quadratics
number theory unsolved
number theory

[quote="harazi"]Find x,y,z such that $ xy^2=z^3+1$ and x does not have prime factors of the form 6k+1.[/quote]
First, $z^3+1=(z+1)(z^2-z+1)$ so, we need to consider two cases:
[list]
1. $(\gcd(z+1,z^2-z+1)=1$ then $z+1=t$ and $z^2-z+1=s$ with $\gcd(t,s)=1$ and $ts=xy^2$,
2. $(\gcd(z+1,z^2-z+1)=3$ then $z+1=3^kt$ and $z^2-z+1=3^ks$ wth $k\in \mathbb{Z}$, $k>0$ and $\gcd(t,s)=1$
[/list]
then I am going to write the solution for the first case, and then extend to the second one. For the expressions we can write the following equation: $t^2-3t+3=s$, so we need that $D=-3+4s$ be a square, namely $\Lambda^2$ then we get that $\Lambda^2-9=4(s-3)$ and $t=\frac {3\pm \Lambda}2$ so $t|(2s-6)$, then $2s-6=E t$, and hence
\[
2z^2-(2+E)z-(4+E)=0 \quad \iff \quad z=-1 \quad {\rm or} \quad z=2+\frac E 2,
\]
in the first case we deduce that $z=-1$ and $xy^2=0$ which give us $(0,y,-1)$ and $(x,0,-1)$, and the second case give us that $E$ must be even, hence $E=2F$, then  $s-3=F t$, $z=2+F$ and now we need to consider this expression $(F+3)((F+3)F+3)=xy^2$, unafortunately I dont know  to solve it :?  but maybe some of us know to do it, so good luck.

Best Regards.
RobertuX TCM

[quote="harazi"]You don't even imagine how difficult this thing is.  ;)[/quote]
Yeah, I know a solution is $((F+3)((F+3)F+3),1,F+2)$ and $((F+3)((F+3)F+3),-1,F+2)$ [taking $y=\pm 1$] but maybe there are more solutions. Well for me the hard work is to probe that there are NOT more solutions in a problem, I hate it.  :blush:

Best regards.

RX

[quote="harazi"]Hint: there are 7 solutions of the problem... ;)  ;)   So, don't try too hard to show what you want. ;)[/quote]
If you see my last comment you can found [b] infinity[/b] solutions taking $y=\pm 1$, for instance $(9,1,2)$, $(28,1,3)$,  and in general $(z^3+1,1,z)$ with $z\ne 0 \ mod \ 6$.  :D  :D

[quote="harazi"]As usual, non-sense. Already, your second example does not work (I applied the modular theory  ;) to think hard and fast and I found that 7 is of the form 6k+1 and divides 28). Now, could you please show me some more of these infinity of solutions?[/quote]
mmm, you are right, every of my examples escept $(0,y,0)$, $(x,0,0)$ and $(2,1,1)$, $(9,1,2)$ are nice solution, but the rest have a [b] bad [/b] divisor. But, can u explain to us, how u probe the bound pf solutions, I remember more or less but I don't use in general modular theory.

Thx and regards.

Robert

[quote="harazi"]It is in the unsolved section, right? I have no solution for it.[/quote]
puffff, so WHY u did the comment of the number of solutions without explanation of it, I hate this kind of comments. Sorry  :( . And I am sure that Modular Theory is NOT trivial in general.

RX

I know some information about this problem, among which the number of solutions. I am sorry you are sorry, so I will just shut up and say absolutely nothing from now on.

we work in $Zw$, we have: $xy^2=(z+1)(z+w)(z+w^2)$. Now every prime of the form $3k+1$ is a product of two irreducibles $(a+bw)(a+bw^2)$, and a prime of the form $3k+2$ is irreducible, and clearly, every prime of this form divides $z+1$ then. so, the only prime that gives us some trouble is the prime $3$. an easy check shows we must have: $z^2-z+1=m^2$ or $z^2-z+1=3m^2$ case 1 can be easily handleed, case 2 is more dificult, we have that $z+1$ is a product of primes of the form $3k+2$ and a product of primes of the form $3k+1$ raised to an even power. I just realised i didnt do anything so forgive me for my nonsense

Edited by pbornsztein to make it readable.

sorry for bringing back an old topic but if anyone is still interested on seeing an elementary solution of this equation, I can post the proof I have.[we happened to discuss this with our proffesor last week...]

Yes, Albanian Eagle, please post an elementary solution.

I found some free time, so I'm posing the solution, finally. Sorry if you were impatient to read this!
Maybe it's a bit long but just take a deep breath !

[color=blue]The problem :[/color]
let x,y,z be three natural numbers and x doesn't contain any prime factor of the form $6k+1$. Show that if $xy^{2}=z^{3}+1$ then
$(x,y,z)$ is equal to precisely one of the following: $(2,1,1)(1,3,2)(9,1,2)(2,78,23)(8,39,23)(18,26,23)(72,13,23)$

Solution:

we will start with some simple lemmas.

[color=red]Lemma 1 :[/color]
$z^{2}-z+1$ can not be a perfect square. (Unless z=1 of course but we can get rid of this trivial case giving the solution $(x,y,z)=(2,1,1)$)
This is trivial and there is no need for proof.

[color=red]Lemma 2 :[/color]
Any odd prime divisor of $3m^{2}+1$ is either equal to 3 or is of the form $6k+1$
[color=orange]proof of lemma 2:[/color]
let $p>2$ be a prime and $p|3m^{2}+1$ we have $(3m)^{2}\equiv-3(\mod p )$. If $\binom{a}{p}$ is the legendre symbol then we know $\binom{-3}{p}=1$ but from the quadratic reciprocity law we see that $\binom{p}{3}=1$ so $p=3$ or $p=6k+1$.

[color=red]Lemma 3 :[/color]
let $p$ be a prime and $z$ is such that $p|z^{2}-z+1$ then either $p=3$ or $p=6k+1$ where $k$ is a positive integer.

[color=orange]proof of lemma 3 :[/color]
$ord_{p}(z)|p-1$ but we can easily see that $z^{6}\equiv 1(\mod p)$
so if $p\not= 6k+1$ then $6\not| p-1$ and $ord_{p}(z)=1,2$ or $3$ but $z^{3}\equiv-1(\mod p)$ so $ord_{p}(z)\not= 1,3$
if it is 2 then $z^{2}\equiv 1(\mod p)$
$0\equiv z^{2}-z+1\equiv 2-z (\mod p)$ so $4\equiv z^{2}\equiv 1(\mod p)$
and $p=3$ so lemma 3 is proved.

[color=red]Lemma 4 :[/color]
$9\not| z^{2}-z+1$ for any natural z.
This also trivial, just check z=3k,3k+1 or 3k+2.

[color=red]Lemma 5 :[/color]
$x^{4}-y^{4}=2z^{2}$ and $x^{4}-y^{4}=3z^{2}$ don't have any solutions in $\mathbb{N}$

[color=orange]proof of lemma 5 :[/color]
I posted this [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=640574#640574]here[/url] but they have no solutions proposed till now.I am going to prove the first of them, the second is similar.
We may assume WLOG that $gcd(x,y,z)=1$ so
$(x^{2}-y^{2})(x^{2}+y^{2})=2z^{2}$ x,y are obviously odd so $4|x^{2}-y^{2}$ but $2\parallel x^{2}+y^{2}$ so
$x^{2}+y^{2}=2m^{2}\; , \; x^{2}-y^{2}=n^{2}$
from the last one we get that $x=a^{2}+b^{2},y=a^{2}-b^{2}$
and putting these in the first equation we get $a^{4}+b^{4}=m^{2}$ well known not to have solutions in $\mathbb{N}$

[color=red]Lemma 6 :[/color]
$x=y=1$ is the only solution in positive integers to the equation $3x^{2}+1=4y^{4}$

[color=orange]proof of lemma 6 :[/color]
$(2y^{2}-1)(2y^{2}+1)=3x^{2}$ as $3\not| 2y^{2}-1$ we must have
$2y^{2}-1=m^{2},\; 2y^{2}+1=3n^{2}$
or $2=3n^{2}-m^{2},\; 4y^{2}=3n^{2}+m^{2}$ from the second equation:
$(2y-m)(2y+m)=3n^{2}$ so we easily deduce that $m=\frac{r^{2}-3s^{2}}{2}$ or $\frac{3s^{2}-r^{2}}{2}$ where $rs=n$
changing the variables of $2=3n^{2}-m^{2}$ to $r,s$ we get :
$2=3r^{2}s^{2}-\left(\frac{r^{2}-3s^{2}}{2}\right)^{2}$
$9s^{4}-18r^{2}s^{2}+r^{4}+8=0$ but the discriminant is $\Delta =12^{2}\cdot 2(r^{4}-1)$ and it should be a perfect square. This is equivalent to saying $r^{4}-1=2q^{2}$ for some $q\in \mathbb{Z}$ but from lemma 5 we see that the only solution is $q=0,r=1$ so also $s=1$ and follows $m=n=1$ and finally $x=y=1$
___________________________________________________________

Now, let's turn back to our problem.
We have $xy^{2}=(z+1)(z^{2}-z+1)$. and we may additionally assume that y has only prime divisors of the form $6k+1$
it is obvious that $gcd(z+1,z^{2}-z+1)\in{1,3}$
but also $gcd(x,z^{2}-z+1)\in{1,3}$ because of the lemmas 3 and 4.
If $gcd(x,z^{2}-z+1)=1$ then $z^{2}-z+1$ would have to be a perfect square and it would contradict lemma 1 (Yes, here's where we use it! :) ).
Thus we have $gcd(x,z^{2}-z+1)=3$ and we build a system:
$y=uv$
$x=9x_{1}$
$z+1=3x_{1}u^{2}$
$z^{2}-z+1=3v^{2}$
from the discriminant of the last equation in z we find that $4v^{2}-1=3m^{2}$ for some $m\in \mathbb{Z}$. now it's clear that v has only factors of the form $6k+1$. So : $2v-1=p^{2},2v+1=3q^{2},m=pq$
$p^{2}-2pq-3q^{2}+4x_{1}u^{2}=0$ and the discriminant with respect to p is :
$\Delta_{p}=16(q^{2}-x_{1}u^{2})$ or $q^{2}=x_{1}u^{2}+a^{2}$
$p^{2}-q^{2}=\frac{4(v-1)}{3}\geq 0$ which implies $p\geq q$ so $p=q+2a$
From the equation $2=3q^{2}-p^{2}$ we get :
$q^{2}-2qa-2a^{2}-1=0$ and we get $q=a+\sqrt{3a^{2}+1}=a+b$
Next, $(a+b)^{2}=x_{1}u^{2}+a^{2}\; x_{1}u^{2}=b(2a+b)$ but $b$ doesn’t have factors of the form $6k-1$ because of lemma 2.
and $b$ is no divisible by 3 so we have wo cases:
1)case $b=2b_{1}^{2}$ so $4b_{1}^{4}=3a^{2}+1$ and $b_{1}=1=a,$ because of lemma 6
$b=2, \; x_{1}u^{2}=8, \; z=23$ as $z+1=3x_{1}u^{2}$
In this case we have he solutions $(2,78,23)(8,39,23)(18,26,23)(72,13,23)$
2)case $b=b_{1}^{2}$  and we get $b_{1}^{4}=3a^{2}+1$ so $a=0,\; b_{1}=b=1$because of lemma 5. and $z=2$ so we finally get the soluions $(1,3,2)(9,1,2)$  :wink:
$\blacksquare$

I like this proof a lot, and I think the ideas used are well known and natural for the solution!

