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special factorizations

In how many different ways can $2/15$ be represented as $1/a +1/b$, if $a$ and $b$ are positive integers with $a \ge b$?

Also, how do I type words in latex on these forums? Dollar signs make the letters run together. =[

Words can be made with \text { blablablba}.

Ex: $\text{blabla blabla}$.

[hide="hint on the problem"] We have calculators don't we?  *cough*graph*cough*.

[hide="hint without calcs"] Is there any way we can split these and factor them into anything? [/hide] [/hide]

One trick nice to know:
[hide]$1/a+1/b=2/15$
or $\frac{a+b} {ab} = \frac 2 {15}$
or $15a+15b= 2ab$
or $4ab - 30a -30b = 0$        ( Add 225 on both sides -  That's the trick! )
or $(2a-15)(2b-15) = 225$
And question reduces to in how many ways you can factor $225$
[hide]The factors are 1,3,5,15 ... etc which will give $b=8,9,10,15$ etc .[/hide].
[/hide]

We just did this problem yesterday in a MathCounts session-but I don't remember what our coach did after this step I spaced out:
[hide]\frac{1}{a}=\frac{2}{15}-\frac{1}{b}[/hide]

Somehow he got to counting the number of numbers that adding or subtracting to 53 would be a prime, and he got 7 numbers.  :?:

[hide]I just did guess and check... $2/15$ - $1/14$ , $2/15$ - $1/13$ etc. down to $2/15$ - $1/8$. You get like 5. When I did it, I FORGOT ABOUT 1/15 + 1/15! SO STUPID![/hide]

Exactly what we did last year at states ( we got it wrong though)

I just discovered how to use Simon's Favorite Factoring Trick on these types of problems today:

[hide]
$\frac 1a + \frac 1b = \frac 2{15}$
$\frac{a+b}{ab} = \frac 2{15}$
$15(a+b)=2ab$
$2ab-15a-15b=0$
$4ab-30a-30b=0$
$4ab-30a-30b+225=225$
$(2a-15)(2b-15)=225$
So this reduces to the number of factor pairs of 225: (1,225), (3,75), (5,45), (9,25)
I hope I got 'em all.
[/hide]

Haha, that's almost the exact same wording as gyan.

isn't there also 15*15

cuz 225 is $15^2$

couldn't you just graph it since this is mathcounts?

Was bringing back a 11-year old post necessary?

Then perhaps Fourth wasn't talking to you ;)


Please can anyone help me??? I have to do this problem for tomorrow!!! i DONT Even know how to start it....I really appreciate it thanks!:)

*A sidewalk of uniform width has an area of 180ft and surrounds a flowerbed that is 11ft wide and 13ft long. Find the width of the sidewalk.

*Boat A and Boat B leave the dock at the same time and each sails a t a constant speed. Boat A sails due north and Boat B due east. After 3 hours, they are 65 miles apart. Boat B sails 7 miles per hour faster than Boat A. Find the speed of each boat to the nearest tenth of a mile per hour.

PLease do anyone know how to do these problems? I really appreciate ur help thanks. ;)

[quote="gurlie_16"]*A sidewalk of uniform width has an area of 180ft and surrounds a flowerbed that is 11ft wide and 13ft long. Find the width of the sidewalk.[/quote]
Try drawing a diagram. Call the width w. What areas do you know? Try to express the total area (flowerbed + sidewalk) in terms of known areas, and also in terms of w. Then you will have an equation to solve to find w.

[quote]*Boat A and Boat B leave the dock at the same time and each sails a t a constant speed. Boat A sails due north and Boat B due east. After 3 hours, they are 65 miles apart. Boat B sails 7 miles per hour faster than Boat A. Find the speed of each boat to the nearest tenth of a mile per hour.[/quote]
Let v be the speed of boat A. Then v+7 is the speed of boat B. Remember that distance = speed * time. You know time so you can find distance, in terms of v. Then use the Pythagorean Theorem to find the distance between the boats. But you know this distance, so you can solve for v.

Thanks a lot! Finally i know how to start! I was completely lost! I felt kinda dumb cause i really have an 'A' in the class...thanks a lot for the help! :)


If $ n(A) \equal{} 3$ and $ n(B) \equal{} 2$, then $ [n(AxB)]{}^{n(A\cap B)}$ is at most:

a) 1
b) 6
c) 12
d) 18
e) 36

ANSWER: E

The largest value is obtained when $ A \cap B$ is as large as possible, that is when $ B$ is contained in $ A$. Thus, for example, $ A$ contains the elements $ a, b$ and $ c$ and $ B$ contians the elements $ a$ and $ b$, so that $ A \cap B \equal{} B$ and since $ n (A X B) \equal{} 3 \times 2 \equal{} 6$, the required answer is $ 6^{2} \equal{} 36$.

Very good ;)
Thanks man. Your explain is perfect.

well since you know that n(A)=3 and n(B)=2

you use those to find the base of the exponential

n(AxB)=3x2=6

now to find the union part you see that n(A) has three elements 1,2,3

and n(B) has 1,2

so the union is 2 because both sets have 1 and 2

so the base of the exponential was 6 and the exponent was 2

so your answer would be $ 6^2$=36


find all p(x,y),such that for all x,y
p(x+y,x-y)=p(x,y)


number theory
prime numbers
number theory unsolved

assume that A is a finite subset of prime numbers, and a is an positive integer.
prove that there are only finitely many positive integers m s.t: prime divisors of a^m-1 are contained in A.

A beautiful problem! Take m_1<m_2<... such that all prime factors of a^m_k-1 are in A for any k.
  Lemma 1:
   The set of prime factors of (m_k)k>=1 is finite.
  Proof:
    Suppose we can find p_1<p_2<... prime factors of the elements m_1,m_2,.... Then for any k the prime factors of a^p_k-1 are in A. Since A has a finite number of subsets, it follows that we can find an infinite set B such that whenever i and j are in B, a^p_i-1 and a^p_j-1 have the same prime factors. Since if i<>J we have (p_i,p_j)=1 ( and thus (a^p_i-1,a^p_j-1)=1), it follows that any prime factor of a^p_i-1 (with i in B) is a divisor of a-1. Take i in B such that p_i>a-1 and take q a prime dividing 1+a+...+a^(p_i-1). Then q|a^p_i-1 and thus q|a-1. Thus, q|p_i and q=p_i>a-1, contradiction with a previous statement.
     Thus, we can find primes p_1,...,p_k (not those from the lemma!) such that each m_i has its prime factors between them. If we denote by x(n,i) the exponent of p_i in m_n, then there is i such that (x(n,i))n>=1 is unbounded. Thus, there is a prime p and an unbounded sequence b_n such that the set of prime factors of a^p^b_n-1 is finite. Take a number s>1 and suppose q is a prime factor of 1+a^p^(s-1)+...+a^p^(s-1)*(p-1). Then q|a^p^s-1 and thus the order of a modulo q is p^r, with r<=s. If r<s, then it easily follows that q|p and thus q=p. Otherwise, p^s|q-1 and thus q>p^s. Since the sequence b_n is unbounded and the set of prime factors of a^p^b_n-1 is bounded, there is n_o such that for all n>n_0  1+a^p^(b_n-1)+...+a^p^(b_n-1)(p-1) is a power of p. But I really don't think this is possible,.

A small missprint: (a^p_i-1,a^p_j-1)=a-1.
  And the solution to the last statement:
  Suppose we have 1+x+...+x^(p-1)=p^t, with t>1. Then (x-1)p^t=x^p-1. From Fermat theorem we have p|x^p-x, thus p|x-1. Write x=1+rp^u, with (r,p)=1. Then rp^(u+t)=rp^(u+1)+Multiple of p^2u. Impossible (it follows that r is divisible ny p!). And the last statement is proved.

:) I have another solution:
We will use two lemmas:
Lemma 1:If a>1 and q is a prime divisior of a p -1 which p is a prime,then p/q-1.
Lemma 2: Every prime divisior of (a p m+1  -1)/(a p m -1) have this form:p m+1 *k+1.(m,k is a positive integer.)
Use little's Fermat theorem,we can easily prove two lemmas.
B is the set of all positive integer m which satisfied:every prime divisior of a m -1  belongs to A.
with this problem,I suppose for contradiction,card B is infinite.We note that :if m belongs to B then every divisior of m belongs to B.So we really needto all the prime and it's power in B.If  in B ,there have finite prime in B:then exits infinite sequences:p,p 2 ,..,p n ,...(which p is a prime belongs B) are belongs to B.By lemma 2,we get contradiction because the number of all divisior of (a p m+1  -1)/(a p m -1) is infinite.So,There have infinite prime in B.Note that: every prime q' in B there exits a prime p' in A which p'/a q' -1 so by lemma 1,q'/p'-1 so card A is infinite,contradiction.
That's all.if I mistake,please sorry to everybody .Thanks! :D


Prove that if $a,b,c$ are positive reals then :

$\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2} \ge \frac{4}{5}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

It's quite trange and I like this kind. :)

i think that we can prove this one even replacing 4/5= 1 by assuming a >= b >= c(WLOG) and then doing some thing we have the problem solved

[quote="darij grinberg"]No, the inequality doesn't hold with 4/5 replaced by 1. Just take a = 1, b = 9, c = 9 for a counterexample. The initial inequality is sharp.

  Darij[/quote]

Yes,It's the reason which I like the inequality.
It can't be prove by old method.

$\frac{4}{5}$ is the best constant.
The equality takes only with $a=b,c=0$

To prove this inequality, I used the following nice ineq.

$\frac{a(b+c)}{b^2+c^2}+\frac{b(c+a)}{c^2+a^2}+\frac{c(a+b)}{a^2+a^2} \ge 2$.

Did you use this ineq. in your proof, hungkhtn ?

[quote="Vasc"]To prove this inequality, I used the following nice ineq.

$\frac{a(b+c)}{b^2+c^2}+\frac{b(c+a)}{c^2+a^2}+\frac{c(a+b)}{a^2+a^2} \ge 2$.

[/quote]

I have expanded it to have

$\sum a^5(b+c) +2\sum a^3b^3 +2abc \sum ab(a+b) \geq 4a^2b^2c^2 +2 \sum a^2b^2(a^2+b^2) $

As $ \sum ab(a+b) \geq 6abc$ we have to prove

$\sum a^5(b+c) +2\sum a^3b^3 +2abc\sum ab(a+b) \geq 2\sum a^2b^2(a^2+b^2) $

but now, Vasc I cannot go on!!!!

I have tried by Tchebychev to have

$\sum ab \sum \frac{1}{a^2+b^2}  \geq \frac 92 $

But this is stronger than Iran 1996.

So  :(

Thank you very much.

I think it should $\geq 2$
for $a=b=1,c=0$

:)
My Solution Only use Cauchy-Swacter Inequality.

similar inequality:
a,b,c>0. prove that:
[tex]\sum{\frac{a}{b^3+c^3}}>= \frac{4}{\sum{a^2}}[/tex]

[quote="stupid_mathematician"]Yes!
Lemma 1:
 $\sqrt {\frac{a^2}{b^2+c^2}}$>= 2>= $\frac{a}{b+c}$
Uses my lemma! you will slove it![/quote]
can you detail your solution? :)

I understand your Idea,Stupid_Mathamatical...

I creat my inequal from:

$\frac{a}{b^2+c^2}\ge \frac{4}{a+b+c}$

(The general is:

$\sum\frac{a}{b^n+c^n} \ge \frac{4}{a^{n-1}+b^{n-1}+c^{n-1}}$    (*)

But for:

$\sum\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum\frac{1}{b+c}$

I don't this this way useful.

(At first,I Invent post this problem (*),but After I found the last is nicer and more difficult. )

This is my solution :
<=> (a+b+c). Left >= (a+b+c). right
So, by B.C.S : [tex](a+b+c). left >= (\sum \sqrt{\frac{a^2}{b^2+c^2}})^2 [/tex] and then it >= 4 by my lemma!
and  (a+b+c) .right <=4 by my lemma!

[quote="hungkhtn"]I understand your Idea,Stupid_Mathamatical...

I creat my inequal from:

$\frac{a}{b^2+c^2}\ge \frac{4}{a+b+c}$

[/quote]
please see my inequality again!
$\frac{a}{b^3+c^3}\ge \frac{4}{a^2+b^2+c^2}$
I think my lemma is useful for you to slove it!

Does anybody remember these four ineq to have a complete solution?

Many thanks.

first one can be written in the form $\sum S_a (b-c)^2 \ge 0$
Second one is nesbitt + vornicu schur
third one is vasc 6 ineq that i proved before. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=a%5E3b&t=26102

Thank you.

But what about fourth ineq ?

I don't remember to have ever seen it in the forum ?

And I cannot prove it. Any help please?

I have an ugly solution
WLOG, let $a \ge b \ge c$, then we have to prove
\[
\begin{gathered}
  \sum {\frac{{a(b + c)}}
{{b^2  + c^2 }}}  \geqslant 2 \hfill \\
   \Leftrightarrow \frac{{ab + ac - b^2  - c^2 }}
{{b^2  + c^2 }} + \frac{{ab + bc - a^2  - c^2 }}
{{a^2  + c^2 }} + \frac{{c(a + b)}}
{{a^2  + b^2 }} \geqslant 0 \hfill \\
   \Leftrightarrow \left( {a - b} \right)\left( {\frac{b}
{{b^2  + c^2 }} - \frac{a}
{{a^2  + c^2 }}} \right) + \frac{{c(a - c)}}
{{b^2  + c^2 }} + \frac{{c(b - c)}}
{{a^2  + c^2 }} + \frac{{c(a + b)}}
{{a^2  + b^2 }} \geqslant 0 \hfill \\
   \Leftrightarrow \left( {a - b} \right)\left( {\frac{{a^2 b + bc^2  - ab^2  - ac^2 }}
{{\left( {b^2  + c^2 } \right)\left( {a^2  + c^2 } \right)}}} \right) + \frac{{c(a - c)}}
{{b^2  + c^2 }} + \frac{{c(b - c)}}
{{a^2  + c^2 }} + \frac{{c(a + b)}}
{{a^2  + b^2 }} \geqslant 0 \hfill \\
   \Leftrightarrow \left( {a - b} \right)^2 \frac{{ab - c^2 }}
{{\left( {b^2  + c^2 } \right)\left( {a^2  + c^2 } \right)}} + \frac{{c(a - c)}}
{{b^2  + c^2 }} + \frac{{c(b - c)}}
{{a^2  + c^2 }} + \frac{{c(a + b)}}
{{a^2  + b^2 }} \geqslant 0 \hfill \\
\end{gathered}
\]
which is true.

Thank you very much.

It is not ugly. :)

How is your solution, hungkhtn ?

Many thanks.

[quote="hungkhtn"]You are false when say:

$\frac{4}{a+b+c} \ge \frac{4}{5}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

I guess It's true <->$a,b,c$ are 3 lengths of triangle.[/quote]
yes! But it is easy to process the case that $a,b,c$ are not! okie?

A nice proof for $ \sum \frac{a(b + c)}{b^2  + c^2 }  \geqslant 2 $ is the following
\[\frac{a(b + c)}{b^2  + c^2 } \ge \frac{2a^2}{a^2+b^2+c^2}\]

Classical and nice your proof, hungk.

[quote="Vasc"]A nice proof for
 $ \sum {\frac{{a(b + c)}}
{{b^2  + c^2 }}}  \geqslant 2 \hfill $

is the following
$ \frac{a(b + c)}{b^2  + c^2 } \ge \frac{2a^2}{a^2+b^2+c^2}$.[/quote]
Using this method, can we extend this ineq in the form

$ \sum {\frac{{a(b + c)}}
{{b^2  + c^2 +mbc}}}  \geqslant 2 \hfill $

for m>0 ?

[quote="Vasc"]Multiplying by $a+b+c$, the ineq. becomes

$\sum\frac{10a^2+10a(b+c)}{b^2+c^2} \ge 24+\sum\frac{8a}{b+c}$.

I obtained this ineq. by summing the following four inequalities:

$\sum\frac{8a^2}{b^2+c^2} \ge \sum\frac{8a}{b+c}$,
$\sum\frac{2a^2+bc}{b^2+c^2} \ge \frac{9}{2}$,
$\sum\frac{a(b+c)-bc}{b^2+c^2} \ge \frac{3}{2}$,
$\sum\frac{9a(b+c)}{b^2+c^2} \ge 18$.

The inequality is not strong, but it is difficult and nice. ;)[/quote]

Oh,I've read your solution...at first I don't see it...It's so difficult,With for inequality very strong!

I think the inequality still true:

$\frac{a}{b^3+c^3}+\frac{b}{a^3+c^3}+\frac{c}{a^3+b^3} \ge \frac{4}{5}({\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2})}$

And :

$\frac{a}{b^{k+1}+c^{k+1}}+\frac{b}{a^{k+1}+c^{k+1}}+\frac{c}{a^{k+1}+b^{k+1}} \ge \frac{4}{5}({\frac{1}{a^k+b^k}+\frac{1}{b^k+c^k}+\frac{1}{c^k+a^k})}$

 :)

[quote="hungkhtn"]Here Is My Solution:

$(\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2})(a(b^2+c^2)+a(b^2+c^2)+a(b^2+c^2)) \ge (a+b+c)^2$

The remain is prove:

$5(a+b+c)^2 \ge 4(ab(a+b)+bc(b+c)+ca(c+a))(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

It's easy after a short expanding![/quote]

Expanding I have

$5\sum a^2  + 6\sum ab \geq \sum ab(a+b)(\frac{1}{a+c}+\frac{1}{b+c})$

but now I cannot to go on.

Can you please complete it, hungkhtn?

Thank you very much. :)

Ok.
Let
$X=a^2+b^2+c^2$
$Y=ab+bc+ca$
$Z=ab(a+b)+bc(b+c)+ca(c+a)$

We need to prove:

$5(X+2Y)(Z+2abc) \ge 4Z(X+3Y)$
$\leftrightarrow XZ+10abcX +20abcY \ge 2YZ$

But by short expanding we have:

$\leftrightarrow XZ+10abcX  \ge 2YZ$

And It complete my solution.

[quote="siuhochung"]first one can be written in the form $\sum S_a (b-c)^2[/quote]

how exactly? i tried, but failed. could detail a little please? (the only soln i could find implied lots of sums and muirhead)

[quote="perfect_radio"][quote="siuhochung"]first one can be written in the form $\sum S_a (b-c)^2$[/quote]

how exactly? i tried, but failed. could detail a little please? (the only soln i could find implied lots of sums and muirhead)[/quote]
Write it in the form, $\sum ab(a-b)(\frac{1}{(b^2+c^2)(b+c)}-\frac{1}{(a^2+c^2)(a+c)}) \ge 0$

[quote="siuhochung"]
Second one is nesbitt + vornicu schur
[/quote]

how does that go? nesbitt applied to what numbers?

ignore that msg, i remember i was wrong at that time.
A nice solution given by darij makes use of gmhm, you can try.

is it somewhere on the forum  :?

i tried to apply gmhm, but i obtained some ugly-looking terms.

[quote="Vasc"]I obtained this ineq. by summing the following four inequalities:

$\sum\frac{8a^2}{b^2+c^2} \ge \sum\frac{8a}{b+c}$,[/quote]

This one was proven in the form $\sum\frac{a^2}{b^2+c^2} \geq \sum\frac{a}{b+c}$ on http://www.mathlinks.ro/Forum/viewtopic.php?t=17207 and http://www.mathlinks.ro/Forum/viewtopic.php?t=34563 ( more general assertion on http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 and http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 , even more general assertion on http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 );

[quote="Vasc"]$\sum\frac{2a^2+bc}{b^2+c^2} \ge \frac{9}{2}$,[/quote]

this one on http://www.mathlinks.ro/Forum/viewtopic.php?t=26104 ;

[quote="Vasc"]$\sum\frac{a(b+c)-bc}{b^2+c^2} \ge \frac{3}{2}$,[/quote]

this one on http://www.mathlinks.ro/Forum/viewtopic.php?t=26102 ;

[quote="Vasc"]$\sum\frac{9a(b+c)}{b^2+c^2} \ge 18$.[/quote]

this one in the form $\sum\frac{a\left(b+c\right)}{b^2+c^2}\geq 2$ in this thread...

[quote="Vasc"]Using this method, can we extend this ineq in the form

$ \sum {\frac{{a(b + c)}}
{{b^2  + c^2 +mbc}}}  \geqslant 2 \hfill \\$

for m>0 ?[/quote]

Indeed, the maximal m for which this inequality will always hold is m = 1. Then, we get the inequality (B) from http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #4.

  Darij


Hi.

Prove that every prime divisor of $2k(k+1)+1$ must be of the form $M(4)+1$.

Now I tried this. Suppose $2k(k+1)+1=pq$ where $p$ is a prime. Then $(2k+1)^2+1=2pq$ So

$(2k+1)^2\equiv -1\mod (p)$, $p$ prime

What I'm not sure is the following: the above implies that $p$ es of the form $M(4)+1$, because of a corollary to a Wilson's Theorem.

(1) Am I right?
(2) Is there a most elemental proof?
(3) I guess I can proove that there are infinitely many primes of the form $M(4)+1$

Thanks

(1) Yes
(2) Most elementary¿ This [u]is[/u] elementary!
(3) I guess too

[quote="hpoincare"]
(3) I guess I can proove that there are infinitely many primes of the form $M(4)+1$[/quote]

This follows from Dirichlet's Theorem on arithmetic progressions.   :D

yeah I know Dirichlet Theorem but I was looking for an elementary proof in THIS case (as for 4n+3).
I thought I've made a good proof but now I see it was wrong.

At first I believed that if p was a prime of the form M(4)+1, then would exist a prime greater than p such that it divides p^2+1.
That's wrong: 41^2+1=2·29^2

No I wonder this: if I suppose there are a finite number of primes of the form M(4)+1, then p^2+1 must have prime factor greater than p_i, where p is the product of all primes p_i of the form M(4)+1.
I couldnt proove that

What can I do?

Well, saying it has a prime factor greater than all $p_i$ or just saying that it has a prime factor different from them doesn't make any difference, since you assume there are no others.

So you want to show:
When we have a finite number of primes, then their product $p$ has the property that $p^2+1$ has a prime divisor $q \equiv 1 \mod 4$ that is different from all the factors of $p$.

But that's just a small step:
Take any prime divisor $q$, it will be $\equiv 1 \mod 4$ by the mentioned theorem (I think it is: $x^2\equiv -1 \mod q$ has a solution iff $q=2$ or $q \equiv 1 \mod 4$; the case $q=2$ is easy to avoid).
Since $p$ and $p^2+1$ are coprime, also $q$ will be coprime with $p$, so is different from all it's divisors.

Another way to prove that there are infinetely many such primes would be to set $k=p$ in the problem of your post above.


function
superior algebra
superior algebra unsolved

I want to show that $ \phi(n) \geq \sqrt{n/2}.$ Here $ \phi$ is Euler's phi function.

Proof: write $ n=2^{k_0}p_1^{k_1}...p_r^{k_r},$
so
\begin{eqnarray*}
\frac{\phi(n)}{\sqrt{n}} & = & \frac{2^{k_0-1}}{2^{k_0/2}}(p_1-1)p_1^{k_1/2 -1}... (p_r-1)p_r^{k_r/2 -1}\\
& \geq \frac{{1}{\sqrt{2}}}
\end{eqnarray*}

Can somebody please tell me that is my argument correct.

Could you repost with the property Tex-ing?  This looks awful

Sorry for the error:

\[ \frac{\phi(n)}{\sqrt{n}} = \frac{2^{k_0-1}}{2^{k_0/2}}(p_1-1)p_1^{k_1/2 -1}... (p_r-1)p_r^{k_r/2 -1}\\
\& \geq \frac {1}{\sqrt{2}}\]

Can somebody please tell me that is my argument correct.

Off the top of my head, a couple questions arise:

One cannot assume that $ k_0 > 0$ (i.e. you are not allowed to assume $ n$ is even!).

Also, you definitely need to explain that $ \geq \frac {1}{\sqrt {2}}$ part=I dont think it follows immediately as, you need to consider what happens
in your argument if $ k_i = 1$ because then $ p_i^{k_i/2 - 1}$ appears in the denominator!

What it all boils down to is that you have:

$ \frac {\phi(n)}{\sqrt {n}} = \frac {2^{k_0 - 1}}{2^{k_0/2}}(p_1 - 1)p_1^{k_1/2 - 1}... (p_r - 1)p_r^{k_r/2 - 1} \geq \prod \frac {p_i - 1}{\sqrt {p_i}}$.

Now, for what values of $ n$ is $ \frac {n - 1}{\sqrt {n}} > 1$?  If you do a quick analysis, it is true for all $ n > 2$ and when $ n = 2$ it is not.  So the product is $ \geq \frac {1}{\sqrt {2}}$


Prove that:
$(ac+bd-1)^{2}\geq (a^{2}+b^{2}-1)(c^{2}+d^{2}-1)$
where $a;b$ are real numbers satisfying $a^{2}+b^{2}\leq 1$

[quote="N.T.TUAN"][quote="andy_tok"]oops，sorry
Then why not just use $f(1)\geq 0$?[/quote]
Because if $f(1)\geq 0$ then not implies  $f(x)=0$ has real root. Ex $f(x)=x^{2}+x+1$.[/quote]

I think it does because $a^{2}+b^{2}-1\leq 0$ means that f(x) is either concave up or a straight line  :wink:

This Azel Inequality for n=2:
$(\sum a_{i}b_{i}-1)^{2}\geq (1-\sum a_{i}^{2})(1-\sum b_{i}^{2})$holds for all reals $a_{i}$ satisfying $\sum a_{i}^{2}\leq 1$


real analysis
calculus
calculus computations

There is the ever popular Principles of Mathematical Analysis by Rudin. This starts where a "beginner" would start, but goes quite quickly. It is a great book, but if it is your first attempt at analysis, it would probably need to be supplemented.

I learned from Thomson, Bruckner, and Bruckner (which I hear is out of print?), but I loved as a beginner. It goes very slowly with lots of examples and nice exercises. It tends toward the concrete side, though. It rarely takes you into the abstract setting that many analysis texts strive towards.

[quote="gauravpatil"]could anyone suggest appropriate begginer and intermediate books for real analysis[/quote]

this is the one that i used when i did my undergrad.

http://www.amazon.com/Methods-Real-Analysis-Richard-Goldberg/dp/0471310654

A combination which was used in my undergraduate real analysis class that worked well for me was to use Strichartz's [i]The Way of Analysis[/i] (a very wordy book that focuses heavily on motivation and on workmanlike as opposed to elegant proofs)  for the first pass through the material, then later going back to a denser and more succinct book (we used Kolmogorov and Fomin's [i]Introductory Real Analysis[/i], which has the bonus of being very inexpensive) as a supplement to review the material and approach it from a more abstract perspective.

I'm a bit hesitant to recommend Rudin's book for beginners -- as HilbertThm90 noted, it starts at the beginning but may not be ideal for your first introduction.


im chinese. u guys are super good at math. just saying i am a fan.

im Chinese too
is everyone hear living in china or just chinese and living somewhere including china

I am Chinese... How can you type in Chinese on Windows Vista?

I just got a Windows Vista and it doesn't have a language bar :(

Go to START -> CONTROL PANEL.
Then click Clock, Language, Region.
There should be a link that says "change keyboard or other input methods".
Click that, and a window should pop up.
Find "Keyboards and Languages" in the top bar and click change keyboard.
Click ADD, find Chinese, and add it.


Diophantine equation
number theory proposed
number theory

Let $ p,b,k \in N$ and $ \varepsilon > 0$
Prove that exist $ \{x_n\}$ such that
[b]i)[/b]$ x_n^2 \plus{} b|[\varepsilon x_n]!$
[b]ii[/b] $ p|x_n \plus{} k,\forall n\in N$
Soure :http://diendantoanhoc.net/forum/index.php?showtopic=32690

We can find $ D>\frac{1}{\epsilon},$ suth that $ \minus{}b$ is square $ mod \ D$, and solve equatian $ x^2\plus{}b\equal{}Dy^2$.

[quote="Rust"]We can find $ D > \frac {1}{\epsilon},$ suth that $ \minus{} b$ is square $ mod \ D$, and solve equatian $ x^2 \plus{} b \equal{} Dy^2$.[/quote]
Are you sure .
For example
$ x^2\minus{}34x^2\equal{}\minus{}1$ has no solution. :wink:
But you can use other method with the Pell eqution.

I can choose x0, then D, suth that $ D\equal{}x_0^2\plus{}b$, (or $ D\equal{}\frac{x_0^2\plus{}b}{y^2}$), then Pell's equation $ x^2\plus{}b\equal{}Dy^2$ had infinetely many solutions.

[quote="Rust"]I can choose x0, then D, suth that $ D \equal{} x_0^2 \plus{} b$, (or $ D \equal{} \frac {x_0^2 \plus{} b}{y^2}$), then Pell's equation $ x^2 \plus{} b \equal{} Dy^2$ had infinetely many solutions.[/quote]
Yes $ d\equal{}m^2\plus{}b$
But $ p|x_n\plus{}k$?

[quote="Rust"]It is easy may be satisfyed if $ p\not |D$.[/quote]
This problem can solve if we use the root sequence of Pell equation and quite easy .Don't need $ p\not |D$

[quote="TTsphn"]This problem can solve if we use the root sequence of Pell equation and quite easy .Don't need $ p\not |D$[/quote]
You are wrong. But we may choose $ x_0$, suth that $ ord_p(x_0^2\plus{}b)$ is even.

Do you remember the root sequence of Pell equation?
$ x^2\minus{}dy^2\equal{}n$ (1)
Call $ (a,b)$ is root of Pell equation $ x^2\minus{}dy^2\equal{}1$ then
$ (c,d)$ is solution of  $ (1)$
$ x_1\equal{}c$,$ y_1\equal{}d$
$ x_{n\plus{}1}\equal{}ax_n\plus{}dby_n$
$ y_{n\plus{}1}\equal{}bx_n\plus{}ay_n$
$ \{x_n\},\{y_n\}$ is a solution of (1)
You can true $ x_1$ then use this result.

More :
$ p,k,a\in N$
$ \gcd(a,p)\equal{}1$
$ \varepsilon >0$
Prove that exist an infinite sequence has property:
[b]1)[/b]$ ax_n^2\plus{}b|[\varepsilon x_n]!$
[b]2[/b]$ p|x_n\plus{}k$

[hide="hint"]
A lemma
$ a,b,c,p\in N$ and $ \gcd(b,p) \equal{} 1$
Consider the sequence
$ x_{n \plus{} 2} \equal{} ax_{n \plus{} 1} \plus{} bx_n$
Consider the sequence
$ r_n\equiv x_n(\mod p)$ and $ 0\leq r_n\leq p \minus{} 1$
Prove that $ \{r_n\}$ is periodic from the first term.
It will help me solve my problem.[/hide]


algebra
polynomial
number theory unsolved
number theory

[quote="ehsan2004"]Let $a,b,c\in Z$ and $a$ be the even number and $b$ be the odd number. Show that for every integer $n$ there exist one positive integer  $x$ such that $2^n\mid ax^2+bx+c$[/quote]

Consider the values of the polynomial $f\left(x\right)=ax^2+bx+c$ for $x\in\left\{0;\;1;\;...;\;2^n-1\right\}$. These are $2^n$ values, and all of them have different remainders $\mod 2^n$; in fact, if, for two different numbers x and y from the set $\left\{0;\;1;\;...;\;2^n-1\right\}$, the values f(x) and f(y) would have the same remainder $\mod 2^n$, then f(x) - f(y) would be divisible by $2^n$, but since

$f\left(x\right)-f\left(y\right)=\left(ax^2+bx+c\right)-\left(ay^2+by+c\right)=\left(x-y\right)\left(a\left(x+y\right)+b\right)$,

and since the right factor a (x + y) + b is odd (since a is even and b is odd), this would mean that the number x - y would be divisible by $2^n$, what is impossible (since x and y are different integers between 0 and $2^n-1$).

So the values of the polynomial $f\left(x\right)=ax^2+bx+c$ have $2^n$ different remainders $\mod 2^n$. But there are only $2^n$ possible remainders $\mod 2^n$, and thus all of these remainders are represented. Particularly, there must be some x such that f(x) has the remainder 0 $\mod 2^n$. But this means that $2^n\mid f\left(x\right)$; in other words, $2^n\mid ax^2+bx+c$. The problem is solved.

  darij


I'll wait to post a solution, I'd like to see other ideas.

[quote="jhcreinhardt"]Let $a,b,c$ be distinct reals. Prove that the following cannot occur.

     $(a-b)^\frac{1}{3}+(b-c)^\frac{1}{3}+(c-a)^\frac{1}{3}=0$[/quote]
We prove the result by contradiction.
Let $x^{3}=a-b, y^{3}=b-c,$ and $z^{3}=c-a$. (This is possible since the cube of a real can be positive, negative or zero.)
Now, if $(a-b)^\frac{1}{3}+(b-c)^\frac{1}{3}+(c-a)^\frac{1}{3}=0$ is true, we have
$x+y+z=0$
$\Rightarrow x^{3}+y^{3}+z^{3}=3xyz$ (This is a fairly elementary result, I believe.)
$\Rightarrow 3xyz = 0$
$\Rightarrow (a-b)^\frac{1}{3}(b-c)^\frac{1}{3}(c-a)^\frac{1}{3}=0$
$\Rightarrow (a-b)(b-c)(c-a)=0$
$\Rightarrow$ At least two of the numbers $a,b,$ and $c$ are equal, which leads to a contradiction since $a,b,$ and $c$ are distinct reals. And, we are done.


modular arithmetic
pigeonhole principle
number theory unsolved
number theory

Let $ n$ be a given positive integer.  Find the smallest positive integer $ u_n$ such that for any positive integer $ d$, in any $ u_n$ consecutive odd positive integers, the number of them that can be divided by $ d$ is not smaller than the number of odd integers among $ 1, 3, 5, \ldots, 2n \minus{} 1$ that can be divided by $ d$.

I think the wording of the problem is quite confusing.. not unclear, just confusing.

I'd reword it:

Let $n$ be a positive integer. Define the set $N_{m,k}$, where $m$ is an odd positive integer, to be the set of the $k$ smallest odd integers that are at least $m$. For an arbitrary positive integer $d$, denote by $N_{m,k}(d)$ the number of integers in the set $N_{m,k}$ that are divisible by $d$. Find the smallest $u_n\in\mathbb{N}$ (in terms of $n$) such that $N_{m,u_n}(d)\ge N_{1,n}(d)$ for every positive integer $d$ and odd positive integer $m$.

Perhaps this is an even more confusing way of putting it! Feel free to choose the version with the better exposition.

Anyway, the first few values for $u_n$ are $1,3,5,7,9$ so we conjecture $u_n=2n-1$.

Clearly $N_{1,n}(2n-1)=1$ because of the presence of $2n-1$ in the set $N_{1,n}$. As the next odd multiple is $3(2n-1)=2(3n-1)-1$, this means that $N_{2n+1,2n-2}(2n-1)=0$. So $u_n\ge 2n-1$. We now need to show $u_n=2n-1$ works. Note that $N_{1,n}(d)=0$ for $d\not\in\{1,3,5,\ldots ,2n-1\}$ so we only need to worry about $d$ in this set. Now let $m$ be an arbitrary odd positive integer. Then the set $N_{m,2n-1}$ consists of the integers $m,m+2,m+4,\ldots ,m+2(2n-1)-2$. One of the first $d$ of these integers is divisible by $d$. Suppose on the other hand none of the first $d$ integers are multiples of $d$. Then none are equal to $0\pmod{d}$ and so by pigeonhole there are two equal $\pmod{d}$, say $m+a$ and $m+b$ where $0\le a<b\le 2d-2$. This means $d|b-a$. Of course, $2d-2\ge b-a>0$ and so the only multiple of $d$ that $b-a$ can possibly be is $d$ itself (the case $d=1$ can be discarded). Now $a$ and $b$ are both even, so $b-a$ is even, meaning $d$ is even, contradiction, as we assumed $d\in\{1,3,5,\ldots ,2n-1\}$. And so the assertion that one of the first $d$ of these integers is divisible by $d$ is correct. So let $m+a$ be the smallest integer in the set $N_{m,2n-1}$ such that $d|m+a$. We have just proven $a\le 2d-2$.

Let $h$ be the largest integer such that $m+a+2hd\le m+4n-4$. Then the integers in $N_{m,2n-1}$ which are divisible by $d$ are $m+a,m+a+2d,m+a+4d,\ldots ,m+a+2hd$. It follows that $N_{m,2n-1}(d)=h+1$. Now let $b$ be the largest odd integer such that $bd\le 2n-1$, and if $b=2c-1$ then it follows that $N_{1,n}(d)=c\ge 1$.

Now $(2c-1)d\le 2n-1$ so $2(2c-1)d\le 4n-2$. Now since $h$ is the largest integer such that $m+a+2hd\le m+4n-4$, we have $a+2(h+1)d\ge 4n-4$. Now $2d-2\ge a$ meaning $2d-2+2(h+1)d\ge a+2(h+1)d\ge 4n-4\ge 2(2c-1)d-2$. Then it follows that $h+1\ge 2c-2\ge c$ for $c>1$. When $c=1$, we have $d=2n-1$ in which case $h=0$ so $h+1=1\ge 1=c$. So $h+1\ge c$. So $N_{m,2n-1}(d)\ge N_{1,n}(d)$ , proving that $u_n=2n-1$.


I heard that some colleges don't consider the student's SAT writing score when looking at his or her application, or at least don't put as much weight on it as they do the other two subjects. Is this true?

Yes. I'll use California public universities as an example:

The UC system has always been very important for the SAT. Ten years ago, the writing exam was an SAT 2. The UCs required it, and counted it equally with the two SAT 1 parts in their admission index. A few years later, they got the writing test changed to an SAT 1.
The CSU system, on the other hand, didn't require the writing exam. After the change, they chose to keep using the math and verbal portions of the SAT while ignoring writing.

This doesn't make a big difference most of the time.

Admissions counselors from MIT, Cornell, and WPI have told me that essays and English class grades give a much better impression of a person's written communication abilities than a score on a 25-minute scribbled out essay and some multiple-choice grammar questions.

This doesn't mean you shouldn't try to do your 100% best on the SAT Writing section.


inequalities
LaTeX
inequalities unsolved

Let a.b.c be positive reals such that their sum is 1. Prove that

$ \frac{a^{2}b^{2}}{c^{3}(a^{2}\minus{}ab\plus{}b^{2})}\plus{}\frac{b^{2}c^{2}}{a^{3}(b^{2}\minus{}bc\plus{}c^{2})}\plus{}\frac{a^{2}c^{2}}{b^{3}(a^{2}\minus{}ac\plus{}c^{2})}\geq \frac{3}{ab\plus{}bc\plus{}ac}$

the inequality is equivalent to :

$ \sum\frac{(ab)^2}{c^3(a^2\minus{}ab\plus{}b^2)}\geq \frac{3(a\plus{}b\plus{}c)}{ab\plus{}ac\plus{}bc}$

$ \iff$  $ \sum\frac{(ab)^5}{a^2\minus{}ab\plus{}b^2}\geq \frac{3(abc)^3(a\plus{}b\plus{}c)}{ab\plus{}ac\plus{}bc}$

but : $ 3(abc)^3(a\plus{}b\plus{}c)\equal{}3abc(\sum a)(abc)^2\leq (\sum ab)^2(abc)^2$  (AM-GM)

hence it suffices to prove that :

$ \sum\frac{(ab)^5}{a^2\minus{}ab\plus{}b^2}\geq (abc)^2(\sum ab)$

$ \iff$ $ \sum\frac{(ab)^3}{c^2(a^2\minus{}ab\plus{}b^2)}\geq \sum ab$

$ \iff$  $ \sum\frac{(ab)^3}{c^2(a^2\minus{}ab\plus{}b^2)}\plus{}\sum c(a\plus{}b)\geq 3\sum ab$

$ \iff$  $ \sum\frac{(ab)^3\plus{}(bc)^3\plus{}(ca)^3}{c^2(a^2\minus{}ab\plus{}b^2)}\geq 3\sum ab$

but $ \sum\frac{(ab)^3\plus{}(bc)^3\plus{}(ca)^3}{c^2(a^2\minus{}ab\plus{}b^2)}\geq 9\frac{(ab)^3\plus{}(ac)^3\plus{}(bc)^3}{2\sum(ab)^2\minus{}abc(\sum a)}$

it suffices to prove that :

$ 9\frac{(ab)^3\plus{}(ac)^3\plus{}(bc)^3}{2\sum(ab)^2\minus{}abc(\sum a)}\geq 3\sum ab$

denote $ x\equal{}ab   , y\equal{}ac$ and $ z\equal{}bc$

$ 3\frac{x^3\plus{}y^3\plus{}z^3}{2(x^2\plus{}y^2\plus{}z^2)\minus{}xy\plus{}yz\plus{}zx}\geq x\plus{}y\plus{}z$

$ \iff$  $ \sum x^3\plus{}3xyz\geq \sum xy(x\plus{}y)$

wich is schur inequality ,and we have done :)

[quote="mashel"]Let a.b.c be positive reals such that their sum is 1. Prove that

$ \frac {a^{2}b^{2}}{c^{3}(a^{2} \minus{} ab \plus{} b^{2})} \plus{} \frac {b^{2}c^{2}}{a^{3}(b^{2} \minus{} bc \plus{} c^{2})} \plus{} \frac {a^{2}c^{2}}{b^{3}(a^{2} \minus{} ac \plus{} c^{2})}\geq \frac {3}{ab \plus{} bc \plus{} ac}$[/quote]
The inequality is equivalent to
\[ \sum \frac{a^2b^2}{c^3(a^2\minus{}ab\plus{}b^2)} \ge \frac{3(a\plus{}b\plus{}c)}{ab\plus{}bc\plus{}ca}\]
Put $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$, then the above inequality becomes
\[ \sum \frac{z^3}{x^2\minus{}xy\plus{}y^2} \ge \frac{3(xy\plus{}yz\plus{}zx)}{x\plus{}y\plus{}z}\]
This is a very known inequality :)

Using CS Engel, we get that
$ (\sum ab)^3 \ge 3 \sum c^3(a^2\minus{}ab\plus{}b^2)$ which can be proved by simple manipulation.

[quote="Agr_94_Math"]Using CS Engel, we get that
$ (\sum ab)^3 \ge 3 \sum c^3(a^2 \minus{} ab \plus{} b^2)$ which can be proved by simple manipulation.[/quote]
I'd like to see how this can be be proved by "simple manipulation" :?:  :maybe:

@can_hang2007,
Could you please give me a link to your known inequality.

[quote="can_hang2007"][quote="mashel"]Let a.b.c be positive reals such that their sum is 1. Prove that

$ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})}\geq \frac {3}{ab + bc + ac}$[/quote]
The inequality is equivalent to
\[ \sum \frac {a^2b^2}{c^3(a^2 - ab + b^2)} \ge \frac {3(a + b + c)}{ab + bc + ca}
\]
Put $ x = \frac {1}{a},y = \frac {1}{b},z = \frac {1}{c}$, then the above inequality becomes
\[ \sum \frac {z^3}{x^2 - xy + y^2} \ge \frac {3(xy + yz + zx)}{x + y + z}
\]
This is a very known inequality :)[/quote]
I will prove the inequality than stronger:
$ \sum \frac {z^3}{x^2-xy+y^2} \ge\ x+y+z$
BY CAUCHY-SCHWARZ inequality ,we have:
$ LHS \ge\ \frac{(x^2+y^2+z^2)^2}{\sum{z(x^2-xy+y^2)}}$
So, we must prove that:
$ (x^2+y^2+z^2)^2\ge\ (x+y+z)({\sum{z(x^2-xy+y^2)})}$
$ <=>x^4+y^4+z^4+xyz(x+y+z) \ge\ x^3(y+z)+y^3(x+z)+z^3(x+y)$
Which is obvious true by Schur inequality :)

[quote] Agr_94_Math wrote:
Using CS Engel, we get that  .. [/quote]

I'm sorry,what is CS Engel  :?:

CS engel is  Cauchy-Schwartz-Bunjakovsky inequality which clears the denominators,sorry for not being specific,but i don't know how to latex that form.

[quote="mashel"]Let a.b.c be positive reals such that their sum is 1. Prove that

$ \frac {a^{2}b^{2}}{c^{3}(a^{2} \minus{} ab \plus{} b^{2})} \plus{} \frac {b^{2}c^{2}}{a^{3}(b^{2} \minus{} bc \plus{} c^{2})} \plus{} \frac {a^{2}c^{2}}{b^{3}(a^{2} \minus{} ac \plus{} c^{2})}\geq \frac {3}{ab \plus{} bc \plus{} ac}$[/quote]
I have two inequality
Pro. let $ a,b,c > 0, a\plus{}b\plus{}c\equal{}1$ and real number $ k$. Prove that:
1/ $ \frac {{{a^{k \minus{} 1}}{b^{k \minus{} 1}}(a \plus{} b)}}{{{c^k}\left( {{a^k} \plus{} {b^k}} \right)}} \plus{} \frac {{{b^{k \minus{} 1}}{c^{k \minus{} 1}}(b \plus{} c)}}{{{a^k}\left( {{b^k} \plus{} {c^k}} \right)}} \plus{} \frac {{{c^{k \minus{} 1}}{a^{k \minus{} 1}}(c \plus{} a)}}{{{b^k}\left( {{c^k} \plus{} {a^k}} \right)}} \ge \frac {3}{{ab \plus{} bc \plus{} ca}}$ with $ k\geq 1$ or $ k < 0$
2/ $ \frac {{{a^{k \minus{} 1}}{b^{k \minus{} 1}}(a \plus{} b)}}{{{c^k}\left( {{a^k} \plus{} {b^k}} \right)}} \plus{} \frac {{{b^{k \minus{} 1}}{c^{k \minus{} 1}}(b \plus{} c)}}{{{a^k}\left( {{b^k} \plus{} {c^k}} \right)}} \plus{} \frac {{{c^{k \minus{} 1}}{a^{k \minus{} 1}}(c \plus{} a)}}{{{b^k}\left( {{c^k} \plus{} {a^k}} \right)}} \le \frac {1}{{3abc}}$ with $ 1 \ge k \ge 0$

[quote="mestav"]First we multiply  $ a + b + c$ Right side and we have

$ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})}\geq \frac {3(a + b + c)}{(ab + ac + bc)}$

This inequality is true for all a,b,c if and only if main inequality is true for $ a + b + c = 1$
This inequality homogeneous and all degree are equal we can $ abc = 1$

$ abc = 1$  $ \implies$  $ \frac {a^{5}b^{5}}{a^{2} - ab + b^{2}} + \frac {b^{5}c^{5}}{b^{2} - bc + c^{2}} + \frac {a^{5}c^{5}}{a^{2} - ac + c^{2}}$

$ \implies$  $ a^{5} + b^{5}\geq ab(a^{3} + b^{3})$

similarly we have
$ ab(a + b) + bc(b + c) + ac(a + c)\geq\frac {3(a + b + c)}{ab + ac + bc}$

from Muirhead inequality $ (2,1,0)(1,1,0)\geq 6(1,0,0)$  $ \Box$[/quote]

Mestav I don't understand ,because $ a^{5} + b^{5}\geq ab(a^{3} + b^{3})$ it is true

But $ a^{5}b^{5}\geq ab(a^{3}+b^{3})$  is it true

Sorry i made huge mistake :roll: but we could prove following ineq.for $ abc\equal{}1$ easily with muirhead inequality
$ \frac{a^{5}b^{5}(a\plus{}b)}{a^{3}\plus{}b^{3}}\plus{}\frac{b^{5}c^{5}(b\plus{}c)}{b^{3}\plus{}c^{3}}\plus{}\frac{a^{5}c^{5}(a\plus{}c)}{a^{3}\plus{}c^{3}} \geq\frac{3(a\plus{}b\plus{}c)}{(ab\plus{}ac\plus{}bc)}$


This is the second time my account has been hacked. Someone changed my password along with my email address registered with my account and I heard some very bad statements had been made with my account. I want to apologize for this but I want to know exactly why and how this is happening so I can stop this from happening again! :mad: What I really can't understand is how someone hacked into my account, I only go on this site through my home computer which only I use and no one but me knows my password :?:

[quote="Iversonfan2005"]I only go on this site through my home computer[/quote]I have to disagree with this statement. There are at least 75 different IP addresses you have posted from.

Anyway the IP address the hacker used made me trace back the computer the hacker used: it'a user of this site. You have previously posted 8 times from that IP address (of the hacker).

To hack an account he must have access to: your email address or guess your password (of the site account and/or of the email account). Try to use non-dictionary words for your password, include both uppercase and lowercase letters and even punctuation signs). Make it at least 6 characters long.

[quote="Iversonfan2005"]Someone changed my password along with my email address registered with my account and I heard some very bad statements had been made with my account. I want to apologize for this but I want to know exactly why and how this is happening so I can stop this from happening again![/quote]

Alright, I'm slightly confused here. How is it that you can still log in (and post) if your password and email address have been changed? That should mean that you currently do not have access to the account (and that it's actually the hacker posting). I'm not all too familiar with hacking, but it seems logical that unless he didn't want you to know that your account has been messed with it's unlikely you'd be able to use the account again. Now if the hacker didn't want you to know that your account had been hacked, why on earth would they change your password and email?

But anyways, you can always ask moderators and admin to delete/edit posts (especially those from your own account)  that you don't like.

Ok, let's recap what happened.

The user (not the person) Iversofan2005 posted something that let's just say it contained a lot of censored words. He then PMed me with a message similar with the post.

I was pretty sure it wasn't the real person behind user Iversonfan2005 (let's call it X), so I temporarily banned the IP address and the account. As it turns out (I've noticed quickly also) the IP address of the person which was posting the censored words (let call him Y) is the same with some other user on the site, call him Z.

I've contacted person Z and he told me that a friend of his (which we called above person Y) used his computer during the time the censored posts were made.

At this point it is not clear to me (person Z failed to clarify this after some emails back and forth)

- if person X = person Y
- if not, how did person Y get his hands on the account information for person X (the usual user of Iversonfan2005).

Ok, so Iversonfan2005's statement above was clearly not entirely accurate. The facts are clear now. I hope this never happens again.
Topic closed.


Find all functions $f: \mathbb R^{+} \to \mathbb R^{+}$ such that
\[ f\left(\frac{2xy}{x\plus{}y}\right) \equal{} \frac{2f(x)f(y)}{f(x)\plus{}f(y)}\]
for all positive reals $x$ and $y$.

[quote="behdad.math.math"]Find all functions $ f: R^ \plus{} \longrightarrow R^ \plus{}$ such that:

$ f(\frac {2xy}{x \plus{} y}) \equal{} \frac {2f(x)f(y)}{f(x) \plus{} f(y)}$[/quote]

I suppose $ 0\notin\mathbb R^\plus{}$

Let $ g(x)\equal{}\frac 1{f(\frac 1x)}$. The equation becomes $ g(\frac{\frac 1x\plus{}\frac 1y}2)\equal{}\frac{g(\frac 1x)\plus{}g(\frac 1y)}2$ and so $ g(\frac {x\plus{}y}2)\equal{}\frac{g(x)\plus{}g(y)}2$ with $ g(x): \mathbb R^\plus{}\to\mathbb R^\plus{}$

This is a very simple and well known equation whose solution (remember the constraint $ \mathbb R^\plus{}\to\mathbb R^\plus{}$) is $ g(x)\equal{}ax\plus{}b$ with $ a,b\ge 0$ and $ a\plus{}b>0$

Hence the solutions of original equation : $ \boxed{f(x)\equal{}\frac{x}{bx\plus{}a}}$ where $ a,b\ge 0$ and $ a\plus{}b>0$


Find all triplets (x,y,z) of natural numbers diferent from 0 so that:

(1+1/x)(1+1/y)(1+1/z)=2

This is basically just doing a lot of cases - not dfifficult and quite boring.

Suppose x \leq y \leq z (without loss of generality).
->If x = 1 then (1 + 1/y)(1 + 1/z) = 1, which is impossible since y, z > 0.
->If x = 2 then (1 + 1/y)(1 + 1/z) = 4/3.
*If y = 2 then 1 + 1/z = 4/3.2/3 < 1 which is impossible.
*If y = 3 then 1 + 1/z = 1 which is impossible.
*If y = 4 then 1 + 1/z = 16/15 which gives z = 15.
*If y = 5 then 1 + 1/z = 10/9 which gives z = 9.
*If y = 6 then 1 + 1/z = 8/7 which gives z = 7.
*If y \geq 7 then we get y \leq z, so there is also no solution.
(If y \geq 7 then 1 + 1/y \leq 8/7, so 1 + 1/z \geq 4/3.8/9 = 7/6. This forces z \leq 6.)
->If x = 3 then (1 + 1/y)(1 + 1/z) = 3/2.
This gives again a lot of different possibilities. One can show in a similar way that the only solutions are (y, z) = (3,8) or (4,5).
->If x \geq 4 then (1 + 1/x)(1 + 1/y)(1 + 1/z)  \leq (5/4) < 2.
So in this cases there are no solutions.

Conclusion:
The only solutions are all possible permutations of the triples
(2, 4, 15), (2, 5, 9), (2, 6, 7), (3, 3, 8), (3, 4, 5).

Yeah, this is the usual way of solving these. Here's a nice problem from the Romanian selection for JBMO 2001 where you can use this: find all quadruples (a, b, c, d) s.t. each number divides the sum of the other 3.


What is the value of $ .375 \plus{} .4$ expressed as a common fraction?


Find smallest and greatest perfect square and perfect cube form with numbers $ 0,1,...,9$ once
Example: 1354267890 (but this is not perfect square or cube)

If I don't misinterpret the problem, I think there are no such number.

$ \sqrt[6]{100000000}\approx21.5$

$ \sqrt[6]{999999999}\approx31.6$

So we just have to observe $ 22^6, 23^6, 24^6, \ldots 31^6$. But none of these numbers uses the numbers 0 to 9 once.

"Perfect square" and "perfect cube" are separate problems.

Yes, t0rajir0u right. That is two different problem
I'm sorry make you confused  :(


arithmetic sequence
number theory proposed
number theory

Find all integrers $n >=3 $ such that if $a_1, a_2, ... a_n$ are integrers in aritmetic progresion, the number $a_1^n+a_2^n+...+a_n^n$ is a multiple of $n$

I think $n$ can be any multiple of $d$ where $d$ is the common differents of the AP

$n$ is fixed and the aritmetic progresion is not. $a_i$ can be any aritmethic progresion. I´ll post mi solution later

I think it can solve this way

step1) m,n be a coprime such that  m,n, satisfying the condition

then mn too.

step2) p be a prime then p^a  satisfying the condition.(when  p is odd)

step3) step2)is true when p=2

Am I Wrong?

[quote="rgep"]The sum of the first $n$ terms of the arithmetic progression $a+kd$ is $na + d.n(n+1)/2$.  This is divisible by $n$ if $n$ is odd or $d$ is even.  The answer is that all odd $n$ have the property you want.[/quote]

You missed a's power of n


Let n,k be natural numbers and p prime such that p>$ (n^2 \plus{} n \plus{} k)^2 \plus{} k$. prove that between the numbers $ n^2,n^2 \plus{} 1,n^2 \plus{} 2,...,n^2 \plus{} m$ where m=$ (n^2 \plus{} n \plus{} k)^2 \minus{} n^2 \plus{} k$ there exist 2 x,y with $ y\minus{}x\equal{}k$ such that $ (x/p) \equal{} (y/p) \equal{} 1$

[quote="Γιώργος"]Let n,k be natural numbers and p prime such that p>$ (n^2 \plus{} n \plus{} k)^2 \plus{} k$. prove that between the numbers $ n^2,n^2 \plus{} 1,n^2 \plus{} 2,...,n^2 \plus{} m$ where m=$ (n^2 \plus{} n \plus{} k)^2 \minus{} n^2 \plus{} k$ there exist 2 x,y with $ y \minus{} x \equal{} k$ such that $ (x/p) \equal{} (y/p) \equal{} 1$[/quote]

I don't think this makes sense

Consider the largest number in $ {n^2,n^2 \plus{} 1,n^2 \plus{} 2,...,n^2 \plus{} m} \equal{} (n^2 \plus{} n \plus{} k)^2 \plus{} k$
Since this [i]is[/i] the prime, there can't be any numbers in between which are equal to that (required for the fractions).

Also, if $ \frac{x}{p}\equal{}\frac{y}{p}\equal{}1 \Rightarrow x\equal{}y$ Therefore $ y\minus{}x\equal{}0 \Rightarrow k \equal{} 0$ That in turn implies our prime number is a square number. Contradiction!

$ \left( \frac{x}{p} \right)$ is the [url=http://en.wikipedia.org/wiki/Legendre_symbol]Legendre symbol[/url].

[quote="t0rajir0u"]$ \left( \frac {x}{p} \right)$ is the [url=http://en.wikipedia.org/wiki/Legendre_symbol]Legendre symbol[/url].[/quote]

Thanks, I wondered why that didn't make much sense

if $ \left( \frac{n^2\plus{}k}{p} \right)\equal{}1$, then $ (x,y)\equal{}(n^2,n^2\plus{}k)$ is a solution. if $ \left( \frac{(n\plus{}1)^2\plus{}k}{p} \right)\equal{}1$, then $ (x,y)\equal{}((n\plus{}1)^2,(n\plus{}1)^2\plus{}k)$ is a solution. otherwise, we have

\[ 1\equal{}(\minus{}1)(\minus{}1)\equal{}\left( \frac{n^2\plus{}k}{p} \right) \left( \frac{(n\plus{}1)^2\plus{}k}{p} \right)\equal{}\left(\frac{(n^2\plus{}n\plus{}k)^2\plus{}k}{p} \right)
\]

so $ (x,y)\equal{}((n^2\plus{}n\plus{}k)^2,(n^2\plus{}n\plus{}k)^2\plus{}k)$ is a solution.


trigonometry
inequalities
inequalities proposed

Prove that $\left(\sin{\pi \over x}\right)^x \ge \left(\sin{\pi (x-1)\over 2}\right)\cdot\left(1+{{(x-2)^2} \over 2}\right),\,\, \forall x\in [1,2]$

:) it was a joke!

for this inequality i know only
computer-powered solution.
with 5000 points.

but without polynom it's very easy.


Let $p$ be a prime number. Prove that there exists an integer $a$ such that $p|(a^2-a+3)$ if and only if there exists an integer $b$ such that $p|(b^2-b+25)$.

$f_c(x)=x^2-x+c = \left( x- \frac{1}{ 2} \right)^2 + \frac{4c-1}{4}$.
Lets assume that $p$ is odd.
Thus $f_c(x) \equiv 0 \mod p$ can be solved iff $-(4c-1)$ is a square $\mod p$.
So we have to show that $-(4 \cdot 3-1) = -11$ is a square iff $-(4 \cdot 25 -1) = -99$ is one. But for $p \neq 3$ thats rather obvious.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Give an example of a noncommutative group in which every nontrivial subgroup is normal :maybe:

The quaternion group $\{\pm1, \pm i, \pm j, \pm k\}.$

Is this quaternion group is from the Complex Analisys $ij=k=-ji , ... i^{2}=j^{2}=k^{2}=1$?

That's correct. It has order 8, so you just need to check that subgroups of order 2 are normal. And there is only one such ie $\{\pm 1\}.$


if we provided glucose marked with radioactive carbon to a cell where we had removed or inactivated the mitochondria, where or in what state would we find the carbon????

still in the cytoplasm as 6-carbon glucose

in the cytoplasm as 3-carbon molecules

inside the mitochondria as 3-carbon molecules

in the cytoplasm as CO2

inside the mitochondria as CO2

GOOD QUESTION HA?,
NEED HELP!

Heres a solution. If the mitochondria was removed, then the glucose would be 2 3-carbon molecules, probably pyruvic acid. The glucose goes through glycolysis before it can enter the mitochondria, which occurs in the cytoplasm with end products of two pyruvic acids.
 :D  wow I feel smart!


Here is the problem:
On a rather unusual day, a bus makes over 13 stops and at each stop it either [b]picks up 4 people or picks up 2 people and lets off 3 people[/b]. If the bus starts off empty, what is the fewest number of stops over 13 that the bus makes so that it has exactly 7 people on board at the end of its run? Do not count the beginning or end of the run as stops.

the bolded part is a bit strange. At first I thought that these are three separate events. But then I saw the "and," so I got confused. How are you supposed to interpret the logic of a sentence like this? (sorry, I don't have the solution to this problem)

At each stop, the bus does one of two things: (a) picks up four people, or (b) picks up 2 people and lets off 3 people. That's how I would interpet it.

Ok, so you should always do the "and" operation before the "or"?

Usually wording doesn't mean ((a or b) and c) but rather (a or (b and c)).

The keyword is "either."  It either (picks up four people) or (picks up 2 people and lets off 3 people).  When you see it that way, it doesn't make sense for the "and" to be separate since, linguistically, it would've been a lot more natural as a separate clause.

I second chess64's interpretation, since otherwise it would be more natural to say "picks up 2 or 4 people and lets off 3 people". Such as it is, a small comma after "picks up 4 people" would solve the problem.

this seems like a problem that you could file an appeal for being ambiguous


Mkay, $ x^2 \minus{} 7x \plus{} 11$ has roots $ \frac {7\pm\sqrt {5}}{2}$ So x can't be either of those values.

Now, either $ x^2 \minus{} 7x \plus{} 11 \equal{} 1$ which has roots 5 and 2, or $ x^2 \minus{} 11x \plus{} 30 \equal{} 0$ which has roots 5,6.

Therefore, 2,5,6 are the solns.

Edit: Whoops, forgot that $ x^2\minus{}7x\plus{}11\equal{}\minus{}1$ will also give the 2 solutions x=3,4.

[hide]$ (x^2\minus{}7x\plus{}11)^{x^2\minus{}11x\plus{}30}\equal{}1$
So in order for an exponent to equal 1 we have to have either $ 1^x\equal{}1$ for any $ x$ or $ y^0\equal{}1$ for any positive $ y$ or $ \minus{}1^z$ for any even $ z$.
Case 1: base is $ 1$
$ x^2\minus{}7x\plus{}11\equal{}1$
$ x^2\minus{}7x\plus{}10\equal{}0 \rightarrow x\equal{} 2$ and $ x\equal{} 5$
Case 2: exponent is $ 0$
$ x^2\minus{}11x\plus{}30\equal{}0 \rightarrow x\equal{} 5$ and $ x\equal{} 6$
Case 3: base is $ \minus{}1$
$ x^2\minus{}7x\plus{}11\equal{}\minus{}1$
$ x^2\minus{}7x\plus{}12\equal{}0 \rightarrow x\equal{} 3$ and $ x\equal{} 4$
Both of this cases gives us even exponents so we are done. The solutions are $ x\equal{}2,3,4,5,6$[/hide]

actually, $ x^2\minus{}11x\plus{}30$ is always even.  If $ x$ is even, it is obvious.  If $ x$ is odd, $ x^2$ and $ 11x$ are also both odd so there difference is even, and then we have even+30=another even.

[quote]actually,$ x^2\minus{}11x\plus{}30$  is always even. If$ x$  is even, it is obvious. If $ x$ is odd, $ x^2$ and$ 11x$  are also both odd so there difference is even, and then we have even+30=another even.[/quote]
it's true only when x is integer

i think this is an old mathcounts problem?

[quote="Pancho Bigotes"][hide]$ (x^2 \minus{} 7x \plus{} 11)^{x^2 \minus{} 11x \plus{} 30} \equal{} 1$
So in order for an exponent to equal 1 we have to have either $ 1^x \equal{} 1$ for any $ x$ or $ y^0 \equal{} 1$ for any positive $ y$ or $ \minus{} 1^z$ for any even $ z$.
Case 1: base is $ 1$
$ x^2 \minus{} 7x \plus{} 11 \equal{} 1$
$ x^2 \minus{} 7x \plus{} 10 \equal{} 0 \rightarrow x \equal{} 2$ and $ x \equal{} 5$
Case 2: exponent is $ 0$
$ x^2 \minus{} 11x \plus{} 30 \equal{} 0 \rightarrow x \equal{} 5$ and $ x \equal{} 6$
Case 3: base is $ \minus{} 1$
$ x^2 \minus{} 7x \plus{} 11 \equal{} \minus{} 1$
$ x^2 \minus{} 7x \plus{} 12 \equal{} 0 \rightarrow x \equal{} 3$ and $ x \equal{} 4$
Both of this cases gives us even exponents so we are done. The solutions are $ x \equal{} 2,3,4,5,6$[/hide][/quote]

wouldn't $ x\equal{}5$ be indeterminate? 'cause we'd have $ 0^0$.

If we are solving $ a^b\equal{}1$ then we should consider all $ a^k\equal{}1$ where $ k$ is an integer... what if $ a\equal{}i$? or any other root of unity for that matter?

But what if for some reason we have that $ k$ isn't an integer, but $ a^k$ is still 1???

Also, would most problems like this say that $ x$ or whatever is an integer???


What non-zero real value for $ x$ satisfies $ (7x)^{14} \equal{} (14x)^7$?

$ \textbf{(A) } \frac 17 \qquad \textbf{(B) } \frac 27 \qquad \textbf{(C) } 1 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 14$


trigonometry
inequalities
real analysis
real analysis solved

One quick reaction and a place not to go: I was thinking about variations on Jensen or AM-GM, but the inequality sign is pointing in the wrong direction for that.

Thank you again, Kent Merryfield! :P

I look forward to your solution.

kunny

I wonder if you get this inequality by comparing

$\cos^{2n}\left(\frac{1}{2}\right)\cos^{2n-1}\left(\frac{1}{3}\right)\cos^{2n-2}\left(\frac{1}{4}\right)\cdot\cdot\cdot\cos^{2}\left(\frac{1}{2n}\right)\cos\left(\frac{1}{2n+1}\right)$

 and

$\frac{\sqrt{n+1}}{2^{n}}$

I tried different ideas, but no proof.
Can you give an hint for this one kunny ?

O.K., Moubinool. :)

The hint is as follows.

You can use the inequality $x\geqq |\sin x|\ (x\geqq 0)$.

kunny

(1) $|\cos x|\geq \sqrt{1-x^2}$ for $x=1/2;...;1/(2n+1)$

Call the product of cosinus,  $P_n$, you get with (1)

$P_n\geq Q_n=\prod_{k=2}^{2n+1} ( 1-\frac{1}{k^2})^{\frac{2n-k+2}{2}}$

From this point simplify $Q_n$ and get the minoration $\frac{\sqrt{n+1}}{2^n}$


search
combinatorics unsolved
combinatorics

In plane are consudered n greater than 2 points, such that the number of segments of length 1 is maximum. Prove that their number is smaller or equal than n^2/3.

Famous Erdös problem. Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=509980709&t=22391

Pierre.


Let $a_{1}, a_{2}, ..., a_{n}$ be positive integers and let $[a_{1},a_{2},...a_{n}]$ denote their least common multiple. Prove that
\[[a_{1},a_{2},...a_{n}] \geq \frac{a_{1}a_{2}...a_{n}}{\prod_{1 \leq i < j \leq n}\gcd(a_{i}, a_{j}) }\]

[hide="By induction"]First, it's easy to verify in the case $n=2$
Suppose the ineq is right with $k-1$ positive integers. Then, let
$lcm(a_{1},a_{2},...,a_{k-1})=m$
$lcm(a_{1},a_{2},...,a_{k})=lcm(m,a_{k})=\frac{ma_{k}}{\gcd(m,a_{k})}$
$\geq\frac{a_{1}a_{2}...a_{k}}{\prod_{1\leq i<j\leq k-1}\gcd(a_{i},a_{j})lcm(m,a_{k})}$
$\Rightarrow \mathbb{QED}$[/hide]


I have a weird query.
pi is a non-terminating & non-terminating number. Now if all the information of the world was to be represented in numbers, is it so that pi can effectively represent the entire information in this world.

No- pi is in a prearranged order, so you would then need to document the number sequences of the information, and their place in pi.  Documenting the place of the information in pi would itself take as much space as merely documenting without pi.

However, the idea of an irrational number (one that changes based on the data being inputted) being able to store infinite data is intriguing.

i think i know how it could work-
if we can find a certain sequence of numbers, we can represent it by (a,b) where a is the distance of the first digit from the beginning and b the distance of the last

[quote="qwang"]Daermon: The idea is very interesting, but having to have such large quantities of data stored (imagine how many digits of pi you'd have to go through just to find the sequence 123456) would sort of kill the point.

This discussion belongs in the round table.[/quote]

Searchstart: 12 The string 123456 occurs at position 2,458,885 counting from the first digit after the decimal point. The 3. is not counted.

or you can just use a search thingy on the internet

[quote="qwang"]Daermon: The idea is very interesting, but having to have such large quantities of data stored (imagine how many digits of pi you'd have to go through just to find the sequence 123456) would sort of kill the point.

This discussion belongs in the round table.[/quote]
its probably not at all useful but its intresting :P

This is actually an argument used against the idea of the real-world existence of infinite-precision numbers.  In other words, the idea that real-world quantities are numbers on a real number line.

Edit:  By the way, if you're interested in this kind of stuff, $\pi$ itself can't really be said to store an infinite amount of information because [url=http://mathworld.wolfram.com/SpigotAlgorithm.html]spigot algorithms[/url] exist for it.

However, in 1927 a man by the name of Borel discussed the possibility of a number that answered all possible yes/no questions.  Simply number all the possible yes/no questions, and the $n^{th}$ bit of Borel's number is $1$ if the answer is yes, and $0$ if the answer is no.  And of course to Borel the idea of the existence of such a number was plainly ridiculous.


[b]Clasa a X-a[/b]
[b]1.[/b] Fie sirul lui Fibbonacci, definit prin $F_{0}=0, F_{1}=1$ si $F_{n+2}=F_{n+1}+F_{n}, \forall n\in \mathbb N$

$a)$Sa se demonstreze relatia $\sum_{k=1}^{n}F_{k}^{2}=F_{n}F_{n+1}, \forall n\in \mathbb N$
$b)$Sa se demonstreze ca pentru orice sir $(x_{n})_{n\geq1}$ de numere reale are loc inegalitatea:

$2(\sum_{k=1}^{n}F_{k}sinx_{k})(\sum_{k=1}^{n}F_{k}cosx_{k}) \leq nF_{n}F_{n+1}$
                                                                                [b]D.M. Batinetu Giurgiu, Bucuresti[/b]

[b]2.[/b] Fie $X$ o multime nevida si finita. Unei functii $f: P(X) \to  P(X)$, unde $P(X)=${$A|A\subset X$}, ii asociem sirul de functii $(f_{k})_{k\geq1}, f_{k}: P(X)\to P(X)$, definit prin $f_{1}=f$ si $f_{k+1}(x)=f_{k}(f(x)), \forall k\in \mathbb N^{*}$. Sa se demonstreze ca daca exista doua numere naturale nenule $p$ si $q$, prime intre ele, pentru care este satisfacuta relatia $f_{p}(A)$ reunit cu $f_{q}(A) = A, \forall A \in P(X)$, atunci $f(A)=A, \forall A\in P(X)$.
[b]Romeo Ilie, Brasov[/b]

[b]3.[/b] Fie $a,b\in (1;\infty)$. Se considera ecuatia:

$a^{8x^{2}+8y}+b^{8y^{2}+8x}=\frac{2}{a^{\log_{b}{a}}b^{\log_{a}{b}}}$, $x,y\in \mathbb R$.

$a)$Sa se rezolve ecuatia in cazul $a=b$.
$b)$Sa se arate ca ecuatia nu are solutii in cazul $a\ne b$.

[b]Cristinel Mortici, Targoviste[/b]

[b]4.[/b]Sa se demonstreze identiatea:
$\sum_{k=0}^{n}\frac{(-1)^{k}}{\binom{n}{k}}= \frac{(1+(-1)^{n})(n+1)}{n+2}, n\in \mathbb N^{*}$

[b]Mihail Bencze, Brasov[/b]

Have fun:) primul la a X-a a fost Svejk. Felicitari domnule:P

Multumiri pentru subiecte...si pentru $LaTeX$ :)

[quote]si pentru $LaTeX$ :)[/quote] :D  :lol:

rezultatele concursului sunt afisate pe undeva?

Am subiectele 9-12 in format jpg daca e interesat cineva.

[quote="Zamfirmihai"]Felicitari domnule:P[/quote]
Merci :lol:

Poti sa atasezi fisiere .jpg in posturi ;)


The two ends of a rope of length $ L$ are tied to two vertical poles, each of height $ h\ge L/2$, that are $ d<L$ distance apart. What will be the shape of the rope in this position?

[hide="I strongly believe"] it will be a conic section, or more specifically, parabolic. But I have no idea how to show this.[/hide]

The shape is that of a [url=http://en.wikipedia.org/wiki/Catenary]Catenary[/url].

how is a catenary different from a parabola?

Try using your graphing calculator or whatever graphing utility you use on your computer to plot the following curves on the same axes:

$ f(x)\equal{}\cosh x\equal{}\frac{e^x\plus{}e^{\minus{}x}}2$

$ g(x)\equal{}1\plus{}\frac{x^2}2$

$ y\equal{}f(x)$ is a catenary. $ y\equal{}g(x)$ is a parabola.

Play around with the graphing "window", looking at it at various zoom levels, in and (especially) out.

can one of you prove that it will be a caternary

In case you are too lazy to scroll down in my original post, [url=http://en.wikipedia.org/wiki/Catenary#Derivation]here[/url] is the exact spot for you!


geometry
3D geometry
sphere
conics
ellipse

A cross section is taken of a solid sphere whose radius is 34. The area of the cross section is $256\pi$. The cross section divides the sphere into two pieces. Find the total surface area of the small piece.

I've figured out it's like halve of a 3-d ellipse, with one radius 4 and the other 16, but I have no idea where to go from there. Help please.

I don't see a good way to do this without calculus.

[quote="Farenhajt"][b]@ragnarok23[/b]: Please, how did you manage to squeeze a "3D ellipse" in a sphere problem?

[b]@t0rajir0u[/b]: Formula for the surface area if a spherical cap is well known and, moreover, surprisingly easy.[/quote]
3-d ellipse was the best way I could describe it, I bet it's just my stupidity again. I made a right triangle using the radius of the cross section and the radius of a sphere and found the vertical part of the hemisphere formed after the cross section.

Hmm.

http://mathworld.wolfram.com/SphericalCap.html

But of course it can't be shown rigorously without calculus...?

I think we had a similar argument before, t0rajir0u:

Please show $V={4\over 3}r^{3}\pi$ for the volume of a sphere [u]without[/u] calculus. No go? Then please submit a petition for removal of that formula from the elementary school curriculum.

@ragnarok23: I still don't get it where you managed to see an ellipse.

[quote="t0rajir0u"]I think he just means that a spherical cap resembles half of an ellipsoid.[/quote]
Yes, this is what I mean.
Thanks for the help.

Oh, I get it know... Sorry for being non-intuitive :)

Anyways, that's not a half of an ellipsoid, but a spherical cap, as both t0rajir0u and I repeated a few times now :) If you're given such a problem, I guess you're expected to know the formula for its surface area. If you don't, check the link given by t0rajir0u.


analytic geometry
calculus
integration
number theory
greatest common divisor
arithmetic sequence
combinatorics unsolved

On the segment $ [0,2008]$, the endpoints and $ n\minus{}1>0$ interior points with integer coordinates are marked so that the lengths of the segments into which the segment $ [0,2008]$ is divided have the greatest common divisor $ 1$. We are allowed to divide a segment with marked endpoints into $ n$ equal parts and to mark the division points,provided that their coordinates are integral. Can we mark all integral points on the segment after several such operations?

Yes. It suffices to show that whenever there are $ k \ge n\plus{}1$ marked points (including the endpoints) such that the segments formed have trivial gcd, either we can mark additional points, or we have already marked all points. Suppose we can not; let our $ k$ points be located in ascending order at $ x_1 \equal{} 0, x_2, \ldots x_k \equal{} 2008$.

For any set $ \{ x_t, x_{t\plus{}1}, \ldots x_{t\plus{}n} \}$, some two of the numbers will be congruent modulo $ n$ so that we may mark intermediate points (possibly already marked) on the segment they form; if all of these $ n\minus{}1$ intermediate points were previously already marked, we must have been dividing the segment $ [x_t, x_{t\plus{}n}]$, and in particular, $ x_t, x_{t\plus{}1}, \ldots x_{t\plus{}n}$ must be an arithmetic sequence. Since $ t$ is arbitrary, it follows that all the $ x_i$ form an arithmetic sequence, and since the gcd of the segments formed is trivial, it follows that the $ x_i$ comprise all the integers from $ 0$ to $ 2008$.

Nice.

What about if we start by only marking the points $ d$ coprime to $ 251$. And we are now allowed to mark the midpoint of any segment whose endpoints are already marked. Can we still mark all integer points?


is it possible for the number 111....1111 to be divisible by 19

is there a solution without actually dividing the number?

Yes.

$111 \ldots 111$ ($n$ times) $= \frac{10^{n}-1}{9}\equiv 0 \pmod{19}$

$10^{n}\equiv 1 \pmod{19}$.

By Euler's Totient Theorem, we know $n = 18$ must satisfy this congruence.

Testing, we see

$111 \ldots 111$ ($18$ times) $= 5847953216374269 \cdot 19$.

For coprime integers $a,n$ where $n>0$ we have $a^{\varphi (n)}\equiv 1\pmod n$.

A nice solution using pigeon-hole principle:
We have infintely many numbers of form $11...11$ so we will find two that will have same remainder when divided by $19$. Now substracting them get the number $11...1100...00$ but $19$ does not ever divide $100...00$ so $111...11$ is divisible by $19$. QED

hi rem, i did not quite get how the pigeonhole principle applies to the problem, can you explain further?

Of the twenty numbers

$\{ 1, 11, 111, ..., 11111111111111111111 \}$

We must have at least two which leave the same remainder upon division by $19$, since there are only $19$ possible remainders.

The number can take the following form:

$111\ldots111=111\ldots108+3$

But $111\ldots108$ is divisible by $4$, that is:
$111\ldots108=4k\implies111\ldots111=4k+3$

But $19$ can take that form too: $4k+3\quad k=4$

So the answer is yes!

But that's not a proof. Just because two numbers are of the form $4k+3$ doesn't mean that they divide eachother (i.e. 3 and 7).

I guess you are right, wrong approach of the problem  :|


invariant
combinatorics unsolved
combinatorics

In the Parliament of Sikinia, each member has at most three(3) enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house

So classical, it's a pity you don't cite the original source (for sure it is in Engel).

Split into two houses such that the split minimizes the sum of the number of enemies within his house for each member. If some member still has at least two enemies in his house, move him in the other house.

Yes it is by Engel .

What I don't understand is that why we consider the "...split minimizes the sum of the number of enemies".

I don't understand why we have to look for "minimizing the sum " ,

Also ,if  A is enemy of B ,does this means that B is enemy of A ?

YES, the relation of "enmity" is supposed to be symmetrical.
The reason we take a split with minimal sum (or equivalently, minimal number of members having each at least two enemies in his house)  is called "extremal element principle" in combinatorics. It is equivalent to Fermat's infinite descent method. Let's apply it here. Start with a random split into houses. If some member has two or more enemies in his house, move him in the other house. What we achieve is that the number of members having each at least two enemies in his house strictly decreases. Since this is a finite descent, we must stop at zero.


I like this better as a problem:

Find $\lim_{x\to-2}\frac{3x^2+ax+2a-12}{x^2+x-2}$.

(But then it's a matter of taste.)

Original problem...

[hide]

The denominator approaches $0$. The numerator approaches $12-2a+a+3=15-a$. So if $a>15$ the limit is $-\infty$, if $a<15$ the limit is $\infty$. If $a=15$ we can apply L'Hopitals rule to get:

$\displaystyle \lim_{x\to-2}\frac{6x+15}{2x+1}=-1$.

[/hide]

As for Kent's... I'm too tired to do it again now... but the same approach should work.

Oops - found the typo. I meant $x\to-2,$ not $x\to2.$

That changes what TheBB said.

[quote="TheBB"]Original problem...

[hide]

The denominator approaches $0$. The numerator approaches $12-2a+a+3=15-a$. So if $a>15$ the limit is $-\infty$, if $a<15$ the limit is $\infty$. If $a=15$ we can apply L'Hopitals rule to get:

$\displaystyle \lim_{x\to-2}\frac{6x+15}{2x+1}=-1$.

[/quote]

[hide]Thanks TheBB,

 I am looking for  differents ways to solve it,Could someone solve without L'Hospital's rule??

Thanks[/hide][/hide]

I note $f(x)\equiv \frac{3x^2+ax+a+3}{x^2+x-2}$.

$3x^2+ax+a+3=(x+2)[3x+(a-6)]+(15-a)$.

$f(x)=\frac{3x+(a-6)}{x-1}+\frac{15-a}{x-1}\cdot \frac{1}{x+2}.$

$a=15\Longrightarrow \lim_{x\to -2} f(x)=\frac{12-a}{3}=-1;$

$a<15\Longrightarrow f(-2-0)=\infty,\ f(-2+0)=-\infty\Longrightarrow \lim_{x\to -2} f(x)$ don't exist;

$a>15\Longrightarrow f(-2-0)=-\infty,\ f(-2+0)=\infty\Longrightarrow \lim_{x\to -2} f(x)$ don't exist.

you  could factor the numerator and denominator.


logarithms
limit
calculus
calculus computations

let $\textbf{(1)}\;\;\;\quad \alpha_{k}\;=\;\frac{1+2+3+4+\cdots+k}{k}$

and $\textbf{(2)}\;\;\;\quad \ln \beta_{k}\;=\;\frac{\ln 1+\ln 2+\ln 3+\ln 4+\cdots+\ln k}{k}$

show, [b][u]without[/u] [/b]applying stirling's formula, $\lim_{k\to\infty}\;\left( \;\frac{2\cdot\alpha_{k}}{\beta_{k}}\;\right) \;=\;\boxed{e}$

 :D

[hide="Click to open the moongate for the Minoc mine"]
Proposition) If $a_{n}> 0$ and $r = \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ exists, then $\lim_{n\to\infty}\sqrt[n]{{a_{n}}}$ exists and its value is $r$.

Not apply it to $\left( \frac{2\cdot\alpha_{n}}{\beta_{n}}\right)^{n}= \frac{(n+1)^{n}}{n!}$. Then you'll have $e$.
[/hide]


$ ABCD$ is a quadrilateral.the line through $ A$ parallel to $ DC$ meets $ BD$ at $ E$ and the line through $ D$ parallel to $ AB$ meets $ AC$ at $ F$ .prove $ EF$ is parallel to $ BC$

Nice problem  :)

Let $ AC$  and $ BD$ intersect at $ O$.

Then we have $ \frac{OE}{OD}\equal{}\frac{OA}{OC}$ and $ \frac{OF}{OA}\equal{}\frac{OD}{OB}$

Thus $ \frac{OE}{OB}\equal{}\frac{OF}{OC}$ and hence $ EF$ and $ BC$ are parallel.


[quote="zeikii"]Let $x, y$ be possitive real numbers such that$0<x, y< 1$.
Prove that
1.$(1+x)x^{y}+(1+y)y^{x}> 1+x+y$
2.$(2x+y)x^{y}+(2y+x)y^{x}> 2(x+y)$
I've posted it but it seems that no one takes care of it.
I need some help.[/quote]
to zeikii:Why do you always post this kind of problem?
I 'll try it .

Thank you Hawk Tiger.
Since i tried to prove the inequality $x^{y}+y^{x}> 1$ (with the same condition), now i'm interested in this kind. And i have some hard problem (as you can see). Thus i post them for a help.

Come on guys! Please help me! I need some suggestion.

libra gold, did you use Bernoulli's Inequality to prove $x^{y}\geq 1-y+xy$ ?

From what I read in my book...
For $a \in \{0,1\}$, the sign is reversed...
Hence, it is $(1+x)^{a}\leq 1+ax$...

So, I think your answer must be wrong...  :maybe:

I thinks these problem can't be proved by your method.


geometry
algebra
system of equations
AMC

The vertices of a $ 3 \minus{} 4 \minus{} 5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
[asy]unitsize(5mm);
defaultpen(fontsize(10pt)+linewidth(.8pt));

pair B=(0,0), C=(5,0);
pair A=intersectionpoints(Circle(B,3),Circle(C,4))[0];

draw(A--B--C--cycle);
draw(Circle(C,3));
draw(Circle(A,1));
draw(Circle(B,2));

label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
label("3",midpoint(B--A),NW);
label("4",midpoint(A--C),NE);
label("5",midpoint(B--C),S);[/asy]$ \textbf{(A) } 12\pi\qquad \textbf{(B) } \frac {25\pi}{2}\qquad \textbf{(C) } 13\pi\qquad \textbf{(D) } \frac {27\pi}{2}\qquad \textbf{(E) } 14\pi$

[hide]
The radii of the circles are 1, 2, and 3.

This gives a total area of $14\pi$
 [/hide]

[hide]
More helpfully, to find the radii of these circles, set up a system of equations.
Let the radius of A be x, the radius of B be y, and the radius of C be z.
This brings you to the following system of equations:
x + y = 3
x + z = 4
y + z = 5
thus solving this system of equations gets you the radii of 1,2, and 3 for x,y, and z respectively.
[/hide]

This way is longer, but could be more useful in some other case and I wanted to show I could find the area without ever finding the individual circle radii.

Say, the radii of the three circles are x, y and z.

$A = \pi x^2 + \pi y^2 + \pi z^2 = Z \pi$
$Z = x^2 + y^2 + z^2$
If we have a given triangle in there with sibes a, b and c. Then we have:
$x + y = a$
$y + z = b$
$x + z = c$
$\Rightarrow x + y + z = \frac{a+b+c}{2} \text{ (1)}$
$\Rightarrow ac = (x+y)(x+z) = x^2 + xy + xz + yz$
$\Rightarrow ab = (y+x)(y+z) = y^2 + yx + yz + xz$
$\Rightarrow bc = (z+y)(z+x) = z^2 + zy + zx + yx$
Now, say $xy + yz + zx = R$.
Adding the three, we get:
$ab + bc + ca = x^2 + y^2 + z^2 + 3R$
$\Rightarrow R = \frac{ab + bc + ca - Z}{3}$
Squaring $\text{(1)}$ we get:
$x^2 + y^2 + z^2 + 2R = \frac{(a+b+c)^2}{4}$
$\Rightarrow Z + 2\frac{ab + bc + ca - Z}{3} = \frac{(a+b+c)^2}{4}$
$\Rightarrow Z = \frac{3}{4}(a+b+c)^2 - 2(ab + bc + ca)$

So, since we are given a = 3, b = 4, and c = 5, we have:
$Z = \frac{3}{4}(3+4+5)^2 - 2(3\cdot 4 + 4\cdot 5 + 5\cdot 3)$
$= \frac{3}{4}(144) - 2(12 + 20 + 15) = 108 - 94 = 14$
$\Rightarrow A = Z\pi = 14\pi$

Long, but I did it anyways. :lol:  (P.S. on the test I just solved the system.)

Yeah, I miss something like this every test...1+4+9 is NOT 13.

I could be  :wallbash_red: -ing like this all day, but here's the interesting part. My chemistry teacher was giving an example of a "wrong" answer when he said, "Now, Sam, if you said the answer was $13\pi$ and the answer really was $14\pi$, then...".  He had never heard of this test, except that I had to skip his class to take it. I told him after he finished, and showed him the test and he was equally spooked out.

Now, considering the fact that that was the only question I missed on that test, and I don't miss questions that often ( :roll: ), that's really weird...


geometry
3D geometry
combinatorics proposed
combinatorics

Each vertex of a cube is assigned an 1 or a -1, and each face is assigned the product of the numbers assigned to its vertices. Determine the possible values the sum of these 14 numbers can attain.

See the product of the eight numbers in the faces of the cube.  :)

[hide="Solution"]
Suppose that for a given configuration, the sum of the vertexes and faces is a number $n$. If we change the number in a vertex from $a$ to $-a$, the total sum of the vertexes can vary by $\pm 2$ and the total sum of the faces by $\pm 6, 2$. This means that in total, $n$ can vary by $\pm 8, 4, 0$. Note that the new result is congruent modulo 4 to the old one.

Now, since all results must be congruent modulo 4, and if we place ones in all the vertexes we obtain 14, the only attainable numbers are:
$\pm 2, 6, 10, 14$

Doing some casework:
-All ones, 14.
-One minus one, 6.
-Using two minus ones:
 *Opposite corners, -2
 *In the same edge, 6
 *Same face, different edges, 2
-We have checked the only three cases that could give 10, a higher number of minus ones will always result in a number smaller than 10, therefore 10 is not attainable.
-Five minus ones, -6
-Two ones, in opposite corners, -10.
-Finally, all minus ones, -2. This means that -14 is not attainable.

All the possible values are: -10, -6, -2, 2, 6, 14.

[/hide]


vector
real analysis
real analysis unsolved

Can someone help me to prove that
$c_{0}(\mathbb{N})^{*}\cong l^{1}(\mathbb{N})$
and
$l^{1}(\mathbb{N})^{*}\cong l^\infty(\mathbb{N})$

One part of the argument: let $P$ be the vector space of finitely supported sequences. The natural basis is a Hamel basis for $P,$ so all linear functionals on $P$ are of the form $L(x)=\sum_{k=1}^{\infty}a_{k}x^{k}$ for some sequence $a.$

The completion of $P$ under the $\infty$ norm is $c_{0},$ and the completion of $P$ under the $1$ norm is $\ell^{1}.$ Find a way of exploiting that.

(You didn't say whether your spaces are real or complex; there are slight technical differences in the argument.)


abstract algebra
group theory
superior algebra
superior algebra unsolved

For each of the following groups, find the smallest integer n such that the group has a faithful operation on a set with $ n$ elements.

(a) the quaternion group $ H$.
(b) $ D_{4}$
(c) $ D_{6}$.

a) n=8
b) If you meant $ D_8$, it is n=4 (also 4 if you meant $ D_4$, the abelian group).
c) n=5 if you meant $ D_{12}$, n=3 if you meant $ D_6 \equal{} S_3$.

The only non-trivial one, I guess, is (a).  But for any rep. from $ H\rightarrow S_n$ for n<8, the kernel will be the intersection of non-trivial subgroups of $ H$, all of which contain $ \minus{} 1$, so the kernel contains $ \minus{} 1$.

Steve


Let $\{x_n\}$ be rationals in $[0,1]$. Can we arrange $x_n$ such that $\sum_k x_k^k$ converges?

I think that given a sequence $(\lambda_n)_n\subset(0,1)$ increasing strictly towards $1$, it's not that hard to see (I think :)) that we can find an enumeration $(x_n)_n$ of the rationals in $(0,1)$ s.t. $x_n<\lambda_n,\ \forall n$, and we can apply this to $\lambda_n=\frac 1{\sqrt[n]{n^2}}$.


Let (c) and (D) be 2 circles externally tangent in A. Let Ac and Ad be points on (C) and (D) respectively s.t. AAc and AAd are perpendicular. What's the locus of:

(a) the projection of A on AcAd;
(b) the midpt of AcAd.

Nice problems!

Let C and D be the centers of the circles (C) and (D), respectively.

[b](a)[/b] The locus of the projection of A on AcAd is an arc of the circle with diameter AS, where S is the external center of similtude of the circles (C) and (D). [Clearly, this circle touches the circles (C) and (D) at A, since it passes through A and has its center on the line CD.]

[i]Proof.[/i] Let Q be the projection of A on AcAd. We want to prove that Q lies on the circle with diameter AS.

Well, the isosceles triangle ACAc gives < ACAc = 180 - 2 * < CAAc, and the isosceles triangle ADAd gives < ADAd = 180 - 2 * < DAAd, so we have

  < ACAc + < ADAd = 360 - 2 * (< CAAc + < DAAd)
                  = 360 - 2 * (180 - < AcAAd)
                  = 360 - 2 * (180 - 90) = 180,

so that CAc || DAd. Now, if the line AcAd meets CD at T, then it follows from this parallelism that T is the external center of similtude of the circles (C) and (D). Hence, T = S, i. e. the line AcAd passes through the point S. Hence, since Q is the projection of A on AcAd, we get < AQS = 90, and Q lies on the circle with diameter AS. Qed..

[b](b)[/b] The locus of the midpoint of AcAd is the circle with diameter CD.

[i]Proof.[/i] If P is the midpoint of AcAd, then P is the circumcenter of the right-angled triangle AcAAd; hence, AcP = AP. Of course, AcC = AC. Hence, the quadrilateral ACAcP is a kite, and the line CP bisects the angle APAc. Similarly, the line DP bisects the angle APAd. But these two angles are complementary; hence, their angle bisectors are perpendicular to each other, i. e., the lines CP and DP are perpendicular to each other, and P lies on the circle with diameter CD. Proof complete.

  Darij

Well, Dan Branzei usually proposes beautiful problems (mainly geometry).


geometry
geometric transformation
rotation
combinatorics proposed
combinatorics

Let $A_{1}A_{2}\ldots A_{2n}$ be a convex polygon and let $P$ be a point in its interior such that it doesn't lie on any of the diagonals of the polygon. Prove that there is a side of the polygon such that none of the lines $PA_{1}$, $\ldots$, $PA_{2n}$ intersects it in its interior.

with this lemma, the problem is easy:

if P is in interior of $A_{1}A_{2}..A_{n+1}$, we have that the lines $PA_{1}$, $PA_{n+1}$ and $PA_{i}$ (where i = n+2, ... , 2n) will intersect the sides $A_{1}A_{2}$, $A_{2}A_{3}$, ... , $A_{n}A_{n+1}$. so, we have n+1 lines intersecting n sides. so, two lines will intersect the same side! (impossible, if we suppose that the problem isn't correct)

We had a little bit harder problem to solve on our TST:

http://www.mathlinks.ro/Forum/viewtopic.php?t=137070

proposed by me :)

another solution i thought of should be by considering the figure formed by $A_{k}A_{k+1}A_{n+k}A_{n+k+1}$, considering just the two areas of the triangles $A_{k}OA_{k+1}$ and $A_{n+k}OA_{n+k+1}$, where $O \in A_{k}A_{n+k}\cap A_{k+1}A_{n+k+1}$.

now it is obvious that these $n$ pairs of two triangles formed by the "big" diagonals and with one common vertices cover the all polygon, therefore $P$ must be in one of the $2n$ triangles, let that be $A_{i}O_{i}A_{i+1}$, therefore the side $A_{i+n}A_{i+n+1}$ is the side which isn't cut by the $PA_{k}$ type lines.

The problem with this second solution is that I didn't know how to prove that the pairs of triangles formed with two consecutive 'big' diagonals cover up the entire polygon. It seems it is not quite an obvious thing. I saw the official solution (I think they had two solutions, in fact, but I can't remember the other one), which involved the rotation of a diagonal and, considering P to be on, let's say, the right, by rotating the diagonal 2n times, the point moved to the left with respect to the diagonal.  Frankly, I don't see how this goes. Why does that mean that the point necessarily belongs to one "papillote"? Someone please enlighten me. Even if my question may be stupid for you :fool:

Can no one comment on this solution to the problem?  :(

Oh, sorry because I didn't see your message at first. This "lemma" is quite well-known since 2006, it is a basic lemma used in an official solution to IMO2006/pb6, as well you can see here, it is mentioned in fedja's solution:
[url]http://www.mathlinks.ro/viewtopic.php?p=572824#572824[/url].


geometry
inequalities
trigonometry
rotation
geometric transformation
homothety
Cauchy Inequality

Let $ABC$ be a triangle with $\angle BAC=60^\circ$. Consider a point $P$ inside the triangle having $PA=1$, $PB=2$ and $PC=3$. Find the maximum possible area of the triangle $ABC$.

A quick remark: it's very similar to the 2'nd problem from the first Romanian TST in 2003: it was the same except for the values, which were PA=1, PB=2, PC=3.

I think it's from Rumania, There is someone ask me to solve it, I got the answer is 36+22 \sqrt 3  is right?

Ehm, were the numerical data in the problem (post #1) changed? Else, I don't understand Grobber's

[quote="grobber"]it was the same except for the values, which were PA=1, PB=2, PC=3.[/quote]

  Darij

Yes, I think they were changed. (In fact, my guess is that sigma copied the wrong values at first, and then, thanks to grobber's post, they were corrected ;))

I think the properties$PA=1,PB=2,PC=3$ can be replace by$PA=x,PB=y,PC=z$for $x,y,z$are given positive number.My solution only use CauChy inequality

maybe rotating the triangle about point $A$ 60 degrees would help...

[quote="mumath"]With my solution the property$\angle BAC=60$can be rewrited $\angle BAC=\alpha$,but the maximum of the area of  triangle must satisfy a 4-degree equation[/quote]

What is your solution?

I think that the Erdos- Mordel inequality will help:
$\ OA+OB+OC \geq 2(OP+OQ+OR)$  ;)

I hope it's correct.

let's take $P$ and $A$ fix. so the locus of $B$ and $C$ will be two circle with center $P$ and radius $2\ (\omega_2)$ and $3\ (\omega_3)$.

we have $S_{ABC}=\frac 12.b.c\sin 60^{\circ}$ so area will be maximum when $b.c$ be maximum.

now rotate $\omega_3$ wrt $A\ \ 120^{\circ}$ name it $\omega_3'$. with this rotaion $C\rightarrow C'$ and $B,A,C'$ are collinear.(SEE FIGURE 1)

now probelm become:

[color=darkblue]We have two intersecting circle$(\omega_2\ ,\omega_3')$ and a point $A$ in these circles.line $\ell$ pass through $A$ and intersect these cicles at $P$ and $Q$.
find line $\ell$ such that $AP.AQ$ be Maximum.[/color](SEE FIGURE 2)

[i]proof[/i].consider a circle $(\mathcal C)$ which is tangent to these circles and the tangency points are collinear with $A$.  name this tangency points $M,N$.
now it's clear for every points $P,Q$ that $A,P,Q$ are collinear we have $AM.AN\ge AP.AQ$.

for drawing this circle it's sufficient that $MN$ pass throuh Ex-homothecy center of these circles.

hence we find triangle $ABC$ .

[color=green]Click on the picture to see larger[/color]

[quote]...now we have that: PA^2+PD^2-PB^2-PC^2=2PO^2+1/2*AD^2-2PQ^2-1/2BC^2=...[/quote]
dear lagrangia
I cannot find points O,Q.May you draw a figure and, may you explain the following equalities:
1)
[quote]PA^2+PD^2-PB^2-PC^2=2PO^2+1/2*AD^2-2PQ^2-1/2BC^2=2*AB*AC*cos(60)[/quote]

and

2)
[quote]PD^2=12+AB*AC[/quote]

thanks.

Can someone explain how we find the area of the triangle after doing what Amir did? I don't know how you would find the circle tangent at points collinear with A, and how you would do the rest.


function
trigonometry
real analysis
real analysis unsolved

Is there exist differentiable function $ f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfying these conditions:

1) $ \forall x \in \mathbb{R}  ; f(x) \geq 0$.

2) $ \forall x \in \mathbb{Q}  ; f(x) > 0$.

3) The equation $ f(x)\equal{}0$ has infinite answer.

You mean $ \cos$ instead of $ \sin$ , because $ \sin(0)\equal{}0$.


One of the Indian olympians is operating with a fake profile...

And it seems he had some hand in trying to gain the postal test solutions...

[b]A moderator can see the IP address and can know if the same person is posting under different names [/b]
And it turns out that we have a person who is using two different profiles...

Till now from unconfirmed sources we know he is a INMO- 2007 awardee...

[b]Please be ethical and no need to hide in a democratic forum...[/b]

Please confirm Mr. pardesi...[/b]

Mr. Pardesi hmmmmmm  :rotfl:    :P
And i happen to know he is your best friend  :wink:

hmmmmmm :D
yes it is and he knows it best

You will be surprised...
Even i happen to know he is my best friend :o

The suspense is building on...

You know that he's your best friend  :o
O god how could you let this happen
One friend knows another  :ninja:
I thought only enemies were very close  :P

madness yen  veena vambu ezhukerai :rotfl:

Yenna vamba naan veena ezhukaren  :D
And by the way are you Vambu?
I thought i was pulling your leg   :rotfl:

no not till he misuses the profile again

We are almost sure about the person but now we cannot find as the person is maybe alert and will no longer use that account...

We will have to keep a  :wink: watch :wink: ...

you know, I left a clue in the above statements for you to guess what the fake profile is named... :wink:

yes a big clue.
soumya see the private message i sent u right above on the right side of the page

Its really obvious if one checks Soumya's (lol a perfect girl name btw) and Ashwath's previous posts.

Maybe you guys should complain about it to some higher-ups. (p.S use gvignesh for this....) The darij guy was totally against this kind of system of postals as it encourages malpractice though most of them are genuinely intelligent.


[quote="Timecat8"]Today was the Hopkins open house????? who went..

the school is so good they have every thing you could think of.....the only things i don't understand was why there was a man outside playing scotland music... with a skirt... :wink:[/quote]
are you talking about the senoir that plays the bagpipes? yeah he always performs whenever there's a big event.

I went, I didn't like it at all.

My tour guide was unfriendly

The teachers there seemed fake

The school seems to only put emphasis on materials

I don't think I'll be considering Hopkins in the list of high schools I'll be going to.

Methinks that Ignite got hacked by the same guy who got anirudh...

[quote="worthawholebean"]Ignite: Stop being an idiot. You go to Hopkins. You sit right above me at assemblies. You are on my bus.[/quote]

Pretty sure it's 2 rows above half the time.

lol.. i was wondering, but why does he preform.. he was walking back and forth and when he saw people walking toward the entrance he would go on the grass and start playing music... :o

doesn't that guy play such good music, and the school was awesome

im going there now
-jorian

how come you've nevr mentioned going to Hopkins

what do you mean, ill let u figure where i sit at the assemblies

lets say the distance between chairs is 1 unit, and 4 make a square

i am $5\sqrt{2}$ units away from ignite

No, you don't go here, so shut up, you aren't even close, it's many layers of platform so front the front you see
 [code]
 ___________
|___________|
|___________|
|___________|
|___________|
|___________|
|___________|[/code]

From the top you see

[code]
 ___________
|           |
|           |
|           |
|           |
|           |
|___________|[/code]

So by 2 rows above, I sit 2 planes above, or 2 "floors" above

[quote="Phelpedo"]Why do you guys even sit in the same building? He's an 8th grader at Smith, and you're a 9th grader at Hopkins...[/quote]

We don't, that's what I've been saying, he's been lying.

[quote="Phelpedo"]Maybe he doesn't even know who you are :ninja:[/quote]
No he knows, he isn't even in the same county as me so he's just being stupid.

ok. i had my laughs. i was just seeing how ignite would react when i did the same thing he did higher on this topic. :rotfl:
-jorian

[quote="Timecat8"]...a man outside playing scotland music... with a skirt... :wink:[/quote]

It's called a kilt, not a skirt, and it's part of traditional Scottish male garb.

[quote="Ihatepie"]im doing the john hopkins test, for u guys who did it recently, wat is in the ACT and SAT?[/quote]

I recommend startnig a new topic in the College forum, this topic is for discussion about Hopkins School and you'll likely find people who know the most about the SAT and ACT (both exams which colleges look at) in there.

The Hopkins school is not the same thing as Johns Hopkins  :wink:   As for the SAT, nothing too hard is on it, just basic Algebra I and Geometry.

isn't Hopkins a high school and John Hopkins a college?

i have a friend who goes to Hopkins, named Alex T. He is Ihatepie's friend too. that is so cool that there are actually people I kin've know on AOPS! (I told IHatePie about AOPS) So do any of you guys no a kid named Lucian I think that's how you spell it... (redcomet46 do you know him, I know a kid named Constantine, Nick and Will, and Cutler. Hopkins-I thought seemed stupid-but now I am kind've interested...there are so many AOPSers and Mathcounts kids AMC from seventh graders to twelfth graders at Hopkins, and college at Jhons Hopkins-it works out....nah! Hopkins probably stinks compared to awesome Fairfield Woods Middle School....er...uh... go wild cats?  :maybe:  :blush:  :|  :(

[quote="ngade"]have a friend who goes to Hopkins, named Alex T.[/quote]

oh god. titinski or whatever you spell it? that kid has the most high pitched voice i have ever heard and it is freakin impossible to get rid of him if he goes around following you

Hey yeah i don't go to Hopkins, but im taking a math course there in the summer..it started june 24 and its called math 5 Functions probability and statistics..my teacher is named Mr. Ayer...just wondering to anyone at Hopkins who might know anything about the course or the teacher..


geometry
geometric transformation
dilation

A colourless solution is known to contain one of these ions. Which ion is present if adding dilute
$ HCl$ produces a white precipitate that dissolves when the solution is warmed?
A $ Ag^\plus{}$
B $ Cu^{2\plus{}}$
C $ Hg_{2}^{2\plus{}}$
D $ Al^{3\plus{}}$
E $ Pb^{2\plus{}}$

I understand that $ Ag^ \plus{}$, mercury(I) cation, and lead(II) will precipitate with Cl. How do we know that lead(II) chloride will dissolve in solution when the solution is warmed and the other two metal with chloride compounds wouldn't?

Maybe you are just supposed to know it.... :rotfl:

Yeah I think you are just supposed to know it. It's pretty common knowledge as far as precipitation stuff go.

Is there anything else we should know about how heating affects solubility for certain ionic compounds?

In general, heating favours the solubility of ionic compounds. However, there are, as usual, some exceptions.

Any reason why the Pb one will dissolve and the other two won't? If heating increases solubility then why can't we heat up the solutions until all of the solutes dissolve?

In my textbook, I cannot find where it says that PbCl2 will dissolve due to heating. What other ionic compounds dissolve due to heating? What should I read to find out about other compounds that dissolve due to heating?


The expression $ 1 \minus{} \frac {1}{1 \plus{} \sqrt {3}} \plus{} \frac {1}{1 \minus{} \sqrt {3}}$ equals:

$ \textbf{(A)}\ 1 \minus{} \sqrt {3} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \minus{} \sqrt {3} \qquad\textbf{(D)}\ \sqrt {3} \qquad\textbf{(E)}\ 1 \plus{} \sqrt {3}$


induction
number theory proposed
number theory

The sequence of integers $ u_0 , u_1 , u_2 , u_3 , . . .$ satisfies
\[ u_0 \equal{} 1\] and
\[ u_{n\plus{}1} u_{n\minus{}1} \equal{} ku_n\]
for each $ n \ge 1$, where $ k$ is some fixed positive integer. If $ u_{2000} \equal{} 2000$, determine all possible values of $ k$.

Let $ u_1 \equal{} x$. If $ x \neq 0$, we have:

$ u_0 \equal{} 1$;    $ u_1 \equal{} x$;    $ u_2 \equal{} kx$;     $ u_3 \equal{} k^2$;     $ u_4 \equal{} \frac{k^2}{x}$;     $ u_5 \equal{} \frac{k}{x}$;     $ u_6 \equal{} 1$;     $ u_7 \equal{} x$; ...

So the sequence is periodic with period $ 6$. Then, $ u_{2000} \equal{} u_2$. Hence $ kx \equal{} 2000$, and $ k \mid 2000$. But also, since $ u_4$ and $ u_5$ are integers, we have $ x \mid k$. Then, $ k \equal{} mx$, where $ m$ is a positive integer, and $ mx^2 \equal{} 2000$. To find the possible values of $ x$ we look at the divisors of $ 2000$ and take the perfect squares. Since $ 2000 \equal{} 2^4 \times 5^3$, the perfect squares are $ 1, 4, 16, 25, 100, 400$, then $ x \equal{} 1, 2, 4, 5, 10, 20$, and $ \boxed{k \equal{} 100, 200, 400, 500, 1000, 2000}$.

If $ x\equal{}0$, we have:

$ u_0 \equal{} 1$;     $ u_1 \equal{} 0$;     $ u_2 \equal{} 0$;     $ u_3 \equal{} y$.

No matter what the values of $ k$ and $ y$ are, we have $ u_1u_3 \equal{} ku_2$. Suppose $ y \neq 0$. Then $ u_2u_4 \equal{} 0 \equal{} ky \rightarrow k \equal{} 0$, a contradiction because $ k$ is a positive integer. Then $ u_3 \equal{} 0$, and by induction, $ u_n \equal{} 0$ $ \forall n \in \mathbb N$.

So the only solutions are those mentioned above.