$ a,b,c$ real number . $ a\plus{}b\plus{}c\equal{}1$ and $ a,b,c\geq\minus{}\frac{3}{4}$              prove:
$ \frac{a}{a^{2}\plus{}1}\plus{}\frac{b}{b^{2}\plus{}1}\plus{}\frac{c}{c^{2}\plus{}1}\leq\frac{9}{10}$


Is there any way I can include enumerate or description environments inside tables?

\begin{tabular}{@{}ll@{}}
2003-2007 & University of XXXXXXX \\
& XXXX Master of XXX (2:1)\\
& [This is where I want my "description" environment"] \\
\end{tabular}

There are 2 alternative ways of dealing with this:
1. Use a minipage inside the cell where you want the environment to be
[code]\begin{tabular}{@{}ll@{}}
2003-2007 & University of XXXXXXX \\
& XXXX Master of XXX (2:1) \\
& \begin{minipage}{6cm}
     \begin{description}
        \item This is where I want
        \item my description environment
     \end{description}
     \begin{enumerate}
        \item Item 1
        \item Item 2
     \end{enumerate}
  \end{minipage}
\end{tabular}[/code] which gives
$\begin{tabular}{@{}ll@{}}2003-2007 & University of XXXXXXX \\ & XXXX Master of XXX (2: 1) \\ & \begin{minipage}{6cm}\begin{description}\item This is where I want \item my description environment \end{description}\begin{enumerate}\item Item 1 \item Item 2 \end{enumerate}\end{minipage}\end{tabular}$

2. Change the table's parameters:
[code]\begin{tabular}{p{2cm}p{6cm}}
2003-2007 & University of XXXXXXX \\
& XXXX Master of XXX (2:1) \\
& \begin{description}
     \item This is where I want
     \item my description environment
    \end{description}
    \begin{enumerate}
       \item Item 1
       \item Item 2
    \end{enumerate}
\end{tabular}$[/code]which gives
$\begin{tabular}{p{2cm}p{6cm}}2003-2007 & University of XXXXXXX \\ & XXXX Master of XXX (2: 1) \\ & \begin{description}\item This is where I want \item my description environment \end{description}\begin{enumerate}\item Item 1 \item Item 2 \end{enumerate}\end{tabular}$

Thanks a lot stevem, I've avoided the issue now that I decided to use a different layout, but it's nice to know that it can be done in case I need to in the future.


$ABCDE$ is a convex pentagon such that $DC = DE$ and $\angle C = \angle E = 90^{\cdot}$. $F$ is a point on the side $AB$ such that $\frac{AF}{BF}= \frac{AE}{BC}$. Show that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.

Let $G$ be a point on extension of $AE$ such that $EC=CB$.And Let $CB$ intersects $EA$ at $O$.we have $EF\parallel BG$. we want to prove $\angle FEC=\angle BDC$.Let $AD$ intersects $EC$ at $K$. Let $\angle AEF=\alpha,\angle CED=\beta,\angle ODB=\theta,\angle KCD=\rho$ so we should prove that $\alpha+\beta=\theta+\rho$.since $DE=DC$ so $\beta = \rho$ so we shold prove $\alpha = \theta$. but since $EF\parallel BG$ so $\alpha = \angle AGB$.so we should prove $GOBD$ is cyclic. or $\angle GOB+\angle GDB=180$ but we have $\angle GOB+\angle CDE=180$and $\angle BDC = \angle EDG$ (because $\triangle EDG = \triangle BDC$). so $\angle GOB+\angle GDB=180$.similarly we can prove $\angle FCE=\angle ADE$. and proof is complete.

[quote="Megus"]$ABCDE$ is a convex pentagon such that $DC = DE$ and $\angle C = \angle E = 90^{\cdot}$. $F$ is a point on the side $AB$ such that $\frac{AF}{BF} = \frac{AE}{BC}$. Show that $\angle FCE = \angle FDE$ and $\angle FEC = \angle BDC$.[/quote]

Here, $\angle FCE = \angle FDE$ should be replaced by $\angle FCE = \angle ADE$ (this is a typo originating from the Kalva website). Anyway, the problem was posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=29178 and solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=22138 .

  Darij


A polynomial $f$ with real coefficients verifies $f(x) f(y) = f(x+y)$. Find $f$.

$f(x)=1$ or $f(x)=0$ everywhere.  Such a function has exponential growth if it is nonconstant, and hence cannot be a polynomial.

There was a mistake in my solution. I had to delete it

Alternately, say that $f(x)$ is a nonconstant polynomial.  Then $f$ has an analytic continuation to the complex plane which obeys that functional equation.  But if $f=0$ anywhere, then $f=0$ everywhere by the functional equation.  So $\text{deg}(f)\in\{0,-\infty\}$, and it's easy to finish from here.

I posted the problem here in order to see some elementary solutions, but instead I got a very nice proof using the Fundamental Theorem of Algebra.

My first solution was the first solution of blahblahblah and my second solution was considering $\lim_{x \to \infty} ( f(x) \cdot f(-x) )$.


A bag contains one marble, either green or yellow. A yellow marble is added to the bag, and one marble is randomly chosen. The chosen marble is yellow. What is the probability that the marble left in the bag is yellow? Express your answer as a common fraction.

The probability that the chosen marble is yellow is $ \frac{1}{2}\cdot \frac{1}{2}\plus{}\frac{1}{2}\cdot 1\equal{}\frac{3}{4}$, and the probability that the other marble is yellow is $ \frac{1}{2}$, so \[ \frac{\frac12}{\frac34}\equal{}\boxed{\frac{2}{3}}\]


algorithm
group theory
abstract algebra
graph theory
combinatorics unsolved
combinatorics

On a table there are 2004 boxes, and in each box a ball lies. I know that some the balls are white and that the number of white balls is even. In each case I may point to two arbitrary boxes and ask whether in the box contains at least a white ball lies. After which minimum number of questions I can indicate two boxes for sure, in which white balls lie?

Does "in each case I may point to two arbitrary boxes and ask whether in the box contains at least a white ball lies." mean

"i can point to 2 boxes and ask if there a white ball in at least one of the 2 boxes"

If so, my best with 5 minutes of thought is within 1004 questions, but I find it hard to believe that 1004 is the minimum.
-Tom

Hm,I don't think 1004 is a big number,since in the case we just have two white balls then the question must be larger or equal 1001.

The answer is: 2*2004-3 = 4005

Proof of lower bound (4004 is not enough):
Suppose the answers to the first 4004 questions are all "YES, at least one of these two boxes contains
a white ball".  Suppose furthermore, the player in the end claims that boxes 1 and 2 both contain white
balls.
  Case A: If the player never queried the pair (1,2), then we make the balls in 1 and 2 black, and all other
balls white. This situation agrees with all answers, and the player cannot be sure that he is right.
  CASE B: The player has queried (1,2). Then there are 2002 pairs (1,x) and 2002 pairs (2,x) with x in
3...2004.  One of these 4004 pairs has never been queried, say the pair (1,z). We make the balls in box 1
and in box z black, and all other balls white. Again, the player cannot be sure that he is right.

Proof of the upper bound (4005 queries are enough):
We describe an algorithm with at most 4005 queries that correctly identifies two white balls. The algorithm
goes through two phases.
  Phase 1: Query the pairs (1,x) with x=2...2004.
  CASE 1.A: If one of the answers is NO, then box 1 has a black ball. Then the YES-answers
identify all boxes with a white ball.  Game is won. We stop.
  CASE 1.B: If all answers are YES, then the ball in box 1 must be white. (If it was black, then
all other 2003 balls are white; contradiction to the even-assumption).  We move on to phase 2.
  Phase 2: Query the pairs (2,y) with y=3...2004.
  CASE 2.A: If one of the answers is NO, then box 2 has a black ball. Then the YES-answers
identify all boxes with a white ball.  Game is won. We stop.
  CASE 2.B: If all answers are YES, then the ball in box 2 is white (same argument as in case 1.B).
We stop with white balls in boxes 1 and 2. Game is won.

I also got 4005 questions, just by thinking of it.
Lets say you pick 2 boxes at random (a and b), remember them, and ask questions about a paired with every other.
Then you ask the questions about b, totally it gives you 2003+2002 = 4005.  By this way you can determine the color of box a and box b (if all questions about box a return a "yes", then a is white, if at least one is no, a is black, and you use the box which returned a "yes" for a.  Same you do with b).  (This will work for any number of white boxes).  Also, all boxes  cannot return "no", since it would imply either one or zero black box, which I think, is impossible.

i think you are thinking wrong about this problem because it says just find 2 white boxes not all of them.
so i think the answer is 1005
lemma 1
if we have 4 boxes which include at least 2 white boxes; we can always find two white boxes in five qustions. this is easy just casework.
the answer is bigger than 1004 because
suppose that there are two white boxes.
lemma 2
we must ask at least one question about every box for the minimum delay
suppose that there is a box that we didn't ask a question about that box (say non-asked) and say that from the questions that we asked we get one "at least one white box" and the other ones are "there aren't white boxes" answers. if want to learn that non-asked box is white or black without asking any questions about non-asked one we must ask 4002 questions(case work also).so we must ask about all the boxes.
now we can get two "at least one boxes" answers in at least 1002 questions(we can get three or four but we are looking for the best answer)now we have at most 2 questions but from the first lemma this isnt always possible to identify two white boxes from four boxes with 4 questions.
the finding algorithm is ask about boxes(with non intersecting two's) when you get exactly two "at least one boxes" answer and after that ask about the four boxes at most 3 questions(2 of the five questions are almost asked) and this is obvious that the number of the answers are smaller than 1005
i think this wasn't a full solution (but it can be also there were parts that i'm so bored to write and pass) but it an take a 5 or 6 points the elegant solutions about this problem can be with graph theory i think
it take a very big 0

[quote="anonymous1173"]if we have 4 boxes which include more than 2 white boxes; we can always find two white boxes in five qustions. this is easy just casework.[/quote]In fact, it's not. That's the point of this question...

No... You don't seem to understand what the problem is about. Do your "easy casework", you will see that it is nonsense. You really need the fact that the number of white balls (and the total number of boxes) is even. This is the crucial point here.

[quote]i think this wasn't a full solution but it an take a 5 or 6 points[/quote]Rather a 0 thus, I'm afraid...

[quote]No... You don't seem to understand what the problem is about. Do your "easy casework", you will see that it is nonsense.[/quote]

His "easy casework" is correct, actually.  What he's trying to do is start from a small case and work his way upwards.

[quote]You really need the fact that the number of white balls (and the total number of boxes) is even. This is the crucial point here.
[/quote]

Actually, it doesn't matter.  The crucial point is that there be at least two white boxes. (An obviously necessary condition)  Then the minimum number of questions you need is 1004, in either case.

[quote="Deedlit"]His "easy casework" is correct, actually. What he's trying to do is start from a small case and work his way upwards.[/quote] It is not. Maybe you should read the problem again. There is no way you can find an answer for a small case. It is simply false for any small case.

[quote="Deedlit"]Actually, it doesn't matter.[/quote]It does matter. For 2005 boxes or without the condition that the number of white balls is even, there is no way you can find two in a finite number of attempts.

Test20's answer is correct, and the only correct answer.

[quote]It is not. Maybe you should read the problem again. There is no way you can find an answer for a small case. It is simply false for any small case.
[/quote]

The answer to "How many questions does it take to find two white balls in four boxes?" is "False"?  Are you saying it is not possible to identify two white balls, no matter how many questions you ask?

[quote]It does matter. For 2005 boxes or without the condition that the number of white balls is even, there is no way you can find two in a finite number of attempts.

Test20's answer is correct, and the only correct answer.[/quote]

You're right, we do need that the number of white balls is even.  My mistake.

Actually, what we need is that there are at least two white balls, and not exactly n-1 white balls.  That's enough to answer the question.  And then the number of boxes does not need to be even.

You're answering the question yourself:

[quote="Deedlit"]What he's trying to do is start from a small case and work his way upwards. [/quote][quote="Deedlit"]Actually, what we need is that there are at least two white balls, and not exactly n-1 white balls.[/quote]

So since you only know that there aren't $n-1$ balls in the total of 2004, not in the smaller subgroup of $4$ boxes (where it can actually happen to be $3$), the conclusion is not true for each set of $4$ boxes. And a wrong conclusion cannot be attained trough a simple case study.

I'm not claiming the problem is wrong, I'm claiming his lemma 1 is wrong, as it misses the condition that there aren't exactly $3$ white balls. And if you ask that for an extra condition, it doesn't help in the proof any further.

i got what you said peter vdd i made a terrible mistake.
not a misunderstooding but a carelessness. thanks pater vdd for showing me my bad mistake and thanks test20 for the right solution

Reworded for clarity.

[quote=Russia 2004] There are $2004$ boxes, each containing a white or a black ball. It is known that a non-zero even number of balls are white. A move consists of picking two boxes and asking whether at least one of them contains a white ball. What minimal number of questions is required, after which one can indicate two boxes both containing a white ball? [/quote]

I claim that $4005$ is the desired minimum. Consider this as a game played by two players, [i]Fate[/i] and [i]Hope,[/i] where [i]Hope[/i] asks the questions and [i]Fate[/i] replies to them truthfully with a "yes" or a "no". For convenience, identify the boxes by the color of the balls they contain.

[b]Strategy for Hope:[/b] Label the boxes as $1,2, \dots, 2004$ and ask questions of the form $(1, i), (1,2), (2,i)$ for all $3 \le i \le 2004$.

For the first $2003$ questions, if any of them is answered by a "no" then we get $1$ is black; so all the "yes" answers tell us exactly which balls are white, and we are done. Now, assume they were all answered by a "yes". If $1$ is black then some other index in $\{2,3,\dots, 2004\}$ is also black as there are an even number of black indices; so $1$ must be white.

