I am Shi-Ke Xue; I used to go to Takoma Park Middle School before I moved. I just thought I might ask/find out how many people posting in the Maryland AoPS forum goes/went there.

I'm not from Takoma, and unfortunately I have a great dislike (no offense to you, or any AoPSer that goes there) for them. Mostly jealousy, since they destroy us in math and chess  :P  but they were pretty arrogant about it.

Fortunately for us, we got back at them in band?

I think everyone in Maryland who doesn't go to Takoma Park hates them.  Their perpetual domination of Mathcounts gets sort of annoying.

As you might guess, I didn't go there, although I met a few of their students when I trained with their coach as the alternate for National MC in 8th grade.  They were less arrogant than I had been led to believe they were.

mchoi815 is a Takoma Park alumni, I believe.

alumnus? or is "alumni" acceptable?

and yes I am.

Yes, yes, I forget that "alumni" is plural . . .

I know this hasn't been posted on in a month, so this should be considered spam, but yeah i currently go to Takoma Park.

I go to TPMS too. Unfortunately, i did not make any competition teams. The basic idea is that a lot of the good people get picked to go to TPMS, they pick out the good people and they come (it is a magnet school).

We also have lots of alumni, and i think that a lot of good, Blair students came from TPMS.

Wow - Takoma's a magnet school? For what subject? I thought it was a public school. Is Montgomery Blair also a magnet school?

Magnet schools, including Takoma and Blair, are public schools.  I think they both have math magnets.

Note that there are actually a [i]lot[/i] of magnet schools around, most of which aren't any good.  Takoma and Blair are unusual because they have such a strong reputation that they get almost everyone in the county who is good at math.

[quote="E^(pi*i)=-1"]Note that there are actually a [i]lot[/i] of magnet schools around, most of which aren't any good.[/quote]

North County is one of them. Their STEM program is supposedly really good, but they perform dismally at math and programming competitions. Ironically, my school has placed 1st at our county competitions for the past two years and managed to attract some of the best (math) students in the county with a non-math program (that I myself succumbed to). (And, on that note, our programming team is made up solely of students that are in that exact same program) :)

[quote="E^(pi*i)=-1"] I think they both have math magnets.
[/quote]

yeah. it has the math, science, adn computer science magnet program

Can Montgomery Blair "magnet" students from outside of Montgomery County?

Sadly, no.  I would go there if they could.

oh that's too bad.
i'm sure they already have a handful with moco though. :)

I might know someone on here cuz I go to a school not named Takoma Park or Montgomery Blair where I go to Montgomery Blair HS from my school so I can ride the bus to go home.

who went to the math competition on the Jan. 11th?

[quote=presto502]Mathalon was trash.

Also yes we (Takoma) got first (some random seventh grade team did).
This was due to the fact that they screwed up the team round scoring. I doubt they will fix it.

A certain person who does not go to Takoma got first in individual.[/quote]

xd that was me
sorry if this is old or something

Gj :P
Also this post was originally from 2008 so you're good

[quote=pie314159265]like
the one at takoma park?[/quote]

yes

Who is ohsomething? I currently go to Eastern and I am on the math team going to MathCounts state round. Are you an alumni or a current student?

current. I will pm you for privacy reasons.

i go to takoma
i can list some other aops users who go if u like

[quote=CatherineAsaro][quote=statman]btw mathalon was troll this year.....   i got 29 and was like 4th LOL (also on 3rd place team)[/quote]

There were three perfect scores, All CMP students, Daniel Yuan (RFMS), Matthew Liu (BMMS), and Brandon Du (EMMS). So you were in good company. On the tie-breakers, Matthew came in first, Daniel came in second, and Brandon came in third because he forgot to write his tie breaker questions on his score sheet instead of the test sheet. Had he put them in the right place, he might have come in first.[/quote]

sorry, that comment i made was from 2016 :D

Oh, yeah, sorry, that was from the previous year. The three perfect scores were this year.


algebra
polynomial
inequalities
function
maximization
IMO Shortlist

Let $ P( \minus{} 1) \equal{} y_1,\ P( \minus{} 1/2) \equal{} y_2,\ P(1/2) \equal{} y_3,\ P(1) \equal{} y_4$. Then from Lagrange's interpolation formula (or from a linear system of equations) we obtain:
$ P(x) \equal{} \frac 1 3 (( \minus{} 2y_1 \plus{} 4y_2 \minus{} 4y_3 \plus{} 2y_4)x^3 \plus{} (2y_1 \minus{} 2y_2 \minus{} 2y_3 \plus{} 2y_4)x^2 \plus{}$
$ (\frac 1 2 y_1 \minus{} 4y_2 \plus{} 4y_3 \minus{} \frac 1 2 y_4)x \plus{} ( \minus{} \frac 1 2y_1 \plus{} 2y_2 \plus{} 2y_3 \minus{} \frac 1 2 y_4)$.
It is easy to check, that if $ y_i \in [ \minus{} 1,1]$, then $ \pm a \pm b \pm c \pm d \le 7$ for any combination of pluses and minuses. So, $ |a| \plus{} |b| \plus{} |c| \plus{} |d| \le 7$.

[b]Remark.[/b] This solution applies the technics, which is common for polynomials, bounded on [-1, 1]. Actually we consider Chebyshev's polynomial $ T_n(x)\equal{}\cos(n\arccos(x))$ of the same degree and compare it with our polynomial in points, where $ T_n(x)\equal{} \pm 1$. In our case it will be a polynomial $ 4x^3\minus{}3x$ and the corresponding points are $ \minus{}1,\ \minus{}1/2,\ 1/2,\ 1.$

[quote="Sasha Rybak"]It is easy to check, that if $ y_i \in [ \minus{} 1,1]$, then $ \pm a \pm b \pm c \pm d \le 7$ for any combination of pluses and minuses. So, $ |a| \plus{} |b| \plus{} |c| \plus{} |d| \le 7$.[/quote]
Can you explain this further?  Why is it "easy to check," and how do you check it?

This problem is very reminiscent of Chebyshev bounds on polynomials. What I have so far is that $|f(x)| \le |C_3(x)|$ for $|x| \ge 1$, where $C_3(x)$ is the 3rd degree Chebyshev polynomial. However, I cannot seem to isolate into $|a| + |b| + |c| + |d|$? Does anyone have a completion?

[quote=orl]Let $ P(x)$ be the real polynomial function, $ P(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d.$ Prove that if $ |P(x)| \leq 1$ for all $ x$ such that $ |x| \leq 1,$ then

\[ |a| \plus{} |b| \plus{} |c| \plus{} |d| \leq 7.\][/quote]
Let $f(x)=a+bx+cx^2+dx^3$ and $|f(0)|\leq 1,|f'(0)|\leq 1,|f''(0)|\leq 1,|f(1)|\leq \frac{1}{2}.$
Then for all $x\in [0,1]$ [url=https://artofproblemsolving.com/community/c6h1425490p8030153] we have [/url]$|f(x)|\leq \frac{11}{8}.$

[quote=sayantanchakraborty]Let $a \le 0, b \le 0$.Then set $Q(x)=-P(x)$
Let $a \le 0,b \ge 0$.Then set $Q(x)=P(-x)$
Let $a \ge 0,b \le 0$.Then set $Q(x)=-P(-x)$

In each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \ge 0,b \ge 0$.

$c \ge 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \le 1$
$c \ge 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \le 3$
$c < 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}P(1)-\frac{1}{3}P(-1)-\frac{8}{3}P(\frac{1}{2})+\frac{8}{3}P(-\frac{1}{2}) \le 7$
$c < 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}P(1)-4P(\frac{1}{2})+\frac{4}{3}P(-\frac{1}{2}) \le 7$

So we are done!![/quote]

What is Q(x) here

Also if |X| ≤ 1 then isn't X only 1and -1

Let $x_1=-1,x_2=-\frac 12,x_3=\frac 12,x_4=1.$ From  Lagrange's interpolation formula$,$
$$\begin{aligned}P(x)&=\sum\limits_{k=1}^4P(x_k)\prod\limits_{j\neq k}\frac {x-x_j}{x_k-x_j}\\&=-\frac 23P(-1)\left(x^3-x^2-\frac x4+\frac 14\right)+\frac 43P\left(-\frac 12\right)\left(x^3-x^2-\frac x4+\frac 14\right)\\&\quad\text{ }-\frac 43P\left(\frac 12\right)\left(x^3+\frac {x^2}2-x-\frac 12\right)+\frac 23P(1)\left(x^3+x^2-\frac x4-\frac 14\right).\end{aligned}$$
For $p,q,r,s\in\mathbb R,$ define $(p,q,r,s)=pP(x_1)+qP(x_2)+rP(x_3)+sP(x_4).$
Then $a=\left(-\frac 23,\frac 43,-\frac 43,\frac 23\right),b=\left(\frac 23,-\frac 23,-\frac 23,\frac 23\right),c=\left(\frac 16,-\frac 43,\frac 43,-\frac 16\right),d=\left(-\frac 16,\frac 23,\frac 23,-\frac 16\right).$
Therefore
$a+b=\left(0,\frac 23,-2,\frac 43\right),a-b=\left(-\frac 43,2,-\frac 23,0\right)\Rightarrow |a|+|b|=\max\{|a+b|,|a-b|\}\leqslant 4.$
$c+d=\left(0,-\frac 23,2,-\frac 13\right),c-d=\left(\frac 13,-2,\frac 23,0\right)\Rightarrow |c|+|d|=\max\{ |c+d|,|c-d|\}\leqslant 3.$
$\therefore  |a| + |b| + |c| + |d|\leqslant 4+3=7.\blacksquare$

[img]https://i.ibb.co/x6SRRwK/1996-ISL-A5.png[/img]
[hide=P(x) is a real polynomial function, why do you subtitute i?]In [url=https://mathworld.wolfram.com/RealPolynomial.html]Wolfram Mathworld[/url], I saw this definition of [b]Real Polynomial[/b]:
[img]https://i.ibb.co/wQ7rSNz/def.png[/img]Also, it's given that $ |P(x)| \le 1$ $\boxed{\text{for all} \; x\; \text{such that}\; |x| \le 1},$ not $\text{for all real}\; x\; \text{such that}\; |x|\le 1.$[/hide]


geometry
power of a point
radical axis
geometry unsolved

For a circle [b]O[/b] with its chord [b]AB[/b] , let two circles [b]P[/b] , [b]Q[/b]  intouch the circle [b]O[/b] at [b]R[/b] , [b]S[/b]
[b]AB[/b] = external common tangent of [b]P[/b] and [b]Q[/b]
Intersection point of internal common tangents of [b]P[/b] and [b]Q[/b] = [b]L[/b]
midpoint of arc [b]AB[/b] = [b]M[/b] , midpoint of arc [b]RS[/b] = [b]N[/b]

[b]L[/b] , [b]M[/b] , [b]N[/b] are collinear

Is this right ?
If so , May I ask  you  a favor ?  ( as synthetic ones as possible )
Thank you in advance  :)
Horrible chores  :(

I am sorry , but this problem maybe wrong :(

Thanks !
I feel that MN is the radical axis of two circles P and Q  from your nice figure.
May I ask you the certification or the proof ?  :blush: sorry to bother you

I don't think so , because PQ and MN are not perpendicularity


logarithms
inequalities unsolved
inequalities

Let $ n \in Z, n \geq 2$ and $ x,y \geq 0$. Prove that:

    $ \sqrt[n]{x^n\plus{}y^n} \ge \sqrt[n\plus{}1]{x^{n\plus{}1}\plus{}y^{n\plus{}1}}$

[quote="ngoclam1088"]Let $ n \in Z, n \geq 2$ and $ x,y \geq 0$. Prove that:

    $ \sqrt [n]{x^n \plus{} y^n} \ge \sqrt [n \plus{} 1]{x^{n \plus{} 1} \plus{} y^{n \plus{} 1}}$[/quote]
If $ x \equal{} 0$ or $ y \equal{} 0$, we obtain the equality. Hence, suppose that $ x > 0$ and $ y > 0$. Then equivalently
$ \sqrt [n]{t^n \plus{} 1}\ge\sqrt [n \plus{} 1]{t^{n \plus{} 1} \plus{} 1},\qquad\text{where}\quad t: \equal{} \frac {x}{y}.$
Now, you can take logarithms to get a more elegant form of this inequality. The rest is easy.


If

\[ f\left(\frac{x}{x-1}\right) = \frac{1}{x}, \ \text{for all} \ x \neq 0,1 \]

and $0 < \theta < \pi/2$, then $f(\sec^2 \theta)=$

$\text{(A)} \ \sin^2 \theta \qquad \text{(B)} \ \cos^2 \theta \qquad \text{(C)} \ \tan^2 \theta \qquad \text{(D)} \ \cot^2 \theta \qquad \text{(E)} \ \csc^2 \theta$

[hide]
$f\left(\frac{1}{\cos^2{\theta}}\right) = f\left(\frac{1}{1-\sin^2{\theta}}\right) = f\left(\frac{\csc^2{\theta}}{\csc^2{\theta}-1}\right) = \frac{1}{\csc^2{\theta}} = \sin^2{\theta}$. So A. [/hide]


algebra
function
domain
topology
real analysis
real analysis unsolved

Let f:U -> R^m be continuously differentiable where m>1 and U is an open set in R^m. Suppose {x in U: det [f'(x)]=0 } only contains isolated points. Prove f is an open map. Use this result to prove the Fundamental Theorem of Algebra.

I have seen this old problem but never manage to solve it. (not even the first part). it is only necessary to prove that if
$ f: \mathbb{R}^n\rightarrow\mathbb{R}^n, f \in C^1, n>1$
$ det(f'(x)) \equal{} 0$ iff $ x \equal{} 0$,

f is injective for some neighborhood around 0.

the theorem is false in $ \mathbb{R}$

sorry I misread the problem... it says prove that it is an open map, injectivity is a stronger result by invariance of domain (i wonder if this stronger result is true though).


There is this ant of mass $m$ that is on a rotating compact disk. Given that the coefficient of friction between the ant and the disk is $\mu$. What is the greatest frequency the disk can be rotating at such that the ant is able to stay on? (The radius of the disk is $R$, the gravitational acceleration is $g$).

