I am in analytical geometry right now and my friend asked me to tutor him and he said he would pay me. He got a 33% on the first test and is failing and I have a 93% in the class. Because I am not a college scholar and am just his friend, I don't know how much to charge him. I plan on going to his house atleast 3 tims a week and just do the homework with him and show him the way I do it. The MAO captain charges 20 dollars per hour but this kid is my friend and he is paying with his own money. I am thinking about charging him around 5 dollars an hour. What do you all think?

make it more than $5
make it profitable at least :)  :D  :)

I think its better to either charge a 'regular value' like 10-20 dollars or just nothing rather than 5... I think anything under 10 dollars would just be weird.

well, is he your friend or is he not your friend...
either accept 10-20 as if it were a standard business contract
or do it free because you want to help a friend

(like what paladin8 said)

Well, it depends on how far away he is and how much time you expect to spend. If he's your neighbor and it's only like 20 minutes each session, then making it free would probably be good. But if you need to drive for quite a while, and then stay for hours, you should probably charge 5-10 dollars.

If it were my friend, I'd do it for free.

But, I'm assuming that you don't want to do that.  So I'd say about 5-10 dollars.

If I were you, I'd do it for free.  You say he's (supposedly) your FRIEND--so why make him pay for assistance?  Friends are supposed to help eachother, it's kinda a no-questions-asked thing in life.  I've been in the same situation before with one of my friends, and I never made him pay money for my help with math.

Around 15 dollars  should be OK , I guess.......

yeah, at our school the regular price for Calculus tutor is 15(that what i charge.. :D )


another solution is to begin with AM-GM, (x^2)+(y^2)+(z^2)>=xy+yz+zx for the correct x,y,z and finish with GM-HM.


WLOG let y be the second largest among x, y, and z.
we have x 2 y+y 2 z+z 2 x \leq (x+z) 2 y.  (Because of LHS-RHS=z(y 2 -yz-2xy+xz) \leq z(y 2 -yz-yx+xz)=z(y-z)(y-x) \leq 0)
but (x+z) 2 y=y(1-y) 2 =4y[(1-y)/2][(1-y)/2] \leq 4({y+[(1-y)/2]+[(1-y)/2]}/3) 3 =4/27.
Q. E. D.


calculus
inequalities
function
inequalities unsolved

x,y,z>=0,x+y+z=1,then show that
x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z)>=0

please prove that with derivation calculus

i often see that you guys prove most of the inequalitys with saying that it's a "convex function".
i just can't catch your meanings
en,so could you say something more about the "convex function"?thanks

[quote="nttu"]Expand all , we get :
 $ 5(x^2+y^2+z^2) \leq 6(x^3+y^3+z^3)+1 $ , which is true by Schur ineq  ;)  (it's just a hint )[/quote]

ah,nttu
Please show me the using of Schur ineq in detail
thanks

Transform the inequality into an homogenous one using $x + y + z = 1$. We get \[5(x^2 + y^2 + z^2)(x + y + z) \leq 6(x^3 + y^3 + z^3) + (x + y + z)^3.\] This is equivalent to \[x^3 + y^3 + z^3 + 3xyz \geq xy(x + y) + yz(y + z) + zx(z + x)\] which is exactly Schur.

ah ,thanks
but how to solve this problem with the viewpoint of convex function
i'm weak in that

Well, I am so lucky at the end. I found another solution of yours but I don't use Schur at all. I think it is a little bit weird but the idea is very nice.
Ok. Here is mine:
$ x+y+z=1$  WOLG   $y\leq\frac{1}{3}$ (It is obviously, isn't it)
So put $x=1-y-z $ in $V(x,y,z)=x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z)$
$\Rightarrow V(x,y,z)=(-18z+8)y^2 + (-18z^2+26z-8)y+(2+8z^2-8z)$
You have to prove $V(x,y,z) \geq0$. Well you can use caculus here. But I think here is a very short way to prove (without caculus):
$V(x,y,z)= (8-18y)(z+\frac{y}{2}-\frac12)^2+\frac92y(y-\frac13)^2 $
Well because $0\leq$$ y \leq\frac{1}{3}$ so It is easy to prove that $8-18y\geq0$ so $V(x,y,z)\geq0$

   We have done it.
But I like to use Schur too. It is a briliant idea.

yes. schur is power
so what is the comman ineq given by schur

and what is Lagrange identity ??
thanks

Lagrange : http://mathworld.wolfram.com/LagrangesIdentity.html
Schur : http://planetmath.org/encyclopedia/SchursInequality.html

[quote="FishDragon"]$x,y,z \geq 0$,$x+y+z=1,$then show that
$x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z) \geq 0$

please prove that with derivation calculus

i often see that you guys prove most of the inequalitys with saying that it's a "convex function".
i just can't catch your meanings
en,so could you say something more about the "convex function"?thanks[/quote]

ineq Churs  :
          $a^k(a-b)(a-c) +b^k(b-c)(b-a) +c^k(c-a)(c-b) \geq 0$


For each positive integer $n\geq 2$ we denote with $p(n)$ the largest prime number less than or equal to $n$, and with $q(n)$ the smallest prime number larger than $n$. Prove that  \[ \sum^n_{k=2} \frac 1{p(k)q(k)} < \frac 12. \]

$\sum_{1}^{\alpha} \frac{1}{p(k)q(k)}=\sum \frac{p_{k}-p_{k-1}}{p_{k}p_{k-1}}=\frac{1}{2}$
so the inequality follows :P

Albanian  Eagle, yours is not an adequate post,

in that you skipped crucial intermediate steps that either a

student or a (potentially grading instructor) would want to see.

Your work doesn't show why the q(k) term is vanishes.  More

generally, you do not show how the different summations are

equal to each other.

Very limiting in its help for this post of yours.


Find all integers $n\geq 2$ and prime numbers $p$ such that
$n^{p^p}+p^p$ is prime.

This is polish MO nr 52, first round, exercice 5. There is no such prime number. For every [tex]n \geq 2[/tex], [tex]n^p^p + p^p[/tex] is composite.

the number is a produt of $(n^p+p)$ I think. This is because every prime is odd number.So There is only one another case. $p=2$ Then we have that
$n^4+4$Inthis case $n$ must be odd number. But I could'nt continue

[quote="silouan"]the number is a produt of $(n+p)$ I think[/quote]

I don't think so...

$n^{p^p}+p^p=(n^{p^{p-1}})^p+p^p$

Which is divisable by $n^{p^{p-1}}+p$
except for p=2.

But then we have $n^4+4=n^4+4.1^4$
Which is the identity of Sophie Germain, which gives us the only possibility is $n=1,p=2$ which indeed works. But $n \geq 2$, so there are no solutions

Here's the solution I was speaking of (sorry for posting it so lately):
If [tex]p[/tex] is an odd prime number then:

<center>[tex]x^p + y^p = ( x + y ) \cdot ( x^{p - 1} - x^{p - 2} y + x^{p - 3} y^2 -\ldots - xy^{p - 2} + y^{p - 1} )[/tex]</center>

by putting [tex]x = n^{p^{p - 1}}[/tex] and [tex]y = p[/tex] we come to the conclusion that [tex]n^{p^p} + p^p[/tex] is composite.

And if [tex]p = 2[/tex] then [tex]n^4 + 4 = ( n^2 + 2 n + 2 ) ( n^2 - 2 n + 2 )[/tex], and because [tex]n \geqslant 2[/tex], both factors are greater than [tex]1[/tex] and thus [tex]n^4 + 4[/tex] is also composite.

Actually it's just a recapitulation of everything that has been said before  ;)


quadratics
function
algebra
domain
linear equation

[hide="General Idea"]
Square both sides and combine like terms.  You get a radical on one side and a linear equation on the other.  Square both sides again and you get a quadratic that can be solved with the quadratic formula.  The numbers get pretty disgusting.
[/hide]

Yup. After working through some pretty large numbers, we get that the only solution is [hide] $\boxed{x=5}$ [/hide].

Is there a shorter way to do this...instead of squaring and squaring until no roots are left..

[hide="Guess that x is an integer"]
The equation is approximately:
$3\sqrt{3x+10}=(16-3\sqrt{10})\sqrt{x}$
$3x+10=(16-3(3.16)/3)^{2}x$ $(16-9.5)/3=6.5/3=2.17$, $\sqrt{5}=2.23$ hints at $3x+10=5x$ in the integers or $x=5$. Checking, this is a solution. Furthermore, there is only one solution since each radical takes only positive values, i.e. the function $f(x)=3\sqrt{3x+10}+3\sqrt{10x-1}-4\sqrt{16x+1}$  is decreasing on all $x$ in domain (easy check with derivative).
[/hide]
Well, you could do that or go through the quadratics...


integration
trigonometry
calculus
calculus computations

Let $I_{n}=\int^{\frac{\pi}{4}}_{0}\tan^{n}xdx$ where $n$ is a positive integer. Prove that:
$I_{n}+I_{n-2}=\frac{1}{n-1}$

[hide="hint"]Recall that $\tan^{2}x = \sec^{2}x-1$[/hide]


geometry
geometric transformation
dilation

Whats the difference between Henry's law and Rault's law? Where should they be respectively used and why?

For a start, Raoult's law refers to ideal solutions, while Henry's law refers to an ideally dilute solution. Now as an homework, you look the diferences between these two types of ideal solutions.

Doesn't Henry's Law discuss the solubilit of a gas in relation to the pressure of the gas?

Sg = k times Pg

A good excercise would be to explain why the solubility of a gas increases as the pressure increases...

And Raolt's Law discusses the colligative property of vapor pressure:

The partial pressure exerted by the solvent vapor equals the mole fraction of the solvent times the pure vapor pressure of the solvent.


Brute force does work:

Define x = c-b and y = b-a; then, $x \geq 0$ and $y \geq 0$, and moreover, b = a+x and c = a+x+y, so that our inequality becomes

$\left( a+3\left( a+x\right) \right) \left( \left( a+x\right) +4\left(a+x+y\right) \right) \left( \left( a+x+y\right) +2a\right) \geq 60a\left(a+x\right) \left( a+x+y\right)$,

or

$\left( 4a+3x\right) \left( 5a+5x+4y\right) \left( 3a+x+y\right) \geq 60a\left( a+x\right) \left( a+x+y\right)$.

Now,

$\left( 4a+3x\right) \left( 5a+5x+4y\right) \left( 3a+x+y\right) -60a\left(a+x\right) \left( a+x+y\right) \\ =5a^{2}x+8a^{2}y+20ax^{2}+27axy+16ay^{2}+15x^{3}+27x^{2}y+12xy^{2}\geq 0$

and the proof is complete.

I know this is not exactly what people here call elegant, but I've really no idea how to find a better solution.

[b]EDIT:[/b] Problem also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=40837 .

  Darij

From  am-gm  we have  a+b+b >=3(ab 2 )^(1/3). Multiply  the  same  inequalities  for  b+2c , c+2a  and  find that (a+2b)(b+2c)(c+2a) >=27abc.
Then  (a+3b)(b+4c)(c+4a) >= (a+a/3 + 8b/3)(b+2b/3+10c/3)(c+2a) =
=20/9 (a+2b)(b+2c)(c+2a)>=60abc.

am-gm solves 90%  of inequalities....


For most places it's 4, 3, 2, 1, 0 for A, B, C, D, and F, with +/- .3 for a +/- indicator. For example, A+ = 4.3 and C- = 1.7.

Also, some schools add extra points for honors and AP courses, but sadly my school doesn't  :(

[quote="Rep123max"]And is a plus from 6-9 and a minus from 0-4. Like a 95 would be an A (4.0) a 96 would be an A+ (4.3) and a 94 would be an A- (3.7), right?[/quote]

It depends on how your teacher grades.

I think that it mostly depends on the school.  I am not sure how colleges do it though.  My school does it weird.  They make it so your gpa can be really high.

[quote="JSRosen3"]Also, some schools add extra points for honors and AP courses, but sadly my school doesn't  :([/quote]

Phooey.

94 is an A in most schools. 90-93 would be an A-.

There are many different ways.  Here are the three big ones I know -

1.   4.0 = 90-100  A
      3.0 = 80-89    B
      2.0 = 70-79    C
      1.0 = 60-69    D
      0.0 = <60       F/E

2.   The same as above, except with +0.3 and -0.3 for +'s and -'s respectively.

3.    4.0 = 96-100
       3.9 = 95
       3.8 = 94
       3.7 = 93 and so on.

I'm pretty sure colleges use the first one, so people aren't penelized that much for only 1 or 2 points.

Like I said before my school makes it so your GPA can be really high.  The GPA depends on your point grade (not letter).  So if you get a 93 (which is the lowest A here) in a CP class you get a 4.0.  The highest GPA that you can get for an AP class is a 5.87, 5.37 for Honors, and 4.87 for CP/TP.  For example, I think the valedictorian had like a 5.2 overall GPA for all four years.  This year with 2 CP classes and 5 Honors classes I got about a 5.1 GPA.

At college most of the time you are not graded with a direct correlation between test grades and final grades.  That is, in any of the systems that created such a correlation, I would have failed math miserably (with a 66/100 test average, give or take), while I in fact ended up with an A.

Generally, it's safe to say that 95, 85, 75 correspond roughly to A, B and C with plusses or minuses filling in the gaps.


A sequence of numbers starts with $ 1$, $ 2$, and $ 3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $ 1\plus{}2\plus{}3\equal{}6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

$ \textbf{(A)}\  11  \qquad
\textbf{(B)}\   20  \qquad
\textbf{(C)}\   37  \qquad
\textbf{(D)}\   68  \qquad
\textbf{(E)}\    99$

[hide="Solution"]
Denote by $ a_n$ the $ n$th term of the sequence.

