geometry
rectangle
combinatorics proposed
combinatorics

We have an $n \times n$ board, and in every square there is an integer. The sum of all integers on the board is $0$. We define an action on a square where the integer in the square is decreased by the number of neighbouring squares, and the number inside each of the neighbouring squares is increased by 1. Determine if there exists $n\geq 2$ such that we can turn all the integers into zeros in a finite number of actions.

What is "neighbouring squares"? Do its share one side? Can its share one vertex?

Uh...
Here's one:
 2 -2  2
-2  4 -2
 2 -2  2
Use action on each corner to get
0 -1 0
-1 4 -1
0 -1 0
Use action on center to get all 0s.
This example fulfills the criteria, so therefore some square like that exists.
Are you sure it was that simple?

Simple... unless neighbors means sharing a vertex. Please clarify on this.

Yeah, you're right, I forgot to add this. The neighbouring squares are the ones that share the side. And I should have also added "Determine if there exists n\geq 2 such that we can turn all the integers into zeros in a finite number of actions [b]for arbitrary combinations of integers on the board[/b]".
It's not that there exzists an n that for some combination we can do this. You must determine if it is possible for all combinations satisfying the assumptions :wink:

Now we can smell the answer is "no"...  :D

I think it is hard but why this problem was No.4-_-;;;
can anyone post solution?

The problem is not that hard. I solved it on the contest, so it should be easy :wink: Well, for now I know 3 complete proofs, and one of them is I think very clever (not mine unfortunately). I'll post 2 hints for the standard solution:

[hide="Small Hint"]
Try to find a good invariable :wink: [/hide]

[hide="Huge Hint"]
Look at the longest diagonal[/hide]

[hide="Full Solution"]
We'll prove that it is not possible for every combination of integers. Let's produce a counterexample. First see that by an action we cannot change the parity of the sum of the numbers on the longest diagonal. To prove this we only need some cases to be done (what's happening when we operate on a square on the diagonal, a square not on the long diagonal etc.). So we can produce a counterexample. Take any odd integer $ k$. Let us have a board with $ k$ on the long diagonal, $ \minus{}k$ which is not on the long diagonal and all the other integers we set as zeros. Then it's obvious that after a finite number of actions the sum of the integers on the long diagonal is still going to be odd, thus a board with only zeros will never occur due to a contradiction to the parity of the sum of the integers on the long diagonal :wink:
[/hide]

[hide="Hint for solution 2"]
Try to find a more complicated invariable mod2 working separetely for $ n\equal{}2k$ and $ n\equal{}2k\plus{}1$[/hide]

[hide="Hint for solution by Piotr Szcześniak"]
Work separetely for $ n\equal{}2k$ and $ n\equal{}2k\plus{}1$. Try to produce a contradiction putting only 1's and -1's into the board (for one case put only one 2 instead of 1)[/hide] :wink:

And can you please tell me why it is so weird for the problem to have the number 4  :lol:

i thought number 5 was much easier than this...but.....

i'm so ashamed that I could not to solve this simple one :blush:

Nice, now try the same for a rectangle $ n$ on $ m$ :P It is solvable, although much harder.

Well, the third solution works also for rectangles :wink:


[quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Prove that:
$ 2.\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {a^2}{c^2 \plus{} ca \plus{} ab} \plus{} \frac {b^2}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {c^2}{b^2 \plus{} bc \plus{} ca} \plus{} 1$[/quote]

It's very interesting  :P

Using cauchy-schwarzt , we get :
$ RHS \equal{} \sum \frac{ a^2(b^2\plus{}ca\plus{}ab) }{ (c^2\plus{}ca\plus{}ab)(b^2\plus{}ca\plus{}ab)  } \le \sum \frac{a^2(b^2\plus{}ca\plus{}ab)  }{ (\sum ab)^2}$
$ \equal{} \frac {\sum a^2b^2 \plus{} \sum a^3(b \plus{} c)}{(\sum ab)^2} \plus{} 1$
Beside ,we have :$ LHS \equal{} 2\frac {\sum a^2 }{\sum bc} \ge \frac {\sum a^2b^2 \plus{} \sum a^3(b \plus{} c)}{(\sum ab)^2} \plus{} 1$ because it's equivalent to $ \sum ab(a \minus{} b)^2 \ge 0$

[quote="Vincent Gilbert"]Using cauchy-schwarzt , we get :
$ RHS \equal{} \sum \frac { a^2(b^2 \plus{} ca \plus{} ab) }{ (c^2 \plus{} ca \plus{} ab)(b^2 \plus{} ca \plus{} ab) } \le \sum \frac {a^2(b^2 \plus{} ca \plus{} ab) }{ (\sum ab)^2}$
$ \equal{} \frac {\sum a^2b^2 \plus{} \sum a^3(b \plus{} c)}{(\sum ab)^2} \plus{} 1$
Beside ,we have :$ LHS \equal{} 2\frac {\sum a^2 }{\sum bc} \ge \frac {\sum a^2b^2 \plus{} \sum a^3(b \plus{} c)}{(\sum ab)^2} \plus{} 1$ because it's equivalent to $ \sum ab(a \minus{} b)^2 \ge 0$[/quote]

It's same my solution  :)


function
limit
real analysis
real analysis unsolved

Prove the following theorem  :D :

" Let $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$ a function with the following property : exists $C>0$ such that \[ \left\|f(x+y)-f(x)-f(y)\right\| \leq C, \; \forall x,y \in \mathbb{R}^m \] then exists an unique $T \in \mathcal{L}(\mathbb{R}^m;\mathbb{R}^n)$  such that $sup\left\{\left\|T(x)-f(x)\right\|; x \in \mathbb{R}^m \right\}< + \infty$ "

(Hint: prove that exists $g(x) = lim_{k \rightarrow \infty} \dfrac{f(2^kx)}{2^k}$.)

:(  please.... $%Error. "HUGE" is a bad command.
Help!!!$

:?   $\|g(x)-f(x)\|\leq\ C$ ?
Someone can prove that $g$ is continous at $0$, I think that it is true.

[quote="tipe"]:?   $\|g(x)-f(x)\|\leq\ C$ ?
[/quote]
You are right. I have modified my post.

[quote="tipe"]then exists an unique $T \in \mathcal{L}(\mathbb{R}^m;\mathbb{R}^n)$[/quote]
What do you mean by $\mathcal L(X,Y)$ here? All linear bounded operators from $X$ to $Y$?

Here $\mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ means the set of all linear operators from $\mathbb{R}^m$ to $\mathbb{R}^n$.

... Which is exactly the same as what liyi thought, in the finite dimensional case at stake.

Uniqueness is easy:
Suppose there are two linear operators $T_1,T_2$ satisfying the condition. Let $\delta(x) = \|T_1(x)-T_2(x)\|$ then $n\delta(x) = \|T_1(nx)-T_2(nx)\|$ which is bounded for all $n$, hence $\delta(x)=0$, which means that $T_1=T_2$.

So the difficulty is the existence. It is not difficult to see that $g(x+y)=g(x)+g(y)$, but I don't know how to show that $g(\alpha x)=\alpha g(x)$.

Yes, I prove this too, the difficult part is that $g(\alpha x)= \alpha x$   :)  , but if we prove that $g$ is continous at the origin the desired result folows?   :huh:


trigonometry
function
vector
algebra unsolved
algebra

[quote="modularmarc101"]Find min $ f$ and max $ f$ where $ x$ and $ y$ are real numbers and

$ f(x,y) \equal{} 2\sin{x}\cos{y} \plus{} 3\sin{x}\sin{y} \plus{} 6\cos{x}$[/quote]

Upper and lower bounds of $ a\sin(x)\plus{}b\cos(x)$ are $ \pm\sqrt{a^2\plus{}b^2}$

So, for a given $ y$, upper and lower bounds of $ f(x,y)$ are $ \pm\sqrt{36\plus{}(2\cos(y)\plus{}3\sin(y))^2}$

And since $ 0\leq (2\cos(y)\plus{}3\sin(y))^2 <(\sqrt{2^2\plus{}3^2})^2\equal{}13$ :

$ \minus{}7\leq f(x,y)\leq 7$
And these bounds may obviously be reached.

[quote="modularmarc101"]I think there's an easier solution with vectors but thanks anyways (even though I can't understand you're solution). But the answer is correct  :)[/quote]

What part of my solution dont you understand ?

oh nvm I see it. Thanks  :)  nice solution.. I just wouldn't have known that

Upper and lower bounds of  are  $ a\sin{x} \plus{} b\cos{x}$ are $ \pm\sqrt{a^2 \plus{} b^2}$.

[quote="modularmarc101"][quote="pco"][quote="modularmarc101"]Find min $ f$ and max $ f$ where $ x$ and $ y$ are real numbers and

$ f(x,y) \equal{} 2\sin{x}\cos{y} \plus{} 3\sin{x}\sin{y} \plus{} 6\cos{x}$[/quote]
...
And since $ 0\leq (2\cos(y) \plus{} 3\sin(y))^2 < (\sqrt {2^2 \plus{} 3^2})^2 \equal{} 13$ :

...
[/quote][/quote]

As I said in the first line, upper and lower bounds of $ a\sin(x)\plus{}b\cos(x)$ are $ \pm\sqrt{a^2\plus{}b^2}$

So lower and upper bounds of $ 2\cos(y) \plus{} 3\sin(y)$ are $ \pm\sqrt{2^2\plus{}3^2}$

So lower and upper bounds of $ (2\cos(y) \plus{} 3\sin(y))^2$ are $ 0$ and $ (\sqrt{2^2\plus{}3^2})^2$


Xy plane.
Starting at the origin, there is a spiral made of 1-unit-long segments joined end to end. The angle between the first two is 179 degrees, the next angle is 178, the next 177, etc. The angles always curve counterclockwise. Where does the curve first intersect itself?

You can find the numbers of the two segments that intersect, or even the exact point if you can!

Say $t_i=\displaystyle\frac{i(i+1)}{2}$ is the $i$th triangular number.
Each point $P_i$ can be denoted by
$P_i=\left(\displaystyle\sum_{i=0}^{n-1}\cos(t_i) , \displaystyle\sum_{i=0}^{n-1}\sin(t_i)\right)$
Now we say $S_i$ is the segment between $P_i$ and $P_{i+1}$
So, the equation of the line $S_i$ is $\displaystyle\frac{y-\displaystyle\sum_{i=0}^{n-1}\sin(t_i)}{x-\displaystyle\sum_{i=0}^{n-1}\cos(t_i)}=\displaystyle\frac{\displaystyle\sum_{i=0}^{n}\sin(t_i)-\displaystyle\sum_{i=0}^{n-1}\sin(t_i)}{\displaystyle\sum_{i=0}^{n}\cos(t_i)-\displaystyle\sum_{i=0}^{n-1}\cos(t_i)}=\displaystyle\frac{\sin(t_n)}{\cos(t_n)}=\tan(t_n)$

$\Rightarrow y = \tan(t_n)\left(x-\displaystyle\sum_{i=0}^{n-1}\cos(t_i)\right) + \displaystyle\sum_{i=0}^{n-1}\sin(t_i)$ given $x\in \left[\displaystyle\sum_{i=0}^{n-1}\cos(t_i),\displaystyle\sum_{i=0}^{n}\cos(t_i)\right]$

This started out a good idea, but its not looking so much anymore. Um, I'll leave it up here, but I can't fix/finish it because I have to go. If it helps anyone you're welcome, if it doesn't, sorry for posting.

Construct the spiral segment by segment.  At each step, consider the polygon formed by the spiral (connecting the beginning and end).

In the nth step, [we count 1 segment as the 0th step] the angles of the polygon are (180-j) for j = 1..n, plus 2 more angles, and we have n+2 sides.  We know that for the polygon to be convex, it must have sum of angles as 180(n-2), so that if it has different, we are done.  The most an angle can be is 180.  So we want to find minimal n so that $|\sum_{j=1}^n 180-j    -  180(n-2) | < 360$.

Expanding, we see that this is $|360 - \frac{n(n+1)}{2}| < 360$, which is true when $n(n+1) > 720$, or when $n = 27$.

--
i am in rush, so sorry if wrong / mis calculate


calculus function algebra polynomial MATHCOUNTS geometry algorithm

hmmm...so i was wondering which calculator to buy. the ti-89 or ti-83 plus silver edition. or what. im leaning towards the 89, even though theyll most likely be giving that away at national mathcounts. just say the pros and cons and stuff. and if you have an 86 or just a regular 83 plus. still post

ive nevar used either for an fyi. but i do know how to use ti-83 plus. im leaning to 89 because i can use it better later on also.

why arent people responding :!:  :?:

P.S.
i really should go practice my violin. hmmm

TI-89 is far more advanced that TI-83 Plus. I use TI-83 myself (since that is my school standard), but I know some ppl with TI-89's. They are particularly helpful for calculus, since I believe they can differentiate/integrate equations in variables and indefinite integrals.

The TI-89 were created specifically for people doing calculus, just as TI-86 leans towards stats. So I would recommend the TI-89. It's more accurate and powerful than the TI-83, silver edition or not. Right now I have a TI-86, but I've used all three before.

[quote="Syntax Error"]who voted for se?[/quote]
I did. I like se more because the calculus functions can be added on to se, plus a lot other things.
I think se is enough for middle schooler to early high schooler.

Syntax Error...

I wouldn't be so sure about getting an '89 next year.

Let me explain:

In 2001, they gave out '89s.  I was expecting that when I went in 2002.  Instead, they gave out '83+ses.  Then, when I went in 2003, they gave out '89s.  So, following the pattern, next year they'll give out '83+ Silver Editions again.

I think they've implemented this "alternating" thing so that people who go two years in a row don't get a duplicate.  Maybe?

