geometry calculus trigonometry function inequalities derivative circumcircle

I recently looked at the Geometry subforum and discovered a couple of problems that resulted in quite long discussions: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209516]Problem 1[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99674]Problem 2[/url], so I decided to take a close look at both. What is done below may be not the simplest solution for either of them but the technique is essentially the same for both problems, so I decided to make this separate post rather than posting in the corresponding threads. A small warning: you need to be comfortable with basic calculus and trigonometry to understand the solutions.

[b]Weak contractions[/b] Suppose that a function $ f$ defined on an interval $ I$ satisfies the inequality $ |f'| < 1$ everywhere on $ I$. Assume that $ x,y,z\in I$ satisfy $ f(x) \equal{} y$, $ f(y) \equal{} z$, and $ f(z) \equal{} x$. Then $ x \equal{} y \equal{} z$. Indeed, the bound for the derivative implies that $ |f(x) \minus{} f(y)|\le|x \minus{} y|$ for all $ x,y$ and the inequality is strict whenever $ x\ne y$, so, assuming that $ x\ne y$, we get $ |x \minus{} y|\le |z \minus{} x|\le |y \minus{} z| < |x \minus{} y|$, which is impossible.

[b]Implicit functions[/b] Sometimes (as it will happen in the solutions below), the function $ f(x)$ is not given explicitly but rather is defined by some relation $ F(x,y) \equal{} 0$. How can we see that it is a weak contraction then? The trick is to use the implicit differentiation theorem according to which $ f' \equal{} \minus{} \frac {\partial F/\partial x}{\partial F/\partial y}$. So, if we can show that $ \left|\frac {\partial F}{\partial x}\right| < \left|\frac {\partial F}{\partial y}\right|$ for all $ x,y$ satisfying $ F(x,y) \equal{} 0$, we shall get our weak contraction property. Of course, we should supplement this with some argument showing that the relation $ F(x,y) \equal{} 0$ indeed defines $ y$ as a function of $ x$.

Now, keeping all that in mind, let's turn to the problems. I'll start with
[b]Problem 2[/b]. It is described by the following picture:
[asy]size(200);
pair A=D("A",(0,0),SW), B=D("B",(20,0),SE), C=D("C",(12,16),N),
     X=D("X",(2*B+C)/3,NE), Y=D("Y",(2*C+A)/3,NW), Z=D("Z",(2*A+B)/3);
DPA(B--Z--C^^C--X--A^^A--Y--B);
MA(B,A,X,3,9);
MA(C,B,Y,3,9);
MA(A,C,Z,3,9);
DPA(A--Z^^B--X^^C--Y,orange+linewidth(1.5));[/asy]
Here 3 orange segments and 3 green angles should be equal. Denote now the length of each orange segment by $ x$ and the size of each green angle by $ \alpha$. The sine theorem yields the equations
\[ 2R \sin\alpha\frac {\sin C}{\sin(B \plus{} \alpha)} \equal{} 2R \sin\alpha\frac {\sin A}{\sin(C \plus{} \alpha)} \equal{} 2R \sin\alpha\frac {\sin B}{\sin(A \plus{} \alpha)} \equal{} x
\]
where $ R$ is the circumradius of the triangle $ ABC$.
Now rewrite that as
\[ \sin B \equal{} \lambda\sin(A \plus{} \alpha),\quad \sin C \equal{} \lambda\sin(B \plus{} \alpha),\quad \sin A \equal{} \lambda\sin(C \plus{} \alpha)
\]
where $ \lambda \equal{} \frac {x}{2R\sin\alpha}$.
We see that these are cyclic equations just as above. There are 2 problems here. The first one is that once we know $ A$, the first equation allows us to determine $ \sin B$ but this is not yet enough to determine $ B$ uniquely. The second one is that, in order to apply our weak contraction idea, we need to know something about $ \lambda$; clearly, when $ \lambda$ is large, then the mapping $ A\mapsto B$ is not going to have small derivative. So, we need to do some

[b]Preliminary work[/b]. Our first goal will be to prove that $ \lambda\in(0,1)$. Clearly, $ 0 < \alpha < \pi$, so positivity is not an issue. To prove that $ \lambda < 1$, assume that $ B$ is the largest angle and consider the following picture:
[asy]size(200);
pair A=D("A",(0,0),SW), B=D("B",(20,0),SE), C=D("C",(12,16),N),
     X=D("X",(2*B+C)/3,N);
path Q=D(circumcircle(A,B,C));
D(A--B--C--A);
pair U=D("U",IP(Q,L(A,X,1)),NE);
D(A--U--B);
MA("\alpha",B,A,X,3,9);
MC("x",B--X,WSW);
MC("y",B--U,W);[/asy]
Note that $ y \equal{} |BU| \equal{} 2R\sin\alpha$. On the other hand, we may be sure that $ x < y$ if $ \angle BXU > \angle BUX$. But these angles are easy to compute: they are $ B \plus{} \alpha$ and $ C$ respectively. According to our assumption $ B\ge C$, so we are done.

So, let us define the function $ f(t)$ as the unique angle between $ \minus{} \pi/2$ and $ \pi/2$ such that $ \sin f(t) \equal{} \lambda\sin(t \plus{} \alpha)$. Since $ \lambda\in(0,1)$, this definition makes sense on the entire real line.

Our next task will be outrule obtuse angles. So suppose that $ A$ is obtuse. Then $ B \equal{} f(A)$, $ C \equal{} f(B)$ and $ A \equal{} \pi \minus{} f(C)$. Note now that $ f(t) < t \plus{} \alpha$ if $ 0\le t\le\pi/2$, so $ f(C) < C \plus{} \alpha\le C \plus{} B$. But then the last equation implies $ A \plus{} B \plus{} C > \pi$ which is impossible.

Now, when we outruled the possibility of an obtuse angle, we can be sure that  $ B \equal{} f(A)$, $ C \equal{} f(B)$ and $ A \equal{} f(C)$, so it is enough to show that $ f$ is a weak contraction. Since it is given by a relation $ \sin s \minus{} \lambda\sin(t \plus{} \alpha)$, it is enough to show that $ |\cos s| > \lambda|\cos(t \plus{} \alpha)|$ for $ s,t$ satisfying the relation. But it is obvious since the smaller sine (in absolute value) implies the larger cosine and $ 0<\lambda<1$. We are done with Problem 2. So, let's go to

[b]Problem 1[/b]
Here 2 diagrams are possible.
[asy]size(250);
picture f(real s)
{
real r=1-s;
pair A=D("A",(0,0),SW), B=D("B",(20,0),SE), C=D("C",(12,16),N),
     N=D("N",s*B+r*C,NE), P=D("P",s*C+r*A,NW), M=D("M",s*A+r*B),
     U=IP(A--N,C--M), V=IP(B--P,A--N), W=IP(C--M,B--P);
fill(U--V--W--cycle,yellow);
D(A--B--C--A); DPA(A--N^^C--M^^B--P);
DPA(A--M^^B--N^^C--P, orange+linewidth(1.5));
return CC();
}
picture a=f(0.3), b=f(0.6);
add(a);
add(shift((30,0))*b);[/asy]
(the orange segments are equal and the yellow triangle is equilateral).

The argument below applies regardless of which picture is used. We shall denote the length of each orange segment by $ \ell$, fix the yellow triangle, and consider $ C$ as a function of $ A$ (so, given $ A$, we define $ C$ as the unique point on the ray extending the appropriate side of the yellow triangle such that $ |PC| \equal{} \ell$. Of course, we need to show that our definition makes sense (i.e., that such $ C$ exists and is unique). We will do that plus, in addition, we will show that the mapping $ A\mapsto C$ is a weak contraction. Then, identifying the rays containing $ A,B,C$ by appropriate rotations around the center of the yellow triangle, we will conclude that $ A$, $ B$, and $ C$ lie at the same distance from the corresponding vertices of the triangle, which immediately yields that the triangle $ ABC$ is equilateral.

Let us look at the following picture:
[asy]size(300);
fill((0,0)--(2,0)--2*dir(60)--cycle,yellow);
D((-20,0)--(0,0)--(10,17), black);
D(L((-8,17),(2,0),0,-0.3),black);
D(L((-8,17),(2,0),-0.7,0.2),dashed);
pair A=D("x",(-15,0)), B=D("y",(8,13.6),SE), C=D(IP(A--B,(-8,17)--(2,0)));
D(L(A,B,0,0.2));
MA("60^\circ",C,(2,0),A,3);
MA("60^\circ",B,(1,sqrt(3)),C,3);
MC(30,"M(x,y)",A--C,N); MC(30,"L(x,y)",C--B,N);
MA("\alpha",(0,0),A,B,2,2,orange);
MA("\beta",B,C,(-8,17),2,2,orange);
MA("\gamma",2*B-A,B,(10,17),2,2,orange);[/asy]
Here the position of the point $ A$ is denoted by $ x$ and the position of the point $ C$ by $ y$. The relation connecting $ x$ and $ y$ is $ L(x,y) \equal{} \ell$. Note that when $ y$ changes from $ 0$ to $ \infty$, so does $ L(x,y)$ (on this diagram; on the other diagram $ L(x,y)$ starts at the length of the side of the yellow triangle and ends at $ \infty$). We need the derivatives $ \frac {\partial L}{\partial x}$ and $ \frac {\partial L}{\partial y}$. The reader needs to be familiar with both calculus and trigonometry to compute them and the computation is a bit boring, but here is the result:
\[ \left|\frac {\partial L}{\partial x}\right| \equal{} \frac {L}{L \plus{} M}\frac {\sin\alpha}{\sin\beta}|\cos\beta|
\]
and
\[ \left|\frac {\partial L}{\partial y}\right| \equal{} \left|\cos\gamma \minus{} \frac {M}{L \plus{} M}\frac {\sin\gamma}{\sin\beta}\cos\beta\right|\ge \cos\gamma \minus{} \frac {M}{L \plus{} M}\frac {\sin\gamma}{\sin\beta}|\cos\beta|
\]
Note now that $ \alpha,\gamma\in(0,\pi/3)$ while $ \beta\in(\pi/3,2\pi/3)$, so the ratios of sines are less than $ 1$ and $ \cos\gamma > |\cos\beta|$. This allows to conclude that $ \left|\frac {\partial L}{\partial x}\right| < \left|\frac {\partial L}{\partial y}\right|$, so $ L(x,y)$ is monotone in $ y$ (thus justifying the existence and uniqueness of $ C$, given $ A$) and the mapping defined by the relation $ L(x,y) \equal{} \ell$ is a weak contraction. Problem 1 is done as well.

[b]Final remarks[/b]. I am very far from claiming that the above is the best solution for either of those problems. All I wanted was to introduce a simple (and very well known) idea of how to deal with cyclic identities arising in triangle geometry. Sometimes it works, sometimes it doesn't. Also, the weak contraction property is not the only one that allows to conclude that $ f(f(f(x))) \equal{} x$ implies $ f(x) \equal{} x$. Another useful property is $ f' < 0$ (so $ f$ is decreasing). Here is a relatively simple exercise which makes use of this observation:

[b]Exercise[/b]. On the picture below the green angles are equal and the orange intervals are equal. Prove that the triangle $ ABC$ is equilateral.
[asy]size(200);
pair A=D("A",(0,0),SW), B=D("B",(20,0),SE), C=D("C",(12,16),N),
     X=D("X",(2*B+C)/3,NE), Y=D("Y",(2*C+A)/3,NW), Z=D("Z",(2*A+B)/3);
DPA(A--Z--C^^B--X--A^^C--Y--B);
MA(B,A,X,3,9);
MA(C,B,Y,3,9);
MA(A,C,Z,3,9);
DPA(B--Z^^C--X^^A--Y,orange+linewidth(1.5));[/asy]

Try to solve :).

I rode above post it seems very interesting. Maybe I'm wrong but I cannot find any gaps. I also think it is the most universal method I've seen for solving such kind of problems. The only thing I can say - these problems are not very easy at least for me ... even with such universal and convinient method they requires some not routine observations and deep math thinking.

I cannot solve the last problem. Here is the place to remember again - math is only my hobby. It is interesting to see a solution for the problem - proposed by fedja as an excercise. I think if it is not solved in a week fedja must show us again some beautiful solution on it using calculus or not.

OK, let it be one week (starting at the time in borislav_mirchev's post time signature). :)

OK, since I'm not certain about my whereabouts at 2PM tomorrow, I'll post the solution today.  Let $ x$ be the length of each orange segment and $ \alpha$ be the size of each green angle. Then, using the Sine theorem, we get
\[ x \equal{} |AY| \equal{} \frac {\sin(B \minus{} \alpha)}{\sin(C \plus{} \alpha)}2R\sin C\,,
\]
so
\[ \sin(B \minus{} \alpha) \equal{} \lambda\frac {\sin(C \plus{} \alpha)}{\sin C}
\]
with $ \lambda \equal{} \frac {x}{2R}$.
Now notice that $ |AY| \equal{} x < |AB|$, so the angle $ B \minus{} \alpha$ cannot be obtuse. Thus, $ B \equal{} f(C)$ where $ f(t) \equal{} \alpha \plus{} \arcsin\left(\lambda\frac {\sin(t \plus{} \alpha)}{\sin t}\right)$. Also, $ C \equal{} f(A)$ and $ A \equal{} f(B)$. The function $ f$ is decreasing on the subinterval of $ (0,\pi)$ on which it is defined (just write $ \frac {\sin(t \plus{} \alpha)}{\sin t} \equal{} \cos\alpha \plus{} \sin\alpha\operatorname{ctg} t$ and recall that $ \operatorname{ctg}$ is decreasing on $ (0,\pi)$ to see that). Thus, $ f(f(f(t))) \equal{} t$ implies $ f(t) \equal{} t$, so all angles are equal. The end.

Short, nice and beautiful.
I'm wondering why no one discovered it.

In the configuration given there may be invented many similar problems. They may be probably solved in the manner, proposed by you. I'm wondering if to publish them. If I publish all such problems - I risk to be boring. But sometimes the manner of solving is similar and no so...some creativity is required there. We also have lots of problems ready for math competitions (in mathlinks there are 4 or 5 similar problems) and we may see some different ideas.

