inequalities
abstract algebra
calculus
calculus computations

I face a problem:

Let $ f: \mathbb{R}---\to  \mathbb{R}$ be differential such that

\[ f(x)x \geq 0, |f'(x)|\leq M, \forall x \in \mathbb{R}, M>0.\]

Prove that $ |f(x)|\leq M|x|, \forall x \in \mathbb{R}.$

Let's do it.

Thanks.

$ f(x)x \geq 0$  $ \forall x \in \mathbb{R}$ implies $ f(x) \geq 0$  $ \forall x >0$ and $ f(x) \leq 0$  $ \forall x<0$.  Since f is continuous, we must have $ f(0)=0$.

Now clearly then $ |f(0)| \leq M \times 0$.

Consider the case where $ x\neq0$.  Then we just use the mean value theorem:

$ \frac{f(x)-f(0)}{x-0}=f'(c)$ for some $ c$ in the interval between $ 0$ and $ x$.  This can be rewritten as:  $ \frac{|f(x)|}{|x|}=|f'(c)|\leq M$ which is equivalent to your result.

[quote="edoo"]$ f(x)x \geq 0$  $ \forall x \in \mathbb{R}$ implies $ f(x) \geq 0$  $ \forall x >0$ and $ f(x) \leq 0$  $ \forall x<0$.  Since f is continuous, we must have $ f(0)=0$.

[/quote]

Thanks. My proof is longer than yours and certainly yours is more interesting than mine.

I proved that $ f$ is an injection due to $ f(x)x\geq 0$ then $ f(0)=0.$ And applying Lagrange's theorem or mean value one, I deduced that result.

Best wishes.

[quote="edoo"]$ f(x)x \geq 0$  $ \forall x \in \mathbb{R}$ implies $ f(x) \geq 0$  $ \forall x >0$ and $ f(x) \leq 0$  $ \forall x<0$.  Since f is continuous, we must have $ f(0)=0$.

.[/quote]

The main step is to prove $ f(0)=0.$ I'd like to make clear what edoo said.

First, since $ f$ is continuous in $ \mathbb{R},$ there exists $ f(0).$
From the given data, we have
\[ x>0\Rightarrow f(x)\geq 0, \]

\[ x<0\Rightarrow f(x)\leq 0, \]
$ \forall x\in \mathbb{R}.$

Now, suppose that $ f(0)\neq 0.$ There are two cases
1. $ f(0)>0\Rightarrow 0>0 (!)$
2. $ f(0)<0\Rightarrow 0<0 (!)$.

Thus $ f(0)=0.$


i converted my latex document to a pdf file with the cutepdf converter that usamts recomended but its kinda blurry so i dont know how to make it look clear again can someone help me with this problem

Why don't you just use the PDF converter that comes with TeXnicCenter?  That's the way I do it and I never have problems.

i just got texnic center like a few weeks ago and all i no how to do is type math i cant do anything else with it so if u can explain the pdf converter in texnic center id be grateful

Look under the "Build" menu, then under the "Current File" submenu. Select either build, or build and view. There should be corresponding icons that will execute that process for you, and I've included an attachment that shows you what they are.

One point to add.  If you're going to use the TeXnicCenter PDF builder (which I recommend), make sure there are no serious errors or else the file will fail.

i figured out how to do it i dont know if its wut u told me but i found something under build called select output profile that did the job :alien:  :spam:  :starwars:  :fourhundredfiftysecond:

thx i used the texnic center pdf converter and it worked perfectly


So, now that 2004 is in the past, what was your favorite movie last year? Mine was probably ROTK.

I started to make a poll, but found there's just a lot of movies that would have to be considered, so just write you opinions here.

well rotk was kinda 03...but that's okay...it's my favorite too...

let's see here

shrek 2
the incredibles
hero
i, robot
king arthur
van helsing
troy
the day after tomorrow
dodgeball
spiderman 2
50 first dates
[hide]the girl next door
eurotrip[/hide]

I must say though...this was a pitiful year for hollywood...

What about national treasure, and princess diaries 2,

Hmm I'd go for
StarTrek N3mesis
Shrek 2
The girl next door
I, robot
Spiderman 2

[quote="eryaman"]Wow....mith has a lot of favorite movies...in only one year, too.[/quote]

haha actually it's a rather short list. i normally have 40+ in a year...

Anybody here seen Finding Neverland? I've heard three really good reviews from different people about it, and I want to know if everyboyd else feels similarly.

[quote="MathFiend"][quote="MithsApprentice"][quote="hatsup123"]well this is off topic, but is pirates of the carribean 2 coming out this year?[/quote]

summer 2006

Keira Knightley *hop hop*[/quote]

they're filming the 2nd and 3rd ones of PoTC right about now i believe...Orlando Bloom *hop hop*  ;)[/quote]

*hop hop*? mathfiend, miths are yall okay?

[quote="BHorseMath"]summer 2006

Keira Knightley *hop hop*[/quote]

they're filming the 2nd and 3rd ones of PoTC right about now i believe...Orlando Bloom *hop hop*  ;)[/quote]

*hop hop*? mathfiend, miths are yall okay?[/quote]
not to make this a ziggurat thread, but i think its just the usual teen fantasy of marrying some person about 20 years older than they
and anyway, i liked ROTK also.
and i only saw commercials for Eurotrip, and it looked like a poorly disguised movie designed for ogling

Harry Potter and the Prisoner of Azkaban because it was the only one I watched...hehe.

[quote="MithsApprentice"]No offense to HP fans here or anything of that sort...but eh...LoTR was indisputably better...and by more than just a bit.[/quote]
Yah i think the LOTR was alot better than Harry Potter, well HP was ok though... I, Robot was a good movie.

so why are [hide] girl next door and eurotrip in hide-a-text[/hide]? Is it cuz they are rated R?

Dunno, but euro-trip? C'mon ... that's rated R? ... very lame movie if you ask me. Everyone should watch the real "trip" movie: [url=http://www.imdb.com/title/tt0215129/?fr=c2l0ZT1kZnxteD0yMHxzZz0xfGxtPTIwMHx0dD1vbnxwbj0wfHE9cm9hZCB0cmlwfGh0bWw9MXxubT1vbg__;fc=1;ft=19;fm=1]Road Trip[/url]. ;) :D

i think i know what you're talking about.....

this is turning into a true forum for adolescents....

[quote="MithsApprentice"]haha well i put them in the hidden text thingy and i was HOPING not to start a conversation...but oh well...[/quote]

lol, wouldn't it just tempt people?

[quote="yif man12"]lol, wouldn't it just tempt people?[/quote]of course  :D

the evil temptations of the hide-a-text's, lol

Well I was hoping that the majority of...uh...young eyes on this forum would not engage in activities involving such curiosity. :-D (Wow I think that was a good excuse.)

But anyways...Troy was rated R but I forgot to include that one...whoops

mine was probably Sponbob Squarepants...no wait, it was Cat in the Hat

Hero is probably the best movie this year then national treasure.
Who didn't need to look at the stupid subtitles for Hero and understood it?  And who understood the story with the subtitles?  Jet Li owns all other martial artists.  (sry no *hop hop*....)
ROTK is my favorite movie overall.

elvenchamp777

on another note, House of Flying Daggers was awesome in the sense of the coreographed fights and the special effects, but the story was so so

House of flying daggers is a rereleased movie from hong kong which is pretty cool (FIGHT SCENES!).  but the story was soso...  That was kinda good.

elvenchamp777

LOL. House of Flying daggers is a good movie, but its loaded with special effects and has a pretty short storyline. I got the chinese dvd 2 and a half months before it came out. LOL! Exact same quality too!

this is REALLY off-topic... but i'm just wondering...

[quote="MithsApprentice"]-.- to scrambled. Actually, I'm...er...yea...I'm kinda obsessed with that girl right there
<<----- that one
and she's barely 18...so it's actually 2 years...not 20...but that's okay :-)

eurotrip was a bad movie in content...but it was realli funny...[/quote]

so ur 16, rite?

i lyked the [b]incredibles[/b]... a lil... mostly cuz i watched it with my friends
hmm... i don't remember which year the movies i lyk came out....

i lyk... Sinbad...
Pirates of the Carribean (hop hop???)
Hary Potter was ok...

what else....

i might post more movies later on

[quote="Valentin Vornicu"]I just saw Meet the Fockers. Very, very funny. Just like the original Meet the Parents :) :D[/quote]

which one do you think is better, meet the fockers or meet the parents???

I think Meet the Parents, but Meet the Fockers is very close. In Meet the Parents most of the stuff was really nice and new. In the sequel we already knew (in large) what was going to happen. But it was still fun. I love the end - the camera viewing :)

Yeah, I'm 16.

The Incredibles was a majorly funny movie. Not quite as good as I expected, but definitely deserving of something nice. :-)

Meet the Parents and Meet the Fockers are...uh...I don't know. I like them both...can't really tell you which is better.

I've never seen any of those two movies

i liked incredibles better than big fish...(that came out in 2004, rite?)

big fish was totally boring for me... i don't know why others think its so good (no offense)


function
geometry
geometric transformation
reflection
algebra
polynomial
integration

Let $ f$ be a function meromorphic in the closed unit disk. Suppose that $ |f|$ is constant on the unit circle.

Prove that $ f$ must be a rational function.

If $ f$ is entire and $ |f| \equal{} 1$ on the unit circle, show that $ f(z) \equal{} kz^n$ for some unimodular constant $ k$ and some integer $ n \geq 0$.

Look at Blaschke product representations

A less explicit approach would be to use a reflection principle to get a meromorphic extension in $ \mathbb{C}$ and show that this extension does not have an essential singularity in infinity

Is there a neat way to show that if a polynomial has modulus 1 on the unit circle, then it must actually be a monomial?

Because I'm still missing that ingredient...

This is not so neat, because it adresses the more general problem, but let the polynomial be $ f$, and let the zeros inside the unit circle (repeated with multiplicity) be $ a_1, \ldots , a_k$.

Now we remove the zeros of $ f$ by dividing away automorphisms of the unit disk. I.e. set $ g(z) \equal{} f(z) \prod_{j \equal{} 1}^k \frac {1 \minus{} \bar{a_k}z}{z \minus{} a}$. $ g$ is analytic in the closed disc and has $ |g(z)| \equal{} 1$ on the boundary. By maximum modulus principle, $ g(z)$ is constant or has $ |g(z)| < 1$ in the open unit disk. On the other hand, the same is true of $ 1/g(z)$. So $ g(z)$ must be constant. This gives a Blaschke product representation of $ f$, and it's clear the only polynomials are the monomials.

[quote="Arne"]Is there a neat way to show that if a polynomial has modulus 1 on the unit circle, then it must actually be a monomial?
[/quote]
If $ P(z)\equal{}a_nz^n\plus{}\dots\plus{}a_Nz^N$, then $ \int_{\mathbb T}|P(z)|^2 z^{N\minus{}n}\,dm(z)\equal{}a_n\overline{a_N}$. ;)

[quote="fedja"][quote="Arne"]Is there a neat way to show that if a polynomial has modulus 1 on the unit circle, then it must actually be a monomial?
[/quote]
If $ P(z) \equal{} a_nz^n \plus{} \dots \plus{} a_Nz^N$, then $ \int_{\mathbb T}|P(z)|^2 z^{N \minus{} n}\,dm(z) \equal{} a_n\overline{a_N}$. ;)[/quote]

err is $ 2 \pi$ missing? Anyway the idea is beautiful!


algebra
polynomial
logarithms
number theory proposed
number theory

Let $f(x)$ be any nonconstant integer polynomial. Show that there is a constant $C > 0$ such that, for any $N>1$, the number of $n$ with $1 \leq n \leq N$ and $|f(n)|$ prime is less than $CN/\log{N}$.

Vess, some hints about this problem? I really hope it doesn't use too sophisticated machinery, not only because it would be highly unfair to post it here in this case.

It was more or less known around 1900 (Nagell had shown at that time that the set of those values has zero density, and later Heilbronn showed how to deduce the refined version with $CN/\log{N}$ from Nagell's proof). The proof does use the prime number theorem in number fields (or some equivalent), but nothing any more sophisticated (in fact, the rest is the simplest possible version of the sieve, and estimates). Does this count as a hint? :)

It would be really great to find a proof without recourse to the prime number theorem, using only sieve and perhaps some clever ideas along the lines of Chebyshev's proof of $\pi(x) < Cx/\log{x}$). I don't know if such proof exists, but I failed to find one... :(


Anything but ceder point. Specifically the height and speed of it, I can't those kind of things on any theme park site.

[quote="Kuku_King"]WTF HELP ME OUT[/quote]We decide whether to help or not, don't make it seem like it's our obligation to do anything whatsover for you. Besides, this forum is NOT a homework-type of site. You're in the wrong forum.


geometry
conics
ellipse
projective geometry

[asy]size(300); draw(unitcircle); pair A,B,C,D; A = (0,-1); B = origin; C = (0,1); D = (0.8,1); draw(A--C); draw((-1,1)--(1,1),Arrows); label("\small{$O$}",B,W); label("\small{$A$}",A,S); label("\small{$B$}",(1,0),E); label("\small{$C$}",C,N); label("\small{$D$}",(0.8,1),N); draw(0.1dir(90)..0.1dir(0)..0.1dir(-90)); dot(B);[/asy]
In this picture, angle $ AOC$ is $ 180^{\circ}$. Can I simply assume that angle $ OCD$  is $ 90^{\circ}$ because of the inscribed angle theorem or is it necessary to prove that angle $ OCD$ is 90 degrees some other way?

Since the tangent line meets the circle at only one point, are we assuming $ C$ is a point where tangent line and diameter intersect?

