integration ratio LaTeX function logarithms trigonometry limit

Give with proof and explicit formula for 	$ \sum_{k\equal{}1}^{\infty} \frac{1}{(4n\minus{}1)(4n\plus{}1)}$.

[quote="primoz2"]Give with proof and explicit formula for 	$ \sum_{k \equal{} 1}^{\infty} \frac {1}{(4n \minus{} 1)(4n \plus{} 1)}$.[/quote]

construct a power series $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {x^{4n \plus{} 1}}{(4n \minus{} 1)(4n \plus{} 1)}$,radius of convergence is 1.Differential it w.r.t x,it follows that

$ \frac {d}{dx}(f(x)) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {x^{4n}}{4n \minus{} 1} \equal{} x\sum_{n \equal{} 1}^{\infty} \frac {x^{4n \minus{} 1}}{4n \minus{} 1}$.Let$ g(x) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {x^{4n \minus{} 1}}{4n \minus{} 1}$,  $ \frac {d}{dx}(g(x)) \equal{} \sum_{n \equal{} 1}^{\infty} x^{4n \minus{} 2} \equal{} \frac {x^{2}}{1 \minus{} x^{4}}$

$ g(x) \equal{} \int_{0}^{x}\frac {x^{2}}{1 \minus{} x^{4}}dx \equal{} \minus{} \frac {1}{4}ln(1 \minus{} x) \plus{} \frac {1}{4}ln(1 \plus{} x) \minus{} \frac {1}{2}arctan(x)$

$ f(x) \equal{} \int_{0}^{x}x( \minus{} \frac {1}{4}ln(1 \minus{} x) \plus{} \frac {1}{4}ln(1 \plus{} x) \minus{} \frac {1}{2}arctan(x))dx \equal{} \frac {1}{8}ln(1 \minus{} x) \minus{} \frac {1}{8}x^{2}ln(1 \minus{} x) \plus{} \frac {1}{8}x^{2}ln(1 \plus{} x) \minus{} \frac {1}{8}ln(1 \plus{} x) \plus{} \frac {x}{2} \minus{} \frac {1}{4}x^{2}arctan(x) \minus{} \frac {1}{4}arctan(x)$
finally,let $ x$ approaches 1,we get the result

$ \sum_{n \equal{} 1}^{\infty} \frac {1}{(4n \minus{} 1)(4n \plus{} 1)} \equal{} \frac {1}{2} \minus{} \frac {\pi}{8}$

Ratio test.

Deng Qunyi, very nice solution.  LaTeX has lots of built-in functions: \ln, \sin, \arctan, etc.  They make everything easier to read :)

Sorry, this isn't a telescoping series at all. So here's a different solution:

Since $ \frac{1}{\left(4n\minus{}1\right)\left(4n\plus{}1\right)}\equal{}\frac{1}{2}\left[\frac{1}{4n\minus{}1}\minus{}\frac{1}{4n\plus{}1}\right]$, we have

$ 2\sum_{k\equal{}1}^{n}\frac{1}{\left(4k\minus{}1\right)\left(4k\plus{}1\right)}\equal{}\left(\frac{1}{3}\plus{}\frac{1}{7}\plus{}\frac{1}{11}\plus{}\cdots\plus{}\frac{1}{4n\minus{}1}\right)\minus{}\left(\frac{1}{5}\plus{}\frac{1}{9}\plus{}\frac{1}{13}\plus{}\cdots\plus{}\frac{1}{4n\plus{}1}\right)$.

Reordering, we get

$ 2\sum_{k\equal{}1}^{n}\frac{1}{\left(4k\minus{}1\right)\left(4k\plus{}1\right)}\equal{}1\minus{}\sum_{k\equal{}0}^{2n}\frac{\left(\minus{}1\right)^{k}}{2k\plus{}1}\equal{}1\minus{}\frac{\pi}{4}\equal{}2\left(\frac{1}{2}\minus{}\frac{\pi}{8}\right)$

by the Leibniz-Gregory formula. Thus the sum is $ \frac{1}{2}\minus{}\frac{\pi}{8}$.

[quote="Mathias_DK"][quote="Deng Qunyi"][quote="primoz2"]Give with proof and explicit formula for 	$ \sum_{k \equal{} 1}^{\infty} \frac {1}{(4n \minus{} 1)(4n \plus{} 1)}$.[/quote]

construct a power series $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {x^{4n \plus{} 1}}{(4n \minus{} 1)(4n \plus{} 1)}$,radius of convergence is 1.[/quote]
Why is that?[/quote]

denote $ u_{n}\equal{}\frac{x^{4n\plus{}1}}{(4n\minus{}1)(4n\plus{}1)}$，$ \lim\limits_{n\to\infty}\frac{|u_{n\plus{}1}|}{|u_{n}|}\equal{}x^{4}<1$,so radius of convergence is 1.

(And it also converges for all $ x$ such that $ |x| \equal{} 1$ by comparison with $ \sum \frac{1}{n^2}$.)


I saw this on here a couple of days ago and the person stated they needed help with it, part C however I wanna know if I could get some help with part A of the following problem:

3. a) Show that a power of 3 can never be a perfect number.

b) More generally, if p is an odd prime, show that a power pk can never be a perfect number.

c) Show that a number of the form 3i5j can never be a perfect number.

d) More generally, if p and q are distinct odd primes, show that the product piqj can never be a perfect number.

Does anyone know why a power of 3 , or any odd prime is never perfect?

I don't understand why the proof I gave here does not satisfy you :
http://www.mathlinks.ro/Forum/viewtopic.php?t=119766
 :wink:

Pierre.

so it works for single primes like three as well?

If there are questions etc., add them to the original thread.
Locked.


For this to be true, either $x^2-5x+6=0$ or $x^2+3x+1=1$, or both equations are equal to $0$, so therefore you get the solutions: $x=3,2,-3,0$

Is this what iversonfan and 236 factorial missed?

[hide]$x^2+3x+1$ could be -1, if $x^2-5x+6$ is even.
So solving for $x^2+3x+1 = -1$, you get x=-1 or x=-2.  If x=-1, then $x^2-5x+6$ is 12, which is even, so it works.  And if x=-2, then $x^2-5x+6$ is 20, which also works.  So, with the solutions that iversonfan and 236facotrial mentioned, x=-3, -2, -1, 0, 2, 3$[/hide]

Yes, it is what I missed  :roll:
I was thinking there were more solutions. But -1^even number=1!!! *hits self on head*

[quote="h_s_potter2002"]Is this what iversonfan and 236 factorial missed?

[hide]$x^2+3x+1$ could be -1, if $x^2-5x+6$ is even.
So solving for $x^2+3x+1 = -1$, you get x=-1 or x=-2.  If x=-1, then $x^2-5x+6$ is 12, which is even, so it works.  And if x=-2, then $x^2-5x+6$ is 20, which also works.  So, with the solutions that iversonfan and 236facotrial mentioned, $x=-3, -2, -1, 0, 2, 3$[/hide][/quote]
That's correct.
People often miss the last case ($x^2+3x+1=-1$) and that was a tricky part :P


Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is

$ \textbf{(A)}\ \frac{3}{5}\qquad\textbf{(B)}\ \frac{\sqrt 5}{3}\qquad\textbf{(C)}\ \frac{2}{\sqrt 5}\qquad\textbf{(D)}\ \frac{\sqrt 5}{2}\qquad\textbf{(E)}\ \frac{5}{3} $

[hide="Answer"]$\sin A=\frac{2}{3}=\frac{opp}{hyp}$, so $adj=\sqrt{3^2-2^2}=\sqrt{5}$ and $\tan A=\frac{opp}{adj}=\frac{2}{\sqrt{5}}\Rightarrow \tan B=\frac{\sqrt{5}}{2}\Rightarrow \boxed{D}$[/hide]

[b]EDIT:[/b] Whoops, fixed.

[quote="JesusFreak197"][hide="Answer"]$\sin A=\frac{2}{3}=\frac{opp}{hyp}$, so $adj=\sqrt{3^2-2^2}=\sqrt{5}$ and $\tan A=\frac{opp}{adj}=\frac{2}{\sqrt{5}}\Rightarrow \boxed{C}$[/hide][/quote]

They ask for $tanB$ JesusFreak197

[hide]We have $AB=3$, $BC=2$, and $AC=\sqrt5$ by Pythagorean theorem.

Then $tanB=\frac{\sqrt5}{2}$

$\boxed{D}$[/hide]

[hide]We can see that $\cos{A}=\frac{\sqrt5}{3}$. Thus $\tan{B}=\frac{\sqrt5}{2}$. [/hide]


trigonometry
calculus
calculus computations

2:
$ \frac{dr}{dt}\equal{}4$

$ \frac{dh}{dt}\equal{}\minus{}10$

$ \frac{dV}{dt}\equal{}2\pi rh\left(\frac{dr}{dt}\right)\plus{}\pi r^{2}\left(\frac{dh}{dt}\right)$

Now plug in:

$ \frac{dV}{dt}\equal{}2\pi (3)(5)(4)\plus{}\pi (3)^{2}(\minus{}10)\equal{}\pi(120\minus{}90)\equal{}30\pi$

The general idea is to list everything that is changing, and make sure you plug everything in and take every derivative.

[quote="magicalorama"]Can you help me figure these out? I only have 4 problems left!

1. A man 6 ft tall walks at the rate of 200 feet per minute towards a street light which is 18 ft above the ground. At what rate is the tip of his shadow moving?
2. Ilsa subjects a cylindrical can to a certain transformation as part of her project for the institute. During this transformation the radius and height vary continuously with time. The radius is increasing at 4 in/min, while the height is decreasing at 10 in/min. Is the volume of the can increasing or decreasing, and at what rate, when the radius is 3 inches and the height is 5 inches?
3. A particle moves along a path described by $ y \equal{} x^{2}$. At what point along the curve are $ x$ and $ y$ changing at the same rate? Find this rate if at that time $ t$ we have $ x \equal{} sin t$ and $ y \equal{} sin^{}2 t.$
4. Two straight roads intersect at right angles near the Krupps factory at Baden-Baden. Fabio drives his scooter towards the intersection at a rate of 50 kph. Ilsa drives her scooter on the other road away from the intersection at a speed of 30 kph. When Fabio is 2 kilometers from the intersection and Ilsa is 4 kilometers from the intersection:
(a) How fast is the distance between them changing?
(b) Are Fabio and Ilsa getting closer together or farther apart?[/quote]
Nr. 1:
Draw a sketch. It is terribly easy really :P
Let $ d$ be the distance from the man to the street light. Then the distance to the tip of the shadow is $ \frac{3}{2}d$ So the tip is moving at speed $ 300$ feet per minute.

[quote="magicalorama"]4. Two straight roads intersect at right angles near the Krupps factory at Baden-Baden. Fabio drives his scooter towards the intersection at a rate of 50 kph. Ilsa drives her scooter on the other road away from the intersection at a speed of 30 kph. When Fabio is 2 kilometers from the intersection and Ilsa is 4 kilometers from the intersection:
(a) How fast is the distance between them changing?
(b) Are Fabio and Ilsa getting closer together or farther apart?[/quote]
Let $ t$ be the time (in hours) after the starting point.
The distance between them is acording to pythagoras: $ \sqrt{(2\minus{}50t)^2 \plus{} (4\plus{}30t)^2} \equal{} \sqrt{3400t^2 \plus{} 40t \plus{} 20}$. Differentiating with $ t$ gives the change: $ \frac{7800t \plus{} 40}{2\sqrt{3400t^2\plus{}40t\plus{}20}} > 0$(in kph) for all $ t > 0$

[quote="magicalorama"]

3. A particle moves along a path described by $ y \equal{} x^{2}$. At what point along the curve are $ x$ and $ y$ changing at the same rate? Find this rate if at that time $ t$ we have $ x \equal{} sin t$ and $ y \equal{} sin^{}2 t.$
[/quote]

$ y\equal{}x^2 \to \frac{dy}{dt}\equal{}2x\frac{dx}{dt}$

$ \frac{dy}{dt}\equal{}\frac{dx}{dt} \to 2x\equal{}1 \to x\equal{}\frac{1}{2} \to y\equal{} \frac{1}{4}$

$ (\frac{1}{2},\frac{1}{4})$

$ x\equal{}\sin t, y\equal{}\sin 2t$
$ \frac{dx}{dt}\equal{}\cos t, \frac{dy}{dt}\equal{}2\cos 2t$

$ \cos t \equal{} 2 \cos 2t \equal{} 2(2\cos^2 t \minus{}1)\equal{} 4\cos^2 t \minus{} 2$

$ 4u^2\minus{}u\minus{}2\equal{}0 \to u\equal{}\frac{1 \pm \sqrt { 33}}{8}\equal{}\cos t$

$ t\equal{}\cos^{\minus{}1} \left(\frac{1 \pm \sqrt { 33}}{8} \right)$  plug in for rate


You haven't answer the question yet  :rotfl:

I know, that's why I said 'My algebra solution' and not 'Answer to your problem'  :P  :rotfl:

:wallbash: [hide]Could it be 171700?[/hide]

The sum of the first $n$ positive integers = $\frac{n(n+1)}{2}$ (triangular numbers)

The sum of the first $n$ triangular numbers = $\frac{n(n+1)(n+2)}{6}$

... = $\frac{n(n+1)...(n+k)}{(k+1)!}$

$5050=\frac{n(n+1)}{2}$
$n=100$

$5050 \cdot \frac{102}{3}=5050 \cdot 34 = 171700$

[quote="anirudh"]how do you do this kind of problem  :([/quote]

look above  :roll:
use a formula.