[quote="Albanian Eagle"]
$ q^{2} \minus{} 2qa \minus{} 2a^{2} \minus{} 1 \equal{} 0$ and we get $ q \equal{} a \plus{} \sqrt {3a^{2} \plus{} 1} \equal{} a \plus{} b$
Next, $ (a \plus{} b)^{2} \equal{} x_{1}u^{2} \plus{} a^{2}\; x_{1}u^{2} \equal{} b(2a \plus{} b)$ but $ b$ doesn’t have factors of the form $ 6k \minus{} 1$ because of lemma 2.
[/quote]

I don't know for the whole solution, but here there is a problem since $ (a\plus{}b)^2\equal{}b(2a\plus{}b)$ doesn't hold.

there is a missing comma, it should read
$ (a\plus{}b)^2\equal{}x_1u^2\plus{}a^2 \rightarrow x_1u^2\equal{}b(2b\plus{}a)$

I still don't know why y has only prime divisors of the form $ 6k\plus{}1$ sorry for my stupid question  :blush:

[quote=Albanian Eagle] If $gcd(x,z^{2}-z+1)=1$ then $z^{2}-z+1$ would have to be a perfect square and it would contradict lemma 1 [/quote]
Sorry, but why is this true? Should it be $gcd(x,z^{2}-z+1)=1$, then $z^2-z+1 | y^2$ and $x | z+1$? But this doesn't imply $z^2-z+1$ is a perfect square and the proof falls apart? Sorry if I am just being dumb. Thanks!

Suppose $z^2-z+1$ is not a perfect square. Then there exists a prime $p\mid y$ such that $p$ divides $z^2-z+1$ an odd number of times. This means $p$ divides $z+1$ at least once, so $p\mid \gcd (z+1, z^2-z+1)\mid 3\implies p=3$. However, we already assumed $y$ only has prime factors of the form $6k+1$ contradiction.

Oh, whoops! Thank you so much for clearing this up for me.


function
algebra
domain
geometry
rectangle
parameterization
advanced fields

Help me to show, that general solution of the equation $ u_{xy}\minus{}\frac{n}{x\minus{}y}u_x\plus{} \frac{m}{x\minus{}y}u_y\equal{}0$, where $ n,m \in \mathbb{N}$ is the function of the following kind: $ \frac{\partial^{n\plus{}m\minus{}2}}{\partial x^{m\minus{}1} \partial y^{n\minus{}1}}[\frac{f(x)\plus{}g(y)}{x\minus{}y}]$, where $ f$ and $ g$ are arbitrary functions of the class $ C^2$.

You should be a bit careful about how you state this problem. First of all, the domain in which you solve the equation matters a lot. Secondly, the $ C^2$ requirement looks very strange: since you differentiate much more than 2 times in your formula, it is much more natural to talk either about $ C^\infty$ or about generalized functions. So, let's restate the problem as follows:

Show that the general $ C^\infty$ solution of that equation in a rectangle $ (a,b)\times(c,d)$ that doesn't intersect the diagonal $ \{x\equal{}y\}$ is given by that formula with $ f,g\in C^\infty$.

The key idea is that if $ U$ satisfies the equation with parameters $ m,n$, then $ U_x$ and $ U_y$ satisfy the equation with the parameters $ m\plus{}1,n$ and $ m,n\plus{}1$ respectively (just multiply the equation by $ x\minus{}y$ and differentiate with respect to $ x$ and $ y$). Conversely, if $ m>1$, you can integrate the function satisfying the equation with parameters $ m,n$ with respect to $ x$ and get the solution of the equation with the parameters $ m\minus{}1,n$ and similarly for $ n$ (here you have to be a bit careful about how you exactly choose your "antiderivative": not all of them will work).

This allows to reduce the question to the case $ m\equal{}n\equal{}1$ but then your equation is equivalent to the equation $ [(x\minus{}y)u]_{xy}\equal{}0$, whose solutions are well-known (just integrate with respect to $ x$ and then with respect to $ y$).


modular arithmetic
algebra proposed
algebra

Find the integers $ p,\ q\ (p>q>1)$ such that $ \frac{2p\minus{}1}{q},\ \frac{2q\minus{}1}{p}$ are integers.

[hide]Clearly $ p$ and $ q$ are odd.

$ 2p\equiv1\pmod{q}\implies p\equiv\frac {q \plus{} 1}2\pmod{q}$

Let $ p \equal{} kq \plus{} \frac {q \plus{} 1}2$

$ p|2q \minus{} 1\implies 2q \minus{} 1\ge p\implies 2q \minus{} 1\ge kq \plus{} \frac {q \plus{} 1}2\implies k < 2$

(i) $ k \equal{} 1$
$ p \equal{} q \plus{} \frac {q \plus{} 1}2 \equal{} \frac {3q \plus{} 1}2$

$ \frac {2q \minus{} 1}{p} \equal{} \frac {4q \minus{} 2}{3q \plus{} 1} \equal{} 1 \plus{} \frac {q \minus{} 3}{3q \plus{} 1}$. So $ q\equal{}3$ or $ q \minus{} 3\ge 3q \plus{} 1\implies 2q\le \minus{} 4$ a contradiction. Thus we get the solution $ (p,q)\equal{}(5,3)$.

(ii) $ k \equal{} 0$
$ p \equal{} \frac {q \plus{} 1}2$

$ p > q\implies \frac {q \plus{} 1}2 > q\implies q \plus{} 1 > 2q\implies q < 1$ a contradiction.

Hence, the only solution is $ (p,q)\equal{}(5,3)$.[/hide]

[quote="Johan Gunardi"]which part is wrong?[/quote]

$ (5,3)$ is a solution ...  :blush:

$ \frac {2p \minus{} 1}q \equal{} m > 0$ and $ \frac {2q \minus{} 1}p \equal{} n > 0$ $ \implies$ $ p \equal{} \frac {mq \plus{} 1}2$ and $ 2q \minus{} 1 \equal{} n\frac {mq \plus{} 1}2$ and so $ q \equal{} \frac {n \plus{} 2}{4 \minus{} mn}$

And so $ mn\in\{1,2,3\}$ and very few trials to find $ (p,q) \equal{} (5,3)$ as unique solution such that $ p > q > 1$

Sorry. I forgot that $ q\minus{}3$ can be zero. :blush:  I have edit my solution.

Now that's correct.

Here is my scheme:

$ 0<\frac {2q \minus{} 1}{p} < \frac {2p \minus{} 1}{p} < 2\Longrightarrow \frac {2q \minus{} 1}{p} \equal{} 1$

[quote="kunny"]Now that's correct.

Here is my scheme:

$ 0 < \frac {2q \minus{} 1}{p} < \frac {2p \minus{} 1}{p} < 2\Longrightarrow \frac {2q \minus{} 1}{p} \equal{} 1$[/quote]

Quick and nice ! :)


modular arithmetic
algorithm
number theory
Euclidean algorithm

I forgot how to use the Chinese Reaminder Theorem. I think it has something using the Euclidean Algorith.

$ S\equiv2\: (mod\: 15)$
$ S\equiv8\: (mod\: 28)$
$ S \equal{} 15x \plus{} 2 \equal{} 28y \plus{} 8$
$ 28 \equal{} 1\cdot15 \plus{} 13$
$ 15 \equal{} 1\cdot13 \plus{} 2$
$ 13 \equal{} 6\cdot2 \plus{} 1$
$ 2 \equal{} 2\cdot1 \plus{} 0$
This is where I get confused.

umm...2 mod 2 simplifes down to 0 mod 2.
therefore S simply = 2x, for some integer x.
Therefore,
2x = 2 mod 8.
The first time that is true is when x = 5. So u plug 5 back into S to get the smallest value for those two statements to be true is 10.

Indeed, $ 2x\equiv 2\pmod 8\implies x\equiv 1\pmod 4$, so your solutions are $ x\equal{}4n\plus{}1$ for any integer $ n$.
Plugging back into $ S$ we get $ S\equal{}2x\equal{}8n\plus{}2$.
This is verified quite easily:
$ 8n\plus{}2\equiv 0\pmod 2$.
$ 8n\plus{}2\equiv 2\pmod 8$.

Srry I messed up in posting it, I have edited it.

Same argument still applies.
$ S\equiv 2\pmod {15}\implies S\equal{}15n\plus{}2$
$ S\equiv 15n\plus{}2\equiv 8\implies 15n\equiv 6\pmod {28}$
Now we want the modular multiplicative inverse of $ 15\bmod {28}$, which can be found with the Euclidean algorithm or by simple inspection. We find that $ 15^2\equiv 1\pmod{28}$.
Therefore, $ n\equiv 90\equiv 6\pmod {28}\implies n\equal{}28m\plus{}6$.
Substituting back, $ S\equal{}15(28m\plus{}6)\plus{}2$.


[quote="ashrafmod"]The sequence 2, 3, 5, 6, 7, 10, 11, ... consists of all positive integers that are not a square or a cube. Find the 500th term.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76478


function
algorithm
limit
number theory
relatively prime
algebra unsolved
algebra

I dont know where to post this , but it seems to be a the most suitable place :) .
Prove that there dont exist a injective ,surjective function from
1.$\mathbb{R} \to \mathbb{Q}$
2.$\mathbb{R} \to \mathbb{N}$

and prove that there exist a injective surjective function from
$\mathbb{Q} \to \mathbb{N}$
[hide]I just find a algoritm for this but never find a strict function.(I dont mean a strict function(like increasing or decreasing:lol:) but just that a direct relation) .[/hide]

anybody knows that Do we have a injective surjective function from
$\mathbb{R} \to \mathbb{C}$.

PS.I have read somewhere in the past that we have different kind of infinities and
later is was showned that the infinity in the $\mathbb{N}$ is the smallest infinity :D (seems exteremly odd).does any body know which one is the biggest ?
[hide]yes i can but not in the math but in the world , the infinity of the universe is the biggest , Or i just think[/hide]

Let's show first that there exists a bijective (both injective and surjective) function
$f: \mathbb{Q} \to \mathbb{N}$
Let $r \in \mathbb{Q}$ then $\exists  (p,q) \in \mathbb{Z} \times \mathbb{N}^{*}$ such that $r=\frac{p}{q}$
This function $g_{1}: \mathbb{Q} \to \mathbb{Z} \times \mathbb{N}^{*}, \frac{p}{q} \mapsto (p,q)$ is a natural bijection.

Considering functions $h_{1}: \mathbb{N} \to \mathbb{N}^{*}, n \mapsto n+1$ and $l_{1}: \mathbb{N} \to \mathbb{Z}, n \mapsto \left\{ \begin{array}{cc}{11}
\frac{n}{2} & \textrm{if n is even}\
- \frac{n+1}{2} & \textrm{if n is odd} \end{array} \right.$
you can easily show that they are bijective.
Thus $g_{2}: \mathbb{Z} \times \mathbb{N}^{*} \to \mathbb{N}^{2}, (p,q) \mapsto (l^{-1}_{1}(p),h_{1}(q))$
is bijective.

Finally, function $g_{3}: \mathbb{N}^2 \to \mathbb{N}, (a,b) \mapsto 2^{a}(2b+1)$ is also bijective and then by composing $g_{3} \circ g_{2} \circ g_{1}: \mathbb{Q} \to \mathbb{N}$ we have what we want :)

Now, we only have to show that there's no bijection between $\mathbb{R}$ and $\mathbb{Q}$ (the other result yields from what we have just shown)....but it is a relatively long proof but it involves one main idea: if A is an non-empty set then there is no surjective function from A to P(A) where P(A) is the set of all the subsets in A (this is not easy to prove...especially when A is infinite).

As for a bijective function from $\mathbb{R} \to \mathbb{C}$....I'm sure it exists but I don't have an example...I'll try to find an easy one.

Sincerely,
                            AlJaber.

Nice attemp for making the function from $\mathbb{Q} \to \math{N}$(you dont use one function but combine of 3 that is intersting )but
i think $g_1$ is not a function cause for each $r \in \mathbb{Q}$
we have infinity of $(p,q)$ such that :
$r=\frac{p}{q}$ like $(p,q),(kp,kq),...,$ but if we add that
$p,q$ be ralativly prime to each other then the $g_1$ is a function.
i really hope that i am wrong cause i liked the solution :)

Indeed...I was too quick with the solution...p and q must be relatively prime :oops: but $g_{1}$ must be modified (because as it is, it will only be injective)...however, the solution will look more ugly :(
Let $B= \{ (p,q) \in \mathbb{Z}^{*} \times \mathbb{N}^{*}, p \wedge q=1 \} \cup \{ (0,1) \}$
Then $g_{1}: \mathbb{Q} \to B, \frac{p}{q} \mapsto (p,q)$ is bijective.
We see that B is an infinite subset of $ H= \mathbb{Z} \times \mathbb{N}^{*}$
we showed (thanks to function $h=g_{3} \circ g_{2}$) that there is a bijection from H to $\mathbb{N}$
Now, we use a useful lemma: Let A be a set such that there is a bijection from A to $\mathbb{N}$, then there is a bijection from $\mathbb{N}$ to any infinite subset of A. Let's prove this in this particular case (the same idea applies to the general proof):
Let $t_{2}:B \to h(B),x \mapsto h(x)$ (this function is bijective by construction) and $t_{1}:h(B) \to \mathbb{N},
min(h(B)) \mapsto 0,min(h(B) \backslash \{t^{-1}(0)\} \mapsto 1, \cdots$ and so on...
Then, finally, by composing $t_{1} \circ t_{2} \circ g_{1}$ we (hopefully!) have what we want :D
Al Jaber.