What you will advice me - to publish or not to publish more problems of this kind?

You mean "to post or not to post?". Well, it seems that even my exercise didn't attract any attention (except from you), and, once you've seen the solution (which is almost a verbatim repetition of what I did in Problem 2, only much simpler), you should agree that one week was plenty of time for this question. So I'm not sure that sufficiently many mathlinkers really like this kind of problems. Also, such problems can be created in bundles, so to post all possible variations would be quite boring. On the other hand, you can try to post one more and see if somebody responds.

In general, I see it like this: there is a certain culture in almost every subforum and the problems that fit with this culture attract attention from many people, including the best ones, and are solved rather quickly if they are solvable at all, while the posts containing much easier (and sometimes even nicer) problems that do not quite fit with the current "standards" are just ignored by those who could really solve them. The dominating culture in the Geometry subforum seems to be what I would call "Classical Synthetic Geometry". The folks here are really good at it and the stuff they do is beautiful, indeed, but if you deviate from the mainstream a little bit, they either get confused, or lose interest, or both. There is nothing unusual about it; the people in "professional mathematics" are hardly any different with a few exceptions (needless to say, i like the exceptions most :)).

So, perhaps, you shouldn't bother posting more than one additional "prove that the triangle is equilateral" problem, if any, but if you have other "non-mainstream" questions that you consider nice and difficult, by all means post them in this subforum. It is always a good idea to attract the attention of people to something that is not completely foreign to them but also not exactly what they are used to.

Anyway, I'm just a fleeting guest in this particular subforum, not one of the regulars. If you want a more informed opinion about your question, ask somebody like Darij or Pohoatza ;).


inequalities
geometry
3D geometry
tetrahedron
sphere
LaTeX
inequalities proposed

Can you solve this problem! :(

Find a least value of x satisfying this condition: for every tetrahedron with volumes V and radius of the circumsphere and sphere inscribed is R,r respectively
we have
\[ V \leq r \cdot [x \cdot R^2+(24\cdot \sqrt 3-9x)\cdot r^2] \]

eum, perhaps can you explain us what this has to do with the IMO in greece? :?

might be a problem for shortlist?  :maybe:

Not be shorlist.

Idea of problem:
Probable every one knew similar inequalityin plane:

(1)For every triangle abc with area S and radius of the circum circle and inscribedcircle is R,r, we have
S <=  [(3.sqrt3)/2].R.r

How to solve (1)?
Put M,N,P are excenter of triangle ABC. Easy to check triangle MNP inscribed in circle with radius R'=2.R and area of MNP =2.R/r. area of ABC.
By the "problem circumference" we have:
area ofMNP<=(3sqrt3)/4.(R'^2) so we have (1)./

What 's the heart of this solution. I thinks it's content of this theorem:

Theorem in plane: OA=OB=OC=OM and 90degree>#CAM=#MAB then OM perpedicular BC.

Theorem is heart of problem 1 in IMO 04.
I don't know is there exist similar theorem in spaces. Can you  :lol:

[quote="nguyenquockhanh"]why don't you solve thispro :D :?  it veryvery hard? :D[/quote]because it's in the wrong place ... watch carefully where you post. The forum I belive has a clear structure :).

[quote="nguyenquockhanh"]sorry for my point, but can you solve it[/quote]Maybe if I could understand it ... I am editing it to make it $\LaTeX$ compatible. Please look at it and tell me if it's what you wanted ...  :?


Let $ABC$ a triangle, $N$ the midpoint of $(AC)$, $AD\bot BC$, $D\in (BC)$ and $P\in (AD)$ such that $\angle ABP \equiv \angle NBC$.
Prove that $m(\angle BAC)=90^0$ if and only if $(AP)\equiv (PD)$.

If $\angle{ABC}=90^o$, then $D=B$, then $P \in AB$, then $\angle{ABP}=0^o \neq \angle{NBC}$.

Sorry for the mistake. Now it's correct.


conics
ellipse
function
symmetry
calculus
calculus computations

Given an ellipse $ D: \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (a > 0,\ b > 0)$ which contains a circle $ C: x^2 \plus{} y^2 \equal{} 1.$ Draw a tangent line from the point $ P(0,\ b)$ on $ D$, let the extented line meets $ D$ at $ Q$. Draw a tangent line from $ Q$, let the extended line meets $ D$ at $ R$, draw a tangent line from $ R$, let the extended line meets $ D$ at $ P$. Express $ a$ as the function of $ b$.

Please translate the context into plain English.

Thanks in advance

kunny

Suppose w.l.o.g $Q=(x_0,y_0)$, $x_0>0$, $y_0<0$. Since $PR$ is also a tangent line, by symmetry, $R=(-x_0,y_0)$. Then that $QR$ is tangent to $C$ gives $y_0=-1$. The line $PR$ thus becomes $x_0y+(1+b)x-x_0b=0$, and the distance from the origin to this line is $\frac{x_0b}{\sqrt{x_0^2+(1+b)^2}}=1$. Since $Q$ lies on $D$, $\frac{x_0^2}{a^2}+\frac{1}{b^2}=1$. Eliminating $x_0$, we get the result:
\[a=\frac{b(1+b)}{b^2-1}.\]

That's a correct answer, that is to say, $a=\frac{b}{b-1}$.

The problem is 1990 Kyoto University entarnce exam/Science.

Hi, kunny.
$\triangle POE\sim\triangle PQF\Longrightarrow\frac{OE}{PE}=\frac{QF}{PF}\Longrightarrow QF=\frac{b+1}{\sqrt{b^{2}-1}}$
$\frac{QF^{2}}{a^{2}}+\frac{OF^{2}}{b^{2}}=1\Longrightarrow a=\frac{b}{b-1}$


How do I get those class reminders to stop coming(to my email)?
I actually like the classes, so there's no way that I could forget that there is a class, and I don't need those reminders(and to be honest, it's kind of annoying).

I'd actually prefer if the reminders were closer to the classes, say 19 hours before class instead of 43 hours?

[quote="junggi"]I actually like the classes, so there's no way that I could forget that there is a class[/quote]

Erm...not everybody remembers everything they like or care about.

[quote="junggi"]But just how can I forget about AoPS classes? :P

[quote="akalra1"]set up a message filter in watev email u have and send all email from classes@artofproblemsolving.com directly to Trash

or u could set up a AoPS folder instead of trash[/quote]
How do I do that in yahoo mail?[/quote]

GO to options (its on the right) and filters. It might not work for you I have a sbcglobal/yahoo accounts which means I might have some options you don't. Just check and see.

If you're using yahoo mail beta it makes you switch back to the old version while you're doing that.


Ya se puede discutir y dar a conocer los problemas verdad?

bueno yo conosco los problemas ( participe en la olimpiada) alguien quiere discutirlos y comparar soluciones?

Por favor, ponga las problemas para discutirlas.


geometry
3D geometry
pyramid
tetrahedron
LaTeX
geometry proposed

It is a not easy problem.

There are four lines (there are no two of them that have two common points or are parallel). Is it possible the angle between [i]any two[/i] of them to be the same?

If you have lines $ l_{1}, l_{2}, l_{3}$ which pairwise all form angles $ \theta$, then either two are parallel, or the triangle they form has all equal angles.  No two of the lines in the problem are parallel, so any three of the lines form an equilateral triangle.  It then follows that two of the lines are parallel: contradiction, and so it's impossible.

[quote="borislav_mirchev"]But please take a look at the problem with the piramid proposed in Vietnam 1999  :)[/quote]

For this one, i'm thinking if you take this regular tetrahedron, ABCD.  WLOG, OX passes through ABC.  Consider the circumsphere of ABCO.  Extend DO to meet this circumsphere again at P.  Then, wLOG, $ X$ also lies on this circumsphere.  Perhaps this makes calculations easier...though I don't know...

[quote="borislav_mirchev"]I took in mind the following problem: In space let be given 4 rays: Ox, Oy, Oz, Ot. The angle between any two is the same. Let Or be a ray rotating around O. Let A, B, C, D are the angles forming by it and rays Ox, Oy, Oz, Ot respectively. Prove that:
i) cosA+cosB+cosC+cosD
ii) (cosA)^2+(cosB)^2+(cosC)^2+(cosD)^2
do not depend on Or.

It is ok but ... you proved that the statement is impossible, because it is impossible
the angle between any two of for lines to be the same ... do you remember? ;)

P.S. I promised to learn Latex ... but I met very nice girl and now I'm very distracted...[/quote]

Well in space, there's exactly one configuration where all pairwise angles are the same...namely where the four rays are the rays from the centroid of a regular tetrahedron through the vertices...

[quote="borislav_mirchev"]I've a question. How you prove that in space there is "exactly one" configuration?[/quote]

Well consider first two rays.  Now there are exactly two directions for the third ray such that all three pairwise angles are the same.  One direction must be the third ray, and the other the fourth, then.  Moreover, the angle between these two rays must be the same.  A continuity argument yields the desired result (create a picture in your head where you vary the angle between the first two rays, and as a result, the other two move in and out).  This isn't rigorous, but it's a sketch of the complete proof.

Another idea is to suppose the angles between the rays are equal... then to take for points M, N, P, Q on the same distance of O. If you take a look at the equal triangles MNO, NPO, PQO, QMO... you can conclude that this is a regular tetrahedron. I remember - similar idea were used somewhere in mathlinks for the Vietnam 1999 problem.


CO anh nao` co bai` ve` day so ko
post len cho em ha
thanks

Cho so thuc $a$ va day {$x_{n}$} xac dinh boi $x_{1}$= $\sqrt{2005}$ va $x_{n+1}$=$\frac{a}{x_{n}^2+1}$.Tim a de day co gioi han huu han khi n dan toi vo cung.

bai nay` sach nao` hong co'
ve mua sach doc.
anh cho em bai` khac hay hon de

ah`
Anh oi; post cho em may bai` luong giac lun nghe
thanks

Thử bài này luôn đi: Cmr với mọi tam giác $ABC$ có $\sum (tg\frac{A}{2})^2 +8\prod sin\frac{A}{2}$ $\geq  2$.

zozozo123  , sao khoong vô http://diendantoanhoc.net
  trong do thi bài tập về dãy số và lượng giác bạt ngàn   . tha hồ làm
          em đừng lên mathlinks ít bài về loại này lắm .

[color=blue]Đúng vậy. [url=http://www.diendantoanhoc.net]Trang nhà[/url] không vào mà lại vào đây làm bài nhỉ?
Kể cũng lạ...[/color]


algebra
polynomial
inequalities proposed
inequalities

http://www.mathlinks.ro/Forum/viewtopic.php?t=324338&sid=d9bdd40acb9494da5b76d5542d047e23
16#.
pay a attention.
BQ

[quote="fjwxcsl"]Suppose that:

$ H_{n + 1}(x_1,x_2,\ldots,x_{n + 1}) = x_1^k[A(x_1,x_2,\ldots,x_{n + 1}) - 2F_n(x_{2},x_{3},\ldots,x_{n + 1})]$

${ + x_2^k[A(x_1,x_2,\ldots,x_{n + 1}) - 2F_n(x_3,x_{4},\ldots,x_1})]$

${ + x_3^k[A(x_1,x_2,\ldots,x_{n + 1}) - 2F_n(x_4,x_{5},\ldots,x_2})]$

$ + ......$

${ + x_{n + 1}^k[A(x_1,x_2,\ldots,x_{n + 1}) - 2F_n(x_1,x_{2},\ldots,x_{n }})],$

where

$ A(x_1,x_2,\ldots,x_{n + 1}) = F_n(x_1,x_2,\ldots,x_n) + F_n(x_2,x_3,\ldots,x_{n + 1}) + \ldots$

$ + F_n(x_{n + 1},x_1,\ldots,x_{n - 1}), (x_1,x_2,\ldots,x_{n + 1})\in{R_{ + }^{n + 1},k\in{N}}$

prove (or disprove) that:

 if   $ F_n(x_1,x_2,\ldots,x_n)$ is positive semi-definite  homogeneous symmetric

polynomial,then $ H_{n + 1}(x_1,x_2,\ldots,x_{n + 1})$ is also  positive semi-definite polynomial.[/quote]

[size=200]More generally,we have proved that :[/size]

[size=150]if $ F_n$ is  cyclic symmetric  positive semi-definite (homogeneous or

inhomogeneous) ,then $ H_{n + 1}$  is also  positive  semi-definite .[/size]


[hide]If x varies inversely with y, then xy will always be the same.

So on Monday, we get $4 \cdot 5 = 20$

On Tuesday, we get $20 \cdot x = 20$
So $x=\boxed{1}$.[/hide]

[quote="Arvind_sn"]Krabby Patties give Sponge Billy Bob bad breath. The number of breath mints he takes varies inversley with the number of minutes he waited after eating a Krabby Patty. On Monday, Sponge Billy Bob ate a Krabby Patty. Exactly four minutes later, he ate five breath mints. On Tuesday, Sponge Billy Bob had to eat twenty breath mints (whoooah!!!!!!!!). How many minutes did Sponge Billy Bob wait?[/quote]

[hide]
monday- 20
tuesday 20x=20
x=1
[/hide]

[quote="Arvind_sn"]Krabby Patties give Sponge Billy Bob bad breath. The number of breath mints he takes varies inversley with the number of minutes he waited after eating a Krabby Patty. On Monday, Sponge Billy Bob ate a Krabby Patty. Exactly four minutes later, he ate five breath mints. On Tuesday, Sponge Billy Bob had to eat twenty breath mints (whoooah!!!!!!!!). How many minutes did Sponge Billy Bob wait?[/quote]

[hide]20x=20
x=1[/hide]

[hide]xy must always be the same, if x varies inversly with y.
$20x=20$
$x=1$[/hide]

[hide]because it varies inversely, after just 1 minute Billy must eat 20 mints... [/hide]


topology
function
real analysis
real analysis unsolved

Could you give an example of separable and Hausdorf space that has no countable local base. (there exists at least a point that does not have a countable local basis)

The topological product $\mathbb R^\mathbb R$ should do the trick.