But why get a Silver Edition?  It's about $50 more, which is a lot for a bit more memory.  I bought a regular TI-83+ for school, and it worked fine.  I thought of Silver as an extravagance, since no one ought to need THAT much memory...
I own a regular 83+, a silver 83+, and an 89.  One I bought for school, and two I got from mathcounts nationals.  I find that, while the TI-89 has a better processor and more advanced functions, in general it's kind of clunky.  Also, it's a lot harder to program on it, and most schools prefer the 83+ anyway.  But if you're going to take Calculus, I've heard it can be helpful.  We'll see.
And then there's the cost difference.  My parents wouldn't shell out $150 (or more?) for an '89, and I would say this is true of most people.  I felt guilty even needing the $80 for an '83+.  So money may also be a factor.

the 89 is way more powerful, but unless you will be taking a calc ap class this year, i dont advise you to buy it. why? First, in most of the pre-calc classes ive taken (algebra 1,2 geometry, trig/precalc/discrete.. etc.) we cannot use the 89 on a majority of tests. i also want to mention the fact that an 89 tends to be to powerful for competitions, ie it can expand, factor polynomials , solve equations and systems of equations, it can find mods, etc. i know some people at my school who use these functions on every competition, not knowing the math behind them. of course, if you already know all the tehniques to solve expressions, algebra and the like it wont help you, but ultimately a lot of people feel encouraged to buy and use the 89 in an unfair way.

[quote] also want to mention the fact that an 89 tends to be to powerful for competitions, ie it can expand, factor polynomials , solve equations and systems of equations[/quote]

whoa there...

so you're saying that it can solve even non-linear system of equations? like x+y+z = 7, xy+yz+zx=-2, and xyz = 12? And it can factor polynomials of degree >= 3? If so, then why even allow them on the AMC?

Personally, I like the TI-89 a lot better.  Although the S.E. isn't bad, it's sort of limited in its capacities even if you download programs that will give you the functions for the TI-89.  The TI-89 has a much higher resolution (100x160 or something close to that) while the TI-83 plus se has a pathetic resolution of 64x96, which definitely can't graph 3-d stuff very well.

-interesting_move

the 89 can factor polynomials of any degree, and also solve any systems of equations with ne amount of variables (as long as the solutions are nice, i.e. there arent an infinite number of sols, etc...). usually the amc questions are 89 proof so that no one with an 89 would really have an advantage. there are even questions where people using a graphing calculator would get the question wrong if just using the calc.

The people writing the AMC exams know the capabilities of calculators and the capabilities of problem solvers, and they make sure that no problem can be solved with calculator skill and not ingenuity. In my opinion, they do a good job of making their problems calculator-proof.

[quote="Mindspa"]Try not using any calculator, there's no reason for doing that actually.[/quote]
Yes, and meke a handicap for yourself...
I think the question ultimately comes down to one's moral and pride in their math skill.

i have an 89 and im in pre-calc, though i am doing calculus as well... as people have said, its really good for calc because of differentiation and integration.... also, the factoring thing is awesome :D . oh ya, and theres just one minor downside... unlike the 83, it does not show your entire expression, because it only has one command line (i think thats what you call it) that extends a long way... but that is easily ignored. TI-89 is a +, especially when MC gives it out for FREE! heh.

how do you solve multivariable non-linear system of equations on the ti-89?

It probably won't do it -- there aren't nice algorithms for doing that, so even if it could (using the solve function just as normal), at its processor speed it would take forever.

You might be surprised how compilicated a system of equations an 89 can solve.
anyway...

solve(eqn1 and eqn2 and eqn3) where eqn1,2,3 are the equations in your system, is the syntax for solving 3 equations, just keep adding "and eqn4 and eqn5" etc. for a larger system.

I have owned an 89 for 5 years.  I have used it very much during those 5 years.  I also know that there are not nice ways of finding solutions for most non-linear sets of equations.  I also know that the processor on an 89 is quite slow, especially compared to a computer.  (I once wrote a program on my 89 and ran it for nearly 8 hours at school, where it calculated 10,000 trials of something.  When I re-wrote the program on Mathematica at home, it did 100,000 trials in under 1 second.)

insert a space, then the letters "a" "n" and "d" and then insert another space.  Then at the end you have to indicate all of the variables you want to solve for.

Or, to do it the hard way, press [2nd] [MATH] 8 8.  (2nd-math-8 is the test menu).

i have a TI and am trying to download some programs but keeping getting the message invalid format, can anyone help?

If any of you know me, you would know that I know how to use calculators fairly well ;)

I have used all the calculators available on the market and I actually HIGHLY RECOMMEND

the HP 49G+

It takes a bit of adjustment from the TI-style of calculators.  The TI-89 is very good for most purposes, but the HP is hands down a far superior calculator once you know how everything works.

(and yes JBL, I have turned my back on the TI-89 since graduation ;), though I still have my trusty TI-89 with me and still develop some alternative programs for them )

I don't believe you. I don't think anything is superior to the 89 unless you're trying to do statistics or somat.

I think if anyone knows what calculators are good for what, Jack is probably him.

Are you sure the programs you're downloading are written for the calculator you're trying to download them to?

Also, if you're using a Ti-83+, you might end up downloading applications in the format .8xk.  Unfortunately, the TI-83+ won't accept those files, so you have to "rename" the program to a .app file.  That is to say, if you want to transfer the application "PUZZPACK", you need to rename "puzzpack.8xk" to "puzzpack.app".  It's weird.

[quote="Tare"]Um you really shouldn't have that problem if you were to use the TI official uplink software, in fact I think it's just much safer to use that then to just go change the extention like *snaps* that. (To send normal applications, I believe go to Link then Applications and Certificates or something like that and if there are two options you're on the right track. Just select the top one (which I believe was also names Applications and Certificates but I'm not a 100% positive on that one) and you shouldn't even worry about it.)

Of course I don't recall ever downloading .8xk files so I may be wrong...I just assumed TI wasn't dumb as that :)[/quote]

No, I've tried sending .8xk files, and the "TI official uplink software" wouldn't accept it.  (It said something like "invalid file format".)  At least, my version of the software wouldn't accept the file.  I noticed the program was looking for .app files, so I figured, "hey, why not?" and I changed the file format by renaming it.  Then I tried it again, and it worked.


function
absolute value
algebra proposed
algebra

Let $ f: \mathbb{Q}\to\mathbb{R}$ satisfy the conditions:

$ \star\;1: f(0) = 0$

$ \star\;2: f(\alpha) > 0$ if $ \alpha\not = 0$

$ \star\;3: f(\alpha.\beta) = f(\alpha)f(\beta)$

$ \star\;4: f(\alpha + \beta)\le f(\alpha) + f(\beta)$

$ \star\;5: f(n)\le 2007,\forall n\in\mathbb{Z}$

Prove that:

$ 1\;,f(\alph + \beta) = \max\{f(\alpha),f(\beta)\}$ if $ \alpha\not = \beta$

$ 2$  Find $ f$

:maybe:

Condition 4 reduces to $ f$ always being nonnegative... was that intentional?

Part 1 is wrong. Take $ f$ to be $ 1$ at all other points besides $ 0$. This is valid, but then $ \alpha \equal{} 1$, $ \beta \equal{} 0$, contradicts what we're supposed to prove.

I'm so sorry :blush: . I have a misstake when i posted this problem [code]\alpha--->\alph[/code]

Now, it was edited, please solve it  :)

I still get part 1 being wrong. With the solution of $ f$ I gave above, which still holds, putting $ \alpha \equal{} \minus{}\beta \neq 0$ gives $ f(0) \equal{} \max\{f(\alpha), f(\minus{}\alpha)\} \equal{} 1$.

OK, i will trying to edit it. It is a problem from my friend, maybe i have some mistake when i copied it to forum  :D

What did you have solution for past 2  :?:

[quote="N.T.TUAN"]Those conditions said that $ f$ is an absolute value on $ \mathbb{Q}$.  :lol:[/quote]

Do you mean $ f(x)\equal{}\text{constant}$ N.T.TUAN. If you think that, i think you are wrong  :)

this is a special case of proof of the Ostrowski theorem for all norms on $ \mathbb Q$.

[quote="N.T.TUAN"]Please see in ''Algebraic Number Theory'' of J. Neukirch.  :lol:[/quote]

Where can i see this book,N.T.TUAN  :D . I don't think $ f(x)\equal{}\text{contsant}$

The first 4 properties say that $ f$ is a multiplicative valuation on $ \mathbb Q$. All of them are known (sometimes called Ostrowski's theorem):
Every such $ f$ fulfills $ f(x) \equal{} \| x \|_p^s$ for some $ s>0$ and $ p \in \{2,3,5,7,11, ... \infty \}$ an extended prime (and $ s\leq 1$ if $ p\equal{}\infty$). Here $ \| \cdot \|_\infty$ is the normal absolute value, and $ \| \cdot \|_p$ is the $ p$-adic absolute value if $ p \neq \infty$.

From that on, the two consequences are really easy. Unlike said in this thread, there are such functions $ f$, in fact all $ s>0$ and alle $ p \neq \infty$ give one.


inequalities
algebra
polynomial
inequalities proposed

[quote="Namdung"]Here is the week variant:

Show that if a, b, c are roots of  x^3 - 3x+ 1 then a^2 -2, b^2 - 2, c^2 - 2 also are roots of this polynomial.

Namdung[/quote]
Just a verification
If a is root of P(x)=x^3 - 3x+ 1 then a^3=3a-1
P(a^2-2)=(a^3)^2-6a.a^3+9a^2-1=(a^2-2)^2-6a(a^2-2)+9a^2-1=0

[quote="Namdung"]Find all polynomial P(x) = x^3 + px + q from R[x], which has only real roots and if x0 is root then x0^2 - 2 is also root.[/quote]
If P(x) satisfy the statement then the real root 2cos(2 \pi /9)

-------------------------------------------------------------------------
For the first problem, here an answer for the case that P has MULTIPLE roots
p(x)=x^3+px+q all roots are REAL
p(x)=(x-a)(x-b)(x-c) a,b,c real root then a-2,b-2,c-2 are also root
p(x)=(x-a^2+2)(x-b^2+2)(x-c^2+2)

Coefficient of x
a+b+c=0
6-a^2-b^2-c^2=0 these two => ab+bc+ca=p=-3

p(x)=x^3-3x+q but p(x) has multiple real root so 4p^3+27q^2=0
q^2=4
q=-2 or 2

x^3-3x+2=(x+2)(x-1)^2,  x^3-3x-2=(x-2)(x+1)^2
-------------------------------------------------
For the 2cos(2 \pi /9), I was trying to find all x^3+p+q with only one
real root and two complex roots s.t if a is root then a^2-2 is also a root.


algebra
polynomial
geometry
circumcircle
number theory
prime numbers
national olympiad

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=140462
Show that the plane cannot be represented as the union of the inner regions of a finite number of parabolas.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=140460
If $b_{1}, b_{2}, \ldots, b_{n}$ are non-negative reals not all zero, then prove that the polynomial
\[x^{n}-b_{1}x^{n-1}-b_{2}x^{n-2}-\ldots-b_{n}=0 \]
has only one positive root $p$, which is simple. Moreover prove that any root of the polynomial does not exceed $p$ in absolute value.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=140464
Let $ABC$ be a triangle. A circle is tangent to sides $AB, AC$ and to the circumcircle of $ABC$ (internally) at points $P, Q, R$ respectively. Let $S$ be the point where $AR$ meets $PQ$. Show that
\[\angle{SBA}\equiv \angle{SCA}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=140463
Show that there are infinitely many prime numbers $p$ having the following property: there exists a natural number $n$, not dividing $p-1$, such that $p|n!+1$.

Thanks to iura  :wink:


inequalities
trigonometry
complex numbers
inequalities unsolved

Let  $a,b,c\geq0  \ a+b+c=abc \ $   Find the minimum value of:

$\sqrt{1+\frac{1}{a^2}}+\sqrt{1+\frac{1}{b^2}}+\sqrt{1+\frac{1}{c^2}}$

I'm not exactly keen on inequalities, but anybody can spot the substitution necessary for this one: $a=\tan A$ etc., with $ABC$ being an acute triangle.

The expression is $\sum \frac 1{\sin A}$, and from the convexity of $\frac 1{\sin x}$ on $(0,\frac \pi 2)$ we can derive the minimum: $2\sqrt 3$, and it's reached when $a=b=c=\tan \frac \pi 3=\sqrt 3$.

Natalio Guersenzvaig gave a really neat solution to this on in Crux.

\begin{eqnarray*}
\sum \sqrt{1 + \frac{1}{a^2}} &=& \sum \left| 1 + \frac{i}{a} \right|  \\
&\geq& \left| 3 + \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) i \right| \\
&=& \sqrt{9 + \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^2 } \\
&\geq& \sqrt{9 + 3\left( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} \right) } \\
&=& \sqrt{12} = 2\sqrt{3}
\end{eqnarray*}

(we used the triangled ineq in the second step)

[quote="billzhao"]
\[
\sum \sqrt{1 + \frac{1}{a^2}} \geq \sqrt{9 + \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^2 }
\]
[/quote]
The use of complex numbers is not necessary here, this is exactly Minkowski's Inequality! Anyway, it's a nice solution.


geometry
combinatorics solved
combinatorics

Let C be a circle of radius 1. Let P1,P2.....,Pn be points incluiding its center. for a point Pj let Xj be the samllest distance to some other Pi i different from j. prove that:

X1 2 +X2 2 ....+Xn 2 <=9

A classical use of the areas...

