geometry
circumcircle
incenter
similar triangles
geometry proposed

Let c be the circumcircle ot triangle ABC and let P be a point on the plain of c. Let's to project from P the points A,B,C to X,Y,Z on c. Let QBC [if P is inside c  Q is in the opposite midplain of A wrt line BC and [i]viceversa[/i]] a triangle  direct(*) similar to XYZ prove that APB is similar to ACQ.

(*) I (would) mean <Q = <X, <B = <Y and consequentely <C = <Z.

In what follows, all similarities are "direct", as this was defined by sprmnt.

If $U,V,T$ are the projections of $P$ on $BC,CA,AB$ respectively, then it's well-known that $XYZ$ and $UVT$ are similar, so we can work with $UVT$ instead of $XYZ$.

The point $Q$ is unique, so it's enough to prove this: if $Q$ is a point s.t. $(APB,ACQ),(APC,ABQ)$ are pairs of similar triangles, then $UVT,QBC$ are similar.

This is, however, a simple angle chase: $\angle QBC=\angle APC-\angle ABC=\angle PAB+\angle PCB=\angle PVT+\angle PVU=\angle UVT$. In exactly the same manner we show that $\angle QCB=\angle UTV$, and that's it.

Nice problem, Sprmnt21! Nice solution, Grobber! A little remark: The triangle QBC is not "directly" similar to triangle XYZ in the usual meaning of the word "directly". What Sprmnt21 meant with "directly" was just the fact that the vertices Q and X, the vertices B and Y, the vertices C and Z are corresponding vertices of the two triangles. In other words, the similitude transformation which maps the triangle QBC to the triangle XYZ must map the points Q, B, C to the points X, Y, Z, respectively. In contrast to this, the usual meaning of the word "directly similar" is that two triangles are called directly similar if and only if they are similar and have the same orientation (either both of them are oriented clockwise, or both are oriented counterclockwise). The opposite is "oppositely similar", when the triangles are similar and have different orientation (one of them is oriented clockwise and the other one is oriented counterclockwise). In your case, the triangle QBC is oppositely similar to triangle XYZ !

Here is my solution of the problem:

Let P' be the isogonal conjugate of P with respect to triangle ABC, and let R be the point of intersection of the circumcircle of triangle BCP' with the line AP' (different from P'). We will first prove that R = Q.

Well, since the triangle QBC is oppositely similar to triangle XYZ, we have < CBQ = - < ZYX. In other words, < CBQ = < XYZ. Now, < XYZ = < XAZ (cyclic), so < CBQ = < XYZ = < XAZ = < XAB + < BAZ. But < BAZ = < BCZ (again by concyclicity), and thus < CBQ = < XAB + < BCZ = < PAB + < BCP.

Since the point P' is the isogonal conjugate of P with respect to triangle ABC, we have < CAP' = < PAB and < P'CA = < BCP. Thus, < CBQ = < PAB + < BCP = < CAP' + < P'CA = < (CA; AP') + < (CP'; CA) = < (CP'; AP') = < CP'R. On the other hand, since the point R lies on the circumcircle of triangle BCP', we have < CP'R = < CBR, and thus < CBQ = < CBR. Similarly, < BCQ = < BCR. These two equations force the points Q and R to coincide.

Thus, the point Q is the point of intersection of the circumcircle of triangle BCP' with the line AP' (different from P'). Well, now everything is easy: Since the point P' is the isogonal conjugate of P, we have < PAB = < CAP' and < ABP = < P'BC. Now, < PAB = < CAP' can be immediately rewritten as < PAB = < CAQ, and < ABP = < P'BC implies < ABP = < P'QC, since < P'BC = < P'QC (remember that the point Q lies on the circumcircle of triangle BCP'), or, equivalently, < ABP = < AQC. Altogether, we have < PAB = < CAQ and < ABP = < AQC, and consequently, the triangles APB and ACQ are directly similar. $\blacksquare$

Sprmnt, I will look at your nice "three lines" generalization later when I have more time.

  Darij

If R=BP^QC, then APCR is Cyclic. Indeed <PRC = <BCQ - <PBC = <YZX - < YZC = <PAC. Now we can prove that ABQR is cyclic too. In fact <BQC + <BAR = <YXZ + <BAC + <CAR, but <CAR = <CPR = <PBC + <PCB = <YAC + <ZAB. Then  <BQC + <BAR = <YXZ + <BAC + <YAC + <ZAB = 180.

From this facts it easy to deduce that ABQ ~ APC and ABP ~ AQC. Furthemore, if we call S=CP^QB, it holds that ABR ~ ASC and ABS ~ ARC.

And finally, if O1,O2,O3 and O4 are the centers of (APB), (APC), (ABQ) and (ACQ), it is easy to see that O1O2O3O4 is cyclic.

PS

Could you discuss the locus of points P s.t. A,R, S are collinears?

Is not difficult to construct (in the euclidean sense) the locus point by point, but I'm not able to recognize any elementar curve.

The incenter of ABC, for example, lies on this locus. Are there  other well-known points?

Very interesting...

[quote="sprmnt21"]And finally, if O1,O2,O3 and O4 are the centers of (APB), (APC), (ABQ) and (ACQ), it is easy to see that O1O2O3O4 is cyclic.[/quote]

One can say even more: The circumcircle of the quadrilateral $O_1O_2O_3O_4$ is orthogonal to the circumcircle of triangle ABC. This is because the points $O_1$ and $O_3$ are mutually inverse with respect to the circumcircle of triangle ABC, and so are the points $O_2$ and $O_4$.

[quote="sprmnt21"]Could you discuss the locus of points P s.t. A,R, S are collinears?[/quote]

I think that this locus is the union of the circles $BCII_a$ and $BCI_bI_c$ (where I is the incenter and $I_a$, $I_b$, $I_c$ are the excenters of triangle ABC). The only thing I can prove is that if the point P lies on the circle $BCII_a$ or on the circle $BCI_bI_c$, then the points A, R and S are collinear.

Here is this [i]proof[/i]: Assume that the point P lies on the circle $BCII_a$ (the proof for the circle $BCI_bI_c$ is completely analogous). We will work with directed angles modulo 180.

I will again denote by P' the isogonal conjugate of the point P with respect to triangle ABC, and use the fact I proved in my previous message, namely the fact that the point Q is the point intersection of the circumcircle of triangle BCP' with the line AP' (different from P'). Additionally, I will use the following formulas holding for any two points P and P' isogonally conjugated to each other with respect to triangle ABC:

  < BP'C = < BAC - < BPC;
  < CP'A = < CBA - < CPA;
  < AP'B = < ACB - < APB.

Of course, all angles are directed angles modulo 180 here. Now, in our case, since the point P lies on the circle $BCII_a$, we have < BPC = < BIC. Now, it doesn't make sense to write things like $\measuredangle BIC=90^{\circ }+\frac{A}{2}$ here, since we are using directed angles modulo 180 and thus the angle $\frac{A}{2}$ isn't defined properly, but we can write

  $2\cdot \measuredangle BIC=2\cdot \measuredangle \left( BI;\;CI\right) =2\cdot \left( \measuredangle \left( BI;\;BC\right) +\measuredangle \left( BC;\;CI\right) \right) $
  $=2\cdot \measuredangle \left( BI;\;BC\right) +2\cdot \measuredangle \left( BC;\;CI\right) $
  $=\measuredangle \left( AB;\;BC\right) +\measuredangle \left( BC;\;CA\right) =\measuredangle \left( AB;\;CA\right) =\measuredangle BAC$.

Thus, $2\cdot \measuredangle BPC=2\cdot \measuredangle BIC=\measuredangle BAC$. Now, since < BP'C = < BAC - < BPC, we have

  $\measuredangle BP^{\prime }C-\measuredangle BPC=\left( \measuredangle BAC-\measuredangle BPC\right) -\measuredangle BPC=\measuredangle BAC-2\cdot \measuredangle BPC=0^{\circ }$,

so that < BP'C = < BPC, what implies that the points B, C, P and P' lie on one circle. The point Q must lie on the same circle, since we know that Q lies on the cirumcircle of triangle BCP'. Hence, < CBQ = < CP'Q. But < CP'Q = < CP'A = < CBA - < CPA; thus, < CBQ = < CBA - < CPA, and

  < ABS = < CBS - < CBA = < CBQ - < CBA
        = (< CBA - < CPA) - < CBA = - < CPA = < APS.

Thus, the points A, B, S and P lie on one circle. Similarly, the points A, C, R and P lie on one circle. [You can also easily show that the points A, C, Q and S lie on one circle, and that the points A, B, Q and R lie on one circle, but we won't need this.]

Now, since the points A, B, S and P lie on one circle, we have < BAS = < BPS. In other words, < (AB; AS) = < BPC. Since the points A, C, R and P lie on one circle, we have < RAC = < RPC. In other words, < (AR; CA) = < BPC. Thus,

  $\measuredangle \left( AR;\;AS\right) =\measuredangle \left( AR;\;CA\right) +\measuredangle \left( CA;\;AB\right) +\measuredangle \left( AB;\;AS\right) $
  $=\measuredangle BPC+\measuredangle CAB+\measuredangle BPC=2\cdot \measuredangle BPC+\measuredangle CAB=\measuredangle BAC+\measuredangle CAB=0^{\circ }$,

and it follows that the points A, R and S are collinear. Proof complete.

Thus, it follows that your locus contains the circles $BCII_a$ and $BCI_bI_c$. Whether it contains any other curves would be a nice question... but I have no time!

  Darij


limit
logarithms
function
trigonometry
Euler
calculus
calculus computations

Evaluate $ \lim_{x\to 0}\left[\frac{\pi}{2}\csc \left(\frac{\pi x}{2}\right)\minus{}\Gamma(x)\right]$, where $ \Gamma(x)$ is the gamma function.

Using the natural logarithm of the Weierstrass form of the $ \Gamma$ function, we get that near zero:

$ \ln \Gamma (z) \sim \minus{} \ln z \minus{} \gamma\,z$

and this gives

$ \Gamma(z) \sim \frac {1}{z} e^{ \minus{} \gamma\,z}$

Also, we have $ \sin z \sim z$ near zero, so the limit becomes:

$ \lim_{z \to 0} \frac {1 \minus{} e^{ \minus{} \gamma\,z}}{z} \equal{} ^{L'H} \gamma \lim _{z \to 0} e^{ \minus{} \gamma \, z} \equal{} \gamma$

where $ \gamma$ is the Euler-Mascheroni's constant and $ \equal{} ^{L'H}$ denotes the application of the L'Hospital's rule.


LaTeX
calculus
integration
limit
trigonometry
real analysis
calculus computations

Homework is homework, even if it's extra credit. Also attachments are discouraged unless absolutely necessary (and I see why, new user, no $ \text{\LaTeX}$ knowledge).

[hide="Hint"]Riemann sum to integral[/hide]

Maybe here should be:
\[ \lim_{n\to \infty} \sum_{k\equal{}1}^{n}{\frac{\sin(\frac{\pi}{k})}{k}}\]
 :wink:

Is the limit as n approaches infintiy equal to zero?

