[quote="Vasc"]If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then

$ \frac 1{1 \plus{} a \plus{} b^2} \plus{} \frac 1{1 \plus{} b \plus{} c^2} \plus{} \frac 1{1 \plus{} c \plus{} a^2}\le 1$.[/quote]
We can rewrite this ineq that:
$ \sum_{cyc} a^2\plus{}\sum_{cyc} a \leq \sum_{cyc} a^3b\plus{}\sum_{cyc} a^2b^3$
By AM-GM we can easy to prove that $ \sum_{cyc} a^3b\equal{}\sum_{cyc} \frac{a^2}{c} \geq \sum_{cyc} a$
Now I am trying proving that $ \sum_{cyc} a^2b^3\equal{}\sum_{cyc} \frac{ab^2}{c} \geq \sum_{cyc} a^2$

[quote="chien than"]
Now I am trying proving that $ \sum_{cyc} a^2b^3 \equal{} \sum_{cyc} \frac {ab^2}{c} \geq \sum_{cyc} a^2$[/quote]
Which obviously wrong! :wink:

[quote="chien than"][quote="Vasc"]If $ a,b,c$ are positive numbers such that $ abc = 1$, then

$ \frac 1{1 + a + b^2} + \frac 1{1 + b + c^2} + \frac 1{1 + c + a^2}\le 1$.[/quote]
We can rewrite this ineq that:
$ \sum_{cyc} a^2 + \sum_{cyc} a \leq \sum_{cyc} a^3b + \sum_{cyc} a^2b^3$
[/quote]

$ \frac{3\sum{a^3b}+\sum{a^2c^3}}{4} \geq \sum{a^2}$

$ \frac{\sum{a^3b}+2\sum{a^3c^2}+\sum{b^3a^2}}{4} \geq \sum{a^{\frac{7}{4}} \geq \sum{a}}$

That's all :)

[quote="chien than"]We can rewrite this ineq that:
$ \sum_{cyc} a^2 \plus{} \sum_{cyc} a \leq \sum_{cyc} a^3b \plus{} \sum_{cyc} a^2b^3$
[/quote]
Can we use SOS ,SS with this inequality?

[quote="cefer"]
$ \frac {3\sum{a^3b} \plus{} \sum{a^2c^3}}{4} \geq \sum{a^2}$
[/quote]
You means
$ \frac {3\sum{a^3b} \plus{} \sum{a^3c^2}}{4} \geq \sum{a^2}$,
which probably is wrong.

[quote="Vasc"]If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then

$ \frac 1{1 \plus{} a \plus{} b^2} \plus{} \frac 1{1 \plus{} b \plus{} c^2} \plus{} \frac 1{1 \plus{} c \plus{} a^2}\le 1$.[/quote]
Substitute $ a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}(x,y,z>0)$, it is equivalent to the following inequality

$ \frac{yz^2}{y^3\plus{}z^2(x\plus{}y)}\plus{}\frac{zx^2}{z^3\plus{}x^2(y\plus{}z)}\plus{}\frac{xy^2}{x^3\plus{}y^2(z\plus{}x)}\leq 1$.

[hide="Proof."]$ 1\minus{}\frac{yz^2}{y^3\plus{}z^2(x\plus{}y)}\plus{}\frac{zx^2}{z^3\plus{}x^2(y\plus{}z)}\plus{}\frac{xy^2}{x^3\plus{}y^2(z\plus{}x)}$

$ \equiv\frac{F(x,y,z)}{\left[y^3\plus{}z^2(x\plus{}y)\right]\left[z^3\plus{}x^2(y\plus{}z)\right]\left[x^3\plus{}y^2(z\plus{}x)\right]}$.

$ F(x,y,z)\equal{}F(x,x\plus{}s,x\plus{}t)$

$ \equal{}9\left(s^2\minus{}st\plus{}t^2\right)x^7\plus{}9\left(2s^3\plus{}s^2t\plus{}2t^3\right)x^6\plus{}3\left(s\plus{}t\right)\left(5s^3\plus{}6s^2t\plus{}5t^3\right)x^5$

$ \plus{}\left(6s^5\plus{}28s^4t\plus{}40s^3t^2\plus{}17s^2t^3\plus{}17st^4\plus{}6t^5\right)x^4$

$ \plus{}\left(s^6\plus{}9s^5t\plus{}29s^4t^2\plus{}26s^3t^3\plus{}7s^2t^4\plus{}9st^5\plus{}t^6\right)x^3$

$ \plus{}st\left(s^5\plus{}6s^4t\plus{}20s^3t^2\plus{}4s^2t^3\plus{}3st^4\plus{}2t^5\right)x^2\plus{}s^2t^3\left(4s^3\plus{}5s^2t\minus{}st^2\plus{}t^3\right)x$

$ \plus{}s^5t^4\geq0$,

which is clear true for $ x\equal{}\min\{x,y,z\}$.[/hide]

We have not a nice solution, yet.

Actually, the inequality below holds for any real $ k\ge 0$ and positive $ a,b,c$ such that $ abc\equal{}1$:

$ \frac 1{1 \plus{} a \plus{} b^k} \plus{} \frac 1{1 \plus{} b \plus{} c^k} \plus{} \frac 1{1 \plus{} c \plus{} a^k}\le 1$

A hint for $ k\equal{}2$: combine Chien than's way and  Cefer's way to give the desired result.

I think it is known that for all positive reals x,y,z that

$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y}\geq x^2 \plus{} y^2 \plus{} z^2$  and now put (x,y,z) -> (b,a,c) to get the inequality .

Have you a complete solution, Silouan?

Thank you very much.

[quote="Vasc"][quote="Vasc"]A hint for $ k \equal{} 2$: combine Chien than's way and  Cefer's way to give the desired result.[/quote]
By the AM-GM inequality, we have

$ 2\sum a^3b \plus{} \sum a^2b^3 \ge 3\sum a^{\frac 7{3}}\ge 3\sum a^2$

and

$ \sum a^2b^3 \ge \sum a$.[/quote]

Thank you very much, Vasc.

To prove $ 3\sum a^{\frac 7{3}}\ge 3\sum a^2$ I used power mean inequality

with $ s \equal{} \frac 7{3}$ and $ t \equal{} \frac 4{3}$ . We have $ (\frac {a^s \plus{} b^s \plus{} c^s}{3}) \geq (\frac {a \plus{} b \plus{} c}{3})^s \equal{} (\frac {a \plus{} b \plus{} c}{3}) (\frac {a \plus{} b \plus{} c}{3})^t \geq (\frac {a \plus{} b \plus{} c}{3})$

Also to prove $ \sum a^2b^3 \ge \sum a$ I used power mean inequality. Is your proof the same?

[b]silouan wrote:[/b]
[quote]I think it is known that for all positive reals x,y,z that
 $ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y}\geq x^2 \plus{} y^2 \plus{} z^2$[/quote]Is a bit confusion,silouan.
Try $ x\equal{}10000,y\equal{}1,z\equal{}10$. :wink:
However, this is true when $ x,y,z$ are sides of triangle or when $ x\geq y\geq z$.

[quote="completely"][quote="Vasc"]We have not a nice solution, yet.

Actually, the inequality below holds for any real $ k\ge 0$ and positive $ a,b,c$ such that $ abc \equal{} 1$:

$ \frac 1{1 \plus{} a \plus{} b^k} \plus{} \frac 1{1 \plus{} b \plus{} c^k} \plus{} \frac 1{1 \plus{} c \plus{} a^k}\le 1$[/quote]

I think may be there are nice solution for it? :blush:[/quote]
Yes, there is. AM-GM do it. :)

[quote=Vasc]If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$ \frac 1{1\plus{}a\plus{}b^2}\plus{}\frac 1{1\plus{}b\plus{}c^2}\plus{}\frac 1{1\plus{}c\plus{}a^2}\le 1$.[/quote]
If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$$\frac 1{1\plus{}a^2\plus{}b^3}\plus{}\frac 1{1\plus{}b^2\plus{}c^3}\plus{}\frac 1{1\plus{}c^2\plus{}a^3}\le 1.$$

So easy.
Muliply first fraction up and down with (c²+1+1/b)
you get
(1+a²+b³)(c²+1+1/b) ≥ (a+b+c)²
Same for other two fraction.
Now you have 2-3 line calculations and in fourth line you have
ab+bc+ca ≥ 3
which is true when abc =1.

Sure, you must use 1/b = ac ^^
and second fraction will be multiplied by (a²+1+1/c) .....

Same technique for the first Vasc's problem above.

[quote=Vasc]If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$ \frac 1{1\plus{}a\plus{}b^2}\plus{}\frac 1{1\plus{}b\plus{}c^2}\plus{}\frac 1{1\plus{}c\plus{}a^2}\le 1$.[/quote]
It remains to prove that
$ \sum_{cyc}\frac 1{1+a^3+b^6}\leq 1,$
$ \sum_{cyc}\frac 1{1+a^3+b^6}\leq \sum_{cyc}\frac{c^4+a+\frac{1}{b^2}}{(a^2+b^2+c^2)^2}=\frac{\sum_{cyc}a^4+\sum_{cyc}a+\sum_{cyc}b^2c^2}{(a^2+b^2+c^2)^2}$
Hence, it remains to prove that
$\sum_{cyc}b^2c^2\geq abc\sum_{cyc}a .$
[quote=sqing] If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$$\frac 1{1\plus{}a^2\plus{}b^3}\plus{}\frac 1{1\plus{}b^2\plus{}c^3}\plus{}\frac 1{1\plus{}c^2\plus{}a^3}\le 1.$$[/quote]
$\sum_{cyc}\frac {1}{1+a^2+b^3}\leq $$\sum_{cyc}\frac{c^2+1+\frac{1}{b}}{(a+b+c)^2}=\frac{\sum_{cyc}a^2+3+\sum_{cyc}bc}{(a+b+c)^2}$
Hence, it remains to prove that
$\sum_{cyc} bc\geq 3\sqrt[3]{(abc)^2}.$
Thank bitrak.

Wow.
Very nice.
If $ a,b,c$ are positive numbers such that $ abc=1$ . Prove that$$ 1\geq \frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\geq\frac{3}{ab+bc+ca}.$$

[quote=oldbeginner][quote=sqing]If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$$\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\leq \frac{3}{4}$$
(Proposed by Marcel Chirita, Bucharest, Romania)[/quote]

The following inequality is also true.
]If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$$\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\geq \frac{(a+b+c)^2}{4(a^2+b^2+c^2)}$$[/quote]
Very nice.
If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then
$$\frac{(a+b+c)^2}{4(a^2+b^2+c^2)}\leq\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\leq \frac{3}{4}
$$

[quote=oldbeginner]The following inequality is also true.
Let $a,b,c>0,abc=1$,prove:
\[\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\ge\frac{3}{ab+bc+ca}\][/quote]
http://www.artofproblemsolving.com/community/c6h1362247p7475264:
Let $a, b, c>0$, and $abc=1$. Prove:
$$\frac{1}{\lambda a^2+b+1}+ \frac{1}{\lambda b^2+c+1}+ \frac{1}{\lambda c^2+a+1}\geq\frac{9}{(\lambda +2)ab+bc+ca}$$where $\lambda\geq 1.$

[quote=Vasc]If $ a,b,c$ are positive numbers such that $ abc\equal{}1$, then

$ \frac 1{1\plus{}a\plus{}b^2}\plus{}\frac 1{1\plus{}b\plus{}c^2}\plus{}\frac 1{1\plus{}c\plus{}a^2}\le 1$.[/quote]
Let $ a,b,c,d>0$ s.t. $ abcd=1$.[url=https://artofproblemsolving.com/community/c6h214901p1188389]Prove that:[/url]
$ \frac{1}{1+a+a^2+a^3}+\frac{1}{1+b+b^2+b^3}+\frac{1}{1+c+c^2+c^3}+\frac{1}{1+d+d^2+d^3}\geq 1$


induction
advanced fields
advanced fields unsolved

simplify $ (\omega \plus{} 1) \cdot \omega^2$.

edit: I was wrong since $ (a\plus{}b)c\equal{}ac\plus{}bc$ does not always hold.

Maybe $ \omega \neq 1$ would mean the complex root of $ z^3 \equal{} 1$.

Maybe it's the least infinite ordinal number.

by definition, $ (\omega \plus{} 1) \omega \equal{} \bigcup_{n < \omega} (\omega \plus{} 1) n$. now an easy induction shows $ (\omega \plus{} 1) n \equal{} \omega n \plus{} 1$. e.g. we have $ (\omega \plus{} 1)2 \equal{} (\omega \plus{} 1) \plus{} (\omega \plus{} 1) \equal{} \omega \plus{} (1 \plus{} \omega) \plus{} 1 \equal{} \omega \plus{} \omega \plus{} 1 \equal{} \omega 2 \plus{} 1$. but $ \omega n \plus{} 1 < \omega n \plus{} \omega \equal{} \omega (n \plus{} 1) < \omega^2$. thus $ (\omega \plus{} 1) \omega \leq \omega^2$. on the other hand, $ \omega^2 \equal{} \bigcup_{n < \omega} \omega n \leq \bigcup_{n < \omega} (\omega n \plus{} 1) \equal{} (\omega \plus{} 1) \omega$.

this shows $ (\omega \plus{} 1) \omega \equal{} \omega^2$ and then $ (\omega \plus{} 1) \omega^2 \equal{} \omega^3$ (this is the cantor normal form, so I guess this is simplified enough).


Hello,

I was wondering if Intermediate Counting & Probability and/or Trig/Complex Numbers would be offered in the winter session.  I had a schedule in mind that looked like this:

Fall:

One of the Intermediate Independent Studies in the Fall
One of the Intermediate Classes

Fall to Winter:

AMC 12 Problem Series
AIME Problem Series

Winter:

The other Intermediate Class...if available.

I could take both Intermediate classes but I don't want to overload and burn out.  Could someone maybe evaluate my schedule?

I'm taking Intermediate Algebra right now.  It's a fun class.

Also, are the Olympiad classes specifically for Olympiad problems and questions?  Would there be any AIME level questions in such classes?

Thank you all for your help.

Our plan is to offer every one of our classes (except possibly the AIME A Problem Series) during the school year.

Your schedule looks pretty packed, but of course, that has a lot to do with the amount of time you plan to devote to mathematics.  The Independent Study should take up 10-15 hours of your time per week.