If this is the case:
[hide]Draw another line connecting $ O$ to the tangent line, naming the intersection $ E$. To determine if $ OCD$ is a right angle, it has to be the shortest distance from the point $ O$ to the line.
If the line we created ($ OE$) is the perpendicular and $ OC$ is not, there's an conflict. $ OC$ is the hypotenuse while $ OE$ is the leg. $ OC$ is within circle radius so it should be  shorter than $ OE$, but $ OE$ is the hypotenuse. There we see the conflict, so $ OCD$ must be $ 90^{o}$.[/hide]

Edit: I'm not sure if that's the type of solution I was looking for. I'll consider it anyway, thanks  :D

Does anyone else have any other methods?

Thanks to all who responded. The solution is pretty simple.

Draw 2 lines perpendicular to each other. Then make their intersection a point on a circle of any radius. And voilà, the solution is blatantly obvious  :lol:

This is probably what minsoens just said but crudely expressed in words.

EDIT:

There's also another method that involves similar triangles. Draw a circle. Then draw a line from the origin to any point on the circle. Drop the perpendicular. Now, simply draw a similar triangle whose perpendicular side is tangent to the circle. Q.E.D.  :P

[quote="minsoens"]hmm... i don't know but can't you just say
$ OC \perp CD$ since $ OC$ is a radius and $ CD$ is a tangent... and $ OC\parallel OP\Longrightarrow CP \perp CD$[/quote]

Actually, you can, because (at least I'm fairly certain - it's in my textbook) it is a recognised property of a circle that a radius (which OC is) meets a tangent at 90 degrees.

[quote="Prime factorization"][quote="BanishedTraitor"]:lol:  Properties of a circle can be proved by proving properties of an ellipse.[/quote]

Can you give me an example of this? I know that the circle is a special case of the ellipse where both focii are located at the center point. Other than that, I don't know very much about their relationship.[/quote]

I will give you more than one! Because I am in a good mood!

So we take an Ellipse $ E: \frac{x^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}= 1$.
$ a$ is half the major-axis and $ b$ is half the minor-axis.

1. Recall that the Cartesian Equation of a circle $ C: x^{2}+y^{2}= r^{2}$. If we let $ a=b=r$ we find that a circle is really just an ellipse in disguise  :ninja: .

2. The area of an ellipse is $ \pi ab$ and if we replace it with $ a=b=r$ we have that $ \pi r^{2}$ is the area of a circle.

3. For an ellipse the tangent at a point $ P \in E$ is equally inclined to the focal chords through $ P$. Now a circle is really just an ellipse where the focii overlap. So we draw a tangent at a point $ P \in C$ and draw the focal chords through $ P$ which is just the radius and since the tangent is equally inclined, two $ 90^{\circ}$ angles are formed! Which is your original problem.

4. Circles and ellipses are both conic sections so theorems like $ \text{ Butterfly Theorem}$ or $ \text{Pascal's Theorem}$ will hold for both of them (in fact they hold for the whole family of conic sections).


Let $ \{x_k\},\{y_n\}$ are two integer number sequences satisfy: $ x_1 < x_2 < ... < x_k < ...$,  $ y_1 < y_2 < ... < y_n < ...$ with $ k\geq5$ and $ n\geq5$, where $ x_i \neq y_j$ for all $ i \equal{} 1,2,...,$ and $ j \equal{} 1,2,...,$.
 Prove that: There exist index $ n_0, m_0$ such that: $ |x_{n_0} \minus{} y_{m_0}|\geq 7$.

Best! :)

it seems that obviously, but can you prove?

WLOG $ x_1<y_1$ (otherwise rename the sequences)

Consider the first $ 5$ terms of each sequence:
$ x_1<x_2<x_3<x_4<x_5$ and $ y_1<y_2<y_3<y_4<y_5$

$ x_3<y_5$ $ \Rightarrow$ $ y_5\minus{}x_1 \ge 7$ because ($ x_2, x_3,y_1,y_2,y_3,y_4$) are between $ x_1$ and $ y_5$

$ y_5<x_3$ $ \Rightarrow$ $ x_5\minus{}y_1 \ge 7$ because ($ y_2,y_3,y_4,y_5,x_3,x_4$) are between $ y_1$ and $ x_5$

but we can solve this problem follow:
we also consider the first 5 elements for each sequence.
WLOG $ x_1<y_1$.
Let assume that there is no pair of $ (x_i,y_i)$ satisfies the condition $ |x_i\minus{}y_j|\geq 7$ then:
+)since is the first number in (A) so the 8th ; 9th and 10th number in (A) must belong to set M (because the difference between the 8th and the 1st )
+)since is the 10th number so that the 2nd ; 3rd in (A) is belong to set M , because the difference between the 10th number and the 3rd number > 7 .
set M only cannot have 6 elements so we see the contradiction.


group theory
abstract algebra
superior algebra
superior algebra unsolved

How to proof:
     Let G be a finitely generated group(not necessarily abelian), and H be a subgroup of G with finite index.
 Show that  H is also finitely generated.

This is an old theorem of Schreier.  Indeed Schreier has shown more: if G is finitely presented, then so is H. It is explained briefly in http://en.wikipedia.org/wiki/Schreier%27s_subgroup_lemma and in more detail in most group theory texts that deal with finitely presented groups at all (Robinson, Rotman, and Scott's textbooks all cover this for instance).

If you read French, there is a complete proof here :
http://fr.wikiversity.org/wiki/Groupe_%28math%C3%A9matiques%29/Classes_modulo_un_sous-groupe#Sous-groupe_d.27indice_fini_d.27un_groupe_de_type_fini

[quote="jmerry"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62075]here[/url] for a constructive solution that tells us something about the number of generators.[/quote]

If it helps, those solutions are called the Schreier graph and the Schreier transversal respectively.


[quote="eraven"]The perpendicular distance [i]O[/i] [i]N[/i] between the origin and the line [i]L[/i] is [i]p[/i], and [i]O[/i] [i]N[/i] makes an angle, alpha, with the positive x-axis.  Show that L has equation x cos(alpha) + y sin(alpha) = p.

Am I right in concluding that the above equation is true only when L is a horizontal line (i.e., y=constant)?[/quote]

You are right only in case alpha= $90^0$ or any odd multiple of it.

In general, when alpha is not a multiple of $90^0$, we have the point at a distance p from origin as (p cos(alpha), p sin(alpha)), the slope of the line ON is tan(alpha) and that of the line perpendicular is -cot(alpha).  Then the eqauation to the line is given by

y-psin(alpha) = -cot(alpha)(x-pcos(alpha))

on simplification, it gives x cos(alpha) + y sin(alpha) = p.


Find the smallest natural number such that when divided by 5 it gives a remainder of 2, when divided by 7 it gives a remainder of 3, and when divided by 11, gives a remainder of 5.

[hide="answer"]192[/hide]

would you care to share with us your solution? :roll:
I just plug values in and out, is there an easier way to do this problem using modular arithmetic?

[hide]Well it ends with 2 or 7, so the multiple of 11 ends with 7 or 2.

try the first ending with 2. 22+ 5 is 27, but not 3 more than a multiple of 7.
77+ 5 is 82, not 3 more than a multiple of 7.
132 + 5 is 137, not 3 more than a multiple of 7.
187 + 5 is 192, 3 more than a multiple of 7.

So the answer is 192.[/hide]

[hide]
We are looking for something that satisfies all three conditions at once.

3(mod 7), 5(mod 11), and 2(mod 5).

We shall continue subtracting backwards each of these things and we get that the answer is the first positive integer congruent to -193(mod385), so the answer is 192.

Or you could just guess and check and see that it has to be greater than 100 and then you will arrive at 192.

$192$[/hide]

i know how you got 385, but where'd u get -193?

[hide]Keep subtracting the numbers in the modulo until you get one number that is common to these three sequences which is -193.  Or you could add and see and this would get you 192.[/hide]

If you want a sure way to solve the problem, use the Chinese Remainder Theorem, that I learned in the Intro to Number Theory Class.

[hide="solution using Chinese Remainder Theorem"]Write out all of the conditions using modular arithmetic, where $N$ is the number:
$N\equiv2\pmod5$
$N\equiv3\pmod7$
$N\equiv5\pmod{11}$
Taking the first two, we have $N=2+5j$ and $N=3+7k$, so
$2+5j=3+7k$
$5j=1+7k$
$5j\equiv1\pmod7$
$j\equiv3\pmod7$
$j=3+7d$
Substituting in for $j$, we have
$N=2+5(3+7d)$
$N=17+35d$
$N\equiv17\pmod{35}$
Now, we can add in the second equation:
$N=17+35j$ and $N=5+11k$
$5+11k=17+35j$
$11k=12+35j$
$k\equiv17\pmod{35}$
$k=17+35d$
Substitute in again to recieve:
$N=5+11(17+35d)$
$N=192+385d$
$N\equiv192\pmod{385}$
The lowest positive value is therefore $\boxed{192}$[/hide]

[hide]
I'm not too good with mods, so I did guess and check.
27 works for everything except for 7.
192 works for all three of them:
192/5=38 r 2
192/7=27 r 3
192/11=17 r 5[/hide]

Of course, that was to my own problem...

Nice, elegant solution to your problem!

Didn't even think of how they were all related  :roll:

[quote="b-flat"][quote="cowpi"]Answer:

[hide]192[/hide]

Solution on my calculator, assuming that:

N=X mod M
N=Y mod N

[hide]ClrHome
Prompt X,M,Y,N
Y-X->Y
If Y=0: Then
Goto 2
Else:
Lbl 1
If iPart(Y/M)=Y/M: Then
Y/M->Y
Goto 2
Else: N+Y->Y
Goto 1
Lbl 2
Disp "Remainder"
Disp MY+X
Disp "Mod"
Disp MN[/hide][/quote]

Wow, that is how I used to do the problems (if a calculator was allowed), until I learned the Chinese Remainder Theorem, or couldn't easily just "see" what the answer was.[/quote]

Irony...I learned the CRT first and stuck the formula into my calculator after.


geometry
cyclic quadrilateral
similar triangles
geometry proposed

Given a quadrilateral $ ABCD$ with $ \angle B\equal{}\angle D\equal{}90^{\circ}$. Point $ M$ is chosen on segment $ AB$ so taht $ AD\equal{}AM$. Rays $ DM$ and $ CB$ intersect at point $ N$. Points $ H$ and $ K$ are feet of perpendiculars from points $ D$ and $ C$ to lines $ AC$ and $ AN$, respectively.
Prove that $ \angle MHN\equal{}\angle MCK$.

I have not a solution yet, so I made a figure ! :blush:

The sketch of solution.
We need 3 steps to compute angles $ \angle MHN$and $ \angle MCK$.
1. $ CD\equal{}CN$. This can be done by a little computing.
2. $ AM^2\equal{}AD^2\equal{}AH*AC$ so $ \angle BMC\equal{}\angle MHC$.
3. $ CN^2\equal{}CD^2\equal{}CH*CA$ so $ \angle CHN\equal{}\angle CNA$.
Finally $ \angle MHN\equal{}\angle MCK$

Hello!

The proposed problem can be generalized:
   Let $ ABCD$ be a cyclic quadrilateral ($ AD < AB$ and $ \angle ADC \equal{} \alpha\leq 90^{\circ}$),
$ M\in\left(AB\right)$ so that $ AM \equal{} AD$ and the points $ E,F\in AC$ so that $ \angle DEA \equal{} \angle DFC \equal{} \alpha$.
The point $ N\in BC$ so that $ \angle MDN \equal{} 90^{\circ} \minus{} \alpha$ ($ \angle MDN$ and $ \angle EDF$ have the same orientation)
and the points $ X,K\in AN$ so that $ \angle AXC \equal{} \alpha$ and $ \angle XCK \equal{} \angle ENF$ ($ \angle XCK$ and $ \angle ENF$
have the same orientation). Prove that $ \angle MEN \equal{} \angle MCK$. :)

Hi Petry,

Nice generalization. Same approach. From $ \angle MDN\equal{}90^{\circ}\minus{}\alpha$, we know $ CD\equal{}CN$. Then use the two similar triangles mentioned in rchlch's solution and some angle chasing to get the result. :)

first,we can get $AD^2=AM^2=AH*AC$;
$\angle MDC=\angle MNC=\frac{A}{2}$
hence $DC=CN$
so $DC^2=CN^2=CH*CA$
then$\angle NHM=\angle NHC-\angle MHC=\angle CNA-\angle CMB=90-\angle KCN-\angle CMB=\angle KCM$
QED


Find the best value of $ k$ such that

$ \sqrt {x \plus{} y \plus{} z \minus{} 2} \minus{} 1\le k(xy \plus{} yz \plus{} zx \minus{} 3)$,

for any positive real numbers $ x,y,z$ satisfying $ xyz \equal{} 1$.

[quote="Vasc"]Find the best value of $ k$ such that

$ \sqrt {x \plus{} y \plus{} z \minus{} 2} \minus{} 1\ge k(xy \plus{} yz \plus{} zx \minus{} 3)$,

for any positive real numbers $ x,y,z$ satisfying $ xyz \equal{} 1$.[/quote]
The best value of $ k$ is zero.

[quote="Vasc"]Sorry, Guest. The correct inequality is reversed.[/quote]
I am sorry, Vasc. The best possible value of $ k$ for the correct inequality is $ \frac{27}{46}$.