Wlog, we may assume that $P_1$ is the center of the disk. Thus $x_i \leq 1$ for each $i$.
For each $i$, let $D_i$ be the disk with center $P_i$ and radius $\frac {x_i} 2$. From the minimality of the $x_i$'s any two of these disks have no interior common points.
Let $A_j$ be any point in $D_j$. Then $A_jP_j \leq \frac {x_j} 2 \leq \frac 1 2$, thus $A_jP_1 \leq A_jP_j + P_jP_1 \leq \frac 3 2 $.
It follows that all the $D_j$ are contained in the disk with center $P_1$ and radius $\frac 3 2 $. Summing over the areas of the $D_j$'s it follows that :
$\sum x_i^2 \cdot \frac {\pi} 4 \leq (\frac 3 2 )^2 \cdot  \frac {\pi} 4$, which leads to the desired inequality.

Pierre.


calculus
integration
inequalities
function
logarithms
real analysis
real analysis unsolved

A function $f$ such that $f(x)>0,\forall x \in [a,b]$ and $f$ satisfy the  Lipschitz condition which means,for $L>0,\forall x_1,x_2 \in [a,b]$ we have $|f(x_1)-f(x_2)\le L|x_1-x_2|$.also we know that for $a\le c \le d \le b$,we have:
\[ \int_c^d \frac{dx}{f(x)}=\alpha,\int_a^b \frac{dx}{f(x)}=\beta  \]
Prove:
\[ \int_a^b f(x)dx \le \frac{e^{2L \beta}-1}{2L \alpha} \int_c^d f(x)dx  \]

[quote="fedja"]As stated, it is trivially true because the inequality $\beta\le \frac{e^{2L\beta}-1}{2L\alpha}\alpha$ holds for all $\alpha,\beta>0$.[/quote]
Yup, Bernoulli's inequality $1+X\le e^X$

[quote="fedja"]As stated, it is trivially true because the inequality $\beta\le \frac{e^{2L\beta}-1}{2L\alpha}\alpha$ holds for all $\alpha,\beta>0$.[/quote]
But I still don't know how to prove $\int_a^b f(x)dx \int_c^d \frac 1{f(x)} dx \le \int_c^d f(x)dx \int_a^b \frac 1{f(x)} dx$
Can you help me again?

Oops.... I misread the problem. Sorry  :blush:

OK, here goes. Let $D=d-c$ and $B=b-a$. Since $\int_c^d \frac{dx}{f(x)}=\alpha$, we must have $f\le \frac D\alpha$ at some point of $[c,d]$. Now, among all $L$-Lipschitz positive functions on $[a,b]$ whose minimum does not exceed $\frac D\alpha$, the function $f(x)=\frac D\alpha+L(x-a)$ gives the largest value to the integral $\int_a^b f(x)dx$ and the least value to the integral $\int_a^b \frac{dx}{f(x)}$. Hence $\int_a^b f(x)dx\le \frac{BD}{\alpha}+\frac{B^2L}2$ and $\beta\ge\frac1L\ln\left(1+LB\frac{\alpha }D\right)$. Also, by Cauchy-Schwartz, $\int_c^d f(x)\,dx\ge \frac{D^2}{\alpha}$. Thus, it suffices to prove that
\[ \frac{BD}\alpha+\frac{B^2L}2\le\frac{D^2}{\alpha} \frac{e^{2L\frac 1L\ln(1+LB\frac\alpha D)}-1}{2L\alpha}\,. \]
But it is an identity.

[quote="fedja"] by Cauchy-Schwartz, $\int_c^d f(x)\,dx\ge \frac{D^2}{\alpha}$. [/quote]

$D\lambda=\int_c^d \lambda\frac{\sqrt{f}}{\sqrt{f}} \le \lambda^2\int_c^d \frac{1}{f} \int_c^d f = \lambda^2\alpha \int_c^d f $
So  $\int_c^d f(x)dx \ge \frac{D}{\lambda \alpha}$, let $\lambda = 1/D$, we get  $\int_c^d f(x)\,dx\ge \frac{D^2}{\alpha}$

Close but not quite correct. Cauchy-Schwarz is $\int fg\le\sqrt{\int f^2}\cdot \sqrt{\int g^2}$. If you forget those square roots, as you did, you can easily prove many interesting things like $\frac 14\le \frac 1{16}$ ($f=g=\frac 12$ on $[0,1]$). ;)


algebra
function
domain
group theory
superior algebra
superior algebra unsolved

Need some help, thx.

$p>2$ is a prime

Prove that there is just one element of order $2$ in $Z_p^{\times}$
and write down the automorphism of order $2$ of cyclic group of order $p$

Since there exists a primitive root, the group $\mathbb{Z}/p\mathbb{Z}^*$ is cyclic, thus isomorphic to $\mathbb{Z}/(p-1)\mathbb{Z}$.

I guess I still confused what you wrote to prove that there is just one element of order $2$ in $Z_p^*$, could u be please to explain more?

uhm..
an element of order $2$ would satisfy the equation $x^2 - 1 = 0$. since $\left(\mathbb{Z}/p\mathbb{Z}\right)^*$ is a commutative group, we can write $x^2 - 1 = (x-1)(x+1)$, and since it's an integrity domain (no divisors of zero), we can conclude that if $x$ has order $2$, it has to be $-1$.
that's all :)

Thx for explaining.

Could u write down the automorphism of order 2 of a cyclic group of order p as well?

i guess you mean an automorphism $f$ such that $f(f(x)) = x$, right?
well, in this case, you need to find an automorphism such that $f(f(1)) = 1$ but $f(1) \neq 1$.
say $f(1) = a \cdot 1$ (meaning $a$ times one). then $f(f(1)) = a^2 \cdot 1$, since $f$ is an automorphism.
now, of course, $a^2 \cdot 1 = a^2 \in \mathbb{Z}/p\mathbb{Z}$, so you're actually looking for an element $a$ of order $2$ in $\left(\mathbb{Z}/p\mathbb{Z}\right)^*$, so you have to send $1 \mapsto -1$, and automorphism you're looking for is $f(x) = -x$.
hope i've been clear...


limit
inequalities
real analysis
real analysis unsolved

Let $p\in\mathcal l N, p\geq 2$. Consider a sequence $(x_n)_{n\geq1}$ such that $x_1>0,x_{n+1}=\sqrt[p]{1+px_n}-1$ for each $n\geq 1$. Prove that $\lim_{n\to\infty} x_n=0$ and $\lim_{n\to\infty} nx_n=\frac{2}{p-1}$.

[hide="help"] I got stuck somewhere at $\frac{x_{n+1}^2}{x_{n}-x_{n+1}}\leq \frac{2}{p-1}$[/hide]

$x_{n+1}=(1+px_n)^{1/n}-1\leq 1+\frac1n(px_n)-1=\frac{p}{n}x_n<x_n$ when n is larger than p.
So, $x_n$ is drecreasing eventually and bounded below from zero.
Therefore, $x_n$ is convergent.
Next, $(1+x_{n+1})^n=1+px_n$ implies $1+px_n=(1+x_{n+1})^n\geq 1+nx_{n+1}$
$x_{n+1}\leq \frac{p}{n}x_n\leq \frac{p}{n}\frac{p}{n-1}x_{n-1}\leq ...\leq \frac{p^n}{n!}x_0$ which tends to zero as n tends to infinity.
By Sandwich Theorem, the given sequence tends to zero as n tends to infinity.

Sorry, I have mistaken the p as n. Too sleepy.....

First we have inductively that $x_{n+1}>0$

From $x_{n+1}=\sqrt[p]{1+px_n}-1$ we get $(x_{n+1}+1)^p=1+p x_n$

and using bernoulli's inequality we get

$1+px_n\geq 1+px_{n+1}$ or $x_n\geq x_{n+1}$.

Since $x_{n}>0$ and is decreasing it converges , $\lim_{n\to\infty}x_{n}=\lambda$

Then $(\lambda+1)^p=1+p\lambda \Longrightarrow \lambda=0$

$\lim_{n\to\infty} nx_n=\frac{2}{p-1}$ this is what I can not prove

By cesaro stoltz the problem reduces to finding the limit $\lim \frac 1 {x_{n+1}}-\frac 1{x_n}$.
After some calculation we get that $px_n=x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-1} x_{n+1}$, Hence
\[ \lim \frac 1 {x_{n+1}}-\frac 1{x_n}=\frac 1 {x_{n+1}}-\frac{p} {x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-1} x_{n+1}}=  \]
\[ =\lim \frac{x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-1} x_{n+1}-p x_{n+1} }{(x_{n+1}) (x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-1} x_{n+1}}=  \]
\[ =\lim \frac{x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-2} x_{n+1}^2}{x_{n+1}^{p+1}+\binom p 1 x_{n+1}^p+\dots+\binom p {p-1} x_{n+1}^2}=  \]
\[ =\lim \frac{x_{n+1}^{p-2}+\binom p 1 x_{n+1}^{p-3}+\dots+\binom p {p-2}}{ x_{n+1}^{p-1}+\binom p 1 x_{n+1}^{p-2}+\dots+\binom p {p-1}}=\frac{p-1}{2}  \]

Hence, the limit has the desired value.

I assumed that $x_n \to 0$. It can be found easily since by bernoulli $x_{n+1}\le x_n,x_n\ge 0$ , hence $x_n$ have limit. The only positive solution of $a=\sqrt[p]{1+ap}$ is 0. Hence the sequence is convergent to 0.

Thx for the solution xirti. You have just a little typo:
[quote="xirti"]$\lim \frac 1 {x_{n+1}}-\frac 1{x_n}=\frac 1 {x_{n+1}}-\frac{p} {x_{n+1}^p+\binom p 1 x_{n+1}^{p-1}+\dots+\binom p {p-1} x_n}$[/quote] it should be $x_{n+1}$.

[quote="xirti"]
\[ =\lim \frac{x_{n+1}^{p-2}+\binom p 1 x_{n+1}^{p-3}+\dots+\binom p {p-2}}{ x_{n+1}^{p-1}+\binom p 1 x_{n+1}^{p-2}+\dots+\binom p {p-1}}=\frac{p-1}{2}  \]

Hence, the limit has the desired value.[/quote]

I don't understand this part  :| ..

Since $x_n\to 0$ every term of the numerator tends to 0, hence their sum tends to 0 (there are finitely many terms: p), except the last one. The same argument holds for the denominator. ;)


number theory
greatest common divisor
modular arithmetic
relatively prime
combinatorics unsolved
combinatorics

Integers $ a,b,c$ have highest common factor $ 1$. The triplet $ a,b,c$ may be altered into another triplet such that in each step one of the numbers in the actual triplet is increased or decreased by an integer multiple of another element of the triplet. Is it true that, no matter how the numbers $ a,b,c$ are given, the triplet $ 1,0,0$ can be obtained in at most $ 10$ steps?

How about let’s shoot for $ 5$ steps?

$ \bullet$ Make $ \text{gcd}\,(a,b) \equal{} 1$; do so by finding $ k_1$ such that $ \text{gcd}\,(a' \equal{} a \plus{} k_1c, b) \equal{} 1$. Let $ \text{gcd}\,(b,c) \equal{} t$, and $ b' \equal{} b/t$; then the $ b'$ numbers in the form $ a \plus{} mt \pmod{b't}$ for $ m \equal{} 0,1,\ldots,b'\minus{}1$ are attainable; of these, by Euler's totient (or similar inclusion-exclusion argument?) at least one is relatively prime to $ b'$, and hence to $ b$.
$ \bullet$ Make $ c' \equal{} c \plus{} k_2b \equiv 1 \pmod{a}$ by letting $ k_2 \equiv (1\minus{}c)b^{\minus{}1} \pmod{a}$.
$ \bullet$ Make $ c' \equal{} c \minus{} k_3a \equal{} 1$.
$ \bullet$ Use $ c \equal{} 1$ to make $ a \equal{} 0$.
$ \bullet$ Use $ c \equal{} 1$ to make $ b \equal{} 0$.


Recently while studying Organic Chemistry, I got really obsessed with the idea of using micro-organisms during organic sythesis process. So I was wondering if there are certain micro-organisms commonly used by Chemists? Also how are the micro-organisms chosen?

There are certain synthetic transformations that can be made through the use of some types of microorganisms (specially reductions and oxidations), but I think this is not commonly used. Some examples:

- Carbonyl groups can be reduced to hydroxyl groups with variable selectivity with several yeasts: Baker's yeast ([i]Saccharomyces cerevisiae[/i]), [i]S. acidifaciens[/i], [i]S. debrueckii[/i], [i]S. fermentati[/i], [i]Candida albicans[/i], [i]Pichia farinosa[/i], [i]Lipomyces starkeyi[/i], [i]Endomycopsis fibligera[/i], [i]Kloeckera saturnus[/i], and others.

- The following transformations can be made with [i]Pseudomonas putida[/i] (where R = H, Me and X = halogen, H):


conics
ellipse
analytic geometry
algebra
function
domain

How do you find the equation of the tilted rings of saturn (supposing that they are ellipses)?
The planet Saturn is composed of rings. Supposing that the "ball" is a Circle and the rings of Saturn are ellipses, how would it be possilbe to find the equation of those rings?
Equation Form Needed

Come on people... Yoou all dont like the challenge or are u just lazzy?

There are two general forms for ellipses.
In Cartesian coordinates
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Where $a, \ b$ are coefficents varying the major and minor axises of the ellipse.
Iff $a=b$, then the graph is a circle.

In polar coordinates
$r= \frac{l}{\left( 1+e \text{cos}\theta\right) }$
Where $l$ is the semi-latus rectum, $l=\frac{b^{2}}{a}$
$a$ is the semi-major axis
$b$ is the semi-minor axis

That pretty good.. But can u go in depth?
Thanks.. We have just learned the ellipses in class.. so I'm not that familiar to the polar form..
LOL

THanks tho.. You were the only one who listen to my prblem

Using polar coordinates, the domian is the $\theta$ and the range is $r$.
The domain, $\theta$, is an angle measured from the positive x-axis counterclockwise.
$r$ is the distance, "radius", from the origin.
So each $\theta$, then gives a $r$ by the function, to define a point; by definition of domain and range that might seem redundant  :wink: .
The main difference is that in polar each points has components of $\theta$ and distance from origin instead of the Cartesian components of displace along the x and y axises.

