Dear MathLinks Community,

as you probably know, the Austrian-Polish Mathematical Competition (APMC) takes place every year, interchangeably once in Austria and once in Poland. This competition can now look back to a 28 year old tradition.

For reasons beyond my understanding the Department of Education in Austria recently decided to cancel the APMC for the year 2006 and it is doubtfully whether it will take place in 2007 or ever again. The official reason mentioned is the EU Presidency in Austria for the year 2006.

As a matter of course the scientific leader of the Austrian Mathematical Olympiad and the APMC, as well as teachers, participants and former participants in Austria are already doing everything to argue the Department of Education out of their intention. We however would appreciate it if we had some support from our international Olympian colleagues - and I am especially adressing our friends from Poland.

If anyone wants to write to the Department of Education in Austria, here is the adress:
[hide]Elisabeth Gehrer
Bundesministerium für Bildung, Wissenschaft und Kultur
Minoritenplatz 5
1014 Vienna, Austria
Tel: +43/1/53 120-0
Fax: +43/1/53 120-3099
E-Mail: ministerium@bmbwk.gv.at
Or if you want to contact the minister personally: elisabeth.gehrer@bmbwk.gv.at[/hide]
As I said, we would strongly appreciate your help. It is not in my interrest to break or even end this tradition and I think neither it is in yours.

Yimin Ge

You should have contacted me so that we can make this an announcement. I will help with what I can.

Thank you for your help, Valentin! And also of course thanks to everyone else who wants to help.
I myself will send a letter to the department tomorrow.
I really hope that they will change their mind.

Yimin Ge

PS: If anyone has the emails of the participants of the previous APMC, please inform them. I do have a participants list but I don't have the emails though.

Go Yimin Ge!!! I will be glad to hear that the Austrian-Polish Mathematical Competition (APMC) will take place, so please go on!!!. Good luck!.

We have had an alike problem in Belgium some years ago. The ministry of education decided it was not necessary for Belgium to compete in "elite contests" as the IMO. The money stopped coming. There was no training stage that year (what could be seen very clearly in our score...) and our team leader had to pay his own plane ticket to the IMO. A lot of effort has been done around it, and finally they did get their money again for the years after that.

Does the APMC have enough funds to organise the competition for one year without governement support? If they have, I think they really should try to keep it going this year. By trying hard, they may be able to change the governement's mind, like the people in our olympiad organisation did.

After all, if the alternative is the end of the competition, you don't have much to lose...

Hello,
I have just received a mail saying that if Austria will be invited by Poland then the invitation will be accepted. Hopefully the invitation really comes after this show by our Ministry.
The mail also said that the department received a lot of mails and letters and that one of our teachers was accused of having started a campaign against the department  :).
And we are asked to "create something new"

I want to express my deepest gratitude to everyone who helped to persuade our ministry/minister.
Yimin Ge

I've just received a message regarding this thread which states [quote]The jury of Polish MO in its official message about the results of current Polish MO (http://www.om.edu.pl/pliki/praso_57.pdf - in Polish) wrote that the AP competitions will be in Poland this year.[/quote]

Well I guess everything is okay now :)


In a 3D rectangular coordinate system with origin $O$, points $A$, $B$, and $C$ are on the $x$-, $y$-, and $z$-axes, respectively. If the areas of triangles $OAB$, $OAC$, and $OBC$ are 4, 6, and 12, respectively, compute the area of triangle $ABC$.

[hide]$AB=8$, $AC=12$, $BC=24$

Solving gives $A=2$, $B=4$, $C=6$.

Take the vectors $\vec{v}=\left(-2,0,6\right)$ and $\vec{w}=\left(-2,4,0\right)$.  Then the area of the triangle $ABC$ is

\[ \textrm{Area}=\frac{1}{2}\left|\vec{v}\times\vec{w}\right|=\frac{1}{2}\sqrt{24^{2}+12^{2}+8^{2}}=\boxed{14}. \]

[/hide]

Wait, I got the same exact thing, except when I took the cross product, I got 22. Maybe I've forgotten how to take cross products properly.

Edit: Nevermind, you did it right.

why did u let the sides be double the area, just losing you on that part

edit* wait nevermind, thought it was one side, not the two legs, got it.


[quote="236factorial"]Sam and Susie are brother and sister. Sam has twice as many sisters as brothers. Susie has twice as many brothers as sisters. How many girls are in the family?[/quote]

[hide]Two girls
$s=2b, 2s=b$
there are just as many boys as girls
since sam is a boy and susie is a girl , susie has to have 2 brothers and one sister, while sam has two sisters and one brother [/hide]

[hide]Let b be the number of boys in the family and g be the number of girls. As Sam is a boy he has twice as many sisters as brothers, 2(b-1)=g. Similarly for Susie, 2(g-1)=b. Solving these two equations yields 2 boys in the family and [b]2 girls[/b].[/hide]


number theory
greatest common divisor
logarithms
number theory unsolved

Find the largest real number $\alpha$, such that there exists an infinite integer sequence $\{a_{n}\}$ which satisifies the following condition:
(1) $a_{n}>2006^{n}$ for any $n$;
(2) for any $n\geq2$, $a_{n}^{\alpha}$ is not larger than the greatest common divisor of the set $\{a_{i}+a_{j}|i+j=n,i,j\in\mathbb{N}\}$.

It is obviosly, that $\alpha\le \frac{1}{2}$, because for infinetely many n $a_{2n}>a_{n}^{2-\epsilon }$ and $gcd (a_{2}n+a_{0},2a_{n})\le 2a_{n}$.

But why is it true that $a_{2n}\geq a_{n}^{2}$ holds for infinitely many $n$?
And also, how to construct an example when $\alpha=\frac{1}{2}$?

This is NOT my solution.

[b]lemma[/b]. For all $\epsilon >0$ exists infty of $n$ such that $a_{2n}\geq a_{n}^{2-\epsilon}$.

In fact, suppose lemma is FALSE, we will find $K$ hold $a_{2n}<a_{n}^{2-\epsilon}\forall n>K$. Therefore $\frac{\ln a_{2n}}{2n}<\frac{2-\epsilon}{2}.\frac{\ln a_{n}}{n}$. Using product telescopic and (1) we have contradition!

[quote="cuopbientcd"]It is from Viet Nam TST [b]19998[/b] :)[/quote]

Are you sure?  :P

You are right! But this problem also can from another contents, or AMM ,or...


inequalities
search
inequalities proposed
111

Let $ a,b,x,y,m,n \in R$,prove that:

$ \sqrt{x^2\plus{}y^2}\plus{}\sqrt{a^2\plus{}b^2}\plus{}\sqrt{m^2\plus{}n^2} \geq \sqrt{(x\plus{}a\plus{}m)^2\plus{}(y\plus{}b\plus{}n)^2}.$

BQ

Let $ a,b,c,x,y,z,m,n,p \in R$,prove that:

$ \sqrt {x^2 \plus{} y^2 \plus{} z^2} \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} \sqrt {m^2 \plus{} n^2 \plus{} p^2}$

$ \geq \sqrt {(x \plus{} a \plus{} m)^2 \plus{} (y \plus{} b \plus{} n)^2 \plus{} (c \plus{} z \plus{} p)^2}.$

BQ

just Minkowski Inequality, do you know? :wink:

[quote="xzlbq"]Let $ a,b,c,x,y,z \in R$,prove that:

$ \sqrt [t] {x^t \plus{} y^t \plus{} z^t} \plus{} \sqrt [t] {a^t \plus{} b^t \plus{} c^t}$

$ \geq \sqrt [t]{(x \plus{} a )^t \plus{} (y \plus{} b )^t \plus{} (c \plus{} z )^t}(t \geq 0).$

$ \sqrt [t] {x^t \plus{} y^t \plus{} z^t} \plus{} \sqrt [t] {a^t \plus{} b^t \plus{} c^t}$

$ \le \sqrt [t]{(x \plus{} a )^t \plus{} (y \plus{} b )^t \plus{} (c \plus{} z )^t}(t < 0).$

BQ[/quote]

please search "Minkowski Inequality" to see that

http://zh.wikipedia.org/wiki/%E9%97%B5%E5%8F%AF%E5%A4%AB%E6%96%AF%E5%9F%BA%E5%AE%9A%E7%90%86

replace it:
$ 27((z\plus{}x)(x\plus{}y)\plus{}(x\plus{}y)(y\plus{}z)\plus{}(y\plus{}z)(z\plus{}x))^4 \geq 256(x\plus{}y\plus{}z)^6(yz\plus{}zx\plus{}xy).$

[quote="xzlbq"]replace it:
$ 27((z \plus{} x)(x \plus{} y) \plus{} (x \plus{} y)(y \plus{} z) \plus{} (y \plus{} z)(z \plus{} x))^4 \geq 256(x \plus{} y \plus{} z)^6(yz \plus{} zx \plus{} xy).$[/quote]

setting $ x\plus{}y\plus{}z\equal{}3u,xy\plus{}yz\plus{}zx\equal{}3v^2$, ineq becomes

$ 3^3((3u)^2\plus{}3v^2) \ge 4^4(3u)^6(3v^2)$

$ \iff (u^2\minus{}v^2)^2(9u^2\plus{}2uv\plus{}v^2)(9u^2\minus{}2uv\plus{}v^2) \ge 0.$

[quote="xzlbq"]$ 3^{16}((x \plus{} 2y \plus{} z)(y \plus{} 2z \plus{} x) \plus{} (y \plus{} 2z \plus{} x)(z \plus{} 2x \plus{} y) \plus{} (z \plus{} 2x \plus{} y)(x \plus{} 2y \plus{} z))^{16}\geq 2^{64}3^{17}(x \plus{} y \plus{} z)^{30}(yz \plus{} zx \plus{} xy)$
BQ[/quote]

I think it should be

$ 3^{15}((x \plus{} 2y \plus{} z)(y \plus{} 2z \plus{} x) \plus{} (y \plus{} 2z \plus{} x)(z \plus{} 2x \plus{} y) \plus{} (z \plus{} 2x \plus{} y)(x \plus{} 2y \plus{} z))^{16}\geq 2^{64}(x \plus{} y \plus{} z)^{30}(yz \plus{} zx \plus{} xy).$

Dear friends,
I hope To put some hard rock,
your solution be  getting well.
BQ


prove that every positive integer can be written as the sum of not more than 53 fourth powers of integers


Suppose 2 distinct integers are chosen from between 1 and 9, inclusive.  What is the probability that they are both odd?

There are $ \binom92\equal{}36$ ways to choose 2 numbers. There are $ \binom52\equal{}10$ ways to choose two odd numbers. So the answer is $ \frac{10}{36}\equal{}\frac{5}{18}$

Alternatively, there is a $ \frac59$ chance for the first integer to be odd and a $ \frac48$ chance for the second to be odd.

Thus, the answer is $ \frac59 \times \frac12$, or $ \boxed{\frac{5}{18}}$.

[quote="ernie"]Alternatively, there is a $ \frac59$ chance for the first integer to be odd and a $ \frac48$ chance for the second to be odd.

Thus, the answer is $ \frac59 \times \frac12$, or $ \boxed{\frac {5}{18}}$.[/quote]

I under stand what your saying but the question dosn't specify if the digits are distinct or not.(Even if you got the correct answer)

[quote="GameBot"]Suppose [b]2 distinct integers[/b] are chosen from between 1 and 9, inclusive.  What is the probability that they are both odd?[/quote]
Actually, it does say that they're distinct  :P


I spent two hours, but I don't know how to solve it.
Maybe who knows inequalities very well will send the solution.
And about problem is 12 is minimum constant that we can
put in our inequality ?

I've made a stupid mistake...ouch. I pressed the "submit" button instead of the "preview" button.

anyway, the proof can be done by
$
|\sqrt{(x+\frac{(y-z)^2}{12})}-\sqrt{(y+\frac{(z-x)^2}{12})}|=\frac{|(x+\frac{(y-z)^2}{12})-(y+\frac{(z-x)^2}{12})|}{\sqrt{(x+\frac{(y-z)^2}{12})}+\sqrt{(y+\frac{(z-x)^2}{12})}}\geq\frac{11|x-y|}{12(\sqrt{(x+\frac{(y-z)^2}{12})}+\sqrt{(y+\frac{(z-x)^2}{12})})}\geq\frac{11|x-y|}{24\sqrt{\frac{7}{12}}}>\frac{|x-y|}{2}
$
where $\frac{7}{12}$ is half the maximum of $(x+\frac{(y-z)^2}{12}+y+\frac{(z-x)^2}{12})$.

It tells that 12 is not the minimum constant which can be put in the inequality. I don't know what value the minimum is.

I find that the inequality :

$\sqrt{x+\frac{(y-z)^2}{p}}+\sqrt{y+\frac{(z-x)^2}{p}}+\sqrt{z+\frac{(x-y)^2}{p}}\leq\sqrt{3}$

is true for $p=4+\sqrt{21}<9<12$.

I found that the inequality :

$\sqrt{x+\frac{(y-z)^2}{p}}+\sqrt{y+\frac{(z-x)^2}{p}}+\sqrt{z+\frac{(x-y)^2}{p}}\leq\sqrt{3}$

is true for $p=4+\sqrt{21}<9<12$. But this one is not the best value of $p$.

I used the Darkseer way. Substitute $a=(y-z)^2/p$,  then square the inequality.
We get
$\sum{[(\sqrt{y+b}-\sqrt{z+c})^2-3a]\geq0}$.
It suffices to show that
$(\sqrt{y+b}-\sqrt{z+c})^2\geq 3a$,
which reduces to
$(\frac{3x+p-1}{(\sqrt{y+b}+\sqrt{z+c}})^2\geq3p$.
Since
	$(\sqrt{y+b}+\sqrt{z+c})^2\leq2(y+z+b+c)$,
it suffices to show that
$(3x+p-1)^2\geq 6p(1-x+b+c)$,
that is
$(3x+p-1)^2-6p(1-x)-6(5x^2-4x+1)+12yz\geq 0$.
It suffices to prove that
$f(x)= (3x+p-1)^2-6p(1-x)-6(5x^2-4x+1) \geq 0$.
Since $f(x)$ is concave, it suffices to show that
	$f(0)\geq 0$ and $f(1)\geq 0$.
These inequalities hold for $p\geq 4+\sqrt{21}$.