We can chug out the numbers: $ a_1\equal{}1$, $ a_2\equal{}2$, $ a_3\equal{}3$, $ a_4\equal{}1\plus{}2\plus{}3\equal{}6$, $ a_5\equal{}2\plus{}3\plus{}6\equal{}11$, $ a_6\equal{}3\plus{}6\plus{}11\equal{}20$, $ a_7\equal{}6\plus{}11\plus{}20\equal{}37$, $ a_8\equal{}11\plus{}20\plus{}37\equal{}68$, which is answer choice $ \boxed{\textbf{(D)}}$.
[/hide]


Μετέφερα από το thread: ΠΡΟΒΛΗΜΑΤΑ ΓΙΑ ΦΟΙΤΗΤΕΣ (από τις 20 Φεβρουαρίου και μετά) ερώτηση του  SorrowMan

[color=blue] SorrowMan wrote:
As valw kai mia akoma.... Na vrethei akolouthia sinartisewn Fn tetoia wste gia kathe x pou anikei sto [0,1] na min sigklinei kai to orio tou olokliromatos apo 0 ews 1 tis Fn na isoute me 0 ...Iparxoun 2 tropoi me tous opoios mporeis na vreis(an kai o enas sxetizete me ton allon) .... [/color]

Στην οποία ο Durandal απάντησε

[color=blue]Heh, δεν ξέρω γιατί λες μόνο για 2 τρόπους... μπορώ να σκεφτώ αρκετούς, και ο καθένας να αναδεικνύει ένα εντελώς διαφορετικό φαινόμενο  [/color]
.......

Μετά προτάθηκαν διάφορες λύσεις στο πρόβλημα του SorrowMan. Είναι όλες  πολύ έξυπνοι τρόποι αντιμετώπισης του προβλήματος αλλά νομίζω ότι όλοι είναι δυσκολότεροι από «όσο πρέπει». Ένας ευκολότερος τρόπος είναι ο εξής: Παίρνουμε διάφορες  συναρτήσεις f, g, h, ... στο [0,1] που έχουν ολοκλήρωμα ίσο με το 0. Π.χ. οι f(x) = x-1/2 ,  g(x) = -x + ½,  h(x) = e^x – (e-1) μας κάνουν. Ορίζουμε τώρα την ακολουθία
[color=white]  ........     [/color]     f, g, h, f, g, h, f, g, h, ....
Είναι σαφές ότι η ακολουθία αυτή απαντά στο πρόβλημα.

Τάκης

[quote="Papadopoulos"]Μετέφερα από το thread: ΠΡΟΒΛΗΜΑΤΑ ΓΙΑ ΦΟΙΤΗΤΕΣ (από τις 20 Φεβρουαρίου και μετά) ερώτηση του  SorrowMan

[color=blue]Μετά προτάθηκαν διάφορες λύσεις στο πρόβλημα του SorrowMan. Είναι όλες  πολύ έξυπνοι τρόποι αντιμετώπισης του προβλήματος αλλά νομίζω ότι όλοι είναι δυσκολότεροι από «όσο πρέπει». Ένας ευκολότερος τρόπος είναι ο εξής: Παίρνουμε διάφορες  συναρτήσεις f, g, h, ... στο [0,1] που έχουν ολοκλήρωμα ίσο με το 0. Π.χ. οι f(x) = x-1/2 ,  g(x) = -x + ½,  h(x) = e^x – (e-1) μας κάνουν. Ορίζουμε τώρα την ακολουθία
[color=white]  ........     [/color]     f, g, h, f, g, h, f, g, h, ....
Είναι σαφές ότι η ακολουθία αυτή απαντά στο πρόβλημα.[/quote]

Well said! For sure, αυτή η ακολουθία πράγματι απαντά στο πρόβλημα όπως διατυπώθηκε :)

Ωστόσο, η ενδιαφέρουσα διατύπωση είναι η ακολουθία $ f_n$ να είναι θετική (ή, ισοδύναμα, να ζητείται το όριο του ολ/τος των [i]απολύτων[/i] να είναι 0). Αν ζητήσεις κάτι τέτοιο (που, πράγματι, το αρχικό πρόβλημα δυστυχώς δεν το ζητούσε...), ένα απλό relabelling όπως αυτό που προτείνεις δε θα δούλευε και θα έπρεπε να βρεις κάτι πιο σύνθετο.

Αλλιώς, όπως λες κι εσύ, το πρόβλημα θα ήταν εντελώς τετριμμένο και δε θα αναδείκνυε κάποιο μαθηματικό φαινόμενο :P Αντίθετα, reading between the lines, το πρόβλημα σε οδηγεί σε μία ακολουθία συναρτήσεων με $ f_n\to 0$ στον $ L^1[0,1]$ (i.e. $ \int_0^1|f_n|\to 0$), χωρίς η $ f_n$ να συγκλίνει κατά σημείο (και μάλιστα πουθενά) :)

Πάντως, ναι, πολύ σωστή η παρατήρηση.

Cheerio,

Durandal 1707

[quote="Durandal"]

Ωστόσο, η ενδιαφέρουσα διατύπωση είναι η ακολουθία $ f_n$ να είναι θετική (ή, ισοδύναμα, να ζητείται το όριο του ολ/τος των [i]απολύτων[/i] να είναι 0). Αν ζητήσεις κάτι τέτοιο (που, πράγματι, το αρχικό πρόβλημα δυστυχώς δεν το ζητούσε...), ένα απλό relabelling όπως αυτό που προτείνεις δε θα δούλευε και θα έπρεπε να βρεις κάτι πιο σύνθετο.

Αλλιώς, όπως λες κι εσύ, το πρόβλημα θα ήταν εντελώς τετριμμένο και δε θα αναδείκνυε κάποιο μαθηματικό φαινόμενο :P Αντίθετα, reading between the lines, το πρόβλημα σε οδηγεί σε μία ακολουθία συναρτήσεων με $ f_n\to 0$ στον $ L^1[0,1]$ (i.e. $ \int_0^1|f_n|\to 0$), χωρίς η $ f_n$ να συγκλίνει κατά σημείο (και μάλιστα πουθενά) :)

[/quote]

Έχεις απόλυτο δικιο. Το μαθηματικά ενδιαφέρον ερώτημα είναι να πάρουμε τις συναρτήσεις θετικές, οπότε τα παραδείγματα είναι πιό λεπτά. Αυτά που δίνεις είναι ευφυέστατα.

Προσοχή όμως σε μία λεπτομέρεια στο παράδειγμα 3 που έδωσες. Έγραψες:
[i]Έστω φ η συνάρτηση Dirichlet, που στους ρητούς είναι 1 και στους άρρητους  0. Βάλε τότε f_n = φ για τα άρτια n και  f_n = 1-φ για τα περιττά.[/i]

Εδώ έχουμε, ουσιαστικά, δύο κατάλληλα επιλεγμένες συναρτήσεις f, g της αρεσκείας μας και η ακολουθία που μελετάμε είναι η f, g, f, g, f,  ... (στο δικό μου παράδειγμα ήσαν τρεις οι περιφερώμενες συναρτήσεις).

Ως εδώ όλα καλά, μόνο που η επιλογή της g δεν είναι καλή, γιατί έχει ολοκλήρωμα 1 (θέλουμε 0). Απλά οι f, g πρέπει να αντικατασταθούν με άλλες.

Ευχαριστώ πάντως για τα post σου. Είναι εμβριθέστατα και ευφυέστατα.

Τ.Π.

[quote="Papadopoulos"]Προσοχή όμως σε μία λεπτομέρεια στο παράδειγμα 3 που έδωσες. Έγραψες:
[i]Έστω φ η συνάρτηση Dirichlet, που στους ρητούς είναι 1 και στους άρρητους  0. Βάλε τότε f_n = φ για τα άρτια n και  f_n = 1-φ για τα περιττά.[/i]

Εδώ έχουμε, ουσιαστικά, δύο κατάλληλα επιλεγμένες συναρτήσεις f, g της αρεσκείας μας και η ακολουθία που μελετάμε είναι η f, g, f, g, f,  ... (στο δικό μου παράδειγμα ήσαν τρεις οι περιφερώμενες συναρτήσεις).

Ως εδώ όλα καλά, μόνο που η επιλογή της g δεν είναι καλή, γιατί έχει ολοκλήρωμα 1 (θέλουμε 0). Απλά οι f, g πρέπει να αντικατασταθούν με άλλες.[/quote]

Yiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiikes!!!!!!!!!!!! :wallbash:

Απροσεξία και βιασύνη... κάποιος να μου δώσει το βραβείο "Δεν Προσέχω τι Γράφω - 2008" :(

Sorry, έχεις απόλυτο δίκιο (έχετε? την πάτησα ήδη μια φορά σήμερα με τους ενικούς  :oops: ). Αυτό που ήθελα να κάνω είναι να μην έχω [i]έναν[/i] παλμό που να περιφέρεται εργοδικά στο [0,1] αλλά, eventually, μια απειρία παλμών που να αλλάζουν θέση κρατώντας αυθαίρετα μικρό ολοκλήρωμα, αυτό ήθελα να αναδείξω. Προφανώς, κάτι τόσο cheap όσο η πρώτη προσπάθεια δεν μπορεί να δουλέψει... για να εξιλεωθώ (και να μη βάζω πάλι κάτι πάνω στους ρητούς ή στο Cantor ;)), κάποιο σήμα της μορφής $ f_n(x) \equal{} \sin_ \plus{} (n(x \plus{} \theta))$ όπου $ 0 < \theta < \pi/2$ και $ \sin_ \plus{}$ είναι το θετικό μέρος του $ \sin$ ($ \sin_ \plus{} \equal{} \max\{\sin, 0\}$) πρέπει να δουλεύει (αν και δεν έχω χρόνο τώρα να δω αν το εμβαδό των λοβών μειώνεται αρκετά γρήγορα). Αν δε δουλεύει ούτε αυτό, apologies, θα το ξανακοιτάξω μόλις βρω λίγο χρόνο...

Cheers,

Pan

[quote="Durandal"]κάποιο σήμα της μορφής $ f_n(x) \equal{} \sin_ \plus{} (n(x \plus{} \theta))$ όπου $ 0 < \theta < \pi/2$ και $ \sin_ \plus{}$ είναι το θετικό μέρος του $ \sin$ ($ \sin_ \plus{} \equal{} \max\{\sin, 0\}$) πρέπει να δουλεύει (αν και δεν έχω χρόνο τώρα να δω αν το εμβαδό των λοβών μειώνεται αρκετά γρήγορα). Αν δε δουλεύει ούτε αυτό, apologies, θα το ξανακοιτάξω μόλις βρω λίγο χρόνο...
[/quote]

Με πρώτη ματιά και εμένα μου φαίνεται ότι θα δουλέψει. Αν δεν δουλέψει αυτό, τότε το
 $ f_n(x) \equal{} \sin_ \plus{} (\bold {n^2}(x \plus{} \theta))$ είναι σίγουρα εντάξει.

Τ.


Prove that the only integers x which make x^4+x^3+x^2+x+1 a perfect square are x = -1,0 and 3.


[quote="quykhtn-qa1"]Let $ a,b,c$ be none-gative numbers ,no two of which are zero.Prove that:
$ \frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \plus{} \frac {(2b \plus{} c)^2}{2a^2 \plus{} bc} \plus{} \frac {(2c \plus{} a)^2}{2b^2 \plus{} ca} \ge 9$[/quote]
An expanding gives
$ \sum_{cyc}(2a^5b\plus{}8a^5c\plus{}16a^4b^2\plus{}4a^4c^2\minus{}20a^3b^3\minus{}2a^4bc\plus{}3a^3b^2c\plus{}12a^3c^2b\minus{}23a^2b^2c^2)\geq0,$
which is obvious.

[quote="arqady"][quote="quykhtn-qa1"]Let $ a,b,c$ be none-gative numbers ,no two of which are zero.Prove that:
$ \frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \plus{} \frac {(2b \plus{} c)^2}{2a^2 \plus{} bc} \plus{} \frac {(2c \plus{} a)^2}{2b^2 \plus{} ca} \ge 9$[/quote]
An expanding gives
$ \sum_{cyc}(2a^5b \plus{} 8a^5c \plus{} 16a^4b^2 \plus{} 4a^4c^2 \minus{} 20a^3b^3 \minus{} 2a^4bc \plus{} 3a^3b^2c \plus{} 12a^3c^2b \minus{} 23a^2b^2c^2)\geq0,$
which is obvious.[/quote]
We can solve it by Cauchy-Schwarz inequality,it very nice :)

[quote="matex (L)(L)(L)"]Could You show us how to apply CS to kill this inequality?[/quote]
By CS we obtain:
$ \sum_{cyc}\frac{(2a\plus{}b)^2}{2c^2\plus{}ab}\equal{}\sum_{cyc}\frac{(2a\plus{}b)^4}{(2a\plus{}b)^2(2c^2\plus{}ab)}\geq\frac{\left(\sum(2a\plus{}b)^2\right)^2}{\sum(2a\plus{}b)^2(2c^2\plus{}ab)}.$
But $ \left(\sum(2a\plus{}b)^2\right)^2\minus{}9\sum(2a\plus{}b)^2(2c^2\plus{}ab)\equal{}\sum_{cyc}(25a^4\plus{}4a^3b\plus{}31ab^3\minus{}60a^2b^2)\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)a(25a^2\plus{}29ab\minus{}31b^2)\equal{}\sum_{cyc}\left((a\minus{}b)a(25a^2\plus{}29ab\minus{}31b^2)\minus{}\frac{23(a^4\minus{}b^4)}{4}\right)\equal{}$
$ \equal{}\frac{1}{4}\sum_{cyc}(a\minus{}b)^2(77a^2\plus{}170ab\plus{}23b^2)\geq0.$

arqady why You transform this in such terrible way:

$ \equal{} \sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \equal{} \sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right) \equal{}$ when $ \left(\sum(2a \plus{} b)^2\right)^2 \minus{} 9\sum(2a \plus{} b)^2(2c^2 \plus{} ab) \equal{} \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2) >\equal{}0$ follows directly from AM=GM?