If the ant rests in the center of the disk it can stay there arbitrarily long...

Djole is right!you must specify the position of the ant OR else the solution to your problem is quite simple I suppose

$mxw^{2}={\mu}mg$
this implies$w^{2}= \frac{{\mu}g}{x}$

thus $w$ is maximised for the position $x = 0$ i.e when thwe ant is situated at the centre of the circle.


function trigonometry calculus derivative

I don't know if this belongs to this forum but oh well...

If $f(x)=\sin x$, find $f^{51}(x)$.

Is $f^{51}(x)$ a common way of writing the $51$-th deriviative of $f$?

I've seen it meaning the derivative a number of times, I've also seen it as the number of function applications, $f^{3}(x)$ being $f(f(f(x)))$.

It's more common to write $f^{(n)}(x)$ for the n-th derivative.


Let x<y be positive integers > 2
p Prime

find all Solutions so that

 $ y\minus{}px | y^2\minus{}xy\minus{}2x$

I guess there are infinitely many.

$ (y \minus{} px)m \equal{} y^2 \minus{} xy \minus{} 2x \\
\implies (x \plus{} m)^2 \minus{} 4x(pm \minus{} 2) \equal{} a^2$ where all symbols represent naturals.

From which we get
$ x^2 \plus{} x(10 \minus{} 4pm) \plus{} (m^2 \minus{} a^2) \equal{} 0$

If
$ x^2 \minus{} x(10 \minus{} 4pm) \equal{} 0 \\
\implies x \equal{} (4pm \minus{} 10)$, then obvious, we get infinitely many solutions.

I think there is an error in your transformation


I'm looking for information on how to combine computers and the internet and handwritten math notes.
My basic idea is to use a graphics tablet and an online canvas that supports collaborative works.

I have few relevant experiences myself, but a number of advantages of combining handwritten notes and computers are obvious:
- Handwritten notes are natural for doing math.
- They are much quicker than TeX etc. and avoid disruptions from coding formulae.
- It's easy to add simple diagrams etc. in handwritten notes.
- Via the internet, collaborative work with all its benefits becomes possible.
- The material can be made available for a large audience.
- Adding audio streams etc. makes the written material easier to understand.
- An electronic canvas or blackboard can be made much larger than a physical one.
- It's easy to manipulate elements on the canvas: remove parts, move them, highlight them, change their size, zoom in and zoom out, add comments (written or audio) ...
- It's easy to keep track of different versions.
(I know there are a number of severe shortcomings - messy handwriting being a major one.)

So - any ideas on this?

[quote="thomasteepe"](I know there are a number of severe shortcomings - messy handwriting being a major one.)[/quote]  Large file sizes is the other: LaTeX documents can be distributed as plaintext, and many people live in places without ready access to broadband internet.

I've seen a few people use tablet computers to very good effect.  They also work very well for presentations, as you can project slides but still have much of the freedom of a chalk talk.


parameterization
geometry
3D geometry
sphere
vector
function
conics

Hey guys,

I'm reading a course in 2nd year calculus. I have some questions. If this isn't the right place to post them let me know.

Find a vector-valued function g(t): R map to R3,  to describe the curve of intersection between x^2 + y^2 + z^2=6, and x+y+z=0.

We want to express $x$, $y$, and $z$ in terms of one variable (say, $t$) called the parameter. First, note that $z=-x-y$, so if we express $x$ and $y$ in terms of $t$, then $z$ will be easy to handle. The first equation tells us that $x^{2}+y^{2}+(-x-y)^{2}=6$. This is an equation of an ellipse, which can be brought into standard form and parametrized.

A more geometric approach is to realize that you have a circle of radius $\sqrt{6}$ in the plane $x+y+z=0$. Find two perpendicular unit vectors in the plane, say $\vec u=\langle 1,-1,0\rangle/\sqrt{2}$ and $\vec v=\langle 1,1,-2\rangle/\sqrt{6}$. Then the equation is $\vec r(t)=\sqrt{6}(\vec u\cos t+\vec v\sin t)$.

Thanks, the geometric approach makes sense, but I'm still not sure how to parametrize 2x^2+2xy+2y^2=6 in the "standard" way.

The axes of this ellipse make the angle of $\pi/4$ with coordinate axes. So we need a new coordinate system $(u,v)$ in which the axes are rotated by $\pi/4$. For example, $x=u+v$ and $y=u-v$ work. In this system the equation becomes $6u^{2}+2v^{2}=6$, which can be parametrized by $u=\cos t$ and $v=\sqrt{3}\sin t$. Finally, return to $x$ and $y$.


function
calculus
integration
real analysis
real analysis unsolved

Let h:[0,1]->R
          h(x)= 0     if x is irrational
                  1/n   if x is rational, i.e. x=m/n , where m and n are
                        natural numbers and coprime.

         Let [a,b] be any interval contained in { x: x>0 }

         Show that h is integrable on [a,b] in Riemann's sense.

Don't assume that you know the characterization of integrable functions in terms of almost-everywhere continuity. This problem would occur in a course just after the raw definition of the Riemann integral, and long before you would be in position to prove that characterization.

A key to the proof: pick an arbitrary $\epsilon>0.$ For how many points in the interval of integration is it true that $h(x)>\epsilon?$

Direct proof using Kent's hint is tricky but informative ($\left\{ f(x) > \epsilon \right\}$ is finite then $N = |\left\{f(x) > \epsilon \right\}|$ then cut with intervals of length at most $\epsilon / N$)

There is a killer argument : $f$ is continuous on irrationals (classical exercice, sure it was solved in this forum) so $f$ is continuous almost-everywhere (= except on a set of measure $0$) so it is Riemann-integrable (Lebesgue's :D theorem), and its integral is $0$


In triangle $ ABC$, $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$. A line $ l$ passes through $ G$, intersecting line $ AB$ at $ P$ and line $ AC$ at $ Q$, where $ P\ne B$ and $ Q\ne C$. If $ [XYZ]$ denotes the area of triangle $ XYZ$, show that $ \frac{[BGM]}{[PAG]}\plus{}\frac{[CMG]}{[QGA]}\equal{}\frac32$.

[quote="Johan Gunardi"]In triangle $ ABC$, $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$. A line $ l$ passes through $ G$, intersecting line $ AB$ at $ P$ and line $ AC$ at $ Q$, where $ P\ne B$ and $ Q\ne C$. If $ [XYZ]$ denotes the area of triangle $ XYZ$, show that $ \frac {[BGM]}{[PAG]} \plus{} \frac {[CMG]}{[QGA]} \equal{} \frac{3}{2}$.[/quote]
Nice problem  :)
Here,
$ [BGM]\equal{}[CMG]\equal{}\frac{1}{6}\cdot[ABC]$
So, We need to prove that,
$ \frac{[ABC]}{9[PAG]}\plus{}\frac{[ABC]}{9[QGA]}\equal{}1......(1)$

Here,
$ \frac{[PBG]}{[PGA]}\equal{}\frac{BP}{PA} \Longrightarrow \frac{[ABC]}{3[PGA]}\equal{}\frac{c}{PA}.....(2)$
[as, $ [ABG]\equal{}\frac{[ABC]}{3}$
Analogously,$ \frac{[QGC]}{[AGQ]}\equal{}\frac{CQ}{QA} \Longrightarrow \frac{[ABC]}{3[QGA]}\equal{}\frac{b}{QA}.....(3)$

So, $ (1)$ becomes,
$ \frac{c}{3PA}\plus{}\frac{b}{3QA}\equal{}1$

We draw $ A'B' \parallel AB$ through $ G$

[Note: Here we shall use directed length]
Using Menelau's Theorem on $ \triangle AB'C'$ and transvarsal $ PQ$ we get,
$ \frac{AP}{PB'}\equal{}\minus{}\frac{QA}{C'Q}
\\
\Longrightarrow \frac{PB'}{AP}\plus{}\frac{C'Q}{QA}\equal{}0
\\
\Longrightarrow \frac{PB'}{AP}\plus{}1\plus{}\frac{C'Q}{QA}\plus{}1\equal{}2
\\
\Longrightarrow \frac{AB'}{AP}\plus{}\frac{C'A}{QA}\equal{}2
\\
\Longrightarrow \frac{\frac{2AB}{3}}{AP}\plus{}\frac{\frac{2CA}{3}}{QA}\equal{}2
\\
\Longrightarrow \frac{AB}{3AP}\plus{}\frac{CA}{3QA}\equal{}1$

So,
$ \frac{c}{3PA}\plus{}\frac{b}{3QA}\equal{}1$
which implies,
$ \frac {[BGM]}{[PAG]} \plus{} \frac {[CMG]}{[QGA]} \equal{} \frac{3}{2}$
$ Q.E.D$

[quote="Johan Gunardi"][color=darkred] In triangle $ ABC$ , $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$ . A line $ d$ passes through $ G$

intersecting sideline $ AB$ at $ P$ and sideline $ AC$ at $ Q$ , where $ P\ne B$ and $ Q\ne C$ . Show that $ \frac {[BGM]}{[PAG]} \plus{} \frac {[CMG]}{[QGA]} \equal{} \frac32$ .[/color] [/quote]

[quote][color=darkred][b][u]Lemma (well-known or you can prove it easily !).[/u][/b] $ P\in AB\ ,\ Q\in AC\ ,\ G\in PQ\ \Longrightarrow\ \frac {\overline {AB}}{\overline {AP}} \plus{} \frac {\overline {AC}}{\overline {AQ}} \equal{} 3$ .[/color][/quote]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote the distance $ \delta (X)$ of the point $ X$ to the line $ AM$ .

Therefore, $ \left\|\begin{array}{c} \frac {[BGM]}{[AGP]} \equal{} \frac {GM\cdot\delta (B)}{AG\cdot \delta (P)} \equal{} \frac 12\cdot\frac {\overline {AB}}{\overline {AP}} \\
 \\
\frac {[CGM]}{[AGQ]} \equal{} \frac {GM\cdot\delta (C)}{AG\cdot\delta (Q)} \equal{} \frac 12\cdot\frac {\overline {AC}}{\overline {AQ}} \end{array}\right\|$ $ \stackrel{LEMMA}{\ \ \ \ \implies\ \ \ \ }$ $ \boxed {\ \frac {[BGM]}{[AGP]} \plus{} \frac {[CMG]}{[AGQ]} \equal{} \frac32\ }$ .[/color]

[quote="Virgil Nicula"][color=darkred][b]Lemma (well-known or you can prove it easily !). $ P\in AB\ ,\ Q\in AC\ ,\ G\in PQ\ \Longrightarrow\ \frac {\overline {AB}}{\overline {AP}} \plus{} \frac {\overline {AC}}{\overline {AQ}} \equal{} 3$ .[/b][/color][/quote]
I have proved it in my sollution :wink: (But it was not known to me....)

[quote="Moonmathpi496"]... So $ \frac {c}{3PA} \plus{} \frac {b}{3QA} \equal{} 1$ which implies ...
I have proved it in my sollution :wink: (But it was not known to me)[/quote]
[color=darkblue][b]Very well ![/b] You passed through the proof of this well-known (remarkable !) property of a line $ d$ , $ G\in d$ . Generally,

$ \left\|\begin{array}{c} M\in (BC)\ ,\ P\in (AM) \\
\ X\in (AB)\ ,\ Y\in (AC)\ ,\ P\in XY\end{array}\right\|$ $ \implies$ $ \boxed {\ \frac {AB}{AX}\cdot MC \plus{} \frac {AC}{AY}\cdot MB \equal{} \frac {AM}{AP}\cdot BC\ }$ .[/color]


inequalities
induction
inequalities proposed

[quote="Zamfirmihai"][quote="mathematica"]Prove that for every $n\in \mathbb{N}$, we have $(n!)^{2}\geq n^{n}$.[/quote]
Maybe is $(n!)^{2}\leq n^{n}$[/quote]
I think the initial problem is ok.Here's a proof using induction:
We only have to prove that:
$((n+1)!)^{2}\geq (n+1)^{n+1}$ $\Leftrightarrow$ $(n!)^{2}(n+1)^{2}\geq (n+1)^{n+1}$,but by the induction hypothesys we have:
$(n!)^{2}(n+1)^{2}\geq n^{n}(n+1)^{2}\geq (n+1)^{n+1}$
And the last one is equivalent to:
$n^{n}(n+1)\geq (n+1)^{n}=\sum_{k=0}^{n}n^{n-k}\cdot C_{n}^{k}$ which is true ,because $n^{n}\geq C_{n}^{k}\cdot n^{n-k},\Leftrightarrow T_{1}\geq T_{k+1}\forall k=\overline{0,n-1}$ ,which is proved by the fact that $\frac{T_{k+2}}{T_{k+1}}<1$ ,where $T_{k+1}=C_{n}^{k}\cdot n^{n-k}$

$(n!)^{2}=\prod_{i=1}^{i=n}((n+1-i)*i)$ so you need to prove that $(n+1-i)*i\geq n$ if $1\leq i\leq n$ which is equivalent to $(n-i)(i-1)\geq 0$ which is true.
this is a very well known proof for this problem.


function
limit
real analysis
real analysis unsolved

Show the Riesz Marcell theorem in case $ p\equal{}2$, with Fourier Transform.
Give a countraexample in $ p\equal{}\infty$.

I know two closely related theorems of Marcel Riesz, and each theorem could be stated either for $ \mathbb{T}$ or for $ \mathbb{R}.$

Here it is on $ \mathbb{T}:$

1. Suppose $ f$ is a "sufficiently nice" function on $ \mathbb{T}.$ Define $ \widetilde{f}(x)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}\minus{}i\text{sgn}\,(k)\widehat{f}(k)e^{ikx}.$ (Note: this can also be obtained by convolving $ f$ with a constant times the principal value of $ \cot(x/2).$) Then  for $ 1<p<\infty$ there exists a constant $ c_p$ such that $ \|\widetilde{f}\|_p\le c_p\|f\|_p$ and hence the operator $ f\mapsto\widetilde{f}$ can be extended to a bounded operator from $ L^p$ to $ L^p.$ The constant $ c_p$ tends to $ \infty$ as $ p\to 1$ or $ p\to \infty$ and this theorem is not true for $ p\equal{}1$ or $ p\equal{}\infty.$

2. Suppose $ f\in L^p(\mathbb{T}).$ Define $ S_Nf(x)\equal{}\sum_{k\equal{}\minus{}N}^N\widehat{f}(k)e^{ikx}.$ Then $ \lim_{N\to\infty}\|S_Nf\minus{}f\|_p\equal{}0.$ That is, Fourier series converge in $ L^p.$ Again, this is false for $ p\equal{}1$ or $ p\equal{}\infty.$

The theorem stated as #1 is used to prove the theorem stated as #2.