It's Hausdorff because it's a product of Hausdorff spaces. In order to exhibit a countable dense subset, consider the functions $: \mathbb R\to\mathbb R$ which are zero everywhere except for several (i.e. a finite number) disjoint intervals with rational endpoints, where they take rational values (we consider such a function for every combination of intervals and values). Finally, let's prove that this space is not first-countable by showing that $f: \mathbb R\to\mathbb R,\ f\equiv 0$ has no countable local basis $\mathcal B$ of open sets (of course, this works equally well for any element of $\mathbb R^\mathbb R$).

Every basis element $B_n\in\mathcal B$ contains a set of the form $S_n=S(x_1,\ldots,x_k;\varepsilon)=\{g: \mathbb R\to\mathbb R\ |\ |g(x_i)|<\varepsilon,\ i=\overline{1,k}\}$, hence every open set around $f$ must contain some $S_n$. Now consider the open set $S=\{g: \mathbb R\to\mathbb R\ |\ |g(x)|<1\}$, where $x$ is a real different from all the $x_i$'s appearing in all the $S_n$'s (it's possible to find such a real number $x$ because there are countably many $x_i$). $S$ is an open set containing $f$, but does not contain any of the sets $S_n$, so we have a contradiction.

I'm don't understand very well your example. Is it separable ($\mathbb{R}^{\mathbb{R}}$ with the product topology ?)

for each $x \in \mathbb{R}$, $V_x$ the set of functions that vanish on $\mathbb{R}-\{x\}$ and such that $|f(x)|< 1$ is an open set. So $\mathbb{R}^{\mathbb{R}}$ is not separable ?

So what's your point? Those $V_x$ are not disjoint. How do you conclude that it's not separable?

OK, but for example if $V_x$ is the set of functions $f$ that satisfy:
1/ $f(t)=x$ on $\mathbb{R}-\{x\}$
2/ $|f(x)-x|<1$

Then the $V_x$ are open disjoint sets, isn't it.

They are disjoint, but not open :). Focus man! :) An open neighborhood of $f$ in $\mathbb R^\mathbb R$ is a set of the form $\{g\in\mathbb R\ |\ |g(x_i)-f(x_i)|<\varepsilon_i,\ \forall i\in\overline{1,n}\}$.

:blush:  my fault .. Sorry for the stupid things I wrote above  :roll:


function
integration
inequalities
real analysis
real analysis unsolved

For $ f \in L^{p}[a,b]$, $ 1 \leq p<\infty$, and a subdivision $ \Delta\equal{}\{ \xi_{0},\ldots,\xi_{m} \}$ of $ [a,b]$, define the $ \Delta$-approximant to $ f$ as the step function on $ [a,b]$ that is constant on each subinterval $ [\xi_{k},\xi_{k\plus{}1})$, taking the value $ \frac{1}{\xi_{k\plus{}1}\minus{}\xi_{k}}\int_{\xi_{k}}^{\xi_{k\plus{}1}} f$ there. Show that the ($ L^{p}$) norm of the $ \Delta$-approximant is at most $ \parallel{}f\parallel{}_{p}$.

Let a $ \delta$-approximant be $ g$.
$ \int_{[a,b]} |g|^p$
$ \equal{} \sum \int_{[a,b]} |g \chi_{[\xi_k, \xi_{k\plus{}1}]}|^p$
$ \equal{} \sum \int_{[\xi_k, \xi_{k\plus{}1}]} \frac{1}{(\xi_{k\plus{}1} \minus{} \xi_k)^p} \left|\int_{\xi_k}^{\xi_{k\plus{}1}} f \right|^p$
$ \equal{} \sum \frac{1}{(\xi_{k\plus{}1} \minus{} \xi_k)^{p\minus{}1}} \left|\int_{\xi_k}^{\xi_{k\plus{}1}} f \right|^p$
$ \leq \sum \int_{\xi_k}^{\xi_{k\plus{}1}} |f|^p$ (Holder's inequality)
$ \equal{} \| f \|p$

The $ \Delta$-approximant converges to $ f$ in $ L^{p}$ as the length of the longest subinterval tends to 0. Does it converge to $ f$ in measure?

I think yes. For continuous functions, the delta approximant actually converges back to it. We know that L^p functions can be approximated by continuous functions. So what you can do is
$ f \in L^p$ approximated by $ g \in C[a,b]$ approximated by $ \delta g \in L^p$ approximated by $ \delta f \in L^p$. I think that works.

In general, does convergence in the $ L^{p}$ norm imply convergence in measure? I think it does by Chebyshev's inequality.


I thought that we get Karma if we answer a problem correctly, even if we've already done the problem before...is this true? Or do we only get Karma the first time around when solving a problem?

you can rate a video or a problem by scrolling down the solution page and clicking the stars to show how much you liked the solution. You can do the same with the videos by looking where the large comment box is and clicking the stars to rate.

[quote="vallon22"]answering a problem correctly does not get you karma
rating the solution does though

here is what gets and loses you karma, taken from the instructions:

[quote]The karma points are currently as follows, but subject to change: Answering a problem Spend 5
Rate problem/solution Receive 5
Submit bug report Spend 10
Bug report determined valid Receive 25
Watch video Spend 5
Rate video Receive 5
Comment on video Receive 5
[/quote][/quote]

       bug report????
idk

[quote="BOGTRO"]If you discover a bug (error) and send it to them, you get karma.[/quote]

You would only get karma if it is a legitimate report about a real problem; this is to prevent spamming of reports.


combinatorial geometry
advanced fields
advanced fields unsolved

In the following, we denote by $ \text{conv}\left(p_1,p_2,...,p_n\right)$ the convex hull of $ n$ given points $ p_1$, $ p_2$, ..., $ p_n$.

[b](a)[/b] Let $ a$, $ b$, $ c$ be three non-collinear points in the plane, and let the set $ \text{conv}\left(a,b,c\right)$ be covered by three open sets $ A$, $ B$, $ C$ such that $ \text{conv}\left(b,c\right)\subseteq A$, $ \text{conv}\left(c,a\right)\subseteq B$, $ \text{conv}\left(a,b\right)\subseteq C$. Prove that $ A\cap B\cap C\neq\emptyset$.

[b](b)[/b] Let $ a_1$, $ a_2$, ..., $ a_{n \plus{} 1}$ be $ n \plus{} 1$ points in $ \mathbb{R}^n$ which form a non-degenerate $ n$-simplex (this is equivalent to the $ n$ differences $ a_1 \minus{} a_2$, $ a_1 \minus{} a_3$, ..., $ a_1 \minus{} a_n$, $ a_1 \minus{} a_{n \plus{} 1}$ being linearly independent). Assume that the set $ \text{conv}\left(a_1,a_2,...,a_{n \plus{} 1}\right)$ is covered by $ n \plus{} 1$ open sets $ A_1$, $ A_2$, ..., $ A_{n \plus{} 1}$ such that $ \text{conv}\left(a_1,a_2,...,a_{i \minus{} 1},a_{i \plus{} 1},...,a_{n \plus{} 1}\right)\subseteq A_i$ for every $ i\in\left\{1,2,...,n \plus{} 1\right\}$. Prove that $ A_1\cap A_2\cap ...\cap A_{n \plus{} 1}\neq\emptyset$.

[b](c)[/b] Derive [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105036]the Brouwer fixed point theorem[/url] from problem [b](b)[/b].

I have edited the above because I noticed that problem [b](b)[/b] follows from the well-known [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=72491]KKM Theorem[/url] (just set $ F_i \equal{} \text{conv}\left(a_1,a_2,...,a_{n \plus{} 1}\right)\setminus A_i$), what made my post pretty much useless. I have thus added part [b](c)[/b], which seems not to have been discussed here yet.

If I remember correctly, [b](a)[/b] was the actual question given to Christian in the 3rd round of the BWM (the west-German mathematical olympiad; note that the 3rd round is a colloquium with a professor to find out whether one is really good at mathematics, so one should expect questions of any possible kind there), and [b](b)[/b] was his generalization. Christian Sattler and me both were pretty proud of our proofs, but it turned out that these proofs are exactly the two ones mentioned by Grobber in http://www.mathlinks.ro/Forum/viewtopic.php?t=72491 , modulo some slight differences in my proof.

Note that there seems to be a similar fact (also proven using Brouwer) called [url=http://www.springerlink.com/content/p411228841781u81/]Scarf's theorem[/url].

  darij


In a sphere of radius $r$, a "truncated" cone (i think this is how it is called in english) of height $r$ is inscribed. The trnc cone has $V_{cone}= \frac{1}{2} \cdot V_{sphere}$.
Find out where it is placed, i.e. what are the distances from its bases to the center of the sphere.

First I want to find out if this is possible. It should be, if X is the largest [i]cylinder[/i] that can fit in this sphere, if V(X) > 1/2V(sphere) then a truncated cone should work.

So the largest such cylinder would have a height equal to its diameter. Let r be the radius of the sphere.
V(sphere) = $ \frac{4\pi r^3}{3} $
V(cylinder) = $ \sqrt{2}r \vdots \frac{\pi r^2}{2} $ = $ \frac{\sqrt{2}\pi r^3}{2} $
The cylinder to sphere ratio is
$ \frac{\sqrt{2}}{2} : \frac{4}{3} $
$ 3\sqrt{2} : 8 $

Which is more than 1/2, so this problem should be a good one. But I'm still not sure there is a unique solution...


To any people who have gone to either of these camps, approximately how long does it take to be notified if u got in or not? I applied to PROMYS and my friend applied to SUMAC, and we are both interested in finding out (especially me, because if i dont get in, then im going to send off an app to ROSS).

I think I got my acceptance letter from SUMaC in early May.


factorial function integration calculus trigonometry logarithms Asymptote

I think that the answer to this question involves the gamma function so I posted it here.
How do you calculate the factorial of a real number that is not an integer?  Can the same be done for complex numbers?

You can compute the "factorial" of any complex number that isn't a negative integer. The accepted conventional defintion of the Gamma function is offset by one from the factorial, so that $x!=\Gamma(x+1).$

If $\text{Real}(z)>0$ we define $\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}dt.$

The reason for the restriction to the right hand half plane is so that the improper integral will converge at the $t\to0^+$ end.

We have an extremely important identity: if $\text{Real}(z)>0,$ then $\Gamma(z+1)=z\Gamma(z).$ This can be proved by using the obvious integration by parts. This identity can now be used to define $\Gamma(z)$ on the rest of the complex plane. That is, since

$\Gamma\left(\frac12\right)=\sqrt{\pi},$ then $\Gamma\left(-\frac12\right)= \frac{\Gamma\left(\frac12\right)}{-\frac12}=-2\sqrt{\pi}.$

Similarly, $\Gamma\left(-\frac32\right)=\frac43\sqrt{\pi}, \Gamma\left(-\frac52\right)=-\frac8{15}\sqrt{\pi},$ and so on.

However, as $z\to0, \Gamma(z)=\frac1z+$ a bounded function. That is, the Gamma function has a simple pole of residue 1 at the origin. This in turn propagates through the recursion formula: Gamma has simple poles at all the nonpositive integers, and the residue at $-n$ is $\frac{(-1)^n}{n!}.$

For example,

$i!=\Gamma(1+i)=\int_0^{\infty}t^ie^{-t}dt =$

$\int_0^{\infty}\cos(\ln t)e^{-t}dt +i\int_0^{\infty}\sin(\ln t)e^{-t}dt.$

If you want to see numerical values for something like this, or graphs, I would suggest a symbolic-manipulation package like Mathematica or Maple.

Thanks for the explanation, it really helped.  I was surprised that my calculator knew $(-1)!$ was undefined.  This made me wonder, is the accepted value of $\frac{1}{0}$ undefined?  My teacher told me that it was $\infty$, but if you look at the graph of $y=\frac{1}{x}$, then it could be $\pm \infty$.  The graph of $y=\frac{1}{x^2}$ however, shows that $\frac{1}{0^2}=\infty$.  My calculator (TI-89) says $\frac{1}{0}=undef$ and $\frac{1}{0^2}=\infty$.  Can you clarify this since $0=0^2$?

Your calculator is making a distinction between $+0$ and $-0$. $0^2=+0$, but it doesn't know which one you meant if you just put in $0$.

A calculator's number system is not one of the standard ones- there are a lot of funny things at the edges.

The 1/0 stuff is a touchy subject.  Your calculator and teacher are both talking about limits.  I'm just going to presume by your presence in this forum that you know what a limit is  :).

For 1/x, the limit as x comes from the right and goes to 0 is +inf, while when it comes from the left and goes to zero, is -inf.  Therefore, the limit is undefined.  For 1/x^2, the limit from both directions is +inf, which is what your calculator is telling you.

However, note that this does not at all mean that 1/0^2=inf, 1/0=inf, or anything of the sort.  The value of a function is not in the least determined by the values of the function near it.  That's precisely what a limit is (especially from a topological viewpoint).  The value of 1/2 has nothing to do with the values of 1/(1.9) or of 1/(2.1), and you would never use those values to calculate it.  Likewise, 1/0 has nothing to do with 1/.1 or 1/-.1.

Thanks, you both verified what I thought about the calculator.
So, just to be sure, $\frac{1}{0}$ and $\frac{1}{0^2}$ is undefined or not a number but NOT $\infty$ right?

Firstly I want to say that infinity is no real no.Its included in the extended number system to provide a more definitive limit notion.
The topic about 1/0 is extremely touchy.If you want the tricky limits, its useful to graph the function(not always its easy!).You will get a clear idea about the function,its limits,asymptotes etc..

Good Luck.......


[quote="hien"]Let $a>2$ be a real number and let $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$. Prove that the equation $f_{n}(x)=a$ always has unique root, for all $n\in\mathbb{N}$, denote by $x_{n}$ and do prove sequence $x_{n}$ has a finite limit when $n$ run at infinity[/quote]

And http://www.mathlinks.ro/Forum/viewtopic.php?p=750368#750368 (problem 7) are different  :maybe:

Yes, this is my mistake typing. I have edited:

[b]Problem 7[/b]
Let $a>2$ be a real number and let $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$. Prove that the equation $f_{n}(x)=a$ always has unique root in $(0,+\infty)$, for all $n\in\mathbb{Z}^{+}$, denote by $x_{n}$ and do prove sequence $x_{n}$ has a finite limit when $n$ run at infinity

so if, this is my solution
[hide]Fix $n$, we have $f_{n}(x)$ is continuous on $[0,\infty)$ , $f_{n}(0)=1<a,f_{n}(1)>a$ and $f$ is increasing, therefore first part of problem solved.