[quote="Vasc"][quote="Guest"] The best possible value of $ k$ for the correct inequality is $ \frac {27}{46}$.[/quote]
I think the same.[/quote]
$ k_{best}\equal{}\max_{(0,\plus{}\infty)}\frac{2t\plus{}1}{(t\plus{}1)(t\plus{}\sqrt{2t^3\minus{}2t^2\plus{}1})}\equal{}\frac{27}{46},$ where $ t_{max}\equal{}\frac{5}{9}.$ :o


Let $k = \sqrt[3]{3}$. Find a polynomial $p(x)$ with rational coefficients and degree as small as possible such that $p(k+k^{2}) = 3+k$. Does there exist a polynomial $q(x)$ with integer coefficients such that $q(k+k^{2}) = 3+k$?

It is obviosly.
All solutions $p(x)\in Q[x]$, suth that $p(k+k^{2}=3+k$ are $p(x)=\frac{x(x-1)}{2}+r(x)(x^{3}-9x-12), r(x)\in Q[x]$.
If $p(x)\in Z[x]$, and $p(k+k^{2})=3+k$, then exist $r(x)\in Z[x]$, suth that $p_{1}(x)=p(x)-r(x)(x^{3}-9x-2)\in Z[x]$ and $deg(p_{1}(x))\le 2, p_{1}(k+k^{2})=3+k$. It give contradition. I don't undestand, what is problem? I give solution in first post.

Solution:
It is easy to understand that $ p(x)\equal{}\frac{x^{2}\minus{}x}{2}$ is the solution of the problem,so now let $ deg(p)<2\Rightarrow deq(p)\equal{}1$.At first we will prove the following lemma:
Let $ r\equal{}\sqrt[3]{3}a\plus{}\sqrt[3]{9}b$ and $ r,a,b\in Q$ then $ a\equal{}b\equal{}0$.Indeed,we have that $ r^{2}\equal{}a^{2}\sqrt[3]{9}\plus{}3b^{2}\sqrt[3]{3}\plus{}6ab\Rightarrow r'\equal{}a^{2}\sqrt[3]{9}\plus{}3b^{2}\sqrt[3]{3}\in Q\Rightarrow$
Suppose $ a\neq 0\Rightarrow\frac{3b^{2}r}{a}\minus{}r'\in Q\Rightarrow\sqrt[3]{9}(\frac{3b^{3}}{a}\minus{}a^{2})\in Q\Rightarrow$condratiction$ \Rightarrow a\equal{}0\Rightarrow b\equal{}0$.
Now it is obvious that $ deq(p)\geq 2$.Using lemma for $ deq(p)\equal{}2$,we will have that $ p(x)\equal{}\frac{x^{2}\minus{}x}{2}$ is the unique 2nd degree polynomial,such that $ p(\sqrt[3]{3}\plus{}\sqrt[3]{9})\equal{}3\plus{}\sqrt[3]{3}$.
My solution to the second part is the as Rust's.

[quote="borislav_mirchev"]What are $ Q[x], r(x)$ in the solutions above? And I also don't understand the second part of the solution. Is it possible to find a polynomial with integer coefficients?[/quote]
$ p(x)\in Q[x]$,means that all coefficients in $ p(x)$ are rational numbers.The answer to the second part is $ NO$.The idea is to consider $ r(x)$,do you need in more details?

In Rust's solution is written:

Let $ p(x)\equal{}a_{0}\plus{}a_{1}x\plus{}a_{2}x^{2}\plus{}r(x)p_{0}(x)\equal{}3\plus{}k,r(x),p_{0}(x)\in Q[x], p_{0}(k\plus{}k^{2})\equal{}0$.
Then $ a_{0}\plus{}a_{1}(k\plus{}k^{2})\plus{}a_{2}k^{2}(k^{2}\plus{}2k\plus{}1)\equal{}(a_{0}\plus{}6a_{2})\plus{}k(a_{1}\plus{}3a_{2})\plus{}k^{2}(a_{1}\plus{}a_{2})$, therefore $ a_{2}\equal{}\minus{}a_{1}\equal{}\frac{1}{2},a_{0}\equal{}0$ .
We can find $ p_{0}(x)\equal{}b_{0}\plus{}b_{1}x\plus{}b_{2}x^{2}\plus{}x^{3}$ , then $ b_{0}\plus{}b_{1}(k\plus{}k^{2})\plus{}b_{2}(6\plus{}3k\plus{}k^{2})\plus{}(12\plus{}9k\plus{}9k^{2})\equal{}0$. It give $ p_{0}(x)\equal{}x^{3}\minus{}9x\minus{}12$

1. Is Rust trying to find all polynomials with degree greater than 3?
2. What are the degrees of $ p_{0}(x)$ and $ r(x)$?
3. Why Rust can be sure that the coefficient before $ x^{3}$ is 1?

P.S. If it easier for you write the solution from scratch it will be easier for undertsnding from new readers. When and if you have time explain in more details second part of your solution for Vietnam 1996 Combinatoric - it is in my post.

I'm understanding far better. Now it is OK.

[quote=Erken]I'll try to explain you:
I hope that from my solution of the first part you understand that there is unique polynomial $ p(x)\in Q[x],deg(p)\equal{}2$,such that $ p(k^{2}\plus{}k)\plus{}3\plus{}k$,but if $ p(k^{2}\plus{}k)\equal{}3\plus{}k$ and $ p(x)\in Z[x]$ then $ deg(p)\geq 3$,so Rust just divide $ p(x)$ to $ g(x)\equal{}x^{3}\minus{}9x\minus{}2$,then $ p(x)\equal{}q(x)(x^{3}\minus{}9x\minus{}2)\plus{}r(x)$,then $ geg(r)\leq 2,r\in Z[x]$ and $ r(k^{2}\plus{}k)\equal{}3\plus{}k$,since $ k^{2}\plus{}k$ is the root of $ x^{3}\minus{}9x\minus{}2$.Contradiction.Now i hope you understand.
P.S.OK,i'll explain solution to Vietnam 1996 in more detalis.[/quote]

I think $r\in Q[x]$


geometry
3D geometry
sphere
real analysis
real analysis solved

Prove that the space can be partitioned in non-degenerate circles. My teacher told me it's an extremely difficult problem. Does anyone have ideas how to solve it?

Do you mean [tex]\mathbb{R}^3[/tex] by saying space?  ;)

They are not points, i.e. circles with zero radii.

OK and I think it's disc instead of circle ?

[quote="alekk"]he gave the same pb last year  :D
1/show that a sphere minus 2 points can be partitioned in non-degenerate circles
2/find a family $F$ of non intersecting circles in the plane $z=0$ such that any sphere with center $O$ cut this
family $F$ in exactly 2 points.
3/conclude[/quote]

 He? Harazi's teacher?  :D
 Anyway, the above solution is due to Halmos, rite? ;)

[quote="fedja"][quote="alekk"]who is Halmos ?[/quote] My guess is that dickchimney meant the famous Hungarian mathematician [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Halmos.html]Paul Halmos[/url].[/quote]

 That's him. :lol:  [url=http://www.geom.uiuc.edu/docs/forum/hoops_links/circle_solution.html]Here[/url] is where I find the solution of the problem.  There're also some other interesting geometry problems.


My professor proved this theorem, but when I write notes I cannot pay attention to what he is saying.

|v*w|<= |v||w|.  He defines "v" as u1 +u2.  where u1 = the vector projection of v onto w. and u2 as v - 1. Towards the end he goes from

||v||^2 = |< v, w >|^2 / < v, w >^2  multiplying <w , w > + ||u2 ||^2

            =  |< v, w>|^2 / ||w||^2.

My question is what happened to  to the " ||u2||^2 in the line second to last.

I can't figure out what you're trying to say (what is the word "multiplying" supposed to be signifying?) -- if you used a bit of [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ), it would make your post much easier to understand.


The USAMTS problems were really hard!

How did you do?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312392]This[/url] topic talks about this issue.

But here there is a poll, so this is probably a slightly better thread.
(Yes, I know I am going against my own thread, but still. I like polls, and so do a bunch of other people.)
So vote here if you wish, and post here if you wish. You may vote on the other thread too, but who doesn't like polls?

Thanks Eric,

I see that there are two 25s on Aops.
Thats amazing!  :D

Of course, they might be lying, but why would one lie when its anonymous?

There should be a "delete" button near where the "quote" and "edit" buttons are.

[quote="KingSmasher3"]How does one delete his/her post?[/quote]

After a certain amount of time, one is unable to delete his/her own post anymore.

Your post must be the last one in the thread for the delete button to be there.


Let $a,b,c$ be side lengths of a triangle. Prove that
$a^2b^4+b^2c^4+c^2a^4+b^2a^4+c^2b^4+a^2c^4 \leq 3a^2b^2c^2+a^6+b^6+c^6$

I don't get this. It is exactly Schur for $ a^2, b^2, c^2$, you don't need any extra-assumption.


Grrrr roadrunner.

Okay, I got disconnected (I have no idea why...my internet connection just severed for a sec I guess) and got kicked out of the classroom. Upon reconnecting, I try to sign in. Everything is normal but I noticed the login button didn't go away. Then I click the classroom link and it gives me the "You must be logged in" page. So I click the login button, and same thing happens. *continues attempting to get into classroom*

When this happens, click logout on the forum.


well, divide both sides by 2 to get $ \log_2 x = 33 $, from which you get $ x = 2^{33} $ (by the definition of $ \log $)

[quote="lamefif"]66 = 2 log2 x

that is 2 log to the base 2 not 2 times log (times 2)
how would i go about finding x
thax  :blush:[/quote]
you can use the latex to help you

$66=\log_2{x}$

[code]66=\log_2{x}
[/code]

where do i get it from can u post a link?
Never mind

Is this an [b]Intermediate Topic[/b] problem?!  :no: And btw Iversonfan2005 i think you should wait to learn $logs$  :P

Here are some things you can learn  ;)

http://mathworld.wolfram.com/Logarithm.html

It's good to know things earlier than you learn at school  :D

[quote="nat mc"][quote="Iversonfan2005"]Hmmm....I haven't really done logs yet, but maybe I can help. If 2log2 of x=66, then log2 of 2x=log2 of 33. So [b]x=log 2 of 33[/b]. Maybe, I did it completely wrong.[/quote]

I don't know what you did
Logs:
If
bx=a
Then  x=logba

And If you have
a log b x
It is the same as
logbxa[/quote]

Iversonfan did $\log_{2} 2x$ instead of $\log_{2} x^2$.

[quote="Slizzel"]Is this an Intermediate Topic problem?![/quote]

Eh, you know how it is. Folks in the Getting Started community don't accept stuff that goes above mathcounts in level, so we'll take any problems with logs here.


What level do you think these books are:

mathematical olympiad treasures
and mathematical olympiad challenges.

Are those books more difficult than PSS.

thanx :lol:

[quote="matt276eagles"]You already posted the same question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104000]here[/url].[/quote]
So they're both olympiad lvl. :)

[quote]So they're both olympiad lvl. [/quote]

Can you be more specific. is it usamo level or imo level?

What about mathematical miniatures
 :?:

well,
-mathematical olympiad treasures: http://www.amazon.com/gp/product/0817643052/sr=1-1/qid=1154359846/ref=pd_bbs_1/002-6308219-2941631?ie=UTF8&s=books

-mathematical olympiad challenges:
http://www.amazon.com/gp/product/0817641556/sr=1-1/qid=1154360006/ref=pd_bbs_1/002-6308219-2941631?ie=UTF8&s=books
See for yourself. :)


inequalities
geometry
circumcircle
trigonometry
geometry unsolved

Prove that in any triangle we have :
a)$ m_{a}+m_{a}+m_{c} \le \sqrt{p^2+\frac{3(a^2+b^2+c^2)}{2}}$
b) $ m_{a}+m_{a}+m_{c} \le 4R+r$

With $ M$ is midpoint of $ BC$, $ O$ is circumcenter of triangle $ ABC$ then
$ m_a\equal{}AM\leq AO\plus{}OM\equal{}R\plus{}R\cos \angle BOM\equal{}R\plus{}R\cos A$
Similarly then
$ m_a\plus{}m_b\plus{}m_c\leq 3R\plus{}R(\cos A\plus{}\cos B\plus{}\cos C)$
$ \equal{}3R\plus{}R(1\plus{}4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})\equal{}4R\plus{}r$

Oh! Slution of [b]tdl[/b] only true when ABC triangle hava $ \angle A$, $ \angle B$, $ \angle C$ are acute angle. Because if $ \angle A$, $ \angle B$, $ \angle C$ are acute angle we have O live in ABC, so we have $ \angle BOM$=$ \angle A$. It's not true if $ \angle A$ is obtuse angle.

If $ \angle A$ $ > 90^0$.
We have $ \angle B$ $ <$ $ \angle BAM$ or $ \angle C$ $ <$ $ \angle CAM$ $ \Longleftrightarrow$ $ AM$ $ <$ $ BM$ or $ AM$ $ <$ $ CM$ $ \Longrightarrow$ $ m_a = AM \le$ $ \frac {BC}{2}$ = $ \frac {a}{2}$.
And we have $ m_b = BN$ $ <$ $ BP + PN$ = $ \frac {a + c}{2}$, $ m_c = CP$ $ <$ $ CN + NP$ = $ \frac {b + c}{2}$.
$ \Longrightarrow$ $ m_a + m_b + m_c <$ $ \frac {3a}{2}$+$ \frac {b + c}{2}$ $ =$ $ 2a + (p - a)$.
But we have $ 2a = 2BC$ $ \le$ $ 2.2R = 4R$.
Let I be center of inscribed circle of $ ABC$ triangle. Let $ IK \perp AB$
In KAI triangle we have $ \|\begin{array}{c}AK = p - a \\
\angle KAI = \frac {A} {2} \geq 45^0 \\
\angle KIA \le 45^0\end{array}$ $ \Longrightarrow$ $ AK = p - a$ and $ AK \le AI$ $ \Longrightarrow$ $ p - a \le r$.
So that $ m_a + m_b + m_c \le 4R + r$.