The correct answer is
[hide]171700  :D [/hide]

[quote="anirudh"]how do you do this kind of problem  :([/quote]

Look at the way frt analyzed it. I think that is a good general approach to this type of problem. Immediately, you can see that it is the sum of a sum. It is the sum of the firsth n triangular numbers, and the kth triangular number is the sum of the first k integers. Now, if you write it as the sum of a sum, and simplify it (since we know what the sum of the first k integers formula ), we get what frt has.

[quote="mathmanman"]Ok, now, can you prove the sum of two consecutive triangular numbers is a perfect square? :)[/quote]

[hide]let n-1 be one number, n be the next.

$\frac{(n-1)(n)}{2}+\frac{(n)(n+1)}{2}=\frac{n^2-n+n^2+n}{2}=\frac{2n^2}{2}=n^2$[/hide]


Let us choose arbitrary $n$ verteces of a regular $2n$-gon and colour them red. The remaining vertices are coloured blue. We arrange all red-red distances into a non-decreasing sequence and do the same with the blue-blue distances. Prove that the sequences are equal.

I have pure combinatorial solution (easy to generalize), but I am pretty sure that there is some algebraic solution involving roots of unity and polynomials (it was student competition), but I can't finish... So, if anyone knows...

I know this is not what you asked for, but I liked the problem, so I'm posting a combinatorial solution :).

For $d$ being a distance between two vertices, which is at least the length of one edge, and at most the length of the diameter of the circle, list all pairs of vertices $(A,B)$ s.t. the distance between $A$ and $B$ is $d$, and $B$ is found by going in a counterclockwise direction from $A$.

Each vertex is listed twice, so the number of red and blue listed vertices is the same. Arcs $AB$ for which $A,B$ have different colors contribute the same number of red and blue vertices ($1$). This means that the number of red-red arcs is the same as the number of blue-blue arcs. We do this for all $d$, and we get the result.

Gde si bre Novakoviću, kako je tamo kod vas? :)

Interesting problem, I'll think about it.


linear algebra
matrix
number theory unsolved
number theory

For every positive integer $n$, let $\phi(n)=\frac{n}{k}$ and $k$ is greatest perfect square such that $k|n$ .
Let $a_1,\dots,a_n$ be $n$ positive integer with factor $p_1,p_2,\dots,p_n$ such that $a_i=p_1^{\alpha_{1i}}\times p_2^{\alpha_{2i}}\times \dots \times p_n^{\alpha_{ni}}\ ,\alpha_{ij}\ge 0$
we know that $p_i|\phi(a_i)$ and $p_i|\phi(a_j)$ then $p_j|\phi(a_i)$.

Prove that there exsist $k_1,k_2,\dots,k_m$ , $1\le k_1<k_2<\dots<k_m\le n$ such that:
$\phi(a_{k_1}\times a_{k_2}\times \dots \times a_{k_m})=p_1\times\dots \times p_n$

[quote="amir2"]
we know that $p_i|\phi(a_i)$ and $p_i|\phi(a_j)$ then $p_j|\phi(a_i)$.
[/quote]

Here, you actually meant "we know that $p_i|\phi(a_i)$ and $p_i|\phi(a_j)$ whenever $p_j|\phi(a_i)$", right?

It's basically the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=characteristic&t=39693]this problem[/url]. The Matrix $A$ over $\mathbb F_2$ is the one in which we put $1$ in position $(i,j)$ if the exponent of $p_i$ in $a_j$ is odd, and $0$ if it's even.


induction
number theory proposed
number theory

$ a,a',b,b',n \in \mathbb{N}$ then prove there exists $ x,y \in \mathbb{N}$ such that:

$ (a^2\plus{}b^2)^n(a'^2\plus{}b'^2)^n\equal{}x^2 \plus{}y^2$

(case $ n\equal{}1$ is trivial)

Case $ n\equal{}1$ is everything you have to do, the rest is induction  :wink:

you're right.

I proved it using complex numbers.


Let $P$ be a polynomial with positive coefficients. Prove that if
$P(\frac{1}{x})\geq \frac{1}{P(x)}$
holds for $x=1$, then it holds for every $x>0$

[hide]If P(1/x) is greater or equal to 1/P(x) when x=1, then the sum of its coefficients is greater than 1.

P(1/x) is greater or equal to 1/P(x) when
P(1/x)P(x) is greater or equal to one.

Let S(x) denote the sum of the coeffecients of P
If you apply Cauchy's inequality you get

P(1/x)P(x) is greater of equal to (S(x))^2 which is greater or equal to 1

since the 1/x and the x cancel out conviently when multiplied.[/hide]


The number of integral pairs $(x,y)$ satisfying the equation $x^2y^2+x^2-3y^2-8=0$ is....? :)

[color=green][size=75]$\LaTeX$ed by the green mod[/size][/color] :D

[quote="dangvh"][quote="riddler"][hide="solution"]
Let $x^2$ be $a$ and $y^2$ be $b$,
So,
$ab+a-3b-8=0$
$a(b+1)=3b+8$
$a=3+\frac{5}{b+1}$
Now $(b+1)$ has to be a factor of $5$ so either $b=0$ or $b=4$
But if $b=0$ then $a=8$
which is not a perfect square.
So, $b=4$ only works and $a=4$,
So,
$\boxed{(x,y)=(2,2)}$
[/hide][/quote]

Riddler you have the right idea, but i think that since $x^2=4$, $x=\pm 2$ and the same goes for $y$. So I think there are four pairs.
So,

[hide]$\boxed{(x,y)=(2,2);(2,-2);(-2,2);(-2,-2)}$[/hide][/quote]
I am soooo dumb!!!!!!!!!!!!

[hide="another way"]

factorise we can get

$(y^2+1)(x^3-3)=5$ . This means there are two ways

(i) $y^2+1=1$ ; $x^3-3=5$ . But these cannot work

(ii)$y^2+1=5$ ; $x^3-3=1$ . We get $x^2=y^2=4$ in this case which yields the same answer as above

 :)
[/hide]

hey guys , dont listen to the rot , riddler's spouting that he's an  amateur ! the guys's got the Grey matter stuffed up in handfuls! I know cos i studied with him when he was over here in india , and boy , XTREME intelligence is what he's gifted with ( and he knows it )!!!

But he hates MUGGING (heehehheheheheheheheh) just like me ..........

I would also like to say about  Shobber, 236 factorial , math92 ,rushil , blahblahh, yetti , 4everwise and of course many more whom i aint remebring that boy, you guys rock !Never seen geniuses like you folks ever, that too many of you are so young !I am still a dumbo  :blush:  but hey i can give credit when due

So keep up the good work guys !

I know we all appreciate riddler's gifted math abilities, but please keep the spam to a minimum. (I just deleted at least 5 posts, and trust me, deleted posts is not fun. That wasted like 5 minutes of my life :P )

Same mistake done here too riddler..... :D

sorry chess.... :blush:  :blush:

it wont happen again ;)  :roll:


Hi,

I am having some trouble with my figure numbering.
I am using the "book" class and i always get my figures numbered this way

Figure 1.2.3.4.5 (depending on depth of section)

those numbers before the last one is always the number of the sections or subsections or even subsubsections if those figures are under one of those.

I want to change it to

Figure <chapter number>.<figure number>

I've tried renewcommand at the beginning of my document, but it still doesn't change it. I've noticed that it is changes back to Figure 1.2.3.4.5 after "section" or "subsection" commands.

When i try to look into the .cls and .sty files that i use, i can't seems to find anything that relates to thefigure that might reset the numbering format there.

can someone please help.

thanks

Kel

I believe that you need to redefine the \thefigure command.

For example:

\renewcommand{\thefigure}{\arabic{chapter}.\arabic{figure}}

would print the first figure in chapter 2 as 2.1


induction
combinatorics proposed
combinatorics

Hi all!

Let [tex]\vert U\vert,\sigma(U),\pi(U)[/tex] be the number, the sum and the product of the elements of [tex]U[/tex]. Further, let [tex]\vert\emptyset\vert =0,\sigma(\emptyset)=0,\pi(\emptyset)=1[/tex]. Now, suppose [tex]S[/tex] is a finite set of positive integers.
Show that [tex]\sum_{U\subseteq S} (-1)^{\vert U\vert} {m-\sigma(U)\choose \vert S\vert} = \pi(S)[/tex] for all [tex]m\geq \sigma(S)[/tex].

Hanno

induct on $\sigma(S)$: if $\sigma(S)=1$ then precisely one of the elements of $S$ is one and the rest are all zero($S\subset\mathbb{N}\cup\{0\}$).but then if say $x_1=0,x_1\in S$ then $\mathcal{P}(S)$ can be partitioned into two classes,those that contain $x_1$ and those that don't.since these can be paired and the corresponding sums are equal but opposite in sign, it follows that the total sum is zero, if $S$ has more than one term and it is trivial if $S=\{1\}$.In fact that same argument works if any one single term in $S$ is zero.

 Now let $S=\{x_1,x_2,\ldots,x_n\},S^*=\{x_1,x_2,\ldots,x_n-1\}$ where without loss of generality, $x_n\geq 1$.
Now
 $\sum_{U\subset S} (-1)^{|U|}\binom{m-\sigma(U)}{|S|} = \sum_{U\subset S^*,x_n-1\notin U} (-1)^{|U|}\binom{m-\sigma(U)}{|S|}+\sum_{U\subset S^*,x_n-1\in U}(-1)^{|U|}\binom{m-1-\sigma(U)}{|S|}$
=
 $\sum_{U\subset S^*,x_n-1\notin U} (-1)^{|U|}\left(\binom{m-1-\sigma(U)}{|S|}+\binom{m-1-\sigma(U)}{|S|-1}\right) + \sum_{U\subset S^*,x_n-1\in U}(-1)^{|U|}\binom{m-1-\sigma(U)}{|S|}$
=
\[\sum_{U\subset S^*,x_n-1\notin U} (-1)^{|U|}\binom{m-1-\sigma(U)}{|S|}+\sum_{U\subset S^*,x_n-1\in U}(-1)^{|U|}\binom{m-1-\sigma(U)}{|S|} + \sum_{U\subset S^*\setminus\{x_n-1\}}(-1)^{|U|}\binom{m-1-\sigma(U)}{|S|-1}\].

now the third sum, by induction is $x_1x_2\cdots x_{n-1}$.and again by induction the first and second sums together give $x_1x_2\cdots x_{n-1}(x_n-1)$. adding these together completes the induction and the proof.

interestingly this is an identity(not just for a set of nonnegative integers but also symbolically).


$ A,B,C,D$ are four "consecutive" points on the circumference of a circle and $ P, Q, R, S$ are points on the circumference which are respectively the midpoints of the arcs $ AB,BC,CD,DA$. Prove that $ PR$ is perpendicular to $ QS$.

[hide]Let $ x\equal{}AP\equal{}PB$, $ y\equal{}BQ\equal{}QC$, $ z\equal{}CR\equal{}RD$, and $ w\equal{}DS\equal{}SA$. (The arc measures, not the segments or their lengths.)
Since there are $ 360^\circ$ in a circle, $ 2x\plus{}2y\plus{}2z\plus{}2w\equal{}360^\circ$, so $ x\plus{}y\plus{}z\plus{}w\equal{}180^\circ$.
Let $ O$ be the intersection of $ PR$ and $ QS$.
Since $ \angle POQ$ and $ \angle ROS$ subtend a total of $ x\plus{}y\plus{}z\plus{}w\equal{}180^\circ$ of the circle, $ \angle POQ\equal{}\angle ROS \equal{} \frac12 \cdot 180^\circ \equal{} 90^\circ$ as desired.[/hide]

Thanks.

Now this is obvious but just to make sure, $ \angle POQ$ and $ \angle ROS$ are same because of vertical angles, right?

[quote="Xantos C. Guin"]Yes. Anyway, CMO is the Canadian Math Olympiad right?[/quote]

Yes.

These problems are actually from earlier years so they're much easier than what they are. I think overall though CMO is easier than USAMO by far. I just do these problems to resolve my writer's block for my contest (and to help Resource page).  :)

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1975Problem5

Vo Duc Dien


Is the $ \ce{AlNi3}$ alloy substitutional or interstitual? Why?

What are the criteria for an alloy to be interstitial or substitutional?

Well, according to [url=http://www.bookrags.com/research/alloy-woc/]this[/url], "In an interstitial alloy, smaller elements fill holes that are in the main metallic structure." and "In substitutional alloys, some of the atoms of the main metal are substituted with atoms of another metal."

But I do not understand how to determine whether $ \ce{AlNi3}$ is substitutional or interstitial.

What you need is the [url=http://en.wikipedia.org/wiki/Hume-Rothery_rules]Hume-Rothery rules[/url].


MATHCOUNTS
AMC
USA(J)MO
USAMO
AMC 8
AMC 10
AMC 10 A

Did you know that, in addition to being MATHCOUNTS National Champion, Kevin Chen also made 3 consecutive years of AMC 8 perfect scores? He also earned a perfect AMC 10 A in his freshman year of high school.

You can view the lists of perfect scorers by viewing the archives under perfect scores at http://www.unl.edu/amc

That is impressive but I do not think this is the place to gossip about other AoPS members.

[quote="gauss1181"]He also earned a perfect AMC 10 A in his freshman year of high school.[/quote]

The AMC 10A of his freshman year in high school has not happened yet.

Wow... Does gossiping about other AoPS members benefit you in any way? Even though being national champion is impressive, who cares about what other people do? If you really want to make USAMO or MOP or whatever, rather than saying how impressive someone else is, just do the work. Honestly, even a super genius at math need to learn the stuff first.

On a side note, AoPS is becoming more and more spammy with topics like this.