[quote="arqady"][quote="matex (L)(L)(L)"]arqady why You transform this in such terrible way[/quote]
Transform this better.  :)[/quote]

how do you transform this at all :( I can never produce these magic factorizations, and you seem to solve 80% of your inequalities that way :O (also, how would you factor things like this under square roots?)

[quote="matex (L)(L)(L)"]arqady why You transform this in such terrible way:

$ \equal{} \sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \equal{} \sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right) \equal{}$ when $ \left(\sum(2a \plus{} b)^2\right)^2 \minus{} 9\sum(2a \plus{} b)^2(2c^2 \plus{} ab) \equal{} \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2) > \equal{} 0$ follows directly from AM=GM?[/quote]
Why does $ \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2) \ge 0$ follow by AM-GM?

Mathias, It follows from Cauchy inequality and Muiheard theorem. :)

[quote="not_trig"]
how do you transform this at all [/quote]
Make concrete, what do you mean? In my proof
[quote="arqady"]
By CS we obtain:
$ \sum_{cyc}\frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \equal{} \sum_{cyc}\frac {(2a \plus{} b)^4}{(2a \plus{} b)^2(2c^2 \plus{} ab)}\geq\frac {\left(\sum(2a \plus{} b)^2\right)^2}{\sum(2a \plus{} b)^2(2c^2 \plus{} ab)}.$
But $ \left(\sum(2a \plus{} b)^2\right)^2 \minus{} 9\sum(2a \plus{} b)^2(2c^2 \plus{} ab) \equal{} \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2) \equal{}$
$ \equal{} \sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \equal{} \sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right) \equal{}$
$ \equal{} \frac {1}{4}\sum_{cyc}(a \minus{} b)^2(77a^2 \plus{} 170ab \plus{} 23b^2)\geq0.$[/quote]
there is five steps:
1) $ \sum_{cyc}\frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \equal{} \sum_{cyc}\frac {(2a \plus{} b)^4}{(2a \plus{} b)^2(2c^2 \plus{} ab)}.$
2) $ \left(\sum(2a \plus{} b)^2\right)^2 \minus{} 9\sum(2a \plus{} b)^2(2c^2 \plus{} ab) \equal{} \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2).$
3) $ \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2)\equal{}\sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2).$
4) $ \sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2)\equal{}\sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right).$
5) $ \sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right)\equal{}$
$ \equal{}\frac {1}{4}\sum_{cyc}(a \minus{} b)^2(77a^2 \plus{} 170ab \plus{} 23b^2).$
About which step you want to ask? Ask me.

[quote="rofler"]I have a question Arqady.

How do you know to magically multiply the top and bottom of each fraction by a certain power for cauchy schwartz in step 1? Was there motivation behind it, or trial and error, because I see many people who destroy inequalities with cauchy schwartz do this with almost no apparent reason.

Thx in advance, Rofler[/quote]
Thank you for your question!
I have write: $ \sum_{cyc}\frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \equal{} \sum_{cyc}\frac {(2a \plus{} b)^2(ka\plus{}b)^2}{(ka \plus{} b)^2(2c^2 \plus{} ab)}\geq\frac{\left(\sum(2a\plus{}b)(ka\plus{}b)\right)^2}{\sum(ka \plus{} b)^2(2c^2 \plus{} ab)}.$
Now we need take a value of $ k$ such that  $ \frac{\left(\sum(2a\plus{}b)(ka\plus{}b)\right)^2}{\sum(ka \plus{} b)^2(2c^2 \plus{} ab)}\geq9$ is true.
You can ask me why you didn't take $ (ka\plus{}b\plus{}lc)^2$ $ ?$
Because we don't want to get $ a^4$ after expanding in the right part of our inequality.
Hence, we assumed $ l\equal{}0.$
Now we need test the last inequality when $ a\rightarrow\plus{}\infty$ and $ c\equal{}0.$
With $ a\rightarrow\plus{}\infty$ it is all right because $ l\equal{}0,$
while an assuming $ c\equal{}0$ and for example $ a\equal{}b\equal{}1$ gives $ \frac{11\minus{}3\sqrt{11}}{2}\leq k\leq\frac{11\plus{}3\sqrt{11}}{2}.$
Let $ k\equal{}2.$ ( because it's nicer  :D )
Now we need prove that
$ \frac{\left(\sum(2a\plus{}b)^2\right)^2}{\sum(2a\plus{}b)^2(2c^2\plus{}ab)}\geq9$ and indeed it is true. :)

[quote="arqady"][quote="matex (L)(L)(L)"]Could You show us how to apply CS to kill this inequality?[/quote]
By CS we obtain:
$ \sum_{cyc}\frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \equal{} \sum_{cyc}\frac {(2a \plus{} b)^4}{(2a \plus{} b)^2(2c^2 \plus{} ab)}\geq\frac {\left(\sum(2a \plus{} b)^2\right)^2}{\sum(2a \plus{} b)^2(2c^2 \plus{} ab)}.$
But $ \left(\sum(2a \plus{} b)^2\right)^2 \minus{} 9\sum(2a \plus{} b)^2(2c^2 \plus{} ab) \equal{} \sum_{cyc}(25a^4 \plus{} 4a^3b \plus{} 31ab^3 \minus{} 60a^2b^2) \equal{}$
$ \equal{} \sum_{cyc}(a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \equal{} \sum_{cyc}\left((a \minus{} b)a(25a^2 \plus{} 29ab \minus{} 31b^2) \minus{} \frac {23(a^4 \minus{} b^4)}{4}\right) \equal{}$
$ \equal{} \frac {1}{4}\sum_{cyc}(a \minus{} b)^2(77a^2 \plus{} 170ab \plus{} 23b^2)\geq0.$[/quote]
Nice solution,arqady :) .Thanks :lol:

[quote="enndb0x"][quote="arqady"][quote="matex (L)(L)(L)"]Could You show us how to apply CS to kill this inequality?[/quote]
By CS we obtain:
$ \sum_{cyc}\frac {(2a \plus{} b)^2}{2c^2 \plus{} ab} \equal{} \sum_{cyc}\frac {(2a \plus{} b)^4}{(2a \plus{} b)^2(2c^2 \plus{} ab)}\geq\frac {\left(\sum(2a \plus{} b)^2\right)^2}{\sum(2a \plus{} b)^2(2c^2 \plus{} ab)}.$
[/quote]
Hi arqady:
Can you help me with this step please ? I don't know how you won  $ \sum_{cyc}\frac {(2a \plus{} b)^4}{(2a \plus{} b)^2(2c^2 \plus{} ab)}\geq\frac {\left(\sum(2a \plus{} b)^2\right)^2}{\sum(2a \plus{} b)^2(2c^2 \plus{} ab)}.$

It seems all these Sigmas are confusing some of us  :blush:[/quote]
Because $ \frac{a^2}{x}\plus{}\frac{b^2}{y}\plus{}\frac{c^2}{z} \ge \frac{(a\plus{}b\plus{}c)^2}{x\plus{}y\plus{}z}$ with $ x,y,z>0$ :)


we define $ ord_{p}(k)\equal{}max\left\{j : p^{j}|k\right\}$

Now prove that

$ \sum^{n}_{i\equal{}1} ord_{p}(i)\equal{}\sum^{\infty}_{l\equal{}1}\left\lfloor \frac{n}{p^{l}}\right\rfloor$

$ ord_p(kp)\equal{}ord_p(k)\plus{}1$
Now,
$ \sum^n_i ord_p(i) \equal{} \sum_{k\equal{}1}^{[n/p]}ord_p(kp)$
recursively
$ \equal{} [n/p] \plus{} \sum_{j\equal{}1}^{[[n/p]/p]} ord_p(jp)$
$ \equal{} [n/p] \plus{} \sum_{j\equal{}1}^{[n/p^2]}ord_p(jp)$
$ \equal{} [n/p] \plus{} [n/p^2] \plus{} \sum_{k\equal{}1}^{[n/p^3]}ord_p(kp)
   \equal{} [n/p] \plus{} [n/p^2] \plus{} [n/p^3] \plus{} ...$


Given an integer $n$, $n\geq 2$, find the largest real number $k_{n}$ such that for any positive numbers $a_{1},a_{2},...,a_{n}$ satisfying $a_{1}a_{2}\cdots a_{n}=1$ one has $a_{1}^{n-1}+a_{2}^{n-1}+\cdots+a_{n}^{n-1}-n\geq k_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n\right)$

I think $k_{n}=n-1$.
This is just Generalization of Popoviciu's Inequality.
See Algebraic Inequality, problem 1 from page 199 and problem 17 from page 242. :lol:


modular arithmetic
number theory unsolved
number theory

[quote="Spribo"]Find the smallest positive integer such that, on moving its last digit to the front, it becomes nine times larger.[/quote]

The required number may be written $ 10x\plus{}a$ where $ 10^{p\minus{}1}\le x<10^p$ has $ p>0$ decimal digits and $ a\in\{0,1,2,3,4,5,7,8,9\}$ and the equation is then :

$ 10^pa\plus{}x\equal{}9(10x\plus{}a)$ and so $ x\equal{}a\frac{10^p\minus{}9}{89}$

The constraint $ 10^{p\minus{}1}\le x<10^p$ becomes $ 89\cdot 10^{p\minus{}1}\le a(10^p\minus{}9)<89\cdot 10^p$ $ \iff$ $ 8.9\le a(1\minus{}\frac 9{10p})<89$ and so $ a\equal{}9$ and $ x\equal{}9\frac{10^p\minus{}9}{89}$

So we need $ 10^p\equal{}9\pmod{89}$ and the least $ p$ is $ 43$ so the required number is $ \boxed{90\frac{10^{43}\minus{}9}{89}\plus{}9}$

Thank you very much, but how did you find that the least $ p$ is 43?


function
induction
algebra solved
algebra

Let $\mathbb{N}^* = \{0,\;1,\;2,\;...\}.$

Give as much information as possible about the function defined by

\[ f(0) = f(1) = 0,\]
\[ f(n) = \max_{1 \leq k \leq n - 1} (f(k) + f(n - k) + \min(k, n - k)),\; n \geq 2, n \in \mathbb{N}^*.\]

This problem is problem 4 in Taiwan olympiad 2000.
i think it is a very hard problem: (this is the sol in 'math olympiads - problems and solutions from around the world - 2000,2001)
for each n consider the binary representation of n. consider the strings u get when u remove some digits from the left, so that the remaining stuff starts with 1. call the sum of the decimal values of these g(n); also for each 1 that appears in binary represention of n, if it reperesents 2^k write it as k*2^(k-1) and sum these; call it h(n). Now it can be proven by induction that f(n) = g(n) + h(n) (but thats not easy and very long, considering 2 cases).
there is like no motivation for the starting part  :D  :D how the hell would you think of doing something crazy like that? btw, is your solution simpler arne?

it's always a good starting point for these tasks to write your numbers in binary representation, because the function depends in most cases on the binary representation. now having written down all f(n) upto n=2^10 you may guess what the function is like :D
but i won't have guessed that weird function anyway...

After some guesses we can try to prove that if we define $f_0: \mathbb{N}^* \rightarrow \mathbb{N}^*$ by \[ f_0(0) = f_0(1) = 0,\;f_0(2m) = 2f_0(m) + m,\;f_0(2m + 1) = f_0(m) + f_0(m + 1) + m\] then for all n: $f(n) = f_0(n)$.
The proof is not trivial. Try it! :)
Later on we define $g(n) = f(n + 1) - f(n)$. Then the link with binary representations becomes clear.

This function $f_0$ is what I found until I gave up because proving $f=f_0$ seemed to be very messy...

Pierre.

I'll give an outline of the proof.

1) Use induction on  $k$ to prove that if $0 \leq m \leq n \leq 2^k$ then \[ f_0(n + m) - f_0(n) - f_0(m) - m \geq 0\] so that for $1 \leq k \leq n - 1$ we have \[ f_0(k) + f_0(n - k) + \min(k,\;n - k) \leq f_0(k + n - k) = f_0(n).\]
2) We also have \[ \max_{1 \leq k \leq n - 1} \{f_0(k) + f_0(n - k) + \min(k,\;n - k)\} \] \[ \geq f_0([n/2]) + f_0(n - [n/2]) + \min([n/2],\;n - [n/2])\] and the right hand side of this expression equals $f_0(n)$ (check the cases $n = 2m$ and $n = 2m + 1$ separately).

So $f(n) = f_0(n)$.


I wrote this program in C,

#include<math.h>
#include<stdio.h>

main()
{
float a,b,c,disc,p,q,x1,x2;
a=getchar();b=getchar();c=getchar();
disc=b*b-4*a*c;
p=-b/(2*a);
q=sqrt(disc)/(2*a);
printf("x1=%f\n x2=%f\n",x1=p+q,x2=p-q);

}

input 1,2,3
it run,and output:
sqrt: DEMAIN erorr
x1=-0.448980
x2=-0.448980
why? what's wrong in my program?

The error message should give it away.. theres a problem with the square root. Namely, you're trying to take the square root of a negative number (2*2 - 4*1*3 = -8). You can't do that. Not all quadratics have real solutions, remember.

