Let $ a$ and $ b$ are positive numbers. Prove that
\[ \left(\sqrt a \plus{} \sqrt b\right)\left(\frac {1}{\sqrt {a \plus{} 3b}} \plus{} \frac {1}{\sqrt {b \plus{} 3a}}\right)\leq2
\]

[quote="arqady"]Let $ a$ and $ b$ are positive numbers. Prove that
\[ (\sqrt a + \sqrt b)(\frac {1}{\sqrt {a + 3b}} + \frac {1}{\sqrt {b + 3a}})\leq2
\]
[/quote]

First post!

[hide]
Take an example where a=b
\[ (2\sqrt a)(\frac {1}{\sqrt {4a}} + \frac {1}{\sqrt {4a}}) \\
(2\sqrt a)(\frac {1}{2\sqrt {a}} + \frac {1}{2\sqrt {a}}) \\
(\frac {2\sqrt a}{2\sqrt {a}} + \frac {2\sqrt a}{2\sqrt {a}}) \\
1+1=2
\]
Therefore equality holds (is how it's said?)
Now let a>b, since both a and b are >0, therefore a=kb, where k is a positive rational

Given
\[ (\sqrt a + \sqrt b)(\frac {1}{\sqrt {a + 3b}} + \frac {1}{\sqrt {b + 3a}}) \\
\mbox{Therefore }(\sqrt ka + \sqrt a)(\frac {1}{\sqrt {a + 3ka}} + \frac {1}{\sqrt {ka + 3a}}) \\
(\sqrt a ( \sqrt {k + 1} ) )(\frac {1}{\sqrt a  ( \sqrt {3k + 1} )} + \frac {1}{ \sqrt a  ( \sqrt {k + 3} )}) \\
(\frac {\sqrt a ( \sqrt {k + 1} )}{\sqrt a  ( \sqrt {3k + 1} )} + \frac {\sqrt a ( \sqrt {k + 1} )}{ \sqrt a  ( \sqrt {k + 3} )}) \\
(\frac {( \sqrt {k + 1} )}{ ( \sqrt {3k + 1} )} + \frac {( \sqrt {k + 1} )}{  ( \sqrt {k + 3} )}) \\
\]
Since k is a positive rational, and since when a/b for b>a>0  a/b<1
Since
\[ ( \sqrt {k + 1} ) <  ( \sqrt {3k + 1} )
\]
and
\[ ( \sqrt {k + 1} ) <   ( \sqrt {k + 3}
\]
Therefore
\[ (\frac {( \sqrt {k + 1} )}{ ( \sqrt {3k + 1} )} + \frac {( \sqrt {k + 1} )}{  ( \sqrt {k + 3} )}) < 2
\]

By putting the two cases together

\[ (\sqrt a + \sqrt b)(\frac {1}{\sqrt {a + 3b}} + \frac {1}{\sqrt {b + 3a}})\leq2
\]

[/hide]

Please don't hesitate to rip it to shreds, its too late for me to check it too deeply

[quote="SimonM"]
Now let a>b, since both a and b are >0, therefore a=kb, where k is a positive rational
[/quote]
Maybe $ b\equal{}ka$ $ ?$ :wink:
[quote="SimonM"]
Given
\[ \left(\sqrt a \plus{} \sqrt b\right)\left(\frac {1}{\sqrt {a \plus{} 3b}} \plus{} \frac {1}{\sqrt {b \plus{} 3a}}\right) \\
\mbox{Therefore }\left(\sqrt ka \plus{} \sqrt a\right)\left(\frac
{1}{\sqrt {a \plus{} 3ka}} \plus{} \frac {1}{\sqrt {ka \plus{} 3a}}\right) \]
[/quote]
Maybe
\[ \left(\sqrt {ka} \plus{} \sqrt a\right)\left(\frac {1}{\sqrt {a \plus{} 3ka}} \plus{} \frac {1}{\sqrt {ka \plus{} 3a}}\right)\]
:wink:
[quote="SimonM"]
\[ \left(\sqrt a \left( \sqrt {k \plus{} 1} \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right) \][/quote]
Maybe
\[ \left(\sqrt a \left( \sqrt k \plus{} 1 \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right) \]
 :wink:

b = ka or a = kb doesnt matter, obviously! approach is proving in both cases!

I was making the assumption that letting a>b being the same as b>a as it doesn't matter which one was larger

[quote]
[quote="SimonM"]
\[ \left(\sqrt a \left( \sqrt {k \plus{} 1} \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right)
\]
[/quote]
Maybe
\[ \left(\sqrt a \left( \sqrt k \plus{} 1 \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right)
\]
[/quote]

SimonM, pay attention: you wrote $ \sqrt{k\plus{}1}$ while it should be $ \sqrt{k}\plus{}1$ wich is similar but you will have to work it more to prove the inequality...

 :)

[quote="ElChapin"][quote]
[quote="SimonM"]
\[ \left(\sqrt a \left( \sqrt {k \plus{} 1} \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right)
\]
[/quote]
Maybe
\[ \left(\sqrt a \left( \sqrt k \plus{} 1 \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right)
\]
[/quote]

SimonM, pay attention: you wrote $ \sqrt {k \plus{} 1}$ while it should be $ \sqrt {k} \plus{} 1$ wich is similar but you will have to work it more to prove the inequality...

 :)[/quote]

Damn, typical of me to make a error like that, easily patched however
\[ \left(\sqrt a \left( \sqrt k \plus{} 1 \right) \right)\left(\frac {1}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {1}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right) \\
\left(\frac {\sqrt a \left( \sqrt k \plus{} 1 \right)}{\sqrt a \left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {\sqrt a \left( \sqrt k \plus{} 1 \right)}{\sqrt a \left ( \sqrt {k \plus{} 3} \right)}\right) \\
\left(\frac {\left( \sqrt k \plus{} 1 \right)}{\left ( \sqrt {3k \plus{} 1} \right)} \plus{} \frac {\left( \sqrt k \plus{} 1 \right)}{\left ( \sqrt {k \plus{} 3} \right)}\right)
\]
Now considering one of the fractions

Let $ \sqrt k \plus{} 1 \equal{} s$ and let $ \sqrt {3k \plus{} 1} \equal{} r$.

Therefore $ 3k \plus{} 1 \equal{} r^2$ and $ k \equal{} (s \minus{} 1)^2$

Substituting
\[ 3 \left ((s \minus{} 1)^2 \right) \plus{} 1 \equal{} r^2 \\
3s^2 \minus{} 6s \plus{} 4 \equal{} r^2 \\
\]
Given $ s > 1$ therefore $ r > s$

Hence, from the same as previously the left hand fraction is less than one. (Am I allowed to leave the other as an exercise?) :P

[hide]If you want to do the otherside bear in mind one of the previous suggestions and what it means for k. From there is should be easy :)[/hide]

Good idea SimonM but it results that $ \sqrt {k \plus{} 3}$ is not always greater than $ \sqrt {k} \plus{} 1$ for $ k > 1$
I have aslo tried many things and failed, but may be we are now closer to solution...

 :)

k is less than one though...

a>b

b=ka

:)

What I mean is this:

$ \sqrt{k}\plus{}1<\sqrt{3k\plus{}1}$                and                             $ \sqrt{k}\plus{}1<\sqrt{k\plus{}3}$
$ k\plus{}2\sqrt{k}\plus{}1<3k\plus{}1$                     and                            $ k\plus{}2\sqrt{k}\plus{}1<k\plus{}3$
$ \sqrt{k}<k$                                    and                              $ \sqrt{k}<2$
$ 1<k$                                            and                                $ k<4$

So both, $ \sqrt{k}\plus{}1<\sqrt{3k\plus{}1}$ and $ \sqrt{k}\plus{}1<\sqrt{k\plus{}3}$, hold only for $ 1<k<4$

 :)

And don't think I am disaproving you with out reason it is just that I have been trying to prove it with out succes so, when someone post a solution a read carefuly in order to understand and learn from the proor.

As our inequality is symmetric, lets assume

$ a\geq b$
$ b\equal{}ak$ for some real $ k\leq 1$
$ \frac 1{\sqrt {a \plus{} 3b}} \plus{} \frac 1{\sqrt {b \plus{} 3a}}\equal{}\frac 1{\sqrt {a \plus{} 3ak}} \plus{} \frac 1{\sqrt {ak \plus{} 3a}}\equal{}\frac 1{\sqrt a (\sqrt { 1\plus{} 3k})} \plus{} \frac 1{\sqrt a (\sqrt {k \plus{} 3})}\equal{}\frac 1{\sqrt a}\left(\frac 1{\sqrt { 1\plus{} 3k}} \plus{} \frac 1{\sqrt {k \plus{} 3}}\right)$
$ \left(\sqrt a \plus{} \sqrt b\right)\left(\frac {1}{\sqrt {a \plus{} 3b}} \plus{} \frac {1}{\sqrt {b \plus{} 3a}}\right)\equal{}\left(\sqrt a \plus{} \sqrt b\right)\frac 1{\sqrt a}\left(\frac 1{\sqrt { 1\plus{} 3k}} \plus{} \frac 1{\sqrt {k \plus{} 3}}\right)\equal{}\left(1\plus{}\sqrt k\right)\left(\frac 1{\sqrt { 1\plus{} 3k}} \plus{} \frac 1{\sqrt {k \plus{} 3}}\right)$

You have to prove the inequality for every pair $ (1,k)$ with $ k\leq 1$, which is probably easier. I don't have a proof for this, I'll post as soon as I find one.

Mithril you've done exactly the same proof as me, except you have changed the order of two items

k by definition is less than one. Since a > b and b = ka, therefore k is less than one :)

$ \sqrt k \plus{} 1 < \sqrt {k \plus{} 3}$
$ k \plus{} 2\sqrt k \plus{} 1 < k \plus{} 3$
$ 2\sqrt k < 2$
$ \sqrt k < 1$


calculus
derivative
calculus computations

This might seem lightweight to some of you, but I can't seem to wrap my head around a problem here...

Differentiate with respect to x.

1.) (x^3+e^(-x)+5)^1/2

Now, using the Power rule,

1/2(x^3+e^-x+5)^-1/2 (3x^2-e^(-x))

...and I'm not sure where to go from there to get the result that my Ti-92 gives me:

[(x^3+3x^2+5)e^(x/2)]/2(x^3e^x+5e^x+1) - {[(x^3e^x+5e^x+1)^1/2]e^(-x/2)}/2

Now, how exactly do I get from point A to point B? I'm lost.

Your silly calculator turned $x^{3}+e^{-x}+5$ into $\frac{x^{3}e^{x}+1+5e^{x}}{e^{x}}$ before differentiating its square root. I would not recommend this method to any human.


Show that for all positive integers n = 1, 2,..., that 14 divides 34^{n+2} + 5^{2n+1).

$ S\equiv 0^{n+2}+1^{2n+1}\equiv 1 \pmod{2}\implies 14 \not | S$, please revise  :)


Suppose [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a297d1227b6683200ab3e7f9075f3c44.gif[/img] are n positive numbers.Let S be the sum of the n numbers.Prove that
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b6b918cedff68e9b63137f0f649daf68.gif[/img] :)

I have forgotten to write that n is an integer which is greater than 2.I should say sorry because my English is poor. :blush:

Well take $s=1$ then $\sum\sqrt{\frac{a_i}{1-a_i}}\ge \sum2a_i=2\Leftrightarrow \sum(2a_i-1)^2\ge0$ .Note that equality holds for example for $a_1=a_2\longrightarrow\frac{1}{2},a_3=a_4=...=a_n\longrightarrow 0$

The proof to my ineq:
Lemma:Assume a,b are two positive numbers such that a+b<2/3,then we have
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a19aafabeba6624026dc850bee3267dc.gif[/img]
  Now suppose S=1,
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/50bc7412a0d7664525fe3efbce935c3f.gif[/img] When n>3,[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6859fcd3b8cf8e14fb3a9fedf8b6ba2f.gif[/img]
 then let a be the sum of  a_n and a_n-1  . Then several times later n will become 3 the following is easy.Since I am weak in English ,my solution may be not fluent or has some grammar mistakes,please forgive me. :blush:
  ehku,your problem is much harder than mine,could you give your solution to us?

[quote="ehku"]Let [tex]a_i, \ t \in R [/tex] satisfying [tex]\frac{a_1}{a_2+a_3+...+ta_1} \ge 0[/tex].

Prove that

[tex]\sum \sqrt[n]{\frac{a_1}{a_2+a_3+...+ta_1}} \ge \frac{n\sqrt[n]{(n-1).(1-t)^{n-1}}}{n-1}[/tex][/quote]

Is $a_i$ positive?

I wrote : $a_i \in R$ !

($n \in N , \ n \ge 2 $)

put s=1
it is easy to prove that
sqrt(a/1-a)  > 2a (use AM-GM)

(What is this, wei_wenzhe9170 ????)

Problem. Let $m, n \in N, \ m,n \ge 2 $ ; $a_i, \ t \in R \ \ i=1,2,...,m ;\ \ t \le 1$ satisfying $\frac{a_i}{S-(t-1)a_i} \ge 0$ where $S=a_1+a_2+...+a_m$ . Prove that $\sum \sqrt[n]{\frac{a_i}{S-(t-1)a_i}} \ge n.\sqrt[n]{(\frac{1-t}{n-1})^{n-1}}$  .