If any of the next $2002$ questions is answered by a "no" then we know that $2$ is black and exactly which of $\{3, \dots, 2004\}$ are white by looking at the "yes" answers. If all are "yes" then $2$ cannot be black since the set $\{2,3, \dots, 2004\}$ contains an even number of black indices. Evidently, we can identify $1,2$ as white. Thus, the strategy indeed works.

[b]Strategy for Fate:[/b] Suppose [i]Hope[/i] has a strategy which allows her a win in $m \le 4004$ questions regardless of what [i]Fate[/i] does. [i]Fate's[/i] strategy now is to say "yes" to all questions. We can assume, in this case, that [i]Hope's[/i] strategy is independent of the feedback she receives; simply because all the answers she receives will be "yes". It is sufficient to show that there exists a coloring of the boxes for which these answers are consistent but [i]Hope's[/i] strategy fails.

Following her strategy; suppose she indicates two boxes $A$ and $B$ at last. If $A, B$ were not inquired about before; then we can color $A, B$ as black and all other balls as white. If $A, B$ were inquired about before, then, say there exists a box $C$ which is not questioned with at least one of $A, B$. WLOG, let it be $A$; then color $A, C$ as black and all other boxes as white. Such a vertex $C$ exists otherwise we have at least $1+2002+2002=4005$ questions asked. Hence, [i]Hope's[/i] strategy does not necessarily work for $m \le 4004$ questions.

The answer is $4005$
Easy to see that $(1,2).....(1.2004),(2,3)......(2,2004)$ satisfies we choose two white balls.
If the answer $\leq 4004$ let the $2$ numbers is $1,2$ and if we say yes for the all questions $WLOG$ he didn't ask the one of the pairs from $(1,2).....(1,2004),(2,3)......(2,2004)$ let this $(1,x)$ and if $1,x$ are black our all answers will be yes too. Contradaction


Hi everyone,

Could someone please help me on this problem?

Hi everyone,

Could someone please help me with this?

1. The problem statement, all variables and given/known data

Newton's law of universal gravitation is represented by the following equation where F is the magnitude of the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance.
F = GMm/r2
Force has the SI units kg·m/s2. What are the SI units of the proportionality constant G

2. The attempt at a solution

Would it be m/kg s^2?

Thank you very much

You need a few more meters: on the right hand side you have two in the [b]denominator[/b] (without $ G$) and on the left hand side you have one in the [b]numerator[/b].

Thank you very much

In order to solve this problem, would I just use cross multiplication to solve for G?

Thank you very much


See here http://www.mathlinks.ro/Forum/viewtopic.php?p=17397#p17397  :wink:


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

[b]Evaluate :

$ \displaystyle\mathop {\lim }\limits_{k \to \infty } (\int\limits_0^k {\frac {{\ln (1 \plus{} \frac {x} {k})^x }} {{k.(1 \plus{} x)}}dx)}$

 :wink: [/b]

let me to see your solution.

Yes , you can use L'Hopital ;

but I just wonder also if there is another solution without using L'Hopital.

wait the guys. :wink:

The first step? Make the obvious substitution $ x\equal{}kt,dx\equal{}k\,dt$. The integral becomes $ \int_0^1 \frac{(\ln(1\plus{}t))^{kt}}{1\plus{}kt}\,dt$. For $ 0<t<1$, the numerator goes monotonically to zero and the denominator goes monotonically to $ \infty$. At zero, the integrand is bounded; $ \lim_{t\to 0^\plus{}}(\ln(1\plus{}t))^{kt}\equal{}\lim_{t\to 0^\plus{}}t^{kt}\equal{}1$, and it is at most 1 pointwise.

In summary, the integrand goes to zero uniformly on $ [\epsilon,1]$, and thus the limit of this part of the integral is zero. On $ [0,\epsilon]$, the integrand is at most $ 1$ and this part of the integral is at most $ \epsilon$. Letting $ \epsilon$ tend to zero, the limit of the integral is zero.

This can be done faster with some more advanced analysis; that estimate near zero is a special case of the argument for the Dominated Convergence Theorem.

[quote="jmerry"]

This can be done faster with some more advanced analysis; that estimate near zero is a special case of the argument for the Dominated Convergence Theorem.[/quote]

thanks jmerry , but could you explain more about this statement plz :)

Statement of the DCT: Let $ f_n$ tend pointwise to $ f$, with each $ f_n$ absolutely integrable. If we also have $ |f_n|<g$ with $ g$ (absolutely) integrable, $ \lim_n \int f_n\equal{}\int f$.
That could be almost any kind of integral, including infinite sums. It's also closely related to Fubini's theorem. In the usual Lebesgue context, we relax "pointwise" to "pointwise almost everywhere". For improper Riemann integrals, that's "pointwise except on a discrete set".
The idea of the proof is to look at $ \int |f_n\minus{}f|$, and split this into two parts. On one part, we make it small with the pointwise convergence. On the other, we make it small by comparing to $ g$ and making the region small. That is exactly what I did for this problem.

Thanks Jmerry ,

that a nice theorem in Lebesgue Integral .

let's wait to see if there is another solution.


[quote="GameBot"]What is the sum of the first 40 positive odd integers?[/quote]

$ 40^2\equal{}1600$.

how do you know that $ 40^2 = 1600 $ is the sum of first 40 positive integers

[hide="Formula"]The first n odd numbers are 1,3,5,7,9,...,2n-1. So, the average number is $\frac{2n-1+1}{2}=n$. Since there are n terms, the sum of the first n numbers is $\boxed{n^2}$[/hide]

[quote="bassamali01"]how is it $n^2$[/quote]

Prove it yourself.

Darn fine [hide="proof"]
The first $n$ odd positive integers are represented by $1, 3, 5, 7, ..., 2n-1$
We notice that we can pair up these integers such that they add up to $2n$. ($2n-1+1=2n-3+3=...=2n$)
There are $\frac{n}{2}$ such pairs, so the sum of the first $n$ positive odd integers is $\frac{n}{2}\cdot2n$.
[/hide]

What a silly answer.

I tried but I got $n=n$


MATHCOUNTS
search
AMC
AMC 8
geometry
geometric transformation
USAMTS

This is a request directed to middle schoolers only.  I am interested in know what middle school students would do if they ran this site.  So, the question:

What would you do with AoPS?  What would make this site more enjoyable and educational for you?

we should have like weekly contest ...
like test preparation....

That might attract more people to this site a

Great suggestion.  Unfortunately we don't have enough time to do something like that weekly.

We do sometimes hold informal MATHCOUNTS-style countdown round contests in Math Jams.  I am in fact preparing to schedule a few of those soon.

Alright

but (another suggestion) I wouldn't suggest making more recreational forum because, that will be the only place where kids would go.( I actually would like it if I could get rid of the fun factor(or whatever ) section for kids)

[quote]I agree with Shinwoo, if ppl doesn't like math, there's no point in making a "fun" forum for them because it doesn't help them in math at all. But if you truly like math, you'll come no matter how not "fun" the site is. The only suggestion is to tell more people about it, I know there must be lots of people who likes math out there.[/quote]

That's a beautiful mind

The purpose of the Games & Fun Factory is not simply to provide a place where students can discuss recreational topics, it is a place where students can discuss recreational topic [b]with other enthusiastic math students[/b].  It also serves to keep such discussion out of the actual math forums, something we had trouble with when the site first opened.

I think it's true that mathematical minds tend toward different sets of recreational activities than other minds.  It's almost certainly true that you will find more students on a site such as this interested in chess or go to a degree that they have become sophisticated players and enjoy seeking out other sophisticated players of their own age.

We might also at some point add entirely mathematical games to the site.  We would design these games to be educational themselves.

[quote="MCrawford"]
We might also at some point add entirely mathematical games to the site.  We would design these games to be educational themselves.[/quote]

yea, but most of these games just repeats over and over again so it gets really boring, it'll be cool if you can design games in which you don't have to do the same type of problems, like the game from the mathcount site, think it's called go figure.

[quote]like the game from the mathcount site, think it's called go figure.[/quote]

It gets really boring after awhile :D

[quote="LuCy4EvA"]I agree with Shinwoo, if ppl doesn't like math, there's no point in making a "fun" forum for them because it doesn't help them in math at all. But if you truly like math, you'll come no matter how not "fun" the site is. The only suggestion is to tell more people about it, I know there must be lots of people who likes math out there.[/quote]

But you have to realize, many people living in Wyoming, for example, might not be able to find any friends that like maths as much as they do. This is their place to socialize. I believe it actually helps peoples maths since it encourages them and makes them feel less lonely while doing maths.

[quote="LuCy4EvA"]
yea, but most of these games just repeats over and over again so it gets really boring, it'll be cool if you can design games in which you don't have to do the same type of problems, like the game from the mathcount site, think it's called go figure.[/quote]

If we ever create a game, you can be sure that it won't simply repeat problem types over and over.  We wouldn't even get started doing something like that if it were going to be that boring.

We have been through the math competition circuits as students and teachers and we feel we have a pretty good handle on how to select problems that are challenging and interesting.

[quote]i don't think so lucy, not quite.. well at least for me, at first....i basically hated math. Until i was really bored one day and decided to search for random sites on the internet. i found this site...but it wasn't the site that actually attracted me...  it was the fact that some of mua friends were posting on this site and i just wanted to keep up with stuff that's going on... so yea....

but then.... after a while, one day.....that moment.... somehow..... i discovered that math was something more than just odd and tiring formulas....it was actually something u could take time to think about and then go " ooo.. i get it now, so it's like this..... ! i actually get what it MEANS now."

from that moment and on... i discovered something new...something that was really fascinating and let's see....beautiful. I learned that math was not just meaningless work that you have to do becuz it's homework (i kno, i'm ashamed of how i thought of some things...), i realized that math was full of mysteries and so many answers that led to more questions. it's like walking down a down a deep hall, opening one door, only to find another, waiting to be opened...even though there always are new challenges and obstacles.

It was kinda like how life is sometimes....[/quote]

This sounds very interesting...

the reason I joined this website was to prepare for AMC8

but

like KBabe, I found the beauty in math, and I am now obssessed with it...

I bougth all kinds of math books(including AoPS text) and started to have fun!

[quote="KBabe"]
from that moment and on... i discovered something new...something that was really fascinating and let's see....beautiful.  I learned that math was not just meaningless work that you have to do becuz it's homework (i kno, i'm ashamed of how i thought of some things...), i realized that math was full of mysteries and so many answers that led to more questions. it's like walking down a down a deep hall, opening one door, only to find another, waiting to be opened...even though there always are new challenges and obstacles.

It was kinda like how life is sometimes....[/quote]

When I read this, two things come to mind.

First, mathematics is about exploration, just like life.  Your hallways of doors analogy is great.  I would add that you never know what shape the next hallway will take and there is no limit to the number of doors in the hallway.

Second, analogies are at the heart of mathematics.  You are right in seeing how mathematics is about the way you think.  It's about connecting ideas and we often do that through analogy.  Every real problem that can be solved by mathematics first undergoes a process of translation from words or ideas into a mathematical form that we can solve with equations or geometry or other tools.  The translation process is exactly what we do in English class when recognize similar relationships between words.  We make a leap of understanding that simplifies the way we think about a problem.

Indeed, we do this everywhere in life.

Since a lot of off-topic posting (and not only from middleschoolers!) was going on, I've split [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23645]that discussion[/url], and put it in the Round table forum.

[quote="MCrawford"]
We might also at some point add entirely mathematical games to the site.  We would design these games to be educational themselves.[/quote]
http://pbskids.org/cyberchase/games.html
That webpage is FULL of math games especially FOR KIDS.

I 'd like some interesting maths problems,not for test ,just for fun

[quote="shinwoo"]We should all turn in to poets

We could make great poems about math!!! :D[/quote]

why not?i don't like the poetry i study at school  :lol:

[quote="KBabe"][quote="LuCy4EvA"]I agree with Shinwoo, if ppl doesn't like math, there's no point in making a "fun" forum for them because it doesn't help them in math at all. But if you truly like math, you'll come no matter how not "fun" the site is. The only suggestion is to tell more people about it, I know there must be lots of people who likes math out there.[/quote]

sry this post is kinda long..

i don't think so lucy, not quite.. well at least for me, at first....i basically hated math. Until i was really bored one day and decided to search for random sites on the internet. i found this site...but it wasn't the site that actually attracted me...  :blush: it was the fact that some of mua friends were posting on this site and i just wanted to keep up with stuff that's going on... so yea....

but then.... after a while, one day.....that moment.... somehow..... i discovered that math was something more than just odd and tiring formulas....it was actually something u could take time to think about and then go " ooo.. i get it now, so it's like this..... ! i actually get what it MEANS now."

from that moment and on... i discovered something new...something that was really fascinating and let's see....beautiful.  I learned that math was not just meaningless work that you have to do becuz it's homework (i kno, i'm ashamed of how i thought of some things...), i realized that math was full of mysteries and so many answers that led to more questions. it's like walking down a down a deep hall, opening one door, only to find another, waiting to be opened...even though there always are new challenges and obstacles.

It was kinda like how life is sometimes....[/quote]

to Kristy:
[hide] How would you feel about your post now? what's with you and hating e-face and stuff?[/hide]

I think you should post a very challenging problem for each week and see who solves it :) Winners get their name posted up on the site for some sort of time. Last week's winners

I guess your question itself made a lot of discussion which in itself is fun.
I recommend you try making sections where students could put their problems and also their answers .
I also recomend youput up loads of puzzles realted to their study cource.