\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\ \\a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\ge 1\\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:

[quote="bbbbbbbbbb"]
\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:[/quote]
We have
$ \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}}  \geq \frac{(a\plus{}b\plus{}c\plus{}d\plus{}e)^2}{ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}be\plus{}cd\plus{}ce\plus{}ca\plus{}de\plus{}da\plus{}db\plus{}ea\plus{}eb\plus{}ec}$

[quote="chien than"][quote="bbbbbbbbbb"]
\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:[/quote]
We have
$ \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \geq \frac {(a \plus{} b \plus{} c \plus{} d \plus{} e)^2}{ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} be \plus{} cd \plus{} ce \plus{} ca \plus{} de \plus{} da \plus{} db \plus{} ea \plus{} eb \plus{} ec}$[/quote]
It's wrong!
I think $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}1$ !

[quote="NguyenDungTN"][quote="chien than"][quote="bbbbbbbbbb"]
\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:[/quote]
We have
$ \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \geq \frac {(a \plus{} b \plus{} c \plus{} d \plus{} e)^2}{ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} be \plus{} cd \plus{} ce \plus{} ca \plus{} de \plus{} da \plus{} db \plus{} ea \plus{} eb \plus{} ec}$[/quote]
It's wrong!
I think $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 1$ ![/quote]
Not wrong.
U must find mind.

no one like this.
Remember $ a,b,c,d,e \in R$

[quote="bbbbbbbbbb"]
\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:[/quote]
The $ \min F$ is not exists. Try $ a\plus{}b\plus{}c\rightarrow0^\minus{}.$ :wink:

[quote="arqady"][quote="bbbbbbbbbb"]
\[ \begin{gathered} Let\,\,a,b,c,d,e \in \mathbb{R} \hfill \\
Find \hfill \\
F_{\min } \equal{} \frac {{a^2 }} {{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }} {{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }} {{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }} {{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }} {{a \plus{} b \plus{} c}} \hfill \\
\end{gathered}
\]
:huh:[/quote]
The $ \min F$ is not exists. Try $ a \plus{} b \plus{} c\rightarrow0^ \minus{} .$ :wink:[/quote]
OMG. SOrry
My condition
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\ge 1$

[quote="bbbbbbbbbb"]
My condition
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2\ge 1$[/quote]
The answer is still the same: the $ \min F$ is not exists.
But if $ a,b,c,d,e\geq0$ then see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63850


function
real analysis
real analysis unsolved

Why is the $ \sup$ of a real continuous function equal to its $ ess\,\sup$?

Because you can stick a box under the bump. If $ f$ is continuous and $ f(x_0)\ge M\minus{}\epsilon$, there is an interval $ (x_0\minus{}\delta,x_0\plus{}\delta)$ on which $ f$ is greater than $ M\minus{}2\epsilon$.


doesn't it follow from the definition?

If not, I would be glad to know the solution...

[quote="brady00n"]Prove that if $x\equiv y \pmod{m}$, the quotient $\displaystyle\frac{x-y}{m}=$ an integral value.[/quote]

[hide]Can you just subtract $y$ from both sides, $km=x-y$ for an integer $k$ since $x-y$ is divisible by $m$ and then $\frac{x-y}{\frac{x-y}{k}}=k$[/hide]

-internet lags and I double-posted accidentally...-

[quote="pkerichang"]kind of easy one to post on intermediate topics...
[hide]
$x \equiv y (\bmod\ m)$ is a short hand of saying that $x = ma + y$ for integer $a$.  so we have $\frac{x-y}{m} \equiv a$, since it is given that $a$ is an integer.  We're done.
[/hide]
maybe a harder one...
proof that if $a|x$ and $a|y$, then:
$a|x+y$
$a|x-y$
$a|px+qy$ for $p$, $q$ integers
$a \ge \gcd(x, y)$[/quote]

[hide="Pretty easy, and i dont know much abt mods :P "]
Just put $x=ka \ , \ y=na$ for some integers $k \ , \ n$.
Then $x-y=a(k-n)$ , $x+y=a(k+n)$, and $px+qy=a(pk+qn)$... so all of those numbers are divisible by $a$. [/hide]

[hide]Let $x=ak_1$ and $y=ak_2$. Then
$x+y=a(k_1+k_2) \implies a|x+y$,
$x-y=a(k_1-k_2) \implies a|x-y$,
$px+qy=a(pk_1+qk_2) \implies a|px+qy$,
and since $a|x$, and $a|y$, a must divide the GCD(x,y), so $a\ge gcd(x,y)$.[/hide]

As xxxyyyy said, it's just the definition......... nothing to prove or disprove there

[quote="Jiang"]Maybe it should be $a\leq\gcd(x,y)$[/quote]

oops, I meant to say proof that $px+qy \ge \gcd(x, y)$


integration
trigonometry
calculus
calculus computations

[quote="eshban"]$\int_{0}^{\frac{1}{\sqrt{2}}}\frac{x\sin^{-1}x}{\sqrt{1-x^{4}}}dx$[/quote]
Use this substitute$y=sin(x)$ I think it's can help you. :wink:

[quote="HilbertThm90"]I don't want to speak for eshban, but I tried this problem yesterday, and the substitution you made led to (indefinite) $\int \frac{y\sin y}{\sqrt{1+sin^{2}y}}dy$, and also I tried letting $x=\sin A$ and ended up with, imagine, $\int\frac{A\sin A}{\sqrt{1+\sin^{2}A}}dA$. Although that substitution seems natural, the next step is not obvious to me. Where would you go from here, or would you start with something different.[/quote]
don't forget interval
[0,pi/4] I think you must use it somewhere


trigonometry
geometry
trig identities
Law of Sines

1. Triangle ABC is a right triangle, with angle C as the right angle. Angle B is 55 degrees. Line AC is 120 INCHES. What is the length of the line BC, in CENTIMETERS??

2. Triangle ABC is a right triangle, with angle C as the right angle. Angle B is 65 degrees. Line AC is 120 INCHES. What is the length of the line BC, in CENTIMETERS??

(see the difference between "inches" and "centimeters"!!)

Simply convert to the same units, namely 2.54 centimeters for every inch.

You can use the Law of Sines to determine your side lengths.

[hide="Solution for 1."]
$ \angle A\equal{}90^o\minus{}\angle B\equal{}90^o\minus{}55^o\equal{}35^o$

By the Law of Sines:

$ \frac{AC}{\sin \angle B}\equal{}\frac{BC}{\sin \angle A}$

$ \frac{120\ \text{in.}}{\sin 55^o}\equal{}\frac{BC}{\sin 35^o}$

$ BC\equal{}\left(120\ \text{in.}\right)\left(\frac{\sin 35^o}{\sin 55^o}\right)\approx 84.02\ \text{in.}$

To convert to centimeters, we note that $ 2.54\ \text{cm}\equal{}1\ \text{in.} \implies \frac{2.54\ \text{cm}}{1\ \text{in.}}\equal{}1$. Since multiplying $ BC$ by $ 1$ does not change the value of $ BC$, we find that:

$ BC\equal{}\left(84.02\ \text{in.}\right)\left(\frac{2.54\ \text{cm}}{1\ \text{in.}}\right)\approx \boxed{213.42\ \text{cm}}$
[/hide]

If you can do 1, you can do 2.

I would not recommend bothering with Law of Sines for this...  The definition of sine and cosine will suffice.

[hide="definition of sine, cosine, and tangent"]The sine of an angle is defined to be the length of the side opposite the angle divided by the length of the hypotenuse.

The cosine of an angle is defined to be the length of the side adjacent to the angle (non-hypotenuse side) divided by the length of the hypotenuse.

$ \tan(\theta)\equal{}\frac{\sin(\theta)}{\cos(\theta)}$

[asy]size(200);
pair A=(0,3), B=(4,0), C=(0,0), BB=(0,1.5), AA=(2,0), CC=(2,1.5);
draw(A--B--C--A);
label("$A$",A,N);
label("$B$",B,E);
label("$C$",C,SW);
label("$\theta$",A,(4,-5));
label("$a$",AA,S);
label("$b$",BB,W);
label("$c$",CC,NE);
MA(C,A,B,0.5);[/asy]

$ \sin(\theta)\equal{}\frac{a}c$

$ \cos(\theta)\equal{}\frac{b}c$

$ \tan(\theta)\equal{}\frac{a}{b}$
[/hide]

[hide="hint"]Rewrite what you know based on definitions of sine and cosine.  You may have to assign variables to what you don't know.  Then see if you can solve.[/hide]


advanced fields
advanced fields unsolved

Let X be a metric compact space. Let f: X->X be an isometric application ( d(x,y) = d( f(x), f(y) ), for every x,y)
Show that f is surjective.

Suppose the contrary: $X \setminus f(X) \neq \varnothing$. Take arbitrary $x_{0}\in X \setminus f(X)$ and consider the sequence $x_{n+1}= f(x_{n})$. One sees that $(x_{n})_{n\in\mathbf N}\subset f(X)$ and since $f(X)$ is compact, one may extract a convergent subsequence of $(x_{n})$ (denote it again $(x_{n})$). But
\[\|x_{n+m}-x_{n}\| = \|x_{m}-x_{0}\| \ge \mathrm{dist}\,(x_{0},f(X)) > 0, \]
since $f(X)$ is a closed subspace of $X$. We have come to a contradiction.

Thus $f$ is a homeomorphism.


How many different integers can be expressed as the sum of two different numbers in the set $ \{9,12,15,18,21,24,27,30,33\}$?

The largest integral value is $ 63$, while the smallest is $ 21$.  Since the commonn difference is $ 3$, our answer is $ \frac{63\minus{}21}{3} \plus{} 1$, or $ \boxed{15}$.

I don't understand where you got 63 from. Could you elaborate a bit?

Thanks!
________________
$\mathfrak{pi man}$

The largest possible sum is 30+33=63 because the 2 largest numbers in the set are 30 and 33.

[quote="ernie"]The largest integral value is $ 63$, while the smallest is $ 21$.  [b]Since the commonn difference is $ 3$, our answer is $ \frac{63\minus{}21}{3} \plus{} 1$, or $ \boxed{15}$.[/b][/quote]

Wait, what formula is this?

There is no formula, or at least none with a specific name @DeathLlama9.

BTW this requires no tricks.
Actually it is quite basic. just think next time.


limit
trigonometry
calculus
calculus computations

hello  :D
\[ \lim_{x\to 0}\frac{\sqrt[5]{x^2\plus{}32}\plus{}\sqrt{x\plus{}36}\minus{}8 }{\sin x}\]

Beautiful? The L'Hopital rule would do it.


Suppose that $ B$ is a $ n\times n$ matrix such that: $ \left(AB\right)^2 \equal{} 0$ for all $ n\times n$ matrix $ A$, then prove that $ B$ is a $ n\times n$ matrix with all entries zero.

Your statement doesn't make sense, for several reasons.
First, "Suppose $ A$ is..." and "for all A" don't belong together. Second, two $ m\times n$ matrices don't have a product. Is the following what you meant?

Suppose $ B$ is a $ m\times n$ matrix such that $ (AB)^2 \equal{} 0$ for all $ n\times m$ matrices $ A$. Show that $ B$ is the zero matrix.

[Edit: grammar fix]

In case [b]jmerry[/b] is correct, the answer is easy to obtain.

Choose $ A$ to be the matrix with exactly one number 1 and the rest 0. Then $ A$ serves as the selection matrix: $ AB$ gives the matrix of exactly one element (which is in $ B$) and the rest 0. Hence, the square of every element of $ B$ is 0. Thus, it must be the zero matrix.

That's not correct. That's not what $ AB$ looks like, and you haven't squared it either.

Oh, I'm sorry for my blindness  :blush: . Now the problem has been edited. Thank you!!!

Well if it is true for all $ n \times n$ matrices A.... doesn't it follow that B is the $ 0$ matrix?
Like... let $ A'$ be a matrix such that A'B = 0.
Now... lets construct $ A''$ where this matrix is the the same as $ A'$ but just add... lets say $ 1$ to any one corner of $ A'$. This is possible since I am not doing a row reduction process, I am just making a new matrix based off of the old one.
Then $ A''B\equal{}0$ and $ A'B\equal{}0$ iff $ B\equal{}0$ matrix since... well it must. Umm the corner where the $ 1$ was added makes it impossible to have $ A''B \equal{}0$ where B is NOT the 0 matrix AND $ A'B \equal{} 0$. (Yes $ AB \equal{}0$ and $ CB \equal{} 0$ work for some distinct matrices $ A$ and $ C$, but not for the given constructed ones).
So... $ B$ must be the 0 matrix.

$ (AB)^2 \equal{} 0$ is also $ (AB)(AB) \equal{} 0$ which by associativity, gives $ (ABA)B \equal{} 0$ and this makes it more clear $ B$ must be the $ 0$ matrix, if you consider $ ABA \equal{} C$.
:S
Hard to explain for some reason, maybe since it's very early.

$ ABAB\equal{}0$ does not imply $ AB\equal{}0$. Try a different argument.
[hide="Hint"]Use $ B$ to construct $ A$ making the product $ (AB)^2$ nonzero. Don't just use a short list of possible $ A$ values.[/hide]

Well the for the $ ABAB\equal{}0$ argument, $ A$ is EVERY $ n \times n$ matrix.... and use my whole argument not just that snippet. I didn't say $ AB\equal{}0$.
Unless you mean $ AB\equal{}0, and CB \equal{}0$. That is a different $ A$. I was just saying two matrices multiplied together can give a $ 0$ even if neither are the $ 0$ matrix.... sorry for the confusion change $ A$ to... $ D$ ? :P

If your whole argument depends on that snippet, it's all I need to shoot down. You don't have a proof. The rest of what you wrote is also a mess, which doesn't really prove anything.

