function
algebra
polynomial
topology
advanced fields
advanced fields unsolved

Hello,

I have the following result about extension of morphisms from curves to projective varieties:

Let $ X$ be an abstract nonsingular curve, let $ P \in X$, let $ Y$ be a projective variety, and let $ \varphi: X \setminus \{P\} \rightarrow Y$ be a morphism.  Then there exists a unique morphism $ \overline{\varphi}: X \rightarrow Y$ extending $ \varphi$.

I'm supposed to show the following:
(a) That this result fails if $ \dim X \geq 2$.  For example, let $ X\equal{}\mathbb{A}^2$, $ Y\equal{}\mathbb{P}^1$, $ P\equal{}(0,0)$, and $ \varphi: X \setminus \{P\} \rightarrow Y$.  Then this cannot be extended to a morphism $ \overline{\varphi}: X \rightarrow Y$ but I'm having trouble showing this precisely.

(b) That this result also fails if $ Y$ is not projective.  I was thinking about letting $ X$ be the blow-up of an affine variety, but then $ \dim X\equal{}2$ at the least and we've not shown that $ Y$ has to be projective.

Any hints/help would be very much appreciated.

For part (a), such a $ \overline{\varphi}$ could not possibly be continuous.  Part (b) is still giving me troubles though.

For part (a), to show that such a $ \overline{\varphi}$ cannot be continuous, does the following argument work?

Suppose there exists such a $ \overline{\varphi}$ extending $ \varphi$.  Let $ U \subset Y$ be an open set with $ \overline{\varphi}(P) \in U$.  Then, $ \overline{\varphi}^{ \minus{} 1}(U) \equal{} \varphi^{ \minus{} 1}(U) \cup \{P\}$, which is not open.

For the second part, the blowing-up doesn't work, but maybe this does?  It's kind of inspired by my answer to part (a).  View $ \mathbb{P}^1$ as having the point $ \infty$, and map it to $ \mathbb{A}^1$ or something.  But how can I map $ \mathbb{P}^1 \rightarrow \mathbb{A}^1$?

Sorry for spamming this topic, but no one's responding and I'm making progress of some sort.

Okay is it true that $ \varphi^{\minus{}1}(U) \cup \{P\}$ is necessarily not open?  I am not sure that that is true.

For (b), would the map $ \mathbb{P}^1 \rightarrow \mathbb{A}^1$, sending $ [x: y] \mapsto x/y$ suffice to show what I want?

Again, sorry to spam.

$ \overline{\varphi}^{ \minus{} 1}(U) \equal{} \varphi^{ \minus{} 1}(U) \cup \{P\}$, whose complement is just $ \mathbb{A}^2 \setminus \varphi^{ \minus{} 1}(U)$, and this is closed right?

Alternatively, could I do this?
Let $ U_y \subset \mathbb{P}^1$ be the open set where $ y \neq 0$, and let $ f: U_y \rightarrow k$ be the regular function given by $ f([x: y]) \equal{} x/y$.  Then, if $ \overline{\varphi}$ were a morphism, we would have to have regular functions pullback to regular functions, i.e., $ f \circ \overline{\varphi}: \overline{\varphi}^{ \minus{} 1}(U_y) \rightarrow k$ would have to be regular.  But, $ f \circ \overline{\varphi}$ is not defined (let alone regular) at $ P \equal{} (0,0)$.

Again, (b) is still giving me trouble though.  Comments/help much appreciated.

For (b), you could let $ X \equal{} \mathbb{A}^1$ and $ Y \equal{} \mathbb{A}^1 \setminus \{0\}$. Then you have the identity morphism $ X \setminus \{0\} \to Y$. Assume this can be extended to a morphism $ \varphi : \mathbb{A}^1 \to Y$.

Of course all morphisms $ \mathbb{A}^1 \to Y$ are classified as maps $ k[x][x^{\minus{}1}] \equal{} A(Y) \to A(\mathbb{A}^1) \equal{} k[y]$. Such a map is determined by the image of $ x$ in $ k[y]$. However its image must be invertible, with inverse the image of $ x^{\minus{}1}$; so the image lies in $ k$. This means the only morphisms $ \mathbb{A}^1 \to Y$ are constants; in particular there are no surjective morphisms, yet $ \varphi$ would be surjective if it existed.

For part (a), consider the usual projection $ p : \mathbb{A}^2 \setminus \{0\} \to \mathbb{P}^1$ given by $ (x,y) \mapsto [x,y]$. Suppose it had an extension $ \varphi$. Let $ q \in \mathbb{P}^1$ be different from $ \varphi(0,0)$. Then $ \varphi^{\minus{}1}(q)$ is a copy of $ \mathbb{A}^1 \setminus \{0\}$ inside $ \mathbb{A}^2$.

This is not closed, for if $ f(tx,ty) \equal{} 0$ for a particular $ (x,y)$ and all $ t \ne 0$, then we may let $ h(t) : \equal{} f(tx,ty)$ and notice that $ h$ is a polynomial in $ t$, while $ h$ has infinitely many roots and so $ h \equiv 0$. Therefore $ 0 \equal{} h(0) \equal{} f(0,0)$.

I'm sorry, but I don't see how that shows that $ \varphi^{\minus{}1}(q)$ is closed.  Could you elaborate?

The way I was initially attacking the problem was trying to show that regular functions don't pullback to regular functions.  Is there a way to do that?

I'm showing that $ \varphi^{\minus{}1}(q)$ is [i]not[/i] closed.  Under the Zariski topology, closed sets are of the form $ Z(I)$ where $ I$ is some ideal of polynomials. I showed that if $ f$ vanishes at all points in $ \mathbb{A}^1 \setminus \{0\} \subset \mathbb{A}^2$, then it necessarily vanishes at $ 0$ as well. Therefore there is no closed set $ Z(I)$ which contains $ \mathbb{A}^1 \setminus \{0\}$ but not 0; so, this set cannot be closed.

I meant to say not closed, but I see now.  Thank you.

In my previous post, is $ f \circ \overline{\varphi}$ regular?  It is well-defined, sort of by assumption.  But it shouldn't be regular, should it?  Or is there another instance of regular functions pulling back to non-regular functions?

It's certainly regular near every point different from $ (0,0)$. The only thing you'd need to worry about is what happens near $ (0,0)$. But that depends where $ (0,0)$ is mapped to by $ \overline{\varphi}$. If its mapped to $ [1,0]$ then the point is moot, since then $ (0,0)$ is not in the preimage of $ U_y$. (Then, of course, you could use $ [x,y] \mapsto y/x$ instead.)

It's almost not even worth considering then, though, because you will have that $ \overline{\varphi}^{\minus{}1}(U_y)$ is not even an open set. Presumably you could give it the subspace structure or something and then inquire as to whether or not its regular, but I don't see the point given that $ \overline{\varphi}$ can't be a morphism because it's not continuous. :).


algebra
system of equations
algebra unsolved

Find all tuples $ (x_1,x_2,x_3,x_4,x_5,x_6)$ of non-negative integers, such that the following system of equations holds:
$ x_1x_2(1\minus{}x_3)\equal{}x_4x_5 \\
x_2x_3(1\minus{}x_4)\equal{}x_5x_6 \\
x_3x_4(1\minus{}x_5)\equal{}x_6x_1 \\
x_4x_5(1\minus{}x_6)\equal{}x_1x_2 \\
x_5x_6(1\minus{}x_1)\equal{}x_2x_3 \\
x_6x_1(1\minus{}x_2)\equal{}x_3x_4$

From the first equation we get there is at least a zero in $ (x_3,x_4,x_5)$, similarly as $ (x_4,x_5,x_6)$, $ (x_5,x_6,x_1)$, $ (x_6,x_1,x_2)$, $ (x_1,x_2,x_3)$  and  $ (x_2,x_3,x_4)$.  WLOG, assume that $ x_3\equal{}0$, then
Case 1:$ x_6\equal{}0$
    we get $ x_1 x_2\equal{}x_4 x_5$ satisfies the conditions.By substituting the conditions in this case into the original equation.
Case 2:$ x_6 \neq 0$ , then $ x_1\equal{}0$ or $ x_5\equal{}0$
    If $ x_1\equal{}0$, by substituting $ x_1\equal{}0$ and $ x_3\equal{}0$ into the original equations, we get that there must be a zero in $ (x_4,x_5)$, and $ (x_5,x_6)$ from 1st and 5th equation , using $ x_6 \neq 0$ ,we get $ x_5\equal{}0$ also, and it's easy to check that this satisfies the equation.Similarly if $ x_5\equal{}0$, $ x_1\equal{}0$ also.
    So the solutions are $ (ab,cd,0,ac,bd,0)$, $ (0,x,0,0,y,0)$,$ (x,0,0,0,y,0)$,$ (0,g,0,h,0,i)$ and any cyclic permutation (may be I use the wrong term  :maybe: ), where $ a,b,c,d$ are any positive integers and $ x,y,g,h,i$ are any non-negative integers.

[quote="FelixD"]Find all tuples $ (x_1,x_2,x_3,x_4,x_5,x_6)$ of non-negative integers, such that the following system of equations holds:
$ x_1x_2(1 \minus{} x_3) \equal{} x_4x_5 \\
x_2x_3(1 \minus{} x_4) \equal{} x_5x_6 \\
x_3x_4(1 \minus{} x_5) \equal{} x_6x_1 \\
x_4x_5(1 \minus{} x_6) \equal{} x_1x_2 \\
x_5x_6(1 \minus{} x_1) \equal{} x_2x_3 \\
x_6x_1(1 \minus{} x_2) \equal{} x_3x_4$[/quote]
Adding all these equations we get,
\[ \sum_{i\equal{}1}^6 x_ix_{i\plus{}1} \minus{} \sum_{i\equal{}1}^6 x_ix_{i\plus{}1}x_{i\plus{}2}\equal{} \sum_{i\equal{}1}^6 x_ix_{i\plus{}1}
\iff S\equal{} \sum_{i\equal{}1}^6 x_ix_{i\plus{}1}x_{i\plus{}2}\equal{}0\]
So, $ (x_1,x_3,x_5)$ or $ (x_3,x_6)$ must be $ 0$. (or other cyclic permutations). Otherwise $ S>0$.
Now,we see that for the second case $ x_1x_2\equal{}x_4x_5$ satisfies the system of equations.

I get this
[quote="Moonmathpi496"]Adding all these equations we get,
$ S \equal{} \sum_{i \equal{} 1}^6 x_ix_{i \plus{} 1}x_{i \plus{} 2} \equal{} 0$
[/quote]

But how do yo get this?
[quote="Moonmathpi496"]
So, $ (x_1,x_3,x_5)$ or $ (x_3,x_6)$ must be $ 0$. (or other cyclic permutations). Otherwise $ S > 0$.
[/quote]
Why would it have to be greater than 0?

$ x_i$ is a non-negative integer for $ i \equal{} 1,2,3,4,5,6$.

Therefore, $ x_i \ge 0$, so $ x_{i}x_{i\plus{}1}x_{i\plus{}2} \ge 0$, and thus, $ S \ge 0$ as well.

So, if $ S \neq 0$, then $ S > 0$.

Whoops, I didn't see non-negative, lol  :blush:


One:

Factoring:
4x^2-1=0

Two:

Completing the square:
4x+2=x^2

Three:

The quadratic Formula:
4+4x=2x^2

[quote="And 1"]Yo whats the sig in problem number one x= (whats this) 1/2 ??[/quote]

Plus or minus. It means 1/2 can be positive or negative.

so like for the answer i can write both - or + right

Both of those are solutions.  The equation has two zeros (points that his the x-axis), so you have to write down both.


does someone can please help me to simplify this expression:

$\frac{1}{a^2-2a-3}+\frac{1}{2a^2-a-3} \cdot (a-\frac{3}{a-2})$

thanks in advance

[color=green][size=75]mod edit: latexed[/size][/color]

I'm feeling a bit slepy but i can tell you that alll you have to do is factorise each denominator by splitting the middle term.


find all triangular numbers that are also perfect squares.
[hide="Remark"]
triangular numbers are: $n=\frac{k(k+1)}{2}$
[/hide]

[hide="The first 4"]
$1,36,1225=\frac{49\cdot 50}{2},41616=\frac{288\cdot 289}{2}$
is there a general formula?
one can easily construct a brute force algorithm.[/hide]

That's a classic one. See e.g. http://en.wikipedia.org/wiki/Triangular_square_number


[quote="feynman_wannabe"]In base-induced elimination of cis & trans-2-methylcyclohexyl bromide, why does cis react fastly while trans reacts slowly?[/quote]

generally, stability is inversely proportional to reactivity.

I'm not entirely sure about this.  But when you draw the cis and trans isomers in the chair diagrams with the bromide in the axial position (required for E2 elimination), there are 2 hydrogens anti-paraplanar to the bromide in the cis isomer, but only 1 that is antiparaplanar in the trans isomer.  So there are two sites for the proton to be pulled off in the cis isomer, but only 1 site in the trans isomer, so I think that's the reason why E2 occurs faster for the cis isomer than the trans isomer.

yes hideaway is rite. the hydrogen must be anti to the leaving group. and also, the product formed due to elimination in case of cis is more stable as the alkene is more substituted.

Just a remark: the hydrogen doesn't [i]must[/i] be antiperiplanar, that is just a preferred pathway, but an elimination can (and will) occur syn.

that is what i was taught. :(  what do you mean 'will'? in this case why shud syn elimination take place, when anti elimiination is possible.

I thought the reason for the antiperiplanar requirement was that the C-H sigma bond had to overlap [i]constructively[/i] with the C-Br sigma* bond.  If E2 were to occur with the two groups synperiplanar, that would mean there would be destructive interference between the two MO's.  It seems like a synperiplanar E2 reaction would go against MO theory then, wouldn't it?  But I did find this site which gives pictures of the MO's interacting in the syn mode [url]http://library.tedankara.k12.tr/carey/ch5-6-1.html[/url].  Seems a bit implausible... I guess I don't understand MO's enough to figure it out.


What is the ratio of the volume of a cube with edge length six
inches to the volume of a cube with edge length one foot? Express
your answer as a common fraction.

Since $ 1\text{ foot}\equal{}12\text{ inches},$ the ratio of the volumes is $ \left(\frac{6}{12}\right)^3\equal{}\left(\frac{1}{2}\right)^3\equal{}\boxed{\frac{1}{8}}.$

Alternatively, we could find the volumes separately. $ 6^3\equal{}216$ and $ 12^3\equal{}1728.$ The ratio of the volumes is $ \frac{216}{1728}\equal{}\boxed{\frac{1}{8}},$ as before.


farz konid A={1,2,..........,n^2} .
tedade zirmajmoehaye n ozvie az A ra peida konid ke mojmoe hich zir majmueie az an bar n+1 bakhspazir nabashad.

mishe sabet kard hame azaye zirmajmoe n ozvi bayad be mod (n+1) hamnehesht bashand.
yani tedad mishe $ \phi(n\plus{}1)$


abstract algebra
logarithms
group theory
puzzles

Here is a problem that I thought is fun to tackle:

for all real numbers excluding 1: the operation * on G by a*b= a^ln b so for all a, b in G prove that G is an abelian group under the operation *.

[hide="One-liner"]$a^{\ln b} = e^{ \ln a \ln b} = b^{\ln a}$[/hide]

[quote="JBL"]That's just the abelian part; you haven't shown it's a group  ;)

What is the set which is supposed to be associated with this group operation?[/quote]
[quote="mikejones"]for all real numbers excluding 1:[/quote]
But isn't $\ln b$ not defined when $b \leq 0$?

Just checking--abelian means it's commutative, right?

[hide="Associativity"] $(a*b)*c = (e^{\ln a \ln b})*c = e^{\ln a \ln b \ln c} = a * (b*c)$ [/hide]
[hide="Neutrality"] $\exists I : I*a = a$
$e^{\ln I \ln a} = e^{\ln a}$
$I = e$ [/hide]
[hide="Inversion"] $\forall a : \exists b : a * b = I$
$e^{\ln a \ln b} = e^{1}$
$\ln b = \frac{1}{\ln a}$

Since $\ln x$ has range across the reals, such a $b$ exists. [/hide]
[hide="Closure (this one I have trouble with)"] $a*b$ is not closed for the reals except $1$!  It is closed for positive reals except $1$, or for $\mathbb{C}$ except $1$.   :huh: [/hide]


How many languages do you speak???

I speak 1 (english) and always feel left out when my Korean friends say stuff in Korean... :(

[quote="dynamo729"]feel left out when my Korean friends say stuff in Korean... :([/quote]

heheee... i'm one of those people who do that to my friends.. xD

Korean and English! w00t!

[quote="D3m0n Shad0w"]You speak as in "fluent"?[/quote]

Basically anything you can hear and understand, even with extended thought. (You dont have to be able to speak it so fast)

Mmm, one language then. I can speak some Mandarin, but not all of it and I certainly can't understand most of what my parents speak.

English for me. Yay... :surrender:

I can speak 0 languages, but I am learning Russian.  :)

100110010010111010

anyone speaking binary??
 :D
lol

What does 156858 have to do with anything? :P

chinese, english, japanese, latin, some spanish, and alien.

that's 6, or 4 if u dont count the last two. taiwanese is basically the same as chinese, but i can speak and understand it, so does it count?

I can speak English, French, Chinese, Alien 1, Alien 2, Alien 3, Alien 4, Alien 5, Alien 6, ..., Alien 798, Alien Omega. That's... 802 languages. Yay! :)

Accidently put 3 when it's actually 0.

[hide]
Public school language classes DO NOT count as fluent.
[/hide]

I'm uberfluent in English, although the grammar often gets me. I'm fluent in Food Tongue. And I'm working toward proficiency in Spanish.

Congrats on your 2000th post!  :D   :D   :D

[quote="1=2"]Based on this poll, hundreds millions of people do not know any languages.[/quote]

Or can't read English
Or are annoyed there was no otter option

Well, i am fluent with English, Bengali,Hindi and Urdu( Hindi is our national language , Urdu is the national language of Pakistan and Bengali is my mother tongue as well as the national language of Bangladesh). I also sometimes speak a mixture of these 4 languages .That's what my friends say.

me fluent in english and mandarin

me understands taiwanese and can kinda speak it

me is learning french in skool

so i put 2 :D

Actually I speak 3 languages
1. Georgian
2. Russian
3. English

I speak 2: English and I can carry on conversations fairly well in French. (Yay for French class!)

I live in the U.S. so I obviously speak English. I also know a bit of French, but I don't think I know enough to actually know it.


function
topology
real analysis
real analysis unsolved

Let f be a single-variable function defined on a closed domain. Prove that f is continuous if and only if the inverse image of every closed set is a closed set.

It's practically the definition. A function $ f$ from a topological space $ X$ to another space $ Y$ is continuous if and only if the inverse image $ f^{\minus{}1}(U)$ of any open set is open. $ f^{\minus{}1}(Y\setminus U)\equal{}X\setminus f^{\minus{}1}(U)$, so the inverse images of closed sets being closed is the same thing.


Choose any two positive numbers a and b. Draw a horizontal line segment of length b and then draw a perpendicular PR of length a/2. Now draw a circle, centre R, radius a/2. Draw the line segment through Q and R and extend it to meet the circle at S. What is the connection between the length of QS and one of the roots of the equation x^2 -ax - b^2 = 0? What, if any, are the constraints on a and b?

[quote="AndrewTom"]Choose any two positive numbers a and b.  Draw a horizontal line segment of length b and then draw a perpendicular PR of length a/2. Now draw a circle, centre R, radius a/2. >> Draw the line segment through Q and R and extend it to meet the circle at S.  << What is the connection between the length of QS and one of the roots of the equation x^2 -ax - b^2 = 0? What, if any, are the constraints on a and b?[/quote]
"Draw the line segment through Q and R and extend it to meet the circle at S."

Andrew Tom,

you abruptly started typing about point Q and point S here.  You did not introduce
them and their setting to the reader so the problem cannot be finished.

Sorry: the second sentence should read "Draw a horizontal line-segment PQ of length b and then draw a perpendicular PR of length a/2".

Well then, your drawing is
 [geogebra]1804fb205972f59abfccf40ce821ed9d74ce0dae[/geogebra]
and we see that $ ST$ has length $ a$ and $ PQ$ has length $ b$. Thus, by power of a point, we have
\[ b^2 \equal{} x(x\plus{}a)
\]
so
\[ x^2 \plus{} ax \minus{} b^2 \equal{} 0.
\]
So we see that one of the roots of $ x^2 \plus{} ax \minus{} b^2$ must be the length of $ QS$. As for restrictions, it looks like this diagram is valid for any positive $ a$ and $ b$, and the equation holds as well.


\[ \sum_{i \equal{} 0}^{n}\left[{n \choose i}^2\right] \equal{} ?\]

A hint:
That sum can be rewritten as $ \sum_{i\equal{}0}^n\binom{n}{i}\cdot\binom{n}{n\minus{}i}$.

$ (1 \plus{} x)^{2n} \equal{} {2n \choose 0} \plus{} {2n \choose 1}x \plus{} \ldots \plus{} {2n \choose n}x^n \plus{} \ldots \plus{} {2n \choose 2n}x^{2n}$,
$ (1 \plus{} x)^n \equal{} {n \choose 0} \plus{} {n \choose 1}x \plus{} \ldots \plus{} {n \choose n}x^n$,
$ (x \plus{} 1)^n \equal{} {n \choose 0} x^n \plus{} {n \choose 1}x^{n \minus{} 1} \plus{} \ldots {n \choose n}$.
$ (1 \plus{} x)^{2n} \equal{} (1 \plus{} x)^n \cdot (x \plus{} 1)^n \Longrightarrow$
\[ \sum_{i \equal{} 0}^{n}\left[{n \choose i}^2\right] \equal{} {2n \choose n} .
\]

Thanks a lot!   :)


You are buying 15 doughnuts for your MATHCOUNTS team and coach. If the doughnuts are packaged singly or in sets of 3 or 4, in how many different ways can you select your order?

here is the same solution, but with a less algebraic wording

if we have 3 4packs, we have 3 doughnuts left to buy (using 3packs or singles). these can be packed in a 1 3pack or 0 3packs (meaning we buy 3 singles instead).  thus, 2 cases so far

if we have 2 4packs, we have 7 doughnuts left to buy. these can be packed in 0,1, or 2 3packs (again, if we have 0 3packs, that means we buy 7 singles, if we have 1 3pack that means we buy 4 singles, etc). we add 3 more cases

if we have 1 4packs, then we have 11 doughnuts  left to buy. we can have 0,1,2, or 3 3packs. 4 more cases

if we have 0 4 packs, then we have 15 doughnuts left to buy. there are 6 cases

thus, the answer is 2cases+3cases+4cases+6cases = 15 (the answer)

[quote=5849206328x]We want to solve $ 3x\plus{}4y\leq 15$ for whole number values of $ (x,y)$.

If $ y\equal{}3$, then $ 3x\leq 3 \Rightarrow x\leq 1$.  There are $ 2$ pairs here.

If $ y\equal{}2$, then $ 3x\leq 7 \Rightarrow x\leq 2$.  $ 3$ pairs.

If $ y\equal{}1$, then $ 3x\leq 11 \Rightarrow x\leq 3$.  $ 4$ pairs.

If $ y\equal{}0$, then $ 3x\leq 15 \Rightarrow x\leq 5$.  $ 6$ pairs.

$ 2\plus{}3\plus{}4\plus{}6\equal{}\boxed{15}$[/quote]

Wait a sec if you need to buy 15 doughnuts for a mathcounts team shouldn't it be 3x+4y = 15?


Which $ n$ have the following properties,

$ a_1,a_2,\cdots,a_n$ is a permutation of $ 1,2,\cdots,n$

$ a_1\plus{}a_2$, $ a_2\plus{}a_3$, $ \cdots$, $ a_{n\minus{}1}\plus{}a_n$ and $ a_n\plus{}a_1$ are all primes ?

Obviosly $ n$ is even.
Let $ p,p\plus{}2$ are primes, and $ n\equal{}p\minus{}2$, then $ a_{2k\minus{}1}\equal{}2k\minus{}1,a_{2k}\equal{}n\plus{}2\minus{}2k$ is solution.

I believe You meant $ n\equal{}p\minus{}1$ ($ n\plus{}1$ and $ n\plus{}3$ primes),
how about other even numbers such as $ n\equal{}6$ or $ n\equal{}8$ ?

$ a_{1\plus{}2k}\equal{}2ak\mod n \plus{}1, a_{2\plus{}2k}\equal{}(p\minus{}2\minus{}2ak)\mod n \plus{}1$ work, if $ p,p\plus{}2a,p\plus{}2a\minus{}n$ are primes.
For example $ n\equal{}6,p\equal{}7,a\equal{}2$.

[quote="Rust"]Obviosly $ n$ is even.
Let $ p,p \plus{} 2$ are primes, and $ n \equal{} p \minus{} 2$, [...][/quote]
So the answer will depend on the twin-prime conjecture ?
May I ask you to write a more detailed proof ?
thanks in advance


superior algebra
superior algebra unsolved

[quote="t0rajir0u"]Here's a cute exercise I just thought up.  Does there exist a field whose additive and multiplicative groups are isomorphic?[/quote]

No, because otherwise your field, $ F,$ would have to be infinite. Now if $ \text{char}(F)\equal{}2,$ then $ x^2\equal{}1$ has only one solution in $ F^{\times}$ but $ 2x\equal{}0$ will have infinitely many solutions in $ F^{\plus{}}.$

If $ \text{char}(F) \neq 2,$ then $ x^2\equal{}1$ has two solutions in $ F^{\times}$ but $ 2x\equal{}0$ has only one solution in $ F^{\plus{}}.$

It has to be noted that this problem is entry [b]10B[/b] of [i]Problems for Mathematicians, Young & Old[/i] by [b]P. R. Halmos[/b].

Best of luck,


I'm doing my research work in the field of 'Climatology
Sciences.' Can anyone show me the proper calculation related to
'volume of water' comimg from both atmosphere,and ocean?
Please show me too 'mathematical descripted calculation' related to December 27,2004 'Tsunami.'

                                                     Respectfully,
                                                    A.Eddington

Can you make a different section for your posts, this has no relevance to economics whatsoever. Please stop spamming. Thank You.


With four votes now, it's looking kinda lopsided.  All of them are [hide="...?"]physics :roll: [/hide]

how can you people like physics? it's applied math! Aaaaaaaaaaaaaarg. It's also just simply awful.

i thought chemistry and comp. science might be close behind. we'll see after we get a more accurate picture with more votes. physics will probably win, but let's see what comes in second.

Computer science... though I haven't taken physics yet.

I like weight training but I also like physics.

Military history is lots of fun to read about  :D (the result of playing too many games of Starcraft :blush:). I've only just begun to pick up programming (Java), but so far, I've found it to be very interesting. The kind of thinking involved seems a lot like what I do when I'm working on math problems. :yup:

Hmm... music should be up there too :?

I would have to go with English.  Just because I want to be a nonconformist.  Actually, it really is my favorite subject (aside from math, of course.)

I consider too many of those subsets of or applications of math, so I picked History.  It's fascinating.  You can use it to explain everything about the world.  :)

Guess I'm the first one to vote for Economics.  I'm only basing this after taking one course of economics...oh well.

I love science as well though...

Haha, yeah, weight lifting is my favorite class this year.  Actually, English might be my favorite core class this year because of the teacher...the teacher really makes or breaks the class.

PHYSICS IS THE BEST!!!!!yOU NEED A REAL BRAIN FOR THAT :P

Physics is more interesting than math,much more and its not a subset of math nor application(justa little)Its pure brain and thinking. :D

I dislike Physics, mainly because over here A-Level physics has had most of the maths ripped out of it to appease medics who only do it for the science parts.

I can't believe you can get an A-Level in Physics without having to know how to integrate.

So I hate physics.

I like Comp Sci a lot though.

History just because I havent been able to take physics yet, but once I get there I will probally enjoy it

Before taking physics, I'd always assumed that it would be my favorite.  But now that I'm in both chem and physics, I'm more of a chemist.  And I kind of actually dislike my physics class..  :?   but I'm thinking it might be because its non-calculus and kinda boring.  AP physics C next year, we'll see if that changes anything.  I'm suprised that there aren't more chemistry people tho.

probably history first. biology is up there too. physics is just ok

wow, history is making a stand. i took AP us. history and got a 5, but that was enough for me. i just want to sail through the rest of my history classes.

Computer science is practically math, just more discrete math and stuff like algorithims and whatnot.

I put other, because mathematics is in there. :lol:  :rotfl:

my favorite subject besides math is german.

[quote="mathjay616"]This poll is pointless, since its obvious that nearly everyone who loves math likes physics. What is physics anyway, just using math to describe real world phenomena? Duh  :P[/quote]

Much to learn this newbie has. :yoda:   (or are you graduated yourself?)