I think perhaps a long term goal would be to establish something like the USAMTS (USAMTS Junior!), except informal, as a small competition. Perhaps just like keeping score in the MathCounts countdown round math jams. There would be no reward, but it would be fun to do. Also perhaps post problems and have us pm you or something to simulate more of a target round or AMC8 or AMC10 type of thing. [hide]
The bottom line is: would you have found these games fun as a student? Keep that in mind, if you choose to create these math games.
[/hide]
Addionally, (in the looonng term) you could set up your own competition that would be at math counts nats level or something. This would be like the USAMTS in the sense that it would take place over a whole year. The winner would get like one free class or something. Whatever. Dont listent to me, you guys know whats best for AoPS and the Middle School math comunity.

Just my 1/50 of a dollar.

Even his user name needs some consideration. He needs to get a life.  ;)

i agree.  and he told us to get a life...

[quote="Lordofhell5"]you are all fauking geeks...... go get a fauking life you should all go out there ... grow your hair then get to FAUK!!!!!!!!! get a fauking decent job all you mutherfauking teaching bastarhds. =) Happy gai mathmatical gainess.[/quote]

What's wrong with being a geek? i think people who hate math are geeks because they are math-hating geeks  :mad:

i just pmed MCrawford...i hope that he will teach him a lesson. ;)


Asymptote
limit
trigonometry
algebra
binomial theorem
real analysis
real analysis solved

as in the topic, how to compute limit of (-1)^n*sin(sqrt(n^2+0.5n+1)*Pi)*sin(Pi*(n+1)/4n)? i know it converges to 1/2. first sin converges to +-1/sqrt(2), second to 1/sqrt(2), but how to prove this? first sin is the main problem for me, we get sin(infinity). can the expression in bracket be replaced by it's asymptote: x+1/4? i think yes, but how to prove this method is correct?

Paulusk, do you want to compute the limit:
\[
\displaystyle \lim_{n\to +\infty} (-1)^n\sin \left (\pi\sqrt{n^2+0.5n+1}\right )\sin \left ( \pi \frac {(n+1)} {4n} \right) ?
\]

yes! i think i found the proof of asymptote approach

Put
\[
(-1)^n \sin \pi \left (\sqrt {n^2+0.5n+1}\right )=(-1)^n \sin \pi \left (\sqrt {n^2+0.5n+1}-n+n \right ) =
\]
\[
=  \sin \pi \left (\sqrt {n^2+0.5n+1}-n \right )=
\sin \pi \displaystyle \frac {0.5n+1} {\sqrt {n^2+0.5n+1}+n}
\]
which goes $ \to \frac {\sqrt {2}} {2} $
Then all is clear.

Time for some more big-O, little-o thinking.

$\sqrt{n^2+0.5n+1}=n\sqrt{1+\frac1{2n}+o\left(\frac1n\right)} = n\left(1+\frac1{4n}+o\left(\frac1n\right)\right)$

(Here I used the binomial theorem: $(1+x)^{\alpha}=1+\alpha x + o(x)$ as $x\to 0$.)

So $\pi\sqrt{n^2+0.5n+1}=\pi n + \frac{\pi}4 + o(1)$

and $\sin\left(\pi\sqrt{n^2+0.5n+1}\right)=(-1)^n\frac{\sqrt{2}}2+o(1).$

Then $\sin\left(\frac{\pi(n+1)}{4n}\right) = \sin\left(\frac{\pi}4+o(1)\right)= \frac{\sqrt{2}}2+o(1)$

So put it all together to get

$(-1)^n\left((-1)^n\frac{\sqrt{2}}2+o(1)\right)\left(\frac{\sqrt{2}}2+o(1)\right) = \frac12+o(1).$

The limit is $\frac12.$


American Scholastic Mathematic Association Competitions... we take em every coupla months... ummm.. im dont think there are any practice ones around... *shrugs*

http://www.asan.com/asa/asma1.htm

Hope that's what you are looking for


Assuming it is drawn to scale:

The large triangle is a $ 9\minus{}12\minus{}15$ triangle while the small one is a $ 6\minus{}8\minus{}10$.  The area of the shaded region is thus $ 54\minus{}24\equal{}\boxed{30}$.


Let $ N$ be a positive integer. How many pairs of $ (m,\ n)$ are there with $ 1\leq m\leq 2N,\ 1\leq n\leq 2N$ such that a quadratic equtaion: $ x^2 \minus{} nx \plus{} m \equal{} 0$ has real root more than or equal to $ N$?

This seems a bit too easy...
[hide]
$ x^2 \minus{} nx \plus{} m \equal{} 0$

$ x \equal{} \frac {n \pm \sqrt {n^2 \minus{} 4m}}{2}$

If one root is greater than $ N$ then

$ n \plus{} \sqrt {n^2 \minus{} 4m} \ge 2N$, But since $ 1\le n \le N$ we can see that,

$ n \plus{} \sqrt {n^2 \minus{} 4m} \le N \plus{} \sqrt {N^2 \minus{} 4m} < 2N$

Hence there are no solutions?[/hide]

[hide="Slightly shorter"] Note that if the polynomial has real roots, they're both positive.  But their sum is $ n \leq N$.  So, this shows that there are no solutions whem $ m > 0$, $ 0 < n \leq N$.[/hide]

Sorry, there are some correction. $ N$ is a positive integer and $ 1\leq m\leq 2N, 1\leq n\leq 2N$.

kunny


Let $a_n$ be the number of nonempty subsets $S$ such that
(i) $S \subseteq {1,2,\ldots,n}$
(ii) all elements of $S$ have the same parity
(iii) each element $k \in S$ satisfies $k \ge 2|S|$, where $|S|$ is the number of elements in $|S|$
Prove that \[ a_{2m-1} = 2(F_{m+1} -1) \] and \[ a_{2m} = F_{m+3}-2 \] for all $m \ge 1$, where $F_n$ is the $n^{th}$ Fibonacci number.


This question comes from an old Senior Challenge UKMT paper ([url=http://www.wpr3.co.uk/UKMT/teachers/SMC05.html]here[/url]) that I was looking through. The question is Q18:

[b]A cube exactly fits inside a sphere and another sphere exactly fits inside this cube. What is the ratio of the volume of the smaller sphere to the ratio of the volume of the larger sphere?[/b]

A) $ 1 : 3\sqrt{3}$
B) $ 1 : 4$
C) $ 1 : 3$
D) $ 2 : 3$
E) $ 1: 2$

[hide="Answer"]A, $ 1 : 3\sqrt{3}$.[/hide]What I don't understand is how they came to this answer. I drew my own diagram and I thought the radius of the larger sphere was:

[hide="Radius of Larger Sphere"]$ \mbox{Radius} \equal{} \sqrt{2} \cdot \mbox{radius of smaller sphere}$[/hide]But this must be wrong, as when put into the ratio it doesn't match their answer. How do you find the radius of the larger sphere in terms of the smaller one, to get the desired ratio? Thanks for any help, I've been struggling with this for a little while.

[hide]Let $ R\equal{}$ radius of large sphere and $ r\equal{}$ radius of small sphere.  Also, $ s\equal{}$ side length of the cube.
For a cube to be inscribed in a sphere, all the vertices of the cube must be on the sphere.  Also, the cube and the sphere are concentric. So, you should be able to see that the space diagonal of the cube is a diameter of the sphere.

Thus, $ 2R\equal{}s\sqrt{3}$, so $ R\equal{} \frac{s \sqrt{3}}{2}$.

For a sphere to be incribed in a cube, the sphere must touch the midpoints of the cube's faces.  Since the sphere and the cube are concentric, we must have the side length of the cube is the diameter of the sphere.

Thus, $ 2r\equal{}s$, or $ r\equal{}\frac{s}{2}$.

Since we want the ratios of the volumes of the spheres, we want $ \frac{r^3}{R^3}\equal{}\left( \frac{r}{R} \right)^3\equal{}1: 3\sqrt{3}$.
[/hide]

OK, with your solution and a little more thought on my part, I now understand where I went wrong and how they came to the final solution. Thanks for the help, this one had me completely buggered for quite some time!


function
calculus
derivative
geometry
3D geometry
sphere
real analysis

Let $ f: R^{3}\longrightarrow R$ be a function such that $ f^{\minus{}1}(0)$$ \equal{}${$ {x\in R^{3}, \left|\left|x\right|\right|\equal{}1}$} Suppose that $ f$ has continuous partial derivatives of orders $ <\equal{}2$. Is there a $ y\in R^{3}$ with $ \left|\left|y\right|\right|\leq 1$ such that $ \frac{\partial^{2}f}{\partial x^{2}_{1}}(y)\plus{}\frac{\partial^{2}f}{\partial x^{2}_{2}}(y)\plus{}\frac{\partial^{2}f}{\partial x^{2}_{3}}(y)\geq 0$?

[quote="Γιώργος"]Let $ f: R^{3}\longrightarrow R$ be a function such that $ f^{ \minus{} 1}(0)$$ \equal{}${$ {x\in R^{3}, \left|\left|x\right|\right| \equal{} 1}$} Suppose that $ f$ has continuous partial derivatives of orders $ < \equal{} 2$. Is there a $ y\in R^{3}$ with $ \left|\left|y\right|\right|\leq 1$ such that $ \frac {\partial^{2}f}{\partial x^{2}_{1}}(y) \plus{} \frac {\partial^{2}f}{\partial x^{2}_{2}}(y) \plus{} \frac {\partial^{2}f}{\partial x^{2}_{3}}(y)\geq 0$?[/quote]
if I understand right your question , yes.
for example : let $ f\equal{}x_{1}^{2}\plus{}x_{2}^{2}\plus{}x_{3}^{2}\minus{}1$ then $ f^{\minus{}1}(0)$ is unit sphere :
then for all $ y$ , $ your sum ...  \equal{} 6 \geq 0$

Hmmmm, i think the answer is no, because consider $ f(x_{1},x_{2},x_{3})\equal{}1\minus{}x^{2}_{1}\minus{}x^{2}_{2}\minus{}x^{2}_{3}$. Then...

yes! hence answer is no. we have two example "my" and "your".
in my example for all $ y$ we have $ \geq$
and in your example for all $ y$ we have $ \leq$ hence answer on your problem depend on $ f$


geometry trapezoid rectangle trigonometry circumcircle symmetry ratio

Let $ ABCD$ be a trapezoid with $ AB\parallel{}CD$, $ AB\equal{}11$, $ BC\equal{}5$, $ CD\equal{}19$, and $ DA\equal{}7$. Bisectors of $ \angle A$ and $ \angle D$ meet at $ P$, and bisectors of $ \angle B$ and $ \angle C$ meet at $ Q$. What is the area of hexagon $ ABQCDP$?

$ \textbf{(A)}\ 28\sqrt{3}\qquad
\textbf{(B)}\ 30\sqrt{3}\qquad
\textbf{(C)}\ 32\sqrt{3}\qquad
\textbf{(D)}\ 35\sqrt{3}\qquad
\textbf{(E)}\ 36\sqrt{3}$

[hide]$ A_{ABQCDP}\equal{}A_{ABCD}\minus{}A_{APD}\minus{}A_{BQC}\equal{}A_{ABCD}\minus{}A_{A'PD}\minus{}A_{B'QC}$
$ \equal{}\frac{11\plus{}19}{2}\times h\minus{}\frac{7}{4}\times h\minus{}\frac{5}{4}\times h\equal{}12h$

Compute $ h$ by Pythagorean Theorem or Heron's Formula:

$ A_\triangle\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$
$ s\equal{}\frac{1}{2}(7\plus{}5\plus{}8)\equal{}10$
$ A_\triangle\equal{}\sqrt{10\cdot 3\cdot 5\cdot 2}\equal{}10\sqrt{3}$
$ h\equal{}\frac{A_\triangle}{4}\equal{}\frac{10\sqrt{3}}{4}\equal{}\frac{5\sqrt{3}}{2}$

$ A_{ABQCDP}\equal{}12h\equal{}30\sqrt{3}$[/hide]

I solved it basically the same way - the key is realizing that extending $ AP$ to intersect $ CD$ at $ A'$ gives you congruent triangles $ ADP$ and $ A'DP$, since $ DP$ bisects $ \angle ADA'$ and since $ \angle APD$ is a right angle (use the fact that $ \angle A \plus{}\angle D\equal{}180$).  And the area of $ A'DP$ is a lot easier to calculate, since you can get the area in terms of just the height.  Then you can get the area of $ BQC$ (the same way) and the area of the whole trapezoid in terms of the height also.  From there all you have to do is calculate the height of the trapezoid and you're good.

rcv, the way you found the height was really nice.  Pythagorean Theorem was a little messier.  :P

If you draw a line AX parallel to BC and intersecting DC at X, you will find an interesting triangle ADC 8-7-5, in which angle AXD (angle C) equals to 60 degree.

Project $ P$ onto $ AB, CD, DA$. The feet of the projections are the points of tangency of a circle with center $ P$. (The triangles formed by the angle bisectors and those perpendiculars come in congruent pairs!)

[hide="brute force"] Solve for the angles (do this by finding the height and dropping perpendiculars). Two miraculously turn out to be 60 and 120, so we get one 30-60-90 triangle (we see the two little triangles are right).

The other...we get arccos11/14 (i think)? Then, we can just use the half angle formula. This wasn't too hard to bash, but it was still ugly.[/hide]

We can solve for the height $ h$ of the trapezoid:

$ \sqrt {49 \minus{} h^2} \plus{} \sqrt {25 \minus{} h^2} \equal{} 19 \minus{} 11 \equal{} 8$.

After solving this equation, we obtain $ h \equal{} \frac {5}{2} \sqrt {3}$ (the negative $ h^2$ must be discarded).

Then we split the trapezoid into a rectangle and two right triangles.

