Does anyone(I know fredbel does..) know english? If so, can we chat in this thread with english to enhance our knowledge? Just so you know I have had sixteen years latin but am just starting college level.

STARTS OUT:

How are you?

I hope my grammar is correct...

Wow, now this is really a parody.  :rotfl:

Are you serious making this topic?, don't we use [b]English[/b] in this forum?  :D
Or you want pure English topic in this topic.  :D

Nein, ich verstehe nicht Englisch! Aber ich möchte viele Englisch Wörter und auch Grammatik (selbstverständlich) lernen. Hilfen Sie mir?

Wait, are you sure about that?  :o

Translation : No, I don't understand English! But, I want to learn a lot of English word and (you should write und, not and) Grammar too. (Wow, my German is quite good :P)

[quote="BoesFX"]Translation : No, I don't understand English! But, I want to learn a lot of English word and ([b]you should write und, not and[/b]) Grammar too. (Wow, my German is quite good :P)[/quote]

Rats, I've been discovered  :blush:

[quote="akalra1"]OMG BUBKA THIS IS spam BECUZ ALL OF US KNO ENGLISH CUZ THIS IS AN ENGLISH FORUM (NOT TO BE RACIST OR ANYTHING) BUT WAT WE DO NEED IS A hindi FORUM![/quote]

*fireforce sighs deppressingly*

Would it hurt to read the rules before you post? First of all, if you want to get the message accross there is no need to write in all caps we can read you just fine normal type.

Anyway, I am fluent in Pig Latin, Ubbi Dubbi, and Gibberish. English is a bit hard.  :P

You are making fun of me!!!! It is absolutely certain! I just like chatting in other languages. *cries and sobs and whines*

Really were you just doing it b/c it is funny?

Okay let me try to use ENGLISH.... Hi everyone. My name is Funcia. The evil child threw stones at the last temple. The brothers gave faith to the bad servants.

Je suis desole.

Nah, no bad feelings funcia, I was simply poking fun at all the obscure (or not so much for those who use them) veranculars of the english language.

I guess I can add basic (yes, the computer programming laguage) and leetspeak to my list of languages too.

it's okay...everyone has gotten a post with their language(it would seem) well i poked fun at my greek curriculum here...

OK OMG THIS IS SPAM CUZ THIS WHOLE TOPIC IS IN ENGLISH SO... YEA

we do need a hindi forum

but it would take too long to read it in hindi for some.

and no one goes to the india forum anyhow.

YEA TAT ONE IS DEAD SO I CREATED MY TOPIC

*ooh*
*ahh*

I no speek teh very good english :blush:

Try doing that wearing a really touristy shirt too. The kind that people buy in airports.

[quote="futuremontrealian"]You know what's fun?  Getting a french-english dictionary and going to restaurants (in the US) and ordering food with a french accent and checking your dictionary constantly.  I once fooled a cashier to ask me what country I was from... it was quite interesting :)[/quote]
I don't have to go to the USA to encounter all sorts of language problems, in fact, I don't even have to leave my country if I would be looking for that :( .

[quote="funcia"]Je suis desole.[/quote]
That should be "désolé" if you are man and "désolée" if you are woman.

[quote="futuremontrealian"]Unless of course you're one of those anti-gender-difference people protesting for equal linguisitic rights for men and women.  Lol.  "Equal Linguistic Rights" is a funny concept :)[/quote]
Je parle
Tu parles
Il/Elle parle

---> your teacher is a man

Je parle
Tu parles
Elle/il parle

---> your teacher is a woman :)

Lemme guess.
Il - he
Elle - she

Incidentally Fred taught me this:
ich bin
du bist
er/sie ist

[quote="bubka"]

er/sie ist[/quote]

er/sie/es ist  :lol:

sie/Sie sind

Maybe we should stop writing German since this is English thread and there is German thread.  :wink:

Warum?
Ich möchte auf Deutsch screiben!!!
 :P
I think you have a point, but as I've seen... that usually dosen't matter in g&ff so...

Hello. My name nutz. I happy. You happy?


ARML geometry calculus integration trigonometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b) : (b+c) : (c+a) = 6 : 7 : 8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

wow, are these actually arml problems?  they seem to be significantly easier than the ones i encountered last year :?

1.
[hide]$a^2+b^2=9$
$(a+b)^2=17=a^2+b^2+2ab$
$2ab=8\rightarrow \frac{ab}{2}=\fbox{2}$[/hide]

2.
[hide]$a+b=6x, b+c=7x, c+a=8x$
$2(a+b+c)=28=21x\rightarrow x=\frac{4}{3}$
$6x+c=14\rightarrow c=14-8=\fbox{6}$[/hide]

3.
[hide]$z=\cis{0+72n}$
the sum any five consecutive powers of $z=0$  because the points are equally distributed about the unit circle
$z^{15}=z^{50}=\fbox{1}$[/hide]

4.[hide]
try numbers and increase sides
square = $200$
hexagon = $300$
octagon = $400$
thus $600 = 8+2(2) = \fbox{12}$ sides[/hide]

5.
[hide]the distance from the center of circles to $XY=3$
they are perpendicular, so we have an isosceles right triangle and distance is $3\sqrt{2}$
unrelated question: why cant i box square roots? :huh:[/hide]

6.
[hide]the numbers desired are of the form $pq$ or $p^3$ where $p,q$ are prime
$p=2$ yields $8$ values for $q, (3,5,7,11,13,17,19,23)$
$p=3$ yields $5$ values for $q, (5,7,11,13,17)$
$p=5$ yields $1$ value for $q (7)$
$p^3$ has only $2$ values, so $8+5+1+2=\fbox{16}$[/hide]

7.
[hide]$n!+(n+1)!+(n+2)!=n!(1+n+1+n^2+3n+2)=n!(n+2)^2$
there must be at least two $7$'s in the expression, so
$n!\ge 14$ or $n+2=7x$
lhs yields $14, 15, 16$ and rhs yields $5, 12$, so the final answer is $\fbox{5, 12, 14, 15, 16}$[/hide]

These are very old ARML problems.  The competition has developed significantly as the students have developed.

fo #8, if anyone wants a hint, use the fact that a = 2RsinA, b = 2RsinB, c = 2RsinC, and the result will follow shortly thereafter.


geometry
3D geometry
geometric transformation
reflection
number theory
complex numbers
number theory unsolved

Bothering me that I can't get this to work out. I think I need to go lie down.

Determine the smallest integer n with the property that in every n-digit number, there exists a string of consecutive digits whose product is a perfect square.

what does perfect square mean? because i think that 2020...0202 isn't right solution, it's too easy for put it into olympiad section... maybe product zero is a perfect square...

[quote="skrecok"]what does perfect square mean?[/quote]

A integer number $n$ is said to be a perfect square if there exists another integer $k$ such that $n=k^2$.

[quote="Rodolphe2005"]I don't understand the problem. If I take 202020202.......02 there is no string of consecutive digits whose product is a perfect square ?! and this number can be as larger as I want[/quote]

Well, in that case why don't you take 11. Read the problem a little more carefully. You will notice that it says that for every string of lenght $n$ the statement is true. Then you have to see that non of the digits is 0, 1, 4, 9 because those are squares by themselves. Therefore there can be only 2,3,5,6,7,8. Now, you can't have two equal digits together, because the product would be a square, you can't have the combinations 28, or 263, etc.

The problem actually seems to be interesting.

Best,

[quote="skrecok"]so zero is perfect square too, thanks... i thought that every square except zero is perfect (btw. why there is always written "perfect square" when "square" is good and shorter? :) )[/quote]

Mainly because of context. Perfect square, as well as perfect cube use to tell to the trained math problem solver that you are talking about number theory, regardless the rest of the context, that some time may be somehow ambiguous. Square or cube alone may refer to several things. It may refer to a geometric figure with four equal sides, four right angles. In some algebraic structures it refer to elements of the form $\sigma^2$ for $\sigma$ in the structure (for example, $-1$ is a square in the integers $\mod 10$), and in the reals, every non-negative number is a square. In the Complex numbers, every number is a square.

Let [tex]d_1d_2...d_n[/tex] be a n-digit number [i]without[/i] the property

Consider the products
[tex]p_1 = d_1[/tex]
[tex]p_2 = d_1d_2[/tex]
[tex]p_3 = d_1d_2d_3[/tex]
...
[tex]p_n = d_1d_2d_3...d_n[/tex]

We are only interested in the parity of the powers of 2,3,5,7.
There are 2^4 = 16 different possible values of the ordered tuple (w(mod 2), x(mod 2), y(mod 2), z(mod 2). Also, it cannot take the value (0,0,0,0) or p_i would be a perfect square.
Therefore, if n >=16 , some pair of tuples for [tex]p_i[/tex] and [tex]p_j (i < j)[/tex] are the same. Then [tex]p_j/p_i[/tex] is a perfect square, and it equals [tex]d_id_{i+1}...d_j[/tex], a contradiction.

Therefore n<16.
So we have proven that 16 is the greater than or equal to smallest number n such that for all n-digit numbers, some substring of the digits has it's product equal to a perfect square.

Now we must find a 15-digit number without the property. This is equivalent to the problem "can we list all binary numbers from 0 to 31 such that only one digit changes from each number to the next?" The answer is yes. Old computers used to do this to save time! There is a simple pattern. Basically you start off with 0000 and 0001, then you add a 1 in the first unused digit, and "reflect" the previous part of the list.
0.) 0000
1.) 0001
2.) 0011
3.) 0010
4.) 0110
5.) 0111
6.) 0101
7.) 0100
8.) 1100
9.) 1101
10.) 1111
11.) 1110
12.) 1010
13.) 1011
14.) 1001
15.) 1000
This corresponds to the number 232523272325232

Hello,

I haven't solve problem, but I don't see the point in your solution. Let's start with

[quote="schulmannerism"]There are 2^4 = 16 different possible values of the ordered tuple (w(mod 2), x(mod 2), y(mod 2), z(mod 2). Also, it cannot take the value (0,0,0,0) or p_i would be a perfect square.
[/quote]

What is the relation between the fourtuples and the $p_i$'s.

[quote="schulmannerism"]Therefore, if n >=16 , some pair of tuples for [tex]p_i[/tex] and [tex]p_j (i < j)[/tex] are the same. Then [tex]p_j/p_i[/tex] is a perfect square, and it equals [tex]d_id_{i+1}...d_j[/tex], a contradiction.
[/quote]

Why does $\frac{p_j}{p_i}$ has to be integer at all?

[quote="schulmannerism"]This corresponds to the number 232523272325232[/quote]

OK, with this number is easy to se that $n\geq16$, but I still don't understand your argument. Could you please reexplain?

[quote="schulmannerism"]
Consider the products
[tex]p_1 = d_1[/tex]
[tex]p_2 = d_1d_2[/tex]
[tex]p_3 = d_1d_2d_3[/tex]
...
[tex]p_n = d_1d_2d_3...d_n[/tex]
[/quote]
[quote="schulmannerism should have"]
Consider the products
[tex]p_1 = d_1[/tex]
[tex]p_2 = d_1*d_2[/tex]
[tex]p_3 = d_1*d_2*d_3[/tex]
...
[tex]p_n = d_1*d_2*d_3*...*d_n[/tex]
[/quote]

Ugh yes it was nice and easy. I do feel stupid now.

[quote="schulmannerism"]I also should have said
[tex]p_i = 2^w3^x5^y7^z[/tex][/quote]

OK, after those two comments, the solution has a lot more sence, and actually is totally correct and very clever. Nice problem.

Best,


In triangle ABC, AB=1, AC=2, Side BC and the median from A to BC have the same length, what is BC?


function
integration
calculus
derivative
real analysis
real analysis unsolved

Suppose $f\in C([a,b])$. All $x \in [a,b]$ except $\{x_n\} \subset [a,b]$ satisfy $f'(x) > 0$. Show that $f$ is strictly increasing.

Is $C([a,b])$ the set of continuous functions from $[a,b]$to $R$?

I think it's continuous functions over the $[a,b]$

I don't think it's hard. Take $a_1<b_1\in [a,b]$. $f(b_1)-f(a_1)=\displaystyle \int_{a_1}^{b_1}f'(x)dx\ge 0$ because when we construct the Riemann sums, given a certain interval, we can always choose points where $f'$ takes positive values (because the set of points where it doesn't take positive values is at most countable). This means that $f$ is increasing (not necessarily strictly, for now).

Now assume $f(a_1)=f(b_1)$. Then $f$ is constant on $[a_1,b_1]$, so its derivative is $0$ on the entire interval, which is clearly false. this means that $f(a_1)<f(b_1)$, and we're done because $a_1<b_1$ were chosen arbitrarily.

Maybe I misunderstood and $\{x_n\}$ isn't a countable sequence?

[quote="grobber"]I don't think it's hard.[/quote]
I don't mean it's hard.
[quote="grobber"]Take $a_1<b_1\in [a,b]$. $f(b_1)-f(a_1)=\displaystyle \int_{a_1}^{b_1}f'(x)dx\ge 0$ because when we construct the Riemann sums, given a certain interval, we can always choose points where $f'$ takes positive values (because the set of points where it doesn't take positive values is at most countable).[/quote]
Can you avoid to use that theorem? This exercise appears before Lebesgue Integral. In Riemann Integral, we can proof that when there are only finite points but not countable points.

I don't think I know what you mean when you say it's ok if there are finitely many points but not if there are countably many. Instead, here's what I mean: the integral is the limit of exressions of the type $\displaystyle \sum_{i=1}^n f'(\xi_i)(a_{i-1}-a_i)$. In each interval $(a_{i-1},a_i)$ we can find $\xi_i$ s.t. $f'(\xi_i)>0$ because otherwise it would mean that $\{x_n\}\supseteq (a_{i-1},a_i)$, which is obviously wrong since the set on the left is countable and the one on the right isn't.