The problem series are easier to handle, taking up maybe 2-5 hours depending on how much time you spend tackling the message board problems, though tackling the practice problems helps students advance more quickly and most completely.

Thanks for replying Mr. Crawford.

I have a weird schedule next year:  I have quite literally, 4 days off per week (weekends, and Tuesdays and Thursdays) with a 50 minute class on Tuesday.  I think I'll have plenty of time for Mathematics enrichment.

I will attempt all message board problems.

I would like a bit of clarification here:  When you say that the AoPs staff plans to offer every of the classes during the school year, does that mean Trig/Complex Numbers and/or Counting will be offered in the Winter as well?  I was worried about overloading, and I think the best course possible is to split the Intermediate courses into two different sessions.

Thanks again.

Our current plan is to begin classes that are not currently listed around January.

[quote="1234567890"]So Olympiad PS may be offered in January?
That's good, since I'm not ready to take it in September.[/quote]

No.  If it's on the schedule now, it [b]won't[/b] be offered again in January.


linear algebra
matrix
vector
combinatorial geometry
linear algebra unsolved

Let $U$ be unitary matrix such that for some vector $x$, $Ux$ is orthogonal to $x$. Prove that any arc on the unit circle $|z| =1$ on which the eigenvalues of $U$ lie has the length at least $\pi$.

Either I'm dumb, or this is obvious: take $v_{1},...,v_{n}$ an orthonormal basis for this matrix and write $x=k_{1}v_{1}+...+k_{n}v_{n}$. The condition shows that $|k_{1}|^{2}r_{1}+...+|k_{n}|^{2}r_{n}=0$ where $r_{i}$ are the eigenvalues and thus the origin is in the convex hull of the eigenvalues. The conclusion follows.

[quote="harazi"]Either I'm dumb, or this is obvious[/quote]

Probably the second and the contrary to the first. (I'm talking about you guesses). You soltuon is ok.


Given two positive sequences $ \{a_n\},\{b_n\}$ s.t.$ a_{1}>a_{2}>a_{3}>...,b_{1}>b_{2}>b_{3}>...$ and $ \sum_{n\equal{}1}^{\infty}a_{n}\equal{}\plus{}\infty,\sum_{n\equal{}1}^{\infty}b_{n}\equal{}\plus{}\infty$.Prove or disprove $ \sum_{n\equal{}1}^{\infty}\min(a_{n},b_{n})\equal{}\plus{}\infty$. :)

Let $ A_n \equal{} \begin{cases}1/n,\quad \textrm{if n is even} \\
1/n^2,\quad \textrm{if n is odd}\end{cases}$ and  $ B_n \equal{} \begin{cases}1/n^2,\quad \textrm{if n is even} \\
1/n,\quad \textrm{if n is odd}\end{cases}$. Clearly $ \sum_{n}\min(A_n,B_n)$ is convergent.

Ah, you want series with decreasing terms? Then replace $ A_n$ with $ a_{n,1} \plus{} \dots \plus{} a_{n,M_n}$ (and $ B_n$ with $ b_{n,1} \plus{} \dots \plus{} b_{n,M_n}$) where the elements $ a_{n,j}$ are of nearly the same size and their sum is $ A_n$. Make sure that the sequence $ (M_n)$ grows fast enough so that $ a_{n,M_n} > a_{n \plus{} 1,1}$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I need a bit of help writing a formal definition of the following:
In a group G, let b be considered a rearrangement of a if there is some way to write a as the product of some elements and a way to rearrange those elements so the product is b.
=Uncool-

You could write: $b$ is a permutation of $a$ if there exist
[list]an integer $n\ge 1$
elements $g_{1},\dots,g_{n}$
and a permutation $\sigma$ on $n$ elements[/list] such that
[list]$g_{1}g_{2}\cdots g_{n}=a$
and $g_{\sigma(1)}g_{\sigma(2)}\cdots g_{\sigma(n)}=b$
[/list]

(Equivalently, $b^{-1}a\in [G,G]$.)

Alright, then can you tell me if this proof works, using that definition: the set of 'permutations of the identity' is a normal group, and the corresponding factor group is commutative:

Let us call the 'permutations of the identity' the set N. Then I assert that N is a normal subgroup.

N is a subgroup: Clearly, N is closed under multiplication (as we can simply use the composition of the two permutations), and N includes inverses, so N is a subgroup.

N is normal: Let's take the coset aN. We wish to prove that it is equal to Na. aN = {a*n | n is a permutation of e}. Let us look at ana^(-1). Clearly, this is also a permutation of e. Therefore, we can say aN = {a*n*a^(-1)*a | n is a permutation of e} = {n2 * a | n2 is a permutation of e} = Na. Therefore, N is normal.

G/N is commutative: Let us take the two cosets aN and bN. I say aN * bN = bN * aN. aN * bN = abN = ab (b^(-1)*a^(-1)*b*a)*N (as b^(-1)*a^(-1)*b*a is a permutation of e) = baN = bN * aN.

And a proposition: If K is a normal group and G/K is commutative, then K must include the set of all permutations of 0.

Yes, your proof works. The normal subgroup already [url=http://en.wikipedia.org/wiki/Commutator_subgroup]has a name[/url].

Ah...I had been looking to see if there was a name for this, couldn't find it and my math teacher didn't know of it...thank you very much.
Just using this because it directly related to my thread on partial rings.
=Uncool-


topology
real analysis
real analysis unsolved

a space $ X$ is  called limit point compact if every infinite subset of $ X$ has a limit point

My question is :  why $ [0,1]$ is not limit point compact as a subset of $ \mathbb{R}_l$ : with lower limit topology?

thanks

Well, $ [0, 1]$ isn't compact (in the sense of open covers) because, for example, we have the open covering $ \left\{\left[0, \frac {1}{2}\right), \left[\frac {1}{2}, \frac {3}{4}\right), \left[\frac {3}{4}, \frac {7}{8}\right), \ldots \right\}$ with no open subcover.  It's not hard to use a similar idea to find a sequence with no limit point.  Just think about what all neighborhoods of a given point (e.g., the point $ 1$) look like.


[quote="umesh123"]>> Which of the following << is the number of Zeros in the product :- 1^1*2^2*3^3*4^........99^99*100^100
my approach
5^5x5^10x5^15x5^20x25^25x5^30x5^35x5^40x5^45x25^50x...........x25^100
total zeros=5+10+15+....+100 +25+25x2+25x3+25x4=5(1+2+3+.....+20)+100=5x10x21+2500=1050+250=1300

it's not perfect solution though got right answer
expecting the perfect way to get the answer from you all[/quote]

You never gave the choices, and there is no "perfect way."

Is there any other way than to count the number of 5 factors?
$ 5^55^{10}5^{15}5^{20}25^{25}5^{30}...25^{100}$ seems good enough to me!

Finding the largest power of $ 5$ that divides the product only gets you the number of zeroes at the end, not the number of zeroes in the product.

[quote="bbgun34"]modular arithmetic or something similar would, though  :D[/quote]

No, as far as I know there is no way to count the number of zeroes that occur in the product with modular arithmetic.

Because for there are obviously more factors of 2 than 5.  And for every 5, it gets paired with an excess two to make a 10
@grn_trtle: But do you know anyway to count the number of zeroes that occur in the product?

Although it is most likely he copied the problem wrong

hi chalao ... just try to understand this stretegy

igot a general solution for the above mentioned question -

5(1+2+3+4+......+[N/5]) +25(1+2+3+4+...+[N/25]) +125(1+2+...+[N/125])+ 5^4(1+2+3+....+[N/5^4])+...

Where [ ] is the greatest integer function.


I have the book "Problem Solving Through Problems"  by Larson, and I've heard that it's excellent prep for the Putnam. Unfortunately, there are no solutions to the end-of-chapter problems, and I was wondering how I could get them. Thanks.

it is a great book, i dont know of any way to get the solutions, but they do post the sources for some questions, so for some you can get answers from kalva.


Why are the disgusting chains like McDonalds, Domino's, and KFC up there but the ok chains like Pizza Hut and Popeye's not?

you should have definitely added chic-fila. but i voted for burger king anyway. i don't go there very often, but if we're going by taste then that's my favorite.

half-pound bean burrito especial with extra jalapeno sauce=deliciosly good.

TACO BELL!!!

[quote="fredbel6"]I voted McDonald's....
.. :maybe:  we don't have any of those other chains here in Flanders.

We do have Pizza Hut, which is very good as well. :wink:[/quote]

Same  :) !

I like McDonald's, i go there once or twice a year so it doesn't really affect me anyways  :D

[quote="goldendomer"][quote="Ignite168"][quote="goldendomer"]Why do people think McDonald's is nasty? I love McDonald's! Cheap and delicious! :icecream:  :icecream:  :icecream:[/quote]

Because it is, that crap is terrible for your body. Have you tried Burger King's burgers? They actually look like some of the pictures and have a nice grilled taste.[/quote]

All fast food resturants' food are bad for you. (except Subway and some other ones)[/quote]

McDonald's is actually pretty good if eaten moderately.

[quote="mdk"][/quote][quote="goldendomer"][quote="Ignite168"][quote="goldendomer"]Why do people think McDonald's is nasty? I love McDonald's! Cheap and delicious! :icecream:  :icecream:  :icecream:[/quote]

Because it is, that crap is terrible for your body. Have you tried Burger King's burgers? They actually look like some of the pictures and have a nice grilled taste.[/quote]

All fast food resturants' food are bad for you. (except Subway and some other ones)[/quote][quote="mdk"]

McDonald's is actually pretty good if eaten moderately.[/quote]

No not really, not only is McDonalds bad for you, it tastes like crap too.

they advertise well, and kids like their kids meal so they get grown ups to go. they also make the world more obese, but yeah...

 :huh:

[quote="mdk"][quote="goldendomer"][quote="now a ranger"]MCDONALDS IS HORRIBLE! I watched that documentary, people become flat out fat! and they puke if they supersize! It is nasty and bad for your body.[/quote]

Alright, I eat McDonald's like 10 times in a year. I WILL NOT BECOME "FLAT OUT FAT AND PUKE". It only happens when you eat it everyday. Duh![/quote]

Duh![/quote]
Duh!

anyways, everything up there, cept taco bell, sucks. like, a lot.

Taco Bell sucks. Denny's and Red Lobster are the best hands down.

too bad both arent fast food restaurants

Do any of you like Arby's? or Shoney's? my sister does

[quote="mdk"][quote="now a ranger"]MCDONALDS IS HORRIBLE! I watched that documentary, people become flat out fat! and they puke if they supersize! It is nasty and bad for your body.[/quote]

Only if you eat it 3 times a day for 30 days. If you eat it moderately, it is quite good. Why do you think they're a multi billion dollar franchise?[/quote]

It's actually quite bad for your health in any quantity. Why do I think they're a multi billion dollar franchise? Certainly not because it's healthy. After all, the cigarette industry is a multi billion dollar franchise but cigarettes certainly aren't good for you.

Bluck. Have you seen the movie "Supersise Me"? apart from the inappropriateness, it's educational about your health.

If you're mexican and you don't say Taco Bell sucks then it is because you're not mexican. (In fact, there aren't many Taco Bells here in Mexico, at least, there weren't many in the cities where I have gone)

I voted for McDonald's

I hate fast food  :D
its horrible for your body and doesn't really taste very good either
if anybody needs convincing, just watch "Supersize Me"

[quote="Max300"]Its true that if you eat McDonalds moderately then you won't get fat or anything, but it is still bad for your health and may affect it in ways not as noticable as weight. Regular or large fries (can't remember) have like 33% DV of sodium, and some people put even more salt on top of it and dip it in ketchup  :what?:[/quote]
what does DV stand for? Darth Vader?

in an earlier post someone said subway is good for you

...thats if you eat the right things!

check this out

[url]http://www.subway.com/applications/NutritionInfo/nutritionlist.aspx?id=sandwich[/url]

(these are for the 6"subs - double for the foot long ones)
-jorian

Seriously, i dont like any burger fast food place, because the burgers always look like they've been flattened by a stack of heavy books. However, KFC is awesome, because I LOVE CHICKEN! Subway is also good, because the subs are okay for you and it actually tastes good. the commercials are also awesome: subway! eat FRESH! That guy is the man!

But yeah, seriously, SUPER SIZE ME is very interesting(and gross) and i only eat McDonalds like once every two months.

Popeyes is als very good, especially the biscuits.


function
induction
algebra proposed
algebra

Find all the functions $f(x),$ continuous on the whole real axis, such that for every real $x$ \[f(3x-2)\leq f(x)\leq f(2x-1).\]

[i]Proposed by A. Golovanov[/i]

Let $g(x)=f(x+1)$, then $g(3x)\le g(x)\le g(2x)$, therefore if g(x) continiosly in 0, then $g(x)\equiv const$.

we can easily show by induction that : $ g(\frac{x}{2^{n}})\leq\ g(x)\leq\ g(\frac{x}{3^{n}}) $ ,  $ \forall\ (x,n)\in\mathbb{R}\times\mathbb{N} $
now for a fixed $x$ take the limit when $n\mapsto\infty$ using the fact that $f$ is continuous we get : $g(x)=g(0)$  ,  $\forall\ x\in\mathbb{R} $
so: $g$ and then $f$ are constant!

Another problem of this type, which includes this particular one as a special case (for the function $g$) has already been posted here:  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=379114&hilit=3x


are they arranged in order of difficulty?  (sorry if this has been asked before)

eh...not really...like, last round, i solved #1 after solving 2,3,4 i think, but #5 is usually the harder than the rest i think...

It really depends on the person.  The directors try to organize the problems in order of increasing difficulty, but they can't account for people being stronger at a certain type of geometry.  For example, someone may find a #4 number theory problem much easier than a #1 geometry problem.  It all depends on your style of solving, really...

i see. and what level are the problems at generally?

CMO lvl?

You could check out some past problems: http://usamts.org/Problems/U_Problems.php


geometry
AMC
AIME
vector
area of a triangle
Heron\u0027s formula
absolute value

Equilateral triangle ABC, with side 2, is in plane P. Segments BX and CY are drawn perpendicular to plane P, both on the same side of the plane. If BX = 1 and CY = 3, compute the area of triangle AXY.

Ah, I remember this problem!  Old AIME, right?

If you don't mind, would you please identify the source of the problem?  Thanks.