The explaination may not make any sense.


For WOOT, it states in the Program Overview that each lecture is given twice and that I "should attend one time or the other" and that I "do not need to attend both times." However, particularly for the more difficult lectures, could I attend both sessions? If yes, and if there is someone that has done so in the past, have you found that attending both classes is worth the extra time?

Yes, you are allowed to attend both classes.  We ask that in the second class, you let others participate rather than answering all the questions immediately.


if every distinct letter stands for a distinct digit, then solve:

two+three+three=eight

[quote="laughinghead505"]Is there supposed to be only one solution? If you replace two with 100t+10w+o and other numbers similarly and combine like terms, [hide="you get"] $20099t+10w+o+1990h+200r-9978e-1000i-100g=0$

That doesn't look very helpful to me... did I do it wrong? :(
[/hide][/quote]
It doesn't really matter what coefficients you get, there are 8 variables. I think there is more than one answer.

Anyway, that eight variables have to be all different.

i think there is only one solution....

you might be able to get it by a lot of scratch work.


Dakle, ako u neki kvadrat pobacamo N nasumičnih (slučajnih) točaka, koliko će njih u prosječnom slučaju pripadati konveksnom omotaču tih točaka (convex hull)?

Jasno mi je da udio takvih teži 0 kad N teži u beskonačno. Ali ima li kakva formula?

Mislim da shvaćam otprilike što tražiš, ali ti je pitanje malo loše definirano i stoga je malo teže i razmišljati o njemu. Zanima me što je prosječni slučaj, tj. kako izgleda prosječni slučaj za, recimo, N=3 ili N=4? Kako izgledaju, po tvom mišljenju ti trokuti ili četverokuti u prosječnom slučaju i kolika je njihova površina?

Prosječni slučaj dobijemo kad napravimo 1000000 testova pomoću random generatora i nađemo prosjek.

OK, kužim... sorry, u pravu si, malo sam se zbunio s tim prosjekom jer mi, očito, ne možemo imati neki "prosječni" trokut (te tri točke su neovisne, pa, barem nekako intuitivno, "u prosjeku" sve tri završe u centru kvadrata i onda uopće nemamo troku  :D). Ali, da, ovak kak si postavio, ima smisla.


probability
function
integration
calculus
derivative
trigonometry
probability and stats

Let $ (\xi_n, n\geq 1)$ be independent, identically distributed random values, that have continious,even in some neighborhood of $ 0$ probability density. Proove weak convergence: $ \frac {\sum \limits_{k \equal{} 1}^\infty \xi_k^{ \minus{} 1}} n \to^{w} \chi, n \to \infty$, where $ \chi$ has Cauchy distribution.
A bit technically hard problem.

Honestly I don't have a clue for this very bizarre law of large numbers  :huh:

but I'd be interested to know from where this comes from.

Sorry not to be able to help

It's rather convergence to a specific distribution, something like Classical Limit Theorem.
The idea of solving is using Levy theorem and some asymptotical things.But I have not managed to solve it completely.

Just play with characteristic functions a bit. Your task is to show that, for any density function $ p$ that is continuous and positive at $ 0$, you have $ \int_{\mathbb R}p(x)e^{it/x}\,dx\equal{}1\minus{}c|t|\plus{}d it\plus{}o(|t|)$. Once this is done, the problem follows immediately. To do it, just change the variable to $ 1/x$ and get the integral $ \int_{\mathbb R}e^{ity}f(y)\,dy$ where $ f(y)y^2$ has a positive limit as $ y\to\infty$. Subtract the Cauchy distribution having the same limit, and you'll get an eventually even integrable function decaying faster than $ y^{\minus{}2}$ at infinity with integral $ 0$. All you need to do now is to show that its Fourier transform is $ d it\plus{}o(|t|)$ as $ t\to 0$ with some $ d\in \mathbb R$. The non-even part gives you an analytic function whose derivative at $ 0$ is purely imaginary, so we are down to showing that the even part gives you $ o(|t|)$ (we can adjust both parts to have integral $ 0$). Now just split the integral into that over $ [\minus{}c/t,c/t]$ with small $ c$ and the rest. Use the estimate $ |1\minus{}\cos ty|\le t^2 y^2$ for the first part and the crude estimate $ |\cos ty|\le 1$ for the second part (taking into account that $ f(y)y^2$ is bounded and tends to $ 0$).

Thx fedja you were very clear (as usual).

I was nearby this solution, but a lack of technic..


Let be the sequence of numbers that: $ a_0\equal{}1;a_1\equal{}6; a_{n\plus{}2}\equal{}6a_{n\plus{}1}\minus{}a_n$
Prove that: $ 2a_n^2\equal{}k(k\plus{}1)$ where $ k \in Z$

Let $ r_{\pm}\equal{}3\pm \sqrt{8}$ be the roots of the characteristic polynomial $ x^2\equal{}6x\minus{}1$, then $ a_n\equal{}\frac{r_\plus{}^{n\plus{}1}\minus{}r_\minus{}^{n\plus{}1}}{r_\plus{}\minus{}r_\minus{}}$ holds. From this we conclude $ 8a_n^2\plus{}1\equal{}\left(\frac{r_\plus{}^{n\plus{}1}\plus{}r_\minus{}^{n\plus{}1}}{2}\right)^2$. If we set $ k_n\equal{}\frac{r_\plus{}^{n\plus{}1}\plus{}r_\minus{}^{n\plus{}1}\minus{}2}{4}$, we have $ k_0\equal{}1,k_1\equal{}8$ and the recursion $ k_{n\plus{}2}\equal{}6k_{n\plus{}1}\minus{}k_n\plus{}2$, i.e. $ k_n\in\mathbb{Z}$. Since $ (2k_n\plus{}1)^2\equal{}8a_n^2\plus{}1\Longrightarrow 2a_n^2\equal{}k_n(k_n\plus{}1)$ holds, we have completed the proof.

[quote="hjbrasch"]Let $ r_{\pm} \equal{} 3\pm \sqrt {8}$ be the roots of the characteristic polynomial $ x^2 \equal{} 6x \minus{} 1$, then $ a_n \equal{} \frac {r_ \plus{} ^{n \plus{} 1} \minus{} r_ \minus{} ^{n \plus{} 1}}{r_ \plus{} \minus{} r_ \minus{} }$ holds. From this we conclude $ 8a_n^2 \plus{} 1 \equal{} \left(\frac {r_ \plus{} ^{n \plus{} 1} \plus{} r_ \minus{} ^{n \plus{} 1}}{2}\right)^2$. If we set $ k_n \equal{} \frac {r_ \plus{} ^{n \plus{} 1} \plus{} r_ \minus{} ^{n \plus{} 1} \minus{} 2}{4}$, we have $ k_0 \equal{} 1,k_1 \equal{} 8$ and the recursion $ k_{n \plus{} 2} \equal{} 6k_{n \plus{} 1} \minus{} k_n \plus{} 2$, i.e. $ k_n\in\mathbb{Z}$. Since $ (2k_n \plus{} 1)^2 \equal{} 8a_n^2 \plus{} 1\Longrightarrow 2a_n^2 \equal{} k_n(k_n \plus{} 1)$ holds, we have completed the proof.[/quote]
Your solution is very brief  :lol: I had a solution, but it's so long and complicated  :(


Diophantine equation
number theory proposed
number theory

Determine all integers $ a$ for which the equation $ x^2\plus{}axy\plus{}y^2\equal{}1$ has infinitely many distinct integer solutions $ x,y$.

$ \Delta: \equal{}(a^2\minus{}4)y^4\plus{}4$ is a square for infinitely many value of $ y$, so $ |a|\ge 2$ is a necessary condition. It is also sufficient since it is a pell equation and (x,y)=(1,0) is a solution.


What is the number of square units in the area of the region determined by $ |x| \plus{} |y| \equal{} 1$?

Graph it (if you can't see it, it's bounded by $ x\plus{}y\equal{}1,x\minus{}y\equal{}1,\minus{}x\plus{}y\equal{}1,\minus{}x\minus{}y\equal{}1$).

This is a square with side $ \sqrt2$, so our answer is $ (\sqrt2)^2\equal{}\boxed2$.


2abc+ab+ac+bc=1 and a,b,c be positive reals.prove that:  (ab)^(1/2)+(bc)^(1/2)+(ac)^(1/2)>=3/2

The inequality is the wrong way round. It should be:

[color=blue][b]Problem.[/b] Let a, b, c be three positive reals such that bc + ca + ab + 2abc = 1. Prove that $\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\leq\frac32$.[/color]

[i]Solution.[/i] Denote $x=\sqrt{bc}$, $y=\sqrt{ca}$, $z=\sqrt{ab}$. Then, x, y, z are three positive reals and satisfy

$x^2+y^2+z^2+2xyz=\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2+\left(\sqrt{ab}\right)^2+2\cdot\sqrt{bc}\cdot\sqrt{ca}\cdot\sqrt{ab}$
$=bc+ca+ab+2abc=1$;

thus, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 (or http://www.mathlinks.ro/Forum/viewtopic.php?t=21387 ), we have $x+y+z\leq\frac32$; in other words, $\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\leq\frac32$, and the problem is solved.

  Darij

i have a document "romanian math olympiad 2005" this question was there and i look again the question and it was true.
can't you see this question in romanian math olympiad 2005 ??? :(

can't there be a typo in the file you've read?
The sign is obviously reversed by the substitution $\sqrt{ab} = cosC$ and so forth, where A+B+C=180

i think my document was false.because the solutions are true.and i haven't got a typo in my file :blush:


[1] Find all functions f: R -> R such that

f(x + y) + f(f(x) - y) = 2f(f(x)) + 2y.

[2] Find all functions f: R+ -> R + such that

f(f(x)) + f(x) = 6x.

[3] For which positive integers k is there a function f: N -> Z such that

f(2003) = 2004, f(xy) = f(x) +  f(y) + k gcd(x, y)?

[i]unknown sources[/i]

For [1]:

We put y=f(x) and we get f(x^2+f(x))+f(0)=2f(f(x))+2(f(x))^2 (#)
We make y=-x^2 and we get f(0)+f(x^2+f(x))=2f(f(x))+2x^4 (##)

From (#) and (##) we find that for any real x f(x) is either x^2 or -x^2 (*). From this we find f(0)=0. We put x=0 to find f(y)+f(-y)=2y^2 (**).

From (*) and (**) we get f(x)=x^2 for all real x.

I think this is the easiest one among the 3 you've posted :D

For [3] :
Let k be such a value and f be such a function. Then, for a = b = 1, we get (k+1)f(1) = 0, that is k = -1 or f(1) = 0.

Suppose that f(1) = 0. Let p = 2003. We deduce that :
f(p 2 ) = (k+2)f(p) thus f(p 3 ) = (2k+3)f(p), which leads to f(p 4 ) = (3k+4)f(p).
In another hand, f(p 4 ) = f(p 2 ) + f(p 2 ) + kf(p 2 ) = (k+2) 2 f(p).
Since f(p) = 2004 =/= 0, we then have 3k+4 = (k+2) 2 , that is k = 0 or k = -1.

Conversely :
- If k = -1, let choose f be constant and equal to 2004.
- If k = 0, the desired condition is f( \pi (p_i) a_i ) =  \sum a_if(p_i) (1) where the p_i's are primes and a_i's are positive integers.
Thus, it suffices to choose f(1) = 0, f(p) = 2004 for each prime p, and define f by (1) on the rest of the integers.

Thus, the 'good' values for k are k = 0  and k = -1.

Pierre.

For [2] :
Let x be a given nonnegative real.
Let a_0 = x, a_1 = f(x), a_2 = f(f(x)), and so on. The sequence (a_n) satisfies the recurrence relation a_(n+2) + a_(n+1) - 6a_n = 0.
The caracteristic equation has two distinct roots, which are 2 and -3.
Thus, for each n : a_n = 2 n (3x+f(x))/5 + (-3) n(2x-f(x))/5. (1)

Note that since a_n = f (n) (x), we must have a_n  \geq 0 for each n.
But, if 2x - f(x) =/=0, then according to its sign and to (1), we will have a_n < 0 or sufficientely large odd n or sufficientely large even n.
It follows that, for each x, we have f(x) = 2x.

Conversely, f(x) = 2x gives indeed a solution.

Pierre.


Is this a correct way of simplifying:

$ \frac {1 \minus{} \sqrt {3x}}{\sqrt {6x}}$

$ \frac {\sqrt {6x} \minus{} 3|x|\sqrt {2}}{6|x|}$

This can't simplified any further?

How about:

$ 3x \minus{} 6|x|$
$ 3(x \minus{} 2|x|)$

Or can the $ x$ be taken out?

How about:

$ \sqrt {(x \minus{} 2)^{2}}$
$ x \minus{} 2$
or
$ |x \minus{} 2|$?

Absolute value signs from radicals are very confusing...

You don't need the absolute value signs for the first one.  You're assuming $ x$ is real so you're not going to get something negative by squaring it.  I would leave it as
$ \frac{\sqrt{6x}\minus{}3\sqrt{2}x}{6x}$

If you take the complex number expression for $ |z|$, then for real numbers $ a$ we have $ |a|\equal{}\sqrt{a^2}$.  So $ \sqrt{(x\minus{}2)^2}\equal{}|x\minus{}2|$.


In a trig problem for sin, I get √12/4 and the book has √3/2. I am guessing that they are both the same, just reduced somehow. What is the procedure for reducing square roots in a fraction?, Thanks

Well it's reduced this way...
$\frac{\sqrt{12}}{4}=\frac{\sqrt{4}\sqrt{3}}{4}=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
In general, you can simplify $\sqrt{an^2}$ to $n\sqrt{a}$.

Ahh, ok Thanks alot that clears that one up.


We are doing a Math Jam as part of a teachers workshop for MATHCOUNTS coaches in Oklahoma. This class will be from 9 to 9:45 AM PT (noon to 12:45 ET) this Friday, Oct 14. We are looking for students to participate. If you (and hopefully some of your friends) can make it at this time, please post here to let me know.

I will be able to come today... looking forward to it! :D

I have a student who wants to join the Jam session.  How does she do this?

Click on Classroom on the left side of the page and follow the instructions.


I have a good solution for this. I wait your solutions.
$x,y,a$ are reals such that $x+y=x^3+y^3=x^5+y^5=a$
Gind all $a$ which stisfay the above condition.

This is pretty boring...

Take $p = xy$. Then $a = x^3 + y^3 = a(a^2 - 3p)$ and hence $a = 0$ or $a^2 - 3p = 1$.

Obviously we get solutions if $a = 0$ so assume that is not the case.

Now, $a = x^5 + y^5 = (x^2 + y^2)(x^3 + y^3) - x^2y^2(x + y) = a(a^2 - 2p) - ap^2 = a(a^2 - p^2 - 2p)$.

Hence $a^2 - p^2 - 2p = 1$.

So, $a^2 = p^2 + 2p + 1 = 3p + 1$. Hence $p = 0$ or $p = -1$.

If $p = -1$ we get $a = 2$ or $a = -2$, if $p = 0$ we get $a = 1$ or $a = -1$.

Hence we can have $a \in \{-2,-1,0,1,2\}$. It's easy to check that these values work.


$2$ persons flip alternatively a coin until tail is obtained. What is the probability that this will happen to the first person who flips the coin?

Pre-olympiad? I thought this was fairly easy.....

Maybe i did it a different way? Infinate series?

it is a 10th grade problem. i think infinite series are quite inadequate. but, anyway, please write it  :(

I im going to bed now, im in the middle of exams :(  Ill write it out later, but goes something like this:
Prob that he gets heads first?
Prob of two tails and one head?
Prob of four tails and a head?

Continue like this, and add em up!

Thats the way I did it too, with sum of an infinite geometric series.  I think this problem was discussed in one of the Mathjams.

as seamusoboyle said,  you have to make an infinite series.
the answer is:
$\sum_{n=0}^{\infty}(\frac{1}{2})^{2n+1}=\frac{2}{3}$

O.o..

So if Joe flips the coin, the chance of a tail is $\frac{1}{2}$

But if he calls his buddy over and tells him to get a coin and then Joe flips the coin, the chance of a tail is $\frac{2}{3}$ ?

Edit: Ok, I understand the problem now, they're talking about the first person not the first flip.

[quote="perfect_radio"][quote="JBL"]It can be done more easily:  let's say the probability that the first person flipping gets the first tail is $p$.  Then $\frac{1}{2} + \frac{(1 - p)}{2} = p$.[/quote]

why?[/quote]
[hide="Hopefully a clear enough explanation"]I will refer to the first person that flips a tail a win. So when I say someone wins, I mean that someone is the first to flip the tail.
Well, $p$ is the probability that the first coin-flipper wins. How do we calculate that?
First, the first person can flip tails immediately, so that's already $\frac12$.
What happens afterward? The other alternative is that the first person flips heads ($\frac12$) and the next person goes.
At this point, it's as if we started the game again, except now, we want the new second person (original first person) to win. Since one person must win and $p$ is the probability that the first person wins, then $1-p$ is the probability that the second person wins.
So to tie it all together:
$\frac12$ for win on the first flip
$\frac12 (1-p)$ for a win in the game after the first flip
These are the two ways the first person can win, so $\frac12 +\frac{1-p}{2}=p$.
Solving gives $p=\frac23$.[/hide]


Hi all !

My name is Issam, I am 16 years old and I live in Tetouan, Morocco ...
I study in a spanish school at Tetouan, and  I speak french, spanish, arabic and english ( not very well ).
I love maths ... computers ... reading ... and bash scripting  :rotfl:

And I wish see more moroccans here  ;)

Hi Issam
Why did you not simply post in the introduce yourself topic? Anyway, nice to see another african here :)

Hey....
u going to pamo?...
does moroco participate?

hI isSAM yOUR aRE fROM tHE nORTH OF MOROCCO Happy to see You in this site we want to attack THe INternational MAthimatics :)  HOw Are you other Nice african peaple :blush:


Let n be a positive integer. How many ways are there to color the n^2 unit squares of a given (n x n)-square with two colors (say red and blue) such that in each row and each column, the number of red squares is even?

The answer is $ 2^{(n\minus{}1)^2}$. We only need to randomly choose the color of each unit square that forms the upper left $ (n\minus{}1)$x$ (n\minus{}1)$-square, and all others are instantly defined. Let's prove it!

The first $ n\minus{}1$ rows are easy to see that are defined, to choose the $ n^{th}$ square painting we need only to see if the number of red painted squares in this row is odd or even, and choose the last in other that the total sum is even. Now there is one last column that for the argument above is also defined, we need only to see if the sum of red squares in this column will always be even. The sum of red squares in all rows already painted is even, so there is an even number os columns with an odd sum of red squares. This way the last column will always have an even number of red squares and the proof is complete!


43110.

now nobody can use leet!
And I lost.

Hola.
spanish

Ni hao.
chinese

0.7734
backwards calculator language

ciao
italian

yo
english

hi
english

hey
english

Bonjour!
Salut!
[i]French[/i]

Guten Tag!
[i]German[/i]

你好!
[i]Chinese[/i]

'e5l'-
r15s11z55y89w21ian pwnage language

olleh

backwards language (sdrawkcab egaugnal)

Γειά σου.

Greek.

안녕하세요

Korean I guess I dunno.

I'm not even Korean :lol:.

Type in 1134209 on your average scientific calculator and flip it around :D.

ok, here I go...

[b]French:[/b]  bonjour, salut

[b]Spanish:[/b] hola

[b]German:[/b] Guten tag

[b]Chinese:[/b] Ni-hao

[b]Japanese:[/b]  Konichiwa (well, that's more of good afternoon) or Ohayou Gozaimasu (formal way of saying good moring) or Ohayou (informal way of saying good morning)

[b]Hindi:[/b] Namaste

[b]English:[/b] (ranging from most formal to least) Hello, Hi, Hey, Yo, 'Sup?

OWNED!!
NO ONE WROTE TAMIL!!
TAMIL=VANAKAM
HA HA!! :lol:

[quote="algebra2000"][quote="Poincare"][quote="nikeballa96"]Namaste-Hindi[/quote]
Darn you.

namaste-gujrati

OH[/quote]

aw copiers...

namaste-punjabi

Also:
 Arabic: مرحبا
 Bulgarian: здрасти
 Catalan(?):hola
 Croatian:bok
 Czech:nazdar
 Danish:hej
 Dutch:hoi
 Estonian:tere
 Filipino:mga pagbati
 Finnish:hei
 French:bonjour
 Galician:ola
 German:Hallo
 Greek:Hallo
 Hebrew:הי
 Hungarian:szia
 Indonesian:hai
 Italian:ciao
 Japanese:やあ
 Korean:야
 Latvian:čau
 Lithuanian:labas
 Maltese:ej
 Norwegian:hei
 Persian:سلام
 Polish:cześć
 Portuguese:oi
 Romanian:salut
 Russian:Гавайи
 Serbian:бок
 Slovak:nazdar
 Thai:สวัสดี
 Turkish:merhaba
 Ukrainian:привет
 Vietnamese:chào[/quote]
Sala chuthia!  (i hope you don't know what that means :P )

[quote="Poincare"]
Sala chuthia!  (i hope you don't know what that means :P )[/quote]

 :wink: Well I think I know,..definitely its not "hello"

"JAMBO"
Its swahili....

Esperanto:

Bonan matenon (Good Morning)

Bonan vesperon (Good Afternoon)

Bonan nokton (Good Night)

fail   >:-(

um...hewo --Bork, Bork, Bork!

:D:D:D

latin:salvete

means hello people!

$ \mathcal{HELLO FROM MATHCAL}$

So is this strictly "hi" or like greetings?

French: Bonjour, salut
Chinese: Ni hao, zhao (I am bad at spelling in pinyin, sorry)
Spanish: Hola

I was going to say it in spanish... :mad:

humm... not knows anything else *loses*

[quote="EggyLv.999"]Konichiwa

Japanese?[/quote]

are you jappanese?

KONICHIWA!!
EH?
Sayounara


We are given a circle $ k$ and nonparallel tangents $ t_1,t_2$ at points $ P_1,P_2$ on $ k$, respectively. Lines $ t_1$ and $ t_2$ meet at $ A_0$. For a point $ A_3$ on the smaller arc $ P_1 P_2,$ the tangent $ t_3$ to $ k$ at $ P_3$ meets $ t_1$ at $ A_1$ and $ t_2$ at $ A_2$. How must $ P_3$ be chosen so that the triangle $ A_0 A_1 A_2$ has maximum area?

Let us change the notation in order to use the standard triangle notation.

[b]Problem:[/b] $ (I_a,r_a)$ is a fixed circle and $ b,c$ are fixed tangents to $ (I_a)$ through $ N,M$ meeting at $ A.$ Variable tangent $\tau$ to $ (I_a)$ through the small arc $ MN$ meets $ AN,AM$ at $ C,B.$ Construct tangent $\tau$ such that $ \triangle ABC$ has maximum area.

Using the well-know relation $ [\triangle ABC] \equal{} r_a(p \minus{} a),$ keeping in mind that $ AN \equal{} AM \equal{} p \equal{} \text{const},$  we deduce that $ [\triangle ABC]$ is maximum $\Longleftrightarrow$ $ a$ has minimum lenght. Let $ h_a$ be the length of altitude issuing from $ A.$ We have then

$ a \equal{} \frac {2r_ap}{h_a \plus{} 2r_a}  \Longrightarrow \ a$ is mimum $\Longleftrightarrow \  h_a$ is maximum.

If $ BC$ cuts $ AI_a$ at $ P'$ and $ P$ is the midpoint of the arc $ MN,$ then $ AP \ge AP' \ge h_a$  $\Longrightarrow$ $h_a$ has maximum lenght $ AP$ $\Longrightarrow$ $\tau$ forms the isosceles triangle $\triangle ABC$ with apex $A.$

[quote="moldovan"][color=darkred]We are given a circle $ w$ and nonparallel tangents $ AE$ , $ AF$ at points $ \{E,F\}\subset w$ . For a point mobile $ M$ on the smaller arc $ EF$

 the tangent to $ w$ at $ M$ meets $ AE$ at $ B$ and $ AF$ at $ C$ . How must $ M$ be chosen so that the triangle $ ABC$ has maximum area ?[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] I'll use the standard notations for the triangle $ ABC$ with the $ A$-exincircle $ w \equal{} C(I_a,r_a)$ . Therefore, $ AE \equal{} AF \equal{} p$ (constant)

and $ \left\|\begin{array}{c} a \equal{} (p \minus{} b) \plus{} (p \minus{} c) \\
 \\
p \minus{} b \equal{} r_a\tan\frac C2 \\
 \\
p \minus{} c \equal{} r_a\tan\frac B2\end{array}\right\|\ \Longrightarrow\ \boxed {\ a \equal{} r_a\left(\tan\frac B2 \plus{} \tan\frac C2\right)\ }$ . In conclusion, $ S \equal{} [ABC] \equal{} r_a(p \minus{} a)\ \minus{} \ \mathrm{maximum}\ \Longleftrightarrow$

$ a\ \minus{} \ \mathrm{minimum}\ \Longleftrightarrow\ \left(\tan\frac B2 \plus{} \tan\frac C2\right)\ \minus{} \ \mathrm{minimum}$ ,  where $ \left(\frac B2 \plus{} \frac C2\right) \equal{} 90^{\circ} \minus{} \frac A2$ (constant) $ \Longleftrightarrow B \equal{} C$ $ \Longleftrightarrow BC\parallel EF$ .[/color]


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Hint: If $ a\in D$, the map $ x\mapsto ax$ is an injective $ F$-linear transformation from $ E$ to itself.

[quote="EvaristeG."]If E is an finite extension field of F, and D is an integral domain such that F < D < E, show that D is a field.[/quote]

$ D$ is algebraic over $ F$ because $ [D: F] < \infty.$ It's very easy now to prove that every nonzero element of $ D$ has an inverse in $ D.$

Or you could note that the ring $ F[d]$ for $ d \in D$ is a quotient of $ F[x]$, and all prime ideals in $ F[x]$ are maximal as it is a principal ideal domain. It follows that $ d$ is invertible in $ D$.


function
linear algebra
linear algebra unsolved

Given: $f(x)=x$ , $V_{3}=\{$functions constant on 1/8 ths$\}$

find $P_{f}$ as a step function by using the standard basis,
and call standard coefficients $s$ and find wavelet coefficients $w$

i know $V_{3}=\{ \phi(8x), \phi(8x-1), \phi(8x-2) ... \phi(8x-7) \}$
but not sure how to get come up with the standard coefficients

What do you mean by "standard basis" - the Haar functions?

yeah, this problem is from a wavelets class, $\phi$ is the haar scaling function


AMC USA(J)MO USAMO college Yale MIT AIME

How much weight do math contest scores and qualifications have on college applications? Is it better to emphasize that or things like summer camps and independent research?