As written, yes. I'm quite sure that what was meant was $ \lim_{n\to\infty}\sum_{k\equal{}1}^n\frac{\sin\frac {\pi k}n}{n}$, though ;)

[quote="fedja"]As written, yes. I'm quite sure that what was meant was $ \lim_{n\to\infty}\sum_{k \equal{} 1}^n\frac {\sin\frac {\pi k}n}{n}$, though ;)[/quote]
 :)
In this case we have:
\[ \lim_{n\to \infty} \sum_{k\equal{}1}^{n}{\frac{\sin(\frac{\pi k}{n})}{n}}\equal{}\lim_{n\to \infty} \frac{1}{n}\sum_{k\equal{}1}^{n}{\sin(\frac{\pi k}{n})}\equal{}\int_{0}^{1}{\sin(\pi x) dx}\equal{}\frac{2}{\pi}\]


geometry
perimeter
inequalities
triangle inequality

$ABCD$ is a convex quadrilateral with perimeter $p$. Prove that \[ \dfrac{1}{2}p < AC + BD < p.  \] (A polygon is convex if all of its interior angles are less than $180^\circ$.)

We consider two possibilities: $AD > BC$ and $BC > AD$

If $AD > BC$:
Triangle inequality:
$AC < AB + BC$ and $BD < BC + CD$
So we see that \[ AC + BD < AB + 2BC + CD < AB + BC + CD + AD = p \]

If $BC > AD$
Triangle inequality:
$AC < AD + CD$ and $BD < AD + AB$
So again, \[ AC + BD < AB + 2AD + CD < AB + BC + CD + AD = p \]

What about the other side of the inequality? :D

We call $S$ the intersection of $AC$ and $BD$

Again triangle inequality:
$AB < AS + BS$, and $CD < DS + CS \Rightarrow AB+CD < AC + BD$ (1)

Triangle inequality:
$AD < AS + DS$ , and $BC < BS + CS \Rightarrow AD + BC < AC + BD$ (2)

We make the sum of (1) and (2):
\[ p < 2AC + 2 BD \iff \dfrac{p}{2} < AC + BD \]

So combine my two posts and get:

\[ \dfrac{p}{2} < AC + BD < p \]

QED !  :D


Αν  $ a\ge1$ , $ b\ge1$ , $ c\ge1$ ,   δείξτε ότι

 $ \frac {ab}{1 \plus{} ab} \plus{} \frac {bc}{1 \plus{} bc} \plus{} \frac {ac}{1 \plus{} ac}\ge$ $ \frac {a^2}{1 \plus{} a^2} \plus{} \frac {b^2}{1 \plus{} b^2} \plus{} \frac {c^2}{1 \plus{} c^2}$

Υ.Σ.  1 Αυτή η όμορφη ανισότητα αφιερωμένη για τους λάτρεις του αντικειμένου   :)
Υ.Σ.  2 Λέγοντας όμορφη δεν σημαίνει απαραίτητα και δύσκολη ...
Υ.Σ.  3 Η δυσκολία ενός θέματος είναι πάντα σχετική ... :lol:

Θέτουμε $ ab \equal{} 1/k$, $ bc \equal{} 1/t$, $ ac \equal{} 1/m$ και παίρνουμε $ a^2 \equal{} t/km$, $ b^2 \equal{} m/kt$, $ c^2 \equal{} k/tm$. Kάνουμε τις πράξεις και φτάνουμε στο αποτέλεσμα $ 1/(k \plus{} 1) \plus{} 1/(t \plus{} 1) \plus{} 1/(m \plus{} 1) >\equal{} 1/(km \plus{} t) \plus{} 1/(kt \plus{} m) \plus{} 1/(tm \plus{} k)$ Τα αντιστρέφουμε και παίρνουμε $ km \plus{} kt \plus{} tm >\equal{} 3$ όπου ισχύει. Άρα ισχύει αι η δεδομένη ανισότητα :)

[quote="vangelismak"]Αν  $ a\ge1$ , $ b\ge1$ , $ c\ge1$ ,   δείξτε ότι

 $ \frac {ab}{1 \plus{} ab} \plus{} \frac {bc}{1 \plus{} bc} \plus{} \frac {ac}{1 \plus{} ac}\ge$ $ \frac {a^2}{1 \plus{} a^2} \plus{} \frac {b^2}{1 \plus{} b^2} \plus{} \frac {c^2}{1 \plus{} c^2}$
[/quote]

Κατ' αρχάς προσθαφαιρόντας το 1 σε όλους τουσ αριθμητές η ανισότητα γίνεται:

$ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ac}\le\frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}$

Τώρα θα αποδείξουμε ότι

$ \frac {2}{1 \plus{} ab}\le\frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2}$.

Φανερά αν ισχύει αυτό προσθέτουμε κυκλικά τις ανισότητες που δημιουργούνται από τα ζεύγη $ (a,b)$ , $ (b,c)$, $ (a,c)$ και προκύπτει η ζητούμενη ανισότητα.

Πολλαπλασιάζοντας την παραπάνω με $ (1 \plus{} ab)(1 \plus{} a^2)(1 \plus{} b^2)$ προκύπτει:

$ 2 \plus{} 2a^2 \plus{} 2b^2 \plus{} 2a^2b^2\leq 1 \plus{} b^2 \plus{} ab \plus{} ab^3 \plus{} 1 \plus{} a^2 \plus{} ab \plus{} a^3b\Longleftrightarrow$

$ ab^3 \plus{} a^3b \plus{} 2ab \minus{} a^2 \minus{} b^2 \minus{} 2a^2b^2\geq0\Longleftrightarrow$

$ (ab \minus{} 1)(a \minus{} b)^2\geq 0$ που ισχύει αφού $ a\ge1$ , $ b\ge1$

[u]@Γιώργος[/u]
Μπορείς να εξηγήσεις τι εννοείς ότι τα "αντιστρέφουμε"; Γιατί αν κάνεις αυτό που νομίζω δεν μου φαίνεται σωστό...

Χίλια συγνώμη, κατάλαβα τη γκάφα μου. Τα αντέστρεψα και απο τη βιασύνη μου δεν πρόσεξα ότι αριθμητής πάει το 1 και παρανομαστής όλα τα υπόλοιπα κλάσματα.

Η λύση σου bilot4 ( Δημήτρης ? αν δεν κάνω λάθος ) είναι ακριβώς ίδια με τη διαδικασία κατασκευής της παραπάνω άσκησης !  :)


trigonometry
geometry
3D geometry
sphere
calculus
integration
analytic geometry

why does (sine)^2(theta) come out to average 2/3 over a sphere?   :huh:

What do you mean by "theta"?

Do you mean that the average of the squared sines of all the angles within a sphere is $ \frac{2}{3}$ or something?

I'm not sure I understand....

Perhaps an integral over the sphere is involved...?

There are two angles used in the spherical coordinate system for $ \mathbb{R}^3:$ the longitude and the co-latitude. As for which letter to use for which angle, conventions vary. The American calculus books I teach this material from typically use $ \theta$ for the longitude and $ \varphi$ for the co-latitude. In that convention, the average of $ \sin^2\theta$ would be the same as the average of $ \sin^2\theta$ over a circle, namely $ \frac12.$ It's clear that the original poster means for $ \theta$ to be the co-latitude, not the longitude.

I'm going to switch over to that American calculus book notation, which makes the original question that of the average of $ \sin^2\varphi.$

We have $ \begin{cases}x\quad \equal{} & r\sin\varphi\cos\theta \\
y\quad \equal{} & r\sin\varphi\sin\theta \\
z\quad \equal{} & r\cos\varphi\end{cases}.$

Now, and this is a very important thing, the surface area element on the unit sphere is $ dS \equal{} \sin\varphi\,d\varphi\,d\theta.$ This is a portion of the whole volume element, which is $ dV \equal{} r^2\sin\varphi\,dr\,d\varphi\,d\theta.$

Let $ \Sigma$ denote the whole sphere, or all possible angles. We note that $ \varphi$ ranges from $ 0$ to $ \pi,$ while $ \theta$ ranges around a circle.

$ \iint_{\Sigma}dS \equal{} \int_0^{\pi}\int_0^{2\pi}\sin\varphi\,d\theta\,d\varphi \equal{} 2\pi\int_0^{\pi}\sin\varphi\,d\varphi \equal{} 4\pi.$

This agrees with what we know about the area of a sphere.

Then the average value of $ \sin^2\varphi$ will be this:

$ \frac1{4\pi}\iint_{\Sigma}\sin^2\varphi\,dS \equal{} \frac1{4\pi}\int_0^{\pi}\int_0^{2\pi}\sin^3\varphi\,d\theta\,d\varphi$

$ \equal{} \frac {2\pi}{4\pi}\int_0^{\pi}\sin^3\varphi\,d\varphi \equal{} \frac12\int_0^{\pi}(1 \minus{} \cos^2\varphi)\sin\varphi\,d\varphi$

$ \equal{} \left.\frac12\left( \minus{} \cos\varphi \plus{} \frac13\cos^3\varphi\right)\right|_0^{\pi} \equal{} \frac23.$

I believe that physicists tend to use the reverse convention- $ \theta$ for colatitude and $ \phi$ for longitude. The inconsistent notation isn't going away anytime soon.

One comment on terminology: while I don't think "co-latitude" or "colatitude" are common terms, they make sense to me. In geography, latitude is $ 0^{\circ}$ at the equator, $ 90^{\circ}\text{N}$ at the north pole, and $ 90^{\circ}\text{S}$ at the south pole. By contrast, the angle usually used in spherical coordinates is $ 0$ at the north pole, $ \frac{\pi}2$ ($ 90^{\circ}$) at the equator, and $ \pi$ ($ 180^{\circ}$) at the south pole. That's the complementary angle to the (north) latitude - hence co-latitude.

thanks, that was pretty neat. I did come across it in a physics text, which didn't say why it was so.

Also, it seems that the average of x-coordinates over a circle is (1/3)(radius of circle)^2 ??


inequalities
function
three variable inequality
convex-concave inequalities

Ok,i will post the solution by use SOS.
First one,we have:
             $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})-9=\sum \frac{(x-y)^2}{xy}$
             $6(\sum \frac{x}{y+z})-9=\sum \frac{(x-y)^2}{(x+z)(y+z)}$
So that the inequality equavilent to prove
               $\sum \frac{(x-y)^2}{xy} \geq \sum \frac{(x-y)^2}{(x+z)(y+z)}$ (when $x,y,z \in [1,2]$)
Put:$S_{x}=\frac{1}{yz}-\frac{3}{(x+y)(z+x)}$
      $S_{y}=\frac{1}{xz}-\frac{3}{(x+y)(z+y)}$
      $S_{z}=\frac{1}{yx}-\frac{3}{(x+z)(z+y)}$
WLOG $x \geq y \geq z$
Se we can easily to prove $S_{x} \geq 0$.Now we're going to prove $S_{y} \geq 0 \Longleftrightarrow xy+zy+y^2-2xz \geq 0$ (1)
we have $xy \geq xz$ (from the condition);$y+z \geq x$ (because $x \leq 2$ and $y,z \geq 1$) so that  $y(y+z) \geq xz$
Hence (1) is proved
If $S_{z} \geq 0$ then the problem is proved
If $S_{z} \leq 0$.You can prove that $S_{y}+2S_{z} \geq 0;S_{x}+2S_{z} \geq 0$
So that from the critetion of SOS we have that $\sum S_{x}(y-z)^2 \geq 0$
Therefore the problem is Proved :D  :D
the equality hold when $x=y=z$ or $y=z=1,x=2$

[quote="Bluesea"]the function is clearly convex for all it'svariable[/quote]

WHY? I DON'T THINK SO!

[quote="hungkhtn"] Only prove $S_x+S_z \ge 0$. [/quote]
Maybe $S_y+S_z \ge 0$ ?

Yes, sorry  :oops: . To prove this, notice that $a \le b+c$.