Great solutions !!!
About p what i wanted to say if we put instead 12 - 4  then noticing
x = p-a, y = p-b, z = p-c.

we get that x + (y-z)2/4 = ma2

where ma is median in triangle ABC so inequality will be

   ma + mb + mc ≤ √3/2 (a+b+c)

which a think this inequality isn't corect because in 1977 was proven
only the following fact using PC :

   la + lb + mc ≤ √3/2 (a+b+c)

where la, lb are bisectors in that triangle.

[quote="prowler"]   ma + mb + mc ≤ √3/2 (a+b+c).[/quote]
It is not correct. For a counterexample, just look at the triangle (0, 0), (0, 0), (1, 0).
It is a great idea anyway^_^.

[quote="Stun"]Let x,y,z be positive numbers and x+y+z=1 . Prove that :
$\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(z-x)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\leq\sqrt{3}$[/quote]
The following inequality is also true.
Let $x$, $y$ and $z$ are non-negative numbers such that $x+y+z=1$. Prove that:
\[\sqrt{x+\frac{(y-z)^2}{8}}+\sqrt{y+\frac{(z-x)^2}{8}}+\sqrt{z+\frac{(x-y)^2}{8}}\leq\sqrt{3}\]
But $\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{7}}\leq\sqrt3$ is wrong already: $x=\frac{10}{11}$ and $y=z=\frac{1}{22}$.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=388225

[quote=Stun]Let x,y,z be positive numbers and x+y+z=1 . Prove that :
$\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(z-x)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\leq\sqrt{3}$[/quote]
http://www.artofproblemsolving.com/community/c6h141836p802407
For  $x,y,z>0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\ge\sqrt{3}$$

Let $a, b, c$ be non negative a+b+c=1 with $k=\frac{2-\sqrt{3}}{2}$ , [url=http://www.artofproblemsolving.com/community/c6h1328440p7148103]prove that[/url]
$$\sqrt{a+k(b-c)^2}+\sqrt{b+k(c-a)^2}+\sqrt{c+k(a-b)^2}\leq \sqrt{3}$$
[url]https://artofproblemsolving.com/community/c6h202647p1114956[/url]
Let $a,b,c\geqq 0$ and $a+b+c=3 .$   [url=https://artofproblemsolving.com/community/c6h2232594p17065421]Find minimum  of  [/url]
$ \sqrt{6a+\frac{(b-c)^2}{2}}+\sqrt{6b+\frac{(a-c)^2}{2}}+\sqrt{6c+\frac{(b-a)^2}{2}}$

[quote=Stun]Let x,y,z be positive numbers and x+y+z=1 . Prove that :
$$\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(z-x)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\leq\sqrt{3}$$[/quote]
Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$. Prove that
$$ \sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(z-x)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\ge  \frac{5\sqrt{3}}{6}.$$

[quote=KaiRain][quote=dragonheart6][hide][quote=sqing][Let $a;b;c>0$ and $a+b+c=1.$ Prove that $$\sqrt{12a+(b-c)^2}+\sqrt{12b+(c-a)^2}+\sqrt{12c+(a-b)^2}\leq 6$$
[url=https://artofproblemsolving.com/community/c6h1965768p13653126]Proof: [/url]Using the idea that $X-Y \ge \frac{X^2-Y^2}{2X}$ for all $X>0$, we have:
$$\sum\left(4a+b+c-\sqrt{12a(a+b+c)+(b-c)^2}\right)\ge \sum \frac{(4a+b+c)^2-\left(12a(a+b+c)+(b-c)^2\right)}{8a+2b+2c}$$
$$=\sum \frac{2(a-b)(a-c)}{4a+b+c}=\sum \frac{2(b+c)(a-b)(a-c)}{a^2+b^2+c^2+3(ab+bc+ca)-(a-b)(a-c)}$$
$$ \ge \sum \frac{2(b+c)(a-b)(a-c)}{a^2+b^2+c^2+3(ab+bc+ca)}=2\frac{\sum bc(b+c)-6abc}{a^2+b^2+c^2+3(ab+bc+ca)}\ge 0.$$[/quote][/hide]

Who gives this very nice proof?

Can this idea work for stronger one? For example, for $p < 12$:
$$\sqrt{x+\frac{(y-z)^2}{p}}+\sqrt{y+\frac{(z-x)^2}{p}}+\sqrt{z+\frac{(x-y)^2}{p}}\leq\sqrt{3}$$[/quote]
Hi, it is my proof. The same idea can help for all $p \ge 8$, but not work for the best constant.[/quote]

You should be pointed out being the author of that proof. For the best constant, I have a ugly proof.

[quote=KaiRain][quote=dragonheart6]
You should be pointed out being the author of that proof. For the best constant, I have a ugly proof.[/quote]
Oh sorry... Mr. sqing attachs the url link in #19, I posted the proof there.[/quote]

In #19, sqing quoted a proof which is given by you in #7 of the link, right? The link: [url]https://artofproblemsolving.com/community/c6h1965768p13653126[/url]


Find the center of the circle described by $ x^2\plus{}2x\plus{}y^2\equal{}8y$. Express your answer as an ordered pair $ (x,y)$.


linear algebra
matrix
superior algebra
superior algebra unsolved

Prove that the set $ A = \{{({\begin{array}{rr} a & 0 \\
0 & 0 \\
\end{array}) | a\in \mathbb{C}}\}}$ is a commutative ring of the matrix ring $ M(2;\mathbb{C})$ and find his unit.

I'll leave it to you to verify the ring axioms, which shouldn't be hard.  We have that $ \begin{pmatrix}
a&0\\
0&0
\end{pmatrix}\cdot\begin{pmatrix}
b&0\\
0&0
\end{pmatrix}\equal{}\begin{pmatrix}
ab&0\\
0&0
\end{pmatrix}$.  When you go in the other direction, you'll get $ ba$ instead, but that's the same as $ ab$ since $ \mathbb{C}$ is commutative.


http://www.trustmeher.net/freeware/free-indian-classical-music.htm
contains information and samples of some of the top ranking artists
both from carnatic and hindustani section

 music.download.com/jayateerthmevundi/3600-8849_32-100914412
.htm[/url]
this contains a good vocal ssmple by jayateerth mevundi.

you can google search the following artists:

VOCAL
M.S.subbulakshami
s.balamuralikrishna
thyagaraja
kumar gandharva
dagar bandhu
bhimsen joshi
bade gulam ali khan
begum akhtar
girja devi
omkarnath thakur
ajoy chakraborty
d.v.paluskar
rashid khan
mallikarjun mansur

PERCUSSION
zakir hussain
ustad allah rakha
rashid mustafa
ahmed jan thirakwa
prabhat kale

FLUTE
s.prasanna
hariprasad chourasia
bismillah khan

STRING instruments
PT.Ravishanker
vilayat khan
vishwa mohan bhatt
shiv kumar sharma
alauddin khan
faiyyaz khan
faiyyazuddin dagar
ali akbar kahn
amjad ali khan

ask sathej or carnatic music section

Firstly,I suppose it is M Balamuralikrishna,not S Balamuralikrishna.
To add to the list,
Vocal:
Search for
Chembai Vaidyantha Bhagavthar
Ariyakudi Ramanuja Iyengar
Semmangudi Srinivasa Iyer
The list is endless
Violin:
MS Gopalakrishnan
Lalgudi Jayraman
TN Krishnan
You can have a look  at the website of Shivkumar Kalyanaraman,a Professor at Rennslaer Polytechnic Institute(RPI).It has a plethora of information on Carnatic music.Just type his name in Google.
PS:Sorry,tokenadult ,if it was sounding like a private conversation.Being a lover of Carnatic music,I get interested if someone talks about it.I agree that in AOPS issues should have a more global perspective.

I think this will be good information for math-liking people around the world to find out more about music from India.


I use FF 3.0 on Leopard.  About 5:20 forum time today, whenever I clicked to view the thread, it would display a completely blank FF window.  When I clicked to view a forum (list of threads) it would work fine...

Haven't you noticed anything in this forum you just posted? ... ;)

I'm fixing a lot of bugs. During this time the forum posting might be affected.


If.

That's the key word. I can sometimes do that with very simple problems. However, I can't keep all the information in my head at one time. Therefore, I need paper and pencil.

I agree with mathnerd. It all depends on the person. I can't spell medium words without using a piece of paper to double check myself. Then again maybe someone can do calculus in their heads.

Also, a great example of this is blindfold chess.

You guys just figured this out? I've been doing it for years...

I use visualization of math problems all the time, and it has advantages and disadvantages. It does require some concentration just to keep the thing in mind, and this is kind of a drain on your mental power. But when you imagine math (esp algebra, in my experience), you can manipulate it much more quickly and easily than you could on paper. This lets you find and test tons of possible approaches very quickly. I end up using visualization in some form in almost every problem.

As a historical figure whom we know must have been very good at mental manipulation without the help of paper, there was Leonhard Euler, whose stupendous mathematical productivity continued unabated even after he became totally blind.

Me, I'm a mere mortal. I do a lot of things in my head, but I also make plenty of mistakes doing things that way. And any scrap of paper loose in my vicinity probably has work pertaining to four or five different problems wedged onto it at various angles.

I find that I can do problems faster and more accurately on paper than in my head. It helps me keep track of all the numbers.

Off-topic: When I looked at my "View your posts", this topic was right by "No Pen". Apparently, no pen and no paper either. :D Interesting how some topic names seem to answer other ones...

I used to like diong the logic spacial problems where you need to do stuff like this. Now that I haven't done it in years, it gives me headaches... :wink:

i can usually visually the actual concept without having to write the details down. It's mainly just the actual numbers that i need to keep track of with pencil and paper

[quote="weiquan"][quote="PSUgradmeteo077"]If one can visualize things very well, then he can make a mental whiteboard to work out problems. Thus, one needs to just read a section, look at a problem, and then create a mental whiteboard showing his work. You think this is possible?[/quote]

I think it is much more than a mental whiteboard that you can create in your mind. It is also possible to create a mental collection of ideas or memories without actually consciously acknowledging any one of them, then putting them all together at once. In terms of solving problems, it is something like solving a problem mentally without knowing much about the solution except that it is correct and I can work it out if I wanted to, but it is extremely difficult. But I often use this "method" for speed-reading so I can read "areas" of words instead of lines.[/quote] Wow, I did a minor form of that the other day in physics. I wasn't paying attention, had never done any of the problems, but they were pretty much complicated conversions of units. The teacher suddenly asked for us to solve a few problems, and I jotted down the arithmetic that needed to be done for the answers without thinking about the conversion factors, about the numbers or objects in the problem, or really about anything at all, and it was right. I mean, I don't need to write down the steps to problems like them, but usually I THINK about what the steps are, or at least about what process I'm doing to solve the problem. Still, I'm paranoid, and I had to go back and figure it all out step by step so that I knew I had done right... and I screwed up the second time.

I have a mental room with memories and emotions and the like that's sort of what you're talking about. Everything's a complex symbol, and I never tried to get them to form or attach memories or ideas to them--objects just started appearing in my mind in place of much more complex ideas. I can break them down into words if I need to (for the most part), but it's kinda strange. The connections between things are pretty wild.

I also do the mental white board thing. Not in general practice, like on homework and stuff, but when I'm not thinking about it. It's faster and I'm more likely to get the right answer and not get bogged down in silly errors, so I'm trying to do it in situations where scratchwork is ingrained. I'm especially better at geometric proofs in my head--not what society'd think, is it?


vector
analytic geometry
trigonometry
linear algebra

Hi

I never took linear algebra so I am unsure of the following:

To find a point on a line (if you know the angle between old and new vector):

New Point (x) = Old Point X + cos(angle);
New Point (y) = Old Point Y + cos(angle);

Graphically this is what I am trying to do (although angle will prob be less than 90 degrees):

   OP (x,y) -------------
    |  angle
    |
    |
    |
   NP (x,y)

Is that correct?  I am doing some 3D programming and I am trying to move a camera.

Thanks

If you want this:
[asy]D((6,-8)--MC("\mbox{horizontal direction}",8,D("A",MA("\alpha",(6,-8),(0,0),(10,0),1,9),W)--(10,0),0.6,N));
MC("r",(0,0)--D("B",(3,-4),SW),SW);[/asy]
then the coordinates of $ B$ are $ X(B)\equal{}X(A)\plus{}r\cos\alpha$, $ Y(B)\equal{}Y(A)\minus{}r\sin\alpha$. If your picture was assumed to mean something else, just tell more precisely what is needed :)


calculus
integration
limit
function
complex numbers
calculus computations

Find

a) $\int_{0}^{\infty}x^{r}(a-x)^{s}dx$

b)Let $u_{n}= \int_{0}^{\infty}e^{-px}sin^{n}x dx$ find $\frac{u_{n}}{u_{n-2}}$ and evaluate $\lim_{x \rightarrow \infty}\frac{u_{n}}{u_{n-1}}$

On the first one, are you sure you want to integrate from $0$ to $\infty$ and not from $0$ to $a$?

I have that doubt too, but the exercise is exactely how i wrote.That is, from $0$ to $\infty$.

