Solve in integers:

x4 + y4 + z4 - 2x2y2 - 2y2z2 - 2z2x2 = 2000.

are you sure it's not from Greece 2000, looking at the numbers :D

Yep, absolutely sure - it's from a selection examination for the JBMO 2001. :D


A circular ceiling fan rotates at a constant speed of $ 80$ revolutions per minute.  A point halfway between the center of the fan and the outer edge of the fan travels $ 97968$ inches in $ 15$ minutes.  How far does a point on the outer edge of the fan travel in $ 30$ minutes?

To find the distance the point on the outer edge of the fan travels in thirty minutes, we need to know the diameter of the fan. Therefore, we need to find the diameter of half the fan then multiply that quantity by two. So $ 97968\div15\div80 \equal{} 81.64$ which is the pi times the diameter of half the fan. To find the pi times the diameter of the whole fan you do $ 81.64\times2 \equal{} 163.28$. And since it travels 80 revolutions in a minute, and it makes 30 revolutions $ 163.28\times80\times30 \equal{} 391872$.


induction
number theory proposed
number theory

Prove that for any $ n\in\mathbb{N}_0$, $ n\ge2$, the equation $ 1\plus{}x_1^2\plus{}x_2^2\plus{}\ldots\plus{}x_n^2\equal{}y^2$ has infinitely many solutions in positive integers with $ 1<x_1<\ldots<x_n$.
[i]Adapted after D. Mihet[/i]

It seems easy modulo 4
Observation:
We know that
$ y^2\equal{} 0,1 [4]$
$ 1\equal{} 1[4]$,
 so,
$ x_1^2 \plus{}x_2^2 \plus{}...\plus{}x_n^2 \equal{}0 [4]$
or
$ x_1^2 \plus{}x_2^2 \plus{}...\plus{}x_n^2 \equal{}3 [4]$
for $ n\equal{}2$
$ 1\plus{}x_1^2 \plus{}x_2^2 \equal{} 1 \plus{}0 [4]$, so $ x_1^2 \equal{}0 [4]$ and $ x_2^2 \equal{}0 [4]$,
$ n\equal{}3$
1 case
 $ 1\plus{}x_1^2 \plus{}x_2^2 \plus{}x_3^2 \equal{} 1\plus{}1\plus{}1\plus{}1 [4]$
$ x_i^2\equal{} 1 [4]$, for  $ i\equal{} 1,2,3$
or
2 case
$ 1\plus{}x_1^2 \plus{}x_2^2 \plus{}x_3^2 \equal{} 1\plus{}0\plus{}0\plus{}0 [4]$
$ x_i^2\equal{} 0 [4]$, for  $ i\equal{} 1,2,3$
So for $ n\equal{}4k\minus{}1$ $ k\in N$ solutions exist, taking $ x_i^2\equal{}1[4]$, $ i\in N$
and for all n, taking $ x_i\equal{}0[4]$.
Since there is an infinite number of perfect squares there is an infinite number of solutions.

[quote="Kabor Suk"]It seems easy modulo 4
Observation:
We know that
$ y^2 \equal{} 0,1 [4]$
$ 1 \equal{} 1[4]$,
 .....etc etc etc
[/quote]

Is just me or this proves nothing related to our problem ?

I manged to prove it for n=2 .Is this helpfull ? I tried using induction but I failed.Can someone provide a hint or a solution ?

We can just take almost any $ x_1 < x_2 < ... < x_{n\minus{}1}$ with $ x_1^2\plus{}...\plus{}x_{n\minus{}1}^2$ even and then $ x_n \equal{} \frac{x_1^2\plus{}...\plus{}x_{n\minus{}1}^2}{2}$ works.


search
number theory theorems
number theory

who can give me a simple proof of Cauchy-Davenport theorem?Now I have got a monstrous proof,which I don't really understand.

You could just search for "Cauchy-Davenport" theorem on ML. You will find threads like

http://www.mathlinks.ro/Forum/viewtopic.php?t=19496 ;
http://www.mathlinks.ro/Forum/viewtopic.php?t=5766

and some others. The second one is supposed to give an elementary proof (but I have not checked it).

  Darij

Thank you,Darij.But I have searched the forum before I posted the topic,and I have also downloaded the attachment,the solution is just from the attachment,but it is too hard for me to understand.What else,I have seen that the theorem was used in Romania or Bulgaria's MO.


limit
logarithms
trigonometry
calculus
calculus computations

[hide="answer"]
$ e$
[/hide]

[hide="solution"]Let $ t\equal{}\sin x$, if $ x\to 0^\plus{}$, then $ t\to 0^\plus{}$. So we can calculate easier limit:

$ \lim_{t\to 0^\plus{}}\frac{1}{(1\plus{}\frac{1}{t})^{\frac{1}{\ln t}}}$

Becouse:

$ \lim_{t\to 0^\plus{}}\left(1\plus{}\frac{1}{t}\right)^{\frac{1}{\ln t}}\equal{}\lim_{t\to 0^\plus{}}\exp\left(1\plus{}\frac{1}{t}\right)^{\frac{1}{\ln t}}\equal{}\lim_{t\to 0^\plus{}}\exp\frac{\ln (1\plus{}\frac{1}{t})}{\ln t}\equal{}$

$ \equal{}\lim_{t\to 0^\plus{}}\exp\frac{\ln (t\plus{}1)\minus{}\ln t}{\ln t}\equal{}\lim_{t\to 0^\plus{}}\exp\left(\frac{\ln (t\plus{}1)}{\ln t}\minus{}1\right)\equal{}\exp (0\minus{}1)\equal{}e^{\minus{}1}$

So:

$ \lim_{t\to 0^\plus{}}\frac{1}{(1\plus{}\frac{1}{t})^{\frac{1}{\ln t}}}\equal{}\frac{1}{e^{\minus{}1}}\equal{}e$[/hide]


IMO scores :

[url]http://www.srcf.ucam.org/%7ejsm28/imo-scores/2004/scores-code.html[/url]

see "LTU6"  :D  :D  :D

how is it possible, lol ? :?

(note : only 11 people has solved the problem 3 and, by example, we were able to get 1 point with "P(x) = P(-x)", then ? ;) )[/url]

An interesting Lithuanian result for an Estonian problem.

  dg

Apparently it was well-known in that region. I think I had seen alike stuff before too...

LTU6 was amazing...  :) But I am almost sure, that LTU6 didn't see that or similar problem before. Such type problems are not so popular here.

Talking about LTU 6, there was a guy, (I think from USA, don't remember)  that used to get the following score: 0-0-7, 0-0-7 and they used to call him James Bond.
Does anyone knows the name of james bond?  ;)
By the way this guys are great. they don't care about the easy problems..heh

I'm wondering if they were doing this on purpose to catch more attention, well. just kidding :D

[quote="wpolly"]I wonder if LTU6 had gone to IMO before.......

In China there was once a guy named Li Xin who got into CMO in grade 7(!) and been choosed into the Chinese Team in grade 9. (That was in 1998 so he didn't went to IMO). He then went to both IMO1999 and 2000 and got two gold medals (36p and 38p). In CMO2001 he got tired of going to IMO and did just problems 3 and 6 of CMO2001. (and CMO2001 is considered by most people to be the second-hardest CMO in the history, the hardest one being CMO2004...............)[/quote]So this means that this year's team (for Mexico) has been already chosen? ...  :?

To make the story more complete - LTU6 ended up with silver two years later.

[quote="Valentin Vornicu"]Who is he?[/quote]

LTU6 = Daumilas Ardickas


[color=darkblue]Find $ n \in N$, know that: $ \boxed{\frac{1}{2}\displaystyle \left ( 1\plus{}\frac{1}{1.3}\right ) \displaystyle \left ( 1\plus{}\frac{1}{3.5}\right ) ... \displaystyle \left ( 1\plus{}\frac{1}{n(n\plus{}2)}\right ) \equal{} \frac{2007}{2008}}$[/color]

Obviosly $ \boxed{\frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right ) \equal{} \frac {n\plus{}1}{n\plus{}2}}$[/color]

[color=darkblue]You can slove:
$ \boxed{\frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right ) \equal{} \frac {n \plus{} 1}{n \plus{} 2}}$ [size=200]?[/size][/color]

[color=darkblue]Hi, Rust. I tried but don't get any solution. The aftermath you post is not true: $ \boxed{\frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right ) \equal{} \frac {n\plus{}1}{n\plus{}2}}$

If $ n \equal{} 3$ $ \Longrightarrow \frac{1}{2}.\frac{4}{3}.\frac{16}{3.5} \not\equal{} \frac{4}{5}$.

Can you help me?[/color]

You're missing the n=2 term. Or you need to fix your notation.

$ a_n\equal{}\frac{1}{2}\prod_k^n(1\plus{}\frac{1}{k(k\plus{}2)})\equal{}\frac 12 \prod_k^n\frac{(k\plus{}1)^2}{k(k\plus{}2)}$.
Chek $ a_1\equal{}\frac 23\equal{}\frac{1\plus{}1}{1\plus{}2}$.
If $ a_{n\minus{}1}\equal{}\frac{n\minus{}1\plus{}1}{n\minus{}1\plus{}2}\equal{}\frac{n}{n\plus{}1}$, then $ a_n\equal{}a_{n\minus{}1}\frac{(n\plus{}1)^2}{n(n\plus{}2)}\equal{}\frac{n\plus{}1}{n\plus{}2}.$

[quote="Rust"]$ a_n \equal{} \frac {1}{2}\prod_k^n(1 \plus{} \frac {1}{k(k \plus{} 2)}) \equal{} \frac 12 \prod_k^n\frac {(k \plus{} 1)^2}{k(k \plus{} 2)}$.
Chek $ a_1 \equal{} \frac 23 \equal{} \frac {1 \plus{} 1}{1 \plus{} 2}$.
If $ a_{n \minus{} 1} \equal{} \frac {n \minus{} 1 \plus{} 1}{n \minus{} 1 \plus{} 2} \equal{} \frac {n}{n \plus{} 1}$, then $ a_n \equal{} a_{n \minus{} 1}\frac {(n \plus{} 1)^2}{n(n \plus{} 2)} \equal{} \frac {n \plus{} 1}{n \plus{} 2}.$[/quote]

Thank Rust very much. :P


HI!

Here is another "difficult" problem:

--------------------------------------------------------------------------------------------------------------
Let ABC be a triangle. Chose X a point onto AB, Y a point onto BC, and Z a point onto AC.
Draw the triangle XYZ. Then the triangle ABC contains 4 triangles (AXZ,BXY,CYZ and XYZ).
Prove that among this 4 triangles, XYZ is never the triangle with minimum area!!!
--------------------------------------------------------------------------------------------------------------

Thank you in advance! This forum is very great

See http://www.mathlinks.ro/Forum/viewtopic.php?t=48583 .

  darij

The proof given here is great

http://www.mathlinks.ro/Forum/viewtopic.php?t=48583

The following step in the proof seems obvious

"If, on the other hand, all three pairs in (*) intersect, then the area of DEF will be at least as large as that of A'B'C' (this should be an easy exercise; just move the points D,E,F around on BC,CA,AB and see what happens)"

But if we wish to prove it, how can we do it?

I mean that I feel this result, but I don't see how to give a real proof...

Thanks !


Studiati convergenta sirului definit prin relatiile $ a_{1}\equal{} 2$ , $ a_{n\plus{}1}\equal{} 2\plus{}\frac{1}{a_{n}}$

[color=darkred]Exista o corespondenta intre siruri recurente de forma $ x_{n\plus{}1}\equal{}\frac{ax_{n}\plus{}b}{cx_{n}\plus{}d}$ si puterile naturale ale unei matrici de ordinul 2 ! Voi reveni curand.[/color]

Domn' profesor am incercat in acest fel ,in ideea de a determina termenul general,dar este mult de calcul.
Am gasit o alta solutie :
Fie $ f(x)\equal{}2\plus{}\frac{1}{x}$
$ |f(x)\minus{}f(y)|\equal{}|\frac{1}{x}\minus{}\frac{1}{y}|\equal{}\frac{|y\minus{}x|}{|xy|}\geq\frac{1}{9}|x\minus{}y|$( deorece x si y sunt din intervalul [2,3) deci functia este o contractie ....deci sirul este convergent,limita aflandu-se trecand la limita in relatia de recurenta
.Va rog sa imi spuneti daca am dreptate

[quote="posabogdan"]Domn' profesor, am gasit o alta solutie :
Fie $ f(x) \equal{} 2\plus{}\frac{1}{x}$ $ |f(x)\minus{}f(y)| \equal{} |\frac{1}{x}\minus{}\frac{1}{y}| \equal{}\frac{|y\minus{}x|}{|xy|}\geq\frac{1}{9}|x\minus{}y|$ deorece x si y sunt din intervalul [2,3). Deci functia este o contractie ....deci sirul este convergent, limita aflandu-se trecand la limita in relatia de recurenta. Va rog sa imi spuneti daca am dreptate.[/quote]
Functia $ f: I\rightarrow\mathcal R$ ar fi o [b]contractie[/b] - o [b]functie Lipschitz[/b] particulara -  daca exista $ k\in (0,1)$ astfel ca $ |f(x)\minus{}f(y)| < k\cdot |x\minus{}y|$ pentru orice $ x$, $ y$ din intervalul $ I$, ceea ce la tine este pe dos ...