Yes ,that is one error.And you can input some other words and have a try .There are some other error.

The problem is in the command getchar()

Try another function for input!

The function getchar() returns the character as a number. For example, if you give $A$ it reads $65$

I didn't know this function, but I put a command

[code]printf("a=%f b=%f c=%f",a,b,c);[/code] after the getchar() commands

When I gave 123 and pressed enter, I saw

[code]a=49.000000 b=50.000000 c=51.000000[/code]

I couldn't give a negative number! When I gave 12-3 I saw

[code]a=49.000000 b=50.000000 c=45.000000[/code]

(The ASCII represantation of 1,2,3 and "-" is 49,50,51 and 45 respectively)

:) My friend amend the program like this.I don't know the particular meaning.

 #include<math.h>
#include<stdio.h>
#include<stdlib.h>

main()
{
float disc,p,q,x1,x2;
char x[10],y[10],z[10];
int a,b,c;
char* str1;
char* str2;
char* str3;
gets(x);
gets(y);
gets(z);
str1=x;str2=y;str3=z;

a=atoi(str1);
b=atoi(str2);
c=atoi(str3);

disc=b*b-4*a*c;
p=-b/(2*a);
q=sqrt(disc)/(2*a);
x1=p+q;
x2=p-q;
printf("x1=%f\n x2=%f\n",x1,x2);
}

:?: Otherwise , how could you input the words in the "Code"?

I cannot understand you

What do you mean you want to input the words in the code?

Which words?

I suppose that you mean, you want a static program which don't ask from user to give the values of a,b,c

for example, there is a command line

a=1; b=2; c=1;

I'm not sure what you mean by the "code" thing but I see what you're doing and how your friend changed the program... you'd be much better off using the [b]scanf[/b] command, it's very similar to [b]printf[/b] but instead of printing the output to screen it reads the user input.

[quote="Liu Ce"]:blush:  Sorry,my English is very pool.You know Chinese is complete different from English.So our learning English is very hard.
 :) I mean that how could you input the expression in the panel above.
Can you get the meaning?[/quote]

我想是語言誤會，所以外國人不懂你的問題而回答一些你不如期的答案。
I think this is the communication problem, so the foreigner does not understand your questions and reply you an unexpected answer.

不如你去翻閱一些C語言的參考書，尤其是基本輸入/輸出的部分，理清楚 getChar()和 scanf(…) 的分別就可以搞懂了。
I suggest you read some C language reference texts, especially the basic Input/Output, in order to understand the different between getChar() and scanf() is fine.


Find $ x,y,z\geq 0$ such that
$ 3x\plus{}2y\plus{}z\equal{}26$ and $ xyz\equal{}2x\plus{}5y\plus{}10z$
 :maybe:

[quote="kunny"]Are $ x,\ y,\ z$ integers?[/quote]
No they are real numbers. Use AM-GM?


group theory
abstract algebra
superior algebra
superior algebra unsolved

How do I prove that a noncyclic group of order $ 21$ has one Sylow 3-subgroup?  I know that by the Third Sylow Theorem, there are either $ 1$ or $ 7$ such groups but I can't seem to rule out the possibility of $ 7$.

Don't prove it, because it's false. The unique (up to isomorphism) nonabelian group of order $ 21$ has $ 6$ elements of order $ 7$ and $ 14$ of order $ 3$.

This group can be realized as a semidirect product, or as a subgroup of a permutation group. For the second version, the elements $ (1234567)$ and $ (235)(476)$ in $ S_7$ generate it.


analytic geometry
geometry
circumcircle
conics
hyperbola
Asymptote
geometry proposed

Let $ ABC$ and $ X$ are triangle and a fixed point, $ AX\cap BC\equal{}D,BX\cap AC\equal{}E$, $ P \in DE$.
$ Q,Y$ is isogonal conjugate point of $ P,X$. $ AQ \cap BC\equal{}D',BQ \cap AC\equal{}E'$.
Prove that $ D',Y,E'$ are conlinear.

*I was prove this problem using barycentric coordinate when I saw it first. But I want see more beutiful solution.

Let $ \mathcal H$ be the circumhyprebola of the $ \triangle ABC,$ which is the isogonal conjugate of the line $ DE.$ $ P \in DE$ $ \Longleftrightarrow$ $ Q \in \mathcal H.$ When $ P \longrightarrow D,$ then $ Q \longrightarrow A$ and when $ P \longrightarrow E,$ then $ Q \longrightarrow B.$ By continuity principle, tangents to $ \mathcal H$ at $ A, B$ are isogonals of $ AD, BE$ WRT the angles $ \angle A, \angle B,$ respectively. They intersect at the isogonal conjugate $ Y$ of $ X.$ By Pascal theorem for the degenerated hexagon $ AAQBBC$ inscribed in $ \mathcal H,$ the intersections $ D' \equiv AQ \cap BC, E' \equiv BQ \cap CA, Y \equiv AA \cap BB$ are collinear.

BTW, what's the circumhyperbola of a triangle :?:  :maybe:

Isogonal conjugate of a line $ d$ intersecting circumcircle $ (O)$ of a $ \triangle ABC$ at $ P, Q$  is a hyperbola $ \mathcal H$ passing through $ A, B, C$ - a triangle circumhyperbola. (Every line intersecting the circumcircle has a different isogonal circumhyperbola.) It intersects the triangle circumcircle again at the pole $ R$ of Simson line perpendicular to the line $ d.$ Its asymptotes are perpendicular to Simson lines with poles $ P, Q.$ See of [url]http://www.mathlinks.ro/viewtopic.php?t=129699[/url] and click on "Show details."

BTW, the original problem does not say that $ DE$ intersects the circumcircle. If not, the isogonal conjugate of the line $ DE$ is a triangle circumellipse. The rest of the proof is the same, for Pascal theorem holds for any conic.

I'm sorry, but anything I know about isogonal conjugates I picked up on Mathlinks; I did not read any books about it.

[quote="NextPeace"]Thank you very much! your proof is very nice, yetti! :lol:

BTW, I really want to studying about isogonal conjugate of lines. Is there anything to study about it?
If there is, can you post(or post the link) please?[/quote]
Probably these will help you: :)
1. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472[/url]
2. [url]http://www.cip.ifi.lmu.de/~grinberg/Isogonal.zip[/url]
3. [url]http://www.cut-the-knot.org/Generalization/ceva.shtml#iso[/url]


calculus
AMC
AIME
MIT
college
linear algebra

How many are you taking this year?

I have 4: Calc BC, Stats, History, and Physics B.

do you have to take the AP test if you take the AP course? (stupid question, but i really don't know and AP courses are a year or 2 away from me)

That's really a question you should ask someone at your school.  But if your AP class is legit and you're doing well in it I see no reason to not take the AP exam.  It would be foolish not to, imo.

I am taking World History and Chemistry.

My school is wierd in that NJ requires two years of US history (a bit stupid, I think), so World History is a junior course.

I took 3 last year as a sophomore: Calculus BC, Macro and Micro Economics.

I know someone who took 6 APs as a sophomore! That was insane. I do not recommend overloading, because it is tough to do well in all. He got 3 $4$s 2 $5$s and a $1$.

Next year is tough though: Physics, English, Euro History, Spanish Language, Gov. and Pol. I do not know if I should take all those AP tests as a senior, because there is really no point if I do not want to place out, and college apps are done before the test.

[quote="mcalderbank"]I know someone who took 6 APs as a sophomore! That was insane. I do not recommend overloading, because it is tough to do well in all. He got 3 $4$s 2 $5$s and a $1$.[/quote]

5 as a sophomore here (Econ counts twice)...
Going through a nervous breakdown / major brain leakage now that they're here, though. AHHHHHHH!

I am taking a mere 2.  But I am a freshman.  I hope to do a lot sophomore year to keep up.  Econ sounds good, I'll be doing calc BC, I think comp. programming too, and chemistry.  Maybe biology.  Any other suggestions for stuff that's easy to study on your own for a mathy person like me?  I enjoy history too, but I don't know how good I am at that.

4 as a soph... 1 self-study, but thats stats. i'm also taking physics B, calc BC, and world history (ugh, im going to fail this one).

Computer Science AB
World History

Bah...

Taking 0 this year...but as a freshman the only AP our school offers is Computer Science.  Next year I will be taking two, AP World History, and AP Comp Sci.

Physics B
Java A
Macro Econ
Micro Econ
Biology

[quote="stupidkid"]World History
only AP i can take as freshmen
darn test is on tuesday maybe i should start studying instead of posting in AoPS[/quote]

 :? Or not... :D

[quote="mcalderbank"]
I know someone who took 6 APs as a sophomore! That was insane. I do not recommend overloading, because it is tough to do well in all. He got 3 $4$s 2 $5$s and a $1$.
[/quote]

What a turkey.

I have English language, US and European history, Physics B, Physics C(both tests), and Chem.  I think I can use up all of my student pack labels :).

BC Calculus
AP English Lit
AP Chemistry

i feel so stupid as a senior with 3 APs while in this thread  :(

Hmm. Well, I took Music Theory this year as a afterschool, noncredit course, just for fun really and to learn more about the subject itself.

i am taking none this year (10th grade) but a couple next year.  i sympathize with all of you who go to schools with lots of requirements about AP classes.  I was not allowed o take BC calc this year after averaging 100 last year in Alg 2.  So, I sat through a year of pre-calc and learned little or no new material.  since i go to a public school, AP's don't have that much importance to administrators in the school.  it is incredibly diffcult to take even 8 AP's at my school because of the many requirements we have to meet before graduating.  we have useless classes like computer apps, where you learn how to type and use microsoft word for a semester. :rotfl:  and also, alot of AP classes in our school are only offered every other year, or conflict with eachother.  I wanted to take AP Chem and physics next year, but AP chem wasn't offered because of lack of demand, and AP physics was the same time as BC calc(possibly the most ridiculous scheduling problem i have heard of).  oh well, maybe senior year.

[quote="mkkool64"]we have useless classes like computer apps, where you learn how to type and use microsoft word for a semester. :rotfl: [/quote]

i had to take that stupid class cause my stupid school wouldn't let me take alg. 2 freshman year.

if you're a senior, could you tell me how many AP classes you are taking? many of the school counselors say if you work hard during first 3 years of HS, then you can slack off a bit senior year, but i also heard that its important to keep up your grades senior year for college. could someone answer my  :?:

for the record, you also learn how to use PowerPoint, Access, and Excel in comp. apps.:D

[quote="plokoon51"]if you're a senior, could you tell me how many AP classes you are taking?[/quote]
I'm a senior and took Physics, Bio(self-studied), Comp Sci, and both Econs.
There is [u]no doubt[/u] that you should take the maximum number of AP classes for the following reasons.

1. $It$ $shows$ $your $ $characteristics$ - you like to learn? you have intellectual curiosity? This doesnt make any sense to the colleges unless you show them. I have heard that taking the most challenging curriculum in your situation is the main admission factor, not 10 points higher on your SAT.
2. $you$ $can$ $save$ $a$ $lot$ $of$ $time$ $/$ $money$ - college classes are expensive! but more importantly, you can pass out baby classes or stupid requirements so that you can take whatever you want to take. This is one of the reason why I took Bio exam self studied after I got into MIT, in order to avoid notorious biology so that I can take more math.
3. $learning$ $for$ $the$ $sake$ $of$ $learning$ - I had alot of fun with java and econ which I never thought was going to be fun. I'm a math person, but taking those classes widen my view definitely. In my opinion, I think this is the most important reason.

What they mean by "slacking off a bit" is that you dont need to stay up late for doing homework and fighting for one point on your grade. They dont mean that you can take minimum classes that are easy and required to graduate.

I know too many smart seniors who took 3-4 classes,(school ends before 12 for them) and [i]almost[/i] all of them failed to go to the colleges that they wanted to go.

[quote="bleumoose"]Computer apps. is a required course in most states. Can't you test out? I tested out. The basic idea behind the class is to get kids who actually have very little access to computers (like in the Black Belt in AL) to HAVE to take a computer course. Which means that the school HAS to provide computers, etc, to the students.[/quote]

No, i couldn't test out although i did try to.  I personally think if they are going to have a required class about computers, it needs to be a lot earlier than high school.  It was scary-alot of kids in my class typed staring at the keyboard, averaging less than 30 wpm., and said that was what they always did.  By high school a lot of kids have developed these habits already (we don't suddenly dicover computers in 9th grade) and have created our own ways of doing things.  They need to teach kids in like 1st or second grade, and then reinforce it, so teachers are not trying to undo 9 years of previous practice in a semester.

I remember learning to type in 3rd grade.  Of course, the public elementary school I went to was one of the better ones in my area.

they taught typing way back in my elementary school, but it was barely enough for kids to practice using it.  although my high school doesnt have a silly computer requirement like that, many many many kids do type the way you're referring to with 2 fingers staring at the keyboard.  it's quite pathetic.

I remember our town's 6th grade school would do a program called "Type to Learn" to practice typing. It was fairly annoying, but useful. It showed what finger to use for each key and stuff like that. Pretty effective.

"Type to Learn" was the exact same program that our elementary school used to teach kids how to type, although it wasn't very effective for most of them

We never really had a typing program although we had to go along with a song.. it's so hilarious

"Press A, Reach for B etc."

Wow, I feel unaccomplished.  :?