A quite easy proof using Jensen's:

(x+y+z) ( 1/x + 1/y + 1/z ) >= 6 ( (x/y+z)+(y/x+z)+(z/x+y) )
<--> ( x+y+z ) ( 1/x + 1/y + 1/z ) + 18 >= 6 ( x+y+z ) ( 1 / (y+z) + 1 / (x+z) + 1 / (x+y) )
Because inequality is homogenous, let x+y+z=1. original condition becomes 1/5 <= {x,y,z} <= 1/2 [since 1/5 and 1/2 are the lower and upper bounds for x/x+y+z since x,y,z are in the interval]
<--> 1/x + 1/y + 1/z + 18 >=  6 ( 1/(1-x) + 1/(1-y) + 1/(1-z) )
<--> [ 1/x - 6/1-x + 6 ] + [ 1/y - 6/(1-y) + 6 ] + [ 1/z - 6/(1-z) + 6 ] >= 0
let f ( x ) = 1/x - 6/(1-x) + 6. It is sufficient to prove f is concave up for all x in the interval [1/5,1/2], since then it would follow by Jensen's that f(x) + f(y) + f(z) >= 3 f (x+y+z/3) = 3f(1/3) = 0 as required.
proving it is concave is a simple matter of proving f''(x) >= 0 for x in that interval, so we are done.

$\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 3^2=9$ -> this is a particular case of $\left(xa+yb+zc\right)\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\geq \left(a+b+c\right)^2$ which can be proved with Cauchy-Buniakowsky-Schwartz. Now it remains to prove that $\sum \frac{x}{y+z} \geq  \frac{3}{2}$ , which is true (Nesbitt inequality). So $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 6\left(\sum \frac{x}{y+z}\right)$ QED

No,you just prove both side is larger than 9,but not the original inequality.

[quote="hungkhtn"]I just see this problem in this moring and i think it's quite easy from SOS. Nhat's solution is nice but i think It's too long. Only prove $S_x+S_z \ge 0$. And the stronger problem is replacing the condition $x,y,z \in [1,2]$ by $x,y,z$ are three lengths of a triangle.[/quote]
Ok, I have same idea with Hungkhtn. :D
Problem very easy. Nhat used Nesbit inequality.
Nhat's solution is nice but i think It's too long. Only prove .
My solution is :
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 6(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})$
 $\Leftrightarrow \frac{a}{b} + \frac{b}{a}+ \frac{b}{c}+ \frac{c}{b}+ \frac{c}{a}+ \frac{a}{c}+3 \ge 6[ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}]$
 $\Leftrightarrow [\frac{a}{b} -2 +\frac{b}{a}]+[\frac{b}{c}-2+\frac{c}{b}]+[\frac{c}{a}-2+\frac{a}{c}] \ge 3[ \frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}-3]$
 $\Leftrightarrow \frac{(a-b)^2}{ab}+ \frac{(b-c)^2}{bc}+ \frac{(c-a)^2}{ca}\ge 3[ \frac{(a-b)^2}{(a+c)(b+c)}+ \frac{(b-c)^2}{(b+a)(c+a)}+ \frac{(c-a)^2}{(c+b)(a+b)}]$
 $\Leftrightarrow \sum (a-b)^2[ \frac{1}{ab}- \frac{3}{(a+c)(b+c)}]\ge 0$
$\Leftrightarrow \sum(a-b)^2(a+b)(c^3+c^2a+b^2a-2abc)\ge 0$
We will SOS method:
$S_a=(b+c)a(a^2+ab+ac-2bc)$
$S_b=(c+a)b(b^2+ba+bc-2ac)$
$S_c=(a+b)c(c^2+ca+cb-2ab)$
WLOG we assume $a\ge b\ge c$.
Se we can easily to prove $S_a, S_b\ge 0$ Now we're going to prove
$S_a+S_c=(b+c)a(a^2+ab+ac-2bc)+(a+b)c(c^2+ca+cb-2ab) \\ \ge (ac+bc)(a^2+ab+ac-2bc+c^2+ca+cb-2ab)\ge 0$
(it true) :D

I think your solution is same as nhat's.Isn't it? :roll:

[quote="duongqua"]Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds:
\[ (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}).  \]
When does the equality occur?[/quote]

Using CBS we obtain that
  \[ \sum x \sum\frac{1}{x} \geq 3^2=9  \]
   Now all we have to do is prove that
   \[ \sum \frac{x}{y+z} >\frac{3}{2}  \]
that is true
.......

Nesbitt's:
$a,b,c>0$
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac32  \]

????
All two side are larger than 9. So you can ???????
Wrong!

That's great if someone can post official solution!

luke999, darkdevil first think then post

Proof using [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108778]Mixing Variables[/url] :D
We may assume that $x\ge y \ge z$.
Let
\[f(x,y,z)=(x+y+z)\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-6\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right) \]
We have

$f(x,y,z)-f(x,t,t)=(y-z)^{2}(x+y+z)\left[\frac{(y+z)(3x^{2}+xy+xz-5yz)+2x(x^{2}+zy)}{yz(x+y)(y+z)(z+x)(2x+y+z)}\right]\ge 0$

where  $\large t=\frac{y+z}{2}$.

It enough to prove that $f(x,y,y) \ge 0$ where $1\le \frac{x}{y}\le 2$, with is equivalent to $(x-y)^{2}(2y-x)\ge 0$. So, the proof is complete.

[quote="duongqua"]Prove that for all real numbers $ x,y,z \in [1,2]$ the following inequality always holds:

$ (x \plus{} y \plus{} z)\left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)\geq 6\left(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y}\right).$[/quote]$ \left(\sum{x}\right)\left(\sum{\frac {1}{x}}\right) \minus{} 6\sum{\frac {x}{y \plus{} z}}$

$ \equal{} \frac{1}{4xyz}\sum{\frac {(y \minus{} z)^2}{(z \plus{} x)(x \plus{} y)}\left[(x \plus{} y \plus{} z)(3x \minus{} y \minus{} z)^2 \plus{} (y \plus{} z \minus{} x)(z \plus{} x \minus{} y)(x \plus{} y \minus{} z)\right]}$

$ \geq0,$ with equality if $ x \equal{} 2,y \equal{} z \equal{} 1.$

This problem is introduced as Exercise 1.22 on page 33 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.bookschina.com/4277335.htm]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

[quote="sqing"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=615194]Strengthen of  Viet Nam TST 2006:[/url]
Let $x,y,z \in[1,2]$.Prove that\[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{18}{x+y+z}\ge 6\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right).\][/quote]
After using Ravi's substitution we need to prove that
$\sum_{cyc}\frac{1}{x+y}+\frac{9}{x+y+z}\geq6\sum_{cyc}\frac{1}{2x+y+z}$ for positives $x$, $y$ and $z$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, it's obvious that our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Thus, by $uvw$ it remains to prove the last inequality in two cases:
1. $y=1$, $z\rightarrow0^+$, which gives $(x-1)^2(2x^2+3x+2)\geq0$;
2. $y=z=1$, which gives $x(x-1)^2\geq0$. Done!

See also here
http://www.artofproblemsolving.com/community/c6h123459p699421

[quote=Ji Chen][quote="duongqua"]Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds:
\[(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}). \]
[/quote][/quote]
http://www.artofproblemsolving.com/community/c275h116646p662466:
Fie $a,b,c$ laturile unui triunghi. Sa se demonstreze inegalitatea:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq  6(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$

[quote=sqing]http://www.artofproblemsolving.com/community/c275h116646p662466:
Fie $a,b,c$ laturile unui triunghi. Sa se demonstreze inegalitatea:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq  6(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$[/quote]


Using k,al,a,m,a,z,o,o  how many words can be written that   o  letter before  k  letter?

At least one $o$ or both $o$'s are to be placed before $k$?

You tape "o" and "k" together as if they were one letter

and then you have ok,al,a,m,a,z,o 7 letters

and the number of ways to arrange those are 7!

however, you have 2 o's

so i think you need to double that number by 2

so the answer would be $2*7! = 10080$

please note you did not give exact clarification of a "word"

if a word can be written with less than all of the letters

then the answer is not 10080


geometry
3D geometry
sphere
geometric transformation
dilation
invariant

[quote="TheHeinrich"]Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of 8, and the largest marble has a radius of 18. What is the radius of the middle marble?[/quote]

You used copy/paste. :wink:

[hide="isn't it"]the average of 18 and 8, which is 13?[/hide]

a) You can't have an apostrophe in the hide tag
b) You should actually try to be rigorous before guessing an answer. It is the geometric mean, not the arithmetic mean; this follows from similar triangles.

Well, if it's like Catalyst said and it's the geometric mean then it would be $ \sqrt {8*18}$ so that would 12?

[b]Visually:[/b] Say you have a cone with a vertex at the origin, and two spheres shoved inside here that are tangent to the walls and to eachother. If we do an origin-centered dilation, the cone will remain invariant because the edges are formed by lines that pass through the origin. However, the spheres will become larger or smaller based on the dilation factor.

So say we have one sphere of radius $ 8$ and another sphere of radius $ r$. Now, let's dilate this setup by a factor of $ \frac{r}{8}$. Thus, the smaller sphere has a radius of $ r$ and is exactly where the previously radius-$ r$ sphere was. The larger sphere has a radius of $ \frac{r^2}{8}$, which we know from the problem to be equal to $ 18$. Thus, $ r^2 \equal{} 8 \cdot 18 \implies r \equal{} 12$.

[quote="TheHeinrich"]how do you rigorously do it?[/quote]

The main idea is similar triangles.


integration
logarithms
calculus
calculus computations

Obtain a reduction formula for $ \int \frac{x^n}{(\log x)^m} dx$.

hello
[hide="Solution"]
$ I_{m,n}\equal{}\int \frac{x^{n}}{(log(x))^{m}}$
$ I_{m,n}\equal{}\int \frac{x^{n\plus{}1}}{x(log(x))^{m}}$
$ I_{m,n}\equal{}\frac{x^{n\plus{}1}}{(1\minus{}m) (log(x))^{m\minus{}1}}\plus{}\frac{n\plus{}1}{m\minus{}1} I_{m\minus{}1,n}$
[/hide]
thank u

That is correct. The solution made use of this fact: $ \int [f(x)]^n f(x)' dx \equal{} \frac {[f(x)]^{n \plus{} 1}}{n \plus{} 1}$, for $ n \neq \minus{} 1$.
Then, in the integration by parts, we set $ u \equal{} x^{n \plus{} 1} \implies u' \equal{} (n \plus{} 1)x^n$ and $ v' \equal{} (\log x )^{ \minus{} m} \frac {d}{dx} \log x \implies v \equal{} \frac {(\log x )^{ \minus{} m \plus{} 1}}{ \minus{} m \plus{} 1}$.


algebra
polynomial
calculus
integration
number theory
relatively prime
prime numbers

excuse me i think there's a fact that "all prime divisors of $ 1\plus{}x\plus{}x^{2}\plus{}...\plus{}x^{n}$ are either $ p$ or $ 1 (mod p)$." could someone explain it to me please? a proof along with it will be more useful. thank you :)

Thats only true for $n=p^{k}-1$ ;)

But that's only a special case (since $\Phi_{p}= x^{p-1}+...+x^{2}+x+1$, $\Phi_{p^{m}}=\Phi_{p}(x^{p^{m-1}})$ and $x^{p^{k}-1}+...+x^{2}+x+1 = \prod_{m=1}^{k}\Phi_{p^{k}}$):

Let $\zeta_{n}= e^\frac{2 \pi i}{n}$ be the smallest $n$-th primitive root of unity and then $\Phi_{n}(x) : = \prod_{\substack{1 \leq k \leq n \\ \gcd(k,n)=1}}(x-\zeta_{n}^{k})$ be the $n$-th cyclotomic polynomial.