Now, for second part, we have $\frac{1}{a}<x_{n}<q$ for $n$ sufficient large, where $q$ satisfy $\frac{q}{q-1}>a,a^{10}q^{9}-a^{10}q^{10}-1>0$ and $f_{n}(x_{n+1})=x_{n+1}^{n+1}(a^{10}x_{n+1}^{9}-1-a^{10}x_{n+1}^{10})+a>a$ for $n$ sufficient large. Therefore $x_{n+1}>x_{n}$ for $n$ sufficient large, and we have done :D[/hide]


[quote="math92"]A soda manufacturer decided to produce one-liter glass bottles of soda.  The retail price of the bottle with the soda is $1.10.  If the bottle costs $1.00 more than the soda, how much does the soda cost without the glass bottle?[/quote]

[hide]  soda=x

(100 + x) + x = 110

100 + 2x= 110

2x= 10

x=5

The soda costs [b] 5 cents [/b] [/hide]

[quote="math92"]A soda manufacturer decided to produce one-liter glass bottles of soda.  The retail price of the bottle with the soda is $1.10$.  If the bottle costs $1.00$ more than the soda, how much does the soda cost without the glass bottle?[/quote]

[hide="And the answer is..."]Let $b=$ $\textrm{price of bottle}$ $\textrm{and}$ $s=$ $\textrm{price of soda}$:

$b=1+s$

$b+s=1.1$

$b+s=1.1$

$1+2s=1.1$

$2s=0.1$

$s=0.05$

[b]5 cents[/b][/hide]

[quote="math92"]A soda manufacturer decided to produce one-liter glass bottles of soda.  The retail price of the bottle with the soda is $1.10. If the bottle costs$1.00 more than the soda, how much does the soda cost without the glass bottle?[/quote]

[hide]
the answer is....
[hide="clck ths"]
$\boxed{\boxed{5\text{ cents}}}$
[/hide][/hide]


geometry
inequalities
trigonometry
function
inequalities proposed

For $m_a,m_b,m_c$ medians of a triangle of area $S$ prove

$\sum \frac{1}{m_bm_c} \leq \frac{\sqrt{3}}{S}$

Using the theorem that the medians $m_a$, $m_b$, $m_c$ form a new triangle $D_m$ with area $\displaystyle S_m = \frac34 S$, it is clear that we can apply the inequality

$\displaystyle \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\leq \frac{3\sqrt{3}}{4S}$

to the triangle $D_m$ and get

$\displaystyle \frac{1}{m_{b}m_{c}}+\frac{1}{m_{c}m_{a}}+\frac{1}{m_{a}m_{b}}\leq \frac{3\sqrt{3}}{4S_{m}}=\frac{3\sqrt{3}}{4\cdot \frac{3}{4}S}=\frac{\sqrt{3}}{S}$.

So it remains to prove the above inequality

$\displaystyle \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\leq \frac{3\sqrt{3}}{4S}$.

In fact, multiplying it with 2S, you get

$\displaystyle \frac{2S}{bc}+\frac{2S}{ca}+\frac{2S}{ab}\leq \frac{3\sqrt{3}}{2}$.

But since $S=\frac{1}{2}bc\sin A$, we have $\displaystyle \frac{2S}{bc}=\sin A$, and similarly $\displaystyle \frac{2S}{ca}=\sin B$ and $\displaystyle \frac{2S}{ab}=\sin C$, hence it remains to show that $\displaystyle \sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}$. But this is clear, since the function sin x is concave on the interval $\left[ 0;\;\pi \right] $, and thus Jensen yields

$\displaystyle \sin A+\sin B+\sin C\leq 3\cdot \sin \frac{A+B+C}{3}=3\cdot \sin \frac{\pi }{3}=3\cdot \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$.

Qed.

  Darij

Very weak inequality! Almost any approach works!


Let ABC be a triangle inscribed in a circle of radius R

and a, b, c be respectly sides of the triangle ABC.

Prove that:

   $\frac{R}{r}\geq\frac{b}{c}+\frac{c}{b}$

Pd.  r: inradio of the triangle ABC.

[color=red][Moderator edit: Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=15807 and at http://www.mathlinks.ro/Forum/viewtopic.php?t=49433 .][/color]

Well pretty nice probem.Here is the solution:
Given inequality is equivalent to the followwing:
[tex]b^{2}+c^{2}\leq\frac{2a}{(a+b-c)(b+c-a)(c+a-b)}\cdot{b^{2}c^{2}} (1)[/tex]
I got it using the square formulas:[tex][ABC]=\frac{abc}{4R},[ABC]=\frac{a+b+c}{2}\cdot{r}[/tex] and [tex][ABC]=\sqrt{p(p-a)(p-b)(p-c)},[/tex]where [tex]p[/tex] is semiperimeter of the triangle [tex]ABC.[/tex]
Next step is to notice that inequality [tex](1)[/tex] is equivalent to
[tex](\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1)(\frac{2ac^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1)\geq{1}[/tex]
And at last we write
[tex]\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1\geq\frac{2a}{c+a-b}-1=\frac{a+b-c}{c+a-b}>0[/tex]
and
[tex]\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1\geq\frac{2a}{a+b-c}-1=\frac{c+a-b}{a+b-c}>0
[/tex]
It remains only to multiple last two inequalities.And equality holds if and only if [tex]a=b=c.[/tex]

Here is a simple method:

Put $x=s-a,y=s-b,z=s-c$ where $s$ is the semiperimeter of the triangle.So it's easy to see that :
\[\frac{R}{r}=\frac{(x+y)(y+z)(z+x)}{4xyz}\] and
\[\frac{b}{c}+\frac{c}{b}=\frac{x+z}{x+y}+\frac{x+y}{x+z}\]
So we have to prove
\[\frac{y+z}{4xyz}\ge \frac{1}{(x+y)^2}+\frac{1}{(x+z)^2}\]
for $x,y,z>0$ . But in fact this follows from that:
   \[\frac{1}{(u+v)^2}\le \frac{1}{4uv}\]for $u,v>0$
This complets the proof.


$ dy/dx \equal{} y/x \plus{} 2x \plus{} 1$

I solved this using integrating factor but am wondering if there's any other way to solve it.

hello, by setting $ u = \frac{y}{x}$ we have $ y = ux$ and by differentiating with respact to $ x$ we got
$ y^' = u^'x + u$, inserting this in our equation we have $ u^' = \frac {2x + 1}{x}$ and from here
$ du = \frac {2x + 1}{x}dx$.
Sonnhard.


Let $ T$ be the set of all $ n$-tuples in the form $ (1,\pm1,\pm1,\dots,\pm1)$.  Let $ x : \equal{} \left(x_1,x_2,x_3,\dots,x_n\right)$.  Define
\[ P(x) : \equal{} \sum\limits_{e \in T} u_e(x\cdot{e})^n\,,
\]
where $ u_e : \equal{} e_1e_2e_3\cdots{e_n}$.  For each $ n$-tuple $ r : \equal{} \left(r_1,r_2,r_3,\dots,r_n\right)$ with $ r_1 \plus{} r_2 \plus{} r_3 \plus{} \cdots \plus{} r_n \equal{} n$, the coefficient of $ x_1^{r_1}x_2^{r_2}x_3^{r_3}\cdots{x_n^{r_n}}$ in  $ u_e(x\cdot{e})^n$ is
\[ c_r(e) : \equal{} \frac {n!}{r_1!r_2!r_3!\cdots{r_n!}}e_1^{r_1 \plus{} 1}e_2^{r_2 \plus{} 1}e_3^{r_3 \plus{} 1}\cdots{e_n^{r_n \plus{} 1}}\,.
\]
If $ r_k \equal{} 0$ for some $ k$, then $ c_r(e) \propto e_k$ and hence,
\[ \sum\limits_{e\in{T}} c_r(e) \propto \sum\limits_{e_k \equal{} \pm1} e_k \equal{} 0\,.
\]
Therefore, the only remaining term is $ x_1x_2x_3\cdots{x_n}$, which has coefficient
\[ \sum\limits_{e\in{T}}\frac {n!}{(1!)^n}e_1^{2}e_2^{2}e_3^{2}\cdots{e_n^{2}} \equal{} \sum\limits_{e\in{T}}n! \equal{} n!|T| \equal{} n!2^{n \minus{} 1}\,.
\]
Therefore, $ P(x) \equal{} n!2^{n \minus{} 1}x_1x_2x_3\cdots{x_n}$.

[quote="kl2836"]Prove that
$ (a \plus{} b \plus{} c \plus{} d)^4 \plus{} (a \plus{} b \minus{} c \minus{} d)^4 \plus{} (a \minus{} b \plus{} c \minus{} d)^4 \plus{} (a \minus{} b \minus{} c \plus{} d)^4 \minus{} (a \plus{} b \plus{} c \minus{} d)^4 \minus{} (a \plus{} b \minus{} c \plus{} d)^4 \minus{} (a \minus{} b \plus{} c \plus{} d)^4 \minus{} ( \minus{} a \plus{} b \plus{} c \plus{} d)^4 \equal{} 192abcd$[/quote]
I have a ideal Let $ LHS\equal{}P(a,b,c,d)$ try with$ a\equal{}0,b\equal{}0,c\equal{}0.d\equal{}0$ hence $ P(a,b,c,d)\equal{}@abcd$ try some values of $ a,b,c,d$ to find $ @\equal{}192$


[quote="i_like_pie"]Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.[/quote]

23 minutes

[quote="i_like_pie"]Does anybody have a solution? And stop rating posts as spam whoever it is.[/quote]

I can try...
[hide]So we have the following information

Alex paints
Red: $ 5 \text{ ft}^{2}/\text{minute}$
Blue: $ 2.5 \text{ ft}^{2}/\text{minute}$

Bob paints
Red: $ 6 \text{ ft}^{2}/\text{minute}$
Blue: $ 3 \text{ ft}^{2}/\text{minute}$

So, we see that together, they paint
Red: $ 11 \text{ ft}^{2}/\text{minute}$
Blue: $ 5.5 \text{ ft}^{2}/\text{minute}$

So, we need $ 125 \text{ ft}^{2}$ red and blue paint. Calculating that, we get
$ \frac{125}{11}+\frac{125}{5.5}\approx 34\text{ minutes}$[/hide]

umm i get 23 minutes too..its actually a little over 23..but you said round..

just scale everything down to 1 minute.

[b]vishalarul, your solution does not take into account once someones already done with their part of the job, they can help with the other person, making the other persons job quicker[/b]

for example,  alex doenst have to paint both blue and red at the same time.  alex can paint all red, and bob all blue (sepcializing on their strengths) to make it quicker.

i_like_pie, it should be whats the minimum time required i assume..

How is this classic?  and theres no twist..
I get $ \frac{375}{16}$ minutes or $ 23.4375$ minutes
You always want to maximize on a strength for both people and then finish off with the weakness for them together (making it stronger since theres 2 people)

The "classic" part is that 2 people paint a wall at the same time. The twist is with the different types of paint.

@ vishal - Each person paints the side that they are fastest at, and whoever finishes first helps the other finish faster. 23 minutes is the correct answer.

[quote="abacadaea"][quote="i_like_pie"]Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.[/quote]

[hide="fast solution"]theres a fast way to do these work problems. i think its $ \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}$. where $ R_{1}$ refers to Alexs minutes per squ ft and $ R_{2}$ refers to Bob's minutes per squ ft. alex paints a squ ft in $ \frac{1}{5}$ minutes with red paint and bob does 1 squ ft in $ \frac{1}{3}$ minutes. pluging the numbers in gets us. $ \frac{1}{8}$ minutes per squ ft if they work together. since half the wall must be red and the other half blue, they must paint 125 squ ft of red wall, so it will take them $ \frac{125}{8}$ minutes to do the red side of the wall. the blue side of the wall is the same ting only different numbers: $ \frac{2}{5}$ and $ \frac{1}{6}$. plugging the numbers in gets us a rate of 1 squ ft every $ \frac{2}{17}$ minutes. then multiply it by 125, you get $ \frac{250}{17}$. add the two together gets us apx. $ 30.33$ minutes.[/hide]

EDIT: after reading the other posts, i think i_like_pie worded the problem badly because you asked "how fast CAN they paint the wall." i think you meant "what is the fastest speed in which they can paint the wall". i just did it the normal way i would do these problems so i didnt get it.[/quote]

That works only if you are given two rates. This problem is different, and that method doesn't work.

That method works perfectly fine on something like this:

Bob drives at 40 mph to work, and drives at 60 mph back home. What is his average speed for the round trip?

Then you can use the harmonic mean to get 48 mph.

I misread the problem at first. I thought that they both specialized in Red. :oops:

Actually, that formula isn't the harmonic mean formula, it works when you are given the time, not rate.


geometry
ratio
AMC
AIME
similar triangles

Given a triangle ABC, mark point P on AB such that AP:PB = 3:4, and mark point Q on AC such that AQ:QC = 3:2. Join and extend PQ to meet BC at Z. Prove that BC = CZ.

Well, a quick and straightforward solution - not particularly beatiful - is to use vectors. I'll think of a purely geometric solution but it seems difficult to find such a solution.

Why is this not just a direct application of Menelaus' Theorem?

Most people don't know about Menelaus' theorem, certainly not a grade 12 Algeo class. ;)

This problem took me about 30 minutes during the said class, which I'm ashamed about.

Can you tell me the Menelaus theorem? It sounds like a good one.