[quote="Pakman2012"]yeah actually, i'm pretty sure creating topics whose subject is an aops member is pretty not allowed.

also, hero-worship, while a time-honored tradition, is not becoming.[/quote]
What? Really? He's an AoPS member? What's his name?

binomquist or something like that... seriously, this thread is an utter waste of time. Who cares if he made IMO in 2nd grade or something, this isn't a place to discuss it.


Suppose that $ \{a_n\}$ and $ \{b_n\}$ are two arithmetic sequences such that

$ \frac{S_n}{T_n}\equal{}\frac{3n\plus{}2}{4n\minus{}5},$

where $ S_n\equal{}a_1\plus{}\cdots\plus{}a_n$ and $ T_n\equal{}b_1\plus{}\cdots\plus{}b_n$. Find $ \frac{a_{2009}}{b_{2009}}$.


Two alcohols, A and B, have the same molecular formula $ C_{9}H_{10}O$ and react with sulfuric acid to give the same hydrocarbon C. Compound A is optically active and compound B is not. Catalytic hydrogenation of C gives a hydrocarbon D, $ C_{9}H_{10}$, which gives two and only two products when nitrated once with $ HNO_{3}/H_{2}SO_{4}$. Give the structures of A, B, C, and D.

A is 1-phenyl-2-propanol
B is 3-phenyl propanol

Obviously, those answers are wrong since they don't even match the given molecular formula.

Let me just post a few thoughts. The last step is nitration which might indicate the presense of a benzene ring but it is possible that other nitration reactions exist that I'm not aware of. Say $ D$ is some form of substitued benzene ring then the ring itself provides $ 4$ degrees of unsaturation leaving $ 1$ more required which is probably a double bond somwhere else. Due to the two nitration product constraint I would say $ D$ is $ \text{p\minus{}ethenyltoluene}$?

Well then what would C be? C is hydrogenated to give D...

[hide="I think..."]If we leave out the nitration part, then D=2,3-dihydro-1H-indene seems logical...[/hide]

In other words Hell_ever, D would be Indane. But why leaving out the nitration part? How many nitration products can be obtained from indane?

BanishedTraitor, Hell_ever is right: if D is obtained from an hydrogenation then how can it contain a double bond?

Now, if D is indane, what are the structures of A, B, and C?

It is easy to go backwards. The optically active alcohol is alpha to the ring, while the inactive one is beta to the ring.

I think we should get 4 nitrated products.....

That's right.

Indane has a plane of simmetry, so there are only two (chemically) non-equivalent aromatic protons. Therefore, mono-nitration is expected to yield only two products. Remember that in nitration only the protons connected to the aromatic ring are substituted, not the ones on alkyl side chains. Are you taking that into account?


function
group theory
abstract algebra
superior algebra
superior algebra solved

I think the following problem is classical, but quite hard (in any case, I could not solve it):
  Let two matrices $a_{11}=0, a_{12}=1, a_{21}=-1, a_{22}=1$ and $ b_{11}=0, b_{12}=-1, b_{21}=1, b_{22}=0$. Prove that a relation of the form $ X_1 B X_2 B... X_n B=I$ or $ B X_1 B... X_n B=I$ or $ X_1 B X_2... X_n=I$ or $ B X_1 B... X_n=I$ where each $ X_i$ is either $A$ or its inverse, is impossible.

I'm having some trouble with the statement. Did the question ask for a proof that the two generate a free group or not? Because the way you've written it, $B$ has order $4$, since it's $\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$ :?.

Ah, sorry, I simply forgot the exact form of the matrices. Here is the equivalent statement, I will let you find the form of the matrices:
  Consider the functions $ f(x)=1/(1-x)$ and $g(x)=-1/x$. Then any composition of these functions which is equal everywhere with the identity is only the trivial combination: that is something of the form $f f^{-1} g^2 g^{-2}$ and so on. Is it correct now? My memory is totally lost...  :blush:

I think the problem text should be like this:

Let $A=\left(\begin{array}{cc} 0&1\\ -1&1\end{array}\right)$, $B=\left(\begin{array}{cc} 0&-1\\ 1&0\end{array}\right)$.

Prove that $\langle A\rangle *\langle B\rangle$ is a subgroup of $GL_2(\mathbb{Z})$. "*" means free product.

Suppose that $ A,B $ are transformations of $ \widehat{\mathbb{C}} = \mathbb{C}
\cup \{ \infty \} $ which are $ Ax = \frac{1}{1-x} $, $ Bx = \frac{-1}{x} $.
If I did the computations correctly, we get
$ Ax = \frac{1}{1-x} $ , $ A^{2}x = \frac{x-1}{x} $, $ A^{3}x = x $, $ Bx = \frac{-1}{x} $,
$ B^{2}x = x $, $ ABx = \frac{x}{x+1} $, $ A^{2}Bx = x+1 $, $ BAx = x-1 $,
$ BA^{2}x = \frac{x}{1-x} $. Denote $ C = AB $, $ D = A^{2}B $. Then any non-trivial
combination of $ A,B $ can be written or as $ X $, or $ BX $, or $ XA $ or $ XA^{2} $
or $ BXA $ or $ BXA^{2} $, where $ X $is a combination of $ C $, $ D $.
Therefore if there is a non-trivial combination of $ A $, $ B $, which is equal
to the identity then for some non-trivial combination $ X $ of $ C,D $ we have
$ X = I $ or for some ( not necesserily trivial combination ) $ X $ of $ C,D $
we have $ X = B $ or $ A $ or $ A^{2} $ or $ BA $ or $ BA^{2} $.
Now observe that transformations $ C,D $ map positive reals to positive reals
and we see that $ B(1) = -1 $, $ A(1) = \infty $, $ A^{2}(1) = 0 $, $ BA(1) = 0 $
and $ BA^{2}(1) = \infty $, which are not positive reals. So we cannot
have the case when some combination $ X $ of $ C,D $ equals to one of
$ B,A,A^{2},BA,BA^{2} $ because then we will have that $ X(1) $ is a positive real.
The remaining case is when some non-trivial combination of $ C,D $ is equal to the
identity. Recall that $ Cx = ABx = \frac{x}{x+1} $ and $ Dx = A^{2}Bx = x+1 $.
Then if we have that $ x $ is some positive rational number $ x = \frac{p}{q} $,
when $ p,q \in \mathbb{N} $ and $ p,q $ are co-prime, then $ Cx = \frac{p}{p+q} $,
when $ p,p+q $ are co-prime and $ Dx = \frac{p+q}{q} $, when  $ p+q,q $ are
co-prime. Therefore if for $ x $ positive rational number we define it's height
as $ p+q $, where $ x = \frac{p}{q} $, $ p,q \in \mathbb{N} $ and $ p,q $
are co-prime, then maps $ C,D $ always increase height. Therefore, if , for example
we start with $ x = \frac{2}{3} $, then applying $ C,D $ some number of times,
the height will increase and we will never come back to $ \frac{2}{3} $.

Thank you, Leva, very nice solution.  :)

To translate the result in Algebra, $[A]*[B]$ is a subgroup of $PSL_2(\mathbb{Z})$, by using the isomorphism $M:PSL_2(\mathbb{Z})\to Aut(\mathbb{\hat{C}})$, $M\left(\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\right)(z)=\frac{az+b}{cz+d}$.

It is easy to see now that $\langle A\rangle * \langle B\rangle<GL_2(\mathbb{Z})$


If $ \alpha$ and $ \beta$ are roots of $ z \plus{} \frac 1z \equal{} 2e^{i\theta}, 0 < \theta < \pi$, then prove that
(1) $ |\alpha \minus{} i| \equal{} |\beta \minus{} i|$ and $ Arg(\alpha \plus{} i) \equal{} Arg(\beta \plus{} i)$
(2) Find locus of $ \alpha$ and $ \beta$ in argand plane as $ \theta$ varies from 0 to $ 2\pi$

Here, $ i \equal{} \sqrt { \minus{} 1}$

I could only prove $ \mid \alpha \minus{} i \mid \equal{} \mid \beta \minus{} i \mid$ ......that too with some ugly manipulations!  :(


An $ 8$ by $ 8$ board is to be tiled with $ 2$ by $ 1$ rectangular tiles, each one black or white. What is the least number of black tiles needed so we can arrange them in a way that no $ 2$ by $ 2$ square on the board is all white?

[hide="Solution"]Each tile can fill at most 6 2x2 squares. There are 49 2x2 squares, so by the pigeonhole principle, there are at least 9 tiles. It is easy to find a placing that has 9 tiles, but I do not know how to show an image.
[/hide]

[quote="Altheman"][quote="pythag011"]Each tile can fill at most 6 2x2 squares. [/quote]

Huh?[/quote]

Originally, we need to partially cover every 2x2 square by at least one black tile. A black tile partially cover at most 6 2x2 squars.

[hide]Okay right I wasn't thinking of the problem correctly. Your proof is nice so far. Make a picture with paint or something so it is complete (I might compile all the solutions at some point).[/hide]

[hide]An example that 9 black tiles are enough:[/hide]


Here were the rules as said by randomdragoon in the original Fight Game:

[b]Match up two characters, real or fictional, to a fight! Choose who would win in a fight (giving explanations if necessary) and then choose a new challenger for the winner!

Base rules:

1) Chuck Norris is not allowed to challenge anyone.[/b]

Here are some modifications:
- God is not allowed to challenge anyone either.

Keep in mind that this is NOT a continuation of the first fight game.

First Round:
Finger VS Toe

fingers are above the ground, so they divebomb to victory

finger vs nailpolish

fingers :they control the nailpolish

Fingers v.s. gangrene

gangrene kills the fingers (in loose scientific terms)

mercurochrome vs. gangrene

nobody wins because i dont know what those are

pikachu vs. daermon

pikachu wins because he shocks deomon

Pikachu V.S. whatever pokemon is on randomdragoons avatar.

Do I even have to say who wins?

Umbreon vs. Jolteon
(and the only reason my avatar is Umbreon rather than Jolteon is because good avatar sized pictures of Jolteon are/were harder to find when I changed my av)

Umbreon and Jolteon vanish because of some supernatural reason....

Which turns out to be the Artifact. It's terrestrial, and you only have 15 seconds to view it, or else you will be breaking exposure limits.

The Artifact versus George Bush.

P.S. The Artifact mutated someone so that he could use 100% of his brain(!).

George Bush of course. 30% of the country is still a lot of support.

George Bush vs. Al Gore

George bush because there were 2 of them

George Bush Senior vs George Bush Jr

George Bush Senior. George Bush Jr is less smart and way younger then Senior

George Bush SEnior vs. SunnyD (the really good orange juice)

George Bush Senior. He drinks all the SunnyD.

George Bush Senior vs. ghost coming for revenge

What kind of ghost gets revenge on George Bush Senior? Probably insignificant.

George Bush Senior vs. Dalai Lama

If a ghost loses, Dalai Lama should also lose.

George Bush Senior vs. Bill Clinton

NO DUR. FIGURE IT OUT YOURSELF.

Bill Clinton V.S. his wife

This may be a hard one, but they both destroy each other.

Chuck norris vs. atomic bomb

Neither because neither of them are Bill Clinton nor his wife.

Bill Clinton or Ernest Hemingway

Bill Clinton. His name sounds cooler :)

Bill Clinton vs Bill Clinton (his evil twin)

the evil twin wins.

Evil twin of bill clinton v.s. Cabbage patch kids

Cabbadge Patch Kids definetely. They use their annoying songs to make evil Bill Clinton implode.

Cabbadge Patch Kids vs. Barbies

barbies!!!

Barbie and Ken v.s. Barney and the rest of his dinosaurs (i think t.j. and baby bop. DON'T ask where i get this!!!

Barney and his Dinosaurs

Barney vs Winnie the Poo

Barney wipes up Winnie the Poo.  So Barney wins.

And now I wait for people to knock me unconscious for such a terrible joke.

Barney vs. Monica Lewinsky

*hopes she gets to face off against Hillary*

[quote="CheeseIsGood"][quote="junggi"][quote="CheeseIsGood"]Barney wins, because he is special.

Barney vs. Bob the Builder[/quote]
alright who are you?
no one could possibly know about bob the builder except ccy.[/quote]

What? I'm not ccy.... Who is ccy?[/quote]
well that's irrelevant. im just curious who you are and how you know of bob the builder
pm me with this or something

[quote="vicseb"]I guess bob the builder because he builds a wall that knocks barney unconsiounce (not that I would know or anything)

Bob the Builder vs. Handy Manny (this television guy that my little sister watches, and he is this guy that fixes things, not that I would know or anything)[/quote]
they both hammer each other
stalemate.

Bob the Builder vs. House (doctor from the show House)

House. Bob the builder cannot repair him

house vs. plasma screen tv

The plasma TV screen stuns the house.

plasma TV screen vs. 16*15 knights vs. 20*12 knights

Plasma TV screen stuns the knights. And it's really, [i]really[/i] hot, as it's a TV screen in the plasma state (har har har!)

plasma TV screen vs. Bose-Einstein condensate!

Plasma since it stuns the bose einstein thingy.

Plasma Screen TV vs educational television

Plasma screen wins.  Educational TV can only show educational TV, while the plasma screen can show educational TV and more.

Plasma screen vs. HDTV

The plasma screen shoots out plasma that destroys the HDTV

Plasma screen tv vs  :rotfl:


Let $\triangle ABC$ be a triangle with unequal sides. Let $D\in[AC]$ and $E\in[AB]$ such that $\widehat{EDB}=\widehat{BCD}$. If $|BC|=|AD|=2$ and $|AE|=|DC|=1$, then what is $|EB|$?