-----------

Both of these could be restated for functions on $ \mathbb{R}.$ On $ \mathbb{R},$ the conjugate function (Hilbert transform) is given by convolution with the principal value of $ \frac1{\pi x}.$

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

As for proving them for $ p\equal{}2:$ that's not really what M. Riesz's theorem is about, and you don't need his arguments. If you look at the conjugate function (Hilbert transform) in its multiplier version  - that is, multiply the Fourier coefficients by $ \minus{}i\,\text{sgn}\,(k)$ - then the boundedness follows immediately from Parseval's identity. As as for #2: the $ L^2$ convergence of Fourier series is a matter of the general theory of orthonormal sets in Hilbert spaces, once we prove that the trigonometric basis is a complete orthonormal set.

As for counterexamples in $ L^{\infty}$ - are you referring the the theorem I've called #1 or the theorem I've called #2? It makes a difference in terms of how hard we need to work.

Thank you!!
You know some book or article?

The Katznelson book, [i]An Introduction to Harmonic Analysis[/i], is a good place to start. It was published by Dover; I haven't checked lately to see whether it's still in print.

Or you could go to the source: Antoni Zygmund's [i]Trigonometric Series[/i].


$construct$ $a$ $triangle$ $given$ $it's$ $A(one$$vertex$),$G(centroid)$,$H(orthocentre)$

Let $A_{1}$ is the midpoint of $BC$ ( the side opposite vertex $A$ ), by $\overrightarrow{GA}=-2\overrightarrow{GA_{1}}$ and $\overrightarrow{GH}=-2\overrightarrow{GO}$ we can construct $O$ (circumcenter) and $A_{1}$. The line which perpendicular $OA_{1}$ at $A_{1}$ intersects $(O,OA)$ ( the circumcircle ) at $B$ and $C$ ( the rest vertexes )

thanks it's a nice solution.can you give me the proof for $A$,$G$,$H$ to be colinear


algebra polynomial induction calculus integration function domain

if we talk about the roots of a polynomial in the sense of when the polinomial =0 mod p, it's obvious that it might have at most n roots where n is the degree, for example $(x-1)(x-2)$ =0 mod 3 for 1 and 2...but if the polinomial cannot be factor, for example $x^{2}+1$ does the theorem is still valid? thanks.

The theorem's still valid though the polynomial cannot be factored.
Here is Lagrange's theorem (according to Sierpinki):
[quote]If $n$ is a natural number and f(x) is a polynomial of degree n with integral coefficient, the coefficient of $x^{n}$ is not divisible by p then the congruence $f(x) \equiv 0$ (mod $p$) has at most $n$ roots[/quote]

[quote="dondigo"]Use induction and Bezout theorem (if $W(a)=0$ then $W(x)=(x-a)P(x)$)[/quote]

       I dont get this stuff, know the Bezout i get bu the other one, there is no proof or work

                     P.S. I NEED WORD TO UNDERSTAND!!!!! :furious:  :wallbash_red:  :spam:

[hide="Discussion"] Lagrange's Theorem holds in prime moduli because they are integral domains; in other words, if $ab \equiv 0 \bmod p$ then either $a \equiv 0 \bmod p$ or $b \equiv 0 \bmod p$.

That is why Lagrange's Theorem fails $\bmod$ composite values.  For example, the quadratic $(x-a)(x-b) \bmod pq$ where $p, q$ are distinct odd primes and $a \neq b$ has four solutions (obvious by Chinese Remainder Theorem). [/hide]

[quote="dondigo"]Use induction and Bezout theorem (if $W(a)=0$ then $W(x)=(x-a)P(x)$)[/quote]what do you mean? if the polynomial cannot be factored...

[hide="Complete proof"]
[b]Theorem[/b]

[color=blue]Polynomial with integer coefficients $W(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$ satisfying $a_{n}\not\equiv{0}\pmod{p}$ has utmost $n$ roots modulo $p$
[/color]

We will perform an induction on $n$

For $n=1$ we have $W(x)=ax+b$. Assume, that it has more that one root.
Let say $r,t$ are roots of $W(x)$ thus:\[ar+b\equiv{at+b}\equiv{0}\pmod{0}\Rightarrow a(r-t)\equiv{0}\pmod{p}\Rightarrow r\equiv{t}\pmod{p}\]
Contradiction.

Now assume, that the theorem holds for $n-1$. I'll show that it holds for $n$
If $W(x)$ has no roots, we are at home, in other case it has at least one root.
Let it be $b$

\[W(x)\equiv{W(x)-W(b)}\equiv{a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}-(a_{n}b^{n}+a_{n-1}b^{n-1}+\ldots+a_{1}b+a_{0})}\pmod{p}\]

\[W(x)\equiv{a_{n}(x^{n}-b^{n})+a_{n-1}(x^{n-1}-b^{n-1})+\ldots+a_{1}(x-b)}\pmod{p}\]
\[W(x)\equiv{(x-b)g(x)}\pmod{p}\]

Polynomial $g(x)$ has integer coefficients and it's leading coefficient if equal to $a_{n}$(one can compute it using the formula $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}$), so it satisfies the conditions of out theorem.
Assume by contradiciton, that $W(x)$ has at least $n+1$ roots. It means that it has has at least $n$ roots different than $b$. All of them are roots of $g(x)$. But from the assumption follows, that it has utmost $n-1$ roots. Contradiciton.

[/hide]

In general, this result holds in any field, but not in an arbitrary integral domain

No, it holds in integral domains, too (assumed integral domains to be commutative and having $1$).

And to srulikbd: You say "...if the polinomial cannot be factor...". Note that factoring should be seen as some property of whatever you work in. Especially, $x^{2}+1$ factors in $\mathbb Z / 5 \mathbb Z$ as $(x-2)(x+2)$.
This makes life easier ;)

Ah yes... of course you are right.
Too much noncommutative ring theory can destroy basic knowledge  :oops:


abstract algebra
group theory
superior algebra
superior algebra unsolved

Suppose $ G$ is a group and that it is isomorphic to $ \textbf{F}_q$. Is it then true that $ \text{Aut}(G)$ is isomporphic to $ (\textbf{F}_q)^*$?

If $ q\equal{}p^n$ then any automorphism of $ F_q$ (as an abelian group) can be viewed as an invertible $ \mathbb{F}_p$-linear transformation of $ V\equal{}(\mathbb{F}_p)^n$, and vice-versa.  So $ Aut(G)\cong GL(n,\mathbb{F}_p)$. (ask for details if you don't understand this step)

But $ |GL(n,\mathbb{F}_p)|\equal{}(p^n\minus{}1)(p^n\minus{}p)\ldots(p^n \minus{} p^{n\minus{}1})$, which is greater than $ q\minus{}1$ if $ n > 1$.  But $ |\mathbb{F}_q^*|\equal{}q\minus{}1$, so it is not true in this case.

When $ n\equal{}1$, however, $ Aut(G) \cong Aut(Z_p) \cong Z_{p\minus{}1} \cong \mathbb{F}_p^*$.

Oh, it was not that hard efter all. I should have had tried harder. Thanks LydianRain!


i know this has been done on the regular amc forum, but who needs all those extra non-hoosier(or whose ear) posts.

who really cares? everyone basically knows everone else's scores, but whatever...

135 on 10a
25 on 8
AIME: i'll edit it in later

too lazy to list seventh grade...


Find the value of x

r and s are parallel.

[img]http://img193.imageshack.us/img193/7902/melhorq.jpg[/img]

From the arcs, you can see that $ 4x \plus{} 30 \equal{} 2x \plus{} 10$, so $ x \equal{} \minus{}10$, unless I'm missing something, which I think I must be...  :huh:

If I interpret that correctly, it's just that the arcs aren't distinguishing noncongruent angles.

Draw a third line parallel to r and s through that point in the middle and then you end up with \[ 2x\plus{}10\plus{}180\minus{}(5x\plus{}20)\equal{}4x\plus{}30\Rightarrow x\equal{}\boxed{20}\]

I don't understand your logic ,5849206328x.

Ok, but What do you do in this example ?
Do you have a site to learn ?

Ah that's ok. Now I understand. Thanks for your help ;)


You are given a string of digits: $718923$. You are allowed to perform the following operation any number of times:

1) Take any two digits that are next to each other.

2) Place their positive difference anywhere in the string.

3) Place their sum anywhere in the string. If their sum is greater than 10, only place its unit digit in the string. After a series of operations, is it possible to transform the original string into $505731218798236$? If not, prove it.

[hide="answer"]
NO because seeing the $1$st $2$ digits $70...$ $0$ can never be formed because $7$ is next to it and $70$ can never be formed by adding $2$ digits
If it is then there would be more digits in the front.
[/hide]

Are you saying that that 70 is too large to be formed by adding two digits? That's true, but 0 can be a result of taking the difference of two equal digits, and 7 can be a result of adding, say, 3 and 4. (remember, after computing the sum and difference, you can place them anywhere you want)

yes I get your point then I think it is Yes It can be formed.

Step 3 should have been fixed (fixed now). Also, the final string is shorter.


induction
inequalities
least common multiple
More Sequences

Does there exist positive integers $a_{1}<a_{2}<\cdots<a_{100}$ such that for $2 \le k \le 100$, the least common multiple of $a_{k-1}$ and $a_{k}$ is greater than the least common multiple of $a_{k}$ and $a_{k+1}$?

Let's prove that for every integer $ n\geq 2$, there exists a sequence $ a_1 < a_2 < \dots < a_n$ of positive integers such that:
1) the sequence $ lcm(a_k,a_{k \minus{} 1})$, $ k \equal{} 2, 3\dots,n$, is strictly decreasing;
2) $ a_n < lcm(a_{n \minus{} 1},a_n).$

Proof by induction on $ n$.

For the induction base case of $ n \equal{} 2$, we can take $ a_1 \equal{} 2 < a_2 \equal{} 3 < lcm(2,3) \equal{} 6$.
Suppose that the required sequence $ a_1 < a_2 < \dots < a_n$ of $ n$ elements exists. Below we will show how to construct a sequence of $ n\plus{}1$ elements.

Taking $ d$-th power of each term and multiplying it by $ a_{n \minus{} 1}^{d \plus{} 1}$, we have $ a_1^d a_{n \minus{} 1}^{d \plus{} 1} < a_2^d a_{n \minus{} 1}^{d \plus{} 1} < \dots < a_n^d a_{n \minus{} 1}^{d \plus{} 1}$. It is clear that this sequence still satisfies the property 1). Let's extend it with an extra term $ a_n^{d \plus{} 1} a_{n \minus{} 1}^d$:
\[ (\star)\qquad a_1^d a_{n \minus{} 1}^{d \plus{} 1} < a_2^d a_{n \minus{} 1}^{d \plus{} 1} < \dots < a_n^d a_{n \minus{} 1}^{d \plus{} 1} < a_n^{d \plus{} 1} a_{n \minus{} 1}^d.
\]
Our goal is to find such $ d$ that the property 1) holds, i.e.:
\[ lcm(a_{n \minus{} 1}^d a_{n \minus{} 1}^{d \plus{} 1}, a_n^d a_{n \minus{} 1}^{d \plus{} 1}) > lcm(a_n^d a_{n \minus{} 1}^{d \plus{} 1},a_n^{d \plus{} 1} a_{n \minus{} 1}^d)
\]
or equivalently
\[ \ell^d a_{n \minus{} 1}^{d \plus{} 1} > a_n^d a_{n \minus{} 1}^d \ell
\]
where $ \ell \equal{} lcm(a_{n \minus{} 1},a_n).$ The last inequality is equivalent to
\[ \left(\frac {\ell}{a_n}\right)^d > \frac {\ell}{a_{n \minus{} 1}}
\]
which has a solution (w.r.t. $ d$) since $ \frac {\ell}{a_n} > 1.$
Therefore, we can select $ d$ such that the sequence (*) of $ n \plus{} 1$ elements satisfies the property 1).
It is also easy to see that the property 2) for the original sequence (i.e., $ a_n < lcm(a_{n \minus{} 1},a_n)$) implies the property 2) for the sequence (*):
\[ a_n^{d \plus{} 1} a_{n \minus{} 1}^d < a_n^d a_{n \minus{} 1}^d lcm(a_{n \minus{} 1},a_n) \equal{} lcm(a_n^d a_{n \minus{} 1}^{d \plus{} 1},a_n^{d \plus{} 1} a_{n \minus{} 1}^d).
\]
Q.E.D.


Shock! is a game is a Mafia-esque game where a group of people stand in a circle.  One of the people is a "shocker" and the aim of the game is to discover who.  The game is played as follows:

- I will PM the shocker a number.
- The shocker will PM one of the people "standing" next to him one less than that number.
- That person will PM me the number they recieved and PM one of the people next to him/her one less than the number.
- Whoever recieves one shock, will be eliminated from the game and I will post in this thread who it was.
- The players converse on the thread.  After the conversation is completed, a vote takes place.  Whoever has the most votes drops dead.  Whether or not they were the shocker is revealed.  The next round begins.

Rules and specifics:
- We need at least 20 people to play.  As soon as I recieve 20 PM's, I will PM one of the 20 people with a message saying that they are the shocker.
- During the "shock period", NO POSTS MAY BE MADE ON THIS THREAD.
- During the conversation period, you may tell people what number you got, but people can lie about this.
- The conversation period is over once everybody has said in BLUE TYPE who they vote for.

Notice that you will know who the shocker is if you are standing next to him/her.  You can vote for this person, but you may not tell people your true reasoning for doing so.