Well,  lets change $\infty$ by $a$ at the first one.
So we do $u=(a-x)^{s}$ and $dv=x^{r}dx$, so $du =s(a-x)^{s-1}$ and $v=\frac{x^{r+1}}{r+1}$.Thus

$\int_{0}^{a}udv=-\int_{0}^{a}vdu=-\frac{s}{r+1}\int_{0}^{a}x^{r+1}(a-x)^{s-1}dx$ (because $uv$ becomes $0$ taken from $0$ to $a$.)
So because of that we get that:
$\int_{0}^{a}x^{r}(a-x)^{s}dx=\frac{(-1)^{s}s!r!}{(r+s)!}\int_{0}^{a}x^{r+s}(a-x)^{s-s}dx=\frac{(-1)^{s}s!r!a^{r+s+1}}{(r+s+1)!}$.

If you note something wrong please tell me, thanks:)

Your sign is off; we're integrating a positive function, so the answer should always be positive. What happens when yoe differentiate a power of $a-x$?

That argument works when $r$ and $s$ are nonnegative integers. In general, we can write down an expression involving the $\Gamma$ function $\Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}\,dx$;$\Gamma(n+1)=n!$ when $n$ is a nonnegative integer, and $\Gamma(s+1)=s\Gamma(s)$ whenever the real part of $s$ is positive.

[hide="General version"]$\int_{0}^{a}x^{r}(a-x)^{s}\,dx=a\int_{0}^{1}a^{r}t^{r}a^{s}(1-t)^{s}\,dt =a^{r+s+1}\int_{0}^{1}t^{r}(1-t)^{s}\,dt$
Consider $I=\int_{0}^{\infty}e^{-t}\int_{0}^{t}x^{r}(t-x)^{s}\,dx\,dt$. By the preceding, $I=\int_{0}^{\infty}t^{r+s+1}e^{-t}\int_{0}^{1}x^{r}(1-x)^{s}\,dx=\Gamma(r+s+2)\cdot\int_{0}^{1}x^{r}(1-x)^{s}\,dx$.

On the other hand, $I=\int_{0}^{\infty}\int_{x}^\infty e^{-t}x^{r}(t-x)^{s}\,dt\,dx$; substituting $y=t-x$,
$I=\int_{0}^{\infty}\int_{0}^{\infty}e^{-x-y}x^{r}y^{s}\,dy\,dx= \left(\int_{0}^{\infty}x^{r}e^{-x}\,dx\right)\cdot \left(\int_{0}^{\infty}y^{s}e^{-y}\,dy\right)$
$=\Gamma(r+1)\Gamma(s+1)$

Equating the two expressions, $\int_{0}^{1}x^{r}(1-x)^{s}\,dx= \frac{\Gamma(r+1)\Gamma(s+1)}{\Gamma(r+s+2)}$ whenever the integral converges.[/hide]

On #2: Integrate by parts (twice) and use the Pythagorean identity to get the recursion.


Would it be feasible to have a math jam on advanced geometry, i.e. all the stuff that is claimed to be "well-known" on the geometry board?(;)) Hopefully, it would pick up where advanced geometry left off - stuff like affine/projective transformations, preservation of cross-ratio, this whole involution business, polar map, and maybe this complex projective plane, because that just seems pretty cool. It would be nice to understand grobber and darij.


My keygenerator generates keys that have a working chance of one in 300. I generate 5000 keys with it. What is the chance that a random 75 keys selected from the 5000 would have at least one working?

[hide]$1-(\frac{299}{300})^{75}$[/hide]???
If it is right I will post a solution...

[quote="SplashD"]is this binomial experimentation?[/quote]
What is that?

[quote="bpms"][quote="ProtestanT"]Correct.  :lol:[/quote]
Who is correct? Me or splash (or both if what I did is binomial experimentation)? Oh snap I made a Latex Error. I'm going to fix it...[/quote]

nvm, forget what i said, i agree that your answer is right


linear algebra
matrix
algebra unsolved
algebra

Let $ A$ be an $ n \times n$ matrix whose elements are non-negative real numbers. Assume that $ A$ is a non-singular matrix and all elements of $ A^{\minus{}1}$ are non-negative real numbers. Prove that every row and every column of $ A$ has exactly one non-zero element.

$\begin{pmatrix}a_{11}&a_{12}& \cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\cdots \\a_{n1}&a_{n2}&\cdots&a_{nn}\end{pmatrix}\begin{pmatrix}b_{11}&b_{12}& \cdots &b_{1n}\\b_{21}&b_{22}&\cdots &b_{2n}\\\cdots \\b_{n1}&b_{n2}&\cdots&b_{nn}\end{pmatrix}=\begin{pmatrix}1&0& \cdots &0\\0&1&\cdots &1\\\cdots \\0&0&\cdots&1\end{pmatrix}$
WLOG $a_{1i}b_{j1}>0$ and $a_{1j}b_{j1}>0$.
That means $b_{ik}=b_{jk}=0$ for all $k>1$.

$A_i$ and $B_j$ be the $i$th and $j$th row and column of matrix $A$ and $B$ respectively.
$A_2\cdot B_i=0$ but $A_2\cdot B_2\neq 0$ that implies there exists $i_0\neq i,j$ such that $a_{2i_0}b_{i_02}>0$. Which thus forces $b_{i_0l}=0$ for all $l\neq 2$.

Extending similar ways, we have an implausibility that $A_n\cdot B_n=0$.


Here are my set of problems from the last Columbia Math Circle. Let's see if we can generate a complete set of solutions for these. I'll try to post some solutions as well as I find time. Note that number 36 had an important piece of information missing which has been added here. If anyone wants to scan Patrick's set of problems and post them also, feel free.

Note that in general:
$ ax \plus{} by \equal{} c$ means
$ x \equal{} \frac {c \minus{} by}{a}$

If $ a$, $ b$, $ c$, $ x$, and $ y$ are all integers, this implies that
$ c \minus{} by\equiv 0 \pmod{a}$ or
$ by\equiv c \pmod{a}$ and
$ ax\equiv c \pmod{b}$

[hide="Number 1"]
Realistic Approach:
There must be $ 3$ pennies, as there are only $ 6$ total coins with a value $ 23\equiv 3 \pmod{5}$.  This leaves $ 3$ coins and $ 20$ cents in nickels and dimes.  The only way this works is if there are $ 2$ nickels and $ 1$ dime.

Systematic Approach:
$ 10d \plus{} 5n \plus{} p \equal{} 23$
$ d \plus{} n \plus{} p \equal{} 6$
$ 9d \plus{} 4n \equal{} 17$

$ 9d\equiv17 \pmod{4}$
$ d\equiv1\pmod{4}$
$ d \equal{} 4t \plus{} 1$

$ n \equal{} \frac {17 \minus{} 9(4t \plus{} 1)}{4}$
$ n \equal{} \minus{} 9t \plus{} 2$

The only reasonable value for $ t$ is $ 0$, so there are $ \minus{} 9(0) \plus{} 2 \equal{} 2$ nickels.[/hide]

[hide="Number 2"]
$ p \plus{} q \equal{} r$ forces all of the primes to be even or for one to be even.  The only even prime is the smallest, $ 2$.  $ p$ is the smallest prime, so it must be $ 2$.
[/hide]

[hide="Number 3"]
$ d\equiv 0 \pmod{2}$
$ d\equal{}2t$

$ 2d\plus{}15\equiv0 \pmod{9}$
$ 2d\equiv3 \pmod{9}$
$ 4t\equiv12\pmod{9}$
$ t\equiv3\pmod{9}$

$ t\equal{}3$; $ d\equal{}6$
[/hide]

[hide="Number 4"]
Factor $ 280$ into $ 2^3 5^1 7^1$.
The number of factors is $ (3 \plus{} 1)(1 \plus{} 1)(1 \plus{} 1) \equal{} 16$.
[/hide]

[hide="Number 5"]
$ 2^9 3^5 7^4$, find all factors with only even exponents for each component.

$ \left(\frac{9\plus{}1}{2}\right)\left(\frac{5\plus{}1}{2}\right)\left(\frac{4\plus{}2}{2}\right)\equal{}45$
[/hide]


$ a,b,c > 0$ then prove:

$ \sum_{cyc} \sqrt {a^2 \plus{} b^2 \plus{} ab} \ge \sqrt {3}.(a \plus{} b \plus{} c)$

$ \sum\sqrt{a^2\plus{}ab\plus{}b^2}\ge \sum\sqrt{\frac 34(a\plus{}b)^2}\equal{}\sqrt3(a\plus{}b\plus{}c)$

oke I think it is easy!!
a²+b²+ab≥$ \frac {3$\{a + b}^2}${ }{4}$
similar .............


Find four perfect fourth power numbers whose is a perfect fourth power.

I assume that you wanted to ask for a sum?
In that case :
$30^4 + 120^4 + 272^4 + 315^4 = 353^4$
found by R. Norrie (1911).

Pierre.


It's not correct -  try $(a,b,c) = (10^{100}, 10^{-100}. 10^{-100})$, and say k = 4.

RS is 25, but LS is atleast $(\frac{ka}{b+c} + 1)$,  which is clearly bigger than 25.

1)If $A>25$ then $A\geq25.$
2)You are right. If $k=4$ then $LS>25.$ All we need now is to prove this inequality. :)

So what is that hard to prove that \[ (1+\frac{4a}{b+c})(1+\frac{4b}{c+a})(1+\frac{4c}{a+b}) >25 \] :P

Nice inequality! :D
$(\frac{ka}{b+c}+1)(\frac{kb}{a+c}+1)(\frac{kc}{a+b}+1)\geq(k+1)^2$
$\Leftrightarrow (ka+b+c)(a+kb+c)(a+b+kc)\geq (k+1)^2(a+b)(b+c)(c+a)$
Expanding, this is equivalent to
$ka^3+kb^3+kc^3+k^3abc-k\sum_{sym}a^2b-kabc-2k^2abc\geq 0$
$\Leftrightarrow a^3+b^3+c^3+k^2abc-\sum_{sym}a^2b-abc-2kabc\geq 0$
By Schur Inequality, $a^3+b^3+c^3+3abc-\sum_{sym}a^2b\geq 0$
$\Rightarrow a^3+b^3+c^3+k^2abc-\sum_{sym}a^2b-abc-2kabc$
$=(a^3+b^3+c^3+3abc-\sum_{sym}a^2b)-3abc+k^2abc-abc-2kabc$
$\geq k^2abc-2kabc-4abc$
$=abc(k^2-2k-4)$
$\geq 0$ ($\because k\geq\sqrt{5}+1, a,b,c>0$)

Let $a, b, c>0$. Prove that
 $$2\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}>\sqrt 2$$
$$\frac{2a}{b+c}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}>\sqrt 2$$
$$\sqrt{\frac{a}{b+c}+\frac{1}{2}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}\geq 2$$
$$\frac{a}{b+c}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}\geq\frac{3}{\sqrt[3]{4}}-\frac{1}{2}$$
$$\frac{a}{b+c}+2\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}\geq\frac{5}{2}$$
$$\frac{a}{b+c}+3\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}\geq3\sqrt[3]{\frac{9}{4}}-\frac{1}{2}$$
$$\frac{a}{b+c}+4\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}\geq3\sqrt[3]{4}-\frac{1}{2}$$
$$a\sqrt{\frac{a}{b+c}+\frac{1}{2}}+b\sqrt{\frac{b}{c+a}+\frac{1}{2}}+c\sqrt{\frac{c}{a+b}+\frac{1}{2}}\geq a+b+c$$
Let $a,\,b,\,c$ are positive real numbers. [url=https://artofproblemsolving.com/community/c6h2030698p15298228]Prove that [/url]for all $-\frac54 \leqslant k \leqslant \frac14,$ then
\[\left(\frac{a}{a+b}+k\right)^2+\left(\frac{b}{b+c}+k\right)^2+\left(\frac{c}{c+a}+k\right)^2 \geqslant 3\left(k+\frac{1}{2}\right)^2.\]
[url]https://artofproblemsolving.com/community/c6h272187p1473871[/url]
[hide=*][quote=xzlbq]My BL24 is(in 2003):

If $x,y,z>0,
\frac{1}{4}(\sqrt{5}-1)\geq k \geq -\frac{3}{4}+\frac{\sqrt{5}}{4}$
then

\[(k+\frac{x}{x+y})(k+\frac{y}{y+z})(k+\frac{z}{z+x}) \le  (k+\frac{1}{2})^3\]

$-\frac{1}{4}-\frac{\sqrt{5}}{4}\geq k \geq -\frac{3}{4}-\frac{\sqrt{5}}{4}$
then

\[(k+\frac{x}{x+y})(k+\frac{y}{y+z})(k+\frac{z}{z+x}) \geq  (k+\frac{1}{2})^3\]

BQ[/quote][/hide]


Euler
LaTeX
function
number theory solved
number theory

Euler had found out that zeta(2) = {\pi ^ 2}/{6}
how is this true?

im going to try latex:

the $\zeta$ function is defined as follows:

\[ \zeta {(x)} = 1 + \dfrac{1}{2^x} + \dfrac{1}{3^x} + \dfrac{1}{4^x} + \dfrac{1}{5^x}+ \ldots \]

Euler had found out that  $\zeta {(2)} = \dfrac{\pi ^ 2}{6}$.

How is this true?

latex worked!