That is not true!   In my case it has always been the same scenario : physics professors don't define their things properly, don't clearly express what is proved and assumed,etc...

In high school, I adored physics as a second course, but now I realised...that was just math, that was not physics, while physics is not math.  Throwing rocks, circling the earth, calculating power outputs.  I absolutely hated physics in university and stay away from it.

Actually the poll is [b]unclear[/b], do you mean subjects to choose yourself, or what alternative study career I would prefer?

I put History.  The science department at my school is atrocious, and despite my efforts to continue to enjoy the ideas of science has actually turned me off science.  I am refusing to take the only AP science course my school offers (Chemistry) because my experiences in Honors Chemistry this year were so awful.  Also, the CS department is, if possible, [i] worse[/i] : it is one guy, with a degree in child psychology, not teaching several classes.  The main class activity is playing Age of Empires II.  I took Intermediate Programming for a week before dropping out and switching to philosophy.  It was in Visual Basic.  What?!
So, I chose History because the other subjects I might have chosen are taught in a way that seems designed to get people that are actually interested in that subject to hate it passionately by the end of the year.  Well, and the fact that History is just 'cool' anyways.

I do agree with fredbel that the poll is unclear, though.  Do you mean favorite subject, outside of how the subjects are taught (in which case I would choose CS), or should we factor in how well the classes are taught?  I think it could be interpreted either way given the title of the poll.

I would put computer science. Biology is one of my least favorite subjects :sleep2:.

Hey, David.  Yeah, I forgot philosophy.  Oops.  First semester was better, though.  Debating existence of God, Cartesian dualism, etc.  Great fun.

I picked CS, although my school doesn't give CS as a subject.  There's a subject called Computer Studies, but we use Visual Basic  :mad:

[i]Moderator's Note:[/i]  Making more posts about how you made to many posts is pointless and creates more work for me.  When your post is the last one on the thread, there is a little box marked "x" in its corner that you can use to delete the post.  Alternatively, if you go into the "edit post" screen, there is a checkbox with the option "Delete this post."

[quote="JavaMan"]I like computer science, I feel the thinking involved is very much problem solving, just like maths :)[/quote]

Yes, what I like about computer science is the logical part: programming, inventing programs, those things.... just magic!

Yeah, doing problems like the ones from [url=http://mathschallenge.net/index.php?section=project]project euler[/url]


trigonometry
limit
algebra proposed
algebra

[quote="Algadin"]Let $ \{x_n\}$ be a real sequence defined by
$ x_1 \equal{} a$
and $ x_{n \plus{} 1} \equal{} 2 \minus{} x_n^2$
For all $ n \equal{} 1,2,3...$ and a is a real number.
Find all $ a$ such that $ \{x_n\}$ has finite limit when $ n\to \plus{} \infty$ and find the finite limit in that cases[/quote]

Let $ y_n\equal{}\minus{}\frac{x_n}2$. The sequence becomes :

$ y_1\equal{}\minus{}\frac a2$
$ y_{n\plus{}1}\equal{}2y_n^2\minus{}1$ and so :

The only possible finite limits for $ y_n$ are roots of $ 2X^2\minus{}X\minus{}1\equal{}0$, so $ 1$ or $ \minus{}\frac 12$. Then :
If $ |y_1|\le 1$ : $ \exists u$ such that $ y_1\equal{}\cos u$ and then $ y_n\equal{}\cos 2^{n\minus{}1}u$ :
Then we need to have either $ u\equal{}k\pi 2^{\minus{}p}$, and the limit will be $ 1$, either $ u\equal{}\frac{\pi}32^{\minus{}p}\plus{}k\pi 2^{\minus{}q}$ and the limit will be $ \minus{}\frac 12$

If $ y_1> 1$ : $ \exists u>0$ such that $ y_1\equal{}\cosh u$ and then $ y_n\equal{}\cosh 2^{n\minus{}1}u$ and the limit is $ \plus{}\infty$

If $ y_1<\minus{}1$ : then $ y_2>1$ and $ \exists u>0$ such that $ y_2\equal{}\cosh u$ and then $ y_n\equal{}\cosh 2^{n\minus{}2}u$ and the limit is $ \plus{}\infty$

Hence the answer : the sequence $ x_n$ may have a finite limit only when :

either $ a\equal{}\minus{}2\cos(\frac{k\pi}{2^p})$ with $ k,p\in\mathbb Z$ and then the limit is $ \minus{}2$

either $ a\equal{}\minus{}2\cos(\frac{\pi}{3\cdot 2^q}\plus{}\frac{k\pi}{2^p})$ with $ k,p,q\in\mathbb Z$ and then the limit is $ 1$

[quote="Algadin"][quote="pco"]

Then we need to have either $ u \equal{} k\pi 2^{ \minus{} p}$, and the limit will be $ 1$, either $ u \equal{} \frac {\pi}32^{ \minus{} p} \plus{} k\pi 2^{ \minus{} q}$ and the limit will be $ \minus{} \frac 12$

[/quote]

Why?????[/quote]

$ \cos(2^nu) \equal{} \cos(2\pi\{\frac {2^nu}{2\pi}\})$ and so, if $ \lim_{n\to \plus{} \infty}\cos(2^nu)$ exists, then the sequence $ a_n \equal{} \{2^nv\}$ (where $ v \equal{} \frac u{2\pi}$) must have at least one and at most two accumulation points in $ [0,1)$.

So, such accumulation points may only be $ (0)$ or $ (\frac 13,\frac 23)$

Single accumumation point $ 0$ means that $ 2^nv$ is integer $ \forall n$ great enough and so $ u \equal{} \frac {k\pi}{2^p}$ for some integers $ k,p$
Double accumulation points $ \frac 13,\frac 23$ means $ 2^nv \equal{} \frac 13 \plus{} k$ or $ \frac 23 \plus{} k$ $ \forall n$ great enough and so $ u \equal{} \frac {\pi}{3\cdot 2^p} \plus{} \frac {k\pi}{2^q}$ for some integers $ k,p,q$


After Sally takes 20 shots, she has made $ 55\%$ of her shots.
After she takes 5 more shots, she raises her percentage to $ 56\%$.
How many of the last 5 shots did she make?

20 shots, 55% = 11 success.
25 shots, 56% = 14 success.
3 successes out of last 5 shots.


linear algebra
matrix
graph theory
combinatorics proposed
combinatorics

let say graph $G$ good ,if for every set of $V$ of vertices with cardinality$m$,and the set $S$ of all neighbours of vertices in$V$ with cardinality $n$,we have$n\geq{m}$.
is there a necessary condition for  a graph being good?

Well, a good graph is equivalent (due to Hall's theorem) to one which has an adjacency matrix with a nonzero permanent.  However, in practice this isn't really any easier to check than the original definition (permanents are notoriously hard to compute).

If your graph has an odd number of vertices, a necessary condition is that it not be 2 colorable, so it must contain at least one cycle.  However, this is a pretty weak condition.

Is there a particular collection of graphs you need to check the goodness/nongoodness of?

i tried to find some conditions,but failed:
e.g if $G$ is bipartite then it should has  a perfect matching.
see that all cycles and all even paths are good.
may be our first graph is union of these three (with some extra edges.)
but all of these criterias fail.

One close thing to your "union of edges and cycles" criterion is this:

A good graph is one that has a spanning subgraph such that each component is either a cycle or an edge.

Construct an auxillary bipartite graph on two copies of $V(G)$, where we connect $x$ in one copy and $y$ in the other iff $(x,y)$ is an edge in the original graph.  Because the graph is good, the conditions of Hall's theorem are satisfied for the auxillary graph, so it has a perfect matching.

Now look at what happens when you pull that matching back to the original graph (connecting $x$ with $y$ if either copy of $x$ is paired with $y$ in the matching).

this might help:
for a graph $G$ we denote by $q(G)$ the number of component with and odd number of vertices.

now prove that we have a perfect matching in a graph $G$ not necesarily bipartite if and only if $q(G-S) \leq |S|$ for every $S$ a subset of $V(G)$.

but kevinatcausa , (due your solution) what about odd cycles?


inequalities algebra polynomial calculus derivative function conics

I studied your solution for

$ \sum a^3 \plus{}3abc \geq \sum a^2\sqrt{a^2\plus{}3bc}$

but there is something it is not clear. I will explain it to you.

$ (\sum a^3 \plus{}3abc)^2 \geq (\sum a^2\sqrt{a^2\plus{}3bc})^2$

$ \sum a^6 \plus{} 2\sum a^3b^3 \plus{}9a^2b^2c^2 \plus{}6abc\sum a^3 \geq \sum a^4(a^2\plus{}3bc) \plus{} 2\sum a^2b^2\sqrt{(a^2\plus{}3bc)(b^2\plus{}3ca)}$

$ 2\sum a^3b^3 \plus{}9a^2b^2c^2 \plus{}3abc\sum a^3 \geq 2\sum a^2b^2\sqrt{(a^2\plus{}3bc)(b^2\plus{}3ca)}$

Now there is a step I don't understand

By AM-GM

$ 2\sum a^2b^2\sqrt{(a^2\plus{}3bc)(b^2\plus{}3ca)} \equal{} 2\sum ab\sqrt{b^2(a^2\plus{}3bc)a^2(b^2\plus{}3ca)} \leq \sum ab[a^2(b^2\plus{}3ca) \plus{} b^2(a^2\plus{}3bc)] \equal{} 2\sum a^3b^3 \plus{}6abc\sum a^3$

and not as you say in solution of problem n° 28 of PDF file nothing.pdf

$ 2\sum a^2b^2\sqrt{(a^2\plus{}3bc)(b^2\plus{}3ca)} \leq 2\sum a^3b^3 \plus{}3abc\sum ab(a\plus{}b)$  ....  $ (*)$

If I am right your solution is wrong. If I am wrong can you please explain me how to obtain $ (*)$

Thank you very much.

P.S. I am checking all your solution of nothing.pdf

Yes, you are right. My solution is wrong, I am sorry about that :(. Let me try it again. By the way, thanks for checking my solution, it is what I need and you are very kind, my friend.:)

We cannot prove this very hard inequality using the theorem:
[b]Theorem.[/b]Let $ f(a,b,c)$ be a symmetric and homogeneous polynomial expression of fifth degree. The inequality
$ f(a,b,c) \ge 0$
holds for any nonnegative real numbers $ a,b,c$, if and only if it holds for $ a \equal{} 0$ and also for $ b \equal{} c \equal{} 1$.

Vacs, can you post the full theorem and proof about above inequality. I see it very interesting.

I cannot, because the theorem above is true only if the degree of $ f$ is smaller than or equal to 5. :oops:

[quote="manlio"]I studied your solution for

$ \sum a^3 \plus{} 3abc \geq \sum a^2\sqrt {a^2 \plus{} 3bc}$

but there is something it is not clear. I will explain it to you.

$ (\sum a^3 \plus{} 3abc)^2 \geq (\sum a^2\sqrt {a^2 \plus{} 3bc})^2$

$ \sum a^6 \plus{} 2\sum a^3b^3 \plus{} 9a^2b^2c^2 \plus{} 6abc\sum a^3 \geq \sum a^4(a^2 \plus{} 3bc) \plus{} 2\sum a^2b^2\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ca)}$

$ 2\sum a^3b^3 \plus{} 9a^2b^2c^2 \plus{} 3abc\sum a^3 \geq 2\sum a^2b^2\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ca)}$

Now there is a step I don't understand

By AM-GM

$ 2\sum a^2b^2\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ca)} \equal{} 2\sum ab\sqrt {b^2(a^2 \plus{} 3bc)a^2(b^2 \plus{} 3ca)} \leq \sum ab[a^2(b^2 \plus{} 3ca) \plus{} b^2(a^2 \plus{} 3bc)] \equal{} 2\sum a^3b^3 \plus{} 6abc\sum a^3$

and not as you say in solution of problem n° 28 of PDF file nothing.pdf

$ 2\sum a^2b^2\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ca)} \leq 2\sum a^3b^3 \plus{} 3abc\sum ab(a \plus{} b)$  ....  $ (*)$

If I am right your solution is wrong. If I am wrong can you please explain me how to obtain $ (*)$

Thank you very much.

P.S. I am checking all your solution of nothing.pdf[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=214409]See my new solution here[/url] :)[/url]

[quote="Vasc"]We can prove this inequality using the following corollary of the EV-Theorem.

[b]Corollary.[/b] Any symmetric and homogeneous polynomial expression of sixth degree can be written in the form
$ f(a,b,c) \equal{} 9Ax^2 \plus{} 2p(Bp^2 \minus{} Cq)x \plus{} D(p,q)$,
where $ A,B,C$ are real numbers, $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$ and $ x \equal{} abc$. If $ A\le 0$ or $ C \minus{} 3B\le A\le B$, then the inequality
$ f(a,b,c) \ge 0$
holds for any nonnegative real numbers $ a,b,c$, if and only if it holds for $ a \equal{} 0$ and for $ b \equal{} c \equal{} 1$.

In our case, I found $ A \equal{} 16$, $ B \equal{} 17$, $ C \equal{} 62$.[/quote]
How can we transform it, VasC? Since RHS has root and...

A more elegant solution follows by the idea below.
The statement reduces to the following:

If $ x,y,z$ are positive real numbers such that $ xyz\equal{}1$, then

$ \sum \frac 1{1\plus{}\sqrt{3x\plus{}1}} \le 1$.

[quote="Vasc"]A more elegant solution follows by the idea below.
The statement reduces to the following:

If $ x,y,z$ are positive real numbers such that $ xyz \equal{} 1$, then

$ \sum \frac 1{1 \plus{} \sqrt {3x \plus{} 1}} \le 1$.[/quote]

Is your solution elementary, Vasc (that is without use of derivative ) ?

Thank you very much.

[quote="Vasc"]By double squaring.....
See[/quote]
Thanks, VasC. Your solution is nicer than mine... :blush:

[quote="Vasc"]We can prove this inequality by the RCF-Theorem.[/quote]

Or by mixing variable method, because
$ f(x)\plus{}f(y)\le f(\sqrt{xy})$
for $ xy\le 1$.

[quote="Vasc"]

Or by mixing variable method, because
$ f(x) \plus{} f(y)\le f(\sqrt {xy})$
for $ xy\le 1$.[/quote]

You discovered a beautiful proof by mixing variable method.

Sorry,but I cannot prove it. A hint, please  :blush:

Thank you very much, Vasc.

