geometry
incenter
trigonometry
ratio
circumcircle
cyclic quadrilateral
geometry proposed

Let  $ABCD$ bicentric quadrilateral, $I$ is the incenter, $X,Y,Z,W$ are the touch points of the incircle $(I)$ and $AB,BC,CD,DA$. Further $E,F$ are the pedal points of the perpendiculars from $I$ to $BD,AC$. Prove that $[XEY]=[WFX]$.

Let $P$ be the common diagonal intersection of the bicentric quadrilateral $ABCD$ and its contact cyclic quadrilateral $XYZW$. The pentagons $AXIFW, BYIEX$ are both cyclic, because the angles $\angle AXI = \angle AFI = \angle AWI = 90^o$ spanning a circle with the diameter $AI$ are right and the angles $\angle BYI = \angle BEI = \angle BXI = 90^o$ spanning a circle with the diameter $BI$ are also right. Hence,

$\angle FXW = \angle FAW \equiv \angle PAD,\ \ \angle FWX = \angle FAX \equiv \angle PAB$

$\angle XFW = 180^o - (\angle FXW + \angle FWX) = 180^o - (\angle PAD + \angle PAB) =$
$= 180^o - \angle A = \angle C$

$\angle EYX = \angle EBX \equiv \angle PBA,\ \ \angle EXY = \angle EBY \equiv \angle PBC$

$\angle YEX = 180^o - (\angle EYX + \angle EXY) = 180^o - (\angle PBA + \angle PBC) =$
$= 180^o - \angle B = \angle D$

where we used the fact that the quadrilateral $ABCD$ is cyclic and $\angle A + \angle C = \angle B + \angle D = 180^o$. Let $r = IX = IY = ...$ be the quadrilateral inradius. From the isosceles triangles $\triangle WXI, \triangle XYI$, the segments $WX, XY$ are equal to

$WX = 2 IX \cos{\widehat{IXW}} = 2r \cos{\frac{\widehat A}{2}}$

$XY = 2 IY \cos{\widehat{IYX}} = 2r \cos{\frac{\widehat B}{2}}$

Using the sine theorem for the triangle $\triangle FWX, \triangle EXY$, their remaining sides $FW, FX, EX, EY$ are

$FW = WX\ \frac{\sin{\widehat{FXW}}}{\sin{\widehat{XFW}}} = 2r \cos{\frac{\widehat A}{2}}\ \frac{\sin{\widehat{PAD}}}{\sin{\widehat C}}$

$FX = WX\ \frac{\sin{\widehat{FWX}}}{\sin{\widehat{XFW}}} = 2r \cos{\frac{\widehat A}{2}}\ \frac{\sin{\widehat{PAB}}}{\sin{\widehat C}}$

$EX = XY\ \frac{\sin{\widehat{EYX}}}{\sin{\widehat{YEX}}} = 2r \cos{\frac{\widehat B}{2}}\ \frac{\sin{\widehat{PBA}}}{\sin{\widehat D}}$

$FW = XY\ \frac{\sin{\widehat{EXY}}}{\sin{\widehat{YEX}}} = 2r \cos{\frac{\widehat B}{2}}\ \frac{\sin{\widehat{PBC}}}{\sin{\widehat D}}$

The areas of the triangles $\triangle FWX, \triangle EXY$ are then

$|\triangle FWX| = \frac 1 2\ FW \cdot FX \sin{\widehat C} = 2r^2 \cos^2{\frac{\widehat A}{2}}\ \frac{\sin{\widehat{PAD}} \sin{\widehat{PAB}}}{\sin{\widehat C}} = r^2\ \frac{\sin{\widehat{PAD}} \sin{\widehat{PAB}}}{ \tan{\frac{\widehat A}{2}}}$

$|\triangle EXY| = \frac 1 2\ EX \cdot EY \sin{\widehat D} = 2r^2 \cos^2{\frac{\widehat B}{2}}\ \frac{\sin{\widehat{PBA}} \sin{\widehat{PBC}}}{\sin{\widehat D}} = r^2\ \frac{\sin{\widehat{PBA}} \sin{\widehat{PBC}}}{\tan{\frac{\widehat B}{2}}}$

where we again used the fact that the quadrilateral $ABCD$ is cyclic, $\angle C = 180^o - \angle A, \angle D = 180^o - \angle B$ and

$\sin{\widehat C} = \sin{\widehat A} = 2 \sin{\frac{\widehat A}{2}} \cos{\frac{\widehat A}{2}},\ \ \sin{\widehat D} = \sin{\widehat B} = 2 \sin{\frac{\widehat B}{2}} \cos{\frac{\widehat B}{2}}$

The ratio of areas of these 2 triangles is then

$\frac{|\triangle FWX|}{|\triangle EXY|} = \frac{\sin{\widehat{PAD}} \sin{\widehat{PAB}}}{\sin{\widehat{PBA}} \sin{\widehat{PBC}}} \cdot \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat A}{2}}} = \frac{\sin{\widehat{PAB}}}{\sin{\widehat{PBA}}} \cdot \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat A}{2}}}$

where we used equality of the angles $\angle PAD \equiv CAD = \angle DBC \equiv \angle PBC$ spanning the same arc $CD$ of the quadrilateral circumcircle $(O)$. To complete the proof, we have to show that

$(?)\ \ \ \frac{\sin{\widehat{PAB}}}{\sin{\widehat{PBA}}} = \frac{\tan{\frac{\widehat A}{2}}}{\tan{\frac{\widehat B}{2}}}$

because then the ratio of areas of the 2 triangles will be equal to 1. Obviously,

$\frac{BX}{AX} = \frac{r}{\tan{\frac{\widehat B}{2}}} \cdot \frac{\tan{\frac{\widehat A}{2}}}{r} = \frac{\tan{\frac{\widehat A}{2}}}{\tan{\frac{\widehat B}{2}}}$

On the other hand, using the sine theorem for the triangle $\triangle ABP$, we get

$\frac{\sin{\widehat{PAB}}}{\sin{\widehat{PBA}}} = \frac{BP}{AP} = \frac{BX}{AX}$

The last equality is due to the fact that the diagonals of the contact quadrilateral of a bicentric quadrilateral are perpendicular to each other, i.e., $XZ \perp YW$ and that they bisect the angles formed by the diagonals of the bicentric quadrilateral. For a proof, see the problem   [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23192]Bicentric quadrilateral[/url], steps 1, 2, and 3. Consequently, the line $PX$ bisects the angle $\angle APB$ and divides the opposite side $AB$ of the triangle $\triangle ABP$ in the ratio of the adjacent sides $AP, BP$. As a result

$\frac{|\triangle FWX|}{|\triangle EXY|} = \frac{BX}{AX} \cdot \frac{AX}{BX} = 1$

From cyclic $IEXBY$ and $IFXAW,$ we get $\widehat{EXY}=\widehat{EBY}=\widehat{FAW}=\widehat{FXW}$ $\Longrightarrow$ $\widehat{FXY}=\widehat{EXW}.$ Moreover $\widehat{BEX}=\widehat{BYX}$ and $\widehat{DEW}=\widehat{DZW}$ $\Longrightarrow$ $\widehat{BEX}+\widehat{DEW}=90^{\circ}-\tfrac{1}{2}\widehat{ABC}+90^{\circ}-\tfrac{1}{2}\widehat{ADC}=90^{\circ}$ $\Longrightarrow$ $\widehat{XEW}=90^{\circ}$ and similarly $\widehat{XFY}=90^{\circ}.$ Therefore, the right $\triangle XFY$ and ${\triangle  XEW}$ are similar $\tfrac{XF}{XE}=\tfrac{XY}{XW},$ i.e. $XF \cdot XW=XE \cdot XY$ and together with $\widehat{EXY}=\widehat{FXW},$ it follows that $[XEY]=[WFX].$

Dear Mathlinkers,
just to remember that the beloved Juan Carlos Salazar of Venezuela was a great contributor to this site.
He died in South Africa the 30 Marz 2008.
Sincerely
Jean-Louis


It's obviously that ab + bc + ca <=1

So 5(ab+bc+ca) <= 5(ab+bc+ca)^2 = 5(a^2b^2+b^2c^2+c^2a^2) + 10abc(a+b+c)

It's remain to show that

          9a^2b^2c^2+ 5(a^2b^2+b^2c^2+c^2a^2) + 3abc(a+b+c) <= 3

But it's also trivial.