Interesting problem.

Hint: use [hide]vectors[/hide]. In particular, [hide]vector formula for area of a triangle[/hide].

How about Heron's formula? All three sides of the triangle AXY can be easily calculated. Oh, wait... It's a right triangle.

[hide]

AX=sqrt(5), AY=sqrt(13), XY=2sqrt(2), <AXY=90,

So the area [AXY]=sqrt(10)

[/hide]

[quote="Osiris"]I think that route is more calculation-intensive.[/quote]
Not really if you read the spoiler in my above post.

More evidence that every AIME problem has an elegent and a computationally complex solution.

[quote="beta"]No,not at all. I used that method too. AX=[tex]\sqrt{1+4}=\sqrt5[/tex]
XY=[tex]\sqrt{4+4}=\sqrt8[/tex]
AY=[tex]\sqrt{13}[/tex]  which is a right triangle.
area=[tex]\frac{\sqrt{40}}{2}  [/tex][/quote]

Good point.  But...typically [tex]\frac{\sqrt{40}}{2}[/tex] is written as [tex]\sqrt{10}[/tex]

Ah, my bad. I didn't realise there was a spoiler. Now in comparison my method is more complex.  In any case, here it is:

[hide] Place A at point (0, 0, 0), B at (-1, :rt3:, 0) and C at (1, :rt3:, 0). Then AB = (-1, :rt3:, 1) and AC = (-1, :rt3:, 3). Cross-multiplying AB and AC and taking its absolute value, then dividing by 2, yields :sqrt: 10.[/hide]

if only i had checked if rt 5, 2rt 2, and rt 13 made a right triangle.. that was easy


I downloaded the solutions.tex and usamts.tex from USAMTS.  I am using MikTeX and TeXnicCenter, with fancyhdr installed.

When I build dvi in solutions.tex, pdf comes out.  I get the error message "Windows API error 2: The system cannot find the file specified. Data: C:\Documents and Settings\[...]\USAMTS\R\solutions.dvi"
***I skipped the long path [...] here.***
TeXnicCenter does not automatically open the pdf file.  However, if I open it manually, it's fine.

When I build pdf in solutions.tex, pdf comes out and is opened automatically Acrobat Reader.

When I build dvi in any other LaTeX files, dvi comes out and is opened automatically in Yap.

When I build pdf in any other LaTeX files, pdf comes out and is opened automatically in Acrobat Reader.

How do I get the dvi file using USAMTS's template?

If you can generate a PDF file, then why worry about a DVI file?  We accept PDF files; in fact, we prefer them.


Find all pairs $ (x,y)$ in $ R^2$ such that:
$ (x^2\plus{}y^2\minus{}x)(xy\minus{}1)^2\plus{} \sqrt{y^2\minus{}x^2} \equal{}0$ ( i proposed it in Pre-olympiad section,but Irecieved no solution :(, so i apologize if this is a spam)

$ R^2$ = set of all pairs of real numbers. For example, say $ A \equal{} \{1, 2\}$. Then $ A^2 \equal{} \{(1, 1), (1, 2), (2, 1), (2, 2)\}$

[quote="anas"]Find all pairs $ (x,y)$ in $ R^2$ such that:
$ (x^2 \plus{} y^2 \minus{} x)(xy \minus{} 1)^2 \plus{} \sqrt {y^2 \minus{} x^2} \equal{} 0$ ( i proposed it in Pre-olympiad section,but Irecieved no solution :(, so i apologize if this is a spam)[/quote]

This problem should be really easy.
Because x must equal y according to the square-root thing,
and also 2x^2-x=0 or x^2=1
so (x,y) = (0,0);(1/2,1/2);(-1,-1);(1,1)

I hope that I'm right... :lol:

@tim26270746: Can you explain for me why did you had that? I'm still working on this problem. Why
Because x must equal y according to the square-root thing, I think $ y^2 \ge x^2$ they don't need to be equal.


limit
inequalities
real analysis
real analysis unsolved

Find:

\[ \lim_{n \to \infty}{ \left( \sqrt{\frac{n\plus{}2}{n\plus{}1}}\plus{} \sqrt{\frac{n\plus{}3}{n\plus{}2}}\plus{} \dots \plus{} \sqrt{\frac{2n\plus{}1}{2n}}\minus{}n \right) } \equal{}?\]

$ \lim_{n \to \infty}{\left( \sqrt {\frac {n \plus{} 2}{n \plus{} 1}} \plus{} \sqrt {\frac {n \plus{} 3}{n \plus{} 2}} \plus{} \dots \plus{} \sqrt {\frac {2n \plus{} 1}{2n}} \minus{} n \right) }$
$ \equal{}\lim_{n\to\infty} \sum_{k \equal{} 1}^n \left(\sqrt {1 \plus{} \frac {1}{n \plus{} k} } \minus{} 1\right)$

And use the fact that in this case $ \sqrt {1 \plus{} t} \sim 1 \plus{} \frac {t}{2} \minus{} \frac {1}{8}t^2$.Or writing it more rigorously :

$ 1 \plus{} \frac {t}{2} \minus{} \frac {1}{8}t^2 \leq \sqrt {1 \plus{} t}\leq 1 \plus{} \frac {t}{2}$

You can always find these kind of convenient inequalities if you look at the function's Taylor expansion.


I can't do this question:

A rigid body rotates about a point of itself with angular velocity $ \mathbf{\omega}$ (what this means is that at any instant, the body is rotating at a rate of $ |\mathbf{\omega}|$ about the line in the instantaneous direction of the vector $ \mathbf{\omega}$). The velocity of a point of the body with the position vector $ \mathbf{r}= x\mathbf{i}+y\mathbb{j}+z\mathbf{k}$ is given by

$ \mathbf{v}=\mathbf{\omega}\times\mathbf{r}$

Explain why every point of the body is moving instantaneously perpendicular to $ \mathbf{\omega}$. Find the matrix $ S$ such that $ \mathbf{v}= S\mathbf{\omega}$. Show that the square of the speed of the point is

$ |\mathbf{v}| ={\mathbf{\omega}}^{T}S^{T}S\mathbf{\omega}$

where

$ S^{T}S =\left(\begin{array}{ccc}y^{2}+z^{2}&-xy &-zx\\ -xy & x^{2}+z^{2}&-yz\\ -zx &-yz & y^{2}+x^{2}\end{array}\right)$

"Every point of the body is moving instantaneously perpendicular to $ \mathbf{\omega}$"
is equivalent to
"for every $ \mathbf{r}$, $ \mathbf{v}$ is perpendicular to $ \mathbf{\omega}$",
and the last statement is true because
$ \mathbf{v}\equal{}\mathbf{\omega}\times\mathbf{r}$, and the cross product is perpendicular to both its factors.

If $ \mathbf{\omega}\equal{}\omega_{x}\mathbf{i}\plus{}\omega_{y}\mathbf{j}\plus{}\omega_{z}\mathbf{k}$, then
$ \mathbf{v}\equal{}\mathbf{\omega}\times\mathbf{r}\equal{}\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \omega_{x}&\omega_{y}&\omega_{z}\\ x & y & z\end{array}\right| \equal{}\left(\begin{array}{ccc}0 & z &\minus{}y\\ \minus{}z & 0 & x\\ y &\minus{}x & 0\end{array}\right)\left(\begin{array}{c}\omega_{x}\\ \omega_{y}\\ \omega_{z}\end{array}\right) \equal{} S\mathbf{\omega}$,
where
$ S \equal{}\left(\begin{array}{ccc}0 & z &\minus{}y\\ \minus{}z & 0 & x\\ y &\minus{}x & 0\end{array}\right)$.
Now $ |\mathbf{v}|^{2}\equal{}\mathbf{v}^{T}\mathbf{v}\equal{} (S\mathbf{\omega})^{T}S\mathbf{\omega}\equal{}{\mathbf{\omega}}^{T}S^{T}S\mathbf{\omega}$.
(sludgethrower, you forgot the square in your formula).
Notice $ S$ is antisymmetric, $ S^{T}\equal{}\minus{}S$ (with odd dimension, therefore singular $ |S| \equal{} 0$).
$ M \equal{} S^{T}S \equal{}\minus{}S^{2}\equal{}\left(\begin{array}{ccc}0 &\minus{}z & y\\ z & 0 &\minus{}x\\ \minus{}y & x & 0\end{array}\right)\left(\begin{array}{ccc}0 & z &\minus{}y\\ \minus{}z & 0 & x\\ y &\minus{}x & 0\end{array}\right) \equal{}\left(\begin{array}{ccc}y^{2}\plus{}z^{2}&\minus{}xy &\minus{}zx\\ \minus{}xy & x^{2}\plus{}z^{2}&\minus{}yz\\ \minus{}zx &\minus{}yz & y^{2}\plus{}x^{2}\end{array}\right)$,
which is symmetric, $ M^{T}\equal{} M$.


Αν $ z\equal{}\prod^{n}_{k\equal{}0}\left[ 1\plus{}\left(\frac{1\plus{}i}{2}\right)^{2^{k}}\right]$ να υπολογίσετε το $ |z|$ και το $ {lim}_{n\rightarrow\infty}|z|$ :)

[quote="Γιώργος"]Αν $ z \equal{} \prod^{n}_{k \equal{} 0}\left[ 1 \plus{} \left(\frac {1 \plus{} i}{2}\right)^{2^{k}}\right]$ να υπολογίσετε το $ |z|$ και το $ {lim}_{n\rightarrow\infty}|z|$ :)[/quote]

Μαλον εχεις βαλει λαθος το προσημο, γιατι οπως το εχεις τωρα για ν>0 κανει 0

Παντος θεωρω οτι ειναι σχολικη ασκηση :)

[quote="Durandal"][quote="NickNafplio"]Μαλον εχεις βαλει λαθος το προσημο, γιατι οπως το εχεις τωρα για ν>0 κανει 0

Παντος θεωρω οτι ειναι σχολικη ασκηση :)[/quote]

Παρακαλώ? :?:  :huh:[/quote]

Οπα, σημειωσατε λαθος :blush:
Ξεχασα να υψωσω και το 2 στο τετραγωνο


Let $ \clubsuit (x)$ denote the sum of the digits of the
positive integer $ x$. For example, $ \clubsuit (8) \equal{} 8$ and
$ \clubsuit (123) \equal{} 1\plus{}2\plus{}3\equal{}6.$ For how many two-digit values of
$ x$ is $ \clubsuit (\clubsuit(x)) \equal{} 3$?

But I think you have to first prove that there are as many numbers that generate 3 as there are that generate 6 or 9...

If we want to check, we could do casework.  Note that $ \clubsuit (x)$ is either one or two digit.

If $ \clubsuit (x)\equal{}3$, then $ x\equal{}12,21,30$, so 3

If $ \clubsuit (x)\equal{}12$, then $ x\equal{}39,48,57,66,75,84,93$, which is 7

If $ \clubsuit (x)\equal{}21$ or $ 30$, then $ x>99$, which is impossible

So $ 3\plus{}7\equal{}\boxed{10}$

No, beastieric123, because the question asks for [i]two-digit[/i] numbers.

[quote=Jwilson1990J]I understand the 7 values of x, 39, 48, 57, 66, 75, 84, and 93, but why is the answer 10?[/quote]

$12,21,$ and $30$ satisfy the function, too.[quote=OlympusHero]Why is y<=18?[/quote]

The sum of the digits in a two-digit number is $\leqslant 18.$[quote=Flame]I found more elegant solution for this problem.

If we think about what this function represents, this is exactly the same algorithm we use to see if a number is divisible by 3. The maximum 2-digit number is 99 so maximum $ f(x)=18 \implies f_{max}(f(x))=9$ so our function will always generate 1-digit number. Now returning to the proposed solution, we will look for the 2-digit numbers divisible by 3:
$ \lfloor{\frac{99}{3}}\rfloor-3=30$
but this would be over counting since for some numbers divisible by 3 the function will generate $ 6$ and $ 9$ as well so we have to divide $ 30$ by $ 3$ to get total of $ \boxed{10}$ numbers for which $ f(f(x))=3$.[/quote]

This is a nice intuition but I do not understand how you can prove that the set of two-digit multiples of $3$ has the same number of numbers whose result in the given function is $3,6,$ and $9$. Can somebody prove the claim? It seems intuitive if you try numbers e.g. $\clubsuit(\clubsuit(33))=6,\clubsuit(\clubsuit(36))=9,\clubsuit(\clubsuit(39)=3$ but can somebody give a rigorous proof?

Ngl but casework is a pretty simple solution as well for this, since you can asy what numbers have digits that add up to 3, and then 12


This one is a particular case of Vasc's problem, also use strong S.O.S method to solve.

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

$\frac{2a^2+5bc}{(b+c)^2}+\frac{2b^2+5ca}{(c+a)^2}+\frac{2c^2+5ab}{(a+b)^2} \ge \frac{21}{4}$

[/quote]
Here is a solution for pedestrians:

The given inequality is equivalent to
$x(b-c)^2+y(c-a)^2+z(a-b)^2 \ge 0$
where
$x = (b+c)^2(4b^2+4c^2+11bc+5ab+5ac-9a^2)$,
$y = (c+a)^2[4c^2+11ac+5bc+5(a-b)b+4(a^2-b^2)]$
and
$z = (a+b)^2[4b^2+11ab+5bc+5(a-c)c+4(a^2-c^2)].$
Without loss of generality, assume that $a\ge b\ge c$. Then clearly $y\ge 0$ and $z\ge 0$.  Further,
$x+y = 8c^4+24(a+b)c^3+(21a^2+21b^2+30ab)c^2+13a^3c+19b^3c+4(a^2-b^2)^2 +5ab(a-b)^2+3ac(2a+b)(a-b) \ge 0$
and the proof is completed.


ARML quadratics calculus function algebra

I looked at the solution for this problem, but I couldn't understand some of the steps they used to reach the final answer:

Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x+4y$ can have?

[hide="Answer"]73[/hide]

I think its from an old AMC...but I might be wrong.

I think i've done this before (AMC?) and it's just the higher tangent line to the circle.

Hmm... I think I can do it this way...