Eu iti propun sa ataci sirul elementar. Se constata usor (prin inductie sau altfel) ca subsirul $ u_{n}\equal{} a_{2n}$ este descrescator si subsirul $ v_{n}\equal{} a_{2n\minus{}1}$ este crescator, unde $ n\in N^{*}$ si ambele satisfac recurenta $ x_{n\plus{}1}\equal{}f(f(x_{n}))\equal{}\frac{5x_{n}\plus{}2}{2x_{n}\plus{}1}$ iar sirul $ a_{n}$ este amestecarea alternativa a acestor doua subsiruri. Mai exact,
$ a_{1}< a_{2}> a_{3}< a_{4}> a_{5}< a_{6}>\ldots$
$ 2\le a_{1}< a_{3}< a_{5}<\ldots$
$ a_{2}> a_{4}> a_{6}>\ldots > 2$
$ u_{n}>\max\{v_{n},v_{n\plus{}1}\}$ si $ v_{n}<\min\{u_{n},u_{n\plus{}1}\}$ etc.

$ \frac{1}{4}|x\minus{}y|\geq\frac{|x\minus{}y|}{|xy|}\equal{}|f(x)\minus{}f(y)|$
Acum ?

Let $ \alpha \equal{} 1\minus{}\sqrt{2},\ \beta \equal{} 1\plus{}\sqrt{2}$, we can rewrite the given recurrence as $ \frac{a_{n\plus{}1}\minus{}\beta }{a_{n\plus{}1}\minus{}\alpha }\equal{}\frac{\alpha }{\beta}\cdot\frac{a_{n}\minus{}\beta}{a_{n}\minus{}\alpha }\Longleftrightarrow\frac{a_{n}\minus{}\beta }{a_{n}\minus{}\alpha }\equal{}\left(\frac{\alpha }{\beta }\right)^{n}$, yielding $ a_{n}\equal{}\frac{\beta\minus{}\alpha r^{n}}{1\minus{}r^{n}}\longrightarrow\beta \equal{} 1\plus{}\sqrt{2}\ (n\rightarrow\infty)$, where $ r\equal{}\frac{\alpha }{\beta}$.

Multumesc domnule Nicula pentru raspuns dar trebuie sa recunosc ca solutia lui kunny este geniala

Mai bine zis cazuta din cer. Presupun ca metoda domnului Nicula din postul #2 (recurenta omografica) da acelasi rezultat.

Pentru mai multe detalii privind recurenta ta, [b]iti recomand urmatorul link[/b]:
http://en.wikipedia.org/wiki/Continued_fraction


vector
linear algebra
matrix
function
linear algebra unsolved

Suppose we have a finite dimensional vector space $V$ equipped with a norm which is given by:  $\|x\|^{2}=(x^{T})X(x)$ where $X$ is a matrix and $x \in V$ is in matrix form.  Further, let us assume that vectors evolve in time $t$ according to the equation:  $\frac{\partial x}{\partial t}= U (x)$.

Consider small perturbations $\delta y$ from a particular initial condition $y(0)$.  How would one calculate the $\delta y(0)$ which evolves into the "largest" small perturbation $\delta y(t)$ after a small time interval $t$?  "Largest" here is taken to be with respect to the norm defined above.  Further, how would one calculate the growth in the "size" of the perturbation?

(Please note that the notation is lousy here.  One could take $U$ to be a function of $x$ or simply a matrix that multiplies $x$.  Well I think the first interpretation is more interesting, I think the second interpretation is the one intended judging by the use of parentheses in the definition of the norm).

Since $\delta y(t)=e^{tU}\delta y(0)$, you should find a vector that realizes the operator norm of $e^{tU}$ with respect to the given vector norm. I think that the largest eigenvalue of $(e^{tU})^{T}Xe^{tU}$ is relevant here.


there's no one online give a solution for this exercice :?:  :?:  :?:  :?:

[quote="silouan"]$ \displaystyle \frac 11+\frac 12+\frac 13+\frac 14+ ...
= \frac 11+\frac 12+(\frac 13 +\frac 14) +(\frac 15+\frac 16+\frac 17+\frac 18)+....
>1+\frac 12+\frac 12+\frac 12+\frac 12+... $
So n=4007[/quote]
the inequality is valid the opposite way.
 the easiest way and definitely the most accurate is to write a program for this(again i doubt that you will get an answer due to the huge numbers). The answer is approximately of the order $10^{870}$

[color=red]silouan[/color] :idea:
[color=green] galeos[/color] :idea:

 [size=150] please explain  [/size]:?:  :!:  :?:  :!:  :?:  :!:  :?:  :!:  :?:  :!:

[color=red]mathematicien@naseej.com[/color]

there is no closed formula for the sum $1+\frac{1}{2}+...+\frac{1}{n}$, which means that there is no "smart" way to solve your problem. Moreover, this sum tends to infinity as $n$ gets bigger but very slow (at least by hand!). However it can be proven that $\sum^N_1 \frac{1}{n}=lnN+\gamma+e_n$, where $e_n$ tends to $0$ for large values of $N$ and $\gamma$ a constant smaller that 1.so, what i did to give you a rough answer was to find the exponent of 2005 and "approximate" the order of the answer. still, you can improve the accuracy by taking into consideration the $\gamma$ constant, but never will you get the exact value of $N$ you ask for till you(or a computer) calculate the sum. i hope i made my point clear!


LaTeX
search
combinatorics proposed
combinatorics

Let A(n) be maximum value of number of acute triangle made by n points on the plane.

Could you prove that  A(n)/(nC3) =< A(n-1)/(n-1C3) ?

Sorry. I don't know how to type a numerical formula.
Could you give me a way to type?  :)  :)

Read about [[LaTeX]].  See also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

The concyclic case, which may help :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1745217168&t=179042

Pierre.

Your initial question is solved in the book of Greitzer "International mathematical olympiads 1959-1977", for IMO problem 1970 #6.

Pierre.


Prove that

$\frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}\geq\frac9{2(a^2+b^2+c^2)}$,

where $a,b,c>0$ and $ab+bc+ca=1$.

$\frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}=\frac{1}{ab+bc+ca-ab}+\frac{1}{ab+bc+ca-bc}+\frac{1}{ab+bc+ca-ca}= \frac{1}{c(a+b)}+\frac{1}{a(b+c)}+\frac{1}{b(a+b)}=A$
Now, from http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=21&year=2002&p=3250
and from$\frac{1}{c(a+b)}+\frac{1}{a(b+c)}+\frac{1}{b(a+b)}\ge\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)}$
We have that $A\ge \frac{27}{2(a+b+c)^2}$
But on the other hand, $\frac{27}{2(a+b+c)^2}=\frac{27}{2(a^2+b^2+c^2+2ab+2bc+2ca)}\ge \frac{27}{3(a^2+b^2+c^2)}=\frac{9}{2(a^2+b^2+c^2)}$.
 :icecream: The problem is solved.

My soln was this : By Vornicu-Schur  , $A\ge \sum_{cyc} \frac1{a^2+bc}$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=368739&highlight=#368739]here[/url])
the RHS of that was shown recently (ref?) to be $\ge \frac92$ if $(a^2+b^2+c^2)=1.$

[quote="campos"]hope i'm not making any mistake....

from cauchy-schwarz we have that

$\frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}\geq\frac9{3-ab-bc-ac}=\frac{9}{2}\geq\frac{9}{2(a^2+b^2+c^2)}$

 :D[/quote]

   Quite funny,but very nice proof Campos...cheer


sorry if i am thick but hey what if x=2 or 3 ? equality wont hold then

can you please clarify your solution?

Andreas's last statement shows that the thing will be greater than 0...but there has to be a paricular value for which it's equal to 0,because ur asking that the thing is greater then $or equal$ to 0....HOpe i i'm not being thicker :blush:

ooooh i think i got it now, for all values of x the last thing is greater than 0 , XCEPT 1

thx andrea

hey cant this be made into another question itself , that find the value too for which the given thing is equal too and not greater than the RHS?? correct me if i'm wrong


LaTeX
inequalities
trigonometry
inequalities unsolved

let a,b,c>0 and 1/ab+1/bc+1/ca=1
prove that:
1/[c^2(1+ab)]+1/[b^2(1+ac)]+1/[a^2(1+bc)]<=1/4

sorry if it was posted before :blush:
 :D  :D

First I rewrite your problem with LaTeX (believe me, it makes the problem much more attractive):

[color=blue][b]Problem.[/b] Let a, b, c be positive reals such that $\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$. Prove that

$\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ca\right)}+\frac{1}{c^2\left(1+ab\right)}\leq\frac14$.[/color]

And here is a solution:

We can transform the given equation $\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$ into the form $\frac{a+b+c}{abc}=1$. Thus, a + b + c = abc. Hence,

$a^2\left(1+bc\right)=a^2+a^2bc=a\cdot a+a\cdot abc$
$=a\cdot a+a\cdot\left(a+b+c\right)=a\cdot\left(b+c+2a\right)$;

similarly,

$b^2\left(1+ca\right)=b\cdot\left(c+a+2b\right)$ and
$c^2\left(1+ab\right)=c\cdot\left(a+b+2c\right)$.

Thus, your inequality

$\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ca\right)}+\frac{1}{c^2\left(1+ab\right)}\leq\frac14$

rewrites as

$\frac{1}{a\cdot\left(b+c+2a\right)}+\frac{1}{b\cdot\left(c+a+2b\right)}+\frac{1}{c\cdot\left(a+b+2c\right)}\leq\frac14$.

Multiplying this by abc, we get

$\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}+\frac{ab}{a+b+2c}\leq\frac{abc}{4}$.

Since a + b + c = abc, this is equivalent to

$\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}+\frac{ab}{a+b+2c}\leq\frac{a+b+c}{4}$.

But this inequality was proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=14289 , posts #2-#4.

  Darij

Here's another solution:

Let $x = 1/bc$, $y=1/ac$, $z=1/ab$. Then the constraint becomes $x+y+z=1$. And

\begin{eqnarray*}
\sum_\mathrm{cyc} \frac{1}{a^2(1+bc)}
&=& \sum_\mathrm{cyc} \frac{yz}{x} \frac{1}{1+1/x}     \\
&=& \sum_\mathrm{cyc} \frac{yz}{1+x}    \\
&=& \sum_\mathrm{cyc} \frac{yz}{(x+y)+(x+z)}    \\
&\leq& \sum_\mathrm{cyc} \frac{1}{4}\left( \frac{yz}{x+y} + \frac{yz}{x+z}\right)    \\
&=& \frac{1}{4} \sum_\mathrm{cyc} \left( \frac{yz}{x+y} + \frac{xz}{x+y}\right)    \\
&=&\frac{1}{4}(x+y+z)    \\
&=&\frac{1}{4}    \\
\end{eqnarray*}

[quote="cuong"]i like your solution very much bilzhao,it looks easy right
i also have another solution:
let a:=cotgA,b:=cotgB,c:=cotgC,it will satisfy the conduction
and translate this inequallities into three sides of triangle
with equivalent translate and USING COSI AND SCHUR TO PROVE IT
I ALSO LIKE DARIJ SOLUTION BUT WITHOUT THE INEQUALITIES YOU MANSION,IT TAKES TOO LONG :P  :(  :D

SORRY FOR MY TERRIBLE ENGLISH :blush:[/quote]
I think you mean let $a=\tan A,b=\tan B,c=\tan C$
but I can't complete the solution.

hei , i terrible sorry i meant let a:=cotgA/2........
using formular to translate to  3 sides of triangle:

you see darij 'S solution right with his last inequality
let a:=x+y; b:=y+z; c:=x+z;

you got my solution after that schur should work :D

i sorry for my mistakes i don't know how to use latex

i can't understand how darij can know a lot of inequalities posted before :(
can you tell me :lol:

Didn't you notice the easy solution for $darij's$ last inequality in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14289]www.mathlinks.ro/Forum/viewtopic.php?t=14289[/url] as you said.and I think it's also like $billzhao's$ solution.
for another question:I think $darij's$ last inequality was sloved by $darij$ :D


geometry
incenter
circumcircle
inequalities
geometry proposed

Let$ G,H,O,I$ denote the  centroid, orthocenter, incenter and circumcenter of a triangle. Prove that: $ OI\plus{}OG \ge IH$
 :)

[quote="quykhtn-qa1"]Let$ G,H,O,I$ denote the  centroid, orthocenter, incenter and circumcenter of a triangle. Prove that: $ OI \plus{} OG \ge IH$
 :)[/quote]
THIS INEQUALITY VERY INTERESTING.PLEASE SOLVE IT.I HAVE A NICE SOLUTION :)

[quote="Virgil Nicula"][u][size=125][color=darkred][b]Nice[/b] inequality ! Thank you, [b]quykhtn-qa1.[/b][/color][/size][/u]

[quote="quykhtn-qa1"][color=darkred]$ G,H,O,I$ - centroid, orthocenter, incenter and circumcenter of $ \triangle\ ABC\ \Longrightarrow\ OI\ \plus{} \ OG\ \ge\ IH$ .[/color][/quote]
[color=darkblue][b][u]Preliminaries[/u].[/b] $ m(\angle HIG)\ge 90^{\circ}\Longleftrightarrow HG^2\ge IH^2 \plus{} IG^2\Longleftrightarrow 4\cdot OG^2\ge IH^2 \plus{} IG^2$ .

