Hi guys !

I'm doing my assigment about the mid point, trapezoid rule and Simpson's rule ! I just can't figure out  in which case, what rule is the best solution ? And I don't know how the mathematicans can find out the formulas of the Errors in the Trapezoid Rule and Simpson's Rule. I need to understand that but I don't know where i can find out these information. May u help me , plz ! I'm appreciated to your help ! thanks! :)

Here's a good site I recommand you to take a look at about the Trapezoidal and Simpson Rules...

http://math.fullerton.edu/mathews/n2003/SimpsonsRule2DMod.html
(I personally don't know these theorems so I just searched on Interent )
and I suggest you not to post these kind of questions in the Classroom Forum...

This is far more advanced than a normal mathclass level....

[quote="shinwoo"]and I suggest you not to post these kind of questions in the Classroom Forum...

This is far more advanced than a normal mathclass level....[/quote]

Oops...I'm so sorri because of posting in wrong place. :blush: thanks for ur help!


slove this equation:

$ \sqrt {1 \plus{} x^2} \equal{} \frac {(1 \plus{} x^2)^3}{6x^5 \minus{} 20x^3 \plus{} 6x}$

Yes. But Can you slove equation: $ (x^2\plus{}1) (5 x^4\minus{}10 x^2\plus{}1) (7 x^6\minus{}35 x^4\plus{}21 x^2\minus{}1)\equal{}0$

Notice that $ 6x^5 \minus{} 20x^3 \plus{} 6x \equal{} Im[ (x\plus{}i)^6 ]$.

I think we can slove this equation by other way. I will try to done it.


[quote="rockymarcianocosmin99"]Cum pot sa ma inscriu si eu la concurs ?[/quote] Please use English outside the Romanian forums.

The translation of the question is: How can I register in the contest?

You should first read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3683]regulations[/url]. If you still have more questions after that, post them here (in English).

For now the 4th Edition has ended, and the 5th Edition will begin in spring, sometime in April / Mai 2005. It will be announced at least one week ahead.


search
combinatorics unsolved
combinatorics

We are given a finite set of lines, no two parallel and all colored either red or blue, such that the intersection of two lines of the same color also lies on a line of the opposite color. Prove that all lines pass through a common point.

I found the following threads by searching for "red blue lines":
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1365001574&t=143627
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198938
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1365001574&t=241670

Well...quite similar to
[url]http://www.mathlinks.ro/viewtopic.php?t=255550[/url].
Let $ S$ be the set of all $ 4$-tuples $ \left(g_{1},g_{2},g_{3},g_{4}\right)$ of lines such that $ g_{1}$, $ g_{2}$ and $ g_{3}$ have a common point $ P$, $ P\notin g_{4}$, $ g_{1}$ and $ g_{3}$ have the same colour, $ g_{2}$ and $ g_{4}$ have the same colour, but not $ g_{1}$ and $ g_{2}$ and such that $ g_{2}$ crosses the triangle formed by $ g_{1}$, $ g_{3}$ and $ g_{4}$. If $ S$ is not empty we can assume that the triangle formed by $ g_{1}$, $ g_{3}$ and $ g_{4}$ has minimal area. But by the initial conditions a line $ g_{5}$ crosses $ g_{2}\cap g_{4}$ such that $ g_{5}$ has a different colour than $ g_{2}$ and $ g_{4}$. Hence either $ \left(g_{2},g_{5},g_{4},g_{1}\right)\in S$ or $ \left(g_{2},g_{5},g_{4},g_{3}\right)\in S$. In both cases the triangle formed by $ g_{2}$, $ g_{4}$ and $ g_{1}$ or $ g_{3}$ has a smaller area. Contradiction. Finally we get $ S \equal{} \emptyset$.
Thus if two lines with one colour and one line with another colour have one common point one now sees easily that all other lines have to pass through the same point.


Post your favorite Mathcounts problems ever on this thread.

Answer the problem before yours first.

Here is one that is really easy, but I liked it a lot last year...

There is a 10 unit by 10 unit square divided into two new squares.  What is the side length of the larger of the two new squares?

the answer is 8.  (pythagorean theorum)

you have 190 marbles.  You have to put a different amount of marbles in each bag.  No bags can contain zero.  What are the most amount of bags you can have?

The answer isnt 19.  It will be too much hassle if someone answers the problem incorrectly and posts a new problem.

[quote="ml2"]190 bags, each one into another , one marble for each... hm i don't have a problem.[/quote]

How did you get that?  Then all bags have only one marble.  They need to have different number of marbles.

Are you sure the answer isn't 19?

If you wanna minimize it, then you go from smallest to biggest... so from 1 to 19 would be 1+2+3...19 = 20*19/2 = 190

no, no put one marble in bag A, then put one in bag B and then put bag A into Bag B, etc. thus incresing the number by 1 each time.

Yeah, this is more of a lateral thinking puzzle than a problem-solving one. Was it actually from a mathcounts exam? Thats interesting..

yes.  The mathcounts hosts initially made it so the answer was 19, but they later realized they made a mistake.

Can someone post another "favorite" past problem?...

Heres another one of my favorites: A circle with area 20 pi had a square inscribed inside it.  Another circle is inscribed inside the square.  What is the area of the smaller circle?  If your smart, you should be able to solve this with countdown speed.

diagonal of square=sqrt 80, side of square=diameter of smaller circle=sqrt 40, sqrt 10=radius of square, area=10pi, this will always be exactly 1/2 of the bigger circle...

ml2, you're clearly missing the point of this exercize -- you are supposed to post another question after you solve one, so that the thread is self-expanding and there is always another problem around.

Oops. sorry for the spam every1... :oops:
One problem that i liked was a problem in the 2001? national sprint.

"A card with a vowel has a prime number on the other side."
You have 6 cards. The letters/numbers on them are a,b,c,3,4,5
How many cards do you have to flip to prove/disprove the statement?

I remember that one.  The answer is 2: the "a" to make sure it has a prime on the other side, and the "4" to make sure it doesn't have a vowel.

I wouldn't say this is my favorite, but there was an interesting one on team last year, just because so many teams got it wrong:

A rat enters the maze whose pathways are diagrammed below.  At each labeled point, it randomly chooses a direction to go, but it never retraces its steps.  It enters the maze at A.  The exits to the maze are at F, G, H, I, J, and K.  What is the probability that it exits the maze at K?

A---B-----C------F
      | \
      |    D--------G
      |      \
      |        H
      E---I
      |  \
      K   J

First i thought of using block walking, if we say how many ways we have to get to each place its 1, to each of the exits so the probability is 1/5,  but probably the right way is to say there is 1/3, to go to B,D and E, thus 1/9 to K.

A problem I like is
"in a single elimination tournament the better player always won. 8 players participated in 3 rounds. round 1 assignments of position were random, what is the probability that the second best player did not finish as runner up?

New one: What percent of 12 :sqrt:12 is 3 :sqrt:3?  Round to nearest tenth.

Edit: Fixed stupid radicals.

For ml2's prob, it's 3/7. You basically have 2 groups of 4 spaces in the first round. If the best takes up one, the chances that the second best is in the same group of 4 is 3/7.
For raccoon's prob, its 12.5% 12/3=4   :sqrt: 12/ :rt3: = :sqrt: 4, which is 2. 2*4=8. 100/8=12.5%

If x isn't  :pm: 1, simplify

((x+1)/(x-1)-(x-1)/(x+1))/((X+1)/(x-1)+(x-1)/(x+1))


How do you get to participate in the APMO? Is it open to everyone? ( which i doubt not!) If not, how does the selection process work?

[quote="fourierseries"]In the USA, only returning MOPpers are allowed to participate in the APMO. In Taiwan, you must be one of the top scorers in the National Exam to be qualified.  These are the two systems I know of, and I'm pretty sure each country has its individual selection criterion.

That being said, the problems on the APMO are usually easier than those on the IMO or the USAMO, but you'll have to see the solution and write it up quicker.

Best of luck![/quote]
In addition to the returning MOPpers, if you are a Canadian citizen living in the USA and have done exceptionally well on the COMC, then you will get an invitation for the APMO.

[quote="smileyboi"]Anyone here knows how to solve the second question of APMO?[/quote]
As per the contest rules no one is allowed to discuss anything pertaining to APMO until the problems/solutions are posted on their website.  I suspect it will be in a few days.

Actually, the solutions HAVE been posted online (this is the official site, I think)

[url]http://www.kms.or.kr/competitions/apmo/data/08APMO-SOL.pdf[/url]

did anyone else think that number 1 was too hard?

There are two ways to join APMO in HK
First, you need to join the IMO Preliminary Selection Contest the year before and than be invited to join the IMO training progamme. The training divides into three phrase and after phrase 3, students can join APMO as one of the selection test for HK IMO team member.
The other way is that you should be qualified for being a exceptionally gifted students in Maths category by the Education Department and you are free to join APMO.

[quote="shaun ong"]I also want to join this. IM frm malaysia so what should i do?[/quote]
Hmm this post is 3 years ago, I hope u are still eligible to participate in APMO.
In Malaysia, u have to be selected by taking a few selection tests, starting with OMK.
You can try out for APMO 2012.

[quote="BigSams"][quote="yunxiu"]Can any one post the problems of 2011 APMO?[/quote]
It cannot be discussed or distributed publicly until posted officially here http://www.mmjp.or.jp/competitions/APMO/.[/quote]
And considering the current event in Japan, I believe it is highly unlikely for the problems to be posted in several weeks. Hopefully they are all O.K.

The problems, solutions, and results are up now.

I just don't understand why China can not send students (even those could not make to the IMO national team) to participate. There are so many good students over there.
[quote="mathq"][quote="xiyuguzu"]Can anybody tell me why China did not participate in APMO?[/quote]

Because China destroys [i]every[/i] country nearly [i]every[/i] year...

jokes, i have no idea.[/quote]


geometry
geometric transformation
reflection
circumcircle
trigonometry
ratio
projective geometry

Can anyone please tell me some references or problems about harmonicity i.e. polarity?

[[color=red]Edited by pohoatza[/color]: Title + content of some following messages for maintaining them (the title + next posts) one related to each other.]

I will start with an example of using it in proving the Butterfly theorem. Here it goes:

[b]let $ AB$ be an arbitary chord of circle $ w$,let $ P$ be the midpoint of segment $ AB$,now let $ KM$ and $ LN$ be two other arbitary chords from $ w$ which pass thruogh $ P$,let $ LM,KN$ intersect $ AB$ at $ X,Y$ respectively,prove that $ PX \equal{} PY$.[/b]

[u]proof:[/u]
note that $ L$ is a point on $ w$ and we know that $ LA,LB,LM,LN$ intersect the circle at $ A,B,X,P$ respectively therefore:

$ (AB,MN) \equal{} (AB,XP)$ (I)
wrt point $ L$.

and similary we get that:
$ (AB,MN) \equal{} (AB,PY)$
wrt point $ K$.

now let $ Y'$ be the reflection of point $ Y$ wrt $ P$,so its sufficient to show that $ Y' \equal{} X$...
the reflection of points $ A,B,P,Y$ wrt point $ P$ are $ B,A,P,Y'$ respectively thus:

$ (AB,PY) \equal{} (BA,PY') \equal{} (AB,Y'P)$ (II)

$ (I),(II)\Rightarrow (AB,Y'P) \equal{} (AB,XP)$
$ \Rightarrow X \equal{} Y'$
as wanted...

also with the same method you can prove one of the generalizations of it:

[b]let $ AB$ be an arbitary chord of circle $ w$ and let $ P$ be an arbitary point on $ AB$,let $ LN,KM$ be two other arbitary chords of $ w$ which pass thruogh $ P$,now assume that $ LM,KN$ intersect $ AB$ at $ X,Y$ respectively,prove that:

$ \frac{1}{PY}\plus{}\frac{1}{PA}\equal{}\frac{1}{PX}\plus{}\frac{1}{PB}$[/b]

[quote="pardesi"]Thanks Babak. Do you know some more problems which can be proved similarly?[/quote]
here you go,the first 5 problems:
[hide="problem 1"]
let $ ,E,F$ be three points on the sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ P$,let $ EF$ intersect $ AD$ at $ K$ prove that $ (AP,KD) \equal{} \minus{} 1$
-------------------------------------------[/hide]
[hide="problem 2"]
let $ ,E,F$ be three points on the sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ K$,let $ FD$ intersect $ BE$ at $ X$,let $ P$ be the midpoint of $ AK$ and let $ EP$ intersect $ AB$ at $ Y$.prove that $ XY\parallel AD$
-------------------------------------------[/hide]
[hide="problem 3"]
let $ D$ be a point on side $ BC$ of triangle $ \triangle ABC$ such that $ AD\perp BC$ and let $ P$ be an arbitary point on $ AD$,let $ BP,CP$ intersect $ AC,AB$ respectively at $ Y,X$ prove that6 $ \angle XDA \equal{} \angle ADY$
-------------------------------------------[/hide]
[hide="problem 4"]
let $ X,Y$ be points on sides $ AC,AB$ of $ \triangle ABC$ such that $ AZ\perp XY$,where $ Z$ is the intersection of $ BX,CY$ and $ BHXC$ is cyclic where $ H$ is the intersection of $ AZ,XY$.prove that $ BX \equal{} CX$
-------------------------------------------[/hide]
[hide="problem 5"]
let $ D$ be a point on side $ BC$ of triangle $ \triangle ABC$ such that $ AC \equal{} CD$ and $ C$ lies between $ B,D$.now let $ E$ be the second intersection of the circle with diameter $ BC$ and the circumcircle of $ \triangle ACD$ (other than C).let $ BE,CE$ intersect $ AC,AB$ at $ X,Y$ respectively.prove that $ X,Y,D$ are colinear.
------------------------------------------[/hide]

hope you enjoy them :D

[quote="BaBaK Ghalebi"]