[hide="Solution"]Standard markings $\alpha,\beta,\gamma$. Denote $\phi=\angle CBD$. Then $\angle ADE=\phi$

Sine Law for $\triangle CBD$: $\frac{2}{\sin(\gamma+\phi)}=\frac{1}{\sin\phi}$

Sine Law for $\triangle ADE$: $\frac{2}{\sin(\alpha+\phi)}=\frac{1}{\sin\phi}$

Therefore $\sin(\alpha+\phi)=\sin(\gamma+\phi)$, which gives either $\alpha+\phi=\gamma+\phi$ or $(\alpha+\phi)+(\gamma+\phi)=\pi$

The first possibility yields $\alpha=\gamma\iff CB=AB$ and that's extraneous since the given triangle is scalene.

The second possibility yields $2\phi=\beta\iff CD: DA=CB: BA\iff BA=4\iff EB=3$[/hide]


Find all $n \in N$ , $n$ is not a perfect square such that : $n^2$ is divisible by $[\sqrt{n}]^3$

Let $l$ be the greatest number such that $l^2<n$ Then $n=l^2+k$ for some $k<2l+1$ Using this representation we obtain $[\sqrt{n}]=l$ So we have divisibility \[ l^3|(l^2+k)^2=l^4+2kl^2+k^2  \] so $l^3|2kl^2+k^2$ in particular $l^2|k^2$ which means that $l|k$ but as $k\leq 2l$, we obtain $k=l$ or $k=2l$. In the first case we reach $l^3|2l^3+l^2$ so $l=1$ and our $n$ is $2$, and in second case $l^3|4l^3+4l^2$, it follows that $l|4$ which gives $n=3, 8,24$ and finaly \[ n=2,3,8,24  \] :)

[quote="jastrzab"]But in your solution you forgot obout $n=24$[/quote]
But $n$ cannot be 24!
$\sqrt{24}=4.8989.....$
$[\sqrt{24}]^3=4^3=64$!
24 doesn't divide 64!

But it seems to me that you misunderstood this problem ...we want so that $[\sqrt{n}]|n^2$ and in your solution you solve the problem with divisibility $[\sqrt{n}]|n$


When taking a log in an inequality, what happens to the sign?

sign not changed: 2^x < 8  <=> log_base2 2^x < log_base2 8 <=> x < 3

sign changed: (0.2)^cos x < 1 <=> log_base.2 (0.2)^cos x  > log_base.2 1 <=> cos x > 0

What's the deal with this?

Think about the following two graphs:

y = n^x

where n is greater than 1 and then the graph where n is less than 1.  The difference between these graphs should help you understand how the inequalities work.

Ah, I see.

When n > 1, the sign stays the same.
When n = 1, the graph is 1 for all x.
When 0 < n < 1, the sign switches.

Ok, thanks for the help Mr. Crawford.

you can also think about like this:

when n>1, log in base n is an increasing function, and thus taking the log of both sides will not change the sign.

when n<1, log in base n is a decreasing function, and thus taking the log of both sides will change the sign.

in general, as long as the function is increasing, the inequality sign will not change. But if the function is decreasing, the inequality sign will change.

[color=cyan]Note on what darkquantum said:  the funtion must be continuous, too.  Otherwise you can get something like f(x) = -1/x, which sometimes reverses the sign of the inequality and sometimes doesn't, even though it's always increacing (except at 0, where it has a discontinuity).[/color]


A triangle $ ABC$ has $ \angle A\equal{}\frac{45}{2}$ and $ \angle C\equal{}90$.  What is $ \cot A$?

I suppose $ cot A\equal{}cot \frac{\pi}{8}$

Also $ cot A\equal{}\frac{AC}{BC}$

yes that is correct skand.  now try to solve it

btw try not to use trig

How can you solve trig problem without trig?

I suppose he means solve geometrically:

[hide="Solution"]Extend BC to point D s.t. AB is the angle bisector of DAC.  That is, triangle DAC is a 45-45-90 right triangle.  By the angle bisector theorem:

$ \frac{DB}{BC}\equal{}\frac{AD}{AC}\equal{}\sqrt{2}$

$ 1\plus{}\sqrt{2}\equal{}1\plus{}\frac{DB}{BC}\equal{}\frac{DB\plus{}BC}{BC}\equal{}\frac{DC}{BC}\equal{}\frac{AC}{BC}\equal{}\cot A$

So the answer is $ \boxed{1\plus{}\sqrt{2}}$[/hide]

Or call a point $ D$ on $ AC$ such that triangle $ DCB$ is an isoceles triangle.  Then if $ BC\equal{}1$, $ CD\equal{}1$.  Note that triangle $ ABD$ is iscoceles too so $ AD\equal{}BD\equal{}\sqrt{2}$.  Thus we get $ \boxed{1\plus{}\sqrt{2}}$

@jili and krsattack,you rock man :)

I couldn't ever think of solving such problems with geometry


Let $ S^2$ be the 2-sphere and let $ D_1,\cdots, D_n$ be $ n$ small open dics on $ S^2$ with disjoint boundaries. Let $ X^\plus{}, X^\minus{}$ be two copies of $ S^2\setminus (D_1\cup \cdots\cup D_n)$ and let $ X$ be the space obtained by identifying, for each $ i\equal{}1,\cdots, n$, the boundary of $ D_i$ on $ X^\plus{}$ with the boundary of $ D_i$ on $ X^\minus{}$, using the identity map. Thus, $ X$ is a "Sphere with $ n\minus{}1$ handles". Compute the homology of $ X$.

Thks in Advance.

Instead of identifying them, which can make for a confusing picture, use the homotopy pushout (which I'll call $ Y$) instead. What this consists of, in this case, is two disjoint copies $ X^\plus{}$ and $ X^\minus{}$, with [i]cylinders[/i] connecting the boundaries of the $ D_{i}$'s (rather than gluing them together. Then $ Y$ is homotopy equivalent to $ X$ by collapsing the cylinders. However, the homology of $ Y$ can be computed via the Mayer-Vietoris sequence. $ Y$ has a cover $ U \cup V$, where $ U$ is a "fattened" copy of $ X^\plus{}$, and $ V$ is a fattened copy of $ X^{\minus{}1}$. Specifically, $ U$ contains $ X^\plus{}$ and just over half of each cylinder, while $ V$ contains $ X^\minus{}$ and the other half of the cylinder (plus a little), so that $ U \cap V$ consists of $ n$ disjoint copies of $ S^1$; and $ U \cup V \equal{} Y$.

Can you finish it from here? :)


function
real analysis
real analysis unsolved

How to show: There is constant $C$ such that:

$log(ex)(loge(log(e(log(ex)))\leq Clog(x)(log(log((log(x)))$ for all $x>1$.

I have a idea now:
maybe enough to show $log(ex)\leq Clog(x), x>1$:
when $x$ large, it is clear. When $x$ near 1, use $log(1+t)=t-t^{2}/2+...$.


projective geometry
geometry
geometry proposed

Let triangle $ABC ,AH\perp BC ,M,N$lie on side$AC,AB$such$\widehat{ACN}=\widehat{ABM}$.$D$ be midpoint of $BC DM,DN$intersect line $\delta$ thruogh $A$and parallel to $BC$ at$P,Q$.Prove that$MN,BP,CQ$and $AH$ are concurent

The concurrence point doesn't necessarily lie on $AH$, I think that's a false statement.
However, all the conditions you state are unnecessary, we can generalise your problem in many ways. $\delta$ can be arbitrary, $D$ can be arbitrary, $M$ and $N$ can be arbitrary too, ...

I can easiley generalise your problem for you:
$ABC$ is triangle and $M$ and $N$ are arbitrary points on $AC$ and $AB$. $D$ is an arbitrary point on $BC$. Now, we draw an arbitrary line $\delta$ through $A$, and we call $P$ and $Q$ the intersections of $DM$ and $DN$ with $\delta$. Now, $BP, CQ$ and $MN$ are concurrent.

Proof:
Call $R$ the intersection point of $BP$ and $CQ$.
$P, A, Q$ are three collinear points, and so are $C,D,B$.
By Pappus-Pascal, we now know that $M, R, N$ are collinear. $\mathbb{QED}$

The proof for your problem now follows, by taking the case where $\delta$ is parallel to $BC$, $B,M,N,C$ are concyclic, and $D$ is the midpoint of $BC$.

Jan Vonk

:)  In my problem the line $\delta$ is parallel to BC so I think my problem is true

If the problem is indeed true, you have a typo. (Assuming there is no misinterpretation by me from the problem as you stated it)  :)
Here is what the configuration looks like as you are describing it. I would say S does not lie on AH!  :wink:

Oh i'm sorry !!
I post my problem is not enough
"" $BM,CN,AH$ are concurent""  :blush:

We can restate your problem in more general formulation, as follows:

A triangle $\bigtriangleup ABC$ is given and let $D,$ $M,$ $N$ be, arbitrary points on it’s sidelines $BC,$ $AC,$ $AB,$ respectively. We denote as $R,$ the intersection point of $BM,$ $CN$ and as $S,$ the one, of the segment lines $MN,$ $AR.$

Prove that the line connecting the points $P\equiv BS\cap DM$ and $Q\equiv CS\cap DN,$ passes through vertex $A$ of $ABC$ and that this line is parallel to $BC,$ when $D,$ is the midpoint of the sidesegment BC.

$\bullet$ It is easy to prove the collinearity of $P,$ $A,$ $Q,$ by  Pappus’s theorem, considering the triads of points, $B,$ $D,$ $C$ and $M,$ $S,$ $N,$ on $BC,$ $MN,$ respectively.

We can prove that $PQ\parallel BC,$ as follows:

We denote as $K,$ $L,$ the intersection points of the sideline $BC,$ from the segment lines $MN,$ $AR,$ respectively and as $N'\equiv AB\cap CQ.$ As well known the points $M, S, N, K$ is in harmonic division and so, we have that the pencil $C.MSNK,$ is in harmonic conjugation.

The points $A, N', N, B,$ is in harmonic division, as the intersection points of the pencil $C.MSNK,$ from the sideline $AB$ of $\bigtriangleup ABC.$  So, the pencil $Q.AN'NB,$ is in harmonic conjugation and now because of $DB = DC,$ we conclude that $BC\parallel PQ$ $($ it is well known that when a pencil is in harmonic conjugation, then a line parallel to one it's ray, is intersected from the other three it's rays, in three points formed two congruent segments $)$ and the proof is completed.

Clearly, it is not necessary the condition of $\angle ACN = \angle ABM$ and the concurrency point of $BM,$ $CN$ $($ here the point $R,$ $),$ lies on the segment line $AS$ and not on the altitude $AH,$ of $ABC.$

Kostas Vittas.

[quote="vittasko"]...Clearly, it is not necessary the condition of $\angle ACN = \angle ABM$ and the concurrency point of $BM,$ $CN$ $($ here the point $R,$ $),$ lies on the segment line $AS$ and not on the altitude $AH,$ of $ABC.$

Kostas Vittas.[/quote]
you dont' use conditions $\angle ACN = \angle ABM$ then R not lie on altitude $AH$ .I think so

[quote="hunter_cvp_ic"]you dont' use conditions $\angle ACN = \angle ABM$ then R not lie on altitude $AH$ .I think so[/quote]

I think that it is bright in Jan's figure, that the concurrency point $R,$ of $BM,$ $CN,$ lies on the segment line $AS$ and not on the altitude $AH.$

By the way, I post again a figure relative to the condition $\angle ACN = \angle ABM,$ because of probably you can't open the Jan's figure.

Kostas Vittas.

Oh!! I'm sorry
Your sol is true but in my problem i construct first $AH \perp BC$  then let point $R$ on $AH$such$\widehat{ABR}=\widehat{ACR}( R$ be concurent point such $BM,CN$ passthruongh on my first post)
hicc
then $R$ be orthocenter of triangle $ABC$ then I will prove that $BP,CQ ,AH$ and $MN$concurent but not use $pappus$'s theorem


Let $ p$ be a prime number. Find remainder from dividing number $ \binom{2p}{p}$ by $ p$.

$ \binom {2p}{p} \equal{} \frac {2p}{p} \frac {2p \minus{} 1}{p \minus{} 1} \cdots \frac {p \plus{} 1}{1} \equiv 2 \cdot 1 \cdot \cdots \cdot 1 \equiv 1 (mod p)$

Sorry, I give you a wrong task.

Problem is:
Let p be a odd prime number. Find remainder from dividing number  $ \binom{2p}{p}$ by $ p^2$.

PS
The answer in task I give first is 2 I think.

@cadiTM: Your solution is not complete..

You forget to check the if $ p\equal{}2$ case, which leaves the remainder to be $ 0$.  :D

Using the fact that $ (p\minus{}k)(p\plus{}k) \equiv \minus{}k^2 (mod p^2 )$
We arrive at
$ \binom{2p}{p} \equal{} \frac{(2p)!}{(p!)^2} \equal{} 2 * \frac{(2p\minus{}1)(2p\minus{}2)\ldots(p\plus{}1)(p\minus{}1)(p\minus{}2)\ldots*2*1}{((p\minus{}1)!)^2} \equiv 2 * \frac{(\minus{}1^2 )(\minus{}2^2 ) \cdots (\minus{}(p\minus{}1)^2 )}{((p\minus{}1)!)^2} \equal{} 2 (mod p^2 )$

Exactly $ ord_p( \binom{ap}{bp}\minus{}\binom{a}{b})\equal{}ord_p(a)\plus{}ord_p(b)\plus{}ord_p(a\minus{}b)\plus{}v_p$, were $ v_2\equal{}1,v_3\equal{}2,v_p\equal{}3,p\ge 5,p\not |B_{p\minus{}3}$. If $ p|B_{p\minus{}3}$, then $ v\plus{}_p\ge 4$.

[quote="perica0001"]Let p be a odd prime number. Find remainder from dividing number  $ \binom{2p}{p}$ by $ p^2$.
[/quote]

$ \binom{2p}{p}\equal{}\binom{p}{0}^2\plus{}\binom{p}{1}^2\plus{}\ldots\plus{}\binom{p}{p\minus{}1}^2\plus{}\binom{p}{p}^2$.

$ \binom{p}{0}\equal{}\binom{p}{p}\equal{}1$;

$ p|\binom{p}{k}$,  $ k\equal{}1,\dots,p\minus{}1$.


algebra
polynomial
integration
function
real analysis
real analysis unsolved

[hide] P any polynomial in $ C[X]$,$ P(z) \equal{} \frac {1}{2\pi}\int_{0}^{2\pi}P(z \plus{} re^{ix})dx$

for any $ z \in C$ and any $ r\in R\plus{}$ [/hide]


integration
function
complex analysis
complex analysis unsolved

Hi all,
I need help in sloving thiscomplex integral.
from 0 to infinity
1/{ [x^3  + 1 ]^2}   dx

Thanks

Let's have a go:

We want to calculate $I = \int_0^\infty \frac{dx}{(x^3 + 1)^2}$.  To do this, I'm going to integrate over the curve $\gamma$ defined by the segment between $0$ and $R$, the circular arc from $R$ to $Re^{2\pi i/3}$ and the segment from $Re^{2\pi i/3}$ to $0$.  Then
$\int_\gamma \frac{dz}{(z^3 + 1)^2} = \int_0^R\frac{dx}{(x^3 + 1)^2} + \int_0^{\frac{2\pi}3} \frac{Re^{i\theta}d\theta}{(R^3e^{3i\theta} + 1)^2} - \int_0^R \frac{e^{2\pi i/3}dx}{(x^3 + 1)^2}$.  Now, as we take the limit of both sides as $R \to \infty$, the right-hand side becomes $(1 - e^{2\pi i/3})I$.  On the left-hand side, we're integrating a nicely behaved function around a simple closed curve, so its value is just given by the residue theorem.  So, the answer is going to be $\frac{2\pi i a_{-1}}{1 - e^{2\pi i/3}}$ where $a_{-1}$ is the residue of $\frac 1{(z^3 + 1)^2}$ at $e^{\pi i / 3}$ (the only pole inside $\gamma$).


Just curious, but who's the currently highest rated FTW player?
Screenshots to back up your rating claim would be good.

Currently, the highest rated player is Unemployed at 1523, and the all-time highest rated player(s) is a tie between e^i penguin and RollandWu at 1552.

[quote="Canton"]If you look back at February 5th's archive, the highest all time rating then, was mathgeniuse^ln(x) with [b]1677![/b] Whoa![/quote]

Unimpossible had 1702, but he was banned and lost his rating.

it was easier before the rating changes because you could leave a game if you were losing and not lose any points

Not only that, but your points would go WAY up when you won. Now, its practically impossible to gain once you are in teh mid 1500s

Sure it is, just play against other 1500s :)

Once there are many people playing, then there will be more 1500's to play against. And some of them won't mem.

Yea, I'm at 1487 right now and I've found that the only way to gain is to countdown people, win and get +1. The only other person who's on and has a rating near mine is TheMiddleSchooler and he doesn't want to cd me for some reason... :mad:

I won superllamel twice when the rating system wasn't working. *Sigh*...

lol

I was much more willing to play people with lower ratings when the rating system wasn't working.  :wink:

LOL super and I are afraid of eachother so neither of us go up.

lol im not afraid of neither super nor mewto lol

[quote="isabella2296"]
[b][i][u][color=red][size=200]GJ SUPERLAMEL!!!!!! AND WHATS A LLAMEL?[/size][/color][/u][/i][/b][/quote]
who knows...

A llamel is a cross between a camel and a llama.  I googled it. :thumbup:

Have you guys ever thought that super might be lying?

i think its |lame|(thanks sheena [guineapiggies])

Yes, I'm a very mysterious person.  :ninja:

Why do so many people post about what a llamel is? It doesn't make me want to tell anybody.

well yeah...but now we have like aime15 whose rating is 2200
also, why did u revive ( i know i shoudnt be talking)

[quote="TwinDracula"]now i am 1500 and i do not mem[/quote]

1. False
2. Revival
3. Pointless

In conclusion, thread is most likely and should be locked.

well...
Phirekalk

I used to be 1500.

What is a revival?

I am pointless?

whatwhatwhat???

@TwinDracula
A revival is when you "awake" a sleeping topic that hasn't been active for a long time. Please do not revive topics unless it's not pointless. Really think before you revive a topic.
Thanks!

A good way to make sure it's not fully asleep yet is to check the last post time,
then compare it to the day your posting's time.
If their is more than a month of a difference, then don't touch the topic, unless it has a point.
But don't go around posting spam in awake topics, not going to help.
-meewheezerez

[quote="TwinDracula"]I used to be 1500.[/quote]

The false part was a different part of that statement.

[quote="TwinDracula"]What is a revival?[/quote]

You have been answered.

[quote="TwinDracula"]I am pointless?[/quote]

No. Your post is.


Putnam
geometry
Princeton
college
CMU
Carnegie Mellon
college contests

Hello,

I'm sorry if this question has been repeated many times but I thought I'd start a new thread because this is somewhat personal.

I'm currently at my college, which is about 40 minutes from Pittsburgh, Pennsylvania. My college at this point does not have Putnam, and I talked with the the chairman of the math department, and he said that due to the lack of the student body, Putnam seems to be quite difficult to carry out. Recently, however, I met the head student of math club here, and he said that he wants to do Putnam this year so maybe, we'll be able to do this.

But, I'm going to hope for the best and plan for the worst so here are my questions.
1) When is the deadline for Putnam registration for student(s) who do not have the exam at their own college and plan to take on other college?
2) Any idea on which colleges around Pittsburgh area have Putnam?
3) Do you think they'll be able to provide vehicles for the exam day (if not, I can ask a friend or just pay my way through taxi or something..)?

So any input'll be great. Thanks!

The deadline for registering for the Putnam is approximately Oct. 15. (I don't know exactly because I haven't received the registration mailing yet.) All registration is through a college or university - there is no provision, anywhere, for unaffiliated students. The registration form includes a line that says something about certifying that the named students are "regularly enrolled" students at the college in question, and calls for the signature of the department chairman.

It's also basically self-policing. Each college's Putnam supervisor is essentially responsible for determining what is proper. The most important thing you'd have to find at your own college is a faculty member willing to be responsible and to take the time required. This faculty member must submit the entries (in October); receive and safeguard the exams when mailed; find an appropriate place in which to take the exam; spend all day on the appropriate Saturday proctoring; and promptly mailing or shipping the papers off to be graded.

As for finding another school: I don't have my Putnam mailing with me (it's in my office and I'm at home right now) - it would have the full list of participating colleges. Off the top of my head, I know that Carnegie Mellon does, and I think also the University of Pittsburgh. But are you a "regularly enrolled student"? I think I'd be willing to accommodate a community college student (who might be in the process of transferring into my institution) but I probably wouldn't accept a student at another 4-year college. You're much better off if you can get your home campus to hold it. (And I really doubt that anyone is going to offer you a ride.)

Thanks for the feedback.

I'm not sure when the next meeting for math club is but I'm pretty confident that it's before October 15 so I'll definitely try to bring this topic up. I'm regularly enrolled in 4-year college so yeah, it seems doubtful then. I just asked this question because I've heard these advices for AMC when I was in high school (funny how everywhere I start, there isn't a math competition  :roll: ).

Oh well.

We had a math meeting, and basically the result was that if I wanted to do it, I could. But I can't because of time.

I wonder if Putnam has any alternate dates? The sad part is that the date that Putnam is given (December 6th I believe? I know it's Sunday) is right around when I have my finals and I do not want to spend 6 hours of my studying time on getting score of 0 or 1. Experience is great but I don't want to fail my classes.

I don't think there's an alternate date (I haven't yet heard anything about it before) but is there? Thanks.

There isn't an alternate date, with one exception. The date is always a Saturday (and yes, it does tend to fall very close to final examinations for the fall semester or quarter - always). The one exception involves religious reasons (presumably mostly affecting orthodox Jews) and allows the student to take it the very next day, the Sunday. This provision also calls for some very strong supervision, probably involving a rabbi.

Thanks for the quick reply. I guess I won't be able to participate in Putnam at all... Oh well.

You really wouldn't take it just because of studying for finals? Study for six more hours earlier in the semester instead of playing games or something, and that's probably more valuable than 6 hours of cramming on a December Saturday anyway!

Um, first of all, I don't really play games or goof off much. I can't because I'm pre-med.  :)

I think that although the exam's 6 hours, the actual impact will be 1 hr or 2 hrs after that so that's like 8 hours... of not studying finals.

I don't know, I'm not math major so yeah.  :|

I think in any case if you study effectively during the course of the semester, you will not need to spend all of your free weekend time in December studying for finals. Anyway I know that at Princeton some students sign up and then just don't show up to the test; if your school is doing Putnam anyway, and students other than you are taking it, it probably won't cost them anything for you to sign up and then take it only if you feel sufficiently good about your classes to take that day off from studying. (Dr. Merryfield or someone else who knows better, please correct me if I'm wrong to suggest this.)

Well, if I signed up 8 people to take the Putnam and 7 of them showed up, that would be OK. But if I signed up 8 people to take it and 2 of them showed up, I'd get pretty annoyed. And for my 1-unit class "Honors Seminar in Problem Solving" it says on the syllabus that taking the Putnam is an obligation of the course.

Fortunately for us, our final exams aren't until the week of Dec. 15-19, so the Putnam won't be the very last weekend.


A ball is tossed into the air from a bridge, and its height, $ y$ (in meters), above the ground $ t$ seconds after it is thrown is given by $ y\equal{}f(t)\equal{}\minus{}5t^2\plus{}11t\plus{}11$.

How high above the ground is the bridge?

What is the average velocity of the ball for the first second?

Approximate the velocity of the ball at $ t\equal{}1$ second. Round the answer to the nearest integer.

Well how do you know the ball is at the height of the bridge?  Answer: When no time has passed.  So $ f(0)\equal{}11$ and the bridge is at 11 m.

If $ v(t)$ is velocity, then average velocity is given by
$ \frac{1}{b\minus{}a}\int_{a}^{b}v(t)\,dt$
But we're not quite there yet.  I'll let you figure out how to get there.

You can find the position of the ball after 1 second, f(1), and since you have f(0) = 11, you know the distance covered during that first second. It's f(1) - f(0). Divide that distance covered by the time (1 second), and you have the average velocity. No need to study calculus.

Actually, notice that $ f(t)$ is not a linear function so $ \left|f(1)\minus{}f(0)\right|$ gives the displacement, not the distance traveled.  That is, it will tell you how high the ball traveled, but not the length of the path it took.

I believe the ball was launched straight up (as there's no indication of horizontal velocity).

[hide="Hint"]
If an object accelerates uniformly, its position is given as a function of time by

$ s(t) \equal{} s_0 \plus{} v_0t \plus{} \frac {at^2}{2}$

Where $ a$ is the acceleration and $ v_0$ is initial velocity. This should allow you to answer all of the questions.
[/hide]

@JRav:As the ball didn't have any horizontal velocity, it's fine to assume that it traveled in a straight line. Even if it went up, then down, the definition of average velocity is not distance/time or displacement/time. It's $ \frac {\Delta s}{\Delta t}$, where $ \Delta s$ (or $ \Delta x$, as some of you use) indicates difference in space (the difference between the three is something I'm not discussing in detail, unless someone asks), so it's irrelevant.

[quote="JRav"]...
If $ v(t)$ is velocity, then average velocity is given by
$ \frac {1}{b - a}\int_{a}^{b}v(t)\,dt$
But we're not quite there yet.  I'll let you figure out how to get there.[/quote]

Since we're given $ y = f(t)$ we don't actually need to calculate that integral.

The average velocity over the first second is given by:
\[ \begin{array*}\overline{v}_y & = \frac {\Delta y}{\Delta t} \\
& = \frac {f(1\, sec) - f(0\, sec)}{1\, sec - 0\, sec} \\
& = \frac {( - 5(1)^2 + 11(1) + 11) - ( - 5(0)^2 + 11(0) + 11)}{1 - 0} \\
& = 6\, m/sec
\]
We can use this same technique to approximate the velocity @ $ t = 1\, sec$. We merely have to assume a very small interval around $ t = 1\, sec$.

Then

$ v(1\, sec)\approx \frac {( - 5(1.01)^2 + 11(1.01) + 11) - ( - 5(0.99)^2 + 11(0.99) + 11)}{1.01 - 0.99} = - 10 + 11 + 0 = 1\, m/sec$.


Let $ f : \mathbb{R^{+}}\longrightarrow\mathbb{R}$ be a function
such that
$ f(x)-f(y)\geq\ln\frac{x}{y}+x-y$ for every $ x,y\in\mathbb{R^{+}}$
and $ f(1)=1$
Find f.

[quote="0714446459923"][hide]
if you flip x and y, you get
$ f(x)-f(y)\leq\ln\frac{x}{y}+x-y$
so therefore,
$ f(x)-f(y)=\ln\frac{x}{y}+x-y$
and f(x)=ln(x)+x
[/hide][/quote]
You are correct  :)

if we take $ g(x)=f(x)-x-ln(x)$
$ f(x)-f(y)\geq\ln\frac{x}{y}+x-y$ gives $ \forall (x,y)\in \mathbb{R}^{+}: \ g(x)\ge g(y)$
so $ \exists c\in\mathbb{R},\forall x\in \mathbb{R}: \ g(x)=c$
then $ \exists c\in\mathbb{R},\forall x\in \mathbb{R}: \ f(x)=x+ln(x)+c$


[quote="pda"]Determine the number of five-digit positive integers whose neighboring digits differ by 3.[/quote]

Im confused at the problem.

Is a number 30303 an example of one that works?

Is it that the positive difference between any two nieghboring digits is 3?

[quote="pda"]Yes, it is. Also examples are 47414 or 63036.[/quote]

altight thanks..Ill work on this problem.

Let me know if this is incorrect:
[hide]
Because each neighboring digit have a difference of 3, the five-digit numbers can only contain digits {$ 1, 4, 7$}, {$ 2, 5, 8$}, or {$ 0, 3, 6, 9$}.

For the first two sets, there is $ 3*2*2*2*2 = 48$ possible numbers.
For the last set, there is $ 3*3*3*3*3 = 243$ possible numbers.

So the total number is $ 2(48)+243 = \boxed{339}$[/hide]

but your solution can contain numbers such as 14711 or 30660

[hide="Big hint"] This is a path-counting exercise in disguise. [/hide]

[quote="p4fn2w"]I dont think it could,
consider $ 3*2*2*2*2$, that means the first digit can be any one of the 3 numbers on the first list. The second digit can be any of the three numbers also, just excluding the one you chose on the first digit (so technically any 2 number on the first list). The third digit can be any number, excluding the one you chose on the second digit (so theres again 2 choices). And so on...

Isnt that how it works?![/quote]

No.  You are allowing, for instance, 17171.

I got $ 45$

Im pretty sure thats correct becasue you just have to consider the cases (0,3,6,9) (1,4,7) (2,5,8)

The last two are easy, its just 4*1*2*1*2+2*2*1*2*1=24

The first case.

say its 303 then there are 3
say its 363, still 3
369, only 2

3+3+2=8

8 cases for the first digit being 6 also

if the first is 9, 963-3 choices, 969-2 choices

24+8+8+3+2=$ 45$

Im pretty sure Astro Physics brute forced it wrong because when its a middle digit of a case, there are more possiblities than an end digit(although his post was 5.5 i think his answers wrong)

[quote="AstroPhys"][hide]
Starting with $ 1$
Possible Digits seperated by $ |$.
$ \{ 1|4|1,7|4|1,7 \}$. Total $ 1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 4$.

Starting with $ 2$
$ \{ 2|5|2,8|5|2,8 \}$.  Total $ 1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 4$.

Starting with $ 3$
$ \{ 3|0,6|3,9|0,6|3,9 \}$.  Total $ 1 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 16$.

Starting with $ 4$
$ \{ 4|1,7|4|1,7|4 \}$.  Total $ 1 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 4$.

Starting with $ 5$
$ \{ 5|2,8|5|2,8|5 \}$.  Total $ 1 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 4$.

Starting with $ 6$
$ \{ 6|3,9|0,6|3,9|0,6 \}$.  Total $ 1 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 16$.

Starting with $ 7$
$ \{ 7|4|1,7|4|1,7 \}$.  Total $ 1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 4$.

Starting with $ 8$
$ \{ 8|5|2,8|5|5,8 \}$.  Total $ 1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 4$.

Starting with $ 9$
$ \{ 9|6|3,9|6|3,9 \}$.  Total $ 1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 4$.

$ \boxed{4+4+16+4+4+16+4+4+4=60}$.
[/hide][/quote]

There is no way thats right.  Look at your overcounts.  You are counting something likek 30930 to work

[quote="t0rajir0u"]Yes.  Using my hint, it wasn't hard to verify.

[hide="The advantage of a path-counting solution"] It extends readily to more digits and longer numbers.  Suppose we asked for the number of seven-digit positive integers whose neighboring digits differed by $ 1$.  This would simply be counting paths on a $ 10 \times 7$ grid instead of on a $ 4 \times 5$ grid, which isn't really any harder. [/hide][/quote]

wait im kinda confused how you did it...I didnt really use your hint I used some easy casework but your idea seems more logical and easier.


function
abstract algebra
algebra unsolved
algebra

Slove this equation: $ 6^x\plus{}2^x\equal{}5^x\plus{}3^x$

Well, I know a really nice solution to these types of algebra problems using analysis... And I'm now referring to Lagarange's theorem.

The equality is equivalent to: $ 6^x\minus{}5^x\equal{}3^x\minus{}2^x$. Let's fix $ x$ and take a function $ f$, $ f(t)\equal{}t^x$. From Lagrange's theorem there are numbers $ c_1, c_2$ with $ c_1\in(5;6)$ and $ c_2 \in (2; 3)$, $ f'(c_1)\equal{}6^x\minus{}5^x$ and $ f'(c_2)\equal{}3^x\minus{}2^x$.
So at last we have that $ x c_1 ^{x\minus{}1}\equal{}x c_2^{x\minus{}1}$. So, we have that either $ x\equal{}0$ or $ x\equal{}1$.

[quote="dina013"]Well, I know a really nice solution to these types of algebra problems using analysis... And I'm now referring to Lagarange's theorem.

The equality is equivalent to: $ 6^x \minus{} 5^x \equal{} 3^x \minus{} 2^x$. Let's fix $ x$ and take a function $ f$, $ f(t) \equal{} t^x$. From Lagrange's theorem there are numbers $ c_1, c_2$ with $ c_1\in(5;6)$ and $ c_2 \in (2; 3)$, $ f'(c_1) \equal{} 6^x \minus{} 5^x$ and $ f'(c_2) \equal{} 3^x \minus{} 2^x$.
So at last we have that $ x c_1 ^{x \minus{} 1} \equal{} x c_2^{x \minus{} 1}$. So, we have that either $ x \equal{} 0$ or $ x \equal{} 1$.[/quote]

Thank you very much.


exist or not exist $ a,b,c,d$ be positive integers such that $ ab \equal{} cd$ and $ a^2 \plus{} b^2 \equal{} c^2 \minus{}d^2$    :D

Assume the contrary. The given conditions imply $ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \plus{} bd)^2$ so $ c^4\minus{}d^4 \equal{} (ac \plus{} bd)^2$. But the equation $ x^4 \minus{}y^4 \equal{} z^2$ has no solution in positive integers ([url=http://www.mathlinks.ro/viewtopic.php?p=1263074#1263074]proof[/url]). Hence such integers do not exist.


quadratics
number theory proposed
number theory

Solve the equation :

$ x^2\plus{}xy\minus{}3y^2\equal{}15$

take the disc. and see that you get $ k^2 \equiv 8 (mod13)$

solve the quadratic in $ y$
the disc. will be $ 2x^2\plus{}180\implies x^2\plus{}90\equal{}2z^2\implies 2x_{1}^{2}\plus{}45\equal{}z^2$.
$ z$ is odd. and so $ 8\mid 2x_{1}^{2}\plus{}44\implies 4\mid x_{1}^{2}\plus{}22\implies 4\mid x_{1}^{2}\plus{}2$ which is not possible.
so no solution is possible.

[quote="Max D.R."]Solve the equation $ x^2 \plus{} xy \minus{} 3y^2 \equal{} 15$[/quote]

We have $ x^2 \plus{} xy \minus{} 3y^2 \equal{} 15$ iff $ (2x \plus{} y)^2 \minus{} 13y^2 \equal{} 60$ that is divisible only one time by $ 5$, but $ \left(\frac {13}{5}\right) \equal{} \left(\frac {3}{5}\right)\equal{} \minus{} 1$,contradiction. []


The points $ (2,\minus{}3)$, $ (4,3)$, and $ (5, k/2)$ are on the same straight line. The value(s) of $ k$ is (are):

$ \textbf{(A)}\ 12\qquad
\textbf{(B)}\ \minus{}12\qquad
\textbf{(C)}\ \pm 12\qquad
\textbf{(D)}\ {12}\text{ or }{6}\qquad
\textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}$

We have $ x_1\equal{}2$, $ y_1\equal{}\minus{}3$, $ x_2\equal{}$, and $ y_2\equal{}3$. We find the slope of the line to be $ \frac{y_2\minus{}y_1}{x_2\minus{}x_1}\equal{}\frac{6}{2}\equal{}3$. Then we form the equation $ y\equal{}3x\minus{}9$. We know $ x\equal{}5$ in the third point, so we find $ y$ to be $ 3(5)\minus{}9\equal{}6\equal{}\frac{12}{2}$. So $ k\equal{}12$. That is answer choice $ \mathbf{(A)}$.


I found this problem [i]really[/i] fun.

"Show for any positive integer $n$ there exists a positive number composed of only $7$'s and $0$'s which is divisible by $n$."

I have two solutions to this one:

[hide=" First One"]
Create the following sequence of positive integers:
$7,77,777,...,\underbrace{777...7}_{n+1}$
We have $n+1$ distinct integers so there exists two who difference is divisible by $n$ by the Pigeonhole Principle (Pigeons = Differece of Numbers, Holes = Remainder mod n).
And if we subtract there we end up with a number composed of only $7$'s and $0$'s
[/hide]

[hide=" Second One"]
We can solve this using the following competition problem (shown in my book):

"Show given $n$ distinct integers then either one is a multiple of $n$ of a sum of several is a multiple of $n$.

[b]Proof:[/b] Given $a_{1},a_{2},...,a_{n}$ consider the integers:
$a_{1}, a_{1}+a_{2}, .... , a_{1}+a_{2}...a_{n}$.
So either one is divisible by $n$ or if not then by first approach there exists two two difference is and hence: $a_{i}+a_{i+1}+...+a_{j}$ is divisible by $n$.

So we construct the following sequence:
$7,70,700,...,7\underbrace{00...0}_{n}$
So either one is divisible by $n$ of a sum of several of them. But the sum of any of them is a number consisiting of only 0's and 7's. So that it.
[/hide]

Here's a solution (I haven't bothered looking at your solutions, because then that will ruin the purpose of your post.  :wink:).

[b]Solution.[/b] Consider the $n$ numbers $7,77,777,\hdots, 77...7\bmod n$.  If any of these terms is divisible by $n$, then we are done.  If not, then we have $n-1$ congruency classes modulo $n$.  Therefore by Dirichlet's Box Principle, there are two (say $p$ and $q$, $p>q$) that are congruent $\bmod n$.  Hence $p-q\equiv 7...70...0\equiv 0\pmod n$, as required. $\Box$


Wat does that double-I notation mean? Give example of usage too plz.

What notation¿¿¿ I don't see anything except a product...

I'd bet he'd never seen a big pi, hence he thinks those are two connected I's.


[quote="epitomy01"]Dear fang-jh,
Apart from the 6 problems in your post, I noticed that you also posted a few problems recently, which you also specified as also from Chinese TST:
http://www.mathlinks.ro/viewtopic.php?t=198128
http://www.mathlinks.ro/viewtopic.php?t=198114
http://www.mathlinks.ro/viewtopic.php?t=198296
http://www.mathlinks.ro/viewtopic.php?p=1090338#1090338
Are those also from the same 2008 Chinese TST, or perhaps from the 'quizzes' leading up to it?[/quote]
Dear epitomy01
You are right! those are from quizzes.

The first quiz  March 17, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=198471[/url]
P2:[url]http://www.mathlinks.ro/viewtopic.php?t=198344[/url]
P3:[url]http://www.mathlinks.ro/viewtopic.php?t=198360[/url]
The second quiz March 18, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=198128[/url]
P2:[url]http://www.mathlinks.ro/viewtopic.php?t=199551[/url]
P3:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=198501[/url]
The third quiz   March 21, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=198114[/url]
P2:[url]http://www.mathlinks.ro/viewtopic.php?t=198299[/url]
P3:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=198296[/url]
The fourth quiz March 22, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=199148[/url]
P2:[url]http://www.mathlinks.ro/viewtopic.php?t=198529[/url]
P3:[url]http://www.mathlinks.ro/viewtopic.php?t=199125[/url]
The fifth quiz March 25, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=198511[/url]
P2:[url]http://www.mathlinks.ro/viewtopic.php?t=198513[/url]
P3:[url]http://www.mathlinks.ro/viewtopic.php?t=198502[/url]
The sixth quiz March 26, 2008
P1:[url]http://www.mathlinks.ro/viewtopic.php?t=199114[/url]
P2:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=199121[/url]
P3:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1097823#1097823[/url]
PS: Of all quizzes, the third is the hardest.


MATHCOUNTS search geometry algebra polynomial function

ok for mathcounts 6th grade we have bases?? i am so confused on it n e one help me? plz help thx

Search for bases in the math forum.  I'm sure someone has asked about this before.  I'm too tired to explain it now.  :D

By the way, the search button is at the top of the page where it says profile, options, etc.

It depends on your goals.  If your aiming for full score (like me XD) you NEED to know.  If your happy with a 40 or so, then you can just let it go.  Same with mod theory, some advanced geometry and algebra (polynomial division and inverse functions) and stuff like that.