There is a mistake.
This is correct:
$ f(x)\plus{}f(y) \ge 2f(\sqrt {xy})$ for $ xy\le 1$.

[quote="Vasc"]There is a mistake.
This is correct:
$ f(x) \plus{} f(y) \ge 2f(\sqrt {xy})$ for $ xy\le 1$.[/quote]

This is a reversed sign, so the proof by MIXED VARIABLES METHOD doesn't work.  :maybe:

Am I right, Vasc?

[quote="manlio"][quote="Vasc"]There is a mistake.
This is correct:
$ f(x) \plus{} f(y) \ge 2f(\sqrt {xy})$ for $ xy\le 1$.[/quote]
This is a reversed sign, so the proof by MIXED VARIABLES METHOD doesn't work.  :maybe:
[/quote]
Of course,
$ f(x) \plus{} f(y) \le 2f(\sqrt {xy})$ for $ xy\le 1$.

[quote="Vasc"]For $ A\le 0$ or $ C \minus{} 3B\le A\le B$,  the inequality
$ f(a,b,c) \ge 0$
holds for any nonnegative real numbers $ a,b,c$ if and only if it holds for $ a \equal{} 0$ and for $ b \equal{} c \equal{} 1$.

In our case, I found $ A \equal{} 16$, $ B \equal{} 17$, $ C \equal{} 62$.[/quote]

And do you think that $ A\equal{}16\le 0$  :?:

[quote="Vasc"]$ f(x) \plus{} f(y) \le 2f(\sqrt {xy})$ for $ xy\le 1$.[/quote]

Has anyone proved this fact. I am very interested to see a complete solution by Mixed Variables Method. I think it is a beautiful proof.  :wink:

Thank you very much.

[quote="Robert Gerbicz"][quote="Vasc"]For $ A\le 0$ or $ C \minus{} 3B\le A\le B$,  the inequality
$ f(a,b,c) \ge 0$
holds for any nonnegative real numbers $ a,b,c$ if and only if it holds for $ a \equal{} 0$ and for $ b \equal{} c \equal{} 1$.

In our case, I found $ A \equal{} 16$, $ B \equal{} 17$, $ C \equal{} 62$.[/quote]
And do you think that $ A \equal{} 16\le 0$  :?:[/quote]
I don't think that $ A \equal{} 16\le 0$. But I think that $ C \minus{} 3B\le A\le B$. :wink:

[quote="Vasc"]We can prove this inequality using the following corollary of the EV-Theorem.

[b]Corollary.[/b] Any symmetric and homogeneous polynomial expression of sixth degree can be written in the form
$ f(a,b,c) \equal{} 9Ax^2 \plus{} 2p(Bp^2 \minus{} Cq)x \plus{} D(p,q)$,
where $ A,B,C$ are real numbers, $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$ and $ x \equal{} abc$. For $ A\le 0$ or $ C \minus{} 3B\le A\le B$,  the inequality
$ f(a,b,c) \ge 0$
holds for any nonnegative real numbers $ a,b,c$ if and only if it holds for $ a \equal{} 0$ and for $ b \equal{} c \equal{} 1$.

In our case, I found $ A \equal{} 16$, $ B \equal{} 17$, $ C \equal{} 62$.[/quote]

Sorry, Vasc but I cannot prove your corollary for case $ A \geq 0$.

We put $ x\equal{}abc$, $ p\equal{}a\plus{}b\plus{}c$ and $ p^2\minus{}q^2\equal{}3 (ab\plus{}bc\plus{}ca)$ so $ p \geq q$.

We know that $ 0 \leq x \leq \frac{(p\minus{}q)^2(p\plus{}2q)}{27} \equal{} M$

$ f(a,b,c) \equal{} 9Ax^2 \plus{} 2p(Bp^2 \minus{} Cq)x \plus{} D(p,q)$

as function of $ x$ ($ A \geq 0$) is a parabola concave up increasing/decreasing in our domain $ [0,M]$ iff $ V \leq 0$ / $ V \geq M$ where $ V$ is the x-coordinate of the vertice of the parabola $ f(x)$

 We have $ V\equal{} \frac{\minus{}p(Bp^2 \minus{} C\frac{(p^2\minus{}q^2)}{3})}{9A}$ and

              $ M\equal{} \frac{p^3\plus{}2q^3\minus{}3pq^2}{27}$

We have to study two cases:

1) $ V \leq 0$ which implies $ 3B\minus{}C \geq 0$ which is false !!!!

2) $ V \geq M$ which implies after simple algebra

  $ p^3(C\minus{}3B\minus{}A) \plus{} pq^2(3A\minus{}C) \minus{}2Aq^3 \geq 0$

As $ p \geq q$ I found this is true if $ C\minus{}3B \geq A \geq C$ which is different from your result $ C\minus{}3B \leq A \leq B$

I cannot find my mistake, Vasc. Can you please explain you corollary.

Thank you very much.


AMC AIME geometry trigonometry USA(J)MO USAMO

Can anyone rank the most important chapters (in preparing for the AIME) in AoPS 2?  What are the most important?  What are the least important?  (etc.)

Thanks

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806&start=20

There's a thread on this same page where a person asked the same exact question  :P

It depends on what you mean by "important." If you want to learn geometry, the geometry chapters are most important. If you want to brush up on trig, read the trig chapters.

When I asked this question, I was told, do every single chapter! Five times!

They are all important, so it makes the most sense to start with stuff that you haven't learned very much of. It helped me to do a few old AIMEs, and figure out what problems I was doing really badly on, and then do the chapters that cover them to start with.

Yeah, but it is still a good thing to know

[quote="Karth"]Well, inversion isn't all that useful on AIMEs...  :wink:[/quote]

I don't think it's on the USAMO either

[quote="Karth"]Well, actually, you can use inversion on a ton of geometry problems on the USAMO and IMO, but it's not exactly REQUIRED. It just makes stuff a whole lot easier  :)[/quote]

Nothing is required on the AIME except for addition and multiplication. More advanced things like algebra just make stuff a whole lot easier  :)

Are you sure, Karth?  I have used it on very few problems (aside from the ones in the inversion lecture at MOP; obviously, there are many problems in which inversion works beautifully, but I don't think they occur too often).  Always be careful not to fall into the trap of using more recondite "advanced" techniques when simpler and more elegant techniques are available.  This is not to say that you will never use the powerful tools :  you just shouldn't be too trigger-happy with them.  Remember that the main point of these contests is to test mathematical ability, not knowledge.

[quote="Boy Soprano II"][quote="chess64"]Yeah it is... (correct me if I am mistaken, Arnav :D)[/quote]
Well, I don't know about USMO, but I don't think it is supposed to be necessary for the IMO.  But it is revelatory, and sometimes even useful, so you should learn it.[/quote]

well if zhai (who is easily the master of synthetic geometry) says he has to use inversion for a problem, I'd say inversion does come in useful every once in a while

granted, he only said that once and it was so momentous an occassion I put the quote in my profile and besides I think he found a synthetic solution later anyway

BUT STILL


Share some methods of problem solving that you use:

1. Work with sample cases and find patterns.
2. Exploit Symmetry
3. Try to prove a stronger statement.
4. Try to find ways to use the given information.
5. Break the problem into different sub-problems that can be proven by lemmas.

Any other things to add to the list?

[quote]Try to prove a stronger statement. [/quote]

What do you mean by this?

By stronger statement, it usually means something like instead of proving the case where you have 2003 objects, you prove the case where you have n objects.

Suppose you were given the problem:

Prove x^2 + y^2 + z^2 = 87 has no solutions in integers x,y,z.

This could be generalized to the much stronger

Prove x^2 + y^2 + z^2 = 8m + 7 where m is an integer has no solutions in integers x,y,z.

Clearly the second statement has a much clearer approach (i.e. quadratic residues modulo 8) and solution, and also proves the first statement.

Although this strengthening may seem artificial, it is a natural consequences of investigation.

Additional "tactic:"

* Consider importance of given numbers.

Specifically, answer the question "why 87?" The answer leads to the solution.

Yeah, strengthing does help at times, gives you a wide range of things to do as well (induction etc). Of course sometimes the hard part is working out what to strengthen it to.. this is where small cases will help.
In fact sometimes making the problem weaker makes it harder to solve.. I heard a good example once where it was real real hard to solve this problem by induction but when you made it stronger it was easier to prove.. I can't remember it though.

Ok, thanks, I see now.

[quote]* Consider importance of given numbers.

Specifically, answer the question "why 87?" The answer leads to the solution.[/quote]

Yeah, and in even more obvious cases. For example, when you see the numbers 7 and 24 in a geometry problem initlally without any mention of right triangles, or Sqrt(2)/2 when no isosceles triangle is around, well...you know what to construct...

This may sound trivial, but I find that substitution is one of the most powerful problem solving tools.  So much of mathematics is finding simpler ways to deal with very complex problems.  Substitution is very often that way of taking a very complicated expression and turning it into something far easier to work with.


Thanks for the help everyone on my last post, I have another proof, just wanted to see how you guys think it sounds.

Let $T$ be a linear operator on $R^n$ and $B = {v_1,v_2,...,v_n}$ be an orthonormal basis for $R^n$ such that ${T(v_1),T(v_2),...,T(v_n)}$ is also an orthonormal basis for $R^n$.  Prove that $T$ is an orthogonal operator. Also let $B$ be in $V$.

Here is what I have so far, I'd like some opinions because it doesn't flow so well.

We know that $[T]_B = ([T(v_1)]_B, [T(v_2)]_B, ... , [T(v_n)]_B)$.  For an orthonormal basis, the dot product of vectors in $V$ is equal to the standard dot product in $R^n$ of their respective column vectors.  If $T$ is orthogonal, the set of vectors $T(v_1), T(v_2), ... , T(v_n)$ is orthonormal.  Hence, the columns of $[T(v_1)]_B, [T(v_2)]_B, ... , [T(v_n)]_B$ are orthogonal so $A = [T]_B$ is an orthogonal matrix.  Which, implies that the vectors $T(v_1), T(v_2), ... , T(v_n)$ is an orthonormal set.  In other words, $T$ preserves all dot products of the elements of basis $B$, $T$ also preserves all dot products of the vectors in $V$.  Therefore, $T$ is an orthogonal operator.

I would appreciate any and all opinions, do I satisfactorily prove this, or is the proof not sufficient.


induction
limit
algebra unsolved
algebra

One easily gets the recurrence relation $ u_{n \plus{} 1} \equal{} \frac {nu_n} {u_n^2 \plus{} n \minus{} 1}$. Now $ u_n < 1$ follows by induction, since $ 1 \minus{} u_{n \plus{} 1} \equal{} \frac {(1 \minus{} u_n)((1 \minus{} u_n) \plus{} (n \minus{} 2))} {u_n^2 \plus{} n \minus{} 1} > 0$.
Then $ u_{n \plus{} 1} \minus{} u_n \equal{} \frac {u_n(1 \minus{} u_n^2)} {u_n^2 \plus{} n \minus{} 1} > 0$, so the sequence is increasing, and being upper bounded, convergent (easily shown $ \lim_{n \to \infty} u_n \equal{} 1$).

I however understand that this is part of a monthly problem at Bilkent University, so we should not delve any further into it

We are given $ u_{n \plus{} 1} \equal{}\frac{n}{u_{1}\plus{}u_{2}\plus{}...\plus{}u_{n\minus{}1}\plus{}u_{n}} \equal{}$ $ \frac{n}{(n\minus{}1)\frac {u_{1}\plus{}u_{2}\plus{}...\plus{}u_{n\minus{}1}} {n\minus{}1} \plus{}u_{n}} \equal{}$ $ \frac {n} {\frac {n\minus{}1} {u_n} \plus{} u_n} \equal{}$ $ \frac {nu_n} {u_n^2 \plus{} n \minus{} 1}$.

nice. Can you explain the second line, and why or how the limit=1?

Yes. Knowing that $ u_{n \plus{} 1} \equal{} \frac {nu_n} {u_n^2 \plus{} n \minus{} 1}$ is increasing, convergent to some limit $ \lim_{n \to \infty} u_{n} \equal{} \sup(u_n) \equal{} \ell \in (0, 1]$, let us assume $ \ell < 1$. Then
$ u_{n \plus{} 1} \minus{} u_n \equal{} \frac {u_n(1\minus{}u_n^2)} {u_n^2 \plus{} n \minus{} 1} \geq \frac {u_3(1\minus{}\ell^2)} {n}$ for $ n\geq 3$, so
$ u_{n \plus{} k} \minus{} u_n \geq u_3(1\minus{}\ell^2) \sum_{i\equal{}0}^{k\minus{}1} \frac {1} {n\plus{}i} \to \infty$ for $ k \to \infty$, absurd.


Ok, so this is this game called 185. It starts out like this:

185______walked into a bar. The bartender says,"We don't serve _______ here". Then, the __________ say _____________________________________.

So this is the main template. You can modify it a little bit. The point is:
The first three blanks are to be filled by a noun, like monkeys, or anything you want to think of. The last blank is filled by something the noun says, which relates to the noun and makes this funny. So, I will start it off.

185 monkeys walked into a bar. The bartender says,"We don't serve monkeys here". Then, the monkeys say,"Oh bananas!". Get it?

Uh, ok?  not that funny thouhg.

185 AIME15's walked into a bar.  the bartender says "we don't sell choo-choo trains here".  then the AIME15's swear loudly and leave.

Ok:

185 iPods walk into a bar. The Bartender says,"we dont serve iPods here. Oh wait . Sorry, we do!" Then, the iPods say, "Well, thats music to my ears."

185 mugs of beer walk into a bar. The Bartender says, "We serve you here." Then the beers pwn the bartender and leave.

185 tennis players walk into a bar. The bartender says, "We don't serve tennis players here." Then the tennis players say, "That's okay.  We can do the serving ourselves."

185 robots walk into a bar. The bartender says, "We don't serve minibars here." One of them says, "I am NOT a minibar!"
(A cookie goes to whoever gets this reference.)

185 Wiis walk into a bar. The bartender says,"We dont serve Wiis here." Then a Wii says," I'm sensing that you do not like me." Get it?

nick42, if you are going to criticize this thread, just do not post here. Anways, my pun was meant to be- Then the Wiis say,"I'm [b]sensing[/b] that you do not like me." Wii have a sensing device. They have a wiimote that senses the signals and the wii registers it. I hope you do know that.

185 "185 "185 "185 "185 "..."s walk into a bar. The bartender says,"OMG IT'S A SELFREFRENTIAL STATEMENT!""s walk into a bar. The bartender says,"OMG IT'S A SELFREFRENTIAL STATEMENT!""s walk into a bar. The bartender says,"OMG IT'S A SELFREFRENTIAL STATEMENT!""s walk into a bar. The bartender says,"OMG IT'S A SELFREFRENTIAL STATEMENT!""s walk into a bar. The bartender says,"OMG IT'S A SELFREFRENTIAL STATEMENT!"
<GENERAL PROTECTION FAULT>

Umm...

185 bears walk into a bar. The bartender says, "I'm sorry, we don't serve bears here." The bears said, "It's surprising you get so many customers with the stock market the way it is."

185 dogs walked into a bar. The bartender said, "I'm sorry, we don't serve dogs here." The dogs said, "Meow!"

185 people walk into a bar. They said, "Ouch."

I got the idea for this from evilhamster's one about the bears.

185 deaf bears walk into a bar.  The bartender says, "I'm sorry, we don't serve bears here."  The bears, being deaf, don't respond.  Then the bartender writes on a sign, "I'm sorry, we don't serve bears here."  The bears, after reading the sign, ask for some wine.  The bartender, furious, writes, YOU ARE BEARS! WE DON'T SERVE YOU!"  The bears say, "Oh, we thought you wrote [i]beers[/i], not [i]bears[/i]."

185 subjects walk into a bar. The bartender says "Sorry, we don't serve subjects here." A subject said the punchline.

185 vacuum cleaners go into a bar. The bartender says, "We don't serve vacuum cleaners here." The vacuum cleaners say, "That sucks."

185 computers walk into a bar. The bartender says,"we dont serve computers here". Then the computers go,"Aw man. This is really gonna make me crash and shut down-forever!".

185 :D s walked into a bar. The bartender says,"We don't serve smilies here". Then, the :bomb: say KABOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOMMM!!!.

Doesn't make sense, but...

185 oceans walk into a bar(they do not hear properly). The bartender says, "We do not serve oceans here." The oceans say, "But we have high waves." The bartender yells, "I SAID 'SERVE' NOT 'SURF'!"

185 spamming jain_harshi's walked into a bar.  The bartender says, "We don't serve spam here".  Then, the jain_harshi's say Dang this thread fails.


invariant
vector
superior algebra
superior algebra solved

[i]A[/i] is an orthogonal transform in an Euclidean space. W is a stable subspace of [i]A[/i]. Prove that the orthogonal complement of W is also stable of [i]A[/i].

does the conclusion hold for infinite dimension?

[quote="liyi"] W is a invariant subspace of [i]A[/i][/quote]

liyi it's me moubinool,

In Paris invariant subspace  means A(W)=W is the same for you  ?

For every two vectors x and y

(Ax,Ay)=(x,A^{*}Ay)=(x,y) since A^{*}=A^{-1}, because A orthogonal.

if x\in W and y\in W^{\perp} then

(Ax,Ay)= (x,y)=0 so A(W^{\perp})=(A(W))^{\perp}

[quote="Moubinool"]
In Paris invariant subspace  means A(W)=W is the same for you?[/quote]

a slightly different from China.
Here we defined it as follow.

W is a linear subspace. For any \alpha \in W, we have A\alpha \in W. W is called an invariant subspace.

Of course, if W has definite dimension, AW=W.

E is euclidean space E=W@W*
where W* is the orthogonal of W, @ is the orthogonal direct sum.

A(E)=E since A is bijective
A(W@W*)=W@W*
A(W)@A(W*)=W@W* but W is invariant
W@A(W*)=W@W* => A(W*)=W* the orthogonal of W is invariant by A

" A is bijective " -- this condition is important for AW=W.

The conclusion does not hold in space with infinite dimension.

Let's have a look at P[x], and we define the inner product for

  a = a0 + a1 x + a2 x^2 + ... + an x^n + ...
  b = b0 + b1 x + b2 x^2 + ... + bn x^n + ...

the inner product

  (a, b) = a0b0 + a1b1 + ... + anbn + ...

and there's a orthogonal transform

  [i]A[/i] f = x * f.

W = L(x,x^2,x^3,...); L(a1,a2,...)means the linear space generated by the basis a1, a2, ...

W is a stable space of [i]A[/i]. But its orthogonal complement, L(1), is not stable.

[quote="liyi"]... there's a orthogonal transform

  [i]A[/i] f = x * f.

W = L(x,x^2,x^3,...); L(a1,a2,...)means the linear space generated by the basis a1, a2, ...

W is an invariant space of [i]A[/i][/quote]

W= Vect (x,x^2,x^3,...)= vector space generated by x, x^2, x^3,...

A is multiplication by x

A(W) is contained in vector space generated by

x^3, x^4, x^5,....,  and so x is not in A(W)=Vect(x^3,x^4,x^5,....)

It seems to me A(W)  is contained in W, but W is not contained in  A(W)

A(W) =/=W

Do I miss something ?

ok. in China, at least on the textbook,

invariant space does not require "W \subseteq AW". It requires only "AW \subseteq W"

Definition: W is stable by A if A(W) is included in W

Definition: W is invariant by A if A(W)=W,
in this case we have A(W) is included in W, and W is included in A(W).

oh..sorry... i never read an english algebra book before...
i will change all the descriptions in the posts.


abstract algebra
group theory
superior algebra
superior algebra unsolved

$ G$-abelian group of finite index
$ p$-prime

$ p/|G| \Longrightarrow \exists x\in G\ (order(x) \equal{} p)$

proof?

If $ H$ is a proper subgroup of $ G$ and p divides its size, then we can induct on $ |G|$ to finish. So assume that for any proper subgroup H of G we have: $ \gcd (|H|, p) \equal{} 1$. Let M be a proper subgroup of G of maximal size.

Pick $ x \neq 1 \in G \minus{} M$. Set $ K \equal{} < x >$. Since $ M$ is maximal, we must have $ G \equal{} MK$.
Then $ |G| \equal{} |MK| \equal{} \frac {|M\parallel{}K|}{|M \cap K|} \equal{} \frac {|M|}{|M \cap K|} \cdot |K| \equal{} |K|\cdot |M: M \cap K|$.

We have $ \gcd (|M|, p) \equal{} \gcd (|M: M \cap K|, p) \equal{} 1$ and $ p| |G| \implies p| |K| \implies pt \equal{} |K|$ for integer t.

We get $ g \equal{} x^t \neq 1$ and $ g^p \equal{} x^{|K|} \equal{} 1 \implies$g has order p.


Hi everyone!

I tried to find numbers with the following properties:

sum of the digits
[list] in base 10 (decimal) = 10
in base 2 (binary) = 2
[/list]
Clearly, the solutions are sum of two powers of 2

Using a simple programme i got the solutions:

$ (136)_{10} \equal{} 2^7 \plus{} 2^3 \\
(514)_{10} \equal{} 2^9 \plus{} 2^1$

and

$ (640)_{10} \equal{} 2^9 \plus{} 2^7 \\
(2080)_{10} \equal{} 2^{11} \plus{} 2^5 \\
(8020)_{10} \equal{} 2^{13} \plus{} 2^3$

There are no other solutions less than $ 2^{20}$

So the question is:
Are there only finitely many solutions? Are this the only 5 solutions?

Well, by intuition, we conjecture that there are only 5 solutions because it`s unlikely that the sum of two powers of 2 gets that close to a power of 10

I already found out, that all solutions are either of the form:

$ 2^{6a \plus{} 3} \plus{} 2^{6b \plus{} 1}$

or

$ 2^{6c \plus{} 5} \plus{} 2^{6d \plus{} 5}$

Note that only one solution (2080) is of the second type.

Another interesting property:
$ (136)_{10} \equal{} 2^7 \plus{} 2^3 ... 7 \plus{} 3 \equal{} 10 \\
(514)_{10} \equal{} 2^9 \plus{} 2^1...9 \plus{} 1 \equal{} 10 \\
(136)_{10} \equal{} (12001)_3...1 \plus{} 2 \plus{} 1 \equal{} 4 \equal{} 2^2 \\
(514)_{10} \equal{} (201001)_3...2 \plus{} 1 \plus{} 1 \equal{} 4 \equal{} 2^2$

$ (640)_{10} \equal{} 2^9 \plus{} 2^7...9 \plus{} 7 \equal{} 16 \\
(2080)_{10} \equal{} 2^{11} \plus{} 2^5 ...11 \plus{} 5 \equal{} 16 \\
(8020)_{10} \equal{} 2^{13} \plus{} 2^3...13 \plus{} 3 \equal{} 16 \\
 \\
(640)_{10} \equal{} (212201)_3...2 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 1 \equal{} 8 \equal{} 2^3 \\
(2080)_{10} \equal{} (2212001)_3...2 \plus{} 2 \plus{} 1 \plus{} 2 \plus{} 1 \equal{} 8 \equal{} 2^3 \\
(8020)_{10} \equal{} (102020201)_3...1 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1 \equal{} 8 \equal{} 2^3$

but not only the sum of the digits is the same-even the number of 2`s and 1`s is the same :huh:

And, of course, there are further questions on this topic :wink: , such as:

Are there infintely many numbers with the following porperties:

sum of the digits
[list] in base 10 (decimal) $ \equiv 0 \ mod \ 10$
in base 2 (binary) = 2
[/list]

I hope, i motivated you to deal with this problem :)
greets

Georg

[quote="Georg-A."]Well, by intuition, we conjecture that there are only 5 solutions because it`s unlikely that the sum of two powers of 2 gets that close to a power of 10[/quote]
It might be reasonable to conjecture that there are finitely many solutions, but I don't think it's reasonable to conjecture that there are only five.


Yes... I don't see that they coincide  :blush:  Sorry,

[quote="toanhocmuonmau"]Using the inequality of mine and ductrung:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}\]
we get the result. :)[/quote]

Can you explain your solution ?

I have a very nice solution without expanding. At the moment, I am waiting for you.

[quote="toanhocmuonmau"]
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}\]
[/quote]

I found an SOS form for this but nothing better , could anyone continue this?
It is equivalent to
$\sum\ (b-c)^{2}(a^{3}+a^{2}b+ba^{2}-\frac{3abc}{2})$

Thank you ,very nice and elementary :D
Now hungkhtn I wait your solution...... :wink:

[quote="svasao_qb111"][quote="toanhocmuonmau"]Now, i will post my solution
The inequality is equivalent to
\[\sum S_{c}(a-b)^{2}\ge 0 \]
where
\[S_{a}=\frac{b}{c}+\frac{a}{b}+\frac{2a}{c}-\frac{5}{2}\]

\[S_{b}=\frac{c}{a}+\frac{b}{c}+\frac{2b}{a}-\frac{5}{2}\]

\[S_{c}=\frac{a}{b}+\frac{c}{a}+\frac{2c}{b}-\frac{5}{2}\]
Note that it suffices to consider the case $a \ge b \ge c$, then we have $S_{a}\ge 0$. We will show that
\[S_{a}+2S_{b}\ge 0,\qquad 2S_{b}+S_{c}\ge 0,\qquad S_{b}+S_{c}\ge 0 \]
Indeed, we have
\[S_{a}+2S_{b}=2\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{4b}{a}\right)+\frac{3b}{c}-\frac{15}{2}\ge 4+4+3-\frac{15}{2}>0 \]

\[S_{c}+2S_{b}=2\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{4b}{a}\right)+\frac{3c}{a}-\frac{15}{2}\ge 4+4+0-\frac{15}{2}>0 \]

\[S_{b}+S_{c}=\left( \frac{a}{b}+\frac{2b}{a}\right)+\left(\frac{b}{2c}+\frac{2c}{b}\right)+\left(\frac{b}{2c}+\frac{2c}{a}\right)-5 \ge \frac{a}{b}+\frac{2b}{a}+2\sqrt{\frac{b}{a}}-3=\left(\frac{a}{2b}+\frac{2b}{a}\right)+\left(\frac{a}{2b}+\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}\right)-3 \ge 2+\frac{3}{\sqrt[3]{2}}-3>0 \]
Now,
+, If $S_{b}\ge 0$, then
\[\sum S_{c}(a-b)^{2}\ge (S_{b}+S_{c})(a-b)^{2}\ge 0 \]
+, If $S_{b}\le 0$, then
\[\sum S_{c}(a-b)^{2}\ge (S_{a}+2S_{b})(b-c)^{2}+(S_{c}+2S_{b})(a-b)^{2}\ge 0 \]
The inequality is proved. Equality holds if and only if $a=b=c.$ :)[/quote]
@Can: very nice  :D
We have
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}\]

\[(a^{2}c+b^{2}a+c^{2}b)(a+b+c)^{2}\ge 9(a^{2}+b^{2}+c^{2})abc \]
Using muiheads and AM-GM.ptdh :lol:[/quote]
This inequality can't be solve by Muirhead theorem, Son! Since the inequality is not symmetric.
We have the sharper
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{10(a^{2}+b^{2}+c^{2})-ab-bc-ca}{(a+b+c)^{2}}\]
:)

To [b]Silouan[/b]
 I thinks some this wrong in your SOS form ??

Dear hungkhtn I cannot use your method to prove the original inequality as I cannot choose $c \geq b \geq a$ and your method works only for this case.  Can I choose   $c \geq b \geq a$ ? I don't think so? A help, please. Thank you very much. :blush:

Dear manlio, we can assume that $a\ge b\ge c$ since if $a\ge b\ge c$ then
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\le \frac{b}{a}+\frac{c}{b}+\frac{a}{c}.\]

Sorry, I want to explain better my problem.I want to prove

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}$

I have written it as

$(a-b)^{2}A+(b-c)(a-c) B \geq 0$

where
$A = \frac{1}{ab}-\frac{6}{(\sum a)^{2}}$

$B = \frac{1}{ac}-\frac{6}{(\sum a)^{2}}$

If $a\geq b \geq c$ we have $A \leq 0$ and now I cannot how to prove the inequality.

Can you please help me? Thank you very much.

Manlio, you can download my solution again. It is very detailed to understand. I consider two cases.

Your solution is very nice and detailed, but it is not stronger than

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}$

so I need a new proof. Am I right?

I dont understand? Two inequality is equivalent? Since
\[9-\frac{18(ab+bc+ca)}{(a+b+c)^{2}}=\frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}?\]

[quote="manlio"]Sorry but the solution in your PDF files is not for

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}$

but for a different inequality. Am I right? :oops:[/quote]

 :blush: It is my stupidity. Sorry, I attachs a wrong file  :(  how stupid am I.

Actually, my solution is to use the same way as 10_maths.

[quote="10maths_tp0609"]OK.
$f(a,b,c)=M(a-b)^{2}+N(a-c)(b-c)$
In that:
$M=\frac{1}{ab}-\frac{6}{(a+b+c)^{2}}-\frac{(a-c)(b-c)}{abc(a+b+c)}$
$N=\frac{1}{ac}-\frac{6}{(a+b+c)^{2}}+\frac{(a-b)^{2}}{abc(a+b+c)}$
c=max(a,b,c).[/quote]

But N it is positive only for $b \geq a$ !!!! How do you solve the case $b\leq a$

[quote=hungkhtn]Let $a,b,c$ be positive real numbers. Prove that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{18(ab+bc+ca)}{(a+b+c)^{2}}\ge 9.\][/quote]
Let $a,b,c$ be positive real numbers.[url=https://artofproblemsolving.com/community/c6h2087245p15058597 ]Prove that[/url]
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+(4\sqrt{2}-3)\frac{ab+bc+ca}{a^2+b^2+c^2}\ge 4\sqrt{2}$$
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{(3\sqrt[3]{4}-2)(ab+bc+ca)}{a^2+b^2+c^2}\ge 3\sqrt[3]{4}+1$$
[url=https://artofproblemsolving.com/community/c6h1892822p12921831]here[/url] [url=https://artofproblemsolving.com/community/c6h146906p831522]here [/url][url=https://artofproblemsolving.com/community/c6h576073p3395487]here[/url]


Determine all natural numbers n for which n divides Binom(2n,n) - 2.

If $n$ is prime, we have: $\binom{2n-1}{n}\equiv1[n]$, so $\binom{2n}{n}\equiv2[n]$.
Proof: $(n-1)!\left( \binom{2n-1}{n}-1\right)=(n+1)\cdots(2n-1)-(n-1)!\equiv(n-1)!-(n-1)!\equiv0[n]$.

But it seems that we have many no prime solutions (ex: $n=9$,$n=25$)

$N.B.$ In fact Wolstenholme's theoreme state that $\binom{2n-1}{n}\equiv1[n^3]$ if $n$ is prime $>3$. I think that we don't know if the converse is true (open question).

I can prove (using Wolstenholme theorem) the following (perhaps well known?) lemma:
$lemma:$ If $p>3$ is a prime and $l\in\mathbb{N}^*$ then $\binom{2p^l-1}{p^l}\equiv1[p^3]$.
$consequence:$ If $n=p^k$ with $p>3$ prime and $k\in\{1,2,3\}$ then $n$ divides $\binom{2n}{n}-2$.

Indeed well-known  :) . Actually, this congruence is itself attributed to Wolstenholme  :P .

Moreover, the congruence holds modulo $p^4$ if and only if $p$ divides the numerator of the $(p-3)$-rd Bernoulli number, and event that seems to occur extremely rarely (this is proven using the $p$-adic Stirling formula, in which the Bernoulli numbers are involved). Thus, it seems that if $n=p^k$ with $k \geq 4$, then $n$ is almost never a solution to $n \mid \binom{2n}{n} - 2$.

One aplication of Lucas Theorem not solve?


geometry
inequalities proposed
inequalities

Find max:
[tex] \sum\frac{1}{ab(sinC)^2}[/tex]

Are a, b, c; A, B, C the sides and angles of a triangle?

[quote="payman_pm"]$\sum\frac{1}{abSin^2{C}}=\frac{1}{2S}\sum\frac{1}{Sin C}$
***S is the area of triangle.[/quote]
ya, so I think there is only minimum but not maximum (take C=90)
The minimum is easy to find.

[quote="kid1412"]don't have max?[/quote]
Take C -> 0?

indeed,I don't understand this problem.
if we take $a=b=c \to 0$
the maximum is not exist. ;)


topology
induction
advanced fields
advanced fields unsolved

Hi all,

I'm supposed to show that any variety of positive dimension over an algebraically closed field $ k$ has the same cardinality as $ k$, and that, consequently, any two curves over $ k$ are homeomorphic.

Now, the second part is easy: any two curves have the same cardinality by the first part, and the finite complement topology, so are homeomorphic.

For the first part, I'm given some hints: Do $ \mathbb{A}^n$ and $ \mathbb{P}^n$ first.  Then for any $ X$, use induction on the dimension $ n$.  Then, $ X$ is birational to a hypersurface $ H \subset \mathbb{P}^n$.  Then, we use the fact that if $ H$ is a hypersurface, we have $ X \cap H \neq \emptyset$ to conclude that the projection of $ H$ to $ \mathbb{P}^n$ from a point not on $ H$ is finite-to-one and surjective.

How can I show that $ | \mathbb{A}^n |\equal{}| \mathbb{P}^n | \equal{}|k|$?  It's hard to understand how that could happen.  Second, how can we conclude anything about the cardinality of $ X$ from $ X$ being birational to a hypersurface?  Does that mean that it has the same cardinality as the hypersurface?  Finally, what is the projection that is referred to in the problem?  I did this problem previously:  Let $ \mathbb{P}^n$ be a hyperplane in $ \mathbb{P}^{n\plus{}1}$ and let $ P \in \mathbb{P}^{n\plus{}1} \setminus \mathbb{P}^n$.  Define $ \varphi: \mathbb{P}^{n\plus{}1} \setminus \{P\} \rightarrow \mathbb{P}^n$ by $ \varphi(Q)\equal{}$ the intersection of the unique line containing $ P$ and $ Q$ with $ \mathbb{P}^n$.  Then $ \varphi$ is a morphism.  Is this the projection that the problem is referring to?

Thank you.

Okay, I realize now that my second question is stupid: a birational map $ X \rightarrow H$ is only defined on some open subset.  Now I'm even at more of a loss as to how this helps me.

[quote="musicmantribalx"]How can I show that $ | \mathbb{A}^n | \equal{} | \mathbb{P}^n | \equal{} |k|$?  It's hard to understand how that could happen. [/quote]
$ A^2$ has the same cardinality as $ A$ for any infinite set $ A$.

[quote="musicmantribalx"]Second, how can we conclude anything about the cardinality of $ X$ from $ X$ being birational to a hypersurface?[/quote]
We can conclude lots.  The set of points at which a birational map isn't defined has cardinality at most that of $ k$, and so does the hypersurface.


I always felt that the standard (scalar) report card was not representative.

Here we go: if you have two competences to grade, a regular (scalar) report card may not be appropriate. A "perfect", orthonormal 2D grade is (100/sqrt 2 I + 100/sqrt 2 J) %. The total grade is (I^2 + J^2)^1/2.

The angle of a grade is arctan J/I. Usually the I-axis will be knowledge and the J-axis, comprehension. However, as long as you have two aspects to grade, you can use an axis as one aspect and the other axis for the second aspect.

However, the standard deviation vector is calculated as follows: (std.dev I + std.dev J)% and the norm, as with all 2D vectors, is calculated by this: (I^2 + J^2)^1/2. The difference between the scalar standard deviation and the norm of the standard deviation vector is what I call an angular dispersion measure. Likewise, the difference between the norm of the average vector and the scalar average is also an angular dispersion measure.

(The average vector is defined as adding up all grade vectors and dividing by the number of vectors to add up)

I'll be happy to discuss any problems that may arise from this kind of report cards.

Huh? I don't get your post at all. Here's a template that you can use:

Grade averages are often calculated using a formula such as (formula). I believe this is inadequate because (list of concerns). Here is a formula/set of formulas that gives a more meaningful representation of performance: (list of formulas). It addresses the issues above in (list of concerns) because (list of reasons).

Why is this useful and/or better than the current grading system in widespread use? Remember, most teachers are not math teachers and shrink away from even basic stuff.

Vectorial report cards would allow more representative grades, because I often see knowledge and comprehension as two different things. And we could more easily find faults with how teachers, well, teach, if the norm of the grade average vector is different from the average of the norms, the average angle is off the "standard" of 45 degrees (or pi/4), the angular dispersion is high, etc...

Sorry but I cannot continue now.


let f is a function from R to R
such that:
f(x) f(1-x)= f(ax+b)

and f is injective .

1) prove that a=0.

2) prove that f(1-b)=1.

3) prove that f isn't surjectif.

f(0)f(1) = f(b)
f(1)f(0)=f(a+b)=f(b)

so a = 0

Now f(1/2-x)f(x-1/2) = f(b) so f(b) cannot be 0 since f is injective. Hence from f(b)f(1-b) = f(b) we see that f(b) = 1.

I guess someone got the third one

[hide="1"]$ x \equal{} 1$ and $ x \equal{} 0$ together give $ f(a\plus{}b) \equal{} f(b)$. Now $ a\plus{}b \equal{} b$ by $ f$ injective, so $ a \equal{} 0$.[/hide]
[hide="2"]$ x \equal{} b$ gives $ f(b)f(1\minus{}b) \equal{} f(b)$. So $ f(1\minus{}b) \equal{} 1$ or $ f(b) \equal{} 0$. But if $ f(b) \equal{} 0$, then plug in, say $ x \equal{} 2$ and $ x \equal{} 3$. We get $ f(2)f(\minus{}1) \equal{} 0$ and $ f(3)f(\minus{}2) \equal{} 0$. So then there are at least two distinct $ f(x)$ that are 0, contradicting injectivity. Therefore $ f(1\minus{}b) \equal{} 1$.[/hide]
[hide="3"]Suppose it is surjective. Then we have a $ c$ such that $ f(c) \equal{} 0$. But plugging in $ x \equal{} c$ gives $ f(b) \equal{} 0$, which we showed was impossible.[/hide]


logarithms
integration
calculus
calculus computations

You may refer to:

http://www.mathlinks.ro/Forum/viewtopic.php?t=143417


να βρεθουν οι φυσικοι αριθμοι $ x,y,z$ οι οποιοι επαληθευουν την εξισωση
$ 2^x \plus{} 4^y \plus{} 8^z\equal{}81$
δικη σας :)

Μήπως εννοείς ακέραιοι ή φυσικοί μαζί με το 0 ή έχεις κάποιο typo? :huh:
Αλλιώς είναι προφανώς αδύνατη αφού $ 2^x \plus{} 4^y \plus{} 8^z \equiv 0(mod2)$ ενώ $ 81\equiv 1(mod2)$ :roll:

Οι φυσικοι τα ''τελευταια'' χρονια εχουν και το μηδεν .Αδυνατη δεν ειναι γιατι $ 2^4\plus{}4^0\plus{}8^2\equal{}81$

Το είπα με την προϋπόθεση ότι δε λαμβάνουμε το 0 στους φυσικούς....είναι πώς το έχει συνηθίσει ο καθένας :)
Τεσπά ας βάλω μία λύση τότε...
[hide]Πρέπει $ x \equal{} 0$ ή $ y \equal{} 0$ ή $ z \equal{} 0$ γιατί αν όλοι είναι μη μηδενικοί τότε το αριστερό μέλος είναι άρτιος ενώ το δεξί περιττός άτοπο.Επίσης δεν μπορεί δύο να είναι μηδέν για τον ίδιο λόγο.Άρα έχουμε 3 περιπτώσεις:

1.$ x \equal{} 0$.Τότε $ 4^y \plus{} 8^z \equal{} 80 \Rightarrow 4^{y \minus{} 1} \plus{} 2 \cdot 8^{z \minus{} 1} \equal{} 20$.Αφού και τα δύο μέλη είναι άρτιοι πρέπει $ y > 1$.Για $ z \equal{} 1$ πρέπει $ 4^{y \minus{} 1} \equal{} 18$ άτοπο.Άρα πρέπει και $ z > 1$.Για $ y \equal{} z \equal{} 2$ έχουμε λύση ενώ για $ y\geq 3$ είναι $ 4^{y \minus{} 1} \plus{} 2 \cdot 8^{z \minus{} 1}\geq 32 > 20$. Άρα η μόνη λύση σε αυτή την περίπτωση είναι η $ (x,y,z) \equal{} (0,2,2)$.

2.$ y \equal{} 0$.Τότε $ 2^x \plus{} 8^z \equal{} 80 \Rightarrow 2^{x \minus{} 1} \plus{} 4\cdot 8^{z \minus{} 1} \equal{} 40$.Αν $ z \equal{} 1$ πρέπει $ 2^{x \minus{} 1} \equal{} 36$ άτοπο.Αφού $ z > 1$ και $ 8|40$ πρέπει $ 2^{x \minus{} 1} \plus{} 4 \cdot 8^{z \minus{} 1} \equiv 0(mod8) \Rightarrow x \minus{} 1\geq 3\Rightarrow x\geq 4$.Μία προφανής λύση τώρα είναι η $ (x,y,z) \equal{} (4,0,2)$.Για $ x\geq 5$ είναι $ 2^{x \minus{} 1} \plus{} 4\cdot 8^{z \minus{} 1} \geq 16 \plus{} 4\cdot8 \equal{} 48 > 40$.Άρα εδώ η μόνη λύση είναι η$ (x,y,z) \equal{} (4,0,2)$.

3.$ z \equal{} 0$.Τότε πρέπει $ 2^x \plus{} 4^y \equal{} 80 \Rightarrow 2^{x \minus{} 1} \plus{} 2\cdot 4^{y \minus{} 1} \equal{} 40$.Αν $ y \equal{} 1$ τότε $ 2^{x \minus{} 1} \equal{} 38$ άτοπο.Άρα $ y\geq 2$ και έτσι $ 2\cdot 4^{y \minus{} 1}\equiv 0(mod8)$ οπότε πάλι $ x\geq4$.Μία προφανής λύση τώρα είναι η $ (x,y,z) \equal{} (4,3,0)$.Για $ x \equal{} 5$ πρέπει $ 2\cdot 4^{y \minus{} 1} \equal{} 24 \Rightarrow 4^{y \minus{} 1} \equal{} 12$ άτοπο.Μία ακόμη λύση εδώ είναι η $ (x,y,z) \equal{} (6,2,0)$ ενώ εύκολα βλέπουμε ότι για $ x\geq7$ δεν έχει λύσεις.