I'm sorry, but could you give me a complete solution, dear jmerry? Thanks in advance  :)

Over $ \mathbb{R}$, we can use $ A\equal{}B^T$. Whenever $ X$ is a nonzero real matrix, $ X^TX$ is symmetric, with at least one positive element on the diagonal. Then $ ABAB\equal{}(B^TB)^T(B^TB)$, and we repeat this. For $ \mathbb{C}$, we use the conjugate of the transpose instead.

Over an arbitrary field, we need a slightly different construction. If $ A$ is the matrix with $ 1$ in the $ ij$ position and zeros elsewhere, $ AB$has the $ j$th row of $ B$ in its $ i$th row, and zeros elsewhere. Choose $ i$ and $ j$ so that the $ ji$ entry of $ B$ is nonzero; this gives $ AB$ a nonzero entry $ c$ in the $ ii$ position. Then $ (AB)^2$ has $ c^2$ in the $ ii$ position, and is nonzero.

By the way, this works for nonsquare matrices with dimensions as I stated.


Does there exist a function $f : \mathbb{R} \rightarrow \mathbb{R}$, which satisfies both conditions :

     [b]a)[/b]	$f( x + y + z) \leq 3(xy + yz + zx)$   for all real numbers $x , y , z$

and

	[b]b)[/b] there exist function $g$  and natural number $n$, such that

             $g(g(x)) = x ^ {2n + 1}$    and         $f(g(x)) = (g(x)) ^2$  for every real number  $x$ ?

I'll see it again , but from the first look it seems to be a correct and nice solution , much better than the official.
 They prove that [hide] $f(x) = x^2$ [/hide]ect

 babis

but if $f(x)=x^2$, then the sign in [b]a)[/b] must be reversed...  :?

That exactly IS the point: the function doesn't exist (as I've proven in my post, but in a much simpler manner)

EDIT: The proof had a grave mistake, but the correct one is given in post #7

Farenhajt you are wrong:

$f(3x) \le 9x^2$ or $f(t) \le t^2$ then $f(g(x))^2 \le g(x)^2$ wich is true.

Oh my God! You're right  :(

Incredible no one so far spotted that moment of stupidity  :blush: myself included...

But here, I'll make up for it:

Putting $y=z=0$ we get $f(x)\leqslant 0$, hence $g(x)^2=f(g(x))\leqslant 0$ implies $g(x)\equiv 0$

To Fahrenhajt:

  Why did you delete your first post ? I Had not read it with detail , but it seemed to me good. Your last solution is , I think the same , and nice.
 Last evening I gave the problem to a student of mine - not contestant , class 12 - and he solved it in the same way , as you. Any way , the problem is now full solved !
 Thanks to all ofyou!
  Babis


limit
function
calculus
calculus computations

That is correct. One way to see it: write this expression in polar coordinates.

Another way to see it: we have a function which is positively homogeneous of degree 1.

We call a function positively homogeneous of degree $k$ if for all $\lambda>0$ we have $f(\lambda x,\lambda y)=\lambda^{k}f(x,y).$

If $f$ is bounded on some circle and positively homogeneous of degree $k$ for some $k>0,$ then $f(x,y)\to0$ as $(x,y)\to(0,0).$

However, if $f$ is positively homoeneous of degree zero and not constant, then it has no limit at $(0,0).$

[quote="sos440"][quote="sylow_theory"]
But, by AM-GM
$\frac{|x|\cdot |y|}{\sqrt{x^{2}+y^{2}}}\leq \frac{\frac{\sqrt{x^{2}+y^{2}}}{2}}{\sqrt{x^{2}+y^{2}}}= \frac{1}{2}$ $(**)$
[/quote]

AM-GM yields $|x\parallel y| \leq \frac{x^{2}+y^{2}}{2}$. Then for $\sqrt{x^{2}+y^{2}}< \delta$,

$\left| \frac{xy}{\sqrt{x^{2}+y^{2}}}\right| = \frac{|x\parallel y|}{\sqrt{x^{2}+y^{2}}}\leq \frac{1}{2}\sqrt{x^{2}+y^{2}}< \frac{\delta}{2}$[/quote]

Thank you for the correct. I wrote that first and then for some reason edited back.  :oops:


How do you figure out rational expressions with word problems, such as

[quote]
 If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
[/quote]

It's about Work Group.

When two people, machines, workers, wanna make a job together, then we gotta use the following formula:

\[\frac1a+\frac1b=\frac1c\]

Where $a$ and $b$ are hours, days, or speed which they use to do the job and $c$ are the hours, days, or speed which they take to do the job together, then, plug the data in the formula:

\[\frac14+\frac16=\frac1c\]

And it follows that $c=2,4 \ \text{hours}$

[quote="MathTeX"]It's about Work Group.

When two people, machines, workers, wanna make a job together, then we gotta use the following formula:
\[\frac1a+\frac1b=\frac1c \]
Where $a$ and $b$ are hours, days, or speed which they use to do the job and $c$ are the hours, days, or speed which they take to do the job together, then, plug the data in the formula:
\[\frac14+\frac16=\frac1c \]
And it follows that $c=2,4 \ \text{hours}$[/quote]

Just so you know, the formula $\frac1a+\frac1b=\frac1c$ comes from a manipulation of the rate formula $d=rt$:

Let the distance in this case be $1$, for [u]one[/u] house to be painted. For other cases, the distance will vary based on the given information. Then, if it takes $t_{A}$ time for person/machine $A$ to complete one job, and it takes $t_{B}$ for person/machine to do the same job, we have that
\[r_{A}=\frac1{t_{A}}\ \wedge \ r_{B}=\frac1{t_{B}}\implies 1=\left( \frac1{t_{A}}+\frac1{t_{B}}\right)\cdot t_{A+B}\]


Here a problem from Kedlaya's Geometry Unbound

Problems for Section 7.5
5. (Dan Moraseski) Let $D$, $E$, $F$ be the feet of the symmedians of triangle $ABC$. Prove
that the Lemoine point of $ABC$ is the centroid of $DEF$.

This problem seems to be wrong but I can't manage to guess what the right one is. Has anyone an idea ?

Actually the Lemoine point in $\triangle{ABC}$ is the Gergonne point in triangle $DEF$, i.e the Gergonne point in its contact triangle $MNP$.  For proving that look at $\triangle{MNP}$ and prove that the Lemoine point in $ABC$ is the Gergonne point in its contact triangle $MNP$, i.e the Gergonne-type-lines in $MNP$ are symmedians in $ABC$.

For that extend $BC$ until touches $MN$ at $Q$, thus the quadruple $(QMAN)$ forms an harmonical division, so $(QBRC)$ is an h.d, but because $MN$ is tangent to the circumcircle of $ABC$, we have that $\frac{RB}{RC}=\frac{QB}{QC}=\frac{AB^{2}}{AC^{2}}$, thus $AR$, i.e $AP$ is a symmedian in the triangle $ABC$. The result follows.

Here's a sketch (the notations differ from the ones I mentioned in the solution)

[quote="pohoatza"]Actually the Lemoine point in $ \triangle{ABC}$ is the Gergonne point in triangle $ DEF$, i.e the Gergonne point in its contact triangle $ MNP$.[/quote]

I think you got slightly confused, since your statement about DEF is wrong, and your picture seems to have the points incorrectly labeled. I guess you meant the facts discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .

I also think the Geometry Unbound problem actually wanted to claim that the symmedian point of triangle ABC is the centroid of its own pedal triangle. See [url=http://www.cip.ifi.lmu.de/~grinberg/SymmedianPedalPDF.zip][i]New Proof of the Symmedian Point to be the centroid of its pedal triangle, and the Converse[/i][/url] on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] for two proofs.

  Darij

Oh thanks, you are right, I got confused with the notations.


Find all solutions to  $2^a+3^b=5^c$  if a,b,c are +ve integers.


geometry
Duke
college
symmetry
3D geometry

We had a practice for the Duke Math Meet this morning, and on the team round, there was a 3D geometry problem that on one knew how to do (my dad didn't have the solutions).

I have the answer, if needed, but what I really want is the solution/explanation.

6.  Find the volume of the solid in $\mathbb{R}^{3}$ defined by the equations
\[x^{2}+y^{2}\leq2\\ x+y+|z|\leq3 \]

Let us calculate the volume of the portion of the region with z>0 and then double that.

This region is bounded by the xy-plane, the cylinder with radius sqrt(2) defined by x^2+y^2=2, and the plane x+y+z=3.

By symmetry, the average height of the plane above the circle x^2+y^2=2 is simply the height at the point x=y=0.  Here the height is 3.

So the volume is pi*(r^2)*h_average=pi*(sqrt(2))^2*3=6pi

Doubling to account for the volume with z<0 we have a total volume of [b]12pi[/b]


If f(x) = (x - B)/(x - A). f(2) = 0 and f(1) is undefined, what are the values of A and B?

[hide="HINTS"]
For that function to be undefined the denominator has to be zero, so $x-A = 0$ let $x=1$, solve for $A$.
To find $B$, let $x=2$ and $f(x)=0$, then sovle for $B$ in $x-B = 0$.
[/hide]

So, basically we are plugging and chugging, right?

Not really. Use logic. For example, in the following f(x)-

$\frac{(x-B)}{(x-A)}$

where if you substitute 1 for x, the solution is undefined, and when you substitute 2, the solution is 0. Well, the only way to have something undefined is to have the denominator set to 0. Therefore since x=1, A must equal 1. To get a fraction to equal 0, the numerator must equal 0. Therefore, B=2.


In $ \triangle ABC$, we have $ \angle C \equal{} 3 \angle A$, $ a \equal{} 27$, and $ c \equal{} 48$.  What is $ b$?

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=origin, B=(14,0), C=(10,6);
draw(A--B--C--cycle);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$a$", B--C, dir(B--C)*dir(-90));
label("$b$", A--C, dir(C--A)*dir(-90));
label("$c$", A--B, dir(A--B)*dir(-90));
[/asy]

$ \textbf{(A)}\ 33 \qquad \textbf{(B)}\ 35 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 39 \qquad \textbf{(E)}\ \text{not uniquely determined}$

[hide="Solution"]
Draw $ \overline{CD}$ so that $ m\angle ACD\equal{}m\angle CAD$. Angle chasing results in the following diagram, and from Stewart's Theorem, $ (21\plus{}27)(21^2\plus{}21\cdot27)\equal{}27^2\cdot21\plus{}b^2\cdot27\implies21(48)^2\minus{}21(27)^2\equal{}27b^2$
$ \implies21(21)(75)\equal{}27b^2\implies21^2\cdot\frac{25}9\equal{}b^2\implies b\equal{}21\cdot\frac53\equal{}\boxed{\textbf{(B)}\ 35}$.
[/hide]

An alternate method?

[hide]Trisect angle C to meet AB at points D and E on line AB where D is closer to A than B. Let $\angle CAB = \angle DCA = x$. By Sine Law we know $\frac{\sin(3x)}{48}=\frac{\sin(x)}{27}$. It is easy to derive that $\sin(3x)=3\sin(x)-4\sin^3(x)$. We can solve for $\sin(x)$ to get $\sin(x)=0,\pm\frac{\sqrt{11}}{6}$. Solve for $\cos(x)$ with the equation $\sin^2(x)+\cos^2(x)$ and I think you get $\cos(x)=\pm1, \pm\frac{5}{6}$. We solve for $\angle B$ to get $180-4x$ and by Sine Law again we have $\frac{\sin(180-4x)}{b}=\frac{\sin(x)}{27}$. We know $\sin(4x)=\sin(180-4x)$ thus $\frac{\sin(4x)}{b}=\frac{\sin(x)}{27}$. You can apply the double angle identity a bunch of times to get $\sin(4x)=4 \sin(x) \cos(x) (2\cos^2(x) -1)$. We now plug in the values we have for $\sin(x)$ and $\cos(x)$. Note that obviously the values of $\sin(x)$ that are negative or 0 obviously won't work because we know this is a triangle! So $\sin(x)=\frac{\sqrt{11}}{6}$. Solving for $b$ here by plugging and chugging we get $b=35$ :).[/hide]


[quote="outback"]Prove that if $ x,y,z > 0$ then

$ xyz \plus{} 2(4 \plus{} x^2 \plus{} y^2 \plus{} z^2)\geq 5(x \plus{} y \plus{} z)$[/quote]

[hide="Hint"]Use mixing variable technique[/hide]

[quote=outback]Prove that if $ x,y,z>0$ then
$ xyz\plus{}2(4\plus{}x^2\plus{}y^2\plus{}z^2)\geq 5(x\plus{}y\plus{}z)$[/quote]
Prove that if $ x,y,z>0$ then$$ xyz+x^2+y^2+z^2+5\geq 3(x+y+z)$$

[quote=sqing]
Prove that if $ x,y,z>0$ then$$ xyz+x^2+y^2+z^2+5\geq 3(x+y+z)$$[/quote]
Since $\prod_{cyc}((x-1)(y-1))=\prod_{cyc}(x-1)^2\geq0$, we can assume $(y-1)(z-1)\geq0$.
Hence, $x(y-1)(z-1)\geq0$ or $xyz\geq xy+xz-x$.
Hence, it remains to prove that $x^2+y^2+z^2+xy+xz-x+5\geq3(x+y+z)$ or
$x^2+(y+z-4)x+y^2+z^2-3(y+z)+5$ for which it remains to prove that $(y+z-4)^2-4(y^2+z^2-3(y+z)+5)\leq0$,
which is $(y-z)^2+2(y-1)^2+2(z-1)^2\geq0$. Done!