We can establish that $ \cos \angle BAD \equal{} \frac {5}{14} \sqrt {3}$ and $ \cos \angle BCD \equal{} \frac {\sqrt {3}}{2}$. The next key observation is that the angle bisectors meet at right angles, and we can use the half-angle formulas to get the sines and obtain the areas of the two triangles that must be subtracted from the trapezoid.

Quite brute-forced, but I couldn't see any other solution =P

Let $ \angle PAD = x$, then $ \angle DAB = 2x$, $ \angle ADC = 180 - 2x$, and $ \angle ADP = 90 - x$. Hence $ \angle APD = 90$. Now let $ M$ and $ N$ be the midpoints of $ BC$ and $ DA$, respectively. It follows that $ M$ is the circumcenter of $ \triangle APD$. Then $ \angle PMD = 2\cdot \angle MAP = 2x$. So $ MP//BC//AB$. But that means that $ MP$ is the mid line of $ ABCD$. But by symmetry $ NQ$ is the mid line as well. So $ M$, $ P$, $ Q$, and $ D$ form a line.

Notice that $ MP$ is equal to the circumradius of $ \triangle APD$, but since it is a right triangle, that value is $ \frac {AD}{2}$. Now it follows that
\[ \frac {[APD]}{[ABCD]} = \frac {\frac{1}{2}\cdot \frac {1}{2} \cdot AD\cdot d(AB,CD)}{ MN\cdot d(AB,CD) } = \frac {AD}{2 AB + 2 CD}
\]
We have a symmetrical result for $ \triangle BCQ$. We can subtract those areas from the total area of the trapezoid to get the desired area. Hence
\[ \frac {[APDCQB]}{[ABCD]} = \frac {2AB + 2CD - BC - AD}{2AB + 2CD} = \frac {2\cdot 11 + 2\cdot 19 - 5 - 7}{2\cdot 11 + 2\cdot 19} = \frac {4}{5}
\]
Now construct $ B'$ on segment $ CD$ such that $ AB = B'C$. Since $ AB//B'C$ as well, $ ABCB'$ is a parallelogram. Hence $ AB' = 5$ and $ DB' = 19 - 11 = 8$. We can use heron's formula to get $ [AB'D] = 10\sqrt {3}$. Now
\[ \frac {[ABCD]}{[AB'D]} = \frac {AB + DC}{ A'D} = \frac {11 + 19}{8} = \frac {15}{4}
\]
We can finish the problem by multiplying those ratios:
\[ [APDCQB] = 10\sqrt {3}\cdot \frac {4}{5}\cdot\frac {15}{4} = 30\sqrt {3}
\]
Hence the answer is $ \boxed{B}$.

whoa lots of crazy trig o.o...

I extended AP and BQ similar to rcv's... Then find the height and you're done because $ \bigtriangleup ADA'$ is isosceles and $ DP\perp AA'$. similarly with other triangle so find half of the area of each of the isosceles triangles and subtract that from the total area... Except rcv has cool lines in his diagram T.T

[quote="rcv"][hide]$ A_{ABQCDP} \equal{} A_{ABCD} \minus{} A_{APD} \minus{} A_{BQC} \equal{} A_{ABCD} \minus{} A_{A'PD} \minus{} A_{B'QC}$
$ \equal{} \frac {11 \plus{} 19}{2}\times h \minus{} \frac {7}{4}\times h \minus{} \frac {5}{4}\times h \equal{} 12h$

Compute $ h$ by Pythagorean Theorem or Heron's Formula:

$ A_\triangle \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
$ s \equal{} \frac {1}{2}(7 \plus{} 5 \plus{} 8) \equal{} 10$
$ A_\triangle \equal{} \sqrt {10\cdot 3\cdot 5\cdot 2} \equal{} 10\sqrt {3}$
$ h \equal{} \frac {A_\triangle}{4} \equal{} \frac {10\sqrt {3}}{4} \equal{} \frac {5\sqrt {3}}{2}$

$ A_{ABQCDP} \equal{} 12h \equal{} 30\sqrt {3}$[/hide][/quote]

Where is the 5-7-8 triangle in the ABCD trapezoid?

[quote="cdymdcool"]
Where is the 5-7-8 triangle in the ABCD trapezoid?[/quote]If you cut out the $ 11\times h$ rectangle in the middle of the trapezoid, you get a $ 5$-$ 7$-$ 8$ triangle with height $ h$.

[quote="rcv"]$ A_{ABQCDP} \equal{} A_{ABCD} \minus{} A_{APD} \minus{} A_{BQC} \equal{} A_{ABCD} \minus{} A_{A'PD} \minus{} A_{B'QC}$
$ \equal{} \frac {11 \plus{} 19}{2}\times h \minus{} \frac {7}{4}\times h \minus{} \frac {5}{4}\times h \equal{} 12h$[/quote][list=1]
[*]Where did he get $ \frac {7}{4}$ and $ \frac {5}{4}$ from?

[*]Also, in the wiki, it says that the area of triangle $ ADP$ is given by $ R \cdot C \sin a \sin b$, where $ a$ is $ \angle ADP$ and $ b$ is $ \angle DAP$. Where did this area formula come from?[/list]

Edit: Never mind, I figured both out.

2. $ S\equal{}\frac{abc}{4R}\equal{}R\frac{a}{2R}\frac{b}{2R}c\equal{}Rc\sin a\sin b$

[quote="aidan"]
Then we split the trapezoid into a rectangle and two right triangles.
[/quote]
You can finish up a little nicer. In any right triangle, the distance from the midpoint of the hypotenuse to the right angle vertex is half the hypotenuse. So if R and S are the midpoints of BC and AD, then RQ=2.5 and PS=3.5, so QP=9, the result follows easily.

We first solve for the height $h$ of the trapezoid. By dropping perpendiculars from $A$ to $DC$ and $B$ to $DC$, we find that
\[ \sqrt{7^2 - h^2} + 1 + \sqrt{25 - h^2} = 19 \implies  \sqrt{49 - h^2} = 8 - \sqrt{25-h^2} \implies \\ 49 - h^2 = 89 - h^2 - 16 \sqrt{25-h^2} \implies  40 = 16 \sqrt{25-h^2} \implies \\ 5 = 2 \sqrt{25-h^2} \implies 25 = 100 - 4h^2 \implies h = \frac{5 \sqrt{3}}{2}.\]
Let $P_1$, $P_2$, $P_3$ be the feet of the perpendiculars from $P$ to $AB$, $AD$, and $DC$, respectively. Because $P$ lies on the bisectors of $\angle A$ and $\angle D$, $PP1 = PP2 = PP3$. But $PP1 + PP3 = h$, so $PP2 = \frac{h}{2}$, whence $[APD] = \frac{7h}{4}$. Similarly, we find that the length of the altitude from $Q$ to $BC$ is $\frac{h}{2}$, so $[QBC] = \frac{5h}{4}$. $[ABCD] = \frac{(11+19)h}{2} = 15h$, so our desired area is $[ABCD] - [APD] - [QBC] = 15h - \frac{7h}{4} - \frac{5h}{4} = 12h = 30 \sqrt{3}$, giving an answer of $\boxed{B}$.

[hide="Solution"]The area of the desired hexagon is just the area of the trapezoid minus the areas of triangles $APD$ and $BQC$. Since the height is $\frac{5\sqrt3}{2}$ as several have stated, the area of the trapezoid is just $15\cdot\frac{5\sqrt3}{2}=\frac{75\sqrt3}{2}$.

Since $\overline{AB} || \overline{CD}$, $\angle{APD}=180^{\circ}-(\angle{DAP}+\angle{ADP})=180^{\circ}-\frac12 (\angle{A}+\angle{D}) = 180^{\circ}-90^{\circ}=90^{\circ}$.
A similar argument yields the same for $\angle{BQC}$.

Let $\angle{ADP}=\alpha$. Then $[ADP]=\frac{AP\cdot DP}{2}=\frac{7\cos\alpha\cdot7\sin\alpha}{2}=
\frac{49\sin{2\alpha}}{4}=\frac{49\sin\angle{D}}{4} =\frac{49}{4}\cdot\frac{\frac{5\sqrt3}{2}}{7}=
\frac{35\sqrt3}{8}$.

[hide="Do the same for $\triangle{BQC}$"]
Let $\angle{BCQ}=\beta$. Then $[BCQ]=\frac{BQ\cdot CQ}{2}=\frac{5\cos\beta\cdot5\sin\beta}{2}=
\frac{25\sin{2\beta}}{4}=\frac{25\sin\angle{C}}{4} =\frac{25}{4}\cdot\frac{\frac{5\sqrt3}{2}}{5}=
\frac{25\sqrt3}{8}$.[/hide]

Therefore $[APD]+[BQC]=\frac{60\sqrt3}{8}=\frac{15\sqrt3}{2}$ and the area of that hexagon is $\frac{75\sqrt3}{2}-\frac{15\sqrt3}{2}=30\sqrt3$ or B.[/hide]

where did the 7 fourth and 5 fourth come from in first sol.

[hide="Synthetic Solution"]Since $P$ lies on the angle bisector of $\angle A$, it is equidistant from $AB$ and $AD$. Since $P$ is also on the angle bisector of $\angle D$, it is equidistant from $AD$ and $CD$. Thus, $P$ is equidistant from $AB$ and $CD$. But $AB$ and $CD$ are parallel, so the distance from $P$ to the bases is just $h/2$, where $h$ is the height. This also means that $P$ is $h/2$ away from $AD$. Similarly, $Q$ is $h/2$ away from the bases and $BC$. Now, all we have to do is compute $h$, find the area of the trapezoid, and subtract the areas of the two right triangles, since we know their altitudes. To find $h$, we drop perpendiculars from $A$ and $B$ to $CD$, hitting $CD$ at points $R$ and $S$, respectively. Letting $RD=x$, we obtain the following system of equations from the Pythagorean Theorem: $$x^2+h^2=49$$$$(8-x)^2+h^2=25.$$ Solving yields $x=\frac{11}{2}$ and $h=\frac{5\sqrt{3}}{2}$. Hence, the area of hexagon $ABQCDP$ is
\begin{align*}
[ABQCDP]=[ABCD]-[ADP]-[BCQ] &=\frac{1}{2}\left(\frac{5\sqrt{3}}{2}\right)\left(11+19\right)-\frac{1}{2}\left(\frac{5\sqrt{3}}{4}\right)\left(7\right)-\frac{1}{2}\left(\frac{5\sqrt{3}}{4}\right)\left(5\right)
\\&=\frac{75\sqrt{3}}{2}-\frac{35\sqrt{3}}{8}-\frac{25\sqrt{3}}{8}
\\&=\frac{75\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}
\\&=\frac{75\sqrt{3}}{2}-\frac{15\sqrt{3}}{2}
\\&=\frac{60\sqrt{3}}{2}
\\&=\boxed{30\sqrt{3}}.
\end{align*}
The answer is $\boxed{\textbf{(B)}}$.[/hide]


geometry
inequalities
integration
calculus
derivative
calculus computations

Let $t$ be a positive constant number.In $x-y$ plane, let $S(t)$ be the area of the figure surrounded by the curve $y=e^{t^2x},$ $x$ axis, $y$ axis and the line $x=\frac{1}{t}.$ For $t>0,$ prove that $S(t)>\frac{4}{3}.$

Of course $S(t)=\int_{0}^{\frac{1}{t}}ydx=\frac{e^t-1}{t^2}$. Now take $f: (0,\infty) \to {\mathbb R},f(s)=e^s-\frac{4}{3}s^2-1$, and after some analysis stuff we get $f(s)>0,\forall s\in(0,\infty)$. So $e^t-\frac{4}{3}t^2-1>0 \iff \frac{e^t-1}{t^2}>\frac{4}{3} \iff S(t)>4/3,t>0$.

Sorry for that how did you find the value of $s$ such that $f'(s)=0.$

who told you i used derivatives? ... i used taylor $e^s = \sum_{k=0}^{\infty}\frac{s^k}{k!}$.
is it good ?  :?

ok now let's make like that:
the left part is: $s(1-\frac{5}{6}s) + blabla ...$ where $blabla... >0$.
[b]1)[/b] if $s \in (0,\frac{6}{5})$ we get that $1-\frac{5}{6}>0$ so all left part is $>0$.
[b]2)[/b] if $s \in [\frac{6}{5},\infty) \Rightarrow s\geq\frac{6}{5}>1$ it is clear that $blabla... > \frac{5}{6}s^2$ so again left part is positive.

So, conclusion .... right part positive i.e. $e^s - \frac{4}{3}s^2 -1 >0,\forall s>0 \Rightarrow S(t)>\frac{4}{3},t>0$.
It's OK now ?

I can understand now.

Multumescu, jianuovidiu. :)

kunny


calculus
integration
trigonometry
calculus computations

For $ p > \minus{}1$, prove that $ \int_{0}^{\frac{\pi}{2}} \cos^{p} t \, \cos (p\minus{}2)t \, dt \equal{} \frac{p \, \pi}{2^{p\plus{}1}}$.

(1) $ I_{m,n} \equal{} \int \cos^m x \cos nx dx \equal{} \frac {\cos^m x \sin nx}{m \plus{} n} \plus{} \frac {m}{m \plus{} n}I_{m \minus{} 1, n \minus{} 1}$.