[quote="Al Jaber"]it is a relatively long proof but it involves one main idea: if A is an non-empty set then there is no surjective function from A to P(A) where P(A) is the set of all the subsets in A (this is not easy to prove...especially when A is infinite).[/quote]
Actually, it is not that difficult: suppose $f:A\to P(A)$ is a bijection. Denote, for each $x\in A$, $f(x)=A_x$. Define
\[
B=\{x\in A| x\notin A_x\}.
\]
Since $B\in P(A)$ and $f$ is surjective, there exists $y\in A$ such that $B=A_y$. But then we see that both alternatives $y\in A_y$ and $y\notin A_y$ lead to a contradiction.

I know this proof is very short yet personally I think it's very tricky since the idea of considering the set B doesn't come very naturally :roll:
Anyway, here is a proof to show there's no bijection from $\mathbb{R}$ to $\mathbb{N}$ based on a homework assignment our teacher gave us this year:
Let $u \in \{0,1 \}^{\mathbb{N}}$ then we see that sequence $v_{n}= \sum_{k=0}^{n} u_{k}3^{-k}$ converges.
Thus, we can define function $f: \{0,1\}^{\mathbb{N}} \to \mathbb{R}, u \mapsto \lim_{n \to \infty}v_{n}$
Let's show that $f$ is an injective function:
Let $(u,t) \in \{0,1 \}^{\mathbb{N}} \times \{0,1 \}^{\mathbb{N}}$ and $u \neq t$
Let $w_{n}=\sum_{k=0}^{n}t_{k}3^{-k}$ and $p=min \{k \in \mathbb{N},u_{k} \neq t_{k} \}$
We can also assume that $u_{p}=1$ then we can see that $\forall n \geqslant p, w_{n}<3^{-p} \leqslant v_{n}$
This shows that $f$ is an injection from $\{0,1 \}^{\mathbb{N}}$ to $\mathbb{R}$ then there exists a surjective function $s_{1}$ from $\mathbb{R}$ to $\{0,1 \}^{\mathbb{N}}$
However, $s_{2}:P(\mathbb{N}) \to \{0,1 \}^{\mathbb{N}}, A \mapsto \varphi_{A}$
Where $\varphi_{A}$ is the function that is 1 if a number is in subset A, 0 if not. This is a well-known bijection.
Therefore, function $s=s_{1} \circ s_{2}^{-1}$ is a surjection from $\mathbb{R}$ to $P(\mathbb{N})$
Finally, this prevents the existence of any surjective function $j$ from $\mathbb{N}$ to $\mathbb{R}$ because composing $s \circ j$ would create a surjective function from $\mathbb{N}$ to $P(\mathbb{N})$ which is contradictory.
AlJaber.

Conected to this, here is an interresting problem:

  Show that any function $f:\mathbb{Q}\to\mathbb{Q}$ is the sum of two injective functions. ;)

Yes, yet the classical proof is Cantor's "diagonal" proof.
BTW, a nice approach to construct a bijection from $\mathbb{N}$ to $\mathbb{Q}$ is the following:
consider the sequence $r_1=1$, $r_{2n}=1+r_n$ and $r_{2n+1}=\frac{1}{r_{2n}}$, for all $n\ge 1$. One can prove that for each $r\in \mathbb{Q},r>0$ there exists a unique positive integer $n$ such that $r_n=r$. This extends immediately to a bijection from $\mathbb{N}$ to $\mathbb{Q}$.


I was wondering if there could be a feature on the report that said $ \frac{x \text{ points}}{y \text{ points possible}}$ or something like that.

Isn't there an infinate number of points possible :huh:

$ y$ points as in possible points out of the questions :wink:


The arrangement below, composed of 32 hexagons, was mounted with rods, all with length equal to the side of the hexagon. How many rods, at least, are required to assemble the arrangement ?

[img]http://img28.imageshack.us/img28/1082/obm.jpg[/img]

A) 113
B) 123
C) 122
D) 132
E) 152

Anyone knows how to do ? Can you explain me step by step ?  :blush:
Thanks.

[geogebra]fcbba89fb8a6f19d87b25404fd0eadaf88d82ec3[/geogebra]
$ 5\plus{}11\times\frac{33}{3}\minus{}3\equal{}123$
Thus $ \boxed{B}$

Hum ... nice method. Very good. Thanks ;)


Το Σάββατο έχουμε το Θαλή. Θα είμαι στο εξεταστικό κέντρο της Χαλκίδας, μια και με έχουν ορίσει πρόεδρο της Μομαρχιακής επιτροπής τέλεσης του διαγωνισμού για την Εύβοια.Θέλω να πιστεύω ότι τα θέματα θα είναι όπως πρέπει και ο διαγωνισμός θα διεξαχθεί καλά , μια και έχουν γίνει φιλόδοξες ενέργειες από την ΕΜΕ. Έχει συσταθεί εκ νέου η επιτροπή διαγωνισμών  που απαρτίζεται από καλούς και άξιους συναδέλφους .Νομίζω ότι θα θελήσουν να δείξουν κάτι καλά οργανωμένο και σωστό , διότι και το θέλουν αλλά και το μπορούν.

 Λοιπόν, μόλις τελειώσει ο διαγωνισμός,  θα βάλουμε τα θέματα στη σελίδα για να τα δουν και όσοι βρίσκονται μακρυά από τα εξεταστικά κέντρα ή τη χώρα μας, είτε για δουλειές είτε για σπουδές. Στο χώρο αυτό μπορούμε να πούμε τις εντυπώσεις , τις λύσεις και άλλα σχόλια .Εύχομαι στην επιτροπή καλές εμπνεύσεις και  σε όλους που θα συμμετάσχουν  :

                   [color=red][b]     ΚΑΛΗ ΕΠΙΤΥΧΙΑ !!![/b][/color]

[b]
[color=red]ΚΑΛΗ ΕΠΙΤΥΧΙΑ κι από εμένα σε όλους!! Ας έχουμε μία καλή ΟΛΥΜΠΙΑΚΗ ΧΡΟΝΙΑ![/color][/b]

Αλέξανδρος Συγκελάκης

ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΚΑΙ ΑΠΟ ΜΕΝΑ

Αντε και καλα μυαλα!!!!!!

[quote="Nick Rapanos"]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΟΥΣ! (Αυτή η ευχή είναι κάπως contradiction από μόνη της, αλλά τι να κάνουμε....)[/quote]

Καλυτερα Καλή Τύχη σε όλους  :wink:

Θα δώσω μια κάπως διαφορετική ευχή στους διαγωνιζόμενους . Χαρείτε το διαγωνισμό και γράψτε αυτό που πραγματικά αξίζετε  :)

εγω βασικα λεω μακαρι ολοι να πιασουν την [color=red]μεγιστη δυνατη τους αποδοση[/color] :)
cu tomorrow :wink:

kurios pros ton moderator alla kai genika...silouan kai kurie stergiou (kai genika opoiodipote allo melos tou forum) to pithanotero einai i eme na argisei na valei ta themata pano, opote anevaste ta osoi ta gnorizete aurio to proi...na ta doume ki emeis pou eimaste stin ksenitia.

kati mou leei oti auth th xronia tha einai kai polu prosegmena ta themata kai polu oraia!

symfono me tin euxh tou silouanou!

Καλη επιτυχια σε ολους τους διαγωνιζομενους

Ειμαι σε νετ-καφε και περιμενω να ανεβουνε τα θεματα :P

Σας στέλνω τα θέματα του διαγωνισμού στο επισυναπτόμενο pdf.

Αλέξανδρος

[quote="NickNafplio"]Απλα ελεος, εχει προβλημα το Acrobat Reader στο net που ειμαι και δεν ανοιγει, ισος να εχουν καποια παλια εκδοση...

Υπαρχει αλλος τροπος να δω τα θεματα??? (Γιατι με εχει φαει η περιεργεια :P)[/quote]

Θα τα ανεβάσω σε λίγο σαν .jpeg. (Ελπίζω να επιτρέπεται)

Μία πολύ εύκολη και εντελώς σχολική λύση για τα θέματα 4 της Β' και Γ' Λυκείου (που για κάποιο λόγο μου έπιασαν το μάτι).

Βάλτε καρτεσιανές συντεταγμένες: $ A(0,0), B(2,0), M(x,y)$.

Τότε, στη διατύπωση της Β' Λυκείου, το $ \Gamma$ θα είναι στο $ ( \minus{} y,x)$ και το $ \Delta$ στο $ (2 \plus{} y, 2 \minus{} x)$, οπότε το μέσο τους θα έχει συντ/νες $ (1,1)$.

Το θέμα της Γ' βγαίνει με ακριβώς την ίδια συλλογιστική και θα έχει συντ/νες $ (1,2)$.

Cheerio,

Durandal 1707

Edit 2: Yikes, ένα typo στη θέση του $ B$.

[quote="Durandal"]Μία πολύ εύκολη και εντελώς σχολική λύση για το 4ο θέμα Β' Λυκείου (που για κάποιο λόγο μου έπιασε το μάτι).

[hide]Βάλτε καρτεσιανές συντεταγμένες: $ A(0,0), B(0,2), M(x,y)$. Τότε το $ \Gamma$ θα είναι στο $ ( \minus{} y,x)$ και το $ \Delta$ στο $ (2 \plus{} y, 2 \minus{} x)$, οπότε το μέσο τους θα έχει συντ/νες $ (1,1)$.[/hide]

Cheerio,

Durandal 1707[/quote]

Έτσι την έλυσα και εγώ. Μόνο που πήρα Β = (1,0) και μου βγήκανε κάτι $ \sqrt {2}/2$.

Edit: Αναφερόμουνα στην άσκηση της Γ' Λυκείου. Τώρα πρόσεξα ότι ήτανε λίγο διαφορετικές.

[quote](kai to allo to kitrino apo tis ekdoseis springer einai arketa kalo alla den boro na thimitho to syggrafea auth ti stigmi)
[/quote]
Christopher G. Small
"Functional equations and how to solve them" :wink:

Του Venkatachala με τις εύκολες ασκήσεις... :)

[quote="filosofimenos"]Καταρχας καλη επιτυχια σε ολους τους διαγωνιζομενους. Εγω εδινα β' λυκειου... Με την παραγοντοποιηση εχω διαζυγιο και συνεπως δεν ελυσα το 3το.
[/quote]

Και εγώ που από βιασύνη δεν κάθισα να δω παραγοντοποίηση το έβγαλα απλώς με πράξεις (αντικαθιστώντας στην 1η εξίσωση έναν από τους τρεις από τη 2η εξίσωση) Βέβαι η διακρίνουσα έβγαινε λίγο [hide="τρομακτική"]3.272.481-3.272.400 περίπου[/hide]  :P

Τελικά τα έλυσα όλα :)  Το 4ο όπως είπαν και άλλοι τσιμπούσε λιγάκι..

"KMN" ( :wink: ) με αντιστροφές την έκανες ή με πιο elementary λύση  τη γεωμετρία στο τέταρτο?

kserete pote the anakoinothoun te apotelesmata??

Όχι η λύση μου ήταν απλούστατη... Αν και ήταν λίγο κουφό γιατί θα έπρεπε να βρίσκεται το Κ σε κύκλο με διάμετρο την ΑΒ..... Για την μεσοκάθετο το ψιλο-μπάνιζες-υπέθετες, όπως θέλετε..

Πιστευω papsr οτι αν ισχυσουν τα δεδομενα προγουμενων ετων τα αποτελεσματα θα βγουν μετα τις γιορτες.Τα γραπτα στη φαση του Θαλη ειναι παρα πολλα.Καλως ηρθες στο Forum και καλη συνεχεια.