By math contest stuff, I mean like AMC perfect scores, USAMO qualifications, MOP qualifications, or high placing at math meets.

For most colleges- the admissions offices have no idea what it means. I don't think my math contest record (2-time MOP+HM before application time) made any difference at all to UC Berkeley; it might have helped for the Regents' Scholar award, but that's it.
It's probably worth some space explaining what your records mean. I used some of the essays for that.

Really?  My assumption was that pretty much every college knew what at least AMC and USAMO were (along with USACO, USAPhO, and all the other ones).

I would say mentioning USAMO would definitely be good, because it sets you apart from 99% of the applicant pool, but if you have other stuff that demonstrates initiative and passion, they might be more worthwhile to write an essay about.

In my year, the Yale and MIT applications asked specifically for AMC and AIME scores, or more accurately, they offered somewhere for you to put your scores. I would be somewhat surprised if the colleges you're interested in do not know about math competitions. They should have had applicants like you in the past and should have an inkling about what it means to qualify for USAMO.
I don't know what you mean by emphasis; I wrote down all of my summer programs and what little research I did. For my personal essay, I avoided the math completely because the rest of my application pretty much summed up how much I love math. I used that space to talk about a unique experience, how it changed me and how my reactions to it reflected the kind of person I am.

[quote]I used that space to talk about a unique experience, how it changed me and how my reactions to it reflected the kind of person I am.[/quote]

Yeah, I don't have any of those.


quadratics
number theory solved
number theory

Assume that $ p$ is prime number with $ p\equiv 3 \mod 4$.
Prove that:
$ \displaystyle \prod_{1\leq x < y\leq \frac {p \minus{} 1}2} \left(x^2 \plus{} y^2\right) \equiv ( \minus{} 1)^{\left[\frac {p \plus{} 1}8\right]}\mod p$.
Here, $ \left[u\right]$ means the integer part of $ u$ for every $ u\in\mathbb{R}$.

Very nice problem. Let eps be a primitive root modulo p. Then the quadratic residues are exactly 1,eps^2,eps^4,..., eps^4k, where p=4k+3. Thus, we have to find product (0 \leq i<j \leq 2k) (eps^2i+eps^2j). Let q=eps^2. Then the product is product( 0<=i<j<=2k) (q^i+q^j)=product (0<=i<j<=2k) (q^2j-q^2i)/ product (o<=i<j<=2k) (q^j-q^i). But product (0<=i<j<=2k) (q^2j-q^2i) and product (0<=i<j<=2k) (q^j-q^i) are Vandermonde's determinants and since q^(2k+1)=1 modulo p, the first determinant is obtained from the second one using k+(k-1)+...+1 transpositions of lines, thus it is (-1)^[k(k+1)]/2* second one. Thus, the product is (-1)^[k(k+1)]/2 modulo p, which is (-1)^[(p+1)/8].


geometry
circumcircle
angle bisector
perpendicular bisector

I am looking for the complete solution (dont refer me to Kalva), to Problem 3 of the 7th JBMO 2003...any tips?

Thanks,

PiDeltaPhi

why not post the question in the forums in its apropriate place, and wait for a solution there ? :)

I will do that, but i thought someone has a solution by hand somewhere...
Are you still in Greece Valentin?  :D

The problem you're asking about is a geometry one so i dont understand how one can find its sdolution at the inequality's forum.
In fact i've posted that problem here before at Begginer's section--Geometry---Why MN is parallel to BC.
But that section doesn't exist anymore and i dont know where you could find it.
Darij replied to it and he supplied a very nice an detailed solution here it follows:

I slightly rewrite your problem:

Let D, E, F be the midpoints of the arcs BC, CA, AB on the circumcircle of a triangle ABC not containing the points A, B, C, respectively. Let the line DE meet BC and CA at G and H, and let M be the midpoint of the segment GH. Let the line FD meet BC and AB at I and J, and let N be the midpoint of the segment IJ. Show that

- the line MN is parallel to BC.
- if P is the point of intersection of the lines AD and EF, then the circumcenter of triangle DMN lies on the circumcircle of triangle PMN.

Indeed, the points D, E, F lie on the angle bisectors of triangle ABC, i. e. on the lines AI, BI, CI, where I is the center of the incircle of triangle ABC. [This is because the angle bisector of an angle in a triangle always passes through the midpoint of the arc cut off from the circumcircle by the opposite side.]

Now, we consider the triangle DEF. We have

< DIB = 180 - < AIB
= < IAB + < IBA (sum of angles in triangle AIB)
= A/2 + B/2,

while

< IDB = < ADB = < ACB (cyclic)
= C = 180 - A - B = 180 - 2 (A/2 + B/2)
= 180 - 2 < DIB.

Thus,

< DBI = 180 - < DIB - < IDB
= 180 - < DIB - (180 - 2 < DIB)
= - < DIB + 2 < DIB = < DIB,

so that triangle BDI is isosceles, i. e. we have BD = ID. Similarly, CD = ID. Hence, BD = ID = CD, and the point D is the circumcenter of triangle BIC. Similarly, E and F are the circumcenters of triangles CIA and AIB. [This is an important fact: that the midpoint of the arc BC on the circumcircle is the circumcenter of triangle BIC.]

Now, since the points E and F are the circumcenters of triangles CIA and AIB, both of them lie on the perpendicular bisector of the segment AI. Hence, the line EF is the perpendicular bisector of the segment AI. Similarly, the lines FD and DE are the perpendicular bisectors of the segments BI and CI. In other words, the lines EF, FD, DE pass through the midpoints of the segments AI, BI, CI, respectively, and are perpendicular to these segments. Hence, the point P, being defined as the point of intersection of EF and AI (you have said AD, but this is the same line), must be the midpoint of AI. Now, since DE is perpendicular to CI, the line CI is an altitude of triangle GCH. On the other hand, it is obviously an angle bisector of this triangle. But if an altitude and an angle bisector of a triangle coincide, then the triangle is isosceles; hence, the triangle GCH is isoscles, and the midpoint M of its base GH must lie on the altitude from the apex C. In other words, the point M lies on the line CI. But since the point M also lies on the line DE, and the line DE is the perpendicular bisector of CI, the point M must be the midpoint of CI. Similarly, N is the midpoint of BI. Thus, we have found:

The points P, N, M are the midpoints of the segments AI, BI, CI, respectively.

Now, at first, MN || BC follows trivially, since N and M are the midpoints of the sides BI and CI of triangle BIC.

Now it remains to prove your second assertion, namely that the circumcenter of triangle DMN lies on the circumcircle of triangle PMN.

Since the lines EF, FD, DE are perpendicular to the lines AI, BI, CI, respectively, the lines AI, BI, CI are the three altitudes of the triangle DEF. Consequently, the point I is the orthocenter of this triangle, and the points P, N and M are the feet of the altitudes of this triangle. Hence, the circumcircle of triangle PMN is the nine-point circle of triangle DEF. Therefore, it remains to show that the circumcenter of triangle DMN lies on the nine-point circle of triangle DEF. But since < DNI = 90 and < DMI = 90, the points N and M lie on the circle with diameter DI; hence, the circumenter of triangle DMN is the center of the circle with diameter DI, i. e. the midpoint of the segment DI, hence the midpoint of the segment joining a vertex of triangle DEF with its orthocenter. But it is well-known that such midpoints always lie on the nine-point circle. Hence, the circumenter of triangle DMN lies on the nine-point circle of triangle DEF, i. e. on the circumcircle of triangle PMN.

I have tried to write up the proof with all details, since you posted the question in the "Beginner's Corner". If you still don't understand something, feel free to ask!

Darij

The nine-point circle of a triangle is quite a famous circle; it is defined as follows:

If we have a triangle ABC, then

- the three midpoints of its sides,
- the three feet of its altitudes,
- and the midpoints of the three segments AH, BH and CH, where H is the orthocenter of triangle ABC

lie on one circle. Since these are nine points altogether, this circle is called the "nine-point circle" of triangle ABC.

For a proof that all of these points lie on one circle, see

http://www.cut-the-knot.org/Curriculum/Geometry/SixPointCircle.shtml

P.S Thanks a lot to Darij for this solution.

[quote="Valentin Vornicu"][quote="PiDeltaPhi"]I found the solution in Inequalities/Solved Problems[/quote]great! can you pin-point a link so that I can add the correct source field? :)[/quote]

[url]http://www.mathlinks.ro/viewtopic.php?t=1045[/url]  :D

[quote]The problem you're asking about is a geometry one so i dont understand how one can find its sdolution at the inequality's forum. [/quote]

Sorry, it is problem , not 3
Thanks for your solution to 3


function
calculus
derivative
calculus computations

I'm given this qn which i could not solve.

V(x, y, z) is E = -gradV

V(x, y, z) = z / (x^2 + y^2 + z^2)^3/2  (x,y,z) is not equal to (0,0,0)

Determine function E

Any help/hint given is much appreciated.

Since you have $V$, all you need to do is take the necessary partial derivatives and make it into a vector.

$\vec{E}=-\vec{\nabla}{V}=-\left(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right)$


The perimeter of a rectangle of width 6 feet is 10 yards. In square
feet, what is the area of the rectangle?

if the perimeter is 10 yards then it is 30 feet so the length and width add to 15. Since we know that the width is 6 then the length is 9 multiplying yields 54.

DONT BE FOOLED. I pity the fool. the perimeter is 10 yards, but one side, the width is 6 feet. How can this be. 10 yards =30 feet. subtract 2 x 6 from it and you get 18/2=9 9x6=54


superior algebra
superior algebra unsolved

I was thinking about this a few days ago, I managed to prove it because it kind off easy. But once I proved it I immediately realized I did that a long time ago, and I was amazed at how I could have forgot this result.

Let $ F\leq E\leq K$ with $ E$ algebraic over $ F$ and $ K$ algebraic over $ E$ show that $ K$ is algebraic over $ F$.

The above result immediately implies a interested theorem:
If $ F\leq E$ with $ E$ algebraic over $ F$ then $ \bar E\simeq \bar F$.

Let $ \alpha$ be any element of $ K$.  Then because $ K$/$ E$ is algebraic, we know that there are $ a_{0},a_{1},\dots,a_{n}\in E$ such that

$ a_{n}\alpha^{n}+\cdots+a_{1}\alpha+a_{0}=0$

where $ n$ is a positive integer.  Now because all the $ a_{i}$'s are algebraic over $ F$, we know that the extension $ F(a_{0},a_{1},\dots,a_{n})$/$ F$ is finite, and moreover, because the extension $ F(\alpha,a_{0},\dots,a_{n})$/$ F(a_{0},\dots,a_{n})$ is at most $ n$, we know that

$ [F(\alpha,a_{0},\dots,a_{n}): F]=[F(\alpha,a_{0},\dots,a_{n}): F(a_{0},\dots,a_{n})]\cdot[F(a_{0},\dots,a_{n}): F]$

must also be finite, and hence $ \alpha$ must be algebraic over $ F$.  Thus $ K$/$ F$ is algebraic.

As for the second part, it is obvious that $ F\le E\Rightarrow \bar{F}\le\bar{E}$.  As for the other direction, using the lemma above for $ F\le E\le \bar{E}$, we have the desired result.

[quote="zgorkster"]Let $ \alpha$ be any element of $ K$.  Then because $ K$/$ E$ is algebraic, we know that there are $ a_{0},a_{1},\dots,a_{n}\in E$ such that

$ a_{n}\alpha^{n}+\cdots+a_{1}\alpha+a_{0}=0$

where $ n$ is a positive integer.  Now because all the $ a_{i}$'s are algebraic over $ F$, we know that the extension $ F(a_{0},a_{1},\dots,a_{n})$/$ F$ is finite, and moreover, because the extension $ F(\alpha,a_{0},\dots,a_{n})$/$ F(a_{0},\dots,a_{n})$ is at most $ n$, we know that

$ [F(\alpha,a_{0},\dots,a_{n}): F]=[F(\alpha,a_{0},\dots,a_{n}): F(a_{0},\dots,a_{n})]\cdot[F(a_{0},\dots,a_{n}): F]$

must also be finite, and hence $ \alpha$ must be algebraic over $ F$.  Thus $ K$/$ F$ is algebraic.
[/quote]
Thats exactly like I proved it. But you got to admit its an elegant result though the prove is easy.

[quote]As for the second part, it is obvious that $ F\le E\Rightarrow \bar{F}\le\bar{E}$.  As for the other direction, using the lemma above for $ F\le E\le \bar{E}$, we have the desired result.[/quote]
Im not exactly sure what you wrote. But I simply said $ \bar E$ is algebraic over $ E$ which is implies $ \bar E$ algebraic over $ F$. So $ \bar E$ is an algebraic closure of $ F$, and using a famous result $ \bar E\simeq \bar F$ where $ \bar F$ is any algebraic closure of $ F$.


How do math contest writers think of contest problems? I'm especially interested in how the olympiad problems are thought up of because they seem so mysterious and complicated to me at the present.

Do writers start pick some basic facts to start with when writing a problem and try to generate unexpected results? Are problems more often written backwards from a result and then disguised as a problem?

Is knowing how to write math problems significant at all to aspiring mathletes? :huh:

I recall asking one of my tutors at Maths Camp once. From what I remember it could be (among other things):

1. A really old problem modified (which is not preferred).

2. A special case of a much more general theorem (which is too hard to prove itself at the Olympiad level).

3. A generalization of an old problem.

4. A small section from the work the tutor is doing at the moment (e.g. the tutor gave us a problem which we later found out was a section from his Algebra paper, but quite doable).

5. Once is a while a completely random idea just comes to him.

6. Start with an obvious fact e.g. $ (a\minus{}b)^{2} \geq 0$ and turn it into a problem e.g. $ a^{2} \plus{} b^{2} \geq 2ab$.