Can't explain it better than Mathworld can:

[url]http://mathworld.wolfram.com/MenelausTheorem.html[/url]

However, this question has a three-line solution that doesn't need any fancy mathematics.  8-)

Menelaos theorem is the dual of Ceva's theorem. In a triangle with vertices A, B, and C and points X,Y,Z on the sides AB,BC,CA respectively, X,Y,Z are collinear iff AX/BX*BY/CY*CZ/AZ=1

Random comment on Menelaus and Ceva: they are both nice theorems because they're useful and easy to remember.  In the same way that we get 16/64 = 1/4 by cancelling 6's, if you cancel letters in Menelaus and Ceva, you get 1 = 1.  They are simple, powerful theorems for proving things about ratios and concurrency and colinearity.

Can someone post a proof of Menelaus?  I don't remember how it goes.

Wow. Thats a cool thing... Do lots of contest problems use Menalaus' Theorem?

*poke* does anyone have a proof to this question yet? Once again I remind you, it's extremely short and does not involve Menelaus' Theorem (or anything of the sort). ;)

i dont know if this is the simple geo soln ur talking about osiris, but heres one with mass points:

[hide] we know that AQ/QC=3/2, so if there is a weight of x at A and y at C, then AQ/QC=3/2=y/x. Also, if there is a weight of z at B, then AP/PB=3/4=z/x. It would be convenient if the weights were all integers, so we can see that x=4, y=6, and z=3. Then, by definition,

3B+nZ=6C (where the number is the weight, the letter is the point). hence n=3, and BC/CZ=3/3=1, so done.[/hide]

Not quite. ^^" My solution involves only parallel lines, similar triangles and proportions of lengths.

Eh, since noone seems to have come up with anything, I'll post my solution:

[hide]Construct CD // PZ, meeting AB at D. Then AP : PD : DB = 3 : 2 : 2. Then PD = DB, therefore BC = CZ.[/hide]

wow. how is it that you gained your geometric intuition, like where and when to add lines and stuff? im horrible at geometry and would really like to know.

i think that can be considered part of one of the proof of menelaos theorem if you take out the numbers. ill post the proof eventually.

I thought of a good way to prove it... If it is 3x:4x, take the 4x and take the midpoint, 2x and call it M.  Then extend it to the point C.  We can say that this line is parallel to PZ because they have the same ratios of 3:2 on both sides.  Triangle BMC and triangle BPZ are similar and they are in the ratio of 2 to 1.  QED! :P

Arrrg...  I just realized that Osiris posted the same solution.

In answer to the bit about geometric intuition - the truly talented geometers do indeed appear to be magicians, but you can get there yourself through practice and constantly asking the question 'How did you think to do that?'  The answer, you'll find, is often 'I've seen it before', which isn't too satisfying, so just ask again, but ask 'How would you have thought of it the first time?'

In this problem, the reason we think 'draw in a parallel' is that this is a problem involving ratios.  Ratios mean similar triangles.  We get similar triangles from parallel lines.  So we just visualize a few parallel lines we could add - the one that solves the problem is easily one of the early candidates to try.

So, now you have another couple tools for your toolbox.  Doesn't always work, but they're tools good geometers reach for frequently.  Specifically:  Ratios -> look for similar triangles.  Need similar triangles -> draw a line through a point in the diagram parallel to some other line.

well the first thing that comes to mind is menelaus... simply because the diagram is all so familiar...

but the second thing that comes to mind is a midline... specifically the midline of BPZ parallel to PZ. If u show it hits C ur finished...

so lets just take the line parallel to PZ through C, since if C is the midpoint of BZ this will be the midline...

SO how do we SHOW this is the midline... well then BD:DP = 2:2... how can we show this?
Well since AQ:QC = 3:2 and AP:PB = 3:4, then PD: DB = 2:2... well look at that we're finished :)


modular arithmetic
number theory proposed
number theory

Prove that the following equation has no solution in integer numbers: \[ x^2 + 4 = y^5.  \] [i]Bulgaria[/i]

We shall prove that $x$ is odd. Indeed, suppose that $x$ is even. Then $y$ is also even, so it's divisible by $16$. It means that $x^2 \equiv 12 \mod{16}$ but that's impossible (it can be checked by hand very easily).
Now, we see that the equation is equivalent to the following:
$(x+2i)(x-2i) = y^5$
Now let $d=(x+2i, x-2i)$
It follows that $d|4i$
But because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers.
$x+2i = (a+bi)^5$ and
$x-2i = (a-bi)^5$
(this is because $x+2i$ and $x-2i$ are conjugates)
Now we substract the second equation from the first and, after doing some boring, bot fortunately not too long, computations, we obtain a equation which has no solutions (after comparing the real and imaginary parts or so...).

I think looking mod 11 works.  We have $y^{10} \equiv 0, 1 \pmod{11}$; thus $y^5 \equiv -1, 0, 1 \pmod{11}$.  Also, $x^2 \equiv 0, 1, 3, 4, 5, 9 \pmod{11}$; thus $x^2 + 4 \equiv 2, 4, 5, 7, 8, 9 \pmod{11}$.  Hence $x^2 + 4$ and $y^{10}$ are different mod 11.

Looking $\mod 16$ gives that $x,y$ are odd (as above) and $y \equiv 1 \mod 4$.
Now the equation is equivalent to $x^2+6^2 = y^5+2^5 = (y+2)(y^4-2y^3+4y^2-8y+16)$, and because of $\gcd(y+2, y^4-2y^3+4y^2-8y+16)|16$ we get that the $\gcd$ is $1$. Now $y+2 \equiv 3 \mod 4$, so there is some prime divisor $q$ of $y+2$ that divides $y$ an odd number of times, then also dividing $y^5+2^5$ an odd number of times, contradicting the fact that this number is sum of two squares.

To show that $m=a^2+b^2$ can't be divided by $p^{2s-1}$ and not by $p^{2s}$ for $p \equiv 3 \mod 4$, just use inductive reasoning:
Let $m$ be the smallest number for that we haven't proved this. Then let $p$ be as above and look $\mod p$: $a^2+b^2 \equiv 0 \mod p$.

1. Case: $b \equiv 0 \mod p$.
This gives that $p|a,b$, thus $m=a^2+b^2=p^2 a'^2 + p^2 b'^2 = p^2 (a'^2+b'^2)$ for some integers $a',b'$, giving that the number $m'=\frac{m}{p^2}=a'^2+b'^2$ is sum of two squares, contradicting the fact that we have shown this as impossible for all smaller $m'$ (note that $p^{2s-3}$ but not $p^{2s-2}$ divides $m'$).

2. Case: $b \not \equiv 0 \mod p$.
Then multiplication with the inverse $b^{-1} \mod p$ of $b$ gives for $c=ab^{-1}$ that $c^2 \equiv a^2 (b^2)^{-1} \equiv -1 \mod p$. But that is impossible since it would lead to $1 \equiv c^{p-1} \equiv (c^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \equiv -1 \mod p$, clearly impossible.

Thus both cases lead to contradictions, thus done.

[quote="TomciO"]
But because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers.
$x+2i = (a+bi)^5$ and
$x-2i = (a-bi)^5$
(this is because $x+2i$ and $x-2i$ are conjugates)
[/quote]

Can you explain why
$x+2i = (a+bi)^5$ and
$x-2i = (a-bi)^5$ but not
$x+2i=\epsilon (a+bi)^5$ and
$x-2i=\overline{\epsilon} (a-bi)^5$, where $N(\epsilon)=1$? You consider the case when $\epsilon=1$ only.

$i = i^5$ :). So for example: $i(a+bi)^5$=$i^5(a+bi)^5=(ai-b^2)^5$
So there's no problem with writing like that (for odd powers of course).

mod 11 :P (standard for integer variables in 5-th power)

This problem is too easy for this competition.The idea is using the little theorm of Fermat and we have y^10=1(mod11)=>y^5=-1,0,1(mod11) and x^2+4=2,4,5,7,8,9(mod11) and we discover a contradiction.

[quote=sttsmet]Clearly something is wrong with this problem.
Taking $ \mod 11$ makes this problem a JBMO P2 rather than a BMO P4.[/quote]

This is a problem from the 1990s.

[quote=sttsmet]@above I don't see your point.[/quote]

Older problems tend to be easier

I agree, I just say it should be a bit more complicated... here only a single idea was needed.


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=72170]example[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69618]doesn't work here[/url]

Why is it that the feature when the mouse icon over latex text becomes a hand sometimes, allowing copying to the clipboard, and sometimes it doesn't

I really don't see a pattern, it would be nice if you could always do that

This has got to be an issue with your browser, since it always works fine for me (and in fact LaTeX is rendered the same way -- as a gif file -- everywhere on the forum).

I can confirm the hand/no hand difference on two different browsers and systems. On the other hand, I still get a menu appropriate to an image when right-clicking.

If my memory doesn't cheat me, the LaTeX image is clickable if and only if the code is over a certain fixed length (so very short codes, like $1+1=2$, are not clickable, while ones like $\sum_{i=1}^{\infty}\frac{1}{2^i}+\sum_{i=1}^{\infty}\frac{1}{2^i}=\sum_{i=0}^{\infty}\frac{1}{2^i}$ are). If you are really desperate to get the code of a short LaTeX string, you can right-click it, go to "Properties", and you see the code as "alternative text" (well, this is how you do it with Firefox). Or you can quote the message, if you are logged in.

[b]EDIT:[/b] Of course, Jmerry is right - simply selecting the text and copying works, too. Of course, provided one has a good browser (FF works, IE doesn't - who would expect -, I haven't checked others).

  Darij

i use internet explorer a lot and rarely opera

i am just trying to see the pattern

so length is the criterium?

i don't see the logic in that, when you need a specific thing in latex it usually pops up in a short string?

but it is true, quoting always works

For me, selecting and copying has always worked regardless of length- it cuts out the dollar signs, but you get the text.

copying?  from the quoted section you mean?

it's just that i myself need to learn some more latex do be a serious member here, and i wanna help a friend who considers joining and learning latex (well he is a member but has zero posts for now)

The point is that each clickable formula is in fact a piece of javascript code. If the formula is less than a certain length limit, there is no need for the code as it is surely well known (so no need to generate further traffic).

You can always however over a formula to see how it is written.


Find the numerical value of
\[ \frac{(\sin 18)(\cos 12) + (\cos 162)(\cos 102)}{(\sin 22)(\cos 8) + (\cos 158)(\cos 98)}, \] where all angles are measured in degrees.

[size=75][color=green]a few modifications by mod[/color][/size]

just use the cofunction identities, and angle addition formulas

[hide][quote="t0rajir0u"][hide]
Recall the identities $\cos (180 - \alpha) = - \cos (\alpha), \cos (90 + \alpha) = - \sin (\alpha)$.  From these we can rewrite the top as $\sin 18 \cos 12 + \cos 18 \sin 12 = \sin 30$ and the bottom as $\sin 22 \cos 8 + \cos 22 \sin 8 = \sin 30$.  Hence the value is $\boxed{1}$.
[/hide][/quote]

Nice solution. I got it right through deciding that the number were ugly, adding or subtracting 2 from each angle, and saying $\frac{n}{n} = 1$  :P [/hide]


If the altitude of an equilateral triangle is $ \sqrt {6}$, then the area is:

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

[hide="Solution"]The altitude cuts it into two 30-60-90 triangles, so one side of the triangle must be $ 2\cdot \frac{\sqrt{6}}{\sqrt{3}}\equal{}2\sqrt{2}$.
Therefore, the area is $ \sqrt{2}\cdot\sqrt{6}\equal{}\boxed{\textbf{(B)}\ 2\sqrt{3}}$.[/hide]


Marian wishes to buy a new computer that will cost her $ \$300$. She receives $ \$5$ per hour for watching her younger brother and $ \$4$ per hour for helping her mother with chores. Each week she watches her brother for four hours and helps her mother with chores for 10 hours. How many full weeks must she work to earn enough money to buy the computer?

Each week she earns $ 5 \times 4 \equal{} 20$ dollars for watching her brother and $ 4 \times 10 \equal{} 40$ dollars for helping her mother, making a total of 60 dollars a week. Thus it will take her $ 300 \div 60 \equal{} 5$ full weeks.


I've been including images in my $ \text{\LaTeX} $ documents by converting those in [i]JPEG[/i] to [b][i]EPS[/i][/b], and then inserting the appropriate code. This website recommends converting them to [b][i]PDF[/i][/b] instead. Does this approach offer any advantage? It seems to me I would not need to use the [i]epstool[/i] program to adjust the [i]BoundingBox[/i]. Anything else?

If your final output is going to be PDF I suggest you plainly use .PNG images (any image editor can convert images to .png) which you can easily insert and compile with pdflatex (won't give you a .dvi output, but it'll be a perfect .pdf).

Otherwise, I'd go with the EPS, as it's easier to work with when you are also creating PS files.

With \includegraphics*, you can include jpg files directly.

Thank you guys for the suggestions.

Using [i]\includegraphics*[/i] I won't need to bother about converting my files to [i]EPS[/i], using [i]epstool[/i], and later having specify the size of picture within the regular [i]\includegraphics[/i] command.

Inserting images in [i]PDF[/i] format through [i].PNG[/i] images and the [i]pdflatex[/i] package also sounds good.


Na apodeiksete oti ean $n\not\equiv 0 \pmod{6}$ tote

$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\equiv 0 \pmod{7}$

Alexandros

Y.G. Na ipenthimisw (gia tous endiaferomenous) oti exei meinei aliti i deuteri askisi apo ton Arximidi tou 94 pou eixa kanei post.

Mpravo Stelio! Poli wraia proseggisi parolo pou egw exw 2 alles sto mialo mou!

Tin askisi auti tin evala gia ena didaktiko skopo ton opoio tha anaferw se epomeno post giati exw ligi doulitsa simera! Tha mporeite meta apo auto na antimetwpisete polles askiseis tetoiou eidous 'h akomi kai diskoloteres tou tipou "Na vreite ola ta dinata ipoloipa tis diairesis tou $2\cdot 3^{n}+3\cdot 7^{2n+1}+5^{3n+1}-7$ me to $11$" (Prosoxi! Den dinw poio einai to ipoloipo tis diairesis gia na ginei me epagwgi (pragma eukolo), alla zitw gia tis diafores times tou $n$, ola ta dinata ipoloipa). Episis tha mporeite na kataskeuasete tis dikes sas askiseis (opws auti pou parethesa egw parapanw) kai na kanete tis dikes sas paratiriseis.