By integration by parts, set $ u \equal{} \cos^m x$ so that $ u' \equal{} \minus{} m \cos^{m \minus{} 1}x \sin x$ and $ v' \equal{} \cos nx$ so that $ v \equal{} \frac {\sin nx}{n}$. Note also that $ \cos (n \minus{} 1)x \minus{} \cos nx \cos x \equal{} \sin nx \sin x$

Hence we get
$ I_{m, n} \equal{} \cos^m x \cdot \frac {\sin nx}{n} \minus{} \int \left( \minus{} m \cos^{m \minus{} 1}x \sin x \right)\frac {\sin nx}{n}dx$

$ \equal{} \cos^m x \cdot \frac {\sin nx}{n} \plus{} \frac {m}{n}I_{m \minus{} 1,n \minus{} 1} \minus{} \frac {m}{n}I_{m,n}$

$ \implies I_{m,n} \equal{} \frac {\cos^m x \sin nx}{m \plus{} n} \plus{} \frac {m}{m \plus{} n}I_{m \minus{} 1, n \minus{} 1}$

(2) $ I_{m, n} \equal{} \int_{0}^{\frac {\pi}{2}} \cos^m x \cos nx dx \equal{} \frac {m(m \minus{} 1)}{m^2 \minus{} n^2}I_{m \minus{} 2, n}$

Integration by parts allows us to write:
$ I_{m,n} \equal{} \left[\cos^m x \cdot \frac {\sin nx}{n} \right]_{0}^{\frac {\pi}{2}} \plus{} \frac {m}{n} \int_{0}^{\frac {\pi}{2}} \cos^{m \minus{} 1} x \sin x \sin nx dx$

$ \equal{} 0 \plus{} \frac {m}{n} \int_{0}^{\frac {\pi}{2}} (\cos^{m \minus{} 1} x \sin x) \sin nx dx$

$ \equal{} \frac {m}{n} \left[\left[(\cos^{m \minus{} 1} x \sin x) \left( \minus{} \frac {\cos nx}{n} \right)\right]_{0}^{\frac {\pi}{2}} \minus{} \int_{0}^{\frac {\pi}{2}}{(m \minus{} 1)\cos^{m \minus{} 2} x ( \minus{} \sin^{2} x) \plus{} \cos^m x }\left( \minus{} \frac {\cos nx}{n}\right) dx \right]$

$ \equal{} \frac {m}{n}\left[0 \plus{} \frac {1}{n} \int_{0}^{\frac {\pi}{2}} {\cos^m x \cos nx \minus{} (m \minus{} 1) \cos^{m \minus{} 2} x \sin^2 x \cos nx}dx \right]$

$ \equal{} \frac {m}{n^2}I_{m,n} \minus{} \frac {m(m \minus{} 1)}{n^2} I_{m \minus{} 2, n} \plus{} \frac {m(m \minus{} 1)}{n^2}I_{m,n}$

Hence we get the result by solving for $ I_{m,n}$.

(3) $ I_{n,n} \equal{} \int_{0}^{\frac {\pi}{2}} \cos^n x \cos nx dx \equal{} \frac {\pi}{2^{n \plus{} 1}}$, whenever n is positive.

By identity (1) we have:
$ I_{n,n} \equal{} \left[\frac {\cos^n x \sin nx}{2n}\right]_{0}^{\frac {\pi}{2}} \plus{} \frac {1}{2}\int_{0}^{\frac {\pi}{2}} \cos^{n \minus{} 1} x \cos (n \minus{} 1)x dx$

$ \equal{} \frac {1}{2}I_{n \minus{} 1, n \minus{} 1}$

$ \equal{} \frac {1}{2} \left[\frac {1}{2} I_{n \minus{} 2, n \minus{} 2} \right]$

$ \equal{} \frac {1}{2^n}I_{0,0}$

$ \equal{} \frac {1}{2^n} \int_{0}^{\frac {\pi}{2}} dx \equal{} \frac {\pi}{2^{n \plus{} 1}}$

Now we have:

$ \int_{0}^{\frac {\pi}{2}}\cos^{p}t\,\cos (p \minus{} 2)t\, dt$

$ \equal{} \frac {p(p \minus{} 1)}{p^2 \minus{} (p \minus{} 2)^2} I_{p \minus{} 2, p \minus{} 2}$         by identity (2)

$ \equal{} \frac {p}{2^2} \cdot \frac {\pi}{2^{p \minus{} 1}}$             by identity (3)

$ \equal{} \frac {p\pi}{2^{p \plus{} 1}}$.

[quote="akech"]Sorry this is old but I am wondering if you had a shorter solution sos440.[/quote]

I have to apologize to you, for I used the title 'Easy Integral' just to attract some interest. In fact, I think this is not an easy problem at all.

Well, I think there is no solution simpler than yours, even when $p$ is an integer. So it would be much harder to prove this integral, concerning that the condition imposed $p$ is only $p > -1$, that is, $p$ need not be integral.

I exploited a very generalized version to solve this:

If $\alpha > -1$ and $\xi \in \mathbb{C}$, then $\binom{\alpha}{\xi} = \int_{-\frac{1}{2}}^{\frac{1}{2}} (1 + e^{2\pi it x})^{\alpha} e^{-2\pi i \xi x} \, dx$, where the binomial coefficient is extended with the identification $z! = \Gamma(1+z)$.

I don't know whether it is well-known or not, for I proved this without any related reference. You can find the proof [url=http://sos440.tistory.com/160]here[/url]. Of course, I guess this must have been known. Anyway, assuming this fact, we may plug $\alpha = p$ and $\xi = p - 1$. This gives

\begin{eqnarray*}
p & = & \int_{-\frac{1}{2}}^{\frac{1}{2}} (1 + e^{2\pi i x})^{p} e^{-2\pi i (p-1) x} \, dx \\
& = & \int_{-\frac{1}{2}}^{\frac{1}{2}} (2 \cos \pi x)^{p} e^{\pi i (2 - p) x} \, dx \\
& = & 2^{p + 1} \int_{0}^{\frac{1}{2}} \cos^{p} (\pi x) \cos (p -2) \pi x \, dx \\
& = & \frac{2^{p + 1}}{\pi} \int_{0}^{\frac{\pi}{2}} \cos^{p} t \cos (p -2) t\, dt.
\end{eqnarray*}

I still wonder if it is possible to find another approach.


Given $ a,b,c,d>0$,and $ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)\equal{}1$
Prove that:$ (2a\plus{}b\plus{}c)^2(2b\plus{}c\plus{}d)^2(2c\plus{}d\plus{}a)^2(2d\plus{}a\plus{}b)^2a^3b^3c^3d^3\le\frac{1}{16}$

This is also valid (same assumptions)
$ (a b c d)^{2m\plus{}1}((2 b\plus{}c\plus{}d)(a\plus{}2 c\plus{}d) (a\plus{}b\plus{}2 d)(2 a\plus{}b\plus{}c))^{2 n} \leq  16^{\minus{}1\minus{}2 m\plus{}2 n}$
for $ 1\leq  n\leq  m$

[quote="EulerIntegral"]This is also valid (same assumptions)
$ (a b c d)^{2m \plus{} 1}((2 b \plus{} c \plus{} d)(a \plus{} 2 c \plus{} d) (a \plus{} b \plus{} 2 d)(2 a \plus{} b \plus{} c))^{2 n} \leq 16^{ \minus{} 1 \minus{} 2 m \plus{} 2 n}$
for $ 1\leq n\leq m$[/quote]

Very nice generalization,could you please post your solution? :)

[quote="EulerIntegral"]This is also valid (same assumptions)
$ (a b c d)^{2m \plus{} 1}((2 b \plus{} c \plus{} d)(a \plus{} 2 c \plus{} d) (a \plus{} b \plus{} 2 d)(2 a \plus{} b \plus{} c))^{2 n} \leq 16^{ \minus{} 1 \minus{} 2 m \plus{} 2 n}$
for $ 1\leq n\leq m$[/quote]
Note that if this is really true for all $ m\equal{}n$, it implies the stronger

\[ abcd\prod(2 a \plus{} b \plus{} c)\leq \prod( a \plus{} b )^2\]

Can you then prove that directly?

:maybe:  :?:

[quote="spanferkel"]\[ abcd\prod(2 a \plus{} b \plus{} c)\leq \prod( a \plus{} b )^2\]
Can you then prove that directly?

:maybe:  :?:[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-97-unknown-author.html

[quote="hxy09"][quote="can_hang2007"][quote="spanferkel"]
\[ abcd\prod(2 a \plus{} b \plus{} c)\leq \prod( a \plus{} b )^2\]
Can you then prove that directly?

:maybe:  :?:[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-97-unknown-author.html[/quote]

I have no more thing to do than applaud :clap:
Thank you so much for your nice solution!!!

P.S.This hard problem was from Ren Qingchun,2006,2007 IMO gold medalist :)[/quote]
There is a little mistake with my typing. The term $ ac(2b\plus{}c\plus{}d)(2d\plus{}a\plus{}c)$ should be $ ac(2b\plus{}c\plus{}d)(2d\plus{}a\plus{}b).$ The other things are still okay. I am very sorry for this mistake. I have fixed it.

[quote="spanferkel"]
Note that if this is really true for all $ m \equal{} n$, it implies the stronger
\[ abcd\prod(2 a \plus{} b \plus{} c)\leq \prod( a \plus{} b )^2\]
[/quote]
After first ingenious Can's step
$ 4abcd\prod_{cyc}(2a \plus{} b \plus{} c)\leq\left(bd(2a \plus{} b \plus{} c)(2c \plus{} d \plus{} a) \plus{} ac(2b \plus{} c \plus{} d)(2d \plus{} a \plus{} b)\right)^2$
we need to prove that
$ 2\prod_{cyc}(a \plus{} b)\geq bd(2a \plus{} b \plus{} c)(2c \plus{} d \plus{} a) \plus{} ac(2b \plus{} c \plus{} d)(2d \plus{} a \plus{} b),$
which is equivalent to $ (ac \minus{} bd)^2 \plus{} \sum_{cyc}(a^2cd \minus{} abcd)\geq0,$ which is obvious.

SImilarly, we can prove
\[ abcd\prod(a \plus{} 2 b \plus{} c)\leq \prod( a \plus{} b )^2\ :\]
After the ingenious first  step (inspired by Can's)  :D
\[ 4abcd\prod_{cyc}(a \plus{} 2 b \plus{} c)\leq\left[ad(a \plus{} 2b \plus{} c)(b \plus{} 2 c \plus{} d) \plus{} bc(c \plus{} 2 d \plus{} a)(d \plus{} 2 a \plus{} b)\right]^2\]
we need to prove that
\[ 2\prod_{cyc}(a \plus{} b)\geq ad(a \plus{} 2b \plus{} c)(b \plus{} 2 c \plus{} d) \plus{} bc(c \plus{} 2 d \plus{} a)(d \plus{} 2 a \plus{} b),\]
which is equivalent to
\[ \sum_{cyc}(a^2bc \minus{} abcd) \plus{} \sum_{cyc}(abc ^2 \minus{} abcd)\equal{}\sum_{cyc}ab(c\minus{}d) ^2\geq0.\]
:lol:


In a chess tournament, every participant played with each other exactly once,
receiving 1 point for a win, 1/2 for a draw and 0 for a loss.
(a) Is it possible that for every player P, the sum of points of the players who were
beaten by P is greater than the sum of points of the players who beat P?
(b) Is it possible that for every player P, the first sum is less than the second one?

[hide="incomplete solution"]
Let the players be $ P_1,P_2,...,P_n$.  Let $ w(P_k)$ be the number of people that $ P_k$ beat, $ l(P_k)$ be the number of players $ P_k$ was beaten by, and let $ t(P_k)$ be the number of players $ P_k$ tied with.  Also, let $ s(P_k)$ be the numerical score of player $ P_k$.

Let $ f(P_k)$ and $ g(P_k)$ be the first sum and second sum, respectively, described in the problem for player $ P_k$.  We will prove that

$ \displaystyle\sum_{k \equal{} 1}^{n}f(P_k) \equal{} \displaystyle\sum_{k \equal{} 1}^{n}g(P_k)$.

First, notice that if all the players tied each other, then

$ \displaystyle\sum_{k \equal{} 1}^{n}f(P_k) \equal{} \displaystyle\sum_{k \equal{} 1}^{n}g(P_k) \equal{} 0$.

Now we show that if

$ \displaystyle\sum_{k \equal{} 1}^{n}f(P_k) \equal{} \displaystyle\sum_{k \equal{} 1}^{n}g(P_k)$

with players $ P_i, P_j$ tied, then changing the game between $ P_i,P_j$ in favor of $ P_i$ (call this transformataion $ A$) will increase both sums by the same value (keeping intact their equivalence).  Notice that $ s(P_i) \equal{} w(P_i) \plus{} 0.5t(P_i)$ and similarly for $ P_j$.  Thus, $ A$ increases $ s(P_i)$ by $ 0.5$ and decreases $ s(P_j)$ by $ 0.5$.  We see what this does to each of the sums.  It affects only $ f(P_k),g(P_k)$ of those who beat or were beaten by $ P_i,P_j$ in addition to $ f,g$ of $ P_i$ and $ P_j$.  For all $ P_k$ who beat $ P_i$, $ f(P_k)$ increases by $ 0.5$.  For all $ P_k$ who were beaten $ P_i$, $ g(P_k)$ increases by $ 0.5$. For all $ P_k$ who beat $ P_j$, $ f(P_k)$ decreases by $ 0.5$.  For all $ P_k$ who were beaten by $ P_j$, $ g(P_k)$ decreases by $ 0.5$.  Additionally, $ f(P_i)$ increases by $ s(P_j) \minus{} 0.5$, and $ g(P_j)$ increases by $ s(P_i) \plus{} 0.5$.