[hide]It's just ALTIMOR arrangements, and all the letters are distinct so it's $7!=5040$.[/hide]

[hide]since b and e are already chosen for us it's just arranging altimor, and there are 7!=5040 arrangements of that[/hide]

[quote="ckck"]If the B and the E must remain in their original positions, in how many ways can the letters in the word BALTIMORE be arranged?[/quote]

[hide="Solution"]It is known that $B$ and $E$ got to stay in their positions. So, the other ones are what you can get. Therefore, teh answer is simply $7!=5040$.

[color=blue]kyyuanmathcount:  NO colors please, I can't keep editing posts, so I might just have to delete if I see much more...[/color][/hide]

[hide]w/o looking at any answers, i believe the answer is just $7!=5040$
[/hide]
-jorian

[quote="ckck"]If the B and the E must remain in their original positions, in how many ways can the letters in the word BALTIMORE be arranged?[/quote]
[hide]
since there are no repeats, the answer is $7!$ since there are 7 different letters. $7!=5040$
[/hide]

[quote="buzzer11"]How about "unitedstates"?  :)[/quote]
[hide]
there are 3 t's, 2 s's, and 2 e's
thats 12!/4!
[/hide]

[quote="buzzer11"]ckck is right.

there are twelve letters: $12!$

and since there are, as ProtestanT said, [quote]3 t's, 2 s's, and 2 e's[/quote],

the denominator should be: $3!2!2!$

and thus the fraction becomes:
$\frac{12!}{3!2!2!}$[/quote]
BTW, $3!2!2!$ is the same as $4!$
$3!2!2!=3*2*1*2*1*2*1=4*3*2*1=4!$

He is talking about the general case, cause it ain't always gonnag get that.

[quote="ckck"]Where did you get 51?[/quote]
I should be the whole number over 1.
Sorry! Will edit...

[hide]there are no same letters in altimor we simply have $7!=5040$

for mississippi:$\frac{11!}{4!*4!*2!}=34560$
This is for any case not for m and i last letters
[/hide]


[quote="andyciup"]If $a,b$ are real numbers which satisfy $a+b+\frac{1}{ab}\geq0$,prove or disprove that $a+b\geq0$
Oh,and sorry if this is really stupid..If you have a counter-example,please send me a PM so I can delete this..[/quote]
Actually, that's not true.  :wink:  Try $a=b=-\frac{1}{100}$.
I don't think it's really necessary to delete this though. It was a good try at making a problem, and this may help other people think of a good similar problem. Besides, I pretty sure it's not possible to delete your first post anyway.

No..actually I needed to know this result for solving another inequality.

$a+b \ge-frac{1}{ab}$

$1>-frac{1}{ab}$ for postivive integers


If $ z$ is a complex number satisfying $ \left|z^3 \plus{} z^{ \minus{} 3}\right|\le 2$, show that $ \left|z \plus{} z^{ \minus{} 1}\right|\le 2$ as well.

Let $ z\equal{}e^{i\theta}$.  Then
$ \left|z^{3}\plus{}z^{\minus{}3}\right|\equal{}\left|e^{3i\theta}\plus{}e^{\minus{}3i\theta}\right|\equal{}2\left|\cos{3\theta}\right|\leq 2\Rightarrow\left|\cos{3\theta}\right|\leq 1$
But the 3 in there just increases frequency-it does nothing to alter amplitude.  Hence $ \max\cos{3\theta}\equal{}\max\cos{\theta}\equal{}1$.
Now note that $ z\plus{}z^{\minus{}1}\equal{}e^{i\theta}\plus{}e^{\minus{}i\theta}\equal{}2\cos{\theta}$.  Then
$ 2\cos{\theta}\leq 2$, as desired.

Your substitution assumes that $ |z|\equal{}1$.


Certainly the following has already been discussed. Who can give references for it?
$a,b,c >0$. Prove
$\sum_{cyc} \frac1{a^2+bc}\leq \sum_{cyc} \frac1{a(b+c)}$.

I don't remember seeing this inequality anywhere, but it can be easily proven:

In the following, we will use the symbol $\sum$ for cyclic sums. Then, we have to prove the inequality $\sum\frac{1}{a\left(b+c\right)}\geq\sum\frac{1}{a^2+bc}$ for any three positive reals a, b, c.

In fact,

$\sum\frac{1}{a\left(b+c\right)}-\sum\frac{1}{a^2+bc}=\sum\left(\frac{1}{a\left(b+c\right)}-\frac{1}{a^2+bc}\right)=\sum\frac{\left(a-b\right)\left(a-c\right)}{a\left(b+c\right)\left(a^2+bc\right)}$.

Now, we WLOG assume that $a\geq b\geq c$. Then, since $a\geq b$ yields $a-b\geq 0$, we have

$a\left(b+c\right)\left(a^2+bc\right)-b\left(c+a\right)\left(b^2+ca\right)=\left(a-b\right)\left(ab^2+ba^2+a^2c+b^2c\right)\geq 0$,

so that $a\left(b+c\right)\left(a^2+bc\right)\geq b\left(c+a\right)\left(b^2+ca\right)$, and thus $\frac{1}{a\left(b+c\right)\left(a^2+bc\right)}\leq\frac{1}{b\left(c+a\right)\left(b^2+ca\right)}$; similarly, $\frac{1}{b\left(c+a\right)\left(b^2+ca\right)}\leq\frac{1}{c\left(a+b\right)\left(c^2+ab\right)}$. So altogether,

$\frac{1}{a\left(b+c\right)\left(a^2+bc\right)}\leq\frac{1}{b\left(c+a\right)\left(b^2+ca\right)}\leq\frac{1}{c\left(a+b\right)\left(c^2+ab\right)}$.

So we can apply the Vornicu-Schur inequality (Lemma 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9[/url]) with $x=\frac{1}{a\left(b+c\right)\left(a^2+bc\right)}$, $y=\frac{1}{b\left(c+a\right)\left(b^2+ca\right)}$ and $z=\frac{1}{c\left(a+b\right)\left(c^2+ab\right)}$, and obtain $\sum\frac{1}{a\left(b+c\right)\left(a^2+bc\right)}\left(a-b\right)\left(a-c\right)\geq 0$. In other words, $\sum\frac{\left(a-b\right)\left(a-c\right)}{a\left(b+c\right)\left(a^2+bc\right)}\geq 0$. According to the above, this is equivalent to $\sum\frac{1}{a\left(b+c\right)}-\sum\frac{1}{a^2+bc}\geq 0$, so that $\sum\frac{1}{a\left(b+c\right)}\geq\sum\frac{1}{a^2+bc}$, and the proof is complete.

  Darij


Hi all, my name is Olf. I just came across the forum and thought that it will be a nice idea to register. I here looking for great new
friends. I like listening to music, some kinds more than others.
That's pretty much it.

Original Science and Math, New information

http://school.discoveryeducation.com/lessonplans/programs/algebra/
http://islamicity.com/science/

You sound like Garry Denke. I suggest you  [b]make a blog[/b]. Seriously, personal stuff is not for this forum. AoPS has a fairly good blog creator here:
[url]http://www.artofproblemsolving.com/Forum/usercp.php?mode=blog&sub=create[/url]

Not realy, I listen to music alot because i feel sometimes lost. I have no faith which is natural and integral part of the human essence.Wheither we like it or not This is an indisputable fact of history.


superior algebra
superior algebra theorems

If $f: G\to G'$ is a isomorphism then $f^{-1}: G'\to G$ has the same propriety.

Since $f$ is an isomorphism $f$ is a bijection, hence is inversable $f^{-1}: G'\to G$. To complete the demonstation of that propriety $f^{-1}$ should be a morphism, right? can anyone show this please :)? Thanks !

Thank you. I'm new to this things.
I have another question concerning morphisms.

If the morphism is $f: GL(n,\mathbb{R})\to(\mathbb{R}^{*},\cdot), f(X)=\det(X)$ then $\ker f=?$

[quote="ZetaX"]$\ker f = SL(n, \mathbb{R})$ by definition.[/quote]

what's that :D?

[quote="ZetaX"]The ones with determinant $1$, the special linear group (of $n \times n$ matrices over $\mathbb{R}$).[/quote]

Okay, I didn't know what SL means  :blush: because it's propriety was obviously. Thanks!


Prove that the area of a quadrilateral $ ABCD$ with sides $ a,b,c,d$ and semiperimeter $ s$ is
\[ \left[ {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d) \minus{} abcd\cos ^2 \left( {\frac{{\angle A \plus{} \angle C}}
{2}} \right)} \right]^{1/2}
\]

Very well known: [url=http://en.wikipedia.org/wiki/Brahmagupta%27s_formula]Brahmagupta's formula[/url]

I think Bretschneider's formula.

[url=http://en.wikipedia.org/wiki/Bretschneider%27s_formula]Bretschneider's formula[/url] - Wikipedia
[url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretschneider's Formula[/url] - Mathworld

Yeah, Bretschneider's formula is an extended form of Brahmagupta's formula.

Hello Guys , for the record ,,Mathematicians still dont known is that formula belows to brahmagupta , ones say yes other say that this formula belows to Heron..But the majority thinks that this is brahmagupta's , This is the same type of Nine point circle controversy , this circle is attributed to Euler, Feuerbach , Morley , Terquen , Ponclet ,Brianchon  :P  ....

I know a lot of proofs to this formula :D  and most are ugly and cumbersome.
I was looking forward to some really cool proof ........ :wink:


Let $ A,B,C$ be three angles of a triangle, show that
\[ \frac {sinA}{sinB \plus{} sinC}(\frac {sinB}{sinC})^2(\frac {1 \minus{} cosC}{1 \plus{} cosB})^2 \plus{} \frac {sinB}{sinC \plus{} sinA}(\frac {sinC}{sinA})^2(\frac {1 \minus{} cosA}{1 \plus{} cosC})^2
\]

\[ \plus{} \frac {sinC}{sinA \plus{} sinB}(\frac {sinA}{sinB})^2(\frac {1 \minus{} cosB}{1 \plus{} cosA})^2 \leq\frac {1}{6}.
\]

I feel this ought to be posted in the Inequalities Open Problems section.

Well it isnt that tough is it? I am ending up with Schur's Inequality with a combination of some Hojoo Lee's Lemma like $ 3\minus{}t \plus{} t(abc)^{\frac{2}{t}}  \plus{}  \sum a^2 \ge 2 \sum ab$. Ya as  alchemist said, it must be in the olympiad section rather than in college playground.


inequalities
integration
algebra solved
algebra

Don't scare it although it seems touch.

Hint: If x+y=1, x>0, y>0 then (1-x^m)^n + (1-y^n)^m>1.

Just one idea and we are done. Come one!

Namdung

Yes, one idea, but it's not a trivial idea anyway! Remember that the hard inequality based on an identity was also based on one idea, as well as the super-inequality!

I dont know for whom it not trivial, but for romanian members this idea is no newer than 2+2=4. Romanians learn much about calculus.

Namdung

Nobody wants to help me to finish this problem. So I have to do:

This inequality is equivalent to

 \int (0..1) (1-x^n)^m dx +  \int (0..1) (1-x^m)^n dx >=1

Change variable, we have  \int (0..1) (1-x^m)^n dx =  \int (0..1) (1-(1-x)^m)^n dx, so we have to show that
 \int (0..1) [(1-x^n)^m+(1-(1-x)^m)^n]dx >=1

But  (1-x^n)m + (1-(1-x)^m)^n >=1 (you can find more general ineq in section solved algebra problem - solved by Pbornzstein) and we are done.


function
algebra
functional equation
algebra proposed

Find all functions $f: \mathbb N \to \mathbb N$ such that \[ f(f(m) \plus{} f(n)) \equal{} m \plus{} n \plus{} 3\] for all positive integers $m$ and $n$.

[quote="theSA"]Find all functions  $ f: N \to N$  such that  $ f(f(m) \plus{} f(n)) \equal{} m \plus{} n \plus{} 3$.[/quote]

The functional equation shows that $ f(n)$ is injective.

Then, $ f(f(m \plus{} 1) \plus{} f(n)) \equal{} m \plus{} n \plus{} 4 \equal{} f(f(m) \plus{} f(n \plus{} 1))$ and so, since $ f$ is injective : $ f(m \plus{} 1) \plus{} f(n) \equal{} f(m) \plus{} f(n \plus{} 1)$, which may be written :
$ f(m \plus{} 1) \minus{} f(m) \equal{} f(n \plus{} 1) \minus{} f(n)$ $ \forall m,n$

So $ f(n \plus{} 1) \minus{} f(n) \equal{} a$, constant and so $ f(n) \equal{} an \plus{} b$

Plugging this back in original equation, we get $ a \equal{} b \equal{} 1$ and so $ \boxed{f(n) \equal{} n \plus{} 1}$
(there is another solution $ a \equal{} \minus{} 1$ and $ b \equal{} \minus{} 3$, and so $ f(n) \equal{} \minus{} n \minus{} 3$, but not from $ \mathbb N\to\mathbb N$)


We wrote down all the integers of 1 to 7 digits we can, using only the two digits 0 and 1. How many 1's did we write?

I am gonna try it and will estimate it, so you mean all 1s from 1 to 1,111,111 - that's quite a lot of ones. Since I'm too lazy to write down all integers and count them, here's my guess: 95

How close am I to the right result?