$f(x)=x^2+y^2-14x-6y-6$
$f'(x)=2x+2y\frac{dy}{dx}-14-6\frac{dy}{dx}$
$\frac{dy}{dx}(2y-6)=14-2x \Rightarrow \frac{dy}{dx}=\frac{14-2x}{2y-6}$

Okay, there's an extra factor of 2 in all terms, so we can simplify that to $\frac{7-x}{y-3}$... well, I guess that doesn't help after all. :blush:

[quote="JesusFreak197"]Hmm... I think I can do it this way...

$f'(x) = 2x+2y=14+6=20 \Rightarrow x+y=10$

So, that would make the answer be that $3x+4y$ reaches a maximum of $40$... is that correct, or can I not use derivatives?[/quote]
i donno how you got that but if you let $f(x)=x^2+y^2-14x-6y-6=0$ then $f'(x)=\frac{14-2x}{2y-6}$ :huuh:

Oh, whoops, I totally screwed that up. I guess I wasn't thinking. :oops: Fixing it now...

[quote="JBL"]This is from ARML, right?

One method:[hide]Draw the sets of points for which 3x + 4y is constant.  Then draw the set of points that satisfy that relation.  Find the intersection of this second set with the set 3x + 4y = k for the largest k for which that intersection exists.  In the end, it is just solving a quadratic equation to see when the discriminant will be 0.[/hide][/quote]

its actually from an old AHSME...

i understand your method somewhat, but wouldn't you have too many variables?

this is a calculus trap: stop applying calc to problems with functions as your first thought!

---------

$3x + 4y$ needs to be maximized, we are given:

$x^2 + y^2 - 14x - 6y - 6 = 0$.

equivalent to:

$(x-7)^2 + (y-3)^2 = 64$

We can try graphing the line and using a formula here: http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html

but i have a transformation approach:  Set x' + 7 = x and y' + 3 = y.  The expression we are trying to maximize is easily found to be $3x' + 4y' + 33$.

Focus on maximizing $3x' + 4y'$.  By cauchy, $(x'^2 + y'^2)(3^2 + 4^2) \geq (3x'+4y')^2$.  It is clear now that the maximum is 40.  So the final answer is 73.  You can check it works.

There are other solutions at these two threads:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23030
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7940


It says "BATT" if your batteries are low in the bottom right hand corner next to the thing that tells you what operation you are doing (ie like 23/30 or something like that).  Or you can always get duracell which have those little power checker strips that you touch to find out how much power is left in your battery.

[quote="mathfanatic"]I know this is silly, but how do you turn the axes on on the TI-89?  When I push GRAPH, the graph comes up, but the axes are missing, and there's no AxesOn or AxesOff function the way there is on the TI-83+.

Thanks.[/quote]
I'll only answer this because it's you, Keone. I'd tell anyone else to RTFM (and you know I would). diamond, f3, f1, up, enter, down, down, down, right, up, enter, enter, Merry Christmas.

Keystrokes and everything.  Thanks, Matt.


group theory
abstract algebra
analytic geometry
superior algebra
superior algebra unsolved

i read this problem on the net- here it is:

Let G be a finite group. Show that there exists a (finite) graph X such that the group of automorphisms of X is isomorphic to G

[quote="splintercelll990"]Show that there exists a (finite) graph X such that the group of automorphisms of X is isomorphic to G[/quote]

Do you mean a finite group or what does automorphism of a graph mean?

actually, i just checked the problem statement and it is correct- if you would like, here is a link to the site where i found the problem (its problem number 7)

http://www.people.fas.harvard.edu/~gcarroll/math/nonelem.pdf

-cheers
 :D  :D

It's easy to imagine what an automorphism of a graph is: it's a bijection f:V->V (V is th set of vertices of the graph) s.t. f(x) and f(y) are connected iff x and y are connected.

[quote="grobber"]It's easy to imagine what an automorphism of a graph is: it's a bijection f:V->V (V is th set of vertices of the graph) s.t. f(x) and f(y) are connected if x and y are connected.[/quote]

then do you know how someone could solve this problem :D

Nope! :D I'd have posted the solution by now. I was merely answering blondu's question.

I think we need to use Cayley's Theorem in this problem.  Cayley's Theorem states that every finite group G is isomorphic to a subgroup of [tex]S_n[/tex] (permutation group on n elements) for n=|G|.  Since we can always construct a graph X such that the group of automorphisms of X is isomorphic to some subgroup of [tex]S_n[/tex] where n is equal to the number of vertices of X, it can be shown that for any finite group, there is a corresponding finite graph with the desired properties.

I thought about this in the very moment in which I saw the problem, but I wasn't sure that for any subgroup of $S_n$ we may find a graph with the desired property. I'm sure that a valid solution would require some sort of explicit construction..

You're right that I didn't prove that.  I still believe that given any subgroup of [tex]S_n[/tex], you can find a graph whose automorphisms are isomorphic to that subgroup.  I just don't know how to show this.  Any ideas?

That's the wrong question. Just try to find a graph with precisely three automorphisms (group cyclic of order 3). It takes quite a bit more than 3 vertices to do it- I found one with 13 vertices, and I'm not sure how much smaller it can be done.

I believe you're right.  Now that I think about it, I don't think we can construct a graph on n vertices such that the group of automorphisms of the graph is isomorphic to a subgroup of [tex]S_n[/tex].  Just look at [tex]S_3[/tex].

Can you always find a graph with precisely x automorphisms so that the group of automorphisms is isomorphic to some subgroup of [tex]S_x[/tex]?  If this can be shown, then we can combine the result with Cayley's theorem to complete the proof.

Could you please tell us what your 13-vertex graph is, jmerry?

The 13-vertex graph, because it looks cool:
Vertices: O, A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4.
Edges: {O, A1}, {O, A4}, {O, B1}, {O, B4}, {O, C1}, {O, C4},
{A1, A2}, {A2, A3}, {A2, A4}, {A3, A4},
{A1, B1}, {A2, B1}, {A3, B1}, {A4, B1},
{B1, B2}, {B2, B3}, {B2, B4}, {B3, B4},
{B1, C1}, {B2, C1}, {B3, C1}, {B4, C1},
{C1, C2}, {C2, C3}, {C2, C4}, {C3, C4},
{C1, A1}, {C2, A1}, {C3, A1}, {C4, A1}.
The automorphisms cycle the letters, preserving the numbers. O is fixed.

Now, a 9-vertex graph for the same group; probably as small as possible:
Vertices: 1, 2, 3, A, B, C, 1A, 2B, 3C
Edges: {1, 2}, {2, 3}, {3, 1},
{1, A}, {2, A}, {2, B}, {3, B}, {3, C}, {1, C},
{1, 1A}, {A, 1A}, {2, 2B}, {B, 2B}, {3, 3C}, {C, 3C}.
The automorphisms cycle $1 \to 2 \to 3, A \to B \to C, 1A \to 2B \to 3C$.
Here A, 1A correspond to the directed edge $1 \to 2$.

This construction gave me an idea for the general case:
Lemma: If there exists a directed graph D with automorphism group G, there is an ordinary graph G with the same automorphism group.
(Notational note: The edges of a directed graph are ordered pairs $(v_1,v_2)$. The edges of an undirected graph are sets $\{ v_1, v_2 \}$.)

Proof: By construction. Let the vertices of the graph be the vertices of the digraph and two new vertices $E_a$ and $E_b$ for each edge $E$ of the digraph. Also, if $v$ is a vertex on only one edge of the digraph, add a vertex $v \prime$ for it.
The edges of the graph: If $(v, w)$ is an edge of the digraph, $\{v,w\}$ is an edge of the graph. If edge $E$ is $(v,w)$, add edges $\{v,E_a\}, \{w,E_a\}, \{v,E_b\}, \{E_a,E_b\}$ to the graph.
Finally, if $v$ had total degree 1 in the digraph, add the edge $\{v,v\prime \}$.
In general, we can distinguish the old vertices from the new by degree; all vertices with degree at least 4 were vertices of the digraph, all vertices of degree 2 are of type $E_b$, all vertices of degree 1 are associated with branch tip vertices, and all vertices of degree 3 are either old branch tips or vertices of type $E_a$. It is fairly easy to see with this that the automorphisms of the digraph induce the automorphisms of the new graph.

Construction: Given a finite group, construct a digraph with that group as its automorphism group.

Let $g_1, g_2, \dots, g_m$ be a set of generators of the group $G$, which has $n$ total elements.
The vertices of the digraph are $(x,1), (x,2), \dots, (x,m)$ for each $x \in G$
The edges are: $((x,1),(x,2)),((x,2),(x,3)), \dots, ((x,m-1),(x,m))$ and $((x,i),(g_ix,i))$ for each $i,x$.
We can distinguish the two types of edges: the first layer of vertices $(x,1)$ has out-degree 2 and in-degree 1, so we can tell which edges stay within and which point to the next layer. Removing those edges, we can do the same in the next layer, and continue up; the second coordinate of each vertex can be deduced from the structure of the graph, and all automorphisms must preserve each layer. Also, since the edges between layers match vertices with the same first coordinate, each automorphism is determined by its action on the first layer, which we can identify with $G$.
Since the path from $(x,1)$ to $(g_kx,1)$ must be sent to a path of the same type (k-1 steps up, one step on the same layer, k-1 steps down), the automorphism maps $g_kx$ to $g_ky$ if it maps $x$ to $y$. Repeating this and using the fact that the $g_k$ generate $G$, we have $f(x)=xf(e)$ when f is an automorphism and $e$ is the identity element of the graph. This gives an identification of the automorphism group with the reverse of the original group; if $p(x)=xa$ and $q(x)=xb$, $p(q(x))=xba$. The reverse group is isomorphic to the original (consider the map $x \to x^{-1}$) and we are done.

Finally, we apply the construction of the lemma, and we have a graph with the given group as its automorphism group.
This construction is not particularly efficient; it generates far larger graphs than necessary for most groups. For example, the alternating group $A_5$ has 60 elements and at least 2 generators, so the construction generates 480 vertices. The group can more easily be found as the automorphisms of the famous Petersen graph on 10 vertices.
The number of vertices in the construction, after converting to an undirected graph, is $n(5m-2)$.

A random graph would most likely have no nontrivial automorphisms- the interesting graphs for this are highly symmetrical.


Let $a, b, c$ be rational numbers with

     $a+b\sqrt[3]{2}+c\sqrt[3]{4}=0$

Show that $a=b=c=0$.

Consider the equation $cx^{2}+bx+a=0.$ By the initial condition, it has a solution $x_{0}=\sqrt[3]{2}$.

Assume $c\neq 0.$ Then

$\frac{-b\pm\sqrt{b^{2}-4ac}}{2c}=\sqrt[3]{2}$

WLOG, we can discuss just one case of $\pm$ (the other one is analogous). From $\frac{-b-\sqrt{b^{2}-4ac}}{2c}=\sqrt[3]{2}$, by cubing and rearranging, we get $\sqrt{b^{2}-4ac}=\frac{-4c^{3}-b^{3}+3abc}{b^{2}-ac}$. This means that $\sqrt{b^{2}-4ac}$ is rational, but then $\sqrt[3]{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$ must also be rational, which is contradiction. Therefore, $c=0$. The equation then becomes $a+b\sqrt[3]{2}=0$, which gives $b=0$ because otherwise we would have $\sqrt[3]{2}=-{a\over b}$ which is contradiction. From there, it directly follows $a=0.$

hay I did this yesterday at hardees

Using $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-xz)$, with $x=a$ and so on, we have \[a^{3}+2b^{3}+3c^{3}=6abc\] . However, by RMS-AM-GM, we have \[\sqrt[3]{2abc}\leq\frac{a+b\sqrt[3]{2}+c\sqrt[3]{4}}{3}\leq\sqrt[3]{\frac{a^{3}+2b^{3}+4c^{3}}{3}}\] But the middle part is $0$ by the initial condition and if you cube and rearrange the left and right parts, you see that they are equal. Since they are equal, they must be zero since that's the middle of the inequality, so $2abc=0$, and one of $a,b,c=0$, and the rest must be zero as follows in that other guys proof.


If a 5x5 cube is painted red and then divided into 125 1x1 cubes, how many 1x1 cubes have exactly one face painted red?

please explain how to do this problem in any way other than drawing and coloring!!!

[hide="hint"]
The only cubes with 1 red face are the ones that are not touching an edge.[/hide]

[quote="DaReaper"]If a 5x5 cube is painted red and then divided into 125 1x1 cubes, how many 1x1 cubes have exactly one face painted red?[/quote]
[hide]The cubes that will be painted on only one side are the ones not touching an edge, which gives them an extra painted face.

$ 6\cdot3^2 \equal{} \boxed{54}$[/hide]

I would do it this way.

[hide]The cubes that only have 1 side painted is $ \ 6(n \minus{} 2)^2$ or you can say the smaller square that is not touching the edges, which would be 9 and there are 6 sides.
So $ \ 9\times6 \equal{} 54$[/hide]

[quote="Elio (n)"]Hey guys tell something about my solution[/quote]

ah... I don't think your solution is right... Draw a 5x5x5 cube....

[quote="Elio (n)"]My calculation:

$ 2(5*5) \plus{} 2(3*5) \plus{} 2(3*3)$
$ 2*25 \plus{} 2*15 \plus{} 2*9$
$ 50 \plus{} 30 \plus{} 18 \equal{} 98$

There are 98 red cubes[/quote]
You calculated the total number of red cubes, which could also have been done by subtracting the inner (3x3) cube from the outer (5x5) one, thus leaving only the outer cubes: $ 5^3 \minus{} 3^3 \equal{} 125 \minus{} 27 \equal{} 98$.

The problem asks for the number of cubes painted red on exactly one face. :wink:

Je suis desole mais je ne comprends pas tres bien l'anglais.
I have misunderstood the question. :oops:

In each face of cube there are 9 cubes one face's colored so easy 1 cubes has 6 faces

$ 6*9\equal{}54$

this is an interesting problem, i've seen it many times, with many variations, and what i just do is do... ((x-2)^2)*6 - just look at one side, and for the area it's x^2, but two edges are taken away from each side, so it's now (x-2)^2.

you know that cubes on any edge will have more than one: therefore, only the cubes in the central 3X3 square of each face will have one coloured side. And since there are 6 sides:

[b]3X3X6=54[/b]


geometry
function
LaTeX
Euler
inequalities
inequalities proposed

Prove that in a convex hexagon ABCDEF we have $\displaystyle AB^2+BC^2+CD^2+DE^2+EF^2+AF^2\geq \frac{1}{2}(AD^2+BE^2+CF^2)$

this was given in a recent contest -gh vranceanu?anyway i found a solution but not beautiful at all.one thing i did not manage to do was to find when equality holds(i saw that we have equality when those points form a regular hexagon but i could not prove that that is the only case).do you have a solution mateivld?