Thus, $ \displaystyle\sum_{k \equal{} 1}^{n}f(P_k)$  changes by $ 0.5l(P_i) \minus{} 0.5l(P_j) \plus{} s(P_j) \minus{} 0.5$ and $ \displaystyle\sum_{k \equal{} 1}^{n}g(P_k)$ changes by $ 0.5w(P_i) \minus{} 0.5w(P_j) \plus{} s(P_i) \plus{} 0.5$[/hide]

I seem to be stuck here, even though the solution worked out in my head.  Did I make a mistake anywhere?


limit
integration
calculus
calculus computations

Evaluate

$ \lim_{m\to\infty}\sum_{i=0}^{m}\frac{i^{n}}{m^{n}}$


topology
real analysis
real analysis unsolved

Does there exist a Hausdorff space which cannot be continiously and bijectively mapped onto compact space?

Edit: By compact space i meant here compact+Hausdorff.

Just to say that in the bicontinuous case, the answer is yes (you need complete regularity and the T1 axiom in order to insure the existence of a compactification). I've not thought about the continuous case yet ...


[quote="mathcountsfreak"]I have looked on the BBCode Guide, but when I wanted to quote someone who quoted another person, I couldn't. Please help!![/quote]

Is now fixed???


trigonometry
inequalities
inequalities unsolved

For any triangle prove that
\[ \frac{\sin{\alpha}}{\alpha}\plus{}\frac{\sin{\beta}}{\beta}\plus{}\frac{\sin{\gamma}}{\gamma}\geq \frac{3\sqrt{3}}{2\pi}\]

It is wrong. \frac{\sin{\alpha}}{\alpha}+\frac{\sin{\beta}}{\beta}+\frac{\sin{\gamma}}{\gamma}\leq \frac{3\sqrt{3}}{2\pi}

it is wrong.
$ \frac{\sin{\alpha}}{\alpha}\plus{}\frac{\sin{\beta}}{\beta}\plus{}\frac{\sin{\gamma}}{\gamma}\leq \frac{3\sqrt{3}}{2\pi}$

[quote="boroo12"]it is wrong.
$ \frac {\sin{\alpha}}{\alpha} \plus{} \frac {\sin{\beta}}{\beta} \plus{} \frac {\sin{\gamma}}{\gamma}\leq \frac {3\sqrt {3}}{2\pi}$[/quote]
So do you have any solutions?


calculus
integration
trigonometry
real analysis
real analysis unsolved

[quote="bookworm_vn"]very easy, it is similar to my post http://www.mathlinks.ro/Forum/viewtopic.php?t=170918[/quote]

really ?  :maybe: i would like to see your solution(s) then, i have not been able to solve these integrals to-date, [b]note [/b]that my upper limit is $ \frac{\pi}{2}$ and not $ \pi$...

[quote="misan"][quote="bookworm_vn"]very easy, it is similar to my post http://www.mathlinks.ro/Forum/viewtopic.php?t=170918[/quote]

really ?  :maybe: i would like to see your solution(s) then, i have not been able to solve these integrals to-date, [b]note [/b]that my upper limit is $ \frac {\pi}{2}$ and not $ \pi$...[/quote]

 :rotfl:  I have not solved them yet, in my post, using $ u\equal{}\tan\frac{x}{2}$....


It's now permitted to discuss the 2009 AIME II contest.

I might as well be the first one to post. LOL.


combinatorics
invariant
discrete system
Process

The numbers $ 1, 2, .. 2008$ are written on the board. Every second, Dr. Math erases some numbers $ a$, $ b$, $ c$, and $ a \plus{} b \plus{} c$, and replaces them with $ a \plus{} b$, $ b \plus{} c$, $ c \plus{} a$,

prove that he must stop after 10 minutes.

I think the idea is to search when a+b, b+c,c+a go over 2008

[quote=gghx][quote=khina]uh sum of squares is an invariant right? i think that was the idea[/quote]

Yes. Two invariants here: sum of terms and sum of squares of terms.

After that, Cauchy will give the result.[/quote]

can you explain a little plz?
i'm stuck  :(

[quote=thrasher64][hide = sol for tommy]
If $X$ is the sum of the terms at any point in the process and $Y$ is the sum of squares at any point then by cauchy we must have \[Y\cdot n\geq X^2,\] where $n$ is the number of terms at any point in the process. Then $n\geq \frac{X^2}{Y} \approx 1506$, so it is not possible for Dr.Math to erase $600$ numbers.
[/hide][/quote]

thanks a lot!


I find that when substituting in like that, using the middle one as "y" works more of the time, but both usually come out nicely, this is an example.

Hmmm look how I have progressed mathematically :lol: I just copied that from my blog entry from a while ago, and I just realized how random my substitution was. I think it would be better to do $x^2-10x-45$. :roll:

[quote="me@home"][quote="4everwise"][hide]Notice that each of the terms in the denominator has $x^2-10x-29$ in it. Thus, we can substitute $n=x^2-10x-29$ to get $\frac 1n + \frac{1}{n-16}=\frac{2}{n-40}$. Solving, $n=10$. Now, we can find $x$. $10=x^2-10x-39\implies (x-13)(x+3)=0$ and the only positive value of $x$ is $\boxed{013}$. [/hide][/quote]

The line that says:
$10=x^2-10x-39\implies (x-13)(x+3)=0$
is this a typo?
I am confused...[/quote]Oops, I typed the 10 again when I already considered it. It should say $10=x^2-10x-29$.  :)

chess64's solution is probably the easiest and best.

[quote="gotztahbeazn"]yea both are fine, although i have seen problems where 4everwise's way leads to no solutions whereas chess64's way will lead to a nice number.... im not sure why theres a difference though[/quote]Because test writers design their problems that way.  :mad:

I usually use that method as well. I just felt like being rebellious this time.  :D

[quote="JesusFreak197"]I don't see how his is better than 4everwise's.[/quote]

Me neither... :?

[hide=Solution]
Let $u=x^2-10x-45$. We want to solve the equation \[\frac{1}{u+16}+\frac{1}{u}=\frac{2}{u-24}.\] By cross-multiplying, we get \[(u+8)(u-24)=(u+16)u.\] We solve to get that $u=-6.$ The equation $x^2-10x-45=-6$ factors as $(x-13)(x+3)=0$ so the positive solution is $x=\boxed{13}.$[/hide]


Given a sequence $ \left(a_1,a_2,...,a_n\right)$ of $ n$ numbers.
Let $ t\leq\frac {n}2$ be an integer.

Let $ \left(b_1,b_2,...,b_n\right)$ be the sequence obtained from $ \left(a_1,a_2,...,a_n\right)$ by sorting the subsequence $ \left(a_1,a_2,...,a_{n \minus{} t}\right)$ in increasing order.
Let $ \left(c_1,c_2,...,c_n\right)$ be the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{n \minus{} 2t\plus{}1},b_{n \minus{} 2t \plus{} 2},...,b_{n}\right)$ in increasing order.
Let $ \left(d_1,d_2,...,d_n\right)$ be the sequence obtained from $ \left(c_1,c_2,...,c_n\right)$ by sorting the subsequence $ \left(c_1,c_2,...,c_{n \minus{} t}\right)$ in increasing order.

Prove that the sequence $ \left(d_1,d_2,...,d_n\right)$ is monotonically increasing.

The only problem is if at the beginning some numbers that should be between
the first $ n\minus{}t$ by size are placed somewhere at the last $ t$ positions. If that
is the case then these numbers will be put in the group of the first $ n\minus{}t$ elements
after the second step. Also in that step the last $ t$ positions are already correct.
After the third step sorting is obviously complete.
Sorry, this could be much more detailed...

I just noticed that this problem generalizes:

[color=blue][b]Theorem 1.[/b] Given a sequence $ \left(a_1,a_2,...,a_n\right)$ of $ n$ numbers.
Let $ t$ and $s$ be nonnegative integers such that $t + s \leq n$.

Let $ \left(b_1,b_2,...,b_n\right)$ be the sequence obtained from $ \left(a_1,a_2,...,a_n\right)$ by sorting the subsequence $ \left(a_1,a_2,...,a_{n-t}\right)$ in increasing order.
Let $ \left(c_1,c_2,...,c_n\right)$ be the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{n-t-s+1},b_{n-t-s+2},...,b_{n}\right)$ in increasing order.
Let $ \left(d_1,d_2,...,d_n\right)$ be the sequence obtained from $ \left(c_1,c_2,...,c_n\right)$ by sorting the subsequence $ \left(c_1,c_2,...,c_{n-s}\right)$ in increasing order.

Then, the sequence $ \left(d_1,d_2,...,d_n\right)$ is monotonically increasing.[/color]

[i]Remark.[/i] When I say that a sequence $\left(p_1,p_2,...,p_k\right)$ is "monotonically increasing", I mean that $p_i\leq p_j$ for any $i$ and $j$ from $\left\{1,2,...,k\right\}$ satisfying $i<j$. Some people call this "non-decreasing" and reserve the "monotonically increasing" notion for the stricter condition $p_i<p_j$.

The following proof is more or less a straightforward modification of shasha's solution, but written less heuristically and more formally. When I started writing it, I hoped for it to become more readable than shasha's, and in a certain sense it did - but at the cost of being 2 pages long.

[hide="Formal proof of Theorem 1"]
[i]Proof of Theorem 1.[/i] Let $i$ and $j$ be two elements of $\left\{1,2,...,n\right\}$ such that $i<j$. We are going to prove that $d_i\leq d_j$.

In order to show this, let us assume that $d_i>d_j$, and try to derive a contradiction.

By the definition of $ \left(d_1,d_2,...,d_n\right)$, the sequence $\left(d_1,d_2,...,d_{n-s}\right)$ is increasing. Hence, if $j\leq n-s$, then $d_i\leq d_j$ (because if $j\leq n-s$, then both $i$ and $j$ lie in $\left\{1,2,...,n-s\right\}$ (since $i<j\leq n-s$) and $i<j$), contradicting $d_i>d_j$. Thus, we cannot have $j\leq n-s$. Hence, $j>n-s$. Thus, $d_j=c_j$ (because the definition of $ \left(d_1,d_2,...,d_n\right)$ yields that the last $s$ elements of the sequence $ \left(d_1,d_2,...,d_n\right)$ are the same as the corresponding elements of the sequence $ \left(c_1,c_2,...,c_n\right)$).

Now let us prove that $i\leq n-s$. In fact, assume the contrary. Then, $i>n-s$. This yields $d_i=c_i$ (because the definition of $ \left(d_1,d_2,...,d_n\right)$ yields that the last $s$ elements of the sequence $ \left(d_1,d_2,...,d_n\right)$ are the same as the corresponding elements of the sequence $ \left(c_1,c_2,...,c_n\right)$). Hence, $c_i=d_i>d_j=c_j$. But by the definition of $ \left(c_1,c_2,...,c_n\right)$, the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ is increasing. Thus, $i\leq j$ yields $c_i\leq c_j$ (since $i>n-s$ yields $i\geq n-t-s+1$ and thus $j>i\geq n-t-s+1$, so that $i$ and $j$ both lie in $\left\{n-t-s+1,n-t-s+2,...,n\right\}$), contradicting $c_i>c_j$. This contradiction shows that our assumption $i>n-s$ is wrong. Hence, $i\leq n-s$ is proven.

By the definition of $ \left(d_1,d_2,...,d_n\right)$, the sequence $\left(d_1,d_2,...,d_{n-s}\right)$ is obtained by sorting the sequence $ \left(c_1,c_2,...,c_{n-s}\right)$ in increasing order. Thus, for every $k\in\left\{1,2,...,n-s\right\}$, the integer $d_k$ is one of the integers $c_1$, $c_2$, ..., $c_{n-s}$. Applying this to $k=i$ (since $i\leq n-s$), we conclude that the integer $d_i$ is one of the integers $c_1$, $c_2$, ..., $c_{n-s}$. In other words, there exists some $p\in\left\{1,2,...,n-s\right\}$ such that $d_i=c_p$. Consider this $p$. We have $c_p=d_i>d_j=c_j$.

Next we will show that $p\leq n-t-s$. In fact, assume the contrary. Then, $p>n-t-s$. But by the definition of $ \left(c_1,c_2,...,c_n\right)$, the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ is increasing. Thus, if $p\leq j$, then $c_p\leq c_j$ (since $p>n-t-s$ and $j>n-s\geq n-t-s$ show that both $p$ and $j$ lie in the set $\left\{n-t-s+1,n-t-s+2,...,n\right\}$), contradicting $c_p>c_j$. Hence, $p\leq j$ cannot be true. We therefore must have $p>j$. But this contradicts $p\leq n-s<j$. This contradiction shows that $p>n-t-s$ is impossible. Thus $p\leq n-t-s$.

The definition of $ \left(c_1,c_2,...,c_n\right)$ yields that the first $n-t-s$ elements of the sequence $ \left(c_1,c_2,...,c_n\right)$ are the same as the corresponding elements of the sequence $ \left(b_1,b_2,...,b_n\right)$. Since $p\leq n-t-s$, this yields $c_p=b_p$.

The definition of $ \left(c_1,c_2,...,c_n\right)$ yields that the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ is obtained by sorting the sequence $ \left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ in increasing order. Thus, the entries of the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ are the same as the entries of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ (up to order).

Now, $c_j$ is the $j-\left(n-t-s\right)$-th entry of the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$. Since this sequence is increasing, this shows that at least $j-\left(n-t-s\right)$ entries of this sequence are smaller or equal to $c_j$. Since $j-\left(n-t-s\right)\geq t+1$ (because $j>n-s$ and thus $j-\left(n-t-s\right)>t$), this yields that at least $t+1$ entries of the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ are smaller or equal to $c_j$. Since the entries of the sequence $\left(c_{n-t-s+1},c_{n-t-s+2},...,c_n\right)$ are the same as the entries of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ (up to order), this rewrites as follows: At least $t+1$ entries of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ are smaller or equal to $c_j$. Since $c_j<c_p=b_p$, all of these entries are smaller than $b_p$. We thus conclude that

[color=green][b](1)[/b][/color] at least $t+1$ entries of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ are smaller than $b_p$.

