geometry
geometric transformation
reflection
circumcircle
trigonometry
geometry proposed

Point $C$ is a midpoint of $AB$. Circle $o_1$ which passes through $A$ and $C$ intersect circle $o_2$ which passes through $B$ and $C$ in two different points $C$ and $D$. Point $P$ is a midpoint of arc $AD$ of circle $o_1$ which doesn't contain $C$. Point $Q$ is a midpoint of arc $BD$ of circle $o_2$ which doesn't contain $C$. Prove that $PQ \perp CD$.

Suppose $CP\cap AD=X,CQ\cap BD=Y$, then the result follows from $\triangle CXY\sim\triangle CQP$.

My solution was based on a well-known:

$CD \perp PQ$ iff $DP^2-DQ^2=CP^2-CQ^2$. Then two Ptolemy's, two cosine's theorem and everything comes out nicely.  :)

[quote="juancarlos"]Take $P',Q'$ by reflection of $P,Q$ w.r.t $O_1,O_2$, now:$O_1O_2//P'Q'//PQ$.As $O_1O_2$ orthogonal $CD$ then $O_1O_2$ orthogonal $CD$. QED[/quote]

  And before you constructed P' and Q'   segment   PQ was not // to O_1O_2  ???

  T.Y.

  M.T.

Dear Armpist:
Sorry, I had a big confusion to try to solve it

I'm sorry, but I can't see why $PQ$ is parallel to $O_1O_2$...

A beaautifull solution using the ortological triangles th can also be done.

Consider the triangles $ABD$ and $PQC$. Perpendiculars from $P,Q,C$ are perpendicular bisectors of sides of the triangle $ABD$ hence intersect each other in it's circumcenter. Thus, perpendiculars from $A,B,D$ on edges of $PQC$ intersect each other in one point.

Taking $T\in (CD$ such that $CT=AC$ we can easily show that perpendiculars on $PC$ and $QC$ through $A$ and $B$ pass through point $T$. Since the three perpendiculars are concurent, follows that $TD\perp PQ$ which finishes the proof.  :lol:

to xirti: your solution is very nice!
i have another solution:
let $M = AD \cap CP, N = BD \cap CQ$.
we know $PD^2 = PM.PC$ and CM.CP = CD.CA; thus $CP^2 - DP^2 = CP^2 - PM.PC = CP.CM = CA.CD$.
analog we have $CQ^2 - DQ^2 = CB.CD$, thus $CP^2 - DP^2 = CQ^2 - DQ^2$, then $CD \perp PQ$

dears ,let me do an inversion with center $C$ and radius $CA^2=CB^2$.

circles $o_1$ and $o_2$ goes through $C$,then the inversive form of them is two line.

$CP$ is bisector of $\angle DCA$. and $CQ$ is bisector of $\angle DCB$. ($\angle DCP+\angle DCQ=180$.

then in our inversive shape, $P'$ and $Q'$ are the intersections of $CP$ and $CQ$ with lines $A'D'$ and $B'D'$

 respectively.

now if $CD$ is prependicular to $PQ, CD'$ should goes through the circumcircle of triangle $P'CQ'$.

and because of $\angle P'CQ'=90$ ,$CD'$ should bisect $P'Q'$.

let me check it...

problem: $D'A'B'$ is a triangle.and $C$ is mid point of $A'B'$ the bisectors of angles $D'CA'$ and $D'CB'$, intersects

$D'A'$ and  $ِDB'$ at$P',Q'$ respectively.

prove that:

               $D'C$ biscts $P'Q'$

ok, now this problem is easier then before.

       $\fac{A'C}: {CD'}$=$\frac{A'P'}{P'D'}$

       $\fac{B'C}: {CD'}$=$\frac{B'Q'}{Q'D'}$

and:$\frac{A'P'}{P'D'}$=$\frac{B'Q'}{Q'D'}$

it means $P'Q'$ is paralell to $A'B'$.

$D'C$ is the median of $A'B'$ .and because of the symilarity of triangles $D'P'Q'$ and $D'A'B'$,

$D'C$ also bisects $P'Q'.$

and the problem is solved.

(NOTE:did u know that the circumcircle of triangle $PCQ$ is tanjent to $AB$?)

megus, what grade is this problem for...?

there are no grade divisions in Polish Mathematical Olympiad, everyone has the same set of problems.

the second round of iran mathematical olympiad will begin at 18 apr 2006

i hope the geometry problems on the exam ,be nice and as enjoyable as poland olympiad. ;)

[hide="Here is a short metrical proof."][quote][color=red][b]Lemma.[/b] Let $ABC$ be a triangle. Denote the circumcircle $w$ of the given triangle and the middlepoint $X$ of the arc $\stackrel{\frown}{BC}\subset w$ which doesn't contain the vertex $A$. Then there is the relation [/color]$XA^2-XB^2=bc\ .$[/quote][b]The first proof.[/b] Denote the intersection $D\in AX\cap BC$. From the similitude $\triangle ABX\sim \triangle ADC$ we obtain the relation $AD\cdot AX=bc$ and from the similitude $\triangle ABX\sim\triangle BDX$ we obtain the relation $XB^2=XA\cdot XD$. Thus, $XA^2-XB^2=$ $XA^2-XA\cdot XD=$ $XA(XA-XD)=$ $AX\cdot AD=bc\ .$ Therefore, $XA^2-XB^2=bc\ .$

[b]The second proof.[/b] In the circle $w=C(O,R)$ we have $XA=2R\sin \left(B+\frac A2\right)$ and $XB=2R\sin \frac A2$. Therefore, $XA^2-XB^2=$ $4R^2\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$ $4R^2\sin B\sin C=bc\ .$
[u][i]Remark.[/i][/u] If denote the middlepoint $Y$ of the arc $\stackrel{\frown}{BC}\subset w$ which contains the vertex $A$, then there is and the relation $YB^2-YA^2=bc$ because $AX^2+AY^2=$ $BX^2+BY^2=4R^2\ .$ In the proof of the lemma I used the well-known relation $\sin^2x-\sin^2y=$ $\sin (x+y)\sin (x-y)$ and the [u]theorem of the Sinus[/u].

[b]The third proof.[/b] Denote the intersection $D\in BC\cap AX$and the incircle $C(I,r)$. Therefore, $XA^2-XB^2=XA^2-XI^2=$ $(XA-XI)(XA+XI)=$ $IA(AX+IX)=$ $AI\cdot AX+\rho (I)=$ $2Rr+AX\cdot \frac{b+c}{2p}AD=$ $2Rr+\frac{bc(b+c)}{2p}=$ $2Rr+bc-\frac{abc}{2p}=$ $2Rr+bc-\frac{4RS}{2p}=$ $2Rr+bc-2Rr=bc\ .$
[u][i]Remark.[/i][/u]. I denoted the power $\rho (I)$ of the point $I$ w.r.t. the circumcircle and I used the well-known relations: $XB=XC=XI$, $\rho (I)=2Rr$, $\frac{IA}{b+c}=\frac{ID}{a}=\frac{AD}{2p}$, $AX\cdot AD=bc$ (from the similitude $\triangle ABX\sim\triangle ADC$), $abc=4RS$ and $S=pr\ .$

[b]The fourth proof.[/b] $XA^2-XB^2=\frac{b^2c^2}{AD^2}-\frac{a^2}{4\cos^2\frac A2}=$ $b^2c^2\cdot \frac{(b+c)^2}{4bcp(p-a)}-\frac{a^2bc}{4p(p-a)}=$ $\frac{bc\left[(b+c)^2-a^2\right]}{4p(p-a)}=bc\ .$
[u][i]Remark.[/i][/u] I used the well-known relations $AX\cdot AD=bc$,
${AD=\frac{2}{b+c}\cdot \sqrt{bcp(p-a)}}$ and $\cos \frac A2=\sqrt{\frac{p(p-a)}{bc}}\ .$

[color=darkred][b][u]The proof of the proposed problem.[/u][/b] Apply the above lemma for the triangles $DAC$ and $CBD$ $\Longrightarrow$ $PC^2-PD^2=CD\cdot CA$ and $QC^2-QD^2=CD\cdot CB$. But $CA=CB$. Thus, [/color]$PC^2-PD^2=QC^2-QD^2\Longleftrightarrow$ $PQ\perp CD\ .$[/hide]

Let $CP$ meet $AD$ at $S$ and $CQ$ meet $BD$ at $K$. $\frac{DS}{SA} = \frac{DC}{CA} = \frac{DC}{CB} = \frac{DK}{KB}$ so $SK || AB$. Note that $\angle PCQ = 90$ so we need to prove $\angle DCQ = \angle BCQ = \angle QPC$. Note that $\angle BCQ = \angle CKS$ so we need to prove $PQKS$ is cyclic. It's well known that $CS.CP = CD.CB = CD.CA = CK.CQ$ so $PQKS$ is cyclic.
we're Done.

A computational solution: it suffices to show $\frac{OO_1}{OO_2} = \frac{r_1}{r_2}$, where $O$ is the circumcenter of $ABD$ and $O_1, O_2$ are the centers of the two circles. Then the result will follow as $\overline{PQ} \perp \overline{O_1O_2}$.

By a computation, $$OO_1 = R_{ABD}\left(\cos \angle DBA - \frac{\cos \angle DCA}{\sin \angle DCA} \cdot \sin \angle DBA \right).$$ Then the desired ratio equals $$\frac{OO_1}{OO_2} = \frac{\cos \angle DBA \sin \angle DCA - \cos \angle DCA \sin \angle DBA}{\cos \angle DAB \sin \angle DCB - \cos \angle DCB \sin \angle DAB} = \frac{\sin \angle BDC}{\sin \angle ADC} = \frac{AB}{BD} = \frac{r_1}{r_2},$$ as needed.


$\text{How come }\sqrt{6\cdot\sum_{j=1}^{\infty}{\left(\frac{1}{j^{2}}\right)}}= \pi \text{ ?}$

This should not be in this forum.  I will post a calculus solution, but would a moderator move this to a suitable place?

First, prove that
\[\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\zeta(2)\]
using a geometric series expansion.
By substituting $x=\frac{\sqrt{2}(\alpha-\beta)}{2}$ and $y=\frac{\sqrt{2}(\alpha+\beta)}{2}$, you compute the integral to obtain $\zeta(2)=\frac{\pi^{2}}{6}$.  Rearranging the identity
\[\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6},\]
the desired result follows.


Are these two terms differnet ...?if yes what are the conditions for a reaction to be feasible and that to be sponataneous :?:

Yes i think so bcos $ \Delta$G can be < 0 but $ \Delta$S can be < than 0  :maybe: thus making it feasible but non spontaneous  :maybe:

the condition for the reaction to be spontaneous is $ \Delta$G < 0

but for every spontaneous process isnt there increase in entropy?  :maybe: that means, $ \Delta$S > 0 doesnt it?  :huh:

EDIT: Maybe they are seen from 2 different perspectives but mean the same  :maybe:

As I have always been taught it:

For a process occuring at an absolute temperature $ T$ with a change in enthalpy $ \Delta H$ and a change in entropy $ \Delta S$, the associated Gibbs Free Energy $ \Delta G$ is given by:

$ \Delta G \equal{} \Delta H \minus{} T\Delta S$

As long as $ \Delta G < 0$ (which means that $ \Delta H > T\Delta S$), the process is spontaneous.

My point is very clear:
Spontaneity=>$ \Delta G < 0$ or thermodynamic feasibility
Feasibility=>$ E_a$ is very small or kinetic feasibility

Yes i just checked out, Kinetics VS Thermodynamics  :)

Feasibility = possible process that can be reversible or irreversible.
Spontaneity = the process is possible and irreversible.

Those arguments above with the Gibbs free energy are valid as long as we are only considering processes at contant T and P. For different conditions, other thermodynamic potentials should be minimized.

is there any specific condition carcul say at constant temperature and pressure by which we can say a reaction is feasible

For a reaction to occur at constant T and P we must have $ \Delta G < 0$.

taht i think refers for it to happen spontaneously :)
what about the feasibility my thought was...
we have $ \Delta G^{0} \equal{} \minus{}RT \ln K$ so for a reaction to be feasible and not only spontaneous in real sense $ K > > 1$ so we  must have $ \Delta G^{0}$ at that temp. $ < < 0$

I think the term feasible refers to the yield. Its more like "Is this reaction feasible for labaratory purpose" and so.

yes that's what i wrote above and the necessary condition for that :)

[quote="Pardesi"]so for a reaction to be feasible and not only spontaneous in real sense[/quote]

That's not also a good criterion. There are reactions for which K >> 1 and that simply do not occur at an appreciable rate. What you are looking for is this: for a reaction to be feasible it must have a reasonable rate constant at room temperature or nearby.

yes had thought of that but...i wanted the feasibilty criterion at a partciular temp...so in some sense the term should not be exavcly 'feasible' rather relatively feasible :)

So feasibility is associated with kinetics (i.e., a feasible process is kinetically favorable)?

And spontaneity is associated with thermodynamics (i.e., a spontaneous process is thermodynamically favorable)?


Well, I have seen any threads here about the US Open Tennis Tournament, so I guess I should start one :)

What have you thought of the matches? What have you enjoyed the most? Have you been there, and if so, what were your experiences? etc, etc

My experiences:

I went there on Friday, September 1.
I first saw the Vaidisova vs. Jankovich, which was a good close match (not to mention sitting in the front row, which you can do without paying any extra in the side courts).

Later, saw the Peer vs. Schiavone match, which was definitely the best one of the day. Peer was down 5 1 in the first set and managed to tie it up, winning *5 match points* against her in the process. She then proceeded to actually win the match in a tie-breaker.

After that, I saw the Navratilova/Bryan vs. Morariu/Bryan match, which was great.

Encounters with players:
- Jankovich almost crashed into me
- Schiavone served a ball and it hit me in the cheek (hard)
- Got Bob Bryan's autograph
- Was within 2 feet of Martina Navratilova, but got no autograph

I must say, it's must more fun to actually be there. :yup:

what channel? i didn't even watch junior nationals in swimming..but i have a guess ESPN?

Well, I was watching it on NBC, USA Network, and CBS.

GRAAAAARGH YOU ARE SO LUCKY
you actually went there????
AHHHH

I only saw one match...my mom wouldn't let me watch them because I had to do homework... but I watched Agassi's second round match!
Definitely the best match I ever saw(but then again, I didn't watch a lot of matches.just part of wimbledon2006 and some usopen series 2006).