I'm a sophomore.  This year, I took (just  ;) ) one AP, Calc AB.  My school has so many requirements for even getting to an AP course-several prerequisites for each course, department approval, etc- that, in general, only seniors take them.  One thing that really annoys me about my school is that the science department is set up such that there is no such thing as AP Physics--freshman year is basic chemistry and physics (an excellent place to catch up on sleep), sophomore is basic biology, junior is chemistry, senior is either AP Chem or physics.  Also, I can't take all the classes that I want to because my schedule is filled with useless classes the school requires--Religions of North America (we spent over a month on Hawaii, and two [i]days[/i] on Christianity.  Uh huh.), Computer Life Skills (learning to use Word and Access, and the Internet, and Age of Empires [the teacher is obsessed]), gym/health .  Aaaah!

Even so, two of my friends and I were the first sophomores ever at our school to take any AP test.  This really surprised me, as the school offers AP Computer Science--why hasn't anyone taken that as a sophomore yet?  I wonder.

[EDIT]  I thought that the AP Calc test was really rather easy--I checked every problem a minimum of 3 times before getting bored and going to sleep (almost) [/EDIT]

[quote="probability1.01"]Hmmm... I didn't have time to check all my problems 3 times.  But I think playing calculator games might have accounted for that...[/quote]

I checked them once... but I didn't have games on my calculator, so I basically slept for an hour on each section.

Earlier, Mr.Rusczyk mentioned that how complaining about "low" score can hurt someone.

Bragging about how much time you have left after Calculus exam would not help you to gain a single extra point, nor the others.

I understand you didnt mean to, but it looks rude to people who might have struggled through the test.


A tennis racquet is elliptical in shape. The formula for the area of an ellipse is $ A \equal{} ab\pi$, where $ a$ is half the width and $ b$ is half the length. If the racquet shown has a hitting surface of $ 35$ square inches, find the width of the hitting surface.
[asy]size(200);
defaultpen(linewidth(0.8));
draw(ellipse((0,0),14,8));
draw((-22,1)--(-12.0,4));
draw((-22,-1)--(-12.0,-4));
draw((-22,1)--(-36,1)--(-36,-1)--(-22,-1));
draw("$14''$",(-13,0)--(13,0),N,Arrows);
draw("width",(16,8)--(16,-8),E,Arrows);[/asy]

Let the width be x.  then 7x/2pi=35, 7xpi=70, xpi=10, so x=10/pi.

Hmmm, on FTW, it says the surface area is $35\pi$ square inches.  I guess they changed it.  So the answer is $\boxed{10}$ instead of $\frac{10}{\pi}$.


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okay, i'm interested in seeing how religious people in AoPS are
these options were taken from 'the god delusion'

i'm personally a #6, leaning towards #7.

[quote="jhredsox"]How the heck is there a 'probability' of believing in God?[/quote]
no, its asking what you think the probability that god exists is

[quote]'How much do you believe in God?'[/quote]

where do you see 'that he exists'?

I might point out that it appears to be logically impossible to prove that God doesn't exist, but it could theoretically be demonstrated that He does exist if the right evidence were there. This is because if the right evidence existed, it could be proven that God exists, while if the evidence didn't exist, it wouldn't mean that it could be proven that God did not exist. The evidence to prove that God doesn't exist would not be possible to ascertain, because we would basically have to know everything there is to know before we could say conclusively that there was no way, place, or form in which God could exist. That, of course, is impossible. If God does exist, however, it would theoretically be possible for Him to give us the evidence to prove that He exists, if He so chose. Thus, it appears that the top poll option is theoretically valid, while the bottom poll option is not a possible state of existence.

Suppose god exists. By definition, he is all-knowing. But by Godel's Incompleteness Theorem, his knowledge must be incomplete; that is, there must be something true that god does not know. This is a contradiction. God does not exist.

It also appears logically impossible to disprove that highly developed little red martians exist, though if we saw one, we would know they exist. I do not believe this is a reason to simply assume that they exist.

However, god certainly exists if you take the meaning very loosely. He could just be the driving force of the universe, without any physical form or conscious decision. At that point, what is being described is simply the universe itself. So sure you can give it a name.

And Arnav, who says there was ever a beginning to the universe? I mean it wasn't much of anything at some point, but there was still empty space there, and there was still all the material to create it. Besides, matter can be created. From tons and tons of energy.

[quote="jb05"]Suppose god exists. By definition, he is all-knowing. But by Godel's Incompleteness Theorem, his knowledge must be incomplete; that is, there must be something true that god does not know. This is a contradiction. God does not exist.

...

However, god certainly exists if you take the meaning very loosely. He could just be the driving force of the universe, without any physical form or conscious decision. At that point, what is being described is simply the universe itself. So sure you can give it a name.[/quote]

all-knowing

yeah I'm referring to the third paragraph

see the problem with most god people

is that you're like

hey guys

god doesn't exist

and then they're all like DUDE HOW DID THE UNIVERSE COME TO BE

and you're like

hmm...that's a tough one

so I have adopted the position that this driving force (which I think is more than simply the totality of the universe; it must have been at some point something external) is god

but that it doesn't necessarily have any of the other qualities one ascribes to god

hmmm it seems pretty suffic to be like eh there was no beginning... because what if there were then nothing could be absolute and things would be all relative... like not even just physics relativity but like BLEARGH IT IS ALL RELATIVE and that sort of sucks because usually you think God is absolute... unless you argue that God was there all along... but then it's like wtf that probably means God is like the theory of evolution and all that junk and you know God and evolution together is [i]not allowed[/i]

[quote="trigonometry1.01"]PLEASE DO NOT CONTRADICT THE LAW OF CONSERVATION OF MASS THANKS[/quote]
The law of conservation of mass is an over-simplification.  Mass can be converted to energy and vice-versa according to Einstein's equation $e=mc^{2}$, where $c$ is the speed of light in a vaccuum (also the maximum speed at which any non-quantum-mechanical influence can propagate).

Please check your facts more carefully, and do not state your claims without defending them.

@Herald of Doom: this is not G&FF.  As such it is meant for [b]serious[/b] discussion, in which you back up you claims logically.

[quote]Suppose god exists. By definition, He is all-knowing. But by Godel's Incompleteness Theorem, His knowledge must be incomplete; that is, there must be something true that God does not know. This is a contradiction. God does not exist. [/quote]

Unfortunently, that is very broken logic. Here's how the argument should run: Suppose God exists: By Gödel's Incompleteness Theorem, all of man's knowledge must be incomplete. Hence, man is not God. Conclusion: Nothing proven about whether God exists, only that man is definitely not God. You have a good argument; it just doesn't prove what you thought it did.

Godel's Incompleteness Theorem would not apply to God in any case, and there is absolutely no logical reason to beleive that it could. By definition, God is above the principles which govern this universe. Hence, trying to prove that he does or doesn't exist by using principles which apply only to us as humans, or to the universe in which we live, is illogical.

(Pssst, he wasn't actually providing a mathematical proof of the non-existence of god, his argument is satirical, similar to the Invisible Pink Unicorn).

[quote]However, god certainly exists if you take the meaning very loosely. He could just be the driving force of the universe, without any physical form or conscious decision. At that point, what is being described is simply the universe itself. So sure you can give it a name. [/quote]

Somehow I think that violates the definition of God about which we are talking in this thread. By your definition, what you are referring to as a god would always have to exist as long as the brain in a vat theory was false.

hm. that's the definition usually used by people in "proving" the existence of God though, as far as I know

any other definition leaves it obviously clear that there is no reason why God should exist

FEEL FREE TO OFFER OTHER DEFINITIONS OH YEAH

God exists. This is so simple and so true that no one can prove it. You see it everyday in your life.

God exists, if he didn't, then we wouldn't be here.

                                                                                Q.E.D

I like how you feel that your case is so strong that it requires no supporting evidence whatsoever.  Please, before you make a similar claim again, be sure to research cosmology, evolution, relativity, quantum physics, etc.  There are plenty of explanations of the universe that do not require appealing to a divine being.  If God's existance were so simple and true, this thread would not be here.  Also, if no one can prove it, why do you end your post with QED.  You claim to have proven the existance of God, which, by your own logic, is impossible.

You may wish to explain why evolution is "just a theory", while gravity is a "law"; the only difference I see is that the latter is a fundamental force, while the former is just an emergent effect of reality. You can't know anything with absolute certainty (see my longer elaboration above), and science is not about that anyway; science is about predicting things, and predictions are measured by how accurately and how often they are correct.

[quote]unless we had a time machine and could "see" the animals' characteristics changing with time through repeated cycles of death and birth.[/quote]

But we [i]can[/i], in a way, observe this. If nothing else, bacteria's life cycles are sufficiently short that we can observe evolution literally going on as we type (it's, for example, the reason that you need a new flu shot every year). As far as I understand, even the most devout (well, the ones who want to try and be taken seriously) creationists have conceded this (with some qualification involving "microevolution").

[quote="EpicSkills32"]quackman is referring to the actual change from old species to new species as opposed to change within species.[/quote]

What exactly is the distinction here (notice you didn't even define what you mean by "species", which is [url=http://en.wikipedia.org/wiki/Species_Problem]not simple[/url])? Regardless, what is the difference between a bacteria adapting to become immune to some form of treatment and speciation? They're the same process.

Species are not very well defined. There is no set line where one species ends and another begins. When some members of a species have changed sufficiently enough they will no longer be part of the old species but will instead have broken away to be part of a different species.

The boring repetition of inane creationist talking points to which the preceding two posts were responding has been removed.

We also can't confirm first hand that gravity always did its job the same way. So what's the point¿
Also, google "abiogenesis"; see that it is not the same as biological evolution. It's like gravity is not accounting for stars being made of hydrogen.

I have proof that God exists. If nobody helped the universe along, it would be a JUMBLED MESS. Also if God doesn't exist, what happens at the end of the world?
BTW strong Catholic, so don't ask

I think you need to learn the difference between "proof" and "incoherent non sequitur."  What does your last sentence even mean?

I believe in creationism. I don't want to say "I'm a creationist," because that implies wild handkerchief-waving gravity-defying fundamentalist. But I guess I am a creationist, because I do *believe* the story that God shaped Adam and Eve from clay, breathed life into them and let them live in a heavenly garden until they ate from the tree of Wisdom.

So far as I can tell, evolution is a fact. A "strong theory," but for my purposes and my lifetime, it's as well as a fact. It's not something I choose to believe in. It's just how homo sapiens developed.

Also so far as I can tell, babies aren't born from virgins, and once people are dead, they don't crawl out of their tombs on the third day. Water can't just turn into wine. People can't walk on water. That's mostly undebatable.
But yes, I *believe* that all of those things happened through Jesus.

I saw an opera last month about a water-fairy that wants to be human. She wants to be with the man she loves, but more than that, she's in love with the idea of love. Her father protests that her life is perfect already; that humans are a mean and imperfect lot, and they will die, while nymphs are immortal. She says: but humans have a *soul,* and their souls live on forever. She sings a heartbreakingly beautiful aria. Souls aren't a scientific fact. Love is just a chemical reaction. Music is just arranged sound. She's singing perched in a tree in a blue fairy-costume on a stage. And yet, the audience cries.

For non-opera folks, I'm sure you've read fiction, maybe science fiction, or a TV series that you were completely captivated by, that enriched your life so much even though it wasn't "real." I *know* water fairies aren't hanging out in the nearest lake, but if I believe in this particular nymph long enough to open my mind to a beautiful new world, I am transformed by it, and when I return to the "real world" I find love a little bit more beautiful, life a little more worthwhile. I find myself loving more, even though there's no reason to. I see a softer world. I become a better person.

Evolution is how it happened. It's the stuff of textbooks, of labs, of universities and PhD's and the educated class. Knowing all about it is a kind of societal badge that elicits respect. It was reasoned, calculated, recorded. It is as real as a rock in the road, and as beautiful, as comforting, as warm.

But what about the homeless who are only cold, alone and hungry day to day: they need no science textbook. What about the child who loses a parent: no all-cells-have-to-die explanation heals their gaping grief. What about the dying man who never could afford college, struggled to support himself, and now dies leaving no legacy? What about the billionaire who cheated thousands out of their savings, and can not live with himself anymore?

There are instinctual human needs throughout life. For love: for unconditional love. For meaning: meaning that is for everyone, everywhere, every time. And we need forgiveness. Forgiveness for the sins we did, but also for the sins we have here on earth: poverty, violence, meaninglessness. God is but a word for love, meaning and forgiveness. Ritual is just a way of reminding ourselves there is more than ourselves. There is candlelight, there is beauty, there is music, there are miracles, and we believe in it all, and our belief is the tip of the iceberg.

Christianity, for one, says blessed are the poor, the needy, the blind, the sick, the hated, for they shall see God. I think at some point in everyone's life, we are hungry for connection, thirsty for meaning, tired of meaningless life, sick of our own cruelty and limitations, homesick in our homes. And then there's no science that can sate us. All we need is love, meaning and forgiveness (and food and shelter, but that is also the work of the church).

So yes, I know evolution happened, but I also believe we are all only made of dust and God, and without being aware of the God, we sometimes feel like we are only dust.

"God" is, for me, a euphemism for what-is-greater-than-us, what is beyond our perception and understanding. I believe in things that aren't "true" because that is my only way of vaguely describing things that are totally outside of my understanding.

As far as I've seen, all arguments against the existence of God are either a.)against the church (understandable, but not all churches are violent and hypocritical...) or b.)against the belief in miracles (which I aimed to refute above), or c.)consist of arguments that define God in our terms (there is no such thing as eternal! etc. however the idea is that time is a human thing, the universe is a human thing, etc.) But I'm open to new ideas to the contrary.

I haven't been following this discussion very closely, so forgive me if what I'm about to state has already been stated.