Note that for every group of 10 numbers, there exists a 1. For every group of 100 numbers there exist 10 1's in the hundreds digit. For every group of 1000 numbers there exist 100 1's in the hundreds place, and so on until every group of 10,000,000 numbers.

[quote="saszs"]I don't think that was a problem but maybe it was.[/quote]

The test is different for different grade levels.

Observe first that, as we're looking for a number of 1s written down, nothing changes if we write 0s in front of all numbers so that each number has exactly 7 written digits. This way, we have written everything from 0000000 to 1111111, a total of 2^7 numbers, each with seven digits, a total of 7 x 2^7 digits.

Fine. But how many of them are 1s? Well, exactly half of this number! Take an arbitrary digit position, fourth for example: then for every number, say abc 1 def, we have that abc 0 def appears in the list as well (and vice versa). Thus, 1s appear in fourth position for exactly half of the written numbers, that is 1s appear 2^6 times in fourth position. Situation is same for each position, and so a total number of 1s written down is 7 x 2^6 = 448.


Let a,b,c be real numbers. Prove that

(1 +(cos 2 a+cos 2 b)/2)(1 +(cos 2 b+cos 2 c)/2)(1 +(cos 2 c+cos 2 a)/2) \geq 8sin^4(a/2)sin^4(b/2)sin^4(c/2)

We will first prove that:

 1+(cos 2 a+ cos2b)/2 \geq  sin 4 (a/2)+sin 4 (b/2). Because cos a=1-2sin 2 (a/2), if we denote x=sin 2 (a/2),  y=sin 2 (b/2), we have to prove that 2+(1-2x) 2 +(1-2y) 2  \geq 2(x 2 +y 2 )  <=> 2[(x-1) 2 +(y-1) 2] \geq 0, which is true.

Back to our problem, if we multiply the above inequality with the analogous ones, and consider the fact that (x+y)(y+z)(z+x) \geq 8xyz, we obtain the desired inequality.

An even easier proof is the following: first prove that sin^4a/2<=(1+cos^2a)/2 and then use the inequality (x+y)(y+z)(z+x)>=8xyz.


If a unit circle of radius $ \sqrt{65}$ is place on a square dot paper so that its center is on  a dot, through how many dots does the circle pass?  Assume the dots are one unit apart, both horizontally and vertically.

[hide="Solution"] In other words, how many pairs of integers $ x, y$ satisfy $ x^2 \plus{} y^2 \equal{} 65$?  By observation,

$ 65 \equal{} 8^2 \plus{} 1^2 \equal{} 7^2 \plus{} 4^2$

and it's not hard to show that no other such pairs are possible.  This gives a total of $ \boxed{16}$ points, given explicitly by

$ (\pm 8, \pm 1)$
$ (\pm 1, \pm 8)$
$ (\pm 7, \pm 4)$
$ (\pm 4, \pm 7)$. [/hide]


function
abstract algebra
modular arithmetic
absolute value
algebra

Find the maximal possible finite number of roots of the equation  $|x-a_1|+\dots+|x-a_{50}|=|x-b_1|+\dots+|x-b_{50}|$, where $a_1,\,a_2,\,\dots,a_{50},\,b_1,\dots,\,b_{50}$ are distinct reals.

Answer: 49.

Firstly we will show that we can't have more than 49 roots. The points $a_i$ and $b_j$ divide the real axis to the intervals in such a way that the function
\[ f(x)=\sum_{i=1}^{50}{|x-a_i|}-\sum_{i=1}^{50}{|x-b_j|}  \]
is linear on each of the intervals (and hence it has at most one root on each interval). Futhermore, these linear functions connect with each other (since the function $f(x)$ is continuous) and the difference of the angular coefficients of the lines from adjacent intervals is equal to 2 (since in every point of the set $\{a_1,\ldots,a_{50},b_1,\ldots,b_{50}\}$ exactly one of the expression under module change their sign). Moreover, observe that on the very left and on the very right intervals (in fact they are rays but not intervals) our function is constant and $f(+\infty)=-f(-\infty)$.

Let $k$ an angular coefficient. Clearly, we may have roots only on the intervals where $k \neq 0$. Obviously, if we have a root on some interval then we may have the next root only on the interval where $k$ has another sign. Therefore, we may not have roots on the adjancent intervals. Altogether we have 101 intervals, but we may ignore the very left and the very right intervals (they coudn't have roots). So we have 99 intervals. Since there are no two adjacent intervals having roots, we have at most 50 roots. But however, if there were exactly 50 roots, the angular coefficients on the second from the right and second from the left intervals would have had different signs, and both of the intervals would have had roots. This is a contradiction, since as we have already observed, $f(-\infty)+f(+\infty)=0$.

So, the number of roots is at most 49. And we have to give an example of equation such that it has precisely 49 roots.

Assume $a_i=2i-r_2(i)$ for $i=1,\ldots,49$; $b_j=2j-(1-r_2(j))$ for $j=1,\ldots,50$; $a_{50}=100-\varepsilon$; where $r_2(x) \in \{0,1\}$ and $r_2(x) \equiv x \pmod{2}$; and finally we may assume $\varepsilon=1/2$.

Obviously,
$a_1<b_1<b_2<a_2<a_3<b_3<b_4<\ldots<a_{49}<b_{49}<b_{50}<a_{50}$. We have $f(x)=-1/2$ on the intervals $(-\infty,a_1)$, $(a_2,a_3)$, $(a_4,a_5)$, ..., $(a_{48},a_{49})$; and we have $f(x)=3/2$ on the intervals $(b_1,b_2)$, $(b_3,b_4)$, ..., $(b_{49},b_{50})$; and finally we have $f(x)=1/2$ on the interval $(a_{50},+\infty)$. So, since the function $f(x)$ is continuous, it has a root on each of 49 intervals $(a_1,b_1)$, $(b_2,a_3)$, ..., $(a_{49},b_{49})$. So, we are done.

P.S. I am sorry about my poor English.

[quote="Remike"]

Let $k$ an angular coefficient. Clearly, we may have roots only on the intervals where $k \neq 0$. Obviously, if we have a root on some interval then we may have the next root only on the interval where $k$ has another sign. Therefore, we may not have roots on the adjancent intervals. Altogether we have 101 intervals, but we may ignore the very left and the very right intervals (they coudn't have roots). So we have 99 intervals. Since there are no two adjacent intervals having roots, we have at most 50 roots. But however, if there were exactly 50 roots, the angular coefficients on the second from the right and second from the left intervals would have had different signs, and both of the intervals would have had roots. This is a contradiction, since as we have already observed, $f(-\infty)+f(+\infty)=0$.
[/quote]

I may be missing something, but I see no contradiction in here. Could you explain it better?

[quote="richard"]I may be missing something, but I see no contradiction in here. Could you explain it better?[/quote]

I suppose it's rather clear...

As I mentioned, the angular coefficients on the second from the right and second from the left intervals would have had different signs, and both of the intervals would have had roots. Therefore the signs of $f(x)$ on the left and on the right intervals are the same. But on the other hand $f(-\infty)+f(+\infty)=0$. This is an obvious contradiction.

Exuse me, I don't know how to explain it more understandable.

[quote="tkhtn"]This problem is easy and with me, only hard at give a example,.[/quote]

I doubted the problem was easy when i participated the olimpiad :)
I've got only 2 points of 7 for this problem  :blush:   :rotfl:


limit
number theory proposed
number theory

This problem I have post but no one answer so I has post again in one topic.
$ f: \ \mathbb{N}\to\mathbb{N}$
$ \lim_{ n\to\infty}\frac {n}{f(n)} < \infty$
For each a odd prime p and a natural x,Prove that there finite n such that
$ p^{f(n)}|x^{n} \minus{} 1$

Something's wrong as $ f(n)\equal{}0$ for all $ n$ shows. Do you want that $ \frac {f(n)}n$ tends to $ \infty$¿

No $ f: N\to N$ it mean that $ f(n) > 0$ :wink:
You can solve that $ f(n)$ is a polynomial.For example $ f(n) \equal{} n \minus{} 1$

Even $ f(n)\equal{}1$ gives counterexamples... As any constant $ f$ does...

[quote="ZetaX"]Even $ f(n) \equal{} 1$ gives counterexamples... As any constant $ f$ does...[/quote]
I am sorry
$ \frac {n}{f(n)} < \infty$


[url]http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm[/url] Just get 2006 on [url]http://www.mathcounts.org[/url]

Or you can PM chess64, and he can hook you up with about 5 years worth of nats tests.

[quote="dragongrl6"]does anyone know a website where I can find previous mathcounts tests from around 2000 to 2006?  :)[/quote]

I thought you werent allowed to publically distribute those test since mathcounts was still selling them?

[quote="anirudh"]Fair use...only for personal purposes...all that good stuff[/quote]
Well look at what happened to the 4shareds :|

[quote="jhredsox"][quote="13375P34K43V312"]Or you can purchase the entire books which have the national tests as well, directly from the mathcounts site.[/quote]

just wondering, but where exactly is this? i'm wondering because they always send me to a pdf file and i don't remember it being there. maybe someone could give a link.
thanx

-jorian[/quote]
Those books are useless, the states and counties are easy to get and it isn't worth 20 bucks for one nationals round. That's just my opinion though...

jhredsox, aren't you still in contact with Mr. Frost anyway?

Umm, well I don't think he likes me very much. I don't really know. I have nothing against him... I just don't want to ask for too much (I believe I already have, I've gotten like 6 states). He's a really nice person.

-jorian

[quote="jhredsox"][quote="13375P34K43V312"]Or you can purchase the entire books which have the national tests as well, directly from the mathcounts site.[/quote]

just wondering, but where exactly is this? i'm wondering because they always send me to a pdf file and i don't remember it being there. maybe someone could give a link.
thanx

-jorian[/quote]
[url]
http://www.mathcounts.org/webarticles/anmviewer.asp?a=183&z=6[/url] Click on Page 7 for the Tests. But I agree with Bpms a nat isnt worth 20 bucks plus they dont include a Masters round.

yah, it's probably not worth it seeing as national tests have a totally different concept than state rounds, which is where i must excel

-jorian


The following are the measures characterizing the distributions of the scores of two parallel English classes :

------------------------- Class 1 ------ Class 2
Mean : ------------------- 78 ------------ 72
Median : ----------------- 65 ------------ 63
Standard Deviation : -- 16 ------------ 6

In which of the two classes should the teacher invest more time in individual tutoring?

Does she want to invest time with the people who are good or those who are bad?

obviously the teacher would invest time with those students who are poor in english.

What's the criterion for deciding which class needs more help?

Class 1 has a lower Pearson skewness coefficient than Class 2, presumably, though not necessarily, indicating that Class 1's scores are more normally distributed. The mean scores for both classes were above their respective medians -- indicating that each class is right-skewed; each class has positive outliers; i.e. overachievers in it.

[quote="mysmartmouth"]Perhaps perform a two sample t-test?  Is this an AP Statistics problem?[/quote]

A two sample t-test is not necessarily appropriate here, since we were not told the number of students in either class. One could conceivably perform a z-test, assuming there were sufficient numbers of students in each class to make the outcome a viable statistic. I ran the numbers for a z-test, assuming 30 students in each class. One would reject the null hypothesis, $ \mu_1 \equal{}\mu_2$, at the $ \alpha \equal{} 0.05$ level for a two-tailed test. In other words, the difference in mean score between the two classes is statistically significant.


geometry
incenter
angle bisector
geometry proposed
U sviti matematyky

Let $ AA_1$ and $ CC_1$ be angle bisectors of triangle $ ABC$ ($ A_1\in BC,$ $ C_1\in AB$). Straight line $ A_1C_1$ intersects ray $ AC$ at point $ D.$ Prove that angle $ ABD$ is obtuse.

Call I the incenter and $ B_1 : BI \cap CA$. Then D is the harmonic cojugate of $ B_1$ w.r.t. AC and then is the point where the external angle bisector from B meet AC. Then $ \angle ABD \equal{} 90 \plus{} \frac{\beta}{2} > 90$


MATHCOUNTS
number theory
AMC 10
\/closed

I have read half of Intro to Algebra, all of Intro to Number Theory and Most of Counting and Probability. All found most of the problem in these books easy to slightly hard when it comes to difficulty. Which Mathcounts course would be for me?

Why didn't you finish Intro Algebra first?

Anyways, if you decide to take a class, Advanced definitely. Since the first half of Intro Algebra pretty much covers, in my opinion, all of the Mathcounts algebra concepts, and you've done NT and most of C&P, I would probably skip the Mathcounts class and do something more difficult, like Intro to Geo. Maybe the AMC 10 class. AMC 10 seems to be nationals level to me.

If you understood all the material in the books you read, Advanced for sure. You probably don't even need the class. Have you taken some MATHCOUNTS tests yet? If so, what are your chapter and state scores?

Hmm yeah if you found everything to be easy to slightly hard..you should probably just get intro to geo.  You can also take the amc 10 class, which is more challenging and would be good for the harder problems on mathcounts. Going through past tests and doing problems is the most important thing.

@cheesyicecream:  It sounds like she's talking about the second half.. or maybe that's how I perceived it?

I think you'd probably be ready for advanced.

[quote="monkeygirl13"]Ok I think I will take advanced.[/quote]
Or you could always start with basic and drop out if it's too easy.

or you could start with advanced, attend one class...if it's too hard, then drop out and attend basics :)
[size=9]That's what I did[/size]


hello   :D
  I'm looking for some help to solve this problem;

f is a function defined on all reals in the interval [0, 1] and satisfies f(0) = 0, f(x/3) = f(x)/2, f(1 - x) = 1 - f(x). Find f(18/1991).
 the whole the solution or any hints will be useful

There should also be the condition that $f$ is non-decreasing.