Namdung

$ab+bc+ca\leq 1\Rightarrow 5(ab+bc+ca)\geq 5(ab+bc+ca)^2$

Is the problem wrong ?I think not but... it is unsolved for one month

I think it's equivalent to: http://www.mathlinks.ro/Forum/viewtopic.php?t=55869

[quote="Vasc"]A hint:
$9a^2b^2c^2 \leq abc(a+b+c)$. ;)[/quote]

I used the hint but nothing. Can anyone help me ?Thanks

Maybe there can be a better way. Many times these "body-naked" inequalities can be obtain from "raw form" inequality. Usually is easier to prove inequality in "raw form", but sometimes it is conversely. In addition is very easy to find the "body-naked" form of an inequality, but conversely it is very diffucuilt.

So everything I wanted to say was an idea to find the "raw form", from the "body naked form". What do you thing about this, silouan?

What I'm thinking is : transfering $\ (a+b+c)^2 = 2(ab+ac+bc) + a^2+b^2+c^2 \geq 3(ab+bc+ca)$
Then $\ 5(ab+bc+ca)+9a^2b^2c^2\leq 3+7abc(a+b+c)$ will become: $\ \frac{5}{3}(a+b+c)^2+9a^2b^2c^2\leq 3+7abc(a+b+c)$
Use the hint(Professor Vasc, would you please tell me how to get this idea?) $\ 9a^2b^2c^2 \leq abc(a+b+c)$
left side = $\ \frac{5}{3}(a+b+c)^2 + abc(a+b+c) = (a+b+c)(abc+\frac{5}{4}(a+b+c))$ :ninja:
$\ \Rightarrow (a+b+c)(abc+\frac{5}{3}(a+b+c)-7abc) \leq 3 \Rightarrow \frac{5}{9}(a+b+c) - 2abc \leq \frac{a^2+b^2+c^2}{a+b+c}$
 :oops: Then what???
I divide both sides by $abc$, then:$\ \rightarrow \frac{5(a+b+c)}{9abc} - 2 \leq \frac{\frac{a^2+b^2+c^2}{abc}}{\frac{a+b+c}{abc}}$
According to the hint, $\ \frac{a+b+c}{abc} \geq 9$...  ;)

[quote="Kalimdor"]I divide both sides by $abc$, then:$\ \rightarrow \frac{5(a+b+c)}{9abc} - 2 \leq \frac{\frac{a^2+b^2+c^2}{abc}}{\frac{a+b+c}{abc}}$
According to the hint, $\ \frac{a+b+c}{abc} \geq 9$...  ;)[/quote]

I think you are wrong. I am sorry.

[quote="Vasc"]A hint:
$9a^2b^2c^2 \leq abc(a+b+c)$. ;)[/quote]

How can we continue?

If we can prove this:$a,b,c>0: 2\sum(a^5b+a^5c)+10\sum b^3c^3+\sum a^4bc+45a^2b^2c^2\ge 15\sum(a^3b^2c+a^3bc^2)$
the ineq will be easily solved.

Well, I try to use $x = ab, y = bc, z = ca$ and finally ended at :

$\frac{x^2+y^2+z^2+xy+xz+yz+x+y+z}{2} \le 1$
(I used Vasc's hint)

Nameless, if you are right, this is easy. Your last ineq can be written as $4[5,1,0]+10[3,3,0]+[4,1,1]+15[2,2,2]\ge30[3,2,1]$.

Now, we have $[4,0]+3[2,2]\ge4[3,1]$ as this is equivalent to $\sum (a-b)^4\ge0$.
Thus, multiplying by  $\frac{10}3[1,1]$ and adding cyclically, we have $\frac{10}3[5,1,0]+10[3,3,0]\ge\frac{40}3[4,2,0]\ge\frac{40}3[4,1,1]$. Adding that to $\frac{2}3[5,1,0]\ge\frac{2}3[4,1,1]$, we have our $LHS\ge 15[4,1,1]+15[2,2,2]\ge30[3,2,1]$ This is Schur multiplied by $15[1,1,1]$.  :roll:

i guess Silouan has a nicer "raw form" for it, allowing for a less 'bunched' proof :?:

Again, I tried to do it this way:
$x = ab, y=bc, z=ca$
$5(x+y+z)+9xyz \le 3 + 7(xy+xz+yz)$
Since $a^2+b^2+c^2 =1 \Rightarrow x+y+z = ab+bc+ca \le a^2+b^2+c^2 = 1$
$\Rightarrow 9xyz+2 \le 7(xy+yz+zx)$

By the hint : $9xyz \le xy+yz+zx$(btw, how to prove this?)
$\Rightarrow 2 \le 6(xy+yz+zx)$
Since $a^2+b^2+c^2 = 1$
$\Rightarrow 2(a^2+b^2+c^2) \le 6abc(a+b+c)$
$\Rightarrow 2(a^2+b^2+c^2) \le 3abc((a+b)+(b+c)+(c+a))$
Since we know $2(a^2+b^2+c^2) \leq \frac{(a+b)^2}{2}+\frac{(a+c)^2}{2}+\frac{(b+c)^2}{2}$
After that, we have to make this prove: $6abc \ge a+b$, $6abc \ge b+c$,$6abc \ge c+a$
However, the hint: $9a^2b^2c^2 \leq abc(a+b+c) \rightarrow 18abc \leq 2(a+b+c)$ disproves this one
Where is wrong? :oops:
Frank

[quote="perfect_radio"]Using Vasc's hint, I ended up with having to prove \[ 5 \sum (a^3b+ab^3) \leq 3 \sum a^4 + 6 \sum a^2 b^2 + \sum a^2 bc , ~ \forall x,y,z \in \mathbb R ^{+} .  \]

I don't think Muirhead will work here. Perhaps some clever trick yet to be discovered shall yield a solution. :D[/quote]
I guess it can proved by SOS. ;)

[quote="arqady"][quote="perfect_radio"]Using Vasc's hint, I ended up with having to prove \[ 5 \sum (a^3b+ab^3) \leq 3 \sum a^4 + 6 \sum a^2 b^2 + \sum a^2 bc , ~ \forall x,y,z \in \mathbb R ^{+} .  \]
[/quote]
$5 \sum (a^3b+ab^3) \leq 3 \sum a^4 + 6 \sum a^2 b^2 + \sum a^2 bc\Leftrightarrow$
$\sum(a^4-a^3b-a^3c+a^2bc)+\sum(a-b)^4\geq0.$ :)
Thank you, Vasile.[/quote]
nice, :)

[quote="arqady"]$\sum(a^4-a^3b-a^3c+a^2bc)+\sum(a-b)^4\geq0.$ :)[/quote]

Indeed, very nice.

I also tried to use Schur, but I didn't realize that the rest was actually $\sum (a-b)^4$. How did you figure it out?


induction
algebra
polynomial
trigonometry
strong induction
number theory
prime numbers

I understand regular induction and can you it in proof, but Im having trouble with strong induction.

All my attempts to find out online did not give me a problem example.

Help please.  :surrender:

Simple induction: Assume the statement is valid for $ n$. Prove that it is valid for $ n\plus{}1$.

Strong induction: Assume the statement is valid for all $ n < k$. Prove that it is valid for $ k$.

You use strong induction when your induction depends on several other results, and not only the previous one.

I googled "strong induction" -- this is a relatively easy problem that came up:

[quote]
Show that every integer greater than 1 may be written as the product of prime numbers.
[/quote]

Induction will lead us nowhere here -- knowing the factorization of n is not very helpful when you're dealing with n+1. Strong induction, however, gives us a simple proof:

[b]Base case[/b]: We can write 2 as the product of prime numbers, as it itself is a prime number.

[b]Induction step[/b]: Suppose all integers $ n$ with $ 1 < n < k$ can be written as the product of prime numbers. We will prove that $ k$ is also expressible as the product of prime numbers.

If $ k$ is prime, then it can be written as itself, which is a product of prime numbers. If $ k$ is composite, then, by definition, it is a product of two smaller numbers (both different from 1). But by our assumption, both of these numbers can be written as the product of prime numbers. So $ k$ is a product of prime numbers as well.

The best strong induction solution I came across recently:
http://www.mathlinks.ro/viewtopic.php?t=101711

Here's a strong inductive proof that there exist polynomials $ T_n$ such that $ T_n(2 \cos x) \equal{} 2 \cos nx$:  let $ q \equal{} e^{ix}$ and note that $ 2 \cos x \equal{} q \plus{} \frac{1}{q}$ and $ 2 \cos nx \equal{} q^n \plus{} \frac{1}{q^n}$.  Then

$ \left( q \plus{} \frac{1}{q} \right)^n \equal{} \sum_{k\equal{}0}^{n} a_k \left( q^k \plus{} \frac{1}{q^k} \right) \Leftrightarrow$
$ q^n \plus{} \frac{1}{q^n} \equal{} \left( q \plus{} \frac{1}{q} \right)^n \minus{} \sum_{k\equal{}0}^{n\minus{}1} a_k T_k \left( q \plus{} \frac{1}{q} \right)$

which is a polynomial in $ 2 \cos x$ by strong induction.  This isn't a great demonstration, since regular induction works just fine here, but it does have the merit of being pretty much automatic.

As a general note about strong induction, you don't need to have a separate mindset when you're trying to use strong induction than when you're trying to use regular induction.  In both cases you're trying to show that a statement follows from a previous result.  Strong induction just lets you know that you can use [i]any[/i] previous result, not just the one immediately before.

Indeed.  Strong induction is really a special case of induction:  a strong induction of the form

$ P(k), k \equal{} 1, 2, ... n \implies P(n\plus{}1)$

is equivalent to a regular induction of the form $ Q(n) \implies Q(n\plus{}1)$ where $ Q(n)$ is the statement $ P(1), P(2), ... P(n)$.


[quote="mmrproblem"]Solve the equation $x+y+(z^2-8z+14)\sqrt{x+y-z}=1$
$2x+5y+\sqrt{xy+z}=3$[/quote]

Are two diferent equations or it is a system?

I've seen it before, but I have no idea about the solution :blush:


I very much enjoyed the "favorite classic books?" thread Silverfalcon put up a week or two ago; however, there were many books that I enjoy that would not fall under the heading of "classic."

Therefore...

What are your favorite modern (e.g., 50 years or less) book?

My favorites are the Hitchhiker's Guide to the Galaxy "trilogy" (Douglas Adams,) the Discworld series (Terry Pratchett,) and the Ender's Game/Ender's Shadow series (Orson Scott Card.)

[i]Dune[/i] by Frank Herbert.  The best science-fiction book I've ever read.
Douglas Adams- the first three Hitchhiker books, the Dirk Gently series, and [i]The Salmon of Doubt[/i], which is a collection of essays, speeches, an incomplete book, and other short works collected after his death.  Excellent.

How many am I allowed to say?

Atlas Shrugged...Surely You're Joking, Mr. Feynman...Ender's Game...The Dark Materials trilogy...the Harry Potter books...

Enchantment, Alvin Maker series, and Ender/Bean books - Orson Scott Card, anything by Tamora Pierce, Hitchhikers Trilogy, Harry Potter, other authors I like are Phillup Pullman, Anne McCaffrey, Garth Nix, Eoin Colfer, Lemony Snicket... the list goes on and on.

Edit: Eragon as well!

I'm not sure books of poetry are included, but Charles Wright's [i]Black Zodiac[/i], is one of my favorites.

Both the Sword of Truth series, and the Ender's Game series are amazing works.

Well right now I'm reading Life of Pi for school (which came out around 2001).  It's a very...interesting...book, is all I have to say.

Seriously though, it explores topics like faith, maybe with a few existentialistic undertones.

A Clockwork Orange by Anthony Burgess
this book is insane, there is a lot of violence in the beginning, but it has a very good theme (it was written in the late '60s in England and the theme is the same as the core of the hippie movement)

these next two books i havent finished:
The Naked Lunch by William Burroughs
and
The Electric Cool Aid Acid Test by Tom Wolfe

The Naked Lunch is experimental literature and the physical setting will change erratically without mention by the author (i've heard that Burroughs wrote the book while experiencing withdrawls from stopping herion cold turkey, but it seems to me he did thought altering drugs while writing.
A Clockwork Orange takes place i think 40 years in the future, so Burgess invented slang (the main character was in a street gang), like ruk meant punch. if you get this book GET THE VERSION WITH 21 CHAPTERS!!! 21 chapters were published in the english version but the publisher only published 20 in the american version, the 21st is the most important chapter, also burgess intentionally wrote 21 (as in 21 years old) chapters because comming of age is involved in expressing the theme

I recently read [i]Surely You're Joking, Mr. Feynman[/i]...great book. It's amusing how you have to have some background in math and physics to understand it.

Of the non-science variety, one of my favorites would have to be [i]Snow Falling On Cedars[/i]. I enjoyed its diversity--although it takes place during WWII, it's not narrative historical fiction. It comprises both mystery and romantic elements, and includes an accurate portrayal of Japanese Americans living in the Orca Islands.

[quote]I recently read Surely You're Joking, Mr. Feynman...great book. It's amusing how you have to have some background in math and physics to understand it.[/quote]

Maybe some math and physics background helps to fully appreciate certain parts, but I don't think you need to be a science person at all in order to get a lot out of that book.  That book is for everyone!

I see Surely You're Joking, Mr. Feynman as a collection of parables that teaches us how to be more scientific people.  He conveys the skeptical, curious, independent attitude of a scientist so well---in my mind that book is one of the pillars that holds up the world's scientific culture.

Okay maybe I love Feynman too much.

The Rule of Four

written by a princton grad and a harvard grad

they both graduated from tjhsst '94 i think

:)

Micheal Crichton, John Grisham, and Dan Brown books.

Prey, Timeline
The Chamber, The Pelican Brief
The Da Vinci Code

from each author, respectively, are my favorites

I love anything by kurzwiel. I really got into 'thinking' as a pass time, once I first read "The age of spiritual machines" ( I reccomend it to everyone ).

Everyone I know loves 1984, maybe I should give it another chance, but I just couldn't be pleased by it.  I MUCH prefured 'brave new world' by huxly, what a much more fun dystopia.

You all seem to like Ender too much! Enders was kind of fun, but really it was just fast paced teenage sci-fi, with a charecter you [u]all identified with[/u] tons I bet. Most thinkers see a bit of themselves in ender, and forget that hes just kind of flat. hopefully this won't start a flame war? I know some people love ender.

faviorite book of all time: 2001 space oddesy, by Arther C Clarke. Clarke was the best of the big three.

When ever I recomend 2001 to people, I tell them " the last quarter will blow your mind, you will not think the same way for the rest of your life ". They don't believe me. Then they hand me back my copy " damn, you were totaly right". richard 5, non-believers 0.

read 2001. maybe childhoods end too?

[i]Skin[/i] by Roald Dahl. Also [i]Shade's Children[/i] (Garth Nix) because it makes me cry and want to be a better person. [i]The Alchemist[/i] is good too, it's nice to read when you are despairing about humanity, because it's hopeful. Paulo Coehlo, or some variation of that spelling. All of the copies of [i]Cadence[/i] that I have are really good, but you wouldn't know what that is, considering that it's a schoolwide publication.

Shade's Children is an excellent book, I should go read that again.

[quote="B-tone_Party"][quote="rzsolt"]
You all seem to like Ender too much! Enders was kind of fun, but really it was just fast paced teenage sci-fi, with a charecter you [u]all identified with[/u] tons I bet. Most thinkers see a bit of themselves in ender, and forget that hes just kind of flat. hopefully this won't start a flame war? I know some people love ender.
[/quote]

Are you speaking of the first book, or the following books?  Ender's game was (in my opinion) the best of the books, but the rest of the series departed quite far from the original work- the Ender series went into a more philosophical tone, and the Shadow books were more war stories than anything else- Ender is not even in the last three Shadow books.[/quote]

I have only read the first one. Kind of fun for teenage science fiction, but more of an adventure book than real science fiction.

The problem i have with ender, is SOME of the fanboys. Most of them are pretty level headed, but I've met a few that idolize ender, and through proxy the book. They do this usualy becuase they identifiy with the flat 'crazy smart, shy, picked on, really special' charecter.

I have to say, the only things I really liked from enders game, was the battle scenes ( in training ), and the very sped up story of colonizing the bug world years and years later.

Great modern book: Zen and the Art of motercycle mantnence.
Exploring the nature of 'quality',  set to a cross country motercycle trip. Very very fun.

[quote="easyas3.14159..."]ERAGON ALL THE WAY!!![/quote]
Yeah! Eldest too!!! Did you know the movie is coming out this december?

Hmmm... the Harry Potter series (especially [i]Prisoner of Azkaban[/i]), and the Discworld books (especially [i]Night Watch[/i], [i]The Fifth Elephant[/i], [i]Interesting Times[/i], [i]Maskerade[/i], and [i]Hogfather[/i]).

I have to say the Dark Materials books are my favorite modern books. They were very intriguing, in a good way. I thought Eragon had a good story, but I don't like it since it COPIED FROM LORD OF THE RINGS AND STAR WARS. And Da Vinci Code was...umm...gripping, though not very well written. I just hope that Harry Potter 7 will be good and answer all our questions, but I doubt it.  :(  Douglas Adams books are very funny, could be genius in disguise. Lots of twists and turns, my mind just isn't big enough to hold all of it. I kept forgetting what had already happened or were some characters were.  :D

The Da Vinci Code is perfect for us mathletes!  ;)  Also, be sure to check out the books about the Da Vinci Code, such as Da Vinci Unraveled, etc.  These are not as light, but can be great if you are really interested in the topic.


Ορίστε μια ωράια ανισότητα (δική μου διασκευή :censored: ) :
Αν $ a,b,c>0$ με $ ab\plus{}bc\plus{}ca \equal{} 9$ να δείξετε ότι : $ a^5 \plus{} b^5 \plus{} c^5>28$

Παίρνω γενικευμένη ανισότητα Andresscu  και έχω ότι

$ a^5 \plus{} b^5 \plus{} c^5\geq\dfrac{(a \plus{} b \plus{} c)^5}{3^4}$ . Άρα αρκεί να δείξω ότι $ \dfrac{(a \plus{} b \plus{} c)^5}{81}>28 \Rightarrow (a \plus{} b \plus{} c)^5>81\cdot28$ (1). Όμως το $ (a \plus{} b \plus{} c)^5 \equal{} (a \plus{} b \plus{} c)^2\cdot(a \plus{} b \plus{} c)^2\cdot (a \plus{} b \plus{} c)$ . Σύμφωνα με γνωστή ανισότητα έχουμε ότι $ (a \plus{} b \plus{} c)^2\geq3(ab \plus{} bc \plus{} ca)\Rightarrow(a \plus{} b \plus{} c)^2\geq27$ και ομοίως $ a \plus{} b \plus{} c\geq \sqrt {3(ab \plus{} bc \plus{} ca)} \Rightarrow a \plus{} b \plus{} c\geq 3\sqrt {3}$  Άρα  έχουμε  $ (a \plus{} b \plus{} c)^5\geq{27^2\cdot3\sqrt {3}}$ . Άρα αρκεί για να ισχύει η (1) να έχουμε ότι $ 27^2\cdot3\sqrt {3} > 28\cdot81 \Rightarrow 27\sqrt {3} > 28 \Rightarrow 27^2\cdot 3 > 28^2 \Rightarrow 2187 > 784$ που σαφώς ισχύει άρα QED.

Κανόνι είσαι ρε!!! Εγώ βέβαια τη βρήκα την ασκηση και δεν ήθελε για $ >28$ αλλά για $ \geq 18$ από το βιβλίο του κυρίου Ρωμανίδη " Η άλγεβρα στις Μαθηματικές Ολυμπιάδες" αλλά τη βελτίωσα ακόμα πιο πολύ και την έδωσα και νομίζω ότι είναι αρκετά ενδιαφέρουσα  :D


geometry Harvard college MIT HMMT symmetry

S26

A rectangular area measuring 3 units by 6 units on a wall is to be covered with 9 tiles each measuring 1 unit by 2 units. In how many ways can this be done?

wtf this question is exactly like Harvard-MIT Mathematics Tournament 1999 Algebra #9, or something (like HMMT Algebra 1999 or 2000)

But this is counting techniques not algebra? How does algebra apply? Anyway my answer was 37, one case has no symmetry so it is odd

Did u use a branching counting method? Or recursion (though i'm not sure how that can be applied here)?


logarithms
analytic geometry
graphing lines
slope

finding some problems  to find the expression of the following problems ....please post only answers

3.
We first have $ f(x) \equal{} \ln{(4\minus{}2x)}$
switch $ x$ and $ f(x)$ to get the inverse

$ f^{\minus{}1}(x) \equal{} \frac{4\minus{}e^{x}}{2}$

5.
$ y' \equal{} 4\minus{}2x$
Slope at $ P(1,3)$ is $ 2$
Slope of Normal Line is $ \minus{}\frac{1}{2}$, and it passes through $ P(1,3)$
Thus Equation of the normal line is $ \frac{y\minus{}3}{x\minus{}1} \equal{} \minus{}\frac{1}{2}$
Set it equal to $ y$ to solve for the co-ordinates of Q.


my silly doubt-

Two circles intersect (not orthogonally) Will the tangent to one of the circle at the point of intersection pass through the centre of the other circle? :maybe:

Only if they intersect orthogonally. But I know what you are heading at since I had the same doubt. The angle between the tangents = angle between the perpendicular to the tangent = angle between the normal drawn from the centre to the points of intersection.

but I have read in books like ML Khanna and arihant that the angle of intersection can be determined using cosine formula as $ \cos \theta \equal{} \frac{r_1^2\plus{}r_2^2\minus{}d^2}{2r_1.r_2}$...is it wrong?

Anand, thats exactly what I'm trying to tell you.

The formula is perfectly valid. Why? Because do you agree that the angle between the tangents at the point of intersection is equal to the angle between the normals at the point of intersection? And the normals pass through the centre. Now take the Triangle $ C_1C_2S$ and apply Cosine Rule.

P.S You deserve special accolades if you are able to read through Arihanth Cogeo.  :rotfl:  :rotfl:

but in that case what is the angle $ C_1SC_2$ be? The book(arihant ) says it is 180 -$ \theta$....I'm unable to make out...Shreyas can you post a diagram please....

Oops. Sorry it is $ \Pi \minus{} \theta$. Actually when 2 lines intersect you get 2 angles. You can either of them. $ \theta$ or $ \Pi \equal{} \theta$, i guess we usually take the smaller angle for sake of convenience.

I apologise for my pathetic diagram. Sorry! I can't draw to save my life.

But see the angle 1 - $ 90 \minus{} \theta$
Angle 2 - $ \theta$
angle 3 - $ 90 \minus{} \theta$

We need 1+2+3.

[quote="Cyber-Shreyas"]

P.S You deserve special accolades if you are able to read through Arihanth Cogeo.  :rotfl:  :rotfl:[/quote]

I ve seen that some of the solutions he gives there arelong and computational rather than straight forward geometry.
Loney s book is better. My favorite :lol:


You can pick whichever version you want, just make sure you make it clear what you mean. Also, it does no harm to add $0$ at the beggining (i.e. $0, 1, 1, 2,\dots$)

Yeah but what if I want to apply formulas like Binet's? There must be some standard agreement.

Binet's formula gives $F_{0}=0$, so I concur with [b]Ravi B[/b] that it's probably the best to stick with Fibonacci Quarterly.

[quote="Ravi B"]The Fibonacci Quarterly follows the second convention.  Namely, $F_{0}= 0$, $F_{1}= 1$, and $F_{2}= 1$.

I would stick to its convention.[/quote]

Only by this convention can we say $F_{n}| F_{mn}\forall m \in \mathbb{N}$.

If anyone cares, what t0rajir0u wrote means that

"$F_{n}$ divides $F_{mn}$ for all natural numbers $m$."

Interestingly, the AoPS books define it the other way.

The fibonnachi sequence goes like this:

$0, 1, 1, 2, 3, 5, 8, 13, 21\dots$
I hope this is what you wanted.

According to this one guy,
$f_{0}=1,f_{1}=2 blabla$ and
$F_{1}=1,F_{2}=2 blabla$ where both are fibonacci.
or maybe it was the other way around


geometry
area of a triangle
calculus
calculus computations

Find the maximum area of a triangle $ APQ$ such that $ A\left(\minus{}\frac{7}{2},\ 0\right)$, two points $ P,\ Q$ are on the circle $ x^2\plus{}y^2\equal{}1$.

[quote="Sara90"]I solved this problem by the method of Lagrange multipliers.
However :maybe: , I think this problem has a more elegant solution.[/quote]
Would you show the details,please.


[color=green]Find the additive inverse of [87] in Z150.[/color]

Thx guys

How do you guys know it's 63?

Just (150-87)?

Yeah.

Let $x$ be the additive inverse of $87$, then:

$x+87\equiv0\;(\mathrm{mod}\;150)$

gives

$x\equiv-87\;(\mathrm{mod}\;150)\equiv150-87\;(\mathrm{mod}\;150)\equiv63\;(\mathrm{mod}\;150)$

So $x=63$, among others.


In the classroom, how do you upload an avatar icon from the internet?

You upload an avatar to the forum with the Profile button atop the forum.  A day later, the avatar will be available in the classroom automatically (if the bug fix I did this morning works!).

[quote="Carbon57"]Well...one way is to just use the avatar from your profile in the classroom.

I think if you click on the little person-ish thing on the bottom left of screen, click the three dots, it could give you a space to put a url...[/quote] i looked...theres no 3 dots!

There is now  :blush: .
This is a month old topic, please lock.


How many four-digit numbers $ N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $ N$?

$ \textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 5 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 7 \qquad
\textbf{(E)}\ 8$

if the four-digit number is abcd, lets say that bcd is q.

Then, $ N\equal{}1000a\plus{}q$

we have $ 1000a\plus{}q\equal{}9q \implies 1000a\equal{}8q \implies q\equal{}125a$

a can be 1, 2, 3, 4, 5, 6, and 7 so that q is a three digit number.

answer: D


Show that if $x$ is a non-zero real number, then
$x^{8}-x^{5}-\frac1{x}+\frac1{x^{4}}\geq 0.$

Multiply all by $x^{4}$, which is positive and won't affect sign.

Factorizes as $(x^{9}-1)(x^{3}-1)$, and both factors are the same sign.

Check:

$x \ge 1$ => both terms non negative.
$x < 1$ => both terms negative.

Thus product is always non-negative.

Multiplying by the positive $x^{4}$  we have to show that
$x^{12}-x^{9}-x^{3}+1\geq 0$ which is equivalent to
$(x-1)^{2}(1+x+x^{2})^{2}(1+x^{3}+x^{6})\geq 0$

which is true....

EDIT :vulgarfraction and Andreas were faster


Let $ f(x)\equal{}a_nx^n\plus{}...\plus{}a_0$ be a polynomial of degree $ \geq 2$ and integer coefficients. Assume that it has only real roots that are all distinct and lie in $ (0,1)$. Prove that: $ |a_0|\geq 2^n\plus{}1$.

What about $ (2x\minus{}1)(3x\minus{}1)$?

Maybe he means $ |a_{n}|\geq 2^{n}\plus{}1$; but in that case, the question becomes trivial.


Solve the inequality

$ \log_{x^2 \minus{}3}(x\minus{}1)\geq 1$

-- edited

sorry i think no solution  :(

Changing to base $ 10$,

$ \frac{\log_{10} (x-1)}{\log_{10} (x^{2}-3)} \ge \log_{10} 10$

that is, $ \frac{\log_{10} (x-1)}{\log_{10} (x^{2}-3)} \ge 1$

We then have two cases, depending on whether ${ \log_{10} (x^{2}-3)}$ is positive or negative.

If it is positive then $ x-1 \ge x^{2} - 3$ and $ x^{2} - 3$ is greater than $ 1$.

If it is negative then $ x-1 \le x^{2} - 3$ and $ x^{2} - 3$ is less than $ 1$.

So there seem to be no solutions.

Perhaps I've made a mistake.

There is NO solution.You are BOTH right.  :D

Yes, I thought I may have made a mistake as it seemed so easy.  But I need to practise my $ \text{\LaTeX}$ at which I am slow and keep making mistakes.

for log to be defined we must have

1) base > 0 and not equal to 1
2) argument > 0

so  x - 1 > 0  $ \Rightarrow$ x > 1

and  $ x^2 - 3 > 0 \ and \ x^2 - 3 \neq 1$

this gives $ x > \sqrt {3} \ , \ x \neq 2$

$ log_{x^2 - 3}(x - 1) \geq log_{x^2 - 3}(x^2 - 3)$

hope you know that when base > 1, inequality as it is for the arguments
but
when 0 < base < 1, inequality is reversed.

[b]Case  1 :[/b]  $ x^2 - 3 > 1 \Rightarrow x > 2$ ... (1)

$ x - 1 \geq x^2 - 3 \\
(x - 2)(x + 1) \leq 0 \\
x \in [ - 1,2]$ ... (2)

No solution from (1) & (2).

[b]Case   2 :[/b] $ x^2 - 3 < 1 \Rightarrow \ \sqrt {3} < x < 2$ ... (3)

NOTE : inequality reverses here

$ x - 1 \leq x^2 - 3 \\
(x - 2)(x + 1) \qeq 0 \\
x \in [2, \infty]$ ... (4)

No solution here too from (3) & (4).

[b]P.S. I am a lil slow at LaTeX tooo [/b]   :blush:

$ \log_{x^2 \minus{} 3}{(x \minus{} 1)}\ge1$
-The solution is from system:

$ x^2 \minus{} 3\ge1$ and $ x \minus{} 1\ge x^2 \minus{} 3$ =>
$ x^2 \minus{} 3\ge1$ , $ x \minus{} x^2 \plus{} 3\ge1$

The equality holds when $ x \equal{} 2$ which is the only solution.

[quote="Obel1x"]
....
The equality holds when $ x \equal{} 2$ which is the only solution.[/quote]

But when x = 2, the logarithmic base becomes 4 - 3 = 1 right ??

[b]And its a property of the logarithmic function that it cannot have the magnitude of base as 0 or 1.[/b]

So, no solutions.

Should logarithms be in High School Basics or High School Intermediate? Personally, they seem to fit more into the Intermediate section but I might be wrong

According to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837]High School Basics Difficulty[/url] post, Algebra 2 topics are completely valid in the High School Basics forum. And this doesn't involve a complex proof or anything; it's a simple one-variable inequality. Multi-variable inequalities, on the other hand, belong in the Intermediate. (I see that enndb0x did so for his other log problems.)


Why doesn't AOPS allow its users to Edit YOUR OWN posts after 2 days or the like... Sometimes, when I've typed a list or  a resource, or a nice solution and want to edit it.. Its kinda frustrating to ask the mod to edit it everytime. I do feel a bit sorry for pardesi, because I've asked him to edit so many of my posts, that he must be waiting to kill me;)

It would be nice if we were allowed to always edit our own posts.

Won't happen. If 2 days have gone, it's time to post a new post.


It was really fun today (not including the fact that my school and I won the team and individual for three years in a row..)

the problems are getting harder and harder, more teams are participating, John Kennedy is a great guy...

I have a feeling that this competition will be bigger and bigger.

[u]Team Winner[/u]
1. Benjamin N. Cardozo HS
2. John Bowne HS
3. forgot  :blush:

[u] Individual Winner - Division 1[/u]
1. Geehoon Hong (A team) / Benjamin N. Cardozo HS $7$
2. He Cui (A team) / Benjamin N. Cardozo HS $5$
3. Qin Feng Chen (Y team) / Benjamin N. Cardozo HS $3$
    Hanpin Yan (A team) / John Bowne HS $3$
    Wensi Yu (R team) / John Bowne HS $3$
    :blush: $3$

[u] Individual Winner - Division 2[/u]
1. :blush:
1. :blush:
1. :blush:

[u] Individual Winner - Division 3[/u]
1. Daniel Vellon / Garden HS
2. Aiden Min / Garden HS
3. Soo Lim Choi / Garden HS

Im realizing I dont remember anything..

well, if we put information from other people together too, maybe we can complete the list...hopefully..

anybody interested in questions / solutions for individual and team rounds?

it was actually pretty hard... at least division 1 was.
out of 10 questions,
mode was definitely 0, mean is between 0 and 1, and i think even median was 1.

if you have gotten 3, youre third place!!

team was relatively easy.
we got 5 out of 6, with stupid $my$ mistake.
bleh.

i don;t know why,but i know my couch never tell us about that,nor did we ever go there in past 10+years.......

I'd like to see the questions...Maybe Tech will participate next year...how do you enter???

[quote="ShadowLord"]in past 10+years.......[/quote]

it's only 4th annual

geehoon blushes too much, it's alright to forget :)
u could have taken notes on winners u kno :D


Let ABC be a triangle, and let D,E,F be points on BC,CA,AB, respectively, such that cevians AD, BE, and CF concur at P. Prove that

$ \dfrac{AP}{PD}\plus{}\dfrac{BP}{PE}\plus{}\dfrac{CP}{PF}\ge6$.

[hide]Denote $ S: \equal{} [ABC]$. Then

$ {AP\over PD} \equal{} {[APB]\over[BPD]} \equal{} {[APC]\over[PCD]} \equal{} {[APB] \plus{} [APC]\over[BPD] \plus{} [PCD]} \equal{} {S \minus{} [PBC]\over[PBC]} \equal{} {S\over[PBC]} \minus{} 1$

Similarly for the other two fractions.

Hence we have to prove

$ {S\over[PAB]} \plus{} {S\over[PBC]} \plus{} {S\over[PCA]}\geqslant 9$

but that follows directly from AM-HM, bearing in mind that $ [PAB] \plus{} [PBC] \plus{} [PCA] \equal{} S$.[/hide]

Put mass points on $ A,B,C$ with weights $ X,Y,Z$ s.t. $ P$ is the center of mass of the system.

$ \frac{AP}{PD}\equal{}\frac{Y\plus{}Z}{X}$ And similarly for the other ones. Then it is simple AM-GM.

[hide="Additional detail"]For the equality to hold, we must have $ [PAB]\equal{}[PBC]\equal{}[PCA]$, hence $ P$ must be the centroid.[/hide]


trigonometry
logarithms
function
algebra
domain
complex numbers
calculus

Hi I was wondering if someone can please help me with this.  Thank you.

Question: Show that $ \displaystyle\rho\sin\phi\plus{}\frac{1}{3}\rho^3\sin 3\phi\plus{}\frac{1}{5}\rho^5\sin 5\phi\plus{}\cdots \equal{}\frac{1}{2}\tan^{\minus{}1}\left(\frac{2\rho\sin\phi}{1\minus{}\rho^2}\right)$

Let $ z\equal{}\rho e^{i\phi}.$ The sum you are looking at is

$ \operatorname{Im}\left(\sum_{k\equal{}0}^{\infty}\frac{z^{2k\plus{}1}}{2k\plus{}1}\right)\equal{} \operatorname{Im}(\tanh^{\minus{}1}z)\equal{} \operatorname{Im}\left(\frac12\ln\left(\frac{1\plus{}z}{1\minus{}z}\right)\right).$

With the note that $ \operatorname{Im}\ln z\equal{}\arg(z),$ I'll leave you to figure out the rest.

Hi Professor Merryfield, thanks for the response.  [quote]I'll leave you to figure out the rest.[/quote]... but I think you are overestimating my abilities.  May you please explain the rest.   :blush:

$ \frac{1 \plus{} z}{1 \minus{} z}$ can be written in the form $ re^{i \theta}$ for some positive real $ r$ and some argument $ \theta$.  Now,

$ \ln (re^{i \theta}) \equal{} \ln r \plus{} (i \theta)$

so the imaginary part of the natural log of a complex number gives its argument.  What is the argument of $ a \plus{} bi$, generically?  What are the components of $ \frac{1 \plus{} z}{1 \minus{} z}$?

Okay, I understand the final part now.  Since $ \displaystyle \log(\omega)\equal{}\ln|\omega|\plus{}i\arg(\omega)$, then $ \displaystyle\frac{1}{2}\log(\omega)\equal{}\frac{\ln|\omega|}{2}\plus{}i\frac{\arg(\omega)}{2}$.  Thus, $ \displaystyle\text{Im}\left[\frac{1}{2}\log(\omega)\right]\equal{}\frac{\arg(\omega)}{2}$.
Here, $ \displaystyle\omega\equal{}\frac{1\plus{}z}{1\minus{}z}$ which we can simplify it by writing it as $ \displaystyle\left[\frac{(\rho\cos\phi\plus{}1)\plus{}i\rho\sin\phi}{(1\minus{}\rho\cos\phi)\minus{}i\rho\sin\phi}\right]\left[\frac{(1\minus{}\rho\cos\phi)\plus{}i\rho\sin\phi}{(1\minus{}\rho\cos\phi)\plus{}i\rho\sin\phi}\right]$
After simplification, $ \displaystyle\frac{1}{2}\arg\left(\frac{1\plus{}z}{1\minus{}z}\right)\equal{}\frac{1}{2}\tan^{\minus{}1}\left(\frac{2\rho\sin\phi}{1\minus{}\rho^2}\right)$.

But I am still not sure how does $ \displaystyle\text{Im}(\tanh^{\minus{}1}z)\equal{}\text{Im}\left[\frac{1}{2}\log\left(\frac{1\plus{}z}{1\minus{}z}\right)\right]$.  Can someone please explain this part to me.  Thanks again.

$ \frac {d}{dz}\tanh^{ \minus{} 1}z \equal{} \frac1{1 \minus{} z^2}.$

$ \frac {d}{dz}\frac12\ln\left(\frac {1 \plus{} z}{1 \minus{} z}\right) \equal{} \frac12\left(\frac1{1 \plus{} z} \plus{} \frac1{1 \minus{} z}\right) \equal{} \frac1{1 \minus{} z^2}.$

Since both functions are zero at zero, then they're the same function - at least within a domain on which we can agree on "standard" or "usual" branches for each.

On second thought, forget that I ever said $ \tanh^{ \minus{} 1}z.$ Just read that the series equals that combination of logarithms, which you can verify directly.


how do you draw a regular polygon, like a hexagon?

polygon(int n) is such a function. See http://asymptote.sourceforge.net/doc/graph.html


an  given sequence $ a_1,a_2,a_2$... .this sequence is increase strictly such that $ a_{gcd(m,n)}\equal{}gcd(a_m,a_n)$ ;
n,k,s are 3 number such that $ {a_k}^2 \equal{} a_n a_s$; n>k>s
prove that s|k|n

[quote="kihe_freety5"]prove that s|k|n[/quote]
Do you mean, $ s\mid k$ and $ k\mid n$ ?
And is it a problem of this year's Indian TST? Too early. :maybe:

ohh , it was appeared in  india team selection test 2001!! and
they are s is a divisor of k and k is a divisor of n!!!
there is a problem in IRAN which is like that a bit!!


probability
algorithm
probability and stats

I was given a problem to find a grid where you can read the squares of numbers $ 1,\ldots ,100$. Here each square contains a single digit and read means that you fix some starting square and the direction (8 possibilities) and you go to that direction as many steps as you want. I heard that one suitable grid is of the size $ 11\times 11$. How long it takes to find that grid? I have no other idea to find the grid than fill the grid by random numbers and check if that grid is suitable. Or does someone has a better idea to solve the problem?

This is reposted from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288507

Correct. But I think it can be still kept here assuming that the implied question is to estimate the probability that a random grid satisfies the condition.

So, please, if you want to discuss such estimates, post here, and if you want to discuss efficient algorithms for solving the problem, post in the thread JBL mentioned. :)


function
integration
abstract algebra
geometry
geometric transformation
dilation
linear algebra

(1)

a)Let K be a commutative field and n>=2 a natural number. Find Z(M_n(K). and show that the ring Z(M_n(K)) is isomorphic to K.
b)Show that M_n(R) and M_n(C) aren't isomorphic.

(2)
Let n>=2 be a natural odd number. Find the continuous functions f:[0,1]->R s.t.  \int {0->1}(f(x 1/k )) n-k dx=k/n for all 1<=k<=n-1.

(3)
Let f:R->R be a continuous function s.t. xf(x)>= \int {0->x}f(t)dt for all real x.

a)Show that the function g:R*->R, g(x)=(1/x)* \int {0->x}f(t)dt is increasing on (-00,0) and on (0,00).
b)If we also have  \int {x->x+1}f(t)dt= \int {x-1->x}f(t)dt, show that f is constant.

(4)
If (G,+) is a finite abelian group with n(G) elements, let i(G) be the number of composition laws (G,*) for which (G,+,*). Show that:

a)i(Z_12)=4
b)i(AxB)>=i(A)*i(B)
c)There exist 2 sequences of finite abelian groups (G_k) and (H_k) s.t.
n(G_k)/i(G_k) -> 0 and n(H_k)/i(H_k) ->00 as k->00.

[quote="grobber"](1)

a)Let K be a commutative field and n>=2 a natural number. Find Z(M_n(K). and show that the ring Z(M_n(K)) is isomorphic to K.
b)Show that M_n(R) and M_n(C) aren't isomorphic as ring.
[/quote]

a) Consider the E(i,j) canonical basis of Mn(K), A is in Z(Mn(K)) means that A commute with any X in Mn(K)
AX=XA choose X=E(i,j) computation => A is a dilation matrix, A=bI_n
The ring Z(Mn(R)) is isomorph to R, Z(Mn(C)) is isomorph to C
b) Suppose there exist f an isomorphism of ring from M_n(R) and M_n(C)
then f(Z(Mn(R)) the image of the center Z(Mn(R)) would be isomorph to the Z(Mn(C)), thus R isomorph to C wich leads that x^2+1=0 has a solution in R .

Really problem in Final Round/RMO/12th grade looks like
MP* exercices, or Oral Examination E.N.S/ X, or Agrgation training exercices.

Could you tell a little more about those examinations you mentioned Moubinool? Thanks :-).

Yes Mindspa have look here

http://www.mathlinks.ro/phpBB/viewtopic.php?p=8166#8166


If $ q$ is an integer that can be expressed as the sum of two integer squares, show that both $ 2q$ and $ 5q$ can also be expressed as the sum of two integer squares.

Thanks in advance!

[quote="t0rajir0u"][hide="Giveaway"] Compute $ (x \plus{} y)^2 \plus{} (x \minus{} y)^2$ and $ (2x \plus{} y)^2 \plus{} (x \minus{} 2y)^2$. [/hide]
[hide="Generalization"] If $ p, q$ are integers that can be expressed as the sum of two integer squares, $ pq$ can be expressed as the sum of two integer squares. [/hide][/quote]

It don't get why $ (x \plus{} y)^2 \plus{} (x \minus{} y)^2$ and $ (2x \plus{} y)^2 \plus{} (x \minus{} 2y)^2$ would help.

Thanks.

If you calculate them, it's basically the answer...

Sorry to be reviving this really old thread, but how did t0rajir0u find
[quote="t0rajir0u"][hide="Giveaway"] Compute $ (x \plus{} y)^2 \plus{} (x \minus{} y)^2$ and $ (2x \plus{} y)^2 \plus{} (x \minus{} 2y)^2$. [/hide]
[/quote]?

Do you just guess and check?

Thanks.

Must I tell you?

[hide="Fine..."]
We see that $(x + y)^2 + (x - y)^2 = 2x^2 + 2y^2$.  This is also $2(x^2 + y^2)$.  So if $q = x^2 + y^2$ then $2q$ can be expressed as $(x + y)^2 + (x - y)^2$.

We see that $(2x + y)^2 + (x - 2y)^2 = 5x^2 + 5y^2$.  This is also $5(x^2 + y^2)$.  So if $q = x^2 + y^2$ then $5q$ can be expressed as $(2x + y)^2 + (x - 2y)^2$.

:)
[/hide]

[quote="exmath89"]No, I meant how did he get $(x+y)^2+(x-y)^2$ and $(2x+y)^2+(x-2y)^2$ in the first place?

Sorry if my question was confusing.[/quote]

i also wonder...

[quote="professordad"]What do you mean where is $p$ coming from?  [b]gaussintraining[/b] just defined $p$ to be $a^2 + b^2$ just to reduce the amount of work needed.[/quote]

How does that work from the begining to the end?


Find  the  maximum  value  of  k  such  that  the  inequality
a^2+b^2+c^2+d^2+e^2>=k(ab+bc+cd+de)
holds  for  all  reals  a,b,c,d,e.
Thanks

bugzpodder wrote:if k(ab+bc+cd+de+ea)  then we can let a=b=c=d=e=1 and k>1 implies that RHS>LHS.  at k=1 we can easily see that the inequality holds

I shall use this as a lemma in my proof, since I don't quite feel like reproving something already stated in the thread...

Spoiler

[hide]

If:

a^2+b^2+c^2+d^2+e^2>=k(ab+bc+cd+de)

for all reals (a,b,c,d,e) then we can permute this set to (e,a,b,c,d), (d,e,a,b,c), and so on such that we account for all cyclic permutations.

in each case,

a^2+b^2+c^2+d^2+e^2>=k(ab+bc+cd+de+ea) - k*ea.

so if we add all these inequalities up (after cycling through), we get:

5*(a^2+b^2+c^2+d^2+e^2) >= 5k(ab+bc+cd+de+ea) - k(ab+bc+cd+de+ea) = 4k(ab+bc+cd+de+ea)

so a^2+b^2+c^2+d^2+e^2 >= 4/5*k(ab+bc+cd+de+ea) and from the above lemma we have that k is at most 5/4, and that this bound is the attainable maximum. QED.

[/hide]

a^2+b^2+c^2+d^2+e^2>=(ab+bc+cd+de+ea)  is a direct application of rearrangement.

it is also a direct application of the trivial inequality, or AM-GM, or possibly even other things. But the problem was not what you stated, it was without the ea term.

Oups, you're right, I'm stupid, I didn't read the terms correctly.


I'm working on a paper and I'm using BiBTeX for the first time; it's really easy, but not entirely trivial. My problem is that BiBTeX only capitalizes the first word in the title, which is kinda not right when the title is Who was Rhee Syngman? and Rhee Syngman has to be capitalized. How do I fix this?

Here's my present ssfpaper.bib file. Neither fareastsurvey nor wwrs capitalizes the title properly.

[code]
@article{fareastsurvey,
	author="Paul S. Dull",
	number="17",
	volume="17",
	title="South Korean Constitution",
	journal={Far Eastern Survey},
	pages="205-207",
	year="1948"}
@misc{wwrs,
	howpublished="\url{www.kimsoft.com/2000/rhett.htm}",
	author="Lee Wha Rang",
	title={Who Was Rhee Syngman?}}[/code]

Thanks!

You can tell BibTeX that you want certain letters capitalised by enclosing those letters in {} so instead of
title={Who Was Rhee Syngman?}
use
title="Who was {R}hee {S}yngman?"
or
title={Who was {R}hee {S}yngman?}
which will come out as
Who was Rhee Syngman?


How many more complete revolutions will a wheel with a circumference of 3 feet make than a wheel with a circumference of 4 feet in traveling one mile?

A mile is 5280 feet. Since a 3 foot wheel would have 1/3 revs per foot and a 4 foot wheel would have 1/4 revs per foot, the difference is 1/12 revs per foot. 1/12*5280=440.


So I have learned about AoPS recently and all those math competitions
and I just took them last month but I did not know that until my school announced it.
I didn't know they would be so hard.

How could I study for it?
Which books should I buy for AMC12?
And AIME?

Is the problem series good?

Thanks guys     :lol:

Even though I'm not a very well experienced problem solver yet, I think it would be helpful to read and consider some statements from this small [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]article[/url], or [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]other articles[/url] of interest once in a while, while preparing, tackling, or beginning the art of solving new problems simultaneously.

In terms of books:

- AoPS: Intro books
- AoPS: Intermediate Books
- Art and Craft of Problem Solving
- Geometry Revisited

Note: These are just a few from the top of my head. Also Kalva (google it) is a good resource for practice problems, and working problems and developing new techniques is just about the best way to get better at math. (For me, anyways)

Doing past problems will also help. They'll give you relative problem difficulties as well as the expected material.

Alright, I'll get those books.
I already found some sites with past problems.

But, are the online classes really good?    :huh:

how do i get the answers to Zeitz's book??

Anyone have a copy?

can someone give me the solutions for Zeitz's book?

Please post the AIME answers.
thanks.


AMC
USA(J)MO
USAJMO
USAMO
AIME
AMC 12
AMC 10

I was trying to decide which test I should take this year. I am in eighth this year. I can score over 130 on the AMC 10 and a (I think) 110+ on the AMC 12. I have some questions:

What are the advantages and disadvantages of the AMC 10?
What are the advantages and disadvantages of the AMC 12?
If I am aiming to make Red MOP, which test should I take?
How should I practice? Should I just practice for the 12 and will my 10 score get better?

Thanks!

EDIT: Ignore the first, second, and last questions.

This is a really hard question to answer, because the system just changed. Now if you take the AMC 10, you are aiming for the USAJMO (Junior), and if you take the AMC 12, you are aiming for the USAMO. The only way to make Red MOP is if you make USAMO and get a good score, or make USAJMO and do good enough on #3 and #6.

I don't know everything about these tests, but here's what I know:

You have to be in 9th grade to make Red MOP. If you're in 8th grade, you must qualify for Blue to go to MOP.

AMC 10:
It's an easier test and you can make USAJMO with it. The downside is that you can't make USAMO with it. Red MOP is based on questions 3/6 on USAJMO and Questions 1/4 on USAMO.

AMC 12:
It's harder, but you have to take this to make USAMO. In my opinion, it's easier to qualify for AIME with this test, but much much harder to get a really high score.

If you're looking to make Red MOP, I think that you should take both since you could qualify for either the USAMO or the USAJMO. Concentrate on getting the questions above right. Also, if you want to make Red, then you also have to be ready for USAJMO/ USAMO so I would recommend getting to the point where you can solve those questions. Then remind yourself the test taking strategies required on the short answer tests (not making silly mistakes, moving through the questions quickly, etc.)

I am in a situation that in my state, me and one other person are sort of competing for first in state (he is in 10th). I think that you have to take the AMC12 to be able to qualify for USA(J)MO through first in state (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294933]here[/url].

I think I got the answer to my first and second question, other than the USAJMO vs. USAMO thing.
For my third question, I am trying to get to Red MOP in 2011, when I am in 9th. If I can make USAJMO or USAMO, from which of those is it easier to make Red MOP? I think I saw somewhere that the 3/6 on the USAJMO are the same as the 1/4 on the USAMO. Does that mean Red MOP qualifications are based on the same problems, no matter which test you take?
Finally, for my last question, I meant: Is it better to take practice AMC10 tests or AMC12 tests if you are going to take 10A and 12B?

So far, that's all I've heard about making Red MOP: No matter what you take, it will not matter since its based on the same problems.

You should probably prepare for the AMC 12 and try to get a really high score. The AMC 10 is easier after that.

How many problems do AMC10 and AMC12 share?

There's no set number, from what I've done, I would guess about 8.