Άρα όλες οι λύσεις είναι:
$ (0,2,2),(4,0,2),(4,3,0),(6,2,0)$[/hide]
Ελπίζω να μην έκανα κάπου λάθος :maybe:

οχι δν εκανες.σωστος :)
παρομοια την πηγα κ εγω μονο
που βασιστηκα περισσοτερο στο $ 8^3\equal{}212$ τοτε$ z<\equal{}2$(και αρχισα να διακρινω περιπτωσεις) οπως:
για $ z\equal{}2$ $ 2^x \plus{}4^y\equal{}17$αρτιος-περιττος αρα $ x\equal{}0$ η $ y\equal{}0$ etc... :)

Όντως έτσι γλυτώνουμε μερικές περιπτώσεις :)

Θετουμε u = 2y και t = 3z και η εξισωση γινεται: $ 2^x \plus{} 2^u \plus{} 2^t \equal{} 81$. Το 2ο μελος ειναι περιτος αριθμος αρα πρεπει και το 1ο να ειναι περιτος. αν κανενας απο τους x,u,t δεν ειναι 0 τοτε το 1ο μελος ειναι αρτιος αριθμος ως αθροισμα αρτιων (ατοπο), αρα ενας απο τους x,y,z ειναι 0, τοτε οι αλλοι 2 αριθμοι πρεπει να ειναι διαφοροι του 0, γιατι αν απο τους 3 αριθμους ακριβως 2 ηταν 0 τοτε το 1ο μελος θα ηταν παλι αρτιος, ενω για x=u=t=0 η εξισωση δεν επαληθευεται, αρα απο τους 3 αριθμους ακριβος ενας ειναι ισος με 0 και η μια απο τις τρεις δυναμεις του 2 που υπαρχουν στο 1ο μελος θα γινεται ιση με 1, ετσι η εξισωση παιρνει τη μορφη $ a \plus{} b \plus{} 1 \equal{} 81 < \equal{} > a \plus{} b \equal{} 80$, οπου a,b ειναι δυναμεις του 2 μεγαλυτερες του 1, επηδη ακομα $ a,b \geq 0$ εχουμε οτι οι αριθμοι a,b ανηκουν στο συνολο {2, 4, 8, 16, 32, 64} και ειναι προφανες οτι το μονο ζευγος αριθμων απο αυτο το συνολο που επαληθευει την $ a \plus{} b \equal{} 80$ ειναι το (a, b) = (16, 64). Οποτε εχουμε:

Αν ειναι x=0 τοτε $ 2^u \equal{} 16 < \equal{} > u \equal{} 4$ και $ 2^t \equal{} 64 < \equal{} > t \equal{} 6$, ή $ 2^u \equal{} 64 < \equal{} > u \equal{} 6$ και $ 2^t \equal{} 16 < \equal{} > t \equal{} 4$ (Αποριπτεται διοτι ειναι t = πολ3), αρα εχουμε $ (x,y,z) \equal{} (x, \frac {u}{2}, \frac {t}{3}) \equal{} (0, 2, 2)$.

Αν ειναι u=0 τοτε $ 2^x \equal{} 16 < \equal{} > x \equal{} 4$ και $ 2^t \equal{} 64 < \equal{} > t \equal{} 6$, ή $ 2^x \equal{} 64 < \equal{} > x \equal{} 6$ και $ 2^t \equal{} 16 < \equal{} > t \equal{} 4$ (Αποριπτεται διοτι ειναι t = πολ3), αρα εχουμε $ (x,y,z) \equal{} (x, \frac {u}{2}, \frac {t}{3}) \equal{} (4, 0, 2)$.

Αν ειναι t=0 τοτε $ 2^x \equal{} 16 < \equal{} > x \equal{} 4$ και $ 2^u \equal{} 64 < \equal{} > u \equal{} 6$, ή $ 2^x \equal{} 64 < \equal{} > x \equal{} 6$ και $ 2^u \equal{} 16 < \equal{} > u \equal{} 4$, αρα εχουμε $ (x,y,z) \equal{} (x, \frac {u}{2}, \frac {t}{3}) \equal{} (4, 3, 0)$, ή $ (x,y,z) \equal{} (x, \frac {u}{2}, \frac {t}{3}) \equal{} (6, 2, 0)$.

Αρα $ (x,y,z) \equal{} (0, 2, 2)$, ή $ (x,y,z) \equal{} (4, 0, 2)$, ή $ (x,y,z) \equal{} (4, 3, 0)$, ή $ (x,y,z) \equal{} (6, 2, 0)$.

qed  :P

[quote="Orfikoss"]Σωστός NickNafplio μόνο, αν θες, αντί για <=> μπορείς να γράφεις \Leftrightarrow στη LaTEX.
Φαίνεται έτσι $ \Leftrightarrow$[/quote]

α οκ, thanx

[quote="Athinaios"]να βρεθουν οι φυσικοι αριθμοι $ x,y,z$ οι οποιοι επαληθευουν την εξισωση
$ 2^x \plus{} 4^y \plus{} 8^z \equal{} 81$
δικη σας :)[/quote]

Αρχικά παρατηρούμε ότι $ x\in\{0,1,2,...,6\}$ και $ y\in\{0,1,2,3\}$ και $ z\in\{0,1,2\}$ διαφορετικά καταλήγουμε σε άτοπο αφού θα ήταν το πρώτο μέλος

μεγαλύτερο από το δεύτερο.

Επιπλέον εύκολα βρίσκουμε ότι

$ 2^{n}\equiv2,4,8,6 mod10$

$ 4^{n}\equiv6,4 mod10$

$ 8^{n}\equiv 4,2,6,8 mod10$

Επειδή τώρα το β΄μέλος είναι περιττός θα πρέπει να έιναι περιττός και το α΄.Οπότε έχουμε τις περιπτώσεις

[u][b]1η Περίπτωση[/b][/u] $ x\equal{}0$

$ 4^{y}\plus{}8^{z}\equal{}80\Longrightarrow4^{y}\plus{}8^{z}\equiv0mod10$. Το οποίο ισχύει μόνο αν ($ 4^{y}\equiv6mod10$ kai $ 8^{z}\equiv4mod10$) ή

($ 4^{y}\equiv4mod10$ kai $ 8^{z}\equiv6mod10$).

Aπό το πρώτο σύστημα παίρνουμε τη λύση $ (x,y,z)\equal{}(0,2,2)$ ενώ στο δεύτερο καταλήγουμε σε άτοπο αφού ο ελάχιστος εκθέτης για τον οποίο ισχύει

$ 8^{z}\equiv4mod10$ είναι ο $ z\equal{}4$

[b][u]2ηΠερίπτωση[/u][/b] $ y\equal{}0$

$ 2^{x}\plus{}8^{z}\equal{}80\Longrightarrow2^{x}\plus{}8^{z}\equiv0mod10$. To οποίο ισχύει μόνο αν ($ 2^{x}\equiv2mod10$ kai $ 8^{z}\equiv8mod10$) ή

($ 2^{x}\equiv4mod10$ kai $ 8^{z}\equiv6mod10$) ή ($ 2^{x}\equiv8mod10$ kai 8^{z}\equivmod10) ή ($ 2^{x}\equiv6mod10$) kai

$ 8^{z}\equiv4mod10$).
Τα τρία πρώτα συστήματα  καταλήγουν σε άτοπο αφού οι ελάχιστοι εκθέτες για τους οποίους ισχύει είναι οι $ z\equal{}5,4,3$ αντίστοιχα ένώ από την τέταρτη περίπτωση παίρνουμε τη λύση $ (x,y,z)\equal{}(4,0,2)$

[u][b]3ηΠερίπτωση[/b][/u] $ z\equal{}0$

$ 2^{x}\plus{}4^{y}\equal{}80\Longrightarrow2^{x}\plus{}4^{y}\equiv0mod10$. To οποίο ισχύει μόνο άν ($ 2^{x}\equiv4mod10$ kai $ 4^{y}\equiv6mod10$) ή

($ 2^{x}\equiv6mod10$ kai $ 4^{y}\equiv4mod10$)

Από το πρώτο σύστημα παίρνουμε τη λύση $ (6,2,0)$ και από το δεύτερο την $ (4,3,0)$

Τελικά όλες οι λύσεις της αρχικής είναι

$ (x,y,z)\equal{}(0,2,2),(4,0,2),(6,2,0),(4,3,0)$

Εύκολα με (mod 2) πρέπει $ xyz \equal{} 0$

το 80 σε δυαδική μορφή είναι

1010000 δηλ. $ 2^6 \plus{} 2^4 \equal{} 2^{3 \cdot 2 \cdot 1} \plus{} 2^{2 \cdot 2 \cdot 1}$

αυτό τώρα γράφεται με τους εξής τρόπους (λαμβάνοντας υπόψιν ότι τα 2, 4, 8 δεν πρέπει να συνυπάρχουν):

$ 80 \equal{} 8^2 \plus{} 2^4$

$ 80 \equal{} 8^2 \plus{} 4^2$

$ 80 \equal{} 4^3 \plus{} 2^4$

$ 80 \equal{} 2^6 \plus{} 4^2$

άρα

$ 81 \equal{} 8^2 \plus{} 2^4 \plus{} 4^0$

ή $ 81 \equal{} 8^2 \plus{} 4^2 \plus{} 2^0$

ή $ 81 \equal{} 4^3 \plus{} 2^4 \plus{} 8^0$

ή $ 81 \equal{} 2^6 \plus{} 4^2 \plus{} 8^0$

Αυτή θα ήταν μια πολύ καλή μέθοδος ανεξάρτητα του αριθμού που θα ήταν στη θέση του 81


If $ 20\%$ of a school was made up of out-of-district students last year, but the number of out-of-district students increased by $ 50\%$ this year, without any change in total enrollment, what percent of the students is out-of-district this year?

pretend there are 100 students total
there would be 20 out-of-district students last year
50% increase would give 30 this year
so it's 30


analytic geometry
geometry
rectangle
similar triangles
Pythagorean Theorem

In quadrilateral $ABCD$, $\angle CAD=\angle CBD=90^\circ$, $CD=10$ and $AD=6$. Let $M$ be the midpoint of $AB$ and let $N$ be the midpoint of $CD$. If $MN=4$, find $BC$.

[hide="answer"]2.8
I used coordinates for all of the points. It's fairly simple, but the big and ugly numbers made it hard and long so I won't post it. Can somebody find a better way to do this?[/hide]

Clearly, $BC // MN // AD$, let $BC=x$

Draw perpendicular from $B$ to $AD$ hitting at $Q$, $BQ$ hits $MN$ at $R$.

It follows that $\angle BRM=90$ and $\triangle BRM \sim \triangle BQA$,

Since $BRNC$ and $BQDC$ are rectangles, $MR=4-x$, and $AD=6-x$. By similar triangles, we have $\frac{5}{10}=\frac{4-x}{6-x}$, so $x=2$

[quote="Altheman"]Clearly, $BC // MN // AD$[/quote]
Hmm... how did you get that? :?
(Maybe you misread the problem?)

By the way, Jiang, your answer is correct.

oh wow...i really misread the problem...woops...it still is not that difficult though

let $AC\cap BD=X$anyway, it is obvious that this inscribes in a semicircle with diameter CD, r=5,

since MN bisects chord AB, $\angle BMN=90$, using the pythagorean theorem, $MB=3$, so $AB=6$, also pythagorean theorem gives $AC=8$

from here we could use ptolemy and get an easy solution, but i rather use the nice similar triangles that occur

note that $AD=AB$, so $\angle ABD=\angle ADB$, by inscribed angles, we have $\angle ADB=\angle ACB=\angle ABD$ so $\triangle ABX \sim \triangle ACB$, then

$\frac{AB}{AC}=\frac{AX}{AB}$
$AX=\frac{9}{2}$, then $CX=\frac{7}{2}$, pythagorean theorem gives $DX=\frac{15}{2}$

Also, $\triangle DAX\sim \triangle CBX$ so $\frac{BC}{AD}=\frac{CX}{DX}$
$BC=\frac{14}{5}$


I'm posting here because my problem is somewhat related to college, and you guys always have great advice. So thanks in advance. It's a long read btw.

[u]about me[/u]: I live in Illinois. I narrowly missed out on applying to IMSA and I regret it. I'm currently a junior taking a precalc-ish class along with calc. My math teacher is great, and I can count on him for help on pretty much everything. I like math, mathematical sciences, and computers. I despise biology.

[u]my problem[/u]: I have no idea what courses I should do senior year. I can take 6-7 classes on a normal schedule. Right now I only want to take spanish 4, english 4, physics, and an advanced history class. I'm trying to get an math independent study going for next year, but this probably wouldn't occupy a class hour. There are no other math classes offered at my school.

[u]my choices[/u]:

[b]1)[/b] take classes at a community college about 45 minutes away. I'll be on the lookout for scholarships; I probably won't consider this option unless I can take what I want for next to nothing. If this seems like a nice choice, should I try to get in classes I would enjoy, or ones for a cheaper price than I would have to pay at college later?

pros: Nice selection of classes. If I take this, I get out of high school an hour early, shortening my required courseload to 5 or 6.

cons: Gas. Credits may not transfer if I get into a (prestigious?) school. Professors may or may not be competent.

[b]2)[/b] take online classes. Right now I'm considering [url=http://www.ivhs.org/index.learn?bhcp=1]Illinois Virtual High School (IVHS)[/url] and EPGY.

pros of IVHS: I can probably get school to pay for it, and the AP courses can be integrated into my GPA as advanced.

cons of IVHS: the [url=http://www2.imsa.edu/programs/ivhs/pdfs/2006-2007_courses.pdf]course offering[/url] seems a little weak, and possibly not of the difficulty I would enjoy.

pros of EPGY: it appears to have an amazing curriculum

cons of EPGY: from searching, it seems the tutors might be, well, bad. I have no idea if my school would pay for or count these as credits. My counselor hadn't even heard of this place, which isn't good in my opinion.

[b]3)[/b] take some AOPS classes.

pros: these really interest me. A lot.

cons: the cost is high considering I'd probably have to pay for all of it. These wouldn't help me solve my credit situation. Truthfully, I don't see that these count for anything besides learning more math.

I'm leaning toward IVHS right now; I could get AP stats through this, and maybe some other things. If I can't get an entire day filled, I might just cave and take Bio 2 and/or Anatomy. I hope it doesn't come to that though... so I'm hoping you guys have advice :maybe:

after rereading that, I need to apologize for not asking what advice I even wanted...and here's a summary if you don't want to read my wall of text

I can't decide on what to do because I don't know what's best for me. You may be thinking, "How can I decide what's best for you?" Well, if you have any experience with high schoolers taking college or online classes, I'd love to hear what you thought about them. If you have experience with both, did one option seem better than another? I think college might be better for me, but then I pay for gas and basically lose any shot at valedictorian. Online classes would be cheaper and easier, but I may feel like I wasted my time.