[color=red]Let $ D,E,F$ be three points on the sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ P$,let $ EF$ intersect $ AD$ at $ K$ prove that $ (A,K,P,D) \equal{} \minus{} 1$[/color]

     [/quote]
[color=blue]Denote $ EF\cap BC \equal{} K'$,since $ (K',B,D,C) \equal{} \minus{} 1$ then the pencil $ F(K',B,D,C)$ is harmonic $ \Rightarrow(A,K,P,D) \equal{} \minus{} 1$[/color]
I think that it will be better if we will write $ (A,B,D,C) \equal{} \minus{} 1$,than $ (AC,BD) \equal{} \minus{} 1$. :wink:

[quote="BaBaK Ghalebi"]
[color=red]Let $ D,E,F$ be three points on the sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ K$,let $ FD$ intersect $ BE$ at $ X$,let $ P$ be the midpoint of $ AK$ and let $ EP$ intersect $ AB$ at $ Y$.prove that $ XY\parallel AD$[/color]
 [/quote]
[color=blue]I'll change the problem a little.Let $ Y$ be the point on $ AB$ such that $ XY\|AD$ and $ Z\in XY\cap AC$.
Then we only need to prove that $ Y$ is the midpoint of $ XZ$,which is equivalent to(from Menelaus theorem):
$ \frac{XB}{BE}\equal{}\frac{AZ}{AE}$,
since $ XY\|AD$ then $ \frac{AZ}{AE}\equal{}\frac{KX}{KE}$,but from the first problem we know that $ (B,X,K,E)\equal{}\minus{}1\Rightarrow\frac{KX}{KE}\equal{}\frac{XB}{BE}$
Thus we are done.[/color]

[quote="nsato"][quote="BaBaK Ghalebi"][hide="problem 4"]
let $ H$ be the orthocenter of triangle $ \triangle ABC$ and let $ X,Y$ be points on $ AC,AB$ respectively such that $ BHXC$ is cyclic.prove that $ BX \equal{} CX$
-------------------------------------------[/hide]
[/quote]

This problem does not use $ Y$.  Is it stated correctly?[/quote]
no actually... :oops:
its corrected now

[quote="Erken"][quote="BaBaK Ghalebi"]
[color=red]Let $ D,E,F$ be three points on the sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ K$,let $ FD$ intersect $ BE$ at $ X$,let $ P$ be the midpoint of $ AK$ and let $ EP$ intersect $ AB$ at $ Y$.prove that $ XY\parallel AD$[/color]
 [/quote]
[color=blue]I'll change the problem a little.Let $ Y$ be the point on $ AB$ such that $ XY\|AD$ and $ Z\in XY\cap AC$.
Then we only need to prove that $ Y$ is the midpoint of $ XZ$,which is equivalent to(from Menelaus theorem):
$ \frac {XB}{BE} \equal{} \frac {AZ}{AE}$,
since $ XY\|AD$ then $ \frac {AZ}{AE} \equal{} \frac {KX}{KE}$,but from the first problem we know that $ (B,X,K,E) \equal{} \minus{} 1\Rightarrow\frac {KX}{KE} \equal{} \frac {XB}{BE}$
Thus we are done.[/color][/quote]
this one has an easier solution:
according to [b]problem 1[/b] we get that $ (BK,XE) \equal{} \minus{} 1$,let $ X'$ be the intersection of $ XY,AD$.now $ YB,YK,YX,YE$ intersect line $ AD$ at $ A,K,X',P$ respectively,so $ (AK,X'P) \equal{} \minus{} 1$,but we know that $ P$ is the midpoint of $ AK$ thus we must have $ X'\to\infty$ i.e. $ XY\parallel AD$.
as wanted...

[quote="Erken"]
I think that it will be better if we will write $ (A,B,D,C) \equal{} \minus{} 1$,than $ (AC,BD) \equal{} \minus{} 1$. :wink:[/quote]
why?
I always write it in that way...
it doesnt really matter,does it? :maybe:
but actually in my notation from $ (AC,BD)\equal{}k$ we mean that [i]segment $ AC$ has been devided internaly and externaly by $ B,D$[/i].therefore we dont put "," between every two points...
anyway I dont think it makes any difference...

[quote="BaBaK Ghalebi"][quote="Erken"]
I think that it will be better if we will write $ (A,B,D,C) \equal{} \minus{} 1$,than $ (AC,BD) \equal{} \minus{} 1$. :wink:[/quote]
why?
I always write it in that way...
it doesnt really matter,does it? :maybe:
but actually in my notation from $ (AC,BD) \equal{} k$ we mean that [i]segment $ AC$ has been devided internaly and externaly by $ B,D$[/i].therefore we dont put "," between every two points...
anyway I dont think it makes any difference...[/quote]
Yes,there is no diffrence.It is the first time,when i saw such notation,so excuse me,because as you said it desn't matter.
P.S Is this problem your owns?(Especcially 5th)
Thank you :)

[quote="Erken"]Is this problem your owns?(Especcially 5th)
Thank you :)[/quote]
no they're not,whats about the 5th one? :maybe:

Oh,just the fifth problem is very beatiful,and i believe that there is a solution without using trig,i haven't still found it,i've found only one ugly solution with metrical  relations. :(
I would like to ask you to not post your harmonic solution to this problem until tommorow. :wink: OK?

[quote="BaBaK Ghalebi"]
let the incircle of triangle $ \triangle ABC$ touch the sides $ BC,CA,AB$ at $ A',B',C'$ respectively,let $ A''$ be a point on $ BC$ such that $ BA' = CA''$ now let $ \ell_A$ be the line passing through $ A'$ and parallel to $ AA''$,define $ \ell_B,\ell_C$ in the same way,prove that $ \ell_A,\ell_B,\ell_C$ are concurrent.
[/quote]
I'm sorry,because i'll not use polarity and harmonic fours theory in my following solution,and also i'm afraid that polarity and "harmonic" solution is much shorter,but anyway here is my proof to this nice problem:
[color=blue]Solution:
Let $ \omega(I,r)$ be the incircle of $ \triangle ABC$.Denote $ B_1\in B'I\cap\omega$(.Define $ A_1,C_1$ in similar way.Let $ l'_A,l'_B,l'_C$ be the tangent lines to $ \omega$ at points $ A_1,B_1,C_1$ respectively.Let $ X,Y,Z$ be the points of their intesection,such that $ X\in l'_C\cap l'_B,Y\in l'_C\cap l'_A,Z\in l'_A\cap l'_B$.
It is well-known that $ B,B_1,B''$ are collinear(similarly for other triplets of points) and it is easy to understand that $ XI = AI,YI = BI,ZI = CI$(from symmetry),
and it is obvious that $ C'I = IC_1 = r$,hence $ ZC'CC_1$ is parallelogramm $ \Rightarrow$ $ ZC'\|CC_1$,so $ ZC'\in l_C$.
Now,all we need to prove is the collinearity of $ XA',YB',ZC'$.But triangles $ \triangle XYZ$ and $ \triangle ABC$ are equal,and $ XY\|AB,YZ\|BC,XZ\|AC$,hence
collinearity of $ XA',YB',ZC'\Longleftrightarrow$ to the collinearity of $ BB_1,CC_1,AA_1$,which is obviously true,and btw their intersection point is called the Nagel point.
Thus we are done :) .[/color]

[color=green][b]Remark[/b]:If we denote $ T\in XA'\cap YB'\cap ZC'$.From my solution it is easy to understand that, the point $ I$ is the homothety center,which turns $ \triangle XYZ$ to $ \triangle ABC$,so the points $ T,I,N$ are collinear,where $ N$ is Nagel point,and $ TI = IN$.
[/color]
Best regards,Erken.
P.S:To BaBaK Ghalebi:
Thank you for your nice problems.And please don't post solutions until next week,OK?
But i'm waiting for your proof of this problem,which uses polarity theory and harmonic fours. :wink:

Proof of Desarqeus Theorem with using :
The Desarques theorem:
Let $ ABC$ and $ A'B'C'$ be the 2 triangles such that $ AA',BB',CC'$ are concurrent(at point $ T$),
then points $ X\in AB\cap A'B',Y\in AC\cap A'C',Z\in BC\cap B'C'$ are collinear.
Let A,B,C,D be the points on a line,which lies in that order,then let
$ (AC,BD) \equal{} \frac {CA}{CB}\cdot\frac {DB}{DA}$.
Proof:
Consider the line $ l$ through the points $ X$ and $ Y$.Denote:
$ P\in AA'\cap l,Q\in BB'\cap l,R\in CC'cap l$,since
$ PQ$,$ AB$,$ A'B'$ are concurrent then
$ (TA,PA') \equal{} (TB,QB')$.
Similarly,since $ PR$,$ AC$,$ A'C'$ are concurrent then
$ (TA,PA') \equal{} (TC,RC')$.
Hence we conclude that
$ (TB,QB') \equal{} (TC,RC')$,so
$ BC,RQ,B'C'$ are concurrent,so we are done. :)

[quote="BaBaK Ghalebi"][quote="unsolved problems"]
[hide="problem 7."]
let $ O$ be the circumcenter of triangle $ \triangle ABC$ and let the perpendicular bisector of $ OA$ meet $ BC$ at $ A'$,define $ B',C'$ in the same way,prove that:
$ i)$points $ A',B',C'$ lie on a line $ \ell$.
$ ii)\ell\perp ON'$
($ N'$ is the isogonal conjugate of the nine-point center)
-------------------------------------[/hide]

[hide="problem 8."]
let $ P$ be an arbitary point in the plane of triangle $ \triangle ABC$,let $ A'$ be a point on $ BC$ such that $ A'P\perp PA$,define $ B',C'$ in the same way,prove that $ A',B',C'$ are collinear.
-------------------------------------[/hide]

[hide="problem 11."]
let $ H$ be the orthocenter of triangle $ \triangle ABC$ and let $ P$ be an arbitary point in the plane of $ \triangle ABC$,let $ A'$ be a point on $ BC$ such that $ A'H\perp AP$,define $ B'C'$ in the same way,prove that $ A',B',C'$ are collinear.
-------------------------------------[/hide]

[hide="problem 12."]
let the incircle of triangle $ \triangle ABC$ touch the sides $ BC,CA,AB$ at $ A',B',C'$ respectively,let $ A''$ be a point on $ BC$ such that $ BA' \equal{} CA''$ now let $ \ell_A$ be the line passing through $ A'$ and parallel to $ AA''$,define $ \ell_B,\ell_C$ in the same way,prove that $ \ell_A,\ell_B,\ell_C$ are concurrent.
-------------------------------------[/hide]

BaBaK[/quote]

these are the unsolved problems of my last set... :)[/quote]

Are you sure you have proofs of all of these using harmonic conjugation? I am wondering what your proof for 12 looks like, since my approach is absolutely straightforward application of homotheties and reflections in points: A' is the reflection of A'' in the midpoint of BC, so if the parallel to AA'' through A'' and cyclically concur (what they surely do - at the Nagel point) and if the parallel to AA'' through the midpoint of BC and cyclically concur (what they also do, because the midpoints of BC, CA, AB form a triangle homothetic to ABC), then the parallel to AA'' through A' and cyclically must also concur. How do you use harmonic conjugation here? If you don't, please don't stuff these problems all into one thread.

  Darij

[quote="darij grinberg"] ....Are you sure you have proofs of all of these using harmonic conjugation? I am wondering what your proof for 12 looks like, since my approach is absolutely straightforward application of homotheties and reflections in points: A' is the reflection of A'' in the midpoint of BC, so if the parallel to AA'' through A'' and cyclically concur (what they surely do - at the Nagel point) and if the parallel to AA'' through the midpoint of BC and cyclically concur (what they also do, because the midpoints of BC, CA, AB form a triangle homothetic to ABC), then the parallel to AA'' through A' and cyclically must also concur....[/quote]
Alternatively, I think it is well known (or easy to prove) that the lines through the midpoints of the side-segments of the given triangle $ \bigtriangleup ABC,$ and parallel to $ AA'',$ $ BB'',$ $ CC'',$ pass through its incenter $ I$ and so, we can easily conclude, by [b][size=100]Thales theoerem[/size][/b], the concurrency of the lines through $ A',$ $ B',$ $ C'$ and parallel to the line segments $ AA'',$ $ BB'',$ $ CC''$ respectively, at one point, as the reflexion of the [b][size=100]Nagel point[/size][/b] $ N\equiv AA''\cap BB''\cap CC'',$ with respect to the incenter $ I$ of $ \bigtriangleup ABC.$

Kostas Vittas.

[quote="BaBaK Ghalebi"][quote="Erken"][quote="BaBaK Ghalebi"][color=red]Let $ w$ be a semicircle with diameter $ AB$ and let $ C$ be a point on $ w$.assume that the tangents from points $ A,C$ to $ w$ intersect at $ D$.let $ H$ be a point on $ AB$ such that $ CH\perp AB$ and let $ CH$ intersect $ BD$ at $ K$.prove that $ HK \equal{} KC$[/color]
[/quote]
[color=blue]Denote $ F\in w\cap BD$,$ G\in AD\cap BC$.Then $ AFCB$ is harmonic quadrilateral,hence $ B(BCFA)\cap AD$ makes a harmonic four,hence $ AD \equal{} DG$.Now using that $ AB$ is diameter,we obtain that $ AG\|CH$,hence $ CK \equal{} KH$.Thus we are done. [/color][/quote]
my solution to this problem is somehow the same,anyway here it goes:
first of all we have the following theorem:
[b]for point $ P$ and circle $ w$,let $ M,N$ be points on $ w$ such that $ PM,PN$ are tangent to $ w$.now let the chord $ AB$ of $ w$ pass through $ P$,then $ (AB,MN) \equal{} \minus{} 1$[/b]
now in our problem let $ BD$ intersect $ w$ at $ P$,then according to the lemma above we get that $ (AC,PB) \equal{} \minus{} 1$ now denote $ \ell$ as the tangent to $ w$ at $ B$.
now we know that $ BA,BC,BP$ intersect line $ CH$ at $ H,C,K$ respectively and also $ B$ lies on $ w$ so $ (HC,KB') \equal{} \minus{} 1$ where $ B' \equal{} \ell\cap CH$ but in the other hand we know that $ \ell\perp AB$ and $ CH\perp AB$ thus $ \ell\parallel CH$ i.e. $ B'\to\infty$ so:
$ (HC,K\infty) \equal{} \minus{} 1$ or in other words $ K$ is the midpoint of $ HC$ i.e. $ HK \equal{} KC$.[/quote]
BaBak i don't understand the part where B is projected about itself...is that possible and why should the line koining B to B we the line $ l$ only it could be any other line through B

pardesi,let $ B$ be a fixed point on circle $ w$ and $ A$ be a variant point on $ w$,now let $ \ell$ be the line passing through $ A,B$ so $ \ell$ variates as $ A$ variates...
but note that $ \ell$ always intersects $ w$ at points $ A,B$
now in the case $ A\to B$ line $ \ell$ intersects circle $ w$ at $ A,B$ where $ A\equiv B$ so $ \ell$ intersects $ w$ at a single point $ B$ i.e. $ \ell$ is tangent to $ w$ at point $ B$...
so the line passing through $ B$ and connecting it to $ B$ is the line tangent to $ w$ at $ B$ i.e. the line $ BB$ is tangent to $ w$ at point $ B$...

thanks BaBaK... :)
also just another doubt when $ A,B,C,D$ are concyclic and we write $ (AB,CD)\equal{}\minus{}1$ ..then do we mean if they are projected about any point $ P$ onto a line then their projections form harmonic :?:

let $ P,A,B,C,D$ be points on circle $ w$,let $ PA,PB,PC,PD$ intersect line $ \ell$ at points $ A',B',C',D'$ now assume that $ (AB,CD) \equal{} k$ i.e. $ \frac {AC}{BC} \equal{} k\cdot\frac {AD}{BD}$...
now note that:

$ \frac {AC}{BC} \equal{} k\cdot\frac {AD}{BD}\iff \frac {\frac {AC}{2R}}{\frac {BC}{2R}} \equal{} k\cdot\frac {\frac {AD}{2R}}{\frac {BD}{2R}}$ (*)

now note that we have $ \frac {AC}{2R} \equal{} \sin \angle APC$ and so on,so the equation (*) is equivalent to:

$ (*)\iff\frac {\sin \angle APC}{\sin \angle CPB} \equal{} k\cdot\frac {\sin \angle APD}{\sin \angle DPB}$

now according to the $ \sin$ law in triangles $ \triangle PA'C',\triangle PC'B'$ we get that:

$ \frac {\sin \angle APC}{\sin \angle CPB} \equal{} \frac {\sin \angle A'PC'}{\sin \angle C'PB'} \equal{} \frac {A'C'}{B'C'}$

similary in triangles $ \triangle PA'D',\triangle PD'B'$ we have:

$ \frac {\sin \angle APD}{\sin \angle DPB} \equal{} \frac {\sin \angle A'PD'}{\sin \angle D'PB'} \equal{} \frac {A'D'}{D'B'}$

hence by (*) we get that:

$ \frac {A'C'}{B'C'} \equal{} k\cdot\frac {A'D'}{B'D'}\Rightarrow (A'B',C'D') \equal{} k$

$ \Rightarrow (AB,CD) \equal{} k\iff (A'B',C'D') \equal{} k$

as wanted...

thank u BaBaK :)
i had a seriuos misconception. :rotfl:

Let $ \omega$ be a circle and $ AB,CD$ be two parallel chords of $ \omega$.Let $ M$ be the midpoint of $ AB$,and $ E$ be the second intersection of $ CM$ with $ \omega$.Prove that $ \angle AKE\equal{}\angle EKB$,where $ K$ is the midpoint of $ DE$.

[quote="Erken"]Let $ \omega$ be a circle and $ AB,CD$ be two parallel chords of $ \omega$.Let $ M$ be the midpoint of $ AB$,and $ E$ be the second intersection of $ CM$ with $ \omega$.Prove that $ \angle AKE = \angle EKB$,where $ K$ is the midpoint of $ DE$.[/quote]
nice problem. :)

let $ S$ be the intersection point of $ DE,AB$,now note that $ M$ is the midpoint of $ AB$ so $ (AB,M\infty)=-1$ and the lines $ CA,CB,CM$ intersects the circle at $ A,B,E$ respectively and also the parallel line with $ AB$,through $ C$,intersects the circle at $ D$ so:

$ (AB,M\infty)=-1\Rightarrow (AB,ED)=-1$

now let the tangent line to the circle at point $ D$ intersect $ AB$ at point $ T$ then we have:

$ (AB,ED)=-1\Rightarrow (AB,ST)=-1$

so in order to prove $ \angle AKE=\angle EKB$ we have to prove that $ \angle TKS=90^\circ$:

as we said before $ (AB,ED)=-1$ so $ AB$ must pass through the intersection of the tangent lines at $ D,E$.so the tangent to $ w$ at $ E$ passes through $ T$ i.e. $ TE$ is tangent to $ w$.now note that $ TE,TD$ are two tangents from point $ T$ to circle $ w$ so $ TE=TD$ so $ \triangle ETD$ is isosceles,but on the other hand $ K$ is the midpoint of $ ED$ thus:

$ TK\mid ED\Rightarrow \angle TKE=90^\circ$.

QED

Great,but another easier solution,could be obtained,if we use the fact,that dioganal is the symmedian in harmonic quadrilateral. :wink:

[quote="Erken"][quote="BaBaK Ghalebi"]
[color=red]Let $ D$ be a point on side $ BC$ of triangle $ \triangle ABC$ such that $ AD\perp BC$ and let $ P$ be an arbitary point on $ AD$,let $ BP,CP$ intersect $ AC,AB$ respectively at $ Y,X$ prove that6 $ \angle XDA \equal{} \angle ADY$[/color]
[/quote]
[color=green][size=150]Lemma[/size][/color]
[color=blue]We will use one famous lemma:
Let $ \angle ABC \equal{} 90^{\circ}$ and $ X,Y\in AC$,then $ (X,A,Y,C) \equal{} \minus{} 1\Longleftrightarrow \angle XBA \equal{} \angle ABY$.
If someone want i could post the proof for this lemma,but it is too easy.
Solution(problem 3):
Let $ XD\cap AC \equal{} Z$ then the division $ (Z,A,Y,C)$ is harmonic,then using lemma we obtain that $ \angle YDA \equal{} \angle ADZ \equal{} \angle ADX$.Thus we are done. :wink:
[/color][/quote]
I want Erken pls

Hi everyone ! I can't understand this "Harmonic Division" what this mean? I have seen about harmonic division at "Mathematical reflections" but Can't understand. :?:

     what this mean (CA/CB=-DA/DB)
 and what this mean (AB,CD)=-1 and (AB,CD)=k ?  :(
please write full about it ! :lol:


A container is filled with 3 cups of lemon juice and 7 cups of water. How many cups of pure water must be added so that the container is 25\% lemon juice?

$ 3$ is $ 25$% of $ 12$. $ \fbox{2}$ cups of water are necessary to be added to the original $ 7$ cups to make $ 9$ cups of water and $ 12$ cups of mixture.


function
integration
real analysis
calculus
derivative
topology
real analysis unsolved

Let $ S$ be a closed set of reals. and $ f$ a function in $ L^1[0,1]$.
Suppose that for all $ E\subset [0,1]$ that are measurable
with $ m(E)>0$
we know that $ \frac{1}{m(E)}\int_{E}f\in S$
Show that it follows that $ f(x)\in S$ for a.e. $ x\in [0,1]$

Let $ E_{n,x}$ be a sequence of sets containing $ x$ which shrink to $ \{x\}$.  Then letting $ \phi(A) \equal{} (m(A))^{\minus{}1}\int_A f(t)dt$, we have $ \phi(E_{n,x})\to f(x)$ a.e. (by the Lebesgue differentiation theorem).  Clearly $ \phi(E_{n,x})\in S$ for all $ n$, so since $ S$ is closed, $ f(x)\in S$ as well.


I saw a online discussion where the question was what is 1/ :inf: . One person said the answer was 0 because 1/ :inf: is the same as how many times  :inf: fits in one. Of course that is incorrect because based on that thinking 1/2 is also 0. :D

What would be a better way to talk about one divided by infinity? Does it even makes sense to talk about dividing by infinity?

[quote]Does it even makes sense to talk about dividing by infinity?[/quote]

Not really...since infinity is technically not a number, but just a representation of an idea. However, it makes sense to talk about 1/n as n approaches infinity.

In some ways it's a shame we don't live in 1750 when you could get away with writing 1/infinity and no one would mind. Euler did things like that plenty of times. In the strictest sense we can't speak about 1/infinity since we can only divide by numbers, and infinity isn't a number. We also can't speak about 1/0 because 0 doesn't have an inverse in R or C or Q or whatever field you're talking about. In terms of intuition, you can divide by whatever you want. I divide by 0 or infinity quite often when I'm trying to solve problems. When I write up solutions formally, I try to leave that stuff out.

Mmm, if you graph y=1/x you get 1/:inf:=0 and 1/0=:inf:, I think. so the inverse of :inf: is 0 and vice versa...?

I've thought about that, and intuitively speaking, I think the inverse of 0 is :pm::inf: and vice versa.,

Sure, intuitively the inverse of 0 is infinity. That doesn't make it true though. Lots of things in mathematics look true but actually aren't.

Can you prove that 1/:inf: isn't 0 and 1/0 isn't :inf:? (Besides the fact that these two are "against the rules")

1/infinity and 1/0 are meaningless statements.

If you aren't allowed to divide by 0, then 1/0 isn't [i]anything[/i], which in particular means that it isn't equal to infinity.  And vice versa.  There isn't anything to be proven.  If you want to talk about what it looks like, well, when you take things that look a lot like 1/:infinity:, they come out looking a lot like 0.  And when you take things that look a lot like 1/0, they end up looking a lot like :infinity:.  So, [i]in a sense[/i] we can say that 1/0 = :infinity:.  But it isn't an actual, valid mathematical statement.

[quote]Again, aside from the fact that those two are undefined by definition, how can people say that 1/infinity isn't 0 and 1/0 isn't infinity?[/quote]

Umm..the statement before your question pretty much answers your question.  8-)

Mmm...I'll get back to you guys in about a month with my proof

Dont numbers get smaller when their denominator gets bigger?

If you have positive rational numbers, then as their denominators increace, they grow smaller (if their numerators don't increace, too).

You should interpret this as a limit !

In fact, those "theorem" mean

limx->01/x = :inf:

limx->:inf:1/x = 0.

Clear ? It is just a "symbolic" notation !

Actually in Complex Analysis is right to say that 1/0 = \infty and 1/\infty = 0, why? The one point compactification of \mathbb{C}.

Limit [size=75]x->0[/size]1/x =  :pm:  :inf:


advanced fields
advanced fields unsolved

Liu writes that a $ \emph{ringed topological space}$ consists of a topological space $ X$ endowed with a sheaf of rings $ \mathcal{O}_{X}$ on $ X$ such that $ \mathcal{O}_{X,x}$ is a local ring for every $ x\in X$. What is the definition of $ \mathcal{O}_{X,x}$?

I guess he means the stalk. It uses the notion of direct limits and should be defined somewhere in the book.

Just a tiny remark meant as a caveat: Some authors distinguish ringed spaces and locally ringed spaces from each other. The former are topological spaces with a sheaf of rings, whereas the latter are ringed spaces whose stalks are local rings.


function integration calculus derivative ratio inequalities

Find all $ x \in \Re$ such that $ 10^x \plus{} 11^x \plus{} 12^x \equal{} 13^x \plus{} 14^x$.

[hide="solution"]

We claim that $ x \equal{} 2$ is the only solution. Dividing the equation by $ 13^x$, we get $ (\frac{10}{13})^x\plus{}(\frac{11}{13})^x\plus{}(\frac{12}{13})^x\equal{}1\plus{}(\frac{14}{13})^x$. The left side is a decreasing function of $ x$ while the right side is an increasing function of $ x$. Therefore there is at most one solution, namely, $ x \equal{} 2$.
[/hide]

What is the proof to the last two statements (that if in an equation, if one side is increasing while the other is decreasing, there is at most one solution)?

[hide="I am not that much sure"]In this problem
If possible,let there be two(or more) solutions.
Let one be $ 2$ and other $ x$.
$ 2$ satisfies the equation(let value of LHS=RHS at $ 2$ be $ y$)
then as we increase values (make them more than $ 2$),
LHS decreases than initial value(LHS $ <y$) but RHS increases(RHS $ >y$)

Thus for no other value,LHS can be equal to RHS

but i am afraid,this is not formal enough[/hide]

Essentially, if you graph it, once the first solution has been found, one side will go down, and one side will go up. They can't "come back" to the x-axis, otherwise it would go from increasing to decreasing, or vice versa. (Another valid, but informal proof.)

[hide="A Rough Sketch"]Suppose the equation is of the form $ f(x)\equal{}g(x)$, and suppose the value $ x\equal{}a$ is a solution. Suppose we are also given that $ f'(x)>0 \; \forall x$ and that $ g'(x) < 0 \implies [\minus{}g]'(x) > 0 \; \forall x$.

Adding, we seek to find other roots of $ [f\minus{}g](x)\equal{}0$, where $ [f\minus{}g]'(x)>0$

Suppose there is another root at $ x\equal{}b$ for $ b>a$

We have that $ [f\minus{}g](b)\minus{}[f\minus{}g](a)\equal{}0$

$ \therefore \int_{a}^{b} [f\minus{}g]'(x) dx \equal{} 0$

However, we know that $ [f\minus{}g]'(x) >0$ over $ [a,b]$. Let $ c>0$ be the minimum value of $ [f\minus{}g]'(x)$ over $ [a,b]$. Then, $ 0 \equal{} \int_{a}^{b} [f\minus{}a]'(x) dx \geq (b\minus{}a)c \geq 0 \implies b\minus{}a\equal{}0 \implies a\equal{}b$.

So there cannot be a larger root.

This means that a value solves the equation, there cannot be a larger value that solves the equation. From here, we can deduce that there is at most one root to the equation.[/hide]

TZF, I really don't think that the calculus is necessary.

Suppose $ f$ is strictly increasing $ \forall x$ and $ g$ is strictly decreasing $ \forall x$.  This means that $ a<b$ implies $ f(a) < f(b)$ and $ g(a) > g(b)$.  If $ f(a)\equal{}g(a)$, then we have $ g(b) < f(b)$ - they cannot be equal.  If, on the other hand, $ f(b) \equal{} g(b)$, then we have $ f(a) < g(a)$  - again the two functions cannot be equal.  So if we have one solution, we cannot have another.

Therefore, for $ f(x)\equal{}g(x)$ where $ f$ is strictly increasing and $ g$ is strictly decreasing, there is at most one solution.

Phrasing it (our arguments are essentially the same) in terms of calculus has advantages and disadvantages - among the advantages is that differentiation is, in many cases, the best way to prove that a function is strictly increasing or decreasing in the first place.

[quote="TZF"][hide="A Rough Sketch"]Suppose the equation is of the form $ f(x) \equal{} g(x)$, and suppose the value $ x \equal{} a$ is a solution. Suppose we are also given that $ f'(x) > 0 \; \forall x$ and that $ g'(x) < 0 \implies [ \minus{} g]'(x) > 0 \; \forall x$.

Adding, we seek to find other roots of $ [f \minus{} g](x) \equal{} 0$, where $ [f \minus{} g]'(x) > 0$

Suppose there is another root at $ x \equal{} b$ for $ b > a$

We have that $ [f \minus{} g](b) \minus{} [f \minus{} g](a) \equal{} 0$

$ \therefore \int_{a}^{b} [f \minus{} g]'(x) dx \equal{} 0$

However, we know that $ [f \minus{} g]'(x) > 0$ over $ [a,b]$. Let $ c > 0$ be the minimum value of $ [f \minus{} g]'(x)$ over $ [a,b]$. Then, $ 0 \equal{} \int_{a}^{b} [f \minus{} a]'(x) dx \geq (b \minus{} a)c \geq 0 \implies b \minus{} a \equal{} 0 \implies a \equal{} b$.

So there cannot be a larger root.

This means that a value solves the equation, there cannot be a larger value that solves the equation. From here, we can deduce that there is at most one root to the equation.[/hide][/quote]

I have no idea what you're trying to prove using calculus! I don't even [i]know[/i] calculus.[/i]

It proves what you asked for - "that if in an equation, if one side is increasing while the other is decreasing, there is at most one solution"

It is also the exact same thing as what [b]facis [/b]did. It is just a more direct computational framework, since you can just take the LHS and the RHS, take the derivatives, and the conclusion follows.

$ (\frac {10}{13})^x \plus{} (\frac {11}{13})^x \plus{} (\frac {12}{13})^x \equal{} 1 \plus{} (\frac {14}{13})^x$

We do not need the derivative (for this problem) to show that the LHS of the equation is strictly decreasing while the RHS is strictly increasing.  All we need to do is take some ratios:

Define $ a(x) \equal{} (10/13)^x$, $ b(x) \equal{} (11/13)^x$, $ c(x) \equal{} (12/13)^x$, and $ d(x) \equal{} (14/13)^x$.  Let $ h$ be any positive number.  Then we have $ a(x \plus{} h)/a(x) \equal{} (10/13)^h < 1$, so $ a(x \plus{} h) < a(x)$ as long as $ a(x) > 0$ (which is always true).  Now since $ x \plus{} h > x$ this proves that $ a(x)$ is strictly decreasing.

The same argument can be made for $ b(x)$ and $ c(x)$, so we have $ h > 0$ implies $ a(x \plus{} h) < a(x)$, $ b(x \plus{} h) < b(x)$, and $ c(x \plus{} h) < c(x)$.  Adding these inequalities, we obtain $ (a \plus{} b \plus{} c)(x \plus{} h) < (a \plus{} b \plus{} c)(x)$ - that the LHS of our equation is strictly decreasing.

Now we also have $ d(x \plus{} h)/d(x) \equal{} (14/13)^h > 1$ which implies $ d(x \plus{} h) > d(x)$ (the multiplication is valid because $ d(x) > 0$ for all $ x$) - that is, that $ d$ is strictly increasing.  Now we can add $ 1$ to both sides: $ d(x \plus{} h) \plus{} 1 > d(x) \plus{} 1$ which tells us that the RHS of our original equation is strictly increasing.

Okay, at the beginning you said that x was a member of (some strange symbol that looks like an R).
What is that strange R?

Oh really, I though it was a different R. Well, I still don't know how to solve it anyway.


articles AMC USA(J)MO USAMO email AIME AoPSwiki \/closed

The early registration deadline for Art of Problem Solving's Worldwide Online Olympiad Training (WOOT) program is [b]July 15[/b].  Register on or before [b]July 15[/b] and save $ \$100$.  WOOT is a 7-month program for the top problem solvers in the world to prepare for high school Olympiad and pre-Olympiad problem solving events.  WOOT includes a series of problem solving classes, articles, practice tests, problem sets, and interaction with many of the top instructors and students from around the world.

In 2007, 10 of the 12 USA Mathematical Olympiad (USAMO) winners were alumni of WOOT.  All participants in the 2007 Math Olympiad Summer Program will be enrolled in WOOT, giving WOOT a truly outstanding peer group.

For more information on WOOT, please visit:

[url]http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php[/url]

To enroll in WOOT, visit:

[url]http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php[/url]

Over the last two years, Art of Problem Solving has sponsored an expansion of the USAMO to 500 students.  Therefore, more students than ever have a chance to participate in this prestigious event.  WOOT offers an excellent opportunity to prepare for the USAMO and for other national qualifying exams.

We will host a free informational online session about WOOT on [b]Wednesday, July 11[/b] at 7:30 PM ET (4:30 PM PT).  For more details on the free session, please visit

[url]http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php[/url]

The registration fee for WOOT before [b]July 15[/b] is $ \$695$; after [b]July 15[/b] is $ \$795$.  If you have any questions about WOOT, please contact us at classes@artofproblemsolving.com.

[quote="Klebian"]So if we sent the letter that we got from MOP, do we receive some acknowledgment that it's been received?[/quote]

We'll send an email to all enrolled WOOTers by the end of July.

Will WOOT be an online program every year?  :D

Yes; we expect to continue running WOOT each year.  It will be different from year to year, though.  (The classes will and assignments will change each year.)

[quote="Omar"]im not sure if im ready for it.  im going to be a sophomore next year, im 99.9% sure ill get at least to aimes, but im not really sure how well ill do there [my goal is like a 7-10ish for now, but i havent done any practice aimes or anything so im not sure how to gauge myself].  do you guys think woot would be beneficial/doable for me, anyway?[/quote]

I think you're ready for it, but if you don't think so, you can do it next year, as it says above, that AOPS tries to do it every year.

If you think you're ready, then go for it!

Just like do a practice test or something.

[quote="Omar"]im not sure if im ready for it.  im going to be a sophomore next year, im 99.9% sure ill get at least to aimes, but im not really sure how well ill do there [my goal is like a 7-10ish for now, but i havent done any practice aimes or anything so im not sure how to gauge myself].  do you guys think woot would be beneficial/doable for me, anyway?[/quote]

You can take some practice AIMEs - you can find some in the AoPSWiki, for example.

Also, we allow students to drop WOOT for a full refund up through October 5, so you can try the first couple classes to see how well it suits you.

I'm pretty excited for WOOT. I believe I signed up back in late April.  :P


Let $ 0<a \le b \le c \le d$ be real numbers. Prove that:

$ 7a^2\plus{}5b^2\plus{}3c^2\plus{}d^2 \le (a\plus{}b\plus{}c\plus{}d)^2 \le a^2\plus{}3b^2\plus{}5c^2\plus{}7d^2$.

we have $ ab\leq b^{2},    ca\plus{}cb\leq2c^{2},    cd\plus{}ad\plus{}bd\leq3d^{2}$
adding these inequalities we get the RH result
we do same with de LH.


linear algebra
matrix
function
quadratics
invariant
induction
pigeonhole principle

I'm just curious how you guys approached the power round.

My main breakthrough came with proving that two forms $ f$ and $ g$ are equivalent if and only if there exists a matrix $ A$ that can be written as a product of $ \begin{bmatrix} 0 & \minus{} 1 \\
1 & 0 \end{bmatrix}$ and $ \begin{bmatrix} 1 & \minus{} 1 \\
1 & 0 \end{bmatrix}$ such that $ f(x, y) \equal{} g(A(x, y)^T)$ for all $ (x, y) \in \mathbb{Z}^2$. In the end, I just got a bunch of systems of equations. That was, for me, a nice way to prove two forms not equivalent (and a nice way to show they are equivalent without drawing them out.)

I used parts from section 3 (including 3.5) to solve problems in section 2. I proved them beforehand. Heheh.

I also realized the day before PUMaC (the day, actually, at around 1:00) that we could show that two forms aren't equivalent if their rivers aren't the same. Ah, whatever.

The first half of section 3 was fairly easy (before the periodic rivers), except for 3.7, which I didn't get.

However, once I got to periodic rivers, I was fairly screwed. Then I realized that a river was periodic if and only if there was a function f with its "center" on the river (the values around the center are $ f(1, 0),$, $ f(1, 1)$, and $ f(0 ,1)$) such that there's a matrix that can be written as a product of the two matrices such that $ f(x, y) \equal{} f(A(x, y)^T)$, i.e., $ f$ is equivalent to itself on a different portion of the landscape. I noticed that a matrix $ A$ with determinant 1 does exist such that $ f(x, y) \equal{} f(A(x, y)^T)$ if the discriminant of $ f$ is not a perfect square (if we do some algebra, we can write the entries of $ A$ in terms of the Pell equation $ x^2 \minus{} Dy^2 \equal{} 1$.) Unfortunately, I couldn't prove that all matrices with determinant 1 can be written as the product of $ \begin{bmatrix} 0 & \minus{} 1 \\
1 & 0 \end{bmatrix}$ and $ \begin{bmatrix} 1 & \minus{} 1 \\
1 & 0 \end{bmatrix}$. (By the way, is that even true?) That was what stopped me cold. That made my proof of everything afterwards nonrigorous.

But I assumed that was true (I stated that in my proof), and from there, 3.8 and 3.9 were trivial. I couldn't do 3.10. What's the answer, anyway?

I got something in section 4, (I gave an example) but I lost my paper with the example, and by the time I realized it, it was about 2:00 am. And I was amazingly sleepy. Gah. Otherwise, I just stopped there.

There were three things I didn't really like about the power round. The first thing is that if we google binary quadratic forms, we get quite a bit of info on it, (especially on Planetmath.) For instance, a different definition of equivalence was on Planetmath (but I think the two definitions are equivalent), and with that in mind, we could simply copy their proofs for some of the early problems in section 3. I also didn't like how the definitions weren't very rigorous. I had to provide my own definitions of landscapes and rivers in order to make my proofs rigorous. Finally, some clarifications were sent on the last day...

Overall, I really enjoyed doing the power round. Unfortunately, I was the only person in my group who really did anything (I was in a group of all freshmen, and I was the only one who knew linear algebra/Pell equations, so everyone got lost past section 2) and I had a bunch of schoolwork that week. Still, it was fun. It was even better since my team got fourth place. =D It really got me going, and before that, I pretty much stopped doing math. Now, I'm back into it again. I can't wait for next year's PUMaC. We really should have more competitions like PUMaC power.

Now, how exactly did you feel about the PUMaC? How'd you work on it? And what were your approaches?
By the way, when will the solutions be posted? And will the 2007 solutions ever be posted? I remember I worked long and hard on that one too. -_-

The Power Test was nice. Here are some approaches that I remember. We didn't do part 4...but we got all of the rest. We probably lost some points for cursory-ness of our proofs though.

3.3: Basically show that if a,b,c are around a vertex, a^2+b^2+c^2-2(ab+bc+ca) is invariant, and show that this has to be the discriminant (this is easy using (0,1),(1,0),(1,1)).
3.6: No. f(x,y)=xy. You basically have two zeroes in the middle of the landscape and four 'binary tree's branching off from it. You can show with strong induction that is pretty darn annoying that these trees are either all positive or all negative, and they like, increase/decrease as you go down the tree. We probably lost points on this.

3.8/3.9 (mostly the work of Matt Superdock): Basically you show that a river is periodic if and only if it doesn't have a 0, which is equivalent to having a non-square discriminant. You can do this by showing that if there is a 0 and it's indefinite, there are two zeroes, and these zeroes bound the river and make it finite and thus non-periodic. If there are no 0's, you can bound (this is M@'s really awesome idea) the values on the river, and by infinite Pigeonhole (it's easy to show that the river goes on forever; assume it's finite, but a bordering region is positive or negative so we can extend it), two vertices have the same arrangement of numbers around it.

3.10: The answer is $ \sqrt {D} \plus{} 1$. You basically start with your (0,a,b) vertex, and show that you can't have (0,a,b) on the other side going in the same direction (this is easy because then you need an a and b on the same side and an a and b on different sides of a river, and this is bad), and this determines the rest of the river so we can just like count the number of (0,a,b) with the appropriate discriminant.

And dude your team came in fourth? How much did you do? Because I was really surprised that we and North Carolina (not North Carolina and us, because you can't have an objective pronoun functioning as a subject) came in as low as 8th and 9th respectively, after getting all of parts 1-3, and they also got like 4.2 or something.

I get what you mean for 3.6; I found the "trees" method for 3.7, but I was too tired/azy to write it down (again, I was working late at night.) However, due to the lack of clarity of the definition, I think my proof counted that 3.6 IS true. If we allow rivers to have length 1 (all it said is, "A river of a form is a connected
chain of edges on its landscape each of which separates a region with a positive value from a region with a negative value, which is not a subset of any other such chain"), then clearly, there must be a river in all indefinite forms. Again, I wasn't too fond of the rigority of the definitions.

What was your general approach for the problems in section 2? I found that analyzing the rivers was helpful, but before I realized that, I just used a bunch of equations. By the way, were the forms in 2.8 equivalent or not? I didn't get that one (I only realized my solutions for 2.7-2.9 wrong at around 11:00; I got 2.7 and 2.9, but not 2.8.)

How'd you show that the values are bounded?

I feel somewhat guilty for transforming everything into an algebra problem. Ah well.

[quote="CatalystOfNostalgia"]And dude your team came in fourth? How much did you do? Because I was really surprised that we and North Carolina (not North Carolina and us, because you can't have an objective pronoun functioning as a subject) came in as low as 8th and 9th respectively, after getting all of parts 1-3, and they also got like 4.2 or something.[/quote]

We placed 7th and we're rather sure we got 4.1-4.3 right. We didn't get 3.10 though since we had $ \sqrt{D}$. It might have been section 2, we did some brute-force testing on some of the forms we weren't sure about and, even an hour before the test was due, we were questioning one of our answers. I'm curious as to where the top two teams came from, I know the top team was "United", but I couldn't remember what the second team was.

Here are most of our (WOOTlings) solutions. We came in 10th.

Argh, I can't believe I missed that the determinants for 2.6 were different. And 3.6, my proof was flawed. Gah. I fail.

[quote]And dude your team came in fourth? How much did you do? Because I was really surprised that we and North Carolina (not North Carolina and us, because you can't have an objective pronoun functioning as a subject) came in as low as 8th and 9th respectively, after getting all of parts 1-3, and they also got like 4.2 or something.[/quote]
I was in division B. Sorry for forgetting to mention that. I suppose that changes everything. :P

[quote]I didn't really like the power test's abstractness, I would've preferred something combinatorial or geometric because they're easier to conceptualize (even though I despise geometry).[/quote]
I did. I work best with abstract things.

I think I'm going to make some edits in my power round and upload it, though I must admit the language is rather dry (even on proof standards) and difficult to understand; I've introduced my own definitions to everything and new notation that have made things difficult.

Judging from the w00tling's solutions, it seems to be that I was the only one who completely algebra-bashed it. They were completely inelegant. Ah well.

Albany Area Math Circle way(4th): 1. do them
2. If they are equivalent, find a triplet that is in both forms in the same order(clockwise or counterclockwise). Otherwise, find one value that is in one form but not the other. I wrote up all of number 2. It was fun.
3-4: I have no idea. Way over my head.

A general comment from the Power Test grading room:

We saw too much reliance on algebra, and even advanced topics in linear algebra.  The teams that were using matrix transformations, etc, quoting advanced books on quadratic forms, and the like were doing too much work.  The people who used mods and number theory were a little more reasonable.  Often, however, the problem can be solved by just drawing the landscapes, which was why they were introduced in the first place.  The point of the power test was to show one area of mathematics in which this seemingly unhelpful notation is actually easier to work with than conventional notation.  Seriously, there have been papers written on quadratic forms that are trivialized by the landscape concept.  More to follow when official solutions are posted.

Do any of you know the results to the power contest?

btw, this is our team's submission.  I drew all the diagrams by hand on a tablet in MS OneNote :D  The language might be a little dry and repetitive, because I was really the only one working on it (the other peeps on my team are students of my math session) and I had to type everything up really quickly.  And it may be confusing in 4 because I introduced a lot of new definitions.

We're [united] from NJ.

Apparently this is the first place power test, though after seeing the other power tests, I'm pretty surprised it is; I couldn't properly bound 3.8, and also pulled off a somewhat hand-wave on 4.3. Though apparently our lost point is from 3.10 in an off-by-one error.

3.3, I thought of rotation of conics.

4.1, 4.4, I luckily (and completely randomly) managed to create an example that worked for both.

After almost completely giving up on 4, I figured that I would completely forget about rivers and just worked with symmetries, and I guess it worked?


Let a,b,c>o.Prove that
$ \frac{a}{a\plus{}b})^3\plus{}( \frac{b}{c\plus{}b})^3\plus{}( \frac{c}{c\plus{}a}) \geq  \frac{3}{8}$

Must is

Let a,b,c>o.Prove that
$ (\frac {a}{a \plus{} b})^3 \plus{} ( \frac {b}{c \plus{} b})^3 \plus{} ( \frac {c}{c \plus{} a})^3 \geq \frac {3}{8}$

Then It is from Vietnam TST 2005.

http://www.mathlinks.ro/Forum/viewtopic.php?p=230046#230046


which is the larger?
$ x \equal{} 4^{321}$ or $ y \equal{} 3^{421}$ .


limit
trigonometry
LaTeX
inequalities
calculus
calculus computations

Find $ lim_{n\to\infty}$ of $ f$ there
$ f = \frac {2^nsin\frac {1 \pi}{n} sin\frac {2 \pi}{n} ...... sin\frac {(n - 1)\pi}{n}}{n + 1}$
I hope some one can kill it without complex number

You might want to specify what $ n$ approaches.

[quote="Dr Sonnhard Graubner"]hello, do you want to calculate
$ \lim_{n\to \infty}\frac {2^n\sin\frac {1\pi}{n}\ sin\frac {2 \pi}{n} \cdots\sin\frac {(n \minus{} 1)\pi}{n}}{n \plus{} 1}$
Sonnhard.[/quote]
Yes,Sonnhard,My proof is used complex number and of course,it is not nice and not suitables for our contest,so I think we can find an easy proof but I failed

I hope the following be true and be what you intend .Let me at first prove the existence of the limit .of course it is not asked and not necessary because of the next stage ,finding the limit but sometimes the existence easier than finding .
f_n+1/f_n = (2*(n+1)/(n+2))*(sin(pi/(n+1))/ sin(pi/(n))* (sin(2pi/(n+1))/ sin(2pi/(n))….
*(sin((n-1)*pi/(n+1))/ sin((n-1)*pi/(n))* sin(n*pi/(n+1))
All above terms except 2 are less than 1 ,specially the last term which is near 0 when n is large so the sequence is decreasing .On the other hand that is lower bounded then it has limit .
If n be even then   n= 2* m     and sin terms are repeating from the half so:
f_n < =  (2^(2*m)/(2m+1))    *  (    (m-1)!  * pi ^ (m-1)  /( (2m)^(m-1)    ) ^2
The limit of right side is 0 when n goes toward infinity ,( it is also true for  odd n )then according to sandwich theorem  lim  f_n = 0 when n ----- > infinity

In $ \text{\LaTeX}:$

$ \frac {f_{n + 1}}{f_n} = 2\cdot\frac {n + 1}{n + 2}\cdot\prod_{k = 1}^{n - 1} \frac {\sin\left(\frac {k\pi}{n + 1}\right)}{\sin\left(\frac {k\pi}{n}\right)}\cdot \sin\left(\frac {n\pi}{n + 1}\right)$

However, there's a problem with the argument.

[quote="christ"]All above terms except 2 are less than 1 [/quote]
That's not true. $ \frac {\sin\left(\frac {k\pi}{n + 1}\right)}{\sin\left(\frac {k\pi}{n}\right)} < 1$ when  $ \frac {k}{n}\le\frac12,$ but the same quotient is greater than $ 1$ when $ \frac {k}{n + 1}\ge\frac12.$ That is, about half of those factors are less than one and about half of them are greater than one.

I’m sorry. I checked last inequality  and limit by MATLAB software so I find those wrong. Indeed  enlarging the f_n  is improper . I’d like to suggest another method to fine the limit, could anyone test  that? The limit ‘2’ is checked by MATLAB  . That is somewhat long but don’t need to complex values.
   What is left to do(which I was unable to do that) is proving of the limit existence .If we would able to prove that the sequence is increasing the existence was proved .It is no necessary to prove that the sequence is upper bounded ,let’s see .
      Let’s suppose  specially   n=2^m , m an positive integer .
g_n = (n+1)f_n/2^n
sin(k* pi /n ) = sin(pi - k* pi /n )= sin((n-k)* pi /n )     , 1<= k <= n-1
1<= n-k <= n-1  then sin’s are repeating after pi/2 so regarding the fact that n-1 is odd:
g_n = (  sin( pi /n )sin(2* pi /n ) sin(3* pi /n ) ….. sin((n/2-1)* pi /n )   )^2
The symmetric sin’s can be joined together and transformed to cos . Only the central sin is left   :
sin((n/4)* pi /n )=1/2^0.5  regarding the fact that n/2-1  is odd .It is necessary to a coefficient equal to
(2^((n/2-2)/2))^2   .
    If we call this stage by 1st  and continue to change product of symmetric cos’s  to sum of cos’s in a stage such as p we will reach to a (cos(pi/4))^2 . In this case
n/2-2^p =n/4  then  p= m-2  .
Powers of  2 which needed to change totally are equal to :
(n/2-2)+(n/4-2) +…. + (n/2^p-2)=n*(1/2)*(1-1/2^p)/(1-1/2) -2*p =2^m -2*p -2^2  = 2^m – 2*m
Coefficient  made by  (cos(pi/4)^2  in p stage is  :
(1/2)^(p+1) = (1/2)^(m-1)
The final yielded coefficient is :
(1/2)^(2^m-m-1) = (2*n)/2^n
Finally    f_n = (2*n)/(n+1)  which results in for  specially n=2^m  ,enlarging n or m justifies  arbitrary vicinity of f_n to 2 .


calculus
integration
geometry
3D geometry
tetrahedron
symmetry
probability

caculate interal :
$ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1}  \max (x,y,z)dxdydz$

It would be very nice to calculate the following integral:

$ I_{n}\equal{}\int_{0}^{1}\cdots \int_{0}^{1}\max(x_{1},x_{2},\cdots, x_{n})dx_{1}\cdots dx_{n}\equal{}?$.

Assuming that I made no mistakes I got that $ I_{3}\equal{}\frac{3}{4}$, is that right?

The set $ \{(x,y,z): 1\ge x\ge y\ge z\ge 0\}$ is a tetrahedron - a simplex. By symmetry, it is one of six such tetrahedra of equal volume whose non-overlapping union is the cube. Utilizing this symmetry, we claim that

$ I_3\equal{}6\int_0^1\int_0^x\int_0^yx\,dz\,dy\,dx$

$ \equal{}6\int_0^1\int_0^xxy\,dy\,dx$

$ \equal{}3\int_0^1x^3\,dx\equal{}\frac34.$

Yes, I agree with didilica. It should be quite possible to generalize this method to cover all of didilica's version of the problem.

Quite interesting, I think that the following formula holds for the general case:

$ I_{n}\equal{}\frac{1}{2}\plus{}\frac{1}{2}\int_{0}^{1}\cdots\int_{0}^{1}\max(x_{1},\cdots,x_{n\minus{}1})^{2}dx_{1}\cdots dx_{n\minus{}1}$.

$ I_n\equal{}n!\int_0^1\int_0^{x_1}\int_0^{x_2}\cdots\int_0^{x_{n\minus{}1}}
x_1\,dx_n\,dx_{n\minus{}1}\cdots\,dx_2\,dx_1$

I claim that $ \int_0^{x_1}\int_0^{x_2}\cdots\int_0^{x_{n\minus{}1}}
dx_n\,dx_{n\minus{}1}\cdots\,dx_2\equal{}\frac1{(n\minus{}1)!}x_1^{n\minus{}1}.$

This is because it's simply the volume of a simplex which is that particular fraction of the volume of a cube.

Then $ I_n\equal{}\frac{n!}{(n\minus{}1)!}\int_0^1x_1^n\,dx_1\equal{}\frac{n}{n\plus{}1}.$

----

There are many other ways to deduce this. This is a result familiar in probability theory under the heading "order statistics." The question can be phased as this: What is the expected value of the maximum of $ n$ independent random variables, each uniformly distributed on the same interval?

How to treat this problem
\[ \int\limits_0^1 {\int\limits_0^1 {\int\limits_0^1 {\max \left\{ {x,y^2 ,z^3 } \right\}dxdydz} } } .
\]

Consider the following

Generalization: Let $ m$, $ n$ and $ p$ be natural numbers:

Find:

$ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\max(x^{m}, y^{n},z^{p})dxdydz$.

Answer, (assuming I made no comp. mistakes):

$ \frac {1}{p \plus{} 1}\left(1 \plus{} \frac {p^{2}(m \plus{} n)}{pmn \plus{} pm \plus{} nm \plus{} pn}\right)$.

My answer is
\[ \int_0^1 {\int_0^1 {\max \left\{ {x^m ,y^n } \right\}dxdy} }  \equal{} \frac{{m \plus{} n}}
{{mn \plus{} m \plus{} m}} \equal{} \frac{{\frac{1}
{m} \plus{} \frac{1}
{n}}}
{{1 \plus{} \frac{1}
{m} \plus{} \frac{1}
{n}}}.
\]

\[ \int_0^1 {\int_0^1 {\int_0^1 {\max \left\{ {x^m ,y^n ,z^p } \right\}dxdydz} } }  \equal{} \frac{{mn \plus{} np \plus{} pm}}
{{mnp \plus{} mn \plus{} np \plus{} pm}} .
\]

\[ \idotsint_0^1 {\max \left\{ {x_1^{\alpha _1 } ,x_n^{\alpha _n } ,...,x_n^{\alpha _n } } \right\}dx_1 dx_2 ...dx_n }  \equal{} \frac{{\sum\limits_{k \equal{} 1}^n {\frac{1}
{{\alpha _i }}} }}
{{1 \minus{} \sum\limits_{k \equal{} 1}^n {\frac{1}
{{\alpha _i }}} }}.
\]

Have fun.

[quote="bookworm_vn"]
\[ \idotsint_0^1 {\max \left\{ {x_1^{\alpha _1 } ,x_2^{\alpha _2 } ,...,x_n^{\alpha _n } } \right\}dx_1 dx_2 ...dx_n } \equal{} \frac {{\sum\limits_{k \equal{} 1}^n {\frac {1} {{\alpha _i }}} }} {{1 \minus{} \sum\limits_{k \equal{} 1}^n {\frac {1} {{\alpha _i }}} }}.
\]
Have fun.[/quote]
But my answer is different from you:
And like Kent Merryfield's way,we can calculus the area that $ x_1^ {\alpha_1}$ is the maximum of $ \left\{ {x_1^{\alpha _1 } ,x_2^{\alpha _2 } ,...,x_n^{\alpha _2} } \right\}$
That:
$ \int_0^1 x_1^ {\alpha_1} \int_0^{{x_1}^{\frac {\alpha_1}{\alpha_2}}}\cdots \int_0^{{x_1}^{\frac {\alpha_1}{\alpha_n}}}dx_1 dx_2 ...dx_n \equal{} \int_0^1 x_1^{\alpha_1 \plus{} \frac {\alpha_1}{\alpha_2} \plus{} \cdots \plus{} \frac {\alpha_1}{\alpha_n}}dx$
We can get the result,Am I wrong?


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

In a domain, do we have that $ a^2|b^2 \Rightarrow a|b$? In a UFD it's trivial, but I don't now about a domain in general...

Thx!


calculus
analytic geometry
graphing lines
slope

I'm trying to figure out how to do instantaneous rates, but it seems like I need to know calculus to do this. I don't know calculus, and it wasn't listed as a prerequisite; I wasn't told the first day I need to know anything beyond basic algebra.

Anyway, I'm suppose to find the "slope of a line tangent to the curve at any point."

I don't really understand that. I tried searching for hours about this stuff, and I couldn't really find anything to help me; well, I tried analyzing stuff, but I couldn't see how it applied to what I was doing.

The book didn't have that information.

I know how to find the slope of something. But drawing this line against the curve... I don't know how to do that.

Therefore, I decided to come here.

Attached to this post is an image.
It's giving an example of instantaneous rate.

Thing is, I don't exactly understand how the authors were able to get that line next to the point (t=600) on the graph.

No equation has been giving to figure out how to do that.

The book also tells me this:

http://img357.imageshack.us/img357/6097/equationpe8.jpg

I'm not very sure about how it obtained those top variables. It seems like it picked them out of thin air. If I'm correct, though, those variables come from the points that the slope comes across. However, I couldn't know those points unless I had done all of this in a graphing calculator, right?

I haven't the slightest clue how those top variables were obtained, nor do I have the slightest clue how to make that line go against the point of the curve.

How do I do these things?

In my opinion, it seems that picture was presented by the authors of the book just to ilustrate the fact that the rate of a reaction is not the same at different instants. So, we define instantaneous rate at any instant t by the slope of the line tangent to the graphic concentration versus time at t. To calculate the slope you need, as you know, the values of concentration and of time, and these can be found by considering two points of the tangent line and its coordinates. To find the coordinates, wether you find them by using milimetric paper, or by using the equation of the graphic. Those top variables given by the authors were calculated by a similar method, and they gave those values very probably just to ilustrate how one can calculate the instantaneous rate. If you are given an exercise to calculate an instantaneous rate, then you must also be given the graphic equation, or otherwise it will be easy to check the values in both axes.


geometry
3D geometry
inequalities open
inequalities

Sorry everyone but I don't know Tex.
Let x,y,z be positive real numbers so that x + 2r(xy) + 3r (xyz) =3/4
notice that nr(x) mean n root of number x (ex. 2r(x) mean square root of x, 3r(x) mean cube root of x).find the maximum of (x+y+z).this is from a mathematical magazines for [b]Secondary school students in Vietnam.
I'm very grateful to s.one who change this one to Tex.thank you[/b]

[quote="Yutzpracachia"]
Let $x,y,z$ be positive real numbers so that $x+\sqrt{xy}+\sqrt[3]{xyz}= \frac{3}{4}$. find the maximum of $x+y+z$. [/quote]

There is no maximum: make $x,y$ very small, then $z$ is big.
For the minimum: see http://www.mathlinks.ro/Forum/viewtopic.php?t=100935
In your case it would be $\frac9{16}$  :wink: Have you quoted the problem from memory??  :lol:

Yes,I forgot two things 4/3 instead of 3/4 and minimum instead of maximum,Thank you very much


Find the distance form the origin to the plane through (1,0,0),(0,2,0), and(0,0,3).

the distance of a point to a line/plane/hiperplane/etc can be easy generalizate to

suppose $g(x,y)-z = 0$ is a plane

and we have the points $(a,b,c)$

the minimum distance is given by the formulae

$d^{2}= \frac{(g(a,b)-c )^{2}}{1^{2}+{\frac{\partial f(x,y)}{\partial x}}^{2}+{\frac{\partial f(x,y)}{\partial 7}}^{2}}$

the proof of this can be found in google or you can make this

Note: Suppose $g(x,y) = a x+b y+c = z$

so we can symplify ${\frac{\partial f(x,y)}{\partial x}}= a$
${\frac{\partial f(x,y)}{\partial y}}= b$

i just put in this form to be more easy to remember or generalize

If you wish to do this without multivariable calculus here is the method of solution:

Make two vectors tail-to-tail out of the three points given on the plane and find the cross product, this will give you the direction vector for the line that is to pass through the plane and the point in question. Define this line and then solve the system between the plane and the line. This will give you the point on the line that also lies on the plane. Proceed to find the distance between this new point and the point in question. Voila!

Things that aren't explained in this post:
-how to define the scalar equation of the plane given 3 points
-how to define the vector equation of the line given a point and a direction vector

[hide="The simplest"] The equation of the plane is $6x+3y+2z = 6$ (make sure you see why this is so).  But by Cauchy,

$(x^{2}+y^{2}+z^{2})(6^{2}+3^{2}+2^{2}) \ge (6x+3y+2z)^{2}= 36 \implies$
$\sqrt{x^{2}+y^{2}+z^{2}}\ge \boxed{ \frac{6}{7}}$ [/hide]

[quote="Gyan"]May have already been pointed out ...[hide]Eq of the plane: $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}= 1$
[/quote]
Why is the equation of the plane that?

[quote="t0rajir0u"]Short answer:  A plane is determined by three non-collinear points (consider the triangle with those points as its vertices), and that equation clearly contains all three given points.[/quote]

Oh, yeah.

[quote="ckck"][quote="Gyan"]May have already been pointed out ...[hide]Eq of the plane: $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}= 1$
or $6x+3y+2z=6$ and the required answer is $\frac{6}{\sqrt{6^{2}+3^{2}+2^{2}}}$
in general distance from point $(x,y,z)$ and the plane $ax+by+cz+d$ is $\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}}$[/hide][/quote]
I'm still confused about how you came to $\frac{x}{1}+\frac{y}{1}+\frac{z}{1}$ to $6x+3y+2z=6$. Isn't the points for the plane $(1,0,0),(0,2,0)$ and $(0,0,3)$? Wouldn't that make it $x+2y+3z=1$?[/quote]Oh yea? Plug in the values for the point (0,0,3) and you will get the false statement 9=1...

[quote="ckck"][quote="Gyan"]May have already been pointed out ...[hide]Eq of the plane: $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}= 1$
or $6x+3y+2z=6$ and the required answer is $\frac{6}{\sqrt{6^{2}+3^{2}+2^{2}}}$
in general distance from point $(x,y,z)$ and the plane $ax+by+cz+d$ is $\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}}$[/hide][/quote]
I'm still confused about how you came to $\frac{x}{1}+\frac{y}{1}+\frac{z}{1}$ to $6x+3y+2z=6$. Isn't the points for the plane $(1,0,0),(0,2,0)$ and $(0,0,3)$? Wouldn't that make it $x+2y+3z=1$?[/quote]

You just multiply by 6. ...

BTW
The equation is  $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}= 1$
and NOT $\frac{x}{1}+\frac{y}{1}+\frac{z}{1}$

(And as explained before,  (1,0,0),(0,2,0), (0,0,3) all are on the plane and it  is  easy to see that
1/1 +0 + 0  =  0 + 2/2 + 0  = 0+0 + 3/3 = 1


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

[quote="pda"]Let ABCDEFGHIJ be a regular decagon with circumradius 1. What is AD-AB?[/quote]

[hide="solution"]$ AD$ is easy to get because it is just the hypotenuse of a isosceles right triangle with leg 1.  So $ AD=\sqrt{2}$.

To calculate $ AB$, consider the triangle $ ABO$ where $ O$ is the center of the circumcircle.  It is a $ 75-75-15$ triangle with $ AO=BO=1$.

Draw the altitude from $ O$.  $ \sin 15=\sin(45-30)=\sin45\cdot \cos45-\sin30\cdot\cos30=\frac{1}{\sqrt{2}\cdot\frac{1}{\sqrt{2}}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{4}$.

Therefore, $ \frac{2-\sqrt{3}}{4}-\sqrt{2}=\boxed{\frac{2-\sqrt{3}-4\sqrt{2}}{4}}$.[/hide]

Um... jli?  The problem asks for a decagon, not a dodecagon.

[quote="vishalarul"][quote="pda"]Let ABCDEFGHIJ be a regular decagon with circumradius 1. What is AD-AB?[/quote]

@jli: I think you thought it was a 12-sided figure...
[hide="Solution"]Basically, all we care is about the quadrilateral shown. With some easy angle chasing, we see that $ m\angle A = m\angle D = 36^{\circ}$ and $ m\angle B = m\angle C = 144^{\circ}$. Draw a perpendicular from the center of the circumscribed circle to determine $ AB$:
$ CD = BC = AB = 2\sin 18^{\circ}$

Draw perpendiculars from B and C to D. Now, we can determine the length of $ AD$:
$ AD = 4\sin 18^{\circ}\cos 36^{\circ}+2\sin 18^{\circ}$

Hence,

$ AD-AB = 4\sin 18^{\circ}\cos 36^{\circ}= 4\cos 72^{\circ}\cos 36^{\circ}= 4\cdot \frac{\sqrt{5}+1}{4}\cdot \frac{\sqrt{5}-1}{4}= \boxed{1}$

Pretty answer?[/hide][/quote]

If you do some shifting stuff, i think the answer is the long side of a 36-36-108 triangle with two sides being equal to AB.

Or you could Law of Cosines bash your way through it... it's honestly not bad at all if you remember that $ \cos{36^\circ}= \frac{1+\sqrt5}{4}$ and that $ \cos{108^\circ}=-\cos{72^\circ}= \frac{1-\sqrt5}{4}$.


geometry
3D geometry
sphere
Ring Theory
circumcircle
projective geometry
power of a point

Who can give me some useful methods or theorems can be used in problems of circles?thank you !
 :oops:  I'm in a hurry
A pdf can be downloaded will be better :roll:

the most basic formula is the area and volume formula's
for circles it's A= (pi)r^2
for a sphere it's
A= 4/3 (pi) r^2
V= 4/3 (pi) r^3
where r is the radius and a is the area and v is the volume.
some others are he diameter s double the radius. and the circumference is
C=2(pi)r OR (pi)d
where d is the diameter
r radius
c circumference
this i all very basic, do you want higher level?
[hide]
don't give me a bad rating for this :D how abotu a good one. Thanx[/hide]

Here are some things you could look up on Google:

General facts about cyclic quadrilaterals (equal angles facing same arc; opposite angles supplementary)
Angle formed by a tangent and a chord is half the measure of the arc facing the chord
Power of a Point (equal tangents too)
Ptolemy's theorem
Pascal's theorem
Area of triangle $ ABC$ is $ \frac{abc}{4R}$ where $ a,b,c$ are side lengths and $ R$ is the circumradius

Then more advanced stuff: (read the threads in the stickies of this forum for a good treatment; Google could be good too)

Inversion in the plane (good for lots of circles, as they are transformed into lines)
Pole-Polar transformation (changes concurrency into collinearity and vice-versa)

yes,sir.
I realy need some higher.
I am familiar with thiese basic knowledge :oops:


Not sure if this is the right spot, but i couldnt find anything that looked similar.

ok i need to find the value of the continuing fraction as follows:

X= 1+1/(2+1/(1+1/(2+1/(1+1/(2+1/(1+........ and so on.

I have read some stuff on how to do these but none conver when the fraction numbers change between 1 and 2.

I read you subtract one from the eqn. so (x-1)= 1/(2+1/(1+1/2+1/.......

after that im lost.  Help would be great.  Please move this thread if its not in the right spot.

Thanks


Find the min value of $ P\equal{}tan^4x\plus{}cot^6x$.


calculus
integration
greatest common divisor
number theory
Diophantine equation
number theory unsolved

[quote="987"]Given that m,n and k are positive integers and 1/m-1/n=1/k, show that mnkh is a perfect square where h is the greatest common divisor of m,n and k.[/quote]

I got to got soon. But here is my approach.

You got,
$\frac{1}{m}-\frac{1}{n}=\frac{1}{k}$
Rewrite it as,
$\frac{1}{m}=\frac{1}{k}+\frac{1}{n}$

And it reduces to the diophantine equation,
$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$
Where $a,b,c$ are positive integers.

Then,
$\frac{a+b}{ab}=\frac{1}{c}$
Then,
$\frac{ab}{a+b}=c$
Meaning find all $a,b$ such that:
$\frac{ab}{a+b}$ is an integer.

Now that happens only when $a|b$ or $b|a$.
So that tells you that one of the denominator numbers got to divide another one.
Doesnt this make it easier  :wink:

[quote="987"]I got that far by myself however it does not really make the problem much easier (it would be a strange national olympiad problem that is that easy to solve).[/quote]
Your right, but it tells us that,
(WLOG $k\leq n$ then by above $k|n$)
$h=\gcd(m,n,k)=\gcd(\gcd(k,n),m)=\gcd(k,m)$

Well,n divides k so an=k for an integer a. Rearranging gives
m=an/(a+1) so n/m =(a+1)/a which is integral iff a=1 and thus iff n=2m. It follows that h=1 if a is anything but 1 (i.e. if n isn't equal 2m). Obviously the mnkh is square if a=1 - otherwise I really have no idea...

This is a well known problem...

You got up to this: $(a+b)c=ab \Leftrightarrow (a+b)c^{2}=abc$. We will show that $a+b$ is a square. Suppose it isn't, then there is some prime $p$ s.t. $p^{2k-1}\parallel a+b$. Then $p^{2k-1}|ab$ so at least one of numbers $a$ and $b$ is divisible with $p^{k}$. But $p^{2k-1}|a+b$ so they are both divisible with $p^{k}$ so $p^{2k}|ab$, and from that we have that $p|c$ so $(a,b,c)\not = 1$. So $a+b$ is a square, as it is $abc$.

Bye


Hi everyone,

Could someone please explain to me how you would use L'hopital's rule on (2/3t^3/2)/(3)-(4t^7/2)/(21), please?

I don't know why anyone would do that.
$ \frac{(2/3)t^{2/3}}{3}\minus{}\frac{4t^{7/2}}{21}\equal{}t^{7/2}\left(\frac{(2/3)t^{\minus{}17/6}}{3}\minus{}\frac{4}{21}\right)$
 where the expression in parentheses tends to $ \minus{}4/21$ as $ t\to\infty$. So your function tends to $ \minus{}\infty$ as $ t\to\infty$, assuming I deciphered it correctly.

Thank you very much

Yes, that's what the equation was. Could you please explain to me how you got negative infinity? Did you use L'hopital's rule?

Thank you

For this problem don't need L'Hospital rule.
You can use some property of limit.
Try to find out solution.If not I will post solution.


geometry
inequalities
calculus
integration
trigonometry
calculus computations

to appear geometry in the first quarter xy level, the R site bounded by the inequalities

a^2<=x^2+y^2<=b^2 and is below the line of equation y = x
and..

to calculate the integral \ iint_ (R) \ frac (y ^ 2) (x ^ 2) dxdy
using the appropriate change VARIABLES
[quote]\ iint_ (R) \ frac (y ^ 2) (x ^ 2)[/quote]dxdy

$ \int_{R} \frac {y^2}{x^2} \ dxdy \equal{} \left| \begin{array}{c} x \equal{} r\cos \varphi \\
y \equal{} r\sin\varphi \end{array} \right| \equal{} \int\int_{\begin{array}{c} a < r < b \\
0\le\varphi \le \frac {\pi}{4} \end{array}} r \cdot (\tan \varphi)^2 \ drd\varphi \equal{} \int_{a}^{b} r\ dr \int_{0}^{\pi/4} \tan^2 \varphi \ d\varphi \equal{} \frac{b^2\minus{}a^2}{2}\left(1\minus{}\frac{\pi}{4}\right)$

I think by 'below the y=x line' the OP meant $ 0 \leq \varphi \leq \frac{\pi}{4}$


Consider two circles $ (C_1),(C_2)$ which their equations are: $ (C_1): x^2\plus{}y^2\equal{}2$, $ (C_2): x^2\plus{}y^2\equal{}5$ and the point $ A(0;1)$. Determine the position of $ M\in (C_1), N\in (C_2)$ when the area of the triangle $ AMN$ attains its greatest value.

Ago some days, I made a nice solution. The greatest value of the area of the triangle AMN is 3, it attains this value when $ M(\minus{}1;\minus{}1),N(2;\minus{}1)$ or $ M(1;1),N(\minus{}2;\minus{}1)$.
Can you solve the problem without analytic ? Good luck!


Kolmogorov`s  competition
Prove that if a, b and c are positive numbers,then
a^2/(2a+b+c) + b^2/(a+2b+c) + c^2/(a+b+2c)≥(a+b+c)/4   thanks :wink:

When was the problem set in the competition?


Please, no discussion of the March 13 2007 AIME (also known as AIME I) until Wednesday, March 14.

Students in many time zones will be taking the contests, so no discussion starting tonight, and continuing until Wednesday, March 14.

Thank you for your cooperation.

Steve Dunbar
Director, American Mathematics Competitions

As requested, please do not discuss the AIME until further notice.  This means no discussion anywhere, including the forum, the wiki, and blogs.

The AMC forum has been locked until Wednesday March 14.

Get a good night's sleep tonight, and good luck tomorrow on the AIME!


Let $ S_{n}=\sum_{k=1}^{n}\frac{F_{k}}{2^{k}}$. Where $F_{1}=F_{2}=1$ and $F_{n}=F_{n-1}+F_{n-2}$ for $n\ge{3}$.

Find a formula for $S_{n}$ without using induction.

\[\\\sum_{k=1}^{n}\frac{F_{k}}{2^{k}}=\sum_{k=1}^{n}\frac{\phi^{k}-(1-\phi)^{k}}{2^{k}\sqrt{5}}\\=\frac{1}{\sqrt{5}}\left[\sum_{k=1}^{n}\left(\frac{\phi}{2}\right)^{k}-\sum_{k=1}^{n}\left(\frac{1-\phi}{2}\right)^{k}\right]\\=\frac{1}{\sqrt{5}}\left[\left(\frac{\phi}{2}\right)\left(\frac{\left(\frac{\phi}{2}\right)^{n}-1}{\left(\frac{\phi}{2}\right)-1}\right)-\left(\frac{1-\phi}{2}\right)\left(\frac{\left(\frac{1-\phi}{2}\right)^{n}-1}{\left(\frac{1-\phi}{2}\right)-1}\right)\right]\] where $\phi=\frac{1+\sqrt{5}}{2}$.  You can prove the substitution for $F_{k}$ using either induction or the characteristic equation of the fibonacci sequence.  The rest requires some simplification which is left to the reader.

Masoud Zargar

the generating function is:
x/(4-2x-x^2)=sum(n=0 to n=infinite)S(n)x^n
sorry but i am not yet able to write in LaTex
yours solutions are very beatiful,if you want you can use my function also

In order to learn LaTex, scroll over other people's work and try and see some patterns.  Also, put dollar signs around your code and it'll appear in the so-called "nice" way. :)

i must download any pragrams? thanks for your advice

You need not download anything, just put dollar signs around your code and the software on the forum does it for you.
For example:

\sum x^2-3x

If you put dollars around what I just typed, you get $\sum x^{2}-3x$.

This thread seems to be diverging from the original topic.

yay for latex tutorials!

I think that my solution is the best, what do you think?

Lol, jk.  I'm not confident about my understanding of generating functions, how do you know that will work and how did you arrive there?

by cross multipling you can verified that.i arrived there with an inverse reasoning.

$\frac{x}{4-x-x^{2}}=\sum_{k=1}^{infinite}S_{k}x^{k}$

(power series)

however your solution is very beatifull with a good idea!

P.S. what is the simbol of "infinite"

[quote="girino"]P.S. what is the simbol of "infinite"[/quote]
It is \infty. For the list of $\LaTeX$ symbols see [url]http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf[/url].


A sum of money is to be divided among Allan, Bill and Carol. Allan receives $\$1$ plus one third of what is left. Bill then receives $\$6$ plus one third of what is left. Carol receives the remaining $\$40$. How much did Bill receive?

[hide="solution"]We could work forwards with a bunch of nasty fractions, but lets not.  Carol recieves the remaining $40$ which is $\frac{2}{3}$ of what had remained.  So $(40)(\frac{3}{2})=60$.  Bill had recieved $6$, so $60+6=66$, which is $\frac{2}{3}$ of what had remained.  so $(66)(\frac{3}{2})=99$.  and $99+1=\boxed{100}$

EDIT: i didnt see it said what did Bill get.  :oops:

so $100-1=99$ and $99/3=33$, so allan recieves $34$ dollars, leaving $66$.  Bill recieves $6$ dollars leaving $60$, and $\frac{60}{3}=20$, so bill recieves a total of $\boxed{26}$

[/hide]

[hide="or, you can do it the even easier way :D "]Carol gets $40$, which is twice the amount of the one third that Bill got. So Bill got $20+6=26$ :D [/hide]

I hate problems where you do a lot of work but then afterwards, you find a much easier way to do it...


A real number m is randomly selected such that 0 is smaller than or equal to m, which is smaller than or equal to 8. What is the probability that m is smaller than or equal to 2. Express your answer as a common fraction.

[hide]So m is greater than or equal to 0, but less than or equal to 8.

m can be: {0, 1, 2, 3, 4, 5, 6, 7, 8}

for m to be less than or equal to 2, it can be: {0, 1, 2}

The probability that m is less than or equal to 2 is the number of elements in the second set / the number of elements in the first set.

Answer is $\frac 13$[/hide]

[hide]a real number is any number, including decimals, which doesn't have a somplex number with in.  So, there are 8 sections (between 0 and 1, between 1 and 2 etc.), and the number of sections under 2 is 2, so the probability is 2/8 or 1/4.[/hide]

[hide="click me!  :D "]0 is < or = to m, which is < or = to m. it is inclusive, so there are 9 possible numbers n could be.
so the prob. of n being between 2 and 0 inclusive is 3. 3/9 is 1/3.[/hide]
Oops. i didn't read real numbers. But i think my answer is still correct.

It is definetely ${\frac 14}$. It never said $m$ has to be an integer.

Actually, if it's a real number, the answer is arbitrarily larger than $\frac{1}{4}$, since the interval chosen from is one greater than $2$ and the entire interval is one greater than $8$. ;) So, if that's the intended answer, it's not written properly; ideally, $0$ should not be included in either interval.

but it said that the real number $m$ must be between 0 and 8, so anything greater than 8 is not and interval

Actually, the answer to this problem is 1/4.

 This is because between and number lets say 0 and 1, there are many, many terms(lets just say 10000000(for example) (doesn't include the o and the 1)) so between 0 and 8 there are 800000008 terms. Then, if m is smaller than 2, there are 200000003 terms. But the answer is about 1/4 becuase that last term is way too small to be of any significance. And, since there are an infinite amount of terms between 2 integers, that one term will really have no importance.

Still, if it didn't include the common endpoint as a possible value, it would definitely be $\frac{1}{4}$, no matter what you said. That's how they should have done it, as there was really no reason not to.

[hide]Notice that we can fit four 2 unit lines in a 8 unit line. THus the probability is 1/4. [/hide]

[quote="Treething"]The answer is 1/4th, after you take a course involving limits then you'll see a more in-depth explanation why.[/quote]

Actually, making a number line is enough.


Mandvil has one standard quarter and one special quarter with heads
on both sides. He selects one of these two coins at random, and
without looking at it first, he flips the coin three times. If he flips a
head three straight times, what is the probability that he selected the
special quarter? Express your answer as a common fraction.

The probability that he picks the regular quarter is $ \frac{1}{2}$ so the probability of also getting heads 3 times is:
\[ \frac{1}{2} \times (\frac{1}{2})^3 \equal{} \frac{1}{16}\]
The probability that he picks the special quarter is also $ \frac{1}{2}$ but if you pick the special quarter you get 3 heads nomatter what.
\[ \frac{\frac{1}{2}}{\frac{1}{16}\plus{}\frac{1}{2}} \equal{} \frac{\frac{1}{2}}{\frac{9}{16}} \equal{} \boxed{\frac{8}{9}}\]

Question: Why do we divide $\frac{1}{2}$ by $\frac{9}{16}$ to find the probability? :help:

This is because we had to specify something:  if he flipped a head three straight times.  The denominator in this case is not 1 because he might not have flipped heads three straight times.  The denominator would be the probability he flipped three heads.

Normal problems would not have any restrictions.


calculus
integration
advanced fields
advanced fields unsolved

Help me to solve
1)$ u_{xx}\plus{}u_{yy}\equal{}0$ in circle $ D\equal{}${$ x^2\plus{}y^2<1$}, $ u(x,y)\equal{}sgn(x\minus{}\frac{1}{2})$ on $ \partial D$, $ sgn{0}$ is not defined.
2)$ u_{xx}\plus{}u_{yy}\equal{}0$ in $ D\equal{}${$ y>0$}, $ u(x,0)\equal{}\frac{1\minus{}cos{x}}{x}$.

1. apply Posion formula. If you are "satisfied" with a solution represented in integral form.
2. apply fourier transform or expand in trigometry series if you are also content to have a series solution.


the top simplifies into $-3(2x-y)(x-2y)(3y-3x)=0$, the system is then easy enough to solve

I don't see how you get that.  Is it just algebra or something else?

EDIT: Never mind, I see... difference of cubes.


inequalities
number theory unsolved
number theory

Let $ a_1\equal{}\frac{1}{2}$and for each $ n>1$let $ a_n\equal{}\frac{1}{d_1\plus{}1}\plus{}\frac{1}{d_2\plus{}1}\plus{}...\plus{}\frac{1}{d_k\plus{}1}$
where $ d_1;d_2;...;d_k$ is the collection of all distinct positive divisorsof $ n$.Prove the inequality:
$ n\minus{}Lnn<a_1\plus{}a_2\plus{}...\plus{}a_n<n$

Remark that

a1+a2+...+an= s(1)/(1+1)  + s(2)/(2+1) +....+ s(n)/(n+1)

where s(k) is the number of integer not exceeding n and is divisible by k

Hence

a1+...+an= sum ( [n/k]/(k+1)]

Not hard to continue

I don't understand! [b]mto[/b], please post your solution! :P

As Mr pvthuan said, [b]this comes from an ongoing contest in Vietnam.[/b]
Mod please lock this topic!

But I'm Vietnamese.. :D So I have right to post my solution here


trace $ A^k$ is 0 for all $ k$ posotive integers. show that $ A$ is nilpotent

Hint: the trace of $ A^k$ is the sum of eigenvalues to $ k$th power.


Find all triples $ (x,y,z)$ of real numbers which satisfy the simultaneous equations

\[ x \equal{} y^3 \plus{} y \minus{} 8\]

\[y \equal{} z^3 \plus{} z \minus{} 8\]

\[ z \equal{} x^3 \plus{} x \minus{} 8.\]

if $ (x \minus{} 2)(y \minus{} 2)(z \minus{} 2) \equal{} 0$ then easlily we get that $ (x,y,z) \equal{} (2,2,2)$ thereforre suppose that
$ (x \minus{} 2)(y \minus{} 2)(z \minus{} 2)$ is not zero.we have:
$ x \minus{} 2 \equal{} (y \minus{} 2)((y \plus{} 1)^2 \plus{} 4)$
$ y \minus{} 2 \equal{} (z \minus{} 2)((z \plus{} 1)^2 \plus{} 4)$
$ z \minus{} 2 \equal{} (x \minus{} 2)((x \plus{} 1)^2 \plus{} 4)$
 by multiply these equation we have:
$ (x \minus{} 2)(y \minus{} 2)(z \minus{} 2) \equal{} (y \minus{} 2)(z \minus{} 2)(x \minus{} 2)((y \plus{} 1)^2 \plus{} 4)((z \plus{} 1)^2 \plus{} 4)((x \plus{} 1)^2 \plus{} 4)$
$ \Rightarrow1 \equal{} ((y \plus{} 1)^2 \plus{} 4)((z \plus{} 1)^2 \plus{} 4)((x \plus{} 1)^2 \plus{} 4)\geq4\times4\times4 \equal{} 64$ which is impossible.therefore the only solution is $ (x,y,z) \equal{} (2,2,2)$

Let $ f(x)\equal{}x^3\plus{}x\minus{}8$.
Then we will have that
$ z\equal{}f(x)$
$ y\equal{}f(z)\equal{}f(f(x))$
$ x\equal{}f(y)\equal{}f(f(f(x)))$
but the function f is increasing (as $ f'(x)\equal{}3*x^2\plus{}1$)
so $ x\equal{}f(x)$ and $ x\equal{}y\equal{}z\equal{}2$ (if for example $ x<f(x) \equal{}> x<f(x)<f(f(x))<f(f(f(x)))$)

$(x-y)=(y-2)(y^2+2y+4)$
$y-z=(z-2)(z^2-2z+4)$
$z-x=(x-2)(x^2+2X+4)$
if y>2,then x-y>3(y-2)
so x>y,x>2,then z>x,z>2
hence y>z>x>y,contradiction!
if y<2,we get a contradictiion analagously
hence $x=y=z=2$


How many ways can 10 people be seated in a row, if 2 certain people cannot sit next to each other?

[hide="solution"]
total number of ways: 10!

possibilities for two people sitting next to each other:
pp--------
-pp-------
--pp------
---pp-----
----pp----
-----pp---
------pp--
-------pp-
--------pp

This gives $9 \cdot 2 \cdot 8! = 2\cdot9!$ ways for the two people to sit together.

$10! - 2\cdot9! = 10\cdot9! - 2\cdot9! = 8\cdot9! = \boxed{2,903,040\text{ ways}}$ they can be seated.

[/hide]

[quote="G-UNIT"]good job, that correct  :10:

Heres another one,

  Prove that ...[/quote]
Proofs are generally considered above the level of MATHCOUNTS, and questions requiring proofs almost never belong here.  [They rarely belong at the High School Getting Started Forum, either.]

Please take such questions to the High School Intermediate or Pre-Olympiad forums.

[color=darkblue][size=75]The proof question has been split and moved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244473#p244473]here[/url] by moderator rcv.[/size][/color]

How come i see some proof problems on problems for younger kids? (Math Olympiad, etc.) Of course, they aren't that hard....

Yeah, that's the correct answer.

Well, younger kids occasionally do proofs, as it helps to demonstrate their full understanding about the subject. That's why some people post them in the sections for younger kids. However, it's still good to keep them out of those sections.


For all $ a,b,c$ be sidelengths of a triangle, show that
\[ \frac {a(b \plus{} c)^2}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)^2}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)^2}{c^2 \plus{} ab} \le a \plus{} b \plus{} c \plus{} \frac {9abc}{ab \plus{} bc \plus{} ca}\]
When equality occurs?

[quote="2424"][quote="dduclam"]For all $ a,b,c$ be sidelengths of a triangle, show that
\[ \frac {a(b \plus{} c)^2}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)^2}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)^2}{c^2 \plus{} ab} \le a \plus{} b \plus{} c \plus{} \frac {9abc}{ab \plus{} bc \plus{} ca}\]
(*)
When equality occurs?[/quote]
$ (*) < \equal{} > \sum{(b \plus{} c)(1 \minus{} \frac {a(b \plus{} c)}{a^2 \plus{} bc})} \plus{} \frac {9abc}{ab \plus{} bc \plus{} ca} \minus{} a \minus{} b \minus{} c \geq 0$
    $ < \equal{} > \sum{\frac {(b \plus{} c)(a \minus{} b)(a \minus{} c)}{a^2 \plus{} bc}} \minus{} \frac {(a \minus{} b)(a \minus{} c)(b \plus{} c) \plus{} (b \minus{} c)(b \minus{} a)(c \plus{} a) \plus{} (c \minus{} a)(c \minus{} b)(a \plus{} b)}{ab \plus{} bc \plus{} ca} \ge 0$
    $ < \equal{} > \sum{\frac {a(b \plus{} c)(b \plus{} c \minus{} a)(a \minus{} b)(a \minus{} c)}{(a^2 \plus{} bc)(ab \plus{} bc \plus{} ca)}} \geq 0$
Let $ x \equal{} \frac {a(b \plus{} c)(b \plus{} c \minus{} a)}{a^2 \plus{} bc} ,y \equal{} ...,z \equal{} ...$
supose $ a \geq b \geq c$
$ \equal{} > a^2 \plus{} bc \minus{} b^2 \minus{} ca \equal{} (a \minus{} b)(a \plus{} b \minus{} c) \geq 0$
$ b^2 \plus{} ca \minus{} c^2 \minus{} ab \equal{} (b \minus{} c)(b \plus{} c \minus{} a) \geq 0$ (Because $ a,b,c$ be sidelengths of a triangle)
$ \equal{} > a^2 \plus{} bc \ge b^2 \plus{} ca \ge c^2 \plus{} ab$
$ a(b \plus{} c \minus{} a) \minus{} b(c \plus{} a \minus{} b) \equal{} (b \minus{} a)(a \plus{} b \minus{} c) \le 0$
$ \equal{} > a(b \plus{} c \minus{} a) \le b(c \plus{} a \minus{} b) \le c(a \plus{} b \minus{} c)$
and $ b \plus{} c \le c \plus{} a \le a \plus{} b$
$ \equal{} > 0 < x \le y \le z$
$ \equal{} > x(a \minus{} b)(a \minus{} c) \plus{} y(b \minus{} c)(b \minus{} a) \plus{} z(c \minus{} a)(c \minus{} b) \ge z(c \minus{} a)(c \minus{} b) \minus{} y(b \minus{} c)(a \minus{} c) \equal{} (z \minus{} y)(c \minus{} a)(c \minus{} b) \ge 0$
So,we are done! :)[/quote]

You are right, 2424. The idea in your proof is same of me. :)  Equality holds for $ (a,b,c)\sim (1,1,1), (1,1,0)$ or $ (1,1,2)$

Besides,  for all $ a,b,c\ge0$ we have $ \sum\frac {a^2(b \plus{} c)^2}{a^2 \plus{} bc}\ge2(ab \plus{} bc \plus{} ca)$

Can you please post your proof, dduclam?

Thank you very much

[quote="dduclam"][quote="dduclam"]

Besides,  for all $ a,b,c\ge0$ we have $ \sum\frac {a^2(b \plus{} c)^2}{a^2 \plus{} bc}\ge2(ab \plus{} bc \plus{} ca)$[/quote]

I am extremely sorry for the mistake of my mind  :oops:  This's problem which I want to post:

For all $ a,b,c$ be sidelengths of a triangle we have $ \sum\frac {a^2(b \plus{} c)^2}{a^2 \plus{} bc}\le2(ab \plus{} bc \plus{} ca)$

(This Inequality is quite evident by Vornicu-Schur).[/quote]

[quote="manlio"]Can you please post your proof, dduclam?

Thank you very much[/quote]

Of course, manlio. :)

The Inequality is equivalent to
\[ \sum_{cyc}\frac {a^2(b \plus{} c)^2}{a^2 \plus{} bc}\le\sum_{cyc} a(b \plus{} c)\]

\[ \Leftrightarrow \sum_{cyc} a(b \plus{} c)\left(1 \minus{} \frac {a(b \plus{} c)}{a^2 \plus{} bc}\right)\ge0\]

\[ \Leftrightarrow x(a \minus{} b)(a \minus{} c) \plus{} y(b \minus{} c)(b \minus{} a) \plus{} z(c \minus{} a)(c \minus{} b)\ge0\]
where $ x \equal{} \frac {a(b \plus{} c)}{a^2 \plus{} bc}, y \equal{} \frac {b(c \plus{} a)}{b^2 \plus{} ca}$ and $ z \equal{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}$.

Without loss of generality, we may assume that $ b \plus{} c\ge a\ge b\ge c$, then $ a \minus{} c\ge\frac b{c}(a \minus{} b)$ and $ bz \equal{} \frac {bc(a \plus{} b)}{c^2 \plus{} ab}\ge\frac {bc(c \plus{} a)}{b^2 \plus{} ca} \equal{} cy$.

Hence,
\[ \sum_{cyc} x(a \minus{} b)(a \minus{} c) \equal{} x(a \minus{} b)(a \minus{} c) \plus{} (b \minus{} c)(z(a \minus{} c) \minus{} y(a \minus{} b))\]

\[ \ge x(a \minus{} b)(a \minus{} c) \plus{} (b \minus{} c)\frac {(a \minus{} b)(bz \minus{} cy)}c\ge0\]
The proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c$, or $ a \equal{} b,c \equal{} 0$, or any cyclic permutation.