Since $ I$ belongs to the disk with diameter $ [HG]$ obtain $ OG\le OI\ .$

[b][u]Proof[/u].[/b] Apply [u][b]Stewart[/b]' relation[/u] to cevian $ IG$ in $ \triangle\ HIO\ : \ \Longrightarrow\ \underline {\underline{3\cdot IG^2}} \equal{} IH^2 \plus{} 2\cdot IO^2 \minus{} 6\cdot OG^2\ (*)$ .

$ 4\cdot OG^2\ge IH^2 \plus{} IG^2\ \Longleftrightarrow\ 12\cdot$ $ OG^2\ \ge\ 3\cdot IH^2 \plus{} \underline{\underline{3\cdot IG^2}}\ \stackrel {(*)}{\Longleftrightarrow}\ \boxed {\ 9\cdot OG^2 \minus{} OI^2\ge 2\cdot IH^2\ }\ (**)$ .

$ (OG \minus{} OI)(7\cdot OG \plus{} 3\cdot OI)\le 0\ \Longleftrightarrow$ $ 2(OG \plus{} OI)^2\ge 9\cdot OG^2 \minus{} OI^2\stackrel {(**)}{\Longrightarrow}2(OG \plus{} OI)^2\ge 2\cdot IH^2$ .[/color][/quote]
Nice proof. :lol:


In the equation 3x^3-7x^2+5x-6=0 the roots are a, b, and c. Find the value of 6/a+6/b+6/c.

Are we just supposed to factor this (in which case, I need to learn how to factor polynomials of degree greater than two)?  Or is there another way?

Find the polynomial whose roots are $ \frac{1}{a}$, etc, then use Vieta's formulas to find the sum of roots, and multiply it by 6.

Let $ f(x)$ be the polynomial you have. What should you make the argument of $ f$ be, in order to get a polynomial whose roots are $ \frac{1}{a}$, $ \frac{1}{b}$, and $ \frac{1}{c}$?

[hide="Useful fact:"] If $ r_1, r_2, r_3, \ldots r_n$ are the roots of the polynomial $ a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} a_{n\minus{}2}x^{n\minus{}2} \plus{} \ldots a_2x^2 \plus{} a_1x \plus{} a_0$, then $ \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}, \ldots \frac{1}{r_n}$ are the roots of $ a_0x^n \plus{} a_1x^{n\minus{}1} \plus{} a_2x^{n\minus{}2} \plus{} \ldots a_{n\minus{}2}x^2 \plus{} a_{n\minus{}1}x \plus{} a_n$ (a polynomial with the coefficients reversed).

Try to prove it.[/hide]

That is what I was hinting at.
To prove that fact, say $ r$ is a root of $ f(x)\equal{}0$. Can you find a root of $ f(x^2)$? Of $ f(x\minus{}1)$?
These are the lines along which you should be thinking.

OK, thanks

grn_trtle: I think I see what you're saying (so sqrt(r)  would be a root of f(x^2) if r is a root of f(x)).

[hide="my solution"]The expression $ \frac{6}{a}\plus{}\frac{6}{b}\plus{}\frac{6}{c}$ can be re-written as follows:

(i) factor out the 6: $ 6\bigg(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\bigg)$

(ii) express as a single fraction with denominator $ abc$: $ 6\bigg(\frac{bc\plus{}ac\plus{}ab}{abc}\bigg)$

Notice that the numerator of the above fraction is the sum of the roots "taken two at a time" and the denominator is the product of the roots.

Using Vieta's Formulas, you can find that these values are $ \frac{5}{3}$ and $ \frac{\minus{}(\minus{}6)}{3}\equal{}2$, respectively.

Substituting these values in the expression in (ii) gives us:
$ 6\bigg(\frac{\frac{5}{3}}{2}\bigg) \equal{} 5$[/hide]


2)Let a be a real number and x1>0 fixed numbers. The sequence (xn) is given by the relation
: x_(n+1)=xn+sqrt(a+xn^2) for any n natural. Compute the limit: lim(n-->oo) 2^n/xn.
(GM 7/8-2003)

x(n+1)=x(n)+sqrt(a+x(n)^2)
  . It's easy to see that x(n+1)>x(n)>0 for al n>=1
1) a>0
   .If x(n) is bounded ,it's converge to some c
        -->   c=c + sqrt(a+c^2) which is impossible
  so,lim{n-->oo} x(n)   =oo
   .  x(n+1)/x(n) =1 +sqrt(a/[x(n)]^2 +1)
   .Hence Lim {ln(x(n+1))-ln(x(n))} =Lim { ln(1+sqrt(a/[x(n)]^2+1)}
   .But lim x(n)=oo  --->  Lim (1/x(n))=0
---> lim{ln(x(n+1))-ln(x(n))} =ln2
   .Use Cesaro 's theorem ,we have lim {ln(x(n))/n }= ln2
--->
      e^[lim (ln(x(n))/n)] =2
<-->
            lim (x_n)^1/n =2
<--->lim x_n/2^n  =1
 2) If a<0
     .If x(1) <sqrt(lal) then the sequence doesn't exist
     .If x(1)=sqrt(lal) x(n)=x(1) for all n>=1 --> lim 2^n /x(n) =0
     .If x(1)>sqrt(lal) x(n)>a for all n>=1
      Let  a= \sqrt (-b), with b>0
      Let ch(y_n)=x_n/b
      {where ch(y)= (e^y+ e^(-y))/2 }
      x_(n+1)=x_n+   \sqrt (x_n^2-b)=x_n+b   \sqrt ((x_n/b)-1)
      From x_n = b.ch(y_n) ,we have
      x_(n+1)=b.ch(y_n) +bsh(y_n)= b.exp(y_n)
      So
         ch(y_(n+1))=exp(y_n)
         y_(n+1) = ln( e^y_n +  \sqrt ( e^(2y_n) -1) )
         y_(n+1) = y_n + ln( 1+   \sqrt (1- e^(-2y_n))
   Hence
         y_(n+1) - y_n = ln( 1+   \sqrt (1- e^(-2y_n)) ---> ln2 when n-->+oo

( y_n -->+oo)

Cesaro theorem => lim ( (y_n- y_(n-1))+...+(y2-y_1))/(n-1) = ln 2
      -->  y_n/n  ~  ln2
  -->
    (2^n/x_n) ~ 2^n.(2/b(2^n + 1/2^n))

so lim 2^n/x_n = 2/b=2/ \sqrt (-a)

Oh,
     I have made a terrible mistake :
     lim (ln x_n)^(1/n)=ln2 doesnot imply
     lim x_n  /2^n =1
     Sorry


I was just wondering, in general are the shortlisted problems harder or easier than the ones that appear on the real IMO?
Thanks

I think it is unsteady from year to year.

If one problem from the shortlist has been non selected for the competition that's because :
- It was considered as too hard
- It was considered as too easy
- It was considered as too well-known
- There was another problem already selected in the same area.

There are other reasons, but I think the two first reasons above answer to your question...:P

Pierre.


The Problem from the book: Graph all the points on the coordinate plan below the graph of y= 3x-5

The Graph from the Answers Section (I made this graph in a program): http://www.yousendit.com/transfer.php?action=download&ufid=D5CBCF570978892E

The Answer from the answers section: "Since the graph only contains points below the graph of y= 3x-5, the boundary is dashed (I know I know, I forgot to dash the line). Thus, the inequality is either y<3x-5 or y>3x-5. The point (3,0) is in the shaded region. Since 0<3(3) - 5, the inequality y<3x-5 describes the region."

My questions: Why do they use (3,0) as a test point instead of using (0,0)?
(0,0) was commonly used in examples in my text book. How do I know which test point to choose?

When the book said "Thus, the inequality is either y<3x-5 or y>3x-5", how do I know which inequality to choose? They are giving me two options, and they haven't done that before so I am confused as to which one I should choose.

[hide]The equation of a graph shaded on one side of a function is either $y>f(x), y<f(x)$ if the function itself is dashed, and $y\ge f(x),y\le f(x)$ if it isn't. So the book said "Thus, the inequality is either y<3x-5 or y>3x-5" because it is dashed. I'm not really sure why they chose $(3,0)$ specifically, but they didn't choose $(0,0)$ because it isn't below the line $y=3x-5$ in the shaded region. Of course, you could evaluate $x=0,y=0$, then reverse the inequality. [/hide]

Do you have an idea as to why to they choose (3,0)? It's pretty confusing, I don't know why they did that.

they just picked a point at random? white people probably wrote the answer key, either that or maybe they want to show that you don't have to pick any special point.

Don't you have better things to do than call people out for things that they're obviously joking about.  You're not his mother.


inequalities
trigonometry
inequalities proposed

Let $ a, b, c$ be the side lenghts of the traingle and $ s$ its perimeter. Show that

      \[ \left(\sum_{cyc}\frac{a}{b+c}\right)\cdot\left(\sum_{cyc}\sqrt{\frac{1}{(s-a)(s-b)}}\right)\geq 6\left(\sum_{cyc}\frac{1}{b+c}\right).\]

wow! nice one, cezar!

rewrite the lhs as

$ \sum_{cyc}\frac{a}{b+c}\sum_{cyc}\frac{1}{\sqrt{bc}\cdot\sin \frac{A}{2}}$.

from am-gm and the famous $ \frac{a}{b+c}\geq \sin\frac{A}{2}$ it follows that

$ \sum_{cyc}\frac{1}{\sqrt{bc}\cdot\sin \frac{A}{2}}\geq \sum_{cyc}\frac{2}{(b+c)\sin\frac{A}{2}}\geq \sum_{cyc}\frac{2}{a}$.

then, we are to show that

$ \sum_{cyc}\frac{a}{b+c}\sum_{cyc}\frac{1}{a}\geq 3\sum_{cyc}\frac{1}{b+c}$,

but the last one follows from chebyschev inequality, and we're done.  :D

[quote="Cezar Lupu"]Let $ a, b, c$ be the side lenghts of the traingle and $ s$ its perimeter. Show that
\[ \left(\sum_{cyc}\frac{a}{b+c}\right)\cdot\left(\sum_{cyc}\sqrt{\frac{1}{(s-a)(s-b)}}\right)\geq 6\left(\sum_{cyc}\frac{1}{b+c}\right). \]
[/quote]
$ \sum_{cyc}\sqrt{\frac{1}{(s-a)(s-b)}}\geq\sum_{cyc}\frac{2}{c}.$
But ${ \sum_{cyc}\frac{a}{b+c}\cdot\sum_{cyc}\frac{2}{c}}\geq 6\sum_{cyc}\frac{1}{b+c}\Leftrightarrow\sum_{cyc}(a^{4}b+a^{4}c+a^{3}b^{2}+a^{3}c^{2}-4a^{2}b^{2}c)\geq0.$

lol.. i didn't see that $ \sum_{cyc}\sqrt{\frac{1}{(s-a)(s-b)}}\geq \sum_{cyc}\frac{2}{c}$ was so immediate  :P


[quote="bui_haiha2000"]Let a1,a2,..an $\ in$ R and f(x)=$\frac{cos(x+ai)}{2^i}$(i=1,2..n) .And f(x1)=f(x2)=0
   Prove that $\frac{x2-x1}{pi}$ in Z[/quote]
I think$f(x)=\sum_{i=1}^{n}\frac{cos(x+a_i)}{2^i},pi=\pi$

I think pi is $\pi$. I don't undestand what is f(x). It is f(x,i), and for all i $f(x_1,i)=f(x_2,i)=0?$


Without using calculus, is there a way to prove that $ \sum^{\infty}_{n\equal{}1}\frac{1}{n^2}\equal{}\frac{\pi^2}{6}$? If so, then how?

Does it look like an elementary result to you?  How would a $ \pi^2$ factor appear by elementary means?

[quote="t0rajir0u"]How would a $ \pi^2$ factor appear by elementary means?[/quote]
Start with the ruler and compass construction that squares a circle ...  :wink:

And then invert it.  Right.  That makes it so much simpler, JBL.

$ \pi$ is transcendental, as Lindemann showed, so I can't think of an elementary proof.  As a matter of fact, the easiest way to show that $ \zeta(4)\equal{}\frac{\pi^4}{90}$ and the further sums involves using Taylor series for complex arguments.  If there were an elementary proof, that would be impressive.

Hmm.  [url=http://en.wikipedia.org/wiki/Basel_problem]The Wiki page[/url] gives a proof using only de Moivre's Theorem, although I can't really justify calling that "not using calculus."

The use of De Moivre's theorem in that proof is not essential.  All that is needed is to express $ \sin nx$ in terms of $ \sin x$ and $ \cos x$.  I consider the proof to be elementary.  I have presented variants of the proof to high school students who have not studied calculus.