To play, you must be an acitve AoPSer, in other words, you will be able to participate whenever need be.

Please submit joining PM's as soon as possible.

Sign-up list:

perfect628
pianoforte
math92

You did not indicate how you shock someone.

This rule makes it stupid:
[quote]Notice that you will know who the shocker is if you are standing next to him/her. You can vote for this person, but you may not tell people your true reasoning for doing so. [/quote]
as we can just say: VOTE FOR THE SHOCKER ONLY IF YOU ARE STANDING NEXT TO HIM
and when two people vote for one person, the game is over.
Or the persons standing next to the shocker might be just the only one who voted without reason.
Or the persons standing next to the shocker could vote for the same person over a period of rounds.
The thing is, you can never actually enforce this rule.

And depending how you shock, it may be too obvious as well considering you can just post you received a number and we know the person is at most 2 spaces from you and if you post who you sent it to or who you received it from then we've narrowed it down to 2 people and the game is quickly over.

Wait a minute.

So we basically give each other shockers.

Do you know how incredibly wrong this sounds?

[quote="Ignite168"]You did not indicate how you shock someone.

This rule makes it stupid:
[quote]Notice that you will know who the shocker is if you are standing next to him/her. You can vote for this person, but you may not tell people your true reasoning for doing so. [/quote]
as we can just say: VOTE FOR THE SHOCKER ONLY IF YOU ARE STANDING NEXT TO HIM
[b]and when two people vote for one person, the game is over.[/b]
[i]Or the persons standing next to the shocker might be just the only one who voted without reason.[/i]
Or the persons standing next to the shocker could vote for the same person over a period of rounds.
The thing is, you can never actually enforce this rule.

And depending how you shock, it may be too obvious as well considering you can just post you received a number and we know the person is at most 2 spaces from you and if you post who you sent it to or who you received it from then we've narrowed it down to 2 people and the game is quickly over.[/quote]

I don't see your reasoning for the bold sentence.
For the italics sentence, you can make up a different reason as to why you are voting for the person.
If it makes the game better in your opinion, I can delete the rule in which I say the number aloud to everybody.
"Shocking" is done by PMing somebody next to you with a number.

You could change the game up slightly so that the people next to the shocker don't know who it is.  Pick a random positive integer to send to the shocker, and let anyone who is shocked shock anyone else.  Furthermore, don't do the shocking by direct PM--have the people PM you with who they want to shock, then PM the shocked person with the number.  That would be a lot harder, though.

Ok then the shocker is obviously the one that's next to the spot in which people keep dying.

Ignite, you misunderstood the rules.  If you are the shocker, let's say you get the number "6".  Then you pass "5" to somebody.  That person passes "4" to somebody.  That person passes "3" to somebody.  And so on until somebody recieves "1", who dies.

Ok then in that case, people claim what number they got and then the shocker has to either claim he never received a number, or he received a number from the person next to him. Then, one of the people next to him will counter whatever he claims and the game is quickly over after throwing two people out the group.

I wrote this incorrectly on the original rules, but you are NOT allowed to claim what number you got.

[quote="miyomiyo"]Sorry to sound like a broken record, but does anybody want to join?  I only have 3 players and I'm going away in a month for a month with no internet, so....[/quote]

In two months, you should get plenty of players I guess ...


Asymptote
limit
calculus
calculus computations

Let $ p,q \in \mathbb{Q}$  such that $ p\plus{}q\equal{}1$. Prove that $ y\equal{}x\plus{}ap\plus{}bq$ is a slant asymptote of the curve $ y\equal{}(a\plus{}x)^p(b\plus{}x)^q$.

This is a basic problem about limits. We are trying to find m and c satisfying

$ \lim_{x\rightarrow \infty }\left( (a\plus{}x)^{p}(b\plus{}x)^{q}\minus{}(mx\plus{}c)\right) \equal{}0$

You will find that $ m\equal{}1,c\equal{}pa\plus{}bq$ will satisfy the criteria.

[quote="IndoChina"]This is a basic problem about limits. We are trying to find m and c satisfying

$ \lim_{x\rightarrow \infty }\left( (a \plus{} x)^{p}(b \plus{} x)^{q} \minus{} (mx \plus{} c)\right) \equal{} 0$

You will find that $ m \equal{} 1,c \equal{} pa \plus{} bq$ will satisfy the criteria.[/quote]How exactly do we prove that?


What is the remainder when the product of the first 25 prime numbers is divided by 4?

Its a multiple of 2 (because of 2) so its an odd number divided by 2 so the answer is 1.

To be a little more clear:

There is only one even prime number ($ 2$).  Then the product of the first $ 25$ numbers is $ 2(\text{some odd junk})$.  Divide by $ 4$, and we get $ \frac{\text{some odd junk}}{2}$, so it has remainder $ \boxed{1}$.

FTW says the correct answer is 2...I dont get it.

AIME15 made a typo at the last part. Since 2(some odd junk) = 2(2n+1) = 4n+2, the answer is 2.

[quote="Mrdavid445"]wait, shouldn't these solutions be hidden?[/quote]

There's a great reason to revive a 3-year old thread.

The answer is 2.  The product of the first three primes is 2*3*5=30.  30/4=7 R2.  You can see that 30*any odd number will result with a remainder of 2 when divided by 4.  For example, 30*7=210.  210/4=52 R2.


What is the sum of all the digits in the sequence 1, 2, 3, 4, 5, 6, 7,...99, 100?

It's asking for the digit's sum.
[hide]
S=1+2+3+...+9=45
Since there are 10 different ten's, 10, 20, 30...,
you do 45*10=450
Since there are the ten's digits, you add 45 more.
Plus 1 for 100=
496[/hide]

[quote="mathmanman"]Huh?  :huh:
Then, I think I misunderstood the question..[/quote]
you did sum of numbers and the question asked for sum of [u]digits[/u]

Oh!  :rotfl:
Yeah, so sorry...
Well, pascal_1623 has solved it :)


geometry
circumcircle
Euler
inequalities
geometry proposed

$O$ is an interior point of a triangle $ABC$. The feet of the perpendiculars from $O$ to $BC$, $CA$ and $AB$ are denoted by $A_{1}$, $B_{1}$, $C_{1}$ respectively. The bisector of  $\widehat{BOC}$ intersects $BC$ at point $A_{2}$, The bisector of $\widehat{COA}$ intersects $CA$ at point $B_{2}$, The bisector of  $\widehat{AOB}$ intersects $AB$ at point $C_{2}$. Prove that $area(A_{1}B_{1}C_{1}) \leq area(A_{2}B_{2}C_{2})$

Yes, it is nice, I change the your notation $O$ by $M$ and circumcenter is $O$
let $MA=x,MB=y,MC=z$ note that $AA_{2},BB_{2},CC_{2}$ are concurrent at $I(MA,MB,MC)$ therefore
$S_{A_{2}B_{2}C_{2}}=\frac{2xyz}{(x+y)(y+z)(z+x)}S_{ABC}$ and by Euler formula
$S_{A_{1}B_{1}C_{1}}=\frac{R^{2}-OM^{2}}{4R^{2}}S_{ABC}$
now we need prove that
$\frac{8xyz}{(x+y)(y+z)(z+x)}\ge(1-\frac{MO^{2}}{R^{2}})$
easily seen the condition $R-OM<x,y,z<R+OM$ by triangle inequality.
Actually there is a problem in inequality theory
If $x,y,z>0$ and $u-v<x,y,z<u+v,u>v>0$ we will get
$\frac{8xyz}{(x+y)(y+z)(z+x)}\ge 1-\frac{v^{2}}{u^{2}}$
I have solution of the last inequality by analysis, it isn't nice therefore I want to find a nice solution from anybody!:D :D

[quote="gemath"]
If $x,y,z>0$ and $u-v<x,y,z<u+v$ we will get
$\frac{8xyz}{(x+y)(y+z)(z+x)}\ge 1-\frac{u^{2}}{v^{2}}$
[/quote]
It doesn't look true - if $u=0$ we have actually reversed inequality by AM-GM.

you are right but I have mistake when I write you can see easily if you read geometric solution part .I edited it!

But still you can take $u$ arbitrarily close to $0$, can't you? So it doesn't change anything

When $u\rightarrow 0 LHS\rightarrow-\infty$ there isn't irrational!
I only forget $u>v$.I edited it.

Ok, I must misunderstand something all the time, so I just leave it as it is with only small comment. If $u>v$ as you claim problem is obvious as LHS is greater than zero whereas RHS is not  :D

no when $u>v\Rightarrow 1-\frac{v^{2}}{u^{2}}>0$ it in't obvious, actually I want to find nice solution!


[quote="N.T.TUAN"]Cac ban cho hoi lam the nao de dang nhap  vao trang nay?
Truoc kia lam rat don gian.[/quote]

De dang nhap vao trang nay`. chi can` vao` trang web nay` http://www.artproblemsolving.com.   Xong roi` Login , sau do' vao` NATIONAL &COMMUNITIES/ASIA/VIETNAM thui !! ma` khong biet toi tra loi co dung' muc. dich' cua cau hoi khong nua~!!  :lol:

Cam on ban,minh lam duoc roi,nhung khong phai bang cach ban noi.


Did you try the "More emoticon" link?  What more could you want?

There are many emoticons:
 :moose:  :ddr:  :spam:

How did you do that spiderboy?   Oops. nvm.. Found out!!!

 :spider:  :starwars:   :gathering:

I'm just having fun with a new discovery.


algebra
polynomial
superior algebra
superior algebra unsolved

if $ f(x)\equal{} a_{0}\plus{}a_{1}x\plus{}a_{2}x^{2}\plus{}....\plus{}a_{n}x^{n} \in R[x]$ with all roots in $ R$(real  number),
show that the roots of $ g(x)\equal{}\binom{n}{0}a_{0}\plus{}\binom{n}{1}a_{1}x\plus{}\binom{n}{2}a_{2}x^{2}\plus{}....\plus{}\binom{n}{n}a_{n}x^{n}$ are also all in $ R$

[quote="mathxxxx"]is there any theorem about real roots of polynomial ?[/quote]
May be [url=http://en.wikipedia.org/wiki/Sturm's_theorem]Sturm's theorem[/url] can help, but I'm not sure.
Whence you took this problem?


algebra
function
domain
polynomial
rational function
superior algebra
superior algebra unsolved

Hello, I would have a hint for this problem:

Let $ K$ be a field. I note $ K[X]$ the domain of polynomials over $ K$
and $ K' \equal{} K(X)$ the field of fractions of $ K[X].$
 $ v : K(X) \rightarrow \mathbb{Z}\cup \plus{} \infty$ defined by $ v(\frac{f}{g}) \equal{} v(f) \minus{} v(g)$ with $ v(f) \equal{} \minus{}deg(f)$ if $ f \neq 0$ and $ v(0) \equal{} \plus{}\infty$.

Show that $ K(X)$ is not  complete for the valuation $ N$ such that $ N(\frac{f}{g})\equal{} \rho^{v(\frac{f}{g})}$ with $ 0 < \rho < 1.$

$ 1, 1\plus{}X, 1\plus{}X\plus{}X^2, \ldots$ is a Cauchy sequence without limit.

[quote="berti"]$ 1, 1 \plus{} X, 1 \plus{} X \plus{} X^2, \ldots$ is a Cauchy sequence without limit.[/quote]

Are you sure this is a Cauchy sequence?

  darij

it's a Cauchy sequence but has a limit $ L\equal{}\frac{1}{1\minus{}X} \in K'.$ So it does not prove that $ K'$ is not complete!

Why is it a Cauchy sequence? For $ n<m$, we have

$ N\left(\underbrace{\left(1\plus{}X\plus{}...\plus{}X^n\right)\minus{}\left(1\plus{}X\plus{}...\plus{}X^m\right)}_{\text{polynomial of degree }m}\right)\equal{}\rho^{\minus{}m}$,

what can become arbitrarily big when $ m$ is big enough (since $ \rho$ is $ <1$).

Whoever composed the problem is pretty good at obfuscation. Why define $ v\left(f\right)$ as [i]minus[/i] the degree of $ f$ and $ \rho$ to be $ <1$ when it could have equally be done the other way round and nothing would change, except that the problem would become more intuitive?

  darij

Anyone still interested in the problem? I have a solution now, but it is pretty long, so I will just sketch it:

It is easy to see that the sequence $ \left(a_k\right)_{k\in\mathbb{N}}$ defined by $ a_k \equal{} \sum_{i \equal{} 0}^k\frac {1}{X^{2^i}}$ is a Cauchy sequence. If we succeed to show that this sequence has no limit in $ K\left(X\right)$, then we are done. Showing this is equivalent to proving that the formal series $ \sum_{i \equal{} 0}^{\infty}\frac {1}{X^{2^i}}$ does not represent a rational function. This is easily seen to be equivalent to the fact that the formal series $ \sum_{i \equal{} 0}^{\infty}X^{2^i}$ does not represent a rational function (in fact, you can substitute $ \frac {1}{X}$ for $ X$).

In order to prove that it does not, assume that it does. Then, there exist two polynomials $ \sum_{u \equal{} 0}^{\infty} p_uX^u$ and $ \sum_{v \equal{} 0}^{\infty} q_vX^v$ such that

[color=green][b](1)[/b][/color] $ \sum_{i \equal{} 0}^{\infty}X^{2^i}\cdot\sum_{v \equal{} 0}^{\infty} q_vX^v \equal{} \sum_{u \equal{} 0}^{\infty} p_uX^u$.

These polynomials should really be polynomials, i. e. all $ p_u$ for sufficiently large $ u$ must be zero, and all $ q_v$ for sufficiently large $ v$ must be zero (this is what distinguishes a polynomial from a formal series). For the sake of simplicity, set $ p_u \equal{} 0$ for all $ u\leq 0$ and $ q_v \equal{} 0$ for all $ v\leq 0$. WLOG assume that $ q_0 \neq 0$ (else, $ p_0$ must also be $ 0$, and we can divide both polynomials polynomials $ \sum_{u \equal{} 0}^{\infty} p_uX^u$ and $ \sum_{v \equal{} 0}^{\infty} q_vX^v$ by $ X$).

Then, for every integer $ u$, we have $ p_u \equal{} \sum_{i \equal{} 0}^{\infty}q_{u \minus{} 2^i}$ (by comparing coefficients in [color=green][b](1)[/b][/color]). Now, for every integer $ s$ big enough, we can take $ u \equal{} 2^s$, and then there is one and only one integer $ i\geq 0$ with $ q_{u \minus{} 2^i}$ being nonzero, namely $ i \equal{} s$ (because $ q_{u \minus{} 2^s} \equal{} q_0\neq 0$, while for $ i > s$ we have $ u \minus{} 2^i < 0$ and thus $ q_{u \minus{} 2^i} \equal{} 0$, and for $ i < s$ we have $ i\leq s \minus{} 1$ and thus $ u \minus{} 2^i \equal{} 2^s \minus{} 2^i\geq 2^s \minus{} 2^{s \minus{} 1} \equal{} 2^{s \minus{} 1}$, so that $ u \minus{} 2^i$ is bounded from below by $ 2^{s \minus{} 1}$, and if $ s$ is big enough, this will mean that $ q_{u \minus{} 2^i} \equal{} 0$). Hence, for every integer $ s$ big enough, we can take $ u \equal{} 2^s$ and then $ p_u \equal{} \sum_{i \equal{} 0}^{\infty}q_{u \minus{} 2^i}\neq 0$ (because a sum with one and only one nonzero term must be $ \neq 0$). Thus, we have found infinitely many $ u$ such that $ p_u\neq 0$, but this contradicts to $ p_u$ being $ 0$ for all sufficiently large $ u$.

The contradiction shows that $ \sum_{i \equal{} 0}^{\infty}X^{2^i}$ does not represent a rational function, and we are done.

A slightly different way to prove this would be to show that, if $ \sum_{u \equal{} 0}^{\infty} p_uX^u$ and $ \sum_{v \equal{} 0}^{\infty} q_vX^v$ are two polynomials, then the coefficients of the formal series $ \frac {\sum_{u \equal{} 0}^{\infty} p_uX^u}{\sum_{v \equal{} 0}^{\infty} q_vX^v}$ (when written as a sum of powers of $ X$) satisfy a linear recurrence, while the coefficients of the formal series $ \sum_{i \equal{} 0}^{\infty}X^{2^i}$ cannot satisfy a linear recurrence (because a sequence satisfying an $ n$-term linear recurrence is zero if it has $ n$ consequent zero entries, while the series of the coefficients of $ \sum_{i \equal{} 0}^{\infty}X^{2^i}$ has arbitrary large blocks of consequent zero entries but are not all zero).

  darij

is there a nice description of the completion?

As Darij already noticed, once you substitute $ X\to X^{\minus{}1}$, $ v(f)$ is just the $ X$-adic valuation. So, the completion (for the original $ v$) is $ K((X^{\minus{}1}))$. owk


trigonometry
inequalities
rearrangement inequality

[quote="TRAN THAI HUNG"][quote="arqady"][quote="TRAN THAI HUNG"]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that
$ \frac {p}{{q \plus{} r}}a^2 \plus{} \frac {q}{{p \plus{} r}}b^2 \plus{} \frac {r}{{p \plus{} q}}c^2 \ge 2\sqrt 3 S$
p,q,r is the three real positive numbers :) .[/quote]
Use
http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf[/quote]How can we use it for this case? :)[/quote]
We know that $ \frac {p}{{q \plus{} r}}(y \plus{} z) \plus{} \frac {q}{{p \plus{} r}}(x \plus{} z) \plus{} \frac {r}{{p \plus{} q}}(x \plus{} y) \ge \sqrt {3(xy \plus{} xz \plus{} yz)},$
where $ p,$ $ q,$ $ r,$ $ x,$ $ y$ and $ z$ are positive numbers.
Let $ y \plus{} z \equal{} a^2,$ $ x \plus{} z \equal{} b^2$ and $ x \plus{} y \equal{} c^2,$ where $ a,$ $ b$ and $ c$ are positive numbers.
Hence, $ a \plus{} b \minus{} c > 0,$ $ a \plus{} c \minus{} b > 0$ and $ b \plus{} c \minus{} a > 0$ and we obtain your inequality for acute-angle triangle.
For other triangles it's obvious:
Let $ c\equal{}max\{a,b,c\}.$ Hence, $ \frac {p}{{q \plus{} r}}a^2 \plus{} \frac {q}{{p \plus{} r}}b^2 \plus{} \frac {r}{{p \plus{} q}}c^2 \ge$
$ \geq\frac {p}{{q \plus{} r}}a^2 \plus{} \frac {q}{{p \plus{} r}}b^2 \plus{} \frac {r}{{p \plus{} q}}(a^2\plus{}b^2) \ge \sqrt 3 ab\geq2\sqrt 3 S.$ :wink:

WoW! I did not see it even after reading it.
Thanks for showing us.

see Topics in Inequalities by Hoojoo Lee. :P

By Rearrangement Inequality, $ LHS \ge \frac {a^2\plus{}b^2\plus{}c^2}{2}$

From Weitzenbock inequality, $ a^2\plus{}b^2\plus{}c^2 \ge 4 \sqrt 3 S$

The given inequality follows from the above two

[quote="hsbhatt"]By Rearrangement Inequality, $ LHS \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{2}$
[/quote]
Explain it please. Thank you!

[quote=TRAN THAI HUNG]Let ABC is a triangle, AB=c,AC=b, BC=a.Prove that
$ \frac{p}{{q \plus{} r}}a^2  \plus{} \frac{q}{{p \plus{} r}}b^2  \plus{} \frac{r}{{p \plus{} q}}c^2  \ge 2\sqrt 3 S$
p,q,r is the three real positive numbers :) .[/quote]
By C-S $\frac{p}{{q \plus{} r}}a^2  \plus{} \frac{q}{{p \plus{} r}}b^2  \plus{} \frac{r}{{p \plus{} q}}c^2=(p+q+r)\left(\frac{a^2}{q+r}+\frac{b^2}{p+r}+\frac{c^2}{p+q}\right)-a^2-b^2-c^2\geq$
$\geq(p+q+r)\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(p+q)}-a^2-b^2-c^2=\frac{1}{2}\sum_{cyc}(2ab-a^2)\geq2S\sqrt3$.


calculus
integration
function
calculus computations

Let $ g: (\alpha,\beta)\rightarrow\Re$ be a differentiable function, with $ g'(x)$ being a continuous function, and let $ f(x)$ be a continuous function for which $ (f\circ g)(x)$ is well-fined as a function on $ (\alpha,\beta)$. Then $ \int^{b}_{a}f(g(x))g'(x)dx \equal{}\int^{g(b)}_{g(a)}f(u)du$ for any $ a,b\in(\alpha,\beta)$.

Verify the substitution method by the Fundamental Theorem of Calculus and the Mean Value Theorems (for Differential Calculus, Integral Calculus).

Thank you for helping. :P

I was told to let $ H(x)\equal{}\int^{b}_{a}f(g(x))g'(x)dx\minus{}\int^{g(b)}_{g(a)}f(u)du$, then prove that $ H'(x)\equiv 0$, and $ H(a)\equal{}0$ for some $ a\in(\alpha,\beta)$.

I don't quite get how to prove $ H'(x)\equiv 0$, can someone please help.
Thank you. :P

[quote]I was told to let $ H(x) \equal{}\int^{b}_{a}f(g(x))g'(x)du\minus{}\int^{g(b)}_{g(a)}f(u)du$ , then prove that $ H(x)\equiv 0$[/quote]
Note that $ H(x) \equal{}\int^{x}_{a}f(g(u))g'(u)du\minus{}\int^{g(x)}_{g(a)}f(u)du$
Let $ F'(x): \equal{} f(x)$ and $ G'(x): \equal{} f(g(x))g'(x)$.So we have
$ H(x) \equal{} G(x)\plus{}F(g(a))\minus{}F(g(x))\minus{}G(a)$.It's easy to see $ H(a) \equal{} 0$ on the other hand:
$ H'(x) \equal{} G'(x)\minus{}(F(g(x)))'\equal{}0$
$ \iff H(x) \equal{} G'(x)\minus{}g'(x)F'(g(x)) \equal{} f(g(x))g'(x)\minus{}g'(x)f(g(x)) \equal{} 0$
So $ H(x)\equiv 0$


real analysis
complex analysis
real analysis unsolved

Hey guys, I have Problem that I can't solve. I would appreciate if someone would help me out with suggestions. It can be found in Rudin's Real and Complex Analysis after the second chapter. It goes like this:

Find a measurable set E, such that for each interval I, you have:

0 < m( E n I) < m(I)

Where m is the Lebesgue measure, and E n I means the intersection of I with E.

More precisely ,the problem is about a borle set E and an open interval I .


calculus
integration
limit
complex analysis

[quote="romy"]Hi
I have tried to evaluate the integral on the unit circle
int {(e^z - 2)/(e^z -2z-1)}
but I end up with ln(0) ie undefined.
Any suggestion?

Thx[/quote]
U mean $\int_{|z|=1}\frac{e^{z}-2}{e^{z}-2z-1}=2 \pi iRes(\frac{e^{z}-2}{e^{z}-2z-1},0)$

$Res(\frac{e^{z}-2}{e^{z}-2z-1},0)=\lim_{z\to 0}{\frac{e^{z}-2}{e^{z}-2z-1}}=\frac{e^{z}-2}{(e^{z}-2z-1)'}|_{z=0}=1$
So ur integral is $2 \pi i$


Let $ a_1, a_2, ..., a_n$ be positive reals and integer $ n>2$. Define
$ S \equal{} \sum a_i$
$ b_i \equal{} S \minus{} a_i$
$ T \equal{} \sum b_i$
Show that:  $ \frac {\prod a_i}{\prod (S\minus{}a_i)} \leq \frac {\prod b_i}{\prod (T\minus{}b_i)}$
(All summations and products range from $ 1$ to $ n$).

Note: Searched and didn't find it. This problem appeared in Amer. Math Monthly about 1 year back. I'm pretty sure the deadline has passed.

this problem is hard, but not nice. oh, can you post a solution for it?
 :D


Tom has six U.S. coins each of a different denomination (penny, nickel, dime, quarter, fifty cent piece, and silver dollar). How many different combinations of money can he formed using at least one coin?
 :lol:  Have fun  :D

For people who try to trial and error this, It'll take much too long   :?  not that I would know about that  :lol:

Weeelllllllllll...
[hide]
This is equivalent to the subset problem, except that the null subset is not allowed. Thus, we have $2^6-1=63$ ways.
[/hide]

is it one of each coin, or 6 of each coin?

I think it's one of each coin. If so, it would be
[hide]$\binom61+\binom62+\binom63+\binom64+\binom65+\binom66=63$ ways.[/hide]


[url=http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg]Picture[/url]

I need to prove that angle D = 90 + .5 (angle A)

Help?

PS. What is this rating under my name? It constantly fluctuates, and I don't know what it's rating. Thanks.

[quote="Icy"]What is this rating under my name? It constantly fluctuates, and I don't know what it's rating. Thanks.[/quote]

This is the average of the ratings that your posts get, on a scale of 0 to 6.

[quote="Icy"][url=http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg]Picture[/url]

I need to prove that angle D = 90 + .5 (angle A)

[/quote]
Well, you forgot to say that $ D$ is incenter.

Define some angles with $ \alpha,\beta$, play with them.

Is not a hard problem.

Sometimes, there are some problems that are incredibly easy. I hit my head against them for an hour or two, and right when I'm about to give up, the answer comes to me and I feel like the dumbest person on the planet. Does this ever go away with geometry proofs? Or is it just for me?

Draw perpediculars from incenter $ D$ to $ AB$ and $ AC$ call the foot of the perpediculars $ R$ and $ S$ respectively. Since we can easily prove by HL that $ \triangle{ADS}\cong\triangle{ADR}$ we can call $ \angle{ADS}\cong\angle{ADR}\equal{}\beta$ so $ \angle{A}\equal{}180\minus{}2\beta$. Now draw lines from $ D$ to $ B$ and $ C$ and a perpendicular from $ D$ to $ BC$ calling it $ T$. Similarly we can prove that $ \angle{RDB}\equal{}\angle{TDB}\equal{}\alpha$ and $ \angle{SDC}\equal{}\angle{TDC}\equal{}\mu$. Now we can show that $ \angle{BDC}\equal{}\alpha\plus{}\mu$ and that $ 2\alpha\plus{}2\mu\equal{}360\minus{}2\beta$ so $ \angle{BDC}\equal{}180\minus{}\beta$.   Now we can easily show that $ \frac{1}{2}(180\minus{}2\beta)\plus{}90\equal{}180\minus{}\beta$

[quote="Icy"][img]http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg[/img]

I need to prove that angle D = 90 + .5 (angle A)[/quote]
Call $ A\equal{}2\alpha,\,B\equal{}2\beta,\,C\equal{}2\gamma$

Since $ \measuredangle\,D\equal{}2\alpha\plus{}\beta\plus{}\gamma$ (very easy to prove), yields $ \measuredangle\,D\equal{}90^{\circ}\plus{}\alpha$

which the asked it's direct $ \blacksquare$

[quote="Hetidemek"][quote="Icy"][img]http://i38.photobucket.com/albums/e146/imdabest112/GeometryProblem.jpg[/img]

I need to prove that angle D = 90 + .5 (angle A)[/quote]
Call $ A \equal{} 2\alpha,\,B \equal{} 2\beta,\,C \equal{} 2\gamma$

Since $ \measuredangle\,D \equal{} 2\alpha\plus{}\beta\plus{}\gamma$ (very easy to prove), yields $ \measuredangle\,D \equal{} 90^{\circ}\plus{}\alpha$

which the asked it's direct $ \blacksquare$[/quote]

Yes, I found this solution about 2 minutes after I read your post. It's what I was referring to in the other post...the solution was so obvious I was ashamed that I asked. But thanks to both of you anyway :D


XY = 10cm, YZ = 14cm, XZ = 16cm. P bisects YZ into equal halves. Many lines can be drawn thru P, meeting XY (possibly extended) at A and XZ at B (possibly extended). Describe how to find the minimum perimeter of triangle XAB.

still cant figure this out. need some help here.

I will sketch my approach:
Consider any point P inside the angle ZXY and draw a line through it. Then draw the excircle to that line. One will notice that the length of the line is equal to the two tangents to this excircle, so the perimeter of the triangle is equal to the two long (equal!) tangents from X to the excircle. Since the angle is constant, it becomes clear that to minimize the perimeter, we have to minimize the radius of the excircle. This radius is minimized when the excircle passes through P (anything smaller and P will be inside it, which is not possible). So to find the min. perimeter, construct the circle tangent to XY and XZ that passes through P and construct the tangent to that circle through P, and this will be the desired line AB. Hopefully I didn't miss anything.  :maybe:


Links to a stupid song. One piece of advice - don't wait for the song to stop, because it won't. Enjoy!

Not worth the effort it took to click the link. Seriously. The animation is meaningless, as is the gibberish being sung. Do geeks spread things like these just so they have a sort of in-joke? Or just because it's the "popular" thing to do? Or do you really think it's funny?

[quote="Klebian"][quote="Ignite168"][quote="junggi"]errr isn't this spam....I mean with the pictures(links) repeated 200 times and stuff...well not spam but....uhhh[/quote]

First of all, no picture is repeated 200 times.

Second of all, I asked Klebian about it first and he said as long as there were 2 different pictures, 4 different links, and the song wasn't in French it was ok.[/quote]

Hm, sure, if you say so.

Don't worry, jmadsen, most people think the song sucks. Ignite thinks it's "awesome" though...[/quote]

Please find sources for both your statements. Wikipedia seems to disagree at least.

well i find the song quite interesting. :blush:
it has something unusual. and that language is very nice. what language is it, ignite?

He says it is in suomi. I won't go as far as actually saying it sucks, but I must admit, as a song, it leaves much to be desired.

The second link has the video of the song. The 5th link is a clock. The 7th link is Star Wars.

You guys didn't even bother looking at all 4 of them did you?

So jmadsen, shut your mouth if you don't know all the facts  :)

I found them amusing...and they took over my mind...*hums eerie song* :blush:

Finnish
More about loituma you have [url=http://en.wikipedia.org/wiki/Loituma]here[/url].


Prove that in any triangle $ ABC$ with angle bisectors $ AP$, $ BQ$, and $ CR$, $ \frac{1}{AR}\plus{}\frac{1}{BP}\plus{}\frac{1}{CQ}\equal{}\frac{1}{AQ}\plus{}\frac{1}{BR}\plus{}\frac{1}{CP}$

Maybe not the coolest proof, but I like the result.

Are $ P,Q,$ and $ R$ on segments $ BC, AC,$ and $ AB$, respectively?

Yes; angle bisectors must be on the side opposite the angle they bisect.

[hide]I will rename $ P, Q, R$ as $ A', B', C'$ if you don't mind, and I will call the lengths of the triangles $ a,b,c$ in the conventional way.  Thus the identity is, in my notation,
\[ \frac{1}{AC'} \plus{} \frac{1}{BA'} \plus{} \frac{1}{CB'} \equal{} \frac{1}{AB'} \plus{} \frac{1}{BC'} \plus{} \frac{1}{CA'} .\]

Notice that by the angle bisector theorem, $ AB' \equal{} b \cdot \frac{c}{a\plus{}c}$, and similarly for the other lengths.  Then we have
\[ \frac{1}{AC'} \plus{} \frac{1}{BA'} \plus{} \frac{1}{CB'} \equal{} \frac{a\plus{}b}{cb} \plus{} \frac{b\plus{}c}{ac} \plus{} \frac{c \plus{} a}{ba} \equal{} \left( \frac{a}{cb} \plus{} \frac{b}{ac} \plus{} \frac{c}{ba} \right) \plus{} \left( \frac{1}{c} \plus{} \frac{1}{a} \plus{} \frac{1}{c} \right) . \]
By symmetry of $ a$, $ b$, and $ c$, the other expression is also equal to this quantity, so the identity is proven.  $ \blacksquare$[/hide]
Nice relation.


A carpenter is preparing to put a roof on a garage that is 30 feet by 50 feet by by 30 feet. A steel support beam 56 feet in length is positioned in the center of the garage. To support the roof, another beam will be attached to the top of the center beam. At what angle of elevation is the new beam? In other words, what is the pitch of the roof?

I can't visulaize this.  Please describe it better.


Prove that in any triangle we have $\frac{R}{2r}\geq\frac{m_a}{h_a}$.

the hour is late so i'm sorry if i'm speaking nonsense in here but if you just use wellknown substitutions like R=abc/4Sand r=S/p and m_a^2=2b^2+2c^2-a^2/4 and a=(x+y)/2},b=y+z/2,c=x+z/2 the ineq turns out to be very weak.equality takes place only when the triangle is degenerate doesn't it?anyway i shall look over it again in the morning cuz now i'm just tooooo sleepy

@Bodom: Of course, an equilateral triangle ABC is an equality case!
See http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 for this problem...

  darij


algebra
polynomial
function
geometry
complex numbers
AoPS Books

hello, does anyone know a website or book or something that can help me with:
polynomial functions (the more indepth stuff)

arguments

de moivres theorem

complex numbers

thanks a bunch!

Mathworld's always a good place to look, and so are the AoPS books.

Yes the [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]AoPS Books[/url] might help you, or this particular [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#inttrig]AoPS Class[/url] might help you even more.

Barbeau's book about Polynomials (Springer-Verlag) is quite good. Hahn's book Complex Numbers and Geometry is a good book too.


Let n be a five digit number and let m be the four digit number obtained by removing
the middle digit of n. Find all such m,n such that n/m is an integer.

[hide="A Start"]
Let N = $ 10000a+1000b+100c+10d+e$
M is then $ 1000a+100b+10d+e$

We need to divide N into M and see what the remainder is.
Then we need to show for what values the remainder is integral, and that's the solution.
[/hide]


$x - 4 = (x + 2) \log 72$
$(1 - \log 72)x = 4 + 2 \log 72$
$x = \frac{4 + 2 \log 72}{1 - \log 72} \approx -8.99845...$

Does not really have a rational solution... (although, with calculator or given the value of $\log 3$ and $\log 2$, this problem is too easy and should not belong in this forum.)


Fill in the number so that this sudoku satisfies the properties:
1. The sum of numbers in the adjacent boxes with the same color must equal to the top left figure.
2. The numbers in the adjacent boxes with same color must be all different.

And it has to be a regular sudoku too!??!?!

[quote="mysmartmouth"]Oh man.  There was a...fun sudoku in the comics section.  Jason is doing it, and instead of numbers, it has "hard" math problems, such as $3\sqrt{9}$ instead of 6.  Here it is:

http://www.gocomics.com/foxtrot/2006/10/15/

[hide="Spoiler - ANSWERS"]
http://adventure.joystiq.com/2006/10/15/foxtrots-sodoku-ometry/
[/hide][/quote]

Unless I'm mistaken $-(2^{-2}) =-1/4$ (square upper-right of lower-middle), so it isn't possible

[quote="besttate"]

Unless I'm mistaken $-(2^{-2}) =-1/4$ (square upper-right of lower-middle), so it isn't possible[/quote]

That says $-(i^{2})$, which would be 1.

[quote="mysmartmouth"]
That says $-(i^{2})$, which would be 1.[/quote]

Ah, the $i$ was just super squiggly so that it looked like a 2 and its "dot" was a negative sign. heh

Yeah, at first my math teacher who showed me this thought it was a 2 as well.


geometry
combinatorics proposed
combinatorics

This one is a very nice one. I have a beautiful proof and hope you do too.
   Let ABCD be a quadrilateral and n be a positive interger > 1. We divide the sides AB, BC, CD, DA each into n equal segments. We draw lines through 2 corresponding points lying on AB and CD, AD and BC and we obtain $n^2$ quadrilaterals. Prove that the sum of areas of n different small quadrilaterals  which pairwisely don't lie on the same row or column is equal to $\frac{1}{n}$ the area of ABCD.
   I note that a corresponding pairs of points are points lying on two opposite sides of ABCD and divide their sides with the same ratio.  :lol:

You might want to look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16137&highlight=]this[/url] :).


real analysis
integration
function
Support
geometry
geometric transformation
real analysis unsolved

Let $ f\in L^{p}$ and $ g\in L^{q}$, where $ p$ and $ q$ are Holder conjugates.  Prove that $ f*g$ is bounded and continuous.

Obviously, $ \left|(f*g)(x)\right|=\left|\int_{\mathbb{R}}f(y)g(x-y)\,dy\right|\leq\left(\int_{\mathbb{R}}\left|f(y)\right|^{p}\right)^{1/p}\left(\int_{\mathbb{R}}\left|g(x-y)\right|^{q}\right)^{1/q}$
$ =\left\|f\right\|_{p}\left\|g\right\|_{q}<\infty$, so we get the bounded claim.  Now for continuity, we get that
$ \left|(f*g)(x+h)-(f*g)(x)\right|\leq\left\|f\right\|_{p}\left\|g(x+h)-g(x)\right\|_{q}$ from above.  However, I don't see why the $ g$ term necessarily goes to 0 without the assumption that $ g$ is continuous (which I never assumed).

In one solution I have seen, the the notation $ \tau_{\xi}(g): =g(\xi-x)\in L^{q}$ is used (not a problem) and it ultimately arrives at the conclusion
$ \cdots\leq\left\|f\right\|_{p}\left\|\tau_{\xi}(g)-\tau_{\xi_{0}}(g)\right\|_{q}=\left\|f\right\|_{p}\left\|\tau_{\xi-\xi_{0}}(\tau_{\xi_{0}}(g))-\tau_{\xi_{0}(g)\right\|_{q}}$, which goes to 0 as $ \xi\rightarrow\xi_{0}$.  Can someone justify this last step and/or my work?

I would start over. Write $ f\equal{}f_1\plus{}f_2,g\equal{}g_1\plus{}g_2,$ with $ f_1,f_2\in C_c$ and $ \|f_2\|_p,\|g_2\|_q<\epsilon.$

Then $ f*g\equal{}f_1*g_1\plus{}f_1*g_2\plus{}f_2*g*1\plus{}f_2*g_2.$

$ f_1*g_1\in C_c.$ As for the other terms, we have $ \|f_1*g_2\|_{\infty}\le \|f_1\|_p\|g_2\|_q<\epsilon\|f_1\|_p,$ and so on.

The fact that $ f*g$ can by uniformly approximated by continuous compactly supported functions means that it is itself continuous and goes to zero at infinity.

I'm reading through Wheeden and Zygmund's book right now and I can't seem to find it, but is there a theorem that guarantees that functions in $ L^{p}$ can be uniformly approximated by compactly supported functions?

Or, does my proof follow from Theorem 8.19, which is continuity in $ L^{p}$?  See [url=http://books.google.com/books?id=YDkDmQ_hdmcC&dq=wheeden+and+zygmund&printsec=frontcover&source=bl&ots=Md_KarLMOl&sig=OcPp2HgnQxW81e2TGfXIquq90JU&hl=en&ei=FLqESqmkI4umMZuahfEE&sa=X&oi=book_result&ct=result&resnum=6#v=onepage&q=&f=false]here[/url] for the statement, bottom of page 134.

Not uniformly approximated, of course, but approximated in the $ L^p$ sense.

Okay, that's just from the fact that continuous functions with compact support are dense in $ L^{p}$, right?

And does Theorem 8.19 justify the final step of my proof?

Yes, theorem 8.19, which is the continuity of translation in $ L^p(\mathbb{R}^n)$ for $ 1\le p<\infty,$ does justify your last step. Of course the proof of that theorem also rests on the density of $ C_c$ functions in $ L^p,$ and the failure of that theorem for $ L^{\infty}$ comes from the failure of that density.


analytic geometry
geometry
3D geometry
prism

Consider the 4x4x4 array of lattice points whose coordinates are either 0, 1, 2, or 3. How many right rectangular solids have eight of these points as vertices and have edges of integer length?

[hide="Solution"]You need to choose two distinct coordinates in each dimension to define a rectangular prism. In a $ 4\times4\times4$ space, there are $ \tbinom{4}{2}$ ways to choose two distinct points in each dimension, so there are $ \tbinom{4}{2}^3\equal{}\boxed{216}$ such rectangular prisms.

In general, there are $ \tbinom{n}{2}^3$ such rectangular prisms in an $ n\times n\times n$ space.[/hide]


geometry
similar triangles
geometry proposed

Two circles with centres C and D intersect at A and B.A st.line through the point A intersects the circle with centre C at P and that with centre D at Q.prove $ \angle PBQ \equal{} \angle CAD$ and $ \angle BPC \equal{} \angle BQD$. Where am I going wrong?

I have never faced a problem like this before.Why am I unable to these sums on circles?.In fact,these sums are supposed to be easy.Please suggest ways to overcome this .E-books can be suggested.( which start right from the basics)

[b]I am actually looking for the reason why I am getting stuck.[/b]

Thank you very much.

Draw in AB, then $ \angle PBQ \equal{} \angle PBA \plus{} \angle ABQ \equal{} 1/2 \angle PCA \plus{} 1/2 \angle ADQ \equal{} 90 \minus{} \angle PAC \plus{} 90 \minus{} \angle QAD$ (since PAC and QAD are isocles)$ \equal{} \angle CAD$.
This forum, in my opinion, is the best way to get better. Learn the basics about cyclic quads and similar triangles and everything (which you seem to know) and then just practice a lot and read others' posts.


Let $ x_1\equal{}1, x_2\equal{}16, x_{n\plus{}2}\equal{}7x_{n\plus{}1}\minus{}x_n\plus{}10$. Prove that $ x_n$ is a perfect square for every positive integer $ n$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ |G| \equal{} 3^n$ and  $ p \equal{} 2\cdot 3^t \plus{} 1$ be prime, where $ t\leq n$. If
the number of cyclic subgroups for  every possible order of $ G$ is a
divisor of $ 3^n\cdot p^l$, then is $ G$ cyclic?

what means "for every possible order of $ G$"?

It means that "the number of  cyclic subgroups of order $ 3^k$ for every possible $ k$", certainly $ k\leq n$. thanks again.


I was trying out a problem but it didn't have the answers so I am wondering if these are the roots to the Polynomial :  $ x^8\plus{}x^4\plus{}1$.

Roots:
$ x\equal{}\dfrac{\sqrt{3}\pm i}{2}$
$ x\equal{}\dfrac{\minus{}\sqrt{3}\pm i}{2}$
$ x\equal{}\dfrac{1\pm i\sqrt{3}}{2}$
$ x\equal{}\dfrac{\minus{}1\pm i\sqrt{3}}{2}$

yes, they are the answers
based on the format of your answers, it looks like (i just said looks like) you factored the polynomial, but you could also multiply the original polynomial by $ x^4\minus{}1$ (with roots 1, -1, i, -i)
and get $ x^12\minus{}1$ whose roots are the 12th roots of unity excluding 1, -1, i, and -i results in your answers

Put braces around the exponent, if it's more than one character, stevenmeow.

e.g.
x^12-1 yields $ x^12\minus{}1$

but

x^{12}-1 gives $ x^{12}\minus{}1$
 :wink:


Source: AMC 12 2000 #14

When the mean, median, and mode of the list: $10,2,5,2,4,2,x$ are arranged in increasing order, they form a non-constant arithemetic progression. What is the sum of all real possible values of $x$?

[hide]The mode has to be 2 since there are already 3 of them.  The median is $x$ if $2 \le x \le 4$.  But x must be greater than 2 because otherwise, both the mode and median would be 2, violating the condition of it being a non-constant arithemetic progression.  When x is greater than 4, the median is always 4.  The mean is: $\frac{25 + x}{7}$.  First, looking at the case where $x>4$, the last term must be 6 because the first two terms of the progression are fixed: 2 and 4.  With 6 as the mean:

$\frac{25 + x}{7} = 6 \rightarrow x = 42 - 25 = 17$

That's the only value for x when x is greater than 4.  The only value that works for x between 2 and 4 is 3.  So the sum is 17 + 3 = 20.[/hide]


The vertices and midpoints of the sides of a regular $2007$-gon are to be numbered with numbers $1,2,...,4014$, so that the sum of three numbers on every side is the same. Show that such a numbering is possible


geometry
trigonometry
circumcircle
trig identities
Law of Cosines
perpendicular bisector

In the triangle ABC, the angle B has 15 degrees, and the angle C has 30 degrees. Let AD=AB with B,A,D colineary.Find the angle ADC.

hint please?

does it require any high-level theorems?

[hide="trig bash"] Drop a perpendicular from $ A$ to $ BC$, and let the length of the perpendicular be 1. We find that $ AC\equal{}2,AB\equal{}\sqrt{6}\plus{}\sqrt{2}\equal{}AD$. By angle chase, $ \angle{DAC}\equal{}45^{\circ}$, so by the law of cosines, $ CD\equal{}2\sqrt{2}$, and by the law of cosines again, $ \angle{CAD}\equal{}30^{\circ}$.[/hide]

$ \triangle XBC$ is equilateral, $ X$ on the same side of $ BC$ as $ A.$ $ \angle BCA \equal{} 30^\circ,$ $ A$ is on bisector of the $ \angle BCX \equal{} 60^\circ,$ $ AX \equal{} AB.$ Circle $ (A)$ with radius $ AB$ passes through $ X$ and cuts $ AB$ again at $ D.$ Perpendicular bisector of $ BC$ cuts the circle $ (A)$ at $ X$ and again at $ Y.$ $ \triangle BYC$ is isosceles with $ YB \equal{} YC.$ $ \angle BDY \equal{} \angle BXY \equal{} 30^\circ.$ $ BD$ is a diameter of $ (A)$ $ \Longrightarrow$ $ \angle BYD \equal{} 90^\circ.$ Then $ \angle ABY \equiv \angle DBY \equal{} 60^\circ$ and $ \angle CBY \equal{} \angle ABY \minus{} \angle ABC \equal{} 45^\circ.$ Since $ \triangle BYC$ is isosceles, $ \angle CBY \equal{} 45^\circ$ and $ \angle BYD \equal{} 90^\circ,$ $ YD$ passes through $ C$ and $ \angle ADC \equal{} \angle BDY \equal{} 30^\circ.$

Let $X$ like in above solution; clearly $AX\bot AB$, hence $DX=BX=CX=BC$, so $X$ is the circumcenter of $\triangle BCD$ and $m(\widehat{BCD})=\frac{m(\widehat{BXC})}{2}=30^\circ$.

Best regards,
sunken rock


linear algebra
matrix
calculus
limit
trigonometry
calculus computations

Hello.

Im reading this book about numerical analysis, Absolute and relative error and there's a part I do not understand. It says as follows:
"
A given size of error is usually more serious when the magnitude of the true value is small.
For example:
$ 1036.62\pm0.010$ is accurate to five significant digits and may be adequate to precision, wheres
$ 0.005\pm0.010$ is a clear disaster.
"

Can someone explain how $ 0.005\pm0.010$ is a clear disaster? not so clear for me. I also didnt understand the first part because 1036.62 itself has 6 significant digits.

Any help would be appreciated.
Thanks a lot.

You're not paying attention to the $ \pm.$

$ 1036.62\pm0.01$ could be $ 1036.61$ or $ 1036.63$ - so while we at least have a hint about that sixth digit, we don't know it with certainty.

As for $ 0.005\pm0.010:$

Let's say I show you an $ n\times n$ matrix $ A$ and ask you whether $ A$ is invertible. You perform Gaussian elimination, doing the arithmetic in a floating point system so that there is roundoff error. At the end of that Gaussian elimination process, the element in the lower right corner is $ 0.005\pm0.010.$ So tell me - is the matrix invertible? Even if it is invertible, is the determinant positive or negative? The point is that your arithmetic is now so screwed up that you can't even answer those questions.

Here's an exercise for you. You know from calculus that $ \lim_{x\to0}\frac {1 \minus{} \cos x}{x^2} \equal{} \frac12.$

Now try to convince yourself of that with a floating point device. Take a calculator (or a spreadsheet on your computer - and make sure you're set to "radians"). Put in various values of $ x$ and record your answer. Then keep pushing - to smaller, and smaller and smaller values of $ x.$ Is smaller always better? If you were doing arithmetic in the real numbers, smaller would be better, but you're not doing arithmetic in the real numbers, you're doing floating point arithmetic.

The error in calculating $ \cos x$ should be roughly constant for any value of $ x,$ and $ 1$ should be an exact value. The problem is that that makes the error in $ 1 \minus{} \cos x$ roughly constant in size, but as $ x\to 0,$ that number (the numerator of your fraction) goes to zero, creating exactly the problem your textbook is talking about.

Added in edit: Using Excel and its standard set of numbers, I seem to get optimum results in that fraction for $ x$ somewhere around $ 0.001$ to $ 0.003.$ For $ x$ smaller than about $ 10^{\minus{}8},$ I just get zero.


Prove that there exists a bijection between [0,1)x[0,1) and [0,1) using properties of R. Please do not use schroder-bernstein theorem or other deep set theoretic facts. My instructor says it is possible to  show a bijection exists using only binary expansion and similar stuff.

Let $ x$ have decimal expansion $ 0.x_1x_2x_3\dots.$ Let $ y$ have decimal expansion $ 0.y_1y_2y_3.dots.$

Since decimal expansions aren't quite in a one-to-one correspondence to real numbers, we must give a rule to resolve the ambiguity. Let it be this rule: if $ x$ or $ y$ is a rational number that can be written with a power of $ 10$ as the denominator, choose the expansion that ends in all zeros. That way, we do not have any expansions that end in all $ 9$s.

Now let $ z\equal{}0.x_1y_1x_2y_2x_3y_3\dots$ Once we note that this cannot end in all $ 9$s, we see that the mapping that sends $ (x,y)$ to $ z$ is one-to-one and maps $ [0,1)\times[0,1)$ onto $ [0,1).$ In particular, it should be apparent what the inverse of this mapping is.

Question: is the mapping I described both measurable and measure-preserving?

Actually, that isn't onto- it's missing numbers like $ 0.9090909090\dots$. The ambiguity still causes trouble, and you need to handle some countable subset separately.

Oops. OK, pull out the entire set of numbers of the form $ \frac{j}{10^m}.$ That countable set can be placed into a one-to-one correspondence with its cartesian product with itself. And then do the decimal expansions splice for all of the other real numbers.

[quote]Oops. OK, pull out the entire set of numbers of the form \frac{j}{10^m}. That countable set can be placed into a one-to-one correspondence with its cartesian product with itself. And then do the decimal expansions splice for all of the other real numbers.[/quote]

When you remove a countable 'A' set from [0,1), you are actually removing Ax[0,1) from the cartesian product so Kent Merryfield's argument is wrong.


For $ x,y \in N$ (x,y>0 )
Find the smallest value of x+y such that $ x^3\plus{}7xy\plus{}y^3$ is a perfect squa :D re

first assume that one of the numbers is one to make the sum as low as possible.
So we get 1^3+7y+y^3=k^2
Perfect squares are either 1mod3 or 0mod3. So lets first look at the case where 0 mod3=k^2 so this means that 1+7y+y^3=0mod3. 7y+y^2=2mod3. y(y^2+7)=2mod3. This splits into two more cases where y^2=0 mod 3 or 1mod3. It does not work when it is 0 mod 3 but y^2 can be 1 mod 3. If k^2 is 1 mod 3. Then y^3+7y would have to be 0 mod 3. Which works. Now we see using inspection to find if y would be greater if it were 0 mod 3 or 1 mod 3. We see that when y is also 1(1 mod 3) it is a perfect square which happens to be 9. So the minimum value of x+y is 2.

Sorry guy what I meant is $ x^{3} \plus{} 6xy \plus{} y^{3}$ :blush: [/quote]


group theory
abstract algebra
inequalities
function
number theory
relatively prime
superior algebra

:?: I have some difficulty with the following problem of Herstein!!
     If Sylow-p subgroup P of a finite group G lies in the center of G, then prove that there exists a normal subgroup N of G such that, P & N intersects at unity & G=PN.

Normality is not a problem,it can also be shown that it suffices to prove for o(G)=pq (q!=0 mod p).It is now required to prove that x->x^q is a homomorphism. Here I got stuck.

Can someone show how you proved the existence of normal subgroup N with PN=G ?

[quote="Subhajit"]it suffices to prove for o(G)=pq (q!=0 mod p).It is now required to prove that x->x^q is a homomorphism. Here I got stuck.[/quote]
This is an odd approach.  In this case you know there is a Sylow $ q$-subgroup $ Q$, which is clearly the required subgroup.

In fact, I don't think you can show $ x\mapsto x^q$ is a homomorphism without also showing $ G$ is abelian, except for the case $ q\equiv1\mod p$: since $ G'\leq P\leq\mathcal{Z}(G)$, $ x^q y^q (xy)^{\minus{}q} \equal{} [x,y]^{\frac{q(q\minus{}1)}2}$.  If $ q\neq 1\mod p$, this is 1 only when $ [x,y]\equal{}1$.

More generally, we have the following two theorems, either of which proves the result in question:

Burnside: Whenever $ C_G(P)\equal{}N_G(P)$, then $ P$ has a normal complement.
Schur-Zassenhaus: Any normal Hall subgroup has a complement.

Is there a more elementary proof without using Schur-Zassenhaus or Burnside's normal complement theorem ?

The proof of Schur-Zassenhaus in the case when a Hall subgroup $ H$ is central goes like this: pick a section $ \sigma: K\to G$ of the SES
\[ 1\to H \to G \to K \to 1,
\]
where $ K \equal{} G/H$, which has order relatively prime to $ H$.  Define a function $ f: K\times K\to H$ by $ \sigma(ab) \equal{} f(a,b)\sigma(a)\sigma(b)$.  Computing $ \sigma(abc)$ in two different ways shows that
\[ (df)(a,b,c) \stackrel{def} \equal{} f(b,c) \minus{} f(ab, c) \plus{} f(a,bc) \minus{} f(a,b) \equal{} 0,
\]
using additive notation for $ H$.  So $ f$ defines an element of $ H^2(K,H)$, which is by definition
\[ H^2(K,H) \equal{} \frac {\{f: K^2\to H | df \equal{} 0\}}{\{f : K^2\to H\,|\,\exists \varphi: K\to H,\,f(a,b) \equal{} \varphi(b) \minus{} \varphi(ab) \plus{} \varphi(a)\}}
\]
It's easy to see that $ H^2(K,H)$ has exponent dividing $ |H|$.  The only non-trivial bit is that since $ (|H|,|K|) \equal{} 1$, we must have $ H^2(K,H) \equal{} 0$.  So by definition, there exists a function $ \varphi: K\to H$ such that $ f(a,b) \equal{} \varphi(b) \minus{} \varphi(ab) \plus{} \varphi(a)$.  Now it is easy to see that this implies that the section $ s(a) \equal{} \varphi(a)\sigma(a)$ is a homomorphism, splitting the above SES, and that $ G \equal{} H\rtimes s(K)$.

(Similarly, the fact that $ H^1(K,H) \equal{} 0$ shows that any two complements of $ H$ are conjugate.)

[quote="Subhajit"]It is now required to prove that x->x^q is a homomorphism. Here I got stuck.[/quote]
Okay, now I see where that's coming from.  Suppose $ P\leq \mathcal{Z}(G)$ is a Sylow $ p$-subgroup of $ G$, with index $ m$, $ p\nmid m$.  We show that $ V(g)\equal{}g^m$ is the transfer homomorphism $ V: G\to P$.