[quote="kid1412"]Ha, this program is easy and i think we need stop . Okie.[/quote]
Then can you show your proof of ehku's inequality?

Who can solve it ? kid1412, can you ?
If no one post the solution, I will post it.

Come on this inequality is easy.It follows from AM-GM.


group theory
abstract algebra
superior algebra
superior algebra unsolved

for A=[b]Z[/b]60 x [b]Z[/b]45 x [b]Z[/b]12 x [b]Z[/b]36. Find the number of elements of order 2 and the number of subgroups of index 2 in A.

this is what I did: after finding the elementary divisors of A,
I obtained : A is isomorphic to [b]Z[/b]4 x [b]Z[/b]4 x [b]Z[/b]4 x [b]Z[/b]3 x [b]Z[/b]3 x [b]Z[/b]9 x [b]Z[/b]9 x [b]Z[/b]5 x [b]Z[/b]5

Only the odd cyclic group has an element of order 2. So the number of elements of order 2 is 6.
And the number of subgroups of index 2 would be 3 ([b]Z[/b]3, [b]Z[/b]9 & [b]Z[/b]5)

Am I correct?????

That's wrong. We get elements of order 2 by taking products of elements of order 1 or 2 in each factor- the subgroup of elements with order dividing 2 is thus isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_{1}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}$. There are 8 elements there- the identity and 7 elements of order 2.
6 is not possible- every group with at least one element of order 2 has an odd number.

There are 7 subgroups of index 2- I won't go into the details there. Your comment on the question is nonsense.


[quote="thanhak5"]For $ x,y,z > 0$ and $ x\plus{}y\plus{}z\equal{} \frac{3}{2}$ . Prove that
\[ \frac{\sqrt{x^2\plus{}xy\plus{}y^2}}{4yz\plus{}1}\plus{}\frac{\sqrt{y^2\plus{}yz\plus{}z^2}}{4zx\plus{}1}\plus{}\frac{\sqrt{z^2\plus{}zx\plus{}x^2}}{4xy\plus{}1} \ge \frac{3\sqrt{3}}{4}\][/quote]
Note that
\[ \sqrt{x^2\plus{}xy\plus{}y^2} \ge \frac{\sqrt{3}}{2}(x\plus{}y)\]
It suffices to prove that
\[ \sum \frac{x\plus{}y}{4yz\plus{}1} \ge \frac{3}{2}\]
By the Cauchy Schwarz Inequality, we get
\[ LHS \ge \frac{4(x\plus{}y\plus{}z)^2}{\sum (x\plus{}y)(4yz\plus{}1)} \equal{}\frac{9}{4\sum (x\plus{}y)yz\plus{} 3}\]
Moreover, using the known
\[ \sum x^2y\plus{} xyz \le \frac{4}{27}(\sum x)^3\]
we get
\[ \sum (x\plus{}y)yz \equal{}(\sum x^2y\plus{} xyz ) \plus{}2xyz \le \frac{4}{27}(\sum x)^3\plus{} \frac{2}{27}(\sum x)^3 \equal{}\frac{1}{2} \plus{}\frac{1}{4} \equal{}\frac{3}{4}\]
Thus
\[ LHS \ge \frac{9}{4\sum (x\plus{}y)yz\plus{} 3} \ge \frac{9}{4\cdot \frac{3}{4}\plus{}3}\equal{}\frac{3}{2}\]
We have done. :)


algorithm
geometry
parameterization
number theory open
number theory

Find the integer solutions of the following two equations:

1) $ 16y^{3}\minus{}x^{2}\equal{}15$.

2) $ 16y^{3}\minus{}x^{2}\equal{}7$.

[hide="does it help for 1)?"]$ 16 (y^3 \minus{} 1) \equal{} (x \plus{}1)(x \minus{} 1)$
$ 16(y \minus{} 1)(y^2 \plus{} y \plus{} 1) \equal{} (x \plus{} 1)(x \minus{} 1)$
[/hide]

I think that the first equation has only that trivial solution but the second one has another solution, namely:

$ (\pm11, 2)$.

and, I do not have a solution for this problems.

[hide="1)"]Let 16y^3 =15 + x^2 so =x^2(mod 5) so y^3 =x^2(mod5). But it only holds when x=y and it yields then x=y=1 inintegers.[/hide]

[hide="2)"]As same with the first of these, y^3 =2 +x^2 (mod 5); more likely we have x=3,y=1[/hide]

[quote="madshiet"]so y^3 =x^2(mod5). But it only holds when x=y and it yields then x=y=1 inintegers.[/hide]
[/quote]
For example,it holds for $ y\equal{}4,x\equal{}2$ too.So your proof is incorrect.

You can try to compute the rank of these curves by John Cremona's program mwrank. However, I was unable to compile that program so I believe I can't help more.

For 2, you can use the unique factorization of $ \mathbb{Z}\left[ \frac{1\plus{}\sqrt{\minus{}7}}{2}\right]$.

1) Its Weierstrass form is $ (x/4)^2 \equal{} y^3 \minus{} 15/16$ or $ (16x)^2 \equal{} (16y)^3 \minus{} 15\cdot 256$. It has a trivial torsion group, Mordell-Weil rank $ 1$, and a generator $ (16,16)$ so the Weierstrass form has only one solution: $ x \equal{} 1, y \equal{} 1$.

2) Its Weierstrass form is $ (16x)^2 \equal{} (16y)^3 \minus{} 7\cdot 256$. It has a trivial torsion group, Mordell-Weil rank $ 1$, and a generator $ (16,48)$ so the Weierstrass form has only one solution: $ x \equal{} 1, y \equal{} 3$.

And what about solutions $ (793,34)$ and $ (11,2)$? Could you give a link to your theory you have used above. Very interesting. :)

See http://en.wikipedia.org/wiki/Elliptic_curve , http://mathworld.wolfram.com/EllipticCurve.html , and http://planetmath.org/encyclopedia/ArithmeticOfEllipticCurves.html .

I found the site http://mathforum.org/dr.math/ask/ where I got an answer. One mathematician used Cremona's program to compute the rank of the curves. I don't know how to compile Cremona's mwrank in my Linux system so I'm unable to do the computations in my home.

But the process of finding the rational points of given elliptic curve is straightforward:
1. Transform the curve into Weierstrass form.
2. Compute its torsion group.
3. Compute its Mordell-Weil rank.

There is a fast algorithm to transform elliptic curves into Weierstrass form. There is a fast algorithm to compute its torsion group. There is some slow algorithms to compute the Mordell-Weil rank of given elliptic curve. I believe that if one proves the Birch-Swinnerton-Dyer conjecture, one can find a fast algorithm to compute the Mordell-Weil rank of given elliptic curve. mwrank uses some relatively fast algorithm to compute the ranks but there are some curves where the algorithm can't compute the rank.

I'm not an expert of elliptic curves. If one wants to start studying those curves, I would like to suggest to read Knapp's Elliptic curves first. It explains the basics of the theory of elliptic curves without messing algebraic geometry into the theory.

Your answers seem to be incorrect, though.  Did you try doubling your generators to see if they gave integer points?

I'm not sure what parameters the mathematician in Dr. Math used while using the program.


blogs AMC AIME trigonometry USAMTS geometry rotation

Share your results here (only if you wish, of course).  I already posted my results in my blog (and many of you probably already know me by name anyways :P ).

Btw, does anyone know what the AIME cutoff was?  It says that I made it...but that doesn't give much insight into what it was :D

[quote="joml88"]Btw, does anyone know what the AIME cutoff was?  It says that I made it...but that doesn't give much insight into what it was :D[/quote]It's 68.

I scored 5/5/5/3/5, no commended.  That's about what I expected, since I found the two solutions to #4, but I couldn't prove that there were no more.  That gives me a three-round score of 69, so I qualify for AIME!  (Not that I need the AIME qualification  :lol: )

Sounds like you got a good score joml  :P

[quote="calc rulz"]5/[b]5[/b]/5/5/3
=23

Total: 56

On question 1 I got "nice generalization" but no commendation. Is this normal?  :huh:[/quote]
Commendations are pretty much at the whim of the grader.  While commendations might be given for an interesting generalization, they also might not be, so it's basically impossible to guess why you may or may not have gotten a commendation.

My scores: 5,[b]5,5,5[/b],5 , which is what I was expecting in terms of score since all my solutions were correct.  I'm happy about the commendations, though  :lol:

5/5/5/5/2

For some reason, they gave me a 2 on #5, saying "Proof is not clear".  Okay, I worked out all possibilities for part b without formally proving that those were all of them, but I thought my solution for part a was pretty clear...

I did abysmally.  Suffice to say that my scores so far form an arithmetic progression in the wrong direction.  I went to New York, worked on them there, got back the day they were due, realized I had omitted a condition on one of them, and ran out of typesetting time.

Then on 2, I messed up somewhere. (And I had been so proud of my infinite sums waltzing across the page).

On 3, my solution was just bizarro.  I made the fractions proper, used Power of Point and the law of cosines, and finished it off with Heron's formula.  I got full credit though.

I actually got a commended on 1.  I didn't think it was really possible to get commended casework.  Apparently, the grader liked my $n=7$.

5 5 5 5 5

75 cumulative

Still no commended though ...

But who am I to complain?

JB

[quote="Boy Soprano II"]I actually got a commended on 1.  I didn't think it was really possible to get commended casework.  Apparently, the grader liked my $n=7$.[/quote]

n=7 was the interesting case for Problem 1/3/17. The other values of n that had solutions either had each digit divisible by n or equal to its opposite mod n. The n=7 case had the sum of the digit and its opposite digit divisible by 7. Anyone who commented on that was viewed as having insight into the problem. Sometimes insight is worth a commendation.

We try to find a standard for commendations on every problem, though commendations are rarer on some types of problems, such as casework.

Erin Schram
Grader for Problem 1/3/17

5 5 3 1 3=17
Which is okay, I guess...current total is 49, so I'm right on track for a bronze medal, which is all I'm aiming for right now...yes, I know that's low. The 3 on #3 is the highest I have ever gotten on a USAMTS geometry problem (got 1's on the first two) which is also kind of sad. So actually I've got another goal--get a 5 on the round 4 geometry problem. Erg.

5 5 [b]5[/b] 5 5 with 75 culimative

I'm happy because I thought I'll get a 3 or 4 on number 5 (4 or 5 pages of brute force...), but the grader comments: "could be more organized at casework but the solution works". Yay! :lol:  :lol:

[quote="salmononpi"]When will the grading rubics be up? For some reason I only got a 2 on problem number 5 despite having a full solution. "Good ideas. Explaination shaky and diffacult to understand"--so I'd like to see exactly why I got so low    :([/quote]

Uh...they've been up ever since the results were posted.

I specifically want some more feedback on my solution to #5, because I answered everything...but my grader said "Proof is unclear" and gave me a 2.  However, I don't know if I can ask for that...

24 this round.  Evidently I missed a case somewhere in #5, which is fair enough, since casework solutions to 5 pretty much dodge the most important aspect of the problem anyway.

That gives me a total of 70, with two commendations (#3 on this round, and #4 on round 2).  Now I just need a 25 on round 4, and I can bask in the glory of knowing that I'm one point from a gold winner ;).

Nice work Eric, Joe, John, and zanntrang (sorry, couldn't easily match you up with your name based on the Leaderboard =p.).

Edit:  Ah hah, so you're name's Erik, zanntrang!  Kudos to you too (and thanks to Joe for filling me in on it)

K81o7:

First, do you know how those two points were distributed (e.g. 1 for part a and 1 for part b)?

For part (a), you end each case with "Bart simply has to rotate the table such that...".  I don't think this is clear/rigorous enough to make the conclusion you did.  They were probably looking for a more concrete method that Bart can follow (you can look at my solution to that one...I attached my solutions).

For part (b):

[quote]Notice that if $m+ n = 8$, then $m\equiv n mod 2$. So if $i$ is even, then $m$ and $n$ will stay the same parity, while if $i$ is odd, then $m$ and $n$ will switch parity.[/quote]

Is that supposed to be $n$?  Based on how you defined the variable in (a) it seems like $l$ would be more logical.

You say: [quote]If we consider one
configuration to be equivalent to its mirror (Ex. 1, 2, 3, 5 equivalent to 1, 3, 4, 5) and equivalent to its reverse (Ex. 1, 2, 4 is equivalent to 3, 5, 6, 7, 8), then we can show that there are exactly 18 configurations.[/quote]

I don't see anything wrong with the first part, but I believe the "equivalent to its reverse" part falls under this category:

[quote="Grading Rubric"]Also, many papers listed vague assertions about "symmetry" which, if unproven, did not receive full credit.[/quote]

Whenever brute forcing a problem, you need to be absolutely certain that you have taken care of [i]everything[/i].  Graders are ruthless (on Olympiads in particular) when they have to grade a massive brute force solution.

5/5/5/5/2.  That's pretty much what I wa expecting, but I hoped for a commendation on #2 or #3.  63 cumulative isn't bad.