The same argument can be used to evaluate zeta(4), and more generally zeta(2j).


function
real analysis
real analysis unsolved

$A \subset \mathbb{R}^{n}$ a subset of $\mathbb{R}^{n}$, $C>0$ a constant and $f: A \to \mathbb{R}^{m}$ a function satisfying:
$|f(x)-f(y)| \leq C |x-y|$
with the usual euclidian norms.
1/ Can one find a Lipschitz function $F: \mathbb{R}^{n}\to \mathbb{R}^{m}$ that satisfies $f(x)=F(x)$ for $x \in A$

2/ Can one find a Lipschitz function $F: \mathbb{R}^{n}\to \mathbb{R}^{m}$ that satisfies $f(x)=F(x)$ for $x \in A$ with $C$ as lipschitz constant. (i.e. $F$ and $f$ have the same lipschitz constant)

The answer to the second question is "yes". The problem has been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=416746&sid=1553dfc3ee27d07c00c51bb00725e0ca#416746]here[/url], but the proof is by no means complete.


trigonometry
geometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

The sides of a triangle are $ 30$, $ 70$, and $ 80$ units. If an altitude is dropped upon the side of length $ 80$, the larger segment cut off on this side is:

$ \textbf{(A)}\ 62\qquad
\textbf{(B)}\ 63\qquad
\textbf{(C)}\ 64\qquad
\textbf{(D)}\ 65\qquad
\textbf{(E)}\ 66$

OMG U HAVE 1001 POST WHICH EQUALS TO 7*11*13!

Any ways... [geogebra]111fe6aa49630e1aac14928ff02435490dc20014[/geogebra]

[hide]$ x^2\plus{}h^2\equal{}4900$
$ (80\minus{}x)^2\plus{}h^2\equal{}900$
Subtracting the 2 equations, we get $ 160x\minus{}6400\equal{}4000$
Thus $ x\equal{}\boxed 65$[/hide]

[hide="Alternatively..."]
We can use Heron's formula to find the altitude. We have

$ s \equal{} \frac{30 \plus{} 70 \plus{} 80}{2} \equal{} 90\implies A \equal{} \sqrt{(90)(90 \minus{} 30)(90 \minus{} 70)(90 \minus{} 80)} \equal{} 600\sqrt{3}.$

We also have $ \frac{80h}{2} \equal{} 600\sqrt{3}\implies h \equal{} 15\sqrt{3}$. Clearly $ 70^2 \minus{} (15\sqrt{3})^2 > 30^2 \minus{} (15\sqrt{3})^2$ so the longer segment has a length of $ \sqrt{70^2 \minus{} (15\sqrt{3})^2} \equal{} \sqrt{4900 \minus{} 675} \equal{} \sqrt{4225} \equal{} \fbox{\textbf{(D) 65}}$.
[/hide]

Here's another one.

[hide]Let the triangle be ABC such that AB = 80, BC = 30, and CA = 70. If D is the projection of C onto AB, then we want the length AD. This is equal to (AC) cos A.

By the Law of Cosines, this equals $ 70 \cdot \frac{80^2 \plus{} 70^2 \minus{} 30^2}{2 \cdot 80 \cdot 70} \equal{} 65$[/hide]


In the figure, circle O of radius 10 fits in the right-angled corner so that AB and BC are tangent lines. What is the radius of the largest circle that fits in the same corner between the larger circle and the corner?
I'd post the picture as well, but I forgot how to upload a pic on here, lol.[/img]

[hide]Draw radii of the smaller circle to hit the lines, to get $30-20\sqrt{2}$ unless I'm being careless[/hide]


[quote="jli"]have you had a really bad teacher before? (really really bad)

i had this science teacher that says using a ruler and protracter is always more accurate than calculating it on paper

more than half the class has dropped now[/quote]

BANANNA!!

My 3rd grade "Science" teacher.  He laughed at my clothes.   :o
My 4th grade teacher...  Tedious divison worksheets every day..
My school's Drama teacher...  evil...Evil...EVIL!!!
My 7th grade 1st semester PE teacher.  We spent more time on the square dancing unit than on the physical fitness unit.  :huh:

Eh..that's it.

the parents even think your science teacher is bad (JLi)
my science "teacher" doen't teach me anything or my math teacher or my music teacher or my odyssey of the mind teacher

My last years team teacher sucked at math. She asked me to show it to the class when she didn't get it! :o And she finally retired this year.

aw man... I thought this was a banana topic :( I LOVE THOSE :lol:

thanks to JLi for almost taking off this topic

My science teacher last year said a whole bunch of weird things.  Like, "God made cows so that we could eat them."  And that for an answer, you should have 13.3 instead of 40/3 because 13.3 is simplified.

my Violin teacher is really really bad.  She yells at me for not playing one note right.

[quote="nonie"][quote="mathclass"]Last year there was this teacher who shall remain "anonymous" let's call him X.

X had stacks of porn magazines... he ate food off the ground... and never taught anything. he never collected homework and if you got a 50% on the final you got an A.

beat that![/quote]

What did X teach???[/quote]

History. well, if you got a 50% you MIGHT not get an A... but as half the class had 120% they probably would.

my freshman ap us history teacher would be a great lecturer, but when we took tests, she would go on e-bay and ask our opinions on what to buy- during the tests.

Well, in 8th grade, I had a teacher who got undescribably pissed at me for anything I said (I guess she really didn't like my first few wisecracks, and even after I stopped, she was on my case). She taught English. For example, once when a girl got back from being sick for a week and got her homework, I said to her that I felt sorry for her since it was going to take a long time. The teacher went off on me for a long time, then the girl whispered to me "What a *err, female dog*".
     Also, I remember a 6th grade history teacher whom I didn't have, but his students told me that he occasionally looked at gay porn during class (during tests).  :o


This is a qualifying round open to anyone who wants to do it. It's ages 16-17 but aimed mainly at people who have never done any problem solving maths before.

Here are some of the tougher questions:

1. The value of 1^2004 + 3^2004 + 5^2004 + 7^2004 + 9^2004 is caluculated using a powerful computer. What is the units digit of the correct answer?

2. The letters S, M and C are whole numbers. S*M*C = 240, S*M + C = 46 and S + M*C = 64. What does S + M + C equal?

3. Which is the lowest positive integer by which 396 must be multipled by to make a perfect cube?

4. The number N is exactly divisible by 7. It has 4008 digits. Reading from left to right, the first 2003 digits are all 2s, the next digit is n and the last 2004 digits are all 8s. What is/are the value/s of n?

5. Positive integers x and y satisfy the equations:

 :sqrt: (x + ( :sqrt: y)/2 ) -  :sqrt: (x - ( :sqrt: y)/2 ) = 1
4<y<10

What are the values of x and y?

Hello, Mikey

Quote:1. The value of 1^2004 + 3^2004 + 5^2004 + 7^2004 + 9^2004
 is calculated using a powerful computer.
What is the units digit of the correct answer?

[hide]

Here's a baby-talk solution which does not involve Modulo Arithmetic.

1^n will always end in 1.

Since 3^4 = 81, then 3^2004 = (3^4)^501 = (81)^501 will end in 1.

5^n will always end in 5.

Since 7^4 = 2401, then (7^4)^501 = (2401)^501 will end in 1.

Since 9^4 = 6561, then (9^4)^501 = (6561)^501 will end in 1.

Therefore, the given sum will end in: 1 + 1 + 5 + 1 + 1 = 9.

[/hide]

Hello, Mikey!

Quote:5. Positive integers x and y satisfy the equation:

What are the values of x and y?[hide]

x = 2, y = 7

[/hide]

Hi there, Mikey!

This is probably the most diabolical problem I've ever seen.

But then, I may be missing a simple and elegant solution.

Quote:4. The number N is exactly divisible by 7. It has 4008 digits.
Reading from left to right, the first 2003 digits are all 2s,
the next digit is n and the last 2004 digits are all 8s.
What is/are the value/s of n?[hide]

   n = 1 or 8

(I will show my work only under the threat of dismemberment or worse.)[/hide]

One easy way to solve the problem is to note that 111,111 is divisible by 7.


abstract algebra
real analysis
topology
calculus
integration
Galois Theory
group theory

Hi all

I am currently just starting my 1st year graduate courses. Despite the syllabus of my algebra course sounding like not too much more then a review of groups, rings, fields and galois theory, our instructor has decide to completely forgo review of any sort and just start teaching entirely new concepts. Since I haven't taken any algebra since my 3rd year course, this is going to be a big challenge for me as I was wholeheartedly looking forward to reviewing these things, especially since some of the stuff he is assuming we know wasn't even covered in my 3rd year course...

Anywho, the course website (http://www.math.mcgill.ca/goren/Math570.08-09/MATH570.html) has some recommended texts, but nothing required. I already have Dummit and Foote, but I was wondering if people here could particularly recommend their favourite of the ones he has listed. You can see the "updated" course syllabus on this page as well so if you know that one of those texts gives a particularly good exposition on the various subjects covered please let me know, after all I am a poor grad student and I need to make sure my money is being well spent haha.

Thanks for your time.

Different people react differently to different textbooks, so it's always best to just spend 20 minutes with a book and see how you like it. Anyway, here's what I think: First of all, if you feel rusty on elementary stuff like groups and rings, you need to catch up quickly. Spend some time with Rotman (Group Theory) and Artin (Algebra) and do some of the exercises to feel comfortable. You mention Galois theory but I wouldn't worry about it at all for now - the course doesn't seem to rely on it in any way.

(I have to wonder: why would a 1st year graduate course be a review of anything? It's time to start learning some real stuff!)

As for the textbooks:

1. Dummit and Foote: I hear it's good but I don't know it. You have it anyway.

2. Fulton and Harris: I'm not a big fan of this is a textbook. It's mostly about Lie theory and I think it takes a somewhat different approach to representation theory than the course.

3. Jacobson: Basic Algebra I,II. Very good text if I remember correctly from my college days, although I never owned it. I'm sure it's a great reference, too.

4. Lang: Algebra. Lang deserves a lot of credit for writing up much of the material of modern algebra and number theory in textbook form, but to be perfectly honest I positively hate that book. It's terrible for self study since it generally insists on the most general and abstract approach with very little motivation, virtually no examples and a total mess in terms of order of presentation. It's strange since the first edition was much shorter, covered much less material and I thought it was much, much better.

5. Rotman: I'm not familiar with these particular texts by Rotman but I really like his Intro to Group Theory so I think those are good bet. He's a very good expositor.

6. Serre: Linear Representations of Finite Groups. Generally a good and useful book, although I somehow like Isaacs' "Character Theory of Finite Groups" better. Isaacs starts of with the group ring right away, considering representations as modules from the very beginning. It's a hurdle, but if you cross it you're in great shape. Moreover, this seems to be the approach of your course. Serre's approach is initially more elementary so that may be easier for you as a first exposure to group representations.

PS Eyal Goren and I were graduate students together. He's a very strong mathematician, but I don't know him as an instructor. Good luck  :)

I love Rotman (Advanced Modern Algebra) as a reference that covers everything, but it is sort of a Rudin in that it is lots is covered in not a lot of space. If you spend the time to work through the details you will be in great shape. I always recommend it, but if you are looking for a thorough covering of the topics, this might not be your best bet.

What is a good book for introduction to rings, groups, etc..?  I mean for 2nd and 3rd year, not grad level.

Fermat's little turtle: I highly recommend Dummit and Foote "Abstract Algebra 2nd ed." (The book I have). It is the book I used in my 3rd year algebra class (For math specialists or maybe your school would call it "honors"). I found it had a great exposition and excellent examples. The problems at the end of each chapter are really nice, usually the first patch being relatively easy and building up to more difficult as you go. I really can't say enough about this book it was great. Just because I can't remember everything I learned from it 2 years ago doesn't mean its not great haha. Seriously get this if you want to learn about groups, rings and fields.

If you need an undergrad-level refresher on lots of aspects of modern algebra, you should definitely try [url=http://www.amazon.com/Elements-Abstract-Algebra-Allan-Clark/dp/0486647250]Elements of Abstract Algebra[/url] by Allan Clarke, which is essentially a problem book with plenty of expository text (but no solutions!). It will take quite a bit of effort to get through all the problems, there are many of them, but I think you will find it worthwhile.

Good luck in grad school :D

I am actually currently using my abstract algebra text for a "refresher" aka. teaching myself modules since we never covered that in my undergrad course, although I definitely want to look over some of the stuff I did learn, I think D & F will be sufficient. More interested in trying to find a good text that will help elaborate on the actual material in this course, but I will definitely take a look at the texts suggested by hilbert and random. So hard to find a grad text that is anything but terse! I mean these guys make Spivak on manifolds look like a Stewart haha.

I just wish that the course listing would have listed the actual syllabus, then I would have started reviewing over the summer instead of now! :maybe:

Seriously this is the course syllabus listed on mcgill website: Review of group theory; free groups and free products of groups. Sylow theorems. The category of R-modules; chain conditions, tensor products, flat, projective and injective modules. Basic commutative algebra; prime ideals and localization, Hilbert Nullstellensatz, integral extensions. Dedekind domains.

I mean I don't mind learning some new challenging stuff, just don't pee on my snowcone and call it sprinkles  :P

I actually just completed a review of basic ring and module theory using Dummit and Foote and found it to be excellent for the purpose.  The exercises are just easy enough that you can work through them all without giving yourself [i]too[/i] much of a headache...


Euler once conjectured that there were no positive integer solutions  to the equation
\[a^5+b^5+c^5+d^5=e^5 \]
But actually there is a solution to the following one: \[133^5+110^5+84^5+27^5=n^5\]
Can you find $n$ without a calculator?

i once did this problem.... [hide]144[/hide] but i am afraid that i do not have a solution to the answer.... maybe something with residues or mods?

nevermind. it has been proven, but with the aids such as calculators.

just some more information:

$2682440^4 + 15365639^4 + 18796760^4 = 20615673^4$
$95800^4 + 217519^4 + 414560^4 = 422481^4$

basically the conjecture was that the sum of $n-1$ numbers of $n$ powers cannot be equal to a single number of $n$ power. however, it has been disproved by the example that jelly gave, for $n=5$, and the ones above for $n=4$. I did some research, and there was no proof---- just calculations with computer/calculator/etc.