On the other hand, the sequence $\left(b_1,b_2,...,b_{n-t}\right)$ is increasing (by the definition of $ \left(b_1,b_2,...,b_n\right)$), and thus $b_p \leq b_w$ for every $w\in\left\{n-t-s+1,n-t-s+2,...,n-t\right\}$ (since $p\leq n-t-s<w$ for every such $w$). In other words, the $s$ entries $b_{n-t-s+1}$, $b_{n-t-s+2}$, ..., $b_{n-t}$ of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ are greater or equal to $b_p$. Thus,

[color=green][b](2)[/b][/color] at least $s$ entries of the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ are greater or equal to $b_p$.

From [color=green][b](1)[/b][/color] and [color=green][b](2)[/b][/color], we conclude that the sequence $\left(b_{n-t-s+1},b_{n-t-s+2},...,b_n\right)$ has at least $t+1+s$ entries (because no entry can be smaller than $b_p$ and greater or equal to $b_p$ at the same time). But this is absurd, since this sequence clearly has only $t+s$ entries. This contradiction proves that our assumption that $d_i>d_j$ was wrong.

Thus, we have $d_i\leq d_j$ whenever $i$ and $j$ are two elements of $\left\{1,2,...,n\right\}$ such that $i<j$. In other words, the sequence $\left(d_1,d_2,...,d_n\right)$ is monotonically increasing.
[/hide]

We can actually do a bit better. First, notice that we have:

[color=blue][b]Theorem 2.[/b] Let $n$, $ t$ and $s$ be nonnegative integers such that $t + s \leq n$.
Given a sequence $ \left(b_1,b_2,...,b_n\right)$ of $ n$ numbers such that the subsequence $\left(b_1,b_2,...,b_{n-t}\right)$ is increasing.

Let $ \left(c_1,c_2,...,c_n\right)$ be the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{n-t-s+1},b_{n-t-s+2},...,b_{n}\right)$ in increasing order.
Let $ \left(d_1,d_2,...,d_n\right)$ be the sequence obtained from $ \left(c_1,c_2,...,c_n\right)$ by sorting the subsequence $ \left(c_1,c_2,...,c_{n-s}\right)$ in increasing order.

Then, the sequence $ \left(d_1,d_2,...,d_n\right)$ is monotonically increasing.[/color]

[hide="Proof of Theorem 2"]
[i]Proof of Theorem 2.[/i] To prove Theorem 2, repeat the proof of Theorem 1 literally.
[/hide]

Now, here is the generalization of Theorem 3:

[color=blue][b]Theorem 3.[/b] Given a sequence $ \left(a_1,a_2,...,a_n\right)$ of $ n$ numbers.
Let $q$, $ t$ and $s$ be nonnegative integers such that $q + t + s \leq n$.

Let $ \left(b_1,b_2,...,b_n\right)$ be the sequence obtained from $ \left(a_1,a_2,...,a_n\right)$ by sorting the subsequence $ \left(a_1,a_2,...,a_{n-t}\right)$ in increasing order.
Let $ \left(c_1,c_2,...,c_n\right)$ be the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{q+1},b_{q+2},...,b_{n}\right)$ in increasing order.
Let $ \left(d_1,d_2,...,d_n\right)$ be the sequence obtained from $ \left(c_1,c_2,...,c_n\right)$ by sorting the subsequence $ \left(c_1,c_2,...,c_{n-s}\right)$ in increasing order.

Then, the sequence $ \left(d_1,d_2,...,d_n\right)$ is monotonically increasing.[/color]

[hide="Proof of Theorem 3"]
[i]Proof of Theorem 3.[/i] Let $\tau = n-q-s$. Then, $\tau + s = n-q \leq n$, so that $\tau \leq \tau+s \leq n$. Also, $\tau = n-q-s\geq t$ (since $q+t+s \leq n$), so that $n-\tau \leq n-t$.
The sequence $\left(b_1,b_2,...,b_{n-t}\right)$ is increasing (by the definition of $ \left(b_1,b_2,...,b_n\right)$). Thus, the sequence $\left(b_1,b_2,...,b_{n-\tau}\right)$ is increasing as well (because $n-\tau \leq n-t$, so that the sequence $\left(b_1,b_2,...,b_{n-\tau}\right)$ is a subsequence of the sequence $\left(b_1,b_2,...,b_{n-t}\right)$).

Also we know that $ \left(c_1,c_2,...,c_n\right)$ is the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{q+1},b_{q+2},...,b_{n}\right)$ in increasing order. Since $q=n-\tau-s$ (since $\tau = n-q-s$), this rewrites as follows: $ \left(c_1,c_2,...,c_n\right)$ is the sequence obtained from $ \left(b_1,b_2,...,b_n\right)$ by sorting the subsequence $ \left(b_{n-t-s+1},b_{n-t-s+2},...,b_{n}\right)$ in increasing order. Thus, we can apply Theorem 2 to $\tau$ instead of $t$, and conclude that the sequence $ \left(d_1,d_2,...,d_n\right)$ is monotonically increasing. Theorem 3 is proven.
[/hide]


induction
Summation
combinatorics
counting
IMO Shortlist

Prove that $\sum \frac{1}{i_1i_2 \ldots i_k} = n$ is taken over all non-empty subsets $\left\{i_1,i_2, \ldots, i_k\right\}$ of $\left\{1,2,\ldots,n\right\}$. (The $k$ is not fixed, so we are summing over all the $2^n-1$ possible nonempty subsets.)

For example, for [tex]n = 2[/tex], the non-empty subsets are {1}, {2}, {1,2}, and the sum is
[tex]\frac{1}{1} + \frac{1}{2} + \frac{1}{1 \cdot 2} = 2.[/tex]

I think mathematical induction works.
If the statement is true for $n=k$, let us consider $n=k+1$.
[color=red][[b]Moderator edit:[/b] The $k$ here is not the $k$ from the original problem, but a new variable. -- Darij][/color]
All non-empty subsets from $\{1,2,\dots,k\}$ are subsets from $\{1,2,\dots,k+1\}$, and the sum over those subsets is $k$.
We need to prove that the sum over the subsets which contain $k+1$ is 1.
For a set $S$ which contains $k+1$, $S\setminus\{k+1\}$ is a subset of $\{1,2,\dots,k\}$. Those kind of sets consist of (i)all non-empty subsets from $\{1,2,\dots,k\}$ and (ii)an empty set. Sum over part (i) is $k/(k+1)$, and part (ii) adds $1/(k+1)$ to the sum.

thus the statement is true of $n=k+1$.

Another approach is to expand the product $\prod_{i = 1}^n (1 + \frac{1}{i})$.


Let $a;b;c \in R$ and satisfying $a^{2}+b^{2}+c^{2}=3$
Find max $P=a+b+c-abc$

actually the cases come down to
1) if $a\geqslant b\geqslant c\geqslant0$ such that $a^{2}+b^{2}+c^{2}=3$ find max
  $P_{1}=a+b+c-abc$
2) if $a\geqslant b\geqslant0,\ c\geqslant0$ such that $a^{2}+b^{2}+c^{2}=3$ find max
  $P_{2}=a+b-c+abc$
3) if $a\geqslant0,\ b\geqslant c\geqslant0$ such that $a^{2}+b^{2}+c^{2}=3$ find max
  $P_{3}=a-b-c+abc$
4) if $a\geqslant b\geqslant c\geqslant0$ such that $a^{2}+b^{2}+c^{2}=3$ find max
  $P_{4}=abc-a-b-c$
and $P=\max\{P_{1},P_{2},P_{3},P_{4}\}$

take $t_{0}\in\left(0,\frac{1}{3}\right)$ such that
$3t_{0}^{3}-11t_{0}^{2}+7t_{0}-1=0$
Now we're going to show that
$P=\sqrt{6-2t_{0}}+\frac{t_{0}}{4}(1-t_{0})=P_{0}=2.5\cdots>\frac{5}{2}$
In fact, take $a=b=\sqrt{\frac{3-t_{0}}{2}}, c=\sqrt{t_{0}}$ we have
$P\geqslant P_{2}=a+b-c+abc=P_{0}$
On the other hand, let's consider the four cases.

[b]Case 1:[/b] with Schur Inequality we have
$a+b+c-abc\\ \leqslant a+b+c-\frac{4}{9}(a+b+c)(ab+bc+ca)+\frac{1}{9}(a+b+c)^{3}\\ =\frac{1}{9}(a+b+c)[15-(a+b+c)^{2}]$
for $\sqrt{3}\leqslant a+b+c\leqslant 3$ we have
$\frac{1}{9}(a+b+c)[15-(a+b+c)^{2}]\leqslant\frac{10}{9}\sqrt{5}<P_{0}$
So we have $P_{1}<P_{0}$

[b]Case 2:[/b]
$a+b-c+abc\\ \leqslant\sqrt{2(a^{2}+b^{2})}-c+\frac{a^{2}+b^{2}}{2}\cdot c\\ =\sqrt{6-2c^{2}}+\frac{c-c^{3}}{2}$
denote
$f(c)=\sqrt{6-2c^{2}}+\frac{c-c^{3}}{2}\ (c\in[0,\sqrt{3}])$
we have
$f'(c)=\frac{1}{2}-\frac{3c^{2}}{2}-\frac{2c}{\sqrt{6-2c^{2}}}$
And it's easy to prove
$f'(c)>0\ (0\leqslant c<\sqrt{t_{0}})\\ f'(\sqrt{t_{0}})=0\\ f'(c)<0\ (\sqrt{t_{0}}<c\leqslant 3)$
so
$f_{\max}(c)=f(\sqrt{t_{0}})=P_{0}$
that means
$P_{1}\leqslant P_{0}$

[b]Case 3:[/b]
$a-b-c-abc\leqslant a\leqslant\sqrt{3}<P_{0}$
so $P_{3}<P_{0}$

[b]Case 4:[/b]
$abc-a-b-c\leqslant abc\leqslant\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{3/2}=1<P_{0}$
so $P_{4}<P_{0}$

Then we'll get
$P=\max\{P_{1},P_{2},P_{3},P_{4}\}\leqslant P_{0}$
Till now we get the conclusion.

[quote="bilarev"]Can't we just use Lagrange multipliers. I think it is easy.[/quote]
That's right. If we use higher method, it's easy. Actually I think we should solve it with computer instead of by ourselves.


analytic geometry
graphing lines
slope
integration
calculus
calculus computations

A slope field is defined by $y'=\frac{1}{2}\left( \frac{y}{x}+\frac{x}{y}\right)$

A curve in this field passes through $(2,1)$. Find its $x$ intercept.

$2yy' = \frac{y^{2}}{x}+x$

Let $v = y^{2}$.  Then $v' = 2yy'$ and we have

$v' = \frac{v}{x}+x \implies$
$\frac{v'}{x}-\frac{v}{x^{2}}= 1 \implies$
$\int \left( \frac{v'}{x}-\frac{v}{x^{2}}\right) \, dx = \int dx \implies$
$\frac{v}{x}= x+C \implies$
$v = x^{2}+Cx \implies$
$y^{2}= x^{2}+Cx$

We are given the piont

$2^{2}= C+1 \implies C = 3$

The $x$ intercepts are then given by $x = \boxed{0,-3}$.


function
algebra
floor function
number theory unsolved
number theory

A prime number $ p$ is given. A sequence of positive integers $ a_1,a_2,...$ satisfies the relation:

$ a_{n\plus{}1}\equal{}a_n\plus{}pf(\sqrt[p]{a_n})$ for $ n\equal{}1,2,3,...$, where $ f(x)$ is the floor function of $ x$.

Prove that there is a term of the sequence which is the $ p$th power of an integer.

we take $ h(a^p + r) = (a,r)$ where $ 0\le r < (a + 1)^{p} - a^p$

we take $ h(a_n) = (b_n,r_n)$
for $ a_n$ such that $ r_n\neq 0$
[quote]if $ r_n + pbn < (b_n + 1)^p - b_n^p$ we have $ (b_{n + 1},r_{n + 1}) = (b_n,r_n + pb_n)$
if not we have $ (b_{n + 1},r_{n + 1}) = (b_n + 1,r_n - 1)$[/quote]
so $ \exists h_n > 0$ such that $ f(a_{n + h_n}) = (b_n + 1,r_n - 1)$
then $ \exists g_n > 0$ such that $ f(a_{n + g_n}) = (b_n + r_n,0)$
so $ a_{n + g_n} = (b_n + r_n)^p$

and we find also that $ \#\{n\in\mathbb{N}: \ f(\sqrt[p]{a_n})\in \mathbb{N}\}=+\infty$
besaus if $ S=\{n\in\mathbb{N}: \ f(\sqrt[p]{a_n})\in \mathbb{N}\}$ is finite and $ m=max(S)$
then $ m+1+g_{m+1}\in S$ and $ m+1+g_{m+1}>max(S)$ (impossible)

[quote="aviateurpilot"]
if not we have $ (b_{n \plus{} 1},r_{n \plus{} 1}) \equal{} (b_n \plus{} 1,r_n \minus{} 1)$[/quote]

Hmm... why is this?


probability
algebra
polynomial
algebra unsolved

probability well known:
let $p= \sum_0^n a_ib^i$ a prime number and its expansion in base $b$.
Prove that the polynomial $P(X)= \sum_0^n a_iX^i$ is irreducible in $Z[X]$

Compare to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32826[/url]  :)


[quote="ehsan2004"]Baron Minchausen talked to a mathematician. Baron said that in his country from any town one can reach any other town by a road. Also, if one makes a circular trip from any town, one passes through an odd number of other towns. By this, as an answer to the mathematician's question, baron said that each town is counted as many times as it is passed through. Baron also added that the same number of roads start at each town in his country, except for the town where he was born, at which a smaller number of roads. Then the mathematician said that baron lied. How did he conclude that ?[/quote]

Okay, what should I say... Consider the graph G whose vertices are the towns of the baron's country; let two vertices be connected by an edge if and only if the two corresponding towns are connected by a road. Then, the condition that from any town one can reach any other town by a road is equivalent to saying that the graph G is connected. The condition that, if one makes a circular trip from any town, one passes through an odd number of other towns, is equivalent to saying that all cycles of G have even length. Hence, the graph G is bipartite, i. e. one can color its vertices with the colors red and blue such that no two adjacent vertices have the same color (this can be easily proved: color the graph step by step, first coloring an arbitrary vertex red, then coloring all vertices adjacent to this vertex blue, then coloring all vertices adjacent to these new vertices red, etc., until all vertices have got colors; it will never happen that one vertex gets two different colors, since all cycles have even length).

Now let the number of all red vertices of the graph G be a and the number of all blue vertices be b.

The last condition of the problem, namely that the same number of roads start at each town in the baron's country, except for the town where he was born, at which a smaller number of roads starts, can be rewritten as follows: All vertices of the graph G have the same degree V, except of one vertex, which has a degree W such that W < V. (This latter vertex corresponds to the town where the baron was born.) We can WLOG assume that this latter vertex with degree W is red. Then, the graph G has a - 1 red vertices with degree V and one red vertex with degree W, so the sum of the degrees of all red vertices is (a - 1) V + W. On the other hand, the graph G has b blue vertices, all of them having the degree V; hence, the sum of the degrees of all blue vertices is bV.

Now, since the graph G is bipartite, every of its edges must connect one red and one blue vertex. Hence, the number of edges equals to the sum of the degrees of all red vertices and to the sum of the degrees of all blue vertices. Since the sum of the degrees of all red vertices is (a - 1) V + W and the sum of the degrees of all blue vertices is bV, we thus have (a - 1) V + W = bV. Hence, W = bV - (a - 1) V = (b - a + 1) V. Therefore, the number W must be divisible by the number V. But this is impossible, since W > 0 (as the graph G is connected), but W < V.

Contradiction. So the baron lied.

  darij

:huh:  :blush:  ;) darij grinberg you are very good  :D thanks very much


geometry
USAMTS
inequalities
triangle inequality

Old USAMTS problem:
Prove that it is impossible for a triangle to have altitudes measuring 4, 7, and 10 units.

[hide="Hint"]Prove the following lemma first: If $h_{a}$, $h_{b}$, $h_{c}$ are the altitudes of a triangle, then $\frac{1}{h_{a}}+\frac{1}{h_{b}}\geq\frac{1}{h_{c}}$[/hide]

[hide] well just set the area constant, at say n/2.  This would then put the side lengths at n/4, n/7, and n/10. and n/10 + n/7 < n/4  so by the triangle inequality this is imposssible.[/hide]

[hide]$[ABC]=ah_{a}/2$, so
$a+b>c \implies 2[ABC]/h_{a}+2[ABC]/h_{b}>2[ABC]/h_{c}$
$\implies 1/h_{a}+1/h_{b}>1/h_{c}$.

$1/10+1/7<1/4$, so it is impossible for a triangle to have altitudes 10,7 and 4.[/hide]


inequalities
Functional Analysis
real analysis
real analysis unsolved

Suppose we have the following:
$X$ is a Banach space and $x_{1},\ldots,x_{n}\in S_{X}$, (i.e. $\parallel x_{i}\parallel = 1$). There exists $K \geq 1$ such that $\parallel \sum_{i=1}^{n}a_{i}x_{i}\parallel \geq (1/K) \sum_{i=1}^{n}|a_{i}|$ for any $(a_{i}) \subset \mathbb{R}$. Furthermore, there exist $z \in S_{X}$ and $\varepsilon>0$ such that $d(z, S_{[x_{i}]_{i=1}^{n}}) > \varepsilon$ where $[x_{i}]_{i=1}^{n}$ denotes the span of $x_{i}$'s.

I am trying to show the following but without success so far:
There exists $L \geq 1$ such that for any $a \in \mathbb{R}$ and $(b_{j}) \subset \mathbb{R}$, $\parallel az+\sum_{j=1}^{n}b_{j}x_{j}\parallel \geq (1/L) (|a|+\sum_{j=1}^{n}|b_{j}|)$.

Please help.

Let $Y$ be the subspace of $X$ with the basis $\{z, x_{1}, \dots, x_{n}\}$. We can define a norm on $Y$ by
\[\left|az+\sum_{j=1}^{n}b_{j}x_{j}\right| = (|a|+\sum_{j=1}^{n}|b_{j}|) \]
Since $Y$ is finite-dimensional, this norm is comparable to the norm inherited from $X$, as required.

Perhaps you meant to add that $L$ should depend only on $K$ and $\varepsilon$, but not on $n$. That would be a more interesting question.

Yes indeed I should have mentioned that $L$ should depend only on $K$ and $\varepsilon$. Any suggestions for handling that case?

Combine the inequalities $\parallel az+\sum_{j=1}^{n}b_{j}x_{j}\parallel \ge \varepsilon |a|$ and $\parallel az+\sum_{j=1}^{n}b_{j}x_{j}\parallel \ge \frac{1}{K}\sum_{j=1}^{n}|b_{j}|-|a|$.

Combining the two inequalities gives me something like $\parallel az+\sum_{j=1}^{n}b_{j}x_{j}\parallel \geq \frac{1}{2K}\sum_{j=1}^{n}|b_{j}|+\frac{\varepsilon-1}{2}|a|$. What if $\varepsilon \leq 1$?

Oh yes...Then I get $\parallel az+\sum_{j=1}^{n}b_{j}x_{j}\parallel \geq (1/L) (|a|+\sum_{j=1}^{n}|b_{j}|)$ where $L = \frac{K(2+\varepsilon)}{\varepsilon}\geq 1$.
Thank you.


linear algebra
matrix
number theory
least common multiple
algebra
polynomial
superior algebra

Is there a 4x4 matrix over the field $ \mathbb{Q}(\omega)$ of order 36?
Prove or disprove.

Here $ \omega$ is the cubic root of unity.

Let K be the algebraic closure of $ \mathbb{Q}(\omega)$. If $ A^{36}\equal{}I_4$ then $ A$ is diagonalizable over K; let $ Spectrum(A)\equal{}\{\lambda_i\}$ ; moreover $ \lambda_i^{\alpha_i}\equal{}1$ where $ lcm((\alpha_i)_i)\equal{}36$. Thus $ \alpha_1\equal{}9,18$ or $ 36$.
Case 1: $ \alpha_1\equal{}9$. the minimal polynomial of $ \lambda_1$  is $ x^3\minus{}\omega$ or $ x^3\minus{}\omega^2$. $ \alpha_1$ has $ 2$ conjugates (which are also eigenvalues of $ A$).
Case 2: $ \alpha_1\equal{}18$. the minimal polynomial of $ \lambda_1$  is $ x^3\plus{}\omega$ or $ x^3\plus{}\omega^2$. $ \alpha_1$ has $ 2$ conjugates.
Case 3: $ \alpha_1\equal{}36$. the minimal polynomial of $ \lambda_1$  is $ x^6\plus{}\omega$ or $ x^6\plus{}\omega^2$. $ \alpha_1$ has $ 5$ conjugates. This case is impossible.
In the first $ 2$ cases $ 4|\alpha_2$ and the minimal polynomial of $ \alpha_2$ is of degree $ \geq{2}$.
Conclusion: necessarily $ n\geq{5}$.
Remark: There exists a solution for $ n\equal{}5$: the companion matrix of the polynomial $ (x^3\minus{}\omega)(x^2\plus{}1)$.


function
number theory proposed
number theory

given $ f(n,k)\equal{} 2^{2n\minus{}1}k^2 \plus{} 2^{n\minus{}1}k \plus{} 1$ for $ n,k$ - positive integers, I need to decide:

- is there any $ m$ s.t. $ f(n,k)\equal{} 5^m$

- is there any $ m$ s.t. $ f(n,k)\equal{} 9^m$

could you give some hints?
thank you

[quote="knoppix"]given $ f(n,k) \equal{} 2^{2n \minus{} 1}k^2 \plus{} 2^{n \minus{} 1}k \plus{} 1$ for $ n,k$ - positive integers, I need to decide:

- is there any $ m$ s.t. $ f(n,k) \equal{} 5^m$

- is there any $ m$ s.t. $ f(n,k) \equal{} 9^m$

could you give some hints?
thank you[/quote]

 [hide="Hint"]
Let $ x\equal{}2^{n\minus{}1}k$[/hide]


geometry
3D geometry
function
Support
linear algebra
matrix
integration

$Q=]0;1[^{n}$ the unit cube of $\mathbb{R}^{n}$, and $B$ the ball with unit volume. Prove that for any $C^{\infty}$ function $f: \mathbb{R}^{n}\to \mathbb{R}$, strictly positive inside $Q$, whose support is $Q$, there exist a $C^{1}$ diffeomorphism $\phi: Q \to B$ such that its jacobian matrix satisfies:
$|J_{\phi}(x)| = \frac{f(x)}{\int_{Q}f(y)dy}$
for all $x \in Q$. Moreover, one can assume that the jacobian matrix is always upper triangular, with positive entries on the diagonal.

PS: with this lemma, (and with a trick), [url=http://planetmath.org/encyclopedia/SobolevInequality.html]sobolev inequality[/url] becomes a piece of cake  :)

"Moreover" gives away the whole idea of construction :) . [hide]Consider $n=2$ to simplify notation. Also assume $\int_{Q}f=1$. If $\phi(x)=(\phi_{1}(x_{1},x_{2}),\phi_{2}(x_{2}))$ is the map we want, then $\phi_{2}'(x_{2})=\int_{0}^{1}f(t_{1},x_{2})dt_{1}$. This tells us what $\phi_{2}$ is. Also, we must have $\phi_{1}'(x_{1},x_{2})=\frac{f(x_{1},x_{2})}{\int_{0}^{1}f(t_{1},x_{2})dt_{1}}$, which determines $\phi_{1}$. [/hide]

I believe J. Moser proved this back in the 1960s. More recently, it was discovered that the conclusion is no longer true if $\phi$ is only required to be bi-Lipschitz and $f\in C(Q)$ is bounded away from $0$ and $\infty$. It is an open problem to characterize the Jacobians of bi-Lipschitz maps.


Have I missed something?  :?  Does it hold at all? And what's the point of such post in the Olympiad Section?  :?

Assume $L>M$ and put $\epsilon=\frac{L-M}{2}$.  Do we get a contradiction?

[quote="Marian Minar"]Show for M and L real numbers, if L = M + e, for all e > 0 (e real number, read "epsilon"), that L is less than or equal to M. I can't belive I can't show this, but I can't! Someone get me started before I brake down in tears. MM.[/quote]

Should it be L<M+e for all e>0?

then it can be proved by contradiction as follows:
Suppose L>M, take e = L-M>0,
then we have L<M+e = M+L-M = L
i.e. L<L, which cannot happen. Hence L=<M

The keyboard was confused: change
L = M + e    to
L <= M + e.

The question reads:

Show for M and L real numbers, if L <= M + e, for all e > 0 (e real number, read "epsilon"), that L is less than or equal to M

Sorry for the confusion.

Marian.

[quote="Marian Minar"]Show for M and L real numbers, if L <= M + e, for all e > 0 (e real number, read "epsilon"), that L is less than or equal to M[/quote]
We have L-M <= e for all positive e
If L-M = d > 0 this contradicts the above (for e=d/2), so we must have L-M<=0, QeD?

I'm afraid there are better forums for questions of this level... :?

Sorry to have wasted everyones' time. The steps in the last proof helped. It really was trivial, huh?

M.


decimal representation
perfect cube
number theory
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Prove that all numbers of the sequence \[ \frac{107811}{3}, \quad \frac{110778111}{3}, \frac{111077781111}{3}, \quad \ldots \] are exact cubes.

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

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[b]Happy Problem-Solving![/b] :)

We let $t_n$ be the $n$th term of the sequence shown to get $t_n=\frac{\overbrace{1\dots1}^n0\overbrace{7\dots7}^n8\overbrace{1\dots1}^{n+1}}{3}$

We know that $10^n-1=\overbrace{9...9}^{n}$ so we can rewrite $t_n$ as $t_n=\frac{\frac{10^n-1}{9}10^{2n+3}+\frac{7(10^n-1)}{9}10^{n+2}+8\cdot10^{n+1}+\frac{10^{n+1}-1}{9}}{3}$

We can get rid of the $9$s to get $t_n=\frac{10^{3n+3}-10^{2n+3}+7\cdot10^{2n+2}-7\cdot10^{n+2}+72\cdot10^{n+1}+10^{n+1}-1}{27}$

We can combine like terms to get $t_n=\frac{1000\cdot10^{3n}-300\cdot10^{2n}+30\cdot10^n-1}{27}$

We can regroup powers of $10$ to get $t_n=\frac{10^{3n+3}-3\cdot10^{2n+2}+3\cdot10^{n+1}-1}{27}$

We take the cube root because we need to prove that $t_n$ is a perfect cube so $t_n=(\frac{10^{n+1}-1}{3})^3$

We need to prove that $\frac{10^{n+1}-1}{3}$ is an integer to make sure that its cube is a perfect cube.

We know that $10^{n+1}-1\equiv 1^{n+1}-1\equiv 1-1\equiv 0\pmod3$ so $\frac{10^{n+1}-1}{3}$ is an integer and $t_n=(\frac{10^{n+1}-1}{3})^3$ is a perfect cube. :)