I need a nice soluition to this.Can somebody help me?
 :)

i think mine is not so beautiful.unless someone does not find a better one in the next few days i will post it

[quote="bodom"][quote="leepakhin"][quote="mateivld"]Prove that in a convex hexagon ABCDEF we have $AB^2+BC^2+CD^2+DE^2+EF^2+AF^2\geq 2(AD^2+BE^2+CF^2)$[/quote]
Should it be $AB^2+BC^2+CD^2+DE^2+EF^2+AF^2\geq\frac{1}{2}(AD^2+BE^2+CF^2)$? :?[/quote]
no it should not. 2 is a good constant[/quote]
excuse me for my stupidity leepakhin.it is indeed 1/2 :blush:  :blush:  :blush:

[quote="leepakhin"]Consider the case of a regular hexagon... Am I missing something? :huh:[/quote]
u are not missing anything.sorry again.mateivld u should correct that

i'm very curious if someone could find a solution using only geometry cuz mine reduces the problem to an "algebric" ineq if i may say so.

Sorry for the mistake, but can someone post a solution no matter how ugly it is?

i will post my solution in the evening.i'm toooo lazy right now in the morning :P  :P

ok.let's solve it.as i said i did not manage yet to prove the equality case(the truth is i lost my interest after i solved the first part) :P .consider a system of orthogonal axis so that all the coord xi,yi are positive.let A(x1,y1);B(x2,y2);C(x3,y3);D(x4,y4);E(x5,y5);f(x6,y6).our ineq becomes 4(sigma(x1^2)-4(sigma(x1x2))>=sigma(x1^2)-2(x1x4+x2x5+x3x6).we see that 2sigma(x1^2)>=2sigma(x1x2) so i suffices to prove that sigma(x1^2)-2(sigma(x1x2)-x1x4+x2x5+x3x6)>=0
consider the function f:R-->R
f(x)=x^2-(x1+x2+x3+x4+x5+x6)x+sigma(x1x2)-1/2(x1x3+x1x5+x2x4+x2x6+x3x5+x4x6)
the determinant of the equation f(x)=0 is delta( :blush: )=(sigmax1)^2-4sigma(x1x6)-2(x1x3+x1x5+x2x4+x2x6+x3x5+x4x6) but delta>=0is exactly our ineq.now if we can find x from R such that f(x)<=0 then delta>=0 and our prob is solved.let's say that max(xi)=x1 and let us take x=x1
now we must show that f(x1)<=0 and that is easy task.do the same replacing xi with yi(of course u will take x in the function above max(yi))and the prob is over.but still it is hard to see when we have equality in this kind of sol.perhaps someone could find a better one.

sorry for not using latex but i have problems when i want to download it.hope i will solve that soon :D

[quote="mateivld"]Well nobody could solve it in 2005,so can somebody can try to solve it this year?:D[/quote]
 :P  i found this sol on the 25th of december 2005 :P

[hide="my solution using the Euler-inequality."][b]Lemma 1.[/b] Let $XYZT$ be a quadrilateral (what isn't necessary convex).
I note the middlepoints $U$,$V$ of the segments $XZ$,$YT$ respectively.
Then there is the following [u]Euler's identity[/u]:
$XY^2+YZ^2+ZT^2+TA^2=XZ^2+YT^2+4\cdot UV^2$

[u]Particularly[/u], $XY^2+YZ^2+ZT^2+TA^2\ge XZ^2+YT^2\ \ (1)$.

We have equality in the relation $(1)$ iff $U\equiv V$.

[b]Solution of the proposed problem.[/b] I note: $AD=x,\ BE=y,\ CF=z$;
$AB=a,\ BC=b,\ CD=c,\ DE=d,\ EF=e,\ FA=f$;
$BF=a_1,\ AC=b_1,\ BD=c_1,\ CE=d_1,\ DF=e_1,\ EA=f_1$;

We will apply the inequality $(1)$ in the following quadrilaterals:
$ABDE\Longrightarrow a^2+d^2+c^2_1+f^2_1\ge x^2+y^2\ ;$
$BCEF\Longrightarrow b^2+e^2+d^2_1+a^2_1\ge y^2+z^2\ ;$
$CDFA\Longrightarrow c^2+f^2+e^2_1+b^2_1\ge z^2+x^2\ .$ a.s.o.
To-morrow.[/hide]

very nice your solution.i have tried with euler but did not see that beautiful way.very nice again ;)  :P

Bodom, I am sorry! I made a small mistake.  I corrected above.

your solution is still nice but now all the magic is lost :(

[hide="my solution using the Euler-inequality and the Leibniz-inequality."][b]Lemma 1.[/b] Let $XYZT$ be a quadrilateral (what isn't necessary convex).
I note the middlepoints $U$,$V$ of the segments $XZ$,$YT$ respectively.
Then there is the following [u]Euler's identity[/u]:
$XY^2+YZ^2+ZT^2+TA^2=XZ^2+YT^2+4\cdot UV^2$

[u]Particularly[/u], $XY^2+YZ^2+ZT^2+TA^2\ge XZ^2+YT^2\ \ (1)$.

We have equality in the relation $(1)$ iff $U\equiv V$.

[b]Lemma 2.[/b] Let $ABC$ be a triangle with the centroid $G$. Then for any point $P$ there is the following [u]Leibniz's identity[/u]:
$PA^2+PB^2+PC^2=3\cdot PG^2+\frac 13\cdot (AB^2+BC^2+CA^2).$

[b]Lemma 3.[/b] Let $A_1A_2A_3$, $B_1B_2B_3$ be two triangles with the centroids $G_a$, $G_b$ respectively. Then exists the following identity:
$\sum\limits_{j=1}^3 \sum\limits_{i=1}^3 A_iB^2_j=9G_aG^2_b+\sum\limits_{1\le i<j\le3} A_iA^2_j\ +\sum\limits_{1\le i<j\le 3}B_iB^2_j.$
Indeed, we will apply the identity from the lemma $(2)$ to the points $B_k$, $k\in \overline {1,3}$ w.r.t. the triangle $A_1A_2A_3$:
$B_kA^2_1+B_kA^2_2+B_kA^2_3=3\cdot B_kG^2_a+$ $\frac 13\cdot \sum\limits_{1\le i<j\le 3}A_iA^2_j$. Thus,
$\sum\limits_{j=1}^3\sum\limits_{i=1}^3 A_iB^2_j=3\cdot \sum\limits_{k=1}^3B_kG^2_a+\sum\limits_{1\le i<j\le 3}A_iA^2_j$. But again
$\sum\limits_{k=1}^3 B_kG^2_a=3\cdot G_aG^2_b+\frac 13\cdot \sum\limits_{1\le i<j\le 3}B_iB^2_j$ a.s.o.

In particularly, $\sum\limits_{j=1}^3\sum\limits_{i=1}^3A_iB^2_j\ge \sum\limits_{1\le i<j\le 3}A_iA^2_j\ +\sum\limits_{1\le i<j\le 3}B_iB^2_j\ \ (2)$.

We have equality in the relation $(2)$ iff $G_a\equiv G_b$.

[b]Solution of the proposed problem.[/b] I note: $AD=x,\ BE=y,\ CF=z$;
$AB=a,\ BC=b,\ CD=c,\ DE=d,\ EF=e,\ FA=f$;
$BF=a_1,\ AC=b_1,\ BD=c_1,\ CE=d_1,\ DF=e_1,\ EA=f_1$;

We will apply the inequality $(1)$ in the following quadrilaterals:
$ABDE\Longrightarrow a^2+d^2+c^2_1+f^2_1\ge x^2+y^2\ ;$
$BCEF\Longrightarrow b^2+e^2+d^2_1+a^2_1\ge y^2+z^2\ ;$
$CDFA\Longrightarrow c^2+f^2+e^2_1+b^2_1\ge z^2+x^2\ .$
From the sum of these relations results $\boxed {\sum a^2+\sum a^2_1\ge 2\cdot \sum x^2}\ \ (3).$

We will apply the inequality $(2)$ for the triangles $ACE$ and $BDF$:
$\left(AB^2+AD^2+AF^2\right)+\left(CB^2+CD^2+CF^2\right)+\left(EB^2+ED^2+EF^2\right)\ge$
$\left(AC^2+CE^2+EA^2\right)+\left(BD^2+DF^2+FB^2\right)\Longleftrightarrow \boxed {\sum a^2+\sum x^2\ge \sum a^2_1}\ \ (4)$.

From the sum of the relations $(1)$ and $(2)$ we obtain the proposed inequality \[ \boxed {2\cdot \sum a^2\ge \sum x^2}\ \ (5)  \] We have equality in the relation $(5)$ iff the triangles $ACE$, $BDF$ have a same centroid and the middlepoints of the diagonals $AD$, $BE$, $CF$ coincide with the centroid of the hexagon. [/hide]


I face some problem in answering these question .Hope someone can explain it for me .Thanks

1)Which of the following species ,doesnt not involve dative bond
A.$HF$  B.$NO_2$ C.$Zn(OH)_4^{2-}$  D.$H_3O^{+}$

2)Which one of the following correctly describes the $Cl-P-Cl$ bond angles in a molecule gaseous phosphorus pentachloride .
A .$120^{\circ}$ only  B.$90^{\circ}$ and $120^{\circ}$  C.$90^{\circ}$ and $180^o$ D.$90^o,120^o,180^o$

3)What is the shape of $I_3^{-}$ ？

4)The chlorine atom present in chlorobenzene is ortho , para directing with respect to further substitution because
A.chlorine withdraws electrons from the ring
B.chlorine activates the ring with respect to the electrophilic substitution
c. the carbon atom combined with the chlorine is sp hybridized
D. chlorine is in electronegative element .

1. B.  A dative bond is one in which one species donates both electrons in the bond.  HF is a combination of H+ and F-.  H+ has no electrons, therefore H-F bond must be dative.  Tetrahydroxyzinc ion is a metal coordination compound.  The OH ion donates two of it's electrons to the Zn ion in order to form the coordination bond.  Hydronium ion is dative for the same reason HF is dative.

2. B.  This follows the trigonal bipyramidal geometry.  There are 3 equatorial bonds each separated by 120 degrees from each other.  The two axial bonds are each separated 90 degress from the 3 equatorial bonds.

3. Lewis structure of the triiodide ion should reveal that it is bent.  The ion is dsp3 hybridized, but the two axial and 1 equatorial position are occupied by lone pairs.

4. uhh... I don't think any of those answers are correct.  Here's why: A. chlorine does withdraw electrons from the ring, but this only explains the fact that chlorine deactivates the benzene ring to electrophilic substitution.  B. As stated above, chlorine deactivates the ring to electrophilic substitution. C.  The carbon atom is clearly not sp hybridized.  D. Though chlorine is an electronegative element, this does not explain why chlorine is ortho, para directing.  This statement is simply a restatement of choice A, that chlorine withdraws electrons from the ring.  The reason that chlorine is ortho, para directing is because it is able to donate two electrons into the ring in order to stabilize the carbocation formed in the process of electrophilic ring substitution.  However, because chlorine is electronegative, it actually deactivates the aromatic compound.

Taylor, I agree with the answer to 4.  However, I am going to have to dispute your answer to the rest of these.

1.  B.  This one I'm not sure about... but it seems that H--F is a combination of neutral H and F, with each atom donating a single electron (it certainly is polar, however, but I don't think that matters in this problem).  And furthermore, draw out one of the Lewis structure of NO2:
/ .. / . /..
:O=N-O:      (ok not the best, but it'll have to do; the slashes are placeholders).  Notice that the singly bonded O
/.. //// ..       has eight electrons.  A neutral O has six electrons.  Hence, both of the electrons came from N.

2.  D.  Don't forget the angle between the axial Cl's, which is 180.

3.  The lone pairs will take the equatorial positions, since this way they are strongly repelled by only two pairs rather than three pairs.  The shape of I3- is linear.

woooow... i'm just dumb... lol.  For 1, i just looked up dative bonding, and HNO3 was one of the examples of dative bonding, and it has similar structural features as NO2, so NO2 should be the correct answer.  For 2, hehe... tribute to reading all of the answer choices... lol.   and number 3, yes... it's supposed to be linear... lol...

thanks for both of your help :D .

Another question our teacher ask is , how to differentiate between propan-1-ol and propan-2-ol without using Iodoform test  :mad: ? Thanks    :)

It's ok i still love you.  Don't worry, that has happened to me more times than I can count.

[quote="gregx007"]shyong:  http://en.wikipedia.org/wiki/Lucas%27_reagent[/quote]

Gosh I forget this one and keep thinking of using oxidation where propan-2-ol stops at propanone but propan-1-ol keeps on oxidation until propanoic acid .... :mad: ....

you could do that as well; add NaOH or something and see if you get a water-soluble result.  Or you could add PCC and do a Tollen's test.  However, I think that Lucas' Test is easiest.

hmm can someone give me an idea of wat the lewis structure of triiodide looks like? drawing it out and giving each of the iodide ions a full ring of 8 still only gives me 20 electrons and not 22 so im not sure where the other 2 are and how it is linear


Given a polynomial [tex]P(x)[/tex] of [tex]2005th[/tex] degree, all its coefficients rational. [tex]P(k)=\frac{k}{k+1},\ k=0,1,...,2005[/tex]. Find [tex]P(2006)[/tex].

I believe this is a standard old olympiad question and this might be the wrong place to ask it.But anyways, here is my solution. :)

$ Consider\ the\ function\ F (\ X )\ =\ (\ X +\ 1 )\ \ P (\ X )\ - X \\ $

Clearly, it has 2005 roots $ (\ 0, 1 , 2, .... , 2005 )\ $ .

Also $ F (\ -1 )\ =\ 1 $ by substitution. Since the polynomial is of $ 2005 $ degree it has 2005 roots all of which we know.

$ \therefore F (\ X )\ =\ (\ (\ X )\ (\ X -1 )\ .... (\ X - 2005 )\ )\ \times C\ $
where C is a constant which can be found by using the fact that $ F (\ -1 )\ =\ 1 $ Once we have determined the polynomial, we can easily find $ F (\ 2006 )\ and\ \therefore P (\ 2006 )\ $

[hide]I am a bit lazy so I haven't given you the final answer Sorry !! [/hide]

Thanks. There's a minor mistake in your solution. [tex]F(x)[/tex] is certainly of [tex]2006th[/tex] degree, so it has [tex]2006[/tex] roots: [tex](0,1,2,...,2005)[/tex]. Everything else clear enough. Thanks once again.


I'm doing Mandelbrot next year.
I don't completely understand the Mandelbrot scoring system. Could someone explain to me?

Please correct me if I'm wrong; I'm not completely sure about the details.

For the individual competition, there are five rounds, each with 7 questions.

Within a round, each problem is assigned a point value, depending on difficulty.

1,1,2,2,2,3,3 for a total of 14 points per round possible, and 70 over all 5 rounds. The regional and national levels have the same point system.

A similar system exists for the team competition. Each question is assigned a point value, and a maximum 28 points can be scored on a round. With 3 rounds, there is a maximum score of 84.

What is the maximum difficulty of proof round? Is that covered by Vol.2?


Give $ m_a,m_b, m_c$ are three madias line of trangle ABC . And r is radius of inside circle
Prove:    $ \frac{m_am_bm_c}{m_a^2\plus{}m_b^2\plus{}m_c^2} \geq r$


inequalities
rotation
geometry
rectangle
symmetry
complex numbers
triangle inequality

Let $ z_1,...,\ z_n$ be complex numbers such that $ z_1\plus{}z_2\plus{}...\plus{}z_n\equal{}0$ and $ \left|{z_1}\right|\equal{}\left|{z_2}\right|\equal{}...\equal{}\left|{z_n}\right|\equal{}1$

Prove (both algebraically and geometrically ) that for a complex number $ z$ it holds
 $ \left|{z\minus{}z_1}\right|\plus{}\left|{z\minus{}z_2}\right|\plus{}...\plus{}\left|{z\minus{}z_n}\right|\geq n$

[hide="Geometrically"] Clearly WLOG $ z_k \equal{} e^{i \frac {2\pi k}{n} }$.  Then geometrically

$ |z \minus{} z_k| \plus{} |z \minus{} z_{n/2 \plus{} k} | \equal{} |z_k \minus{} z| \plus{} |z_k \plus{} z| \ge |2z_k| \equal{} 2$

(the Triangle inequality), so the conclusion follows (for even $ n$; for odd $ n$, rotate the point farthest away from $ z$ to be $ \minus{}1$ without loss of generality (this distance is at least $ 1$), and pair up the remaining points accordingly) . [/hide]

Those conditions don't imply that it is a regular polygon...

Just take two sets of points that are on the unit circle, that have a centroid at 0 and union of those two sets clearly has the same property, but it does not necessarily have to be regular.

Just take any rectangle, for example.

I am not sure if you need to have symmetry or not. Or must we decompose things into regular polygons.

Hint for another solution wirite the inquality in the equivalent version
$ \sum |\bar{z_i}|\cdot |z \minus{} z_i|\geq n$  and then use the triangle inequality


Αν a,b,c θετικοί αριθμοί ν.δ.ο.

$ {(a\plus{}\dfrac{1}{b}\minus{}1)(b\plus{}\dfrac{1}{c}\minus{}1)\plus{}(b\plus{}\dfrac{1}{c}\minus{}1)(c\plus{}\dfrac{1}{a}\minus{}1)\plus{}(c\plus{}\dfrac{1}{a}\minus{}1)(a\plus{}\dfrac{1}{b}\minus{}1)}\geq{3}$

Να μια λύση που έχω βρει ...
Έστω $ x=\frac a+\frac{1}{b}-1$ κτλ  τότε

$ (x+1)(y+1)(z+1)\geq 2+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5+x+y+z$

άρα $ xyz+xy+yz+zx\geq 4$ (1) .Προφανώς αν x είναι ο max τότε $ y+z>0$ άρα $ x>0$ και $ yz\leq 0$ ή $ x>,\ y>0,\ z>0$

Αν   $ x>0$ και $ yz\leq 0$ τότε $ xy+yz+zx>3$ το ζητούμενο .

Αν ;ola είναι θετικα τότε $ xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2}$ άρα $ xyz\leq\sqrt{\frac{(xy+yz+zx)^3}{27}}$  (2)

και αν $ xy+yz+zx=q$ τότε από (1) ,(2) έχουμε

$ \sqrt{\frac{q^3}{27}}+q\geq 4$  που ισοδυναμεί με

$ \frac{(q-3)(q-12)^2}{27}\geq 0$ άρα πρέπει $ q\geq 3$ άρα $ xy+yz+zx\geq 3$  qed :D

Ωραία λύση Σιλουανέ

exw apo AM-GM
 $ (a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1) \plus{} (b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1) \plus{} (c \plus{} 1/a \minus{} 1)(a \plus{} 1/b \minus{} 1) > \equal{} 3*riza tritis( (a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1) )^2$
arkei dld $ riza tritis ((a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1))^2 > \equal{} 1$
simfwna me tn anisotita holder exw:
$ (a \plus{} 1/b \minus{} 1)^{1/3} (b \plus{} 1/c \minus{} 1)^{1/3} (c \plus{} 1/a \minus{} 1)^{1/3}>\equal{}(abc)^{1/3} \plus{}(1/abc)^{1/3} \minus{} 1$
thetw $ (abc)^{1/3} \equal{} x > 0$ kai exw na apodeiksw $ x \plus{} 1/x \minus{}1 >\equal{} 1$ <=> $ x \plus{} 1/x>\equal{} 2$  ,poy isxyei
ara exw $ ((a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1))^ {1/3}>\equal{} 1$
<=>
$ (a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1)>\equal{}1^3\equal{}1$
<=>
$ riza tritis ((a \plus{} 1/b \minus{} 1)(b \plus{} 1/c \minus{} 1)(c \plus{} 1/a \minus{} 1))^2 > \equal{} 1$
xrisimopoiw tin anisotita x + 1/x >=2 gia x thetiko. an exw kanei kapoio lathos peite to mou.

[quote="Athinaios"]
ipothetw xwris vlavi tis genikotitas $ a > \equal{} b > \equal{} c$ kai exw $ 1/a \minus{} 1 < \equal{} 1/b \minus{} 1 < \equal{} 1/c \minus{} 1$
[/quote]

Φίλε Athinaios κατ' αρχήν η συγκεκριμένη ανισότητα είναι [i]κυκλική[/i] αλλά όχι [i][u]συμμετρική[/u][/i]. Ετσι μπορείς να υποθέσεις π.χ. ότι το $ a$ είναι το μεγαλύτερο στοιχείο απ' τα 3 αλλά δεν μπορείς να υποθέσεις χωρίς βλάβη της γενικότητας αυτή την διάταξη των $ a,b,c$ . Για να είναι συμμετρική πρέπει αν αλλάξεις 2 οποιαδήποτε γράμματα μεταξύ τους η ανισότητα να μέινει ίδια πράγμα το οποίο δεν συμβαίνει εδώ.

to diorthwsa allazwntas ena meros tis lysis.  :)

[quote="Athinaios"]simfwna me tn anisotita holder exw:
$ (a \plus{} 1/b \minus{} 1)^{1/3} (b \plus{} 1/c \minus{} 1)^{1/3} (c \plus{} 1/a \minus{} 1)^{1/3} > \equal{} (abc)^{1/3} \plus{} (1/abc)^{1/3} \minus{} 1$
[/quote]

Καλή η προσπάθεια σου αλλά εδώ κάνεις λάθος .Όταν εφαρμόζεις Holder πρέπει οι αριθμοί σου να είναι θετικοί (και το -1 δεν είναι θετικός)  :wink:

ap;oti fenetai dn m thelei ayti i anisotita...alla tha tn palepsw ligo akoma :oops:


[quote="Myth"]I want Mathlinks shirts!  ;)  :D[/quote]

Unfortunately we can't do everything at once.  We asked if Santa would lease out his elves so that everything could get done in a year, but the rates were too high.

[quote="Valentin Vornicu"]There will be MathLinks shirts available, just that the request is heavier at this moment for AoPS T-shirts. There is a MathLinks T-- in planning ;)[/quote]

First you need a thread for it.  ;)

Well, I suspect that a lot of MathLinkers want theirs by July. ;) (More specifically, they will purchase it personally from Valentin in July and get it autographed by lots of other MathLinkers there. :D )

That would be pretty much the plan :) :P

[quote="Valentin Vornicu"][quote="shobber"]Well, i am in China, can i also get one??[/quote]You will be able to get one, once we get it done.[/quote]
Thanks! :D I am looking forward to it... ...

[quote="Moubinool"]Would be a nice present for moderators to receive a free mathlink T-shirt, isn't it ?[/quote]

Well, that would be nice, but I don't think Valentin would like to pay of his own pocket a T-shirt for you, me and every single moderator on the forum (there are quite a lot actually), specially that this would mean that he would have to pay for the shipping too  :D , and well, if the t-shirts are produced in Romania, I suppose it would be expensive to make them reach every single  spot in the planet!

[quote="shobber"]how much will it be??[/quote]It will cost around 15-20$\$ depending on the location and where we will make them (ie in Romania or in US) ...
As djimenez pointed a t-shirt is around 10-15$\$ to which you add the transportation price and you get to the final price of 15-20$. I want them to be available around the end of May, but we will see  :?

[quote="shobber"]not cheap for me...
after all, i am a high school student and perhaps won't be able to afford 150RMB for just a Tshirt...[/quote]Those were estimative prices. The final price may be lower, depending on how many T-shirts we make at once.

[quote="shobber"]not cheap for me...
after all, i am a high school student and perhaps won't be able to afford 150RMB for just a Tshirt...[/quote]

You could go into debt for your favorite community!  :D  Just kidding...

members in all countries can have Mathlinks T-shirts??

T-shirts will soon be available for buying. Unfortunately, their price, for a single item, with shipping is around 20-30$ depending on your country of origin. This is because the shipping costs much in case of one item.
So to minimize the costs I suggest people from the same cities submit an order together. This might lower the costs up to 15$ (with shipping) per t-shirt (in case of 5 people ordering each one a shirt).

I will give more details later on, but I wanted to know what you guys feel about this  :?  :?

[quote="Valentin Vornicu"]T-shirts will soon be available for buying. Unfortunately, their price, for a single item, with shipping is around Dollars 20-30 depending on your country of origin. This is because the shipping costs much in case of one item.

So to minimize the costs I suggest people from the same cities submit an order together. This might lower the costs up to 15 Dollars (with shipping) per t-shirt (in case of 5 people ordering each one a shirt).

I will give more details later on, but I wanted to know what you guys feel about this  :?  :?[/quote]

Valentin, are they going to be produced on the US or in Romania (or somewhere else :? ).

Now, if there is any interest in the profits to sustain the forum, well, most of us like the forum a lot and wouldn't care to buy it that way, but if you are giving them basically at cost, well, they could be produced in several places at once (Just an idea).

"You Have Read At Least One Post by Grobber."
 :D Impressive!

Really this the simplest way to handle things, rather than to produce them in all over the world.
The t-shirt's price is the cut price. I prefere these people for many reasons, the most important being the quality of their fabric and printing. The t-shirt's gotta last long, and these ones are guaranteed 100 washes. In plus you can return them for a full-refund in case you don't like them in the end.

[quote="kueh"]hmm will there be someone at the imo?  ;)[/quote]I will put up the details sometime this week. I certainly hope so, given the shipping time of 5 days :)

There's a mistake in number 5 (the first one):

"You know that a correct solution may worth only 6..."

Correct version:

"You know that a correct solution may be worth only 6..."

Only the text ? AoPS looks nice with the little "cube".. :P

I just noticed that "Olympiad" is spelled "Olympad" in the t-shirt...

[quote="shobber"][quote="dts"]Another correction: "You already have a shirt identical with this one" should be "You already have a shirt identical to this one".[/quote]
what's the difference?[/quote]

One has with, the other has to.  In this context, to is the correct word.

IMO is approaching more and more and I think we should revive this topics. Otherwise T-Shirts are not very likely and it is hard for MLer to recognize each other. On the hand we should recycle some of the respective AoPS thread as they are obviously some similarities.

You are a MLer if ...

- you are not scared by IMO Shortlists.
- you wonder whether Orlando is the German IMO Team Leader.
- you know that Inequalities are the most discussed topic in the olympiad sections
- you know that Gabriel has proposed most of the ML contest problems
- you know that i-th ML contest results are not expected earlier than (i+2)-th ML contest
- you know that ML has the biggest number of people connected to IMO in the subspace of
forums worldwide
- you wonder whether it is possible to teach Darij a new important theorem in Euclidean geometry
- you know that Pierre has the best record of combinatorics references ever
- you wonder whether Vesselin is a math professor
- you wonder whether Mikhail can solve problems faster than you can typeset
- you wonder whether Alexandru solves math problems 24 hours a day
- you wonder how many people will wear a ML T-Shirt during IMO

cool, i want one.  What will the design look like?


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Hallo,

Consider R is a commutative ring with 1, O is the structure sheaf defined in page 741 of the "Abstract Algebra" book by Foote,

I need to find the stalk at P in spec(R) in the case of:

R= Z / 11 Z

P= (2) ideal generated by 2.

All what I know up to now is stalk at (2) is isomorphic to the localization (Z/11Z) with respect to the prime ideal (2).

Now I need to find what is localization of (Z/11Z) with respect to the prime ideal (2) ??

I guess it will be zero but I couldn't move any step in the definition of the localization.

Thanks.

$ P \equal{} R$ is not prime. in fact, $ 0$ is the only prime ideal of $ \mathbb{Z}/11$ and the corresponding stalk is $ \mathbb{Z}/11$.

if you meant $ R \equal{} \mathbb{Z}/10$, we have $ (\mathbb{Z}/10)_{(2)} \cong \mathbb{Z}_{(2)} / 10 \mathbb{Z}_{(2)} \equal{} \mathbb{Z}_{(2)} /2 \mathbb{Z}_{(2)} \cong \mathbb{Z}/2$.

Thanks for the explaination, but I was regarding Z/11Z as Z-module and I want to find the localization with respect to the multiplicative set S=Z/(2).

Sorry for the incomplete question.

Thanks

in other words, you want to find the stalk of the associated $ O_{Spec(\mathbb{Z})}$-algebra $ \widetilde{\mathbb{Z}/11}$ at $ (2)$.

every element of the $ \mathbb{Z}$-module $ \mathbb{Z}/11$ is annihilated by some element of the multiplicative subset $ \mathbb{Z} \backslash (2)$, namely $ 11$; thus the localization is zero.


Is there any $ t > 0$ such that
\[ 4e^{16t^3 \plus{} t^2 \plus{} 20t}\left(t^3 \plus{} t\right) < 1?
\]
And find $ \inf\{t > 0| t \text{ is the solution of the above inequality }\}$ if any? :oops:

$ \lim_{t\rightarrow 0} 4e^{16t^3 \plus{} t^2 \plus{} 20t}\left(t^3 \plus{} t\right) \equal{} 0$, so some interval $ (0; \epsilon )$ is solution of this inequality. Hense, the $ \inf$ is 0.


From Holder's inequality, $ \prod_{cyc}(1 \plus{} a^3)(1 \plus{} b^3)(1 \plus{} b^3)\ge \prod_{cyc}(1 \plus{} ab^2)^3$ which implies the above inequality.


Hello

Any ideas on this problem?
Find the last two digits of 2^(1998)
Find the last three digits of 2^(1998)
Find the last five digits of 2^(1998)

Regards

I am not sure if this helps in finding a solution. But if you look at the Sequence of remainders of $2^ n$ upon 10, 100, 1000, etc., it is periodic. I.e. for 10, it is 2 - 4 - 8 - 6 -2 - 4 - 8 - 6.

So the last digit of $2^{1998}$ is 4, since $1998 \equiv 2 \mod 4$.

I think, there is some form to determine the Length of this period (but I cannot remember it).

I hope, this helps you.

OK. Its the same Idea I had. But you have to guess, which power of 2 to use, dont you?

The best way to solve this type of problem is to use the euler's phi function.
$ a^{\phi (n)} \equiv 1$  mod  n

so for the a) $\phi (100) = 40$ so $2^{1998} \equiv 2^{(40*49)}*2^{38} \equiv 1^{49}* 2^{38} \equiv 44$ $mod$ $100$
b) $\phi (1000) = 400$ so $2^{1998} \equiv 2^{398} \equiv  344$ $mod$ $1000$
c) $\phi (100000) = 40000$  for this one just try use the doubling method that Schoppenhauer use.  so if my calculation is correct you should get 57344 for the last 5 digits.

Thank you all for your reply,
mtphaovibu: when you use the fact that a^(phi(n))=1 mod n, you need (a,n)=1 right? and (2,100) = 2.

In fact using a computer: 2^(40*49) = 76 mod 100.

2)
Notice that (2,125)=1 so 2^100 = 1 ( mod 125) . Hence 2^1998 =1/4 =94 ( mod 125 )
Since 2^1998 = 0 (mod 8 ) , by Chinese remainder theorem , solve for

x = 0 ( mod 8)
x= 94 ( mod 125)            where x = 2^1998

You will get  x=344 (mod 1000)

Clo you are making a mistake...
$a^{p-1}\equiv 1(mod$ $p) => (p,a)=1$
$a^{\phi(p)}\equiv 1(mod$ $p) => (p,a)=S$ and $S$ is not equal to $1$  :D

oops! I forgot about the G.C.D  well the answer turn out to be correct some how
well another way to approach would list the residue until it repeat, or using the doubling method


Hi,
Can someone please help me with the following problem:
http://www.oi.edu.pl/php/show.php?ac=e171100&module=show&file=zadania/oi12/aut
I found a K^2 dp algorithm, but that's too much (max K = $10^5$)..

Thanks in advance..

Well, I think the problem can be converted to shortest path problem. and we know dijskarta's algorithm which takes VlogV. So our algo is KlogK.

By the way, i am not sure how efficient it will be, but what's ur K^2 soln KPO?

Here's my K^2 recursive dp solution:

First I create a "fictional" bus stop at position (0,0) with 0 passengers. This bus stop will be identified as bus stop 0.
maxPassengers[i] denotes the maximum number of passengers I can take starting from bus stop i (of coordinates (x,y)).
To calculate maxPassengers[i]:
[code]
int max = 0;
For each bus stop j whose coordinates (x', y') verify x' >= x, and y' >= y
   if(maxPassengers[j] > max)
      max = maxPassengers[j]

maxPassengers[i] = busStop[i].numPassengers + max
[/code]

Then the solution is maxPassengers[0].

Also, can you please explain a bit more in detail about your shortest path solution?

Thanks,

Ok.

Assert 1,2,...,k to the k vertices.

Then make a path from i to j, if j is right-up of i. (may be at same x or y axis but not below). Assert the people number of i to the weight of that path. Now the Longest path (u may make it shortest path by multiplying each edge by -1) will be the answer. :lol:

Does anyone have better solution than this? :oops:

But you will also need K^2 time to build the graph, that is to determine, for each i, all the "right-ups" of i. Or is there is a more efficient way than to loop through all j and check if x(j) >= x(i) and y(j) >= y(i)?

well, the most efficient dijkstra uses heap, and it will make your code KlogK instead of K^2 :D

well i think for craeting the graph it will take K^2. So my algo is actually K^2+KlogK.

So does anyone have any better algo?

I finally found a solution that runs in O(NlgN) time:
The algorithm is based on the previous dp algorithm i gave above, but with some modifications:
First, sort the stations from bottom to top (increasing y), from left to right (increasing x).
Now, for each station i, instead of looping over all other stations whose x <= x(i) and y <= y(i) and taking the best value for maxPassengers, we will use a binary indexed tree (see "Mobiles" problem, IOI 2001):
This data structure will allow us to get the best value for maxPassengers for all stations whose x position is in the interval [1;p].
So now, for each station i, maxPassengers[i] = tree.get_max(x[i]) + busStop[i].numPassengers
Then, you just add the value of maxPassengers[i] in the tree.

So the final code will look like this:
[code]
sort stations bottom to top, left to right
for i=0 to K
   maxPassengers[i] = tree.get_max(x[i]) + busStop[i].numPassengers
   tree.insert(busStop[i].x, maxPassengers[i])
endfor;
ans = tree.get_max(maxX)
[/code]

Hope that helps,


inequalities
geometry
trigonometry
triangle inequality

I know that you can always construct a new triangle using medians of any triangle.Is it always possible the same construction for heights and bisectors(internal or external) any triangle? I know not always,but why?Any proof?Any criterion?

Well, there are formulae for the bisectors and altitudes in terms of the sides, so one could just plug those equations into the triangle inequality and call it a criterion, but that's not very enlightening.  The "why" has to do with the types of relationships the altitudes and angle bisectors have with respect to the sides.  For example, if we make two sides very big and one very small, because of how altitudes, area and sides interact we get two altitudes very small and one very large, so no triangle.  So I guess the criterion for altitudes can be expressed algebraicly as $\frac{1}{a} + \frac{1}{b} > \frac{1}{c}$ where in triangle ABC, $a \geq b \geq c$.  You can also do things with trig on the angles, and probably millions of other things.  I don't know any that is particularly beautiful or enlightening.


hello, we have $ 70^{\circ}\plus{}40^{\circ}\plus{}\angle1\equal{}180^{\circ}$, frome here we get $ \angle1\equal{}70^{\circ}$. Further we get $ 70^{\circ}\plus{}\angle2\equal{}180^{\circ}$, $ \angle2\equal{}110^{\circ}$. Further we get $ 2\cdot \angle3\plus{}110^{\circ}\equal{}180^{\circ}$, from here follows $ \angle3\equal{}35^{\circ}$.
Sonnhard.

[hide]40+70=110

180-110=70  >1=70 >2=110

4 and 3 are = so and the +=70 /2=35

so the answer is 35 or $\boxed{35}$[/hide]

[b][i][u]please rate my post thank you so much[/u][/i][/b]

please be fair though! thanks [=

 :spider:  :spider:  :spider:  :spider:


Bob's student ID number consists of 10 digits. Based on the clues below, what is Bob's number?
1.Each digit is different
2.The product of the sixth and seventh digits is equal to the third digit
3.The fourh, eighth and tenth digits are multiples of 3.
4.The sum of the fourth and sixth digits is the same as the sum of the fifth and eighth digits.
5.the second, third, sixth an dsevent digits are powers of 2.
6.The first, fifth, seventh and tenth digits are prime

[hide]
Note: Commas represent "or" - you'll see what I mean.

OK, lets call the digits a through j. Here's what's given:
2. $fg = c$
3. $d,h,j=3,6,9$
4. $d+f=e+h$
5. $b,c,f,g = 1,2,4,8$
6. $a,e,g,j = 2,3,5,7$

Since $g$ is on both of the last two lists, the only one shared by the two is 2, so $g=2$, leaving $b,c,f=1,4,8$ and $a,e,j=3,5,7$. Now, because $fg = c$, $2f=1,4,8$, and since g is already 2, the only remaining option for $f$ is 4. Then, $c=8$. In equation 5, all that remains is $b=1$. Looking at equations 3 and 6, we find that $j$ and $3$ are the only shared ones, so $j=3$. We now experiment with possible options for equation 4, and we find that: $d=9$, $h=6$, and $e=7$. In equation 6, the last remaining pair is $a=5$, and the only remaining value for $i$ is $0$. Thus, the final number is $abcdefghij=\fbox{5189742603}$ [/hide]

[quote="indianamath"]I dont think its possible. The only prime numbers are 2,3,5,7    but there is not a power of two that coincides with a prime.[/quote]

2 works...

oh sorry i didnt see that 7 was in both of them. :oops:


Μηπως ξερει καποιος καποια αλλη αποδειξη για το θεωρημα του Wilson κανοντας χρηση οτι η ομαδα $ (\mathbb{Z}^{*}_{p},\odot)$ ειναι κυκλικη. Το θεωρημα αυτο μας δινει ικανη και αναγκαια συνθηκη για να ειναι καποιος αριθμος πρωτος και ειναι
\[ (p - 1)!\equiv - 1modp\]

Στα στοιχεία $ 1,2,...,p \minus{} 1$ εμφανίζεται κάθε ζεύγος της μορφής $ \{x,x^{ \minus{} 1}\}$ (με $ x\neq x^{ \minus{} 1}$), το 1 και το -1 (η περίπτωση $ p \equal{} 2$ θέλει ειδική μεταχείριση). Συνεπώς το γινόμενο είναι -1.

Ωραιος Jondrette.
chr1855 αν θες και καποια αλλη αποδειξη με group theory σου κανω attach ενα σχετικο pdf που εχει αρκετα τετοια αποτελεσματα.

Jondrette sorry αλλα δεν καταλαβαινω που χρησιμοποιεις οτι η ομαδα ειναι κυκλικη?? athinaie πολυ ωραιο το pdf..

'Εχεις δίκιο. Δεν το χρησιμοποίησα. Μπορεί να χρησιμοποιηθεί ως εξής σε μια πιο περίπλοκη απόδειξη:

Αφού η $ (\mathbb{Z}/p\mathbb{Z})^*$ είναι κυκλική τα στοιχεία της έχουν τη μορφή:
\[ 1,a,...,a^{p \minus{} 2}\]
Άρα το γινόμενό τους είναι:
\[ x \equal{} a^{(p \minus{} 1)(p \minus{} 2)/2}\]
Η απόδειξη ολοκληρώνεται με δύο παρατηρήσεις:

ι) $ x\neq 1$ γιατί αλλιώς το $ a$ θα είχε τάξη μικρότερη από $ p \minus{} 1$

ιι) $ x^2 \equal{} 1$ το οποίο έπεται από το $ a^{p \minus{} 1} \equal{} 1$

Άρα $ x \equal{} \minus{} 1$. Παρατήρηστε ότι για το ι) χρησιμοποιούμε το γεγονός ότι το $ p \minus{} 1$ είναι σχετικά πρώτο με το $ p \minus{} 2$.

Κοντολογής, πάμε Μοναστηράκι - Ομόνοια μέσω Λαμίας ...

thanx καπως ετσι το σκεφτομουνα και εγω αλλα καπου το εχανα μετα..

Επειδή η ομάδα είναι κυκλική με περιττό αριθμό στοιχείων, η εξίσωση $ x^2 \equal{} 1$ έχει ακριβώς δύο λύσεις. Μετά συνεχίζουμε όπως στο πρώτο post του Jondrette.


limit
logarithms
function
inequalities
continued fraction
real analysis
real analysis unsolved

This one is very easy, but I find it rather cute :).

Is it possible to find an enumeration $(a_n)_n$ of the positive rationals  s.t. $(\sqrt[n]{a_n})_n$ is convergent?

Since it's so easy, how about a nice explicit enumeration with this property:

Write $n$ in binary, drop the leading 1, and append a terminal 1. Convert this binary string into a finite sequence of positive integers (except for the first, which may be zero) $b_0,b_1,b_2,\dots,b_{2k}$ by counting the alternating runs of ones and zeros. Convert this sequence into a continued fraction $b_0+\frac1{b_1+\frac1{b_2+\frac1{\dots+\frac1{b_{2k}}}}}.$ Interpret this continued fraction as a positive rational number $a(n)$.

Some properties of this enumeration: $a(2^n)=\frac1{n+1}$, $a(2^n-1)=n$. The subsequences $a(2^n),a(2^n+1),\dots,a(2^{n+1}-1)$ are increasing. This is enough to prove the problem asked, since $\lim_{n\to\infty}\frac{\ln a(n)}{n}=0$.

We even have that $\lim_{n\to\infty}\frac{\ln a(n)}{\ln n}=0$.

Now, we can define a function $f(x)$ for $0<x<1$ by letting $f\left(\frac{n+\frac12}{2^r}-1\right)=a(n)$ when $2^r\le n\le 2^{r+1}-1$, and extending to the whole interval by continuity- $f(x)$ is strictly increasing on the dense set of points we know, and the values it takes there are also dense. In general, $f$ converts the binary expansion of its argument into a continued fraction.
The inverse of $f$, which I will call $g$, takes a positive real number and encodes its continued fraction as a binary string, which is then interpreted as a real number. $g$ has the intriguing property that for all rational $r$ and all exponents $n$, $\lim_{x\to r}\frac{f(x)-f(r)}{(x-r)^n}=0$- but $g$ is still a strictly increasing function.