This was an old AIME problem, so no calculators should be used.

[hide="hint"]
use mods ;)[/hide]

Or you can use the fact that the unit's digit of n^5 is the unit's digit of n.  So it has to be 4, in the case of n.  Then you can set boundaries for it...obviously it's above 134^5, and it's below 154^5( because the ratio between 154 and 133 is slightly more than 8/7, and when one takes that to the fifth power, one gets a number greater than 2...so 154^5>(134^5)2), so then it's gotta be 144^5.
Or you could use mods to find that it must be a multiple of 3.

Looking at the units digit [i]is[/i] looking at it mod 10.

Yeah, if you look at it that way...I was thinking of it as, for anyone digit number, if you take the sequence which the units digit undergoes as you raise it to successively higher powers, any number which is 1mod4 will be the number, any which is 2mod4 will be another number, any 3mod4 another, and 0mod4 another.  So in a sense it is mods...I just never defined my approach.  After all, there are more mod approaches.


It give Pell's equation $3m^{2}+5m+7=n^{2}+8n$. Let $x=6m+5,y=2n+8$, then $x^{2}-3y^{2}=-251=7^{2}-3*10^{2}\Longrightarrow x+y\sqrt 3=(\pm 7+10\sqrt 3)(2+\sqrt 3)^{k}.$ From y is even we get k is even, from x=-1(mod 6) we get $x+y\sqrt 3 =(-7+10\sqrt 3)(7+4\sqrt 3)^{l},l=1,2,...$
It give needed result.

[quote="mayhem"]$m$ and $n$ are denote any system of numerations. If  $(357)_{m}$=$(180)_{n}$ then find $m+n$.[/quote]
What's the meaning of "system of numerations"?
Thank you.

Yes Rust you are right. There are infinetely many solutions for m+n. Your solution is very good but are there any solution without using Pell's equation?


Just today I was working on this for case $n=4$ and got close but couldn't solve with rigor... any ideas would be appreciated

I think the inequality has been appeared in the forum many time before.One of the methods is to consider the equivalent inequality:
if
\[x_{1}x_{2}\cdots x_{n}=(n-1)^{n}\]
,then
\[\sum_{i=1}^{n}\frac 1{x_{i}+1}\le 1 \]
try to find $a$ such that:
\[\frac 1{1+x_{1}}\le \frac{x_{1}^{a}}{x_{1}^{a}+x_{2}^{a}+\cdots+x_{n}^{a}}\]

Wow, very nice... However, I have a question for zhaobin:

Can you explain to me how those inequalities are equivalent?
I tried the following on the $n=4$ problem I was working on:

Given $\sum \frac{1}{1+a^{4}}=1$ prove that $abcd\geq 3$.

All I got was, $\sum \frac{1}{1+a^{4}}=1\implies \sum 1+a^{4}\geq 16\implies (\sum a^{4})/4 \geq 3$
which is weaker than the result. But can it be patched using your logic?

(notice, that $abcd=3\implies \sum a^{4}+1\geq 16$ which is close but I don't think enough to prove the rest... :( )

[quote="vietnam2007"]no one of the moroccan contests had done that exercise.I had used all the theorems but no one was useful.the professor suggests that :

we put $y_{i}=\frac{1}{x_{i}+1}$ , $\sum_{i=1}^{n}{y_{i}}=1$ continue...     so it becomes more elegant and more easy![/quote]

Read my post , please!  :D

Another solution goes as follows:

First make the substitution $y_{i}=1/(x_{i}+1),$ so that $x_{i}=(1-y_{i})/y_{i}$ and $0<y_{i}<1$  and $y_{1}+y_{2}+\cdots+y_{n}=1.$  Then we need to prove that $\frac{1-y_{1}}{y_{1}}\frac{1-y_{2}}{y_{2}}\cdots \frac{1-y_{n}}{y_{n}}\geq (n-1)^{n}.$  We do this by smoothing.  Take any two of the $y_{i}$--say $y_{1}$ and $y_{2}$--and replace them by their arithmetic mean $(y_{1}+y_{2})/2.$  This keeps the relation $y_{1}+y_{2}+\cdots+y_{n}=1$ intact.  Then we claim that

\[\frac{1-y_{1}}{y_{1}}\frac{1-y_{2}}{y_{2}}\geq \frac{1-\frac{y_{1}+y_{2}}{2}}{\frac{y_{1}+y_{2}}{2}}\frac{1-\frac{y_{1}+y_{2}}{2}}{\frac{y_{1}+y_{2}}{2}}.\]

This can be seen by clearing the fractions, simplifying, and factoring, writing the inequality in the equivalent form

\[(y_{1}-y_{2})^{2}(1-y_{1}-y_{2}) \geq 0,\]

which is clearly true.  Thus the product $\frac{1-y_{1}}{y_{1}}\frac{1-y_{2}}{y_{2}}\cdots \frac{1-y_{n}}{y_{n}}$ is decreased by smoothing, and thus its minimum occurs when all the $y_{i}$ are equal.  But $y_{1}=y_{2}= \cdots = y_{n}=1/n$ gives

\[\frac{1-y_{1}}{y_{1}}\frac{1-y_{2}}{y_{2}}\cdots \frac{1-y_{n}}{y_{n}}=(n-1)^{n},\]

and thus our proof is complete.

[quote="zhaobin"][quote="me@home"]Wow, very nice... However, I have a question for zhaobin:

 )[/quote]
This because the inequality is obvious equivalent to:
\[x_{1}x_{2}\cdots x_{n}<(n-1)^{n}\]
then
\[\sum_{i=1}^{n}\frac 1{1+x_{i}}<1 \]
Which is also equivalent the inequality I wrote.[/quote]

I see what you are saying but there is a small issue... it is not truly "equivalent" as you say, in fact if we can prove either (1) or (2) below then the problem is solved, but does it hold the other way around? (I.e. problem $\implies$ (1), like you claim)

(1) ... the conditions you wrote

(2) ... replace second line of yours with a $>$ sign

Also, having read Adalbert's proof I realized that I already did this problem before, but under that different formulation... I was wondering why it looked so familiar (actually, is that solution truly correct? Becuase I tried something like it and concluded that Am-Gm wasn't strong enough...)


Hello,

Answering a problem of a guy in the forum, I typed this (REMARK: I didn't wrote the dollar sign in the code, because otherwise it does weird things!)

[code]Now, note that if x\neq0 (x\in\mathbb{C}^n), then, if Ax=0, then (A^2-2A+I_n)x=x\neq0, hence, null(A) \leq rank(A^2-2A+I_n). Also, if (A^2-2A+I_n)x=0\Rightarrow (A-2I_n)Ax=x\neq0\Rightarrow Ax\neq0\Rightarrow  null(A^2-2A+I_n) \leq rank(A), hence [/code]

And got this

[quote="The Code Above"]Now, note that if $x\neq0$ ($x\in\mathbb{C}^n$), then, if $Ax=0$, then $(A^2-2A+I_n)x=x\neq0$, hence, $null(A) \leq rank(A^2-2A+I_n)$. Also, if $(A^2-2A+I_n)x=0\Rightarrow (A-2I_n)Ax=x\neq0\Rightarrow Ax\neq0\Rightarrow  null(A^2-2A+I_n) \leq rank(A)$, hence [/quote]

at least in my browser the equation is like broken into two lines, but the word hence is in the first line, then it seems awkward! am I doing something wrong?

Best,

There is no problem with that. The equation is being split (a width limit is imposed so that large equations to not create horizontal scroll bars on small resolutions). The simplest way to "fix" this would be to split your latex formula into two consecutive formulae:
$ \displaystyle (A^2-2A+I_n)x=0\Rightarrow (A-2I_n)Ax=x\neq0\Rightarrow Ax\neq0\Rightarrow$$ null(A^2-2A+I_n) \leq rank(A) $


geometry
inequalities
Geometric Inequalities
geometric inequality

Prove that in every triangle $ABC$ with sides $a,b,c$ we have
\[a^2+b^2+c^2 \ge 4S\sqrt3\max\left\{\frac{m_a}{h_a}, \frac{m_b}{h_b}, \frac{m_c}{h_c} \right\}\]
where $S$ is the area of $\triangle ABC$ and $m_a, m_b, m_c$ and $h_a, h_b, h_c$ are the medians and altitudes of the triangle corresponding to the sides $a,b,c$ respectively.

WLOG consider that $\max\left\{\frac{m_a}{h_a},\frac{m_b}{h_b},\frac{m_c}{h_c}\right\}=\frac{m_a}{h_a}$.
Hence it is enough to prove that
\[a^2+b^2+c^2 \ge 4S\sqrt3\frac{m_a}{h_a}\]
We replace $S$ with $\frac{ah_a}2$ to get that the inequality is equivalent to
\[\frac{a^2+b^2+c^2}{2ah_a} \ge \sqrt3\frac{m_a}{h_a} \iff (a^2+b^2+c^2)^2 \ge 12a^2m_a^2\] \[(a^2+b^2+c^2)^2 \ge 3a^2(2b^2+2c^2-a^2) \iff  (2a^2-b^2-c^2)^2 \ge 0\] which is obvious.


How do we derive this formula for the gravittional potentital energy of two point masses separated by a distance $ r$:

$ U \equal{} \minus{}\frac{Gm_1m_2}r$?

It seems to me that it's just the gravitational force between the masses multiplied by the distance between them, but what about the negative sign?

(Btw, why is the formula for the gravitational force between two masses $ F_g\equal{}\frac{Gm_1m_2}{r^2}$?)

Let's start with your second question.  This is an experimentally observed fact, not a matter of mathematics.

Given that fact, recall that the work due to a force $ F$ is given by $ \int F \cdot dr$, hence the work due to the gravitational force as we move two masses from a distance $ r_0$ to infinity is

$ \int_{r_0}^{\infty} \frac {Gm_1 m_2}{r^2} \, dr \equal{} \minus{} \frac {Gm_1 m_2}{r_0}$.

The sign is negative because moving two masses closer together is a more favorable situation (has less energy) than two masses further apart; moving two masses further apart increases $ r_0$, which increases their potential energy and requires work.

I don't think there should be a negative sign here
$ \int_{r_0}^{\infty} \frac {Gm_1 m_2}{r^2} \, dr \equal{} \minus{} \frac {Gm_1 m_2}{r_0}$

My apologies - of course the integral of a positive quantity can't be negative.  What I mean is this:  the gravitational force is always attractive, so when we move two masses apart the force always points opposite the direction of motion, so $ F \cdot dr$ is negative and the relevant integral is

$ \int_{r_0}^{\infty} \minus{} \frac{Gm_1 m_2}{r^2} \, dr$.


The nine point circle passes through the midpoints of the sides and the feet of the altitudes, so if it's tangent to a side, the altitude to that side is the perpendicular bisector of that side. I think you can go from there.

[quote="MNrule"]If a side of Triangle ABC is tangent to its nine- point circle then triangle ABC is isosceles.

How do I prove this?[/quote]

this is direct from the nuine ppoint circle definition.


Prove that:
$a+b+c \leq \frac{3}{2}$
where $a;b;c$ are positivereal numbers satisfying $a^{2}+b^{2}+c^{2}+2abc=1$

Quite easy!!!!
$a=cosA,b=cosB,then c=cos(\pi-A-B)$


trigonometry
trig identities
Law of Cosines

A triangle with consecutive integer side lengths also has the property that its largest angle is twice as large as its smallest angle. Find the cosine of its smallest angle.

it is a 4,5,6 triangle, the other angle is $\frac{4^2-5^2-6^2}{-2*5*6}=\cos\theta$

[quote="arkmastermind"]how did you know it would be a 4,5,6 triangle? (your answer is correct, btw)[/quote]Because we've seen this before.  :D

I think this is how you find the side lengths, not entirely sure since I can't work through it all right now.

[hide="how to find side lengths"]
Call the side lengths of the triangle $a$, $a+1$, and $a+2$. We know that the largest angle  (call it $\angle M$) is opposite the side $a+2$, so using law of cosines, we have:
$\cos M=\frac{(a+1)^2+a^2-2a(a+1)\cosh}{2a(a+1)}$
$\cos M=\frac{a-3}{2a}$

Similarly, we can find $\cos N$ (where $N$ is the smallest angle, which we know is opposite side $a$).

$\cos N=\frac{a+5}{2(a+2)}$

We know that $M=2N$, and $\cos2\theta=2\cos^2\theta-1$

So: $\frac{a-3}{2a}=2\displaystyle \left(\frac{a+5}{2(a+2)} \right)^2-1$

You can then solve for $a$ to get the side lengths.

[/hide]


Εύκολη: 2^x +3^x = 5^x

Δύσκολη (messy): 2^x +3^y = 5^z

the easy one can be rewritten as $(\frac{2}{3})^x+(\frac{3}{5})^x=1$

and finally the only solution is x=1 due to monotony.

for the other i'm thinking right now....

i've just found that y must be even.