[b]Lemma 1:[/b] $\prod_{d|n}\Phi_{d}(x)=x^{n}-1$
Proof: compare the roots.

[b]Lemma 2:[/b] $\Phi_{n}(x)$ is a polynomial with integers as coefficients.
Proof: use Lemma 1 and polynomial division.

[b]Lemma 3:[/b] Let $p$ be prime, $a$ be an integer, $p \nmid a$, and $k: =\mathrm{ord}_{p}(a)$ be as always the smallest positive integer with $a^{k}\equiv 1 \mod n$. If $a^{m}\equiv 1 \mod p$, then $k|m$.
Proof: Use division with remainder to write $m=qk+r$, $0 \leq r <k$ and show that $r=0$.

[b]Corollary:[/b] $k=\mathrm{ord}_{p}(a)$ divides $p-1$.
Proof: use Lemma 3 and Fermat's little theorem.

[b]Lemma 4:[/b] Let $p$ be prime. $x^{n}-1$ can only have a double root $a \mod p$ if $p|n$, in other words: if there is an integer $a$ and an integral polynomial $h(x)$ with $x^{n}-1\equiv (x-a)^{2}h(x) \mod p$, then $p|n$.
Proof: substitute $y=x-a$, then simple calculation using the binomial theorem.

[u][b]Theorem:[/b][/u] For all integers $a$ and prime divisors $p$ of $\Phi_{n}(a)$ we have $p \equiv 1 \mod n$ or $p|n$.
Proof: We clearly have $\Phi_{n}(a) \equiv 0 \mod p$ by definition of $p$, thus $\Phi_{n}(x) \equiv (x-a)g(x) \mod p$ for some polynomial $g(x)$.
1. case: $k: =\mathrm{ord}_{p}(a) < n$, giving $k|n$ using Lemma 3.
Then $a^{k}\equiv 0 \mod p$, thus, $x^{k}-1 \equiv (x-a)h(x) \mod p$ for some polynomial $h(x)$.
Using Lemma 1 twice, we get $x^{n}-1 \equiv (x-a)^{2}g(x)h(x) \mod p$, thus $p|n$ by Lemma 4.
2. case: $k: =\mathrm{ord}_{p}(a) = n$.
Then $n|p-1$ by the corollary, thus $p \equiv 1 \mod n$ as it should be.
3. case: $k: =\mathrm{ord}_{p}(a) > n$.
This cannot happen since $a^{n}\equiv 1 \mod p$ be Lemma 1.

[b]Corollary:[/b] There are $\infty$ many primes $p \equiv 1 \mod n$.
Proof: easy, try yourself.

[b]Corollary:[/b] There are $\infty$ many primes $p \equiv 1 \mod n$.
Proof: easy, try yourself.[/quote]

Doesn't it come directly from Dirichlet's theorem?

[quote="ZetaX"]Tried to prove the full Dirichlet theorem yourself¿ ;)[/quote]

[b]I[/b], prove [b]Dirichlet's theorem[/b]? are you nutts? :P

okey, here's the proof of the corollary:(compare with Euclid's)

assume that there are only finitely many primes of the form $tn+1$.
Let $P_{k}=kn+1$ be the largest prime of that form.
consider the number $P=P_{1}P_{2}...P_{k}+1=(n+1)(2n+1)...(kn+1)+1$. note that $P$ is a composite number of the form $tn+1$. and a number of this form must have a prime divisor of this form. but it is relatively prime to the rest of the prime numbers of the form. so we can conclude that a prime > $P_{k}$ divides $P$ and we are done.$\blacksquare$

Please tell me the solution's correct!

[quote="nayel"][quote="ZetaX"]Tried to prove the full Dirichlet theorem yourself¿ ;)[/quote]

note that $P$ is a composite number of the form $tn+1$. and a number of this form must have a prime divisor of this form. [/quote]

can someone explain this fact? how about you ZetaX?

I don't know what you mean...
- $P$ is not of the form $kn+1$, otherwise $P-1 = P_{1}\cdots P_{k}$ would be a multiple of $n$...
- If $P$ is of the form $kn+1$, then not necessarily every prime factor of $P$ is of that form! Take $n=4$, $9 = 3 \cdot 3$

ZetaX did not write all that very interesting stuff for nothing... I think you shoud use those facts to prove che "corollary"   :roll:
bye!

[quote="edriv"]ZetaX did not write all that very interesting stuff for nothing... I think you shoud use those facts to prove che "corollary"   :roll:
[/quote]

well i'll try to study them soon but not rigth now. i have important exams. but don't worry i saved them in my file. :D

[quote="edriv"]ZetaX did not write all that very interesting stuff for nothing... I think you shoud use those facts to prove che "corollary"   :roll:
bye![/quote]
Yes, that's how it was intended. I know no easier way to prove Dirichlet's theorem for $1 \mod m$, $m$ arbitrary. The case $-1 \mod m$ is still possible, but harder (same for some other types of special cases).

Sorry by the way for not responding earlier, I was away the next day and didn't see this thread when I returned...

[b]Lemma 1[/b] the set of prime divisors of the image set of a polynomial with integer coeffcients is infinite. (well, we suppose deg(P) > 0)
By contradiction: if $p_{1},p_{2},\ldots,p_{k}$ are the elements of this finite set and $P(x) = a_{n}x^{n}+\ldots+a_{1}x+a_{0}$, then $P(p_{1}p_{2}\cdots p_{k}a_{0}) = a_{0}(a_{0}p_{1}\cdots p_{k}(\mbox{stuff})+1)$, where stuff can be very big and $p_{i}\nmid a_{0}p_{1}\cdots p_{k}(\mbox{stuff})+1$ for i=1,2,...,k.

Now just apply Lemma 1 to $\Phi_{n}(x)$. Since the prime divisors of n are finite, and every prime p such that $p \mid \Phi_{n}(a), p \nmid n$ is 1 modulo n, there are infinite primes of the form $kn+1$.

Good night $\zeta x$!

Or else take a prime divisor of $ \Phi_{n}(n)$ . Prime p divides $ \Phi_{n}(n)$ divides $ n^{n}\minus{}1$ implies $ (p,n)\equal{}1$ and $ p\equal{}1(modn)$ and there are infinitely many of them


Are there any useful properties of polynomials that have positive values for all integers? Or all real numbers?

What do you mean by "useful"?  I could make up some weird properties that are true, like, if
$ f(x)\equal{}a_{n}x^{n}\plus{}\cdots\plus{}a_{1}x\plus{}a_{0}$
is a polynomial, then $ a_{n}^{2}\plus{}\cdots\plus{}a_{1}^{2}\plus{}a_{0}^{2}\geq 0$.  Is this "useful"?  Probably not, but it fulfills your requirement.

This is a very vague question, but perhaps you have something like [url=http://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem]Hilbert's 17th problem[/url] in mind. The solution to this problem (one of the first to be solved) says in particular that a polynomial which takes only non-negative values over the reals is a sum of squares of rational functions.


AIME problem #1 from 1997, a sample problem on the AMC website:

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

I have failed to solve this problem. This question #1 from an AIME seems considerably more difficult than the #1's from most AIME's. Any insight on the solution would be appreciated, thanks.

Fierytycoon

The difference of squares = a:^2:-b:^2: = (a-b)(a+b). What can you tell about parity (even/odd) of these two brackets?

All the odd numbers would be acheivable, because if x and y are consecutive, (x+y)(x-y) = 1 x (x+y) which would be odd because x+y = x+x+1 = 2x+1 which will be a odd number. Hope I was a help.

Thanks for your responses. Hmm, well, I have gotten to a certain point:

Let us say that 1 <= n <= 1000, and let n = a :^2: - b :^2: = (a+b)(a-b).
Now let us take two cases.

1) Let a = b + 1. Substituting, we have n = 2b + 1, and so for b = 0, 1, 2, ..., all odd numbers n can be satisfied.

2) Let a = b + 2. Substituting, we have n = 4(b + 1), and so for b = 0, 1, 2, ..., all even numbers n that are divisible by 4 can be satisfied.

Thus, 3/4 of the cases can be satisfied, leading to an answer of 3/4 * 1000 = 750, which is the correct answer. However, I am still a bit vexed. I can show that 3/4 of the cases can be satisfied, but I have NOT shown that the other 1/4 of the cases CANNOT be satisfied. How do I know that if I let a = f(b), where f(b) is some function of b that I have not yet tried (such as f(b) = b + k), then more n could be satisfied?

Fierytycoon

Two proofs:

Proof one: All perfect squares are congruent to 0 or 1 (mod 4), so you can't get to 2 (mod 4) by subtracting them.

Proof two: When you factor the difference as (a-b)(a+b), either both factors are even or both are odd. If they're both odd, you're not going to get 2 (mod 4). If they're both even you'll get a number divisible by 4, so it will be 0 (mod 4).

ComplexZeta,

Thanks for your reply. However, since I am not too proficient at number theory (currently reading up on it right now), I do not understand how you came up with your statement:

"If they're both odd, you're not going to get 2 (mod 4)."

Is there a quick proof?

Fierytycoon

If a and b are both odd, a+b and a-b are both even. So (a+b)(a-b) is a multiple of 4, and thus can't be 2 mod 4.

TripleM,

Thanks for yoru response. However, I was confused about the case when the factors, not a and b, were both odd. as I understand, the "they're" in "they're both odd" refer to the factors, not a and b. I think you thought that it was a reference to a and b, and thus, took the case when the factors were even.  :D

Fierytycoon

Oh whoops, your right. Well if both brackets are odd, then their product is odd but 2 (mod 4) is always even.

lol, whoops, sorta missed that...that was sorta simple.  :lol: Anyways, thanks for your response TripleM.

Fierytycoon

Ahhh.. i remember this. It wasn't that hard, just think about how to express all odds and evens as diff of squares. [i]Especially the evens.[/i]

This problem is made simple using parity. All of the numbers from 1-1000 can be expressed as the difference of two squares if and only if (a+b)(a-b) both have the same parity. All numbers work that are odd and even, but there are special cases when the number is in the form, 4n+2. Because this factors as 2(2n+1) which have different parities. This happens 1/4 of the time. So 1000-1000/4= 1000-250=750 which is your answer hope this helped.


Let x, y, z, t be acute angles with sum  \pi. Show that
1/cos(x) + 1/cos(y) + 1/cos(z) + 1/cos(t) <=  \sqrt (2)(tan(x)+tan(y)+tan(z) + tan(t))

[i]unknown source[/i]

Namdung

I mus admit that this is a better variant of the inequality proposed by me in the Proposed problems section. But it is exactly the same problem. I won't post my solution now, I prefer to let all the others think a little bit about the problem.

Yes, and solution based one Harazi indentity
Sum (1+tg(x)^2)/(tg(x)+t(y)) = Sum tg(x)  :)

Namdung


function
integration
real analysis
real analysis unsolved

Let $ f: [0,1]\to\mathbb{R}$ be a continuous function such that
\[ \int_{0}^{1}f(x)dx\equal{}\int_{0}^{1}xf(x)dx.\]
Prove that there exists $ c\in (0,1)$ such that
\[ \int_{0}^{c}xf(x)dx\equal{}0.\]

Yet another typical Cezar Lupu problem.