Episis h askisi auti einai enas kalos tropos (3os tropos ektos apo ekeinon tou Steliou kai ekeinon pou tha anaferw se allo post) gia na thimithoume ena sigekrimeno sxima tis epagwgis me to opoio, se kathe epagwgiko vima den diwxneis mono mia periptwsi alla parapanw apo mia! [Xaraktiristiko einai to paradeigma tou provlimatos "Na apodeixthei oti gia opoiondipote fisiko arithmo $n$ megalitero tou 6 isxiei to eksis: Kathe tetragwno mporei na diameristhei, me ti voitheia eutheiwn parallilwn pros tis pleures tou, se $n$ to plithos tetragwna (anisa en genei metaksi tous)"].

O logos omws gia ton opoio stelnw auto to post einai allos: Paratirisa me vasi tin parapanw askisi pou evala kai me ti voitheia tou ipologisti oti isws i parapanw protasi na isxiei kai genika. Den exw omws kathisei na tin apodeiksw 'h na tin aporripsw. Den kserw an episis anaferetai ws askisi kapou.

[size=125]Ean $p$ prwtos $>2$ kai $n$ fisikos me $n\not\equiv 0 \pmod{p-1}$
tote $1^{n}+2^{n}+\ldots+(p-1)^{n}\equiv 0 \pmod{p}$[/size]

Ti lete? Elkistikotato etsi?
Alexandros

πραγματι ελκυστικοτατη ασκηση.μια ερωτηση μηπως η συνθηκη για το γενικο προβλημα που θετεις αλεξανδρε ειναι οτι ο p δεν διαρει το 2^n-1? και κατι αλλο .για την λυση μηπως πρεπει να ξεκινησω με ενα πληρες συστυμα υπολοιπων? :)

Kwsta exw elenksei to provlima gia olous tous prwtous tous mikroterous 'h isous tou $29$ kai den exei parousiastei kapoio provlima! Gi'auto pisteuw oti isxiei genika etsi opws to edwsa!

Den xreiazetai na dwsw tin sinthiki $2^{n}\not\equiv 1 \pmod{p}$ pou anafereis Kwsta. Opws to edwsa me $n\not\equiv 0 \pmod{(p-1)}$ eimaste kalimenoi. Exeis kapoio antiparadeigma mipws?

Oso gia to pws mporei na lithei agnwsto kai se mena pros to paron... Isws na apotelei kali skepsi na pareis ena plires sistima ipoloipwn.

Alexandros

Kwsta molis twra teleiwsa tin dokimi me tous prwtous tou Mersenne pou anafereis mexri ton $2^{13}-1$ (diladi tous $5$ prwtous) kai pali isxiei. Exw ipopsi mou tin idia protasi pou anafereis kai apaitei to $p\not|2^{n}-1$ (kai tin opoia ap'oti thimamai pragmati douleueis me plires sistima ipoloipwn) alla fainetai na isxiei genika...

Alexandros

μαλλον εχεις δικιο

Καλη....
Αν ν ειναι περιττος,τοτε $1^{n}+6^{n},2^{n}+5^{n},3^{n}+4^$ ειναι πολ7.......
Αν n=αρτιος τοτε θα ειναι της μορφης 6κ+2ή 6κ+4(θα αποδειξω μονο την μια περιπτωση και[b] εντελως ομοια [/b]αποδυκνειετε και η αλλη...)
τοτε το αμελος γινεται $1^{3k+1}+4^{3k+1}+9^{3k+1}+16^{3k+1}+25^{3k+1}+36^{3k+1}$ που ειναι ισουπολοιπο με $2[1^{3k+1}+4^{3k+1}+9^{3k+1}]$και επειδη$(2,7)=1$ θα πρεπει νδο$1^{3k+1}+4^{3k+1}+9^{3k+1}$ =πολ7
 Ε,και που το προβλημα? Επαγωγικα....
Για κ=ο, 14=πολ7
Εστω οτι ισχυει $1^{3k+1}+4^{3k+1}+9^{3k+1}$=πολ7
Για κ+1 εχουμε...α μελος=$1^{3k+4}+4^{3k+4}+9^{3k+4}=1^{3k+1}+64*4^{3k+1}+729*9^{3k+1}$που ειναι ισουπολοιπο με $1^{3k+1}+4^{3k+1}+9^{3k+1}$που ειναι πολ7 απο την υποθεση...
Εντελως ομοια και η περιπτωση6κ+4.....
Creatanman,μαλλον εχεις δικιο....Δοκιμασα και τον 6ο αριθμο του Mersenne,και ισχυει..
Α,να σε ρωτησω και κατι...Η σκηση που δινεις στο 3ο post απο την αρχη...(οχι η γενικευση,αλλα η ασκηση...)Δεν λυνετε και αν θεωρησουμε ολες τις μορφες ν=10κ+υ...?[παιρννω ως προς 10 για να χρησημοποιησω fermat...,(οχι το τελευταιο θεωρημα :lol: )]

 :D  :D  :D

Vevaiws RAKAR! Auto akrivws prepei na kaneis! Oi askiseis tetoiou eidous exoun na kanoun me to $\phi(n)$ pou stin prokeimeni periptwsi $\phi(11)=10$. Tin afinw liges imeres akomi gia na tin dei opoios thelei prin kanw tin geniki paratirisi gia oles autes tis askiseis. Parolo pou tha pareis oles tis periptwseis $10k+u$, me $0\leq u<9$ einai poli eukolo ean to organwseis se ena omorfo pinakaki. Gia koitakste to kai pali kai kanete tis paratiriseis sas (RAKAR eisai mesa...)

Pame gia tin genikeusi tou allou twra...

Alexandros

Katapliktiki skepsi Stelio!!  Sigxaritiria...

Nomizw oti tha mporouse sigoura na einai ena kalo thema olimpiadas ean i lisi tou einai auti pou parethese o stelios kai den iparxei sintomoteri... Ti lete ki eseis?

Kai pali mpravo!

Alexandros

Ωραία Στέλιο  :)

Δείτε και εδώ για μια άλλη λύση . Όπως αναφέρεται και εκεί η άσκηση ονομάζεται ως θεώρημα Chevalley-Warning

http://www.mathlinks.ro/Forum/viewtopic.php?t=40171

The Chevalley-Warning Theorem can be found [url=http://en.wikipedia.org/wiki/Chevalley-Warning_theorem]here[/url].

Note that the proof given in silouan's link is not completely elementary, since it uses the concept of primitive roots.

PS: It's very nice to see that the Greek sub-forum is hyper-active.


Let G and I be respectively the centroid and the incenter of a give triangle ABC.Prove that if $ AB^2 \plus{} 2.IC^2 \equal{} 2.IB^2 \plus{} AC^2$ then GI is parallel to BC.

In order to get $ GI \parallel{} BC$ we need $ AB \plus{} AC \equal{} 2\cdot BC$ ( * ) but, the given relation leads straight to (*), hence the result.

Best regards,
sunken rock

This is an ongoing problem on The M&Y magazine in Vietnam .

Well , I dont know whether it is just accident but all two posts by the author mkt are both on going problems on the M&Y magazine .


calculus
integration
LaTeX
calculus computations

HI, can someone help me evaluate this definite integral using the substitution method?

definite integral of  40/(x^2+25) dx from x = 0 to x = 5. I can't figure out what to make for the substitution for du and u (or any other corresponding dummy variables) as to simplify the expression. Could anyone help out? Thanks a bunch.

Sorry, I don't know how to work latex yet.

I don't really know where textbook problems like this go in, but I would post them in Pre-Olympiad because they aren't really challenging.

Let $x=5u, dx=5du$

Then we have

$\int^5_0 f(x)dx={8}\int^1_0 \frac {du}{u^2+1}=8(\arctan 1-\arctan 0)=2\pi$

Note the change of limits on the definite integral, alternately, you get that:

$\int \frac {40}{x^2+25}=8\arctan \left(\frac {x}{5}\right)+C$

blahblahblah, thanks so much for helping. No, the problem did come out of a textbook, was not meant to be challenging. Just that I couldn't figure it out.  ;)


[hide="ramblings"]
hmmm... always try to get 4 on your turn, so obviously never give the number 4 to your opponent. Number four on your turn guarantees victory. 4 - 1 = 3, 3 - 1 = 2, 2 - 1 = 1.

If n is prime on your turn, your opponent will have n-1, is that a good thing?

If n is 10 on your turn, you have won (congrats)
[/hide]

[hide="definitive"]
[code]
1 <- 2 <- 3

..........^
..........|

1 <- 2 <- 4 <- 5
[/code]

This is the table of forced moves. As you can see, having a four on your turn guarantees a victory if you know what you are doing. However, if your opponent does not know what he is doing, it is possible to win even though the opponent has a four, as shown.

The number is generally irrelevant (read on) if it is not one of the chart numbers OR a number that can be turned into a number on the chart. There is only one other beneficial number, and it is 10. If you have 10 on your turn, take five away so your opponent is forced to give you a four.

Another beneficial strategy is leaving your opponent with a prime number. If your opponent is left with a prime number, then they only have one choice, which can be manipulated. Say you have a 22. You take away 11, and your opponent is forced to take away one. You now have 10, and you can turn that 10 into a 5, which all leads to my chart.
[/hide]

Thanks treething for your reply!!! I have some questions: please help me asap

1)
How did you came up with 10?
I mean... how do you know that the other numbers will not have the 10 property (that quickly left you or your opponent with 4)?

2)
Does leaving your opponent with any other prime number (besides 11) guarantee you the win? or just primes that will help you to 10? and how do you prove that?

3)
Are there winning strategies for starting players? .... starting number?

4)
I got to answer this question:
clearly explains whether a longest and a shortest game exist for particular starting numbers  :?

many thanks   :)

THANK YOU SO MUCH treething!!! I finished the write-up on time!
:)


integration
trigonometry
function
calculus
calculus computations

I found this one on a paper, but it doesn't look correct. Can anyone prove me wrong?
$ \int_0^\pi\cos(m\theta) \sin(x_{ml}|\cos(\theta) \minus{} p|)\,d\theta \equal{} 2\int_0^{\cos^{ \minus{} 1}p}\cos(m\theta) \sin[x_{ml}(\cos(\theta) \minus{} p)]\,d\theta$

$ p \in [0,1]$
$ x_{ml} \in \Re$
$ m$ is integer

Thanks

It doesn't seem to hold for, for example, $ m \equal{} 1$, $ p \equal{} 0$, and $ 0 < |x_{ml}| \leq \pi$.  The left hand side is $ 0$ but the right is strictly positive for $ 0 < x_{ml} \leq \pi$ and negative for $ \minus{} \pi \leq x_{ml} < 0$.  Given the subscript, I would guess that it is meant that $ x_{ml}$ is not arbitrary, but depends in some very specific way on $ m$.

You're right about $ x_{ml}$. It's the $ (l \plus{} 1)$th zero of the Bessel function of the first kind $ J_m(x)$, where $ l$ is a non-negative integer. $ m$ is also non-negative.
$ x_{ml}$ not arbitrary, but for all purposes in this integral it can be assumed to be.

$ \int_0^\pi\cos(m\theta) \sin(x_{ml}|\cos(\theta)\minus{}p|)\,d\theta \minus{} \int_0^{\cos^{\minus{}1}p}\cos(m\theta) \sin [x_{ml}(\cos(\theta)\minus{}p)] d\theta$

$ \equal{} \int_{\cos^{\minus{}1}p}^\pi\cos(m\theta) \sin(x_{ml}(p\minus{}\cos(\theta)))\,d\theta \equal{} 0$ when?

when?
In the paper, it looks as though "always" would be the answer. Look at how they make the statement:
[img]http://i191.photobucket.com/albums/z226/pocaracas/eq8_9.gif[/img]

ooops.. I forgot the 2 on the right hand side... it's not too relevant because $ 2*0$ should the $ 0$.

$ [0,\pi] \equal{} [0,\cos^{ \minus{} 1}p] \cup (\cos^{ \minus{} 1}p, \pi \minus{} \cos^{ \minus{} 1}p) \cup [\pi \minus{} \cos^{ \minus{} 1}p,\pi]$

$ I \equal{} \int_0^\pi\cos(m\theta) \sin(x_{ml}|\cos(\theta) \minus{} p|)\,d\theta \equal{} I_1 \plus{} I_2 \plus{} I_3$

$ I_1 \equal{} \int_0^{\cos^{ \minus{} 1}p}\cos(m\theta) \sin [x_{ml}(\cos(\theta) \minus{} p)] d\theta$

$ I_2 \equal{} \int_{\cos^{ \minus{} 1}p}^{\pi \minus{} \cos^{ \minus{} 1}p} \cos (m\theta) \sin [x_{ml}(p \minus{} \cos(\theta)] d\theta$

$ I_3 \equal{} \int_{\pi \minus{} \cos^{ \minus{} 1}p}^\pi \cos(m\theta) \sin [x_{ml}(\cos(\theta) \minus{} p)] d\theta$

i think what we do is show $ I_2 \equal{} 0, I_1 \equal{} I_3$

$ I_3=\int_{\pi-\cos^{-1}p}^\pi \cos(m\theta) \sin [x_{ml}(\cos(\theta)-p)] d\theta$
$ \chi=\pi-\theta$
$ =\int_{\cos^{-1}p}^0 -\cos(m(\pi-\chi)) \sin [x_{ml}(\cos(\pi-\chi)-p)] d\chi$

$ =\int_0^{\cos^{-1}p} -\cos(m\pi) \cos(m\chi) \sin [x_{ml} (\cos \chi + p )] %Error. "dchi" is a bad command.
$

$ =(-1)^{m+1} \int_0^{\cos^{-1}p}  \cos(m\chi) \sin [x_{ml}(\cos \chi -p)+2px_{ml}]$

$ =(-1)^{m+1} \cos (2px_{ml})\int_0^{\cos^{-1}p}  \cos(m\chi) \sin [x_{ml}(\cos \chi -p)]d\chi$
$ +(-1)^{m+1} \sin(2px_{ml}) \int_0^{\cos^{-1}p}\cos[x_{ml}(\cos \chi -p)]d\chi$

maybe $ x_{ml}$ matters here?