Γεαι!!!Τα x,y,z μαλλον ειναι φυσικοι,ε????
Το θεμα αυτο ειναι θεμα διαγωνισμου,απλως δεν θυμαμαι απο ποιον......
Αν θυμαμαι καλα ειναι (χ.y,z)=(2,0,1),(1,1,1),(4,2,2)
Αποδειξη αυριο η και μεθαυριο....

ΡΑΛΛΗΣ :D

Μπράβο Σιλουανέ!

   Για να βρεις το κατάλληλο mod όμως χρειάζεσαι αρκετή ώρα. Το πρόβλημα αυτό είναι η αρχή μιάς (άπειρης)ακολουθίας Διοφαντικών εκθετικών εξισώσεων: Έστω (α,β,γ) μια Πυθαγόρεια τριάδα. Να βρεθούν οι ακέραιες λύσεις της εξίσωσης

   α^x + β^y = γ^z

Για κάθε τριάδα και διαφερετικό mod!

Ποια είναι η λύση σας στο $2^x+3^y=5^z$   ??
Μπορούμε να βρούμε μια τεχνική στην εύρεση του κατάλληλου mod ώστε να μην έχουμε τόσες πολλές δοκιμές ?? ;)

Σιλουανέ,

Η συνταγή έχει ως εξής: Ξεκινάς με μικρές δυνάμεις του 2 (ή του α στη γενική περίπτψση, όσες βαστάει η ψυχή σου), μετά το ίδιο με δυνάμεις του 3 (β) και του 5 (γ). Μετά διαλέγεις κατάλληλους πρώτους για να αποκλείσεις εκθέτες μεγαλύτερους του 2 (ή του 4 για τον x στην περίπτωση που το 4 έγινε 2). Μετά κάθεσαι πίσω και παίρνεις το γινόμενο των mods και dumbfound το ακροατήριο! Για πιο γενικές εκθετικές Διοφαντικές βλ. Leo Alex (Λεωνίδας Αλεξάκης) & Loraine Foster εργασίες στο Pacific. Πάντως τύπος δεν υπάρχει!

Lance

Πολύ ωραία ευχαριστώ πολύ . :D   .Καλό θα ήταν που και που να βάζετε ασκήσεις σαν και αυτή...... ;)
Και όχι μόνο στη θεωρία αριθμών  :o


[This turned into a discussion about something else]

[quote="bugzpodder"]http://mathworld.wolfram.com/SkolemSequence.html

Show that a necessary condition for a Skolem Sequence of order n to exist is n=4k or n=4k+1[/quote]

http://mcs.open.ac.uk/mjg47/papers/skol.pdf

How did you do on the Euclid?

[quote="bugzpodder"]eek?  who are you?? how did you know i took the euclid?[/quote]

Because I did this morning as well.  How else would I know?

You've PMed me before, anyways.  I figure I got around 90 - you?

[quote="bugzpodder"]I did?? I am completely out of it!! yeah I did.  same here, ~90.

either I am really slow (which is probably the case), or that I do not see the proof for this theorem in the pdf...[/quote]

Nope, that was my fault.  They just cite some book, I didn't read the pdf close enough.

But anyways, I made one stupid computation error, for which I think I lost 1-2, and didn't answer 9c and 10c.  I figure that this should give me (78-79)+(5-7)+(5-7) --> (88-93)

[quote="bugzpodder"]obviously i didnt get 10c, but also i missed the drawer question as well as the f(g(X))  function.  John, from my school i believe got everything correctly cept for 10c[/quote]

How did you do 9c?

[quote="bugzpodder"]long story... my method is far from elegant anywayz, i think i got it down to z^2/x^2=(p^2-4q^2)/3q^2, where y=(p/q)x, gcd(p,q)=1, p,q are integers

z^2=(p^2-4q^2)/3  (you can analyze that z^2=(p^2-4q^2) doesnt work)
from the fact that z is a prime and gcd(p,q)=1
we get for which p+2q=3z^2 and p-2q=1  also you want p/q to be as close to 1.98 as possible.  solving you get (37,73,11) or something close to that.[/quote]

Interesting.  I wish I had more time to do that, as a minor mistake on 10b meant that it took me about half an hour too long.

I didn't take the Euclid but someone gave me the question sheet.

BTW...is 8a) b=-3 c=0

just wondering...

Yes. (by the way, I've heard as many answers for 9c as there were people doing it - which ain't many.)

*unintelligible rant*

This year's #10 is a wild goose chase. I figure I'd get high 80s. Bleh.

John claims that all you need is the relationship between x, y,z which is

3z^2+4x^2=y^2 (which is equilvalent to z^2/x^2=(p^2-4q^2)/3q^2)

then we have 3z^2=y^2-4x^2, or 3z^2=(y-2x)(y+2x)  similarly y+2x=3z^2, y-2x=1,   then you do the same thing to test for 0.99
so what I did was pretty much redundant (introducing p,q)
I did it because thats how I handled b part...  8 a) yes, thats correct

btw I am 99% sure that my solution to 9c) is correct...  our school has the habit of collecting everyone's papers, premark them and distribute the marks (of course, non-official marks)... so the answers we come up with are mostly correct  (only once had there being a mistake in the last 4 years for all the contests)  but i think they'll mail out the solution booklet (at least to my teacher) in 3 days

Well, you're from the [b]Honourable[/b] (take that to John for a laugh?) Vincent Massey, whose math program is second to none in Canada.

As opposed to AY Jackson whose sole reason for not participating in the American Math Competitions is that there are not enough teachers to supervise.The last good math student AY has had was Keon Choi, and that was a good 5 years ago...

Anyway. So, yeah, I messed up on Euclid. Didn't even do 10 b).

Um ... what are all of you talking about?

Euclid, the grade 12 Canadian math competition.

thats too bad Tom...

well for one our math director Dr. White is retiring this year... so I have no idea how our school is gonna be after the grade nines this year graduate four years later.  Hopefully Mr. Ing will develop an equally good program...

I have always been wondering what it takes to have a good track record in contests.  In my opinion, a good program isnt everything - as we are not a magnet school (Well okay, we are a magnet school in the city of Windsor, which only has 200,000 population).  Just speaking of the graduates this year, in my opinion there is generally a lack of enthusiasm - hence bad results for Euclid today - last year our team was 98+96+96, this year its probably at most 96+92+91 (For the same reason our ARML results will probably drop three notches this year -- although somehow we were 3rd in the Power Contest --- cant believe we beat TJ and Stuy).   Most of them just dont care enough...  Maybe this is true for other "math schools" also, but I dont know, I am just speaking from my persepective.  We still do well in ARML with great teamwork which is usually not present in American Teams, as they tend to focus on the individual aspects more.  Of course there are always Jacob and Janos from UTS who are just too good.

I dont know why I said all this, guess I am just bored

Jacob from UTS...is he the guy that has probably been ranked number one or close to that in many of the Canadian math contests?

he did not get #1 in CMO last year and unoffical* USAMO last year (used for Canadian IMO selection), as well as Fermat this year.  he has scored perfect or at least first in canada on all other contests, including a gold medal from IMO last year and 150 on AMC 12 this year...  this guy is probably almost as good as reid barton or tiankai at the least.

OMG tom you are right, my answer for 9c is wrong... I am soo pathetic, I should just kill myself right now


ratio geometry trapezoid AMC AIME rectangle trigonometry

An angle is drawn on a set of equally spaced parallel lines as shown.  The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$.  Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$.

[asy]
defaultpen(linewidth(0.7)+fontsize(10));
for(int i=0; i<4; i=i+1) {
fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);
}
pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
draw(B--A--C);
fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white);
clip(B--A--C--cycle);
for(int i=0; i<9; i=i+1) {
draw((i,1)--(i,6));
}
label("$\mathcal{A}$", A+0.2*dir(-17), S);
label("$\mathcal{B}$", A+2.3*dir(-17), S);
label("$\mathcal{C}$", A+4.4*dir(-17), S);
label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

[hide="hint"]
areas of trapezoids are in arithmetic progression.
[/hide]

[hide="hint"]
WLOG parallel lines are 1 apart.
[/hide]

[hide="hint"]
base of each triangle/trapezoid is also in arithmetic progression
[/hide]

[hide="Solution Outline"]
If we let $x$ be the height of the triangular region A, $b$ be the base of region A, and $h$ be the distance between parallel lines, then we can use similar triangles to show that the longest base of the $n$th trapezoid from region A is given by \[ \frac{x+nh}{x}\cdot b \]
From this, you can relate $x$ and $h$ using [C]/[B] = 11/5 and use that relation to solve for [D]/[A]
[/hide]

The areas of the trapezoids follow an arithmetic progression.  WLOG, we can assign areas, $A_B=5, \thinspace A_C=11.$  Then $A_D=17.$  Including the unshaded trapezoids, the areas are $2,\thinspace 5,\thinspace 8,\thinspace 11, \thinspace\ldots.$  The average base of each trapezoid also follows an arithmetic progression.

The AIME allows graph paper, and I think this is one of those questions where a redrawn sketch will really help.  A careful sketch shows that triangle A has height, $\frac{1}{2},$ and base, $\frac{1}{6}.$  So the area of triangle A is $\frac{1}{24}.$  $A_D/A_A=408$

[quote="zanttrang"][quote="ccy"]I had an ugly solution...

[hide]
Let the area of region A be $A$, B be $B$, C be $C$ and D be $D$.  Let the distance between each of the line segments intercepted by the parallel lines be $x$.  Let $a$ equal the distance between the vertex of the angle and the first parallel line it hits.

By similar triangles, $B=A\left(\frac{a+2x}{a}\right)^2-A\left(\frac{a+x}{a}\right)^2=\frac{A}{a^2}(3x^2+2ax)$
$C=A\left(\frac{a+4x}{a}\right)^2-A\left(\frac{a+3x}{a}\right)^2=\frac{A}{a^2}(7x^2+2ax)$
So
$\frac{C}{B}=\frac{7x^2+2ax}{3x^2+2ax}=\frac{11}{5}$, so $x^2=6ax$.  Since it doesn't really matter, let $a=1$, so $x=6$.
Doing the same thing for $D$,
$D=A\left(\frac{a+6x}{a}\right)^2-A\left(\frac{a+5x}{a}\right)^2$
$=A\left(1+6x\right)^2-A\left(1+5x\right)^2=A(37^2-31^2)=408A$
Dividing by $A$, you get $408$.
[/hide][/quote]
I did exactly the same thing - and it's not that messy a solution once you see it (to get there, I had to do about a page of algebra that got nowhere).[/quote]

same here
it only took like 4 mins

What I did to solve it fairly simply and quickly was, after finding that the bases are 1,7,13,19,etc, was to draw a rectangle from the two vertexes of the small triangle that are intersected by the parallel line.  Knowing that the smaller base of trapezoid B is 7, the two segments of the base not part of the rectangle are 3 and 3.  I then set the length between each parallel line = to 4.  There are now two 3,4,5 triangles between the first triangle and the first trapezoid, one above and one below the 1X4 rectangle.  Using the upper triangle,  I then formed a similar triangle on the other side of the first parallel line and above the small, shaded triangle.  The smallest side is 1/2, since its half of the triangles base, so the triangles height must be 2/3.  Area of trapezoid D=68*4/2 and area of triangle A = (1*2/3)/2.

For some reason i thought that it would be in geometric progression...

I did it the same way rcv did it, almost exactly.

ccy, could you please explain where you got the A((a + 2x)/a)^2 and A((a +x)/a) ^ 2
Thanks for the help

[quote="solafidefarms"]One can simply assign a measure to the angle given, say 90 degrees, and thus it is pretty easy to find the area ratios.[/quote]
We have a winner!

how would you go about doing this? obviously the angle isn't 90 degrees, wouldn't you have to first solve for the angle?
or could you use in general the law of cosines for any angle?

you can assume the angle to be any you like. the ratio will still be the same. you can assume the distance between the parallel lines is 1 or something too

[hide="Solution"]Let $b$ be the length of the median of trapezoid $\mathcal{B}$, let $c$ be the length of the median of trapezoid $\mathcal{C}$, and let $d$ be the length of the length of the median of trapezoid $\mathcal{D}$.  Define a step as the distance between two adjacent parallel lines.  Let $s$ be the length of a step.

All three trapezoids have the same height, so the ratios of their areas is equal to the ratios of their medians.  Therefore, $\frac{c}{b}=\frac{11}{5}$.  Let $b=5x$ and $c=11x$.  There are two steps between the medians of trapezoids $\mathcal{B}$ and $\mathcal{C}$.  Therefore, in each step, the length between the two rays increases by $3x$.

There are two steps between the medians of trapezoids $\mathcal{C}$ and $\mathcal{D}$.  Therefore, $d=17x$, and the area of $\mathcal{D}$ is $17sx$.

There are one and a half steps between the median of trapezoid $\mathcal{B}$ and the base of triangle $\mathcal{A}$.  Therefore, the base of $\mathcal{A}$ has a length of $\frac{1}{2}x$.

The length between the two rays decreases by $3x$ in each step, but it can only decrease by $\frac{1}{2}x$ more before the rays intersect.  Therefore, the height of triangle $\mathcal{A}$ is $\frac{1}{6}s$, and the area of $\mathcal{A}$ is $\frac{1}{2}\left(\frac{1}{2}x\right)\left(\frac{1}{6}s\right)=\frac{1}{24}sx$.