[color=blue]Lemma. Suppose that $ g : [0,1]\to\mathbb R$ is differentiable and that $ \frac{g(1)\minus{}g(0)}{1\minus{}0}\equal{} g^{\prime}(0)$. Then, there is $ \xi\in (0,1)$ such that
\[ g^{\prime}\left(\xi\right)\equal{}\frac{g\left(\xi\right)\minus{}g(0)}{\xi\minus{}0}.\]
[/color]

[i]Proof.[/i] Let $ h(x) \equal{}\frac{g(x)\minus{}g(0)}{x\minus{}0}$. If there is $ \xi\in (0,1)$ such that $ h^{\prime}\left(\xi\right) \equal{} 0$, then we're done.
Else, WLOG $ h$ is strictly increasing, i.e. $ \frac{g(1)\minus{}g(0)}{1\minus{}0}>\frac{g\left(\epsilon\right)\minus{}g(0)}{\epsilon\minus{}0}$.
Letting $ \epsilon\to 0$, we see that $ \frac{g(1)\minus{}g(0)}{1\minus{}0}> g^{\prime}(0)$, contradiction.

Now just apply this lemma to $ G$ such that $ G^{\prime\prime}\equal{} f$.


let $l,m,n$ be three lines ,wich we draw on earth and intersect each other in point $s$ prove that these three lines

also have another common point.

What do you mean by a line? How do you define straight on a sphere?
Isn't that like showing a "line" is a great circle of a sphere? If you define a line to be a great circle, then the result is trivial.

But if they aren't great circles, isn't the result untrue? :huh:

ure right guys ...

and because of this i said:we draw them on $earth...$

i want say that a line is a circle wich its radius is very very geat... [/list]


Let $ABCD$ be a cyclic quadrilateral. Let $X$ be the foot of the perpendicular from the point $A$ to the line $BC$, let $Y$ be the foot of the perpendicular from the point $B$ to the line $AC$, let $Z$ be the foot of the perpendicular from the point $A$ to the line $CD$, let $W$ be the foot of the perpendicular from the point $D$ to the line $AC$.

Prove that $XY\parallel ZW$.

  Darij

The QED is a German mathematics association with about 150 members, most of them young (middle school, high school university students), many of them participants in regional,national or even international mathematical competitions.

The QEDMO is some sort of mathematical competition inside the QED. It has taken place twice so far, and works somewhat like this: When there's a meeting of the QED with a sufficient number of participants, the QEDMO Jury proposes a list of 14 problems. The participants are now split into two teams. Both teams now try to solve as many problems as possible (as QED meetings are approx. 3-4 days, problems are given to the teams on the first evening, solutions will be presented on the last day of the meeting), then present their solutions as follows:

One team starts and asks the other team to present their solution to a particular problem. The other team may accept or decline the challenge. If they decline, the first team has to present the solution. The team which presents their solution chooses one of their team members which will do that, the other team chooses a critic who may ask questions during the presentations and point out mistakes in the solution. If there are mistakes in the solutions that the presenting team's "speaker" cannot resolve, the critic is given the opportunity to correct the mistake or give his own proof. At the end, the jury awards 12 points distributed among the teams: For example, if the presenting team has solved the problem without a flaw, they will get 12 points. If the presenting team has made a minor mistake which they can't correct but the critic can, the critic's team receives 2 points, the presenting team receives 10; you get the idea. If the problem can't be solved by either the presenter or the critic, the jury will be awarded all remaining points. After points are distributed, the other team now asks for a solution to another problem.

There are some additional rules which I don't remember or can't explain correctly, but I think you get the general idea. If there are mistakes in my description (which is quite possible because I just attended 1 of the 2 QEDMOs so far), I'm sure the jury members will point them out.

[quote="darij grinberg"]Let $ABCD$ be a cyclic quadrilateral. Let $X$ be the foot of the perpendicular from the point $A$ to the line $BC$, let $Y$ be the foot of the perpendicular from the point $B$ to the line $AC$, let $Z$ be the foot of the perpendicular from the point $A$ to the line $CD$, let $W$ be the foot of the perpendicular from the point $D$ to the line $AC$.

Prove that $XY\parallel ZW$.

[/quote]
Indeed very easy.  :)

Since $\angle{AXB}+\angle{AYB}=\frac{\pi}{2}+\frac{\pi}{2}=\pi$, so $A, X, B, Y$ are cyclic. Hence $\angle{ABX}=\angle{AYX}$.

Since $\angle{AWD}=\angle{AZD}=\frac{\pi}{2}$, so $A, W, Z, D$ are cyclic. Therefore $\angle{AWZ}+\angle{ADZ}=\pi$.

Because $A, B, C, D$ lie on the same circle, so we have: $\angle{ABX}=\angle{ADZ}$, which means:$\angle{ABX}+\angle{AWD}=\pi$, so $\angle{AYX}+\angle{AWD}=\pi$, thus $\angle{XYW}=\angle{YWZ}$, which means $XY || WZ$.


Last year I was new to math, and applied to the only two summer math programs I knew of, however this year I know of a bunch and I was thinking about applying to 4 or 5, but that seems somewhat excessive. I mean the applications are fun so I will do them anyway, but I was wondering how many other people are applying to.

Perhaps a poll would help show this. Me personally probably two or three.

What happened to me was I found a couple I was really interested in, and decided to start applying to all of them. However, the response (and acceptance) from one of the first two camps I applied to came before the deadline for the application to the rest, so I only ended up applying to two.

At my level and what I was looking for in a math camp, I considered several but ended up applying to only Awesomemath.  If I didn't get in, then I would have considered other options.  However, I am glad that it didn't come to that.  Apply to as many as you want, but make sure that you know what each offers and for what you are looking.


Find the finished digit of this number \[ 9999^{999^{99^9}}\]

You should not post such a problem in this forum.
Please make sure where you should post some problem?

Japanese Communities Moderator

kunny


I have lots of questions:
1. Is it possible to take 2 classes in one day: AIME/ Intermediate Trig?
2. I understand 70% of Intro Counting, but only 10% of Intermediate Counting. Should I take intro, or can I learn it through the aops book?
3. How much will I be affected if I miss the wednesday geometry class, but am there for fridays? I really want to learn geometry. Can I catch up?
4. How much classes it possible to take?
5. Does this schedule look good?

Tuesday: Intermediate Trig or/and AIME
Wed: Geometry
Thursday: Intro to Counting
Friday: Geometry

6. Also: will more courses be offered in the spring of 2005?

Some answers:

1. Is it possible to take 2 classes in one day: AIME/ Intermediate Trig?

If the classes are scheduled at different hours, there is not reason why you couldn't. However if the classes are at the same hour, even if you now can actually log in into two instances of classes, I belive it is hard to catch up on two different subjects online at the same time.

What you could do is to alternatively participate in each class, obtaining the transcript for the other one.

2. I understand 70% of Intro Counting, but only 10% of Intermediate Counting. Should I take intro, or can I learn it through the aops book?

You should definitely take a class if you feel that you only handle 70% of the stuff the class is covering. In plus, almost all intermediate classes require matters discussed previously in the intro class.

3. How much will I be affected if I miss the wednesday geometry class, but am there for fridays? I really want to learn geometry. Can I catch up?

You can use transcripts to see what happened in the class. Catching up differs from individual to individual, but usually it's not much of a problem.

4. How much classes it possible to take?

You can take as many classes as you please. You can even register for classes just to take the transcripts, there is no need for you to actually be present in each class.

5. Does this schedule look good?

Tuesday: Intermediate Trig or/and AIME
Wed: Geometry
Thursday: Intro to Counting
Friday: Geometry

Richard, Mathew or David can answer this one :)

6. Also: will more courses be offered in the spring of 2005?

Yes, AoPS plans to offer more courses in the future. Just stay tooned :)

Hope this helps.

When you say you understand 70% of the material, does that mean you can solve 70% of the problems? If so, that class will probably move too slowly for you. On the other hand, if you know 70% of the topics, it would still be wise for you to take the class. Some of the problems covered in class are quite difficult, and it will be useful for you to see how to apply the theorems to solving challenging problems.

I'll second what ComplexZeta said. It is possible to be familiar with essentially ALL the topics in an AoPS course (from previous curricular math study in school) and still need lots practice in actually solving problems on those topics. My advice, when in doubt, is to register for the easier of two AoPS courses if you think your own level is somewhere in between the level of each course.

Good luck in your course. I think the AoPS courses are an excellent, challenging look beyond what is usually taught in school math courses.


abstract algebra
linear algebra
matrix
geometry
projective geometry
superior algebra
superior algebra solved

Hello all mathematicians out there!.

Lef u:L->F be a homomorphism of finitely generated free A-modules (A=commutative local ring with identity), fix a pair of bases for L and F respectively, and let M be the associated matrix of u. If n=rank(F) and N is the annihilator of the cokernel of u, show that the radical of N equals the radical of I, where I is the ideal of A generated by all nxn minors of M.
(We assume rank(L)>rank(F)).

Remark: I found this problem in a commutative algebra book by C.Peskine (An algebraic introduction to complex projective geometry), the author claims this is an important result, so i would appreciate any hint on how to prove the above statement.

Thanks!

I'll prove that $I\subset (Imu:F)=(0:F/Imu)=(0:coker u)=Ann(coker u)$. Actually I'll only prove the inclusion because the equalities are obvious  :D

To prove the inclusion we only have to prove that the generators of $I$ are in $(Imu:F)$.
In Peskine, $M$ is defined as an $n\times l$ matrix, $u(e_i)=\sum_{j=1}^na_{ji}f_j\ \forall i=\overline{1,l}$, where $rank(L)=l$.

The generators of $I$ are $n\times n$ minors. Choose such a minor $x$. There exists a matrix $B$ such that $MB=xI_n$. Just put the comatrix of the matrix yielding that minor on the corresponding positions in $B$ and put $0$ in the rest of $B$.
So $x\cdot f_i=\sum_{j=\overline{1,n},k=\overline{1,l}}f_j\cdot a_{jk}b_{ki}=\sum_{k=\overline{1,l}}u(e_k)b_{ki}\in Imu\ \forall i=\overline{1,n}\Rightarrow x\in Ann(coker u)\Rightarrow I\subset (Imu:F)$

Now I prove $(Imu:F)\subset\sqrt I$.

$x\in (Imu:F)\Leftrightarrow \exists B=(b_{ij})_{i=\overline{1,n}j=\overline{1,l}}$ such that $x\cdot f_i=\sum_{j=1}^lu(e_j)b_{ji}$
This is easily seen, using the fact that $F$ is free, to be equivalent to $xI_n=AB$.
This means $x^n=\det AB$. Using the Binet-Cauchy Formula, we get that $x^n\in I$, meaning $x\in\sqrt I$

With this, the problem is solved.


maybe you meant
$ x^4 \minus{} 4$

cause I'm pretty sure you can factor that.
As for $ x^4 \plus{} 4$..i cant find a way..

I know there is an answer. Because I was worked handing out the questions in a math competition in my school and that problem showed up.. They showed the answer .In the answer they put something like $ \minus{} 2i$ the $ i$ is an imaginary number.

So I think $ x^4 \plus{} 4$
$ \equal{} x^4 \minus{} ( \minus{} 4)$
$ \equal{} x^4 \minus{} ( \minus{} 2i)^2$
$ \equal{} (x^2 \plus{} ( \minus{} 2i))(x^2 \minus{} ( \minus{} 2i)$

I am sure that the question is $ x^4 \plus{} 4$ and that the answer has to do with imaginary numbers.

If you wanted to factor this expression over integers you can add $ 4x^2$ and subtract $ 4x^2$ to obtain $ (x^4 \plus{} 4x^2 \plus{} 4) \minus{} (4x^2)$ which equals $ (x^2 \plus{} 2)^2 \minus{} 4x^2$ which equals $ (x^2 \plus{} 2x \plus{} 2)(x^2 \minus{} 2x \plus{} 2)$.