$ I_2=\int_{\cos^{-1}p}^{\pi-\cos^{-1}p}\sin [x_{ml}(p-\cos(\theta)] d\theta$

$ \xi=\theta - \frac{\pi}{2}$, $ \alpha = \frac{\pi}{2}-\cos^{-1}p$

$ =\int_{-\alpha}^\alpha  \cos( m\xi +\frac{m\pi}{2}) \sin [ x_{ml} (cos(\xi+\frac{\pi}{2}) - p)] d\xi$

$ =-\int_{-\alpha}^\alpha  [\cos( m\xi)\cos \frac{m\pi}{2}-\sin( m\xi)\sin \frac{m\pi}{2}]\sin [ x_{ml} (\sin(\xi) + p)] d\xi$

$ =\sin \frac{m\pi}{2}\int_{-\alpha}^\alpha  \sin( m\xi)[\sin ( x_{ml}\sin\xi)\cos(x_{ml}p)+\cos (x_{ml}\sin\xi )\sin(x_{ml}p) ] d\xi$
$ - \cos \frac{m\pi}{2}\int_{-\alpha}^\alpha \cos( m\xi)[\sin ( x_{ml}\sin\xi)\cos(x_{ml}p)+\cos (x_{ml}\sin\xi )\sin(x_{ml}p) ]d\xi$

$ =\sin \frac{m\pi}{2}\cos(x_{ml}p)\int_{-\alpha}^\alpha \sin( m\xi)\sin ( x_{ml}\sin\xi)$
$ + \sin \frac{m\pi}{2}\sin(x_{ml}p)\int_{-\alpha}^\alpha \sin( m\xi)\cos ( x_{ml}\sin\xi)$
$ - \cos \frac{m\pi}{2}\cos(x_{ml}p)\int_{-\alpha}^\alpha \cos( m\xi)\sin ( x_{ml}\sin\xi)$
$ - \cos \frac{m\pi}{2}\sin(x_{ml}p)\int_{-\alpha}^\alpha \cos( m\xi)\cos ( x_{ml}\sin\xi)$

$ =2\sin \frac{m\pi}{2}\cos(x_{ml}p)\int_0^\alpha \sin( m\xi)\sin ( x_{ml}\sin\xi)$
$ -2\cos \frac{m\pi}{2}\sin(x_{ml}p)\int_0^\alpha \cos( m\xi)\cos ( x_{ml}\sin\xi)$

or maybe im not getting anywhere =(

Thanks for the effort Ken4000, but I also think it's not leading anywhere... :(


counting
derangement
geometry
geometric transformation
rotation

JACK, AND, JILL, WENT, UP, THE, and HILL are sitting at a round table. They want to reorder themselves such that JACK isn't in JACK's spot, AND isn't in AND's spot,... (everybody isn't in each other's original spot)How many possible arrangements are there?

what if they were originally just sitting in a row of chairs? thx!  :)

For the second problem, this becomes the number of derangements of $ 7$ objects.  For $ n$ objects this is \[ \sum_{i\equal{}0}^{n}(\minus{}1)^i \frac{n!}{i!}\] Plugging in $ 7$ we get $ \boxed{1854}$.

For the first problem I suppose you can take the derivation of this formula and modify it for the problem?

How do you derive that formula? PIE? I keep getting 4320 from epic casework... :maybe:

Yeah it's PIE with the sets $ A_i$, $ 1\leq i\leq n$, where $ A_i$ is the set of all arrangements where person $ i$ is in the right place.

i think for the first problem that it would be 1854*6! cuz for each row of seats, u could put it in a circle and for each circle there are 6! arrangements. am i right?  :|

5...x, what's a sideways, capital "M" supposed to mean????

@akhil0422:

$ \sum_{k\equal{}a}^{b}f(k)\equal{}f(a)\plus{}f(a\plus{}1)\plus{}\cdots\plus{}f(b\minus{}1)\plus{}f(b)$.

So do we have a final answer to the round circle question? To reiterate,
[quote]I don't know if this problem applies to a case of a round table where rotations are indistinct, since you can't say that any particular seat is Jack's seat. If we do say that seat 1 is Jack's seat, etc. then rotations are distinct and our answer is the same.[/quote]


without use of any calculator,determine:
sin(pi/14)+6sin²(pi/14)-8sin^4(pi/14)


algebra
polynomial
calculus
integration
geometry
3D geometry
number theory

[hide]$P1(x)=\frac{x^{101}-1}{x-1}$
$P2(x)=\frac{x^{101}+1}{x+1}$
$P3(x)=\frac{x^{68}-1}{x-1}$

so $P1(x)$ has 100 roots (don't include 1 because then the fraction's undefined), $P2(x)$ has 100, and $P3(x)$ has 67. i guess none of them are the same, because the ones for $P1(x)$ are in the form $e^{\frac{2k\pi}{101}}$, for integral k from 1 to 100, the roots of $P2(x)$ are in the form $-e^{\frac{2k\pi}{101}}$, and the roots of $P3(x)$ are in the form $e^{\frac{2k\pi}{68}}$.

so there should be 267 distinct roots.  [/hide]

You didn't look carefully enough at $P_{1}$.

[hide]266; the difference in $P1$ still has the same number of roots (100,) but this time, there is a double root (-1.)  Therefore, you simply subtract 1 from perfectnumber628's original answer (267) to get the correct answer.[/hide]

You've also done something wrong (there are far more overlapping roots than you've listed) but since you didn't actually show any work, I can't be sure what it was.

P1 is missing every third power, $x^{2}, x^{5}, \ldots, x^{9}8$.

[hide]well look at that, it [b]is[/b] missing every third power.  so factor out a $1+x$ to get $P1(x)=(x+1)(1+x^{3}+x^{6}...x^{99})=(x+1)\left(\frac{x^{102}-1}{x^{3}-1}\right)$.  so $P1$ has 100 roots, -1 and the 102nd roots of unity, not counting the cube roots of unity.  so -1 is a double root, so there are only 99 distinct ones.

now some of those will be the same as the 68th roots from that other polynomial.  the roots of $P1$ are in the form $e^{\frac{2k\pi}{102}}$ (k is an integer, $1\leq k\leq 101$, k is not 34 or 68), the roots of $P2$ are in the form $e^{\frac{2k\pi}{101}}$ (k is an integer from 1 to 100), and the roots of $P3$ are $e^{\frac{2k\pi}{68}}$ (k is an integer from 1 to 67).

since 101 is relatively prime to 102 and 68, none of the roots of $P2$ overlap the other ones. let $k_{1}$ be the k from a root of $P1$ and $k_{3}$ be a k from $P3$.  so the roots that aren't distinct are the ones where $\frac{k_{1}}{102}=\frac{k_{3}}{68}$

mulitplying by 34, we get $\frac{k_{1}}{3}=\frac{k_{3}}{2}$, which means $2k_{1}=3k_{3}$.  so $k_{3}$ is divisible by 2, and $k_{1}$ is divisible by 3.  i think any even number from 2 to 66 works for $k_{3}$, so there's 33 of them.

so we have $99+100+67-33=233$.  so hopefully 233 is the right answer  :lol: [/hide]


What is the largest integer $x$ for which $\frac{1}{x}$ is larger than $\frac{4}{49}$?

The answer should be [hide="answer"]$12$[/hide]

Yes that is.
[hide="brief explanation"]4/49 is close (but smaller than) 4/48, which is also 1/12. Common sense would tell us that 1/12 would be greater than 4/49.
1/13 is 4/52, obviously bigger than 4/49.
$x = 12$[/hide]

[quote]1/13 is 4/52, obviously bigger than 4/49[/quote]

I think you mean smaller. ;)

A little bit more of a proof... $\frac{4}{49}=\frac{1}{x} \Rightarrow 4x=49 \Rightarrow x=12.25$
Therefore, for $\frac{1}{x}>\frac{4}{49}$, $x<12.25$. And since the problem specifies the largest integer, the answer is $12$.

[quote="JesusFreak197"][quote]1/13 is 4/52, obviously bigger than 4/49[/quote]

I think you mean smaller. ;)[/quote]

Whoops, yes I did. I hope you all realize that I knew that already  :blush:

$\frac{1}{x} = \frac{4}{4x} < \frac{4}{49}$
$4x<49 \rightarrow x<12.25$ so the greatest integral $x$ is 12.


How does one get titu's lemma from cauchy's inequality

I'll prove this for the 3 variable case. Let $ u \equal{} <\sqrt{a}, \sqrt{b}, \sqrt{c}>$, and let $ v \equal{} <\frac{x}{\sqrt{a}}, \frac{y}{\sqrt{b}}, \frac{z}{\sqrt{c}}>$. We have that $ |u|^2|v|^2 \geq |u \cdot v|^2$, that is, $ (a \plus{} b \plus{} c)(\frac{x^2}{a} \plus{} \frac{y^2}{b} \plus{} \frac{z^2}{c}) \geq (x \plus{} y \plus{} z)^2$. Divide both sides by $ (a \plus{} b \plus{} c)$ to get the desired result.

Or you could just prove it for the two variable case: $ \frac{a_1^2}{b_1}\plus{}\frac{a_2^2}{b_2}\geq \frac{(a_1\plus{}a_2)^2}{b_1\plus{}b_2}$, which is true if you clear denominators and multiply out (or you can use Cauchy.)
Then, if $ \sum_{i\equal{}1}^n\frac{a_i^2}{b_i}\geq \frac{\left(\sum_{i\equal{}1}^n a_i\right)^2}{\sum_{i\equal{}1}^n b_i}$, then $ \sum_{i\equal{}1}^{n\plus{}1}\frac{a_i^2}{b_i}\geq \frac{\left(\sum_{i\equal{}1}^n a_i\right)^2}{\sum_{i\equal{}1}^n b_i}\plus{}\frac{a_{n\plus{}1}^2}{b_{n\plus{}1}}\geq \frac{\left(\sum_{i\equal{}1}^{n\plus{}1} a_i\right)^2}{\sum_{i\equal{}1}^{n\plus{}1} b_i}$.

He wanted to see how is T2's lemma a special case of CS.

Like this:

CS: $ (x_1^2\plus{}...\plus{}x_n^2)(y_1^2\plus{}...\plus{}y_n^2)\ge(x_1y_1\plus{}...\plus{}x_ny_n)^2$

Take $ x_i\equal{}\frac{a_i^2}{b_i},y_i\equal{}b_i$, and T2's lemma is there.

Simply,
If there is $ a,b,c$ and real $ x,y,z>0$ then: \[ \frac{\left(a\plus{}b\plus{}c\right)^2}{x\plus{}y\plus{}z} \geq \frac{a^2}{x}\plus{} \frac{b^2}{y}\plus{}\frac{c^2}{z}\]

Proof: All of it is choosing good values for Cauchy then some algebra. Notice:
\[ \left(a\plus{}b\plus{}c\right)^2\equal{}\left(\frac{a}{\sqrt{x}}\sqrt{x}\plus{}\frac{b}{\sqrt{y}}\sqrt{y}\plus{}\frac{c}{\sqrt{z}}\sqrt{z}\right)^2\]
Now make: $ \left(a_1,a_2,a_3\right)\equal{}\left(\frac{a}{\sqrt{x}},\frac{b}{\sqrt{y}},\frac{c}{\sqrt{z}}\right)$ and $ \left(b_1,b_2,b_3\right)\equal{}\left(\sqrt{x},\sqrt{y},\sqrt{z}\right)$
Now apply CS
\[ \left(a\plus{}b\plus{}c\right)^2\equal{}\left(\frac{a}{\sqrt{x}}\sqrt{x}\plus{}\frac{b}{\sqrt{y}}\sqrt{y}\plus{}\frac{c}{\sqrt{z}}\sqrt{z}\right)^2 \geq \left(\frac{a^2}{x}\plus{} \frac{b^2}{y}\plus{}\frac{c^2}{z}\right)\left(x\plus{}y\plus{}z\right)\]
By dividing we get \[ \frac{\left(a\plus{}b\plus{}c\right)^2}{x\plus{}y\plus{}z} \geq \frac{a^2}{x}\plus{} \frac{b^2}{y}\plus{}\frac{c^2}{z}\]

[quote="Bugi"]He wanted to see how is T2's lemma a special case of CS.

Like this:

CS: $ (x_1^2 \plus{} ... \plus{} x_n^2)(y_1^2 \plus{} ... \plus{} y_n^2)\ge(x_1y_1 \plus{} ... \plus{} x_ny_n)^2$

Take $ x_i \equal{} \frac {a_i^2}{b_i},y_i \equal{} b_i$, and T2's lemma is there.[/quote]

Just a small typo: $ x_i \equal{} \dfrac{a_i}{\sqrt{b_i}}$, $ y_i\equal{}\sqrt{b_i}$

[hide="proof "]
Since

$ \frac {a_1 ^2}{b_1} \plus{} \frac {a_2 ^2 }{b_2} \geq \frac {(a_1 \plus{} a_2) ^2}{b_1 \plus{} b_2} \iff (a_1 b_2 \minus{} a_2 b_1)^2 \geq 0$

We have that the inequality is true for $ n \equal{} 2$

Suppose it is true for $ n$

$ \sum_{k \equal{} 1}^{n} {\frac {a_k ^2}{b_k}} \geq \frac {\left(\sum {a_k}\right)^2}{ {b_k}}$

Then we add $ \frac {a_{k \plus{} 1}^2}{b_{k \plus{} 1}}$ to both sides .