By the way, I've decided I'll enroll in an AOPS class(es) if I have the time and money.


let's see what countries are active on the forum :)
sorry about those countries that I didn't name in the poll, but the list was getting too long :)

this poll runs for 14 days.

too bad. but the poll was made to check the "nations" that actively are present on the forum. it seems that the results are somewhat real. :)
hopefully at the next one USA will get some more votes.

[quote="integral0"]Too bad I missed the poll. I am from the US of A!

 :)[/quote]

I am from the USA, too.  No votes?  Uy.

Where is Hong Kong? Then again Hong Kong is not a nation  :)

is the reason that there is always a please wait loading... thing on this website becasue it is hosted somewhere else in the world and is slower?

what do you mean "somewhere else in the world"?
the page loading page appears each time (it's the script). if it wouldn't be for that, users with low connections such as dial-up would have to see a white screen for some while which is not quite pleasant.

I am assuring you that for a free-of-charge, personal made with my own resources math web site, with no adds and everything, the speed is quite good (also counting the more and more increasing visitators).  :D


induction
number theory unsolved
number theory

Find all pairs in nonnegative integer $ (x,y)$ that solve the equation:
$ p^{x}-y^{p}=1$,
where $ p$ is a given odd prime.

Use that for $ p \ge 3$,
$ p^{x}=y^{p}+1=(y+1)(y^{p-1}-y^{p-2}+\cdots-p+1)$.
Notice that this forces $ y+1=p^{x'}$ or $ y=0$.  In the first solve for $ y$.
For the case $ p=2$, I think you could just work in $ \mathbb{Z}[i]$ or an [url=http://links.jstor.org/sici?sici=0025-570X%28197409%2947%3A4%3C231%3AANOMN%3E2.0.CO%3B2-M&size=LARGE&origin=JSTOR-enlargePage]elementary argument exists.[/url]
Let me know if this is enough detail  :) .

[quote="PhilAndrew"]In fact, the problem is trivial by Mihailescu's Theorem: either $ (x, y) = (0, 0)$ (and $ p$ is any odd prime), either $ p = 3, x = y = 2$. The single nontrivial case reduces to $ p \ge 2^{p}+1$ for $ p \ge 3$, which by induction is false.[/quote]
Yes, you are correct.  But, that theorem is very deep and took years and many mathematicians to finish it.  Nevertheless, the Tijdeman-Mihailescu's Theorem is a very powerful, beautiful theorem.  :)


[color=blue][i]Αυτή τη φορά η άσκηση στο περιοδικό RMT προτείνεται για την VI !  Μου πήρε κανα εικοσάλεπτο να βρω μια συνθετική λύση. Δόστε την , όταν σας δίνεται ευκαιρία , και στους Junior. Είναι ωραία ασκησούλα για όσους θέλουν να αγαπήσουν τη γεωμετρία , αλλά ακόμα δεν βρήκαν την κατάλληλη άσκηση ![/i][/color]

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

 Σε ένα ισοσκελές τρίγωνο $ ABC$ με  $ AB\equal{}AC$ είναι   $ \angle A \equal{} 20$.

Στην πλευρά $ AB$ παίρνουμε τμήμα   $ AD\equal{}BC$ . Στο εξωτερικό του τριγώνου   $ ABC$

 κατασκευάζουμε το ισόπλευρο τρίγωνο $ ADE$.

 Να αποδείξετε ότι η   $ CD$  διχοτομεί τη γωνία  $ \angle ACE$.

 [u]Μπάμπης[/u]

Nμίζω ότι είναι κάπως έτσι :
[hide] Tα τρίγωνα  ABC  και   CEA είναι ίσα

1. AC  κοινή
2. ΑΕ = ΒC
3. και οι γωνίες  EAC , BCA είναι ίσες από 80 μοίρες η κάθε μία.

Άρα  CE = CA.  Στη συνέχεια η σύγκριση των τριγώνων DAC και DEC είναι εύκολη

και προκύπτει το ζητούμενο.[/hide]

[quote="bilot4"]κ. Στεργίου είστε σίγουρος ότι είναι έτσι η διατύπωση :?:

Γιατί εμένα η $ CD$ μου βγαίνει πάντα εκτός της γωνίας $ \angle{ACE}$ πράγμα που σημαίνει ότι δεν μπορεί να είναι διχοτόμος της :huh: . Είναι κάπως αλλιώς το ερώτημα ή απλά κάνω τελείως λάθος σχήμα?...

Με την ευκαιρία, σας ευχαριστώ πολύ και για τα καλά σας λόγια στο άλλο πρόβλημα που προτείνατε (το γράφω εδώ για οικονομία στα posts)[/quote]

Ετσι ακριβώς την έχει το περιοδικό και με αυτά τα γράμματα την έλυσα. Το ίδιο πρόβλημα είχα και γω , αλλά προσάρμοσα το σχήμα σε αυτό που μας ζητάει. Δυστυχώς έχω ξεπεράσει το όριο στα MB του φόρουμ και δε μπορώ να στε'ιλλω εικόνα για το σχήμα.
 Ας μας το φτιάξει  , αν έχει χρόνο , ο κος Βήττας.

Βάζω το σχήμα που αντιστοιχεί στη λύση του [b][size=100]vangelismak[/size][/b], συνδυασμένο με το κανονικό [b][size=100]9-γωνο[/size][/b] που είναι πολύ εξυπηρετικό (για το σχήμα αλλά και για τη λύση), σε προβλήματα όπου εμφανίζεται ισοσκελές τρίγωνο με γωνία κορυφής $ 20^{o}.$

Με το υπολογιστή είναι πολυ εύκολο να σχεδιάσουμε ένα κανονικό [b][size=100]9-γωνο[/size][/b] και να πάρουμε το ισοσκελές τρίγωνο $ \bigtriangleup ABC$ με $ \angle A \equal{} 20^{o}.$

Αν δεν έχουμε υπολογιστή, ένα ισοσκελές τρίγωνο με βάση $ BC \equal{} 2$ και ύψος $ h_{a} \equal{} 5,65$, είναι μία καλή προσέγγιση, για να έχουμε γωνία κορυφής $ 20^{o}.$

Κώστας Βήττας.

Γενικά τα τρίγωνα 80-80-20 προσφέρουν "διαβολεμένες" ασκήσεις  :(
Τις πιο πολλές φορές για να βρεθεί το ζητούμενο είναι απαραίτητη η κατασκευή ισοπλέυρου τριγώνου ή η κατασκευή kite "χαρταετού". Ευτυχώς εδώ το ισόπλευρο το έδινε η άσκηση. Επίσης Δημήτρη η κατασκευή του ιδίου τριγώνου είναι λίγο δύσκολη. Να σας πω ότι η πρώτη άσκηση που έπεσε στα χέρια μου και ήταν τέτοιου είδους μου έφαγε μια βδομάδα. Θυμάμαι είχα κατασκευάσει το σχήμα μετά από πολλές προσπάθειες. Ευχαριστούμε κύριε Κώστα για την παρατήρηση με το 9-γωνο.  :)


Find $ x, y \in R$ of this equation: $ \begin {cases} y^2\plus{}2y\sqrt{x^2\minus{}1}\equal{}26\minus{}x^2 \\ y^2\plus{}y\sqrt{x^2\minus{}1}\equal{}10\end {cases}$

Let $ u \equal{} \sqrt{x^2 \minus{} 1}$. Then the equations are:

$ \begin{cases} y^2 \plus{} uy \minus{} 10 \equal{} 0 \\
u^2 \plus{} uy \minus{} 15 \equal{} 0 \end{cases}$

This is a second degree system whose solution is $ (3, 2)$.

$ \sqrt{x^2 \minus{} 1} \equal{} 3 \implies x^2 \minus{} 1 \equal{} 9 \implies x^2 \equal{} 10 \implies x \equal{} \pm\sqrt{10}$.

Final solution: $ (x, y) \equal{} (\pm\sqrt{10}, 2)$

[quote="ffao"]Let $ u \equal{} \sqrt {x^2 \minus{} 1}$. Then the equations are:

$ \begin{cases} y^2 \plus{} uy \minus{} 10 \equal{} 0 \\
u^2 \plus{} uy \minus{} 15 \equal{} 0 \end{cases}$

This is a second degree system whose solution is $ (3, 2)$.

$ \sqrt {x^2 \minus{} 1} \equal{} 3 \implies x^2 \minus{} 1 \equal{} 9 \implies x^2 \equal{} 10 \implies x \equal{} \pm\sqrt {10}$.

Final solution: $ (x, y) \equal{} (\pm\sqrt {10}, 2)$[/quote]
Thank ffao! This system can be done by other ways! :P

I believe your way is let $ u\equal{}\sqrt{x^2\minus{}1}$ then
     $ (y\plus{}u)^2\equal{}25$
     $ y^2\plus{}yu\equal{}10$  :D
and it is not really different than the way of [b]ffao[/b]


[quote="spider_boy"][quote="quangtrung"]find all functions$f,g: R\to R$ satisfy
                 $f(x+g(y))=xf(x)-yf(x)+g(x)$  $\forall x,y\in R$[/quote]
Is the question like this one? find all functions$f,g: R\to R$ satisfy
                 $f(x+g(y))=xf(y)-yf(x)+g(x)$
If so see     [url=http://www.kalva.demon.co.uk/short/soln/sh00a3.html]www.kalva.demon.co.uk/short/soln/sh00a3.html[/url]
I remember this shortlist problem because we have solved it together with my teacher at class last year and it was very hard.[/quote]
    yes! the right condittion of problem is like what you say .I  have wrong problem  from my friend ( I know that it is wrong.I have solved it too) but  why not we try to solve it .I thing it is  as interesting as the original problem


Prove that $\overline{\exists} x,y,z,t \in N^*$ such that:
   $x^2+y^2=z^2 - t^2, xy=zt$

This was clearly posted before or anyway it is closely related to an already discussed bulgarian diofantine equation and also to Mathlinks Contest 4-th edition, first round. I do not feel necessary to post again a solution.  ;)


superior algebra
superior algebra solved

$G$ a group and $f \in Aut(G)$ such that $fof=f$ and $f$ has only one fixed point. Prove that $G$ is abelian.

Julien, i think you meant $f(f(x)) = x$ since you said "involutive". Then you may look here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=eugene&t=25871

Sorry, sorry, sorry. Yeah, I really meant involutive: $fof=id$.

Thank you very much, all the more that probably I was trying to solve something impossible. I had a look at your link, and it seems that you assumed $G$ to be finite. I had found the second part of your proof, but for the first part, finiteness is needed. Surely the result does not hold for infinite groups.


[color=darkred]Let $ ABC$ be a triangle for which exists a point $ M\in (BC)$ so that $ AM < AB$ and $ AM < AC$.

Consider the circles $ w_{b}$, $ w_{c}$ which are tangent to the line $ AM$ in the point $ M$ and $ B\in w_{b}$, $ C\in w_{c}$.

$ 1\blacktriangleright$ Denote the points $ \{\begin{array}{c}P\in (AB)\cap w_{b}\\ \ R\in (AC)\cap w_{c}\end{array}$. [b]Prove that the quadrilateral $ APMR$ is cyclically[/b] (let $ w$ be the its circumcircle).

$ 2\blacktriangleright$ Denote the points $ \{\begin{array}{c}\{M,N\}= BC\cap w\\ \\ X\in PM\cap RN\ ;\ S\in AX\cap BC\end{array}$. [b]Prove that[/b] $ \frac{SM}{SN}=\frac{SC}{SB}$.[/color]

[hide="1"]
Clealry $ m\angle{AMP}\equal{}B$, and $ m\angle{AMR}\equal{}C$, by arc chasing.  Therefore, summing,
$ A\equal{}180^{\circ}\minus{}m\angle{PMR}$, and so $ APMR$ is cyclic, as desired.  This holds for any point $ M$ on segment $ BC$.[/hide]

And the second point ?! The first point appears only for understanding the second point.

What does $ \{M,N\}\equal{}BC\cap w$ mean? Does $ w$ refer to both circles?   :huh:

[quote="Anaxerzia"]What does $ \{M,N\}\equal{} BC\cap w$ mean? Does $ w$ refer to both circles?   :huh:[/quote]
Sorry, You didn't read all !
[quote="Virgil Nicula"].... [b]Prove that the quadrilateral $ APMR$ is cyclically[/b] (let $ w$ be the its circumcircle).[/quote]

It is fantastically ! A set with only two points !

[quote="Anaxerzia"]What about $ \{M,N\}$? What does that mean?[/quote]
$ M$ and $ N$ are the points of intersection.


do there exist $ k$ such that $ \frac{2x^2\plus{}3}{x\plus{}2}\equal{}k$
where $ k$ is integral. :?:

I think there would be an infinite amount of solutions if $ x$ is not an integer. If $ x$ is an integer,  here is the solution:

[hide="Solution"]If $ \frac {2x^{2} \plus{} 3}{x \plus{} 2}$ is some integer, then $ \frac {2x^{2} \plus{} 3 \minus{} 2x(x \plus{} 2)}{x \plus{} 2}$ is an integer as well. Simplifying, we get $ \frac {3 \minus{} 4x}{x \plus{} 2}$ is an integer. Then if $ \frac {3 \minus{} 4x}{x \plus{} 2}$ is an integer, $ \frac {3 \minus{} 4x \plus{} 4(x \plus{} 2)}{x \plus{} 2}$ is an integer as well. So $ \frac {11}{x \plus{} 2}$ must be an integer. From there, we can easily find all values of $ x$, then plug them back into our original equation to find $ k$.[/hide]

from what you did it appears that $ x\equal{}9$.
substituting back into my question does not make $ k$ integral
my problem remains is $ k$ integral.

Er yeah it does...

$ k \equal{} \frac {2 \cdot 9^2 \plus{} 3}{11} \equal{} \frac {165}{11} \equal{} 15$

You can also just use synthetic division to rewrite it as $ 2x\minus{}4\plus{}\frac{11}{x\plus{}2}$. Then $ x\plus{}2|11$, so $ x\plus{}2\equal{}\pm 1,\pm 11$, so $ x\equal{}\minus{}1,\minus{}3,9,\minus{}13$ are all solutions that give integral $ k$.

[quote="21 7 15"]well its correct and your method is the same as that used by naiko sato in 1996 to teach students[/quote]

Hmm...can someone explain this method (used by Thunder365) or point me to an explanation? Thanks.

@mismatchtea: look at the second post.

Say you have an $ x$ term in the numerator, and you have an expression containing $ x$ in the denominator. Take that denominator and multiply it by anything needed to cancel with the numerator and add it on, and it will still be an integer.