One argument that I have seen used in this thread and that is a very popular argument in favor of the existence of God is that "there must have been something originally. The universe cannot have been created out of nothing." Somehow, yet, it seems entirely intuitive/plausible that the universe will exist forever. So why couldn't time and the universe have existed forever? If we can accept that the universe will exist forever into the future, what argument can be made that it's impossible for the universe to have existed forever into the past?

@AbsoluteFriend:
You seem to say you *believe* this stuff, but then you say you think it couldn't happen. Can Jesus perform miracles or not? You also say that religion provides comfort. Credo consolans isn't a reason to believe in something. We have to accept what we can see to be the truth, regardless of how discomforting it is. You say that God is a euphemism for what is greater than us. You're just turning God into something it isn't. God, as far as I can tell, is an omnipotent, omniscient, and fundamentally good being. You have no evidence this being actually exists other than that it would be nice if it did.

Omnipotent and omniscient and good are human characteristics. In my belief (I don't speak for all Christians) the Bible is a set of human stories we use to *try* and describe God, using human characteristics, but that doesn't mean that God actually fits within all those guidelines.

I definitely have no "evidence." The point of faith is that it has no evidence, because it is a belief in something that makes our logic/science/life meaningful, and thus is greater than and outside of our own logic/science/life.

Jesus is a story about a literal human embodiment of God, and thus he could perform miracles because God himself/itself is fundamentally above our own logic and experience, so the story goes that in sending a human form of himself, the human acted as God could on earth. Jesus is 100% human AND 100% God. Again, this isn't "possible," but the point is that "impossibility" is a way of describing a God we cannot understand.

@EricMathPath09
There is evidence that the universe had a finite beginning, though of course this doesn't prove that God exists.
http://www.hawking.org.uk/the-beginning-of-time.html


I have recently become familiar with contests, so I'll be writing AMC and hopefully AIME soon.  What topics would you recommend I study, and how would you say I prepare for both of those contests.  Would you recommend that I go straight to preparing for AIME?  Thanks in advance.

Masoud Zargar

Are you confident in your abilities to make the AIME? If you are, go ahead and start studying for that. I never studied for the AMC until this year, when I'm hoping I get 140+ (hopefully 150). As long as you are sure you'll qualify for the AIME, you're going to benefit far more by studying for it instead of the AMC.

[quote="boxedexe"]Thanks.  I have another questions.  What is the benefit of signing up for both AMC 12 A and AMC 12 B, other than the fact that it is fun.  Will I be invited for AIME even if I just take AMC 12 A?[/quote]

Yes, if you qualify through either test, you will be invited to take the AIME. It's just to increase your chances or yeah, have fun.

Thanks again.  This is my last question.  Can we just write down the answer and get full marks in AIME?

Yes, your method for getting the answer is irrelevant on the AIME.

Well then, I shall be happy; I usually write and elaborate too much, so I lose time. :P


Can we have a general method to express any positive integer in the form $ a^2 \plus{} b^2 \minus{} c^2$, where $ a$, $ b$, $ c$ are positive integers and $ a < b < c$.

[hide]Let $ n$ be the desired positive integer. Take a high integer $ a$ such that $ n,a$ have different parities. Let $ b \equal{} c \minus{} 1$, and pick $ c$ such that $ a^2 \minus{} 2c \plus{} 1 \equal{} n$. Increasing $ a$ by $ 1$ increases $ c$ by about $ a$, so with a sufficiently high $ a$ we have $ a < c \minus{} 1 \equal{} b < c$.[/hide]


Find all integers solution to the equation $2b^2=a^2+c^2$.

This is quite basic. Use Pythagorean triad formulae.

Bomb

This is not basic, this is posted before! http://www.mathlinks.ro/Forum/viewtopic.php?t=72960


$ AD,BE,CF$ are the altitudes of the triangle ABC $ (D\in BC, E\in CA, F\in AB)$. Prove that $ AB\plus{}BC\plus{}CA\geq 2(DE\plus{}DF\plus{}FE)$

clearly must be an acute triangle

$ 2(DE\plus{}EF\plus{}FE) \equal{} 2(a \cos A \plus{} b \cos B \plus{} c \cos C) \equal{} \frac{abc}{R^2} \equal{} \frac{4S}{R} \equal{} \frac{4pr}{R}$

so it become $ \frac{4pr}{R} \le 2p \ \Longleftrightarrow \ R \ge 2r$ wich is know


Hey,

I am pretty new to logarithms, so I want to start this thread to help myself and others out with basic logarithms.  Please post any logarithmic rules and tips, as well as sample problems.

Thanks.

do you have AOPS book2? The second chapter, I believe, has extremely good hints and tricks for logs.

Yes, chapter 1 of AoPS Volume is an excellent introduction to logs.

To start with,
\log_a b stands  for "log b base a".
let we write: $b=a^x$ than we can write it as $\log_a b=x$.

There are some properties related to logarithms:
(1) $\log_a b=\frac{\log_c b}{\log_c a}$
(2) We generally eliminate base, $\log x$, for base 10.
     Also with $\ln x$ we mean base is "e".

Assuming any base c
(3) $\log {(ab)} = \log a + \log b$
(4) $\log {\frac{a}{b}} = \log a - \log b$
(5) $\log {a^x} = x\cdot\log a$

Someone can post related problems.

[quote="prototypex24"]Napier's Formula (I think):

(6)                 $a^{\log_{a}{b}}=b$

Also, (1) is called change of base[/quote]

A proof for that is very easy.
Just consider  $a^{\log_{a}{b}}=k$
$\Rightarrow \log_{a}{b}=\log_{a}{k}$
$\Rightarrow b=k$
$\Rightarrow a^{\log_{a}{b}}=b$


Find all $ x \in \mathbb{R}$ such that $ \frac {|x^2 \minus{} 1|}{x \minus{} 2} \equal{} x.$

We have two cases :
 for x > 1

¦x^2-1¦ = x^2 - 1

And
For x<1
¦x^2-1¦= 1 - x^2

* solving two cases:
[b]For x> 1[/b]
x^2-1= x^2-2x

*** No Solution ***
[b]For x<1[/b]
1-x^2=x^2-2x
so,
x= [1+sqrt(3)] / 2
or
x= [1-sqrt(3)] / 2

[quote="pbatra"] [b]... * solving two cases:[/b]

[b]IF[/b] $ x\ \ge\ 1$ [b]then[/b] $ x^2\minus{}1\equal{} x^2\minus{}2x$ *** [b]No Solution[/b] ***

[b]IF[/b] $ x\ <\ 1$ [b]then[/b] $ 1\minus{}x^2\equal{}x^2\minus{}2x$ [b]so[/b] $ \boxed{x\equal{} \frac {1\plus{}\sqrt 3}{2}}\ \ \vee\ \ x\equal{} \frac {1\minus{}\sqrt 3}{2}$[/quote]
$ x\equal{} \frac {1\plus{}\sqrt 3}{2}\not < 1$ ...

hello, @pbatra, there is only one real solution, namely $ x\equal{}\frac{1}{2}(1\minus{}\sqrt{3})$.
Sonnhard.

Since $ x \equal{} 0$ is not the solution, we can rewrite thr given equation as
$ \left|1 \minus{} \frac {1}{x^2}\right| \equal{} 1 \minus{} \frac {2}{x}$. Let $ \frac {1}{x} \equal{} t$, we have $ |1 \minus{} t^2| \equal{} 1 \minus{} 2t\ (t\neq 0)$.
Drawing graphs of the curve $ y \equal{} |1 \minus{} t^2|$ and the line $ y \equal{} 1 \minus{} 2t$ on $ ty$ plane,
we can get one solution $ t \equal{} \minus{} 1 \minus{} \sqrt {3}$, yielding $ x \equal{} \frac {1 \minus{} \sqrt {3}}{2}$.

Alternative Solution:

Let $ x\equal{}e^{i\theta}$, we have .....

I'm bumping this because I'd really like to see this solution...

If we let $ x \equal{} e^{i\theta}$ we have$ \left|1 \minus{} \frac {1}{x^{2}}\right| \equal{} 1 \minus{} \frac {2}{x}$

$ |2\sin\theta| \equal{} 1 \minus{} 2(\cos\theta \minus{} i\sin\theta)$
$ \sin\theta \equal{} 0$ but this implies $ 1 \minus{} 2\cos\theta \equal{} 0$ which cannot be true...

of course I don't think we can assume x to have a norm of 1...

Setting $ x\equal{}re^{i\theta}$ we have
$ |1\plus{}r^2\minus{}2r\cos2\theta|\equal{}1\minus{}2\cos\theta$ with $ \theta\equal{}n\pi$
$ \cos2\theta\equal{}1$ and $ \cos\theta\equal{}\pm1$
Now just solve for $ r$.

Can you explain it a little more, like where do the sine and cosine come from and stuff, im guessing polar form of a point or vector? I duno..

Try searching for "complex numbers" in the aops wiki; it has some decent explanations. also feel free to click on some related links (like euler's formula, polar form, etc)

hello,
see here: http://www.wolframalpha.com/input/?i=%7Cx%5E2-1%7C%2F%28x-2%29+%3D+x
Cody.

EDIT: I thought this was the format for the German National Olympiad, sorry.

 $ |x^2 - 1| = x /cdot (x-2)$
$implies  x /cdot (x-2) /ge 0  /implies x not accepts [0;2]$
$ x^2 - 1=x^2 - 2x /implies x=0.5 $not solution
$ 1- x^2 = x^2 - 2x  /implies  x=\frac{1}{2}(1-\sqrt{3})$
   Solved by $O.Y.SH. :D $

[hide=Fixed Latex][quote=MerlinLegion07]$ |x^2 - 1| = x \cdot (x-2)$
$\implies  x \cdot (x-2) \ge 0  \implies x$ not accepts $[0;2]$
$ x^2 - 1=x^2 - 2x \implies x=0.5 $ not solution
$ 1- x^2 = x^2 - 2x  \implies  x=\frac{1}{2}(1-\sqrt{3})$
   Solved by $O.Y.SH$. :D [/quote][/hide]

[quote=Virgil Nicula][hide="Here is a little entertainment."]$\frac{|x^2-1|}{x-2}=x\Longleftrightarrow |x^2-1|=x(x-2)$, where $x\ne 2,\ x(x-2)\ge 0$, i.e. $x\not \in (0,2]$.
$\Longleftrightarrow \left(x^2-1\right)^2-\left(x^2-2x\right)^2=0\Longleftrightarrow$$(2x^2-2x-1)(2x-1)=0,\ \frac 12\in (0,2]\Longleftrightarrow$
$f(x)\equiv2x^2-2x-1=0,\ x\not \in (0,2]$. But $f(0)=-1<0$, $f(2)=3>0\Longrightarrow$
$x_1<0<x_2<2$, where $f(x_1)=f(x_2)=0$. Therefore, $\boxed {\ x=\frac{1-\sqrt 3}{2}\ }\ .$[/hide][/quote]

entertainment?


with $ a,b,c,x,y,z\geq 0$ and $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$ . prove that :
$ a^{2}x\plus{}b^{2}y\plus{}c^{2}z\plus{}4(ax^{2}\plus{}by^{2}\plus{}cz^{2})\geq 15abc$

[quote="iqtimo"]with $ a,b,c,x,y,z\geq 0$ and $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$ . prove that :
$ a^{2}x \plus{} b^{2}y \plus{} c^{2}z \plus{} 4(ax^{2} \plus{} by^{2} \plus{} cz^{2})\geq 15abc$[/quote]
Use http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31344


topology
geometry
real analysis
real analysis unsolved

Prove that there exists a simple continuous curve in the plane with positive 2-dimensional measure.

(Note: the standard Peano curve is not a simple curve.)

Does this one do it? http://mathworld.wolfram.com/Peano-GosperCurve.html

It doesn't, precisely because it is space-filling. A one-to-one continuous map between compact Hausdorff spaces is a homeomorphism. A line segment (or a circle) is not homeomorphic to a subset of $\mathbb R^{2}$ with nonempty interior. Indeed, you can remove a lot of points out of such subset without disconnecting it. Thus, a simple continuous curve must have empty interior.

You can find two constructions of such curves in the book "Counterexamples in analysis" by Gelbaum and Olmsted. Or go to jstor.org and look up the paper "A Jordan Curve of Positive Area" by William F. Osgood, in Transactions of the AMS, vol.4 [color=red](1903)[/color], pp.107-112.


This is supposed to be really easy, but I just can't figure out how to do it  :wallbash:

Bina and Andreas start at the same point on a circular track and begin running in
opposite directions. After 35 minutes they are both back at the starting point. Bina
has run 12 laps, while Andreas has run 17 laps. How many times did they pass each
other during their run, not including the times they were together at the start and
finish?

Thank you!

I'm probably wrong but this seems like a 'think about it' question, so isnt the answer just 17?

Nope. The answer is 28. But thanks anyways!  :D

I tried using my fingers to trace their paths around the track. It was an epic fail.  :cursing:

[quote="FullMetalNaruto"]

I tried using my fingers to trace their paths around the track. It was an epic fail.  :cursing:[/quote]

LOL, but isn't it just 12+17-1?

[quote="FullMetalNaruto"]But why would it be 12+17-1? I can't really picture it in my mind.  :?:[/quote]

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1149322399&t=218400]here[/url] at dragon's solution. I think he does a reasonably well job of explaining, if you still don't understand, ask again.


I know there is some well known formula and some well known technique for solving this type of problem, but I've forgotten it.
The problem is:
$\prod_{n=1}^9 \sin{\frac{\pi n}{18}}$
Any input, even on proving a well known formula concerning this will be appreciated.

List them out, then convert the larger angles from sin x into cos x, and hten you should have pairs of sinx cosx, which equals 1/2 sin2x. You'll get 1/2^4 sin pi/9 sin 2pi/9 sin 4pi/9 sin pi/6 sin pi/2.