$ \sum_{k}^{n} {\frac {a_k ^2}{b_k}} \plus{} \frac {a_{k \plus{} 1}^2}{b_{k \plus{} 1}} \geq \frac {\left(\left(\sum {a_k}\right)^2 \plus{} a_{k \plus{} 1} \right)^2}{\sum {{b_k}} \plus{} b_{k \plus{} 1}} \iff \left((\sum a_k)^2 \cdot b_{k \plus{} 1} \minus{} a_{k \plus{} 1} \sum b_{k} \right)^2 \geq 0$

We  see the inequality is true for $ n \equal{} k \plus{} 1$ ,then it's true for every $ n \in \mathbb{N}$ ,as desired

That was the proof of Titu's Lemma by induction  :

Now by Cauchy Schwartz inequality

$ \left(\frac{a_1 ^2}{b_1} \plus{} ...\plus{}\frac{a_n ^2}{b_n}\right)\cdot (b_1 \plus{}...\plus{}b_n) \geq {(a_1 \plus{} ...\plus{}a_n)^2}$ ,done [/hide]

Please change the sign in my post from $ \geq$ to $ \le$ I made a typo but I can't edit it now  :|


[quote="perfect_radio"]I think we should keep this forum up-to-date. Does anybody agree?
[img]7129[/img][/quote]

ROFL its just a minor detail....but if u want Mr. Vornicu can change it  :D

[quote="7h3.D3m0n.117"]ROFL its just a minor detail...[/quote]

For some reason, I have a feeling it is more like a subtle allusion...

  darij

It's skin-dependent: the date is 2007 in AOPS style and 2004-2006 in Mathlinks style.


Let $ P(x)$ be a sixth degree polynomial so that the coefficient of $ x^6$ is $ 1$.
If $ P(1) \equal{} 1, P(2) \equal{} 2, P(3) \equal{} 3, P(4) \equal{} 4, P(5) \equal{} 5$ and $ P(6) \equal{} 6$, what's the value of $ P(7)$?

hello, solving the linear system which we get by the conditions we have
$ P(x)\equal{}x^6\minus{}21x^5\plus{}175x^4\minus{}735x^3\plus{}1624x^2\minus{}1763x\plus{}720$
and we get $ P(7)\equal{}727$
Sonnhard.

We have:
$ P(x)\minus{}x \equal{} c(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)(x\minus{}6)$ where $ c\equal{}1$ because P(x) is a monic polynomial, which gives
$ P(7) \equal{} 6! \plus{} 7 \equal{} 727$


function
limit
real analysis
real analysis unsolved

Let $ f: [0,\infty)\to\mathbb{R}$ be a continuous function on $ (0,\infty)$.Prove that :

$ f$ has The Intermediate Value Property $ \Leftrightarrow$ $ \exists$ a sequence $ (a_n)_n$  , $ a_n>0 \forall n \in \mathbb{N}$ such that $ \lim_{n\to\infty}a_n\equal{}0$ and $ \lim_{n\to\infty}f(a_n)\equal{}f(0)$

[i]author :C.Mortici[/i]

The question is not really that interesting. I don't even know why I posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75435]here[/url] :?:


function trigonometry algebra domain Support integration calculus

I dont really get how to show that a function is of exponential order.  Here are some functions from the book:

$ t^{n}$

$ \sin (at)$

$ e^{\sqrt}{t}$

Could someone just explain how its done or do one of these as an example?

Thanks

The question is what you mean by exponential order. What I'm going to assume, since you mention differential equations, is that you're looking for function to which you can apply Laplace transform techniques.

So assume we have a function $ f$ whose domain is $ [0,\infty).$ (Sometimes it's useful to consider functions with domain $ \mathbb{R}$ and support $ [0,\infty),$ but that doesn't really change anything.

What we need is for $ \int_{0}^{\infty}f(t)e^{-st}\,dt$ to converge for some real $ \s_{0}.$ Then the integral will converge for all $ s$ such that $ \text{Re}(s)>s_{0}.$

A sufficient (but not necessary) condition for this is that $ f$ is locally integrable and that there is some $ s$ such that $ f(t)e^{-st}$ remains bounded as $ s\to\infty.$ That's probably what you mean by "exponential order."

$ t^{n}$ is of exponential order; for every $ s>0,\ t^{n}e^{-st}\to0$ as $ t\to\infty.$

$ \sin(at)$ is of exponential order since $ \sin(at)e^{0t}$ is bounded. You can say this about any bounded function.

$ e^{\sqrt{t}}$ is of exponential order. For any $ s>0,\ \sqrt{t}-st\to-\infty$ as $ t\to\infty,$ therefore $ e^{\sqrt{t}}e^{-st}\to0.$ [It doesn't have a "nice" Laplace transform, but that's a different question. It does have some Laplace transform.]

$ t^{m}e^{at}\cos{bt}$ is of exponential order. And has a nice Laplace transform.

So what isn't of exponential order? We need $ e^{b(t)}$ where $ b(t)$ increases at a faster-than-linear rate. For instance, $ e^{t^{2}}.$ Or $ t^{t}.$ Or $ \Gamma(t).$


Could you,please send me the full solutions of the fourth and sixth Junior Balkan Olympiad.

if you post the questions yourself on the forum in the apropriate places you will probably get some answers. otherwise I really doubt whether anyone would write soln to problems whoose statements they don't know :D


Find the maximum value of $ f(x)$, if $ f(x\plus{}\pi)\equal{}12sin(3x)\plus{}5cos(3x\plus{}\frac{\pi}{2})\plus{}3$

[hide="Solution (involves calculus)"]We know $ f(n\plus{}\pi)\equal{}12\sin 3n\plus{}5\cos\left(3n\plus{}\frac{\pi}{2}\right)\plus{}3$. Plug in $ n\equal{}x\minus{}\pi$ to get $ f(x)\equal{}12\sin(3x\minus{}3\pi)\plus{}5\cos\left(3x\minus{}\frac{5\pi}{2}\right)\plus{}3$.

So, $ f^\prime(x)\equal{}36\cos(3x\plus{}3\pi)\minus{}15\sin\left(3x\plus{}\frac{5\pi}{2}\right)$.

$ 36\cos(3x\minus{}3\pi)\minus{}15\sin\left(3x\minus{}\frac{5\pi}{2}\right)\equal{}0$

$ 36\cos 3x\minus{}15\sin\left(3x\plus{}\frac{\pi}{2}\right)\equal{}0$

$ 36\cos 3x\plus{}15\cos 3x\equal{}0$

$ \cos 3x\equal{}0\implies x\equal{}\frac{\pi}{6}\plus{}2\pi n,\;n\in\mathbb{Z}^\plus{}$.

Note that half of these values will be minima and the other half maxima. The first derivative test shows that $ x\equal{}\frac{\pi}{6}$ is a minimum, and $ x\equal{}\frac{\pi}{2}$ is a maximum.

Finally, we find $ f\left(\frac{\pi}{2}\right)\equal{}12\sin\left(\frac{3\pi}{2}\minus{}3\pi\right)\plus{}5\sin\left(\frac{3\pi}{2}\right)\plus{}3\equal{}12\minus{}5\plus{}3\equal{}\boxed{10}$[/hide]

Non-Calculus solution:

[hide]$ f(x \plus{} \pi) \equal{} 12\sin(3x) \plus{} 5\cos(3x \plus{} \frac {\pi}{2}) \plus{} 3$

$ f(x \plus{} \pi) \equal{} 12\sin(3x) \plus{} 5\left(\cos3x\cos\frac {\pi}{2} \minus{} \sin3x\sin\frac {\pi}{2}\right) \plus{} 3$

$ f(x \plus{} \pi) \equal{} 7\sin(3x) \plus{} 3$

Now we can already see that the maximum will be $ 10$, but we'll proceed formally:

$ x \plus{} \pi \equal{} t$

$ f(t) \equal{} 7\sin(3t \minus{} 3\pi) \plus{} 3$

$ f(t) \equal{} 7(\sin 3t \cos 3\pi \minus{} \cos 3t \sin 3\pi) \plus{} 3$

$ f(t) \equal{} \minus{} 7\sin 3t \plus{} 3$

$ \sin \varphi \geq \minus{} 1$ so $ f(t)\geq 7 \plus{} 3 \equal{} 10$

which is the maximum and it is reached for $ \sin 3t \equal{} \minus{} 1$.

[/hide]


[quote="hungkhtn"]OK. It seems too hard.  My solution is quite long and I wait someone solve it easier.[/quote]

is the question changed?!

Yeah, I have strugle with Hung puzzle too. Did you change from"4" to "12"?
Because if it is "4" then I have no idea where the inequa hold.
Anyway, if it is "12" then probably I will try again.

Ok. Sorry. I used to post solution in hurry then sometimes It wasn't like of mines at first. :(  I change the constant $4$ to $12$.

Hi... It's very hard. I like it because $n-1 EV$ can't kill it!   :lol:  :lol:

  I hope before a month, there's a solution here.

[quote=hungkhtn]Prove that if $a,b,c$ are positive real then:
$(a+\frac{b^2}{c})^2+(b+\frac{c^2}{a})^2+(c+\frac{a^2}{b})^2\ge 12\frac{a^3+b^3+c^3}{a+b+c}$[/quote]
We have
\[\left(a+\frac{b^2}{c}\right)^2+\left(b+\frac{c^2}{a}\right)^2+\left(c+\frac{a^2}{b}\right)^2 - \frac{12(a^3+b^3+c^3)}{a+b+c} = \frac{F(a,b,c)}{a^2b^2c^2(a+b+c)},\]
for
\[(a+b+c)F(a,b,c) = \sum abc^2 (2a^2+7ab+2b^2)(a^2+bc-2ab)^2\]
\[\displaystyle +\sum c^2a^3\left[(2a+4b+c)ca+2b(b-2c)^2+2b(a-b)^2\right](a-b)^2\]
\[+\sum a^2b^2(b^3+c^3+2abc-b^2c-2bc^2-ca^2)^2 \geqslant 0.\]
The following inequality is also true
\[\left(a+\frac{b^2}{c}\right)^2+\left(b+\frac{c^2}{a}\right)^2+\left(c+\frac{a^2}{b}\right)^2 \geqslant \frac{73(a^3+b^3+c^3)-39abc}{5(a+b+c)}\]


Let be $ A\equal{}\{(x,y)\in \mathbb{R}^2 \ |\ x^2\plus{}y^2>2\}$ and $ B\equal{}\{(x,y)\in \mathbb{R}^2\ |\ x\plus{}y\plus{}xy<3\}$ . Determine $ A\cup B$ .

[quote="dgreenb801"]I think you mean A intersect B, not A union B. Anyways, there would still be many solutions, take y=-x, x>10.[/quote]

It is $ A\cup B$ .

Oh I see, you intend for there to be infinitely many solutions, but you have to describe them. Sorry.

The answer is all possible ordered pairs but (1,1). Assume an ordered pair doesn't satisfy either. Then
$ 2xy < x^2 \plus{} y^2 < 2$, so $ xy < 1$
Since $ x\plus{}y\plus{}xy>3$, we find $ x\plus{}y >2$
Squaring, $ x^2 \plus{} 2xy \plus{} y^2 > 4$. Adding $ 2 > 2xy$, we get
$ x^2 \plus{} y^2 > 2$
Contradiction
(1,1) works for the equality case, which I left out

[quote="dgreenb801"]Oh I see, you intend for there to be infinitely many solutions, but you have to describe them. Sorry.[/quote]

The solution is $ \mathbb{R}\backslash \{(1,1)\}$ .

[quote="alex2008"]Let be $ A \equal{} \{(x,y)\in \mathbb{R}^2 \ |\ x^2 \plus{} y^2 > 2\}$ and $ B \equal{} \{(x,y)\in \mathbb{R}^2\ |\ x \plus{} y \plus{} xy < 3\}$ . Determine $ A\cup B$ .[/quote]

[b][u]Proof[/u].[/b] Observe that $ \overline {A\cup B} \equal{} \overline A\cap\overline B$ and $ (x,y)\in A\cup B\Longleftrightarrow (y,x)\in A\cup B$ . Therefore,

$ (x,y)\in \overline {A\cup B}\Longleftrightarrow \left\|\begin{array}{c} x^2 \plus{} y^2 \le 2 \\
 \\
3\le x \plus{} y \plus{} xy\end{array}\right\|^{(*)}\ \bigoplus\ \ \Longrightarrow\ x^2$ $ \plus{} y^2 \plus{} 3\le x \plus{} y \plus{} xy \plus{} 2\ \Longleftrightarrow$

$ x^2 \plus{} y^2 \minus{} xy \minus{} xy$ $ \plus{} 1\le 0\ \ \|\ \bigodot\ 2\ \Longleftrightarrow\ (x \minus{} y)^2$ $ \plus{} (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2\le 0\ \Longleftrightarrow\ x \equal{} y \equal{} 1$ .

Since $ (1,1)$ verify both relations from $ (*)$ obtain $ \overline {A\cup B} \equal{} \{(1,1)\}$ . In conclusion, $ A\cup B \equal{} \mathbb R^2 \minus{} \{(1,1)\}$ .

To find $ (x,y)$ not belong to $ A \cup B$.

$ \begin{cases} x^2\plus{}y^2 \le 2 \\
(|x|\plus{}1)(|y|\plus{}1|) \ge (x\plus{}1)(y\plus{}1) \ge 4 \end{cases}$

Now $ 2 \ge \frac{(x^2\plus{}1)\plus{}(y^2\plus{}1)}{2} \ge \frac{(|x|\plus{}1)^2}{4}\plus{}\frac{(|y|\plus{}1)^2}{4} \ge \frac{2(|x|\plus{}1)(|y|\plus{}1)}{4} \ge 2$

Equality holds $ \Leftrightarrow x\equal{}y\equal{}1$

So $ A \cup B\equal{}\mathbb{R}^{2}\backslash\{(1,1)\}$


Find all reals $x$ such thaat $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$.

Should it be : $x = \sqrt{2+\sqrt{2-\sqrt{2+x}}}$? ;)

.

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 through half angle and double angle formulas.

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geometry
inequalities
trigonometry
geometry proposed

For any quadrilateral with the side lengths $a,$ $b,$ $c,$ $d$ and the area $S,$ prove the inequality$S\leq \frac{a+c}{2}\cdot \frac{b+d}{2}.$

Let $A,B,C,D$ be the vertices of the quadrilateral s.t. $a=AB,b=BC,c=CD,d=DA$. We then have $\frac{a+c}2\cdot\frac{b+d}2=\frac{ab+cd+ad+bc}4$, which is $\ge \frac{ab\sin\angle B+cd\sin\angle D+ad\sin\angle A+bc\sin\angle C}4=S$.