14 proofs here : http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf  :wow:

Enjoy! ;)


Let $a_1 < a_2 < .... < a_n < 2n$ ne $n$ positive integers such that $a_j$ does not divide $a_k$ or $j \not= k$. Prove that $a_1 \geq 2^{k}$ where $k$ is defined by the condition $3^{k} < 2n < 3^{k+1}$ and show that it is the best estimate for $a_1$

represent all the numbers in form $2^q.l$, where $l$ is odd. If there were two numbers $a_i$ with the same $l$, one of them would divide the other. So each of $n$ possible values of $l$ is among the numbers $a_i$ exactly once.  If $a_i=2^{q_1}3^{m_1}$,  $a_j=2^{q_2}3^{m_2}$, $q_1\leq q_2$, $m_1\leq m_2$, then $a_i|a_j$, contradiction. Because of that for the numbers of the form
$2^q3^m$ it holds $q+m\geq k$.
the number with l=1 has $q\geq k$. Now it's enough to show that for $a_1$ $l=1$.
This is quite easy when we suppose the number in form $2^q3^mp$ and use similar technique as above.


calculus
integration
trigonometry
function
logarithms
calculus computations

Caculate length of the curve $ r\equal{} \cos^2 \phi$  where $ \phi  \in [0, \pi/6]$

I   try to solve it ...  in   the end   i received this integral
$ \int_{0}^{1/2} \sqrt {1 \plus{} 3t^2} dt$

[quote="AndrewTom"]Unless I'm mistaken, this is

$ \int_{0}^{\frac {\pi}{6}} \sqrt {1 \plus{} (f\prime(\theta))^{2}}\, d\theta$

$ \equal{} \int_{0}^{\frac {\pi}{6}} \sqrt {1 \plus{} \sin^{2}(2\theta)}\,d\theta$

which is an elliptic integral of the second kind.

P.S. Our posts crossed, so I see we kind of agree.[/quote]
  If  this  problem  in polar  system,  i think that we  must  use  formula:
$ L\equal{} \int_{\phi_1}^{\phi_2} \sqrt{ (r')^2\plus{}r^2} d \phi$

You're right.

This gives $ \int_{0}^{\frac {\pi}{6}} \sqrt {\cos^{4}\theta + 4\cos^{2}\theta\sin^{2}\theta}\, d\theta$

$ = \int_{0}^{\frac {\pi}{6}} \cos\theta \sqrt {\cos^{2}\theta + 4\sin ^{2} \theta}\,d\theta$

$ = \int_{0}^{\frac {\PI}{6}} \sqrt {1 - 3\sin^{2}\theta}\,\cos\theta\,d\theta$

$ = \int_{0}^{\frac {1}{2}} \sqrt {1 - 3u^{2}}\,du$

$ = \frac {1}{2}x\sqrt {1 - 3u^{2}} + \frac {1}{2\sqrt {3}}\arcsin{\sqrt {3}u}$

$ = \frac {1}{8} + \frac {\pi}{18}$.

(Have rushed and probably made more mistakes.)

[quote="AndrewTom"]You're right.

This gives $ \int_{0}^{\frac {\pi}{6}} \sqrt {\cos^{4}\theta + 4\cos^{2}\theta\sin^{2}\theta}\, d\theta$

$ = \int_{0}^{\frac {\pi}{6}} \cos\theta \sqrt {\cos^{2}\theta + 4\sin ^{2} \theta}\,d\theta$

$ = \int_{0}^{\frac {\pi}{6}} \sqrt { \cos^{2}\theta +\sin^2 \theta+ 3\sin^{2}\theta}\,\cos\theta\,d\theta$

$ = \int_{0}^{\frac {1}{2}} \sqrt {1 + 3u^{2}}\,du$

$ = \frac {1}{2}x\sqrt {1 - 3u^{2}} + \frac {1}{2\sqrt {3}}\arcsin{\sqrt {3}u}$

$ = \frac {1}{8} + \frac {\pi}{18}$.

(Have rushed and probably made more mistakes.)[/quote]
 you are wrong!    In  the integral  signs - plus  a not minus

$ = \int_{0}^{\frac {\PI}{6}} \sqrt { \cos^{2}\theta +\sin^2 \theta+ 3\sin^{2}\theta}\,\cos\theta\,d\theta$

$ = \int_{0}^{\frac {1}{2}} \sqrt {1 + 3u^{2}}\,du$

[quote="Dr Sonnhard Graubner"]hello, your result is wrong, in a few minutes i will send you my solution.
Sonnhard.[/quote]
My result

I don't think so original fuction is: $ L \equal{}\frac{1}{2\sqrt{3}} \ln\left( \sqrt{3}t \plus{}\sqrt{1\plus{}3t^2}\right) \plus{} \frac{t \sqrt{1\plus{}3t^2}}{2}$
$ L \equal{} \frac {1}{2\sqrt {3}} \ln \left(\frac {\sqrt {3} \plus{} \sqrt {7}}{2} \right) \plus{} \frac {\sqrt {7}}{8}$

This gives the same value as mine.

But I look forward to Sonnhard's solution.

Gotta rush.

hello, i have derived $ \int\cos(\phi)\sqrt{1\plus{}3\sin(\phi)^2}\,d\phi\equal{}\frac{1}{2}\sin(\phi)\sqrt{1\plus{}3\sin(\phi)^2}\plus{}\frac{1}{6}\sqrt{3}arsinh(\sqrt{3}\sin(\phi))$
Sonnhard.

hello, my result is $ \frac{\sqrt{7}}{8}\plus{}\frac{\sqrt{3}}{6}\ln\left(\frac{\sqrt{3}\plus{}\sqrt{7}}{2}\right)$.
Sonnhard.


geometry
circumcircle
incenter
geometry unsolved

consider a triangle $ ABC$ let us invert the triangle about it's circumcentre $ O$ and radius of inversion being $ R$ now the triangle along with it;s angular points is preserved...now consider the orthocentre of the triangle before and after inversion..it has changed positions but the configuration of the triangle is the same :maybe:

Sorry, but I don't quite get what you are doing. The "triangle" is not preserved, only its vertices A, B, C are preserved. Its sidelines BC, CA, AB are mapped to circles - namely, to the circumcircles of triangles BCO, CAO, ABO. Perpendicular lines map to orthogonal circles. There is no reason why the orthocenter should remain on its place.

  darij

well actually was seeing the solution to iran 1995 problem...
Problem:Let M, N and P be points of intersection of the incircle of triangle ABC with sides AB, BC and CA respectively. Prove that the orthocenter of ∆MNP, the incenter of ∆ABC and the circumcenter of ∆ABC are collinear.

the solution was an inversion about the incentre of radius equal to inradius.then the points A,B,C were mapped to the mid point of $ MN$...so the circumcircle mapped to the nine point circle of MNP.hence the $ O'$ image of circumcentr under inversion,circumcentre of MNP(incentre of ABC) and orthocentre of MNP are collinear hence $ O,I$ and orthocentre of MNP are collinear...
is this the right reasoning...i think it is

I had misunderstood this and taken fro granted that the orthocentre under inversion doesn't change position sorry
and THANKS :)

There is just a slight lack in your reasoning...It seems that you work as if the image under inversion of a center of a circle is a
center of a circle-image. But, in general inversion doesn't preserve center of circle!

So O' is not center of circumcircle of $ \triangle$ A'B'C'. Still, we can say that O', center of circumcircle of $ \triangle$ A'B'C' and I are collinear. Since center of circumcircle of $ \triangle$ A'B'C', I and orthocentre of $ \triangle$ MNP lie on the Euler's line of the $ \triangle$ MNP, the desired follows.

no i don't work under that assumption  :huh:
just got tooo carried away by this beautiful question :wink:


Prove that $ \sum^{k}_{n\equal{}0} 5^n$ can be represented as the sum of two (not necessarily positive) triangular numbers for all positive integers $ k$.

[hide="Lemma"] $ n$ is the sum of two triangular numbers if and only if $ 4n \plus{} 1$ is a sum of two squares.

[i]Proof.[/i]  $ n \equal{} T_a \plus{} T_b \Leftrightarrow$
$ 4n \plus{} 1 \equal{} \left( a \plus{} b \plus{} 1 \right)^2 \plus{} \left( a \minus{} b \right)^2$. [/hide]
[hide="Solution"] We therefore desire to show that $ 5^{k \plus{} 1}$ can be written as the sum of two squares, but this is obvious from some [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]well-known lemmas[/url] about representations as sums of squares.  Explicitly, one computes the real and imaginary parts of $ (1 \plus{} 2 \imath)^{k \plus{} 1}$. [/hide]
Interesting questions to ask yourself:  when is a number the sum of a triangular number and a square?  What about combinations of higher polygonal numbers?


Call $ r$ the inradius. We have $ \frac{r}{\sin \frac{A}{2}} \minus{} r \equal{} \frac{r_1}{\sin \frac{A}{2}} \plus{} r_1$ so $ r_1 \equal{}r \frac{1\minus{} \sin \frac{A}{2}}{1\plus{} \sin \frac{A}{2}}$ and cyclicaly $ r_2$ and $ r_3$. Then $ \sqrt{r_1 r_2} \equal{} r \sqrt{ \frac{\left ( 1\minus{} \sin \frac{A}{2} \right ) \left ( 1\minus{} \sin \frac{B}{2} \right )}{\left ( 1\plus{} \sin \frac{A}{2} \right ) \left ( 1\plus{} \sin \frac{B}{2} \right )}} \equal{} r \tan \left ( 45 \minus{} \frac{A}{4} \right ) \tan \left ( 45 \minus{} \frac{B}{4} \right )$

We know that $ R\ge 2r$ so we have to prove  $ \tan \left ( 45 \minus{} \frac{A}{4} \right ) \tan \left ( 45 \minus{} \frac{B}{4} \right ) \plus{} \tan \left ( 45 \minus{} \frac{B}{4} \right ) \tan \left ( 45 \minus{} \frac{C}{4} \right ) \plus{}$ $ \tan \left ( 45 \minus{} \frac{C}{4} \right ) \tan \left ( 45 \minus{} \frac{A}{4} \right ) \ge 1$

but is well-know i think that $ \tan \left ( 45 \minus{} \frac{A}{4} \right ) \tan \left ( 45 \minus{} \frac{B}{4} \right ) \plus{} \tan \left ( 45 \minus{} \frac{B}{4} \right ) \tan \left ( 45 \minus{} \frac{C}{4} \right ) \plus{}$ $ \tan \left ( 45 \minus{} \frac{C}{4} \right ) \tan \left ( 45 \minus{} \frac{A}{4} \right ) \equal{}1$

so we have also $ \sqrt{r_1r_2}\plus{}\sqrt{r_2r_3}\plus{}\sqrt{r_3r_1}\equal{}r$


inequalities
superior algebra
superior algebra unsolved

Let L/K be a field extension of degree [L:K] and Galois group G. Prove that if [L:K]=|G|, then the extension is Galois.

I guess L/K is finite.

Let E = Inv(G). Of course is $ K \leq E \leq L$.
By Artin's lemma (you might know it as an inequality, but considering independence of automorphisms and using a theorem due to dedekind it can be proven that the equality holds)
you get:

$ [L : E] \equal{} |G| \equal{} [L: K]$ where the second equality is by hypotesis and the first one is artin.

This means $ [E: K] \equal{} 1$ and $ K \leq E$ hence $ E \equal{} K$. We proved that $ Inv(G) \equal{} K$ and since $ L|K$ is finite, we have it is Galois.

bye


Why are the following not contradictory?
$ r\equal{}\sqrt{x^2\plus{}y^2}$
$ \frac{\partial r}{\partial x}\equal{}\frac{x}{\sqrt{x^2\plus{}y^2}}\equal{}\frac{x}{r}\equal{}cos{\theta}$

and
$ r\equal{}\frac{x}{cos{\theta}}$

$ \frac{\partial r}{\partial x}\equal{}\frac{1}{cos{\theta}}$

I understand that we got these results from differentiating two separate relations, but I don't see why they're different, and seemingly contradictory.

They're contradictory because the second one is wrong.  $ \theta$ is not independent of $ x$.


function
quadratics
algebra
polynomial
inequalities theorems
inequalities

[quote="djuro"]Can somebody post ABC theorem and / or explain ABC method?[/quote]
OK.I'll only start it,maybe someone would like to continue.
[color=blue][size=150]
[u]ABC THEOREM[/u]
[/size][/color]
I'm going to post here three theorems without proof,i think that everyone can prove it.
At first we'll consider the function $ f(abc,ab\plus{}bc\plus{}ca,a\plus{}b\plus{}c)$,as one one variable function,with variable $ abc$ on $ R$ or $ R^{\plus{}}$.
[size=100]
[u][color=blue]
First Theorem:[/u][/color]
If the function $ f(abc,ab\plus{}bc\plus{}ca,a\plus{}b\plus{}c)$ is a monotonous function then it gets max. and min. values:
1.on $ R$ if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.on $ R^{\plus{}}$ if $ abc\equal{}0$ or $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
[/size]
[size=100]
[u][color=blue]
Second Theorem:[/u][/color]
If the function $ f(abc,ab\plus{}bc\plus{}ca,a\plus{}b\plus{}c)$ is a convex function then it gets maximum values:
1.on $ R$ if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.on $ R^{\plus{}}$ if $ abc\equal{}0$ or $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
[/size]
[size=100]
[u][color=blue]
Third Theorem:[/u][/color]
If the function $ f(abc,ab\plus{}bc\plus{}ca,a\plus{}b\plus{}c)$ is a concave function then it gets minimum values:
1.on $ R$ if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.on $ R^{\plus{}}$ if $ abc\equal{}0$ or $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
 [/size]
Maybe later i'll post consequences,that could be obtained from three theorems above. :wink:
Thank you.