The ratio between the areas of $\mathcal{D}$ and $\mathcal{A}$ is $\frac{17sx}{\frac{1}{24}sx}=17(24)=\boxed{408}$[/hide]

[quote="calc rulz"]Hmm, I though this problem was much easier than everyone made it out to be.

Just notice that the squares of any 1 dimensional length (ratio of any sides, inradii, altitude, etc, you name it) are in proportion with the areas of any two similar triangles. As well, for a trapezoid between two such similar triangles, the difference between the squares of some 1 dimensional length for each similar triangle will be in proportion with the area.

So here is my really short solution based on that concept:
[hide]So so just use altitude here. Call the height of triangle A to be x. Then we have: $\displaystyle \frac{(x+4)^2 - (x+3)^2}{(x+2)^2 - (x+1)^2} = \frac{11}{5}$. Solving gives $x= \frac{1}{6}$. Then we have the answer is $\displaystyle \frac{(x+6)^2-(x+5)^2}{x^2} = \frac{(\frac{1}{6}+6)^2-(\frac{1}{6}+5)^2}{\frac{1}{6}^2} = 408$.[/hide][/quote]
Wow, that's my exact solution, pretty much verbatim

[quote=RoastBeef][quote=lazysnorlax]Sorry for the bump, but what happens if the leftmost point (is that called the apex?) is on one of those parallel lines? Confused on why this would change the problem, could someone explain? Thanks.[/quote]

cuz then the distance from the apex to regions B and C would change, causing the areas of regions B and C to change. To be fair it doesn't really matter you're given the ratio of the areas of B to C anyway.[/quote]

So it wouldn't change it? Then, why did [quote=Altheman]i thought the vertex was on one of the parallel lines... :wallbash_red: , otherwise, this is so easy...[/quote] (not to point out) Altheman say that it's easier if the vertex was not on the parallel line? Wouldn't be the exact same (maybe the diagram might change)

I am just really confused, I think even the coordinate solution stays the same, since that vertex would be on the line x =0, and so on (here it's not, it's in between x  =0 and x = 1

[hide=Solution]
Refer to the following picture for lengths (and assume distance between parallel lines is $1$ WLOG)
[img]https://i.postimg.cc/QdmSwSDb/image.png[/img]

We substitute side lengths into the equation $\frac{\mathcal{B}}{\mathcal{C}}=\frac{5}{11}$ to get
\begin{align*}
\frac{a+d+a+2d}{a+3d+a+4d} &= \frac{5}{11} \\
\frac{2a+3d}{2a+7d} &= \frac{5}{11} \\
22a+33d &= 10a+35d \\
12a &= 2d \\
6a &= d.
\end{align*}

We use similar triangles to get the ratio $\frac{h}{a}=\frac{h+6}{a+6d}$.

We substitute $6a=d$ into this equation to get
\begin{align*}
\frac{h}{a} &= \frac{h+6}{a+6\cdot6a} \\
\frac{h}{a} &= \frac{h+6}{37a} \\
37ah &= ah+6a \\
36ah &= 6a \\
6h &= 1 \\
h &= \frac16.
\end{align*}

We substitute all this in to $\frac{\mathcal{D}}{\mathcal{A}}=\frac{a+5d+a+6d}{ah}$ to get
\begin{align*}
\frac{a+5d+a+6d}{ah} &= \frac{2a+11d}{ah} \\
&= \frac{2a+11\cdot6a}{a\frac16} \\
&= \frac{68a}{a\frac16} \\
&= 68\cdot6 \\
&= \boxed{408}.
\end{align*}
[/hide]

[hide=quick]Let the area of $\mathcal{B}$ be $5k$, $\mathcal{C}$ be $11k$, $\mathcal{D}$ be $17k$.
Extend the rays to intersect the leftmost line. Then let the vertical segments formed by the rays have lengths $-2.5, 0.5, 3.5, 6.5, 9.5, 12.5, 15.5, 18.5$; this satisfies the area ratios because the medians of the trapezoids are in the ratio $5:11:17$.
For the self-intersecting trapezoid bounded by the two leftmost lines, its "signed" area is $-k$, so the difference between the left triangle's area and the right triangle's area is $k$. Additionally, the left triangle's area is $25$ times the right triangle's area because they are similar. Thus, $[\mathcal{A}]=\frac{k}{24}$.
The desired ratio is $\frac{17k}{k/24}=\boxed{408}$.

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLscelesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

Scale the diagram so that the distance between parallel lines is 1.

Note that the diagram has seven nested similar triangles $T_{0}\dots T_{6}$ in order of increasing size. If $T_{0}$ has height $x$, $T_{k}$ has height $k+x$. So the ratio
$$
\frac{[T_{k}]}{[T_{\ell}]}=\frac{(k+x)^{2}}{(l+x)^{2}}.
$$

We are given that $\frac{[T_{4}]-[T_{3}]}{[T_{2}]-[T_{1}]}=\frac{11}{5}$.

\begin{align*}
\frac{11}{5}&=\frac{(4+x)^{2}-(3+x)^{2}}{(2+x)^{2}-(1+x)^{2}} \\
&=\frac{7+2x}{3+2x} \\
33+22x&=35+10x \\
12x&=2 \\
x&=\frac{1}{6} \\
\implies \frac{[T_{6}]-[T_{5}]}{[T_{0}]}&=\frac{11+2x}{x^{2}}=396+12=\boxed{408}.
\end{align*}


Find real solutions of the system

$ \sin x \plus{} 2 \sin(x\plus{}y\plus{}z) \equal{} 0$,
$ \sin y \plus{} 3 \sin(x\plus{}y\plus{}z) \equal{} 0$,
$ \sin z \plus{} 4 \sin(x\plus{}y\plus{}z) \equal{} 0$

Let  $ x_n \equal{} \pi \, C_n$ with $ C$ integer and we are donne

[quote="ashrafmod"]Find real solutions of the system

$ \sin x \plus{} 2 \sin(x \plus{} y \plus{} z) \equal{} 0$,
$ \sin y \plus{} 3 \sin(x \plus{} y \plus{} z) \equal{} 0$,
$ \sin z \plus{} 4 \sin(x \plus{} y \plus{} z) \equal{} 0$[/quote]
It follows that:
$ sinx \plus{} siny \minus{} sinz \plus{} sin{(x \plus{} y \plus{} z)} \equal{} 0$
$ \Leftrightarrow 2sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} y}{2}} \plus{} 2sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} y\plus{}2z}{2}} \equal{} 0$ (from [b]sum-to-product fomulas[/b])
$ \Leftrightarrow sin{\frac {x \plus{} y}{2}}(cos{\frac {x \minus{} y}{2}} \plus{} cos{\frac {x \plus{} y \plus{} 2z}{2}}) \equal{} 0$
$ \Leftrightarrow sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} z}{2}}cos{\frac {y \plus{} z}{2}} \equal{} 0$

[quote="NguyenDungTN"][quote="ashrafmod"]Find real solutions of the system

$ \sin x \plus{} 2 \sin(x \plus{} y \plus{} z) \equal{} 0$,
$ \sin y \plus{} 3 \sin(x \plus{} y \plus{} z) \equal{} 0$,
$ \sin z \plus{} 4 \sin(x \plus{} y \plus{} z) \equal{} 0$[/quote]
It follows that:
$ sinx \plus{} siny \minus{} sinz \plus{} sin{(x \plus{} y \plus{} z)} \equal{} 0$
$ \Leftrightarrow 2sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} y}{2}} \plus{} 2sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} y \plus{} 2z}{2}} \equal{} 0$ (from [b]sum-to-product fomulas[/b])
$ \Leftrightarrow sin{\frac {x \plus{} y}{2}}(cos{\frac {x \minus{} y}{2}} \plus{} cos{\frac {x \plus{} y \plus{} 2z}{2}}) \equal{} 0$
$ \Leftrightarrow sin{\frac {x \plus{} y}{2}}cos{\frac {x \plus{} z}{2}}cos{\frac {y \plus{} z}{2}} \equal{} 0$[/quote]
You are careless NDung! Cause sina+sinb=2sin(a+b)/2.cos(a-b)/2 sO THE FINAL RESULT IS: sin(x+y/2).cos(x+z).cos(z-y)=0


function
trigonometry
limit
complex analysis
complex numbers
calculus
calculus computations

I couldn't find a way to sum this series:

$f(n) = \sum_{m}\frac{1}{n^{2}-m^{2}}$
where n and m are odd, positive integers. m should run over all positive integers, except the case $m=n$.

I need this to complete the solution of a quantum-mechanical problem. It arises in the 2nd order perturbation of a Dirac delta potential to infinite square well.

Thanks in advance!

[quote="rayne"]where n and m are odd, positive integers. m should run over all positive integers, except the case $m=n$.[/quote]

These two sentences are mutually contradictory.  Does $m$ run over all integers or all odd integers?

In the following, I guessed that $m$ runs over the positive odd integers; the details change if I guessed wrong.

The tools I know for doing something like this are complex analysis.
[hide="Answer"]$\frac{-1}{4n^{2}}$[/hide]
[hide="Solution (long)"]The function $F(z)=\sum_{k=0}^{\infty}\frac{1}{z^{2}-(2k+1)^{2}}= \sum_{k=0}^{\infty}\frac{1}{(4k+2)(z-(2k+1))}-\frac{1}{(4k+2)(z+(2k+1))}$ is a meromorphic function with simple poles at each odd integer and no others. If we set $g(z)=2zF(z)$, $g$ has poles in the same location with all residues equal to 1; we can write $g(z)=\sum_{k=0}^{\infty}\frac{1}{z-2k-1}+\frac{1}{z+2k+1}=\sum_{j=-\infty}^{\infty}\frac1{1-(2j+1)}$, interpreting the latter sum as a periodic function. This last sum is clearly periodic in $z$, so $g$ is periodic with period 2. Since $F$ is clearly even, $g$ is odd.

A simple estimate of size: If $|a|\le 1$ and $|b|$ is large, $|g(a+ib)|\le 2|a+ib|\sum_{k=0}^{\infty}\left|\frac{1}{a^{2}-b^{2}-(2k+1)^{2}+2abi}\right|\le 2(b+1)\sum_{k=0}^{\infty}\frac1{4k^{2}+4k+b^{2}}$
If we split that sum into terms with $k<|b|$ and terms with $k>b$, the first have sum at most $2(|b|+1)\sum_{k\le |b|}\frac{1}{b^{2}}\approx 2$ and the second have sum at most $2(|b|+1)\sum_{k>|b|}\frac1{4k(k+1)}\approx \frac12$. The total is bounded (as long as stay away from the poles on the real axis).
Consider $h(z)=g(z)+\frac{\pi}{2}\tan(\frac{\pi}2z)$; the residues cancel at each odd integer, and the singularities are removable. This $h$ is therefore an entire function with period 2. Since $\tan z$ tends to 1 as the imaginary part of $z$ tends to $\pm\infty$, $h$ is bounded.  By Liouville's theorem, $h$ is constant. Since $g(0)=0$, $g(z)=-\frac{\pi}{2}\tan(\frac{\pi}2z)$, and $F(z)=-\frac{\pi}{4z}\tan\frac{\pi z}2$.

Now, your $f(n)$ is $\lim_{z\to n}F(z)-\frac1{z^{2}-n^{2}}= \lim_{t\to 0}\frac{-\pi}{4(t+n)}\tan\frac{\pi(t+n)}{2}-\frac1{2nt}+\frac1{2n(t+2n)}$
We have $\tan\frac{\pi(t+n)}{2}= \frac{1-\frac{\pi^{2}t^{2}}8+\cdots}{-\frac{\pi t}2+\frac{\pi^{3}t^{3}}{48}+\cdots}=-\frac2{\pi t}+O(t)$; multiplying by $\frac{-\pi}{4(n+t)}=-\frac{\pi}{4n}+\frac{\pi t}{4n^{2}}+O(t^{2})$, we get $\frac1{2nt}-\frac1{2n^{2}}+O(t)$. This balances the $\frac{-1}{2nt}$ term, leaving the limit as $\frac{-1}{2n^{2}}+\frac1{4n^{2}}=\frac{-1}{4n^{2}}$[/hide]

There are probably other ways of getting at this.

Sorry, I was supposed to write "all odd integers"!
Huge thanks, jmerry! The solution now agrees with the correct answer! You just don't know how much you've helped me! I better check out how you've managed to sum it. It'd be great if there was a shorter solution, though  :huh:

It is very likely that there's a shorter solution. I don't feel like looking for it myself.

That doesn't quite work; the problem with your claim is that both sums diverge. There is an argument there, but it needs careful attention to convergence issues.
The only sum I ever wrote down in my argument that wasn't absolutely convergent could be handled as a principal value (I said something else; that's a mistake). The exact value of that sum was also unimportant- I only used that form to argue periodicity.

[Edited under the response]

[quote="jmerry"]The problem with your claim is that both sums diverge. There may be an argument there, but it needs careful attention to convergence issues.[/quote]
yes that's true that's whhy my post is so small further since this was reqd. in quantum physics not giving much mathematical considerations i derived the sum

[quote="jmerry"]There's just no reason to write down a nonsensical equality of two infinities when you could put together a mathematically correct statement that does what you want (consider partial sums- it's a principal value). Physics is not an excuse for bad math.[/quote]
ya that's true  :oops:


Hello

Lets say someone masters multivariable calculus. Is it safe to say then they will know single-variable calculus well? In other words, if a student had a rusty knowledge of single-variable calculus, and then he studies multivariable calculus, will most of the concepts in single-variable calculus be reinforced (i.e. student didnt learn about single-variable delta-epsilon proofs) but then goes and learns it from scratch in multivariable calculus (involving 2 or more variables).

Thanks

Generally, the answer is no. For example, the multivariable calculus class that I teach does not include infinite series or differential equations. Special techniques of integrations like trigonometric substitution are not emphasized. The art of lifting a leaky bucket with a rope of variable density is not practiced at all.

Aside from all of that: differential equations, infinite series, or techniques of integration. I am talking about the basic theorems (i.e. mean value theorem, extreme value theorem, finding maxima and minima, related rates, etc...).

Thanks

Usually it is assumed that the students who take multivariable calculus know the one-variable calculus if not perfectly, then reasonably well. The point is that many proofs of the theorems in multivariable calculus are just reductions to the similar theorems in the one-variable one and even those that are not are somehow analogous to their one-dimensional versions. When I gave lectures in multivariable calculus last semester, I built the entire course around the idea that every statement about functions of several variables is just an extension of some "well-known" statement about functions of one variable and every statement about one variable has a multidimensional analogue.  Another thing I consider a necessary prerequisite for the multivariable calculus is linear algebra. Unfortunately, the university where I teach doesn't require the students to take it before multivariable calculus and I had to spend 5 lectures giving a crush course. If I had to remind the basics of one-variable theory as well, I would never be able to finish the course on time. So, most likely, taking a multivariable class without taking a single variable one first (or without learning it by yourself) will be a bad idea. It doesn't mean you'll not be able to survive but your life won't be easy. As to $\epsilon-\delta$ proofs, they are most easily mastered on the level of limits and continuity and quite often the courses in multivariable calculus start with differentiability, where everything is harder. Of course, a lot depends on how your professor will build the course and the best idea would be to address your question directly to him/her.

Thanks a lot for all your replies. If a student self-studies a math subject, can he obtain credit for the course (is this true in many colleges)? Like if a student follows a syllabus, does the required problems, reads and studies the material thorougly, takes timed sample tests from previous semesters and does well, do colleges allow him to obtain credit? Or is this generally not the case?

Thanks

drat, I was hoping to save some money! I feel that I am learning it more thoroughly by myself, then attending  lecture (i.e. took practice tests).

Thanks for your replies!


geometry
geometric transformation
reflection
trigonometry

Let ABC be a triangle and [AD]  be median of [BC].If   angle m(BAD)/m(DAC)=2 then find [AC]  lengths as a trigonometric functions.

I note the reflection $A'$ of the point $A$ w.r.t. the point $D$ and $x=m(\widehat {DAC})$.

Theorem of Sinus in the $\triangle ACA'\Longrightarrow \frac{c}{\sin x}=\frac{b}{\sin 2x}=\frac{2m_a}{\sin 3x}\Longrightarrow$

$c=\frac{b}{2\cos x}=\frac{2m_a}{4\cos^2x-1}\Longrightarrow b^2=c(c+2m_a)$.

But you don't explain [AC] as  only trigonometric function...

What are the data in this problem ? What know we ? What is entrance and what is exit ?


What is the center of the circle passing through the point $(6,0)$ and tangent to the circle $x^{2}+y^{2}=4$ at $(0,2)$.

[hide]
[asy]import graph; pair A,B,C,D; A=(0,0); B=(0,2); C=(6,0); D=(0,-8); dot(A); dot(C); dot(D); draw(Circle(A,2)); draw(Circle(D,10)); draw(B--A--C); draw(A--D--C,dashed); label("$2$",(0,1),W); label("$6$",(3.5,0),S); label("$(6,0)$",C,NE); label("$x$",(0,-5),W); label("$x+2$",(3,-4),SE);[/asy]
The diagram pretty much says everything.  We have a right triangle:
$6^{2}+x^{2}=(x+2)^{2}\Rightarrow x=8$
Since the centers of the circles and the point of tangency are collinear, the point we need lies on the $y$-axis.  So the center is at $\boxed{(0,-8)}$.
[/hide]


For $ 1\le m\le n$ and the set $ X$ with $ n$ elements, find the number of all  m-tuples $ (X_1, X_2, ...,X_m)$  such that $ X_i\subset X, |X_i|\ge 2$ and $ X_i\cap X_j \equal{} \emptyset$ for $ i\neq j$ and $ i, j, \equal{} 1,2,...,m$.

Enjoy! :oops:

Nobody likes to solve it?  :huh:  Try using the Principle of Inclusion-Exclusion


Rezolvati in $\mathbb C$ ecuatia \[x^{4}+135x-26=0\].Am incercat sa scriu ca un produs de 2 expresii de gradul 2 dar nu am reusit.

[quote="Svejk"]Rezolvati in $\mathbb C$ ecuatia
\[x^{4}+135x-26=0 \]
.Am incercat sa scriu ca un produs de 2 expresii de gradul 2 dar nu am reusit.[/quote]
$(x^{2}-5x+26)(x^{2}+5x-1)$


The sum of three integers \textit{a},\textit{b} and \textit{c} is 0. Prove that $2a^4+2b^4+2c^4$ is the square of an integer.


vector
inequalities
real analysis
real analysis unsolved

$X$ is an $n$-dimensional vector space and $\{e_1,\dots,e_n\}$ is a basis. $(t_{\mu \nu})$ is an $n\times n$ matrix. Define an operator from $X$ to $X$ as follows: let $x=\sum_{\nu=1}^n x_\nu e_\nu$, define
$y=Tx=\sum_{\mu=1}^n y_\mu e_\mu$, where $y_\mu = \sum_{\nu = 1}^n t_{\mu \nu} x_\nu, \quad \mu=1,2,\dots, n$. We define the norm of $x$ as $\|x\| = \left(\sum_{\nu=1}^n |x_\nu|^2\right)^\frac{1}{2}$. Prove the norm of the operator satisfy
\[\max_\nu \left(\sum_{\mu=1}^n |t_{\mu\nu}|^2\right)^\frac{1}{2} \leq \|T\| \leq \left(\sum_{\mu=1}^n\sum_{\nu=1}^n |x_{\mu\nu}|^2\right)^\frac{1}{2}.\]

I can prove that right-hand side inequality, how I don't know how to prove the left-hand side one.

It's a vector space over $\mathbb C$, right? If you assume, for simplicity, that $\max_\nu \left(\sum_{\mu=1}^n |t_{\mu\nu}|^2\right)^\frac12$ is equal to $\left(\sum_{\mu=1}^n |t_{\mu 1}|^2\right)^\frac12$, we can simply apply our operator to $1\cdot e_1$ (or we may represent it as $(1,0,0,\ldots,0)$). I believe this proves the inequality on the left.

But I think you prove the Equality, not the Inequality

So? :) If there is an $x$ s.t. $\|Tx\|=\lambda\|x\|$ (for some $\lambda\ge 0$), then $\|T\|\ge \lambda$, because $\|T\|=\sup_{x\ne 0}\frac{\|Tx\|}{\|x\|}$. This means that finding equality for a vector implies the desired inequality.

I did not realise it...   :blush:  :blush:


Generally, how would you find the number of rectangles enclosed in an N by M dot grid when all of the vertices of the rectangles are vertices inside the region?

Is this where all sides of the rectangle are parallel to sides of the array? If so...

[hide="hint"] A rectangle has two vertical sides and two horizontal ones. Thus, you need to find the number of ways to choose two vertical sides and two horizontal ones.[/hide]

[hide="solution"] Out of the M points, you need to choose two to form part of the rectangle. You also need to choose two of the other N to form the rest of it. Thus there are $ \binom{M}{2}\cdot\binom{N}{2}$ rectangles.[/hide]

oh oops i actually meant squares instead of rectangles  :oops:

If we want the number of squares with sides parallel to the edges and $ m\ge n$ (where $ m$ and $ n$ are the number of dots on each side), then there are $ (m\minus{}1)(n\minus{}1)\ \ 1\times1$ squares, $ (m\minus{}2)(n\minus{}2)\ \ 2\times2$ squares, etc., $ (m\minus{}k)(n\minus{}k)\ \ k\times k$ squares, etc., and $ (m\minus{}n\plus{}1)\ \ (n\minus{}1)\times(n\minus{}1)$ squares. Thus there are
$ \sum_{k\equal{}1}^{n\minus{}1}(m\minus{}k)(n\minus{}k)\equal{}\sum_{k\equal{}1}^{n\minus{}1}mn\minus{}k(m\plus{}n)\plus{}k^2\equal{}mn\sum_{k\equal{}1}^{n\minus{}1}1\minus{}(m\plus{}n)\sum_{k\equal{}1}^{n\minus{}1}k\plus{}\sum_{k\equal{}1}^{n\minus{}1}k^2$
$ \equal{}mn(n\minus{}1)\minus{}(m\plus{}n)\cdot\frac{n(n\minus{}1)}2\plus{}\frac{n(n\minus{}1)(2n\minus{}1)}6$
$ \equal{}\frac{n(n\minus{}1)}6(6m\minus{}3(m\plus{}n)\plus{}(2n\minus{}1))\equal{}\frac{n(n\minus{}1)}6(3m\minus{}n\minus{}1)$
$ \equal{}\boxed{\frac{n(n\minus{}1)(3m\minus{}n\minus{}1)}6}$ squares total.

If we want the number of squares with all four vertices on the points in the array of $ m\times n$ dots with $ m\ge n$, then it's somehow (the rather elegant) $ \boxed{\frac{n(n^2\minus{}1)(2m\minus{}n)}{12}}$. It is beyond me to derive this, though...

Normally you wouldn't even need the formulas anyways :wink:

The logic used in the first part of the above post is sufficient for MathCounts purposes


algebra
polynomial
number theory
modular arithmetic

I've been thinking about doing modular arithmetic (perhaps not arithmetic now?) with polynomials. I've searched the internet, but I don't really seem to understand the results there. Can someone just give me an introduction to it, or give me a link with a nice introduction?

Thanks.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]You're in luck![/url]  I can understand why it would be difficult to find a low-level introduction; the first place it seems the notion of modding out by a polynomial occurs in the standard undergraduate curriculum is in algebraic number theory courses when you want to explicitly construct field extensions, but it is possible to approach the idea from the same perspective that you approach normal modular arithmetic with the important difference that you can "substitute" values for $ x$, which makes your life a lot easier.


inequalities
geometry
inequalities proposed

:lol: :)  Let ABC be a triangle and a real number k>0. Prove that:
$(ka)^2\ + 4m_a^2\ \ge\ 8kS.\ \ \ (*)$ When is equality ?   .A particular case:
$\ k=\sqrt{3}\ \ ===>\ \   a^2+b^2 +c^2\ge 4\sqrt{3}\ (The\ Weitzenbock's\ inequality).$ :rotfl:

We note that the area $S$ is equivalent to $\frac{1}{2}(a)(m)sin(x)$, where $x$ is the angle between the median m and the side $a$, and thus $(m)sin(x)$ is the height.

Thus $8kS$ is equivalent to $4(ka)(m)sin(x)$.

By the Trivial Inequality and the fact that $sin(x)$ never exceeds 1, we have $(ka-2m)^{2}+4(ka)(m)(1-sin(x)) \geq 0.$ This implies $(ka)^{2}-4(ka)(m)+4m^{2}+4(ka)(m)(1-sin(x)) \geq 0,$ which further implies that $(ka)^{2}+4m^{2} \geq 4(ka)(m)sin(x).$ Substituting the equivalent expression for the right hand side, we yield the desired inequality $(ka)^{2}+4m^{2} \geq 8kS.$ We note that the equality case occurs only when both $(ka-2m) = 0$ and $4(ka)(m)(1-sin(x)) = 0.$ This is when $k = \frac{2m}{a}$ and $x$ is a right angle, implying the triangle is isosceles. QED


Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 3.

[hide]By Pigeon Hole Principle, there are three "boxes" and five objects, so at least two "boxes" must have at least one object; either there are two numbers $\equiv 0, 1, \text { or } 2$. Write the list mod 3: $\{ 0, 1, 2, x, y \}$ for some values $0 \le x,y \le 2$. Therefore we are done because $0 + 1+ 2 \equiv 0 \mod {3}$.[/hide]

Did I misinterpret the question? That was easy...


pigeonhole principle
combinatorics unsolved
combinatorics

Suppose that $ \mathbb{X}$ is an $ n$-element set,$ n\geq 12$ and suppose $ \mathbb{F}$ is a family of $ 4$-element subsets of $ \mathbb{X}$ such that the intersection of each pair of distinct sets in $ \mathbb{F}$ has at most two elements. Prove that there is a subset $ \mathbb{S}$ of $ \mathbb{X}$ containing at least $ (6n\minus{}3)^{ \frac 13}$ elements such that none of the $ 4$-element subsets of $ \mathbb{S}$ is in the family $ \mathbb{F}$.

I don't think it's that hard, although I didn't work through the numbers. But probably you can just take S to have maximal cardinality while satisfying the conditions and consider each $ x \in X, x \not\in S$. There must be a four-element subset in $ F$ which consists of $ x$ and three elements of $ S$... the numbers work out so that in all likelihood, you will be able to argue by pigeonhole that some two subsets in $ F$ will cover the same three elements of $ S$, leading to contradiction, unless $ |S|$ is sufficiently large.