Let $ \{x_1,\ldots,x_m\}$ be a set of coset representatives of $ P$, i.e., $ G\equal{}\coprod x_i P$.  Let $ g\in G$.  For each $ i$, there is a unique $ j\equal{}\sigma_g(i)$ such that $ x_{\sigma_g(i)}^{\minus{}1}gx_i \in P$.  The transfer from $ G$ to $ P$ is defined as
\[ V(g) \equal{} \prod x_{\sigma_g(i)}^{\minus{}1}gx_i \in P.\]
Since
\[ x_{\sigma_{gh}(i)}^{\minus{}1} gh x_i \equal{} \left(x_{\sigma_g(\sigma_h(i))}^{\minus{}1}g x_{\sigma_h(i)}\right)\left(x_{\sigma_h(i)}^{\minus{}1}hx_i\right),\]
and $ P$ is abelian, it is easy to see that $ V$ is a homomorphism.  (If $ P$ were not abelian, we would only have a homomorphism $ G\to P/P'$.)

Now consider an orbit of size $ k_i$, $ \{x_i,gx_i,\ldots,g^{k_i\minus{}1}x_i\}$.  Since $ P$ is abelian, the product of the factors corresponding to this orbit can be rearranged to give $ x_i^{\minus{}1}g^{k_i}x_i \equal{} g^{k_i}$, since $ g^{k_i}\in P$ is central.  Taking the product over each orbit, we just get $ V(g)\equal{}g^m$.

Now if $ N\equal{}\ker V$, then because $ (m,|P|)\equal{}1$, we have $ N\cap P\equal{}1$.  But also $ G/N \leq  P$ is a $ p$-group, so in fact $ G\equal{}NP$.


Hallo!
Damit es auf der Zugfahrt nach und von Rostock nicht so langweilig ist, fände ich es cool, wenn möglichst viele Teilnehmer in einem Zug fahren.
So gibt es z.B. am 10.2.05 um 7:44 ab München eine Verbindung, mit der man um 17:20 in Lütten-Klein ankommt.

München Hbf 	10.02.05 	ab 	07:44 	18 		ICE 882 	InterCityExpress
BordRestaurant
Hamburg Hbf 	10.02.05 	an 	13:54 	12a/b
Hamburg Hbf 	10.02.05 	ab 	14:19 	6a 		RE 33013 	RegionalExpress
Fahrradmitnahme begrenzt möglich, Fahrzeuggebundene Einstiegshilfe: Anmeldung 01805-512512 (12 ct/Min.)
Rostock Hbf 	10.02.05 	an 	16:54 	8
Rostock Hbf 	10.02.05 	ab 	17:07 	1 		S 	S-Bahn
Fahrradmitnahme begrenzt möglich

Als Rückfahrt bietet sich bieten sich beispielsweise folgende Züge an:

Lütten Klein 	17.02.05 	ab 	08:46 	2 		S 	S-Bahn
Fahrradmitnahme begrenzt möglich
Rostock Hbf 	17.02.05 	an 	08:59 	1
Rostock Hbf 	17.02.05 	ab 	09:06 	9 		RE 33006 	RegionalExpress
Fahrradmitnahme begrenzt möglich, Fahrzeuggebundene Einstiegshilfe: Anmeldung 01805-512512 (12 ct/Min.)
Hamburg Hbf 	17.02.05 	an 	11:38 	6b
Hamburg Hbf 	17.02.05 	ab 	12:05 	14a/b 		ICE 789 	InterCityExpress
BordRestaurant
München Hbf 	17.02.05 	an 	18:13 	18
Dauer: 9:27; fährt täglich, nicht 21. Feb bis 8. Mär 2005, 30. Mär bis 20. Apr 2005

Lütten Klein 	17.02.05 	ab 	09:46 	2 		S 	S-Bahn
Fahrradmitnahme begrenzt möglich
Rostock Hbf 	17.02.05 	an 	09:59 	1
Rostock Hbf 	17.02.05 	ab 	10:07 	3 		IC 2373 	InterCity
Fahrradmitnahme reservierungspflichtig, Fahrradmitnahme begrenzt möglich, BordBistro
Hamburg-Harburg 	17.02.05 	an 	12:37 	5
Hamburg-Harburg 	17.02.05 	ab 	13:12 	4 		ICE 589 	InterCityExpress
BordRestaurant
München Hbf 	17.02.05 	an 	19:01 	13
Dauer: 9:15; fährt täglich, nicht 25. Feb, 4. Mär

Ich schätze mal, die Leute aus Süddeutschland fahren alle entlang dieser Linie. Was meint ihr zu den Vorschlägen?

Ich bin dabei!

Der erste Zug ist gut. Bei den Rückfahrtszügen würde ich eher mit dem zweiten rechnen, ich glaube nicht, dass wir so wahnsinnig schnell frühstücken und dann das durchaus nicht kurze Stück zur S-Bahn rennen wollen.

ich wollte nach dem bwm-kolloquium (7.-8.2.) mehr oder weniger direkt nach rostock weiterreisen...
bei der rückfahrt kann ich dann aber auf jeden fall mitfahren. :)

Komme natürlich (wie bereits gsagt) auch mit!  :)


How do you find out on what intervals $ y \equal{} x^5 \minus{} 5x$ is both increasing and concave up?

$ \frac{dy}{dx}\geq 0 \Rightarrow$ increasing.
$ \frac{d^{2}y}{dx^{2}} \geq 0\Rightarrow$ concave up.

[hide]$ {y}'\equal{}5x^4\minus{}5$
$ {y}''\equal{}20x^3$
Therefore $ \boxed{x\geq1}$.[/hide]


linear algebra
matrix
linear algebra unsolved

Ahoj :)

Let A be the n x n matrix, the [tex](k,l)^{th}[/tex] complex entry is [tex](-1)^{\frac{2 k l}{n+1}}[/tex] for k,l=1,...,n . Find and prove the formula for [tex]det(A^2)[/tex] .

Do you mean for the entries to be $e^{\frac{2\pi ikl}{n+1}}$ ?

Yes, that what it is, sorry if it isn't clear.

Hello

"Vandermonde" is not any "closed formed" solution for me :roll:

p.s. some german journal states that he was [b]van der Monde[/b] ;)


Let $ X$ be a subset of $ R^n$, such that the distance between any pair of points $ x$ and $ y$ is in given set $ \{ d_1, d_2, \ldots, d_s \}$, i.e. $ \parallel{}x \minus{} y\parallel{} \equal{} \sqrt { (x_1 \minus{} y_1)^2 \plus{} (x_2 \minus{} y_2)^2 \plus{} \ldots \plus{} (x_n \minus{} y_n)^2} \equal{} d_k$, where $ 1 \leqslant k \leqslant s$.
Prove that
\[ |X| \leqslant \binom{n \plus{} s \plus{} 1}{s}.\]

The current best bound is $ {n\plus{}s} \choose{s}$ by Bannai, Bannai and Stanton.

The paper is free here :
http://www.springerlink.com/content/9th7t0l322644p94/

Pierre.


I need to prove n^4+64 is composite (non-prime) for all positive integer values of n.

It is straightforward enough except in the cases where n is of the form 5(2p+1). Can anyone sort this out?

Too easy and too well known for Pre-Olympiad.

$n^{4}+64 = n^{4}+4(2)^{4}= (n^{2}-4n+8)(n^{2}+4n+8)=((n-2)^{2}+4)((n+2)^{2}+4)$

Clearly none of the factors can be $1$ so it is composite.
This is called the Sophie-Germain identity


Let $f(z)\equiv az^2+bz+c=0$ be a ecuation with complex coeficients and $f(z_1)=f(z_2)=0$. Ascertain $|z_1|+|z_2|.$

You square and get it????..

[hide]I'll post my soln later!!![/hide]


ratio
probability
expected value
real analysis
real analysis unsolved

I don't know if this post is in the right thread or not but I'll ask anyway.

This is the problem i need to solve:

Team A is playing Team B. Team A offense averages 29.5 points against teams that on average give up 25.76 points on defense. Team B gives up on defense an average of 32.1 points against teams that average 25.48 points on offense. What is the best estimate of how many points Team A will score???

I have 2 ways to solve this problem but I'm getting different answers. I always set these up as fractions. Like 29.5/25.76 and 32.1/25.48 as actual on top and opponents average on bottom.

Method 1. Find out the differences. Team A outscores their oppontents averages by (29.5 - 25.76) = 3.74 and then Team B defense the same thing (32.1 - 25.48) = 6.62 Add these two together 3.74 + 6.62 = 10.36 and then add 10.36 to the average of the denominators which is (average of 25.76 and 25.48=25.62) which gives you 35.98

Method 2. Cross Multiply and take averages. Team A averages 29.5 against teams that give up 25.76 But Team B doesn't give up 25.76 they give up 32.1 so Team A has 36.76 But then Team B gives up 32.1 against teams that put up 25.48 but Team A doesn't put up 25.48 they put up 29.5 so Team B defense has 37.17 Take the average of the two (36.76 and 37.17) and you get 36.96

35.98 and 36.96 are too far apart for me. Maybe there is another more accurate way to solve this than what i'm doing. there shouldn't be two different answers for 1 problem. I understand it's averages, so maybe a range of numbers are possible. Please help me out i need the most accurate way to find out the solution whether it's an actual number or a range.

The problem with both methods is that you have two sets of statistical data to work with:  one that tells you how Team A performs on average and one that tells you how Team B performs on average.  You can't assume that both averages will hold in a game between the two, so you have to take some kind of average of the two averages.

For example, in Method 1 you assume that the spread of the game will be the sum of the spread of an average game that A plays and the spread of an average game that B plays, which is obviously absurd (consider what happens if Team A and Team B are the only teams in the league and always play each other, and Team A always scores $ 30$ points and Team B zero.  Your method will then predict a $ 60$-point victory).  You need to average $ 3.74$ and $ 6.62$, not add them, if you're going to do it this way.

In Method 2 you appear to be assuming that the ratio between points scored and points given up (rather than the difference) is constant.  I'm not sure what you're doing here, but this is a different statistical assumption than the first method.

You shouldn't expect there to be one answer to this; the question isn't well-defined enough yet.  It should be phrased as an expected-value question, but we are given two separate expected values.


On the chess board we want to put bishops such that no bishop threatens another one but each of the unoccupied field is at least threatened by one bishop.

a.) Find such a configuration of 8 bishops.

b.) Determine the maximum number of bishops that can be put on the chess board satisfying above rules.

c.) Analyse for which number $n$ there is such a configuration with $8 < n < m$.

b) Probably more than this, but this is what I got.

Sorry about the double post, but it won't let you attach more than one document.

i figure that there is a maximum of 14 bishops on the board.  The 8 on the first rank and then 6 on the 8th rank but none in either of the corners

[hide="a start"]well when a bishop is placed on any of the sides, it eliminates the position for 8 squares and another bishop can be placed anywhere and they can only have one intersecting "line of fire" and every other square gives us a solvable equation so we just occupy all of the squares available to us then[/hide]

i think that that is the max.  Its kinda like putting the 8 queens on the board in such a way that none is attacking the other.  In that case you put them a knight's move away from each other... Trust me i solved it at chess camp this summer

[quote="Treething"]a) The red squares are the ones occupied by bishops.[/quote]

Well, that's not correct.

Look at some of the top squares not occupied by bishops. They aren't threatened by bishops either, and the problem asks for an arrangement where other squares are threatened also. So, that doesn't work.

Let's say the chessboard is facing you, so that the bottom left square is a1, and the top right square is h8.

Place the bishops on a1 through h1, and then place bishops on b8 through g8. You will find that this is the maximum arrangement as demji has mentioned already.

There are 14 total, and every unoccupied square is being attacked by a bishop.[/img]

Try pigeonhole principle (the floor variant).


[quote="Quevvy"]A jar contains 6 red balls, 5 green balls and 9 yellow balls. Danny draws a ball at random from the jar and does not return it. Celine draws another ball at random from the jar. What is the probability that they drew a ball of the same color? Give your answer as a common fraction in lowest terms.[/quote]

[hide="solution"]The number of ways to draw 2 red balls is ${6\choose 2}=15$

The number of ways to draw 2 green balls is ${5\choose 2}=10$

The number of ways to draw 2 yellow balls is ${9\choose 2}=36$

The total number of ways to draw 2 balls is ${6+5+9\choose 2}=190$

$\frac{15+10+36}{190}=\boxed{\frac{61}{190}}$[/hide]


topology
geometry
3D geometry
sphere
combinatorial geometry
advanced fields
advanced fields unsolved

Let $ X\equal{}l^2(\mathbb{N})$. Prove that the closure of the unit sphere in the weak topology is the unit ball.

* I think one can prove this using Krein-Milman: the closed(in the norm topology) unit ball is the closure of the convex hull of it's extreme points. Here, since we are dealing with the convex set (unit ball in particular), there is no difference between weak closure and norm closure(Hahn-Banach theorem consequence). And it is only left to notice that extreme points of closed unit ball of $ l^2(\mathbb{N}$  is the unit sphere.

I'm interested whether there is a straight way, without invoking lots of theorems.

$ a\plus{}cf_n\to a$ weakly, where $ (f_n)$ is an ONB for $ a^{\perp}$ and $ c\equal{}\sqrt{1\minus{}\|a\|^2}$.


A man walks $x$ miles due west, turns $150^\circ$ to his left and walks 3 miles in the new direction. If he finishes a a point $\sqrt{3}$ from his starting point, then $x$ is

$\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$

[hide]A triangle is formed with side lengths $x$, $3$, and $\sqrt{3}$. Angle opposite side of length $3$ is  uniquely determined. Law of Sine shows this is not an ambiguous case as the sin of the angle is $\frac{\sqrt{3}}{2}$, which denotes either $60^\circ$ or $150^\circ$, and the latter's sum with $30^\circ$ is over $180^\circ$. The angle is $60^\circ$ and thus the it is a $30-60-90$ triangle, $x$ is thus $2\sqrt{3}$

$\boxed{E}$[/hide]

[hide="Solution"]$\text{Answer: (E)}$

This problem is easy to figure out with a drawing. Let's say he started on some line $l$. The man walks for some distance x until he reaches a point $A$ and turns 150 degrees, forming a 30 degree angle. Next, he walks 3 miles to a point $B$.

Now, there are two such points located on line $l$ that are of a distance $\sqrt{3}$ from point $B$, so the answer is not uniquely determined.[/hide]

[hide]The triangle formed is not uniquely determined (SSA), as two triangles can be formed using the Door-Swinging Theorem. So the answer is $\boxed{E}$[/hide]