[quote="quangpbc"]Uhm, you can find it on mathnfriend .org but not English :lol:[/quote]
Maybe you could translate it to english?Not for me,i know ABC theorem and examples,that were given there.I could also give here some examples and consequences. :wink:

This post is just a continue if ABC theorem :) .I know that there are 4 consequences:
[color=red][u] [size=150]
First Consequence:[/size][/u][/color]
[color=blue]
If $ f(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)$ is linear function,then $ f$ get maximum,minimum
1.On $ R$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.On $ R^{\plus{}}$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ or $ abc\equal{}0$.
[/color]
[color=red][u] [size=150]
Second Consequence:[/size][/u][/color]
[color=blue]
If $ f(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)$ is quadratic trinomial,then $ f$ get maximum,minimum
1.On $ R$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.On $ R^{\plus{}}$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ or $ abc\equal{}0$.
[/color]
[color=red][u] [size=150]
Third Consequence:[/size][/u][/color]
[color=blue]
If $ f$ is symmetric three variable polynomial,such that $ deg(f)\leq 5$,then $ f$ get maximum,minimum
1.On $ R$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.On $ R^{\plus{}}$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ or $ abc\equal{}0$.
[/color]
[color=red][u] [size=150]
Fourth Consequence:[/size][/u][/color]
[color=blue]
If $ f(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)$ is symmetric three variables polynomial which degrees $ \leq 8$,with nonnegative coefficient of $ a^{2}b^{2}c^{2}$,then $ f$ gets maximum
1.On $ R$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$.
2.On $ R^{\plus{}}$ if and only if $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ or $ abc\equal{}0$.
[/color]
First and Second Consequences could be easily obtained from 1st and 2nd theorems.The Third and Fourh are quite hard,but if you know vietnam language,then you can read their proofs in quangpbc's file.(download it)
I have not finished it yet.I'll come back soon,i promise. :wink:I would like to give here some examples,of using ABC theorem.

[quote="djuro"]Erken, thank you very much!
Well, you can write a pdf paper about ABC :)[/quote]
OK,i'll post it later  :wink:

[quote="Erken"]
Maybe you could translate it to english?Not for me,i know ABC theorem and examples,that were given there.I could also give here some examples and consequences. :wink:[/quote]
I think it have a different version by English, i don't have it now.But i will find it , then upload to forum :roll:

[quote="Luba1123"]May i use ABC theorem in IMO? :oops:[/quote]

I'm almost sure, without outlining the proof you aren't allowed - well you may provide a link, but you never know if this will be accepted.

[quote=Supercali]Isn't this very similar to $uvw$?[/quote]
In $uvw$ we can work also with $u$ and $v^2$.


a,b,c>0. prove that
3(a/b+b/c+c/a) \geq 9+ \sum (a-b) 2 /(a+c)(b+c)

[color=red][ Admin Edit: Corrected Version of the Ineq.][/color]

Sorry arsisi,but maybe your problem is not right completely.Let a=b and we will see this.Ok,maybe a=b=2 and c=3,your problem is not true. :?

[quote="arosisi"]the true problem is
3(a/b+b/c+c/a) \geq 9+ \sum (a-b) 2 /(a+c)(b+c).
thanks for your idea, anhcuong.[/quote]

This is my solution
After transforming the beginning inequality is equivalent with
 \sum cyc (a ^3 +3 (a ^3 c/b + a 2 c 2 /b +2a 2 c) )  \geq 5 \sum sym(a 2 b)+ 9abc  (*)

By the AM-GM inequality we have
a ^3 c/b + a 2 c 2 /b + ab 2  \geq 3a 2 c
a ^3 c/b + a 2 b 2 /c + ab 2  \geq 3a 2 b
...
(Similarly)
Thus  \sum cyc( 3(a ^3 c/b + a 2 c 2 /b +a 2 c)) \geq 9/2  \sum sym(a 2 b)  (1)
Another way a ^3 +b ^3 +c^3  \geq 1/2  \sum sym(a 2 b)  (2)
                    \sum cyc(3a 2 c) \geq 9abc  (3)
(1),(2),(3)=> (*) is correct

you can prove it by
let x=a/b,y=b/c,z=c/a then x*y*z=1
we have
 \sum x 2 y+3 \sum x 2 z+3( \sum x 2 )>= ...........

i don't think my problem has so many solutions. thank for yours idea
and here's my solution
3(a/b+b/c+c/a)  \geq  9+  \sum (a-b)  2  /(a+c)(b+c)
<=> 3(a/b+b/c+c/a) \geq 3 + (a+b)/(b+c)+(b+c)/(a+b)+(a+c)/(b+c)+(b+c)/(a+c)+(a+b)/(a+c)+(a+c)/(a+b).
we have another inequality
a/b+b/c+c/a \geq 1 + (a+b)/(b+c)+(b+c)/(a+b).
so we have our result  :D

Arosisi ,you prove your beautiful inequality from another inequality

a/b+b/c+c/a \geq 1 +(a+b)/(b+c) +(b+c)/(a+b)

but how do you prove this last?

Thank you, very much.

Can anyone give me, please, a help to solve Arosisi's inequality

a/b+b/c+c/a  \geq  1 +(a+b)/(b+c) +(b+c)/(a+b)      ?

Thank you very much.

i think we also have the inequality
a/b+b/c+c/a+1
\geq (a+b)/(b+c)+(b+c)/(a+b)+(a+c)/(b+c)+(b+c)/(a+c)
i'm not sure it's right
can you help me
thanks.  :D

let x=a/b,y=b/c,z=c/a then x*y*z=1
    =>3(a/b+b/c+c/a)  \geq  9+  (a-b) 2 /(a+c)(b+c).
     x ^2 y+3 \sum   x ^2 z+3(  \sum  x 2 )>= ....
because this I boring thanhliem for writer so I haven't time .sorry bacause prove  :D


Hee hee...interesting.  One of the former authors of Mandelbrot is The Great Rusczyk.  Your question is the equivalent of asking Titu Andreescu where you can find free AMC questions that aren't on his site (i.e. cheating).

They can be ordered from mandelbrot.org individually and now also in 2 books. In general if you want one problem from a contest, you can find it on my website in the math contests section, but that's mainly to give a sample of the level of difficulty rather than to give people large collections of problems to work on.

The two books are relatively cheap - nowhere near the $130 you say you spent on Educaide MCC (a think I used to lust for back when I was in mathcounts...but then I got a bunch of old nationals from the state coordinator).  If you're planning on doing mandelbrot, then go for it.  Or look up Midlevels, the middle school version, at http://www.midlevels.org.  Or at least it used to be.  Now all I can find is [url]http://mathforum.org/library/view/12238.html[/url].


Prove that for all positive integers n,

1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2

[hide="Solution"] [b]0. Dirty Work:[/b]
$ k^3$
$ \equal{} k \left( 2 \cdot \frac{k (k\plus{}1)}{2} \minus{} k \right)$
$ \equal{} \left( (1\plus{} \ldots \plus{} k) \plus{} (1\plus{} \ldots \plus{} (k\minus{}1)) \right) \cdot \left( (1\plus{} \ldots \plus{} k) \minus{} (1\plus{} \ldots \plus{} (k\minus{}1)) \right)$
$ \equal{} (1\plus{} \ldots \plus{} k) ^2 \minus{} (1\plus{} \ldots \plus{} (k\minus{}1)) ^2$

[b]1. Base Case:[/b]
$ 1^3 \equal{} (1)^2$

[b]2. Inductive Step:[/b]
Suppose $ 1^3 \plus{} \ldots \plus{} (k\minus{}1)^3 \equal{} (1\plus{} \ldots \plus{} (k\minus{}1))^2$

Add $ k^3$ to both sides: $ 1^3 \plus{} \ldots \plus{} k^3 \equal{} (1\plus{} \ldots \plus{} (k\minus{}1))^2 \plus{} k^3$

Use the identity $ k^3 \equal{} (1\plus{} \ldots \plus{} k) ^2 \minus{} (1\plus{} \ldots \plus{} (k\minus{}1)) ^2$ on the term $ k^3$ on the RHS of the above. You get:

$ 1^3 \plus{} \ldots \plus{} k^3 \equal{} (1\plus{} \ldots \plus{} (k\minus{}1))^2 \plus{} (1\plus{} \ldots \plus{} k) ^2 \minus{} (1\plus{} \ldots \plus{} (k\minus{}1)) ^2$

Simplify the RHS: $ 1^3 \plus{} \ldots \plus{} k^3 \equal{} (1\plus{} \ldots \plus{} k) ^2$

[b]3. Conclusion:[/b]
Because the statement is true for $ n\equal{}1$, and the truth of the statement for $ n\equal{}k\minus{}1$ implies the truth of the statement for $ n\equal{}k$, the statement must be true for all $ n \in \mathbb{N}$

[b]Note:[/b]
It is generally bad practice to induct from $ k\minus{}1$ to $ k$ -- it is better to induct from $ k$ to $ k\plus{}1$. I chose to induct this way here because the algebra looks cleaner in part [b]0[/b].
[/hide]


group theory
abstract algebra
superior algebra
superior algebra theorems

Let $ G$ be a group with $ N\lhd G$. Show that if $ M\lhd G/N$, then there is a normal subgroup $ M'\lhd G$ such that $ M'/N\cong M$.

$ M' \equal{} \{x \in G : xN \in M\}$ works.


geometry
geometric transformation
reflection
linear algebra
matrix
vector
linear algebra unsolved

I am trying to find a matrix transform that reflects a point P across the line y = mx + b in 2 dimensional space.

Based on this question, can we assume that the line goes through the origin, and requires a linear transformation?  If so, I think I can solve it.  Otherwise, an affine transformation would be necessary and I would be confused and not sure how to proceed.

Simply by rearranging the equation of the line, can I represent it as a 2x1 vector, and work it into the following equation:
[img]http://upload.wikimedia.org/math/c/5/a/c5a0e087e3e04bbc7d89724150fdc96c.png[/img]

Giving me a series of linear transformations...?

Any help, ideas or etc. would be greatly appreciated.

The line goes through the origin if and only if $ b \equal{} 0$, so no.  You do in fact need an affine transformation.  I feel as if this exact question has been posted recently.

In any case, what you've got there looks like (the negative of) a [url=http://en.wikipedia.org/wiki/Householder_transformation]Householder transformation[/url], which is one way to think about reflection, but you don't need to know what a Householder transformation looks like to work out what a reflection looks like.  Try to figure out what the translation part of the affine transformation is so that you reduce to the case of a linear transformation.

No clue...?

I don't know much about linear algebra.  So what your asking me to do is very difficult from my perspective.

Since a point is just a point (and has none of the properties of a vector), What am I supposed to do to it to project it onto the line?  Is the translation supposed to accomplish this?

Do I just use the values y = mx+b, x = y-b/m as my values in a translation matrix?

[quote="frankiecostello"]Since a point is just a point (and has none of the properties of a vector)[/quote]
I'm not sure why you think this.  For the purposes of this type of question you can regard points and vectors as equivalent.  (The equivalence is somewhat subtle, but we'll ignore this for now.)

The transformation you want is of the form $ \mathbf{x} \mapsto \mathbf{Ax} \plus{} \mathbf{t}$ where $ \mathbf{t}$ is a vector (the "affine" part of the affine transformation) and $ \mathbf{A}$ is a matrix.  You know what happens to the origin when it's reflected across $ y \equal{} mx \plus{} b$, hence you know what

$ \mathbf{0} \mapsto \mathbf{A0} \plus{} \mathbf{t} \equal{} \mathbf{t}$

is.  As for figuring out what $ \mathbf{A}$ is, it turns out that $ \mathbf{A}$ shouldn't depend on $ b$, but only on $ m$, so now you can assume that the line goes through the origin.  Can you figure out what $ \mathbf{A}$ looks like from here?

I'm not sure i get it.  The matrix A, should have the vector t (transformation vector) as a part of it...right?

[code]
                                           A              t
Ax + t => Ax + t(a,b) =>             { 1 0 a }
                                      { 0 1 b }   x   + t(a,b)
                                      { 0 0 1 }
                                        matrix          translation vector

All that remains now is solving a and b?  To do this we will represent the line segment, the point in question as matrices of homogeneous coordinates.  Than using projections we can solve T(a,b) by projecting the point onto the line, giving us values to represent a and b within the translation.  Is this correct?
[/code]


I have just installed MiKTeX and TexnicCenter, using Protext. The first thing I'd like to do is to open John Scholes's .tex file of IMO questions, which I got from [url=http://web.archive.org/web/20050216020605/http://www.kalva.demon.co.uk/imo/tex/imo-all.tex]here[/url].

When I open TexNicCenter, open the file, and then try to output the file in .pdf, .ps, or .dvi format, it tells me there are hundreds of errors with the Latex code. It's impossible to believe the code is so bad, so I must be doing something wrong! Any advice would be appreciated! Many thanks, Carrie

It's a TeX file not LaTeX so that's why you are getting the errors. There are a couple of ways to compile it:

1. Use the command prompt and type
tex imo-all.tex
and you will get imo-all.dvi whch you can view

[i]or[/i]

2. open imo-all.tex in TexnicCenter and at the start put

\documentclass{article}
\usepackage[margin=1in]{geometry}
\begin{document}

At the end replace

\bye

by

\end{document}

Then compile your new document to dvi, ps or pdf as you please.


I want to get a normal subscript to come after a labeled \underbrace.

Let's say I want to produce this:

$ \underbrace{\ldots}_{x}$ followed by a $ _2$

Normally, you'd think you'd use something like this:

[code]$\underbrace{\ldots}_{x}_2$[/code]

But this gives me a double subscript, and is therefore not processable.

$ {\underbrace{\ldots}_{x}}_2$ with {} round the underbrace part works, but is it the effect you want?

Is this what you are looking for????

\[ \underbrace{\ldots}_{x^2}\]

How about this?

$ \underbrace{\ldots}_{x}\,_2$

or

$ \underbrace{\cdots}_{x}\,_2$

[quote="googology101"][quote="stevem"]but is it the effect you want?[/quote]

Not really, but it's better than not getting it processed at all. ;)[/quote]As you see lots of ideas here but it's difficult to work out exactly what you want. If no-one has shown you what you want could you show us what you are looking for, even if you have to scan handwritten text?

What I wanted to write was $ \underbrace{\parallel{}\ldots\parallel{}}_ab\underbrace{\parallel{}\ldots\parallel{}}_a{}_c(d)$ (thanks to your answers, I can write it properly now.) My problem was that I couldn't get the subscript $ c$ to operate properly after the second $ \underbrace{\parallel{}\ldots\parallel{}}_a$.


Silly question, I know, but how does one say "AIME"?  (this is the problem with not knowing anyone /in person/ who has an interest in this sort of thing!)

I'm tempted to say "eye-may" b/c Spanish influence...
but I think most people say "amy"

I say the letters seperately.  Nobody pronounces AMC or USAMO as a single word, so why should AIME be pronounced that way?

I pronouce it as "amy". But I sort of don't like it.

u-sa-mo.  :D

I pronounce USAMO, USAMTS, AHSME, AJHSME as a single word. The only one I don't pronounce as a word is AMC, because its just three letters. Like the others, I say AIME as Amy.

AMC = A-M-C
AIME = Amy
USAMO = Use-Uh-Moe
AHSME = As-Me
AJHSME = Ah-Jiz-Me
MOSP = Mop (like the broomish thing)
IMO = I-M-O

A-M-C

A-I-M-E or Amy, I use both

U-S-A-Mo

M-O-P

T-S-T

I-M-O

BTW is there a standard, formal pronouciation for these abbreviations?

Ay Em See

Aim (I guess it's 6 years of studying French there)

You Ess Ay Em Oh

Yeah, I usually hear "amy" but i dunno, i don't really like it.  One of my teachers says "aim," but I usually stick to "amy" or just "A-i-M-E"

I always say "amy" but some of my friends say "I Me". But they are obvious self-centered =P.

We say:

AMC: A-M-C
AIME: A-I-M-E
USAMO: A-M-O (No need for US...)

AMY? as opposed to AIM one syllable. Knowing how lazy  :P  I am, I just go with one syllable.

I say:
AMC=A-M-C
AIME=Amy
USAMO=U-S-A-M-O or use-a-mo (soft "s" as in "in use")

I say:
AMC: A-M-C
Aime: Amy
USAMO: U-S-Aimo


[quote="ragnarok23"]Consider a painted $6\times 6$ cube which is broken completely into unit cubes. Let

 A= The number of unit cubes with one side painted.

 B= The number of unit cubes with two sides painted.

 C= The number of unit cubes with three sides painted.

Find A+B+C.[/quote]
[hide]Note that all of the unit cubes can be described this way other than the unit cubes on the inside. If you cut of the "edges" than you have a 4x4x4 cube with volume 64 giving you 152 cubes on the outside.[/hide]

[quote="ragnarok23"]Consider a painted $6\times 6$ cube which is broken completely into unit cubes. Let

 A= The number of unit cubes with one side painted.

 B= The number of unit cubes with two sides painted.

 C= The number of unit cubes with three sides painted.

Find A+B+C.[/quote]
[hide]$A=96$

$B=48$

$C=8$

$A+B+C=152$[/hide]

[quote="jhredsox"]i know how to do this, but is there a good formula?
-jorian[/quote]
The answer for an NxNxN cube would be [hide]$6n^{2}-12n+8$[/hide]

wow. ty. how do you know this stuff?
-jorian

[quote="13375P34K43V312"]Formula's are useless unless you understand them. I could recite the fundamental theorem of calculus, but do I know why it is, absolutely not!

bpms, are you sure about your formula, though? I think that's the number of cubes on the inside, not A+B+C.[/hide][/quote]
No, I am right, I just checked it. For the number of cubes on the inside it is [hide]$(n-2)^{3}$[/hide]. To jhredsox, leetspeak is right. These formulas are easy to derive when you understand the formula. What would you do if you memorized the formula I made, but then came across a problem that asked for B+C?

yah, i know, thats why i only know basic and above average formulas, but sometimes figure it out during a test, because its hard to remember too many variables
-jorian
for example, i know heron's, pyth, areas, perimeters, etc.

[quote="ragnarok23"]Consider a painted $6\times 6$ cube which is broken completely into unit cubes. Let

 A= The number of unit cubes with one side painted.

 B= The number of unit cubes with two sides painted.

 C= The number of unit cubes with three sides painted.

Find A+B+C.[/quote]

[hide="Answer"]A=96; B=48; C=8. When you add them up, you get 152.[/hide]

why are you answering the question if several people above already have?
-jorian

Because he did the problem, found a solution, and wants to post it? There's nothing wrong with it, the whole point of putting in hide tags is for when other people post their soloutions.

[quote="Walk Around The River"]ninja.   the formula for this is...

[hide]n^3-((n-2)^3)

so 6^3-((6-2)^3)=216-64=[b]152[/b][/hide][/quote]
I expanded mine...


Is it okay if I did part b for a certain problem first in order to solve part a? I'm sorry if this question has already been asked.

[quote="13375P34K43V312"]I think he means are you allowed to do part b then use some of the results from part b to solve part a, like 18.1.3.[/quote]
The answer is the same.  What else would he have meant?

I don't know...he might have meant the order in which he did the problems on his own?


Let $ a,b,c$ be non negative real numbers. Prove that
$ \sqrt{3(a^4\plus{}b^4\plus{}c^4)}\plus{}\frac{9(\sqrt6\minus{}1)abc}{a\plus{}b\plus{}c}\ge \sqrt6(ab\plus{}bc\plus{}ca)$

Let $ a,b,c \ge0$. Prove that
$ \sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}\frac{9kabc}{ab\plus{}bc\plus{}ca}\ge (k\plus{}1)(a\plus{}b\plus{}c)$ where $ k\equal{}\frac{\sqrt6}2\minus{}1$

Note that equality holds (both problem) for $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$  :)


geometry
rhombus
perimeter
symmetry
geometry solved

Diagonals AC and BD of a convex quadrilateral ABCD intersect at point O. Prove that if triangles OAB, OBC, OCD, ODA have the same perimeter, then ABCD is a rhombus. What happens if O is some other point inside the quadrilateral?

Very nice problem!

[quote="amir2"]Diagonals AC and BD of a convex quadrilateral ABCD intersect at point O. Prove that if triangles OAB, OBC, OCD, ODA have the same perimeter, then ABCD is a rhombus.[/quote]

The triangles OAB, OBC, OCD, ODA have the same perimeter. Denote this perimeter by p. Then, p = OA + OB + AB = OB + OC + BC = OC + OD + CD = OD + OA + DA. Hence, AB = p - OA - OB and CD = p - OC - OD, so that

  AB + CD = (p - OA - OB) + (p - OC - OD) = 2p - (OA + OB + OC + OD).

Similarly, BC + DA = 2p - (OA + OB + OC + OD). Hence, AB + CD = BC + DA, so that the quadrilateral ABCD has an incircle. Let this incircle touch the sides DA, AB, BC, CD of the quadrilateral ABCD at the points X, Y, Z, W, respectively. Denote by a = AX = AY, b = BY = BZ, c = CZ = CW and d = DW = DX the lengths of the tangents from the points A, B, C, D to the incircle of the quadrilateral ABCD. Then, since O is the point of intersection of the diagonals AC and BD of the circumscribed quadrilateral ABCD, we have $\dfrac{OA}{OC}=\dfrac{a}{c}$ (as I proved in http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/419.shtml ). Hence, there exists a real number k such that $OA=k\cdot a$ and $OC=k\cdot c$. On the other hand, OA + OB + AB = OB + OC + BC yields OA + AB = OC + BC, so that

  OA - OC = BC - AB = (BZ + CZ) - (AY + BY) = (b + c) - (a + b) = c - a.

In other words, $k\cdot a-k\cdot c=c-a$. In other words, $k\left(a-c\right)=-\left(a-c\right)$. Thus, either a - c = 0, or k = -1. Well, if k = -1, then $OA=k\cdot a=-a$, what is impossible since OA and a are both positive numbers. Hence, we must have a - c = 0, so that a = c. Hence, AB = AY + BY = a + b = c + b = CZ + BZ = BC. Similarly, BC = CD and CD = DA. Thus, AB = BC = CD = DA, so that the quadrilateral ABCD has all its four sides equal, i. e. it is a rhombus. This solves the problem.

[quote="amir2"]What happens if O is some other point inside the quadrilateral?[/quote]

Then, we cannot say anymore that the quadrilateral ABCD must always be a rhombus. In fact, let ABCD be a kite symmetric with respect to its diagoanl AC, and let O be a point on this diagonal such that the triangles OAB and OBC have the same perimeter (this point can be easily constructed: it must satisfy OA + OB + AB = OB + OC + BC, what is equivalent to OA - OC = BC - AB). Then, by symmetry, the triangles OCD and ODA also have the same perimeter, but not every kite is a rhombus. In general, a point O such that the triangles OAB, OBC, OCD, ODA have the same perimeter seems to exist for many circumscribed quadrilaterals ABCD (at least, one can find examples of such quadrilaterals ABCD which are neither rhombi nor kites).

  Darij

[quote="amir2"]and here is my solution.

we can consider that$OC\geq AO,OD\geq BO$let $P,Q$ be two points on $OC,OD$ such that $OP=OA,OQ=OB$.

so $ABPQ$ is parallelgoram hence $P\equiv C,Q\equiv D$ now we get that  $ABCD$  is paralelgoram.

from primeter equality of$\triangle ABO\ \text{and}\ \triangle BOC$ we can get that $ABCD$ is rhumbos.[/quote]

[b]CUTE!!![/b]


[quote="Together"]Find all $f: N^* \to N^*$ :
 $f(m+f(n) = f(f(m)) + f(n)$[/quote]
[quote="Fermat -Euler"]it's OIM 1996 day 1 problème 3  :?:[/quote]
sorry it's like OIM 1996
 (Find all $f: N \to N$ : $f(m+f(n) = f(f(m)) + f(n)$ )

It likes IMO 1996 but the solution for this problem like IMO 1996 ??
I don't think so.

[quote="Labiliau"]$m=0$ and $n=0$=> $f(f(0))=f(f(0))+f(0)$ => $f(0)=0$

$m=0$ => $f(f(n))=f(f(0))+f(n)=f(n)$

So, we have $f(x)=x$ for $x \in Im(f)$

But I have no idea to follow[/quote]

but you can't take $m=0$ and $n=0$
because $f: N^* \to N^*$ :


Hi, I  was just wondering if there is any high school math competition/contests between school math teams in Raleigh.  Something that's not as intense as Duke thing (which isn't really between SCHOOLS necessarily, I guess).

Also, I'm guessing you guys are in your school math club... what do you do in math club? just practice problems?

well, there's a mock-mathcounts competition btween 4 schools in Raleigh... :maybe:


for positif numbers x, y, z, show that

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\geq \frac{3}{2}$

come on  :lol:
$\sum_{cyc}\frac{x}{y+z}=\sum_{cyc}\frac{x^{2}}{xy+xz}\geq \sum_{cyc}\frac{(x+y+z)^{2}}{2(xy+yz+zx)}\geq \frac{3}{2}$ This is the famous Nesbitt. don`t forget it;)

U can also use Chebychev then C-S.
Or : Use the convexity of 1/x
Another method is to put: $y+x=c$   $x+z=b$   $z+y=a$ and it 'll become very easy  :wink:


function
trigonometry
analytic geometry
geometry
rectangle
vector
special factorizations

For a function $y=3\cos \theta+4\sin \theta \ (0^\circ \leq \theta \leq 90^\circ)$

(1) Find the range of $y.$
(2) Find the values of $\sin \theta ,\ \cos \theta$ when $y$ is maximized.
(3) Find the values of $z=3\sin 2\theta+4\cos 2\theta$ when $y$ is maximized.

[hide] instead of maximizing y, I'm going to change it to r so i can use polar coordinates.

1)
r = 3cos + 4sin
r = 3x/r + 4y/r
r^2 = 3x + 4y
x^2 + y^2 = 3x + 4y

well from that we can see it is a circle, but rather than completing the square we can just note that from the first equation the rectangular coordinates (3,0) and (0,4) are on the graph, and from that last equation so is (0,0) and since they are in a right angle, then (3,0),(0,4) is a diameter. So now we know that the diameter has length 5.  Ok now since we are trying to maximize r we just look for the point farthest from the origen,  but since its a circle the farthest distance is the diameter, so 5 is our maximum value. And we see that with 0<x<90 we have a minimum at 3 so the range is from 3 to 5

2) Ok well now we just draw in that diamater from the origen, and we can simply draw in the right trangle, with that hypotenuse which happens have (3,0) or (0,4) as its other vertex as it makes a rectangle, but well look at the one with (3,0). its clear to see that this is a 3-4-5 triangle so we get sin = 4/5 and cos = 3/4

3) ok well sin2x = 2cosxsinx, and cos2x = cosx^2 - sinx^2, from  this we get sin2x = 24/25 and cos2x = -7/25[/hide]

sorry its messy

[hide="Shortly..."] $(3^{2}+4^{2})(\cos^{2}\theta+\sin^{2}\theta) \ge (3 \cos \theta+4 \sin \theta)^{2}$ by Cauchy; our max is $5$ and our min is $-5$.  The rest is computation from the equality case. [/hide]

[quote="nateharman1234"]haha t0rajir0u's solution makes mine look ridiculous[/quote]

Yes, $y$ can't be take negative value.
This is Cauchy-Schwarz's trap.

Anyway you can use the transformation such as $a\cos \theta+b\sin \theta =r\sin (\theta+\alpha)$ or $r\cos (\theta-\beta)\ (r>0)$ or inner product.

What do you call this formula?

[hide=" Another Solution"]Consider scaling the equation by some factor $\ S$  $\ Sy= 3S \cos \theta+4S \sin \theta$.  Now we try to write the RHS as $\ \cos ( \theta+y )$ where $\ \sin y = 3S$ and $\ \cos y = 4S$.  So $\ 25 S^{2}= 1$ so $\ S= \frac{1}{5}$ and we can write $\ y= 5 \cos ( \theta+y )$ which has a range of $\ 3 \leq y \leq 5$ because we see $\ y$ must also fall in the range $\ 0 \leq y \leq 90$[/hide]

A useful formula (what [b]kunny[/b] was talking about, I believe):

For $a, b \geq 0$ we have $a\cdot \sin \theta+b \cdot \cos \theta = (\sqrt{a^{2}+b^{2}}) \cdot \sin( \theta+\arctan \frac{b}{a})$.

For this formula, if $a < 0$ then i think you add $\pi$ to the angle in the sine function.

So, for this problem we would have $4 \sin x+3\cos x = \sqrt{4^{2}+3^{2}}\cdot \sin(x+\arctan \frac{3}{4})$,  which has range $[-5, 5]$.

No one seems to be noticing $0^\circ\leqslant \theta\leqslant 90^\circ$

[hide="which gives"](1) $y\in[3,5]$
(2) $y_{max}=5\implies 5=5\sin(\theta_{0}+\arctan{3\over 4})\implies\theta_{0}={\pi\over 2}-\arctan{3\over 4}\implies\sin\theta_{0}={4\over 5},\cos\theta_{0}={3\over 5}$
(3) $z=5\sin(2\theta_{0}+\arctan{3\over 4})=5\sin\arctan{3\over 4}=5\frac{{3\over 4}}{\sqrt{1+{3^{2}\over 4^{2}}}}=3$[/hide]

[quote="mna851"]Both nateharman and myself gave that answer.[/quote]

That's right, sorry. Haven't noticed final answers.

[quote="kunny"]you can use the transformation such as $a\cos \theta+b\sin \theta =r\sin (\theta+\alpha)$ or $r\cos (\theta-\beta)\ (r>0)$ or inner product.

What do you call this formula?[/quote]

It's simply two ways of calculating the inner product.

Sorry, Farenhajt, I didn't notice $0^{\circ}\le \theta \le 90^{\circ}$!   :(

[hide="In any case..."] Then what we want to do is minimize the dot product of a unit vector and $\left< 3, 4 \right>$ on that particular interval.  This occurs at $\theta = 0$ giving our minimum of $3$. [/hide]

O.K. but no one has obtained a correct answer for [color=blue](3)[/color].

[hide="3"] By Cauchy's equality case we have

$\frac{ \cos \theta}{3}= \frac{ \sin \theta}{4}$

And of course this requires $\cos \theta = \frac{3}{5}, \sin \theta = \frac{4}{5}$.  Then

$\sin 2 \theta = 2 \cos \theta \sin \theta = \frac{24}{25}$
$\cos 2 \theta = \sqrt{1-( \sin 2 \theta)^{2}}= \frac{7}{25}$

And

$z = \frac{72+28}{25}= \boxed{4}$. [/hide]

err for 3...
[hide] I think that that should be -7/25 not 7/25, as 0<x<90  and since cosx = 3/5 < (sqrt2)/2  implies x > 45 so 2x > 90 and cos2x must be negative, so I get (72-28)/25 = 44/25[/hide]

Oh, of course.  Sorry.

[hide="In fact..."] $\cos 2\theta = \cos^{2}\theta-\sin^{2}\theta =-\frac{7}{25}$. [/hide]

I apologize for overlooking quadrants so often while solving this problem  :P

O.K. At last this problem has just finished! :)

To Japanese junior, highschool studets:

$\tan^{-1}\frac{3}{4},$ which is read by arc tangent $\frac{3}{4},$ means by $\alpha$ such that $\tan \alpha =\frac{3}{4}.$

You may not use this notation in Japanese university entrance exam.


Solve the following for $ x$

$ x(\sqrt {3 - 2x + \sqrt {5(1 - x^{2})}} + \sqrt {\frac {3}{2}}) = \sqrt {\frac {2}{3}}$

[hide]Note that we can take $ x = \cos (a)$, make some transformations and then use $ 4 \cos^{3}(a) - 3\cos(a) = \cos(3a)$[/hide]


In a certain year, January had exactly four Tuesdays and four Saturdays.  On what day did January $ 1$ fall that year?

\[ \textbf{(A)}\ \text{Monday} \qquad
\textbf{(B)}\ \text{Tuesday} \qquad
\textbf{(C)}\ \text{Wednesday} \qquad
\textbf{(D)}\ \text{Friday} \qquad
\textbf{(E)}\ \text{Saturday}
\]

[hide="Solution"]
To have $ 4$ Tuesdays and $ 4$ Saturdays in a month with $ 31$ days, we must get rid of the $ 3$ extra days, and we can only do this by going Wednesday-Thursday-Friday, so the month began on Wednesday, or answer choice $ \boxed{\textbf{(C)}}$.
[/hide]

C: To get exactly 4 Tuesdays and Saturdays, we need to get rid of 3 extra days. Two possibilities are Wednesday to Friday and Saturday to Monday, however the latter means an extra Saturday. Thus we can only start on [b]C) Wednesday[/b]

[quote=RoyalCactus][quote=PlaneGod]You can only do Wednesday-Thursday-Friday because if you tried to go the other way, it would be Saturday-Sunday-Monday, giving us an extra Saturday, which contradicts the problem. This is a good problem.[/quote]

Why did you bump an 11 year old thread :([/quote]

wait, its that old?


modular arithmetic
number theory proposed
number theory

Could  anyone  prove  that  for  every  prime   p  there  exists x , y , z  and \[ 0<w<p \] such  that

        \[ x^2 + y^2 + z^2 = wp \]

Is every number $x$ with $x \neq 7 \mod{8}$ not the sum of three squares?

But there ARE primes $\equiv 7\pmod{8}$, such as $23, 31, 47$ etc.

ok   this  problem  has  been  proposed  at  korea  2000   ;)  ;)
 but  i  think  that  it is not  so  easy  problem .I have  found  a  strange  proof .If  you think  that  the  problem  is easy  please  post  your  solution ...................... :P  :P

[quote="seamusoboyle"]Whats wrong with mine?[/quote]

  I think  that  you  wrote  $w=2$  but  there  exist  primes  like 17   $3*17 = 5^2 + 5^2 + 1^2$   ;)  ;)  :?:

  then  $w=3$ .................

@gdeskos: what is your post intended to say.

@seamusoboyle: prove the statement in your first post (in elementary way) or stop using it.

@dna: stop spamming "this is an easy problem"!

It is a theorem of LaGrange.
Ill look for a link.

Edit: Here it is.

http://mathworld.wolfram.com/LagrangesFour-SquareTheorem.html

I can see i might have to do a bit more with it, i didnt get it completely right.

The Four-Squares-Theorem is not what you need and use here...
But I wanted to say that you shouldn't use such hard theorems as the Three-Squares-Theorem for a lot easier problems.

Indeed, see http://www.mathlinks.ro/Forum/viewtopic.php?t=49271 for my prove of this.


Determine whether the series $ \sum_{n\equal{}1}^{\infty} \frac{n\minus{}1}{5n\plus{}1}$ converges or diverges. If it converges, find the sum. If it diverges, explain why.

n-th member of series does not tend to zero as n approaches infinity, so it diverges. It's kind of obvious, isn't it?


linear algebra
matrix
linear algebra theorems

I would like to prove this theorem:

[u]Theorem.[/u] Suppose T:V->W is a linear transformation and B, B' are bases' for V and W respectively. Let A the matrix of representation for T. Then:

                                                                                dim(ImT)=rk(A)

Any help would be handy.

Let $ crk(A), rrk(A), rk(A)$ denote column rank, row rank and rank of a matrix $ A$, respectively. But $ dim(im(T))\equal{}crk(A)$ by definition (just note that $ T(X)\equal{}AX$) and since $ crk(A)\equal{}rrk(A)\equal{}rk(A)$, we have $ dim(im(T))\equal{}rk(A)$.


geometry
real analysis
real analysis unsolved

Here's a neat covering lemma in $\mathbb{R}^{d}$ from Stein's book:

There exists an integer $N$ depending only on the dimension $d$ with the following property. Suppose $E$ is any bounded set in $\mathbb{R}^{d}$ that is covered by a collection $\mathcal{B}$ of balls in the (strong) sense that for each $x \in E$, there is a $B \in \mathcal{B}$ whose center is $x$. Then, there are $N$ sub-collections $\mathcal{B}_{1}, \mathcal{B}_{2}, \ldots, \mathcal{B}_{N}$ of the original collection $\mathcal{B}$, such that each $\mathcal{B}_{j}$ is a collection of disjoint balls, and moreover,

$E \subset \bigcup_{B \in \mathcal{B}'}B$, where $\mathcal{B}' = \mathcal{B}_{1}\cup \mathcal{B}_{2}\cup \cdots \cup \mathcal{B}_{N}$

My additional question: Can we get any kind of estimate for $N$ in terms of $d$?

[quote="mlok"]Don't remember the bound, but it is in the book by P. Mattila titled "Geometry of sets and measures in Euclidean spaces". An excellent book, by the way.[/quote]
I am now reading that book, and it is pretty good. Thanks for the recommendation! Anyway, the bound is in fact
$64^{n}N(n)+1$
where $N(n)$ is the constant [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112415]here[/url].

I am sure that 64 can be improved. Mattila gives some references about improving the bound, but I don't want to check too many books out of the library, and it's fun to at least try this kind of thing before giving in and reading someone else's solution.


The squares of a $2000*2000$ board are filled with $1$ and $-1$ in such a way that the sum of the numbers in the board is not negative. Prove there exist $1000$ columns, and $1000$ rows such that the sum of the numbers in their intersection is at least $1000$.


inequalities
trigonometry
inequalities proposed

Let $ x,y\in R; x \plus{} y\ne 0$ and $ x^4 \plus{} y^4\le \frac {1}{2}$

Prove $ x^2 \plus{} y^2\ge \frac {2xy}{xy \plus{} \sqrt {(x^2 \minus{} 1)(y^2 \minus{} 1)}}$

Sorry for my bad english

Very easy :) .

Inequality is equivalent with

 $ x^2(1 \minus{} x^2) \plus{} y^2(1 \minus{} y^2)\ge 2xy\sqrt {(1 \minus{} x^2)(1 \minus{} y^2)}$ wich is true.

[quote="gigaman"]Let $ x,y\in R; x \plus{} y\ne 0$ and $ x^4 \plus{} y^4\le \frac {1}{4}$

Prove $ x^2 \plus{} y^2\ge \frac {2xy}{xy \plus{} \sqrt {(x^2 \minus{} 1)(y^2 \minus{} 1)}}$

Sorry for my bad english[/quote]
I will prove it when $ x^4 \plus{} y^4 \le \frac{1}{2}$ ! :D
Then $ \frac{1}{4}\ge \frac{x^4\plus{}y^4}{2} \ge \left ( \frac{x^2\plus{}y^2}{2} \right)^2 \iff x^2\plus{}y^2 \le 1$.
Let $ x^2 \plus{} y^2 \equal{} t^2 \in [0;1]$. Then $ (\frac{x}{t}, \frac{y}{t})$ is a point on the unit circle. Hence $ \exists \theta$ such that:
$ x \equal{} t \cdot \cos \theta, y \equal{} t \cdot \sin \theta$. Inserting gives:
$ t^2 \ge \frac{t^2 2\sin 2\theta}{t^2 \sin 2\theta \plus{} 2\sqrt{(1\minus{}t^2\cos^2\theta)(1\minus{}t^2\sin^2\theta)}} \iff$
$ 2\sqrt{(1\minus{}t^2\cos^2\theta)(1\minus{}t^2\sin^2\theta)} \ge (2\minus{}t^2)\sin 2\theta \iff$
$ 4(1\minus{}t^2\cos^2\theta)(1\minus{}t^2\sin^2\theta) \ge (2\minus{}t^2)^2\sin^2 2\theta \iff$
$ 4 \minus{} 4t^2 (\cos^2\theta \plus{} \sin^2\theta) \plus{} t^4 \sin^2 2\theta \ge (2\minus{}t^2)^2\sin^2 2\theta \iff$
$ 4(1\minus{}t^2)(1\minus{} \sin 2\theta) \ge 0$, which is obviously true.

@Marius Mainea: Can you explain how you got the equivalence?

[quote="Mathias_DK"]I will prove it when $ x^4 \plus{} y^4 \le \frac {1}{2}$ ! :D
[/quote]

Sorry, I just fix it  :lol:

Anyone have nice solution for this problem


analytic geometry
linear algebra
matrix
real analysis
real analysis theorems

If S is a connected k-dimensional surface in R(n) then my book says we can orient S by choosing any local chart on S or fixing a frame in any of its tangent planes.

However from the definition of orienting a surface as fixing one of its orientation classes of atlases (or, equivalently, choosing an orienting atlas) I don't see how this is equivalent.

[u]If the surface is orientable[/u], then its orientation can be determined by looking at just one chart. Indeed, the two possible orientations would produce opposite orientations of that chart.

Thanks mlok. I was able to prove it yesterday but my proof is not so simple. I got a hint that we are supposed to consider connected components of the surface but I don't know what a connected component is.

You could [url=http://mathworld.wolfram.com/ConnectedComponent.html]look it up[/url].

I forgot to note that S is oriented, as mlok pointed out.

If a local chart $ \varphi _1$is specified for $ TS_x$ then designate the orientation it defines to be the equivalence class of orienting atlases such that the local charts of each of its elements are consistent with $ \varphi _1$. It is clear that this orientation class exists and is unique since S has exactly two orientations.

If a frame in $ TS_x$is fixed then choose a coordinate chart $ \varphi _1$containing x. If the columns of the Jacobi matrix of $ \varphi _1$are consistent with the fixed frame then we designate that $ \varphi _1$defines the same orientation as the frame. If not we designate that it defines the opposite orientation to that given by the frame. Now apply the above convention.