Here's a hint: Write $x$ in base 3, and write $f(x)$ in base 2.

BTW, this function is known as the Cantor Function:
[url]http://mathworld.wolfram.com/CantorFunction.html[/url]
(don't look this up if you don't want to see the answer!)


How do you save class transcripts without making them look all messed up? Whenever I save a class transcript, it looks weird and messy. I would like to save my transcripts for my class but I can't figure out how. Can someone please help me?

Thanks  :)

Well how do you save them? Do you copy it onto a Word Doc or something?

I just save the HTML page entirely. When you view a transcript, you can see a print view. Just save the entire webpage.

I don't like that though, is there a way to make it a PDF directly?
Also, when you open up a class transcript, does it look the exact same as a math jam transcript? (When you just look at it online, not after you save it)?

idk, it seems more cluttered and I just like PDF's better.

Well I print the transcript as a pdf using cute pdf printer.
[url=http://www.cutepdf.com/products/cutepdf/Writer.asp]here is a link to the website.[/url]

So do you just copy+paste the whole thing to a word document, then convert it to a PDF?

[quote="monkeygirl13"]I highlight the transcript and then leftclick and the click print.[/quote]

Just in case you didn't get it cutepdf acts like a "printer" just like microsoft onenote so you click print to cutepdf.

When I try copying a transcript onto Microsoft Word, it copies fine, but there is no color and no spaces between each separate speaker. I know that other people don't have this problem, so I was wondering if there was a way to convert the transcript to PDF form without copying it to a Word document first?

[quote]I highlight the transcript and then leftclick and the click print.[/quote]

That's how you make it a PDF without even copying and pasting to word  :lol:

is there a way you can save the link, and then go back to it, even when the transcripts aren't on the "My Classes" page anymore?

EDIT: Never mind got it.

The HTML document will stop working when the links are removed, correct?

you have to just do control s and save it? It saved fine on my comp..

I wanted to make it into a PDF and my question was already answered. Please read the posts above yours. Thanks.

Sorry to bring up the topic again, but is there a way we can save it, without a loss in quality?  PDF LaTeX graphics look horrible.

[quote="AIME15"][quote="AIME15USAMO"]I don't like that though, is there a way to make it a PDF directly? [/quote]

What's wrong with a HTML or MHTML...they save the images nicely and not blurry like a PDF.[/quote]

:)

[quote="AIME15"]What's wrong with a HTML or MHTML...they save the images nicely and not blurry like a PDF.[/quote]

How do you do this on a Mac?

I have the Firefox add-on [url=https://addons.mozilla.org/en-US/firefox/addon/unmht/]UnMHT[/url] which allows you to save and open MHT files in Firefox. If you have the add-on installed, then the menu that appears upon right-clicking on a page will have the option "Save as MHT."

I'm trying to save one of my class transcripts as a .HTML but the Latex doesn't show up, and I don't know how to make it show.
I know [i]why[/i], though. When you save the transcripts to your computer, a new folder is created with the images and Latex, but for some reason my computer is only saving the images and not the Latex.

I recommend saving as a PDF.  There's too much going on in some of the transcripts in terms of other files that are needed to render a transcript properly to trust that other methods will properly save the transcript.

Do: File Print (or [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=90347]CMD[/url]/CTRL P), then at the bottom left of the pop up box, select 'OPEN AS PDF'.  Then save the PDF.


Let:$ a;b;c>0$.Find k maximum to satisfy:
$ (a\plus{}b\plus{}c)(a\plus{}b)(a\plus{}c) \ge kabc$

[quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find k maximum to satisfy:
$ (a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c) \ge kabc$[/quote]

I find $ k_{max}\equal{}12$,is it ?

hello, i have found $ k_{max}\equal{}\frac{1}{2}(11\plus{}5\sqrt{5})$ which is attained by $ a\equal{}\frac{1}{2}(\sqrt{5}\minus{}1),b\equal{}1,c\equal{}1$.
Sonnhard.

[quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find k maximum to satisfy:
$ (a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c) \ge kabc$[/quote]
Dr Sonnhard Graubner, I agree with you. :)
$ (a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c) \ge kabc\Leftrightarrow$
$ \Leftrightarrow a^3\plus{}2(b\plus{}c)a^2\plus{}(b^2\minus{}(k\minus{}3)bc\plus{}c^2)a\plus{}(b\plus{}c)bc\geq0.$
But $ a^3\plus{}2(b\plus{}c)a^2\plus{}(b^2\minus{}(k\minus{}3)bc\plus{}c^2)a\plus{}(b\plus{}c)bc\geq$
$ \geq a^3\plus{}4\sqrt{bc}a^2\minus{}(k\minus{}5)bca\plus{}2\sqrt{b^3c^3}\equal{}\sqrt{b^3c^3}(t^3\plus{}4t^2\minus{}(k\minus{}5)t\plus{}2),$ where $ t\equal{}\frac{a}{\sqrt{bc}}.$
We'll find the maximal $ k$ for which $ t^3\plus{}4t^2\minus{}(k\minus{}5)t\plus{}2\geq0$ holds for all $ t>0.$
But $ t^3\plus{}4t^2\minus{}(k\minus{}5)t\plus{}2\geq0\Leftrightarrow k\leq t^2\plus{}4t\plus{}5\plus{}\frac{2}{t}.$
Thus, $ \max k\equal{}\min_{t>0}\left(t^2\plus{}4t\plus{}5\plus{}\frac{2}{t}\right).$
Let $ f(t)\equal{}t^2\plus{}4t\plus{}5\plus{}\frac{2}{t}.$
Hence, $ f'(t)\equal{}\frac{2(t\plus{}1)(t^2\plus{}t\minus{}1)}{t^2},$ which gives $ t_{min}\equal{}\frac{\sqrt5\minus{}1}{2}$ and
$ \max k\equal{}f\left(\frac{\sqrt5\minus{}1}{2}\right)\equal{}\frac{1}{2}\left(11\plus{}5\sqrt5\right).$ Done! :)

[quote="arqady"][quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find k maximum to satisfy:
$ (a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c) \ge kabc$[/quote]
Dr Sonnhard Graubner, I agree with you. :)
$ (a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c) \ge kabc\Leftrightarrow$
$ \Leftrightarrow a^3 \plus{} 2(b \plus{} c)a^2 \plus{} (b^2 \minus{} (k \minus{} 3)bc \plus{} c^2)a \plus{} (b \plus{} c)bc\geq0.$
But $ a^3 \plus{} 2(b \plus{} c)a^2 \plus{} (b^2 \minus{} (k \minus{} 3)bc \plus{} c^2)a \plus{} (b \plus{} c)bc\geq$
$ \geq a^3 \plus{} 4\sqrt {bc}a^2 \minus{} (k \minus{} 5)bca \plus{} 2\sqrt {b^3c^3} \equal{} \sqrt {b^3c^3}(t^3 \plus{} 4t^2 \minus{} (k \minus{} 5)t \plus{} 2),$ where $ t \equal{} \frac {a}{\sqrt {bc}}.$
We'll find the maximal $ k$ for which $ t^3 \plus{} 4t^2 \minus{} (k \minus{} 5)t \plus{} 2\geq0$ holds for all $ t > 0.$
But $ t^3 \plus{} 4t^2 \minus{} (k \minus{} 5)t \plus{} 2\geq0\Leftrightarrow k\leq t^2 \plus{} 4t \plus{} 5 \plus{} \frac {2}{t}.$
Thus, $ \max k \equal{} \min_{t > 0}\left(t^2 \plus{} 4t \plus{} 5 \plus{} \frac {2}{t}\right).$
Let $ f(t) \equal{} t^2 \plus{} 4t \plus{} 5 \plus{} \frac {2}{t}.$
Hence, $ f'(t) \equal{} \frac {2(t \plus{} 1)(t^2 \plus{} t \minus{} 1)}{t^2},$ which gives $ t_{min} \equal{} \frac {\sqrt5 \minus{} 1}{2}$ and
$ \max k \equal{} f\left(\frac {\sqrt5 \minus{} 1}{2}\right) \equal{} \frac {1}{2}\left(11 \plus{} 5\sqrt5\right).$ Done! :)[/quote]
I will check where I am wrong.Thank you arqady.


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

PLease help me to prove the following statments. I have searched the topic similar to this one but get nothing. Thanks in advance.

Let H be a nonempty subset of a group G. The set N(H)={a $\in G | aHa^{-1}=H$} is called normalizer of H in G.
(i)  Prove that  N(H) is a subgroup of G.
           Suppose  H is a subgroup of G.
(ii) Prove that H is normal in G iff N(H)=G.
(iii)  Prove that H is normal in N(H).
(iv)  Prove that N(H) is the largest subgroup of G in which  H is  normal, i.e., if H is normal in a subgroup K of G , then K is a subset of N(H).
Is the last one is straight forward from the (ii) and (iii)?

All of this is not very hard.
Please type your definition of normal group, and of subgroup.  You should solve these problems almost immediately then.

Today i have thought a lot about it and here what i got:

 [hide="  For (i):"]
I see that e $\in N(H)$. But How to show that the inverse of the element belongs to N(H)?
If $a\in N(H)$ then  $\forall h' \in H \text{ we have }aha^{-1}=h' \implies a^{-1}h'a=h \Rightarrow a^{-1}Ha=H$ and therefore  $a^{-1}\in N(H)$ too. Now let $a,b \in N(H)$. Then $abHb^{-1}a^{-1}=aHa^{-1}=H$ that's why $ab \in N(H)$ and hence N(H) is a group.
Is that correct? :lol: [/hide]
[hide="For (ii):"]
[color=blue]Theorem: Let H be a subgroup of a group G.Then H is a normal in G iff $aHa^{-1}$ is a subset of H for all $a \in G$.[/color]

Let H be a normal subgroup ,then $aHa^{-1}$ is a subset of H for all $a \in G$. Since  $|aHa^{-1}|=|H|$ we have $aHa^{-1}=H$. (this result i have got today!) and we get that  N(H)=G.
Conversely, let  N(H)=G. Then  $aHa^{-1}=H$ for all $a \in G$ and that's why $aHa^{-1}$ is a subset of H for all $a \in G$ . Hence H is normal.
 :lol: [/hide]
[hide="For (iii):"]
Using the definition of normalizer we get that $aHa^{-1}$ is a subset of H for all $a \in N(H)$ and by the theorem the result follows. :oops: [/hide]
[hide="For (iv):"] Let H be a normal subgroup in a subset K of G ,then $aHa^{-1}$ is a subset of H for all $a \in K$. Since  $|aHa^{-1}|=|H|$ we have $aHa^{-1}=H$ for all $a \in K$. Therefore Kis a subset of N(H). :oops:
[/hide]
When i first saw this problem i did not realized that we can extend the theorem above given in my book. That was why i couldn't solve it. I am studying Abstract Algebra in summer without anybody's help eccept Mathlinkers, thank you all a lot!

Hmm,
some corrections or comments

you are frequently using the size |H| of a group.  What makes you sure they are finite.
$(\mathbb{Z},+)$ is an infinite subgroup of $(\mathbb{Q},+)$
I can assure you they have the same "size" but obviously the latter is strictly bigger!

I would advise you to prove this little lemma first, it will HELP you  :wink:

The following statements about a subgroup $H$ in $G$ are equivalent :

1.  $H$ is a normal subgroup of $G$
2.  $a^{-1}H a=H \ \forall a\in G$  (you may find it interesting to know that $a^{-1}H a$ is often written as $H^{a}$)
3. $a^{-1}H a\subseteq H \ \forall a\in G$
4.  $\forall a\in G: a H = H a$

[quote="fredbel6"]...you are frequently using the size |H| of a group.  What makes you sure they are finite.....I can assure you they have the same "size" but obviously the latter is strictly bigger![/quote]

If i change the word the same size to  equipolent, will the rest be correct?

Hm, what exactly would you mean by equipollent?

Is there one-one fuction from $a^{-1}Ha$ onto H?

You mean a bijection

Yes, just map every $h\in H$ onto $h^{a}=a^{-1}h a$
But that will NOT be a correct proof.  you cannot say one is in the other and there is a bijection :thus the sets are equal.
Just check my example with $\mathbb{Q}$ I gave up here.

Then how can i prove the $H^{a}=H$ iff H is normal?
Z and Q are equipolent. Sorry I think i cant get your question sorry :(

Two comments :

1. what is your definition of normal subgroup?? That would really clear things up!  :o

2. Once we have that sorted out, you should try this again :
[quote] The following statements about a subgroup $H$ in $G$ are equivalent :

1.  $H$ is a normal subgroup of $G$
2.  $a^{-1}H a=H \ \forall a\in G$  (you may find it interesting to know that $a^{-1}H a$ is often written as $H^{a}$)
3. $a^{-1}H a\subseteq H \ \forall a\in G$
4.  $\forall a\in G: a H = H a$[/quote]

My defenition:
  $H$ is a normal subgroup of $G$ if $\forall a\in G: a H = H a$.
Then there is a theorem:
$a^{-1}H a\subseteq H \ \forall a\in G$ iff $H$ is a normal subgroup of $G$.
And I haven't seen  $a^{-1}H a=H \ \forall a\in G$ ,that's what i am trying to prove  proving incorrectly the |H|=|H^a|

Assume that :

$H^{x}\subseteq H \forall x \in G$

Well then we also have :

$H^{a^{-1}}\subseteq H\forall a\in G$

now if $H^{a^{-1}}\subseteq H$
then $(H^{a^{-1}})^{a}\subseteq (H)^{a}$

or    $H \subseteq (H)^{a}$ as well

Oh, thanx a lot. It was easy again .  :blush:  :ewpu: