Juhan wants to order by weight five balls of pairwise different weight, using only a balance scale. First, he labels the balls with numbers 1 to 5 and creates a list of weighings, such that each element in the list is a pair of two balls. Then, for every pair in the list, he weighs the two balls against each other. Can Juhan sort the balls by weight, using a list with less than 10 pairs?

[hide="Solution"]Well, we take 3 balls and in 3 moves we can see ther sortings(e.g. A>B>C-the ball with the most weight is first). Then we take the last ball from those 3(C) and in 3 moves we compare that ball and remaining two(D and E). That's six moves(we can't have more than 3 moves more).WLOG D>E, and A>B>C
1) D>E>C
1. move: B v. D
If B>D then it' s over. If B<D then:
2. move: A and D
3. move B and E

2) D>C>E
In 3 moves we compare A,B and D

3) C>D>E
It's over(A>B>C>D>E)!

The weightings are moves for me(it's easier to write :rotfl: )[/hide]

Bugi - I believe the idea in this problem is to define a static list of "moves", that is a sequence of weighings that does not depend on the outcome of previous ones. This has to be like "compare A and B, then compare B and C, then A and D, ..." where in the end you're able to sort the five balls completely with the information collected in the process.

The relevant term is [url=http://en.wikipedia.org/wiki/Sorting_network]Sorting Network[/url], and the problem is asking for a network of 5 inputs with 9 comparators (see more info [url=http://www.cs.brandeis.edu/~hugues/sorting_networks.html]here[/url]). This is well known to be optimal (i.e. there is no way to sort 5 balls with just 8 weighings).


Find the remainder when 10988065 is divided by 2103.

No calculator, naturally.

It was on a problem set, so I'm assuming there is an elegeant solution, I'm not sure if there is one. Yes, I just took out some scrap and divided it out. What I tried was to factor 2103 into some more manageable numbers, but it's 3*701, and 701 is prime. At least I hope.

if you had a link to the prime pages you would know! :lol:
ANYWAY THIS IS 701: prime number, sum of three consecutive primes (229 + 233 + 239), Chen prime, Eisenstein prime with no imaginary part, the area code for North Dakota(wiki is ur friend wtf)


Euler
geometry
circumcircle
geometry proposed

Prove that $ d^2\le R(R\minus{}2r)$ where $ d$ is the distance between the circumcenter and centroid of a triangle, $ R$ is its circumradius and $ r$ is its inradius.

[quote="manjil"]This is a standard result and any good book will be able to give you a proof![/quote]
Of course, it is well know. Many problems that are refered by books still be discuss on Mathlinks because they have many synthetic proofs. In addtion, I am introducing some problems about Balkan MO.

http://www.mathlinks.ro/viewtopic.php?t=85162
There is no need to post problems from Balkan MO. All of them [url=http://www.mathlinks.ro/resources.php?c=1&cid=19]have been posted[/url]. :)

What we need is Balkan and Junior Balkan Shortlist for many years!


I can't post the picture, but i'll try my best.

Debbie has an ample supply of tiles with the patterns shown below. She intends to create a horizantal, rectangular wall decoration by cementing three of the squares to the wall in a row. How many different designs can she create in this way that have at least one line of symmetry?

X, +, /, \

Thanks,
LHM

P.S.- I can only find 8.

Are they $ 3 \times 3$ tiles?  That is, do they look like these:

[code]
x-x -x- --x x--
-x- xxx -x- -x-
x-x -x- x-- --x
[/code]

[url]http://www.yashar.biz/mathcounts/mathcounts_files/2002/2002%20State%20Target.pdf[/url]

there is a picture there

No they are 1*3 rectangles.

the answer is

[hide="answer"]
12
[/hide]

but how do you get that answer?
please show your solutions.

Thanks,
LHM

I'm not 100% sure about this but here goes.
[hide]Since all designs using only the first two tiles will have a horizontal line of symmetry, the possibilities using these are $ 2\times2\times2\equal{}8$.  Using the two tiles on the right, there are 4 more designs, as follows.  (1 is the first tile, 2 the second, and so on).
314
324
413
423 [/hide]


ratio
geometry
incenter
geometry unsolved

In a triangle ABC, AB=AC. Point M is on side AB, & AM=MC; Point N is on side AC & BC=CN.
The ratio of the size of angle NBA to BAC is 3:2.
Find the size of angle NMC.

This problem seems simple but I just got stuck somewhere.
I can figure out that angle NBA=30 degrees; & angle BAC=20 degrees.

Let $\angle{A}=2 \theta$, then $\angle{ABC}=\frac{180^{o}-2\theta}{2}=180^{o}-10 \theta$, yielding that $\theta = 10^{o}$.

For the rest see:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=angle+solutions&t=39387
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=incentre+reflect&t=64298
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=angle+reflect&t=71052

[quote="shobber"]Let $\angle{A}=2 \theta$, then $\angle{ABC}=\frac{180^{o}-2\theta}{2}=180^{o}-10 \theta$, yielding that $\theta = 10^{o}$.[/quote]

thanks


The medians to the legs of an isosceles triangle are perpendicular to each other.  If the base of the triangle is 4, find its area.

Yeah, that is the right answer. Good job!


@Mods: Please put this on autolock when it's November 1st (at 12 am :lol: )

The point of this will be to write and solve as many problems involving halloween until its over, I'll start:

Bob and Joe were going trick-or-treating on a snowy October night in their cute pumpking costumes :o. Then, they went by ol' bob joe's house and would only give them candy if they found out the sum of the  area of thevtwo yellow circles on their costumes in sq. inches. Knowing that their costumes were circular (we're just measuring the front, not the sphere) and had an area of 2 sq. ft. and had a diameter 2x bigger than the yellow circles. What would of been the correct answer to tell ol' bob joe?

I know the word construction in this problem is horrible

-jorian

If I remember correctly, there will be no more marathons created for a while. Thus I go on and lock this one. It does guarantee that it'll be locked Nov. 1, 12:01 am, though :P.

CLARIFICATION: Someone asked why there weren't to be any new marathons. Here's why: Some time ago there was a spate of a great many marathons which were annoying. Thus there was an announcement of no new marathons for a while, and after that no new ones without good reason. Well, I haven't heard that we're allowed to make more marathons, and this proposed marathon doesn't have a good enough reason to continue to exist. Thus this one is locked.


[quote="rickyricardo"]Problem:
Can any rational number x be represented as

$x = q_1r_1 + q_2r_2 +....+q_nr_n$

For rational $q_i$ and integer $r_i$?[/quote]

What's the problem with q_1=x and r_1=1?

Noooo, ANY x needs to be representable by a finite set ${q_1,q_2,...,q_n}$ and any r_n's.  The statement is false but I don't know why.  I think it is because x can be chosen to be q_i/10, where q_i is the smallest element of the set of q's, but I'm not sure

Do you mean to say that q_1 to q_n is a given set, and you have to construct x in such a way?  If x=m/n, for sufficiently large n it will be impossible.

The problem worded magnificently is:

Prove that the following statement is false:  There is a finite set of rational numbers S = {q1, q2,....,qn} such that ever x in Q can be written as

x = r1q1 + ..... r_n*q_n

for integer r's.

[quote="Kent Merryfield"]Problems about rational numbers are often at heart number theory problems. Here's a theorem of number theory:

If $\{a_i\}_{i=1}^k$ is a set of nonzero integers, then one can find integers $n_i$ such that $m=n_1a_1+n_2a_n+\cdots n_ka_k$ iff $m$ is divisible by the greatest common divisor of $a_1,a_2,\dots,a_k$.

Can you see how this theorem applies to this problem?[/quote]

Here's some more about it.

The case: ax + by = d iff gcd(a,b)|d is handled by the Euclidean Algorithm.

The case ax + by +cz = d iff gcd(a,b,c)|d is handled by noting that gcd(a,b,c) = gcd((a,b),c)

We can then prove the rest by induction.

Can't we just say that x is less than the smallest element out of $q_1,q_2,q_3,...,q_n$ because then we won't be able to get it with all the r's we want.

Note that the r's are integers - they may be [i]negative[/i] integers. So we can get values of x that are smaller than the smallest q - but we cannot get all rationals.

For example, if the set of q's is {1/2, 1/3}, we can make 1/6 because

1/6 = 1/2 - 1/3

but we cannot make 1/7.

let  q_i =a_i/b_i    in reduced form so that r_i and s_i are coprime to each other,

Now let N = b_1*b_2*b_3* .....b_n +1

and x= 1/N  if  it is possible to write  in the form above than we have

b_1*b_2*b_3......*b_n =  (Integer)*N

bun N is coprime to all  b_1,b_2.. etc .. so there is a cotradiction.
QED

[quote="rickyricardo"]Sorry, I don't understand your answer completely.

how did you get $(b_1)(b_2)(b_3)...(b_n) = (integer)(N)$?

I understand that you made x equal to the product of the b's plus one but not the proof for how it cannot be expressed as the sum of products of rationals with integers.  Can you clarify?  Thanks.[/quote]

look at
x=1/N = (r1*a1)/b1+(r2*a2)/b2 .... etc
            = (1/(b1*b2*......))(r1*a1*b2*b3*....+r2*a2*b1*b3..+ ----- etc but  sum of all the terms would be an integer)
            =   (some_integer)/(b1*b2*b3.....)

or b1*b2*b3*..... =N*(some_integer)

HTH.


geometry
circumcircle
trigonometry
geometry unsolved

[quote="Elementaryyy"][quote="math10"]Let $ D$ be the midpoint of the base $ AB$ of an isosceles acute triangle $ ABC$. Let $ E$ be a point on $ AB$ and let $ O$ be the circumcenter of $ \triangle ACE$. Prove that the perpendicular to $ DO$ through $ D$, the perpendicular to $ BC$ through $ E$, and the line through $ B$ parallel to $ AC$ are concurrent[/quote]
It ís not true.
See:
[img]http://i614.photobucket.com/albums/tt221/hung_dhsp/untitled.jpg?t=1258369959[/img]
[color=red]Mathematics infinitely[/color]
[color=yellow]Return is the bank[/color][/quote]
No, it is right  :wink: .

PS:Em xem lai hinh di,tam giac nay can tai C

[quote="math10"]Let $ D$ be the midpoint of the base $ AB$ of an isosceles acute triangle $ ABC$. Let $ E$ be a point on $ AB$ and let $ O$ be the circumcenter of $ \triangle ACE$. Prove that the perpendicular to $ DO$ through $ D$, the perpendicular to $ BC$ through $ E$, and the line through $ B$ parallel to $ AC$ are concurrent[/quote]
$ AB=AC$,$ D$ is midpoint $ BC.K$ is midpoint $ AB$.F is circumcenter of $ \triangle ABE$.$ FH\perp BC,KI%Error. "PERP" is a bad command.
BC$.
Apply $ Carnot$ theorem. We have to prove that: $ DF^2-DD^2+ED^2-EK^2+CK^2-CF^2=0$
$ DF^2-CF^2+CK^2-EK^2+ED^2=0$
$ DH^2-HC^2+CI^2-EI^2+ED^2=0$
$ DH^2-(CD+DH)^2+(CE+EI)^2-EI^2+ED^2=0$
$ CE(2DI-CD)=0$
True.
P/s:Em thay 1 số kí hiệu. Trong bài này 1 số chỗ dùng độ dài đại số nhưng em ko biết gõ.

Draw a line through $ D$ and perpendicular to $ CB$, it cuts $ BG$ at $ K (BG//AC)$
Let $ J$ be the midpoint of $ AC \Rightarrow DJ//CB \Rightarrow \angle JDK\equal{}90^o$.
By angle chasing, $ \angle OJD\equal{}\angle DKG$.
We will show that $ \Delta OJD\sim \Delta GKD. (1)$
$ \frac{JO}{JD}\equal{}\frac{JO}{JA}\equal{}\frac{ED}{CD}$.
So we need to prove $ \frac{ED}{CD}\equal{}\frac{GK}{DK} \Leftrightarrow \frac{DB}{BK}\equal{}\frac{ED}{GK}\equal{}\frac{CD}{DK}$
$ \Leftrightarrow \tan C/2\equal{}\frac{DB}{CD}\equal{}\frac{BK}{DK}\equal{}\frac{\sin \angle KDB}{\sin \angle DBK}\equal{}\frac{\sin C/2}{\cos C/2}$ (right)
$ \Rightarrow (1)$ is true $ \Rightarrow \angle JDO\equal{}\angle KDG \Rightarrow \angle ODG\equal{}\angle JDK\equal{}90^o$. We are done.


How many pairs of perpendicular line segments can be drawn using the points of this $ 3$ by $ 3$ grid as endpoints?  (Segments must intersect at one point; either at an endpoint or within the segment.)

.         .         .
.         .         .
.         .         .

Answer: [hide]$ 78$[/hide]

Generally, if someone post a problem and an answer, then they don't know the solution and they want to know why the answer is what it is.

Three cases.  Two segments of the same length and one of each.  I'm assuming it isn't a square.
There are overcounts in each of those cases.  Try it. XD

Hmm..I must have done something wrong..

I did case work.
[u]Corners[/u]- I found 12 possible perpendicular lines and 4 corners, so $ 12\times4\equal{}48$

[u]Midpoints[/u]- I found 12 possible perpendicular lines on these as well, and 4 of them, so $ 12\times4\equal{}48$

[u]One in the middle[/u]- I found 16 possible perpendicular lines, so $ 16$

I added those and divided by 2, but it gave me 56..

[asy]size(150); pair A,B,C,E,F,G,I,J,K; A=(-3,3); B=(-1,3); C=(1,3); draw(A--C); E=(-3,1); F=(-1,1); G=(1,1); draw(E--G); I=(-3,-1); J=(-1,-1); K=(1,-1); draw(I--K); draw(A--I); draw(B--J); draw(C--K);[/asy]

Oh, BTW, is there a way you can do this by setting up an equation like you do in finding the all possible rectangles in $ m$ by $ n$ grid?

[quote="alanchou"]Did you include diagonals?[/quote]

Yeah, I am pretty sure I did...Don't know...I will think again..


Fill ingrid Such a way that every row and column and every 3*3 boxes contains number 1-9.
Payman Morteza


The Original Master Mind is a game for two players, and the rules are thus:

Master Mind includes a board and colored pegs. The codemaker secretly makes a code out of those six colors(red, blue, yellow, white, black, green), and the code breaker uses logic to guess the code.

The codebreaker gives what (s)he thins is the code to the codemaker, and the codemaker gives a combination of white and black keys, he gives a black key for every peg that is in the correct position, and a white key for every peg that is the same color as one of the pegs, but in the incorrect position. It is possible to receive no keys.

The codebreaker can only use ten guesses. :evilgrin: If the codebreaker uses all of his guesses, the codemaker wins and shows the codebreaker the code. If the codebreaker guesses the code, the codemaker gives four black keys and gives the code.

Maybe we can make a tournament on this. Thoughts?

So [i]who[/i] wants to be in the tournament?

Here is a Wikipedia link: http://en.wikipedia.org/wiki/Mastermind_%28board_game%29

We will be using 6 colors, repetition allowed, maximum of 10 turns, code is four pegs long.

Feel free to post in this topic.

Isn't this kind of useless? All mastermind games can be solved within 8 tries.

[quote="zolojetto"]Why would you want to delete it, id be interested in playing, and just becuase a code can be solved in 8 turns or less doesnt mean that it will.[/quote]

Theres a strategy to ALWAYS solve it in 8 tries. It's why most games end at 10 tries (to give noobs 2 chances)

[quote="azjps"]Isn't it actually always solvable in like 6 steps or so? I usually take 5.[/quote]

Knuth (1976-77) showed that the codebreaker can always succeed in five or fewer moves (i.e., knows the code after four guesses). His technique uses a greedy strategy that minimizes the number of remaining possibilities at each step, and requires 4.478 guesses on average, assuming equally likely code choice. Irving (1978-79) subsequently found a strategy with slightly smaller average length. Koyama and Lai (1993) described a strategy that minimizes the average number of guesses, requiring on average 4.340 guesses, although may require up to six in the worst case. A slight modification also described by Koyama and Lai (1993) increases the average to 4.341, but reduces the maximum number of guesses required to five.

I am not sure if your numbers are right, but I want to try this. Sounds like fun!

I want to try it, but if masterminds solve it instantly then I'll have no chance to solve and actually progress in the game.

Why don't you try changing the rules a bit?  Maybe make the code longer or use more colors.


Let $ M$ be a point on $ BC$ and $ N$ be a point on $ AB$ such that $ AM$ and $ CN$ are angle bisectors of the triangle $ ABC$. Given that $ \frac {\angle BNM}{\angle MNC} \equal{} \frac {\angle BMN}{\angle NMA}$, prove that the triangle $ ABC$ is isosceles.

Let $ \frac {\angle BNM}{\angle MNC} \equal{} k$ and $ \angle NMA \plus{} \angle MNC \equal{} h$, then we have $ 180 \minus{} k \cdot h \equal{} 2 (180 \minus{} h \minus{} 90)$ then $ k\equal{}2$, so $ \angle BNM \equal{} 2 \angle MNC$.
Now let I the incenter of ABC and J of BMN. Then by angle J is the symmetric of I wrt MN and B,J,I are collinear, then $ NB \equal{} MB \ \Longleftrightarrow \ \frac{ca}{a\plus{}b} \equal{} \frac{ca}{b\plus{}c} \ \Longleftrightarrow \ c\equal{}a$.

Let $ \angle BNM=\alpha$, $ \angle MNC=k \alpha$, $ \angle BMN=\beta$, $ \angle NMA=k \beta$. Let also $ I$ be the incircle of $ \triangle ABC$. Then

$ \angle AIC = \angle NIM = 180^{\circ} - \angle MNC - \angle NMA = 180^{\circ} - k \alpha - k \beta$

and also

$ \angle MAC = \angle BAM = 180^{\circ} - \angle ABC - \angle BMA = 180^{\circ} - (180^{\circ} - \alpha - \beta) - (\beta + k \beta) = \alpha - k \beta$
and similarly $ \angle ACN = \beta - k \alpha$.

We know that $ \angle AIC + \angle ACN + \angle CAM = 180^{\circ}$, thus
$ 180^{\circ} - k \alpha - k \beta + \alpha - k \beta + \beta - k \alpha = 180^{\circ}$ which gives us
$ \alpha + \beta = 2k(\alpha + \beta)$
$ 2k=1$
$ k=\frac{1}{2}$

Now draw $ \angle BNM$ and $ \angle BMN$ bisectors $ NX$ and $ NY$ respectively. Let $ O$ be the incircle of the $ \triangle NBM$ (also the $ NX$ and $ NY$ common point). Now we see that $ \angle ONM = \frac{1}{2} \angle BNM = k \alpha = \frac{1}{2} \alpha = \angle MNI$. Similarly we obtain that $ \angle OMN = \angle IMN$. Thus $ INOM$ is a deltoid and its diagonals are perpendicular, i.e. $ BI\bot NM$. But it means that $ \triangle NBM$ is isosceles ($ NB=MB$) because $ BI$ is both altitude and bisector at the same time. Thus $ \alpha = \beta$. Now we show that
$ \angle BAC = 2\alpha - 2k \beta = 2\alpha - \frac{1}{2} \cdot 2\alpha = \alpha$ and similarly $ \angle ACB = \alpha$, which means that $ \angle BAC = \angle BCA$, q.e.d.


Hello. I am going into Grade 10 and am working through some books. My goal is not super ambitious, all I want to do is make the USAMO and CMO (Canadian Math Olympiad) before my Grade 12 year and hopefully get a respectable score on each. I own a few books which I outlined below.

AOPS V1
AOPS V2
ACOPS
Mathematical Circles
Challenging Problems in Geometry

So far, I have been going way too slowly. I have almost finished AOPS V1, but I have spent a lot of time skimming some parts of the other books that interest me. When I work, sometimes I find that I am spending too much time on one chapter, trying to get down the formulas once and for all. How do you guys study, and do you guys go back to old chapters and sections later or get it down once and for all? Any help is greatly appreciated, and any suggestions for practice or further reading would be helpful. (as a reference, I can get hopefully around 5 questions right on the AIME, maybe more if I am lucky). Also, I am willing to work hard if that means anything.

Thanks.  :)

Personally I would recommend that you take an Aops class. Those are really helpful. If next year you simply want to qualify for USAMO, I recommend an AIME class, but you could take WOOT if you really want a challenge.

I think you have the right books, but what's important isn't speed, but understanding the material.

I suggest: try to finish a chapter(regardless of how long) of whatever book you're currently studying and work maybe half/all of the exercises and maybe 1/2 or 3/4 of the review and challenge problems. That should hopefully get you farther than your current speed. Certain chapters encompass a lot of chapters, so it's O.K. if you can't finish some chapters in a day.

Hope that helps! :lol:

I recommend that you don't speed up too much. Just go at the pace at which you feel best suits you. And as battered said, try to do as many of the review and challenge problems as possible (I recommend giving a serious effort to ALL the review and challenge problems. You are not expected to get the difficult ones straightaway, you may need to come back and if you truly understood your mistake and corrected yourself, should be able to do it then.) But since you kind of do have to hurry a bit, just make sure you take loads and loads of practice materials. That should get you to your destination. And then eventually read some more of the advanced books such as Mathematical Olympiad Challenges, Problem Solving Stragies, and such books.

Good luck on your mathematical conquests. :)

[quote="limac"] And then eventually read some more of the advanced books such as Mathematical Olympiad Challenges, Problem Solving Stragies, and such books.

[/quote]

ACoPS and AoPS volume 2 [b]are[/b] advanced books such as the ones you mentioned... They are both pre-olympiad level... :wink:

Thanks guys! Oh and I am signed up for 3 AOPS classes already (one of which is AIME). I don't think I am good enough for WOOT though (I might try to be by the end of the summer though).


Let $ a,b,c,d$ are real numbers such that $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} 13.$

Find the maximum value of $ S \equal{} (13 \minus{} 2a \minus{} 3b)(13 \minus{} 2c \minus{} 3d)(13 \minus{} ac \minus{} bd).$

[b]Here is my solution[/b]
Consider $ (C): x^2 \plus{} y^2 \equal{} 13$. It is so obvious that $ A(a,b)$, $ B(c,d)$, $ C(2,3) \in (C)$.
Therefore $ 8P \equal{} ((a \minus{} 2)^2 \plus{} (b \minus{} 3)^2)((c \minus{} 2)^2 \plus{} (d \minus{} 3)^2)((a \minus{} c)^2 \plus{} (b \minus{} d)^2) \equal{} AB.BC.CA$
$ \Longrightarrow 8P \equal{} 8(\sqrt {13})^3.sinA.sinB.sinC\le 8(\sqrt {13})^3.(\frac {\frac {3\sqrt {3}}{2}}{3})^3$.
$ \Longrightarrow P\le \frac {39\sqrt {39}}{8} \Longrightarrow maxP \equal{} \frac {39\sqrt {39}}{8}$.
Then we are done!! :lol:


Is there any way we can get him to stop clogging up HSB with textbook/homework problems? Because it's been going on for a good 8 months now, and it's quite irritating.

But remember, we are a loving caring community so let's still be nice, whatever we do.

Yes, but still, it has been emphasized that this is a place of problem solving, and that textbook problems and homework problems are generally not welcome.

Just ask the moderator to remove the topics he puts on.

Okay, 1337, your signature goes way too far...  we are not trying to kill fellow AoPSers.

If I, a random person, looked on this site and the first thing I saw was "KILL SHARKMAN/INTERVAL" then I don't know what I would think.

[quote="bpms"]I distinctly remember a sticky being made specially for him to post in, that way the boards aren't clogged.[/quote]

I just made another one but for some reason, I don't think its going to work.


complex numbers
calculus
calculus computations

If $ a_1, \ldots a_n$ and $ b_1, \ldots b_n$ are complex numbers, then $ \left|\sum_{j = 1}^{n} a_{j}\bar{b}_{j} \right|^2 \leq \sum_{j = 1}^{n} |a_j|^2 \sum_{j = 1}^{n} |b_j|^{2}$.

I do not understand why ${ \sum |Ba_{j} - Cb_{j}|^{2} = \sum (Ba_{j} - Cb_{j})(B \bar{a}_{j} - \bar{Cb_{j}}})$

The bar on the $ Cb_j$ should be over the whole thing.

This is from p 15 of Rudin.

$ |z|^2 \equal{} z \cdot \bar{z}$ for any complex number $ z$.  Just let $ z \equal{} Ba_j \minus{} Cb_j$.  By the way, Cauchy-Schwarz holds for any inner product; are you familiar with the general proof?


function
integration
linear algebra
matrix
real analysis
real analysis unsolved

Prove or Disprove :

if $ f_n$ is any sequence of measurable function on the measurable set $ E$, then :

if $ \int_{E} |f_n - f| dm \rightarrow \ 0 \Longrightarrow f_n \longrightarrow f \ ( a.e)$, where $ f$ is any measurable function .

I'm thinking about this :

let $ n = k + 2^t$ , define $ f_n$ as follows :

$ f_n(x) = \{\begin{matrix} 1 & ,x\in [\frac {k}{2^t}, \frac {k + 1}{2^t}] \\
0 & , O.W \end{matrix}$.

Let $ f = 0$ , then :

$ \int_{E} |f_n - f| dm \rightarrow \ 0$  but $ f_n \not\longrightarrow 0 (a.e)$ .

Does what I said true ? also , do u have another nice example ?

 :)

That's pretty much the standard example. I usually refer to it as the "rotating tooth" example. It's also the standard example that shows that convergence in measure need not imply convergence [i]a.e.[/i]

Note that there is a subsequence that converges [i]a.e.[/i] - and that's a theorem. A sequence with converges in $ L^1$ must have a subsequence that converges [i]a.e.[/i]

You could vary many of the details of the example while keeping the same basic idea, but there's not a lot of motivation for doing so.


algebra
polynomial
real analysis
real analysis solved

Find all complex polynomials such that if $|z|=1$ then $|P(z)|=1$

It seems to me that it has already been discussed http://www.mathlinks.ro/Forum/viewtopic.php?highlight=S+stable&t=25258


Hmm according to weather forecasts it's going to snow in Massachusetts on Monday/Tuesday next week, so there's a chance of a snow day on the day of AMC A. Now what happens to AMC then? Do I get to take AMC B for free  :) (I wish)?

[quote="probability1.01"][quote="DPatrick"]If your school is closed on the "A" day, you can still take the 10B or 12B.  Not for free, though.[/quote]

If your school is snowed out both days, or it only takes the B and it's closed then, does that mean you just don't get to take the AMC?[/quote]

Indeed it does. My school would never do the B (it would take away football money, lord forbid), so if there's a snow day, my MOP hopes are screwed.

[quote="probability1.01"][quote="DPatrick"]If your school is closed on the "A" day, you can still take the 10B or 12B.  Not for free, though.[/quote]

If your school is snowed out both days, or it only takes the B and it's closed then, does that mean you just don't get to take the AMC?[/quote]

Go to Florida!(looking at plane tickets now :D )

Oh wait hopefully no hurricanes.

Fine I'll go to Texas and take it at  A & M Consolidated High School (and stalk some AoPSers there).

[quote]Indeed it does. My school would never do the B (it would take away football money, lord forbid), so if there's a snow day, my MOP hopes are screwed.[/quote]

Offering the AMC is not very expensive. My son's math circle is offering the A test for $\$$5 a person. If your parents are supportive, maybe they can approach the school and offer to pay for the A test. Then you need a supportive teacher who will proctor it.

Of course, this is obviously for next year.

Haha I tried to get my school to do the B.  I asked my math teacher who said he'd ask the division head but didn't.  So then I emailed her and she didn't reply, and so I talked to her a week ago and she said it was too late to worry about setting it up...

JB

Yeah, Western MA is screwed too.

Worc Acad might offer it anyway, and just let who from school wants to come come.


Prove $x^3 - 2xz^2 + 2z^3 \ge 0$ for $x \ge -2z$.   ($x,z$ are reals, $z$ is a constant)


Evaluate:
\[ \sum_{n=1}^{100} fib (n) \]
Masoud Zargar

$\sum_{k=1}^{n}F_n = F_{n+2}-1$ (show this by induction)

then we want the $F_{102}$ and just use the explicit formula for the fibbonacci numbers (i forgot what it is at the moment) but it is in one of your formula topics

Thanks.  I haven't seen the formula in one of my formula topics, however, are you talking about the following formula?
\[ F_n=\frac{\phi^n-\left(\frac{-1}{\phi}\right)^n}{\sqrt{5}}  \]
EDIT: Yes, that's it.  Thanks, I get it now.

One more thing, do you have any idea how they derived the Fibonacci Number formula?

Masoud Zargar

When you solve a linear recurrence (e.g, one of the form)

$F_{n+i}=\sum^{i-1}_{j=0} \alpha_j F_{n+j}$

it is evident that if $t$ is a root of $x^{i}-\sum^{i-1}_{j=0} \alpha_j x^j$ (the characteristic equation)

then $F_{n}=t^n$ solves the original recurrence.

Now use linearity.  There will be $i$ roots (this is the content of the fundamental theorem of algebra) and also $i$ initial conditions.  Call these roots $t_1,t_2,\ldots$.  Thus we can pick $\beta_1,\beta_2,\ldots$ such that

$F_{n}=\sum_i \beta_it_i^n$ (as long as $t_i\not=t_j$ for $i\not=j$)

and this will solve our recurrence AND the initial conditions.

In the case of the fibonacci sequence, the characteristic equation is $x^2=x+1$.


There was a bug previously (the url('...') part wasn't added automatically.) Refresh your browser on the page to change the CSS code cache. Now it should work.

PS For those of you that had added the url part automatically you don't need to change anything, as I will automatically detect when you use the url( tag in the background image.

It does. Refresh the page. Also note that if you want to see the full background you would have to use extra CSS to make the midpart transparent ...


algebra
polynomial
Galois Theory
superior algebra
superior algebra solved

Anyone know how to show that t^4 + 1 is not irreducible over a field F with characteristic p > 0?

If $p\equiv 1(mod\ 4)$, then $-1$ is a square in $\mathbb F_p=\mathbb Z/p\mathbb Z$ which is a subfield of $F$. Then there exists $a\in\mathbb F_p$ such that $a^2+1=0$. From this it follows $X^4+1=(X^2-a)(X^2+a)$.

If $p\equiv 3(mod\ 4)$, then $-1$ is not a square in $\mathbb F_p$.
There exist $a,b\in \mathbb F_p$ such that $X^4+1=(X^2+ax+b)(X^2-aX+b^{-1})$ iff $b=\pm1$ and $a^2=2b$.
Since $-1$ is not a square in $\mathbb F_p$, we have that $2$ is not a square iff $-2$ is a square. The conclusion is that there are $a,b$ such that $b=\pm1$ and $a^2=2b$.

Basically, it factors when $-1$ is a square, when $2$ is a square, and when $-2$ is a square. At least one of those is always true in characteristic > 0, since the product of two nonsquares is a square in $F_p$.
The factorizations:
$t^4+1=(t^2+\sqrt{-1})(t^2-\sqrt{-1})$
$t^4+1=(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)$
$t^4+1=(t^2+\sqrt{-2}t-1)(t^2-\sqrt{-2}t-1)$

You could also use that the Galois group of $X^4 + 1 \in \mathbb{Q}[X]$ is the Klein four group, so the result follows from the same arguments as in http://www.mathlinks.ro/Forum/viewtopic.php?t=76707

Or you can use smth like: A polynomial $P$ of degree $n$ over $\mathbb F_p$ is irreducible iff $\min\{m\in\mathbb N^*|\ P\ |\ (X^{p^m}-X)\}=n$.

Prove that $X^4+1|X^8-1|X^{p^2}-X$ if $p$ is an odd prime number and apply the above. For $p=2$ we have $X^4+1=(X+1)^4$.


I have been teaching math for about 4 years. Students in my algebra class have informed me that the textbooks do not have enough examples and lack detail. I am currently working on an example book just for ALGEBRA. The idea is to have a book where different type of problems are solved. I do plan on marketing this book for teachers, students as well home-schoolers. This book will provide students with detailed examples at different degrees of difficulty.

I realize there are some books out there such as Schaum's outlines, but they leave out some important examples and are sometimes not detailed enough.

It should not be mistaken for a textbook, but rather an example book to be used along with the required textbook.

I would like to hear from some of you and what you think. Any suggestions are appreciated. Thank you.

Samples:

http://www.scribd.com/doc/3813757/orderofoperations

http://issuu.com/lazyeightbooks/docs/lcm?mode=embed&documentId=080706032526-9a1e87f1469d482daf911f6be1d3d823&layout=grey

You've already posted exactly this content ([i]sans[/i] links) in your other post.  If you like, I suggest you post those links as a comment in that thread.


geometry
3D geometry
tetrahedron
counting
distinguishability
rotation
group theory

Each face of a regular tetrahedron is painted either red, white, or blue.  Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical.  How many distinguishable colorings are possible?

One solution is in the archive...can anyone provide another one, or a different way of looking at/explaining it?

[hide="Massive epic extreme overkill"]
The identity contributes one transformation in which $ 3^4$ subgroups of the group of rotations is preserved, obviously.

We can now consider 8 rotations where we rotate around the axis perpendicular to each face by $ 60^\circ$. The face whose axis we're using obviously doesn't change, and the only members of the group that will not change will have the other three sides the same, so we have $ 3\cdot 3$ group members in this case. (And there are three of these cases)

By [url=http://en.wikipedia.org/wiki/Burnside's_lemma]Burnside's lemma[/url], the answer is $ \frac{3^4\plus{}2\cdot 3^2\cdot 3}{8\plus{}1}\equal{}\boxed{15}$.
[/hide]


Princeton college articles vector calculus integration probability

Hi all, I would like to know whether Princeton's Finance (Bendheim Center for Finance) program is[b] good or 'great'.[/b] I have already perused the website; I know the 'prerequisites'. I intend to pursue an M.S. (after some years) in Finance from Princeton. Right now, I'm 20 years old.

Do anyone from this forum have any acquaintances/friends who're pursuing an M.S. in Finance from Princeton university? How good is the program? Can I assume given Princeton's reputation its finance program is also 'great'?

Do give your opinion or suggestion/advice.

Thank you!

:huuh: This is one of the vaguest questions I've seen on this forum in a while. If you take the time to think about and articulate what you're looking for out of your Finance program, you might actually get an answer to your thread.

[quote="TZF"]:huuh:This is one of the vaguest questions I've seen on this forum in a while.[/quote]
:D I think I have articulated my queries pretty well. I find your 'above observation' to be overly exaggerated.

[quote="TZF"]If you take the time to think about and articulate what you're looking for out of your Finance program, you might actually get an answer to your thread.[/quote]
Like I have already said, I intend to pursue an M.S. in Finance from Princeton (The Bendheim Center for Finance). I want to know certain things like: The quality of the program (i.e. Ranking), placements, etc. (I already know the prerequisites, curriculum)

[url=http://grad-schools.usnews.rankingsandreviews.com/best-graduate-schools/top-business-schools/finance]US News and World Report rankings for best finance programs[/url]. Where's Bendheim?? As you must've already realized from browsing the website, Bendheim offers an MFin degree (probably what you meant when you said MS in Finance), which is not that mainstream. It focuses on quantitative aspects of finance. As a result, it clearly won't make it on to your mainstream rankings of finance MBA programs.

This is exactly what I meant when I said that your question was vague -- we have absolutely no idea what 'good' means to you. If good means going with MBA, then this clearly isn't the right choice for you. On the other hand, if you are specifically interested in the quantitative aspect, then Bendheim may very well be a good fit.

Like I said, the decision is about thinking about what you want out of your finance program. One of the most basic things you need to ask yourself is what degree you want and what kind of training you're looking for. You can check out [url=http://www.ederman.com/new/docs/risk-job_market.pdf]this article here[/url], but I'm sure there's lots of other things out there.

As for placement afterward, I'd imagine it's pretty good, since the school is highly selective and fairly specialized (and also, because it's Princeton).

Thank you [b]TZF[/b]!  :)

[quote="TZF"] It focuses on quantitative aspects of finance. As a result, it clearly won't make it on to your mainstream rankings of finance MBA programs. [/quote]
Yes, Clearly Princeton's Master in Finance program is highly quantitative-- It  necessitates people to take some advanced 'math' courses. At the moment, I just manage to conform to their minimum prerequisites for Math (Vector calculus, ODE, PDE, Linear Algebra etc). I'm looking forward to taking more Math courses, and also honing my math abilities.

[quote="TZF"] This is exactly what I meant when I said that your question was vague -- we have absolutely no idea what 'good' means to you. [/quote]
According to me, [b] good [/b] in the present context was -- A finance (Master of Finance) program that culminates in a good placement.

[quote="TZF"] Like I said, the decision is about thinking about what you want out of your finance program. One of the most basic things you need to ask yourself is what degree you want and what kind of training you're looking for. [/quote]

Actually, I'm interested in both the quantitative and 'general' aspects of Finance (M.B.A. stuff). So, I'm contemplating of pursuing both an M.B.A. and M.S. in Finance. I can't apply for Princeton's program right now because it will inevitably be rejected due to lack of 'stellar' recommendations (My recos are pretty good, though), etc. But, I have excellent extra-curricular activities.

After perusing the site it appeared to me that 'extra-curricular activities' aren't accorded much importance.

So, I have decided to apply to Princeton's program after some years. (Most probably after pursuing an M.B.A. degree)

Hello, I just thought I'd respond with my thoughts. I am a rising sophomore at Wharton, taking a bunch of advanced math (but mostly in the form of statistics) courses. I am not very familiar with Princeton's MS, I could probably tell you no more than their website can. At Wharton, a big focus is placed on (a) Leadership and participation (b) The Wharton Core, which includes Finance, Management, Marketing, Accounting, etc...

I'd say it's a different environment from Princeton. Princeton is more focused on the quantitative aspects, and less on the general/well rounded business education aspect. If you wanted to go into Financial engineering, you'd go to Princeton. If you wanted to do mainstream Finance (ie. I-banking, investment management, private wealth, etc..), Wharton is a better place for you. I've tried to cheat this whole thing by enrolling in PhD level quantitative courses here at Wharton, but that's not the norm.

Both are fantastic universities and if you do well in your classes, and can present yourself well on paper, you will get fantastic placements (seeing as how the economy seems to be bottoming out).

Going to MS after MBA is..well, if you are very interested in MS, then perhaps a good choice. It's just that many of our MBA grads accept 300K salaries after graduation, and that's a very large opportunity cost.

Thank you [b] ddziabenko [/b]!

[quote="ddziabenko"]Hello, I just thought I'd respond with my thoughts. I am a rising sophomore at Wharton, taking a bunch of advanced math (but mostly in the form of statistics) courses. [/quote]
Oh! That's nice. If you don't mind, may I know some details pertaining to your academics like: your under-graduate/graduate GPA, GRE/GMAT scores, quality of letters of recommendation, extra-curricular activities(if any). If you don't like to tell me over here, would you mind telling me via a PM? What probably facilitated your admission into Wharton?

[quote="ddziabenko"] I'd say it's a different environment from Princeton.  Princeton is more focused on the quantitative aspects, and less on the general/well rounded business education aspect. If you wanted to go into Financial engineering, you'd go to Princeton.[/quote]
Right now, I'm contemplating of significantly honing my math ability. At present, I just conform to Princeton's Master of Finance program's minimum Math prerequisites-- I had just taken 3 math courses at the under-graduate level until now. These three courses encompassed vector calculus, vector integral theorems, ODE, PDE, Curve Tracing, Probability and Statistics, Double and Triple Integrals, Laplace Transforms, Linear Algebra, Interpolation: numerical,cubic spline, etc. I have good grades in these courses.

Princeton's Finance program almost explicitly underscores the necessity of significant math ability in prospective applicants. Right now, given the number of math courses that I had taken -- I wouldn't consider myself to possess the math ability that Princeton generally seeks out of its prospective students.

I'm going to turn 21 in some time now. I had attempted the General GRE and managed to secure a pretty decent score.(1400)

Though it might sound silly, Would it be better if I attempt the Math subject GRE (after some years)? The reason I want to attempt the Math subject GRE is that provided I manage to score in the 60-90 percentile range on the Math subject GRE -- Would it in any way convince the admissions committee at Princeton that my math abilities are on par with rest of the prospective students/applicants?

I'm an international student-- A non-native speaker of English. But I consider myself to be very well acquainted with the admission procedures, etc of most of the 'good' universities in the U.S. (Many of my friends/relatives/acquaintances are currently pursuing their graduate studies in the U.S.)

My grades aren't extraordinary (but good though), my letters of recommendation are also pretty good but they aren't stellar. In some time, I'm also going to complete my under-graduation. I'm also highly inclined towards pursuing an M.B.A. -- After which I want to pursue an M.S. in Finance from a renowned school like Princeton; I personally consider the tuition fee of Princeton's Finance program to be pretty high.

Can you or for that matter anyone over here suggest something which might increase the probability of me getting admitted to Princeton ? (Please keep in mind that I will most probably be applying to Princeton with an M.B.A. 'stapled' to my B.S. degree, GRE/GMAT score, Math subject GRE score (if possible), good letters of recommendation, work experience, very good extra-curricular activities)

Thank you!

Hi [b]all[/b], Aren't there anyone who is willing or interested to give me some suggestion/advice? :(

I acknowledge that most of these queries are hypothetical in nature, but any relevant suggestion/advice is being eagerly awaited.


vector
analytic geometry
geometry unsolved
geometry

Let ABC is triangle and let I, J, K be the midpoint of AC, CB, BA respectively. Let P is change point. The line parallel to CA through P meets CB at M, the line parallel to CB through P meets CA at N. Let S be the intersection of MI and NJ.
Prove that three points P, K, S are collinear.

---------------------------------------------------------------
Maths is my life. I want is good friend with every body!

This can be easily done with vectors. Let A,B,C be the points, then I,J,K are the averages of those, and take N and M on AC and CB ex. N=tA+(1-t)C where t is a real number. Then use that CNPM is a parallelogram.

You can post a full answer by Vector? I don't understand symbol "N = tA + (1-t)C, where t is a real number", because N, A, C are point.

It's easy and nice with Menelaos:

First observe, that $ CIKJ$ is a paralellogram. Let the line $ IK$ intersect $ PM$ at point $ Z$. From Menelaos in triangle $ IMC$ for the line $ NSJ$ we have $ \frac{IN}{NC} \cdot \frac{CJ}{JM} \cdot \frac{MS}{SI} \equal{} 1$. But since $ \frac{CJ}{JM} \equal{} \frac{IK}{KZ}$ and $ \frac{IN}{NC} \equal{} \frac{ZP}{PM}$ we have from Menelaos in triangle $ IZM$ that points $ P , K , S$ are colinear.

Thank you very much. I wish you good luck.

[quote][color=darkred][b][u]An easy extension (old).[/u][/b] Let $ ABCD$ be a parallelogram. For a point $ M$

define $ \{\begin{array}{c} X\in AB\ ,\ MX\parallel AD \\
 \\
Y\in AD\ ,\ MY\parallel AB \\
 \\
N\in BY\cap DX\end{array}$ . Prove that $ C\in MN$ .[/color][/quote]

[color=darkblue][b]Remark.[/b] We can use the same nice [b]Myszon11[/b]'s proof. Indeed, here is it !

[b][u]Proof.[/u][/b] Denote the intersection $ Z\in CD\cap MX$ . Observe that $ \begin{array}{c} BX = CZ\ ,\ BA = CD \\
\ YA = MX\ ,\ YD = MZ\end{array}$ .

Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {BNY}$ and the triangle $ AXD$ :

$ \frac {BX}{BA}\cdot\frac {YA}{YD}\cdot\frac {ND}{NX} = 1$ $ \implies$ $ \frac {CZ}{CD}\cdot\frac {MX}{MZ}\cdot \frac {ND}{NX} = 1$ $ \implies$ $ C\in MN$ .[/color]

[quote][color=darkred][b][u]An "strong" extension (new).[/u][/b] Let $ ABCD$ be a convex quadrilateral.

Consider  the points $ \{\begin{array}{c} X\in AB\ ,\ Y\in CD\ : \ \frac {\overline {XA}}{\overline {XB}} = \frac {\overline {YD}}{\overline {YC}} \\
 \\
M\in XY\ ,\ Z\in AD\ : \ \frac {\overline {MX}}{\overline {MY}} = \frac {\overline {ZA}}{\overline {ZD}}\end{array}$ .

Denote the intersection $ N\in BZ\cap DX$ . Prove that $ C\in MN$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ \frac {\overline {BX}}{\overline {BA}} = \frac {\overline {CY}}{\overline {CD}}$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {BNZ}$ and $ \triangle\ AXD$ :

$ \frac {\overline {BX}}{\overline {BA}}\cdot\frac {\overline {ZA}}{\overline {ZD}}\cdot\frac {\overline {ND}}{\overline {NX}} = 1$ $ \implies$ $ \frac {\overline {CY}}{\overline {CD}}\cdot\frac {\overline {MX}}{\overline {MY}}\cdot \frac {\overline {ND}}{\overline {NX}} = 1$ $ \implies$ $ C\in MN$ .[/color]

[color=darkred][b]Thank you, [u]Thanhnam2902[/u][/b], for the your nice proposed problem !

[b]Thank you, [u]Myszon11[/u][/b],  for the your nice proof ![/color]

[quote="Virgil Nicula"][color=darkred][b][u]An "strong" extension (new).[/u][/b] Let $ ABCD$ be a convex quadrilateral.

Consider  the points $ \{\begin{array}{c} X\in AB\ ,\ Y\in CD\ : \ \frac {\overline {XA}}{\overline {XB}} = \frac {\overline {YD}}{\overline {YC}} \\
 \\
M\in XY\ ,\ Z\in AD\ : \ \frac {\overline {MX}}{\overline {MY}} = \frac {\overline {ZA}}{\overline {ZD}}\end{array}$ .

Denote the intersection $ N\in BZ\cap DX$ . Prove that $ C\in MN$ .[/color]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ \frac {\overline {BX}}{\overline {BA}} = \frac {\overline {CY}}{\overline {CD}}$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {BNZ}$ and $ \triangle\ AXD$ :

$ \frac {\overline {BX}}{\overline {BA}}\cdot\frac {\overline {ZA}}{\overline {ZD}}\cdot\frac {\overline {ND}}{\overline {NX}} = 1$ $ \implies$ $ \frac {\overline {CY}}{\overline {CD}}\cdot\frac {\overline {MX}}{\overline {MY}}\cdot \frac {\overline {ND}}{\overline {NX}} = 1$ $ \implies$ $ C\in MN$ .[/color][/quote]
[color=darkblue][b][u]A short and interesting commentary.[/u][/b] In this nice extension, in the given hypothesis, the [b]direct Menelaus' theorem[/b] applied to the transversal $ \overline {BNZ}$ and the triangle $ ADX$ is equivalently with the [b]reciprocal Menelaus' theorem[/b] applied to the points $ C\in DY$, $ M\in YX$ , $ N\in XD$ and the triangle $ DXY$ .[/color]

Thank you [b]Myszon11[/b] very muck and Thank you [b]Virgil Nicula [/b]very muck.
Your post are very nice. I have a slove this problem by coordinate Afin. Now i'm not free, i will post it tomorrow
----------------------------------------------------------------------

Wow, [u][b]Virgil Nicula[/b][/u] said that I solved a problem ina  nice way. I am really honoured  :lol: Now i feel encouraged to post more solutions on Mathlinks Forum  :)

Thank you again for the your nice and short proof, [b]Myszon11[/b] ! See at the my problem from

(*) http://www.mathlinks.ro/Forum/viewtopic.php?t=183271 whose proof is identically

with the your proof in the particular case. I dedicate you the all these extensions.

From this moment, you can consider that these three succesive extensions  belong to you !

I"ll send you soon (private message) the proof of  (*) .


Prove that in a simple graph with at least two vertices,there must be at least two vertices with the same degree. How to do this question?  :maybe:

Let $ n$ be the number of vertices.
Thus, the possible values for the degrees are $ 0,1,\cdot,n\minus{}1$. In another hand, it is impossible to have a vertex with degree $ 0$ and another one with degree $ n\minus{}1$, then you just have to use the pigeon-hole principle.

Pierre.

[quote="pbornsztein"]Let $ n$ be the number of vertices.
Thus, the possible values for the degrees are $ 0,1,\cdot,n \minus{} 1$. In another hand, it is impossible to have a vertex with degree $ 0$ and another one with degree $ n \minus{} 1$, then you just have to use the pigeon-hole principle.

Pierre.[/quote]

Thanks, I got it now.


In a right triangle with integral length sides, the hypotenuse has length 39 units. How many units is the length of the shorter leg?

This is similar to a 5-12-13 triangle, so the shortest leg is 15 units long.


inequalities
trigonometry
inequalities proposed
111

I inspired this inequality from my friend - NguyenVanDung's result:

Let $ ABC$ be a triangle. Prove that
$ \frac{\cos A}{\cos\frac{B}{2}\cos \frac{C}{2}}\plus{}\frac{\cos B}{\cos \frac{C}{2}\cos \frac{A}{2}}\plus{}\frac{\cos C}{\cos \frac{A}{2}\cos \frac{B}{2}} \ge 2.$

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. Prove that
$ 2\sum \frac {b \plus{} c \minus{} a}{\sqrt {(a \plus{} b)(a \plus{} c)}} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 4.$
Equality holds if and only if $ a \equal{} b \equal{} c,$ or $ a \equal{} b$ and $ c \equal{} 0,$ or any cyclic permutation.[/quote]

By Am-Gm and CS We have:
$ 2\sum \frac {b \plus{} c \minus{} a}{\sqrt {(a \plus{} b)(a \plus{} c)}} \ge 4\sum \frac{b\plus{}c\minus{}a}{2a\plus{}b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}\equal{}2\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$
So we need prove that:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 2$
It's true by Shur  :P

[quote="caubetoanhoc94"]

By Am-Gm and CS We have:
$ 2\sum \frac {b \plus{} c \minus{} a}{\sqrt {(a \plus{} b)(a \plus{} c)}} \ge 4\sum \frac {b \plus{} c \minus{} a}{2a \plus{} b \plus{} c} \ge \frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \equal{} 2 \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$
So we need prove that:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 2$
It's true by Shur  :P[/quote]

@Dang: The condition is $ a, b, c > 0$, so we don't have $ b \plus{} c \minus{} a > 0$. And we can't use Cauchy Schwarts like this  :wink:

Let ABC be a triangle. Prove that
$ \frac{3\sqrt{3}}{2} \geq \frac{3}{8}\sum\frac{\sin{B}\plus{}\sin{B}}{\cos\frac{B}{2}\cos\frac{C}{2}} \geq \sum\cos\frac{A}{2}$.

BQ

Can someone find a nice proof for the initial inequality? I am really interested in seeing it. Thanks.

[quote="can_hang2007"]I inspired this inequality from my friend - NguyenVanDung's result:

Let $ ABC$ be a triangle. Prove that
$ \frac {\cos A}{\cos\frac {B}{2}\cos \frac {C}{2}} \plus{} \frac {\cos B}{\cos \frac {C}{2}\cos \frac {A}{2}} \plus{} \frac {\cos C}{\cos \frac {A}{2}\cos \frac {B}{2}} \ge 2.$[/quote]

If triangle is acute we have

$ 2(\sum \cos A\cos\frac {A}{2})\ge 2(\sum\cos A\cos\frac {A}{2}\cos\frac {B \minus{} C}{2}) \equal{} \sum \cos A(\sin B \plus{} \sin C) \equal{} \sin A \plus{} \sin B \plus{} \sin C\equal{}4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

We are done

If $ A\ge 90^\circ$ I haven't solved :P

[quote="can_hang2007"][quote="geniusbliss"]@Can,
i have given up on this inequality.. its too hard for me.. what i your solution? i want to see it..[/quote]
My proof is rather long. Please help me to check it: http://can-hang2007.blogspot.com/2009/12/inequality-15-n-v-dung-v-q-b-can.html[/quote]
i think your solution is correct and nice.. anyway how did you think of this term- $ \frac{9c(a\plus{}b)^2}{4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}$ ?? it really wonders me...

[quote="geniusbliss"][quote="can_hang2007"][quote="geniusbliss"]@Can,
i have given up on this inequality.. its too hard for me.. what i your solution? i want to see it..[/quote]
My proof is rather long. Please help me to check it: http://can-hang2007.blogspot.com/2009/12/inequality-15-n-v-dung-v-q-b-can.html[/quote]
i think your solution is correct and nice.. anyway how did you think of this term- $ \frac {9c(a \plus{} b)^2}{4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$ ?? it really wonders me...[/quote]
Well, it is not any secret, dear friend. First, I see the inequality has equality for $ a\equal{}b$ anc $ c\equal{}0.$ That suggested me to using "Mixing Variables Method" to reduce the number of variables, that is to try to mix $ a\equal{}b.$ And the above term is the main step in the mixing variables step.

[quote=can_hang2007]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. Prove that
$ 2\sum \frac {b \plus{} c \minus{} a}{\sqrt {(a \plus{} b)(a \plus{} c)}} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 4.$
[/quote]
We need to prove that $\sum_{cyc}\frac{2(a+b-c)}{\sqrt{(a+c)(b+c)}}\geq4-\frac{9abc}{(a+b+c)(ab+ac+bc)}$ or
$\sum_{cyc}\frac{2(a+b-c)}{\sqrt{(a+c)(b+c)}}\geq\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}$ or
$\sum_{cyc}\left(\frac{a+b-c}{a+c}+\frac{a+b-c}{b+c}\right)-\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}(a+b-c)\left(\frac{1}{\sqrt{a+c}}-\frac{1}{\sqrt{b+c}}\right)^2$ or
$\sum_{cyc}\left(\frac{a+c-b}{b+c}+\frac{a+b-c}{b+c}\right)-\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}\frac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$ or
$\sum_{cyc}\frac{2a}{b+c}-\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}\frac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$ or
$\sum_{cyc}\tfrac{2(a^3+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b)}-\tfrac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}\tfrac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$ or
$\sum_{cyc}\tfrac{2(a^3-a^2b-a^2c+abc)}{\prod\limits_{cyc}(a+b)}+\tfrac{4\sum\limits_{cyc}(a^2b+a^2c)}{\prod\limits_{cyc}(a+b)}-\tfrac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}\tfrac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$ or
$\sum_{cyc}\tfrac{(a-b)^2(a+b-c)}{\prod\limits_{cyc}(a+b)}+\tfrac{4\sum\limits_{cyc}(a^2b+a^2c)}{\sum\limits_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)}-\tfrac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{\sum\limits_{cyc}(a^2b+a^2c+abc)}\geq\sum_{cyc}\tfrac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$ or
$\sum_{cyc}\tfrac{(a-b)^2(a+b-c)}{\prod\limits_{cyc}(a+b)}+\tfrac{abc\sum\limits_{cyc}(a^2b+a^2c-2abc)}{\prod\limits_{cyc}(a+b)(a+b+c)(ab+ac+bc)}\geq\sum_{cyc}\tfrac{(a+b-c)(a-b)^2}{(a+c)(b+c)\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}$.
Thus, it remains to prove that $\sum_{cyc}(a-b)^2(a+b-c)\left(1-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}\right)\geq0$.
Let $a\geq b\geq c$.
Hence, $\sum_{cyc}(a-b)^2(a+b-c)\left(1-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}\right)\geq$
$\geq(a-c)^2(a+c-b)\left(1-\frac{a+c}{\left(\sqrt{a+b}+\sqrt{b+c}\right)^2}\right)+(b-c)^2(b+c-a)\left(1-\frac{b+c}{\left(\sqrt{a+b}+\sqrt{a+c}\right)^2}\right)\geq$
$\geq\frac{a^2}{b^2}(b-c)^2(a-b)\left(1-\frac{a+c}{\left(\sqrt{a+b}+\sqrt{b+c}\right)^2}\right)+(b-c)^2(b-a)\left(1-\frac{b+c}{\left(\sqrt{a+b}+\sqrt{a+c}\right)^2}\right)=$
$=\frac{2(b-c)^2(a-b)}{b^2}\left(\frac{a^2b+a^2\sqrt{(a+b)(b+c)}}{\left(\sqrt{a+b}+\sqrt{b+c}\right)^2}-\frac{b^2a+b^2\sqrt{(a+b)(a+c)}}{\left(\sqrt{a+b}+\sqrt{a+c}\right)^2}\right)\geq0$ because
$a^2b\geq b^2a$;
$a^2\sqrt{(a+b)(b+c)}\geq b^2\sqrt{(a+b)(a+c)}$ and
$\frac{1}{\left(\sqrt{a+b}+\sqrt{b+c}\right)^2}\geq\frac{1}{\left(\sqrt{a+b}+\sqrt{a+c}\right)^2}$.  Done! :)

[quote=can_hang2007]I inspired this inequality from my friend - NguyenVanDung's result:

Let $ ABC$ be a triangle. Prove that
$ \frac{\cos A}{\cos\frac{B}{2}\cos \frac{C}{2}}\plus{}\frac{\cos B}{\cos \frac{C}{2}\cos \frac{A}{2}}\plus{}\frac{\cos C}{\cos \frac{A}{2}\cos \frac{B}{2}} \ge 2.-----(1)$[/quote]

I just now try it,found it equality to walker inequality:

in any triangle ,(1)<=>

in acute triangle,have

\[M=\sum{\frac{-2\cos^2(A)+1}{\sin{B}\sin{C}}}-2\geq 0\]

But

\[M={\frac {{s}^{2}-2\,{R}^{2}-8\,Rr-3\,{r}^{2}}{Rr}}\]

[quote=can_hang2007]I inspired this inequality from my friend - NguyenVanDung's result:

Let $ ABC$ be a triangle. Prove that
$ \frac{\cos A}{\cos\frac{B}{2}\cos \frac{C}{2}}\plus{}\frac{\cos B}{\cos \frac{C}{2}\cos \frac{A}{2}}\plus{}\frac{\cos C}{\cos \frac{A}{2}\cos \frac{B}{2}} \ge 2.----(1)$[/quote]

Arqady,
can put a this result,
<=>
walker inequality.

and
(1)<=>

$x,y,z>0,$
\[{\frac { \left(  \left( z+x \right) ^{2}+ \left( x+y \right) ^{2}-
 \left( y+z \right) ^{2} \right)  \left( y+z \right) }{\sqrt {yz
 \left( x+y \right)  \left( z+x \right) }}}+{\frac { \left( - \left( z
+x \right) ^{2}+ \left( y+z \right) ^{2}+ \left( x+y \right) ^{2}
 \right)  \left( z+x \right) }{\sqrt {zx \left( y+z \right)  \left( x+
y \right) }}}+{\frac { \left(  \left( z+x \right) ^{2}+ \left( y+z
 \right) ^{2}- \left( x+y \right) ^{2} \right)  \left( x+y \right) }{
\sqrt {xy \left( z+x \right)  \left( y+z \right) }}}\]
\[\geq 4(x+y+z)\]

Who can do it?

[quote=can_hang2007][quote="geniusbliss"]@Can,
i have given up on this inequality.. its too hard for me.. what i your solution? i want to see it..[/quote]
My proof is rather long. Please help me to check it: http://can-hang2007.blogspot.com/2009/12/inequality-15-n-v-dung-v-q-b-can.html[/quote]

Nguyenhuyen_AG,please,

move can http://can-hang2007.blogspot.com/2009/12/inequality-15-n-v-dung-v-q-b-can.html,

copy can_hang2007 link here.

[quote=xzlbq][quote="can_hang2007"]I inspired this inequality from my friend - NguyenVanDung's result:

Let $ ABC$ be a triangle. Prove that
$ \frac {\cos A}{\cos\frac {B}{2}\cos \frac {C}{2}} \plus{} \frac {\cos B}{\cos \frac {C}{2}\cos \frac {A}{2}} \plus{} \frac {\cos C}{\cos \frac {A}{2}\cos \frac {B}{2}} \ge 2.$[/quote]

only prove that:
 $ \minus{} 3r^2 \plus{} s^2 \minus{} 8Rr \minus{} 2R^2 \geq 0$
in acute triangle,
but easy.
BQ[/quote]

Now see it,

I saying is not true!
BQ

[quote=xzlbq]$x,y,z>0$,prove that

\[\sqrt {1+4\,{\frac {xyz}{ \left( y+z \right)  \left( z+x \right)
 \left( x+y \right) }}}\geq {\frac {y+z-x}{\sqrt { \left( y+z \right)
 \left( x+y+z \right) }}}+{\frac {z+x-y}{\sqrt { \left( z+x \right)
 \left( x+y+z \right) }}}+{\frac {x+y-z}{\sqrt { \left( x+y \right)
 \left( x+y+z \right) }}}\][/quote]
We need to prove that $\sqrt{(x+y+z)\sum_{cyc}(x^2y+x^2z+2xyz)}\geq\sum_{cyc}(x+y-z)\sqrt{x+z)(y+z)}$.
By C-S $\sqrt{(x+y+z)\sum_{cyc}(x^2y+x^2z+2xyz)}=\sqrt{(x+y+z)\sum_{cyc}x(y+z)^2}\geq\sum_{cyc}x(y+z)=2(xy+xz+yz)$.
Thus, it remains to prove that $2(xy+xz+yz)\geq\sum_{cyc}(x+y-z)\sqrt{x+z)(y+z)}$ or
$\sum_{cyc}(x+y-z)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2\geq\sum_{cyc}(x+y-z)(x+z+y+z)-4(xy+xz+yz)$ or
$\sum_{cyc}\frac{(x-y)^2(x+y-z)}{\left(\sqrt{x+z}+\sqrt{y+z}\right)^2}\geq0$.
Let $x\geq y\geq z$.
Hence, $\sum_{cyc}\frac{(x-y)^2(x+y-z)}{\left(\sqrt{x+z}+\sqrt{y+z}\right)^2}\geq\frac{(x-z)^2(x+z-y)}{\left(\sqrt{x+y}+\sqrt{y+z}\right)^2}+\frac{(y-z)^2(y+z-x)}{\left(\sqrt{x+y}+\sqrt{x+z}\right)^2}\geq$
$\geq\frac{(y-z)^2(x-y)}{\left(\sqrt{x+y}+\sqrt{y+z}\right)^2}+\frac{(y-z)^2(y-x)}{\left(\sqrt{x+y}+\sqrt{x+z}\right)^2}=(y-z)^2(x-y)\left(\frac{1}{\left(\sqrt{x+y}+\sqrt{y+z}\right)^2}-\frac{1}{\left(\sqrt{x+y}+\sqrt{x+z}\right)^2}\right)\geq0$.

[quote=Nguyenhuyen_AG][quote=xzlbq]Nguyenhuyen_AG,please,
move can http://can-hang2007.blogspot.com/2009/12/inequality-15-n-v-dung-v-q-b-can.html,
copy can_hang2007 link here.[/quote]
You can download here.[/quote]

very good and thank you!
Nice to all!

[quote=can_hang2007]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. Prove that
$ 2\sum \frac {b \plus{} c \minus{} a}{\sqrt {(a \plus{} b)(a \plus{} c)}} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 4.$
Equality holds if and only if $ a \equal{} b \equal{} c,$ or $ a \equal{} b$ and $ c \equal{} 0,$ or any cyclic permutation.[/quote]

we have another general method.

[quote=zdyzhj]we have another general method.[/quote]
Can you post your solution. :D

Please someone know where we can find the C-S book of can-hang2007 ?

[quote=xzlbq]$x,y,z>0,x+y>\frac{1}{5}z,,y+z>\frac{1}{5}x,,z+x>\frac{1}{5}y,$,prove that

\[{\frac {y+z-x}{\sqrt {y+z}}}+{\frac {z+x-y}{\sqrt {z+x}}}+{\frac {x+y-
z}{\sqrt {x+y}}}>0\][/quote]
Karamata helps! :-D


The number 2^48 −1 is exactly divisible by what two numbers between 60 and 70?


linear algebra
matrix
calculus
integration
algebra
polynomial
linear algebra unsolved

Suppse $C$ is an $n$ square  matrix with integer entries, and integer eigenvalues. Show there exists a $n$ square matrix $T$ such that $det(T) = +/- 1$, with integer entries such that the matrix $T^{-1}CT$ is uppertriangluar.

Apparently this is a lemma in the book $Integral Matricies$ by $M.Newman$, but I dont have the book and would appriciate if i can get a proof for this lemma.

Thanks for your time

it is false because in this case, every matrix with integer entries would have a characteristic polynomial that has all its roots in $\mathbb{R}$, and it is straightforward to find a counterexample. (try in $M_2{\mathbb{Z}}$)

my apologies, i just real I forgot to add a necessary condition. please see above


geometry
trapezoid
power of a point
radical axis
geometry unsolved

Given are 2 circles in a plane, and their 2 common tangents. These 2 tangents touch the circles in A,B,C and D. Prove that you can inscribe a circle in ABCD if and only if the 2 circles touch each other.

It wasn't answered on http://www.mathlinks.ro/viewtopic.php?t=250560

Never solved. Maybe a preformulation?
We have 2 circles in a same plane. They have 2 common exterior tangents. Those tangents touch first circle in A and D, and they touch the other circle in B and C. Prove that ABCD is tangential quadrilateral if and only if the two circles touch each other in one point.

I've found hint in a book:
If AD and BC are common exterior tangents then bisectors of the angles A and B intersect at the midpoint of arc AB, and the bisectors of the angles D and C - at the midpoint of arc CD and those midpoints of arcs coincide with the common point of the circles.
Is it of any help?

(P), (Q) are the 2 circles with common external tangents AB, DC => AB = DC. ABCD is isosceles trapezoid with AD || BC (both perpendicular to PQ). Radical axis of (P), (Q) cuts the center line PQ at K and AB, DC at their midpoints M, N. K is the midpoint of MN and MN is the trapezoid midline, 2 MN = AD + BC. ABCD is tangential <=> AD + BC = AB + DC <=> MN = AB <=> MK = MA = MB <=> MN is tangent to both (P), (Q) at K.


induction
binomial coefficients
number theory unsolved
number theory

Let $ a>b$ and $ n$ be natural numbers such that $ n$ divides $ a^n\minus{}b^n$.

Prove that $ n$ also divides $ \frac{a^n\minus{}b^n}{a\minus{}b}$

First we show that the theorem holds for primes, then for prime-powers, and then for all integers $ n$. In case $ n = 2$, $ 2 \mid a^2-b^2$ iff $ a$ and $ b$ have the same parity, in which case $ 2 \mid a+b$. If $ n$ is a prime $ p > 2$ and $ p \mid a^p - b^p$, we have

$ 0\equiv a^p - b^p \equiv a^p + ( - b)^p \equiv (a - b)^p\: (mod\: p)$
hence
$ a - b \equiv 0 \: (mod\: p)$

$ a = sp + r$
$ b = tp + r$

If $ s = p^\alpha s', \: t = p^\beta t'$, then $ a - b = p^{\alpha + 1}s' - p^{\beta + 1}t'$ is divisible by $ p^\min\{\alpha + 1, \beta + 1\}$. Moreover

$ a^p - b^p = (sp + r)^p - (tp + r)^p$

The terms in $ r^p$ cancel, and every other term is divisible by $ p^\min\{\alpha + 2, \beta + 2\}$ because the binomial coefficients of the form $ p \choose j$ are all divisible by $ p$ for $ 1 < j < p$.

Now we show that the theorem holds for $ n = p^\alpha$, $ \alpha = 1,2,3,...$. For $ \alpha = 1$ that is what we have just shown. Suppose the theorem holds up to $ \alpha - 1$. Then if
$ p^\alpha | a^{p^\alpha} - b^{p^\alpha}$, we have
$ p^{\alpha - 1}\: |\: a^{p^\alpha} - b^{p^\alpha}$
$ p^{\alpha - 1}\: |\: (a^p)^{p^{\alpha - 1}} - (b^p)^{p^{\alpha - 1}}$
so, by induction,
$ p^{\alpha - 1}\: |\: \frac {(a^p)^{p^{\alpha - 1}} - (b^p)^{p^{\alpha - 1}}}{a^p - b^p}.$

From here there are two possibilities. Either $ p\: | \: a^p - b^p$, but then, since $ p \: |\: \frac {a^p - b^p}{a - b}$, we have

$ p^\alpha\: | \: \frac {(a^p)^{p^{\alpha - 1}} - (b^p)^{p^{\alpha - 1}}}{a^p - b^p} \frac {a^p - b^p}{a - b} = \frac {a^{p^\alpha} - b^{p^\alpha}}{a - b}$.

The other possibility is $ p\: \nmid \: a^p - b^p$, but then $ p \nmid a - b$ and

$ p^\alpha | \frac {a^{p^\alpha} - b^{p^\alpha}}{a - b}$

since the numerator is divisible by $ p^\alpha$ and the denominator is not divisible by $ p$.

See the next post for the final blow!

Final step : we show that if the theorem holds for two coprime integers $ x,y$ then it holds for $ xy$. This, along with the fact that the theorem holds for prime powers, proves the theorem. Suppose the theorem holds for $ x$ and $ y$ and $ (x,y)\equal{}1$. Then if

$ xy \mid a^{xy}\minus{}b^{xy}$, clearly
$ y \mid (a^{x})^y\minus{}(b^{x})^y$

and since the theorem holds for $ y$,
$ y \mid \frac{(a^{x})^y\minus{}(b^{x})^y}{a^x\minus{}b^x}$

so that also
$ y \mid \frac{(a^{x})^y\minus{}(b^{x})^y}{a^x\minus{}b^x}\frac{a^x\minus{}b^x}{a\minus{}b} \equal{} \frac{a^{xy}\minus{}b^{xy}}{a\minus{}b}.$

Similarily,

$ x \mid \frac{(a^{y})^x\minus{}(b^{y})^x}{a^y\minus{}b^y}$

$ x \mid \frac{(a^{y})^x\minus{}(b^{y})^x}{a^y\minus{}b^y}\frac{a^y\minus{}b^y}{a\minus{}b}\equal{}\frac{a^{xy}\minus{}b^{xy}}{a\minus{}b}$

So $ y \mid \frac{a^{xy}\minus{}b^{xy}}{a\minus{}b}$, $ x \mid \frac{a^{xy}\minus{}b^{xy}}{a\minus{}b}$, and since $ (x,y)\equal{}1$, $ xy \mid \frac{a^{xy}\minus{}b^{xy}}{a\minus{}b}$.

Anyone care to verify my proof? I had much fun writing it.

For any prime $ p$ and positive integer $ k$ such that $ p^k|n$ and $ p^{k\plus{}1} \nmid n$, let $ x$ be a positive integer such that $ (p^k)^x|a\minus{}b$ and $ (p^k)^{x\plus{}1} \nmid a\minus{}b$

Then $ a^n\minus{}b^n\equal{}(m(p^k)^x\plus{}b)^n\minus{}b^n\equal{}nbm(p^k)^x\equal{}0(mod (p^k)^{x\plus{}1})$ because $ p^k|n$, done.

P.S. The case that n=1 is easy to check and sorry for letting too many variables.

Much better what you've got there. I should always look for the one-liner!
Oh well. I still like my proof somewhat.

[quote="stephencheng"]For any prime $ p$ and positive integer $ k$ such that $ p^k|n$ and $ p^{k \plus{} 1} \nmid n$, let $ x$ be a positive integer such that $ (p^k)^x|a \minus{} b$ and $ (p^k)^{x \plus{} 1} \nmid a \minus{} b$

Then $ a^n \minus{} b^n \equal{} (m(p^k)^x \plus{} b)^n \minus{} b^n \equal{} nbm(p^k)^x \equal{} 0(mod (p^k)^{x \plus{} 1})$ because $ p^k|n$, done.

P.S. The case that n=1 is easy to check and sorry for letting too many variables.[/quote]

Nice proof. Only it should be $ b^{n \minus{} 1}$ instead of $ b$ (sorry for the pettiness but this kind of stuff confuse people  :P )


induction
linear algebra
matrix
linear algebra solved

Let $A \in M_n(\mathbb R)$ invertible such that there exist $B\in M_n(\mathbb Z)$, $A+A^{-1}=B$

Prove that for any $n \in \mathbb N-(0)$, $A^{n}+A^{-n} \in M_n(\mathbb Z)$

I've seen this problem for real numbers rather than for matrices. It's the same problem and has the same solution. The proof is by induction, and uses that the products and sums of integer matrices are integer matrices.

First note that $A^2+A^{-2}=\left(A+A^{-1}\right)^2-2I$ is an integer matrix.

For $n\ge 2$, $A^{n+1}+A^{-(n+1)}=\left(A^n+A^{-n}\right) \left(A+A^{-1}\right)-\left(A^{n-1}+A^{-(n-1)}\right).$

Since all the pieces that make up the right hand side are integer matrices by inductive hypothesis, the result is an integer matrix.

We did make full use of the fact that powers of the same matrix commute.


Να λυθεί στο $\mathbb{R}$ η εξίσωση

                                         $\frac{2^x-1}{2^x+1}=3^{-x}$

 :)

Η αριστερή (f) είναι γν. αύξουσα , η δεξιά(g) είναι γν.φθίνουσα άρα το πολύ να τέμνονται η f,g σε ένα σημείο
που είναι το χ=1.

Χαλαρά έπεφτε σε πανελλήνιες αυτή η άσκηση  :P


How many numbers have the sum of their digits equal to 14, between 1 and 1000?


2000  is the sum of 40 consecutive odd integers.  The smallest  of these is

A.9    B.11   C.27	D.29	E.31

[hide]I'm sorry that I can't postan explanation right now, but I belive the answer is C[/hide]

[hide]The average is $ \frac{2000}{40}\equal{}50$.
40 consecutive odd numbers have a span of $ (40\minus{}1)*2\equal{}78$ or 39 on each side.
Therefore, the smallest is 11. The answer is B.

We could also search for $ n$ in $ (40\plus{}n)^2\minus{}n^2\equal{}2000$. $ n$ happens to be 5.
$ 45^2\minus{}5^2\equal{}2000$, so the answer is B (11).[/hide]


[quote="mathmen"]let $ a_1,a_2..........,a_n$ reals numbers such that: $ a_1 \plus{} a_2 \plus{} ........ \plus{} a_n \equal{} n$
prove that

$ \frac {a_1}{a_{1}^2 \plus{} 1} \plus{} \frac {a_2}{a_{2}^2 \plus{} 1} \plus{} ............. \plus{} \frac {a_n}{a_{n}^2 \plus{} 1} \le \frac {1}{a_1 \plus{} 1} \plus{} \frac {1}{a_2 \plus{} 1} \plus{} ...... \plus{} \frac {1}{a_n \plus{} 1}$[/quote]
It's wrong. Try $ a\rightarrow\minus{}1^\minus{}$ :wink:

[quote="mathmen"]let $ a_1,a_2..........,a_n$ reals numbers such that: $ a_1 \plus{} a_2 \plus{} ........ \plus{} a_n \equal{} n$
and $ a_i > 0$
prove that

$ \frac {a_1}{a_{1}^2 \plus{} 1} \plus{} \frac {a_2}{a_{2}^2 \plus{} 1} \plus{} ............. \plus{} \frac {a_n}{a_{n}^2 \plus{} 1} \le \frac {1}{a_1 \plus{} 1} \plus{} \frac {1}{a_2 \plus{} 1} \plus{} ...... \plus{} \frac {1}{a_n \plus{} 1}$[/quote]
$ \sum_{i\equal{}1}^n\frac{a_i}{a_i^2\plus{}1}\leq\sum_{i\equal{}1}^n\frac{1}{a_i\plus{}1}\Leftrightarrow\sum_{i\equal{}1}^n\left(\frac{1}{a_i\plus{}1}\minus{}\frac{a_i}{a_i^2\plus{}1}\plus{}\frac{a_i\minus{}1}{4}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{i\equal{}1}^n\frac{(a_i\minus{}1)^2(a_i^2\plus{}2a_i\plus{}3)}{(a_i\plus{}1)(a_i^2\plus{}1)}\geq0.$

The condition implies $ \sum_{i \equal{} 1}^{n}{(a_i \minus{} 1)} \equal{} 0$

We have to prove that :
\[ \sum_{i \equal{} 1}^{n} {\frac {a_i \minus{} 1}{(a_i ^2 \plus{} 1)(a_i \plus{} 1)}} \leq 0\]
Assume that $ a_1 \geq a_2 \geq ...\geq a_n$ .Observe that $ a_i \minus{} 1$ and $ \frac{1}{(a_i ^2 \plus{} 1)(a_i \plus{} 1)}$ are oppositely sorted,so by Chebyshev inequality the conclusion follows


My teacher gave me this really hard question today that gave me a really bad headache once i thought about it a lot. Heres the question.

If a, b and c are real numbers such that a2 + b2 + c2 = 1, find the miniumum value of ab + bc + ca.

The 2s are suppose to be squares.

This is what i thought at first. I thought that maybe a and b would = to 0 and c would equal to one and the question would make sense but the teacher said i was wrong. I'm not sure about this questions so if anyone knows the answer to the question could you lead me to an answer plz. Appreciate it.

[quote="king_23"]If $a, b$ and $c$ are real numbers such that $a^2 + b^2 + c^2 = 1$, find the miniumum value of $ab + bc + ca$.[/quote]

TeX'd.  Now.

[hide="Solution"]
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1 + 2(ab + bc + ca) \ge 0$
$2(ab + bc + ca) \ge -1$
$ab + bc + ca \ge \boxed{\frac{-1}{2}}$
[/hide]

i'm assuming that by a2, you mean $a^2$.

In this case, we have
$a^2+b^2+c^2=1$
Note that $(a+b+c)^2-2(ab+bc+ac)=a^2+b^2+c^2$

So we have $1=(a+b+c)^2-2(ab+bc+ac)$
$ab+bc+ac=\displaystyle\frac{(a+b+c)^2-1}{2}$
$ab+bc+ac=\displaystyle\frac{(a+b+c-1)(a+b+c+1)}{2}$

A minimum value would be where $a+b+c=0$, so
$ab+bc+ac=\frac{-1}{2}$


A shiny red racecar drove around a 2-mile track 10 times exactly at 250 mph. He then drove 8 more laps at exactly 'a' mph. If it took the racecar 2.4 more minutes to drive around the track 10 times than it did 8 times, what is the speed the racecar went (in mph) the second time? Edit: Round your answer to the nearest whole number, if applicable.

[quote="ckck"][quote="superwizard188"]A shiny red racecar drove around a 2-mile track 10 times exactly at 250 mph. He then drove 8 more laps at exactly 'a' mph. If it took the racecar 2.4 more minutes to drive around the track 10 times than it did 8 times, what is the speed the racecar went (in mph) the second time?[/quote]
[hide="really really not sure because i did this in my head"]The distance for the first track is 20 miles. His rate was 250 mph. Solving for the time, we get that the time is $\frac{2}{25}$ of an hour. Then the distance again is 20. His rate was a mph. So his time his $\frac{20}{a}$. But this time is $\frac{12}{5}$ longer. So we have the equation $\frac{2}{25}=\frac{20}{a}+\frac{12}{5}$. Multiply everything by 25a to get $2a=500+60$. Solving for a we get that a is 280. [/hide][/quote]
[hide]
umm, I don't think that should be right because 2.4 is in min., and the rest of the equation is in hrs, and I think the track for the second time should be 8*2=16. Also, when you multiplied by 25a, I think you forgot about the second part of the addition on the left side. (12/5). When I did it basically by the same way, but with pencil and paper and calc. (my mental math is horrible, and I  am lazy on paper) I got 400.
I might have completely misunderstood the problem or did some sort of calculation mistake though...[/hide]

Oh, minutes. Right. I misread the problem again.  :wallbash_red:  :wallbash_red:  :wallbash_red:

[quote="sharkboy"]No, I dont think 400 is correct because that would mean he is taking longer to go a shorter distance with a faster speed. Does not add up.

[hide]I got 133.333333333... which rounds out to 133. It took him 2/25 of an hour to complete the first track.  This comes out to 4.8 minutes.  Add 2.4 and you get 7.2. Then you get 16 miles= 7.2 minutes*rate.   solving you get 2.22222222... miles per minute, then multiplying by 60 you get 133.33333333333...[/hide]

I could be wrong so tell me if I am.[/quote]
You don't [i]add[/i] 2.4 minutes, you [i]subtract[/i] them, because the first 10 laps took longer than the last 8.

[hide="Solution"]$\frac{20}{250}=\frac1{25}+\frac{16}a$

$\frac1{25}=\frac{16}a$

$a=\boxed{400\text{ mph}}$

Unrealistic... :o[/hide]


AMC AIME MATHCOUNTS ARML USA(J)MO USAMO probability

If you took AIME and is a MC contestant, tell me how you did!

you should really be posting this on the amc forum. :roll:

I think this is OK here, he's asking for a kind of "Mathcounts Vs. AIME."  Its interesting, because MATHCOUNTS and AIME have such different problem styles.

Would you like me to add a poll anyone?

3...with my stupid errors, i think a 5 was well-attainable if i only had a few flaws

Coming out of it, I thought I had a nine.

At least three of them were DUMB errors. The other two -- I'm pending on whether or not they were dumb errors or just errors.

4.

[quote="kstan013"]yes please. anway I got a 6. I hope the floor is 6!!!![/quote]

I got a 6, too.

If I hadn't been an idiot on #1, I would have gotten a 7.

Wow... you guys are making me feel extremely bad... I got 3 out of 8 answered, all of which i was sure i did in the correct way.
This year's hasn't been very good for me :P (in fact missed 4th in state by 1 point, though i went nats last year) And didn't even make ARML team (also by 1 point on the tryouts). I might get even madder if the floor value was 4!!!!
Is everyone from mathcounts getting scores like you guys'? (which would lower my self esteem to the extremes  :(

Are you the kid who got 6th in MA(I'm not sure you want your name disclosed)

[quote="1two3"]gj to all those who got 7 or higher!
almost certainly, the floor value for usamo will be lower than that. (im expecting a 6 because everyone seem to think this one was a particularly hard one)

just to make myself feel better, i have to remind myself ----> "i have four more years!"

also, perfect what score did you get?[/quote]
Leave that discussion to pms, we aren't allowed to discuss state scores yet.

lol is it even possible for a mathlete to get a 15?

jorian

8 15's. That's really possible from Mathcounts participants.

I got a 4. Missed 7 by 1 and obviously missed #4. :P Also missed #5 because I did something dumb. Oh well.

I got a 4. Not making excuses either  :wink:

I got a 2.  I reached my goal of 1!   :lol:

i doubt that there is much correlation between mc and aime.  i do pretty well in mc and did horribly in aime.  there is a big difference in difficulty, i'd say.  i dont think i even improved from last year.

5; my goal before it was 6; right after the exam, it was 8...

Can you please explain how more middle school kids on AoPS got a perfect score on the AIME than there should be amongst all the high school math geniuses in the US?

[quote="xscapezaer"]
Can you please explain how more middle school kids on AoPS got a perfect score on the AIME than there should be amongst all the high school math geniuses in the US?[/quote]

people lie...apparently way more often than what you and i believe...

[quote="Phelpedo"][quote="jli"][quote="13375P34K43V312"]My index was 190 (sigh), the 12B was so much harder than the 12A...[/quote]

yeah i know!

i got like a 126 on the 12B  :ninja:

omg...the poll is so pointless now :mad: .  im willing to bet $5{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}$ dollars that nobody in mathcounts got a 15[/quote]

Most scores on the 12b were higher...just look at the statistics.[/quote]

True, but a lot less people take the 12B contest, and most of the people that do are generally more serious; they either take both 12A and 12B for their own benefit, or want to take 10 and 12 and have no choice but to take the B contest (most people I've seen on the forum do 10A/12B).

[quote="inyoung"]I got a 9.
I have a theory. I do mediocre in contests that are a little bit harder. AND UTTERLY FAIL the easier contests. i.e. why I suck so much at mathcounts.[/quote]

How is USAMO + National MC Qualifier two years in a row in MIDDLE SCHOOL mediocre or utterly failing?  People these days...psh  :D .

I'm trying to decide whether I should show up for the USAMO...I did a practice and for #1, it was an iff proof, and I did about half of the "if" and got stuck.

[hide="DONT LAUGH"]
4
[/hide]
[hide="If u laughed"]
i'm a 6th grader dont blame me
[/hide] :rotfl:

[quote="jli"][quote="13375P34K43V312"]My index was 190 (sigh), the 12B was so much harder than the 12A...[/quote]

yeah i know!

i got like a 126 on the 12B  :ninja:

omg...the poll is so pointless now :mad: .  im willing to bet $5{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}$ dollars that nobody in mathcounts got a 15[/quote]
Hey remember... 8 people got a 15! pay the bucks :P

[quote="mgao"]
[quote="jli"]omg...the poll is so pointless now :mad: .  im willing to bet $5{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}{}^{5}$ dollars that nobody in mathcounts got a 15[/quote]
Hey remember... 8 people got a 15! pay the bucks :P[/quote]
obviously a conspiracy ignited by some freshman

I got a 7. =)
This year (for me) seems to be filled with 7's. I also like how 18% of people put they got a 15. =D


Let $ a,b,c$ be non-negative such that $ abc \equal{} 1$.Prove that
$ \frac1{a} \plus{} \frac1{b} \plus{} \frac1{c} \plus{} \frac {a^2}2 \plus{} \frac {b^2}2 \plus{} \frac {c^2}2\ge\frac3{2}(\frac a{\sqrt c} \plus{} \frac c{\sqrt b}\plus{}\frac b{\sqrt a})$

The inequality is equivalent to
\begin{align*} \frac {1}{2}\left(\frac {abc}{ab} + \frac {abc}{bc} + \frac {abc}{ca} + a^2 + b^2 + c^2\right) & \ge \frac {3}{2}(abc)^\frac {1}{2}\left(\frac {a}{\sqrt {c}} + \frac {b}{\sqrt {a}} + \frac {c}{\sqrt {b}}\right) \\
(a + b + c)^2 & \ge 3\left(a^\frac {3}{2}c^\frac {1}{2} + b^\frac {3}{2}a^\frac {1}{2} + c^\frac {3}{2}b^\frac {1}{2}\right) \\
\sum_{cyc}\frac {1}{2}\left(a - 2\sqrt {ab} + \sqrt {bc} - c + \sqrt {ca}\right)^2 & \ge 0 \end{align*}
which is obviously true :)

Yes,bertram. Essence,this problem is solved very easy by inequality'Vasc :D


Robyn counted by 6's to 400 beginning with 6. Matthew counted to 400 by 4's beginning with 4. How many numbers were counted by both? :?:

[hide="it wasnt THAT tricky..."]
So, any number that is divisble by both 6 and 4 is divisible by the LCM, which is 12. In other words, we are just finding how many numbers less than or equal to 400 that are divisible by 12. So we have

12, 24, 36, ....... 396

Now we just divide every number by 12 (because it doesnt actually affect how many numbers are in the list).

1, 2, 3, .......... 33

So there are

$ \boxed{33\text{ numbers}}$[/hide]

the official answer to this problem is:

[hide]33 numbers. Nerd_of_the_Ages was correct.[/hide]


topology
real analysis
real analysis unsolved

1.
if $0 < \epsilon < 1$ construct an open set $E \subset [0,1]$ which is dense in $[0,1]$ such that $m(E) = \epsilon$

2.
Construct a totally disconnected compact set $K \subset R$ such that $m(k) > 0$

[quote="windowlistener"]1.
if $0 < \epsilon < 1$ construct an open set $E \subset [0,1]$ which is dense in $[0,1]$ such that $m(E) = \epsilon$
[/quote]

Take the rationals of $[0,1]$ and add balls of small radii around each of them.

2- Any Cantor set of measure > 0 would do it (http://mathworld.wolfram.com/CantorSet.html)


Έστω $a$ ένας φυσικός αριθμός ώστε    $3^a-2^a=p^b$ με   $b>1$ και  $p$ ένας πρώτος αριθμός.Δείξτε ότι ο $a$
είναι επίσης πρώτος

Έστω ότι ο $a$ είναι σύνθετος τότε $a=mn$ για κάποια $m,n>1$και

$3^a -2^a = p^b \Longleftrightarrow (3^m)^n - (2^m)^n =p^b \Longleftrightarrow \\ (3^m-2^m)(3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)})=p^b \ \ (1) ,$

άρα
$3^m \equiv 2^m (mod p) \Longrightarrow 3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)} \equiv n3^{m(n-1)} (mod p)$,

όμως είναι  $3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)} \equiv 0 (mod p)$  (από (1))

άρα $n3^{m(n-1)} \equiv 0 (mod p) \Longrightarrow p=3$ (άτοπο) ή $p|n \Longrightarrow p|a$.

Συνεπώς $a=kp$για κάποιο $k>1$ και όμοια βγαίνει$p|k \Longrightarrow p^2|a$.Συνεχίζοντας με αυτή τη λογική βγαίνει $p^x|a$για κάθε $x\geq 1$ (άτοπο) ή ότι $a=p^k$ για κάποιο $k\geq 2$ {επίσης άτοπο αφού λόγω της (1) για $m=p$ βγαίνει $3^p\equiv2^p(mod p) \Longrightarrow 3\equiv2(mod p)$ (αδύνατο) }.

Άρα ο $a$ είναι πρώτος.

Στέλιος... :!:


calculus
integration
function
real analysis
real analysis unsolved

hello, here it is what i have got
$ \minus{}1.429915780956231133195607666$
Sonnhard.

Thanks for your care to my problem,but what I need is how to calc the integral exactly by mathematical procedures not any approximation by any maths software  :blush:
I am pondering of seperating the original function under the integral sign into one even function and one odd function,thus we can eliminate x^2.sin(2x) since it compounding with  the denominator makes an odd function and the bounds of our integral are symmetric over 0.But it seems still difficult for the rest .


function
induction
number theory unsolved
number theory

Find all fuctions $ f: N\to N$
$ f(f(n))\plus{}f(n)\equal{}2n\plus{}6$


geometry
cyclic quadrilateral
geometry unsolved

Let $ABCD$ be a cyclic quadrilateral $\angle{BAD}< 90^\circ$ and $\angle BCA = \angle DCA$. Point $E$ is taken on segment $DA$ such that $BD=2DE$. The line through $E$ parallel to $CD$ intersects the diagonal $AC$ at $F$. Prove that \[ \frac{AC\cdot BD}{AB\cdot FC}=2.\]

Because $\angle BCA=\angle DCA$, we have $AD=AB$. Hence \[FC/ED=AF/AE=AC/AD=AC/AB.\] So we have \[FC\cdot AB=DE\cdot AC=AC\cdot BD/2.\]

Let angle BCA= angle ACD=x
Since ABCD is cyclic,
angle BCA= angle ACD=angle ABD= angle ADB= x
So, AB= AD we get
∆AEF~∆ADC
ED/ AD=FC/ AC
DE=BD/2
BD.AC= 2AD.FC
since , AB= AD,
BD.AC= 2AB.FC


Two interior angles $ A$ and $ B$ of pentagon $ ABCDE$ are $ 60^{\circ}$ and $ 85^{\circ}$. Two of the remaining angles, $ C$ and $ D$, are equal and the fifth angle $ E$ is $ 15^{\circ}$ more than twice $ C$. Find the measure of the largest angle.


topology
group theory
abstract algebra
limit
advanced fields
advanced fields unsolved

A group $G$ is called [i]profinite[/i] if it is the [url=http://en.wikipedia.org/wiki/Inverse_limit]projective limit[/url] of an inverse system $(G_{i})_{i\in I}$ of finite groups. It can be given a topology as follows: consider each $G_{i}$ as being a topological group with the discrete topology; $G$ can be identified with a subgroup of the direct product of the $G_{i}$, and hence we can give it the topology inherited from this direct product.

Prove that a topological group is profinite iff it is compact and totally disconnected.

(topological groups are assumed to be Hausdorff here)

I guess the followinng works. I'm not very familiar with topological or profinite groups.

Let us prove the Lemma:

Lemma 1: The inverse limit $S$ of a partially ordered system of nonempty compact Hausdorff spaces $S_{i}$, $i\in I$, is a nonempty compact Hausdorff space.

Proof. We have only to prove that $S$ is nonempty. Indeed, $S=\cap R_{kj}$, where $R_{kj}=\{s\in\prod S_{i}|\pi_{kj}(s_{k})=s_{j}\}$ and the intersection is over all $j,k\in I$ with $j\leq k$. The natural map $\text{pr}_{k}\times\text{pr}_{j}: S\to S_{k}\times S_{j}$ is continuous. Since $T=\{(s_{k},s_{j})\in S_{k}\times S_{j}|\pi_{kj}(s_{k})=s_{j}\}$ is a closed subset of $S_{k}\times S$ (by Hausdorff property), $R_{kj}=(\text{pr}_{k}\times \text{pr}_{j})^{-1}(T)$ is a closed subset of $\prod S$. As $\prod S_{i}$ is compact, we have only to show that the intersection of finitely many of the $R_{kj}$ is nonempty. Indeed, let $j(1)\leq k(1),\cdots,j(n)\leq k(n)$ be $n$ pairs in $I$. Choose $l\in I$ with $k(i)\leq l$, $i=1,\ldots,n$, and $s_{i}\in S_{i}$. Define $s_{j(i)}=\pi_{i,j(i)}(s_{l})$ and $s_{k(i)}=\pi_{l,k(i)}(s_{i})$, for $i=1,\cdots,n$. For each $r\in I\setminus \{j(1),\cdots,j(n),k(1),\cdots,k(n)\}$ pick $s_{r}$ to be an arbitrary element of $S_{r}$. The element $s$ belongs to $\cap_{i=1}^{n}R_{k(i),j(i)}$.

Similarly

Lemma 2: A basis for the topology of $S=\lim_{\longleftarrow}S_{i}$ is given by the sets $\pi_{1}^{-1}(U_{i})$, where $U_{i}$ is open in $S_{i}$.

Now by Lemma 1, $G$ is a compact, totally disconnect group. By Lemma 2, the sets $\pi_{i}^{-1}(g_{i}), g_{i}\in G,i\in I$, form a basis for the topology on $G$. Note that each of the sets $\pi^{-1}(g_{i})$ is open and closed. Therefore each element $g\in G$ is its own connected component, i.e. $G$ is totally disconnected.

For the converse it suffices to show that the normal open subgroups form a base for the neighborhood of $e\in G$. Then use the following Lemma

Lemma 3. Let $F$ be a collection of open normal subgroups of $G$ such that $F$ is a base for the neighborhoods of $e\in G$. Then, we may order $F$ by inclusion,and for normal subgroup $N\supset N'$ we have a projection $G/N'\to G/N$. This makes the system of quotients $G/N$ into an inverse system, with \[G\cong \lim_{{\longleftarrow}\atop{{N\in\mathcal{F}}}}G/N,\] where the inverse limit and isomorphism are in sense of topological group.


Solve the inequality $ x>log_6 11$

I came across this problem, I don't know what its asking for or what the answer is? please help.

If you put it in exponential form it is what value of x satisfies $ 6^x>11$

The question is solved already. unless it wants you to convert the RHS to decimal form, there's nothing more you can do (and that requires a calculator)


If $ A, B\in C^{m\times n}$, and $ rank(A)\equal{}rank(B),$whether $ A^\plus{}AB^\plus{}\equal{}B^\plus{}AA^\plus{}$?
Where $ A^\plus{}$ denotes the Moore-Penrose inverse.

[quote="miwalin"]If $ A, B\in C^{m\times n}$, and $ rank(A) \equal{} rank(B),$whether $ A^ \plus{} AB^ \plus{} \equal{} B^ \plus{} AA^ \plus{}$?
Where $ A^ \plus{}$ denotes the Moore-Penrose inverse.[/quote]

This cannot generally hold - not even for $ m\equal{}n$. In fact, assume this would hold. First replace $ B$ by $ B^\plus{}$ (remember that $ \left(B^\plus{}\right)^\plus{}\equal{}B$ and $ \mathrm{rank}B^\plus{}\equal{}\mathrm{rank}B$ for every matrix $ B$), and your assertion becomes: If $ A,B\in\mathbb{C}^{n\times n}$ and $ \mathrm{rank}A\equal{}\mathrm{rank}B$, then $ A^\plus{} AB\equal{}BAA^\plus{}$.
Now, the equation $ A^\plus{} AB\equal{}BAA^\plus{}$ is linear in $ B$, so if it holds for all matrices $ B$ with $ \mathrm{rank}B\equal{}\mathrm{rank}A$, then it also holds for all linear combinations of such matrices. But every matrix in $ \mathbb{C}^{n\times n}$ can be written as such a combination, unless $ A\equal{}0$ (since for any $ k\in\left\{1,2,...,n\right\}$, the linear span of all matrices in $ \mathbb{C}^{n\times n}$ which have rank $ k$ is the whole $ \mathbb{C}^{n\times n}$). Hence, if your assertion was right, we would have $ A^\plus{} AB\equal{}BAA^\plus{}$ for any two matrices $ A,B\in\mathbb{C}^{n\times n}$ satisfying $ A\neq 0$. This means that the space $ \left\{B\in\mathbb{C}^n\mid A^\plus{} AB\equal{}BAA^\plus{}\right\}$ must be the whole $ \mathbb{C}^{n\times n}$, so its dimension is $ n^2$. But as a byproduct of http://www.mathlinks.ro/viewtopic.php?t=256194 , we know that this is only possible if both matrices $ A^\plus{}A$ and $ AA^\plus{}$ have the form $ \lambda I_n$ for some $ \lambda\in\mathbb{C}$. But there are plenty of matrices $ A\in\mathbb{C}^{n\times n}$ such that $ AA^\plus{}$ is not of this form. Contradiction.

I guess it would be easy to snipe the conjecture using computer algebra software, but I don't have anything that computes pseudoinverses (or don't know how to use it).

  darij


How does one prove that (cosA)^2 +(cosB)^2 +(cosC)^2 + 2cosAcosBcosC=1, if A,B,C are the angles of an acute triangle using the sum to product formulas?

Thanks,
mathgeek93

This relation is true for all triangles .

$ \cos^2 A \plus{} \cos^2 B \plus{} \cos^2 C \plus{} 2\cos A\cos B\cos C \equal{} 1\ (*)$ .

Case 1)

The triangle is right-angled (suppose that $ A\equal{}\frac{\pi}{2}$) and the identity reduces to $ \cos^2 B\plus{}\cos^2 C\equal{}1\ (1)$ . Because $ B\plus{}C\equal{}\frac{\pi}{2}$ we get $ \cos B\equal{}\sin C$ and relation $ (1)$ becomes $ \sin^2 C\plus{}\cos^2 C\equal{}1$ , which is true .

Case 2)

The triangle isn't right-angled :

$ \cos A \equal{} \cos[\pi \minus{} (B \plus{} C)] \equal{} \cos\pi\cdot\cos(B \plus{} C) \plus{} \sin\pi\cdot\sin(B \plus{} C)$

$ \Longleftrightarrow\ \cos A \equal{} \minus{} \cos(B \plus{} C) \equal{} \sin B\cdot\sin C \minus{} \cos B\cdot \cos C$ . Multiply by $ \cos A\ (\ne 0)$

$ \implies\ \cos^2 A \equal{} \cos A\sin B\sin C \minus{} \cos A\cos B\cos C$ .

Similarly we get $ \cos^2 B \equal{} \sin A\cos B\sin C \minus{} \cos A\cos B\cos C$  and $ \cos^2 C \equal{} \sin A\sin B\cos C \minus{} \cos A\cos B\cos C$ .

So relation $ (*)$ becomes : $ \cos A\sin B\sin C \plus{} \sin A\cos B\sin C \plus{} \sin A\sin B\cos C \minus{} \cos A\cos B\cos C \equal{} 1\ (**)$ .

Now, $ \cos (A \plus{} B \plus{} C) \equal{} \minus{} 1\ \Longleftrightarrow\ \cos(A \plus{} B)\cos C \minus{} \sin (A \plus{} B)\sin C \equal{} \minus{} 1$

$ \Longleftrightarrow\ \cos A\cos B\cos C \minus{} \sin A\sin B\cos C \minus{} \sin A\cos B\sin C \minus{} \cos A\sin B\sin C \equal{}$ $ \minus{} 1$ , which is just relation $ (**)$ , so we are done .

[quote="Mateescu Constantin"]This relation is true for all triangles .[/quote]
Your remark is incorrect.
If suppose that $ A\equal{}\frac{\pi}{2}$ then you can't multiply both hand sides by $ \cos A$
[quote="Mateescu Constantin"]
$ \Longleftrightarrow\ \cos A \equal{} \minus{} \cos(B \plus{} C) \equal{} \sin B\cdot\sin C \minus{} \cos B\cdot \cos C$ . Multiply by $ \cos A$
[/quote]

Well in that case, $ \cos A\equal{}0$ so the question reduces to $ \cos^2 B\plus{}\cos^2 C\equal{}1$.  Since $ B\plus{}C\equal{}\frac{\pi}{2}$, $ \cos B\equal{}\sin C$ so we replace this above to get $ \cos^2 C\plus{}\sin^2 C\equal{}1$ which is obviously true.

[quote="tenniskidperson3"]Well in that case, $ \cos A \equal{} 0$ so the question reduces to $ \cos^2 B \plus{} \cos^2 C \equal{} 1$.  Since $ B \plus{} C \equal{} \frac {\pi}{2}$, $ \cos B \equal{} \sin C$ so we replace this above to get $ \cos^2 C \plus{} \sin^2 C \equal{} 1$ which is obviously true.[/quote]
Oh, I see...thank you !
But I think Mateescu Constantin need to complement his solution.

[quote="Mateescu Constantin"]Prove that in any triangle $ ABC$ exists the identity $ \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = 1$ .[/quote]
[b]Proof.[/b] It is O.K. ! Thus, $ \underline{4\cos A\cos C\cos C} =$ $ 2\cos A\cdot 2\cos B\cos C =$ $ 2\cos A\cdot [\cos (B + C) + \cos (B - C)] =$

$ 2\cos A\cdot [ - \cos A + \cos (B - C)] =$  $ - 2\cos^2A - 2\cos (B + C)\cos (B - C) =$ $ - 2\cos^2 A - \cos 2B - \cos 2C =$

$ - 2\cos^2A - (2\cos^2B - 1) - (2\cos^2 C - 1) =$ $ \underline{2(1 - \cos^2A - \cos^2B - \cos^2C)}\Longleftrightarrow \sum\cos^2A + 2\prod\cos A = 1$ .


calculus
integration
real analysis
function
real analysis unsolved

$ f(x) = \begin{cases}1 \quad \mbox{if}\ x \ \mbox{is irrational}\\ 0 \quad \mbox{if}\ x \ \mbox{is rational}\end{cases}\\ \int_{0}^{1}f(x) \, \mbox{d}x = \ldots ?$
Is it possible to compute that? I know that it isn't possible using Riemann integral, but maybe other?

with Lebesgue integral it is equal with
$ \mu(\mathbb Q\cap[0,1])\cdot 0+\mu(\mathbb [0,1]-\mathbb Q)\cdot 1=1$.

In the Riemann sense, this function is 1 on a dense set and 0 on another dense set, so it's not integrable. In the Lebesgue sense, it's 1 almost everywhere, so it's integrable with integral $ \mu(X)$ on any measurable set $ X$.

OK, thanks.

But Dirichlet's function, defined as
$ \mathbf{1}_{\mathbb{Q}}(x) = \begin{cases}1, & \mbox{when }x \in \mathbb Q \\ 0, & \mbox{when }x \notin \mathbb Q \end{cases}$
is integrable in Lebesgue sense, but whatever the integration range is, the result is 0.
So... :?:

$ \mu(\mathbb{Q})=0$, so this function is $ 0$ almost everywhere. The integral for any measurable set $ X$ is $ \mu(X\cap\mathbb{Q})\leq\mu(\mathbb{Q})=0$. This implies that the integral is $ 0$, since the function is nonnegative. With your first function, it was $ 1$ almost everywhere.


What are all ordered pairs of integers (x,y) which satisfy
$x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+6=0?$

[hide="my solution"]
that would equal
$(x-1)^{3}+(y+2)^{3}-1=0$
$(x-1)^{3}+(y+2)^{3}=1$
it's a circle! *wishes I knew Asymptote*
center= (1,2), radius 1
therefore, the integer pairs are
(0,2), (2,2), (1,3), (1,1)
[/hide]
nice avatar, Kouichi

mgao, I think you made a small error. The center of the circle is really (1,-2)
This makes the points (0,-2) (2,-2) (1,-1) and (1,-3) solutions, although (0,-2) and (1,-3) turn out to be extraneous.


Hello,

The problem is this: given a multiset containing 900 elements of 10 types, how many ways can you partition the set into groups of 3 so that all 900 are used?  e.g. partitioning such as |CEF|FFE| would be the same as |FCE|FEF| (although we are concerned with 300 groups of 3).

I understand that the number of group types is 10 multichoose 3 = 220.  I think that the number of arrangements of 300 groups with replacement from that set of 10 types would be 220 multichoose 300, but how do I consider that we are drawing from a finite multiset of 900?

I'm kind of stuck here so any help would be very appreciated.

Thank you,
-Morgan

You seem to like questions with some NP-hard flavor :)

A question: by a multiset of 900 elements of 10 types, do you mean that there are 90 elements of each of the 10 types, or might the distribution of types be more complicated?


Let $ ABC$ be a triangle and the points $ M,N \in AB$, $ P,Q \in BC$, $ R,S \in AC$ suche as $ \frac{\vec{AM}}{\vec{AB}}\equal{}\frac{\vec{NB}}{\vec{AB}}\equal{}k$,$ \frac{\displaystyle\vec{BP}}{\vec{BC}}\equal{}\frac{\vec{QC}}{\vec{BC}}\equal{}p$, $ \frac{\vec{CR}}{\vec{CA}}\equal{}\frac{\vec{SA}}{\vec{CA}}\equal{}q$,where $ k,p,q \in R^{*}$.Consider $ X \in MS,Y\in NP, Z \in QR$ such as $ \frac{\vec{MX}}{\vec{MS}}\equal{}\frac{\vec{PY}}{\vec{PN}}\equal{}\frac{\vec{RZ}}{\vec{RQ}}$.prove that  $ \Delta XYZ$ and $ \Delta ABC$ have the same centroid $ \Leftrightarrow$ $ X,Y,Z$ are the midles of the segments $ MS,NP,QR$ or $ k\equal{}p\equal{}q$.

[color=blue][quote="balan razvan"]Let $ ABC$ be a triangle and the points $ M,N \in AB$, $ P,Q \in BC$, $ R,S \in AC$ suche as $ \frac {\vec{AM}}{\vec{AB}} = \frac {\vec{NB}}{\vec{AB}} = k$,$ \frac {\displaystyle\vec{BP}}{\vec{BC}} = \frac {\vec{QC}}{\vec{BC}} = p$, $ \frac {\vec{CR}}{\vec{CA}} = \frac {\vec{SA}}{\vec{CA}} = q$,where $ k,p,q \in R^{*}$.Consider $ X \in MS,Y\in NP, Z \in QR$ such as $ \frac {\vec{MX}}{\vec{MS}} = \frac {\vec{PY}}{\vec{PN}} = \frac {\vec{RZ}}{\vec{RQ}}$.prove that  $ \Delta XYZ$ and $ \Delta ABC$ have the same centroid $ \Leftrightarrow$ $ X,Y,Z$ are the midles of the segments $ MS,NP,QR$ or $ k = p = q$.[/quote][/color]

Let $ G$ be the centroid of $ \Delta ABC$ abd we heve:
\[ \vec{GA}+\vec{GB}+\vec{GC}=0\]
Let $ \frac{\vec{MX}}{\vec{MS}}=\frac{\vec{PY}}{\vec{PN}}=\frac{\vec{RZ}}{\vec{RQ}}=x$.From $ \vec{GX}=x\vec{GS}+(a-x)\vec{GM}$,$ \vec{GY}=x\vec{GN}+(1-x)\vec{GP}$,$ \vec{GZ}=x\vec{GQ}+(1-x)\vec{GR}$ $ \implies$ $ \vec{GX}+\vec{GY}+\vec{GZ}=x(\vec{GS}+\vec{GN}+\vec{CQ})+(1-x)(\vec{GM}+\vec{GP}+\vec{GR})=x(\vec{AS}+\vec{BN}+\vec{CQ})+(1-x)(\vec{AM}+\vec{BP}+\vec{CR})$.
While $ \vec{AS}=\vec{RC}$,$ \vec{BN}=\vec{MA}$,$ \vec{CQ}=\vec{PB}$ eh have that $ \vec{GX}+\vec{GY}+\vec{GZ}=(1-2x)(\vec{AM}+\vec{BP}+\vec{CR})$
On the other side $ \vec{AM}=k\vec{AB}$,$ \vec{BP}=p\vec{BC}$,$ \vec{CR}=q\vec{CA}$ $ \implies$ $ G$ is the centroid of $ \Delta XYZ$ $ \equiv$ $ x=1\2$ or $ (k-q)\vec{AB}+(p-q)\vec{BC}=0$ $ \equiv$ $ X,Y,Z$ are the middles of the segments $ MS,NP,QR$ or $ k=p=q$.


geometry
rectangle
real analysis
real analysis unsolved

is it possible to pack all the rectangles of side length 1/k by 1/(k+1), k=1, 2, 3,...., without overlap into a square of side length 1?

how will I prove this?

[hide="OK, forget this"]It's only half the area, so total area is not a problem and we have the room to do many different arrangements.

Divide the square into strips that are $ 1\times \frac1{k + 1}$. We'll fit all of the the rectangles into these strips.

Start with the $ 1\times\frac1{k + 1}$ rectangle in the first strip.

Put the $ \frac1{k}\times\frac1{k + 1}$ and the $ \frac {k - 1}k\times\frac1{k + 1}$ so that they fill the second strip.

The $ \frac2k\times\frac1{k + 1}$ and the $ \frac {k - 2}k\times \frac1{k + 1}$ rectangles fill the third strip.

If $ k$ is odd, this all works out perfectly, and we actually pack the rectangles into a $ 1\times\frac12$ rectangle.

If $ k$ is even, we fill a $ 1\times\frac {k}{2(k + 1)}$ rectangle and have one $ \frac12\time\frac1{k + 1}$ piece left over, which easily fits into the space left over.

The fit into the $ 1\times 1$ square is easy. As I said, for $ k$ odd, we can actually fit these rectangles into a $ 1\times \frac12$ rectangle. That does not seem to be possible for even $ k;$ in particular, it fails for $ k = 2.$[/hide]

That looks like a misreading; we have a $ 1\times \frac12$, a $ \frac12\times\frac13$, a $ \frac13\times \frac14$, and so on. It's all the area, and dividing into strips is unlikely to help.

well it seems to be a very hard problem.....
can we argue along the following line?
the total area of the rectangles is
$ \sum \frac{1}{k(k\plus{}1)} \equal{} 1$
which is equal to the area of the unit square.... :maybe:
does this help in some way?

No, that was clear from the very beginning. What Kent wrote is similar to a famous packing problem, but here the two areas being equal, the chance of this being true (by my lights) is almost nil.