One other approach for the latter is to note that the continuous fields of frames on S divide into two equivalence class if we introduce a relation between any two fields of frames which are consistent at every point. Hence S could be oriented by specifying a field of frames and providing a convention. It follows even that S could be oriented by specifying a frame at one point.


Hello friends!
I just completed my two years here at mathlinks!
It was like, um, a remarkable first one month (I fell in love with AoPS then, the board style still lingers in my fond memory), followed by a struggling few months or so probably, then a surrender when I saw it was hopeless to try and return Games and Fun Factory to the way it was in the spring of 2006.
For a long time I boasted the highest number of posts in the G&FF (even ahead of valentin himself lol) though probably the babbit kid (1=2 for you) overtook me, and then I grew kinda old for these games i guess.
Sigh.
Anyway, congratulate me guys!
It's a birthday of sorts you know...

Haha nice! I'm only about a year and 4 or so months  :) . My join date was Friday, October 13th, 2006  :| .

I'm celebrating my 2-year this May.  Happy birthday!


induction
combinatorics solved
combinatorics

Consider w a permutation of the finite set {1,2,...n}.
For each w we perform the following operation: if the first term is k, we reverse the first k terms. Prove that after finitely many operations we will reach a set with first term 1 (so no more operations can be performed).
just interested, are there any valid generalizations when dealing with other sets rather than {1,2...n} - maybe {1,3,5...2n+1}?

For the initial question :
First note that if the number $n$ is on the last place then it will never be used by any operation.
Thus we prove the result by induction on $n$.
It is trivial for $n=1$, now suppose that the result holds for $n-$.

Consider the sequence of operations (the sequence does not depend from us...).

- If the number $n$ appears on the first place during the sequance of operations :
Then, it will be at the last place at the end of the considered operation. Thus, from now, we only have to consider the first $n-1$ terms of the permutation, since from above $n$ will stay at the last place forever.
The induction hypothesis ensures us that $1$ will appear at the first place at some time.

- If the number $n$ never appear at the first place. Let $p$ be the number which appears to the last place of the sequence at the beginning. In this case, the number $p$ will never move (it moves only if $n$ appears at the first place). In particular, it will never reach the first place.
Since the value of a number has only interest for the sequence if it appears at the first place after some operations, it follows that we may interchange the values of $p$ and $n$ without disturbing the sequence. The induction hypothesis now ensures that the conclusion holds again.

Pierre.


Just wondering, what's your rating?

my rating's 2.9... :( People rate my aweseome posts spam for some reason...although sometimes I cannot resist the urge to spam when I have a reall brilliant thing to say.

3.2.

>_> I remember that on Treething, my rating was 1.0 for a very very VERY long time.

[quote="13375P34K43V312"]my rating's 2.9... :( People rate my aweseome posts spam for some reason...although sometimes I cannot resist the urge to spam when I have a reall brilliant thing to say.[/quote]

=P 3.4....it used to be 3.9 but then someone rated my post (i don't know which post nor what rating) and then it dropped to 3.4

4.2

One-tenth of the Answer to Life, the Universe, and Everything!!! :w00tb:

ok what the heck this is EXACTLY what i'm talking about someone rated my original post a 1...

mine's 2.9 and hasn't changed since like the start of the ratings.

3.2... Hasn't changed for a while.

@ Leet - Ratings in G&FF don't count.

[quote="jmerry"]5.6. Well-explained solutions in the college forum tend to get high ratings.[/quote]
:winner_first:

I wonder if anybody has a higher rating than that...?

From the college forum, both Kent Merryfield and fedja should have quite high ratings. Grobber and harazi also.
Anyway, the people with high ratings are not that hard to spot.

When the ratings system began, I had a rating of 5. After some time, it dropped to a stunning
[hide="(fasten your seatbelts)"]4.9[/hide]
and has remained there ever since.

[quote="13375P34K43V312"]my rating's 2.9... :( People rate my aweseome posts spam for some reason...although sometimes I cannot resist the urge to spam when I have a reall brilliant thing to say.[/quote]
Maybe people got used to your spams, eventually getting sick of them. And when they see a post of yours...

[i]Who's that? 13375P34K43V312? Aahhh... that bloody spammer!
*** gives your awesome post a not so awesome rating of 1 ***
I feel much better now :D[/i]

You see what I mean?

when you post, look at your avatar, then look below your "location" , the "rating:" should be somewhere below

[quote="spell it out 11"]my rating is 6.9[/quote]

the max rating you can have is a 6 :roll:

[quote="#H34N1"]the max rating you can have is a 6 :roll:[/quote]
Not really, I once had 6.5 ... those were the good days.

[quote="#H34N1"]No. How can you have a rating of 6.5 if there are only 6 cubes? That means that there is some weird bug...[/quote]
Could you count them* again?

* the cubes

For those people that are going to rate this post as spam

Guess what spam in the games and factory section doenst count. :P

[quote="#H34N1"]No. How can you have a rating of 6.5 if there are only 6 cubes? That means that there is some weird bug...[/quote]

BECAUSE HES SPECIAL

[quote="mz94"]dingzhou has a 7.4 rating...

mine is 2.0 :D

and im proud of it!!!!!!!![/quote]

you beat me: 1.7

._.
Why did you revive a 7-year-old thread?!

Also, they don't mean Alcumus rating, they are talking about a very old thing that got removed from AoPS.


A is the set of positive integers and B is $ A\cup {0}$. Prove that no bijection $ f: A\rightarrow B$ can satisfy $ f(mn) \equal{} f(m) \plus{} f(n) \plus{} 3 f(m) f(n)$ for all $ m, n$.

A hint :
$ g(n)\equal{}3f(n)\plus{}1$ then
$ g(mn)\equal{}g(m)g(n)$
Prove $ g(n)$ can't be a biject function from N to $ A_{3k\plus{}1}$ where
$ A_{3k\plus{}1}\equal{}\{x\in N |x\equiv 1 (\mod 3)\}$

Using the hint: First, $ g(1)\equal{}g(1)^2\Rightarrow g(1)\equal{}1$.

Assuming $ g$ onto, there exists $ p$ such that $ g(p)\equal{}4$. If $ p$ were composite, $ p\equal{}ab$ with $ a,b>1$, then $ g(a)$ or $ g(b)$ would have to equal 1, a contradiction. So $ p$ prime.

Similarly, there exists $ q,r$ such that $ g(q)\equal{}10,g(r)\equal{}25$ and $ q,r$ prime. So $ 100\equal{}\cdots$ giving $ q^2\equal{}pr$ and then $ p\equal{}q\equal{}r$, an obvious contradiction.


Support
probability
expected value
probability and stats

A student is growing algae in five different beakers with water from each of the Great Lakes. He wants to know if algae grows  better in some lakes... or is the rate of growth basically the same.

Support or refute the null hypothesis: "There will be no significant difference in algae growth among the five samples of water from the Great Lakes."

An equal mass of algae (beginning with 5g) is grown in each of the five beakers for a period of 12 weeks. The amount of algae per
cubic centimeter (cc) is recorded for each beaker below:

Huron: 43g
Ontario: 58g
Michigan: 72g
Erie: 64g
Superior: 63g

My question is, how do you compute the chi-square here? I don't really know what the expected outcome and I'm a little confused about the problem statement.

Any insights or advice would be appreciated.

Hi mathemagician1729.

I'm not sure about this one, but perhaps it makes sense to take the mean average of the observed outcomes as the expected outcome. This works out nicely as $ 60$. Then use $ X^{2} \equal{} \sum \frac{(O\minus{}E)^{2}}{E}$.

Thanks, I was sort of thinking towards that end because the expected value should be calculated by assuming the null hypothesis to be true, (e.g, the water in all the lakes do [i]not[/i] significantly affect the growth of algae.) Although I would like to say we could simply average it, I'm not completely certain we can do that because the problem is referring to growth.

Sorry, I didn't read the question carefully: perhaps subtract 5 from each value first (to get growth) and proceed as before.


Let $a,b,c$ be positive real numbers such that $21ab+2ac+8bc\leq 12$ . Find the minimum value of:
 $f(a,b,c)=\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$


$ \mathop x\nolimits^y  \plus{} \mathop y\nolimits^x  > 1(0 < x,y < 1)$
 :D

I've seen this problem a few times before. It works for $ x,y\ge1$ as well. :D

That has probably been posted here before, it looks familiar.

Yes, it's very clearly if x,y>1. :)But can you solve it when x,y<0

It looks familiar much because it's from France . I don't remember exactly the year .
However , Bernouli's inequality works  :)

[quote="Yongyi781"]I'm not quite sure this works...
With $ x \equal{} 3$ and $ y \equal{} \minus{} 3$...
\[ 3^{ \minus{} 3} \plus{} ( \minus{} 3)^3 \equal{} \frac 1{27} \minus{} 27 \equal{} \minus{} \frac {728}{27} < 1
\]
:maybe: And with $ x, y < 0$, if we have $ x \equal{} \minus{} 3$ and $ y \equal{} \minus{} 3$...
\[ ( \minus{} 3)^{ \minus{} 3} \plus{} ( \minus{} 3)^{ \minus{} 3} \equal{} \minus{} \frac 2{27} < 1
\]
[/quote]

$ 0<x,y<1$ means $ 0<x<1$ and $ 0<y<1$

Lazarus, TRAN THAI HUNG posted another problem which is that we solve the inequality when $ x,y<0$.

No, TRAN attempted to suggest that people actually solve the original question (since solving the question for either of $ x, y > 1$ is trivial) but accidentally wrote $ < 0$ instead of $ < 1$.  This has been an awful lot of thread with no attempt at a solution so far :) (unless you count HTA's comment).  Luckily, the two words "bernoulli" and "france" from HTA's post are enough to do a search and find http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204371


[quote="mathnerd314"]How many ordered pairs (x, y) are there such that both x and y are prime, and x+y=999?[/quote]

[hide]
Only 2. All primes are odd, except for the number 2. Therefore, the only ordered pairs are (2,997) and (997,2), because $\text {odd + odd = even}$.
[/hide]

:thumbup:  :10:  :clap2:

Well now, that wasn't so bad, was it? Does anyone else have another easy problem if one thinks about it? I'm bored.

How about this?
Taken from the 2005 handbook warm-ups:
->What is the probability of randomly picking a vowel out of the word "CAT"?<-
 :D

That's pretty hard if you don't know what vowels are.

Even and odd numbers solutions are so familiar to me because I prove stuff with that a lot.

[quote="shinwoo"]

chess64's solution is absolutely correct but one flaw

when he said all primes are odd, that's not true

2 is an even number and a prime number at the same time

though the answers are correct[/quote]

He said that didn't he?

[quote="chess64"]
[hide]
Only 2. All primes are odd, except for the number 2. Therefore, the only ordered pairs are (2,997) and (997,2), because $\text {odd + odd = even}$.
[/hide][/quote]

Yep. There is a 0 in between all this. The answer is 0.

[quote="dwx314"][quote="shinwoo"]:spam:  :spam:  :spam:  :spam:  :spam:  :spam:  :spam:  :spam:  :spam:  :spam:

you guys are insane!

you guys are spamming!!!

noone actually answered the problem except chess64

chess64's solution is absolutely correct but one flaw

when he said all primes are odd, that's not true

2 is an even number and a prime number at the same time

though the answers are correct[/quote]hey, I was not trying to spam. He asked for an easy prob, I gave him one. (though it's easy from the start, and you don't have to think about it at all). As for the guy before me though...

Also, he did said that all primes were odd, EXCEPT FOR 2.

As for an actually easy prob when you kinda think about it:

40*39*38*...*-38*-39*-40[/quote]
0

Not necessarily 0. The sequence

40*39*38*37*...*-37*-38*-39*-40

could or could not include 0. If it doesn't for some reason, then the answer is

(40!)^2

which is a number so ugly I don't want to even think about it.

As for :spam: ming, sorry Shinwoo, but I did ask for easy problems if one thinks about it because I'm bored, that is, problems that require a "trick", are solved in five minutes, and cause people's palms to strike the front of their heads.

Try this one.

Evaluate WITHOUT A CALCULATOR:

(1 - 1/4)(1 - 1/9)(1 - 1/16)...(1 - 1/400)

and SHOW YOUR WORK.

that would be:

$\frac 34 \cdot \frac 89 \cdot \frac {15}{16} \cdot \frac {24}{25} \cdot \frac {35}{36} \cdot \frac {48}{49} \cdots \frac {360}{361} \cdot \frac {399}{400}$

Each term is in the form $\frac {n^2-1}{n^2}$ which simplifies to $\frac{(n-1)(n+1)}{n^2}$

we start at $n=2$ and we see that each $n^2$ in the denominators is cancelled out by the $n+1$ on the left side and the $n-1$ on the right. This leaves us with:

$\frac 12 \cdot \frac{21}{20} = \boxed{\frac{21}{40}}$

I think that's right...

:10: :10: :10:

So you get the idea. Any other problems?

[quote="dwx314"]what is the sum of the following geometric sequence:
1/x+1/x^2+1/x^3+...+1/x^(infinity)[/quote]

$S=\frac{1/x}{1-1/x}=\frac{1}{x[(x-1)/x]}=\frac{1}{x-1}$

all prime numbers above 2 are odd. Odd + odd is even. So the answer is 1. [/code]

that was solved a while ago...
but its 2 because you can reverse it


if$ABCD$is as convex quadrilateral with $<ADC=30$°and $AB+BC+CA=BD$prove that $BD$bissects$<ABC$

Let $O$ the circuncentre of the triangle $ACD$. The triangle $ACO$ is equilateral.
For Ptolomeo's inequality in the cuadrilateral $ABCO$ we obtain: $AB+BC\leq BO$.
So $BD=AB+BC+CA\leq BO+OD\leq BD$.
Then, $ABCO$ is cyclic and $B, O , D$ are collinar.
In particular, $<ABD= 60^{0}=<DBC$.