Καταρχας καλα αποτελεσματα σε ολους..
Μηπως γνωριζει καποιος που μπορουμε να βρουμε το μπλε βιβλιο?
Εψαξα για ebook αλλα δεν βρηκα τιποτα

STIN ISTOSELIDA TIS MATHIMATIKIS ETAIREIAS EXOUN DIMIOURGISEI FAKELO GIA TIN KATAXORISI TON ONOMATON TON EPITYXONTON KAI LEEI POS SUNTOMA THA MPOUN K TA ONOMATA!!!!!AUTO SIMAINEI MIPOS OTI TA APOTELESMATA THA VGOUN PRIN TA XRISTOUGENNA??

Αν διατηρησουν την παραδοση ψιλο χλωμο... :(

Αρχές Ιανουαρίου είπαν.

Mιας και εδω ειχε αρχισει η συζητηση για τους διαγωνισμους ας βαλω το λινκ με τους επιτυχοντες http://www.hms.gr/ (hard to guess :P ).

Αντε και καλη επιτυχια να 'χουμε στον "Ευκλειδη". :first:

Καλησπερίζω! Και εγώ μαθητής της Β' Λυκείου είμαι, Θα ήθελα να μοιραστώ την λύση που έδωσα εγώ στο 2ο θέμα. (δεν ξέρω αν είναι καλή λύση, αν έκανα κάτι πρωτότυπο η συνηθισμένο αλλά απλός θα πω ότι σκέφτηκα εκείνη την ώρα)

ΑΒΓΕ: τετράπλευρο εγγράψιμο σε κύκλο (εφόσον είναι ορθογώνιο παρ/μο)
ΖΒΓΕ: τετράπλευρο εγγράψιμο σε κύκλο (εφόσον απέναντι πλευρές κάθετες)

3 σημεία χαρακτηρίζουν έναν κύκλο άρα οι δύο κύκλοι ταυτίζονται (ΒΓΕ, ΒΓΕ)

άρα το σημέιο Ζ βρίσκεται πάνω στον κύκλο που περνάει από τα ΑΒΓΕ

ΑΓ είναι διάμετρος εφόσον Β εγγεγραμμένη και είναι 90 μοίρες
άρα εφόσον η γωνία ΑΖΓ είναι και αυτή εγγεγραμμένη στην διάμετρο θα είναι και αυτή 90 μοίρες.

Ευχαριστώ!!!

Σημερα ειχαμε τον ευκλειδη βαζω τη γεωμετρια που επεσε. Τα αλλα 3 εβγαιναν αν και ηθελαν προσοχη. Χαζοιδως δεν ολοκληρωσα το 4ο(η τελικη σκεψη μου ηρθε οταν εφυγα) αν και το εχω προχωρησει αρκετα. Οριστε η γεωμετρια.

Δινεται ορθογωνιο τριγωνο ΑΒΓ με Α=90. Αν ΑΔ το υψος και Κ1, Κ2, Κ3 τα κεντρα των εγεγραμμενων κυκλων των τριγωνων ΑΒΔ, ΑΓΔ, ΑΒΓ, αντιστοιχα Ν.Δ.Ο. ΑΚ3=Κ1Κ2.

[u][b]Edit:[/b][/u] Ανεβασαν τα θεματα http://www.hms.gr/images/stories/themataeukleidi%202008-09final.pdf

[b][size=150][color=red]ΑΣΚΗΣΗ[/color][/size][/b]
Δίνεται ορθογώνιο τρίγωνο $ ABC$ με $ \angle{A} \equal{} 90$. Aν $ AD$ το ύψος του τριγώνου και $ K_1,K_2,K_3$ τα έγκεντρα των τριγώνων $ ABD, ACD,ABC$ αντίστοιχα, ν.δ.ο. $ AK_3 \equal{} K_1K_2$

[b][size=150][color=red]ΛΥΣΗ[/color][/size][/b]
Η ονοματολογία είναι σύμφωνα με το σχήμα μου. Τα σημεία $ L,M,E,H,T$ όπως φαίνεται και από το σχήμα είναι σημεία επαφής των κύκλων με τις πλευρές του τριγώνου.

Θα δείξουμε ότι τα τρίγωνα $ AK_3H$ και $ K_1MK_2$ είναι ίσα. Παρατηρούμε ότι $ K_3H \equal{} AH \equal{} \rho$ όπου $ \rho$ η ακτίνα του εγγεγραμμένου κύκλου στο τρίγωνο $ ABC$. Συνεπώς το τρίγωνο $ AK_3H$ είναι ορθογώνιο και ισοσκελές. Επομένως αρκεί ν.δ.ο. $ K_1M\equal{}MK_2\equal{}\rho$ και $ \angle{K_1MK_2}\equal{}90$. Επίσης γνωρίζουμε ότι $ K_3M\equal{}K_3T\equal{}K_3H\equal{}\rho$ και ως ακτίνες του εγγεγραμμένου κύκλου του $ ABC$ είναι αντιστοίχως κάθετες στις πλευρές $ AB,BC,CA$.

[i]Παρατηρήστε τώρα ότι αν αποδείξουμε πως τα τετράπλευρα $ K_1MK_3T$ και $ K_3MK_2H$ είναι παραλληλόγραμμα τότε προφανώς θα είναι και ρόμβοι και  θα έχουμε $ K_1M\equal{}MK_2\equal{}\rho$ και $ \angle{K_1MK_3}\equal{}\angle{K_1TK_3}\equal{}90\minus{}\angle{BTK_1}$ και επειδή $ TK_1//K_3M\Longrightarrow TK_1\perp BC$ δηλαδή $ T,K_1,L$ συνευθειακά και $ TK_1//AD$ άρα $ \angle{BTK_1}\equal{}\angle{BAD}\equal{}\angle{C}\Longrightarrow \angle{K_1MK_3}\equal{}\angle{B}$. Όμοια βρίσκουμε ότι $ \angle{K_2MK_3}\equal{}\angle{C}$ επομένως $ \angle{K_1MK_2}\equal{}90$.
[/i]

Συνεπώς αρκεί να δείξουμε ότι $ TK_1//K_3M$ και $ HK_2//K_3M$. Είναι όμως προφανές ότι αυτό είναι ισοδύναμο με το να δείξει κανείς ότι $ TL//AD$ και$ HE//AD$. Θ.δ.ο. $ HE//AD$ και με όμοιο τρόπο μπορεί κανείς να δείξει ότι και $ TL//AD$. Εύκολα υπολογίζει κανείς τα εξής:

1.$ CH \equal{} \frac {a \plus{} b}{2}$

2.$ CE \equal{} \frac {b(a \plus{} b)}{2a}$

3.$ CD \equal{} \frac {b^2}{a}$

Επομένως $ \frac{CH}{CA}\equal{}\frac{a\plus{}b}{2b}$ και $ \frac{CE}{CD}\equal{}\frac{\frac{b(a\plus{}b)}{2a}}{\frac{b^2}{a}}\equal{}\frac{a\plus{}b}{2b}$, άρα $ \frac{CH}{CA}\equal{}\frac{CE}{CD}$ και η απόδειξη ολοκληρώθηκε $ q.e.d.$  $ \blacksquare$


Side $ AC$ of right triangle $ ABC$ is divide into $ 8$ equal parts. Seven line segments parallel to $ BC$ are drawn to $ AB$ from the points of division. If $ BC \equal{} 10$, then the sum of the lengths of the seven line segments:

$ \textbf{(A)}\ \text{cannot be found from the given information} \qquad
\textbf{(B)}\ \text{is }{33}\qquad
\textbf{(C)}\ \text{is }{34}\qquad
\textbf{(D)}\ \text{is }{35}\qquad
\textbf{(E)}\ \text{is }{45}$

The answer is (A).

There is not enough information because the sides AC and AB could change while the triangle still fits the parameter. In other words, the triangle, with BC at the bottom, could be REALLY pointy or REALLY flat, and this would change the lines drawn.

Note: It is assumed in the problem that $ \angle C \equal{} 90^\circ$.

@PowerOfPi: Nope; you can use similar triangles to find it.

This is probably the best geo problem on the AHSME 1957. (But maybe not since I haven't looked at the last 15 problems.)

[hide="My Solution"]
There are so many similar triangles to use to find the lengths. First, we attempt to find the length of the shortest line segment. The ratio is as follows:

$ \frac{x_1}{10} \equal{} \frac{1}{8}\implies x_1  \equal{} \frac{10}{8}.$

In general, we have $ \frac{x_n}{10} \equal{} \frac{n}{8} \implies x_n \equal{} \frac{10n}{8}$, where $ x_n$ denotes the $ n$th line from $ A$. So we need to find

$ \sum_{k \equal{} 1}^{7}{x_n} \equal{} \frac{1}{8}\sum_{k \equal{} 1}^{7}{10k} \equal{} \frac{10}{8}\sum_{k \equal{} 1}^{7}{k} \equal{} \frac{10\cdot 7\cdot 8}{2\cdot 8} \equal{} \boxed{35 (D)}.$
[/hide]

whoa this is a really old thread why are you bumping it

(but why does C have to be right?)

the problem says it is but it doesnt have to be so the problem is giving unnecessary information


Here is a good problem.

Show that a unit square cannot be covered by five squares with side less than 1/2.

Sathej

[quote="cm"]consider 9 points (0,0) (1/2, 0) (1,0) (0,1/2) (1/2,1/2) (1,1/2) (0,1) (1/2,1) (1,1), any square with side length smaller than 1/2 can only cover one of these 9 points, so actually it's true for 8 such squares. hope it's right.[/quote]
It's easy to cover two of that points with a square... But at least it works for four squares since it's not possible to cover three points at once.


function
algebra solved
algebra
Function equations
Fixed point

1)  $x,y$ are positive integers.
     find all $f$ functions such that
     $f(x+f(y)) = f(x)+y$ and     $f: \ Z^{+}\longrightarrow Z^{+}$

2)  $x,y$ are real numbers.
     find all $f: \R \longrightarrow R$ functions such that
     $x?+y?+2f(x.y) = f(x+y).(f(x)+f(y))$

The first one is easy. Take y=1 and find that $ f(x+f(1))=1+f(x) $. Next, take $ x=f(1) $ and use the first fact. We find $ 1+f(f(x))=x+f(f(1)) $. If we take $x=y$, we see that f must jave a fixed point. Thus, $ f(f)$ has a fixed point and from the above we find that f is in fact involutive. Thus, if we put $ f(y$ instead of y, we find that f is also aditive. From now on, we are on a known field.

i didnt think a lot but i have an idea

if we put x to y and y to x we get f(y+f(x))=f(y)+x
let x+f(y)=a  and y+f(x)=b   then f(a)=b and f(b)=a  so f(f(a))=a
after that we put f(y) to y in main equation we get f(x+y)=f(x)+f(y) then f(x)=nx

is this idea true?

As for the second one, after $x=y=0$ we get $f(0)\in\{0,1\}$.

If $f(0)=0$, then for all $x$ we either have $f(x)=x$ or $f(x)=-x$. This means that $x^2+y^2+2f(xy)\in\{(x-y)^2,(x+y)^2\}$, and from here it follows easily that we either have $f(x)=-x,\ \forall x$ or $f(x)=x,\ \forall x$. Since the former doesn't work, it must be the latter: $f(x)=x,\ \forall x$.

On the other hand, if $f(0)=1$, we get $\forall x,\ f(x)=\frac{-1\pm\sqrt{9+4x^2}}2\ (*)$. This means that $f(2)\in\{-3,2\}$. From $x^2+f(x^2)=f(2x)f(x),\ x=2$ we get $f(4)\in\mathbb Q$, but that's not true if we look at $(*)$. This means that it's impossible to have $f(0)=1$, so $f=1_{\mathbb R}$ is the only solution.


Support
geometry
3D geometry
ratio
trigonometry

Write the name of a bizarre thing you saw today!  :D  :D

A topic that belongs in G&FF in RT. [color=green][i]Since moved by moderator.[/i][/color]  No, actually, while strange, it's not unusual enough to be bizarre.

I saw many snowflakes today!

I saw a drop of water suspended in motion today! :D

I saw snow today, which made me see that meteorologists can actually be right!!! :o

i saw a human today!
they look very queer..
jkjk..

Today I saw the temperature drop below 20 degrees.

Today I saw the Jahn-Teller Effect explained. That was slightly confusing.

Today I saw a rocket launcher... on the TV, in Xbox Halo.

i saw my own breath today.

in t3h car... :(

I saw the letters r, o, s, e, t, a, and n today.

I like Knex. Everybody tells me I pronounce it wrong.

I saw an error in my reasoning for 2003 ACM Problem F. *sigh*

i saw a lot of bad grammar

[hide="click to find out where, although its pretty darn obvious"]een thees thred[/hide]

I saw an English Language Olympiad today which was pretty weird :wink:

I SAW *GASP* AN EMO PENGUIN!!!!!!!noooooo!!!!!!! :O

I saw a fragment of Lego bend itself and make loud whirring noises.

I saw the creation of my AoPS account (hurray!) today. No one else witnessed it.

I saw a trash truck take out a telephone pole by my house today. XP Stupid driver!

I saw a strange expression. I simplified it and got 916.

I saw a button above me that did not work.

[quote="Rafaelloaa"]I saw a button below me that said "Click here for free popcorn!!" I clicked it and so should you.[/quote]

Nope. Sorry. My brain has been eaten too much already and I'm not going to let it be eaten again.

I saw a vast proof that $ \sum_{cyc}\cos{A}{2}\equal{}\frac{3}{2}$

Where A is an angle in a triangle, and so is B and C.


Dear Mr. Crawford,

I have a short message to ask you.  How far are you going to explain on sequences, series, and sigma math jam? To geometric sequences? or just to arithmetic sequences?

Dear Mr. Silverfalcon,

There are many other types of series and sequences other than arithmetic and geometric.

I personally think he intends to go further than the two you mentioned.

I am currently writing some scripts for my algebra class.  After I see how long it takes to explain certain ideas I will decide how far I can go in the Math Jam.  Currently my thoughts are that I will be examining the way that differences can be used in summations and build toward the use of difference equations to help derive and evaluate polynomials.

Arithmetic and geometric series will be briefly touched on and we will tackle a couple of warm up problems that related to such series and that give a natural direction toward developing the ideas on difference equations.

A more complete version of the lesson will appear in the AMC/AIME Algebra course.


geometry
graph theory
combinatorics proposed
combinatorics

H is called an induced subgraph of G if H contains all edges xy of G such that x and y are vertices of H.

Let P(G) be the number of vertex disjoint paths of 3 vertices that are induced subgraphs of G.

Problem: Let G be cubic (all vertices are of degree 3), then P(G) >= 2/11n, where n is the order or the number of vertices of G.

I solved it, but i would like to see how some you would approach it.
Thanks

sorry, the problem is to prove the claim that I stated above...

If G is cubic then P(G) >= 2/11n

Disjoint in terms of edges I presume, am I right?

In that case, I've found a trivial lower bound of $\frac n 2$...

First, since each vertex has degree $3$, it follows that the total number of edges is $\frac {3n} 2$.
Now, let $p$ be the maximal number of pairwise disjoint paths of length $2$ (so that each of them involves $3$ vertices).

Clearly, for each vertex $A$, at most one of the edges with endpoint $A$ is not used by the $p$ paths above, otherwise we could add such a path, contradicting the maximality of $p$. Thus, at most $\frac n 2$ edges (each edge is counted twice in the above reasoning) do not belong to the $p$ paths. it follows that the $p$ paths use at least $n$ edges. Since each path uses exactly two edges, we deduce that there are at least $\frac n 2$ paths, as claimed.

What did I miss?

Pierre.

Yes edge disjoint too...

I am going to post my proof in some minutes.

You should try considering to subgraphs A(P) and B(P). Let P be the 2-path packing in G. Let A(P) be the graph induced by P and let B(P)= G-A(P).
You should try to prove that |A(P), B(P)|.=  :varepsilon: |P|.

I am going to post it the complete proof.

By the way, do you like graph theory Pierre?

I do not understand.
Is my proof wrong?

Pierre.

Yes it is wrong.

I can give you a trivial counterexample. K_4 is cubic an only has 1 disjoint 2-path.

This is what is wrong.
"Clearly, for each vertex A, at most one of the edges with endpoint  A is not used by the  paths above"

that is not true for every vertex of G.

Do you mean that the paths must be disjoint in terms of vertices too?

Pierre.

Yes I said it in the first message.

When you say disjoint it is the same as vertex disjoint. What I missed to say was that the 2-paths have to be edge dijsoint.

Ok, it was unclear for me. I will try to think about it.

Pierre.

This one is not that hard.
Furthermore, the bound can be improved to n/4. I am thinking on this better bound, but i do not know how to prove it. After you think on this one, you can maybe help me..

Hi everyone!

I have proved estimate n/5. Though I can't say that my proof is short and easy.

First step.
Consider bipartite graph G', which can be obtained by splitting each node of G into two nodes (blue and red) in such way that if A is connected to B in graph G then A(blue) is connected to B(red) and A(red) is connected to B(blue) in graph G'. It is clear G' is cubic bipartite graph. It is easy to show that G' satisfies Holl's theorem. Hence we can find complete matching. Now if we return back to G then complete matching for G' means that we found pairwise disjoint cycles in G and each node of G lies in one of these cycles.

Second step.
It is possible that some of founded cycles have length 2, i.e. it is two connected nodes. For each such "cycle" we remove connecting edge. All remaining cycles are "true" cycles.
So what do we have? Our graph G consist of pairwise disjoint cycles and some set of nodes with degree 2 (!). It follows that all nodes of degree 2 can be organized to cycles or to paths, which connect cycles. (See Figure 1: dashed lineas are paths and solid lines are cycles).

Third step.
It is clear that each node in cycles is connected to at most 1 path (since degree of nodes is 3).
Let us to orient each cycles in clockwise order. Nodes in which paths connect to cycles break cycle into "segments". We add to each path two segments which precede to break-points. In this way we obtain "extended" paths (see Figure 2: path AB is extended).
Finally, whole graph is organized into extended paths and cycles. And it is easy to see that extended path or cycle with m nodes gives us at least m/5 disjoint paths of length 2.

P.S. if there are no paths or cycles of length 5 then we can find n/4 desired paths. Maybe it is way to improve my solution.

Myth, I have not read the proof I am going to read it know carefully,
but by the way, who are you? where do you study?

[quote="Myth"] Now if we return back to G then complete matching for G' means that we found pairwise disjoint cycles in G and each node of G lies in one of these cycles.
[/quote]

I don't see clearly this part. Assuming this is true the proof makes sense, but can you help me se this part.

thanks

Hi!
Consider complete matching in G'.
Suppose A(red) is connected to B(blue) and B(red) is connected to A(blue). In this case we obtain degenerate cycles AB in G.
Now suppose A1(red)->A2(blue); A2(red)->A3(blue); ...; Ak(red)->A1(blue). In this case we have "true" cycle in G: A1A2...Ak.

I am post-graduate.

Oh i see..
thanks,

Also I proved that the number of disjoint 3-vertex paths in a 2-connected and claw free graph is n/3.
It has a nice inductive proof.
and i've heard you can relate this number with the number of end blocks in a graph.

BTW, oin what area of mathematics are you going to work in, Myth?


Let $ a,b,c,d,e>0$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2 \geq 1$. Find minimum of :
$ A\equal{}\frac{a^2}{b\plus{}c\plus{}d}\plus{}\frac{b^2}{c\plus{}d\plus{}e}\plus{}\frac{c^2}{d\plus{}e\plus{}a}\plus{}\frac{d^2}{e\plus{}a\plus{}b}\plus{}\frac{e^2}{a\plus{}b\plus{}c}$

[quote="conan_naruto236"]Let $ a,b,c,d,e > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \geq 1$. Find minimum of :
$ A \equal{} \frac {a^2}{b \plus{} c \plus{} d} \plus{} \frac {b^2}{c \plus{} d \plus{} e} \plus{} \frac {c^2}{d \plus{} e \plus{} a} \plus{} \frac {d^2}{e \plus{} a \plus{} b} \plus{} \frac {e^2}{a \plus{} b \plus{} c}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63850


Let
$ P(x) \equal{} x^{2n} \plus{} a(x^{2n \minus{} 1} \plus{} x^{2n \minus{} 2} \plus{} ... \plus{} x) \plus{} 1 \equal{} (x^2 \plus{} b_1x \plus{} c_1)(x^2 \plus{} b_2x \plus{} c_2)...(x^2 \plus{} b_nx\plus{}c_n)$
$ a,b_i,c_i \in R$
prove that $ c_1 \equal{} c_2 \equal{} ... \equal{} c_n \equal{} 1$
I need your help  :lol:

See here : http://www.mathlinks.ro/viewtopic.php?t=188682

Rust i don't understand your solution can you write it again with more detail please  :blush:

If $ x_0$ is root $ P(x)$, then $ \frac{1}{x_0}$ is root too. If $ x_0$ is complex root, then $ Rex_0\minus{}2Imx_0$ is root too.
Let $ y\equal{}x\plus{}\frac 1x$, then $ P(x)\equal{}x^n(y^n\plus{}d_1y^{n\minus{}1}\plus{}...\plus{}d_n)$. If all roots $ y_1,...,y_n$ are real, then
$ P(x)\equal{}x^n(y\minus{}y_1)...(y\minus{}y_n)\equal{}(x^2\minus{}y_1x\plus{}1)...(x^2\minus{}y_nx\plus{}1)$. We must prove, that all $ y_i$ real.
Consider $ x\equal{}exp(iz), y\equal{}2cosz$, then $ P(x)\equal{}x^n(cosnz\plus{}a(\frac 12 \plus{}\sum_{k\equal{}1}^{n\minus{}1}coskz ))$.
Consider number of roots $ 1 \plus{}2\sum_{k\equal{}1}^{n\minus{}1}coskz\plus{}bcosnz\equal{}\frac{sin(n\minus{}1/2)z}{sinz/2}\plus{}bcosnz\equal{}0,b\equal{}\frac 2a$.
It equavalent to $ sin(n\plus{}\frac 12 )z\equal{}(1\minus{}\frac a2 )sin(n\minus{}\frac 12 )z, \ z\not \equal{}2k\pi$.
It give at least $ n\minus{}1$ real roots for $ y$, therefore n-th root real too.


algebra
polynomial
superior algebra
superior algebra unsolved

Take the ring R=Z/4Z. Find a polynomial of degree 1 in R[X] such that its square is the constant 1.


let $ f: R \to R$ such that :
\[ f(x)\equal{}f(2x)\equal{}f(1\minus{}x)\].
prove that $ f$ is periodic

$ f(x\plus{}1)\equal{}f(1\minus{}(x\plus{}1))\equal{}f(\minus{}x)$

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

so $ f(\minus{}x)\equal{}f(\minus{}\frac{x}{2}\plus{}\frac{1}{2})$

 then $ f(2x)\equal{}f(x\plus{}\frac{1}{2})\equal{}f(x)$

[quote="stvs_f"]let $ f: R \to R$ such that :
\[ f(x) \equal{} f(2x) \equal{} f(1 \minus{} x)\]
.
prove that $ f$ is periodic[/quote]
$ x \equal{} 1\minus{}y$:
$ f(1\minus{}y) \equal{} f(2\minus{}2y) \equal{} f(y)$. But $ f(y) \equal{} f(2y)$, so $ f(2y) \equal{} f(2\minus{}2y)$, and hence with $ 2y \equal{} z$: $ f(1\minus{}z) \equal{} f(z) \equal{} f(2\minus{}z)$. Letting $ 1\minus{}z \equal{} n$: $ f(n) \equal{} f(n\plus{}1)$, and hence $ f$ is periodic.

Does this also imply $ f$ is constant, by making the period arbitrarily small? I think it does: Suppose $ f$ is periodic and nonconstant, with minimal period $ k$; but then we just pick $ t$ so that $ \frac{1}{2^t}<k$; contradiction.

If $ f$ is continiosly, then $ f \equal{} const$.
For example $ f(x) \equal{} 0$ if $ x$ irrational, $ f(\frac {a}{b}) \equal{} \frac {1}{b_1}$ were $ b_1$ is maximal odd divisor of $ b$ satisfyed your condition and for any k $ 2^{\minus{}k}$ is period. All periods are $ \frac{n}{2^k}$.

Okay; thanks. Yeah, at first I was sure, but then I realized that a pathological function like that might defy it.


function
calculus
derivative
real analysis
real analysis unsolved

i wich that some one help me to solve this problem , i have many solution but i think they are obvious , so i'm not shur :
fund all function that are C°°  and 2pi peridic  so that all there derivative are bounded

Well, one ought to be able to write them as Fourier series:

$f(x)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)e^{ikx}.$

$f$ is $C^{\infty}$ iff $\widehat{f}(k)$ goes to zero as $k\to\infty$ factor than any power of $k.$

The set of all derivatives of $f$ is bounded iff $\exists\,M$ such that for all $k$ and all $m,$ $\left|\widehat{f}(k)k^m\right|\le M.$

Here's one direction of the proof. Suppose the Fourier coefficients satisfy the above inequality. Then

$\sup_x\left|f^{[m]}(x)\right|\le\sum_{k=-\infty}^{\infty} \left|\widehat{f}(k)k^m\right|$

$\le\sum_{k=-\infty}^{\infty}\left|\widehat{f}(k)k^{m+2}\right|\frac1{k^2}$ (assuming that $m>0$)

$\le2\sum_{k=1}^{\infty}\frac M{k^2}$

(The $m=0$ case can be handled separately.)

think you kent , that what i have proved , but it seems to be obvious ,i hope that to be what they were looking for .

but we will deduce that for |K|>1  we will have from
 \left\widehad{f}(k)k^n\right\leM
by tending m to infinity that \widehade{f}(k)=zero\
so f must be of the forme a+bsin(x)+ccos(x) wich clearly satisfy our conditions


All the vertices of a 15-gon, not necessarily regular, lie on the circumference of a circle and the center of this circle is inside the 15-gon. What is the largest possible number of obtused-angled triangles where the vertices of each triangle are vertices of the 15-gon?

the requirement:
3 points on a circle define an obtuse triangle
is equivalent to:
it is possible to choose a 180 degree arc such that all 3 points are on it but not on the endpoints (double angle theorem)

so we want to minimize the number of combinations of 3 points that do not follow this rule
because the center of the circle is in the 15-gon, all 15 points cannot be on the same semicircle.
i think the optimal arrangement is 13 points on a semicircle, none near the endpoint of the semicircle, and 2 other points, one on each side, an infinitesimal distance beyond the semicircle so that:
the only combination of 3 points that do not describe a an obtuse triangle are the combinations that include the 2 other points
so $ \binom{15}{3} \minus{} 13 \equal{} 442$
edit: lol i made that same mistake at the MC marathon
i cannot find a rigorous proof though

We can choose any combination of 3 of vertices, $ \binom{15}{3} \equal{} 455$.
Now we see how many triangles are not obtuse.

Say we have a circle. The center of the circle lies inside the 15-gon means that not all the vertices of the 15-gon lie in one hemi-circle. Also, an acute triangle is formed if 3 points selected on the circle does not lie in one hemi-circle; an obtuse triangle is formed if the 3 points do lie in one hemi-circle. Since we want to maximize, we assume the least possible number of triangles to exclude. We move 1 vertex out of the hemi-circle, and thus remove all the triangles it forms [b]with the vertex farthest away from it[/b]. However, we may still use that vertex with any other two vertices to form an obtuse triangle.

We subtract the $ 13$ acute triangles that the two opposite vertices make. Our answer is $ 442$ triangles.

Steven: $ \binom{15}{3} \equal{} 455$, not $ 715$.


trigonometry
function
inequalities
geometry
inequalities unsolved

Prove

$\frac{x}{\left(y+z\right)^2}+\frac{y}{\left(z+x\right)^2}+\frac{z}{\left(x+y\right)^2} \geq \frac{9}{4\left(x+y+z\right)}$

for any three positive reals x, y, z.

Equivalently, prove

$\frac{s-a}{a^2}+\frac{s-b}{b^2}+\frac{s-c}{c^2} \geq \frac{9}{4s}$,

where a, b, c are the sides of a triangle and $s=\frac{a+b+c}{2}$.

Equivalently, show

$\frac{\sin B \sin C}{\sin^2 \frac{A}{2}} + \frac{\sin C \sin A}{\sin^2 \frac{B}{2}} + \frac{\sin A \sin B}{\sin^2 \frac{C}{2}} \geq 9$,

where A, B, C are the angles of a triangle.

Thanks!
  Darij

This one can be proved by Cauchy:
(x+y+z)[x/(y+z)^2+y/(z+x)^2+z/(x+y)^2]   \geq (  x/ (y+z)+y/(z+x)+z/(x+y)  )^ 2 .
Then use Nesbitt, we have the result.

take s=x+y+z. we have to prove:
 \sum sx/(s-x) 2  \geq 9/4. function

f(x)=sx/(s-x) 2 is convex. apply Jensen.  :)

Klamkin Ine solve it!we have
(s-a)/a^2+....>=s(a^3+...)/(ab+...)*abc
Klamkin Ine:(x+...)(xPA^2+yPB^2+zPC^2)>=yza^2+....
who know Klamkin Ine please help me to write in detail

We can use homogeneity to let x+y+z=1

Then we have to prove that $ \sum_{cyc} \frac{x}{(x\minus{}1)^2} \ge \frac{9}{4}$ which is true by Jensen

hello Darij, here is my proof, after a short algebraic computation we have to prove
$ \minus{}12a^2b^2c^2\plus{}b^4c^2\plus{}b^2c^4\plus{}a^4c^2\plus{}a^2c^4\plus{}a^4b^2\plus{}a^2b^4\plus{}2(a^3b^3\plus{}b^3c^3\plus{}a^3c^3) \geq0$
This is equivalent with
$ c^2(a^2\minus{}b^2)^2\plus{}a^2(b^2\minus{}c^2)^2\plus{}b^2(c^2\minus{}a^2)^2\plus{}2\left(a^3b^3\plus{}b^3c^3\plus{}a^3c^3\minus{}3a^2b^2c^2\right)\geq0$, which is obvioulsy true because of AM-GM.
Sonnhard.

[quote="darij grinberg"]Prove

$ \frac {x}{\left(y \plus{} z\right)^2} \plus{} \frac {y}{\left(z \plus{} x\right)^2} \plus{} \frac {z}{\left(x \plus{} y\right)^2} \geq \frac {9}{4\left(x \plus{} y \plus{} z\right)}$

for any three positive reals x, y, z.

Equivalently, prove

$ \frac {s \minus{} a}{a^2} \plus{} \frac {s \minus{} b}{b^2} \plus{} \frac {s \minus{} c}{c^2} \geq \frac {9}{4s}$,

where a, b, c are the sides of a triangle and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$.

Equivalently, show

$ \frac {\sin B \sin C}{\sin^2 \frac {A}{2}} \plus{} \frac {\sin C \sin A}{\sin^2 \frac {B}{2}} \plus{} \frac {\sin A \sin B}{\sin^2 \frac {C}{2}} \geq 9$,

where A, B, C are the angles of a triangle.

Thanks!
  Darij[/quote]
Let $ y\plus{}z\equal{}2a,z\plus{}x\equal{}2b,x\plus{}y\equal{}2c$ then the inequality equivalent to
$ (a\plus{}b\plus{}c)\left(\frac{b\plus{}c\minus{}a}{a^2}\plus{}\frac{c\plus{}a\minus{}b}{b^2}\plus{}\frac{a\plus{}b\minus{}c}{c^2}\right)\geq 9$
let $ a\plus{}b\plus{}c\equal{}1$ equivalent to
$ \sum \frac{1\minus{}2a}{a^2}\geq 9$
By AM-GM inequality,we have
$ \frac{1\minus{}2a}{a^2}\plus{}81a^2(1\minus{}2a)\geq 18(1\minus{}2a)$,thus
$ \frac{1\minus{}2a}{a^2}\geq 18(1\minus{}2a)\minus{}81a^2(1\minus{}2a)$
on the other hand
$ a^2(1\minus{}2a)\leq (\frac{a\plus{}a\plus{}1\minus{}2a}{3})^3\equal{}\frac{1}{27}$
$ \frac{1\minus{}2a}{a^2}\geq 18(1\minus{}2a)\minus{}3$
and other simillary results,done

another equivelent form is:
$ \sum\tan\frac{A}{2}\sec^2\frac{B\minus{}C}{2}\geq\frac{9R}{2S}$
S is the area of triangle.

As we have that $ \frac{x(x\plus{}y\plus{}z)}{(y\plus{}z)^2}\ge \frac{9x^2}{4(x^2\plus{}y^2\plus{}z^2)}$ we can write an inequality what we need to prove.


Points $ a$ and $ b$ are in nonadjacent vertices in a 8 by 8 grid square. This square is then sliced in half by cutting through the line ab. Find the number of ways to get from a to b in this new triangle.

How do I approach this problem? Where do I start?

Hint: Find the number of ways to get from Point A to another certain point, and build off that.


A triangle and quadrilateral intersect in finetly many points.  What is the most number of intersection points they can have?

Don't feel bad Braves, I was about to come to your defense as I was convinced you had it right.  Mathgeek's question is preying on our assumptions :)

Just so no one feels bad, when I was given this problem, I missed it.

8 is the highest it can go!
CLUE: dont use square for the quardilateral there are more shapes! :D

Now that two people have gotten it, I'll show the diagram.

[hide]A quadrilateral is made up of 4 lines. A line may interect the triangle in 2 places. Therefore 4 lines may intersect the trianlge in 8 places.[/hide]

And actually I will work on a picture of a square or a  rectangle interecting 8 times, because I think it can be done. :P

Billy

I think I know what you're thinking, and I doubt it can be done.


Consider the plynomial $f(x)$ with degree $n$ and integer coefficients.  Given that:

$f(1)=f(2)=f(3)=f(4)=2004$,

prove that there is no integer $m$ such that $f(m)=1$

[hide]
We have $f(x)-2004 = (x-1)(x-2)(x-3)(x-4)g(x)$ for some $g(x)$ which also has integer coefficients.

So $f(x) = (x-1)(x-2)(x-3)(x-4)g(x)+2004$. If $m$ exists, then

$f(m) = (m-1)(m-2)(m-3)(m-4)g(m)+2004 = 1$

so

$(m-1)(m-2)(m-3)(m-4)g(m) = -2003$.

But since $m-1$, $m-2$, $m-3$, $m-4$ and $g(m)$ are all integers and either $m-1$ or $m-2$ is even, their product must be even and therefore cannot be $-2003$.

BTW, shouldn't this work for $f(1) = f(2) = 2004$ too?[/hide]

Or any pair of consecutive integers then (not including 0)?

you're right paladin, but i guess that 1,2,3,4 are used only to hide the basic solution...

[quote]why $g(x)$ has integer coefficients?[/quote] :)

[quote="Vutanga"]I don't think so.
If $f(x)=h(x)g(x)$, $f(x)$ and $h(x)$ have integer coeffictions then we can't imply $g(x)$ has integer coeffictios. For examble:
$x =2x\frac{1}{2}$  [/quote]

- Why are you putting what you want to say inside quotes?

- $2x\frac{1}{2}=x$ and $1\in\mathbb{Z}$ so it does have integer coefficients.

$g(x)$ must have integer coefficients because of the way it was factored. The other terms are $(x-4), (x-3), (x-2), (x-1)$ The coefficients of the $x$ terms are all $1$ which won't be able to cancel out any fractions such as the example you gave.

I can prove why $g(x)$ must has integer coeffictions, thank you! :)

[quote="Petoria"]Does this mean that $f(m)$ can never equal an odd integer?

What can we generalize this to?

$f(1) = a, f(2) = a, f(3) = a, f(4) = a$ given $a$ is an even integer, then $f(m)$ is never an odd integer.

Would that be a correct statement?[/quote]

yes, it's right.


In a triangle $ ABC$, $ \widehat{B}\equal{}90^\circ$ $ D\in [AC]$ and let $ K$,$ M$ be the feet of perpendiculars from $ D$ to $ [AB]$ and $ [BC]$.

$ |DK|\equal{}x$,$ |DM|\equal{}y$
$ (x\plus{}y)_{min}\equal{}3$
$ (x\plus{}y)_{max}\equal{}8$

S(ABC)=?

[quote="sinankaral53"]In a triangle $ ABC$, $ \widehat{B} \equal{} 90^\circ$ $ D\in [AC]$ and let $ K$,$ M$ be the feet of perpendiculars from $ D$ to $ [AB]$ and $ [BC]$.

$ |DK| \equal{} x$,$ |DM| \equal{} y$
$ (x \plus{} y)_{min} \equal{} 3$
$ (x \plus{} y)_{max} \equal{} 8$

S(ABC)=?[/quote]
 suppose AB<BC => $ \widehat{A}$ > $ \widehat{C}$ , $ \widehat{KDA}$=$ \widehat{C}$ => $ \widehat{A}$>$ \widehat{KDA}$ => x>AK, BK=DM=y so $ x \plus{} y \geq AK \plus{} BK \equal{} AB$ similar we can demonstrate $ x \plus{} y \leq BC$
(x+y)max=8 and (x+y)min=3 so AC=8, BC=3 so SABC=12


[quote="joejia"]Four consecutive positive integers have a product of 840. What is the largest of the four integers?[/quote]

[hide]$x(x+1)(x+2)(x+3)=840$

$(x^2+x)(x^2+5x+6)=840$

$x^4+6x^3+11x^2+6x=840$

$x^4+6x^3+11x^2+6x-840=0$

$\text{The root of the equation is 4, so the largest number must be 7}$[/hide]

Algebra is completely unessecary for such a simple problem.

[hide="Answer"]  The prime factorization of $840$ is $ 2\cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 7 $

The number 3, 5, and 7 are so close together were probably dealing with integers in that range.

  Notice how $ 2\cdot 2=4$ and $ 2 \cdot 3=6$.  So our 4 consecutive integers are $4,5,6$ and $7$, among thse $\boxed{7}$ is the largest. [/hide]

[hide]
i set up an algebraic equation that would take me a while to write but the answer to the problem is 7.[/hide]

[hide] You can use guess and check or prime factorization, either way, you get 7; 4 * 5 * 6 *7 = 840. [/hide]

[hide]the fourth root of 840 is 5.38... which is somewhere around the average of the 4 numbers. 4, 5, 6, and 7 seem to work. 7 is the answer. [/hide]

[hide]Prime Factorisation. I get $4 \cdot 5 \cdot 6 \cdot 7$. Largest is 7[/hide]


i wish i could go...

did u not go? and why are u asking for results before it started?

i wish i could've went... :(  :(  :(

i could've saw xinke

i'm really sad...

we really need some spain park ppl on aops

yao why didnt you go?

hmmm gene got like 6th? Sisi got 10th, I think somebody got 4th and 5th, girls.. and Farhan got... 14th? We got 1, 2, 3, 12, 13. I was on the end of the stage and they didnt see me. Then Zaki pointed out "calculator boy"... :'(

Raisa decided she was going to put my face in a heart and have it say calculator boy on a tshirt and she said ppl would buy it.

Hmmm like 20 points maybe? If u got perfect on written (very probable), I think that would only raise u guys by like 5 or so on written bcuz of close scores, then u would have to pwn ciphering.

[quote="dynamo729"]I got my first ever perfect :) They scored everybodys wrong though, they had an answer on the answer key that they corrected.
Challenger won, w00t. And even if chestnut would have been there, we still would have won, i think.
And a bunch of people followed me around all day. Apparently, I'm Sisi's bff? And Genes? And some people were claiming they kniew me and I'm like uhhhh I've never seen you and someone was talking behind some people I DO know and they were saying German stuff and my name  :o

This makes me feel better from my MC/ AIME failures. :)[/quote]

That happens to me at some tournaments…. Ppl  are like “hi Xinke!” and im thinking “who are you…and how do you know my name…should I know you??..” but I say “..uhh..hi..” back as soon as I can. Fortunately I’ve never noticed ppl following me.
Oh and LOL! Nick you were a good 7 or  8 inches  taller than every one of the 14 others that got trophies. I thought that was pretty hilarious.
Anyways, yea grissom test is always really easy (which means that the good ppl can never place as well since its more about not making stupid mistakes as compared to mathematical ability).

[quote="dynamo729"]I  thought it was easy, but as I've said before all of our skills are increasing, so we cant really judge what easy is, right now at easy.

Xinke got 4th, i think. Spain Park go 3rd.[/quote]

I got 2nd

[quote="chestnut"]uh k how about the people from pizitz....
yes i'm that difference maker...
didn't go to ceder ridge, hoover, berry-simmons, and grissom.....
thats my commitment to math team....[/quote]

dont worry i personally think your time is probably better spent on soccer, since those tourneys often arent worth the time. the only thing that should really matter (after mc nats) for you is making red mop next year.

[quote="yaofan"]they followed me around everywhere at berry's tournament.

i wish i could go...

was it easy?

did xinke go? and did spain park go?[/quote]

spain park went.

[quote="hurdler"][quote="dynamo729"]I got my first ever perfect :) They scored everybodys wrong though, they had an answer on the answer key that they corrected.
Challenger won, w00t. And even if chestnut would have been there, we still would have won, i think.
And a bunch of people followed me around all day. Apparently, I'm Sisi's bff? And Genes? And some people were claiming they kniew me and I'm like uhhhh I've never seen you and someone was talking behind some people I DO know and they were saying German stuff and my name  :o

This makes me feel better from my MC/ AIME failures. :)[/quote]

That happens to me at some tournaments…. Ppl  are like “hi Xinke!” and im thinking “who are you…and how do you know my name…should I know you??..” but I say “..uhh..hi..” back as soon as I can. Fortunately I’ve never noticed ppl following me.
Oh and LOL! Nick you were a good 7 or  8 inches  taller than every one of the 14 others that got trophies. I thought that was pretty hilarious.
Anyways, yea grissom test is always really easy (which means that the good ppl can never place as well since its more about not making stupid mistakes as compared to mathematical ability).[/quote]

funny this happens to me TOO

[quote="hurdler"][quote="chestnut"]uh k how about the people from pizitz....
yes i'm that difference maker...
didn't go to ceder ridge, hoover, berry-simmons, and grissom.....
thats my commitment to math team....[/quote]

dont worry i personally think your time is probably better spent on soccer, since those tourneys often arent worth the time. the only thing that should really matter (after mc nats) for you is making red mop next year.

[/quote]
oh i was being sarcastic hurdler....
lol hurdler... i missed this one for a track meet..... connection?
Thats pretty much my mindset on local mt tournys. if i have something, just skip it since all of us Aops ppl are good enough that the typical 25 question 1 hour test doesn't make you better.(we're just that cool)
so thats my math life in a nutshell
do good at nats
make usamo  :D
and next year
and hopefully make red mop :D  :D
dynamo you should try and make it too :D

[quote="chestnut"]uh k how about the people from pizitz....
yes i'm that difference maker...
didn't go to ceder ridge, hoover, berry-simmons, and grissom.....
thats my commitment to math team....[/quote]
pizitz is overrated...

[quote="dynamo729"]yao why didnt you go?[/quote]
mrs.clopton never said anything

i wish i could go... :(  :(  :(

[quote="chestnut"]how much did yall win by?[/quote]
i didn't know chinese people said yall! i dont say it cuz it's too southern...

and there's no more tournaments left..............................
i'll never learn the pencil spinning trick now...

Uh, I dont care what the score sheet said,. They messed number 5 up and forgot to grade it, maybe? But Bo asked me what I got and I said perfect and she was like "Oh..." which means she did not get perfect.
[quote]
and hopefully make red mop :D :D
dynamo you should try and make it too :D[/quote]
Actually I'm more into the broom making business VBWAHAHAH get it make mop... whatever.
What is mop?

[quote="weird888"]BTW, third to gfirst were all challenger kids....
I SEE BIASED[/quote]
maybe it's because challenger people don't have to wake up early in the morning and drive to huntsville..

pizitz was biased...
they don't like berry or simmons so they made a school from georgia into division I. otherwise I WOULD'VE BEEN 1ST!!!!!!!!!!!

[quote="weird888"]grissom is so BIASED!!!!
look, nick did not really get "first"
on the scor sheet, it said nick ahd a 95.31, and bo had a 95.3??!?!?!
WTF?????
ok, there tottaly is something wrong...
just like last year.....
asutosh and i tied...
and, for some reason...
they said i got a 96, and third got a 96.1, second got 96.15?!?!, and asutosh got a 96.2 (the score that i got)

BTW, third to gfirst were all challenger kids....
I SEE BIASED[/quote]

this may not be the case, but sometimes the .01 or .05 is to break the tie in placement in the computer when they look at who missed the first question backwards...like i tied Kyle but he missed 24 and I didn't so my score said like 86.31 and his was 86.3 (actually we forgot to protest one of the tiebreakers they had wrong and it was 86.21 and 86.2)

MOP the math olympiad program, where you go (for 3 weeks, and for free) if you do well on usamo. The US IMO team is chosen from the 12 usamo winners. The top 12 scores from usamo are invited (they're in black mop), then the top 12 or 18 or something (i forget) nonseniors (that aren't already a usamo winner) (they're in blue mop). However, in recent years they have upped the number of ppl invited for freshmen -  they invite 25 or 30 freshman (they're in red mop...for some reason, if you are below 9th grade, you can't get into red mop. Red mop is exclusively for freshman, but you're reallly really really really good as a middle schooler, you can get into blue or black), so that you only need to get about 9 or 10 points to get in. 9 or 10 points is the equivalent of getting 1 1/2 questions right, since each question is scored on a scale of 0 to 7.

number 1 is usually easiest, and number 4 usually the second easiest. So like most ppl spend their first 4 1/2 hours on number 1 and their second 4 1/2 hours on number 4.

But yea, making usamo and mop is pretty much the most useful use of your time if you capable of getting perfect or near perfects on pretty much all local math tourneys, then you should spend your time working on aime and proofs and stuff, since its a big jump to get to level of being able to get perfect/near perfect on aime and an even bigger one to get from consistently solving 1,4 on usamo to solving at least one of 2,3,5,6. Too bad I will never make those last 2 jumps.

I say yall too

[quote="hurdler"]
MOP the math olympiad program, where you go (for 3 weeks, and for free) if you do well on usamo.

I say yall too[/quote]

well i have no chance of making it in that case...
got a 50% on AMC10... :rotfl:  :rotfl:

yall is so weird

[quote="weird888"]no, hurdler, if bo and nick both got perfects, than how does nick get 95.31, and bo gets 95.3
just like last year, one question was wrong so the ythrow out, i get 96.2 for real... but it said 96 ?!?!?[/quote]

i said "sometimes"....other times...WHO KNOWS

gene, you know what you got, and the only difference anything would make is a number on a trophy. You can't worry about specifics all the time.z

i wish i could talk about this too...
but i didn't go...
sigh...


function
parameterization
logarithms
topology
real analysis
inequalities
triangle inequality

Hi,

Do you think that there's a continuous fonction $ f: [0,1] \rightarrow \mathbb{C}$ such that
$ \forall x,y\in [0,1] \quad |f(x)\minus{}f(y)|\geq \sqrt{|x\minus{}y|}$

I've sent this question in
http://www.mathlinks.ro/Forum/viewtopic.php?t=294174

thanks

I think that the [url=http://en.wikipedia.org/wiki/Dragon_curve]dragon curve[/url] (with obvious parametrization) is an example of such a function.

UPD: No, it is obviously not :(

Obviously -- because the dragon curve has selfintersections.

Further thoughts:

For every $ \alpha > 1/2$ there exists a function $ f: [0,1]\to \mathbb R^2$ such that $ |f(x) \minus{} f(y)|\ge |x \minus{} y|^\alpha$. For instance, for $ \alpha \equal{} \log 3/\log 4$, a possible example is properly scaled and parametrized part of the [url=http://en.wikipedia.org/wiki/Koch_snowflake]Koch snowflake[/url] (among what I write here this is the only statement I can prove  :) ).

But for $ \alpha \equal{} 1/2$, there is no such function.

The reason is that (heuristically) the graph of this function should have Hausdorff dimension $ 1/\alpha$. For $ \alpha \equal{} 1/2$ this curve should have dimension $ 2$, so (highly heuristically) it should have selfintersections.

Having dimension 2 does not mean it must interesect.
You are looking for a space filling curve ( see [url]http://en.wikipedia.org/wiki/Space-filling_curve[/url] )
I believe this solves the original problem.

hi,

I agree that a such curve should have Haussdorf dim =2, but I still don't understand why it's impossible to find a such function $ f$.

Thanks,

[quote="PhilG"]
You are looking for a space filling curve ( see [url]http://en.wikipedia.org/wiki/Space-filling_curve[/url] )
I believe this solves the original problem.[/quote]
But this curve IS self-intersecting. And, of course, it does not solve the problem. By the way, can you give me a real example of non-intersecting curve in the plane with graph dimension 2?

I discussed this problem with a very smart guy, he pointed me where I should look for a solution (and indeed the answer is that such bounded - not even continuous - function does not exist).

But the interesting question what happens if we drop the boundedness assumption.

you are right, it does intersect, my mistake

[quote]
I discussed this problem with a very smart guy, he pointed me where I should look for a solution (and indeed the answer is that such bounded - not even continuous - function does not exist).
[/quote]

Could you be more precise, please ?

Now, I couldn't because I don't know the solution. I have the idea how to prove it, but there is a lot of technical work and I'm too lazy to do it.

Nevertheless, if someone gives me a reference to the 239 St Petersburg school olympiad of 2002 (I didn't find), I can translate the solution.

Ok, this is really nice problem, I'll try to ask my friend to send me the solution in russian and forward it to zhoraster :wink:

[quote="leshik"]Ok, I've got the response from my friend (with attached problems and solutions), but I could not find any problem there, related to the topic we are discussing here. :|[/quote]
Then we must beat Slobodianiuk. He told it should be there.

I know that pupils of 239 school are very strong and have advanced techniqal knowledgies but it's still strange that the problem of such type has been proposed on school olympiad.

No-no, the formulation was different.
Something like: we have $ n^2$ points in a disk of radius $ 100n$ and distance between $ i$th point and $ j$th point is not less than $ \sqrt{i\minus{}j}$. Prove that $ n$ cannot be too large.

Ok I'm sending the files I've got, maybe zhoraster will be able to find the problem that we need. :)

Oops. I can't compile it. Could you also give me rotpr02.sty and ris.tex? And where are the 11th grade problems? Aha, the same as 10th. Anyway, probably Slobodianiuk confused the year.

I don't have these files.Just  comment-out them, then it should be at least readable :lol:

Actually, folks, by switching to the discrete version, you make the problem harder, not easier.

The original one (about continuous function) is not hard at all. Let $ F$ be the (compact) image of $ [0,1]$. Then the diameter of the pre-image of every open disk of radius $ r$ is at most $ Cr^2$. Also, if $ x\in F$ is a density point of $ F$ (with respect to the 2-dimensional Lebesgue measure), then the diameter of the pre-image of $ D(x,r)$ is $ o(r^2)$ as $ r\to 0$ (this is the only non-routine part in the whole solution, but if you think of it for a few minites, you'll realize that it is next to obvious ;) ). These two observations together immediately imply that the 1-dimensional Lebesgue measure of the total pre-image must be $ 0$, which is absurd.

But, of course, the version quoted by zhoraster is more interesting and certainly deserves the extra effort :).

I am trying to understand your reasoning. Can you elaborate these points?

[quote="fedja"]
Let $ F$ be the (compact) image of $ [0,1]$.
[/quote]

Is it an extra assumption that image is compact, or can it be deduced?

[quote="fedja"]
Then the diameter of the pre-image of every open disk of radius $ r$ is at most $ Cr^2$.
[/quote]

OK I get this with C = 4.

[quote="fedja"]
Also, if $ x\in F$ is a density point of $ F$ (with respect to the 2-dimensional Lebesgue measure), then the diameter of the pre-image of $ D(x,r)$ is $ o(r^2)$ as $ r\to 0$
[/quote]

$ D(x,r)$ is the disk centred on a of radius r, right? So the pre-image has diameter at most $ 4r^2$ which is $ O(r^2)$.
How do you get $ o(r^2)$ (little o rather than big O)

[quote="fedja"]
These two observations together immediately imply that the 1-dimensional Lebesgue measure of the total pre-image must be $ 0$, which is absurd.
[/quote]

I will agree, if you can show that the image is compact and that the diameter of the pre-image of $ D(x,r)$ is $ o(r^2)$ rather than $ O(r^2)$
You have not used the fact that the curve is continuous, are you saying that this result is true even without the continuity condition?

1) The image of a compact set under a continuous mapping is compact.

2) Yes, $ C\equal{}4$ works.

3) Fix $ \delta>0$. Since $ x$ is a density point of $ F$, for small $ r>0$, the disk $ D(x,r)$ cannot contain a disk of radius $ \delta r$ free of points of $ F$. So, every 2 points $ y,z\in D(x,r)\cap F$ can be connected by a chain of points $ y\equal{}t_0,t_1,\dots,t_m\equal{}z$ in $ F$ such that $ |t_j\minus{}t_{j\plus{}1}|\le C\delta r$ and $ \sum_j |t_j\minus{}t_{j\plus{}1}|\le Cr$ whence $ \sum_j |t_j\minus{}t_{j\plus{}1}|^2\le C^2\delta r^2$. But this sum dominates $ |f^{\minus{}1}(y)\minus{}f^{\minus{}1}(z)|$ by the triangle inequality.

4) I used continuity to get the compactness of the image (actually I needed just measurability of the image) but I, indeed, claim that it is not necessary (though, of course, my solution should be modified to cover the general case).

hi

thank you fedja.

But can you explain the last part :

[quote]These two observations together immediately imply that the 1-dimensional Lebesgue measure of the total pre-image must be 0, which is absurd. [/quote]

The set $ F$ consists of two parts: the set of density points $ G$ and the set of other points $ H$.

Note that $ H$ has measure $ 0$, so it can be covered by open disks with arbitrarily small sum of squares of radii, so $ f^{\minus{}1}(H)$ can be covered by sets with arbitrarily small sum of diameters whence it has linear measure $ 0$.

Now, fix $ \delta>0$. For every $ x\in G$, and all small enough $ r>0$, the pre-image of $ D(x,r)$ has the diameter less than $ \delta r_j^2$. Now just use your favorite covering lemma (Vitali or Besicovich, doesn't matter). Then you'll get the covering of $ G$ by disks $ D(x_j,r_j)$ with uniformly bounded $ \sum_j r_j^2$. The rest should be clear.

Hi fedja, you wrote

Fix $ \delta > 0$. Since $ x$ is a density point of $ F$, for small $ r > 0$, the disk $ D(x,r)$ cannot contain a disk of radius $ \delta r$ free of points of $ F$. So, every 2 points $ y,z \in D(x,r)\cap F$ can be connected by a chain of points $ y \equal{} t_0,t_1,\dots,t_m \equal{} z \in F$ such that $ |t_j \minus{} t_{j \plus{} 1}|\leq C\delta r$ and $ \sum_j |t_j \minus{} t_{j \plus{} 1}|\leq Cr$ whence $ \sum_j |t_j \minus{} t_{j \plus{} 1}|^2\leq C^2\delta r^2$

it's ok for me (I think), but then you said :

But this sum dominates $ |f^{ \minus{} 1}(y) \minus{} f^{ \minus{} 1}(z)|$ by the triangle inequality.

I don't get it, in fact I've
\[ |f^{ \minus{} 1}(y) \minus{} f^{ \minus{} 1}(z)| \leq |y \minus{} z|^2 \leq m \sum_j |t_j \minus{} t_{j \plus{} 1}|^2 \leq m C^2\delta r^2\]

You use the triangle inequality on the wrong side. What you should write instead is
$ |f^{\minus{}1}(y)\minus{}f^{\minus{}1}(z)|\le \sum_j |f^{\minus{}1}(t_j)\minus{}f^{\minus{}1}(t_{j\plus{}1})|\le \sum_j|t_j\minus{}t_{j\plus{}1}|^2\le\dots$.


FYI: I noticed that today's Google home page has pictures of fractals and chaos and associated math.  I think that's because today (Feb. 3) is the birthday of Gaston Julia (1893-1978), one of the founders of fractals.

Actually, if you click on the logo, it send you to a link with pictures of various fractals (I think the result of doing a google image search for "Gaston Julia.")  Go figure -- I wouldn't have thought that this made their "special days" list -- I wonder if they do anything for March 14.

I'm pretty sure they've done March 14th in the past. They do different special days each year.

i clicked on the link and i saw this french guy weiring a mask what's up with that

Maybe he didn't have a nose like Van Gogh...oh wait that was his [b]ear[/b] :lol:

well he probably was sane and got shot in the nose by the germans (i'm guessing) during world war 1 or 2.
 bullet proof mathematician (the movie starring Julia)

Yes, Julia was severely wounded in World War I, losing his nose.  He had to wear a leather strap across his face for the rest of his life.

Maybe they'll do something for towel day.

[quote="churchilljrhigh"]then confuted how does he breathe? i think he still has the cute hole that goes from the nostrils to the windpiple.[/quote]
Yes. Define hole to be "the absence of anything else"


How would you find the maximum number of distinct regions formed by the intersection of five coplanar lines? Of $ n$ coplanar lines? Assuming unbounded regions count as reginos as well.

I'm not sure how to go about this...At first I thought about somehow counting the regions with 2 edges, 3, 4, and 5, but I don't know how to count that. I thought there might be some sequence or something...but...

Help please?

Observation:
#lines    #regions
0          1
1          2
2          4
3          7
4          11
...         ...

If we say $ S_n$ is the maximal number of regions that $ n$ intersecting lines can make, then you can see: $ S_n \equal{} S_{n \minus{} 1} \plus{} n$. From that, we can easily get the formula: $ S_n \equal{} S_0 \plus{} 1 \plus{} 2 \plus{} ... \plus{} n \equal{} \frac {n(n \plus{} 1)}{2} \plus{} 1$
This can be explained in pictures: imagine there's $ n \minus{} 1$ intersecting lines, adding one more, we can at most intersect $ n \minus{} 1$ times (straight lines at most have one intersection); so this line is cut into $ n$ segments, and each segment exactly parts a former entire region.Thus, $ S_n \equal{} S_{n \minus{} 1} \plus{} n$.

[hide="A direct approach"] $ n$ lines in a plane make at most $ {n \choose 2}$ distinct intersection points.  Those intersection points are the "bottom leftmost point" of exactly one region each, so we have at least $ {n \choose 2}$ regions.

Now, there are regions with no "bottom leftmost point."  These are the (unbounded) regions formed below all of the other regions.  Because we have $ n$ lines, exactly $ {n \choose 1}$ of these regions have an "upper rightmost point" (formed by each of the $ n$ lines) and exactly $ {n \choose 0} \equal{} 1$ region has neither (it will be on the bottom right).  Hence the total number of regions is

$ {n \choose 2} \plus{} {n \choose 1} \plus{} {n \choose 0}$

I phrase the solution this way because this proof is easily generalized:  for example, the maximum number of distinct regions formed by $ n$ planes in space is

$ {n \choose 3} \plus{} {n \choose 2} \plus{} {n \choose 1} \plus{} {n \choose 0}$

An interesting consequence:  the maximum number of distinct regions formed by $ n$ hyperplanes in dimension $ n$ is given by

$ {n \choose n} \plus{} {n \choose n\minus{}1} \plus{} ... \plus{} {n \choose 1} \plus{} {n \choose 0} \equal{} 2^n$

In other words, each hyperplane divides $ n$-space in half in a different dimension.  Consider the cases $ n \equal{} 2, 3$ if you are confused.  :) [/hide]


Let $a_n$ be a sequence of positive integers, such that $a_{n+1}=a_n+b_n$ where $b_n$ is the same as $a_n$ but written backwards. Can $a_7$ be a prime?

The original solution shws that among the first six nubers there must be one with an even number of digits and all the next are divisible by 11. I just wonder  if this solution can be improved by some nice tricks or if there is a completely different solution.