10000y=50000$
$ x+y=5$

Subtract the two equations to get:
$ 1504x-1504y=1504$
$ x-y=1$

The rest is simple... Add the two equations formed now:
$ 2x=6$
$ x=3$.

Plug into $ x+y=5$ and get $ y=2$. So $ (x,y)=(3,2)$[/hide][/quote]
That's the way I solved it too  :)

But to solve the problem, the easiest way is to solve the equations when you add them together AND subtract one from the other.  Its a lot easier than doing long substitutions. :idea:

[hide]you probably won't be very mean, so you can add them up.

5752x + 4248y = 25752
4248 + 5752y = 24248

10000x+10000y=50000
x+y=5

Then you substitute and get x=3 y=2.

how.do.you.get.that.kind of bold text that looks better?
[/hide]

You put dollar signs around the equation. So the two linear equations you do to get:
[code]
$5752x+4248y=25752$
$4248x+5752y=24248$
[/code]
To give you:
$ 5752x+4248y=25752$
$ 4248x+5752y=24248$

But I personally like to this:
[code]
\begin{eqnarray*}
5752x+4248y&=&25752\\
4248x+5752y&=&24248
\end{eqnarray*}
[/code]
To give you:
\begin{eqnarray*}5752x+4248y&=&25752\\ 4248x+5752y&=&24248 \end{eqnarray*}

[quote="galois01"]Solve the system of the equations:
$ 5752x\plus{}4248y \equal{} 25752$
$ 4248x\plus{}5752y \equal{} 24248$[/quote]

$ \text{Add the two. }10000x\plus{}10000y\equal{}50000\rightarrow x\plus{}y\equal{}5.\text{Then subtract. }1504x\minus{}1504y\equal{}1504\rightarrow x\minus{}y\equal{}1.$
$ 2x\equal{}6$ $ x\equal{}3.$ $ y\equal{}2.$

Please do not revive old topics.

Locked


Havent done math so long...but ill try that one

So (Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)*R(x) meaning Q(x) is second degree function since P(x) and R(x) are both third degrees.

So plugging in x=1, (Q(1)-1)(Q(1)-2)(Q(1)-3)=0, notice that Q(1)=1,2,3 (cant be both 1 and 2 because its only 1 y value for x value)
Same for Q(2)=1,2,3 and Q(3)=1,2,3 (pretty easy, same idea of plugging in x=2 and x=3)

So there are 3*3*3=27 values for Q(1),Q(2),Q(3) combination, all different functions.  But notice that you cannot have (1,1,1),(2,2,2),(3,3,3),(1,2,3), or (3,2,1) because they give non 2nd degree functions.

So 3^3-5=22.

Uh, yeah, no need to spam like that.

EDIT: sorry, never mind.

^huh i wanst spamming i was seeing if i could double post...
i ended up editing it to give him the correct solution


Kevin will start with the integers 1, 2, 3 and 4 each used exactly once and written in a row in any order. Then he will find the sum of the adjacent pairs of integers in each row to make a new row, until one integer is left. For example, if he starts with 3, 2, 1, 4, and then takes sums to get 5, 3, 5, followed by 8, 8, he ends with the final sum 16. Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4, how many possible final sums are there?

The only possible way (I think) is by trying out each possibility.

So, I won't list it out, but the answer is $ \boxed{5}$.  I believe the sums are the even numbers from $ 14\minus{}22$?

Let the numbers be $ a,b,c,d$ in order, so that the sum is $ a\plus{}3b\plus{}3c\plus{}d$, so that we only need to worry about $ 2b\plus{}2c$ since $ a\plus{}b\plus{}c\plus{}d$ is constant. $ b\plus{}c$ (this determines $ 2b\plus{}2c$) can take on the values $ 3,4,5,6,7$ for an answer of $ \boxed5$.


I was looking through my history folder studying for my final, when I saw my math team schedule, which said something about a state math contest at the end of February. Does anyone know what this is?

I think people who do well at two or more of the madison math meets do the state math meet, but I'm not sure how others qualify.

Well, there is the Wisconsin Mathematics League (WML) once a month. It's split up into 6th grade, 7th grade, 8th grade, and high school. Different leveled tests with everything being 30 minutes. 40 questions for everything except high school, which has six. Calculators [i]are[/i] allowed as long as they don't have a QWERTY keyboard.