[quote="joml88"][quote="eyefragment"]
Nice work Eric, Joe, John, and zanntrang (sorry, couldn't easily match you up with your name based on the Leaderboard =p.).[/quote]

Thanks. (Btw, zanttrang is Erik...that should help you match him up :D )[/quote]

Wow, so you are the other guy from CA...
I was wondering because I felt California curriculum is not that good...(I dominate my school simply because they don't teach advance math...)

[quote="joml88"]
Thanks. (Btw, zanttrang is Erik...that should help you match him up :D )[/quote]
Yep, you caught me  ;)

[quote="kperichang"]Wow, so you are the other guy from CA...
I was wondering because I felt California curriculum is not that good...(I dominate my school simply because they don't teach advance math...)[/quote]
California does have really bad public schools.  I think most people who get advanced math do it from programs outside of school.  I go to a private school anyway, so I don't have to deal with public school curricula, but on the other hand, our small size means we have fewer outstanding math students, so we tend to have a harder time scoring at the top in team events (like the CML).  But in any system, once you start doing the sort of math covered in contest math programs (like really advanced geometry, number theory, and algebra), you really have to get the training from outside of school.  That's why I rely so heavily on places like AoPS, as well as on a ton of math books (some topic-based, some contest-oriented), that I've been trying to work my way through.  I think next year I'll also join the Bay Area MC, since I live relatively close to Berkeley, since it looks like a lot of fun.

Oh, I don't know if CA schools are that bad. I live in the bay area and our schools are excellent. I go to private, but the public is good too. Also, how bad can they be if 2 perfect scores are from CA? :lol:

I don't know much about the California school system, but I'll take their word that the state curriculum is bad.  On the other hand, you will see a lot of mathematical talent from California because

1. They have a HUGE population...you're bound to have talented students pop up regardless of what their schools are like.
2. There are certain areas where the math culture is exceptionally good (SFBA and San Diego to name two).

[quote="pkerichang"]Well...I will kinda take my word back a little and say that "in general" (no offense), Southern CA have a bad curriculum on mathematics (well...bad for the people who desire to learn more).  For my school, we are excellent in science and computer science, but only slightly "better" than the schools around us on mathematics.

For AMC score, in our school's entire history the highest is 138 and only one guy in 40 or 50 years qualify for USAMO (look at last year's qualifiers for USAMO from CA, none came from Southern California (or a few if I miss someone)), and last year I got 6 out of 15 on AIME and this is the best score out of the whole school...

Not to blame schools or anything, because not much schools are willing to offer or teach classes that goes beyond the curriculum.  Everyone at my school is rushing to take Calc, Linear Alegebra, etc to gain college credits or do good on AP.  And even a Freshmen who is learning Geometry knows more about Geometry than those rushing people.

Seriously, I think we should have AP PreCalc with advance Geometry, number theory, etc.  That will be REALLY nice[/quote]

Well, the inadequacies you describe are not just characteristic of Southern California.  I believe that most of the country is that way.

well...Seeing that most of the people on IMO are from east coast and the difference between Northern CA and Sothern CA made me feel that way, and I don't really know about other places' standard (my friends who moved from East coast said CA is easier too)

[quote="pkerichang"](look at last year's qualifiers for USAMO from CA, none came from Southern California (or a few if I miss someone)),[/quote]
I can't let that pass without a response: there were six USAMO qualifiers from Southern California last year, two from San Diego (Michael Viscardi, Henry Tung) and four from Los Angeles County (Zeb Brady, Tedrick Leung, Roger Li, and David Philipson).

Not that that changes your larger point, including the discrepancy between Southern and Northern Calfornia. But even at that - the large cluster of USAMO qualifiers from Northern California mostly come from about a dozen high schools in the southern half of the Bay Area. It's not that the mathematical problem-solving culture permeates the entire area - it's concentrated in relatively few places.

You attend a high school that, by reputation, is one of the "best" schools in the area, especially for science. Despite your complaints about it, at least the AMC exams are available at your school. At 80% or 90% of the high schools in Southern California, no one ever takes the AMC or any other problem-solving competition. At the few schools that do participate in the AMC, the impetus for doing so usually comes from one or two math teachers, not the principal, creating the risk that the retirement of one teacher could end the program.

Thank you for clarifying.  I was reading the qualifiers for USAMO from CA and found out it is mostly from Northern CA, but I could be wrong because of my lack of geography knowledge (For example, I always thought San Diego is Northern CA until you points out...) :blush:

Also, my response could also be bias because I found my AMC and AIME score in the lower range among the AoPSers, while it is consider high in my school.  It shocked me with the difference and naturally I want to learn more math from my school (It may also result from my discontentment that I am stuck with trig as a junior...)

Lastly, I must clarify that I like my school's science and computer science classes.  They are great and I enjoy them, and I only wish that advance math can be more available to the talented students.  (I found independent studies difficult due to my lack of knowledge of formal mathematics notations and vocabularies..)

[quote="Kent Merryfield"][quote="pkerichang"]Also, my response could also be bias because I found my AMC and AIME score in the lower range among the AoPSers, while it is consider high in my school.[/quote]
That's one issue with AoPS as a community: any topic in which posters volunteer their scores in some contest, or a poll is taken on those scores, can be [b]very[/b] intimidating. Always bear in mind when you read such a topic or such a poll that there are many, many registered AoPS users who aren't going to answer that poll, and if they did, the scores would skew a lot lower. Don't let the exploits of an elite few dampen your own sense of accomplishment. How many people live in Orange County? Certainly more than 2 million, maybe more like 3 million. In all of that, there were only four people in the whole county with higher AIME scores than you last year, and four others with the same score. (And I know most of them.)[/quote]
Agreed - AoPS tends to give the illusion that very respectable scores are in fact very low, because the people willing to share their scores are generally the ones with the highest scores (also, the most active ones are also the most experienced ones, leading to the same problem).  I think, though, that the problem you point out about most people not even having access to the AMC contests is not merely a northern vs. southern California problem - it's actually a nationwide problem.  A lot of schools just don't have teachers who have the knowledge or interest, and that means the school isn't going to look into the contests and sponsor them.  I think northern California has a few clusters of strong math interest primarily because of the San Jose and Berkeley/Oakland areas (where San Jose is a huge technology center, and UCB is located in the latter location; in addition, there are large Asian populations in both these areas, and not to stereotype too much, but Asian parents often push their children more to excel in math and science, meaning they would be more likely oftentimes to participate in math contests).  Elsewhere, I think that the level of interest is about on part with southern California.

I'd also like to note that while only two Californians are at the top of the leaderboard, we are also the ONLY state with two people at the top.  Also, if you look at the rankings immediately below the top, you'll see a very large number of Californians as compared to other states.  So I don't think we're underrepresented in any way on the leaderboard (go CA  :lol: ).

[quote="zanttrang"]Also, if you look at the rankings immediately below the top, you'll see a very large number of Californians as compared to other states.[/quote]
Two of the seven people listed at 74 (and they both have lots of commendations) are specifically Southern Californians. Hooray for my guys!

[quote="zanttrang"][quote="Philip_Leszczynski"]I wish I lived in southern California.

I got 5/5/5/5/4. That's 73 cumulative.[/quote]
Why?  It's too hot in SoCal.  I prefer the more temperate weather of the Bay Area, especially during the summer - plus we have two Math Circles (Berkeley and San Jose)![/quote]
I resent that! I attend the Stanford Math Circle, which, though in its first year, is excellent

I used to live in south CA...and moved to NY about 2.5 years ago.  Well...I only got better at math...

Well...I think it depends on the area and the school, I like here (CA) because my school has good science olympiad and programming teams and some events are fun. (Engineering ones and programs).  Other school either does not participate or never know about them.  For a guy like me who will never like english, science and math competitions are the only reasons I go to school...(or else I'll home school)

[quote="asquared"]
I resent that! I attend the Stanford Math Circle, which, though in its first year, is excellent[/quote]
Sorry, I didn't mean to be exclusive - so we have more than two Math Circles (I'd just heard of the two I'd named)!

Ya, the emphasis on good math education here is pretty good.

I got 5/[b]5[/b]/[b]5[/b]/3/[b]5[/b], which was kinda expected. I knew I was at the mercy of the grader on #4, because I didn't really prove what I got but just tried to sneak that fact by :P My cumulative score is 72, but I've gotten 6 commendations, so that's good.

5/5/5/5/4

I am pleased, but I got a 67 cumulative.  AAARRRG!  If only I had a 68 and could be in the AIME...
I haven't got my scores back from the AMC-10, but since I didn't take the AMC-12, I have a lower chance of actually making it.

5/5/5/5/4

I lost a point because my solution "wasn't clear."  Yea, I guess it wasn't.

I have a 67 cumulative.  Darn it!  If only I could have gotten a 68 and been able to go to the AIME...
I haven't gotten my scores for the AMC-10; maybe I could still qualify...


A sphere is inside a circumscribing cube.  What fraction of the volume of the cube is inside the sphere?
\[ \text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{\pi}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{\pi}{4} \qquad \text{(E)}\ \frac{\pi}{6} \]

[hide](E) If you pick a random number for the length of  a side of your cube, you could pick 4. The volume of the cube is 64. The diameter of your sphere will be 4, and the volume will be 4/3 :pi: r :^3: = 32 :pi: /3. The fraction of the volume of the cube inside the square is (32 :pi: /3)/64, which equals 32 :pi: /192. This simplifies into  :pi: /6. [/hide]

It will be pi/6, I just remember that from a problem I did a while back. pi/4 is for a circle circumscribed in a square. :blush:


integration
function
topology
real analysis
real analysis unsolved

Let $K \in C \left( \left[a,b\right] \times \left[a,b\right] \right)$ such that $\left|K \left(t,x \right)\right|\leq C$ ,$\forall t,x \in \left[a,b\right].$ If $f\in C\left( \left[a,b\right] \right)$ y $r\in \mathbb{R}$ is such that $\left|r \right|<\dfrac{1}{C(b-a)}$, prove that exists an unique $g\in C \left( \left[a,b\right]\right)$ such that \[ f(x)=g(x)+r\int^{b}_{a}K(t,x)g(t)dt, \ \forall x \in \left[a,b\right]. \]

P.D.:Here $C\left( \left[a,b\right] \right)$ denotes the set of all functions $f: [a,b] \rightarrow \mathbb{R}$ that are continous.

:blush:  I don´t understand ... you use the Fixed point´s theorem ?and why $g$ is in the left an in the right?

[quote="tipe"]you use the Fixed point´s theorem ?and why $g$ is in the left an in the right?[/quote]
Yes, I use the fixed point theorem. The notation with $\mapsto$ may be a bit confusing, though I thought it is fairly standard. What it means is "the mapping that maps whatever is to the left  of $\mapsto$ ( in our case, a continuous function $g$) to whatever is to the right of $\mapsto$ (in our case, a fixed function $f(x)$ minus the integral)".
Sorry for the confusion :blush:

[quote="tipe"]So this mapping its a contraction on the set of functions... right?   :P[/quote]
Right you are :P

[quote="fedja"]$g(x) \mapsto f(x)-r\int_a^bK(t,x)g(t)\,dt$ is a contraction on $C[a,b]$.[/quote]
Or more elementary, start with $g_0\in C[a,b]$. Then let $g_{n+1} = f(x) - r\int_a^bK(t,x)g_n(t)\,dt$. Prove that $\{g_n\}$ converges in $C[a,b]$.

But in the fixed point theorem you need a closed set, does $C[a,b]$ is closed?

Since $C[a,b]$ is the entire space in question, it is automatically closed. What we really need is that it is a complete metric space, and it is under the uniform norm. If we are talking about subsets of some larger space, completeness of the set implies that the set is closed.


The square \[ \begin{array}{|c|c|c|}\hline 50 & b & c\\\hline d & e & f\\\hline g & h & 2\\\hline\end{array}  \] is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$? \[ \text {(A) } 10\qquad \text {(B) } 25\qquad \text {(C) } 35\qquad \text {(D) } 62\qquad \text {(E) } 136  \]

[hide]ok, because the products of the numbers in each columns, diagonals, and rows are equal then we have, 50*2*e=g*e*c, and thus gc=100.  Also we have 50bc=2fc, so 25b=f and its easy to see the same is true for d and h by symmetry, 25d=h.  Furthermore we have gec=beh=def, gc=bh=df, 100=bh=df.  this gives us, 1,2,4,5,10,20,25,50,100 as possibilities.  checking for different possibilities, that satisfy the above conditions, we see that b must equal 1,2,4 because anything higher will make f greater than 100 and thus df can't equal 100.  Now we have four possibilities
50 1   c     50 2 c     50 4 c
4   e  25   2  e 50    1  e 100
g 100 2    g  50 2     g 25 2
solving 50dg=50bc and 100=cg, we have g = 5, 10, 20
thus C.. whew, much shorter when done in head.  :lol: [/hide]

[hide]All the unknown entries can be expressed in terms of $b$. Since $100e = beh = ceg = def$, it follows that $h = \frac{100}{b}, g = \frac{100}{c}$, and $f = \frac{100}{d}$.

Comparing rows $1$ and $3$ gives us:

$50bc = 2\frac{100}{b}\frac{100}{c}$, from which $c = \frac{20}{b}$.