Am I missing something?

I'm terribly sorry for the stupid mistake I've made: we can't know that $f'$ is integrable. I think this is what liyi meant, but I didn't understand. I guess the problem is still open.

This gives a new meaning to a problem posted by Harazi: give an example of a function and an interval s.t. the derivative of the function is positive on a set which is dense in the iterval, but which isn't strictly increasing on the interval. The problem posted by liyi shows that the set of points where $f'$ isn't positive should be uncountable.

I'll have to ask again: is this the problem, liyi:

Given $f:[a,b]\to R$ a derivable function s.t. $f'>0$ everywhere except for a countable set of points, show that it's strictly increasing on $[a,b]$ ?

This seems to be a remarkable result, and I was wondering why it isn't better known (maybe it is, and it's only me who has never heard of it :)).

You know, I have no idea why we left this unsolved:

$f'$ has Darboux property, so if $f'(x_0)<0$ for some $x_0$, then $f'([a,b])$ will include an entire interval included in $(-\infty,0)$, $f'(x)<0$ only for countably many values of $x$, so $f'$ will take a countable set to an interval, which is obviously false, so $f'\ge 0$, and the rest is easy.

I hope there are no mistakes this time.

[quote="liyi"][quote="grobber"]
I'll have to ask again: is this the problem, liyi:

Given $f:[a,b]\to R$ a derivable function s.t. $f'>0$ everywhere except for a countable set of points, show that it's strictly increasing on $[a,b]$ ?
[/quote]
Yes, it is the problem.[/quote]
sorrrry, i didn't see your description clearly.
The original problem doesn't say that $f$ is derivable.

But they mention $f'$ in the text of the problem! Maybe it's differentiable everywhere except for $\{x_n\}$?

yes. $f'$ may or may not exist on $\{x_n\}$


If $e$ is prime you can use the quadratic formula.  (Modular division and modular roots!  This is equivalent to completing the square, of course.)

Otherwise, solve for every prime factor of $e$ and use the Chinese Remainder Theorem.


algorithm
linear algebra
linear algebra theorems

Can someone give me a link where algorithm for svd is described and maybe a proof of the algorithm ? (i tried google but haven't found)

It's chapter 2.6 of [url=http://www.library.cornell.edu/nr/cbookcpdf.html]Numerical Recipes in C[/url].


[quote="usaha"]In a bag, there are 4 yellow marbles, 5 red, 3 green, and 9 black.  What is the probability that I draw atleast one red or yellow marble if I draw 3 marbles from the bag without replacement?[/quote]
[hide] Complimentary Counting: $1-\frac{12}{21}\frac{11}{20}\frac{10}{19}=1-\frac{22}{133}=\frac{111}{133}$ [/hide]

[hide]It is actually complementary counting but same difference.

You still get the same answer of $\frac{111}{133}$.[/hide]


Anyone have a short proof of Muirhead's Inequality lying around? Or knows where I can find a proof of it online? Kedlaya's notes were incomplete/confusing...

Thanks.

It's always nice if you state the inequalities you are asking about, because most people may not know them by their names.

Isn't this what KK calls the "bunching theorem"?

It is just "weighted AM-GM"...remember that in symmetric sums you can replace something like a^4*b^3 by 2/3 * a^4*b^3+ 1/3 * a^4*c^3, which is greater than  a^4*b^2*c (for example).

By the way, is there a better method to solve those big homogenous inequalities than multiplying out and using Schur and bunching?

[quote="Celeborn"]Isn't this what KK calls the "bunching theorem"?[/quote]

Yes, this is 'bunching.'

[quote="Celeborn"]By the way, is there a better method to solve those big homogenous inequalities than multiplying out and using Schur and bunching?[/quote]

Yes, there are usually better ways to do it, but these methods are usually less obvious. However, schur+bunching is very effective if you're willing to give the algebra enough time (and if you only have 3 variables so you [i][b]can[/b][/i] use schur :)).

speaking of Schur, why does it come up frequently?  i mean, if its proof is so obvious (just compare the absolute values of terms) why does it appear so often after expanding inequalities?

thank you!

There's a version of Schur that works in general: 3d_1 + d_3 >= 4d_1 d_2, where the d's are the symmetric averages as defined by Kedlaya.  I've never actually used this (terrible at inequalities) but it might come in handy for >3-variable inequalities sometimes.

[quote="zabelman"]
Yes, there are usually better ways to do it, but these methods are usually less obvious. However, schur+bunching is very effective if you're willing to give the algebra enough time (and if you only have 3 variables so you [i][b]can[/b][/i] use schur :)).[/quote]

[quote]remember that in symmetric sums you can replace something like a^4*b^3 by 2/3 * a^4*b^3+ 1/3 * a^4*c^3[/quote]

Hmm, could you clarify this please? Are you saying that SymmetricSum(a^4*b^3) = SymmetricSum(2/3 * a^4*b^3 + 1/3 * a^4*c^3)?

Yes, because if you symmetric that, you get 2/3rds of what you had before in the first term, and the third third in the second term.

[quote="Fierytycoon"]Anyone have a short proof of Muirhead's Inequality lying around? Or knows where I can find a proof of it online? Kedlaya's notes were incomplete/confusing...[/quote]
Not sure any proof online. I am sure it's in the book "Inequalities" by G. H. Hardy, J. E. Littlewood, G. Plya. See  [url=http://www.amazon.com/exec/obidos/tg/detail/-/0521358809/qid=1084555441/sr=1-2/ref=sr_1_2/104-0048912-6303167?v=glance&s=books]here[/url].

The whole point of the proof is what Kiran Kedlaya doesn't give: to prove that if {a} majorizes {b}, then {b} can be represented as a linear combination of permutations of {a} such that the sum of the coefficients is 1.  After that, the result follows immediately by weighted AM-GM (don't worry about those scary subscripts; just try a few examples).


let 1/1+rt(x) be a and similarly define b
lhs is a^2+b^2
but from definition,x=(1/a -1)^2 and similarly y=(1/b -1)^2

from AM-HM

a^2+b^2>=4/(1/a^2+1/b^2)
it is sufficient to show that given RHS is less than the 2HM
ie 4/(1/a^2 +1/b^2)>=2/{(1/a -1)^2+(1/b -1)^2 +2}

on cross multiplying
,
     2/a^2-4/a+2+2/b^2-4/b+2>=1/a^2+1/b^2
ie   (1/a -2)^2+(1/b -2)^2>=0

hence it is proved....
equality iff a=b=1/2  ie  x=y=1

[quote="Dimitris X"][quote="linda2005"]hello,

  Let, $ ;x,y \,\geq 0$

prove that : $ \;\; \frac {1}{(1 \plus{} \sqrt {x})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {y})^2} \,\geq \, \frac {2}{x \plus{} y \plus{} 2}$[/quote]
From Adreescu inequality we have:
$ RHS \geq \frac {4}{x \plus{} y \plus{} 2(\sqrt {x} \plus{} \sqrt {y}) \plus{} 2}$
So we have to prove that
$ \frac {4}{x \plus{} y \plus{} 2(\sqrt {x} \plus{} \sqrt {y}) \plus{} 2} \geq \frac {2}{x \plus{} y \plus{} 2}$
Which is
$ 2(x \plus{} y) \plus{} 4\geq x \plus{} y \plus{} 2(\sqrt {x} \plus{} \sqrt {y}) \plus{} 2 \rightarrow x \plus{} y \plus{} 2\geq 2(\sqrt {x} \plus{} \sqrt {y}) \rightarrow (\sqrt {x} \minus{} 1)^2 \plus{} (\sqrt {y} \minus{} 1)^2 \geq 0$
So we prove it..Equality occurs when $ x \equal{} y \equal{} 1$[/quote]
A little typo, I think you meant $ LHS\geq\frac {4}{x \plus{} y \plus{} 2(\sqrt {x} \plus{} \sqrt {y}) \plus{} 2}$............ :oops:

Anyways, a VERY NICE solution! :lol:


Find all real solutions of the equation: $ [x]\plus{}[x\plus{}\dfrac{1}{2}]\plus{}[x\plus{}\dfrac{2}{3}]\equal{}2002$

Let $ \{x\}\equal{}x\minus{}[x]$.

Clearly, $ 0 \leq \{x\} < 1$. We have three cases:

a) If $ 0 \leq \{x\} < \frac{1}{3}$, then $ [x]\equal{}\left[x\plus{}\frac{1}{2}\right]\equal{}\left[x\plus{}\frac{2}{3}\right]\equal{}r \rightarrow 3r\equal{}2002 \rightarrow r$ is not an integer, contradiction.
b) If $ \frac{1}{3} \leq \{x\} < \frac{1}{2}$, then $ [x]\equal{}\left[x\plus{}\frac{1}{2}\right]\equal{}r$ and $ \left[x\plus{}\frac{2}{3}\right]\equal{}r\plus{}1$, so $ 3r\plus{}1\equal{}2002 \rightarrow r\equal{}667$. Hence all $ x \in \mathbb R$ such that $ 667\plus{}\frac{1}{3} \leq x <667\plus{}\frac{1}{2}$ satisfy the conditions.
c) If $ \frac{1}{2} \leq \{x\} < 1$, then $ [x]\equal{}r$ and $ \left[x\plus{}\frac{1}{2}\right]\equal{}\left[x\plus{}\frac{2}{3}\right]\equal{}r\plus{}1$, hence $ 3r\plus{}2\equal{}2002 \rightarrow r$ is not an integer, contradiction.

So the only solutions are the ones we found on b).

See [url=http://en.wikipedia.org/wiki/Hermite_identity]Hermite's identity[/url].  :)


Show that, for some $ n \in \mathbb{N}$:
$ \sum_{n\equal{}1}^{p\minus{}1} n^k \equiv 0 \bmod p$
when $ (p\minus{}1) \nmid k$.

[hide="Solution"]
We consider the equation $ x^{p \minus{} 1} \minus{} 1\equiv 0\bmod p$.  By Fermat's Theorem, we see that $ x \equal{} 1, 2, 3, \cdots , p \minus{} 1$ are roots of this equation.  Thus,\[ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)\cdots (x \minus{} p \plus{} 1)\equiv x^{p \minus{} 1} \minus{} 1\bmod p\]Now, letting\[ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)\cdots (x \minus{} p \plus{} 1) \equal{} x^{p \minus{} 1} \minus{} a_2x^{p \minus{} 2} \plus{} a_3x^{p \minus{} 3}\cdots \plus{} a_p\]we notice that $ a_i\equiv 0\bmod p$ when $ 2\le i\le p \minus{} 1$ and $ a_p\equiv \minus{}1\bmod p$ by the first equation.  Now, since $ a^{k}\equiv a^{k \plus{} p \minus{} 1}\bmod p$, it suffices to consider $ 1\le k\le p \minus{} 2$.  Notice that $ \sum_{n \equal{} 1}^{p \minus{} 1} {n^k}$ can be put only in terms of the symmetric sums of $ 1, 2,\cdots , n$ with degree less than or equal to degree $ k$.  Since $ 1\le k\le p \minus{} 2$, we have that the symmetric sum expression is congruent to a sum of $ 0$s in $ \bmod p$, meaning that $ \sum_{n \equal{} 1}^{p \minus{} 1} {n^k}\equiv 0\bmod p$.

Here is an example ($ k \equal{} 2$), just to make the symmetric thing clear:\[ \sum_{n \equal{} 1}^{p \minus{} 1} n^2\equiv a_2^2 \minus{} 2a_3\equiv 0^2 \minus{} 0\bmod p\equiv 0\bmod p\][/hide]


Suppose there are n red and n blue points on a plane. You are to connect every red point to all the blue points with a [u][b]straight line segment[/b] [/u](so it will be a K(n, n) graph but with straight lines). Prove that there will be at least one  line segment  which will not intersect with the others.

I do not know the solution.

Best regards,

Kcefle

I'm sure it's been discussed before here. Consider the configuration which minimizes the sum of the lengths of the segments and show this satisfies the condition.

You're right, that's what I meant. Sorry about that: as soon as I read the first few words, I automatically asumed that was the problem.  :blush: I guess this really is a brand new problem.

[quote="Myth"]In this way you will show that it is possible to choose $n$ segments which don't intersect each other.[/quote]

Actually I "found" this problem while trying to solve  how to pair each of the n blue points to one of the n red ones with a line segment so that none of them wil intersect with the others.

[quote="Myth"]Actually, I think there is a simple counterexample for the problem from post 1.

Consider two concentric circles of radii $1$ and $2$. Choose very large $n$ and place blue points on the small circle and red points on the big circle in such a way that. they form regular $n$-gons.

Does it work?[/quote]

Mmm.... I am afraid it appears to work. :blush:

kcefle


[quote="mathwizarddude"]Find integers a, b, c, d, such that $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} 4 \equal{} 0$ has four distinct integers solutions.[/quote]
[hide]Applying Vieta's formula, we have: $ x_1.x_2.x_3.x_4 \equal{} 4$. Since $ x_1\neq x_2\neq x_3\neq x_4$, we see that the only 4-tuple$ (x_1,x_2,x_3,x_4)$ is: $ (1,2, \minus{} 2, \minus{} 1)$. So, we have: $ x^4\plus{}ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}(x^2\minus{}1)(x^2\minus{}4)$, which means $ b \equal{} \minus{} 5$ and $ a \equal{} c \equal{} 0$
[Edit: Thank JBL's remind. I will try to be careful on my LATEX][/hide]

Please review your posts after you post them -- your attempts to guess at latex syntax result in incomprehensible sentences.  Use \cdot for a multiplication dot, \neq for the not-equals sign (which is not transitive, so it's not enough to write $ x_1 \neq x_2 \neq x_3 \neq x_4$), and use dollar signs around all your equations!  And you can always look up LaTeX commands here: http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols


geometry
3D geometry
sphere
integration
trigonometry
calculus
geometry unsolved

What is the surface area of the Earth between 23.5 degrees north (Tropic of Cancer) and 23.5 degrees south (Tropic of Capricorn) latitude, if Earth is a perfect sphere with radius 3963 miles and if the straight-line distance between the tropics is 3250 miles?

Picture Here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200881

this is repied on the on the other one, but here's the answer again

the equation is:
2 * 2 * $ {\pi}$ * radius of earth *$ \int^{23}_{0} \sin x\minus{}90\,dx$ <= note, that this is in degree

basically "2 * radius of earth *$ \int^{23}_{0} \sin x\minus{}90\,dx$" breaks the part of the earth down to its individual pieces along the arc with the new radius at each point.

the "* $ {\pi}$" part gives the circumference of each point along the arc. after adding them all up (the defination of an integral) you should get the area of the top section.

the last "*2" would double it, giving you the whole solution.

sorry, that this is a bit confusing


Given a,b,c are prositive real numbers ; $ a^2\plus{}b^2\plus{}c^2\equal{}1$
Find min of
$ M\equal{}\sum \frac{a}{1\plus{}a^2}$

This's one problem in one junior high school ..........

[hide]We will require some facts:
$ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca\Longrightarrow ab \plus{} bc \plus{} ca\le 1$.......(1)
$ ab \plus{} bc \plus{} ca\ge 3\sqrt [3]{a^2b^2c^2}\Longrightarrow abc\ge\sqrt {\frac {1}{27}} \equal{} \frac {1}{{3\cdot\sqrt {3}}}$ .........(2)
From (1) and (2), $ \sum\frac {1}{a}\le\frac{1}{3\sqrt {3}}$......(3).
From QM AM, $ a \plus{} b \plus{} c\le 3\sqrt {\left(\frac {a^2 \plus{} b^2 \plus{} c^2}{3}\right)}$
or, $ a \plus{} b \plus{} c\le \sqrt {3}$.............(4).

[b]The final step:[/b]
Using CS, we have:
$ M\ge\frac {9}{\sum\frac {1 \plus{} a^2}{a}} \equal{} \frac {9}{\sum\frac {1}{a} \plus{} \sum a}$
From (3) and (4),
$ M\ge\frac {9}{3\sqrt {3} \plus{} \sqrt {3}} \equal{} \frac {3\sqrt {3}}{4}$
Hence maximum value of $ M \equal{} \frac {3\sqrt {3}}{4}$

Hope this is all right, I feel as if it is not QUITE RIGHT....  :huh: [/hide]

[quote="Potla"][hide]We will require some facts:
$ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca\Longrightarrow ab \plus{} bc \plus{} ca\le 1$.......(1)
$ ab \plus{} bc \plus{} ca\ge 3\sqrt [3]{a^2b^2c^2}\Longrightarrow abc\ge\sqrt {\frac {1}{27}} \equal{} \frac {1}{{3\cdot\sqrt {3}}}$ .

Hope this is all right, I feel as if it is not QUITE RIGHT....  :huh: [/hide][/quote]
This step is wrong :wink:
Because: $ 1\equal{}a^2\plus{}b^2\plus{}c^2 \ge 3\sqrt[3]{a^2b^2c^2}$

[quote="quykhtn-qa1"][quote="Potla"][hide]We will require some facts:
$ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca\Longrightarrow ab \plus{} bc \plus{} ca\le 1$.......(1)
$ ab \plus{} bc \plus{} ca\ge 3\sqrt [3]{a^2b^2c^2}\Longrightarrow abc\ge\sqrt {\frac {1}{27}} \equal{} \frac {1}{{3\cdot\sqrt {3}}}$ .

Hope this is all right, I feel as if it is not QUITE RIGHT....  :huh: [/hide][/quote]
This step is wrong :wink:
Because: $ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2 \ge 3\sqrt [3]{a^2b^2c^2}$[/quote].
Yeah, you are right.....

My solution is incorrect...

[quote="foraus"]Given a,b,c are prositive real numbers ; $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Find min of
$ M \equal{} \sum \frac {a}{1 \plus{} a^2}$

This's one problem in one junior high school ..........[/quote]
Predict that equality occurs when $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt {3}}$.
We try to find $ k$ satisfy:
$ \frac {a}{a^2 \plus{} 1} \ge \frac {\sqrt {3}}{4} \plus{} k\left(\frac {1}{3} \minus{} a^2\right)$
$ \Leftrightarrow 12a \ge 3\sqrt {3}\left(a^2 \plus{} 1\right) \plus{} 4k\left(1 \minus{} 3a^2\right)\left(1 \plus{} a^2\right)$
$ \Leftrightarrow k\left(\sqrt{3}a\minus{}1\right)\left(\sqrt{3}a\plus{}1\right)\left(1\plus{}a^2\right) \minus{} 3\sqrt {3}a^2 \plus{} 12a \minus{} 3\sqrt {3} \ge 0$
$ \Leftrightarrow \left(\sqrt {3}a \minus{} 1\right)\left(k\left(\sqrt {3}a^3 \plus{} a^2 \plus{} \sqrt {3}a \plus{} 1\right) \plus{} 3 \minus{} 3\sqrt {3}\right) \ge 0 \text{ (1)}$
We want the Left hand side of (1) have $ \left(\sqrt {3}a \minus{} 1\right)^2$, so we'll find $ k$ satisfy that $ a \equal{} \frac {1}{\sqrt {3}}$ is a root of the following equation:
$ k\left(\sqrt {3}a^3 \plus{} a^2 \plus{} \sqrt {3}a \plus{} 1\right) \plus{} 3 \minus{} 3\sqrt {3} \equal{} 0 \newline \Rightarrow k \equal{} \frac { \minus{} 9 \plus{} 9\sqrt {3}}{8}$
If $ k \equal{} \frac { \minus{} 9 \plus{} 9\sqrt {3}}{8}$, then:
$ LHS(1) \equal{} \left(\sqrt {3}a \minus{} 1\right)\left(\frac { \minus{} 9 \plus{} 9\sqrt {3}}{8}\left(\sqrt {3}a^3 \plus{} a^2 \plus{} \sqrt {3}a \plus{} 1\right) \plus{} 3 \minus{} 3\sqrt {3}\right)$
$ \equal{} \frac {1}{8}\left(3\sqrt {3} \minus{} 3\right)\left(\sqrt {3}a \minus{} 1\right)\left(3\sqrt {3}a^3 \plus{} 3a^2 \plus{} 3\sqrt {3}a \minus{} 5\right)$
$ \equal{} \frac {1}{8}\left(3\sqrt {3} \minus{} 3\right)\left(\sqrt {3}a \minus{} 1\right)^2\left(3a^2 \plus{} \frac {6}{\sqrt {3}}a \plus{} 5\right)$
$ \equal{} \frac {1}{8}\left(3\sqrt {3} \minus{} 3\right)\left(\sqrt {3}a \minus{} 1\right)^2\left(3\left(a \plus{} \frac {1}{\sqrt {3}}\right)^2 \plus{} 4\right) \ge 0$
So, we've proved: $ \frac {a}{a^2 \plus{} 1} \ge \frac {\sqrt {3}}{4} \plus{} k\left(\frac {1}{3} \minus{} a^2\right)$ where $ k \equal{} \frac { \minus{} 9 \plus{} 9\sqrt {3}}{8}$
Hence, $ M \equal{} \sum \frac {a}{1 \plus{} a^2} \ge \frac {3\sqrt {3}}{4} \plus{} k\left(1 \minus{} a^2 \minus{} b^2 \minus{} c^2\right) \equal{} \frac {3\sqrt {3}}{4}$
$ \Rightarrow \min M \equal{} \frac {3\sqrt {3}}{4}$, equality occurs if and only if $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt {3}}$.

Do anyone can give me a solution by AM-GM for this problem?  :?:

Or titus lemma immediately gives

$ M\le\frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{4}\equal{}\frac{9(\frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{3})^2}{4}$

$ \le\frac{9\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}}}{4}\equal{}\frac{3\sqrt{3}}{4}$

[quote="rofler"]Or titus lemma immediately gives

$ M\le\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{4} \equal{} \frac {9(\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{3})^2}{4}$

$ \le\frac {9\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}}{4} \equal{} \frac {3\sqrt {3}}{4}$[/quote]
No, I do not agree. Titu's lemma gives:
$ M \equal{} \sum\frac {(\sqrt {a})^2}{1 \plus{} a^2}\ge \frac {(\sum\sqrt {a})^2}{4}$ :wink:
Thus your application does not work here.
[hide]
For reference of titu's lemma, visit my blog at:
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=264227[/hide]

[quote="geniusbliss"]for the sum to be to find the max,

we know from AM-GM,
$ a \plus{} \frac {1}{a} \ge 2$
or,
$ \frac {a^2 \plus{} 1}{a} \ge 2$
or $ 2 \ge \frac {a}{1 \plus{} a^2 }$
similarly for $ b,c$
we get $ 6 \ge \sum_{cyclic}(\frac {a}{1 \plus{} a^2})$[/hide][/quote]

Wrong genius bliss.
If $ a \plus{} \frac {1}{a} \ge 2$, then $ \frac {1}{2} \ge \frac {a}{1 \plus{} a^2 }$, not $ 2$.

And Potla Rofler's solution is correct.

[quote="geniusbliss"]for the sum to be to find the max,
[hide="solution"]
we know from AM-GM,
$ a \plus{} \frac {1}{a} \ge 2$
or,
$ \frac {a^2 \plus{} 1}{a} \ge 2$
or $ 2 \ge \frac {a}{1 \plus{} a^2 }$
similarly for $ b,c$
we get $ 6 \ge \sum_{cyclic}(\frac {a}{1 \plus{} a^2})$[/hide][/quote]
You're wrong because the equality hold if and only if $ a\equal{}b\equal{}c\equal{}1$. So, this result doesn't satisfy the condition  :P

[quote="geniusbliss"]it is $ \frac {3}{2}$, not $ 6$
sry for the dum error[/quote]
Anyway, you're wrong because the equality can't hold   :lol:

wat do u mean $ equality$ wont hold mr.bikrone?
when $ a\equal{}b\equal{}c\equal{}1$
$ \sum_{cyclic}\frac{a}{a^2 \plus{} 1} \equal{} \frac{3}{2}$ !!!!

[quote="geniusbliss"]wat do u mean $ equality$ wont hold mr.bikrone?
when $ a \equal{} b \equal{} c \equal{} 1$
$ \sum_{cyclic}\frac {a}{a^2 \plus{} 1} \equal{} \frac {3}{2}$ !!!![/quote]
Look at the condition: $ a^2\plus{}b^2\plus{}c^2\equal{}1$  :lol:

[quote="Mr.Bikrone"][quote="foraus"]Given a,b,c are prositive real numbers ; $ a^2 + b^2 + c^2 = 1$
Find min of
$ M = \sum \frac {a}{1 + a^2}$[/quote]
I was wrong in previous solution  :( , I think this problem is
[quote="foraus"]Given a,b,c are prositive real numbers ; $ a^2 + b^2 + c^2 = 1$
Find max of
$ M = \sum \frac {a}{1 + a^2}$
This's one problem in one junior high school ..........[/quote]

[hide="And this is my solution"]
Predict that equality occurs when $ a = b = c = \frac {1}{\sqrt {3}}$.
We try to find $ k$ satisfy:
$ \frac {a}{a^2 + 1} \le \frac {\sqrt {3}}{4} + k(\frac {1}{3} - a^2)$
$ \Leftrightarrow 12a \le 3\sqrt {3}(a^2 + 1) + 4k(1 - 3a^2)(1 + a^2)$
$ \Leftrightarrow 4k(\sqrt {3}a - 1)(\sqrt {3}a + 1)(1 + a^2) - 3\sqrt {3}a^2 + 12a - 3\sqrt {3} \le 0$
$ \Leftrightarrow (\sqrt {3}a - 1)(4k(\sqrt {3}a^3 + a^2 + \sqrt {3}a + 1) - 3a + 3\sqrt {3}) \le 0 \text{ (1)}$
We want the Left hand side of (1) have $ (\sqrt {3}a - 1)^2$, so we'll find $ k$ satisfy that $ a = \frac {1}{\sqrt {3}}$ is a root of the following equation:
$ k(\sqrt {3}a^3 + a^2 + \sqrt {3}a + 1) - 3a + 3\sqrt {3} = 0 \newline \Rightarrow k = \frac { - 3\sqrt {3}}{16}$
If $ k = \frac { - 3\sqrt {3}}{16}$, then:
$ LHS(1) = (\sqrt {3}a - 1)(\frac { - 3\sqrt {3}}{4}(\sqrt {3}a^3 + a^2 + \sqrt {3}a + 1) - 3a + 3\sqrt {3})$
${ = \frac { - 3\sqrt {3}}{4}(\sqrt {3}a - 1)^2(a^2 + \frac {2}{\sqrt {3}} + 3})$
${ = \frac { - 3\sqrt {3}}{4}(\sqrt {3}a - 1)^2((a + \frac {1}{\sqrt {3}})^2 + \frac {8}{3}}) \le 0$
So, we've proved: $ \frac {a}{a^2 + 1} \le \frac {\sqrt {3}}{4} + k(\frac {1}{3} - a^2)$ where $ k = \frac { - 3\sqrt {3}}{16}$
Hence, $ M = \sum \frac {a}{1 + a^2} \le \frac {3\sqrt {3}}{4} + k(1 - a^2 - b^2 - c^2) = \frac {3\sqrt {3}}{4}$
$ \Rightarrow \max M = \frac {3\sqrt {3}}{4}$, equality occurs if and only if $ a = b = c = \frac {1}{\sqrt {3}}$.[/hide][/quote]
we have $ \sum {\frac{a}{{1 + a^2 }}}  = \sum {a - \sum {\frac{{a^3 }}{{1 + a^2 }}} }$
By AM-GM we gives $ \frac{{a^3 }}{{1 + a^2 }} + \frac{{\sqrt 3 ( {1 + a^2 } )}}{{16}} + \frac{1}{{4\sqrt 3 }} \ge \frac{3}{4}a$
similarity, by adding up we've: $ \sum {\frac{{a^3 }}{{1 + a^2 }} \ge \frac{3}{4}\sum a  - \frac{{\sqrt 3 }}{4} - \frac{{\sqrt 3 }}{{16}}\sum {( {1 + a^2 } )}  = \frac{3}{4}\sum a  - \frac{{\sqrt 3 }}{2}}$
Hence $ \sum {\frac{a}{{1 + a^2 }} \le \frac{{a + b + c}}{4} + \frac{{\sqrt 3 }}{2} \le \frac{{3\sqrt 3 }}{4}.Done!!}$

$ M \equal{} \sum_{cyc} \frac{a}{1\plus{}a^2} \leqslant \sum_{cyc} \frac{a}{2a} \equal{} \frac{3}{2}$

[quote="KevinB"]$ M \equal{} \sum_{cyc} \frac {a}{1 \plus{} a^2} \leqslant \sum_{cyc} \frac {a}{2a} \equal{} \frac {3}{2}$[/quote]
hello KevinB you're finding the maximum not the minimum

No, not 'finding' the maximum because this estimation is too weak (see post#11). As shown by leedt26 in his nice solution post#17, the max is indeed $ \frac {{3\sqrt 3 }}{4}$.  :lol:


The side lengths of a right triangle are integers. If its perimeter and area are the same, find the side lengths.

Actually there are more than one answer, but the number of answers is finite, of course :)

For a triangle with legs a, b, the equation is:
$\frac{ab}{2}=a+b+\sqrt{a^2+b^2}$
$2a+2b+2\sqrt{a^2+b^2}-ab=0$

Also, a, b, $\sqrt{a^2+b^2}$ are all integers.

2 solutions I found:
6-8-10 triangle
5-12-13 triangle

Any more?

Pythagorean triples are of the form: $x = m^2 - n^2, y = 2mn, and z = m^2 + n^2$.

So we set $m^2 - n^2+2mn+m^2 + n^2=2m^2+2mn equal to 1/2*mn(m^2 - n^2)$

Simplifying gives: $2=n(m-n)$

So, n equals to either 1 or 2, and m equals to 3 in both cases. Plugging these values in gives the triples: 8,6,10; 5,12,13

This "similar" problem appeared in IMTS. Only two are right triangles so if you want just the triangles that have equal perimeter and area, look at the link.  :)

There are 5 sets of answers and this solution was kindly provided by MysticTerminator.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Heronian+Triples&t=27256


abstract algebra
category theory
group theory
superior algebra
superior algebra unsolved

Suppose $ Grp$ is the category of all groups equipped with all homomorphisms, and $ Ab$ is the subcategory of abelian groups equipped with all homomorphisms.
Let $ I: Ab \to Grp$ be a functor.
Show that if G is any group, then a universal morphism from G to I is $ (G/[G,G], v)$, where $ v: G \to G/ [G,G]$.

I am not very clear on how to proceed.

Do I choose $ A\in Ab$ and suppose $ f: G \to I(A)$ is a group homomorphism, and then find a unique g, which is a homomorphism with $ g: G/[G,G] \to A$ such that $ I(g)v \equal{} f$?

$ I$ is not [i]a[/i] functor, it's the forgetful functor which forgets the commutativity of the abelian groups, i.e. just an inclusion.

your definition of a universal morphism in this case is correct, and this is verified by the homomorphism theorem and that commutator subgroups are mapped into commutator subgroups by homomorphisms.


Ring Theory
Galois Theory
superior algebra
superior algebra unsolved

Proof that there is subrings $I$ and $J$, such that $I$ is an ideal of ring $R$ and $J$ is an ideal of $I$,But $J$ isn't an ideal of ring $R$.

[quote="julien_santini"]It needs clarification: if $I$ and $J$ are unitary subrings, and $I$ is an ideal of $J$, then $I=J$ since $1 \in I$. Or maybe you are considering pseudorings?[/quote]
Of course it holds in certain cases eg see :  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26305#
But he says, prove that if $R$ is a ring and $I$ is an ideal of $R$ and $J$ is an ideal of $I$, it is not necessarily true that $J$ is an ideal of $R$.

Well, one of the things I did not understand is that if $I$ is seen as a subring, then it should contain the unit, and being an ideal of $R$, $I=R$. That's why I was inquiring about pseudorings.

Well,It seems that there are some mistakes;it's better to say that:

Proof that [b]there is a ring $R$[/b] and subrings $I$ and $J$, such that $I$ is an ideal of ring $R$ and $J$ is an ideal of $I$,But $J$ isn't an ideal of ring $R$.

So we have different definitions of "ring".
Julien sees them as having a unit (as I had when reading the problem), but as follows from juliens remark, the problem must be interpreted with rings not necessarily having a unit.

So we have to find a ring without unit R, an ideal I of R, and an ideal J of I (regarded as a ring without unit), s.th. J is not an ideal of R.

LiMa

tanx LiMa.u r right. :wink:
I didn't know anything about different definitions of rings :P

Since we hear "ideal" and "ring", is the ring commutative ? Usually you use the term ideal for a commutative ring, and left/right ideal otherwise ...

Please look up all definitions in a standard book, eg [i]Algebra[/i] by [i]Hungerford[/i], before sending your questions/replies. In general, rings are not assumed to be commutative or unitary unless stated otherwise.

Look up Birkhoff, Milne (Galois theory) or even better Bourbaki. Do you think Hungerford contributed a hundredth of what Bourbaki did ?

Besides, I've never said a ring is supposed commutative. But the standard way is to talk about ideals when we are dealing with a left ideal which is also a right ideal. Most of the times, this naturally happens in commutative rings. So i thought maybe the hypothesis was left out ...

Well we may have different understandings of the term "standard". Let me tell me what I mean by it: [i]widespread[/i]. And please don't tell me that the definitions you thought of, are more common than those I mentioned. :wink:
[quote="julien_santini"]Look up Birkhoff, Milne (Galois theory) or even better Bourbaki. Do you think Hungerford contributed a hundredth of what Bourbaki did ?
[/quote]
I'm not interested in arguing about these things in the forum, but let me remind you that Bourbaki was a group, and Two heads are better than one. :lol: (Just kidding.)

[u]Finally I have a suggestion: Since this is not the first and the last time that we have this problem, I suggest that we have the definitions used in this forum as an announcement or something.[/u]

Well back at the problem, ...

Well, if you want you can have such an announcement (for this subforum).
But I suggest putting it into some wiki page at http://www.artofproblemsolving.com/Wiki/index.php/Main_Page (and then making an announcement giving a link there).

Good suggestion, ZetaX. I hope we can save a lot of time and energy by doing this.

Hint: I could prove the existence of a ring $R$ with order $16$ and having two subrings $S,T$ such that $S$ is a two-sided ideal of $R$, $T$ is a two-sided ideal of $S$, but $T$ is not an ideal (neither left nor right) of $R$. With such a cardinal, there are not so many things you can think of ...


[quote="trung92"]let a,b,c ,d>0.prove that
$ \sum\limits_{a,b,c,d} {} \frac {1}{{4a}} \plus{} \frac {2}{{a \plus{} b \plus{} c \plus{} d}} \ge \frac {1}{{2a \plus{} 2b}} \plus{} \frac {1}{{2c \plus{} 2d}} \plus{} \frac {1}{{2a \plus{} 2c}} \plus{} \frac {1}{{2b \plus{} 2d}} \plus{} \\
\plus{} \frac {1}{{2a \plus{} 2d}} \plus{} \frac {1}{{2b \plus{} 2c}}$[/quote]
Yes! It follows from Turcevici's inequality. :wink:


How many real roots does the equation

$ \frac{1}{x\minus{}a} \plus{} \frac{1}{x\minus{}b} \plus{} \frac{1}{x\minus{}c} \plus{} \frac{1}{x\minus{}d} \equal{} 0$

have, where $ a<b<c<d$?

Show that the sum of these roots is exactly $ \frac{3(a\plus{}b\plus{}c\plus{}d)}{4}$.

For the second part, getting a common denominator and collecting terms, I get

$ 4x^{3} \minus{}3(a\plus{}b\plus{}c\plus{}d)x^{2} \plus{}2(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)x\minus{}(abc\plus{}abd\plus{}acd\plus{}bcd)\equal{}0$,

so that the sum of the roots is $ \frac{3(a\plus{}b\plus{}c\plus{}d)}{4}$.

And the first part?


inequalities
function
logarithms
search
inequalities unsolved

Let $(a)\succ(b)$
Prove or give counterexamples for this inequality
$a_{1}a_{2}...a_{n}\ge b_{1}b_{2}...b_{n}$ ($a_{i},b_{i}>0 \forall i\in N^{*}$)

The reverse inequality is true because of the concavity of $\log.$

This is the result of Karamata inequality for the function $\log$. Search Karamata inequality or majorization inequality.


geometry
search
circumcircle
geometric transformation
homothety
3D geometry
tetrahedron

I'll be glad to see the proof of following statement:

[quote="darij grinberg"]
Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle.
[/quote]

I just started playin with projective geometry so all help welcome.

BTW. Do you know where I can find some useful theorems and lemmas for projective geometry with proofs  ?

Anybody thinking about it? I gave it to my math teachers but none of them can solve it quickly [they told me that they going to think about it] - though I don't know if I understand it correctly. So here is expanded version of my understanding of the problem:

Given a circle with inscribed arbitrary triangle, there always exists a projective transformation which tranforms the circle into itself and the triangle into equilateral triangle which is still inscribed in the circle.

Is it correct?

can you introduce sth about the Projective Geometry??
I have been looking for it's introductions but didn't find anything.

Up!

I think this theorem would be powerful... if I knew the proof!   :lol:

this topic is "old", but i find it very interesting (i "discovered" it only now - i mostly stay on the romanian forum).
indeed, the result is very nice. is it *really* true ? i am not very good at geometry, and i would really like to see a simple proof of it. it is simply "too nice to be true".
i attempted a proof, however i could bet that it has at least 3-4 errors in it. anyways, i hope that this topic will "come back to life", and maybe someone who knows this kind of things well can "bring the light" on the topic.
(ps : sorry for my bad english, especially words in quotes. i really don`t know english very well).
enough of the blabla for me. below is the "proof" :

[hide="proof ?"]
firstly, i search for a projective transformation that maps a triangle into another and the circumcircle of the first triangle into the circumcircle of the second triangle. i also exclude the homothethy out of this, because i want to obtain, in the end, an equliateral triangle.
for that, i take a circle $\omega$ and a point $T$ outside the plane of $\omega$ such that the orthogonal projection of $T$ on the plane of $\omega$ it`s not the center. (the cone formed by $T$ and $\omega$ is not a right-cone). now invert through $T$. (can i call it that way?). let $A,B,C$ be points on $\omega$, and $A',B',C'$ be their respective images. now if $D$ is another point of $\omega$, then, using the formula $X'Y'=\frac{XY}{TX\cdot TY}$, it is easy to see that ptolemy`s relation is true for points $A',B',C',D'$ (the common denominator $TA\cdot TB\cdot TC\cdot TD$ vanishes, and all that`s left is ptolemy`s relation for points $A,B,C,D$ which are the vertices of a cyclic quadrilateral).
thus, for any point $D\in \omega$, the image of $D$, $D'$, is on $\omega '$, the circumcircle of $\triangle A'B'C'$. therefore $\omega$ maps to $\omega '$.
now if i regard the inversion as a projective transformation, i found what i was searching for. $\omega$ maps to $\omega '$, and $\triangle ABC$ maps to $\triangle A'B'C'$.
also, this transformation is invertible. therefore, it will be easier for me to look for a transformation that maps an equilateral triangle into some triangle $\triangle MNP$ (with the above property).
if i find such a map, then it`s inverse will be the projective map from the lemma (because it maps some triangle $\triangle MNP$ into an equilateral one, but keeps the circumcircles).

let $\triangle ABC$ be equilateral of side-length $l$, $T$ a point as above, such that $TA=x$, $TB=y$, $TC=z$. i invert through $T$ and power $k$. i want to find $x,y,z,l,k$ such that the side-lengths of triangle $\triangle A'B'C'$ are some $a,b,c$ (fixed).
because of the inversion, using the formula for $A'B'$ i get :
$A'B'=\frac{kl}{TA\cdot TB}$, thus, $c=\frac{kl}{xy}$.
now, i see that if i take $x=a,y=b,z=c$, it is ok, because the three relations are all equivalent with $kl=abc$ (constant).
all that`s left now is to show that there exists a tetrahedron $TABC$ with sides $TA=a,TB=b,TC=c,AB=BC=CA=l$, where $a,b,c$ are fixed and $l$ can be chosen convenably.
i don`t know if this is true or if this is an easy part. i will try to finish it tommorow (the computations are somewhat ugly - maybe someone will find a nice method).

i only hope i am on the right way.
[/hide]

i really wait for some opinions about this, because it is a very nice result.

[quote="darij grinberg"]
Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle.
[/quote]

Just for the sake of completeness: This follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=795373#795373]http://www.mathlinks.ro/Forum/viewtopic.php?t=138671 post #4[/url] Theorem 1.

[quote="maky"]for that, i take a circle $\omega$ and a point $T$ outside the plane of $\omega$ such that the orthogonal projection of $T$ on the plane of $\omega$ it`s not the center. (the cone formed by $T$ and $\omega$ is not a right-cone). now invert through $T$. (can i call it that way?). let $A,B,C$ be points on $\omega$, and $A',B',C'$ be their respective images. now if $D$ is another point of $\omega$, then, using the formula $X'Y'=\frac{XY}{TX\cdot TY}$, it is easy to see that ptolemy`s relation is true for points $A',B',C',D'$ (the common denominator $TA\cdot TB\cdot TC\cdot TD$ vanishes, and all that`s left is ptolemy`s relation for points $A,B,C,D$ which are the vertices of a cyclic quadrilateral).[/quote]

Are you sure? I think your calculation is mistaken.

[quote="maky"]thus, for any point $D\in \omega$, the image of $D$, $D'$, is on $\omega '$, the circumcircle of $\triangle A'B'C'$. therefore $\omega$ maps to $\omega '$.[/quote]

In fact, I doubt that $D^{\prime}$ will lie on the plane $A^{\prime}B^{\prime}C^{\prime}$ at all.

Your ansatz sounds interesting, but I don't really see how you want to make a projective transformation out of the inversion... However, I am not sure whether I understand your argument at all. Do you happen to mean stereographic projection rather than inversion?

[b]PS.[/b] In case this is of any help to you:

[b]In the plane[/b], you can transform any triangle $ABC$ into an equilateral triangle by a suitable inversion - actually, any inversion with respect to a circle centered at one of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6489]two isodynamic points[/url] of triangle $ABC$ will do! In space, there will be even more inversions satisfying that.

  Darij


how difficult is APMO compared to other national Olympiads?  is it well know for any specific types of problems, or is it basically all round decent?

how are its number theory problems?

[quote="iYOA"]APMO= asian pasific math olympiad.[/quote]

Looks like a BMO 1 from first glance.

and how hard is that?

is it AIME level? is it harder?

APMO is definitely not AIME level.

I would say more like the easiest problems are around USAMO #1 level and the hardest ones are USAMO #5-#3 level.

No, the easiest problems are <usamo 1 level. I can solve most APMO 1/2s but i cant solve half the USAMO 1/4s(well, maybe cuz i havent tried in a while...)

It's about USAMO level, but more "technical," like many Olympiads involving Asian countries.

better idea: try the problems?


Find the maximum value of $ Q(x)$ subject to the constraint: $ x^Tx\equal{}1$.

   $ Q(x)\equal{}3x_1^2\plus{}2x_2^2\plus{}2x_3^2\plus{}2x_1x_2\plus{}2x_1x_3\plus{}4x_2x_3$

$ Q(x)\equal{}x^TAx$ where $ A$ is the symmetric matrix associated to $ Q$. You can diagonalize $ A$ in an orthonormal base,......

It doesn't take a lot of computation.

The matrix loup blanc is talking about is $ A\equal{}\begin{bmatrix}3&1&1\\1&2&2\\1&2&2\end{bmatrix}.$

I'll name two things about that matrix that are clear by inspection:
(1) The matrix is clearly singular, since it has two identical rows.
(2) The sums of every row are the same. Each row adds up to $ 5.$

From (1), we know that $ 0$ is an eigenvalue of $ A.$ From (2), we know that $ [1,1,1]^T$ is an eigenvector of $ A$ with eigenvalue $ 5$ (the row sum). The trace of $ A$ is $ 7,$ and that's the sum of the eigenvalues. Hence, the three eigenvalues are $ 0,2,5.$

The maximum of $ x^TAx$ occurs in the direction of the largest eigenvalue. If you let $ x$ be that eigenvector, normalized (we know what it is, but we don't need to write down too many details), then $ x^TAx\equal{}x^T(5x)\equal{}5x^Tx\equal{}5.$

This quadratic form is positive semidefinite but not positive definite. That is, we have $ Q(x)\ge 0$ for all $ x$ but $ Q(x)\equal{}0$ whenever $ x$ is any scalar multiple of $ [0,\minus{}1,1]^T.$ We know that just from the list of eigenvalues given above; the direction I just named is the direction of the eigenvector associated to the eigenvalue $ 0.$

The set $ \{x\in\mathbb{R}^3: Q(x)\equal{}1\}$ is a cylinder with an elliptical cross-section.

In still another words, we are to maximize the Rayleigh quotient $ \frac{x^TAx}{x^Tx}$ for $ A$, which we know to be the maximum eigenvalue of $ A$. As Professor Merryfield pointed out, pure inspection yields 0, 5, and 2 as the eigenvalues of A. Therefore, $ 0 \leq Q(x) \leq 5$.


I took part in Round1 and Round2 but i missed the submission deadline of Round3. OOOOPS :(
I also didn't do well in both Round1 and Round2. For both i got a 15. 30 in total right now :(
Assume i still only can get a 15 for Round 4. Then it would be a 45 in total. :(
So is score for USAMTS quiet important when applying for colleges?
Consider i only take 3 rounds instead of 4? :(

I dunno about usefulness when applying to colleges (though getting a perfect helps :P), but I think the most important part is just learning to solve problems.  Even if you don't get the score you want, you can learn a lot from just doing the rounds, and that can benefit you in the long run.  Oh, and having fun is also a goal for most who do the USAMTS. :)

[quote="CatalystOfNostalgia"]If the only reason you're doing USAMTS is for college, you shouldn't be doing it at all.[/quote]
this is true for almost everything.

Also if your sole goal is to do stuff for college, there are much more time-efficient things to do than USAMTS.

well...it kind of depends
are you a senior or a junior?
If you're a senior, then i presume that college apps. were due long ago, and if you're a junior, then you could try something other that USAMTS, like trying to qualify for Blue MOP for example.
If you're a sophomore, then you still have two years ahead of you, so don't worry about how you do this year

Most (if not all) colleges don't require USAMTS participation and won't look at USAMTS scores/results during the admissions process (unless you mention it). They are more likely to care about other things (such as your GPA, SAT scores, ACT scores, extracurricular, etc.) since most students that apply to college have these, but haven't heard about the USAMTS. Doing well in this might be worth noting on an application, but if you don't do so well (or you have enough other achievements, activities, etc.) then you could just not mention that your application.

Your reason for doing the USAMTS should not be to get into college, since it probably won't get you into college by itself, although it might help you develop math skills that would help you getting into college.

I guarantee most admission officers don't know what USAMTS is. Possible exceptions are MIT/Caltech etc. but I wouldn't beat yourself upp.

I can say that even if you get a perfect score on the USAMTS from 6th to 12th grade it won't mean a single thing to any college anywhere. That being said, USAMTS is one of the most enjoyable competitions out there (in terms of the problems), with the added benefit that you can do it anywhere. It's worth every minute you spend on it.

I think that if you explain what it is, and give a few sample problems, it will help you. I mean, if you can consistently solve problems like year 21, problem 5, round 3, that means a lot.

But if you can solve those kinds of problems, you will probably have done good in something more prestigious, like USAMO.

In general, your USAMTS scores won't get you very far in colleges.  The important thing is that you learn from participating in the rounds, and get experience on solving difficult problems.  (Well, they're hard to me-- I dunno about the rest of you).

If you're participating just because its name has the word "mathematics" in it (okay, maybe not that extreme), then this is for you.  If you are more worried about college apps, then you should mainly focus on other competitions (the Olympiads) and just use this as practice, if at all.

Just my two cents there.

[quote="Frank94"]even if you dont do well you should still put it on ur application... it shows interest in math[/quote]
Colleges don't give a rats behind about what you're interested in, they only care about what you're good at. Come on now.


http://www.mathlinks.ro/viewtopic.php?t=6360

and the problem is sooooo recent..  :?  :?  :?


A chord of the larger of two concentric circles is tangent to the smaller circle and measures 18 inches. Find the number of square inches in the area of the shaded region. Express your answer in terms of $ \pi$.

[asy]pair O; path A = Circle(O,1); path B = Circle(O,2);
fill(B,grey); fill(A,white); draw(2*dir(30)--2*dir(150),black);
label("18''",dir(90),N);[/asy]

$ R^2 \minus{} r^2 \equal{} \left(\frac {18}{2}\right)^2 \equal{} 81$.
Thus, $ \pi R^2 \minus{} \pi r^2 \equal{} \pi(R^2 \minus{} r^2) \equal{} \boxed{81\pi}$.


advanced fields
advanced fields unsolved

Let for $ j\equal{}1,\ldots,n$ Bernoulli variables $ D_{j}\equal{}Ber(p_{j})$ with $ 0<p_{j}<1$ be given and denote the (number of jump) probabilities for all $ k\equal{}0,1,\ldots,n$ by $ q_{k}\equal{}\mathbb{P}(\sum_{j\equal{}1}^{n}D_{j}\equal{}k)$ and let this be the only information we have about the joint behaviour.

For positive numbers $ b_{1},\ldots,b_{n}$ with $ 1\equal{}\sum_{j\equal{}1}^{n}b_{j}$ define for all $ k\equal{}1,\ldots,n$ and $ j\equal{}1,\ldots,n$ the expression $ P(k,j)$ via

$ PP(n,j)\equal{}1$ and for all $ k\equal{}1,\ldots,n\minus{}1: P(k,j)\equal{}\sum_{i_{1}\not\equal{} j}\sum_{i_{2}\not\equal{}j,i_{1}}\ldots\sum_{i_{k}\not\equal{}j,i_{1},\ldots,i_{k\minus{}1}}\left[\prod_{v\equal{}1}^{k}\frac{b_{i_{v}}}{\sum_{m\not\equal{} i_{1},\ldots,i_{v\minus{}1}}b_{m}}\right]$

(Remark: the $ P(k,j)$ can be interpreted as probabilities associated to an $ n$-numbered balls-urn model. In this urn model drawn balls are removed and the initial likelihood of drawing ball no. $ j$ is given by $ b_{j}$, i.e. $ P(k,j)$ is the likelihood that ball no. $ j$ is not among the first $ k$ draws)

Question: do positive numbers $ b_{1},\ldots,b_{n}$ exist such that for all $ j\equal{}1,\ldots,n$ the representation $ p_{j}\equal{}\sum_{k\equal{}1}^{n}q_{k}[1\minus{}P(k,j)]$ holds ?

Three comments/amendments

1. all $ q_{k}$ are assumed to be non-zero

2. $ P(n,j)\equal{}0$ (a typo  :blush: )

2. for the $ n\equal{}2$-case, note that $ q_{2}\leq p_{j}\leq q_{1}\plus{}q_{2}$ - now set $ b_{j}\equal{}\frac{p_{j}\minus{}q_{2}}{q_{1}}\in[0,1]$ and from $ \mathbb{E}\left[\sum_{j\equal{}1}^{n}D_{j}\right]\equal{}\sum_{j\equal{}1}^{n}p_{j}\equal{}\sum_{k\equal{}1}^{n}kq_{k}$ we verify $ b_{1}\plus{}b_{2}\equal{}1$, i.e. the $ b_{j}$ are well-defined and since $ P(1,j_{1})\equal{}b_{j_{2}}$ with $ \{j_{1},j_{2}\}\equal{}\{1,2\}$, the problem can be solved for this case


As of now, whenever I want to plot two functions, I am only able to do so by naming one, plotting it, naming the other, plotting it, and then asking for both to be shown (example:
gj0 = Plot[{21 x + 9 x^2 - x^3}, {x, 0, 9}]
Graph
gy1 = Plot[{21 + 18 x - 3 x^2}, {x, 0, 9}]
Graph
gjy = Show[gj0, gy1]  )

How would I be able to do so within one command?

I also would like to know how to use the 'Manipulate' function well.

And could anyone please explain how (or where to find out) to use dashes in order to accentuate points of intersection (or inflection, critical points, etc.) with the x and y values labeled?

I realize that all these questions and more could be answered by purchasing the newest Mathematica book; however, I am a college student and considering my department already purchased my copy of Mathematica, I really don't feel like asking them to buy the newest book.

Thanks for any info.

I CAN tell you a bit about manipulate, though I'm not sure how to plot two graphs at once. You may have figured out the most efficient method.

In any case, the Manipulate command basically acts as a dynamic parametrization of any function of $ variable$ you wrap it around.

So the syntax for it is Manipulate[(fuctinon of some variable($ variable$), {$ variable$, start, end, interval between steps}]

That interval between steps is important for several things, notably if you want to step by integers. If you leave it out, Mathematica assumes you want the step size to be a real, and so that can mess up some things with number theory manipulates.

Thank you very much, but how can I control the y-range?


abstract algebra
group theory
superior algebra
superior algebra solved

Let R be a commutative ring  and $ I_1,$ $ I_2$ and $ I_3$ three Ideals such that
$ I_1 \plus{} I_2 \equal{} R$, $ I_1 \plus{} I_3 \equal{} R$ and $ I_2 \plus{} I_3 \equal{} R$.
Prove or disprove:
$ I_1 \plus{} (I_2 \cap I_3) \equal{} R$

it's true, and we don't need $ I_2 \plus{} I_3 \equal{} R$.