I'm in awe of Agassi. and Baghdatis.

WOW.

[quote="junggi"]:(  :(
Agassi just lost :([/quote]

Yes, his last match. Agassi is oficially retired from tennis.

davenport vs. S. Williams

should be a good match.

[quote="José"]What about Hewitt vs. Gasquet? Really interesting match tomorrow[/quote]
that match is today.
I wanna see the Murray v. Davydenko match and Haas v. Safin

I went to the US open for the first weekend (kid's day/practice day) and opening monday and tuesday. AMAZING experience. I saw the Agassi-Pavel match which was really good. Got lots of autographs from players such as Mauresmo and Sharapova, and i am looking forward to going back for the finals weekend.

I was wondering who everyone is cheering for.
My favorites are Blake, Nadal, Mauresmo, and Sharapova.

OH MY GOD
I am so jealous of everyone who got to go to US Open and get autographs and stuff... :(

My favorites are
men's:Federer,Murray,Haas and Ferrero.
women's:Henin-Hardenne and Sharapova.

Henin-Hardenne! Go Belgium! Go Belgium! :)

I was soooooooooooo sad when Agassi lost. :( He also gave a very touching speech.
I am actually quite surprised how well Janovic is doing. I hope she continues to do well.

Henin-Hardenne will lose to Davenport.
My favourites:
men: Federrer, Safin
women: Sharapova, Davenport

Do you really think henin will lose to Davenport? I hope so, but most people are expecting henin to win. I hope they will at least Play today. There are WAY to many rain delays. Who will win the Sharapova vs. Golovin? I think that will be a very good match. I like them both a lot, but think Sharapova will prevail in three sets.

I think henin will beat davenport and go on to win the open :ninja:

Federer and Mauresmo or Sharapova will win the tournament, that's my bet!

omfg Nadal lost today to Youzhny.
Wow Youzhnyc took out the 3 top spaniards in a row :o  :o

I can't believe Nadal LOST! Have you seen Youzhny's ritual after winning? It is cool

nooo....I just saw the news, b/c I was at school.. :(
oh wow I don't think I ever saw Robredo or Ferrer play :(
anyways, I was kinda hoping Nadal will lose. Now Federrer will win for sure :D

haha I knew Henin would beat davenport... :ninja:

Have you heard of the new stuffed toy called the feder-bear? It is sme fundraiser for Unicef or something.
Anyway..... Davydenko beat Haas! I am pretty sad, because i was hoping Haas would play federer in the semi's (unless of course Blake beats him :P ) and beat him. He has a fairly good chance. I heard the Roddick- Hewitt was a good match- but, i couldn't watch it because of HOMEWORK! I hope Roddick beats Youzhny.

Sharapova v. Henin-Hardenne in the finals  :)
should be a good match...but I was hoping for Mauresmo Henin rematch for some reason. Probably b/c I didn't get to see the wimbledon finals :(

I REALLY hope Maria (the pretty girl) will beat henin. I will be cheering her on, but i have the WORST seats. Section v in Promenade or something. I hope Navratilova and Bob win also.

[quote="Arne"]Of course Henin will win... :D[/quote]
of course... :D

federrer won  :D

Now Rodick vs. Youzhny....I want Rodick to win. Big Serve ftw!  :P

Unfortunately, Rodick had a mere 70% first serve the first set...

[quote="calc rulz"]Unfortunately, Rodick had a mere 70% first serve the first set...[/quote] Is that bad??  :maybe:
PS. It's Roddick. :D

It's not THAT bad but it would be a lot better if it's higher(huge first serves :D )edit:isnt 70% good?

Wow tv broadcast says henin will become #1 with usopen win! :D
go henin!

whoahoho...
sharapova wins US open! :clap:  :clap2:
Now how does that change the rankings....
$\#4\rightarrow\#2$??

Yes! Sharapova won! Now, she's number 3, number 2 is Henin, and 1 is Mauresmo.

I stayed up late last night and watched Federer vs. Roddick (the match started at 23:30 or so in my country). I was very sad that Federer won. Again. Roddick made two very good sets (2 & 3), but I guess he couldn't handle Federer all the way.

Now only the Masters are left. Will he win there also? I hope not.

He'll win again too. He's, far far far away, the best player nowadays (maybe the greatest ever?)


vector
linear algebra
matrix
linear algebra unsolved

Show algebraically that if u and v are non-zero vectors in R^2 which are not
parallel to each other, then any vector 〈x, y) ∈ R^2 can be written as a linear
combination of these two vectors.

It just seems obvious to me but i don't know how to prove

Rephrasing this: let $ u$ and $ v$ be two given linearly independent vectors in $ F^2$, and let $ w$ be any other vector in $ F^2$. Show that $ u,v,w$ are linearly dependent.

This is a case of the theorem that says dimension makes sense; we want to show that there can't be three linearly independent vectors in the two-dimensional space $ F^2$.
The standard tool is Gaussian elimination. You're working with a non-square matrix, so there isn't room for enough pivots and you must get a zero row/column. If you track the operations to get there, you can see how that zero is a linear combination of the original vectors.

(Note: that $ F$ is an arbitrary field. You're only doing arithmetic on the entries, so the only relevant properties of $ \mathbb{R}$ are that it's a field. I prefer stating these basic theorems in the general form, so you don't have to redo them when you deal with other fields later.)


algorithm
logarithms
advanced fields
advanced fields unsolved

Take a look at the following algorithm that approximates the Graph coloring algorithm (in which we should color vertices with  minimal number of colors possible, and where a legal coloring is such that no two adjacent vertices have the same color).

input - graph G
1. find the I -  biggest independent set in G and color it with some unique color.
2. G <- G - I
3. goto 1

Can you please help me finding the relation between the optimal solution and the approximation? what is the guarantee factor?

thanx

Given this 3-colorable graph

[img]http://www.math.tamu.edu/~chillar/images/chaograph.gif[/img]

your algorithm may choose $ I\equal{}\{3,4,5,7\}$ on the first step and end up using four colors. So the guarantee factor is at least $ 4/3$.

(Image taken from [url=http://www.math.tamu.edu/~chillar/]Chris Hillar's webpage[/url])

If the graph is $ k$-colorable and has $ n \ge 2$ vertices, then your algorithm will use at most $ k \ln n$ colors.  That's because, at each step, the independent set will remove at least a fraction $ \frac{1}{k}$ of the vertices.

And the $ \ln n$ factor is inavoidable unless you improve your algorithm somehow. Indeed, consider the regular $ n \equal{} 2^m$-gon and number its vertices from $ 0$ to $ n$. Inscribe the regular $ 2^{m \minus{} 1}$-gon into it joining all even vertices in a cycle. Then inscribe the regular $ 2^{m \minus{} 2}$-gon into it joining all vertices with indices divisible by $ 4$ and so on. Here is the picture for $ n \equal{} 16$.
[asy]size(150); int k,m; for(k=1; k<16; k*=2) for(m=0; m<16; m+=k) D(dir(360/16*m)--dir(360/16*(m+k))); for(k=1; k<16; k+=2) D(dir(360/16*k));[/asy]
Suppose that you chose the red vertices for the first step (clearly, it is [b]the[/b] maximal independent set). Then, after the first step, you are down to the same picture with $ m \minus{} 1$ and you are welcome to do this unwise choice again. So, you'll get about $ m$ colors. On the other hand, 3 colors will suffice for this graph (just use induction).


What is the ration of the area of the square to the area of the circle drawn in the diagram? Kindly explain the answer. Thanks.

[hide="I got"]
$\frac{64}{25 \pi}$

(See graph)

Set EC=x, FE=y.
FB=FE(Both are radii), FG=EC, BG=2x-y. So
\[y^2=x^2+(2x-y)^2\]
Solve it and we have 4y=5x. The area of circle is $y^2 \pi$ and the area of square is $4a^2$.
[/hide]

I did it a similar way you did BornforMath but I got $64/100pi$. Here's how I did it: I let the radius equal to $2$ and 1/2 the side length of the square $x$. Then I said that the right triangle would have hypotenuse of $2$ and one side length of $x$ so the other would be the square root of $4-x^2$. So then $2x=sqrt4-x^2+2$ since it's a square and I got $x=4/5$ so the area of the square would be $64/25$. Maybe I made an error in finding $x$?

[quote="dakyru"]Let's start from here:

$2x = \sqrt{4 - x^2} + 2$

$(2x - 2)^2 = 4 - x^2$

$4x^2 - 8x + 4 = 4 - x^2$

$5x^2 - 8x = 0$

$x(5x - 8) = 0$

$x = \frac{8}{5}$
$A_{square} = 4x^2 = \frac{256}{25}$[/quote]

Yup, I screwed up on my calculations, $x$ should have been $8/5$ not $4/5$, thanks darkyru


MATHCOUNTS calculus trigonometry articles

If an arrow is shot at a target, it is moving right.
But if you pause time at any given moment how is the arrow
moving :?:
Discuss the question.
You can also ask other questions or post other problems
Best of Luck to all you ppl going to states! :lol:

wasn't that from the art of problem solving book 1?
i remember seeing that somewhere...i thought about it... :D

yeah
I'm pretty sure that it isn't moving
it all is based on you perception of
movement, and the different forms of physics.
You do Mathcounts?

yeah, im taking it tomorrow (i live in California)

Awww this is such a basic paradoxical question. Shows up in any random book that talks about paradoxes or puzzling stuff. I think it has something to do with an infinite number of actions of zero length can be done in a finite time or something but you don't have to listen to that I'm not good at this philosophical stuff.

Anyway hey are you Kevin Li from Texas? You know like in 7th grade and went to San Marcos last summer

Nope wrong guy dude
Nationals are in Texas I think though
I've lived in PA my whole life.

the answer has to do with something you can't picture but doesn't exist like ripping a hole so fast that you puncture the atoms and it creates a vacuum, bad example. :blush:

[quote="Walk Around The River"][quote="Kevinli99"]States?
Mine are on March 23rd :thumbup:
Our coach had us learn everything
we did everything we got Calculus, Trigonometry, Algebra 2, etc.[/quote]

is it just me, or does that seem a tad pointless? well the trig could come in handy...but calculus??[/quote]

Hey you never know when calculus can come in handy in Mathcounts. Just another way to brute force an easy problem to check if it's right

We can not possibly take a picture at every moment, and we are not responsible for explaining what cannot be observed. Took that argument from a Scientific American article on Zeno's paradoxes.

kevin li is from college station, texas he's my best friend and he's going to nationals.are you kevin chen from sugar land?[/hide]


A large cube is dipped into red paint and then divided into 125 smaller congruent cubes. One of the smaller dubes is then randomly selected. What is the probability that the cube selected will have at least $ 25\%$ of its surface area painted red? Express your answer as a common fraction.

This will mean it has at least 2 red faces.

Lets use complementary counting. WE have 125 total cubes and we want to eliminate ones with 0 or 1 face painted.

125-27(inside 3x3x3)-6x9(6 faces, each with 9 central cubes with only one painted face)=98-54=44.

Assuming that's correct, the answer is then $ \frac{44}{125}$ :)


dude people wear your ARML shirt

(no not the dirty ARML shirt or the inexplicably red ARML shirt, the real ARML shirt)

why wouldn't I?

Good luck to everyone going!

W00T!!!  I ordered turkey?  anyone else?

ham ftw

oh and spread the word about ARML shirtingness tis an awesome ARML shirt

oh im wearing a shirt even better than an ARML shirt. Ima wear a DUKE shirt. yea i know i'm so original  :rotfl:

dude no, that's silly

wear the ARML shirt

(were you on the ARML teams)

nah I wasn't that great at math last year. Hopefully I can make it this year  :)

yes in fact I am wearing the ARML shirt RIGHT NOW

[quote="MysticTerminator"]oh ok this is an acceptable reason for not wearing the ARML shirt[/quote]

What are you talking about? That's a horrible reason.

Sorry guys! When you leave for Duke at 6am and go to sleep at 9 the night before, it's kind of hard to get messages. Trust me, I seriously considered wearing that shirt. And I [i]obviously[/i] will for Princeton.

dude I did not tell jeremy nor noah yet they both wore it

it is just something you KNOW to do

also

wearing the red arml shirt is simply NOT ACCEPTABLE

yeah I know

We've got to wear the same thing again this year


[quote="SplashD"]In year N, the 300th day of the year is a Tuesday. In year N+1, the 200th day is also a Tuesday. On what day of the week did the 100th day of year N-1 occur?

(A) Thursday
(B) Friday
(C) Saturday
(D) Sunday
(E) Monday

Please post a solution as well as an answer.

[hide]I first got (B) Friday, then got (D) Sunday, and then (A) Thursday. I still don't know what I did.[/hide][/quote]

[hide] The 300th day of year N and the 200th day of year N+1 are either 265 or 266 days apart. Since $266\equiv0\pmod7$, year N is a leap year. Therefore, year N-1 is not a leap year and 300th day of N and the 100th day of N-1 are 565 days apart. $565\equiv5\pmod7$. $\boxed{Sunday}$ is the 5th day from Tuesday. :D [/hide]

[quote="surge"][hide]
The first part of the problem shows that year $N$ is a leap year, so year $N-1$ is not a leap year.

Therefore, $565 \mod {7}\equiv 5 \mod {7}$

$5$ days before Tuesday is $\boxed {Thursday}$[/hide][/quote]

Oh. Yeah.  :blush:

[quote="surge"][hide]
The first part of the problem shows that year $N$ is a leap year, so year $N-1$ is not a leap year.

Therefore, $565 \mod {7}\equiv 5 \mod {7}$

$5$ days before Tuesday is $\boxed {Thursday}$[/hide][/quote]

actually year N+1 is a leap year
but it doesnt change the answer
i think that's right

No... we wouldn't be able to tell if $N+1$ is a leap year...

Remember it's going by days, not months (so February 29th cannot be distinguished).

actually you can tell N+1 is a leap year

(365-100)=265 which is not a multiple of seven but (366-100)=266 is :D

Right, but going from the 300th day of year $N$ to the 200th day of year $N+1$ you cannot pass the 366th day of year $N+1$, but rather you pass the 366th day of $N$. Therefore, $N$ is a leap year and $N+1$ is not.

[quote="surge"]No... we wouldn't be able to tell if $N+1$ is a leap year...[/quote]

you just contradicted yourself by saying year $N$ is a leap year and year $N+1$ isn't :P

That isn't contradiction (tsk.. tsk... an integral word in the realm of math). If I said year $N+1$ is a leap year and it is not a leap year then it would be contradiction.

Year $N+1$ IS NOT a leap year. Year $N$ is a leap year, check lotr's solution and my original solution.


If $n$ is positive integer, prove that $(n+1)(n+2)...(2n)$ is divisible by $2^n$

Base case: $n=1$
Then the product is just $(1+1)=2$, which is divisible by $2^1=2$.

Inductive step:
Assume $2^{n-1}|(n-1+1)(n-1+2)...(2(n-1))$.
Then $2^{n-1}|n(n+1)...(2n-2)$.
Want to show: $2^n|(n+1)(n+2)...(2n)$.
The product can also be written $(2n)(n+1)(n+2)...(2n-2)(2n-1)$.
Since $2^{n-1}|n(n+1)...(2n-2)$, then $2^{n-1}|n(n+1)...(2n-2)(2n-1)$.
However, the product still has an extra factor of two at the very beginning, so $2^n|(2n)(n+1)...(2n-2)(2n-1)$ and we are done.


linear algebra
matrix
linear algebra unsolved

$ \left|\begin{array}{cccc}0 & x & y & z \\
x & 0 & z & y \\
y & z & 0 & x \\
z & y & x & 0\end{array}\right| \equal{} (xyz) \left|\begin{array}{cccc}0 & x & y & z \\
1 & 0 & z/x & y/x \\
1 & z/y & 0 & xy \\
1 & y/z & x/z & 0\end{array}\right| \equal{} \left|\begin{array}{cccc}0 & xz & xy & yz \\
1 & 0 & z & y^2/x \\
1 & z^2/y & 0 & x \\
1 & y & x^2/z & 0\end{array}\right|$

$ \equal{} \frac {1}{xyz} \left|\begin{array}{cccc}0 & xyz & xyz & xyz \\
1 & 0 & z^2 & y^2 \\
1 & z^2 & 0 & x^2 \\
1 & y^2 & x^2 & 0\end{array}\right| \equal{} \left|\begin{array}{cccc}0 & 1 & 1 & 1 \\
1 & 0 & z^2 & y^2 \\
1 & z^2 & 0 & x^2 \\
1 & y^2 & x^2 & 0\end{array}\right|$, for $ x,y,z \neq 0$.


Determine all pairs of positive integers such that
$ \frac{1}{x}\plus{}\frac{x}{y}\plus{}\frac{1}{xy}\equal{}1$

[quote="galois01"]Determine all pairs of positive integers such that
$ \frac{1}{x}\plus{}\frac{x}{y}\plus{}\frac{1}{xy}\equal{} 1$[/quote]

[hide="solution"]
Combine all the fractions to get:

$ \frac{y\plus{}x^{2}\plus{}1}{xy}\equal{}1$.

Solving for $ y$ gives:

$ y\equal{}\frac{x^{2}\plus{}1}{x\minus{}1}$.

We can also rewrite the fraction:

$ y\equal{}x\plus{}1\plus{}\frac{2}{x\minus{}1}$.

$ x$ is an integer so $ x\plus{}1$ is also an integer.  Thus $ \frac{2}{x\minus{}1}$ must also be an integer.  However the only values for $ x$ that will make it an integer is when $ x\equal{}2$ or $ 3$.

Therefore the only solutions are $ \boxed{(2,5)\text{ and }(3,5)}$.[/hide]

Kondr and jli your solutions are correct  :D


Let $ABDC$ be a tangential quadrilateral with an incircle $(I)$ and $X$ the intersection of the diagonals $AC, BD$. A normal to the diagonal $BD$ through the point $X$ intersects the angle bisectors $AI, CI$ at points $P, R$. Show that $XP = XR$.

I think it's been solved before on the forum.

I'll use something which I won't prove. It's a pretty nice problem in itself: Let $M$ be the projection of $I$ on $BD$. Then $\angle AMB=\angle CMD$. Let's refer to this result as $(*)$.

From Menelaus, we see that our conclusion is equivalent to $\frac{PA}{AI}\cdot\frac{IC}{CR}=1\ (\#)$. Now let $U=MA\cap PR,V=MC\cap PR$. It's easy to see, from the fact that $IM\|PR$, that $(\#)$ is equivalent to $\frac{UA}{AM}\cdot\frac{MC}{CV}=1$. This, however, is easy to derive from $(*)$.

LOL, grobber - looks like it's easier to you to write down a solution than to find the link...  :D

Actually, the previous discussion of the problem was at http://www.mathlinks.ro/Forum/viewtopic.php?t=20538 .

  darij


What kind of cell phone do you like/have/would like to have :)? I'm satisfied about my Sagem MyX-7. Lately I use it more than my PC :)

I have one, but I don't know anything about it. I rarely use it.

You have a ton of good topics today  :D

i like all of the nextel phones because they look cool.  I wish i had one of them or  the motorolla razor.

i have a funny looking t mobile with horrible connection becuz t mobile sucks

Yeah! That's a nice phone. I would like a Motorola mpx300 :D

i have a Siemens CF62. its a very convenient phone. you can use it in any country. the motorola RAZR looks awesome. but, what happens when you drop a thin phone like that? ;)

Today I went out with some friends and I saw one. It's so thin ... I took it in my hand but that didn't last too much `cuz I was afraid I could broke it :P. Nokia N90 is awesome  :o

[quote="MithsApprentice"][quote="aznmonkey1992"]i dont want a RAZR since the battery sucks :P[/quote]

The put it on the charger when you go to sleep like every other day. You have to do that with most phones anyways.[/quote]

My phone is strange. When I use it, it sort of "recharges" itself. wHen I dont' use it, the batteries drain.

Eh, I only use my phone at school. And sometimes on vacations and stuff. Not too good, but it's light weight and small, which is just fine for me.

[quote="yif man12"]Eh, I only use my phone at school. [/quote]

I thought cell phones were illegal in school (if they ring or something). They are in mine.

Not really in school. Just calling my parents about pick-up times and things.

I want one of those awesome Nextel phones...so awesome ;)

I share a phone with my mom, it's a siemens C56 I think... A candy bar phone, I like it. I'm not all that picky though, my brother gave me his old cellphone that I love!!!! I'ts a nokia color screen and you can open it up and it's like a keyboard to send e-mails and txt messages and such. I love it!

[quote="236factorial"]I thought cell phones were illegal in school (if they ring or something). They are in mine.[/quote]

In our distrcit, you can have them as long as they don't ring. i.e. put it on silent and send text messages.

I have the really basic nextel phone, the one they're offering free with a contract. It's not really that special, but it's battery lasts for a while if I don't do anything like go on the internet.

Course, we should probably be calling nextel something like Sprint Nextel, because that's the name of the new company.

[quote="noneoftheabove"]whats a razr[/quote]

This is a Motorola v3 Razr  :D

[img]http://www.cacell.co.za/images/v3-large.gif[/img]

[quote="noneoftheabove"]what are the pros and cons cuz from what i can understand all it does is look cool[/quote]
pros: looks cool
cons: price

i have a motorola v180 flip phone.

lol i have a something phone that i barely know how to work...
i mean...all you need it for is calling...

i wouldnt want a phone just because it looked cool i want a fun cuz it works well and the features are cool

[quote="furious"]pros: looks cool
cons: price

i have a motorola v180 flip phone.[/quote]

Apparently it also gets dirty quite easily. But I guarantee you there's more pros than that. For one, it has some fairly nice features. And that amazing keypad isn't bad at all.

how much does it cost becuz u could get some cell phones for free as long as u buy a certain plan

Some Cingular agents offer it for free. T-Mobile also has the phone and you can also get it free with them. But you have to get one of the more expensive plans with T-Mobile.

[quote="MithsApprentice"]

 And that amazing keypad isn't bad at all.[/quote]

Um, sorry... Amazing? How does it work, it looks like a piece of paper. There's a nokia that I like, I have no idea what the model is...

[quote="Black_stripe*36"]How does it work, it looks like a piece of paper. [/quote]

Touchpad may have been a more appropriate description. But you know what I mean.

we bought a certain type of phone only so that we could have 3 phones for free but the connection is horrible


In how many ways can eight horses finish a race provided that there can be ties?


rotation
number theory
modular arithmetic

The integers from 1 to 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?

Sigh, I'm dumb. :D

OK....well there is actually a solution for this.  Look at problem 15.  I think you can solve this with modular arithmetic but I don't know how.  But here's a stopgap solution I guess(Problem 15):

This is a particular case of the famous [b]Josephus Flavius problem[/b], see [url]http://en.wikipedia.org/wiki/Josephus_problem[/url] and many other links (enter the bold words in Google).

Hint: Work Backwards

[hide="Solution"]
I don't know why this works, but if you start at 11 and do whatever the problem says, you arrive at the answer, which is 9.
[/hide]

alternatively, you can just pick any number to erase first, let's pick 3
after you are done, you will end up with a number that would be the answer if our first number erased was 3
in this case, it is 5

we know the actual number left is 11, which is 6 more than our answer

thus, the first number the student erased must be 3+6, or 9

[quote="vallon22"]we know the actual number left is 11, which is 6 more than our answer

thus, the first number the student erased must be 3+6, or 9[/quote]

Is this always true? I'm a bit skeptical at how convenient that is. :huh:

it's actually the way i solved it the first time i saw this problem

i'm pretty sure it always works

let's call the first number we erase "x"
then the second number is "x+3"
the third number is "x+6", etc (all mod 15, in this case)

no matter what number we start with, we should get the same answer, since the intervals between the numbers are always the same

Think of it this way. You assign letters to the spots (ABC...O), so if you start at C, you end up at E. With the numbers, rotate the whole thing 6 numbers counterclockwise such that 9 is in the C spot and 11 is in the E spot. It will now work.


geometry
3D geometry
tetrahedron
sphere
algebra unsolved
algebra

This is from a Brazilian Olympiad.
Find the maximum volume of a tetrahedron that can be subscribed in a ellipsoid of equation $ \dfrac{x^2}{9} \plus{} \dfrac{y^2}{16} \plus{} \dfrac{z^2}{25} \equal{} 1$
I have the solution here. But I can't understand. The man calculates the volume of the tetrahedron in a sphere of unitary radius. Then he finds the volume of this tethahedron subscribed in this sphere. Soon after, he does a linear transformation to transform the sphere into the ellipsoid. Like this:

T=
\[ \left( \begin{array}{ccc} 3 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 5 \end{array} \right)
\]
Then he multiplies the volume of the tethahedron by the determinant of the transformation. Why?
I didn't understand the meaning of this transformation. Is that $ T: \Re^3 \minus{} > \Re^3$ ?

Maybe you should see this: http://en.wikipedia.org/wiki/Determinant

look at the applications section.  It gives a geometric definition for the determinant of a linear transformation.


Given [tex]A=\{1;2;...;m\}[/tex]
and [tex]A_1;A_2;..;A_n\subset A ;A_i\neq A_j\forall i\neq j[/tex]
Prove that :exists [tex] x_1;x_2;..;x_n [/tex] such that [tex]x_i\in A_i \forall i=i;2;..;n[/tex]  and [tex]x_i\neq x_j \forall i\neq j [/tex]
if only if [tex]\forall I\subset \{1;2;...;n\}[/tex] then [tex]|\cup\limit_{i\in I}A_i|\ge |I|[/tex]

This is Hall's marriage theorem. You could Google it. I'm sure tons of links will turn up.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group and intersection of its p-Sylow subgroups is trivial($p\,|\,|G|$). Let $G$ has abelian p-Sylow subgroups. Prove that there are $R$ and $Q$ such that $P\bigcap Q=1$ where $P$ and $Q$ are p-Sylow subgroups of $G$.

Don't you mean :

Let $G$ be a finite group and intersection of its p-Sylow subgroups is trivial($p\,|\,|G|$). Let $G$ has abelian p-Sylow subgroups. Prove that there are $P$ and $Q$ such that $P\bigcap Q=1$ where $P$ and $Q$ are p-Sylow subgroups of $G$?

 :maybe:


calculus
LaTeX
videos
geometry
3D geometry
blogs
MATHCOUNTS

This game is simple.

First, I'll say *inserts coin* and then you say what comes out of the vending machine

here i go (hope i dont die)

*inserts coin*

lots of math homework pops out and you spend your time not on the good computer

*inserts coin*

Receives an alternate reality portal.

Inserts a mouse.

a soda comes out and sprays you with root beer

*inserts coin*

Receives a vending machine that dispenses vending machines that dispense...

Inserts 999 Hyrule Rupees.

I remember this game!

recieves 999 root beers that were shaken a little too much.

inserts an elephant and a donkey.

Recieves elephent and donkey ah........ waste.

Puts in teacher.

Receives a death-metal listening, circuit debator, sri lankan, serial killer(for all of you people satisfying some of these, don't be insulted, I'm just trying to describe my friend).

*inserts TWO coins*

Recuevers TWO root beers that spray all over you.

*Inserts a computer.*

Receives the "I love you" virus and 3 keys from the keyboard
ALT CTRL DEL

*inserts coin*

recieves a pressed penny. :]

inserts history homework.

Receives C+

Inserts 99 mario coins, and a 1-up mushroom.

recieves nintendo ds that is broken

*inserts coin*

Receives a black hole.

Inserts MOOFINS! (fake muffins!!)

Receive a moderator who locks this thread.

Inserts a request PM to demon that asks to lock this hate-filled thread.

Receives a "NO!" for 1=2

Inserts 7h3.D3m0n.117

Receives a pile of poo.

Inserts a coin that I pull out after I insert it.

Receive a request PM to demon that asks to lock this hate-filled thread.

Receives nothing.
Inserts a lock on the thread.

Receives a key to unlock it.

Inserts icy pi

receives error message..
inserts a PM for this to be locked.

Receives a shut up sign for kimmystar94.

Inserts kimmystar94's heart, lungs, and brain. :D

Receives kimmystar94.

Inserts a shameful guy :blush:

Yeah, this is getting kind of spammy, sorry I didn't see the posts earlier.

Sigh. It is with great regret that I lock G&FF's current longest running marathon (not to mention it was my sister's creation).

 :(


Let $ a_1,a_2,..,a_n\in R$. and $ \max |a_i \minus{} a_j| \equal{} 1$ ($ 1\leq i,j\leq n$)
Find the minimum of $ A \equal{} \sum^n_{i \equal{} 1}|a_i|^3$


find all functions from the nonegative integers into themselves, such that: $2f(m^2+n^2)=f^2(m)+f^2(n)$ and for $m\geq n$ $f(m^2)\geq f(n^2)$.

In fact, you can solve it without using the second condition (that's what I did for Crux some years ago  :)  )

Pierre.


induction
ratio
number theory
relatively prime
number theory unsolved

$ F_{n}$ is the Fibonacci sequence $ F_{0}= F_{1}= 1$, $ F_{n+2}= F_{n+1}+F_{n}$. Find all pairs $ m > k \geq 0$ such that the sequence $ x_{0}, x_{1}, x_{2}, ...$ defined by $ x_{0}= \frac{F_{k}}{F_{m}}$ and $ x_{n+1}= \frac{2x_{n}-1}{1-x_{n}}$ for $ x_{n}\not = 1$, or $ 1$ if $ x_{n}= 1$, contains the number $ 1$[/quote] :D

assume $ x_{0}\not=1$ otherwise we are done.
then $ x_{0}>1$ and $ x_{1}<0$. then a trivial induciton gives us $ x_{n}<0$ so the only solution is $ x_{0}=1$ which directly implies $ m=1$ and $ k=0$...

[quote="goc"]Then $ x_{0}>1$ and $ x_{1}<0$.[/quote]

No, $ x_{0}<1$.

m=2n+1, k=2n

We see that $ x_{n}=\frac{x_{n+1}+1}{x_{n+1}+2}$. Then we can easily show by induction that if $ x_{n}=1$, $ x_{n-i}=\frac{F_{2i}}{F_{2i+1}}$. In particular, $ x_{0}=\frac{F_{2n}}{F_{2n+1}}$ for any n.

Now suppose that some other ratio of Fibonacci numbers is of this form, i.e. that $ \frac{F_{2n}}{F_{2n+1}}=\frac{F_{k}}{F_{m}}$. Since $ F_{2n}$ and $ F_{2n+1}$ are relatively prime, $ F_{k}>F_{2n}$. Note that the ratio of consecutive terms $ \frac{F_{n+1}}{F_{n}}$ is strictly increasing. Then $ F_{m}=F_{k}\frac{F_{2n+1}}{F_{2n}}<F_{k}\frac{F_{k+1}}{F_{k}}=F_{k+1}$, so $ k<m<k+1$, a contradiction. Hence the only solutions are (m,k)=(2n+1,2n).


Find the real value of $\frac{\log_{3}i}{i}$ where $i=\sqrt{-1}$.

[hide="Solution"] $\frac{\ln i}{i \ln 3}= \frac{i \left( \frac{\pi}{2}+2 \pi k \right) }{i \ln 3}= \boxed{ \frac{\pi+4 \pi k}{2 \ln 3}}$ [/hide]

[hide="I highly doubt that my crazy answer is right:"]Rationalizing the denominator (multiplying num and denom by $i$):
\[\frac{\log_{3}i}{i}=-i \log_{3}i =-\log_{3}i^{i}\]
Since for any integer $i$, $i^{i}= e^{\frac{(2n-1)\pi}{2}}$,
\[=-\log_{3}(e^{\frac{(2n-1)\pi}{2}}) =-\frac{(2n-1)(\log_{3}e)\pi}{2}\]
[/hide]

EDIT: t0ra beat me to it... What am I doing wrong? We are getting different answers...

EDIT: Thanks, paladin8

[quote="vishalarul"]$i^{i}= e^{\frac{(2n+1)\pi}{2}}$[/quote]

I think you want $i^{i}= e^{-\frac{\pi}{2}+2 \pi n}$ instead.

[quote="t0rajir0u"]$\frac{\ln i}{i \ln 3}= \frac{i \left( \frac{\pi}{2}+2 \pi k \right) }{i \ln 3}= \boxed{ \frac{\pi+4 \pi k}{2 \ln 3}}$[/quote]
I don't think that the $k$ term is generally included, but I may be mistaken.  Also, the problem asked for the real [b]value[/b] (singular), so I wouldn't think that you need the $k$ term.

The subject actually appears to be fairly interesting.  Consider: http://en.wikipedia.org/wiki/Complex_logarithm


Prove that do not exist any pair of natural numbers $ (x;y)$ such $ x^{2}\plus{}y$ and $ y^{2}\plus{}x$ be total squares.

We must have for some m,n natural numbers $ y = 2mx + m^2$ and $ x = 2ny + n^2$
Solving the simultaneous equations we get $ x = \frac {m(3m + m^2)}{3(9 - 4mn)}$

since x and y are positive, we have the following possibilities for (m,n): (1,1), (1,2), (2,1).

The corresponding (x,y) are $ (\frac {1}{3},\frac {11}{3}), ({\frac {16}{3}, \frac {11}{3}), (\frac {11}{3}, \frac {16}{3})}$, none of them natural numbers

[hide="Alternately"] If $ y \equal{} 2mx \plus{} x^2$ then $ y^2 \plus{} x \equal{} (x^2 \plus{} 2mx)^2 \plus{} x$.  But the smallest square larger than $ y^2$ is $ (y \plus{} 1)^2 > (x^2 \plus{} 2mx)^2 \plus{} 4x$.  Contradiction. [/hide]

WLOG assume that $ x\geq y$ then we have:

$ x^2<x^2\plus{}y\leq x^2\plus{}x<(x\plus{}1)^2$

$ \Rightarrow x^2<x^2\plus{}y<(x\plus{}1)^2$

contradiction...

hsbhatt, I think something went massively awry in your solution. Your basis, the fact that y = 2*m*x + m^2 (for natural numbers m and n) and the other equation for x imply that,

y > 2*x     and

x > 2*y
which doesn't make sense. Also, I have no idea how you solved for x. I got something completely different. Moreover, the solutions that you "happened upon", when plugged back into the equation don't return any of the values of y and x that you posted.

Anyway, I thought maybe you mistyped and that m,n could be any integer. but even that doesn't make sense. Every number y cannot be represented as

y = 2*m*x  + m^2 (with m,n any integer)

Take y = 6

We get

m^2 + 2*m*x -6 = 0

Solving this as a quadratic in m, we get (after simplification)

m = -x +- sqrt(x^2 + 6)

but the quantity x^2 + 6 never is equal to a perfect square because this quantity oscillates between 2 and 3 mod 4, while perfect squares must be 0, or 1 mod 4.

I have no idea what happened.

[quote="daydrummer347"]hsbhatt, I think something went massively awry in your solution. Your basis, the fact that y = 2*m*x + m^2 (for natural numbers m and n) and the other equation for x imply that,

y > 2*x     and

x > 2*y
which doesn't make sense.[/quote]

He got his two initial equations by letting $ x^{2}\plus{}y\equal{}(x\plus{}m)^{2}$ and $ y^{2}\plus{}x\equal{}(y\plus{}n)^{2}$ and then simplifying.  The contradiction you point out is true, but it doesn't invalidate the solution - actually it provides a quicker way to finish the problem.  He could have just noticed the contradiction and finished there without solving the equations.  Then the solution would be basically equivalent to Babek's.

I didn't check whether he solved the equation correctly, so you may be right about that.

hey sorry abt that.

yeah. i had done a similar problem recently, the question being find x,y such that $ x^2\plus{}3y$ and $ y^2\plus{}3x$ are both perfect squares. I seem to have typed it out wrong. But otherwise its fine


I noticed that not so many people like geometry... I think it's more fun that other types of math, because you don't have to go squaring both sides of an equation a million times to solve for x. SO I wonder why people hate Geometry so much... I think I know why.

The main reason why I don't like geometry is because there are way too many proofs. Other than that, I like geometry.

Same.

I hate geometry.

Although I got an AMC 10 question right by measuring out with my graph paper, b/c it was draw to scale :D

I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's. I hate geometry. I hate proofs. Appears too much on the AMC's.

What do you mean? We don't hate geometry!

I chose "Some of the above" for everything except proofs. Because proofs are part of every Olympiad-level problem.

And now we can't bring out programmable graphing calculators with all our formulas :(

I am not a good artist. I suck at graphing. I suck at diagrams.

Whoa. I can't believe that the major reason why people hate geometry (not icluding those who love it) is because they can't draw a perfect diagram. I thought proofs were the main reason.

The only possible reason I could have for disliking geometry is that I'm not particularly good at it contest proofs in geometry, but that's hardly a reason. I admire those who can do them well :)

Depends what you mean by Geometry. I like geometry when it's taught/self-taught the AoPS way. It also depends what you mean by proofs - two-column proofs or real proofs? I despise two-column proofs but I like real proofs. They're hard, I'm bad at them, but they're also fun.

[quote="pythag011"]Proofs = GOOD. PROOFS = EXTREMELY GOOD. Just not Geometry proofs that teachers make you do. Teachers are bad at making you do proofs in good ways. So everything in geometry is taught the wrong way. I don't hate geometry...I hate school geometry.[/quote]

I completely agree!

If you do a two-column proof of an IMO problem using basic stuff you would have to use quite a number of pages. The two-column proof is inefficient.

The two column proof, in my opinion, is an excellent way to initiate a person into proofs. I also think that, when studying basic geometry, it is important to see how the fundamental postulates and theorems are rigorously used to prove new results. It's worth entertaining even if you already know the stuff.

I like geometry, so there's obviously nothing I hate about it! :)


hi
I have a large (too large) table built with tabular.  I centered it using \begin{center} and in ajusting the margins and it worked.  Now I want to add a caption to my table so...

\newenvironment{changemargin}[2]{%
 \begin{list}{}{%
   \setlength{\topsep}{0pt}%
   \setlength{\leftmargin}{#1}%
   \setlength{\rightmargin}{#2}%
   \setlength{\listparindent}{\parindent}%
   \setlength{\itemindent}{\parindent}%
   \setlength{\parsep}{\parskip}%
 }%
\item[]}{\end{list}}

\begin{changemargin}{-4cm}{-4cm}
\begin{table}[htb]
\caption{Matrice des reponsabilités}
\begin{center}
\begin{tabular}{c|l|c|c|c|c|}
...
\end{...bla}

but now my table is left align with no consideration to my margin changes and it doesn't fit in the page on the right.

thank you

ledurt

well, i have an answer that worked really well.

\begin{table}
\caption{...}
\end{table}
\begin{tabular}
...
\end{tabular}

so case closed!

ledurt


[color=red][b]Ο χώρος αυτός θα μείνει μόνο για Ολυμπιάδες και διαγωνισμούς.[/b][/color] .

Άλλες τυχόν μαθηματικές απορίες μπορούν να μπαίνουν εμβόλημα στα μηνύματα.

 Για λυκειακές - σχολικές ασκήσεις , όσοι θέλουν μπορούν να γράφουν στο :

http://clubs.pathfinder.gr/MATHEMATICA

 Δώσαμε την ευκαιρία για αρκετούς μήνες με τα σχετικά στίκις για να πλατύνουμε το κοινό , αλλά η ανταπόκριση ήταν μικρή. Φαίνεται ότι η διάθεση για μελέτη μαθηματικών με βιβλία και επικοινωνία είναι ακόμα μικρή στην πατρίδα μας. Αυτό είναι  φυσικό, αφού υπάρχει το εύκολο σχολείο και το φροντιστήριο.

 Εδώ όμως είμαστε για να ακούμε τις ανησυχίες του κοινού και να περεμβαίνουμε. Η προτεραιότητα είναι στις ολυμπιάδες. Να σας θυμίσω ωστόσο ότι τα στίκις γίναν για να μη γεμίζει το φόρουμ με διάσπαρτες ασκήσεις και έτσι να γίνεται αδύνατη η εύρεση μιας καλής άσκησης τη στιγμή που τη θέλεις. Και είχαμε όλοι σχεδόν συμφωνήσει , αυτές τις ασκήσεις , έστω λίγες , να τις συγκεντρώσουμε σε ξεχωριστό χώρο.
Το ξέρετε ότι αυτό που με ενδιαφέρει είναι η φωνή και η άποψη σας .  Ο δικός μου ρόλος είναι πρωτίστως να κρατήσω το χώρο μακρυά από έριδες και ανεπίτρεπετες ενέργειες καθώς και να παρακολουθώ διακριτικά την αλληλογραφία.

[b]Αγαπητοί φίλοι , [/b]
   τα μαθηματικά , οι διαγωνισμοί και το μέλλον ανήκουν σε σας. Τραβάτε μπρός  και χαράξτε το δικό σας δρόμο !

    [color=red][b]    Καλή συνέχεια  !!![/b][/color]

 [i]( Α! Αν συμφωνείτε , να κάνουμε στικις τους κλάδους των ολυμπιάδων , για να συγκεντρώνονται οι ασκήσεις σε ομάδες.Πχ: [b]ΟΛΥΜΠΙΑΔΕΣ - ΑΝΙΣΟΤΗΤΕΣ [/b],   [b]ΟΛΥΜΠΙΑΔΕΣ - ΑΡΙΘΜΟΙ  [/b] κλπ.  Δεν το προτείνω , ούτε το θεωρώ φοβερό , αλλά λύνει κάποια προβλήματα που μας βασάνισαν αρκετά στο παρελθόν και φτιάξαμε , θυμάστε , τις κατηγορίες.  Εκτός αυτού, για αυτό ζητάγαμε υποφόρουμ από τον Βορνίκου και διαχειριστή.
 Σκεφτείτε και πείτε τη γνώμη σας.)[/i]

Είμαι καθηγητή;ς Μαθηματικών σε σχολείο Β/θμιας εκπαίδευσης στην Εύβοια.Παρακολουθώ την κίνηση του forum οπότε και οσο μπορώ και με αφορμή την απόπειρα "γκετοποίησης" του Ελληνικού fοrum  με το σβήσιμο φακέλων με μη ολυμπιακό περιεχόμενο και κάποια μηνύματα μελών του , θέλω να εκφράσω μία αντίθετη  άποψη στο θέμα αυτό.
 Θα ήθελα τα μέλη να ξανασκεφτούν την απόφασή τους αυτή για το περιεχόμενο  της  παρούσας ιστοσελίδας , η οποία αποτελεί ένα μοναδικό φωτεινό παράθυρο   προς τον κόσμο των μαθηματικών .
 Σας παροτρύνω ο καθένας σας να γυρίσει μερικά χρόνια πίσω στη ζωή του  όταν βρισκόσασταν στο Δημοτικό ή στα πρώτα γυμνασιακά χρόνια και ανακαλύψατε ότι υπάρχει και ο κόσμος των μαθηματικών .΄Οταν ο καθένας σας  είδε ότι έχει μια κλίση προς αυτά τα δύσκολα και απλησίαστα για τους πολλούς Μαθηματικά, όταν γοητευτήκατε από μία έξυπνη άσκηση που σας έδωσε ο καθηγητής σας , όταν βρήκατε μία δεύτερη λύση (διαφορετική από το λυσάρι ) σε μια άσκηση γεωμετρίας , όταν ακούσατε ή όταν( από περιέργεια ίσως ) συμμετείχατε σε έναν μαθηματικό διαγωνισμό για πρώτη φορά. Και τότε γοητευτήκατε από τα Μαθηματικά και αρχίσατε να ψάχνετε και οι απορίες διαδέχονταν η μία την άλλη και βοήθεια ίσως να μην υπήρχε , ενώ απόψεις του στυλ " τι θες και ασχολείσαι με αυτά  , είναι δύσκολα, που θα σου χρησιμεύσουν; ' κλπ  σας έκαναν να αισθάνεστε  σαν      " εξωγήινοι" .
 Προσπαθώ , φλυαρώντας ίσως , να σας υπενθυμίσω πως και σήμερα κάπως έτσι ξεκινάει κάποιος να περπατάει στο δρόμο που εσείς έχετε πορευθεί και για να γίνει αυτό χρειάζεται αρωγούς που δεν τους έχει πάντα. Ένας τέτοιος αρωγός  είναι και αυτό το forum  και πιστεύω ότι πρέπει να μείνει αρωγός για τον καθένα προς αυτήν την κατεύθυνση. Για να γίνει αυτό καλύτερα και να μην δημιουργούνται εμπόδια σε όσους είναι προχωρημένοι , ίσως πρέπει να δομηθεί η ιστοσελίδα καλύτερα σε επίπεδα όπως το διεθνές forum, ώστε ο καθένας (δεν αναμένεται βέβαια και κοσμοπλημμύρα , αν και οι συμμετοχές σε διαγωνισμούς σε τάξεις του Δημοτικού αφήνει αρκετές ελπίδες για το μέλλον ) να βρίσκει αυτό που θέλει . Γιατί οι σχολικές ασκήσεις και η συζήτησή τους ρίχνει το επίπεδο , αν αυτό γίνεται χωριστά από την Ολυμπιακή σελίδα;
Παιδιά γιατί να μην γίνουν γνωστές σε όσο το δυνατόν περισσότερους οι επιτυχίες σας για τις οποίες σας συγχαίρουμε και η πορεία σας να αποτελέσει κίνητρο για κάποιους; Γιατί να μην μπορεί κάποιος να γνωρίσει μέσα από το παρόν forum σταδιακά τον όμορφο κόσμο και τα άλματα της μαθηματικής σκέψης στη χώρα μας , τα οποία δεν προβαλλονται όπως τα άλματα ενός καλαθοσφαιριστή;
 Συγχωρέστε με , ξέρω ότι  σας κούρασα λίγο παραπάνω με όλα αυτά , κάνοντας κατάχρηση της φιλοξενίας σας , αλλά ήθελα να τονίσω ότι κανείς δεν θα χάσει στο τέλος , ακόμα κι αν μπουν π.χ. και ανέκδοτα στη σελίδα ή αθλητικά , αρκεί το καθένα να μην μπερδεύεται με τα υπόλοιπα .
Ελπίζω η ιστο σελίδα να εξακολουθήσει να είναι και ΟΛΥΜΠΙΑΚΟΥ ΕΠΙΠΕΔΟΥ  αλλά και ΟΛΥΜΠΙΑΚΟΥ ΠΝΕΥΜΑΤΟΣ.
 Ευχαριστώ

Καταρχήν σας ευχαριστούμε που γράφετε εδώ και εμένα προσοπικά με χαροποιεί ιδιαίτερα που ένας καθηγητής Β/βαθμιας εκπαίδευσης ασχολείται με το φόρουμ αυτό γιατί όπως πολύ καλά ξέρετε οι περισσότεροι συνάδελφοί σας δεν παροτρύνουν τα παιδιά να πάνε στους διαγωνισμούς της ΕΜΕ (τους σνομπάρουν...γιατί άραγε ?(ξέρω την απάντηση αλλά δεν είναι της παρούσης  :wink: ))

Θα σας εξηγήσω γιατί αποφασίστηκε να γίνει αυτό το <<ξεδιάλλεγμα>> στο φόρουμ .
Καταρχήν δεν μπορούμε να δημιουργήσουμε υποφόρουμ (όπως στο άγγλικό ) , το μόνο που μπορούμε να κάνουμε είναι τα stickies . Αλλά τα stickies μπαίνουν πάνω πάνω και δεν μπορούμε εκεί  να έχουμε σχολικά μαθηματικά ή αθλητικά κτλ .
Αυτά δεν είναι σε καμιά περίπτωση άσχημα αλλά χαλάνε την συνοχή του φόρουμ (μιλάμε με τα δεδομένα που έχουμε που δεν μπορούμε να φτιάξουμε υποφόρουμ ) και επιπλεόν χτυπάνε πολύ άσχημα . (δηλαδή όταν μπαίνεις σε ένα καθαρά μαθηματικό φόρουμ να βλέπεις αν πήρε η τάδε ομάδα τον τάδε παίχτη ή αστειακια κτλ) .
Αυτό  πιστεύω είναι φυσιολογικό . Δεν μπορεί ούτε το αθλητικό σάιτ να μιλάει για μαθηματικά ούτε το ανάποδο .
Επιπλέον πρέπει να καταλάβουμε όλοι ποιος είναι ο σκοπός αυτού του φόρουμ .
Ο σκοπός αυτού φόρουμ όπως τον αντιλαμβάνομε εγώ τουλάχιστον είναι να βοηθήσει τα παιδιά που [color=red]ασχολούνται [/color] ήδη με διαγωνισμούς να πάνε καλύτερα σε αυτούς . Να συζητηθούν δηλαδή κυρίως δύσκολα θέματα που όλοι μαζί να βοηθήσουν ώστε να λυθούν απορίες να μαθευτούν νέες τεχνικές , να συζητηθούν ασκήσεις από ξένα περιοδικά ,ξένους διαγωνισμούς και όλα τα συναφή .
Το φόρουμ αυτό δυστυχώς δεν μπορεί να πάρει τη μορφή για να παρακινήσει άτομα να ασχοληθούνε με τα μαθηματικά . Αυτός ο ρόλος είναι του σχολείου και της οικογενείας .Και δεν μπορεί κάποιος να βρει αυτό το φόρουμ αν δεν ασχολείται με μαθηματικά (εκτός από κάποιους που έμαθαν ότι υπάρχει ένα λυσάρι που τα λύνει όλα )
Το πιο απεχθές πράγμα για μένα στο φόρουμ ήταν ότι έμπαιναν διάφοροι για να λύσουν την άσκηση για το φροντηστήριο ,το σχολείο ,για να περάσουν το μάθημα στο πανεπηστήμιο .Αυτά δεν είναι σωστά πράγματα και το φόρουμ δεν θα πρέπει να είναι αρμόδιο ώστε ο τάδε να μάθει διανύσματα ή να μάθει ταυτότητες ή ολοκληρώματα κατά Rieman απέξω για να πάει να τα δώσει και να περάσει το μάθημα .
Είχαμε καταντήσει σε αυτή την κατάσταση γιαυτό έγινε ότι έγινε .Και ότι έγινε έγινε γιατί από το φόρουμ έφυγαν οι καλοί , αυτοί που θα μπορούσαν να προσφέρουν ....
Εξαλλου για το σχολείο , το πανεπηστήμιο και όλα τα παραπάνω υπάρχουν άπειρα βοηθητικά λυσάρια κτλ
Για τις ολυμπιάδες δεν υπάρχει τίποτα σε υψηλό επίπεδο (ΒΜΟ ,ΙΜΟ) σε ελληνική βιβλιογραφία και γιαυτό αυτό το φόρουμ προσπαθεί να καλύψει αυτό το κενό που υπάρχει όσο μπορεί .
Τέλος για μένα τουλάχιστον αυτό το φόρουμ είναι ένα μεράκι .
Δηλαδή , εδώ που ζω δεν μπορώ να μιλήσω σε κανέναν γιαυτά που διαβάζω γιαυτά που λύνω για τις απορίες μου κτλ γιατί ή θα με πάρει για τρέλό ήθα μου λέει διάφορα άλλα σπαστικά . Αυτό που γίνεται εδώ είναι μοναδικό !
Μπορείς να συζητήσεις πάνω σε τέτοια θέματα με άτομα που κατέχουν καλύτερα από σένα το αντικείμενο και να μάθεις από αυτούς και επιτέλους κάποιος να εκτιμήσει τη μαγεία και την ομορφιά μιας λύσης σου . Όχι όπως λένε κάποιοι
<<Ε, συντρέχουν και τι έγινε .... :!: >>
Πιστεύω πως γιαυτό υπάρχει αυτό το φόρουμ εδώ (γιατί δεν υπάρχουν ειδικά σχολεία για να είμαστε όλοι μαζι κτλ)
Και επειδή είναι μοναδικό στο είδος του δεν πρέπει να το χαλάσουμε .(Για όλα τα προαναφερθέντα υπάρχει και δεύτερο και 100οστό στο είδος τους )
Και όταν παραλαμβάνεις κάτι αναλοίωτο ,μια παράδοση από τους παλαιότερους , δεν πρέπει να το αλλοιώσεις ...
Ήμουνα κουραστικός το ξέρω αλλά πρέπει να ξεκαθαρίζονται οι θέσεις για να μην έχουμε ερωτηματικά και παρεξηγήσεις .
Ευχαριστώ για τον κόπο να το διαβάσετε (αν και αυτά τα μπουρου μπούρου χαλάνε το χαρακτήρα του φόρουμ  :P  . Αλλά αφού δεν έχουμε κατασταλάξει ακόμα τέλος πάντων)

Βασικά, θέλω να ρωτήσω κάτι.

Τα topics που είχαν δημιουργηθεί ως stickies δεν υπάρχουν πλέον καθόλου στο φόρουμ; (Δηλαδή οκ, τα βγάλαμε από την "περίοπτη" θέση, αλλά, καταργήθηκαν τελείως; )

Τέλος, με συγχωρείτε για την απουσία από το φόρουμ, αλλά έχω αναλάβει το στήσιμο του ασύρματου μητροπολιτικού δικτύου στο Κιλκίς (Kilkis wireless metropolitan network - http://www.kilwmn.gr για πληροφορίες ) και δεν έχω καθόλου χρόνο  :(

Πάντως το φόρουμ εμφανώς πάει πολύ καλύτερα ! Μπράβο σε αυτούς που το κατάφεραν και ανεβάζουν τον πήχυ συνεχώς !  :)

Επίσης , αν κάποιος έχει να προτείνει κάποια [b]άσκηση του μήνα[/b] για να μπεί στο site για τον [b]Οκτώβριο[/b], ας μου στείλει ένα pm ή e-mail στο stfu.gr@gmail.com !

Ευχαριστώ,
Στέλιος

[quote="mostel"]Βασικά, θέλω να ρωτήσω κάτι.

Τα topics που είχαν δημιουργηθεί ως stickies δεν υπάρχουν πλέον καθόλου στο φόρουμ; (Δηλαδή οκ, τα βγάλαμε από την "περίοπτη" θέση, αλλά, καταργήθηκαν τελείως; )
Ευχαριστώ,
Στέλιος[/quote]

Στέλιο ΝΑΙ !

Αυτό είπατε στα μηνύματά σας και ήδη το έπραξα. Εδώ είμαστε ωστόσο να το ξανασκεφτούμε , αν χρειαστεί.
 Θα φτιάξω όμως τα στίκις για ολυμπιακά κεφάλαια , ένα ανά κεφάλαιο ,  διότι διαφορετικά το site θα γεμίσει με σκόρπιες ασκήσεις , παρεμβολές , ερωτήσεις , άσχετα πράγματα κλπ και όλο το καλό και αξιοποιήσιμο υλικό θα εξαφανιστεί . Το πάθαμε ήδη πέρισυ αυτό και θυμάστε τι αγώνα κάναμε μέχρι να φτιάξουμε εκέινα τα '' Γεωμετρία , Ασκήσεις 1 - 20 '' , ώστε να μην πάει η δουλειά χαμένη.
 Για την αλληλογραφία πάνω στο σχετικό θέμα (δες  τα δύο προηγούμενα μηνύματα) , θέλω να ακούσω πρώτα τις απόψεις σας και μετά να  σας πώ την άποψή μου.Προσωπικά δε βλέπω να υπάρχει κανένα πρόβλημα , αρκεί όλοι μας να θέλουμε να βοηθήσουμε και να μην σνομπάρουμε αυτούς που τώρα κάνουν τα πρώτα τους βήματα ή μας απευθύνουν ρηχές για μας ερωτήσεις.
 Σας το είχα ξαναγράψει : όταν ξεκίνησα να μελετώ ανισότητες , αλληλογραφούσα με έναν τρομερό λύτη στα μαθηματικά , τον Καθηγητή στο Πανεπιστήμιο του Brandenburg  Eckard Spect. Τώρα που ξανακοιτάζω τα μηνύματα και βλέπω τι τον ρωτούσα , κοκκινίζω από ντροπή. Και όμως ο άνθρωπος αυτός , χωρίς να με ξέρει , μου απαντούσε σε κάθε ερώτηση ( τώρα τις βλέπω τελείως χαζές) που του έκανα. Αν δεν μου απαντούσε , ή με αντιμετώπιζε αλλαζονικά , θα είχα εγκαταλείψει κάθε  σχέση με το αντικείμενο. Από αυτόν τον άνθρωπο - ήμουν ήδη 44 χρονών - διδάχτηκα πολλά  πάνω στην ανθρώπινη επικοινωνία και συμπεριφορά και τον ευχαριστώ.
 Βλέπω και την αλληλογραφία μερικών στο αγγλικό φόρουμ. Ξεκινάνε από πολύ στοιχειώδεις ερωτήσεις και σε τρία χρόνια δεν πιάνονται !
  Λοιπόν , συνεχίστε εσείς τον ολυμπιακό σας δρόμο και ανοίξτε την αγκαλιά σας σε όλους, όσοι δείχνουν αγάπη στα μαθηματικά. Μην ξεχνάτε ότι στην ΙΜΟ θα πάνε 6 , ενώ τα μαθηματικά πρέπει να τα χαρούνε χιλιάδες . Αν οι μαθητές καταλάβουν ότι μόνο για τους έξι ενδιαφερόμαστε , τότε όλοι θα αποχωρήσουν και πρώτα πρώτα οι έξι , γιατί θα νοιώθουν μοναξιά. Αν δημιουργήσουμε όμως ολυμπιακό κίνημα , τότε οι έξι θα φέρνουν χρυσά και οι υπόλοιποι θα έχουν μια βαθειά αγάπη για τα μαθηματικά, όντας συγχρόνως και πολύ καλοί επιστήμονες.

 Στο φόρουμ υπάρχει χώρος για όλους ! Αλλά για να πάει μπροστά , πρέπει οι μικρότεροι να βλέπουν τους μεγαλύτερους να έχουν διάθεση να τους βοηθήσουν. Έτσι δημιουργούν σωστά πρότυπα και δεν εγκαταλείπουν στη μέση τον δύσκολο δρόμο της μαθηματικής επιστήμης. Συνοπτικά , το φόρουμ θα συνεχίσει να υπάρχει ,
 [b]α) [/b]αν δεν φύγετε εσείς που το φτιάξατε με το ενδιαφέρον και το μεράκι σας
 [b]β) [/b]αν με τη συμπεριφορά σας παρακινήσετε και τους μικρότερους να έρθουν κοντά σας και να διδαχθούν από εσας
 [b]γ) [/b]αν οι ολυμπιονίκες και κυρίως αυτοί που σπουδάζουν μαθηματικά ή σε πολυτεχνία μπαίνουν στο φόρουμ , είτε προτείνοντας είτε λύνοντας και προτρέπουν τους νεότερους να ασχοληθούν όλο και πιο βαθειά με το θέμα ολυμπιάδες.
 [b] δ)[/b] Αν καταφέρουμε να βγάλουμε το βιβλίο και προβάλλουμε το έργο μας.

  Αυτά για τώρα . Τα ξαναλέμε

   Μπάμπης

[quote="silouan"]Για να μην παρεξηγηθώ γράφω αυτό το μύνημα .
Εγώ είμαι υπέρ του stickies που φτιάξατε για Ολυμπιάδα Νέων και θα είμαι δίπλα σε κάθε απορία μικρού που προσπαθεί πάνω στα μαθηματικά και που η ερώτηση αυτή αναφέρεται σε διαγωνισμούς (έστω και στην πιο απλή ανισότητα του Cauchy)
Αυτό που έγραψα είναι ότι εδώ δεν μπορούν (εκ των πραγμάτων ) να λύνονται οι σχολικές απορίες για γ λυκείου και να λύνουμε ασκήσεις για φοιτητές που θέλουν να περάσουν το μάθημα .
Ελπίαω να ξεκαθάρισα τη θέση μου[/quote]

[b]Σιλουανέ, [/b]
μην ανησυχείς γιατί ήσουνα σαφής από την αρχή. Δεν σε παρεξηγεί κανένας και όλοι μας καταλαβαίνουμε το βαθύ περιεχόμενο των προβληματισμών σου. Όλοι θέλουμε η σελίδα να έχει [b]ολυμπιακό και μαθηματικό προσανατολισμό[/b].Φυσικά , οι ανάγκες , είναι πολλές και διαφορετικές για τον καθένα και για την κάθε στιγμή.Είδες ότι και στο αγγλικό πεδίο υπάρχει χώρος όχι μόνο για φοιτητές αλλά και για μεταπτυχιακούς.
Μπορούμε εύκολα να ξεχωρίσουμε ποιος μπαίνει για τις ανάγκες του σχολείου και ποιος επειδή θέλει να μάθει και βρίσκει δυσκολίες. Αυτός που βλέπει το φόρουμ ως τον δωρεάν δάσκαλο για το σχολείο θα καταλάβει από την αρχή , μετά από δύο μηνύματα , ότι εδώ δεν έχει θέση. Αλλά σε κανένα δεν μπορούμε να απαγορεύσουμε να γράφει. Αν κάποιος έχει ανάγκη έστω και για σχολική βοήθεια  - φαντάσου πχ κάποιον που του κάηκε το σπίτι ή η περιουσία του στις φωτιές - είναι δυνατόν να μην τον βοηθήσουμε ; Φυσικά δεν χρειάζεται να το κάνετε εσείς αυτό . Εσείς έχετε άλλο σκοπό για τη στιγμή. Υπάρχουν όμως άλλοι , φοιτητές ή καθηγητές που ίσως θέλουν  - μπορεί και να τους αρέσει - να βοηθήσουν.Με αυτή την έννοια πρέπει να είμαστε ανοιχτοί . Ούτε είναι επερκής λόγος να αποχωρήσουμε από το φόρουμ επειδή που και που μπορεί να μπει κάποιος για του λύσουμε και δυο κλασματικές εξισώσεις. Στο φόρουμ οι ρόλοι χρειάζεται να είναι διακριτοί και ξεκάθαροι :
   [b]α) [/b]Αυτοί που πάνε δυνατά για την ΙΜΟ και θέλουν να κάνουν καριέρα στα μαθηματικά.
  [b] β)[/b] Αυτοί που ασχολούνται γενικότερα με τους διαγωνισμούς - μικροί και μεγάλοι -  και τους αρέσει να παλεύουν , χωρίς όμως να έχουν σώνει και καλά στόχο το μετάλλιο.
   [b]γ) [/b]ΑΥτοί που αγαπούν τα μαθηματικά και θέλουν από οποιαδήποτε θέση να συμμετέχουν θετικά για την προώθηση της μαθηματικής σκέψης.

 Για να έχει μεγάλο κοινό η πρώτη ομάδα , πρέπει οι άλλες δύο να έχουν κόσμο. Διαφορετικά και η πρώτη ομάδα θα συρικνωθεί. Δεν σου κρύβω ότι η καθοριστική χρονιά για το φόρουμ θα είναι η επόμενη που θα έχετε τελειώσει το σχολείο εσείς , η ιδρυτική ομάδα.Για αυτό θα προσπαθήσουμε από τον Νοέμβριο που αρχίζει η νέα περίοδος των διαγωνισμών να προτρέψουμε με κάθε τρόπο το νέα παιδιά  να γίνουν ενεργά μέλη . Σίγουρα χάνουν χρόνο , κερδίζουν όμως εμπειρία.

 Λοιπόν , παρακαλώ τα περισινά παιδιά που φτάσαν στον Αρχιμήδη  , αλλά  και όλους εσας που στελεχώσατε τις ολυμπιακές μας όμάδες , junior και senior , να ξεκινήσετε σιγά σιγά να ζεστένεστε ! Ο Σάββας ήδη ξεκίνησε και τον ευχαριστώ . Ζητείστε ο ένας τον άλλο και γράψτε κάτι . Μία άσκηση και η αρχή έγινε .Ας μην κολήσουμε σε διαδικαστικά και τέτοια θέματα. Τα έχουμε όλα ξεκαθαρίσει . Η ζωή είναι απλή και τα μαθηματικά δύσκολα . Η πατρίδα πρέπει να βγει από το τέλμα. Χωρίς μαθηματική σκέψη , αυτό δε θα γίνει ποτέ !
 Πάμε λοιπόν !

[b] Μπάμπης[/b]

 [b]Άντε και κάτι καλό για να σας ευχαριστήσω:[/b]

 [color=blue]Σε μερικές μέρες θα βάλουμε στο χώρο μας την βίβλο της θεωρίας αριθμών. Ένα σπάνιο βιβλίο (σε pdf) καθαρά ολυμπιακού χαρακτήρα. Θα λατρέψετε τους αριθμούς ! Θα το στείλλω  στο Στέλλιο.[/color]


find all the Functions,  for that all  x and y are real,

f(x.f(y))=f(x.y)+x

when f:R---> R

Taking $ x \equal{} 1$ yields $ f(a) \equal{} a \plus{} 1$ for $ a\in f\left( \mathbb{R}\right)$.  Taking $ y \equal{} 0$ gives $ f(cx) \equal{} x \plus{} c$ for some constant $ c$ and all real $ x$.  Since $ f\left( \mathbb{R} \right)\neq \emptyset$, we must have $ c \equal{} 1$ and hence $ f(x) \equal{} x \plus{} 1$ for all real $ x$.

A problem of this level would more likely belong in the pre-olympiad forum.


What is the greatest number of points of intersections can be made with two circles and a square?
[hide="CLUE"][size=59] Use a pattern :cool:  [/size][/hide]

Two circles cut at at most two places. (Can i assume they arn't the same circle?) The maximum crossing points of a square and a circle is 8. Therefore, the absolute maximum is $8+8+2=18$ A quick sketch shows this will work. (draw a square and a circle inside it that cuts it in eight places. Draw another circle that is nearly exactly the same place, of equal radius. QED)

Since nobody is answering I'll Just post the answer.

What seamusoboyle said is the practicly the same thing I was going to say you just need to put the two circles where they cross over one another but they still make 8 oints of intersection with the sqaure

It looks like an illusion

looking from the outer side of the square, it's bent

but

if you follow the lines, they are all straight

my eyes hurt now....

by the way,

how do you use patterns for this problem?

Woah, that illusion is so cool. The circles look like ovals, the square looks like it's curved. Wow! :rotfl:

Its not supposed to be an Illusion and I was wrong about  using a pattern


function
real analysis
real analysis solved

Here go three famous problem:
1. Characterize all functions f: R -> R such that f(x) + f(y) = f(x+y).

2. Does there exist a function f: R -> R which is continuous at all irrationals but discontinuous at the rationals?

3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?

And here is a problem posed by Gabriel Carroll. I would imagine it has been considered before, but I have not seen it:
Does there exist a continuous f: R -> R such that f(x) is rational iff x is irrational?

[quote="gauss202"]

3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?
[/quote]

Van Der Waerden function
For t \in R , <t>=d(t;Z)=inf{|t-m|; m\in Z}
Consider f(t)=\sum_{n \geq 0}<(10^n).t>/10^n  f is defined R->R
continous, 1-periodic, nowhere differentiable.

Another example Weierstrass function
f(t)=\sum_{n \geq 0}2^(-n).cos(((2004)^n)t)

Mr.Omarjee
STT teacher

[quote="gauss202"]Here go three famous problem:
3. Does there exist a function f: R -> R which is everywhere continuous but nowhere differentiable?
[/quote]
you can find the answer to this question nearly in all textbooks on mathematical analysis.

I never knew the Weierstrass function was so simple in form. I remember reading that he shocked the entire mathematical world by discovering this.

I think problem 2 was solved b4 on this forum.

For problem 2, yes, there is :
x irrational ==> f(x)=0
x=p/q rational , (p,q) coprimes ===> f(p/q) = 1/q

I think that's called Riemann's function. I was wrong, I thought it was another problem. I was just wondering: what is f(0)? Are you sure the soln works?

Yes, Diogene's function does in deed work! Though he needs to make an appropriate definition for f(0). Can you see why this function works? It is very neat! I don't know what the name of this function is though, if it has a name. I've never heard it called the Riemann function, but then again there are so many functions known as the Riemann function it's hard to keep track!

As for the Weierstrass function, it is in deed very interesting. I did my senior thesis as an undergraduate on this subject. Using wavelets to study the regularity (i.e. degree of "non-differentiability") of these types of functions.

There was actually an even more famous function before this one, called the Riemann function (yes another one!! lol!!) defined by f(t) =  \sum {n \geq 1}[sin (pi*n^2*t)/n^2], which was shown to be continuous everywhere, but differentiable only at rational points, p/q, with both odd numerator and denominator!! This remarkable result was shown by Gerver in 1971 when he was still an undergraduate!!

So all of the questions have been appropriately solved except for the first one - which is the easiest, but also very interesting in my opinion.

Congradulations to Mindspa on the Carroll one! That was a very elegant and clever solution.

Gerver also discovered the function, or it was known before, but he was the first to prove (or to even try?!) that it is continous everywhere and differentiable only on rationals?  :?

No, it was known before Gerver. It was proposed by Riemann as a function not differentiable on a set dense in the reals. But this was first proven and published about by Weirstrass (hence it is also sometimes known as the Weierstrass function - how confusing!). However his proof did not say at which points, if any, it was differentiable.

The problem was since worked on by many people, including Hardy, who showed that it was not differentiable at any irrational points. However it was not until Gerver that the problem of it's differentiability was completely solved. His proof was long, but more or less elementary (meaning it did not require any hard results). Quite a remarkable feat for an undergraduate! I have no idea what he has gone on to do in his career.

But it just shows that some unsolved problems are not out of the reach of undergraduate students!

Yeah, I think I was hasty about the function Diogene posted. I thought it wouldn't work because it wasn't defined properly in 0, but all we need to do is make f(0)=a=/=0 and it's done.

The problem that was solved b4 and which I was speaking abt was this: does there exist a function f:[0,1]->R s.t the pts of discontinuity are the irrational points of [0,1] ? I think the answer for this one was no.


calculus
derivative
function
real analysis
real analysis unsolved

[quote="drapt@"]suppose that a differentiable real function has a periodical nonconstant derivative f:R--->R .prove that f has at least positive period[/quote]
I dont understand the problem. Consider $ .5x^2$ its derivative is $ x$ which is a non-periodic function?  :?:

[quote="sylow_theory"]Consider $ .5x^2$ its derivative is $ x$ which is a non-periodic function?  :?:[/quote]

and so this is not a counterexample for this :wink:


The rocket MaoMao is on the point (3, -5) and needs to get to (10,1) by only moving up and right. How many possible paths are there along gridlines?

I've seen all kinds of these problems, with pictures too, but do not know how to approach these.

[hide]he goes to the right 7 times and up 6 times, and since any order works thats 13!/(6!7!) = 13C6[/hide]

extension: what if he cannot pass through the point (6,3)?

[quote="nateharman1234"]

extension: what if he cannot pass through the point (6,3)?[/quote]

Did you mean some other coordinate? If so, I guess you would just subtract the triangle formed from that point you can go to, and 10,1.

[quote="nateharman1234"]he goes to the right 7 times and up 6 times, and since any order works thats 13!/(6!7!) = 13C6 [/quote]

We can think of this as having to arrange 7 R's and 6 U's to form a 13 letter word where R represents a right and U represents a up.

[quote="nateharman1234"]extension: what if he cannot pass through the point (6,3)?[/quote]
The ans will be the same as he will never go to (6,3) as the y coord > given y coord and he is taking the shortest path.


I am having trouble with asymptote, and I was hoping someone could help me.  I have installed TeXnic, GSview, GhostScript, and Asymptote.  I have done what it says in the asymptote getting started in AoPSWiki.  When I paste the commands into a new file (which I saved as picture.asy) and run asymptote, a black box pops up, stays blank, flashes an error message (which I think says: error: could not load module picture.asy) and immediately closes.  If I open asymptote and tell it to draw a line, a GSview window opens and I see the line.  Will someone please help me be able to use Asymptote in Texnic? Thank you very much.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557]this thread[/url] for ways to view error messages.


Let $\triangle ABC$.
The inequality $\sum \frac{1}{a}\cdot \sum a^{2}\ge 3\cdot \sum a$ is abviously.
Prove that a reinforcement of this inequality:
$\sum \frac{1}{a}\cdot \sum a^{2}\ge 3\cdot \sum a+\sum \frac{a}{bc}\cdot (b-c)^{2}\ge \sum \frac{1}{a}\cdot \sum ab\ge 3\cdot \sum a.$

The original inequality is equivalent to $\sum a(b+c-a)(b-c)^2 \ge 0$ which I think may not be true.. :huh:

[quote="levi"]I am sorry, you didn't watch condition: $\triangle ABC$ ! This inequality is a manufacture of the my inequality (1) from the post "uncommon". See
http://www.mathlinks.ro/Forum/viewtopic.php?t=44897[/quote]
oops, I really missed the condition a,b,c are sidelengths of a triangle.
That means I've solved it.  :D

Actually,Levi about the first one,here is better one,but trivial:

         $\sum\frac{1}{a}\cdot \sum ab \geq 3\cdot \sum a $ ;)  so now you may reconsider a new improvment for the second one. ;)

 By  the way, this is true for any $a,b,c>0$

May be true the following inequality [b](Try you !)[/b]:
$\sum \frac{1}{a}\cdot \sum ab\le 3\cdot \sum a+\sum \frac{a}{bc}\cdot (b-c)^{2}.$

Well,if we assume that it is true, then it's for $a,b,c$ sides of a triangle.I don't know,I'll give it a try, but not now.
 Tomorrow I am going to do that. Now I'm very sleepy. :sleeping:

I improved the initial message. Caesar, I solve the your assumpsion.

[quote="Cezar Lupu"][size=117][color=darkred]Well,if we assume that it is true, then it's for $a,b,c$ sides of a triangle.
I don't know,I'll give it a try, but not now. Tomorrow I am going to do that. Now I'm very sleepy.[/color][/size] :sleeping:[/quote]

[color=darkblue]See http://www.mathlinks.ro/Forum/viewtopic.php?t=47305

The my $\triangle ABC$-inequality (proposed in Aug. 07 2005)

$\sum\frac{1}{a}\cdot\sum bc\le3\sum a+\sum\frac{a}{bc}\cdot(b-c)^{2}$

you didn't  prove so far, Aug 17 2006 ! [b]Please, try again ... to-morrow ![/b][/color]

Actually, just expand it out. \[3\sum{a}+\sum(\frac{a}{bc}(b-c)^{2})=3\sum{a}+\sum(\frac{ab}{c}+\frac{ac}{b}-2a)\]  \[=\sum{a}+\sum(\frac{ab}{c}+\frac{ac}{b})=\sum{a}+2\sum\frac{ab}{c}\] Whereas \[\sum{\frac{1}{a}}\sum ab=2\sum{a}+\sum\frac{ab}{c}\] So we then are talking about a comparison between $a+b+c$ and $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$.  But then \[(ab)^{2}+(bc)^{2}+(ca)^{2}\ge (ab)(bc)+(bc)(ca)+(ca)(ab)=abc(a+b+c)\]  \[\implies \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\] which tells us \[3\sum{a}+\sum(\frac{a}{bc}(b-c)^{2}) \ge\sum{\frac{1}{a}}\sum ab\] ...

After that, the last of the three inequalities should be easy, by the same principle, and the first should be easy also.  Essenaitlly, it all reduces to \[\sum(a+\frac{a^{2}}{b}+\frac{a^{2}}{c}) \ge \sum(a+2\frac{ab}{c}) \ge \sum(2a+\frac{ab}{c}) \ge \sum(3a)\] which all follows from the fact that $\sum{a}\le \sum{\frac{ab}{c}}$.


If $a,b,c$ are the three sides of a right triangle , and $a^{n}+b^{n}+c^{n}=2(a^{2n}+b^{2n}+c^{2n})$ . And $n \geq 2$ , $n$ is a integer , can you tell me $n = ?$

I think you can use subst $a^{n}=x^{n}+y^{n}$,etc.


After watching the show THE HAPPENING, thers this maths teacher who gave a maths quiz, its about if u get 2 dollars on the first day of April, and the 2nd u day u get 2 times the amount of the first day and continue doing that for the next 30 days.
I found out its actualy 2^1+2^2+2^3+2^4................+2^30. How do i sum up 2^1+2^2+2^3+2^4...................+2^30[/code]

No need for that. Notice that it's actually $ 2^{31} \minus{} 1$.

Why? $ a^n \minus{} b^n \equal{} (a \minus{} b)(a^{n \minus{} 1} \plus{} a^{n \minus{} 2}b \plus{} ... \plus{} ab^{n \minus{} 1} \plus{} b^{n \minus{} 1})$. Now plug in $ a \equal{} 2$, $ b \equal{} 1$, $ n \equal{} 31$.

Edit: Actually, you have an extra dollar in this calculation, (can you see why-it's the $ b^{n \minus{} 1}$ member) so the result is actually $ 2^{31}\minus{}2$.

i dun under stand the a^n-b^n formula.....so deep

There are unfortunately only two ways to simplify this calculation: first one is this ($ a^n\minus{}b^n$), and the second one is by geometric series (probably too advanced for your level), so I used this approach, as it is how the [url=http://en.wikipedia.org/wiki/Second_half_of_the_chessboard]wheat and chessboard problem[/url] (more famous version of your problem) was explained to me in middle school.

I don't see a nice approach to the explaining of the formula. I hope someone else can explain it...

Well, if you don't want to be so rigorous, you can notice that:

\begin{align*}
2 &= 2 = 2^2 - 2 \\
2 + 4 &= 6 = 2^3 - 2 \\
2 + 4 + 8 &= 14 = 2^4 - 2 \\
2 + 4 + 8 + 16 &= 30 = 2^5 - 2 \\
2 + 4 + 8 + 16 + 32 &= 62 = 2^6 - 2 \\
2 + 4 + 8 + 16 + 32 + 64 &= 126 = 2^7 - 2
\end{align*}

You might notice that pattern and deduce that $ 2 + 4 + 8 + \dots + 2^n = 2^{n + 1} - 2$.

If you want to be more rigorous, you can use induction:

[b]Base case[/b]
$ n = 1$: $ 2 = 2^2 - 2$. Check.

[b]Inductive hypothesis[/b]
$ 2 + 4 + 8 + \dots + 2^n = 2^{n + 1} - 2$

[b]Inductive step[/b]
Assume $ k = n$. Then for $ k + 1$:

$ 2 + 4 + 8 + \dots + 2^k + 2^{k + 1} = 2^{k + 1} - 2 + 2^{k + 1} = 2 \cdot 2^{k + 1} - 2 = 2^{k + 2} - 2$

Therefore, if we proved that IF it works for $ n$, then it works for $ n + 1$. Since we proved our base case for $ n = 1$, then it works for 2, 3, 4, etc...

Geometric series is nice to learn btw too... it comes up a lot even in the middle school forums (although some of the stuff in the MC forum arguably could go in the Intermediate HS section lol...) Here is one case:

$ \displaystyle 1\plus{} r\plus{}r^2\plus{}r^3\plus{}r^4\plus{}\ldots\plus{}r^n \equal{} \frac{r^{n\plus{}1} \minus{} 1}{r\minus{}1}$

This is simply a specific case of the finite geometric series formula, with $ a\equal{}2$ and $ r\equal{}2$. Then, just plug in:
\[ S\equal{}\frac{a\left(1\minus{}r^{n\plus{}1}\right)}{1\minus{}r}\]

For a proof of this specific case define the sum to be [b]S[/b].
Then [b]Sr=r+r^2+r^3+...+r^n+r^(n+1)[/b]....1
Subtracting [b]S[/b] from 1 we get,
[b]Sr-r=S(r-1)=r+r^2+r^3+...+r^n+r^(n+1)-(1+r+r^2+r^3+...r^n)
                   =r^(n-1)-1.[/b].
Therefore we get [b]S=r^(n-1)-1/(r-1)[/b]

Here's another way to solve it:

Let $ x \equal{} 2^0 \plus{}2^1 \plus{} 2^2 \plus{} 2^3 \plus{} ... \plus{} 2^n$

We multiply by $ 2$ on both sides.

$ 2x \equal{} 2(2^0 \plus{} 2^1 \plus{} 2^2 \plus{} 2^3 \plus{} ... \plus{} 2^n$

This is the same as:

$ 2x \equal{} 2(2^0) \plus{} 2(2^1) \plus{} ... \plus{} 2(2^n)$

When you multiply numbers with the same bases, you add the exponents. $ a^x \times a^y \equal{} a^{x \plus{} y}$.

$ 2x \equal{} 2^{0\plus{}1} \plus{} 2^{1 \plus{} 1} \plus{} 2^{2 \plus{} 1} \plus{} ... \plus{} 2^{n \plus{} 1}$

which simplifies to:

$ 2x \equal{} 2^1 \plus{} 2^2 \plus{} 2^3 \plus{} 2^4 \plus{} ... \plus{} 2^{n \plus{} 1}$

What's the difference between this and our first equation? We notice that $ 2^0$ is missing, and we have an extra term - $ 2^{n \plus{} 2}$. So we can make this a new equation.

$ x \minus{} 2^0 \plus{} 2^{n \plus{} 1} \equal{} 2x$

Any number to the $ 0$th power is equal to $ 1$(except 0.)

$ x \minus{} 1 \plus{} 2^{n \plus{} 1} \equal{} 2x$

Now we can solve.

$ 2^{n \plus{} 1} \minus{} 1 \equal{} x$

Great! So now we know the answer! The original problem was up to $ 2^{30}$, so we can substitute this.

$ 2^{30 \plus{} 1} \minus{} 1 \equal{} x$

$ 2^{31} \minus{} 1 \equal{} x$

And that is our answer.

My favorite way of looking at it (very easy to understand):
1+1+2+4+8+16+32...
The first 1+1 makes 2, plus the next 2 makes 4, plus the next 4 makes 8, etc. This pattern keeps continuing and as you can see, it always makes double the last number. So for your problem, you'd have to subtract the first two ones from the final answer.
Using this method, find $ 3 \plus{} 2\cdot3 \plus{} 2\cdot9 \plus{} 2\cdot27$...

Starting out with 2^n and successively subtracting all the powers of 2 below it is another way to see it. Let's say you start out with 16. Subtracting 8, you have the other 8 left. Subtracting 4, you have the other 4 left. Subtracting 2, you have the other two left. Subtracting the last one gives 1 left over. So 8+4+2+1=16-1.

Another way to look at it is using base.
1+2+4+8+16...=1111111... (base 2)=1000000...-1(base 2)=$ 2^n \minus{} 1$

I believe there is a combinatorial argument too, since $ 2^n$ is the number of subsets of n distinct objects, since each can either be included or not. Can anyone come up with a combinatorial proof?

This is the basis of the double-or-nothing strategy at the casino. First you bet 1. If you lose, you bet 2, net 1 in return if you win. If you lose again, you bet 4, again netting 1 if you win. Then you bet 8, 16, etc until you win. Once you win, no matter how long it takes, you will always end up netting 1 in the end (unless you keep losing and run out of money and go broke!).

[quote="moogra"]Unfortunately, it doesn't equal $ 2147483646$ which is the right answer.
It equals $ 2147483647$ (this also happens to be the max int)[/quote]

Oh, because I thought the problem was $ 2^0 \plus{} \text{etc.}$. It really doesn't have the $ 2^0$, I misread it. That's why it's 1 off.


function
superior algebra
superior algebra unsolved

Hello everybody!

I have this exercise with a hint, in elementary group theory. I proved the hint quite easily but can't, for the life of me, see how it helps prove the statement in the exercise. So I'll be really grateful for any further hints or general direction.

Let G be a final group with the property $ (ab)^3 \equal{} a^3b^3$ for all $ a,b \in G$. Prove that if $ gcd(o(G),3)\equal{}1$ (where $ o(G)$ is the order of the group G), than G is abelian.
Hint: Prove that $ a^2b^3 \equal{} b^3a^2$ for all $ a,b \in G$ and that the function $ f: G \to G$ defined by $ f(a) \equal{} a^3$ is isomorphism.

I can't see how the hint helps here, I'm sure I'm missing something simple  :(

Many thanks,
Perrin

Let $ |G|\equal{}m$.Since $ (m,3)\equal{}1$ it follows that $ f$ is izomorphism (1).Since $ f(x)\equal{}x^3$ is surjective morfism we have that $ x^2 \in Z(G)$ $ \forall x\in G$ (2). ( where $ Z(G)\equal{}\{g \in G | gx\equal{}xg ,\forall x \in G\}$)

(You said you managed to prove (1) and (2), so I think you are interested in the next part)

Since $ (ba)^3\equal{}b(ab)^2a\equal{}ba(ab)^2$ (since $ (ab)^2$ commutes with any element of $ G$, in particular with $ a$ ), from where we get that $ (ba)^2\equal{}(ab)^2$.From this we have that $ (ba)^3\equal{}(ab)(ab)ba\equal{}abab^2a\equal{}aba^2b^2\equal{}a^3b^3\equal{}(ab)^3$.So we got that $ (ba)^3\equal{}(ab)^3$.But since $ f$ is injective we can conclude that $ ab\equal{}ba$, for any arbitrary $ a,b \in G$.

Thanks! All that was needed is playing a bit with those equalities ($ (ab)^3$ etc.). I always get stuck on the last, most simple step. Thank you very much for your help!  :)


For each $ n\in N$ find the largest number $ C(n)$ such that the inequality

$ (n \plus{} 1)\sum_{j \equal{} 1}^{n}a_j^2 \minus{} (\sum_{j \equal{} 1}^{n}a_j)^2\geq C(n)$

holds for all $ n$-tuples $ (a_1,a_2,...,a_n)$ of pairwise different integers.

:maybe:

I think it is: $ \frac{n(n\plus{}1)^2(n\plus{}2)}{12}$

 :)

ElChapin, I think you would be right if the problem was about nonzero integers.

Hint:

Assume WLOG that $ a_1<a_2<...<a_n$. Then, $ a_i\minus{}a_j\geq i\minus{}j$ for any $ i$ and $ j$ with $ j<i$. Now,

$ \left(n \plus{} 1\right)\sum_{j \equal{} 1}^{n}a_j^2 \minus{} \left(\sum_{j \equal{} 1}^{n}a_j\right)^2\equal{}\sum_{j\equal{}1}^na_j^2\plus{}\sum_{1\leq j<i\leq n}\left(a_i\minus{}a_j\right)^2$.

Estimating $ \sum_{1\leq j<i\leq n}\left(a_i\minus{}a_j\right)^2$ is easy; $ \sum_{j\equal{}1}^na_j^2$ needs some case distinction.

  Darij


induction
probability
logarithms
expected value
combinatorics unsolved
combinatorics

can u please prove that $R(3,q) \leq \frac {q^2+3} 2$      and $ \frac  {p \cdot 2^{\frac p 2}} {e \cdot 2^{\frac 1 2} } <R(p,p) \leq 4R(p-2,p)+2.$

Edited by pbornsztein, to make it readable. I hope it is what you meant.

At first sight, the first upper bound is an inductive one, and the lower bound is a probabilistic one.
I will try to think about it later.

Pierre.

For first problem.

Base $q=2$ and $q=3$.

Suppose contrary, $q\geq 4$. Consider the node $A$ of degree $k$ and its neighbourhoods $A_1$, ..., $A_k$. It is clear $A_i$ and $A_j$ are unconnected for all $i\neq j$, and $k<q$.

If $k\leq q-2$ then exclude $A$, $A_1$, ..., $A_k$ from the graph. New graph will contain at least $\frac{(q-1)^2+4}{2}$ nodes. So by induction supposition we will find either triangle (impossible) or $q-1$ disconnected nodes, but it is impossible too, since they are not connected to $A$, so we will find $q$ disconnected nodes.

So if there is a node of degree $\neq q-1$ we are done. Therefore we are to consider case all nodes have the same degree $q-1$.

Besides of nodes  $A$, $A_1$, ..., $A_{q-1}$  there are $\frac{(q-1)^2+2}{2}$ nodes and there are $(q-1)(q-2)$ edges from $A_1$, ..., $A_{q-1}$ leading to these nodes. Since $(q-1)(q-2)>\frac{(q-1)^2+2}{2}$ for $q\geq 4$ we conclude there exists node $B$ which connected to at least two nodes from  $A_1$, ..., $A_{q-1}$. It follows that by excluding nodes $A$ and $B$ and all their neighbourhoods we will remove at most $2q-2$ nodes. So remaining graph will consist of at least $\frac{(q-2)^2+3}{2}$ nodes. Thus we will be able to find a triangle or $q-2$ disconnected nodes. Both variants leads to contradiction.

Ok, now for $\frac {p \cdot 2^{\frac p 2}} {e \cdot 2^{\frac 1 2} } <R(p,p)$.

Consider the graph $G$ of $n$ nodes. We will include edges to $G$ with probability $\frac{1}{2}$. Let's count the expected value of bad $p$-gons, where $p$-gon is called bad iff it is complete $K_p$ or it is fully empty $K_p$, $p$-gon is a subgraph of $G$ of $p$ nodes.
Indeed, each $p$-gon has probability $2\cdot 2^{-p(p-1)/2}$ to be bad and there are $C_n^p$ $p$-gons in graph $G$. So expected number of bad $p$-gons is $A_n=C_n^p\cdot 2^{-p(p-1)/2+1}$. If $A_n< 1$ then there exists graph $G$ of $n$ nodes s.t. it doesn't contain bad $p$-gons.
It is clear $C_n^p< \frac{n^p}{p!}\leq \frac{n^p}{2p^pe^{-p}}$ due to Stirling's formulae. Therefore, \[ A_n< \exp\left(p\ln n-p\ln p+p-\frac{p(p-1)\ln 2}{2}\right).\]
Substituting $n=\frac {p \cdot 2^{\frac p 2}} {e \cdot 2^{\frac 1 2} }$ we obtain $A_n< \exp(0) =1$.

For upper bound.
First of all, $R(p-1,p)\leq 2R(p-2,p)$. Indeed, consider graph $G$ of $2R(p-2,p)$ nodes, $X\in G$. Then all remaining nodes divide into two sets: set $A$ -- connected to $X$ and set $B$ -- unconnected with $X$. We have $|A|\geq R(p-2,p)$ or $|B|\geq R(p-2,p)$. Now it is clear how to complete the proof.

Secondly, consider the graph $G$ of $4R(p-2,p)+1$ nodes, $X,Y\in G$. Then all remaining nodes divide into four sets: $A$ -- connected to $X$ and $Y$; $B$ -- connected to $X$ and disconnected to $Y$; $C$ -- disconnected to $X$ and connected to $Y$, $D$ -- disconnected to $X$ and $Y$.
If $|A|+|B|\geq R(p-1,p)$ or $|A|+|C|\geq R(p-1,p)$ or $|B|+|D|\geq R(p-1,p)$ or $|C|+|D|\geq R(p-1,p)$ then we are done :) Otherwise we will obtain \[4R(p-2,p)-1=|A|+|B|+|C|+|D|\leq 2R(p-1,p)-2\leq 4R(p-2,p)-2.\] Contradiction.

As we see Pierre was right.

Hope my solutions are correct :)

You are great Myth! I have nothing to write  :)
At first sight your solutions are the same as mine in the two first answers. I will read the third one, even if I trust you.

Pierre.


vector
combinatorics solved
combinatorics

assume that X is a set of n prime number and m is a positive integer . how many subset like A with m  members exist s.t:
a)the members of A are empty of per fect square (means that we can't show them k 2 x )
b)the product of the members of any subset of A isn't square .
c)if n be in  A  and p be prime number s.t n is divisible by p then p is in X
( could u post splendid solution beacaus eI saw boring one)

The answer is (2n-1)(2n-2) ... (2n-2m-1)/m! for m \leq n, and 0 otherwise.

Suppose m\leq n. Because all the elements of A are square free, we may write a vector in Z2n for each of them, where i-th position is 1 iff the i-th prime in X divides the number.

Since no product of the numbers in a subset of A amounts to a perfect square, this translates to no sum of vectors in a subset is 0. Therefore they are all linearly independent. Also, any m linearly independent vectors satisfy this. If m>n it's clear that we have 0 possibilities (since the m vectors form a basis of a subspace of Z2n, which has dimension n).

The first vector can be chosen in 2n-1 ways. Assume that we have chosen the first k. They span a subspace of dimension k of Z2n, with 2k elements. There are 2n-2k vectors left and we can choose any of them. In the end we divide by m! because we count the same m-uple of linearly independent vectors m! times, once for each permutation.


1. Find all positive integers m, n, where n is odd, that satisfy
1/m+4/n=1/12.

Some help please, I started by letting n=2k+1, this gives an equation that can be continually cancelled down working mod 2, but now i can't see where to go.

Just solve it as if there wasn't any restriction, and then only include the allowed.

Yes.  :D
I was hurrying.
But the method is the same...


geometry
rectangle
perpendicular bisector
geometry proposed

Let $ABCD$ be a  rectangle.Let $E$ be a point on $BD$ such that $\hat{DAE}=15^{o}$.$EF \perp AB (F \in AB)$$EF =\frac{1}{2}AB;AD= \alpha$.Caculate $\hat{EAC}=?;EC=?$

[quote="chien than"][color=darkred]Let $ABCD$ be a  rectangle. Let $E$ be a point on $BD$ such that $\widehat{DAE}=15^{\circ}.$ Let $F$ be the projection of $E$ on $AB$ and assume that $EF =\frac{1}{2}AB.$ Caculate $\widehat{EAC}$[/color] [/quote]
[color=indigo][b]Solution. [/b] Denote $B'$ be the intersection of perpendicular bisector of $AE$ with $AB$

We have: $\angle EAB'=\angle AEB'=75^\circ\Longrightarrow\angle AB'E=30^\circ$
Thus $AB'=EB'=2EF=AB,$ so $B\equiv B'$

And hence: $\angle EAC=\angle DAC-\angle DAE=60^\circ-15^\circ =45^\circ$[/color]


Quite easy.
The key is that a divides c 3 which divides b 9
Thus, a divides b 9 . Simplifying the (a+b+c) 13 under modulo a, we get that it is congruent to 0. Similarly for the others. It remains to consider the cases of common factors...

It is easy to see that a,b,c have the same prime factors so
a=p_1 a_1...p_n a_n
b=p_1 b_1...p_n b_n
c=p_1 c_1...p_n c_n
where a_i,b_i,c_i>0
We have from the conditions b|a 3 c|b 3 a|c 3 that
b_i \leq 3a_i
c_i \leq 3b_i
a_i \leq 3c_i

(a+b+c) 13 is a sum of repeating terms of the form a xb y c z where x+y+z=13.
There is no use considering the case xyz>0 because the term in the sum is divisible by abc.
I want to prove that abc divides each term in the sum.
This goes to showing that a|b yc z where y+z=11, and the others. it is easy to see that z \geq 3 or y \geq 9 or else z+y \leq 10 (false). Prove the others similarly.

[quote="orl"]If positive integers a,b,c are such that b divides a^3, c divides b^3, and a divides c^3, prove that abc divides (a+b+c)^13.[/quote]
If a prime $\forall p\mid a$ then $p\mid a \mid b^3 \mid c^9$, and it implies $\text{rad}(a)=\text{rad}(b)=\text{rad}(c)$.
Then $\upsilon_p\left((a+b+c)^{13}\right)\ge 13\min\{\upsilon_p(a),\upsilon_p(b),\upsilon_p(c)\}$ $=(3^2+3+1)\min\{\upsilon_p(a),\upsilon_p(b),\upsilon_p(c)\}\ge \upsilon_p(a)+\upsilon_p(b)+\upsilon_p(c)$. []

[quote="orl"]If positive integers a,b,c are such that b divides a^3, c divides b^3, and a divides c^3, prove that abc divides (a+b+c)^13.[/quote]
in fact always this is true: if $  b|a^n  $,$  c|b^n  $,$  a|c^n  $ then we have $  abc|(a+b+c)^{n^2+n+1}  $

Consider the expansion of $(a+b+c)^{13}$.
It consists of 3 types of terms - i) terms consisting of one variable only like $a^{31}$
                                                  ii) terms consisting of 2 variables like $Ka^{m}b^{n}$ with $(m+n) = 31$
                                                  iii) terms consisting of 3 variables like $La^{p}b^{q}c^{r}$ with $(p+q+r+) = 31$
Its is easy to show that abc divides each of these.
So abc divides $(a+b+c)^{13}$

[quote="TheBernuli"]What means $rad(a)$ ?[/quote]
It's the [url=http://en.wikipedia.org/wiki/Radical_of_an_integer]radical[/url] ;)


Hey,

The AP coordinator at a high school near me has blocked some of my friends from taking AP tests because they haven't taken the AP class for that subject and other reasons.

Is there a way to bypass the AP coordinator at the high school somehow? I have a friend who has been studying calculus on his own for a year and he was denied being able to take the test.

Go to a different school?  Unfortunately, at a given school, it's more or less up to the school.

According to the AP site:

"If you are a homeschooled student or attend a school that doesn't offer AP, you can still participate. Each year hundreds of students participate through independent study. Some states even sponsor online AP courses."

http://www.collegeboard.com/student/testing/ap/reg.html

I guess it depends on the coordinator.

I know for math contests, I've competed at different schools because mine wouldn't offer it.

Call around AP schools in your area, maybe one of them would be willing to allow you to write with their students.

What I've heard is that it depends on whether your school offers the test.
If they don't offer the test, then it's sorta their fault that they didn't open that test to you guys.  On the other hand, they also get some money for proctoring but they would probably rather not proctor 2 people or something.
If your school doesn't offer the test, you probably could take it at some other school, but you have to ask that other school and turn in their form or something to make sure that Collegeboard knows that you are taking it there.

[quote="aufha"]Is there anyway to bypass the high school AP coordinators?[/quote]If you are willing to rock the boat a little, go over their head.  Talk to your school's principal.  If that doesn't work, talk to somebody in your school's superintendent's office or your representative on the school board.  (I'm not saying this is going to work, but it's something to try if you feel strongly enough.)

Ultimately, an AP exam has to be administered through a school.

My son (homeschooled) has had a very good experience taking AP tests in ninth grade and tenth grade at public schools in our town that he has never attended. Both schools are willing to offer ANY test in the AP program to students who sign up in advance, even if the school doesn't offer that course. The College Board website

http://www.collegeboard.com/student/testing/ap/reg.html

makes reasonably clear how to sign up for AP tests at a different high school. To go take the tests, you would have to be excused from attendance at your regular school, but that is a reasonable schedule conflict for missing class, if your own school won't help you register.

When I took AP tests without taking the class, I just signed up for them.  I didn't tell anyone anything, and no one checked.  Of course, they found out afterwards, but they couldn't do anything about it then.  I've heard of schools doing this before, and it seems to usually not work out well.  But you should just keep pushing it... go to the counselor, principal, higher people, w/e.  It's your risk, not theirs.

oh yeah. forgot to mention - like what do you do if they give you trouble because you are in middle school? A middle school kid I know wants to pass out of calc...

so what do you do if they reject you because you're a middle school student?

Young people of any age can take AP tests. Each year there a few eighth graders who take various AP tests, and even a very few who win AP scholar awards, which means they took multiple AP tests.


Δείτε στο συνημμένο αρχείο .

Οk,thanks.Το κατεβασα.....

ΡΑΛΛΗΣ :D