Take the part sin pi/9 sin 2pi/9 sin 4pi/9. Multiply top and bottom by cos pi/9, then use the same formula earlier to get it up to 1/8 * cos 8pi/9 / cos pi/9 = -cos pi/9 / cos pi/9 = -1. Thus, you'll be left with -1/128 sin pi/6 sin pi/2 = -1/256.

Well , $\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}$ is positive for sure  :) so it cant be -1

What I know is

For $\theta=\frac{\pi}{2n+1}$ $\forall n\in\mathbb{N}$

Then $\prod^n_{i=1}\sin i\theta =\frac{\sqrt{2n+1}}{2^n}$

 :)


[quote="romano"]Let $ a,b,c > 0 $ such that : $ a^2+b^2+c^2 = 3 $ . Prove that :
 a) $ \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} \geq \frac{3}{2} $
 b) $ \frac{a}{2+b} + \frac{b}{2+c} + \frac{c}{2+a} \leq 1 $[/quote]

First one
$L.H.S. \ge \frac{(a+b+c)^2}{\sum a + \sum ab} = \frac{2(a+b+c)^2}{2 \sum a + (\sum a)^2 - 3}$
We have to prove $\frac{2(a+b+c)^2}{2 \sum a + (\sum a)^2 - 3} \ge \frac{3}{2}$ so it's equivalent to $4(a+b+c)^2 \ge 3(a+b+c)^2+ 6(a+b+c)-9$ which is equivalent to $(a+b+c-3)^2 \ge 0$ so it's true.


Let $ BQNR$ be a quadrilateral, WLOG convex, and let $ M \equiv BQ \cap NR,$ $ A \equiv QN \cap RB.$ Internal bisectors of $ \angle NAR, \angle RMB$ intersect at $ I.$ Perpendicular to $ AI$ through $ B$ cuts $ AN$ at $ C$ and perpendicular to $ MI$ through $ N$ cuts $ MB$ at $ P.$ $ \triangle ABC, \triangle MNP$ are then isosceles with $ AB \equal{} AC, MN \equal{} MP$ and $ N \in AC, B \in MP.$

$ BP \equal{} CN$ $ \Longleftrightarrow$ $ BQNR$ is tangential with an incircle.
[hide]Assume $ BQNR$ is tangential with an incircle $ (I),$ touching $ BQ, QN, NR, RB$ at $ X, Y, Z, T.$ Then $ BP \equal{} BX \plus{} XP \equal{} BT \plus{} ZN \equal{} CY \plus{} YN \equal{} CN.$ Conversely, let $ BP \equal{} CN.$ Since also $ IB \equal{} IC$ and $ IN \equal{} IP,$ it follows that $ \triangle IBP \cong \triangle ICN$ are congruent by SSS $ \Longrightarrow$
$ \angle IBR \equal{} \angle IBA \equal{} \angle ICA \equal{} \angle ICN \equal{} \angle IBP \equal{} \angle IBQ,$
$ \angle INB \equal{} \angle INM \equal{} \angle IPN \equal{} \angle IPB \equal{} \angle INC \equal{} \angle INQ$
$ \Longrightarrow$ $ BI, NI$ bisect $ \angle RBQ, \angle QNR,$ respectively $ \Longrightarrow$ $ BQNR$ is tangential.[/hide]

$ RS$ bisects $ \angle BRN$ $ \Longleftrightarrow$ $ BQNR$ is tangential with an incircle.
[hide]As before, assume $ BQNR$ is tangential with an incircle $ (I),$ touching $ BQ, QN, NR, RB$ at $ X, Y, Z, T.$ Let perpendiculars $ BC \perp AI$ intersect at $ U$ and let perpendiculars $ NP \perp MI$ intersect at $ V.$ Since the angles $ \angle AUB, \angle MVN$ are right, $ U, V \in XY \perp QI.$ $ IUSV$ is cyclic on account of the right angles at $ U, V$ and $ UV \perp QI$ $ \Longrightarrow$ $ \angle UIQ \equal{} \angle SUV \equal{} \angle SIV$ $ \Longrightarrow$ $ IS, IQ$ are isogonals WRT angles $ \angle UIV, \angle AIM$ $ \Longrightarrow$ $ R \in IS$ and $ RS$ bisects $ \angle BRN.$ Conversely, let $ RS$ bisects $ \angle BRN$ and assume $ BQNR$ is not tangential. Tangentof the A-excircle $ (I')$ of $ \triangle ARN$ parallel to $ MBQ$ cuts $ MRN, ARB, ANQ, NP$ at $ M', B', Q', P',$ such that $ (I')$ is incircle of $ B'Q'NR.$ Perpendicular to $ AI'$ through $ B'$ cuts $ AN$ at $ C'$ and $ B'C', NP'$ intersect at $ S' \not\in RS.$ $ B'Q'NR$ is tangential with incircle $ (I')$ $ \Longrightarrow$ both $ RS' \not\equiv RS$ internally bisect the same angle $ \angle B'RN \equiv \angle BRN,$ which is a contradiction. [/hide]

[quote="Virgil Nicula"] [color=darkred]Let $ ABC$ , $ MNP$ be two isosceles triangles so that $ \begin{array}{c} AB \equal{} AC \\
\ MN \equal{} MP\end{array}$ and $ \begin{array}{c} N \in (AC \\
\ B \in (MP\end{array}$ . Define

$ \begin{array}{c} R \in AB\cap MN \\
\ S \in BC\cap NP\end{array}$ . Prove that $ BP \equal{} CN\ \Longleftrightarrow$ the ray $ [RS$ is one of the bisectors of the angle $ \widehat {BRN}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote  $ X\in BC\cap MN$ , $ Y\in AB\cap NP$ , $ Z\in NB\cap RS$ . Apply [u][b]Menelaus[/b]' theorem[/u] to :

$ \begin{array}{cccc} \overline {XBC}/\triangle ANR\ : & \frac {XR}{XN}\cdot\frac {CN}{CA}\cdot\frac {BA}{BR} \equal{} 1 & \implies & \frac {XN}{XR} \equal{} \frac {CN}{BR} \\
 \\
\overline {YNP}/\triangle BMR\ : & \frac {YR}{YB}\cdot\frac {PB}{PM}\cdot\frac {NM}{NR} \equal{} 1 & \implies & \frac {YR}{YB} \equal{} \frac {NR}{PB}\end{array}\ .$  Apply [u][b]Ceva[/b]'s theorem [/u] to $ S/\triangle BRN\ :$

$ \frac {ZB}{ZN}\cdot\frac {XN}{XR}\cdot\frac {YR}{YB} \equal{} 1$ $ \implies$ $ \frac {ZN}{ZB} \equal{} \frac {XN}{XR}\cdot\frac {YR}{YB}$ $ \implies$ $ \frac {ZN}{ZB} \equal{} \frac {CN}{BP}\cdot\frac {RN}{RB}$ . In conclusion,

$ BP \equal{} CN$ $ \Longleftrightarrow$ $ \frac {ZN}{ZB} \equal{} \frac {RN}{RB}$ $ \Longleftrightarrow$ the ray $ [RS$ is one of the bisectors of angle  $ \widehat {BRN}$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An equivalent enunciation[/u].[/b] Let $ ABCD$ be a convex quadrilateral for which denote $ R\in AC\cap BD$ .

Consider two points $ \begin{array}{c} E\in (AD\ ,\ AE \equal{} AC \\
 \\
F\in (BC\ ,\ BF \equal{} BD\end{array}$ and define the point $ S\in CE\cap DF$ .

Prove that $ DE \equal{} CF\ \Longleftrightarrow$ the ray $ [RS$ is one of the bisectors of the angle $ \widehat {CRD}$ .[/color] [/quote]


geometry
parallelogram
geometry proposed

Try this one:
Prove that if OI is parallel AB then CO is parallel to HK where K is point of tangency of the incircle with AB.(The points are noted as usual)

Let $L$ be the antipode of $K$. We quickly see that $CHKL$ is a parallelogram, so $CL\|HK$. If we could show that $C,L,O$ are collinear, then we would be done.

This, however, is easy to prove. Assume $O'$ is the intersection between $IO(\|AB)$ and $CL$. If $M$ is $CL\cap AB$, then $O'$ is the midpoint of $LM$, so the projection of $O'$ on $AB$ is the midpoint of $KM$, which is also the midpoint of $AB$, meaning that $O'=O$.

Yes, the only things to observe are: r=RcosC and CH=2RcosC.


Imaginary unit by definition is: $i^2=-1$

or $i=\sqrt{-1}$

Then:
$i^2=i\cdot i=\sqrt{-1}\sqrt{-1}=\sqrt{-1\cdot -1}=\sqrt{(-1)^2}=\sqrt{1}=1$

So we got $i^2=1 \Rightarrow i=1,i=-1$

Where is mistake in the chain of calculation? :roll:  :)

[quote="Eros88"]
$i^2=-1$

or $i=\sqrt{-1}$
[/quote]
First mistake (or at least lack of definition): by that, $a=-\sqrt{2}$ fulfills $a^2=2$, thus $a = \sqrt{2}$.

[quote="Eros88"]
$\sqrt{-1}\sqrt{-1}=\sqrt{-1\cdot\ -1}$[/quote]
$\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't hold, either, even when using a "working" definition of $\sqrt{\cdot}$.

[quote="ZetaX"]
$\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't hold, either, even when using a "working" definition of $\sqrt{\cdot}$.[/quote]
I know.
I was just joking  ;)


Hello! Can you help me with this problem please?

Let X,T be a topological space and let p a point in X.
Let {B(p,(m_i))} an uncountable base of neighbourhoods of p, indexed by the set of all increasing sequences (m_i) of natural numbers.
Now suppose that if B and B' belong to this base,then B(p,(m_i)) is subset of B'(p,(m'_i)) iff m'_i =< m_i for every i.
I want to show that we can't find a countable neighbourhood of p.

Thank you in advance!![/quote][/code][/list]

I want to show that we can't find a countable BASE of neighbourhoods


induction
number theory unsolved
number theory

Prove that there exists infinite number of positive integers $n$ with the following property: from the numbers

$n^2+1,n^2+2,n^2+3,...,n^2+2n$

we can choose non-empty set of numbers which product is a square of integer.

find any $n$ that satisfies conditions and then by induction - if $q_1,..q_i$ are the numbers from $\{n_0^2+1,...,n_0^2+2n_0\}$ which product is a square  then there exists - what's easy to prove distinct numbers
$t_1,...,t_i$ such that $q_1t_1^2,...,q_i,t_i^2$ are disitnct numbers in an interval $\{n^2+1,...,n^2+2n\}$ where $n$ is sufficiently big.


USAMO qualifiers are listed at [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoqual.shtml#canada[/url].
Congratulations to all qualifiers on their success  :coolspeak:
I'm wondering what the index cutoff was this year: I can't seem to find it on the website. Does anyone know?

P.S. I'm the only one from SK. Go me :P

[quote="Steve Dunbar"]* The index is calculated as follows: AMC score + 10 x(AIME score).

USAMO qualifers must attend a US or Canadian school or be a US Citizen. Students who wrote the AMC 12 and achieved an index of at least 197.5 qualified for the USAMO. This was a total of 325 students.

Students in 10th grade or below who scored 6 or higher on AIME
AND
have an index of at least 183.5 also qualified for the USAMO.
This was a total of an additional 171 students.

A student with the highest index in a state not already represented was also invited to write the USAMO. This was 9 students (inlcuding one pair of tied students) from 8 states.

Total number of invited students is 505.

The list of invited stdents will be on our webpage, http://www.unl.edu/amc
by sometime on Friday, April 6, 2007.

Emails to invited students are now being sent, emails to teachers of invited students will be going out later today, Thursday, April 5, 2007. [/quote]

The discussion can be found here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291[/url]

Yayyy!!! Congrats Bobby, and all!! Hope we all do well this year!!

Let the pwnage begin!!!!!!!

Yes, congratulations everyone. :D

btw I thought the cutoff index would be 210+ lol. Good thing it's not. :lol:

did you notice something?
Alex Song, grade 4, Sandowne Public School, Waterloo, ON. qualified for usamo. GRADE 4!!!!!
I wonder if he's going to write IMO 6 times.

glad to see that the cutoff is around 195...just like what i hoped :)

I am very glad to see that. Even though I thought it would be around 210 LOL.

Speaking of Terry Tao, he came to MIT a few days ago to give a special guest lecture series. It was awesome!  :D

Dude I predict no USAMO winners from Canada this year  :ninja: ...

[quote="blootzboy"]sorry um who's  terry tao  :blush:[/quote]

Terry Tao is this genius guy who is the youngest ever gold medallist at the IMO (age 13) and i think he is one of two ppl to get 700+ on the SAT math section at the age of 8..(he got 760) just look him up on wiki :)

to probability1.01: umm ya i think canada will have a few winners  :P

let's invite Terry Tao to our Forum HOHO

[quote="SM4RT"]
to probability1.01: umm ya i think canada will have a few winners  :P[/quote]
EH, who?

i'm sure we'll have a few ppl on the list..hopefully..

[quote="blootzboy"]is there any harm in writing usamo and geting 0?[/quote]
yes...if u get a 0...u will be punished to learn to spell "geting" correctly... :rotfl:

jk jk...i dont think there is any harm...as it says in the teacher's manual, getting into usamo is already very honorable...

remember that only 500 ppl qualified, the worst u can get is No. 500 in this continent.