A lot of what you need is based on what you want.

learn them. its really useful and important if you ever want to do mathcounts

Yes, definately learn them.
Even if you didn't need them, I suggest you learn them anyway.
Because I said so.


calculus
induction
geometry
trigonometry
AMC
AIME
number theory

-      Next year in the fall, I can either take differential equations at Oakland Community College or  a proof writing course at Oakland University.  I've already taken Calculus III. The thing with the differential equations course is that it is free because school will pay for all of it, while the proof course will cost 1000 dollars.
-      The differential equations course is a fairly standard applied college class, while the proof course covers basics of proofs; induction, contrapositives, etc... (20%, all of which I know already), includes many proofs, some which are helpful in math competitions, from elementary geometry, trignometry, and number theory (30%), then has some fairly standard proofs of calculus (20%), and the rest is commonly used proof techniques in higher math (30%). It is a course that will enable me to take the proof-based courses of linear algebra in the winter of my junior year, real analysis in the fall and spring in my senior year (its a full year course), which is all I have room for because I'm taking 5 classes at school, with one free period.
-         Anyways, I can't take both because of stress, and the proof class [b]is [/b]the prerequisite for the classes mentioned above, which I really want to take, because I'd like to be a theoretical math major in college. I'd rather not take any courses over the summer because: 1) I'd like to prepare for math competitions, improving on my pathetic AIME score of 4 this year. and 2) only a few courses are offered in that time period, making them very hard to schedule.
-          Yes, I've looked into EPGY and other distance education courses, but I have bad habits of procrastinating unless I'm in a physical class, so I've completely ruled that out.
-          I'm leaning towards the proof course because I'd like to take that as its suited towards my interest, but two factors are preventing me from enrolling. One is that differential equations is considered a "core" course in mathematics by nearly all the professors I've talked to, and they suggest that I take it while calculus is fresh in my mind. The other is the cost, which is a lot, but I can pay with money from my bank account.

-So, what should I take? Any ideas?

In the circumstances you describe, I would take the proof course, to get more of the mental tools to accomplish the other personal goals you mentioned.

Hello -  My advise would be  take diffeq.

Unless the  prof for the  other course is REALLY good or
( Unless) you really want to take the  other course  Then the above advice  can be ignored :)

Few things to think about  (My train of thought )

[b]Diffeq is a core course [/b] (No question about it)  and would help in    all   the later  Engineering type courses. Having luxury to do it  earlier would be nice (You would have more time to practice)

And yes  I have to  tend to suggest that you listen to ..(as I would say the same)
[quote] One is that differential equations is considered a "core" course in mathematics by nearly all the professors I've talked to, and they suggest that I take it while calculus is fresh in my mind.
[/quote]
Proof course could be enjoyable  but if the teacher  is  not good it can be really  make you  dislike the course.
(Not having the  best Diffeq teacher  is  less risky than not having the best  teacher for the other course)

Hopefully that is somewhat helpful.[quote][/quote]

[quote="Gyan"][b]Diffeq is a core course [/b] (No question about it)  and would help in    all   the later  Engineering type courses. Having luxury to do it  earlier would be nice (You would have more time to practice) [/quote]

This is not true.  It is a core course only for applied mathematicians and for engineers and physicists.  Of course, since this makes up a large portion of mathematicians and for people who use mathematics in their everyday lines of work, I guess the difference is a bit pedantic.  But it's not really important for pure mathematicians (this has been discussed around here before).

All of that being said, I would take diffeqs, mainly because its free and then you've completed pretty much all of 'basic, plug and chug university math' at quite a young age after you're done with linalg.

[quote]Diffeq is a core course (No question about it) and would help in all the later [b]Engineering type courses.[/b] Having luxury to do it earlier would be nice (You would have more time to practice)   [/quote] [quote]
This is not true. [b]It is a core course only for applied mathematicians and for engineers and physicists.[/b] Of course, since this makes up a large portion of mathematicians and for people who use mathematics in their everyday lines of work, ...[/quote]..
Seems like we are saying the samething :) (Yes I do agree that it  is not the core course for pure math  ... even for Theoratical physics , IMO,  good background in, say,  abstract algebra, group representation etc   are  more "core"/helpful  than Diffeq -- but that's my opinion...)


Let $ABC$ be a triangle inscribed in circle $R$ .Also let the angle bisector line of $A,B,C$ intersect the circumference of circle $R$ at$A',B,'C'$ respectively . Let $AA' \cap BC=N$, $C'A'\cap BB'=P$. Also denote the orthocenter as $I$ . Now , let $O$ be the circumcenter of triangle $IPC'$ , and $OP\cap BC=M$. If $BM = MN$ , and $\angle BAC = 2\angle ABC$ , find all the angle $\angle A ,\angle B,\angle C$ in the triangle $ABC$ .


We call a sequence $a_1 , ... , a_n \in {0 ; 1}$ good if for each $ 0 \leq k leq n-1 $
the sum $ a_1 a_{k+1} + a_2 a_{k+2} ... + a_{n-k} a_n $ is odd.
Prove that:
a. There exist a good sequence with n=25.
b. There exist a good sequence with abritary large length.
c. (My result ) Prove that if there exist a good sequence with length n then the order of 2 mod 2n-1 must be odd.
(I believe that this is also the sufficent condition but I have some difficulties in proving this.)
(Part a and b was given in an old All-Russian Olympiad).

I don't think (c) is correct. We have to show that $n=25$ works, so the order of $2$ mod $2\cdot 25+1=51$ should be odd. However, if $d$ is the order, then $3|51|2^d-1$, so $d$ must be even. Am I missing something? Maybe by order you mean something else? (I mean the order of $2$ in the multiplicative group of invertible elements from $\mathbb Z_{2n+1}$).

can you please post the solution? I remember I saw it and its solution in the book with Soviet Union Olympiads. And I remember the idea, but I am so stupid I can't go further..

I have just looked solution, and, in my opinion, it is quite standart idea. In general, I don't consider this problem as interesting one.

Idea is following:
if $A_n$ (n elements) and $A_m$ (m elements) are two sequences which satisfy condition then
1) we replace each '1' in $A_n$ to block $A_m0...0$ (m-1 zeros),
2) each '0' in $A_n$ we replace to block 0...0 (2m-1 zeros)
3) exlude last (m-1) zeros from obtained sequence.
As a result we obtain sequence $A_{2mn-m-n+1}$ which satisfies condition (though it is nontrivial exercise).

Initially we have $A_4=1101$. Using our construction from pair $A_4$ and $A_4$ we obtain $A_25$.


function
trigonometry
topology
real analysis
real analysis solved

[quote="Anto"]Let E be a dense subset of the metric space X. Let Y be an other metric space, and f a omiomorphical (probably i have translated wrongly ...i mean that for every e>0 there is a d>0 s.t. for all p,q in E with dx(p,q)<d we have dy(f(p),f(q))<e) continuos function from E to Y. Prove that we can find a unique continuos function g from X to Y s.t. f(p)=g(p) for all p in E.

The uniqueness is trivial, i have proved that the result is true if Y=$R^k$ or any compact space. But does the general case hold too?[/quote]

 I think you meant Uniformly continuous :D
 The result holds whenever $Y$ is a complete metric space. A counter example in the case $Y$ is not complete is provided by the following case : $E = Y = \mathbb{Q}$ , $X = \mathbb{R}$, and $f$  is the identity mapping. Then there is no continous continuation of $f$ onto $X= \mathbb{R}$

Indeed both $R^k$ and compact sets are complete spaces.
Thanks you  :)
I have realized that it is exactly the same (the proof of above result) as the proof on compact spaces :)
And the counterexample is very clear...

As far as the translation...
Obviously I meant Uniformly < Uni + form which has the same origin with the one i used...
This the way we spell uniformly continuous function in Greece, and i imagine that it is the same and out of here.

dickchimney's example missed the point. No one said "onto"; at issue is the existence of a unique continuous continuation.

If $E=X$, then there is nothing to continue, so there is no issue. If $E=\mathbb{Q}$, $X=\mathbb{R}$ and $f$ is the identity mapping, then it is uniformly continuous and does admit the unique continuation to the indentiy mapping on $\mathbb{R}.$

For a cheap example, consider this: $(0,1]$ is dense in $[0,1].$ Define $f: (0,1]\mapsto\mathbb{R}$ by $f(x)=\sin\left(\frac1x\right).$ This $f$ is continuous  but not uniformly continous and cannot be extended to a continous function on $[0,1].$

The key issue, of course, is the completeness of $Y$. The reason you say it works for compact metric spaces is that a compact metric space must be complete.

Heres another example f defined on R-{0} by the following
f(x)=1/x .Clearly R-{0} is dense in R but f has no continuous extension on R though it is continuous on R-{0}


Let $x,y,z\in\ [0;1],n\in\mathbb{N},n\geq1$,Prove that the following inequality occurs: \[x^{2n}+y^{2n}+z^{2n}\geq\left(xyz\right)^{n}+2\]

[hide="Solution"]
Let $x^{n}=x, y^{n}=y, z^{n}=z$.
We prove: $x^{2}+y^{2}+z^{2}-xyz \le 2$.
Keep y, z fixed and consider $f(x) = x^{2}+y^{2}+z^{2}-xyz, \forall x \in [0,1]$.
$f(x) \le f(1) \Leftrightarrow x(x-yz) \le 1-yz$.
But $x(x-yz) \le x(1-yz) \le 1-yz$.
So $f$ has a maximum for $x=1$.
We have $g(x,y,z) \le g(1, y, z)$, where $g(x, y, z) = x^{2}+y^{2}+z^{2}-xyz, \forall x,y,z \in [0,1]$.
Now we keep $y$ fixed and we get:
$g(x,y,z) \le 2+y^{2}-y = 2+y(y-1) \le 2$, QED.
[/hide]

Very nice!My solution is the same..Are there other ways to prove this?


[quote="mathVNpro"]Let $ f: (0,\infty)\longrightarrow (0,\infty)$ such that:
$ f(x \plus{} \frac {1}{x}) \equal{} \frac {\sqrt [3]{2009}}{f(1)}$[/quote]

Choose $ f(x)$ such that :
$ f(x)$ takes any value for $ x\in(0,2]$
$ f(x)\equal{}\frac {\sqrt [3]{2009}}{f(1)}$ $ \forall x>2$

[quote="pco"][quote="mathVNpro"]Let $ f: (0,\infty)\longrightarrow (0,\infty)$ such that:
$ f(x \plus{} \frac {1}{x}) \equal{} \frac {\sqrt [3]{2009}}{f(1)}$[/quote]

Choose $ f(x)$ such that :
$ f(x)$ takes any value for $ x\in(0,2]$
$ f(x) \equal{} \frac {\sqrt [3]{2009}}{f(1)}$ $ \forall x > 2$[/quote]
Can you tell me why $ f(x)$ takes any value for $ x\in(0,2]$?? :blush:

[quote="mathVNpro"][quote="pco"] Can you tell me why $ f(x)$ takes any value for $ x\in(0,2]$?? :blush:[/quote][/quote]

Because, $ \forall x\in(0,\plus{}\infty)$, $ x\plus{}\frac{1}{x}>2$ and so the equation implies no constraint on $ f(x)$ for $ x\in(0,2]$

Please stop solving and the lock the topic, since the problem was the one on the Mathematics and Youth magazine of Vietnam after some small work.

The original problem is:

[b]Problem:[/b] Find every functions $ f: (0,\plus{}\infty) \to (0;\plus{}\infty)$ satisfies:
$ f(xy)f(yz)f(zx)f(x\plus{}y)f(y\plus{}z)f(z\plus{}x)\equal{}2009, \forall x,y,z \in (0;\plus{}\infty)$.

After some simple substitutions, it will lead to the problem posted in this topic.

@[b]mathVNpro:[/b] I don't think this way of substituting could solve the problem :D


Can any of you guys provide me with a copy of the following paper?

[b]E. M. Wright[/b], A class of representing functions, [i]J. London Math. Soc.[/i] 29 (1954), 63-71.

I thank you in advance for you support.

[quote="coquitao"][i]Peoples[/i], 'tis a matter of death and lifeee![/quote]

I did your Google search for you.  This was the first result:

http://jlms.oxfordjournals.org/cgi/reprint/s1-29/1/63.pdf

Can you see an actual [i]pdf[/i] file there?  :(


1.$\sqrt[3]{2^{\frac{1}{2}}}*\sqrt[3]{6^{\frac{1}{5}}}$ can be written in the form of $2^p*3^q$. Find the value of $p+q$.

2. Let a,b, and c be real numbers such that $a+\frac{1}{b}=5$, $b+\frac{1}{c}=12$, and $c+\frac{1}{a}=13$. Find the value of $abc+\frac{1}{abc}$.

[quote="lkryptonicl"]2. Let a,b, and c be real numbers such that $a+\frac{1}{b}=5$, $b+\frac{1}{c}=12$, and $c+\frac{1}{a}=13$. Find the value of $abc+\frac{1}{abc}$.[/quote]

[hide="Solution"]Notice the $abc$ and the $\frac{1}{abc}$ terms.  One way would be the multiply the three equations together:

$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=13\cdot 12\cdot 5=780=abc+\frac{1}{abc}+a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}=abc+\frac1{abc}+30\implies abc+\frac1{abc}=750$ [/hide]


Let a, b, c be three positive reals. Then, prove or disprove the inequality

$\frac{2a^2+b^2+c^2}{2b^2+c^2+a^2}+\frac{2b^2+c^2+a^2}{2c^2+a^2+b^2}+\frac{2c^2+a^2+b^2}{2a^2+b^2+c^2}$
$\geq \frac{2a+b+c}{2b+c+a}+\frac{2b+c+a}{2c+a+b}+\frac{2c+a+b}{2a+b+c}$.

[b]EDIT:[/b] Well, this is wrong. As a counterexample, you may take a = 11, b = 10 and c = 0 (or $c=\epsilon$ with a very small $\epsilon>0$, if you really want to have the numbers a, b, c strictly positive). Sorry for not properly checking the inequality before posting it...

  darij

The following inequality is true.
Let $a, b, c>0$. Prove that
\[\sum_{cyc}{\frac{3a^2+b^2+c^2}{3b^2+c^2+a^2}}\ge\sum_{cyc}{\frac{2a+b+c}{2b+c+a}}\]


Functional Analysis
real analysis
real analysis unsolved

How to prove that set $ A \equal{} \{x\in \mathcal{H}: \sum_{n \equal{} 1}^{ \plus{} \infty} (1 \plus{} \frac {1}{n})^2|(x|e_n)|^2\leq 1\}$, doesn't have element of maximum norm, where $ \mathcal{H}$ separable Hilbert space, $ (e_n)_{n\in\mathbb{N}}$ orthonormed base. I tried everything I could remember but I had no luck  :blush: .

Prove that the norm of any such $ x$ is always less than $ 1$ but can be made arbitrarily close to $ 1$.


function
trigonometry
real analysis
real analysis theorems

Hey guys,

can anybody explain to me what exactly a damped cyclic phenomena is? and if i am given a set of points which specifically identify the peaks of the graph, how can i form an equation from them?

more specifically:

i'm given 4 sets of points for $t$ and $y$ which are the measurements of 4 consecutive peaks, and am being asked to determine an equation for $y$ as a function of $t$.

Damped cyclic phenomena are any variables that take the form of a $\sin$ or $\cos$ graph, but whose amplitudes decrease over time.  For example:

[img]http://img.sparknotes.com/figures/1/103aba3edae345165f4823058d6a1e6f/damped.gif[/img]

The equation will typically take the form

$y = b^{t}\sin ( \omega t+\gamma)$

Where $0 < b < 1$.


inequalities
vector
linear algebra
linear algebra unsolved

Let V,W be Euclidean vector spaces over $R$ and $\phi : V \rightarrow W$ be a linear map. Show that there exists such a constant C that for every $x \in V$ $\|\mathbf{\phi(x)}\| \leq C\|\mathbf{x}\|$.

Hint: $C$ is the maximum of $\|\phi(x)\|$ on the closed unit ball of $V$.


function
probability
binomial coefficients

Say we have a basic dice problem: What is the probability of rolling a $ 25$ with $ 4$ standard dice?

I want to generate functions to solve this, but since I'm new to this please check if the below is correct.
\[ (x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6)^4\]
Now, we must find the coefficient of the $ x^{25}$ term.

How do I go about doing that? The AoPS Wiki contains nothing and I couldn't find anything online.

Please don't tell me to get the book; I was about to get it, but now I'm going to get Intermediate Algebra.

I'm pretty sure that equation doesn't have a x^25 term (6*4=24, too low) so while I don't know about generating functions, I'm pretty sure you did it wrong.

Also, what book would that stuff be in?

It'd be in Interm. CP., but there is a section about generating functions in Volume 2.

Also, there is no way to roll 25 with 4 standard dice :) because, like Ihatepie said, 6*4 is too low.

For dice a fast way is to use [url=http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_to_negative_integers]negative binomial coefficients[/url] (I believe this technique is in Volume 2); write the generating function as

$ x^4 \left( \frac{x^6 \minus{} 1}{x \minus{} 1} \right)^4$.

Then you know the coefficients of $ (x^6 \minus{} 1)^4$ and the coefficient of $ x^k$ of $ (1 \minus{} x)^{\minus{}4}$ is given by $ (\minus{}1)^k {\minus{}4 \choose k} \equal{} {k\plus{}3 \choose 3}$.  This computation is equivalent to a certain inclusion-exclusion argument.


Solve in $ \mathbb{N}^*$ the inequation :

$ (1\plus{}xy)(1 \plus{} x^3)(1 \plus{} y^3)\ge 3(1 \plus{} x)(1 \plus{} y)(1 \plus{} x^3y^3)$

Note that $ (1\minus{}x^3) \equal{} (1\minus{}x)(1\plus{}x\plus{}x^2)$ and similarly for $ 1\minus{}y$ and $ 1\minus{}xy$.  With $ x,y >1$ (the inequality is clearly true when either $ x$ or $ y$ is $ 1$) we have

$ (1\plus{}x\plus{}x^2)(1\plus{}y\plus{}y^2) \ge 3(1\plus{}xy\plus{}(xy)^2)$
$ 1\plus{}y\plus{}y^2 \plus{}x\plus{}xy\plus{}xy^2 \plus{} x^2\plus{}x^2y\plus{}x^2y^2 \ge 3 \plus{} 3xy\plus{}3x^2y^2$
$ x\plus{}y\plus{}x^2\plus{}y^2 \plus{}x^2y\plus{}xy^2 \ge 2 \plus{} 2xy\plus{}2x^2y^2$
$ (x\minus{}y)^2 \plus{}(xy\plus{}1)(x\plus{}y)\ge 2(1\plus{}x^2y^2)$

not getting anywhere...

Ok , i think i was wrong ... It is a new type of exercise for me

Try this :

$ (1 \plus{} xy)(1 \plus{} x^3)(1 \plus{} y^3)\ge 3(1 \plus{} x)(1 \plus{} y)(1 \plus{} x^3y^3)$

I think this should work .

Ok so the given inequality is equivalent to

$ \frac{1\plus{}x^3}{1\plus{}x} \frac{1\plus{}y^3}{1\plus{}y} \le 3 \frac{1\plus{}x^3y^3}{1\plus{}xy}$
$ (1\minus{}x\plus{}x^2)(1\minus{}y\plus{}y^2) \le 3(1\plus{}xy\plus{}x^2y^2)$
$ 1\minus{}x\plus{}x^2 \minus{}y\plus{}xy\minus{}yx^2\plus{}y^2\minus{}xy^2\plus{}x^2y^2 \le 3\plus{}3xy\plus{}3x^2y^2$
$ \minus{}x\plus{}x^2 \minus{}y\minus{}yx^2\plus{}y^2\minus{}xy^2 \le 2\plus{}2xy\plus{}2x^2y^2$
$ x^2\plus{}y^2 \le x\plus{}y\plus{}x^2y\plus{}xy^2\plus{}2\plus{}2xy\plus{}2x^2y^2$
$ 0 \le x\plus{}y\plus{}x^2(y\minus{}1) \plus{} y^2(x\minus{}1) \plus{} 2xy\plus{}2x^2y^2$

This is clearly true when $ x$ and $ y$ are greater than $ 1$.

Let $ x\equal{}1$.  Then
$ 0 \le 1\plus{}y\plus{}(y\minus{}1) \plus{}2y\plus{}2y^2$
$ 0 \le 2y^2\plus{}4y$
$ 0 \le y(y\plus{}2)$
which is true for $ y \ge 0$ and $ y \le \minus{}2$.  The case where $ y\equal{}1$ is similar.

Let $ x\equal{}0$.  Then
$ 0 \le y \minus{}y^2$
$ 0 \le y(1\minus{}y)$
which is only true for $ 0 \le y \le 1$.  The case where $ y\equal{}0$ is analogous.

I guess my only question is what do you mean by $ \mathbb{N^*}$?  If this is the set of natural numbers (integers starting with $ 1$) then the inequality is clearly always true.  If this is supposed to include $ 0$, the inequality is not always true.

If we wanted to solve this in the set of real numbers, we could play with discriminants of our quadratics

I'm sorry , facis , for my big wrong . I changed . Now i think it is the correct form . And by $ \mathbb{N}^*$ i meant $ x,y\ge 1$

that's okay.  Anyway, I just realized I made a pretty bad mistake.  I should have had $ 1 \minus{} xy \plus{} x^2y^2$ instead of $ 1 \plus{} xy \plus{} x^2y^2$

so my second step should be:
$ (1 \minus{} x \plus{} x^2)(1 \minus{} y \plus{} y^2) \le 3(1 \minus{} xy \plus{} x^2y^2)$
$ 1 \minus{} x \plus{} x^2 \minus{} y \plus{} xy \minus{} yx^2 \plus{} y^2 \minus{} xy^2 \plus{} x^2y^2 \le 3 \minus{} 3xy \plus{} 3x^2y^2$
$ x^2 \plus{} y^2 \le 2 \minus{} 4xy \plus{} 2x^2y^2 \plus{} xy \plus{} yx^2 \plus{} xy^2 \plus{}x\plus{}y$
$ 0 \le 2(xy \minus{} 1)^2 \plus{} xy \plus{} x^2(y \minus{} 1) \plus{} y^2(x \minus{} 1)\plus{} x\plus{}y$

this is clearly true when $ x,y > 1$.  Now if $ x \equal{} 1$
$ 0 \le 2(y \minus{} 1)^2 \plus{} 2y \minus{} 1\plus{}1\plus{}y$
$ 0 \le 2y^2 \minus{}3y \plus{} 2$
$ 0 \le 2(y \minus{} 3/4)^2 \plus{} 7/8$
which is always true.  The case where $ y \equal{} 1$ is analogous.

Thus the inequality is true for all $ x,y \ge 1$ (not just for positive integers)

Ok , facis , you proved that one , but i changed the sign ...

my last two responses were both for your second problem - my last response was a correction to the one before it, since $ (1 \plus{} x^3y^2)/(1 \plus{} xy) \equal{} 1 \minus{} xy \plus{} x^2y^2 \neq 1 \plus{} xy \plus{} x^2y^2$ which was my mistake.

EDIT: OH I didn't realize what you meant by the sign change.  All my math still works except flip the inequality signs.  We then are trying to find
$ 0 \ge 2(xy \minus{} 1)^2 \plus{} xy \plus{} x^2(y \minus{} 1) \plus{} y^2(x \minus{} 1)\plus{} x\plus{}y$
and with $ x,y \ge 1$, this is impossible.

Anyway, some general rules for inequalities:

if $ a < b$ and $ c < d$ (or $ c \equal{} d$) then $ a \plus{} c < b \plus{} d$

if $ k > 0$ and $ a < b$, then $ ak < bk$.  If $ k < 0$ and $ a < b$ then $ ka > kb$.  So when we are multiplying or dividing we need to take into account the sign of what we are multiplying or dividing by.

Putting these two rules together, we can find similar results for subtraction, division,  exponentiation, etc.

After these rules it is pretty much just play with the algebra.  Various inequalities are well known:

trivial inequality: $ x^2 \ge 0$ with equality when $ x \equal{} 0$

the power mean inequality: let $ X$ is a set of $ N$ positive reals, and for $ k \neq 0$, $ f(k) \equal{} \sqrt [k]{\frac {\sum X^k}{k}}$ be the $ k$th power mean.  Then $ a < b$ implies $ f(a) \le f(b)$ with equality iff all the numbers in $ X$ are equal.  Note that we can define $ f(0) \equal{} \lim_{k\to 0}f(k) \equal{} \sqrt [N]{\prod X}$ and the inequality then holds for all $ k$.  this gives results such as HM-GM-AM-RMS ($ 2/(1/x\plus{}1/y) \le \sqrt{xy} \le (x\plus{}y)/2 \le \sqrt{\frac{x^2\plus{}y^2}{2}}$ with equality only when $ x\equal{}y$)

there are numerous other inequalities,  as well as plenty of articles about them.  For instance, http://www.imomath.com has an article about inequalities, and AOPS has 2 in their resources section (under Articles).


What is the difference and purpose of the sentbox and outbox   :?:  :huh:
(in the messaging system on the site)

Thanks.

The outbox is for messages that you have sent, which the other person has yet to open.  These can still be edited.

The sentbox is for messages that you have sent, which the other person has already opened.  You can[i]not[/i] edit these.


There is an 8x8 square with a starting point on the bottom left 1x1 square.  By moving horizontally and/or vertically 5 units backwards or forwards, how many locations can you have from the starting point?  This is a 2003 Nats Target #3. Feel free to give your [b]own[/b] solutions as well as those from the MC solutions book.  :jump:

I still can't believe this was a Nat's Target.

[hide]If the checkerboard were turned into an $xy$ plane, and each space were one unit (starting with 0), then you can travel to any (even, odd) and (odd, even) space, bounded by the axes and the line $x+y=6$.

You can travel to $12$ different places in 5 moves.[/hide]


Deterrmine all naturals n such that f(n)=3n where f denoted the sum of the positive divisors of n

Bomb

PS Should look familiar to someone.....

[quote="bomb"]PS Should look familiar to someone.....[/quote] Do you mean Euclid?

Euclid, not quite. But close to someone of Euclidean proportions of intellect.
Can anyone solve it...

Bomb

Born to be wild....


group theory
abstract algebra
superior algebra
superior algebra unsolved

let G be a finite group with K be the normal subgroup of G. if (|K|,[G:K])=1, prove that K is the unique subgroup of G having order |K|.

[hide="Hint"]Look at the Syllow subgroups of order a divisor of $ |K|$.[/hide]

Sorry i am the beginner in group theory. What is the Syllow subgroup??

Alternatively, assume the existence of another subgroup $ K_1$ of order $ |K|$. Then consider the subgroup $ KK_1$ and look at its order.

I think I get the solution:
Since $ {K \subseteq KK_1}$, it is clear that $ {K \le KK_1}$. Let $ |G|\equal{}|K|st$, where $ |KK_1|\equal{}|K|s$. By product formula, $ |KK_1\parallel{}K \cap K_1|\equal{}|K\parallel{}K_1| \Rightarrow (|KK_1|/|K|)|K \cap K_1|\equal{}s|K \cap K_1|\equal{}|K|$ and $ (|K|,[G: K])\equal{}1 \Rightarrow (|K|,s)\equal{}1$, contradiction.
Is that true??
But, I still don't know how to use Sylow Theorem to do it. :?:

A stronger assertion was proven by -oo- at http://www.mathlinks.ro/Forum/viewtopic.php?t=223605 .

  darij