The enumeration is possible,
Let $a_n$ be an enumeration of the positive rationals, then we can reconstruct this sequence to get a new one, say $b_n$ such that $n > b_n > \frac{1}{2n}$ for all  big $n$.
Because, If $a_{n_0} \geq  n_0$ then put some rationals of $(0,1)$ in the new sequence before we get the value of $a_{n_0}$ ... e.t.c

The idea is that if we have establish that $b_n < n$ and then shift the sequence $b_n$ to the right then $b_n < n$ will still hold, so we can also achieve the second inequality $ b_n > \frac{1}{2n}$.

[quote="jmerry"]Some properties of this enumeration: $a(2^n)=\frac1{n+1}$, $a(2^n-1)=n$.[/quote]

I don't get this. If I do what you said prior to that, I'd get $\displaystyle a \left( 2^2 \right) = 1 + \frac{1}{2}$.

Maybe it would be clearer if I gave that case as an example:

$2^2=100_2$. This becomes the string $001$, which gives the continued fraction $0;2,1=0+\frac1{2+\frac11}=\frac13$.

The integer part of the continued fraction is the number of leading ones in the binary represtentation after removing one. In this case, that's zero.
It looks like you forgot the "drop the leading 1, append a terminal 1" step. If we don't drop the leading 1, we don't get the rationals between zero and 1. If we don't append a terminal 1, the function is 2-1; flipping the last digit wouldn't change $a(n)$.

Thanks for the reply... It is clear now.

As a side remark, I didn't forget to drop/append an one... what I did was to count the ones and the zeros from the end to the begining, not the other way around. The string $001$ ends with one $1$, then it has two $0$'s.
 :blush:


The lengths of the sides of a triangle are 9, 15, and 18. What are the lengths of the sides of a similar triangle with area 1/9 that of the given triangle?

[hide] Since area is a qudratic measure, the similarilty ratio for sides is $ \frac13$. Thus, the side lengths are $ 3$, $ 5$, and $ 6$.[/hide]


Problem 1
$a, b$ are natural numbers such that $\frac{a+1}{b}+\frac{b+1}{a}$ is an integer.
Prove that $gcd(a,b) \leq \sqrt{a+b}$

Problem 2
$a, b, c, d$ are positive reals such that $a+b+c+d \leq 1$. Prove that
$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\leq \frac{1}{64abcd}$

Problem 3
$a, b$ are positive reals such that $a+b = 1$. Prove that
$\frac{1}{3}\leq \frac{a^{2}}{a+1}+\frac{b^{2}}{b+1}< \frac{1}{2}$

Hmm... Never done an inequality before, so this won't seem very rigid:

Problem 2:

The problem is equivalent to $cd(a^{2}+b^{2})+ab(c^{2}+d^{2}) \leq \frac{1}{64}$.

If we maximize the left side by setting $a=b=c=d=0.25$, then we end up getting $\frac{1}{64}$.  Intuition tells me then that there is no way it could be greater than that since I maximized it, but I can't prove that.

[quote="Jan"]Problem 3:
\[\frac{a^{2}}{a+1}+\frac{b^{2}}{b+1}\geq \frac{(a+b)^{2}}{a+b+2}= \frac{1}{3}\]

[/quote]

Umm..did you manipulate the equation before applying cauchy....or do i have the definition wrong for cauchy because i dont see how you got that..

Problem 1 was a pretty nice problem:

Let $gcd(a,b)=x$ with $a=xa', b=xb'$ and $k=\frac{a+1}{b}+\frac{b+1}{a}=\frac{a^{2}+b^{2}+a+b}{ab}$.

Therefore, $(a+b)(a+b+1)=(k+2)a'b'x^{2}$
Now notice that $a+b+1=a'x+b'x+1$ is not divisible by $x$, unless $x=1$. If $x=1$ then clearly and $\sqrt{a+b}\geq gcd(a,b)=x=1$. Otherwise, the result follows from the fact that $x^{2}$ divides $a+b$, and thus ${a+b}\geq x^{2}=(gcd(a,b))^{2}$

I kinda brute forced problem 2 by using only AM-GM. I would like to see a more elegant solution.

[quote="euvtl"]
Problem 2
$a, b, c, d$ are positive reals such that $a+b+c+d \leq 1$. Prove that
$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\leq \frac{1}{64abcd}$
[/quote]

Expand out the LHS and cancel out $abcd$ in the denominator. Then we get

$a^{2}bc+b^{2}cd+c^{2}ab+d^{2}ab$
$= (ad+bc)(ac+bd)$
$\leq \frac{(ad+bc+ac+bd)^{2}}{4}$
$\leq \frac{1}{4}((a+b)(c+d))^{2}$
$\leq \frac{1}{4}((\frac{a+b+c+d}{2})^{2})^{2}$
$\leq \frac{1}{64}$


A point $O$ inside a convex quadrilateral is joined to its vertices .Find the area of the quadrilateral with vertices in the centroids $S_{1},\ S_{2},\ S_{3},\ S_{4}$ of the four triangles $ABO,\ BCO,\ CDO,\ DAO$ if the area of the $ABCD$  is $(ABCD)=81$

See the attached diagram.

$B'$ and $D'$ are, of course, the midpoints of $OB$ and $OD$ respectively. Then

\begin{eqnarray*}[AOB'] &=&{1\over 2}[AOB]\\ \lbrack COB'] &=&{1\over 2}[COB]\\ \lbrack COD'] &=&{1\over 2}[COD]\\ \lbrack AOD'] &=&{1\over 2}[AOD]\end{eqnarray*}

Adding those up, we get

$[AB'CD']={1\over 2}[ABCD]$

Next, from $AS_{1}={2\over 3}AB'$ and $AS_{4}={2\over 3}AD'$ it follows

$[AS_{1}S_{4}]={4\over 9}[AB'D']\qquad(1)$

Similarly, from $CS_{2}={2\over 3}CB'$ and $CS_{3}={2\over 3}CD'$ it follows

$[CS_{2}S_{3}]={4\over 9}[CB'D']\qquad(2)$

Adding up $(1)$ and $(2)$ we get

$[AS_{1}S_{4}]+[CS_{2}S_{3}]={4\over 9}([AB'D']+[CB'D'])={4\over 9}[AB'CD']$

In exactly the same manner, we find

$[B'S_{1}S_{2}]+[D'S_{3}S_{4}]={1\over 9}([ACB']+[ACD'])={1\over 9}[AB'CD']$

Therefore

\begin{eqnarray*}[S_{1}S_{2}S_{3}S_{4}] &=& [AB'CD']-([AS_{1}S_{4}]+[CS_{2}S_{3}])-([B'S_{1}S_{2}]+[D'S_{3}S_{4}])\\ &=& [AB'CD']-{5\over 9}[AB'CD']\\ &=&{4\over 9}[AB'CD'] \\ &=&{2\over 9}[ABCD]\end{eqnarray*}

For the given numeric data, we get $[S_{1}S_{2}S_{3}S_{4}]=18$


induction
inequalities unsolved
inequalities

[quote="hungkhtn"]Of course, Use induction we only need prove when $n-1$ variables equal.

And I think that $a=\sqrt{\frac{a_1^2+a_2^2+...+a_n^2}{n}}$.[/quote]
Yes,sorry for my error type.i've edited it!


algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Suppsose [b]A[/b]  = $<x^3 + x + 1>$ is a principal ideal in [b]$Z_2$[/b] generated by $x^3 + x + 1$.

How do you find the distinct cosets in [b]$Z_2 / A$[/b]? How many are they?

[quote="ZetaX"]I think you meant a principal ideal in $\mathbb{Z}/2\mathbb{Z}[x]$, not in $\mathbb{Z}/2\mathbb{Z}$.
Take division with remainder to get that any coset is (unique) represented by a polynomial of degree $2$ or less. Thus there are $8$ of them.[/quote]

I meant the principal ideal in $Z_2[x]$ generated by $x^3 + x + 1$. Can you give me an example of one of the cosets. My main problem is I am not sure what they look like and my book is not explaining well.

Some examples of 'smallest' representants are $1,x,x^2+x,x^2+1$ etc..
E.g. by definition you have that $1,1+(1+x+x^3)=x+x^3$ and $1+(x+1)(1+x+x^3)=x^2+x^3+x^4$ are in the same coset. In general, all polynomials of the form $1+p(x)(1+x+x^3)$ are in that coset and only those ($p(x)$ runs through all polynomials). The same works when replacing $1$ with another polynomial.
To see the structure of multiplication, always reduce to a polynomial with degree $<3$:
$(x^2+x)(x^2+1)=x^4+x^3+x^2+x$ is the same as $x^4+x^3+x^2+x-(x+1)(x^3+x+1)=x^4+x^3+x^2+x-x^4-x^2-x-x^3-x-1=x+1$ seen modulo $x^3+x+1$.


http://www.nus.edu.sg/oam/doc/uee/math.pdf

Thanks in advance.

#1 is B.

x-1 cannot equal 0, so x cannot equal 1.

#2 is B.

You use the bottom expression. 2^2 +1 =5.

AHHH there are too many of them.

I'll do the last one for fun.

#20 is B

10/18 x 9/17 = .294...


Prove that any 55 element subset of the integers in [1,100] contains elements that differ
by 10, 12 and 13 but need not contain elements differing by 11.

Proof for $ 10$: By the pigeonhole principle, there must exist a subset of six numbers ending in the same digit. It is then clear that two numbers must differ by $ 10$ since no subset of six integers can satisfy the condition.

Proof for $ 12$: Using the pigeonhole principle; there must exist a subset of at least five integers that are equivalent to each other modulo $ 12$. If there exists a subset with six numbers, the above reasoning holds. If there are only subsets of five integers, then for the classes of integers equivalent to $ 5$ through $ 11$ modulo $ 12$, the constraints cannot be satisfied yet again.

Proof for $ 13$: Again, by pigeonhole, there must exist a subset of five integers equivalent to each other modulo $ 13$. Since there are at most eight integers in a given modulo class, the constraints cannot be satisfied via the above reasoning.

Proof for $ 11$: Subset $ 1$ through $ 11$, $ 23$ through $ 33$, $ 45$ through $ 55$, $ 67$ through $ 77$, $ 89$ through $ 99$. is a valid subset that has no elements differing by 11.


[quote="Tiks"]$ a^7 \plus{} a^2 \plus{} 1 \equal{} (a^2 \plus{} a \plus{} 1)(a^5 \minus{} a^4 \plus{} a^2 \minus{} a \plus{} 1)$[/quote]

How did you realize this factorization?

I think you mean a complete set of residue classes (a complete set of residues), not a reduced set of residue classes.

[quote="bambaman"]It's a known and useful 'trick': let $ p(x) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} x^{a_i} \in Z[x]$, where $ a_i$ are non-negative integers. If they form a reduced residue system, $ p(x)$ is divisible by $ g(x) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} x^{i}$.
In our case, $ 0, 7, 2 \equiv 0, 1, 2 \pmod 3$, thus $ 1 \plus{} x \plus{} x^2 | x^7 \plus{} x^2 \plus{} 1$, and long division yields the factorization.

Proof of the lemma - $ g$'s roots are the n'th roots of unity, except for 1, and are all distinct (follows from $ g(x)(x \minus{} 1) \equal{} x^n \minus{} 1$). Thus, to show divisibility, one just have to plug the roots in $ p$ and check whether it yields 0. Because the roots satisfy $ x^n \equal{} 1$, we get that $ x^i$ is only dependent on the residue of $ i$ modulo $ n$. It follows, from the fact that $ 0,1,\cdots,n \minus{} 1$ is a reduced residue system mod $ n$ (and because the n'th roots of unity, excluding 1, are of the form $ a,a^2,\cdots,a^{n \minus{} 1}$ where $ a \equal{} e^{\frac {2i\pi}{n}}$), that $ p(x) \equal{} g(x) \equal{} 0$.[/quote]This is really helpful! I wonder why I've never seen this beautiful lemma before. :D

Then tell us what exactly is unclear instead of expecting us to explain everything.

[quote="persya"][quote="bambaman"]It's a known and useful 'trick': let $ p(x)\equal{}\sum_{i\equal{}0}^{n\minus{}1} x^{a_i} \in Z[x]$, where $ a_i$ are non-negative integers. If they form a reduced residue system, $ p(x)$ is divisible by $ g(x)\equal{}\sum_{i\equal{}0}^{n\minus{}1} x^{i}$.
In our case, $ 0, 7, 2 \equiv 0, 1, 2 \pmod 3$, thus $ 1\plus{}x\plus{}x^2 | x^7\plus{}x^2\plus{}1$, and long division yields the factorization.

Proof of the lemma - $ g$'s roots are the n'th roots of unity, except for 1, and are all distinct (follows from $ g(x)(x\minus{}1)\equal{}x^n\minus{}1$). Thus, to show divisibility, one just have to plug the roots in $ p$ and check whether it yields 0. Because the roots satisfy $ x^n\equal{}1$, we get that $ x^i$ is only dependent on the residue of $ i$ modulo $ n$. It follows, from the fact that $ 0,1,\cdots,n\minus{}1$ is a reduced residue system mod $ n$ (and because the n'th roots of unity, excluding 1, are of the form $ a,a^2,\cdots,a^{n\minus{}1}$ where $ a\equal{}e^{\frac{2i\pi}{n}}$), that $ p(x)\equal{}g(x)\equal{}0$.[/quote]
Can somebody post an other proof or explain that proof for me?
I can't understand the above proof :([/quote]

Note that $x - \omega$ and $x - \omega^2$ are the roots of $P_1(x)=x^2 + x + 1=0$ at $x= \omega, \omega^2$, where $\omega^3 = 1$. So those 2 linear polynomials are the factors of $P_1$. Also, $\omega \neq 1$ and the $x - \omega$ and $x - \omega^2$ also divide $P_2(x)=x^7 + x^2 + 1$. Thus, all the factors of $P_1$ are also factors of $P_2$, so $P_1$ itself must divide $P_2$.

Is there another way to solve this problem without using this lemma and the factorization from it?

[quote=Tiks][quote="Peter"]Prove that $1280000401$ is composite.[/quote]

$a^7+a^2+1=(a^2+a+1)(a^5-a^4+a^2-a+1)$ so every number that can be expresed in this form is composite.

Therefore,$1280000401=20^7+20^2+1$ is composite! :wink:[/quote]

How???