Comparing columns $1$ and $3$ gives us:

$50d\frac{100}{c} = 2c\frac{100}{d}$, from which $d=\frac{c}{5} = \frac{4}{b}$.

Finally, $f= 25b, g = 5b$, and $e=10$. All of the entries are positive integers iff $b = 1, 2$ or $4$. The corresponding values of $g$ are $5, 10, 20$, and their sum is $35$, which means our answer is $\boxed{C}.$

[/hide]


integration
limit
algebra
polynomial
real analysis
real analysis unsolved

I think the following formula is true:

$\frac{\sum_{i=1}^{n}~i^{k}}{n^{k+1}}\rightarrow \frac{1}{k+1}$

e.g. for $k=1$: $\frac{\sum_{i=1}^{n}~i }{n^{2}}= \frac{\frac{n(n+1)}{2}}{n^{2}}= \frac{n^{2}+n}{2n^{2}}= \frac{1+\frac{1}{n}}{2}=\frac{1}{2}~(n \rightarrow \infty)$

How can I prove it?

Simply apply Cesaro Stolz lemma and you should get it.

The limit of Riemman's sum, $\frac{\sum_{i=1}^{n}i^{k}}{n^{k+1}}=\frac{1}{n}\sum_{i=1}^{n}\left(\frac{i}{n}\right)^{k}\ \longrightarrow \int_{0}^{1}x^{k}\ dx=\frac{1}{k+1}\ (k\neq-1)$.

For nonnegative integer $k$, write $i^{k}= k! \binom{i}{k}+c_{1}\binom{i}{k-1}+\ldots+c_{k}\binom{i}{0}$.  Then
\[\sum_{i = 0}^{n}i^{k}= k!\sum_{i = 0}^{n}\binom{i}{k}+\ldots = k!\binom{n+1}{k+1}+\ldots, \]
where "$\ldots$" are $k$ polynomials of lower degree.  $n^{k+1}= (k+1)!\binom{n+1}{k+1}$.  Thus the limit is $\frac{k!}{(k+1)!}= \frac{1}{k+1}$.


geometry
counting
distinguishability
probability
AMC
AMC 12
AMC 12 B

I didn't take the 12B, but I'm just curious to see how others thought of it.

Sorry... I sort of messed up on the polls. :(

I thought it was harder than the A, mostly in terms of time pressure.  I especially disliked problem 11 (I think -- whichever one was the seeds problem), as in trying to avoid stupid mistakes I must've spent 5 minutes multiplying out the numbers in that one, quite a lot of time for such a short test.  Also, the problems were a lot more combinatorics and geometry-oriented, compared to the relative nice and simple algebra and number theory problems on the A

I think it was much, much easier - the last 10 or so on the AMC 12A were annoying, but the last 10 or so on the AMC 12B were pretty much all easy (except #18 which was really silly computation and #22 which was simply annoying).

I did like #17 on the 12B, though - instead of just counting all (not just distinguishable; that would be somewhat annoying) cases where there was a stripe and put that over the number of all cases in general, (which was what I believe most people did) you could solve it in a few seconds with some pseudo-tricky probability multiplying and adding.

[quote="perfect628"][quote="Temperal"]I did like #17 on the 12B, though - instead of just counting all (not just distinguishable; that would be somewhat annoying) cases where there was a stripe and put that over the number of all cases in general, (which was what I believe most people did) you could solve it in a few seconds with some pseudo-tricky probability multiplying and adding.[/quote]
Umm what? How is there are 3 ways each with 1/16 probability tricky? Three other people at my school (Everyone I talked to about it) did this.[/quote]

uhh, okay. I personally found it tricky (since I took a few seconds to see that the probabilities could be treated as independent) but I probably have a bloated sense of self-worth or something.

Hmm, I thought that the 12B was harder.  I took the 12A unofficially and got a 141, but I got a 129 on the 12B, so yeah...  I sorta panicked after the first 20 minutes and thought that all of the last 10 questions were unsolvable.  So that was bad...  The 12A was, imo, pretty easy, and the 12B was slightly harder.

As for #17, I don't think it was tricky, but it's all right, Temperal.  :)

I got a 138 because I thought 3*4 = 16 (#17) and sin comes in periods of pi (not 2pi) (#24).  :(

this was a bad AMC year for me, but i defintely think the amc12b was a lot easier

during the test, i thought it felt harder, but i realized that it was a personal block rather than the test itself. i didn't do some of the problems just because i THOUGHT they looked hard

[b]i truly think that problems 1-21 are accessible to most problem solvers - none of them required any real ingenuity beyond just taking the first step. some of them were frustrating computationally, but very doable. if 1-21 are answered, thats a score of 132 already.[/b]

of course i'm not considering the possibility of making dumb mistakes, which is very real. i've sort of lost my rigor in doing competition problems, so i made a lot of dumb mistakes

yes you are right about the computational part. especially number 11 T.T

I think this test was easier in content, but more difficult in the contest setting--that is, with time constraints and just the test environment, the test was much more intimidating.

Earlier I said I thought this was the easiest AMC 12 I've ever seen, but I think a lot of that perception comes from just feeling better that day (missed school to take it, slept well, got a bit of morning exercise, and, of course, loaded up on caffeine and sugar prior to taking).

Add my name to the list of people who hate #11 and #17. However, I think most of these problems are pretty straightforward, including #11 and #17, and also especially, among the later ones, #24, #23, and #20. #20 in particular was astoundingly easy for something that late in the test.

I didn't get it on the test, but I really like problem #22.

I thought it was pretty hard. Maybe I'm just ranting because I somehow managed to get three wrong and ended up with my lowest score since eighth grade. But yeah, I guess the problems themselves weren't too hard, but they just took a while. That millet one, the circular track problem, and the one with the inverse sine (which I got wrong) probably ate up 20+ minutes together. As a result I didn't get to doing 22 (which was pretty easy), and messed up on 23 due to time pressure.

I think the AMC needs more nice problems where you sit a while before you think of the trick, then you're basically done immediately afterward, like the trapezoid one where you construct a bisector and the 3/4+3/4i one. But all those time-wasters like 11, 18, 24, 25 - blah.

i missed #1. don't ask.
i thought #23 and #24 were nice and easy
#20 kinda required the seeing the word "pyramids" to solve (it took a while)
and i agree with you about #11.

Time consuming and *much* easier to make mistakes than the 12A, especially because of the tons of counting problems, which might of helped some people, but not others. 136.5 on A to 97.5 on B.

it is harder, because i practised the 12A paper and get 130s, but in 12B i messed up and only get 97.5


[quote="felixmann"]Let $a,b,c$ be [i]positive[/i] numbers satisfying $ab+bc+1\geq 3ac$. Find the largest number $P$ such that $a+abc+c\geq Pac$.[/quote]

Nobody?

Does somebody want to find out how this inequality can be proved?

[quote="piever"]We can suppose $b=\frac{3ac-1}{a+c}$

The initial inequality becomes $a+c+ac\frac{3ac-1}{a+c}\geq Pac$

Let's suppose $a=c=1$

We get $3\geq P$

Now if we prove $a+c+ac\frac{3ac-1}{a+c}\geq 3ac$ we are ended.

$a+c+ac\frac{3ac-1}{a+c}\geq 3ac\leftrightarrow c^{2}(3a^{2}+1-3a)+c(a-3a^{2})+a^{2}\geq 0$

We have that $3a^{2}+1-3a\geq 0$ so we have to proof that $\Delta\leq 0$

$\Delta =(a-3a^{2})^{2}-4a^{2}(3a^{2}-3a+1)=-3a^{2}(a-1)^{2}\leq 0$[/quote]
Yes, I think it's correct! I have another solution:
$\frac{a}{b}+\frac{c}{b}+ac\geq \sqrt{3\frac{ac}{b}(a+c+\frac{1}{b})}$
further it is easy


What is the product of all real numbers that are doubled when added to their reciprocals?

$ x\plus{}\frac{1}{x}\equal{}2x$
$ \frac{1}{x}\equal{}x$
$ x\equal{}\pm 1$

Product is -1.


What is the greatest three-digit multiple of 33 that can be written
using three different digits?

Just start at the largest and go down
$ 990$, nope
$ \boxed{957}$, yay


Find all triples $(x,y,z)$ of real numbers satisfying: $x^2+y^2+z^2\le 6+min\left(x^2-\frac{8}{x^4},y^2-\frac{8}{y^4},z^2-\frac{8}{z^4}\right)$.

Easy.
Another:
Case 1: $|x|=|y|=|z|$
Case 2: wlog suppose $x>y>z$
...


linear algebra
matrix
algebra
polynomial

[b]Moreover ![/b]
$ A \equal{} E_{ij},i \neq j$ then
$ e^{E_{ij}} \equal{} I_{n} \plus{} E_{ij}$
$ e^{E_{ii}} \equal{} e.E_{ii} ,\forall i \in$ [b]{1,2,3,....,n}[/b]
$ sin{E_{ii}} \equal{} sin1.E_{ii} ,\forall i \in$ [b]{1,2,3,....,n}[/b]
$ cos{E_{ii}} \equal{} cos1.E_{ii} ,\forall i \in$ [b]{1,2,3,....,n}[/b]
$ Sin{E_{ij}} \equal{} E_{ij} ,\forall i \neq j$
$ Cos{E_{ij}} \equal{} I_{n} ,\forall i \neq j$
[b](check for me !)[/b]
[b]"..Do math love math to love native place , love country ..."[/b]

[b]When do i write   i had  log out !![/b]
We can difine :
$ Ch(A)=\frac {e^{A}+e^{-A}}{2}=\sum_{k=0}^{+\infty} {\frac {A^{2k}}{(2k)!}}$
$ Sh(A)=\frac {e^{A}-e^{-A}}{2}=\sum_{k=0}^{+\infty} {\frac {A^{2k+1}}{(2k+1)!}}$
Now , you can prove that :
 $ Cos(A)=\frac {e^{iA}+e^{-iA}}{2},Sin(A)=\frac {e^{iA}-e^{-iA}}{2}$(i is maginary number )
 From you can prove that :
 $ Cos^{2}(A)+Sin^{2}(A)=I_{n},Ch^{2}(A)-Sh^{2}(A)=I_{n}$
and $ Ch(I_{n})=Ch1.I_{n},Sh(I_{n})=Sh1.I_{n}$
as we have exsecise :
   $ Lim_{m \to +\infty} Sin(2.\pi.(2m+1)!.Sh(I_{n}))=0$
 $ Lim_{m \to +\infty}  Cos(2.\pi.(2m+1)!.Sh(I_{n}))=I_{n}$
  $ Lim_{m \to +\infty} Sin(2.\pi.(2m)!.Ch(I_{n}))=0$
 $ Lim_{m \to \+infty} Cos(2.\pi.(2m)!Ch(I_{n}))=I_{n}$
Very much .....
 :roll:  :roll:  :roll:
[b]"...Do math , love math to love native place , love country ..."[/b]

[b]I am sorry , i have edited !!![/b]$ Sin(A)\equal{}\frac {e^{iA}\minus{}e^{\minus{}i
A}}{2i}$
$ ( \frac {e^{iA} \plus{} e^{ \minus{} iA}}{2})(\frac {e^{iB} \plus{} e^{ \minus{} iB}}{2}) \plus{} (\frac {e^{iA} \minus{} e^{ \minus{} iA}}{2i})(\frac {e^{iB} \minus{} e^{ \minus{} iB}}{2i}) \equal{} \frac {e^{i(A \plus{} B)} \minus{} e^{ \minus{} i(A \plus{} B)}}{2}$
$ ( \frac {e^{iA} \plus{} e^{ \minus{} iA}}{2})(\frac {e^{iB} \plus{} e^{ \minus{} iB}}{2}) \minus{} (\frac {e^{iA} \minus{} e^{ \minus{} iA}}{2i})(\frac {e^{iB} \minus{} e^{ \minus{} iB}}{2i}) \equal{} \frac {e^{i(A \minus{} B)} \plus{} e^{ \minus{} i(A \minus{} B)}}{2}$
$ ( \frac {e^{iA} \minus{} e^{ \minus{} iA}}{2i})(\frac {e^{iB} \plus{} e^{ \minus{} iB}}{2}) \plus{} (\frac {e^{iA} \plus{} e^{ \minus{} iA}}{2})(\frac {e^{iB} \minus{} e^{ \minus{} iB}}{2i}) \equal{} \frac {e^{i(A \plus{} B)} \minus{} e^{ \minus{} i(A \plus{} B)}}{2}$
$ ( \frac {e^{iA} \minus{} e^{ \minus{} iA}}{2i})(\frac {e^{iB} \plus{} e^{ \minus{} iB}}{2}) \minus{} (\frac {e^{iA} \plus{} e^{ \minus{} iA}}{2})(\frac {e^{iB} \minus{} e^{ \minus{} iB}}{2i}) \equal{} \frac {e^{i(A \minus{} B)} \minus{} e^{ \minus{} i(A \minus{} B)}}{2}$
$ Cos(A \plus{} B) \equal{} cos(A)cos(B) \minus{} Sin(A)Sin(B) ,Cos(A \minus{} B) \equal{} cos(A)cos(B) \plus{} Sin(A)Sin(B),Sin(A \plus{} B) \equal{} Sin(A)cos(B) \plus{} Cos(A).Sin(B), Sin(A \minus{} B) \equal{} Sin(A)Cos(B) \minus{} Cos(A)Sin(B)$

[b]$ A,B,C \in M_{n} \mathbb{(R)}$,[b]AB=BA=CA[/b]
We have  :[/b]
$ \frac {e^\frac {i(A + B)}{2} - e^\frac { - i(A + B)}{2}}{2i}\frac {e^\frac {i(A - B)}{2} + e^\frac { - i.(A - B)}{2}}{2} = \frac {e^{i.(A + B)} - e^{ - i(A + B)}}{2i}$
Then $ Sin(A + B) = 2Sin(\frac {A + B}{2})Cos(\frac {A - B}{2}) = 2Cos(\frac {A - B}{2})Sin(\frac {A + B}{2})$
Now , Give $ A + B + C = \pi.I_{n}$
from you can find very problem of math :
If $ A + B = \frac {\pi}{2}.I_{n}$ then  ,$ Sin(A) = Cos(B)$
$ A + B + C = \pi.I_{n}$
$ Sin(A) + Sin(B) + Sin(C) = 4Cos(\frac {A}{2})cos(\frac {B}{2})cos(\frac {C}{2})$,
,......,
[b]....We  creat to  find the truth ....[/b]
[b]....Very much ....[/b]

Let we have :
$ \parallel{}e^{A}\parallel{}=\parallel{}\sum_{k=0}^{+\infty} {\frac {A^{k}}{k!}}\parallel{} \leq \sum_{k=0}^{+\infty} {\frac {\parallel{}A\parallel{}^{k}}{k!}} =e^{\parallel{}A\parallel{}}$
$ \parallel{}Sin(A)\parallel{} \leq \sum_{k=0}^{+\infty}{\frac {\parallel{}A\parallel{}^{2k+1}{(2k+1)!}}=Sh(\parallel{}A\parallel{}) ,
\parallel{}Cos(A)\parallel{} \leq \sum_{k=0}^{+\infty}{\frac {\parallel{}A\parallel{}^{2k}{(2k)!}}=Ch(\parallel{}A\parallel{})}}$(I will check after ... when ||.||<1)
....A=B then :
$ cos(2A)=2cos^{2}(A)-I_{n}$
Now $ m \in N^{*}$
use $ Cos((m-1)A)+cos((m+1)A)=2cos(A)cos(mA)$
Let $ T_{0}=I_{n},T_{1}(A)=A$
Then we have chebyshep  polynomical type 1 :
  $ T_{m+1}(A)=2AT_{m}-T_{m-1}(A)$ (check for me !!)
You can do math as :
  $ A_{i} \in M_{n} \mathbb{(R)} ,i \in$ [b]{1,2,3,.....}[/b]
 $ A_{i}A_{j}=A_{j}A_{i}$ $ \forall i,j \in$ {1,2,...}
and $ A_{m+1}=2A^{2}_{m}-I_{n},A_{1}=B \in M_{n} \mathbb{(R)}$
Find $ A_{m}=?$

If $ \parallel{}B\parallel{} \leq 1$ then let $ Cos(C)\equal{}B$ and
$ A_{m}\equal{}cos(2^{m}C)$
 If $ \parallel{}B\parallel{}>1$ then let $ Ch(C)\equal{}B$
and $ A_{m}\equal{}Ch(2^{m}C)$
 :P
[b]"..Do math love math to love native place , love country ..."[/b]
 :roll:

[quote="studen of jean"]
 :P
[b]"..Do math love math to love native place , love country ..."[/b]
 :roll:[/quote]

i am going to delete all of your posts after this one that have silly phrases like this at the end.


See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266215

This time, Peter really sucks, and cannot solve math problems. Peter correctly solves the problem with probability 0. Alan does not have to worry about Peter and solves a problem correctly in the time limit with probability 2/3.  Clearly Alan wins with probability 1. Find the expected number of problems that will be given in the countdown.

I'm just gonna throw a number out there and say -3

wow, i just realized how easy this problem is...

its just $ \frac{4}{2/3}\equal{}\boxed{6}$, right?

yeah....The message is too small. Please make the message longer before submitting.


M is a union of finitely many disjoint intervals with total length > 1. Show that there are two distinct points in M whose difference is an integer.

Just a remark to say that I solved the 2-dimensional version of that problem, with a method which can be generalized to higher dimensions as well than to the linear case.

http://mathlinks.ro/viewtopic.php?t=4916&highlight=torus

Pierre.


Gauss function integration calculus algebra polynomial inequalities

Here's the problem:

Let $f$ be an analytic function in the disk $|z|<1$. Prove that for any $0<r<1$,

$|f(0)|^{2}\leq \frac{1}{\pi r^{2}}\int \int_{\substack{x^{2}+y^{2}\leq r^{2}}}|f(x+iy)|^{2}dx dy$

I'm guessing using Gauss mean value theorem, I don't know if it is the right way to solve it.

There are many approaches, but here's one I like:
Consider this space of analytic functions as a complex inner product space: $\langle f,g\rangle=\frac1{\pi r^{2}}\int\int_{x^{2}+y^{2}\le r^{2}}\overline{f(x+iy)}g(x+iy)\,dx\,dy$
We first show that powers of $z$ are orthogonal: $\langle z^{k},f(z)\rangle= \frac1{\pi r^{2}}\int_{0}^{r}\int_{0}^{2\pi}f(se^{i\theta})s^{k}e^{-ik\theta}\cdot s\,d\theta\,ds$
$=\frac1{\pi r^{2}}\int_{0}^{r}\frac{2\pi s^{2k+1}}{k!}\cdot\frac{k!}{2\pi i}\int_{0}^{2\pi}f(se^{i\theta})\left(se^{i\theta}\right)^{-k-1}\cdot ise^{i\theta}\,d\theta=\frac1{\pi r^{2}}\int_{0}^{r}\frac{2\pi s^{2k+1}}{k!}f^{(k)}(0)\,ds$
$=\frac{r^{2k}}{(k+1)!}f^{(k)}(0)$, using the Cauchy integral formula to evaluate the inner integral. Specializing to make $f$ a power of $z$, $\langle z^{k},z^{k}\rangle=\frac{r^{2k}}{k+1}$ and $\langle z^{k},z^{m}\rangle=0$ if $k\neq m$.

The orthogonal projection of $f$ onto the space of polynomials with degree at most $m$ is its $m$th Taylor polynomial, and we get (a form of) Bessel's inequality: $\|f\|^{2}\ge \sum_{k=0}^{\infty}|f^{(k)}(0)|^{2}\cdot\frac{r^{2k}}{(k+1)(k!)^{2}}$. Your desired inequality comes from the first term $k=0$.
Since the polynomials are dense (the power series of any function $f$ in the space converges to $f$), the inequality is actually an equality.


[quote="ywl"]when a+b+c=3 positive reals
prove that (a^2-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2) obtains maximum on (0,1,2)
this was solved by arqady in the forum
I have come across a four variable inequality
prove that (a^2-ab+b^2)(b^2-bc+c^2)(c^2-dc+d^2)(d^2-ad+a^2)(a^2-ac+c^2)(b^2-bd+d^2)
for a+b+c+d=7 positive reals  obtains maximum on (0,1,3,3)[/quote]

The first problem is posted by a student of Vietnam who get my book. This problem is created by me  :blush:  The second one is very nice, and my first idea is to suppose that $ a=0$ (as what we do for the first one).

The second part is also problem 5.68 in my book.

Are you the real Alain Connes, the Fields medailler ?  :o

EDIT : I saw your last messages, and I don't think so in fact ...  :P

Well.. although references to the problems are nice,
I posted because I could not find a solution.
I do not think I can buy either of the books that are mentioned in my country for
at least a year.
Could anybody solve the second problem??


algorithm
algebra
polynomial
number theory
prime numbers
superior algebra
superior algebra unsolved

I have no idea whether the following is well-known, unsolved or trivial (although I have some serious doubts about the last possibility...): is there a caracterization of those $ n$ for which the ring of algebraic integers of $ \mathbb{Q}(\sqrt[n]{2})$ is $ \mathbb{Z}[\sqrt[n]{2}]$? I know this is the case for $ n$ at most equal to $ 5$, but the proofs I have surely cannot adapt to the general case. Does anyone have some information about this?

Can you show us your proofs for $ n \leq 5$?

You can write down the discriminant of the polynomial. Then you have finitely many prime numbers for which you can locally check normality. Perhaps this simplifies the problem.

There is also an algorithm for determining the ring of integers, see Cohen, Algorithmic Algebraic Number Theory

Oh, sorry, i forgot to answer this. For $ n=3$ it's well known. For n=4, one computes the determinant and finds that it is a power of $ 2$. Now, we use the following useful fact: if $ x$ is an algebraic integer whose minimal polynomial is Eisenstein at $ p$, then if $ x_{0}+x_{1}x+...+x_{n-1}x^{n-1}$ is an algebraic integer and all $ x_{i}$ are rational, then $ p$ does not appear in their denominators. I will leave this as an (easy) exercise. The conclusion is clear now. The same argument works for $ n=5$ two.

Maybe the following is usefull.....

Let $ p$ be a prime number and let $ a$ be squarefree and coprime to $ p$. Then the ring of algebraic integers of $ \mathbb{Q}(\sqrt[p]{a})$ is $ \mathbb{Z}[\sqrt[p]{a}]$ if and only if $ a^{p\minus{}1}\minus{}1$ isn´t divisible by $ p^2$.


Every time i try to open the thread "hacked satilites" in the round table, the computer fails to load the page for about a minute. Ive tried this on three computers and none of them can open the page. I think theres an attachment limit to each post, so it probably wouldnt contain too much data to be that slow. Is this just my computers?

It takes minutes for me to load it (probably the pictures that make it take so long).  It seems like a bunch of garbage if you ask me.  I'll quote his message (without the pictures):

[quote="cansouth"]
1. A private person has hacked several satellites and sends out illegal signals probably all over the world. The signals make people behave physical different than normal.

People use their hands more to touch their faces, cheeks, chins, mouths and their heads in general.

2. Here are some examples of hand movements which have increased in frequency:

The person on the left on the photo and the person on the right on the photo make these kind of hand movements which have increased because of illegal satellite signals:

http://gfx.dagbladet.no/pub/artikkel/4/45/450/450643/saddam1.jpg

For more pictures you can look here:
http://forums.perspectives.com/forums/view_topic.php?id=74734&forum_id=73

3. These kinds of hand movements are not necessarily unnatural, but they appear more often now, because of the illegal satellite signals.

It seems to be that people who relax physically, for example Saddam Hussein and the other accused in the Bagdad trial, react more visibly on the satellite signals which lead to more hand movements to the face.

Also for example politicians which sit in parliaments and listen to speeches, seem to react more visibly on the signals which increase the frequency of these mentioned hand movements.

4. One satellite is enough to cover a large parte of the planet with satellite signals. The person who has hacked satellites, can send much more dangerous signals than he does now. By using satellites, he can probably kill a large part of the population on the earth in a few seconds. To have satellites in space is extremely riscy and dangerous.

5. The person which has hacked satellites, probably uses private equipment to reach the satellites directly. Then the companies, which are technical responsible for the satellites, do not observe the hacking.

6. Normal satellite signals are so large that they do not get into the brains of human beings. But the hacker sends very fine small signals which enter inside peoples brains. These signals then hit nerve systems or functions systems in peoples brains. Then the hacker can make human beings behave different than normal.

7. The illegal satellite signals are so fine (high-resolution) that they enter through the brain cranium and metal.

8. If you observe the increase in the mentioned hand movements, you can report it to:

Interpol
http://www.interpol.net/

The Swedish government
http://www.regeringen.se/

The Norwegian government
http://www.odin.no/

The Spanish government
http://www.la-moncloa.es/default.htm[/quote]

Looks like spam to me... (I can't load it either.)


My little bro has just started learning how to do math... I set him up with AoPS vol. 1 and he is going through it very slowly... I am getting the feeling that he is getting discouraged... How do I encourage him without being to square about it?

IW.OAV

Depends on how old your brother is.  Idea #1: slap him whenever he stops reading.

Okay, that was a joke.  Sort of an ice breaker, if you could call it that.  But here's an idea.  Read the lessons yourself, then teach him without actually having him read the text that much.  I found that worked with teaching my 5th grade brother Algebra, since he found it hard to understand exactly what the text is saying.

The fact is, no matter how good the author is at writing math texts, it's hard to explain something just by writing something down.  You need someone who understands the material, or you need the reader to have all the necessary skills for reading textbooks.

Also, make sure he works out the examples himself, without really looking at the book's solution.  Reading the book and teaching him yourself will help with this because then he won't see the solutions.  By doing the examples instead of just looking and memorizing how to do them, anyone is sure to learn the methods better.

If your concern is just his [i]speed[/i], then don't worry about it.  Just keep trying!  Good luck.

[quote="mathfanatic"]The fact is, no matter how good the author is at writing math texts, it's hard to explain something just by writing something down.  [/quote]

True. It's much harder to understand something just by reading it than it is when you're actually shown the steps one by one. I would recommend you sitting down with him whenever possible.

*gasp* Mathfanatic! And you said I was violent...let's PM the rest of this =P

True, True but I find AoPS almost as good as sitting down with him... I mean I do know how teach kids math (at least at through the calculus level) fairly easily the only problem is that it takes me a lot of tiime to teach him... I basically told him to read AoPS and if he has trouble with it to ask me any questions. And for those bio freaks out there he's also reading the God of Bio texts Biology 6th ed. by Campbell which is also quite tough if I do say so myself. How do you explain electron orbitals and wave numbers and things like that to an 8th grader?

BTW he is in the 8th grade...

*shrug* I learnt orbitals in 9th grade from reading about it online. But that's off-topic. You'll simply have to invest some time for him to really understand. I know - I have a little brother too.

[quote="Idiot_without_a_village"]True, True but I find AoPS almost as good as sitting down with him... I mean I do know how teach kids math (at least at through the calculus level) fairly easily the only problem is that it takes me a lot of tiime to teach him... I basically told him to read AoPS and if he has trouble with it to ask me any questions.[/quote]

You said that he's not doing that well with you just telling him to learn it.  If you want him to learn it right, then you don't have much of a choice - you just have to put in the time and help him out.  If you don't have the time, then I don't know what to say.  I can't think of any strategies that work [i]just like that[/i], without you having to get involved.

And how to explain electron orbitals to an 8th grader?  Well, my science teacher managed to do it (when I was an 8th grader).  Just take your time.  It's pretty easy to understand the concept that electrons aren't in specific paths, but you can find them in certain areas called orbitals.  It's not too bad to say that there are several shapes.  Explain that 2 electrons go to each orbital.  Explain that there are 4 different levels of orbitals (the s, the p, the d, the f).  Explain that the s has only 1 orbital (so 2 electrons), the p has 3 orbitals (so 6 electrons), etc.  Just explain.  It doesn't matter if it takes a little time...it takes everyone a little bit to grasp all that info.  Later, you can relate this to the periodic table.

[quote="Eve"]This probably won't help him get through AoPS (or Campbell!) more easily, but what I did with my sister a couple years ago was to get these powerpoint presentations of mathcounts problems and "play mathcounts" with her.  We raced and kept score and all that, and I think she had a lot of fun with it.  When she didn't know how to do a problem, I'd go over it with her and then usually refer her to a section of AoPS for more practice later.  It didn't last for more than a couple weeks, but it really got her interested and motivated.[/quote]

Sounds like fun...that's what I've been doing with my brother...for science though. We practice Science Bowl questions and he tries to get as many as possible. On that note...I wonder who's done Science Bowl here...

When I was little I used to play mathcounts with myself.  Same idea, only I'd time myself on countdowns, and then I'd count the number of seconds I had left.  After 10 questions or so, I'd add up my score and compare it to my previous scores.   :D

I also did this with Quiz Bowl (mini-College Bowl for middle schoolers).  But I've never done Science Bowl, sadly.  :(

BTW, I like your avatar, Eve.  It reminds me of how I feel some days.  Maybe I said that already?  Yeah, I did...

[quote="mathfanatic"]BTW, I like your avatar, Eve.  It reminds me of how I feel some days.  Maybe I said that already?  Yeah, I did...[/quote]

Did you?  Good thing you said it again because I didn't read it the first time. =)

It's actually my end of proof symbol, except "Hi!" usually says "QED!"  Long story...

[quote="Eve"]Did you?  Good thing you said it again because I didn't read it the first time. =)

It's actually my end of proof symbol, except "Hi!" usually says "QED!"  Long story...[/quote]

That's a nice touch.  After reading a long, mathematically complex proof, one comes to the end and finds a delightful little potato figure announcing joyfully to the world "Quod Erat Demonstratum!"  I like it.

Now you need one that says "Hi!  I like cabbage" or something equally random.  Oh, darn.  The bubble isn't big enough...

mathfanatic -- I'm almost certain it's a pear.

Back to math...

When I try to teach people to enjoy mathematics, I like to give them just the most elegant solutions to problems to start with. Take a look in [i]The Joy of Mathematics[/i] by Theoni Pappas for some inspiration. The mathematics is very simple (I was able to get through it in elementary school), but it is a wonderful way of introducing a love of mathematics to someone who is not so sure yet. I hope it works out well.

[quote="ComplexZeta"]mathfanatic -- I'm almost certain it's a pear.[/quote]

Uhh, yeah - that's what I meant.

Er, okay, I did think it was a potato.  A funny-shaped one.

Anyway, back to the original topic:

I'm not sure if IW.OAV has checked this thread in a while (a while = 2 days) but maybe he's using that time to help out his brother.  Yay!  That would be nice.  My little brother is a social animal, and he does much better when he's with someone rather than when he's sitting alone at a desk puzzling over a book.  Some people are that way, some people aren't.  Some people are both.  Anyway, just some thoughts on what might help your brother to learn.

Yeah Yeah I've been checkin' the thread just haven't had any creative thang to say or not enough time to post a reply.

I appreciate the suggestion, I guess I'll have to make time for to do math and Bio..

[b]We'll see what happens![/b]

IW.OAV

[quote="mathfanatic"]After reading a long, mathematically complex proof, one comes to the end and finds a delightful little potato figure announcing joyfully to the world "Quod Erat Demonstratum!" I like it.[/quote]

Hehe.  You made me laugh. =)

[quote="ComplexZeta"]I'm almost certain it's a pear.[/quote]

Yep, that's correct.  It's a [i]pear on a chair[/i].

[quote="ComplexZeta"] Take a look in [i]The Joy of Mathematics[/i] by Theoni Pappas for some inspiration.[/quote]

Never seen it.  Hmm... now I want it.

[quote]Never seen it. Hmm... now I want it.[/quote]

Isn't it always that way...you hear about something new...you want it... you read it...you add it to your ceiling-high stack...

On a side note, you need a pear that says "Wow!", "Sorry", and other sorts of phrases.  You could even expand this into a giant greeting card market or something.

[quote="ElectricSnowman"]
On a side note, you need a pear that says "Wow!", "Sorry", and other sorts of phrases.  You could even expand this into a giant greeting card market or something.[/quote]

Dude, I should get in on this business deal.... Let's do it! We'd become moghuls!!!

IW.OAV

Okay.  Sounds good to me, IW.OAV.  Only one thing: we have to make it a potato!

[quote="ElectricSnowman"]Only one thing: we have to make it a potato![/quote]

Absolutely not!!  You will do no such thing to my poor little pear!  (Though he does like the idea of being marketed...)

=)

After all, a pear does taste far better than a potato if my opinion has any influence on anything (which it probably doesn't).

[quote="Eve"]Absolutely not!!  You will do no such thing to my poor little pear!  (Though he does like the idea of being marketed...)[/quote]

Awww...well, okay, we'll keep him a pear.

Simon - while [i]raw[/i] pears taste better than raw potatoes, have you ever tried French Fries?  You will definitely change your mind if you ever have...French Fries rule!  In fact, so does all junk food.

Oh, and Eve, you're lucky it's a pear, because all this talk of French Fries is making me hungry enough to fry meself up a batch or two... *snarl* *gulp*

Actually I do not like French fries. But both raw and cooked pears are very tasty.

Oh, too bad.  *slurp* *crunch*  This McDonald's is mightly tasty.  Or as they would say, "Hot, Fresh, and Just For You", a phrase that might be applicable to other objects.

Actually, I'm not eating McDonalds.  I haven't in about 5 years.  But French Fries are pretty good...

after the joy of mathematics there is a follow-up. the more joy of mathematics. or even more joy of mathematics. i forgot. but i havent read either of them. i did glance through both though.

Yes. He wrote a bunch of other books, such as [i]More Joy of Mathematics[/i], [i]The Magic of Mathematics[/i], and a whole bunch more. But in my opinion, [i]The Joy of Mathematics[/i] is the best.

[quote="ComplexZeta"]He[/quote]

im [i]sure [/i]theoni pappas is 'she'.

she also does the calendar thing, but i think thats unfortunately been discontinued

Yeah, I think its a she, I have two of her books.. they're pretty cool, heaps of one or two page facts about fun stuff and things like that.

*nods* Yep. Definitely a she.

http://wideworldpublishing.com/math.html <- List of books she authored.

Oops. Sorry about that. I should try to find some time to learn more about Greek names it appears.


quadratics
function
calculus
derivative
blogs
search
real analysis

Hello,

Say I have a sequence $ b_n \equal{} \binom{3n}{n}$ and the series $ P(x) \equal{} b_0 \plus{} b_1x^2 \plus{} b_2x^4 \plus{} ... \plus{} b_nx^{2n} \plus{} ...$
(it's quadratic generating function)

I want to find a short / closed form of $ P(x)$, and I found that it satisfied this DE:

$ (4\minus{}27x^2)P'' \minus{} 81xP' \minus{}24P \equal{} 0$

(one can check that it's correct by taking derivatives and calculating the coefficient of $ x^{2n}$)

How do I solve this DE?
Or is there a way to derive the expression simply from the definition of $ b_n$ and some clever combinatorial arguments?

The generating function satisfies a cubic equation; see [url=http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/]this blog post[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=920316419&t=281653]these computations[/url].  So one can apply the cubic formula.

hello, the solution is
$ y(x)\equal{}\frac{C_1(12x\sqrt{3}\plus{}4\sqrt{\minus{}4\plus{}27x^2})^{1/3}}{\sqrt{\minus{}4\plus{}27x^2}}\plus{}\frac{C_2}{\sqrt{27x^2\minus{}4}(12x\sqrt{3}\plus{}4\sqrt{\minus{}4\plus{}27x^2})^{1/3}}$
Sonnhard.

Hello,

Thank you. How did you arrive with that solution? If you used some kind of program, which one is it?

Any computer algebra system can solve cubic equations.  I believe Dr Graubner uses Mathematica.


Duis te feugifacilisi. Duis autem dolor in hendrerit in vulputate

 velit esse molestie consequat, vel illum dolore eu feugiat nulla
       to be?          he/ she/ it follows?
 facilisis at vero eros et accumsan et iusto odio dignissim qui

blandit praesent luptatum zzril delenit au gue duis dolore te

feugat nulla facilisi.

Ut wisi(could it be visi?) enim ad minim veniam, quis nostrud exerci taion
How to see about to little indulgence, who....
ullamcorper suscipit lobortis nisl ut aliquip ex en commodo

consequat. Duis te feugifacilisi per suscipit lobortis nisl ut aliquip

ex en commodo consequat. Ut wisi enim ad minim veniam, quis

nostrud exerci taion ullamcorper suscipit lobortis nisl ut aliquip

ex en commodo consequat.

Duis te feugifacilisi per suscipit lobortis nisl ut aliquip ex en

commodo consequat.Lorem ipsum dolor sit amet, consectetuer

adipiscing elit, sed diem nonummy nibh euismod tincidunt ut

 lacreet dolore magna aliguam erat volutpat. Ut wisis enim ad

minim veniam, quis nostrud exerci tution ullamcorper suscipit

lobortis nisl ut aliquip ex ea commodo consequat

EDIT: I edited it since if it is latin...i admit i have to get out my lexicon for this...

well that is what i thought with quis but all the alphabet lacked j and w? and z was to the end of the republic! i thought it was one of the languages which was taken from latin...maybe w and v are the same?

People ARE capable of making typographical errors, you know. Sometimes, foreign language speakers are perplexed when this happens. A good example is the internet slang term "por k" for "Por que" in spanish. (Maybe Spanglish?) (Confused the heck out of me, this one did)

</offtopic>

:maybe:

[quote] Duis te feugifacilisi per suscipit lobortis nisl ut aliquip [/quote]

Try this in Google and you find some suspicious hits.

This looks like fake Latin.

labortis? (laboratis = you work?)
nlis?
aliquip?
duis(deus =god?)


geometry
rectangle
combinatorics open
combinatorics

I came up with this new problem when I try to select icons on my desktop. I myself isn't a nice combinatorician, so can anybody help me to solve this?

For a $ n\times n$ table, find the maximum number $ a$ of cells that can be coloured such that the following is satisfied:
For all $ 0\leq x\leq a$, there always exist a rectangular region in the $ n\times n$ table, that contains $ x$ coloured cells.

I have some very weak result for lower bound:
If $ n\equal{}2k\plus{}1$, then $ a \ge 2k^2\plus{}4k\plus{}1$;
if $ n\equal{}2k$, then $ a \ge 2k^2\plus{}k$.

The construction is as follow (I only do the case of odd $ n$):
Colour the first $ k\plus{}1$ th row and the first column of the table, notice that for every $ 0\le x \le 2k^2\plus{}4k\plus{}1$, $ x\equal{}(k\plus{}1)m\plus{}r$ where $ 0 \le r < k\plus{}1$. If $ m>0$, then we can draw the rectangle from row $ 1$ to row $ k\plus{}r\plus{}1$ and from column $ 1$ to $ m$. If $ m\equal{}0$, simply draw a $ x \times 1$ rectangle in the first column.

[quote="mszew"]May it helps some boundaries:

1	1					a=3 (clearly the max)
1

1	1	1				a=7 (max, because there is no way for (x,a)=(7,8 or 9)
1	1	1
1

1	1	1	1
1	1	1	1			a=12
1	1	1	1
1	1

1	1	1	1	1
1	1	1	1	1
1	1	1	1	1		a=18
1	1	1	1	1
1	1[/quote]

It seems to me that you've misunderstood the question. The number of coloured cell also has to be $ a$.


$ x^{81}$ is divided by $ (x^2\plus{}1)(x^2\plus{}x\plus{}1)$, find the remainder.

:whistling:

[hide]$ x^4\minus{}1 \equiv 0 \implies x^4 \equiv 1 \implies x^{81} \equiv x \pmod{x^2\plus{}1}$
$ x^3\minus{}1 \equiv 0 \implies x^3 \equiv 1 \implies x^{81} \equiv 1 \pmod{x^2\plus{}x\plus{}1}$

The remainder $ R(x)$ is a polynomial of degree at most $ 3$, and therefore satisfies:

$ R(x) \equal{} (ax\plus{}b)(x^2\plus{}1) \plus{}x$
$ R(x) \equal{} (cx\plus{}d) (x^2\plus{}x\plus{}1)\plus{}1$

Expanding and equating: $ (a)x^3 \plus{} (b) x^2 \plus{} (a\plus{}1) x\plus{}(b) \equal{} (c) x^3 \plus{} (c\plus{}d) x^2\plus{}(c\plus{}d) x\plus{}(d\plus{}1)$

Equate coefficients: $ a\equal{}c\; , \; b\equal{}d\plus{}1\; , \; b\equal{}c\plus{}d\; , \; a\plus{}1\equal{}c\plus{}d$

$ \therefore (a,b,c,d)\equal{}(1,2,1,1)$.

$ \implies \boxed{R(x) \equal{} (x\plus{}2)(x^2\plus{}1)\plus{}x \equal{} x^3\plus{}2x^2\plus{}2x\plus{}2}$[/hide]

[quote="vinskman"]$ x^{81}$ is divided by $ (x^2 \plus{} 1)(x^2 \plus{} x \plus{} 1)$, find the remainder. [/quote]
[hide="May be another approach"]

consider $ x^{81} \minus{} 1 \equal{} (x^2 \plus{} 1)(x^2 \plus{} x \plus{} 1)q(x) \plus{} R(x)$
            $ R(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$

suppose $ p$ is a complex root of $ x^2 \plus{} x \plus{} 1$ , then
            $ p^2 \plus{} p \plus{} 1 \equal{} 0 \implies p^3 \equal{} 1$ & $ p^2 \equal{} \minus{} p \minus{} 1$
Hence $ p^{81} \minus{} 1 \equal{} 0 \plus{} ap^3 \plus{} bp^2 \plus{} cp \plus{} d(x)\implies 0 \equal{} a \plus{} b( \minus{} p \minus{} 1) \plus{} cp \plus{} d$
$ \implies a \plus{} d \minus{} b \equal{} 0 , c \minus{} b \equal{} 0\cdots(1)$

suppose $ t$ is a complex root of $ x^2 \plus{} 1$ , then
 $ t^2 \plus{} 1 \equal{} 0 \implies t^4 \equal{} 1$ & $ t^2 \equal{} \minus{} 1 ,t^3 \equal{} \minus{} t$
Hence $ t^{81} \minus{} 1 \equal{} 0 \plus{} at^3 \plus{} bt^2 \plus{} ct \plus{} d(x)\implies t \minus{} 1 \equal{} \minus{} at \plus{} \minus{} b \plus{} ct \plus{} d$

$ \implies c \minus{} a \equal{} 1 , d \minus{} b \equal{} \minus{} 1\cdots(2)$

From (1),(2) : $ a \equal{} 1,b \equal{} 2,c \equal{} 2,d \equal{} 1,R(x) \equal{} x^3 \plus{} 2x^2 \plus{} 2x \plus{} 1$

Thus, the remainder for $ x^{81} \equal{} R(x) \plus{} 1 \equal{} \boxed{x^3 \plus{} 2x^2 \plus{} 2x \plus{} 2}$
[/hide] :welcomeani:


Somehow, I can't access the latex help and downloadable software. Is it just me or is there something wrong with most of the links above? [b](LaTeX Help, Classes, Books, Classroom)[/b]

Does anyone know where I could get some free latex software?

Try using IE if you are using mozilla.  Otherwise, if you follow directions [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php[/url] to the word, then you should be okay.

hope this helps


inequalities
algorithm
limit
rearrangement inequality

Find the least positive integer whose sum of digits is $ 2009$.

To minimize the number of digits, we set all but the first digits to be 9. Next, we find that 2007 is a multiple of 9. So there is 2 left over. To minimize the value of the number, we set the first digit equal to 2 and the rest equal to 9. Then we get 2 followed by (2007/9) 223 nines.

2999999999999999...9999999 is our answer with a total of 224 digits.

I encourage you to think more carefully about your proof. Why does setting as many digits as possible make the number as small as possible?

Think about it. Which one is bigger, 999 or 1000? The sum of the digits of 999 is a lot bigger, but the value is only one less than 1000. Setting the digits to 9 gets the most possible sum from each digit, while minimizing the number of digits and thus the value.

I would argue along the lines of: show that the least number of digits that we can have is $ 224$ (and showing that is attainable). Then using the rearrangement inequality to show that 299...999 is the smallest one.

It is like squeezing the juice out of a lemon. You can't get anymore than 9 from each digit. Also consider this:

$ a\times b \equal{} 2009$. If a is the number of digits, and b is the digit value, as b goes to infinity, a goes to 0. So b has to be the largest.

:D

I guess the idea here is to get your mind around the fact that the greedy algorithm isn't always best: http://en.wikipedia.org/wiki/Greedy_algorithm

Is it wrong? No.

Is it rigorous? No.

What Altheman is trying to get you to do is be rigorous. A solution with the words "think about it" is extremely hand-wavy.

Okay. I will try my best:

There are two parts to this explanation/proof/solution. Number one, minimizing the number of digits minimizes the value. Number two, setting as many digits as possible to 9 minimizes the number of digits.

$ \mathbf{1)}$ Suppose a number, Number 1, has $ n$ digits. The number is in the form $ abcdefghi...$ where the letters are digits. The value of the number is $ a(10^{n\minus{}1})\plus{}b(10^{n\minus{}2})\plus{}c(10^{n\minus{}3})\plus{}\dots\plus{}n(10^0)$ (by definition of base ten). Say there is another number, Number 2, that has $ n\plus{}x$ digits, in the form $ opqrstu...$. The value of the number is $ o(10^{n\plus{}x\minus{}1}\plus{}p(10^{n\plus{}x\minus{}2}\plus{}\dots$. We have to prove that Number 1 is smaller than Number 2. If we prove this when x is 1, we can prove it for x is 2, when x is 3, ... by induction. We will look at just the first few digits. Number 1$ \equal{}a(10^{n\minus{}1})\plus{}b(10^{n\minus{}2})\plus{}...$ and Number 2$ \equal{}o(10^{n\plus{}1\minus{}1}\plus{}p(10^{n\plus{}1\minus{}2}\plus{}q(10^{n\plus{}1\minus{}3}\plus{}...$ (we set $ x$ equal to 1). For Number 1 to be bigger, $ a(10^{n\minus{}1})$ has to be bigger than $ o(10^{n\plus{}x\minus{}1}\plus{}p(10^{n\plus{}x\minus{}2}$. We know $ a$ and $ o$ are nonzero (or else it wouldn't be an $ n$ or $ n\plus{}x$ digit number). So we know the smallest value of $ o$ is 1. Now, since $ p$ could be zero (it doesn't matter), we have to see if $ a(10^{n\minus{}1})$ can be bigger than than $ 10^n$. For this to be true, $ a$ must be 10. It can not, because a single digit has to be a one-digit number. Therefore, it is proved.

$ \mathbf{2)}$ Suppose a number has $ n$ digits and each digit is $ x$. The sum of the digits of the number is $ x\plus{}x\plus{}x\plus{}\dots\plus{}x\equal{}n\times x$. We will set this to $ a$. So $ a\equal{}nx$. We want to minimize $ n$ while keeping $ a$ the same. So $ n$ and $ x$ are inversely proportional. This means $ \displaystyle\lim_{x\to\infty}n\equal{}0$. In other words, the bigger $ x$ gets, the smaller $ n$ gets. This is exactly what we want to do, make $ n$ the smallest. So we have to maximize $ x$. The largest digit is $ 9$, so $ x\equal{}9$. This proves that, if we want a sum (of digits) that is a multiple of 9 ($ a$), we will have a minimum of $ \dfrac{a}{9}$ digits.

Phew! That took me about half an hour.

[hide="how about..."]
Suppose a number had <224 digits. Then the maximum possible sum is 223*9=2007. Contradiction. Now assume the number has 224 digits. we see that 2999...9999 has a digital sum of 2009. We show there is no smaller number. suppose there was. If the first digit were 2, the sum would be less than 2+223*9=2009. Contradiction. Now suppose the first digit is 1. Then the sum is less than or equal to 1+223*9=2008. Contradiction. Thus, 2999...999 is the smallest.
[/hide]


1337... are you going to organize another one any time soon, because the competition is a week away. o_O. . . Also, maybe you can get some previous national rounds, and make up some hard ones.  Thanks !! :D

If you want to send a message directly to a specific user, please use the Private Message feature.  There is an option for it toward the top left hand corner of the website.

This seems like a very interesting type of Math Jam I would love to participate.

I feel kind of stupid asking this, but what's the AIM thing? (Not the Math Jam)

http://en.wikipedia.org/wiki/AOL_Instant_Messenger

so many people ask this now...


[quote="fskilnik"]Hi there!

For any given rational N/M, is it possible to decide (without doing the obvious calculation) whether it is a terminating decimal or not?

Example: $ 1/2$ is a terminating decimal, $ 1/3$ is not.

[You could argue that the 2 possible remainders when a certain integer is divided by 2 are zero and one and both of them result in terminating decimals; as far as the division by three is concerned, there are 3 possible remainders and just one of them will result in terminating decimals, i.e., zero!  My question is, what can be done/said "in general"?][/quote]

It will only terminate it if the primes in the denominator are only 2 and 5.

To generalize, the fraction $ \frac{m}{n}$ terminates in base $ b$ if and only if there exists a positive integer $ k$ such that $ n\mid b^{k}$ (equivalently, all prime divisors of $ n$ are also prime divisors of $ b$).

Follow-up: Determine the criteria for a fraction to be purely periodic (i.e. the period starts immediately after the point) and impurely periodic (i.e. the period starts after a one-time-occuring sequence of digits).


[b]hi!  :D  :lol:  :) happy 8-3 [/b]
[hide]many wish for you :) [/hide]

Its already June but cheers to all the women(or girlz) in our community  :rotfl:


Let $ a$ , $ b$ , $ c$ be three positive constants and let $ x$ , $ y$ be two positive variables for which $ x \plus{} y \equal{} a$ .

Ascertain ([b]without derivatives[/b] !) the minimum of the expression $ E(x,y) \equal{} b^2x^3 \plus{} c^2y^3$ .

See and [url=http://www.mathlinks.ro/viewtopic.php?t=193496]here[/url] .

By Holder's Inequality, $(b^2x^3+c^2y^3)\left(\frac{1}{b}+\frac{1}{c}\right)^2\geq (x+y)^3=a^3$. Thus, the minimum value of $b^2x^3+c^2y^3$ is $a^3\left(\frac{1}{b}+\frac{1}{c}\right)^{-2}=\frac{a^3b^2c^2}{(b+c)^2}$.


1. Simplify the following expression for $ x\neq \pm{1}$. Express your answer as a common fraction in terms of $ x$.
$ \frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{\frac{x+1}{x-1}+\frac{x-1}{x+1}}$
[hide="Answer"]$ \frac{2x}{x^2+1}$[/hide]
2.Suppose that $ ^*(n)={[{n-2,n+2,2n,\frac n2}]}$. For example, $ ^*(6)={[{3,4,8,12}]}$ For how many distinct integers $ n$ does $ ^*(n)$ have exactly three distinct elements?
[hide="Answer"]5 integers[/hide]
3.The sum of two fraction is $ \frac{11}{12}$ and their product is $ \frac16$. What is the lesser of the two fractions? Express your answer as a common fraction.
[hide="Answer"]$ \frac{1}{4}$[/hide]

[hide="3"]Let them be x and y. $ x \plus{} y \equal{} \frac {11}{12}$ and $ xy \equal{} \frac {1}{6}$, and y<x.
Now, $ x \equal{} \frac {11}{12} \minus{} y$, so we substitute that in. $ y(\frac {11}{12} \minus{} y)$$ \equal{} \frac {1}{6}$. By multiplying by 12, we have $ y(11 \minus{} 12y) \equal{} 2$, so we solve and get $ \frac {1}{4}, \frac {2}{3}$. Since we want the smaller, it's $ \frac {1}{4}$.[/hide]

[hide="1"]
$ \frac {\frac {x \plus{} 1}{x \minus{} 1} \minus{} \frac {x \minus{} 1}{x \plus{} 1}}{\frac {x \plus{} 1}{x \minus{} 1} \plus{} \frac {x \minus{} 1}{x \plus{} 1}}$

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

$ \frac {x^2 \plus{} 2x \plus{} 1 \minus{} x^2 \plus{} 2x \minus{} 1}{x^2 \plus{} 2x \plus{} 1 \plus{} x^2 \minus{} 2x \plus{} 1}$

$ \frac {4x}{2x^2 \plus{} 2}$

$ \boxed{\frac {2x}{x^2 \plus{} 1}}$
[/hide]

[hide="2"]
In order for there to be not 4 terms, but exactly 3 terms, one of the following must be true:
$ n \minus{} 2 \equal{} n \plus{} 2$, Impossible
$ n \minus{} 2 \equal{} 2n$, $ n \equal{} \minus{} 2$
$ n \minus{} 2 \equal{} 2$, $ n \equal{} 4$
$ n \plus{} 2 \equal{} 2n$, $ n \equal{} 2$
$ n \plus{} 2 \equal{} 2$, $ n \equal{} 0$
$ 2n \equal{} 2$, $ n \equal{} 1$

None of the solutions overlap, so there are $ \boxed{5}$ integers.
[/hide]

[hide-faster solution for 1]$ (x \plus{} 1)^2 \minus{} (x \minus{} 1)^2$ factors into $ ((x \plus{} 1) \plus{} (x \minus{} 1))((x \plus{} 1) \minus{} (x \minus{} 1)) \equal{} (2x)(2) \equal{} 4x$.[/hide]


function
inequalities proposed
inequalities

Let $ x_1,x_2,...,x_n$ ($ n\ge 3$) be positive real numbers such that $ x_1,x_2,...,x_n \ge \frac {n\minus{}3}{n\minus{}2}$ and $ x_1x_2...x_n\equal{}1$. Prove that

$ \frac 1{x_1\plus{}2}\plus{}\frac 1{x_2\plus{}2}\plus{}...\plus{}\frac 1{x_n\plus{}2} \le \frac n{3}$.

I think we can apply your LCF theorem , just consider the function $ f(x) \equal{} \frac {1}{3} \minus{} \frac {1}{e^ {\minus{} x} \plus{} 2}$ , am I right ? :(


calculus
integration
function
trigonometry
complex analysis

Here's a cool problem from my book (Complex Variables and Applications, Brown and Churchill):

By considering the function $f(z) = \frac{e^{az}}{z}$, and the contour $C: z=e^{i\theta}, \ \ (-\pi\le\theta\le\pi)$, calculate $\int_{0}^\pi e^{a\cos\theta}\cos(a\sin\theta) \ d\theta$

[quote="PTynan89"]Is the answer $\pi$?[/quote]

 :yup:

Indeed, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70544]this topic[/url].

[i]Edit.[/i] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58309]Here's[/url] another topic.

Thanks, I guess this is a pretty famous result.  Anyway, I really liked Kent's solution, very "outside the box."


Hey,
Let's try to see how many people we can get on FTW at once! Just post a link to this page in your sig and be on FTW at 9:00 EST (8:00 Central, 7:00 Mountain, 6:00 pacific) on Friday October 17.
That's it!


inequalities
blogs
inequalities proposed

If $ a,b,c$ are nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}1,$ then
$ \frac{1}{\sqrt{b^2\plus{}bc\plus{}c^2}}\plus{}\frac{1}{\sqrt{c^2\plus{}ca\plus{}a^2}}\plus{}\frac{1}{\sqrt{a^2\plus{}ab\plus{}b^2}} \ge 2\plus{}\frac{1}{\sqrt{3}}.$

[quote="can_hang2007"]If $ a,b,c$ are nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1,$ then
$ \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt {c^2 \plus{} ca \plus{} a^2}} \plus{} \frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}} \ge 2 \plus{} \frac {1}{\sqrt {3}}.$[/quote]
See here for my proof: http://canhang2007.wordpress.com/2009/11/09/inequality-58-v-q-b-can/

:( Could someone help to type Can's proof here,I still can't open the hint.Thanks very much.

Sure no problem.

Firstly, we claim that for any $ \displaystyle x,y\ge z\geq 0$, the inequalities holds

$ \displaystyle \frac {xy \plus{} yz \plus{} zx}{y^2 \plus{} yz \plus{} z^2} \ge \frac {x \plus{} z}{y \plus{} z}$,

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

The first inequality is equivalent to

$ \displaystyle \frac {xy \plus{} yz \plus{} zx}{x \plus{} z} \minus{} y\geq \frac {y^{2} \plus{} yz \plus{} z^{2}}{y \plus{} z} \minus{} y$,

or

$ \displaystyle \frac {xz}{x \plus{} z}\geq \frac {z^{2}}{y \plus{} z}$,

which is true because $ \displaystyle x(y \plus{} z) \minus{} z(x \plus{} z) \equal{} x(y \minus{} z) \plus{} z(x \minus{} z)\geq 0$.

The second inequality is equivalent to

$ \displaystyle \frac {(x \plus{} z)^{2} \plus{} (x \plus{} z)(y \plus{} z) \plus{} (y \plus{} z)^{2}}{x^{2} \plus{} xy \plus{} y^{2}} \minus{} 1\geq \frac {(x \plus{} z)(y \plus{} z)}{xy \plus{} yz \plus{} zx} \minus{} 1$,

or

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

which is true because

$ \begin{aligned} \displaystyle (xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z) & \geq z(x \plus{} y)(x \plus{} y \plus{} z)\geq z(x \plus{} y)^{2} \\
& \geq z(x^{2} \plus{} xy \plus{} y^{2}). \end{aligned}$

Turning back to our problem, assume that $ \displaystyle c\equal{}\min \{a,b,c\}$. By the above two inequalities, we have

$ \displaystyle \frac {1}{\sqrt {b^{2} \plus{} bc \plus{} c^{2}}} \equal{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{b^{2} \plus{} bc \plus{} c^{2}}}\geq \sqrt {\frac {a \plus{} c}{b \plus{} c}}$,

$ \displaystyle \frac {1}{\sqrt {a^{2} \plus{} ac \plus{} c^{2}}} \equal{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^{2} \plus{} ac \plus{} c^{2}}}\geq \sqrt {\frac {b \plus{} c}{a \plus{} c}}$,

and

$ \displaystyle \frac {1}{\sqrt {a^{2} \plus{} ab \plus{} b^{2}}} \equal{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^{2} \plus{} ab \plus{} b^{2}}}\geq \sqrt {\frac {(a \plus{} c)(b \plus{} c)}{(a \plus{} c)^{2} \plus{} (a \plus{} c)(b \plus{} c) \plus{} (b \plus{} c)^{2}}}$.

Therefore, it suffices to show that

$ \displaystyle \sqrt {\frac {a \plus{} c}{b \plus{} c}} \plus{} \sqrt {\frac {b \plus{} c}{a \plus{} c}} \plus{} \sqrt {\frac {(a \plus{} c)(b \plus{} c)}{(a \plus{} c)^{2} \plus{} (a \plus{} c)(b \plus{} c) \plus{} (b \plus{} c)^{2}}}\geq 2 \plus{} \frac {1}{\sqrt {3}}$.

Setting $ \displaystyle u \equal{} \sqrt {\dfrac{a \plus{} c}{b \plus{} c}} \plus{} \sqrt {\dfrac{b \plus{} c}{a \plus{} c}}\geq 2$, this inequality becomes

$ \displaystyle u \plus{} \frac {1}{\sqrt {u^{2} \minus{} 1}}\geq 2 \plus{} \frac {1}{\sqrt {3}}$.

We have

$ \displaystyle \begin{aligned} u \plus{} \frac {1}{\sqrt {u^{2} \minus{} 1}} \minus{} 2 \minus{} \frac {1}{\sqrt {3}} & \equal{} u \plus{} \frac {\sqrt {u^{2} \minus{} 1}}{u^{2} \minus{} 1} \minus{} 2 \minus{} \frac {1}{\sqrt {3}}\geq u \plus{} \frac {\sqrt {3}}{u^{2} \minus{} 1} \minus{} 2 \minus{} \frac {1}{\sqrt {3}} \\
& \equal{} (u \minus{} 2) \minus{} \frac {u^{2} \minus{} 4}{\sqrt {3}(u^{2} \minus{} 1)}\geq (u \minus{} 2) \minus{} \frac {2(u^{2} \minus{} 4)}{3(u^{2} \minus{} 1)} \\
& \equal{} \frac {(u \minus{} 2)(3u^{2} \minus{} 2u \minus{} 7)}{3(u^{2} \minus{} 1)}\geq 0.\end{aligned}$

So, the last inequality is true and our proof is completed. On the assumption $ \displaystyle c \equal{} \min \{x,y,z\}$, equality holds for $ \displaystyle a \equal{} b \equal{} 1 and \displaystyle c \equal{} 0.$

A similar result
$ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$,then
\[ \frac{1}{\sqrt{a^2\plus{}ab\plus{}b^2}}\plus{}\frac{1}{\sqrt{b^2\plus{}bc\plus{}c^2}}\plus{}\frac{1}{\sqrt{c^2\plus{}ca\plus{}a^2}}\geq 4\plus{}\frac{2}{\sqrt{3}}\]

[quote="Tourish"]A similar result
$ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$,then
\[ \frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt {c^2 \plus{} ca \plus{} a^2}}\geq 4 \plus{} \frac {2}{\sqrt {3}}\]
[/quote]
This inequality was presented in my seminars (about inequalities) one year ago. I proved it by mixing variables method. You can download my seminar lecture here: http://canhang2007.files.wordpress.com/2009/10/donbienthuatru.pdf, where this inequality was presented as Example 1.

In case, you cannot visit my blog and download the file. I upload my file here:

I am doubtful about this, but can we WLOG assume $ a\leq b\leq c$ (since the inequality is cyclic)? If so, then here is my proof:
Let $ a\leq b\leq c$. Then we have that
\[ a^2\plus{}ab\plus{}b^2 \equal{} 1\plus{}a^2\minus{}bc\plus{}b^2\minus{}ca \equal{} 1\plus{}a(a\minus{}c)\plus{}b(b\minus{}c) \leq 1\]
\[ c^2\plus{}ca\plus{}a^2 \equal{} 1\plus{}c^2\minus{}ab\plus{}a^2\minus{}bc \equal{} 1\plus{}a(a\minus{}b)\plus{}c(c\minus{}b) \leq 1\]
So we have $ \frac 1{\sqrt{a^2\plus{}ab\plus{}b^2}}\plus{}\frac 1{\sqrt{c^2\plus{}ca\plus{}a^2}}\geq 2$
So we are required to prove that $ 3\geq b^2\plus{}bc\plus{}c^2$
Which is obvious since $ ab\plus{}bc\plus{}ca\equal{}1 \implies a,b,c\leq 1 \implies b^2\plus{}bc\plus{}c^2\leq 3$
?  :?:  :?:  Can anyone please tell me where I got wrong?   :?:
( Sorry for spam ) :(

[quote="Mr.Bikrone"][quote="Potla"]
Which is obvious since $ {ab \plus{} bc \plus{} ca \equal{} 1 \implies a,b,c\leq 1}$
[/quote]
you got wrong at this :D[/quote]
Well, yes, maybe. I thought the condition was $ a\plus{}b\plus{}c\equal{}1$; but in case of $ ab\plus{}bc\plus{}ca\equal{}1$, I think this is not forced :(
Say, $ a\geq 1$ and $ a(b\plus{}c)\plus{}bc\equal{}1$; we might let $ (a,b,c)\equal{}\left(\frac 32, \frac 23 , 0 \right)$ which satisfies the condition $ ab\plus{}bc\plus{}ca\equal{}1$  :(
Really sorry for this mistake  :( :( .


I have got the following inequality:

if $ a$, $ b > 0$ and $ a^2 \plus{} b^2 \equal{} 2$ then

$ a^{a} \plus{} b^{b} \geq 2$

[quote="kunny"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=243050]Simillar problem[/url][/quote]
also I have:
if $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ and $ a,b,c > 0$ then

$ a^{a} \plus{} b^{b} \plus{} c^{c} \geq 3$.

[quote="Nak"]
also I have:
if $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ and $ a,b,c > 0$ then

$ a^{a} \plus{} b^{b} \plus{} c^{c} \geq 3$.[/quote]

I have a similar problem: Given $ a, b, c \geq\ 0$ satisfy $ a^2\plus{}b^2\plus{}c^2\equal{}a\plus{}b\plus{}c$
Prove that: $ a^ab^bc^c \leq\ 1$
 :)

To your problem, nguoivn
$ a^ab^bc^c \le \left(\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}\right)^{a\plus{}b\plus{}c}$
We are done!


let $ A_n \equal{} 4^{n\minus{}1}\minus{}n$

we have $ (A_2,A_3,A_5)\equal{}(2, 13, 251 )$ - primes.
are there any other primes in this sequence?
is the number of primes in $ (A_n)$ finite?

Yes there are other primes, e.g. for n=17, 945, 33479 (thanks to Mathematica).
It's the sort of sequence where you would "expect" infinitely many primes but there seems at present no hope of proving this.