$ R \equal{} (I_1 \plus{} I_2) (I_1 \plus{} I_3) \subseteq I_1 \plus{} I_1 I_3 \plus{} I_2 I_1 \plus{} I_2 I_3 \subseteq I_1 \plus{} (I_2 \cap I_3)$.

Why is $ R\equal{}(I_1\plus{}I_2)(I_1\plus{}I_3)?$
That is only true if every element of R is a produkt of two other elements.
For example, if $ R\equal{}2 \mathbb{Z}$, it's wrong.

yes, I assumed that your ring is unital (which is usual ;)).

Then, any abelian group becomes a ring with multiplication defined by $ ab \equal{} 0$ for any $ a$ and $ b$... and the assertion is wrong.

  darij

Is this proposition wrong in Abel group?  If A+B=A+C=G, where G is an abel group and A,B,C its subgroup, then $ A\plus{}B\cap C\equal{}G$ ?
According to structure theorem of abel group it needs only checking for a "single term", which seem to be true since the subgroups are linear ordered.

Counterexample: $ G\equal{}(\mathbb{Z}/p\mathbb{Z})^2$, and pick any three different subgroups of order $ p$.


induction
algebra
polynomial
combinatorics unsolved
combinatorics

Prove that for $\forall r\in \mathbb{R}$ ,and forevery $n\in\mathbb{N}$
$\sum_{k=1}^{n}\frac{r(k^{2}-n^{2})+k(2k-1)}{r+k}{r+k \choose k}=n^{2}$

I tried induction,but it seems to long and complicated.

What is $r+k \choose{k}$ with $k$ is real?

${r+k \choose{k}}=\frac{(r+k)(r+k-1)...(r+1)}{k!}$? Am I true?
If so that , we have LHS is the polynomial has degree equal $n-1$. And you only need check with $n$ value of $r$. I think  :wink:

What about r = -1.

Then all binomial coefficients are [b]0[/b].and so [b]LHS = 0[/b]


Help me solve this problem:
There is a small beetle living on infinite checkered sheet. Size of each cell is 1x1, the axes of Cartesian coordinate system are lying on the border of cells. Early in the morning beetle travels from his house at point (x1,y1) to the point (x2,y2). All cells that he passed are numbered in order of passing (borders and corners of cells are not considered to be parts of cells). You need to find coordinates of left lower corner of n-th cell on the way of Mr Beetle.

Start by assuming x2-x1 and y2-y1 are coprime, and x1<x2. Consider what happens as the beetle passes from x=x1 to x=x1+1, and then x1+1 to x1+2, .. etc. You should be able to get some expression in terms of the gradient and floor/ceiling symbols for how many cells the beetle passes over, then its just checking where the nth one occurs.


i send here a geo. problem, i do not know how to draw the picture and add it in here .
Problem: Let $\Delta ABC$ inscribed in the circle $(O)$ (Both are given). A moving point $M$ on the circle $(O)$. Find the least possible value of the following expressions:
1) $\sum_{cyclic}\frac{MA}{BC}$
2) $\sum_{cyclic}\frac{MA^{2}}{BC^{2}}$

They are two distinct questions if Myth appears again :D

[Moderator edit: The first one has been sufficiently discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=103823 , so let's use this topic for the second one.]

One question:
You sent it to Unsolved Section. Is it true that you haven't solution?

I found the first when i was manging to solve the second. So in fact i solved the first but I still do not know how to solve The second. Amazing that this seems both attain the least value at $2$.
By the way for the first , if $M$ any on the plan of the triangle then the least value is $\sqrt{3}$.

I invented short and nice solution for a).

I use Ptolemy's inequa.  and really he stays everywhere in my solution :D . But it's nice if you post your solution in here.  ;)

I use Ptolemei theorem (not inequality :) ).


integration
real analysis
real analysis unsolved

If $ f,f'$ are continuous in $ [a,b]$,$ f'(a)\equal{}f'(b)\equal{}0$ and there is $ f''$ in $ (a,b)$ prove that there is $ c\in(a,b)$:

$ f''(c)>\equal{}\frac{4(f(b)\minus{}f(a))}{(b\minus{}a)^{2}}$

By mean of Lagrange theorem, there exist $ \theta_1$ and $ \theta_2$ such that
\[ \begin{gathered}
  f''\left( {\theta _1 } \right) \equal{} \frac{{f'\left( x \right) \minus{} f'\left( a \right)}}
{{x \minus{} a}}, \hfill \\
  f''\left( {\theta _2 } \right) \equal{} \frac{{f'\left( x \right) \minus{} f'\left( b \right)}}
{{x \minus{} b}}, \hfill \\
\end{gathered}
\]
where $ \theta _1  \in \left( {a,x} \right)$, $ \theta _2  \in \left( {x,b} \right)$. Since $ f'(a)\equal{}f'(b)\equal{}0$, then
\[ \begin{gathered}
  f'\left( x \right) \equal{} f''\left( {\theta _1 } \right)\left( {x \minus{} a} \right), \hfill \\
  f'\left( x \right) \equal{} f''\left( {\theta _2 } \right)\left( {x \minus{} b} \right). \hfill \\
\end{gathered}
\]
On the other hand, $ f''$ is continuos in $ [a,b]$ then
\[ \left| {f''\left( x \right)} \right| \leqslant M \equal{} \mathop {\max }\limits_{\left[ {a,b} \right]} \left| {f''\left( x \right)} \right|.
\]
Therefore
\[ \begin{gathered}
  \left| {f'\left( x \right)} \right| \leqslant M\left( {x \minus{} a} \right), \hfill \\
  \left| {f'\left( x \right)} \right| \leqslant M\left( {x \minus{} b} \right). \hfill \\
\end{gathered}
\]
Thus
\[ \begin{gathered}
  \frac{{4\left( {f\left( b \right) \minus{} f\left( a \right)} \right)}}
{{\left( {b \minus{} a} \right)^2 }} \equal{} \frac{4}
{{\left( {b \minus{} a} \right)^2 }}\int_a^b {f'\left( x \right)dx}  \hfill \\
   \leqslant \frac{4}
{{\left( {b \minus{} a} \right)^2 }}\int_a^b {\left| {f'\left( x \right)} \right|dx}  \hfill \\
   \leqslant \frac{4}
{{\left( {b \minus{} a} \right)^2 }}\left( {\int_a^{\frac{{a \plus{} b}}
{2}} {M\left( {x \minus{} a} \right)dx}  \plus{} \int_{\frac{{a \plus{} b}}
{2}}^b {M\left( {b \minus{} x} \right)dx} } \right) \equal{} M. \hfill \\
\end{gathered}
\]
This means that
\[ \mathop {\max }\limits_{\left[ {a,b} \right]} \left| {f''\left( x \right)} \right| \geqslant \frac{{4\left( {f\left( b \right) \minus{} f\left( a \right)} \right)}}
{{\left( {b \minus{} a} \right)^2 }}.
\]

I think that we don't know if $ f''$ is continuous.... :wink:


How do you prove Ceva's Theorem without using mass points?

http://www.cut-the-knot.org/Generalization/ceva.shtml

one of the proofs is also by our own Darij Grinberg :!:

Darij Grinberg writes often here in AoPS. Try Search and Search for Author.

[quote="shobber"]Well I am curious to know: Why a lot of American students like to talk about and use Mass point? Is that a kind of must-learn lesson in the US Math Olympiad Course?[/quote]

You're right! a lot of US students love talking about mass point... maybe a lot of problems of their national olympiad need it...

Alot of geo problems on US tests are built using only ratios, so Ceva's/Menelaus work great on those, but many of them are weak enough that knowing mass points (which honestly is alot easier to use than ceva/menelaus) will solve the problem

thus everyone knows mass points and uses it alot

here is a proof that i posted elsewhere:

[quote="Altheman"][hide="solution"]Notation: Let there be a $\triangle ABC$ with $D$, $E$, and $F$ existing on $AB$, $BC$, and $CA$, respectively such that $\frac{BD}{DA}=d$, $\frac{CE}{EB}=e$, and $\frac{AF}{FC}=f$. Let $AE\cap BF=X$, $BF\cap CD =Y$, and $CD \cap AE=Z$.

Just for now, ignore $CD$, using mass points, Let $w_C=1$, then $w_B=e$, $w_A=\frac{1}{f}$, $w_E=1+e$, $w_f=1+\frac{1}{f}$ and $w_X=1+e+\frac{1}{f}$, then using the masses, $\frac{BF}{BX}=\frac{ 1+e+\frac{1}{f} }{1+\frac{1}{f} }$, also $\frac{BC}{BE}=\frac{ 1+e}{1}$

Then through the law of sines, $\frac{ [BFC] }{ [BEX] }=\frac{ 1+e+\frac{1}{f} }{1+\frac{1}{f} }*\frac{ 1+e}{1}$ then looking at the bases, $\frac{ [ABC] }{ [BFC] }=\frac{1+f}{1}$, multiplying the 2 we get
$\frac{ [ABC] }{ [BEX] }=\frac{ (ef+f+1)(1+e)}{1}$

Then $\frac{ [BFC] }{ [ABC] }-\frac{ [BEX] }{ [ABC] }=\frac{ [FCEX] }{ [ABC] }=\frac{1}{1+f}-\frac{1}{(1+e)(ef+f+1)}$

Then by symmetry, we have

$\frac{[ABC]}{[ABC]}-\frac{[CEXF]}{[ABC]}-\frac{[FYDA]}{[ABC]}-\frac{[DBEZ]}{[ABC]}=\frac{[XYZ]}{[ABC]}$
$=1-\sum_{cyc}\frac{1}{1+e}+\sum_{cyc}\frac{1}{(1+e)(ef+f+1)}=1-\sum_{cyc}\frac{f}{ef+f+1}$

setting that equal ot $0$, we get

$1=\frac{1}{ e+1+\frac{1}{f} }+\frac{1}{f+1+\frac{1}{d}}+\frac{1}{d+1+\frac{1}{e}}$

which after lots of algebra leaves $def=1$[/hide][/quote]

Proof from the planet math: http://planetmath.org/encyclopedia/ProofOfCevasTheorem.html
Proof by myself when I first see this theorem:  http://www.mathlinks.ro/Forum/viewtopic.php?t=14581
A proof similar to Altheman's: http://www.mathlinks.ro/Forum/viewtopic.php?t=40828

[hide="I shall present here the oldest (in the my opinion) proofs of the Menelaus and Ceva theorems."]
[color=blue]Given are the triangle $ABC$ and the points $M\in BC$, $N\in CA$, $P\in AB\ .$
Denote the line $d$ for which $A\in d$ and $d\parallel BC$. I shall use orientated segments.[/color]

[color=red][b]The Menelaus' theorem.[/color][/b]$\boxed {P\in MN\Longleftrightarrow \frac{MB}{MC}\cdot \frac{NC}{NA}\cdot \frac{PA}{PB}=1}\ \ (1)\ .$
$\boxed {\blacktriangleright \Longrightarrow}$ [color=darkred]Denote $S\in d\cap \overline {MNP}$. Thus, from the product of the relations
$\frac{MB}{AS}=\frac{PB}{AP}\ ,\ \frac{AS}{MC}=\frac{NA}{CN}$ obtain $\frac{MB}{MC}=\frac{PB}{PA}\cdot \frac{NA}{NC}\ ,$ i.e. $(1)\ .$

$\boxed {\Longleftarrow\blacktriangleleft}$ Denote $M'\in BC\cap NP$. From (direct Menelaus' theorem proved previously applied to the points $M',\ N,\ P$) the relations $\frac{M'B}{M'C}\cdot \frac{NC}{NB}\cdot \frac{PA}{PB}=1$ and $(1)$ obtain $\frac{M'B}{M'C}=\frac{MB}{MC}$, i.e. $M'\equiv M$ what means $M\in NP\ .$[/color]

[color=red][b]The Ceva's theorem.[/b][/color] $\boxed {AM\cap BN\cap CP\ne \emptyset\ \ \vee\ \ AM\parallel BN\parallel CP\Longleftrightarrow \frac{MB}{MC}\cdot \frac{NC}{NA}\cdot \frac{PA}{PB}=-1}\ \ (2)\ .$
$\boxed {\blacktriangleright\Longrightarrow}\ .$
$1^{\circ}.$ [color=darkred][b]The case[/b] $I\in AM\cap BN\cap CP\ .$Denote the intersections $U\in BN\cap d$, $V\in CP\cap d$. Thus, from the product of the relations $\frac{MB}{UA}=\frac{IM}{AI}$, $\frac{VA}{MC}=\frac{AI}{IM}$, $\frac{NC}{NA}=\frac{BC}{UA}$, $\frac{PA}{PB}=\frac{AV}{BC}$ obtain the relation $(2)\ .$ Remark the relation $\frac{MB}{MC}=\frac{UA}{VA}\ (\triangle IBC\sim \triangle IUV$ and the rays $[IM\ ,\ [IA$ are homologously$)\ .$
$2^{\circ}.$ [b]The case[/b] $AM\parallel BN\parallel CP\ .$ The product of $\frac{NC}{NA}=\frac{CB}{MB}\ ,\ \frac{PA}{PB}=\frac{MC}{BC}\Longrightarrow (2)\ .$

$\boxed {\Longleftarrow\blacktriangleleft}$ Suppose that $BN\cap CP\not =\emptyset$ and denote the intersections $I\in BN\cap CP$, $M'\in BC\cap AI$. From (direct Ceva's theorem proved previously applied for $AM'\cap BN\cap CP\ne\emptyset$) the relations $\frac{M'B}{M'C}\cdot \frac{NC}{NA}\cdot \frac{PA}{PB}=-1$ and $(2)$ obtain the relation $\frac{M'B}{M'C}=\frac{MB}{MC}$, i.e. $M'\equiv M$ what means $AM\cap BN\cap CP\ne\emptyset\ .$[/color]

[color=blue][b]Remark.[/b] [u]The trigonometrical form of the Ceva's theorem.[/u]

For the points $M\in BC$, $N\in CA$, $P\in AB$ denote :
$m(\widehat {MAB})=x_1\ ,\ m(\widehat {NBC})=y_1\ ,\ m(\widehat {PCA})=z_1\ ;$
$m(\widehat {MAC})=x_2\ ,\ m(\widehat {NBA})=y_2\ ,\ m(\widehat {PCB})=z_2\ .$
Then the cevians $AM$, $BN$, $CP$ are concurrently or parallelly if and only if \[ \boxed {\sin x_1\cdot \sin y_1\cdot \sin z_1=\sin x_2\cdot \sin y_2\cdot \sin z_2}\ .  \][/color][/hide]

[hide="Here is the Euler relation."][color=red][b]The Euler's relation.[/b] Let $ABC$ be a triangle. For a interior point $I$ denote the intersections $M\in AI\cap BC$, $N\in BI\cap CA$, $P\in PI\cap AB$. Then there is the following relation: \[ \boxed {\frac{IA}{IM}+\frac{IB}{IN}+\frac{IC}{IP}+2=\frac{IA}{IM}\cdot \frac{IB}{IN}\cdot \frac{IC}{IP}}\ .  \][b]Remark.[/b] If the segments are orientated, then is truly the [u]Euler's relation[/u] for any point $I$ from the plane of the triangle $ABC$ ?[/color][/hide]

[quote="maokid7"]quick question...what are mass points?[/quote]

Can anybody answer???


here is a Division where someone has removed all digits leaving
just one digit...

Can you find the divisor, dividend and quotient?

[quote="computer_nuke"]Do all the X represent the same digit?[/quote]

no, there is no condition on digits. They may be same, they may not be same.

Have you found a solution?


FTW
AMC
AIME
MATHCOUNTS
AMC 8
USA(J)MO
USAMO

OK guys, we know how bad the chat has been. It recently got disabled (AGAIN). It hasn't been this bad since last summer!

WE all want to be able to chat, right?

So what do we do?

We start a brand-new organization, the Coalition for Good Behavior. CGB for short.

People sign some agreement to join the CGB.

People get CGB points for every week they spend followign the rules of the CGB. These points can be traded for various things I have not decided upon yet.

If someone breaks the CGB code, it will be brought to the attention of the members of the CGB. Then there will be action taken. If found guilty, the person will be temporarily ejected from CGB.

If you would like to join this revolutionary group that will save FTW (we are aiming for everyone to be members at some point) please post: I, (username) agree to follow the CGB code.

[hide="The CGB Code"]
No inappropriate language should be used (this includes acronyms and substituting letters to avoid language filter)
No multiple accounts.
No excessive spam or usage of all CAPS
No discussion of any type of illegal activity
Do not reveal your or others' name or location
Do not harrass other users
Do not concern yourself with the actions of others or why they were muted/banned
Act responsibly and be respectful at all times
[/hide]

Questions? Comments? Concerns? Please post here!

I, THECHAMP, agree to follow the CGB code.

[quote="SuperMathBoy"]Will CGB work? Who will regulate it? Levans?

Still, I will adhere to the CGB code.[/quote]

That's whats brilliant. There is no regulation. That leads to strife and fighting. Just behave well, and you will be rewarded.

I, webmath, agree to follow the CGB code.

KNOCKOUT, did you mean I Knockout agree to folly CGB rules or I Knockout agree to follow CGB rules? Because first one isn't that good?

Heh, isn't CGB rules just ToC?

Whatever, I agree to follow CGB rules...

[quote="pythag011"]
Heh, isn't CGB rules just ToC?
quote]

You mean ToS, but yes. Keep it coming guys. If all of you do what you agreed to do, that is 6 less people who will contribute ot bad behavior.

I truly agree to follow the CBG rules...

I, EBob agree to follow the CGB code.

I hope everyone else will to. :yoda:  :starwars:  :welcome:

I do solemnly swear on oath to defend the CGB, adhere to it, and promptly try to execute this oath onto others.

It doesn't help if you spam up the chat with posts like

[quote]
[19:41] _____________: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1418272#1418272
[19:41] _____________: sign up for the CGB
[19:42] _____________: start by our one and only ex-spammer mewto55555!!!!!
[19:42] _____________: (contains atoned ex-spammers like mew and thechamp)
[/quote]

I, (AIME_is_hard), agree to follow the CGB code.

This isn't gonna work. levans never checks this forum anyway. and if he sees it, he won';t just go ohh people pleged to be good. I will just put chat back on. there will always be peopel who are bad on ftw.

Add your name to the hide tag (hopefully future vice-prez for signing up first, must approve with romanian). If you have any complaints, comment here!!! (I mean about CGB abusers) Also, if levans temporarily grants chat, make sure to spread the message of CGB at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1418272#1418272

Also contains exspammers like mewto55555 and TheChamp
[hide] Romanian Genius (mewto55555)
SuperMathBoy
TheChamp
Knockout
Webmath
Pythag011
pingala
ebob
mz94
[/hide]

I can offer prizes:

CD me day: happens when chat is completely clean from [b]everyone[/b] for 1 month. everyone who sees me on (and yes, i will go on) gets to cd me. once.
Chance to gain quite a bit of points!
AMC 8: if someone is clean for 1 month s/he gets an AMC 8
AMC 10: if someone is clean for 2 monts and takes the 10s s/he gets an AMC 10
AMC 12: same as above but takes the 12s
AIME: Clean for 3 months
USAMO: Clean for 6 monts
MC Chapter: clean for 1 month
state: clean for 3 months
nats: clean for 4 months
all of the above is subject to change or be rejected without reason. to earn these I need to see you on at least a couple times and not abusing chat. This stuff is valuable, so i reserve the right to give this to only a certain number of people. all of the the above is also individual except for the CDing. What most likely will happen with a large number of people is I will find a trustworthy person that has a website and PM the link to people.
Note that these will probably change. Sometimes I don't see why rewards need to be given, sometimes i feel that it helps.

Somebody else better start putting prizes

oh and BTW if I give you something you already have, tough luck.

~BOGTRO~

I'm not sure it's not legal, but it's definitely a bit sketchy. I don't see the point of a materialistic system.

Because it is the only thing that motivates people.

BOGTRO, very generous, but I was thinking more like a "CGB Store" where you trade points you get for items/privileges.

Now what would these privilides and "items" be.

I, jain_harshi, agree to follow CGB code

@RomanianGenius: Seriously, this thread has really gone astray. levans is the boss of AoPS FTW, not you. Please refrain from this pretense of a "coalition" to promote behavior.

If people get a lot of rewards for this, then do you think they've really learned? :wink:

we are really taking on a *legalist point of view...

* an old chinese teaching, emperor qin shuangdi used it to rule his government...

That's a bit of an exaggeration. FTW is not that strict at all.

[quote="ReseesPieces"]wow the CGB has gone far.......[/quote]

That's right. Far off track.

If this arguing continues, this topic will be locked

THis thread being locked would be a supremely ironic gesture. The fact that a mod would find the need to lock a forum telling AOPSers to behave better speaks volumes.

Anyway though guys stop arguing. If you aren't part of the solution you are part of the problem.

Just lock it, jjx. This entire thread is absolutely redundant, at best.


geometry
function
trigonometry
geometry unsolved

to express the value of the shadowed area using trigonometric functions  it's quite easy.
The question is to prove ( or disprove ) that this value can be expressed using geometrical  methods ( ruler and compass ) only.

If $ \alpha\equal{}2$ then the area is $ A\equal{}\minus{}2\plus{}\arcsin(2/\sqrt{5})\plus{}2\arcsin(4/5)$.
Your question is (if I understand it): we consider $ A$ as a real number; is this number constructible by rule and compass ?
Let $ u\equal{}\arcsin(4/5)$; then $ \arcsin(2/\sqrt{5})\equal{}\pi/2\minus{}u/2$ and $ A\equal{}\minus{}2\plus{}\pi/2\plus{}3u/2$ and the question becomes: is $ \pi/3\plus{}u$ constructible ?
$ \cos(\pi/3\plus{}u)\equal{}(3\minus{}4\sqrt{3})/10$ is algebraic.
If $ v$ is algebraic then $ \arccos(v)$ is not algebraic because $ \arccos$ is a complex logarithm.
Then the answer is NO.
Example: $ \cos(2\pi/5)$ is algebraic but $ 2\pi/5$ is not.


Forgive me if you guys have all already seen this, but I have an interesting logic question:

You have 12 coins, 11 of which are exactly the same, 1 of which has a different weight from the others (can be higher or lower,) you have to find out which coin it is using a scale.  The tricky part is, you only get three weighings.

note: the scale only tells you which side is heavier, not by how much or anything like that, also, just to clarify, the weight of the odd coin is completely unknown.

hint:
[hide]You can place separate piles on the same side of the scale[/hide]

solution:
[hide]
call the different coin the Odd Coin, and the rest Normal.

Start with weighing 4 against 4.
      If they are even, they do not contain the Odd Coin, so weigh 3 of the remaining ones against 3 of the Normal coins.  If they are equal, the remaining coin is the Odd Coin, if the 3 coins are heavier, weigh 2 of the 3 against each other. If one is heavier, that one is the Odd Coin, if neither are, the other of the three is the Odd Coin.  If the three coins are lighter, weigh 2 against each other, if they are equal it is the other of the 3; if they are different, the lighter of the coins is the Odd Coin.

      If your original piles of 4 are not equal, one is heavier, and the reamining 4 coins are Normal.  Take 2 coins from the heavy pile, and 3 coins from the light pile and weigh them against the 4 Normal coins, and one coin from the lighter pile, leaving 2 coins from the heavy pile off of the scales.
If the 2 sides are even, then weigh the 2 remaining coins, the heavy one is the Odd Coin.
If the side containing the 4 Normal coins, and the one from the lighter pile is heavier, then the Odd coin must be lighter than the Normal coins, and must be on eof the 3 coins from the lighte pile of the other side of the scale.  In this case, weight 2 of the 3, and if they are equal, the remaining coin is the Odd Coin, if one is lighter, that one is the Odd Coin.
If the side with the the 4 normal coins and the coin from the lighter pile is lighter, than either the coin from the lighter pile is the Odd Coin, or one of the 2 coins from the heavier pile, on the other side of the scale is the Odd Coin.  Weigh the two coins on the scale, from the heavier pile, against each other.  IF they are even, the coin on the other side, from the lighter pile is the Odd Coin.  If one of them is heavier, that is the Odd Coin.[/hide]

posted by me before: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=coin&t=34138]here[/url]


geometry
3D geometry
tetrahedron
MATHCOUNTS
trigonometry
vector
analytic geometry

is there a formula for this: calculating volume of tetrahedron

for calculating the volume of a tetrahedron :D

Trig/vectors don't belong in the MathCounts forum, so I'm moving these posts to Pre-Olympiad.

Say you have four vertices. Find three vectors from one vertex. The volume is half the volume of the parallelopiped formed by the three vectors.

Too lazy to put down formula. It's just a dot and a cross product. :D

I'm pretty sure the original poster wanted the volume of a [b]regular[/b] tetrahedron in terms of its side length $s$.

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108559[/url] for somewhat easier formula (resembling Heron's).

If you have all the sides of a tetrahedron, map it out onto a three-dimensional coordinate plane and use determinants to figure out the volume. Read up AoPS Volume 2 on Analytic Geometry.  :) But then again... mapping stuff out onto coordinate planes is quite annoying.

[quote="Franz Joseph"]Sorry, a kind of stupid question. Why don't you just make use of the fact that a tetrahedron is a form of pyramid?

$\frac{1}{3}\times Base Area \times Height$ which should then easily yield a simple formula expressing the volume in terms of a side length.[/quote]
But I don't think there's a nice and simple formula expressing height in terms of side lengths.  :maybe:

[quote="i_like_pie"][quote="bpms"][quote="i_like_pie"]Vertices A, B, C, D: $V=\frac{|det(A-B,\ B-C,\ C-D)|}{6}$

Vertices A, B, C, D: $V = \frac{AD \cdot BD \cdot CD}{6}\cdot \sqrt{1+2 \cdot \cos ATB \cdot \cos BTC \cdot \cos CTA-\cos^{2}ATB-\cos^{2}BTC-\cos^{2}CTA}$[/quote]
Have fun memorizing that...[/quote]
I don't think I want to. :ninja:[/quote]
I was referring to anirudh, who wanted a formula, although generally complex formulas aren't as good as the derivation...

[quote="bpms"][/quote][quote="i_like_pie"][quote="bpms"][quote="i_like_pie"]Vertices A, B, C, D: $V=\frac{|det(A-B,\ B-C,\ C-D)|}{6}$

Vertices A, B, C, D: $V = \frac{AD \cdot BD \cdot CD}{6}\cdot \sqrt{1+2 \cdot \cos ATB \cdot \cos BTC \cdot \cos CTA-\cos^{2}ATB-\cos^{2}BTC-\cos^{2}CTA}$[/quote]
Have fun memorizing that...[/quote]
I don't think I want to. :ninja:[/quote][quote="bpms"]
I was referring to anirudh, who wanted a formula, although generally complex formulas aren't as good as the derivation...[/quote]
I know. I don't think anybody would want/need to memorize that.

I could come up with one in terms of side lengths... but I'd have to do it in the coordinate plane, which would not be pretty.  :(

Are there any alternative methods to derive it???


inequalities
search
inequalities unsolved

Let $a,b,c$ be nonegative real numbers such that $ab+bc+ca=1$, prove that \[\frac1{1+a^{2}}+\frac1{1+b^{2}}+\frac1{1+c^{2}}+\frac1{1+abc}\geq3.\]

Maybe prove an strogner inequlaity :maybe:  $\frac{4}{3}\le a^{2}+b^{2}+c^{2}+abc$

I will work on it somemore tommorow :|

Sorry for my nonsence post in #2.
I find this inequality is wrong.
Because this inequality is equal to: \[\frac{1}{(a+b)(a+c)}+\frac{1}{(b+a)(b+c)}+\frac{1}{(c+a)(c+b)}+\frac{1}{1+abc}\ge 3\]  \[\iff \frac{2(ab+bc+ca)(a+b+c)}{(a+b)(a+c)(b+c)}+\frac{1}{1+abc}\ge 3\]  \[\iff \frac{2abc}{(a+b)(a+c)(b+c)}+\frac{1}{1+abc}\ge 1\]  \[\iff \frac{2abc}{(a+b)(a+c)(b+c)}\ge \frac{abc}{1+abc}\]  \[\iff \frac{2}{(a+b)(a+c)(b+c)}\ge \frac{1}{1+abc}\]  \[2+2abc \ge (a+b)(a+c)(b+c)\] But this is not always true,I think.
But I think it is hard to find the minimin of the left hand :(

Actually it interests me because it seems like we have equality whenever one of the variables is 0.

I notice this when I see $a=0, b=2, c = \frac{1}{2}$ is the counterexample that zhaobin expected, but in this case equality actually holds in the original guy. I don't understand how to go from line 2 to line 3 in zhaobin's post so maybe this is where it is a problem.

In fact the same example shows DPL one is wrong, unless he meant the inequality sign to be the other way. Maybe I misunderstand his somehow, but I can't see when equality even holds, and since it is not the same as in pvthuan's, I am not sure how it is related.

Unfortunately I don't have a solution to go with all of this talking. :P :(

[quote="zhaobin"]Xevarion,I think you should use this: \[(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc\] :)
For an example,you can try $b=c=k,a=\frac{1-k^{2}}{2k}$
When $k$ is  small enough. :lol:[/quote]
I see.  :blush:  :blush:  :blush:

[quote="Xevarion"]$a=0, b=2, c = \frac{1}{2}$ :([/quote]
You shouldn't search a counterexample for one of the variables is zero.Because I reduce $abc$ in both side.When I do $\iff$ I assume $abc>0$,If $abc=0$ then the equation is hold.

[quote="Xevarion"][quote="zhaobin"][quote="Xevarion"]$a=0, b=2, c = \frac{1}{2}$ :([/quote]
You should search a counterexample for one of the variables is zero.Because I reduce $abc$ in both side.When I do $\iff$ I assume $abc>0$,If $abc=0$ then the equation is hold.[/quote]
When we say a variable is 0 usually it suffices that the variable is just close to 0.[/quote]
I think you still didn't understand my meaning. :D
I mean just for this problem if a variable is 0.Then the equaiton is hold.
So you can't search a counterexample for one of the variables is zero.
Because all of them mustn't be counterexamples. :)
(Yes,in many other inequality we search a variable is 0 usually it suffices that the variable is just close to 0.But it has a requiment that it must be a conterexample when the variable is 0)


. 375. Find all continuous functions , such that

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

for all x,y R.


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AMC 8

So AMC season has begun. How did we do on our AMC 10/12's? I took the 10A and think I got all of them  :)

121.5 on AMC12. The last five problems were too hard for me to even try.

The AMC 10 was really easy this year though, so I still think you'll make USAMO, based on your AIME score last year.

126 on the 12A.  Who all is taking the 12B?

[color=red][b]EDIT:[/b] Please [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260631]don't post AMC B scores[/url] until the AMC forum is unlocked.  Thanks! - Moderator[/color]

How did everyone do on the AIME?
I almost failed it.  :oops:

gthogsgtythna I did fail it. I made a silly mistake on 1,5, and 9 (and I found out later a silly mistake on 8 also), for a total of [b]6[/b]. Yes, two points lower than last year 6. Clearly a very bad day, I suppose. I probably won't make it this year, which would be rather depressing, though I would have made it last year with that score with my AMC. Idk, I'm very annoyed...

Dude you will most definitely make USAMO if you have a 210 index.

Maybe, it depends on the floor and that depends on the AIME II. Supposedly there was someone with a 144/6 AMC 12/AIME I, and since the AMC 12 was so hard this year that will likely let the floor be 6. Since the AMC 10 was so easy, the index will probably be around  210, which means I make it borderline. Again, all speculation. The point was my goal was twice of what I got, and the very real possibility of not making it (which did not exist last year) makes me very annoyed. I find it pretty funny though, that while my AIME tanked, my index actually improved.

Hmm I bombed the AMC 8 so I sorta have to get a 11 on the AIME which won't happen both due to my inability of pwning AIME problems and making stupid mistakes >.>

Oh darn I should learn combo

My index is about the same as yours, only slightly higher, so we're pretty much on the same boat here. I hope that you make it because you can do so much better on Olympiad-type problems than me, it would be a shame to see you not to. If we both qualified, chances are you will make Red Mop and I will get another zero again.

I need to get better at combo ARGH

Combo will own me on the usamo

But your index is with an AMC12, not an AMC10. When you take the AMC 10, you have to have a minimum floor score on AIME no matter what your index is, and mine may not make the cut (Its very likely you made it if you have an index of 211.5). I certainly hope I make USAMO too, ESPECIALLY since I would have technically qualified for Red MOP last year. I'm still studying in case I do make it.


Find the primes $ p$ such that $ p^{2} \plus{} 11$ has exactly $ 6$ divisors (including $ 1$ and itself )


Use the distributive property.
x(y)-z(y)=y(x-z).
Thus, 26(64)-24(64)=2(64)=128.


inequalities
algebra
polynomial
inequalities proposed

If $ a , b , c$ are different numbers between them  , prove that :

  $ (\frac {a \plus{} b}{a \minus{} b})^2 \plus{} (\frac {b \plus{} c}{b \minus{} c})^2 \plus{} (\frac {c \plus{} a}{c \minus{} a})^2 \geq 2$

[quote="stergiu"]If $ a , b , c$ are different numbers between them  , prove that :

  $ (\frac {a \plus{} b}{a \minus{} b})^2 \plus{} (\frac {b \plus{} c}{b \minus{} d})^2 \plus{} (\frac {c \plus{} a}{c \minus{} a})^2 \geq 2$[/quote]
Maybe
$ (\frac {a \plus{} b}{a \minus{} b})^2 \plus{} (\frac {b \plus{} c}{b \minus{} c})^2 \plus{} (\frac {c \plus{} a}{c \minus{} a})^2 \geq 2$

If so, it's equivalent to the following inequality:
$ \left(\sum_{cyc}(a^2b \plus{} a^2c \minus{} 2abc)\right)^2\geq0.$

[quote="epitomy01"]Another solution:
let x=(a+b)/(a-b); then a/b = (x+1)/(x-1)
let y=(b+c)/(b-c), then b/c = (y+1)/(y-1)
let z=(c+a)/(c-a), then c/a = (z+1)/(z-1)
Since a/b * b/c * c/a = 1, (x+1)(y+1)(z+1) = (x-1)(y-1)(z-1),
so xy+yz+xz = -2.
Since x^2 + y^2 >= 2|xy|,
adding up (for x^2+z^2, y^2+z^2), gives:
x^2 + y^2 + z^2 >= |xy + yz +xz| = 2,
which is the required inequality.[/quote]

 Great ! This was also the official solution !!!

 Babis

[b] NEW SOLUTION[/b]

 [size=150][b]Silouan [/b]( a greek student ) gave this solution !!![/size]

 Obviously

 $ (\frac {a \plus{} b}{a \minus{} b} \plus{} \frac {b \plus{} c}{b \minus{} c} \plus{} \frac {c \plus{} a}{c \minus{} a})^2\geq 0$

I will prove that
 $ \prod\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

[color=red][b]Proof[/b][/color]

 I consinde the polynomia

 $ f(x) \equal{} \prod\frac {(x \minus{} a)(x \minus{} b)}{(c \minus{} a)(c \minus{} b)} \minus{} 1$

We have
  $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$ , so  $ f(x) \equal{} 0$ for every  χ
Thus  $ f(a \plus{} b \plus{} c) \equal{} 0$ , hence

 $ \prod\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

See also here

http://www.mathlinks.ro/viewtopic.php?t=172976&start=20

It's an extremely classical problem although it's still nice! I think nobody can miss this problem in the competition! :)

Some corrections on Sympols : On the above solution by Silouan, please put $ \sum$ instead of $ \prod$.

[quote="stergiu"][quote="epitomy01"]Another solution:
let x=(a+b)/(a-b); then a/b = (x+1)/(x-1)
let y=(b+c)/(b-c), then b/c = (y+1)/(y-1)
let z=(c+a)/(c-a), then c/a = (z+1)/(z-1)
Since a/b * b/c * c/a = 1, (x+1)(y+1)(z+1) = (x-1)(y-1)(z-1),
so xy+yz+xz = -2.
Since x^2 + y^2 >= 2|xy|,
adding up (for x^2+z^2, y^2+z^2), gives:
x^2 + y^2 + z^2 >= |xy + yz +xz| = 2,
which is the required inequality.[/quote]

 Great ! This was also the official solution !!!

 Babis

[b] NEW SOLUTION[/b]

 [size=150][b]Silouan [/b]( a greek student ) gave this solution !!![/size]

 Obviously

 $ (\frac {a \plus{} b}{a \minus{} b} \plus{} \frac {b \plus{} c}{b \minus{} c} \plus{} \frac {c \plus{} a}{c \minus{} a})^2\geq 0$

I will prove that
 $ \sum \frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

[color=red][b]Proof[/b][/color]

 I consinde the polynomia

 $ f(x) \equal{} \sum \frac {(x \minus{} a)(x \minus{} b)}{(c \minus{} a)(c \minus{} b)} \minus{} 1$

We have
  $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$ , so  $ f(x) \equal{} 0$ for every  χ
Thus  $ f(a \plus{} b \plus{} c) \equal{} 0$ , hence

 $ \sum \frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$[/quote]


Let ABC be a triangle.Let take a D point on [AC], an E point on [BC], and a F point on [DC].Point M is the intersection between AE and BD, N is the intersection between AE and BF, P is the intersection between MF and DE, Q is the intersection between BF and DE, and R is the intersection between MF and BC, suposing that MF is not parallel to BC. If EF and BD are parallel, prove that AP, CQ and RN have a common point.

I'm interested about the source of this problem and its autor.
May this problem is good enough for Virgil Nicula if he don't know it source in advance  :wink:

What means "interesting problem" ? Where is the subject of this problem or the subject description ?

Please answer me. Thanks.

it's very simple: Appling ceva on CEF and point Q we get that CQ pass through the midpoint of FE; appling ceva of AEF and point P we get that AP pass through the midpoint of FE; appling ceva of REF and point N we get that RN pass through the midpoint of FE. That's all.

[quote="borislav_mirchev"]You may see so many problems with similar description on mathlinks so why not  :wink: ... I think it is not an easy problem and it may be interesting to some of experienced matematicians it was the reason to describe the problem in such a way.

If you want you may try it. There are far more difficult problems but it is not easy at all ... you may see I post problems that vary in their difficulty and it is human to err.[/quote]

You didn't understand me. You aren't right, [b]Borislav[/b] ! It is no correctly. Often I find a problem by its subject or its subject description.

Virgil Nicula there is n.p. in the future I'll try to find better subjects. But it is not easy. When I post many problems at once it requires time. It is the reason.

Gabriel you always give excellent solutions to the problems posted by me. I think you have excellent skills. You may try the following problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=209514 . It have no clear written solution from long time. Virgil Nicula have good skills with analytical solutions so he may have success on it. Ohter people on mathlinks are also welcome. I think Analytical geometry may be required and lots of unusual calculations. I'm not so good in math - it is the reason to not have solution from so many time.

Thank you!


Solve the equation 2^a+2^b=3^c+3^d in natural numbers. Are there infinitely many solutions ? :? If not write all the solutions :D ( i found 5 or 6 solutions for the equation but only for small nb; after that i could not find any :( )

[quote="ciprian"]Solve the equation $2^a+2^b=3^c+3^d$ in natural numbers. Are there infinitely many solutions ? :? If not write all the solutions :D ( i found $5$ or $6$ solutions for the equation but only for small nb; after that i could not find any :( )[/quote]

$1+3^x=2^y$ has few solutions. It is surely true, but how do you prove it?

WLOG $a \ge b, c \ge d$ and we can rewrite $2^a (2^{a-b} - 1) = 3^c (3^{c-d} - 1)$ and we see we have to solve the related congruences

$2^{a-b} \equiv 1 \bmod 3^c$
$3^{c-d} \equiv 1 \bmod 2^a$

Note that for $a, c \gg 1$ the minimum solution should be of the form $\phi(3^c) = 2(3^{c-1}) | a-b$ and $\phi(2^a) = 2^{a-1} | c-d$ - however, noting that these divisors become exponentially larger than the numbers that they must divide tells us intuitively that there are a finite number of solutions, but I am far too lazy to find them all  :)