@phi-unit
That im preatty sure is also an answer.. there were two answers in the answer sheet they gave me at the competition and one had imaginary numbers and the other one had something like that.

$ a^4\plus{}4b^4 \equal{}(a^2\plus{}2b\plus{}2b^2)(a^2\minus{}2b\plus{}2b^2)$ is also known as the Sophie Germain identity, I believe.

Yes, and also, it turns out that in order for $ a^{4}\plus{}n$ to be factorable, $ n$ must be of the form $ 4b^{4}$ for some integer $ b$. This was proved by Sophie Germain, and is why the identity is named for her (because the factorization, compared to the proof that $ n\equal{}4b^{4}$, is quite obvious).


Hello,

I am a USAMTS participant (USAMTS ID: 12048) and today is the deadline for the Round 1 Solutions. I attempted to submit my solutions from the home page, however, I recieve the following message: Sorry, the deadline has passed for submission. The due date was October 13, 2009.

Why is this? Am I not allowed to submit my solutions on the due date?

Any feedback would be helpful.

Thanks,

--
-SB

Sorry dude... the deadline for uploading was at 3 pm pacific...

SEE IF YOU CAN DRIVE TO THE POST OFFICE REALLY QUICK AND MAIL YOUR SOLUTIONS!!!

To be clear, the deadline was 3 PM [b]Eastern Time[/b] on the due date.  This information is on the first page of the problem document, and in several places on the USAMTS website (including the page where you download the problems).


What is the Int. Math. Union logo exactly? I know it represents something in topology..

see http://mathunion.org/

it's a picture of the (actually, a) borromean link..
i think wikipedia has something to say about it :)
the first thing you can actually notice it that they unknot (unlink, if you prefer) as soon as you remove one component.. the second thing you could prove is that they're knotted (linked), but that's not as easy as the first ;)

anyway, it seems a ropelength-minimizing representation: just take three "real" ropes with a constant section and try to knot them up like that using the least rope you can. what you seem to obtain (by computer approximations) should have a shape like that, just thickened as much as possible.
by the way, the theory who tries to solve this kind of problems is called physical knot theory, in case you're interested.

thanks for the fast answer :-) I see why this would be used as the 'MAIN' logo for an international mathematical union..kind of


For number 6, it says we can always remove 7 magazines but does it mean that we know the areas of each magazines or are they random and we have to explain how to remove in such a way.[/quote]

you would know the areas of the magazines you remove. But the magazines could be any size, shape, area, arrangement, etc. as long as they entirely cover the table.

Im pretty sure thats all the clarification I can give without breaking the rules.


[quote="tdl"]The real sequence is defined by $ x_1 \equal{} 1, x_{n \plus{} 1} \equal{} \frac {1}{x_1 \plus{} x_2 \plus{} ... \plus{} x_n}$
Show that $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n > 1989$ for sufficiently large $ n$.[/quote]
[hide]
$ \frac{1}{x_n}\equal{}x_1\plus{}x_2\plus{}\ldots\plus{}x_{n\minus{}1}$

$ x_{n\plus{}1}\equal{}\frac{1}{\frac{1}{x_n}\plus{}x_n}\equal{}\frac{x_n}{x_n^2\plus{}1}$ and the partial sums diverge...[/hide]

All terms are positive, so if we can find one $ N$ such that $ \sum_{k\equal{}1}^N x_k >1989$, we'll be done. Suppose no such $ N$ exists. Then $ x_k\geq \frac{1}{1989}$ for all $ k$, and we get
\[ \sum_{k\equal{}1}^{1989^2\plus{}1} x_k \geq \frac{1989^2\plus{}1}{1989}>1989\]
anyway, a contradiction.


Let $ X$ be a topological space. Prove that:

$ X$ is Hausdorff if and only if the diagonal $ \{ (x,x) | x \in X \}$ is closed in $ X \times X$.

A set is closed if its complement is open. A set is open if we can fit one of the base open sets around any given point in it. A base for the product topology on $ X\times X$ is the products of open sets.

Now, what does it mean if for any $ (a,b)$ not on the diagonal, we can find $ U,V$ with $ (a,b)\in U\times V$ and $ U\times V$ not intersecting the diagonal?


find funcition f N -> N for all n is integer n \geq 0
       f(n)+f(n+1)=f(n+3)f(n+4)+2004 :D  :?


calculus
integration
number theory
prime numbers

What is the least odd integer greater than 4 that cannot be written as the sum of an integral power of 2 and a prime number?

hmm dont really know how to do this.. is it infinity?

maybe try finding odd numbers that are not prime when you subtract powers of two?  :maybe:

Yes.. and the answer is?  I still cant get it, but I probably got careless and missed something

whats the answer?

[quote="lotrgreengrapes7926"][quote="Astronum"]Is it $25$ ?[/quote]
$25=17+8=23+2$[/quote]

Ops, my bad  :blush: ... that's right. Thank you.   :)

[quote="mathgeniuse^ln(x)"]I can't find an answer to this question, are you sure it is possible?[/quote]
Yes, it is possible. May be I will give some more time before posting the answer.

[quote="perfect628"][hide]149[/hide][/quote]
You should have given a hint.  :roll:
Any solution?

My only solution was trial and error, by taking odds, subtracting the greatest power of 2, and seeing if it was prime. If not, I continued until I found a prime. 149 never yielded a prime.

You could always try to write all the prime numbers and see what happens to the numbers you get.  Maybe a pattern will occur. :?:

[quote="perfect628"][hide]149[/hide][/quote]
Unfortunately, I dont have a method for this. I also did it by trial and error.

perfect628's answer is incorrect. My only hint is that it is less than 149. If anyone knows a nice method for this, that will be nice.

[quote="lotrgreengrapes7926"]You should have given a hint.  :roll: [/quote] Oops, sorry. I thought you were the one who posted the problem.  :blush:

[hide]I get 127 by Trial and Error.[/hide]

[hide]95-16=79, so no, its not 95.[/hide]

[quote="mathgeniuse^ln(x)"]I meant 97, don't know where 95 came from. :?:

I edited it in the previous post.[/quote]

89+8=97

I edited both of my previous posts.  I don't know where 97 came from.

Yeah, I think you actually finally found the answer.

[quote="mathgeniuse^ln(x)"][hide]I meant 127, don't know where 95 came from. :?:

I edited it in the previous post.[/hide][/quote]
Correct! 127 it is.

I am not sure if there is a sleek way to do this problem quickly without resorting to trial and error :?:

Yeah, I just wrote all the primes and powers of 2 up to 150 and realized that 127 is impossible to achieve.

I was just wondering that because on some test I remember seeing a question that asked to find all positive prime numbers.

[quote="mathgeniuse^ln(x)"]Yes.

If primes weren't positive then it would be really hard to check the smallest integer...if any.[/quote]

I thought there were negative primes too?

http://www.mazes.com/primes/negative.html

Or did you mean for this particular problem?

I know it is possible for negative primes, but if they ever ask that, they would usually say positive or negative primes.  If they asked for both it would be a little more trickier, but maybe doable.


vector
function
limit
inequalities
real analysis
real analysis unsolved

Let $1 \leq p <\infty$.
a).Prove that $(l^p)^*=l^q$ in a sense that there is a bijection between $(l^p)^*$ and $l^q$, provided $\frac{1}{p}+\frac{1}{q}=1$.
b).Prove that for $1<p<\infty$, the space $l^p$ contains weakly convergent sequences, which are not convergent strongly(in the norm).

Let $P$ be the set of finitely supported sequences. The natural basis for the vector space $P$ is a countable Hamel basis. Hence, any linear functional is uniquely determined by its action on the basis elements. From that, if $L$ is a linear functional on $P,$ then there exists a sequence $y$ such that for all $x\in P,\,L(x)=\sum_{n=1}^{\infty}x_ny_n.$

Now, note that $P$ is dense in $\ell^p$ for $1\le p<\infty.$ Suppose that $L$ is a bounded linear functional on $\ell^p.$ Then $L$ restricts to a linear function on $P,$ generating the sequence $y$ as above.

For $x\in \ell^p,$ let $x^{(k)}$ be the sequence of sequences defined by $x_n^{(k)}=x_n$ for $n\le k$ and $x_n^{(k)}=0$ otherwise. Since $L$ is a bounded (that is, uniformly continuous) linear function on $\ell^p,$ we have that

$L(x)=\lim_{k\to\infty}L(x^{(k)})=\lim_{k\to\infty}\sum_{n=1}^k x_ny_n= \sum_{n=1}^{\infty} x_ny_n$

We never said whether our vector spaces were real of complex, did we? Just to make my life easier, let's assume real (although we could fix it for complex if we wanted to.)

On the one hand, by Hölder's inequality, if $y\in \ell^{p'},$ where $\frac1p+\frac1{p'}=1,$ then

$|L(x)|=\left|\sum_{n=1}^{\infty} x_ny_n\right|\le\|x\|_p\|y\|_{p'}.$

On the other hand, let $x_n=|y_n|^{p'-1}\text{sgn}(y_n).$ Then,

$|L(x)|=\sum_{n=1}^{\infty}|y_n|^{p'}=\|y\|_{p'}^{p'}.$

But $\|x\|_p=\left(\sum_{n=1}^{\infty}|y_n|^{(p'-1)p}\right)^{\frac1p}.$

Note that $(p'-1)p=p'.$ Thus $\|x\|_p=\|y\|_{p'}^{\frac{p'}p}.$

Then $\frac{|L(x)|}{\|x\|_p}=\frac{\|y\|_{p'}^{p'}}{\|y\|_{p'}^{\frac{p'}p}} =\|y\|_{p'}.$

Thus the operator norm of the functional $L$ is $\|y\|_{p'}$ and we have described the bijection called for in part (a). With suitable modifications, this includes the case $p=1,\,p'=\infty.$

For part (b). Consider the basis sequence $e^{(k)}$ defined by $e_n^{(k)}=1$ if $n=k$ and zero otherwise. Then for all $L\in (\ell^p)',$ if $y$ is the element of $\ell^{p'}$ associated with $L$ as above, then $\lim_{k\to\infty}L(e^{(k)})= \lim_{k\to\infty}y_k=0.$

That argument works for $1<p<\infty.$ The particular example fails for $p=1,$ since in that case, $\lim_{k\to\infty}y_k$ may not equal zero, if it exists at all, which cannot be guaranteed.

Thank you,Kent. But for the question that alekk posted, i think you havn't answered fully(or maybe i didn't understand the last lines in you post), i'll write what i think:
we need to prove that every sequence which converges weakly in $l^1$ converges strongly(in the norm of $l^1$). Let $x^{(k)} \to 0$ weakly, then since the dual of $l^1$ is isomorphic to $l^{\infty}$(with it's norm $||x||=sup_n|x_n|$) and since the identity functional($1(x)=x$) is in $(l^1)^*$ we can say that $lim_{k \to \infty} sup_n |x^{(k)}_n|=0$, so for every $\epsilon>0$ we can find such $K$ that for all $k>K$ we have that $|x^{(k)}_n|<\frac{\epsilon}{2^n}$, so $||x^{(k)}||_{l^1} = \sum |x^{(k)}_n|<\epsilon$, which means that $||x^{(k)}||_{l^1} \to 0$, when $k \to \infty$.


If A is a symmetric nxn mx of rank<=r>=1 and X is a unit eigenvector of A, with eigenvalue a not= 0, let B=A - aXX^T. Show that B is symmetric and that N(A) is a proper subspace of N(B). Conclude that rank B=<r-1.

(  someone said, show X is in N(B) but not in N(A)  ).

Thanks for your help!