[quote="RAKAR"]Να προσδιορισετε τους μη αρνητικους ακεραιουςx,y,z ωστε να ισχυει

$2^{x}+3^{y}=z^{2}$

 :D  :D  :D[/quote]

Πολυ ωραια ασκηση.... :)

Θα δωσω μονο την αρχη(δηλαδη το μισο...)

[hide]χ αρτιος[/hide]

(Diagrafike, logw anixneusis lathous... Koitakste parakatw)

Alexandros

Δηλώνω ότι δεν είμαι καλός σ' αυτά. Βρήκα όμως ( οπτικά ) μία λύση.

χ = 4,  ψ = 2, z = 5.

Κώστας Βήττας.

Αλεξανδρε στην αρχή της απόδειξής σου στις περιπτώσεις που πέρνεις λές ότι εξετάζεις αν $x=4k$ kai $y=4k+1$ παραδείγματος χάρη . Νομίζω όμως ότι θα έπρεπε να διακρίνεις την περίπτωση $x=4k$  kai  $y=4m+1$  για να μην δημιουργείται σύγχηση .

Silouan,

Asfalws!! Einai teleiws lathos i apodeiksi stin arxi... enw to elava ipopsin mou otan kataskeuaza ton pinaka twn periptwsewn twn x kai y den to elava ipopsi mou parakatw pairnontas gia paradeigma x=4k+3 kai y=4k anti tou y=4m pou einai to swsto.

Kwsta,

Exeis apolito dikio!

Tha tin diorthwsw teleiws kai tha tin prosthesw kai pali apo tin arxi limeni... Pros to paron tin afairw gia na tin meletisw kalitera...

Alexandros

Grafw tin diorthwmeni askisi:

Ean $x=2k+1$  kai $y\geq 1$, tote $2^{2k+1}+3^{y}\equiv-1 \pmod{3}$ diladi $z^{2}\equiv-1 \pmod{3}$ pou den exei lisi [dioti to $-1$ den einai tetragwniko ipoloipo $\pmod{3}$].

Ean $x=2k+1$ kai $y=0$ tote i dosmeni sxesi ginetai $(z-1)(z+1)=2^{2k+1}$ pou gia $k=0$ den exei lisi enw ean $k\geq 1$ tote $\frac{z-1}{2}\cdot\frac{z+1}{2}=2^{2k-1}$ ap'opou logw tou oti $\Big(\frac{z-1}{2},\frac{z+1}{2}\Big)=1$ pairnoume $\frac{z-1}{2}=1$ kai $\frac{z+1}{2}=2^{2k-1}$ ap'opou $z=3$ kai $k=1$ ara $x=3$ kai etsi $y=0$ opote pairnoume tin lisi $(x,y,z)=(3,0,3)$.
Ara iparxei mia lisi gia x peritto.

An $x=0$ tote h eksiswsi ginetai $3^{y}=(z-1)(z+1)$ kai o $z$ einai artios sinepws $(z-1,z+1)=1$ ap'opou $z-1=1$ kai $z+1=3^{y}$ kai pairnoume $y=1$ me afairesi kata meli kai etsi $z=2$. Ara to zeugos $(x,y,z)=(0,1,2)$ einai mia lisi.

Ean $x\neq 0$ kai $x=2k$ tote sigoura o $z$ einai perittos kai h eksiswsi ginetai $2^{2k}+3^{y}=z^{2}$ ap'opou $3^{y}=(z-2^{k})(z+2^{k})$.

Sigoura $k\geq 1$ afou an $k=0$ tote $x=0$, atopo

Ara sigoura an $k\geq 1$ kai $z$ peritto, exoume $(z,2^{k})=1$ opote $(z-2^{k},z+2^{k})=1$ 'h $2$ kai epeidi $z-2^{k},z+2^{k}$ perittoi ara $(z-2^{k},z+2^{k})=1$. Ara $z-2^{k}=1$ kai $z+2^{k}=3^{y}$ ap'opou me afairesi kata meli exoume $2^{k+1}=3^{y}-1 \ \ (1)$.

An $k=1$ tote i teleutaia den exei lisi ws pros $y$.
Ean $k\geq 2$ tote pairnontas $\pmod{8}$ kai sta dio meli pairnoume oti to $y$ prepei na einai artios. Ara $y=2l$ gia kapoio $l\in \mathbb{N}$.
Sinepws h $(1)$ ginetai $2^{k+1}=(3^{l}-1)(3^{l}+1)$. Profanws $l\neq 0$ kai etsi $(3^{l}-1,3^{l}+1)=2$ ara $\Big(\frac{3^{l}-1}{2},\frac{3^{l}+1}{2}\Big)=1$ ap'opou $2^{k-1}=\frac{3^{l}-1}{2}\cdot \frac{3^{l}+1}{2}$ ara $\frac{3^{l}-1}{2}=1$ kai $\frac{3^{l}+1}{2}=2^{k-1}$.

Etsi pairnoume $3^{l}-1=2$ ap'opou $l=1$ kai $y=2$ ara $k=2, x=4$ kai $z=5$. Opote to zeugos $(x,y,z)=(4,2,5)$ einai alli mia lisi tis arxikis.

Sinopsizontas oi monadikes liseis tis arxikis einai $(x,y,z)=(3,0,3),(0,1,2),(4,2,5)$
Alexandros

Y.G. 1: Euxaristw tous silouano kai Kwsta Vhtta gia tis paratiriseis tous.
Y.G. 2: Xrisimes protaseis pou xrisimopoiithikan ligo poli (edw vriskontai pio genikeumenes) kathara kai mono gia osous ksekinoun twra...

$(i)$ Ean $(a,b)=1$ tote $(a-b,a+b)=1$ 'h $2$.
$(ii)$ Ean $a,b,c,d$ thetikoi akeraioi me $(a,b)=1$ kai $ab=dc^{n}$ tote $a=d_{1}c_{1}^{n}, b=d_{2}c_{2}^{n}$, opou $c_{1},c_{2},d_{1},d_{2}\in \mathbb{N}$ me $(d_{1},d_{2})=1$ kai $(c_{1},c_{2})=1$, $d_{1}d_{2}=d$ kai $c_{1}c_{2}=c$. (POLY XRHSIMH PROTASH)
$(iii)$ Ean $a,b,c$ thetikoi akeraioi $(a,b)=2$ kai $ab=c^{n}(n\geq 2)$, tote $a=2c_{1}^{n}, b=2^{n-1}c_{2}^{n}$ eite $a=2^{n-1}c_{2}^{n}, b=2c_{1}^{n}$, opou $c_{1},c_{2}$ einai thetikoi akeraioi, prwtoi metaksi tous, o $c_{1}$ einai perittos kai $c=2c_{1}c_{2}$.

[quote="cretanman"]
Etsi pairnoume $3^{l}-1=2$ ap'opou $l=1$ kai $y=2$ ara $k=2, x=4$ kai $z=2$. Opote to zeugos $(x,y,z)=(4,2,2)$ einai alli mia lisi tis arxikis.
[/quote]

Ένα μικρό τυπογραφικό λάθος , $z=5$ .

Nai exeis dikio! Euxaristw poli! To allazw amesws!

Alexandros

μπραβο Αλεξαντρε. πολυ καλη λυση.προσπαθησα και εγω αλλα δεν την καταφερα

Kwsta prospathisa poli tin askisi ekeini pou exeis valei prin merikes meres me to $n!$ pou itheles na ksekinaei me $1995$ alla den tin exw kataferei para tis diafores paratiriseis pou ekana! Ean exeis kapoia analitiki lisi grapse tin se parakalw gia na tis riksw mia matia giati eimai poli periergos na dw ti efarmozei... Egw empleksa kai logarithmous mesa alla den kserw ean bgainei kati! Isws na thelei kapoies prosegiseis.

Ean pali theleis prin tin dimosieuseis na sou grapsw 2-3 pragmata pou exw skeftei pes mou...

Alexandros

Αλεξαντρε εισαι στο σωστο δρομο...αν λες λογαριθμους. :roll:  φυσικα και αν εχεις χρονο πες μου τις παρατηρησεις σου.ειναι δυσκολο προβλημα..προσπαθησε την και αλλο.αφου πλησιαζεις


geometry
analytic geometry
area of a triangle
geometry proposed

let $M$ and $N$ be points in interior of square $ABCD$ such that the line $MN$ contains no vertex of the square.
Denote by $s(M,N)$ the smallest area of a triangle with vertices in the set $\{A,B,C,D,M,N\}$.
Find the smallest real number $k$ such that for any such points $M,N$ it holds that $s(M,N)\le k$

Suppose ABCD is a unit square. Denote the intersection of diagonals S. Let's split the squareinto triangles ABS, BCS, CDS and ADS. If M and N are in triangles with comon side, these two triangles of total area 1/2 are split into 5 triangles (by connecting A, B, C, D, M and N), one of them is surely smaller than 1/10.  If points M and N are in opposite triangles, the square is split into 6 triangles (by connecting A, B, C, D, M and N). Two of these have total area 1/2, so among the 4 others ther must be one with area smaller than 1/8, so k=1/8 is sufficient.
If $M=[\frac 1 2,\frac 1 4]$ $N=[\frac 1 2,\frac 3 4]$ (one of square vertices having coordinates [0,0] and the oposite one [1,1]), it's obvious that k=1/8  is also necessary.


probability counting distinguishability AMC

Three red beads, two white beads, and one blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?

$ \textbf{(A)}\ \frac{1}{12} \qquad
\textbf{(B)}\ \frac{1}{10} \qquad
\textbf{(C)}\ \frac{1}{6} \qquad
\textbf{(D)}\ \frac{1}{3} \qquad
\textbf{(E)}\ \frac{1}{2}$

[hide]
Since there are as many red beads as both the white and blue beads combined,

R _ R _ R _

there are $ \frac{3!}{2!1!} \equal{} 3$ ways to put a bead in each blank, and flipping over the sequence like:

_ R _ R _ R

$ 3 \times 2 \equal{} 6$ ways.

there's also

R _ R _ R

where the white beads [u]must[/u] be separate. The blue beads can be in either position, and in either position it could be white-blue or blue-white, so

so $ 2 \times 2 \equal{} 4$ ways here.

thus the total is $ 6 \plus{} 4 \equal{} 10$ out of $ \frac{6!}{3!2!1!} \equal{} 60$ possible arrangements,

so $ \frac{10}{60} \equal{} \boxed{\frac{1}{6} \Rightarrow C}$

(I have a feeling I'm being a bit too (annoyingly) detailed with these solutions XD)
[/hide]

Wait, I'm confused. I got 6/60 or 1/10.
Where did the additional 4 come from?
I understand that there could be:
R W R W R B
R W R B R W
R B R W R W
B R W R W R
W R B R W R
W R W R B R
Just like what undefined117 said for the first 6.

But aren't the other 4 already included in these 6?
Can someone clarify where the other 4 came from?

[hide="The other four"]
R W R W B R
R W R B W R

R W B R W R
R B W R W R
[/hide]

Sorry.. to prevent confusion I should've made my post above something like R _ R _ _ R instead of R _ R _ R

[hide]Well since two beads of the same color are indistinguishable, we really only have to care about the order that the colors are arranged. The total number of arrangements is $ \frac{6!}{3!2!}$ or $ 60$ possibilities. Next, we just list out the total number of arrangements and find that there are $ 10$ of those. Thus our answer is $ \boxed{\frac{1}{6}}$ [/hide]

I'm sorry, but I'm not very good at probability.
Could someone explain [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/e/7/ee729d059a8e34a7167e66c7e9bbabfaddaa65b4.gif[/img]
to get the 60 total possibilities.  I thought it would just be 6!.

[quote="xpmath"]The 3 red and 2 white can't be told apart.[/quote]
Yeah.. and just to elaborate, you divide by 3!2! because:

Let's assume that the red beads can be told apart. There would then be 3! ways to order the red beads in the following arrangement:

R _ R _ R _

or any arrangement, for that matter. However, since they _can't_ be told apart, you divide the total 6! by 3! to get rid of the double (or rather, sextuple) counts. It works similarly for the white beads.

please see the problem:
http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_10B_Problems/Problem_22

I understand the solution, but why can't we use balls and urns?

Here is my attempted solution.

We can make the following diagram: \[-R-R-R-\]

There are 4 places we can put the 3 other beads. Thus the answer is $4+3-1 \choose 3$, which is $20$. Since there are 60 ways to arrange all the beads the probability is $\frac{1}{3}$, but this is incorrect.

What is wrong with my solution?

[color=#00CCA7]Mod: Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=506457]this sticky about reposting AMC problems[/url]. Merging with the appropriate thread.[/color]

You can't put two white balls in one of the four urns.  You might be able to salvage this method if you manually subtract out the number of ways you can put two white balls into the same urn.

I did not understand the solution lol why he multiply 10 by 2?
And there is not 6! possibilities, but 6! / 3!.2!   because 3 are reds and 2 are whites....why he did 6! ?


Find all positive integers $ n$ for which the numbers $ 1,2,3,...,n$ can be colored with $ n$ colors such that each color is used exactly twice and each of the numbers  appear exactly once as the difference between two numbers with the same color.

Someone help me out here -- is this question just a garbled version of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=406829 ?

Not really, if I interpreted this problem correctly. However, with my interpretation, this problem seems too easy.

So we need to color integers $1,2,3,\cdots,n$ with $n$ colors, such that each color is used exactly twice (a number can be colored more than once, or even none), so that if for each color, we list all the absolute differences between the two numbers that are colored with that color, the list is (a permutation of) $1,2,3,\cdots,n$.

However, since the largest number is $n$, while the smallest number is $1$, so the largest possible difference is $n-1$, so $n$ can't appear in the list. So there's no $n$ satisfying the criteria.