I can't remember any other ones at the moment.


Let be $ a,b,c>0$ . Prove that :

$ 3\left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\right)\plus{}\frac{2ab\plus{}2bc\plus{}2ca}{a^2\plus{}b^2\plus{}c^2}\ge 11$

From Cauchy, $ \frac ab \plus{} \frac bc \plus{} \frac ca\ge\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}$.

Let $ x \equal{} a^2 \plus{} b^2 \plus{} c^2,y \equal{} ab \plus{} bc \plus{} ca$. Then it remains to show that
\[ \frac {3(x \plus{} 2y)}{y} \plus{} \frac {2y}{x}\ge 11,
\]
which in turn is equivalent to
\[ (x \minus{} y)(3x \minus{} 2y)\ge 0
\]
which is obviously true since $ x\ge y$.

[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ 3\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right) \plus{} \frac {2ab \plus{} 2bc \plus{} 2ca}{a^2 \plus{} b^2 \plus{} c^2}\ge 11$[/quote]
$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge 3\frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$ comes from $ \frac{a}{b} \plus{} \frac{a}{b} \plus{} \frac{b}{c} \ge \frac{3a}{\sqrt[3]{abc}}$.
$ 9\frac{a\plus{}b\plus{}c}{3\sqrt[3]{abc}} \plus{} 2 \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 11$
Using the uvw method we can wlog assume $ b\equal{}c\equal{}1, a\equal{}t^3$:
$ 9\frac{t^3\plus{}2}{3t} \plus{} 2\frac{2t^3\plus{}1}{t^6\plus{}2} \ge 11 \iff$
$ 3t^9\minus{}11t^7\plus{}6t^6\plus{}4t^4\plus{}6t^3 \minus{} 20t\plus{}12 \ge 0 \iff$
$ (t\minus{}1)^2(3t^7 \plus{} 6t^6 \minus{} 2t^5 \minus{} 4t^4 \minus{} 6t^3 \minus{} 4t^2 \plus{} 4t \plus{} 12) \ge 0 \iff$
$ (t\minus{}1)^4(3t^5 \plus{} 12t^4 \plus{} 19t^3 \plus{} 22t^2 \plus{} 19t \plus{} 12)\plus{} 9t(t\minus{}1)^2 \ge 0$. QED :)

This result is very,very weak.We can prove the stronger
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {k(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge  k \plus{} 3$
$ k \equal{} \frac {2}{3}$,we get ineq of you.
Can_hang2007 and nguoivn have a nice proof for $ k \equal{} \frac {8}{3}$
for the best k,we can use pqr,but very long.
For your ineq,we can prove easily.For example:
$ 2(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}\ge \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} 4 \ge 6$ by AM-GM.:)

we can use the lemma:
$ 2\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right) \plus{} 1 \ge \frac{{21({a^2} \plus{} {b^2} \plus{} {c^2})}}{{{{(a \plus{} b \plus{} c)}^2}}}$(*)
(*) is very strong,we can prove $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{8(ab\plus{}bc\plus{}ca)}{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\ge \frac{17}{3}$($ k\equal{}\frac{8}{3}$) with (*)  :)

[quote="nguoivn"][quote="Toan_VN_LC"]we can use the lemma:
$ 2\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \plus{} 1 \ge \frac {{21({a^2} \plus{} {b^2} \plus{} {c^2})}}{{{{(a \plus{} b \plus{} c)}^2}}}$(*)
(*) is very strong,we can prove $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {8(ab \plus{} bc \plus{} ca)}{3(a^{2} \plus{} b^{2} \plus{} c^{2})}\ge \frac {17}{3}$($ k \equal{} \frac {8}{3}$) with (*)  :)[/quote]
(*) is not stronger than the ineq $ \frac {8}{3}$, Toan. We can't compare 2 ineqs  :wink:
Indeed,  $ k \equal{} 21$ is easy and we can prove it by some ways. But $ k \equal{} \frac {45}{2}$ is very hard, I haven't found solution for it.[/quote]
I see,I'm sorry for my post

[hide="@nguoivn"]em có một cách chứng minh khác bằng S.O.S cho k=21,cách này rất mạnh thì phải,cách này biểu diễn về S.O.S khác với cách của anh Cẩn,em nghĩ là có thể nó còn đúng với k=27,em đang thử chứng minh nó bằng kiểu biểu diễn về S.O.S đặc biệt này,nhưng với k=27 thì quá mạnh thì phải :(,thậm chí em còn chưa bấm máy tính để thử tìm phản ví dụ nữa :P ,cứ thể mà chứng minh với k=27 biết đâu lại tăng thêm được kinh nghiệm giải :) [/hide]

With $ k \geq 23$, this ineq 'll be wrong, Toan. $ k \equal{} \frac {45}{2}$ is the best :wink:


What is the probability that, when Eve selects a positive even integer less than twenty and Theo picks a positive multiple of 3 less than thirty, they pick the same number? Express your answer as a common fraction.

First, list out the possibilities.

List 1: 2,4,6,8,10,12,14,16,18
List 2: 3,6,9,12,15,18,21,24,27

Notice that only 6, 12, and 18 are on both lists. These are the possibilities for the same number. The possibility of choosing one of these from the first list is $ \frac{3}{9}\equal{}\frac{1}{3}$, and the possibility of choosing the same one if the secondd list is $ \frac{1}{9}$. Multiplying the probabilities togother, we get $ \frac{1}{3}\times\frac{1}{9}\equal{}\fbox{1/27}$.

Instead of listing, we know the numbers that satisfy this will be the multiples of $ 6$.  There are $ 3$ of these less than twenty, and each of them picks out of $ 9$ integers, so we do $ \frac{3}{9} \times \frac{1}{9}\equal{}\boxed{\frac{1}{27}}$.


geometry
3D geometry
combinatorics unsolved
combinatorics

Let $n$ be a positive integer and let $\displaystyle S_n = \{(a_1,a_2,\dots,a_{2^n}) \vert a_i \in \{0,1\} \text{ for } 1 \leq i \leq 2^n\}$. We call a subset $A$ of $S_n$ "good" if any two elements of $A$ differ in at least $2^{n-1}$ places. What is the maximum size of a good subset of $S_n$?

The answer is $2^{n+1}$. First, let's prove necesity. Denote by $d(a,b)$ the number of positions where $a$ and $b$ differ. $X$ is the set of those elements of $A$ starting with $0$ and $Y$ are those starting with $0$. Let $N=|X|$. I'll prove that $N \leq 2^{n}$. This follows by a standard double-counting argument: Let $ S=\sum_{\{a,b\} \subseteq X} d(a,b). $. On one hand, $d(a,b) \geq 2^{n-1}$ for all $a,b \in X$, $a \neq b$, so $ S \geq {N \choose 2}2^{n-1} $. On the other hand, on each of the $2^{n}-1$ remaining positions we have at most $N^{2}/4$ "noncoincidences", so
$ S \leq (2^{n}-1)\frac{N^{2}}{4}$. Hence $ (2^{n}-1)\frac{N^{2}}{4} \geq {N \choose 2}2^{n-1} \Rightarrow \frac{N}{N-1} \geq \frac{2^{n}}{2^{n}-1} \Rightarrow N \leq 2^{n} $ so $|X| \leq 2^{n}$. In the same way, $|Y| \leq 2^{n}$, so $|A| \leq 2^{n+1}$.

Now we have to build a good set $A$ with $2^{n+1}$ elements. For any binary sequence $s=(s_{1},s_{2},...,s_{k})$, let $\overline{s}=(1-s_{1},1-s_{2},...,1-s_{k})$; $sr$ means sequences $s$ and $r$ concatenated. I will build good sets inductively. Let $A_{0}=\{0,1\}$ and $A_{k+1}=\{aa:a \in A_{k}\} \cup \{a\overline{a}: a \in A_{k}\}$ for all $k \geq 0$. It is easy to check that $A=A_{n}$ is a good example.

We restate the problem in terms $2^{n}$ digit binaries. Let $B_{n}$ be the set of all $2^{n}$-digit binaries. For any binaries $X, Y \in B_{n}$, let $H(X, Y)$ denote the Hamming distance of $X$ and $Y$. (The number of bits in which $X$ and $Y$ differ).

Let $C_{0}$ and $C_{1}$ be subsets of $A$ such that $C_{0}$ has only elements of $A$ that begin with the bit 0, and $C_{1}$ has only elements of $A$ that begin with the bit 1. (Clearly $\left|A\right|=2\left|C_{0}\right|=2\left|C_{1}\right|$.). Let $\displaystyle \mathcal{K}=\sum_{X,Y\in C_0} H(X, Y)$. By the problem statement $H(X, Y) \geqslant 2^{n-1}$, for all $X\neq Y \in C_{0}$. And we have $\dbinom{|C_0|}{2}$ such pairs. This gives $\mathcal{K} \geq \dbinom{|C_0|}{2}\cdot 2^{n-1}$.

Now considering the remaining positions we can have at most $\dfrac{|C_0|^2}{4}$ bits which do not differ. This gives the upper bound on $\mathcal{K}$. Combining the two we get $$\left(\frac{|C_0|^2}{4}\right)\cdot\left(2^{n}-1\right) \geqslant \binom{|C_0|}{2}\cdot 2^{n-1} \Rightarrow \left|C_{0}\right| \leqslant 2^{n}.$$ Thus $|A|=2\left|C_{0}\right| \leqslant 2 \cdot 2^{n}=2^{n+1}$.

[hide=Construction (Same as above)] For any binary $X$, let $\overline{X}$ denote the binary that has all bits of $X$ flipped and let $X+Y$ denote $Y$ appended to $X$. We can get a construction for $n$ recursively. Let $A_{n}$ be the set $A$ for $n$. Then $$A_{n}=\left\{X+X \mid X \in A_{n-1}\right\} \cup\left\{X+\overline{X} \mid X \in A_{n-1}\right\} \textrm{ for } n \geq 1.$$ Here $A_{0}=\{0,1\}$. So for example $$A_1=\{0+0=00,1+1=11\} \cup\{0+1=01,1+0=10\}=\{00,11,01,10\}.$$ Clearly this construction satisfies.[/hide]


Suppose there is a function $ f$ such that $ |f(x)|<\equal{}1$, why do I can say that f has a least upper bound M, such that
$ |f(x)|<\equal{}M$ ? M is real.

this is an axiom! it can't be proven without stepping out into ZFC territory.

Ok. But what can I say about this M? And I still didn't know why of this M, the function is already limited. And let's suppose exists this M how can I garantee that there is not an $ a$ such $ M < a < 1$?
and what is this ZFC territory?

ZFC i.e Zermelo-Fraenkel set theory with the Axiom of Choice

how can you guarantee? well, you said the function was bounded above by 1, so the least upper bound $ M$ must be $ \leq 1$.


geometry
parallelogram
analytic geometry
graphing lines
slope
vector

[b]Three vertices of parallelogram PQRS are P(-3,-2), Q(1,-5), R(9,1) with P and R diagonally opposite. What is coordinate S?[/b]

[hide="Approach 1"] Diagonals bisect each other [/hide]
[hide="Approach 2, longer"] Slopes of opposite sides equal [/hide]

[hide="Approach 3"] $ \overline{QP} \plus{} \overline{QR} \equal{} \overline{QS}$ [/hide]

[quote="KMST"]I’d consider vectors (1X2 matrices to write it easier). Each vector represents direction (slope) and distance. Let O be the origin. Note that I'm making up my own notation.
O→P = [-3  2]
O→Q = [1  -5]
O→R = [9  1]
P→S = Q→R = O→R  -   O→Q  = [9  1] – [1  -5] = [8  6]
O→S = O→P  +   P→S = [-3  2] + [8  6] = [5  8][/quote]

wait.. so your answer is (5,8)?

Somehow I get (5,4)..

oh I know what you did wrong, you miscopied P its (-3,-2) not (-3,2)

oh well yea I basically did your way except I didnt write all the terms out.


As 7h3.D3m0n.117 suggested, I've made a seperate thread for this. The Mole is a fun game to watch(originally on TV), and even more so to play. It works well on forums, and I could see it being popular on AoPS. Would anyone else play?

Here's a link to the wikipedia article on it:[url]http://en.wikipedia.org/wiki/The_Mole[/url]

Sounds interesting.  :)
How would it be played?

Basically, there would be a total known only to the mod and the mole. The mole would win if the players did not meet this goal by the end, [i]and[/i] if at least one of the final two players doesn't get a perfect score on the final test. Otherwise, the player with the highest score on the final test wins.

Even through the uncertainty, it sounds cool...
i'm in  :lol:

In the tasks, the players earn money. On the real show, the winner gets all the money at the end. Of course, this is unrealistic online. So, to give the players an incentive to actually win money, they need to reach a certain monetary total to win. (the same goes for the mole-if total wasn't important, why should they even sabotage anything?)

I'll play, although I can't think of any possible challenges.

I have a bunch(I've played in several forum versions, and have some ideas of my own). This isn't a sign up-just something to gauge interest.

If anyone wants to look at some of the tasks in the real show, click [url=http://en.wikipedia.org/wiki/The_Mole_%28US_TV_series%29]here[/url] or [url=http://en.wikipedia.org/wiki/Mole_2:_The_Next_Betrayal]here[/url]

OK-I guess this'll generate enough interest. I'm now accepting applications. To apply, simply fill out the form below, and pm it to me.

Birthday(mm/dd only):
Favorite (strictly) two digit integer:
Favorite color:
Are you a leader or a follower?:
Have you seen the mole:
If so, do you like it? Why or why not?:
Would you like to be the mole? Why or why not?:
What is your strategy to win(assuming you're a player)?:

I don't want novels, just a few sentences for the why and why nots. I will only accept people whom I see as committed, and who won't just stop playing. I'm formatting this now-I have many of the tasks already set up.
I'd like to start sometime next week.

How long will the game take?

I don't know exactly how long I can be committed...

How do I know I'm in?
I PMed him, but haven't heard anything yet...

I'm anticipating the game lasting about, I'd say, a month and a half-2 months or so. The tasks are decided-there will be 10 players(including the mole).

Now, the player list has yet to be determined. Applications are due 3:00 EST(or whenever after that I'm on...)
I've already gotten more than 10 applications, so I'll have a good replacement list. :)

I will be hand picking the players. I will select you based on previous behavior, quality of your app, and, of course, how well I feel you will play the game.

This is just a note: the tests will be slightly different than on the show. Since I can't stop you from looking at the thread during the tests, I've decided just to post the tests, and people can use whatever resources they want in completing them. The player with the lowest score, and latest submission in the case of a tie, will be executed. Plus I couldn't write code for my life... :blush: This method works well, so don't worry. (If any non players are willing to write this code, don't hesitate to pm me!)

Remember, get those apps in today and tomorrow, I won't wait...

I might join the next one, just so I can see how it goes first.  :)

I will join the replacement list.

Or the next game.


LaTeX
function
limit
algebra
domain
inequalities
real analysis

If $ X\subset R\wedge X\neq\emptyset$ and v belongs to R ,THEN v=SupX if and only if:

1)$ x\leq v$, for all x belonging to X

2) FOR all  ε>0  there exists an x belonging to X AND SUCH that x>v-ε

FORMALIZE the above theorem and then give a formal proof

[quote="JBL"][quote="τρικλινο"]$ X\subset R\wedge X\neq\emptyset$[/quote]  This is an example of a phrase that has been made less readable and more ambiguous by formalization.  Also, if you were more consistent in your use of [[LaTeX]], it would be much easier to read your posts.[/quote]

[b]THAT is the way it is written in analysis books. Ι copied it from one[/b].

BUT to help you in formalizing the above AND then give a formal proof ,here is the formalization in words:

FOR any non empty subset X OF the real Nos ,R AND any real No ,v ,then v=SupX if and only if :

1) FOR all x belonging to X ,x is less or equal to v

2) FOR all ε>0 there exists an , x belonging to X and such that x>v-ε

I am sorry if you don't know the meaning of
"$ \wedge$" which is: and

[quote="t0rajir0u"][quote="τρικλινο"]I am sorry if you don't know the meaning of
"$ \wedge$" which is: and[/quote]
There's a world of difference between "this is difficult to read" and "I don't understand this notation."  JBL's point is that in making your notation more compact you have only made this thread more of a hassle to respond to.  It is your responsibility to phrase your questions in a legible way, and it is also your responsibility to do it [b]politely.[/b][/quote]

[b] I DO APOLOGIZE ,IF my respond was not done in a polite way[/b].

MY intentions are not to cause any mis comfort to anybody

as a rule of thumb, I try to make sure that using a mathematical symbol in place of a more easily understood word or phrase is only done when it actually saves time.

For example, '$ X$ is nonempty' is essentially the same amount of time and space as '$ X\not\equal{}\varnothing$', so the former is preferable.  Using $ \wedge$ for 'and' should pretty much never be done in mathematical writing, since $ \wedge$ has many possible meanings and it doesn't save any time or space!

In general I don't like using symbols for quantifiers in mathematical writing, I find that things flow much better if you just write things out in english.  Of course in logic this may be unavoidable but even most graduate texts in analysis or mathematics don't usually inundate you with unnecessary symbols.

suppose you write the limit of a function at an accumulation point ,a in a semi formalized form like most books in analysis  do,as you mentioned:

And $ \lim_{x\rightarrow a}{f(x)} \equal{} m\Longleftrightarrow \forall\epsilon>0.\exists\delta>0$ such that:

|f(x)-m|<ε ,whenever xεD(f) and 0<|x-a|<δ

where D(F) is the domain of the function

Now let us write the limit in fully formalized form:

$ \lim_{x\rightarrow a}{f(x)} \equal{} m\Longleftrightarrow\forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x( x\in D(f)\wedge 0<|x\minus{}a|<\delta\Longrightarrow |f(x)\minus{}m|<\epsilon))]$

 In trying to [b] PROVE  what the negation of the definition is [/b],which of the two definitions would you use??

I definitely wouldn't use the latter. It's over-formalized, redundant, and basically unreadable.

Mathematics has grammar, and is thus human-readable as a language. The version of formal logic you're using ignores and obscures this grammar, making it not human-readable except in certain simple cases.
In fact, I'll also find fault with your first version of the definition: the period in "$ \forall \epsilon>0.$" should be a comma, because that's not a complete sentence. It has no verb, and needs to be connected to the next clause.

So how would you [b]prove what the negation of the limit is ??[/b]

You want the negation of the statement:

For all $ \epsilon > 0$ there is some $ \delta > 0$ such that for all $ x \in D(f)$ with $ 0 < |x \minus{} a| < \delta$, we have $ |f(x) \minus{} m| < \epsilon$?

That would be:

There is some $ \epsilon > 0$ such that for all $ \delta > 0$ there is some $ x \in D(f)$ with $ 0 < |x \minus{} a| < \delta$, but also having $ |f(x) \minus{} m| \geq \epsilon$.

I don't see what you would want to prove, really.

("Formal")
$ \forall \epsilon > 0 : \exists \delta > 0 : (0 < |x \minus{} a| < \delta \Longrightarrow |f(x) \minus{} m| < \epsilon)$ negates to
$ \exists \epsilon > 0 : \forall \delta > 0 : (0 < |x \minus{} a| < \delta \not\Longrightarrow |f(x) \minus{} m| < \epsilon)$, the last part of that statement indeed implying that there is some $ x$ satisfying the inequalities $ 0 < |x \minus{} a| < \delta$, but not the last one.

That is not a proof

It is an intuitive conclusion.

How do you know that the formula you have written is true

His statement is the correct negation of your statement as it follows the formal rules of negation of [url=http://en.wikipedia.org/wiki/Universal_quantification#Negation]universal[/url] and [url=http://en.wikipedia.org/wiki/Existential_quantification#Negation]existential[/url] quantifiers.

I think "prove" in your question was taken to mean "find" since the "proof" follows directly from the laws of negation. Maybe you could have phrased your question as: provide the negation of the given statement, and justify why it is the correct negation. Prove sounds awkward in this instance.

When you are asked to show that:

$ \sim \forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x( x\in D(f)\wedge 0<|x\minus{}a|<\delta\Longrightarrow |f(x)\minus{}m|<\epsilon))]$

Logically implies :

$ \exists\epsilon[\epsilon>0\wedge\forall\delta(\delta>0\Longrightarrow\exists x( x\in D(f)\wedge 0<|x\minus{}a|<\delta\wedge |f(x)\minus{}m|\geq\epsilon))]$.

[b]is not that a proof???[/b]

There is unfortunately not a lot of content to negating this formalization beyond
applying simple formal rules. This could easily be done by computer, for example,
which some would say is part of the reason for creating this formalizations in
the first place --- the idea of a computerized "proof-checker" can be appealing.
On the other hand, formal logic manipulation by people doesn't accomplish much
in terms of content, and does not appropriately represent knowledge: we tend
to think in pictures and concepts when we understand things, and resort to
symbolic manipulation when we don't know any better. This is why I would say
that good mathematical writing makes leaps when showing the details would
otherwise distract from the main idea.

I am, however, completely turned off by the amount of bold and all-caps in this
thread.

Learning mathematics without learning formal proofs at the same time is like learning medicine without learning Anatomy

No, it's like doing medicine without insisting that every treatment be justified beginning with quantum physics and building up.  Justifying in symbols what can more easily be justified in words does not increase correctness, and it increases rigor only in a very trivial sense that isn't important when working far enough away from the axioms (as one does in nearly all of mathematics outside of logic and set theory).


The letters $A,B,$ and $C$ represent different digits, $A$ is a prime, and $A - B = 4$. If the number $AAABBBC$ is a prime, compute the ordered triple $(A,B,C)$.

This is an ARML problem, isn't it?  I remember getting it wrong because I forgot a case :(

[hide="Hint 1"] What can't $C$ be?[/hide]
[hide="Hint 2"] $3$ divisibility rule ;) [/hide]
[hide="Hint 3"] $11$ divisibility rule ;) [/hide]

well I figured that A = 5 or A = 7.

If A = 5, then you have 5551117 as the only possibility.

If A = 7 then you have 7773331 as the only possibility.

Where I got confused was how to eliminate one of these. I endede up putting (5,1,7) but it was actually (7,3,1)

Oh doh. I didn't realize that $5+5+1 + 7 - 5 - 1 -1 = 11$  :blush:

I did this problem last Sunday as homework. Yes, agolsme, you wer right.


Let  a,b,c be positive real numbers . Prove that

$ \frac{a\plus{}3c}{a\plus{}2b\plus{}c}\plus{}\frac{4b}{a\plus{}b\plus{}2c}\minus{}\frac{8c}{a\plus{}b\plus{}3c}\ge12\sqrt{2}\minus{}17$

[quote="teerawat"]Let  a,b,c be positive real numbers . Prove that

$ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c}\ge12\sqrt {2} \minus{} 17$[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147142


inequalities
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inequalities proposed

Sorry  i  dont  have  time  to  learn  Latex....

i have an idea for this inequality
since $\displaystyle \sum_{i=1}^n \frac 1{1+x_i} =1$ we may assume
$x_i=\frac{S-a_i}{a_i}$ where $S=\sum_{i=1}^n a_i$ and $a_i \geq 0$
and the inequality is equivalent to
$\sum^n_{i=1} \sqrt{\frac{S-x_i}{x_i}} \geq \sum^n_{i=1}  (n-1)\sqrt{\frac{x_i}{S-x_i}}$
we will prove it for $n=4$, $n=k$ similarly
$\sum \sqrt{\frac{a+b+c}{d}} \geq \sum \frac {\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{3d}}$
but $\frac {\sqrt{a}}{\sqrt{3}}(\frac 1{\sqrt{b}}+\frac 1{\sqrt{c}}+\frac 1{\sqrt{d}})\geq 3\sqrt{3}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}+\sqrt{d}} \geq 3\sqrt{\frac{a}{b+c+d}}$
write down similar inequalities and add them up, we will have what we need
 :D

Try!!!!!!!!!! :D
This problem is not hard

I solved this 7 months ago
$\sum^n_{i=1} \frac {x_i+1}{\sqrt x_i} \geq n\sum^n_{i=1} \frac{1}{\sqrt{x_i}}$
WLOG suppose:
a)$x_1 \geq x_2 \geq ... \geq x_n \geq 1$ or
b)$x_1 \geq x_2 \geq ... \geq x_{n-1} \geq 1 \geq x_n$
Now for a and b use chebishev inequality and proof is complete.

[b]W.L.O.G[/b], we can suppose that: $ 0<x_{1}\leq x_{2}\leq\dots\leq x_{n},(n\geq 2).$

It's enough to prove that: $ \left(\displaystyle\sum_{i\equal{}1}^n\frac{x_{i}\plus{}1}{\sqrt{x_{i}}}\right)\left(\displaystyle\sum_{i\equal{}1}^n\frac{1}{x_{i}\plus{}1}\right)\geq n\cdot \left(\displaystyle\sum_{i\equal{}1}^n\frac{1}{\sqrt{x_{i}}}\right).$

Consider the function $ f(x)\equal{}\frac{x\plus{}1}{\sqrt{x}},x\in (0,\infty).$ It's easy to prove that $ f$ is [u]nondecreasing[/u] on $ (1,\infty)$ and $ f(x)\equal{}f\left(\frac{1}{x}\right),$ for all $ x>0.$

Now, we'll prove that only $ x_{1}$ can be less than $ 1.$ From the given condition we get that: $ \frac{1}{x_{2}\plus{}1}\leq 1\minus{}\frac{1}{x_{1}\plus{}1}\equal{}\frac{x_{1}}{x_{1}\plus{}1}.$ Hence $ x_{2}\geq\frac{1}{x_{1}}.$

So $ f(x_{1})\equal{}f\left(\frac{1}{x_{1}}\right)\leq f(x_{2})\leq\dots\leq f(x_{n}).$ Thus according to [color=blue][b]Chebyshev inequality[/b][/color] we have:

$ \left(\displaystyle\sum_{i\equal{}1}^n\frac{x_{i}\plus{}1}{\sqrt{x_{i}}}\right)\left(\displaystyle\sum_{i\equal{}1}^n\frac{1}{x_{i}\plus{}1}\right)\geq n\cdot \left(\displaystyle\sum_{i\equal{}1}^n\frac{1}{\sqrt{x_{i}}}\right).$ Equality occurs if and only if $ x_{1}\equal{}x_{2}\equal{}\dots \equal{}x_{n}\equal{}n\minus{}1.$

[quote=nickolas]Let $x_i$ be positive reals such that $\displaystyle \sum_{i=1}^n \frac 1{1+x_i}=1$. Prove that
\[ \sum^n_{i=1} \sqrt {x_i} \geq (n-1)\sum^n_{i=1} \frac 1{\sqrt {x_i}}. \]

[Admin Edit: Made Latex compliant].[/quote]

Another proof :

[quote=nickolas]Let $x_i$ be positive reals such that $\displaystyle \sum_{i=1}^n \frac 1{1+x_i}=1$. Prove that
\[ \sum^n_{i=1} \sqrt {x_i} \geq (n-1)\sum^n_{i=1} \frac 1{\sqrt {x_i}}. \]

[Admin Edit: Made Latex compliant].[/quote]

Published by the professor Bogomolny too.
http://www.cut-the-knot.org/arithmetic/algebra/ConstraintInequality4.shtml