I think you didn't proof N(A) is a proper subspace of N(B). Also, how can you tell B is symmetric accordinng $A-B$ is $A^T-B^T$ and $(AB)^T=B^TA^T$?
Thanks agian!
[quote="fredbel6"]i had trouble figuring out what exactly you meant so i hope this is correct

1) keep in mind that the transpose of $A-B$ is $A^T-B^T$

and that $(AB)^T=B^TA^T$
that proofs that B is also symmetric

2)when v is in $N(A)$ we have $Av=0$ and thus also $x^T Av=0$  or $(Ax)^T v=0$ or $a x^Tv=0$
(this may seem farfetched but when you know a thing or two about symmetric matrices and their eigenspaces it geometrically becomes very clear)

so this gives $Bv=Av-ax x^Tv=0-0=0$
now here is a vector that is in $N(B)$ but not in $N(A)$ : x !

because $Ax=ax$ and $ax x^T x=ax$ (in the last equality i  used that x is a unit vector)

i hope this helps,

fred[/quote]

If I know that is working in general, i could show N(A) is a proper subspace of N(B). [quote="xyidan"]Does anyone know how to prove it in general if I can show X is in N(B) but not in N(A). Thanks![/quote]

Sorry about it. you are right. I misunderstanded the second part because I didn't understand the question clearly. So how can you Conclude that rank B=<r-1 finally?
 [quote="fredbel6"]I do not understand your questions?
What is wrong with my proof?

First I proved every vector in $N(B)$ is in $N(A)$ , then I gave you one that is only in the first and not in the latter?[/quote]

there is this theorem for square matrices $A$ :

rank(A)+dim(N(A))=n

here we have found that $N(A)\subset N(B)$ and thus $rank(B)\leq rank(A)$

One more question, why ax x^Tv is equal to 0?
[quote="fredbel6"]i had trouble figuring out what exactly you meant so i hope this is correct

1) keep in mind that the transpose of $A-B$ is $A^T-B^T$

and that $(AB)^T=B^TA^T$
that proofs that B is also symmetric

2)when v is in $N(A)$ we have $Av=0$ and thus also $x^T Av=0$  or $(Ax)^T v=0$ or $a x^Tv=0$
(this may seem farfetched but when you know a thing or two about symmetric matrices and their eigenspaces it geometrically becomes very clear)

so this gives $Bv=Av-ax x^Tv=0-0=0$
now here is a vector that is in $N(B)$ but not in $N(A)$ : x !

because $Ax=ax$ and $ax x^T x=ax$ (in the last equality i  used that x is a unit vector)

i hope this helps,

fred[/quote]

Great! Now, It made perfert sence to me. thanks for your help! [quote="fredbel6"]$a x x^T v=x(a x^T v)$  and in the beginning of my part 2) i explicitly found $a x^T v=0$[/quote]


Find all continuous function f:R+ --> [1, infinity) satisfying

  f(x).f(2x)...f(nx) < 2008.n^2009 for all any positive real number x and positive integer n

we suppose that $ \exists a\in\mathbb{R}^\plus{}: \ f(a)>1$
$ f$ continous gives $ \exists b>a,\forall x\in[a,b]: \ f(x)\ge \frac{f(a)\plus{}1}{2}$ (we take $ c\equal{}\frac{f(a)\plus{}1}{2}>1$)

now for $ e>0$ we take $ n\in\mathbb{N}$ such that $ ne\le b<(n\plus{}1)e$
we have $ |\{k\in[|1,n|]: \ f(ke)>c\}|\ge  |\{k\in[|1,n|]: \ ke\in[a,b]\}|\ge \frac{b\minus{}a}{2b}n$
then $ 2008>\frac{\prod_{k\equal{}1}^{n}f(ke)}{n^{2009}}\ge \frac{A^n}{n^{2009}}$ where $ A\equal{}c^{\frac{b\minus{}a}{2b}}>1$
but we have $ \lim_{n\to \plus{}\infty} \frac{A^n}{n^{2009}}\equal{}\plus{}\infty$ (impossible)

then $ \forall a\in\mathbb{R}^\plus{}: \ f(a)\equal{}1$


Points $E$ and $F$ are chosen on the sides $AB$ and $BC$ respectively of a square $ABCD$ such that $BE = BF$. Let $BN$ be an altitude of the triangle $BCE$. Prove that the triangle $DNF$ is right-angled.


[quote="eli140"]Well, i was going through some problems and i was having trouble figuring this one out:

Simplify:

$ \sqrt{3\plus{}\sqrt8}\minus{}\sqrt{3\minus{}\sqrt8}$

I think it has something to do with the binomial theorem, i don't know how to get it though.[/quote]

Well, if we square it, we get $ 3 \plus{} \sqrt{8} \minus{} 3 \sqrt8 \equal{} x^2 \implies x \equal{} 0$

I think it is $ 0$, should belong in HSB.

--------------------------------------------------------------------------------

eli140 wrote:
Well, i was going through some problems and i was having trouble figuring this one out:

Simplify:

I think it has something to do with the binomial theorem, i don't know how to get it though.

Well, if we square it, we get

I think it is , should belong in HSB.

Same thing from me. eli is right

@user above me:  The "quote" button near the bottom right most part of every post will quote the entire thing a user said and you can post your own words along with it.  that will help retain the format of the post.

Um...I don't really know what you said, but just let $ x\equal{}\sqrt{3\plus{}\sqrt8}\minus{}\sqrt{3\minus{}\sqrt8}\implies x^2\equal{}3\plus{}\sqrt8\plus{}3\minus{}\sqrt8\minus{}2\sqrt{9\minus{}8}\equal{}4\implies x\equal{}\boxed2$ (since the first radical is larger than the second, and both are positive).

007 u know me and anyway sorry meant to say you were right. and thanks!

[quote="math154"]Um...I don't really know what you said, but just let $ x \equal{} \sqrt {3 \plus{} \sqrt8} \minus{} \sqrt {3 \minus{} \sqrt8}\implies x^2 \equal{} 3 \plus{} \sqrt8 \plus{} 3 \minus{} \sqrt8 \minus{} 2\sqrt {9 \minus{} 8} \equal{} 4\implies x \equal{} \boxed2$ (since the first radical is larger than the second, and both are positive).[/quote]

ARGH!!!!  I keep on forgetting that i must multiply each thingie by all the other thingie's...phailure.....

Obviously, it cannot be zero.  If it were zero, it would imply that $ \sqrt{3\plus{}\sqrt{8}}\equal{}\sqrt{3\minus{}\sqrt{8}} \implies \sqrt{8}\equal{}0$, which is false.

As I said earlier (why don't people read? xD), the answer is $ 2$.

Umm, I get where the two 3s come from. How do you get the 2sqrt(9-8)?

[quote="eli140"]Well, i was going through some problems and i was having trouble figuring this one out:

Simplify:

sqrt(3+sqrt8)-sqrt(3-sqrt8)

I think it has something to do with the binomial theorem, i don't know how to get it though.[/quote]

[hide]$ \begin{array}{l}\sqrt {3 \plus{} \sqrt 8 }  \minus{} \sqrt {3 \plus{} \sqrt 8 }  \equal{} \sqrt {3 \plus{} 2\sqrt 2 }  \minus{} \sqrt {3 \minus{} 2\sqrt 2 }  \equal{} \sqrt {\left( {1 \plus{} \sqrt 2 } \right)^2 }  \minus{} \sqrt {\left( {1 \minus{} \sqrt 2 } \right)^2 }  \\\equal{} 1 \plus{} \sqrt 2  \minus{} \left( { \minus{} 1 \plus{} \sqrt 2 } \right) \equal{} 2 \\\end{array}$[/hide]

maybe this might help
$ \sqrt{(a\plus{}b)\pm2\sqrt{ab}}\equal{}\sqrt{a}\pm\sqrt{b}$

[quote="Fibonacci11235"]Wait why wouldn't it be:

$ \begin{array}{l}\sqrt {3 \plus{} \sqrt 8 } \minus{} \sqrt {3 \minus{} \sqrt 8 } \equal{} \sqrt {3 \plus{} 2\sqrt 2 } \minus{} \sqrt {3 \minus{} 2\sqrt 2 } \equal{} \sqrt {\left( {1 \plus{} \sqrt 2 } \right)^2 } \minus{} \sqrt {\left( {1 \minus{} \sqrt 2 } \right)^2 } \\\equal{} 1 \plus{} \sqrt 2 \minus{} \left( {1 \minus{} \sqrt 2 } \right) \equal{} 2\sqrt 2 \\\end{array}$[/quote]

Because $ \sqrt{(1\minus{}\sqrt2)^2}\equal{}|1\minus{}\sqrt2|\equal{}\sqrt2\minus{}1$. Always check that--but personally, I think my method is easier for this problem.

Oh, thanks.  :oops:

I would have totally gotten that wrong.


[quote="mathkid"]This may be a silly question.

You pick 5 cards from 52, and at least two of them have to be aces. This can be done in how many ways?

I know the answer: ${4\choose2}{48\choose3}+{4\choose3}{48\choose2}+{4\choose4}{48\choose1}$

This makes sense, and I understand it.

However, I need someone to explain why ${4\choose2}{50\choose3}$ is incorrect.

You choose 2 aces from the 4, and then you choose three cards from the remaining 50 cards. Makes sense to me.

Anyone? I'm tripping over a simple concept. :oops:[/quote]
$\binom{4}{2}\binom{50}{3}$ is incorrect because you aren't counting the possibility that you get 3 or 4 aces.

[quote="mathkid"]Why not? Of those 50 cards, two will be aces, so choosing three could include two of those.

And my incorrect answer is larger than the correct answer, if that means anything.[/quote]
In that case... I think they're both correct. :rotfl:
But how? :rotfl:

You only want to count a combination like As Ac Ah Ks (lolmaha) once, but you count it several times.

For example, you count it as (As Ac)(Ah Ks), (Ah As)(Ac Ks), (Ac Ah)(As Ks) in the quantity \[\binom{4}{2}\binom{50}{3}\]


[size=150]已知$ \Delta ABC$是等边三角形，$ O$是$ \Delta ABC$内的任意一点、$ OD\perp BC$，$ OE\perp AC$，$ OF\perp AB$，
$ D$、$ E$、$ F$分别是垂足，连接$ OA$、$ OB$、$ OC$.
求证：$ S_{\Delta OAF}\plus{}S_{\Delta OBD}\plus{}S_{\Delta OCE}\equal{}\dfrac{1}{2}S_{\Delta ABC}$.[/size]

[quote="wxd686"][size=150]已知$ \Delta ABC$是等边三角形，$ O$是$ \Delta ABC$内的任意一点、$ OD\perp BC$，$ OE\perp AC$，$ OF\perp AB$，
$ D$、$ E$、$ F$分别是垂足，连接$ OA$、$ OB$、$ OC$.
求证：$ S_{\Delta OAF} \plus{} S_{\Delta OBD} \plus{} S_{\Delta OCE} \equal{} \dfrac{1}{2}S_{\Delta ABC}$.[/size][/quote]

提示图:
(有点误差)


[color=darkred] Let $ABC$ be a triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$. Denote the middlepoint $M$ of the side $[BC]$. Prove that the $A$- simmedian and the lines $EF$, $MI$ are concurrently if and only if $\boxed{\cos B+\cos C=1}$.[/color]

[color=darkred][b]Generalization I.[/b] Let  be $ABC$ an acute triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$. Denote the middlepoint $M$ of the side $[BC]$. Let $S\in [BC]$ for which $\frac{SB}{SC}=s$. Prove that the lines $AS$, $EF$, $MI$ are concurrently if and only if $\boxed{\frac{c-sb}{1-s}=\frac{2(p-b)(p-c)}{a}}$.

[b]Particular cases.[/b]

$1.\blacktriangleright$ For $S\in BC$, $AS\perp BC$, i.e. $s=\frac{c\cdot\cos B}{b\cdot\cos C}$ obtain

http://www.mathlinks.ro/Forum/viewtopic.php?t=141487

$2.\blacktriangleright$ If the point $S$ is the foot of the $A$- simmedian, i.e. $\frac{SB}{SC}=\left(\frac{c}{b}\right)^{2}$, then

$AS\cap EF\cap MI\ne\emptyset$ $\Longleftrightarrow$ $\cos B+\cos C=1$.[/color]

[color=darkred][b]Generalization II.[/b] Let  be $ABC$ an acute triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$. Let $\{M,S\}\subset (BC)$ be two points for which $\frac{MB}{MC}=m$ and $\frac{SB}{SC}=s$. Prove that the lines $AS$, $EF$, $MI$ are concurrently if and only if $\boxed{(s-m)a^{2}+(m+1)(c-sb)a=(b-c)(s-1)(mb-c)}$.

[b]Particular case.[/b] If $m=1$, i.e. the point $M$ is

the middlepoint of the side $[BC]$, then obtain the [u]generalization I[/u].[/color]


calculus
integration
calculus computations

In the $ xy$-plane, if $ C$ is the circle $ x^2\plus{}y^2\equal{}9$, oriented counterclockwise, then calculate $ \int_C \minus{}2ydx \plus{} x^2dy$.

I tried Green's Theorem:
$ \int\int (2x \plus{} 2) dA$ where that is through the circle of radius $ 3$ centered at the origin. This isn't all that pleasant without a nice polar sub (in fact I can't get the correct answer). Is there a better way?

Green's theorem is completely fine and the resulting integral is not hard at all if you notice that the integral of $ x$ over the circle is $ 0$ due to antisymmetry. So I don't understand what your difficulty is :?. Can you show the details of what you did?

$ 2x\plus{}2$ is not even with respect to the $ x$ variable, so the integral over the circle is not $ 4$ times the integral over the quarter-circle (it seems like you computed the latter integral correctly but, of course, since it was a wrong integral, you got a wrong value).


If you view this person's profile, it says they have 146 posts.  But click "find all posts by pifreak" and there are none.  Now, how do 146 posts disappear just like that?

[url]http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=515[/url]

maybe all his/her posts come from those online classes?

Did you PM him?

I agree, either that or they were all in Fun and Games which is not likely since I keep a close eye there. In fact it's probably the earlier because I've never seen his post anywhere.

P.S. Please excuse me for calling you he if you are a girl.

Right, Syntax - posts in the classes wouldn't show up in a search (unless, perhaps, you are also in that class).  pifreak's posts are all in online classes.

pifreak is a she. She's in my class, so if you look at mine message a lot will be missing. Anyways, who's Vanessa Rusczyk? Her name is in the AoPS 2 books for the cover. And Claire and Stanley?

Oh, wow.  She must be very disciplined to not have ever posted in F&G...

Vanessa = my wife; Claire = my mother; Stanley = my grandfather.


algebra
polynomial
Vieta
product of roots

A polynomial with roots -4,3, and 24 is of the form $x^3+ax^2-bx+c$. Which of the following statements is true?

A) $a$ is the sum of the roots
B) $b = -36$
C) $c$ is the product of roots
D) $a$ is the product of the roots
E) None of these

Use Vieta's formulas for third-polynomial degree

[hide]$e$[/hide]

[hide="Answer"]If the cubic was in the form $x^3-ax^2+bx-c$, then the first three would be true, but since the signs are reversed, the answer is $E$.[/hide]

[quote="236factorial"]What's vieta's formulas for the third degree? (I can solve this by multiplying the expression out, but I doubt that's the best way.[/quote]It's not hard to see why the formulas are true intuitively, so there's no need to memorize them. The coefficient for $ x^{n-k} $ in the expansion of $ (x-r_1)\cdots(x-r_n) $ is just the sum of the products of all $ \binom{n}k $ groups of $ k $ of the $ -r_i $.

So the coefficent of $ x^n $ is $ 1 $, the coefficient of $ x^{n-1} $ is $ -(r_1+\ldots+r_n) $, etc.

[quote="Silverfalcon"]A polynomial with roots -4,3, and 24 is of the form $ x^3+ax^2-bx+c $. Which of the following statements is true?

A) $ a $ is the sum of the roots
B) $ b = -36 $
C) $ c $ is the product of roots
D) $ a $ is the product of the roots
E) None of these
[/quote]

[hide]$ 16a+4b+c=64\\ 9a-3b+c=-27\\ 576a-24b+c=-1384\\\\ \text {Just solve with calculator...}\\ \text {Problem is trivial after this.} $[/hide]

:D :D

[quote="JesusFreak197"]Also in simpler terms:

For any monic polynomial $ ax^3-bx^2+cx-d $ (monic just means that $ a=1 $), then $ b=r+s+t $, $ c=rs+rt+st $, and $ d=rst $. Note that the signs in the cubic are necessary. If we had, for example, $ ax^3+bx^2+cx+d $, then $ b=-(r+s+t) $ and $ d=-rst $. ($ r $, $ s $, and $ t $ are the roots of the equation, and $ ax^3-bx^2-cx-d $ is the equation $ (x-r)(x-s)(x-t) $ multiplied out.)[/quote]One more thing to note, if the leading coefficient $ a $ is not $ 1 $, then you can just divide the equation by $ a $ and use what JesusFreak said.  This works because you can divide an equation by anything $ \neq 0 $ and have the same roots.

Right, I should have mentioned that. The general formula would actually be: $ \frac{b}{a} =r+s+t $, $ \frac{c}{a} =rs+rt+st $, and $ \frac{d}{a} =rst $.

[quote="236factorial"]What's vieta's formulas for the third degree? (I can solve this by multiplying the expression out, but I doubt that's the best way.[/quote]

Vieta's forlumas for the third degree states that:

If you have polynomial: $ ax^3+bx^2+cx+d=0 $ with 3 real roots say $ x1, x2, x3 $

Then: $ x1 + x2 + x3 =\frac{-b}{a} $
         $ x1*x2+x2*x3+x3*x1=\frac{c}{a} $
          $ x1*x2*x3 =\frac{-d}{a} $


function
integration
calculus
inequalities
real analysis
calculus computations

Suppose $ f$ and $ g$ are Riemann Integrable Functions on $ [a,b]$ and $ f(x)\equal{}g(x)$ if $ x$ not equal to $ c$.
Show that: $ \int ^a_b f\equal{} \int ^a_b g$

A rough sketch, using the Darboux theory (upper and lower sums). Find a partition so that the upper and lower sums for the integral of $ f$ are within $ \epsilon$ of each other. (Maybe $ \frac{\epsilon}2$ - that can be worked out later.) Take the common refinement. Refine that further to make the subinterval containing $ c$ extremely narrow. How narrow? We can figure that out once we know the size of $ |f(c)\minus{}g(c)|$ and the $ \epsilon$ we want to match. The upper and lower sums for $ f$ and $ g$ are identical except for the one term for the subinterval containing $ c,$ and we can control how big that is.

There's a straightforward inductive extension to the case in which $ f$ and $ g$ differ at finitely many points.

We can show $ \int^{a}_{b}f = \int^{a}_{b}g$ also directly from Riemman sums.

[b]Definition:[/b]
$ f$ is Riemann integrable on $ [a,b]$ if and only if for each $ \epsilon > 0$ there exists $ \delta > 0$ such that for all partitions $ D$ of $ [a,b]$, where $ \max{\delta_k} < \delta$ $ (\delta_k$ is the length of $ k$-th interval of $ D  \right)$, and all choices of $ \xi_k$, we have $ |\sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \int^{b}_{a}f| < \epsilon$.

[b]For the problem:[/b]

We choose $ \epsilon > 0$.

Now we find $ \delta_f$, for which $ |\sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \int^{b}_{a}f| < \frac {\epsilon}{2}$,
whenever $ \max{\delta_k} < \delta_f$.

Analoguously $ \delta_g$, for which $ |\sum_{k = 1}^{n}(g(\xi^{\prime}_k)\delta^{\prime}_k) - \int^{b}_{a}g| < \frac {\epsilon}{2}$,
whenever $ \max{\delta^{\prime}_k} < \delta_g$.

Since the upper inequalities must be true for all partitions and choices of points, they must be true also for a partition, where $ \delta_k < \min{\delta_f, \delta_g}$ and where $ \xi$-s are the same for $ g$ and $ f$ and $ c$ is not among them. Therefore

$ |\sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \sum_{k = 1}^{n}(g(\xi_k)\delta_k) - \int^{b}_{a}f + \int^{b}_{a}g | \leq |\sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \int^{b}_{a}f| + |\sum_{k = 1}^{n}(g(\xi_k)\delta_k) - \int^{b}_{a}g| < \epsilon$

Now: $ \sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \sum_{k = 1}^{n}(g(\xi_k)\delta_k) = \sum_{k = 1}^{n}(f(\xi_k) - g(\xi_k))\delta_k = 0$

So $ |\sum_{k = 1}^{n}(f(\xi_k)\delta_k) - \sum_{k = 1}^{n}(g(\xi_k)\delta_k) - \int^{b}_{a}f + \int^{b}_{a}g | = |\int^{b}_{a}f + \int^{b}_{a}g | < \epsilon$.

Since $ \epsilon$ was arbitrary, we have $ \int^{b}_{a}f = \int^{b}_{a}g$.

[quote="Kent Merryfield"]There's a straightforward inductive extension to the case in which $ f$ and $ g$ differ at finitely many points.[/quote]

Using my approach, I think, that the result is valid even if $ f$ and $ g$ differ at infinitely many points, just that they do not form any closed interval $ [c,d]\subset [a,b].$

[quote="tmrfea"]Using my approach, I think, that the result is valid even if $ f$ and $ g$ differ at infinitely many points, just that they do not form any closed interval $ [c,d]\subset [a,b].$[/quote]
It's more subtle than that.  For example, the result isn't valid if $ f$ and $ g$ differ by a constant on $ \mathbb{Q}$ - in that case, at least one of $ f$ and $ g$ aren't Riemann integrable.  The most general result I know of for the Riemann integral is that $ f$ and $ g$ differ on a compact set of [url=http://mathworld.wolfram.com/MeasureZero.html]measure zero[/url] (and note that this is well-defined even if you haven't defined the Lebesgue measure).  Extending this result to all sets of measure zero is one of the motivations for the theory of the Lebesgue integral.


$ a,b,c>0\ ,\ a\plus{}b\plus{}c\equal{}1\Rightarrow 1\plus{}\sum_{cyc}\frac{(a\plus{}b)(b\plus{}c)}{c\plus{}a}\ge \sum_{cyc}\frac{1\plus{}9ab}{3(1\minus{}c)}$

[quote="alex2008"]$ a,b,c > 0\ ,\ a \plus{} b \plus{} c \equal{} 1\Rightarrow 1 \plus{} \sum_{cyc}\frac {(a \plus{} b)(b \plus{} c)}{c \plus{} a}\ge \sum_{cyc}\frac {1 \plus{} 9ab}{3(1 \minus{} c)}(*)$[/quote]
IT's very easy and not interesting!
$ (*)<\equal{}> \sum_{cyc}\frac{a^2}{b\plus{}c}\plus{}a\plus{}b\plus{}c \geq \sum_{cyc}\frac{3ab}{a\plus{}b}$