[quote="digger"]Given any integer $n\ne 0$ can you find an integer $x$ for which $1+nx$ is composite integer?[/quote]

what does the word "composite" mean?

[hide]
Yes.  $1+n(n+2)=n^{2}+2n+1=(n+1)^{2}$, which is obviously composite.
[/hide]

[hide="Another approach"] By Bezout's Lemma we can find $y$ such that

$py-nx = 1$

For an arbitrary $p$ such that $gcd(n, p) = 1$.  By a slight extension of Bezout's Lemma, we can find infinite such $y$. [/hide]

[quote="amirhtlusa"]not prime. :wink:[/quote]
1 is not prime.  :wink:

[quote="anirudh"]A composite number, according to mathworld is:

[quote]A composite number n is a positive integer n>1 which is not prime (i.e., which has factors other than 1 and itself)[/quote]

Google is your friend.  :)[/quote]
No, it's not.
Wikipedia is your only online friend. :D
edit: sorry that is not true. Anything AoPS is your best online friend


You probably think that somewhere in the net there are some great websites you never had the opportunity to discover, and in this topic we can exchange a looooooot by just asking : what's your favorite websites?
(of course it includes (http://www.mathlinks.ro));)

http://www.myspace.com
http://www.myyearbook.com
http://www.dogpile.com
http://www.blogtop100.com
http://www.miniclip.com
http://www.artistdirect.com
http://www.abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijk.com
http://www.funnyjunk.com
etc etc....i dunt really know what wud interest nerds tho
im just on this site as an xcuse 4 studying cuz of my mom lol...bet some others r too

[url=http://www.skyscrapercity.com]www.skyscrapercity.com[/url]
very nice, not only have they pictures of all skyscrapers, you can ask questions, show your own pictures, and they often show pictures of parks, bridges, rivers etc, you also get to see a lot of cities like in eastern europe or africa or north korea, where you rarely catch a glimpse of

[url=http://www.imdb.com]imdb[/url]
mother of all movie and tv sites, pages about everything, links to all actors and directors, trivia, fora, etc

[url=http://www.space.com]space.com[/url]
covers everything about space : unmanned, manned, astronomy

[url=http://www.nasa.gov]www.nasa.gov[/url]
very nice site

[url=http://en.wikipedia.org/wiki/Main_Page]wikipedia[/url] well you all know this encyclopedia right

Some websites I like a lot:

http://www.scienceblogs.com/pharyngula

A weblog written by a developmental biologist.

http://www.nws.noaa.gov/

The National Weather Service.  (If you don't like Senator Rick Santorum of Pennsylvania, you should use their website instead of commercial weather companies to check the forecast.  Besides, weather is cool.)

http://zenoferox.blogspot.com/

A weblog written by a math instructor at the college level.

http://www.pestiside.hu

A news/weblog site for English-speakers in Budapest.

[url=http://en.wikipedia.org/wiki/Main_Page]wikipedia[/url]
[url=http://google.com]google[/url] (random googling and going on the sites i get)
[url=http://www.artofproblemsolving.com/Forum/index.php]AoPS[/url]
[url=http://facebook.com]facebook[/url]
[url=http://http://ocw.mit.edu/index.html]MIT OpenCourseWare[/url]
i mostly surf the net for time wasting purposes, not much interest in knowledge :(
its amazing how many AoPSers have blog sites/myspace-like-sites as their favorites, is that true of most people?

[url=http://www.mugglenet.com]Mugglenet[/url]
[url=http://www.jkrowling.com]JK Rowling's site[/url]
and ofcourse this site.

I'll post more once I think of them.

I can't believe nobody has mentioned it yet....

http://www.theonion.com

I [i]can't[/i] be the only AoPS'er that loves that site :D

[url=http://www.mathlinks.ro]www.mathlinks.ro[/url] : for math things obiouvsly :D
[url=http://www.hi5.com]www.hi5.com[/url] : keep in touch with friends, collegues, etc. I check it regularily but I don't use it as a mail box at all.
[url=http://www.imdb.com]www.imdb.com[/url] : i watch a movie only if is well rated ($\geq 6.0$). This is very useful and true in the same time.
[url=http://www.explodingdog.com/drawonpaper/]www.explodingdog.com/drawonpaper/[/url] : nice drawings. I like them very much. I used on as an avatar :P.

I didn't mentioned but it's a popular thing to use Google or Y! Mail :)

The most hilarious clip:
http://www.endofworld.net

What was the adress of that page, whose owner won 1 million dollars????

[url=http://Wikipedia.org]Wikipedia, of course[/url]

[url=http://myspace.com]And Myspace[/url]

[url=http://www.research.att.com/~njas/sequences/]OEIS[/url] is pretty fun, too.

And, to round it off, [url=http://rinkworks.com]Rinkworks[/url]

Post number thousand!

if you just want a quick look

[url]http://www.skyscraperpicture.com/index1.htm[/url]

for closer inspection :

[url]http://www.skyscrapercity.com[/url]
[url]http://www.emporis.com[/url]

http://www.kurzweilai.net is fascinating.  Among other things, it contains links to a lot of really interesting science news stories.  (Mostly related to biotech, nanotech, and AI/robotics.)


algebra
polynomial
geometry
analytic geometry

Any guesses?

... I've always screwed up on the last problem.. I'm placing my bet on either polynomial problem, or messy analytic geometry!  :D

I think it'll be a polynomial this year.
Last year I got the last problem. but screwed up the one before it. lol


function
limit
real analysis
real analysis unsolved

Find all functions  such that |f(x)-f(y)|<=|x-y| for all x,y>0 and there exist positive constants a, b, c satisfying

                                                    f(ax)+f(b/x)=f(cx) for all x.

(Edgár Dobribán, Cluj-Napoca)

Lipschitz functions are uniformly continuous and can be extended to the limit points of their domains. Hence, $ \lim_{x\to0^{\plus{}}}f(x)$ exists. Just to have a notation, call this limit $ f(0).$

$ \lim_{x\to0^{\plus{}}}f\left(\frac{b}{x}\right)\equal{}\lim_{x\to0^{}}(f(cx)\minus{}f(ax))\equal{}f(0)\minus{}f(0)\equal{}0.$

Hence, $ \lim_{x\to\infty}f(x)\equal{}0.$

By taking the limit of the same expression as $ x\to\infty,$ we conclude that $ f(0)\equal{}0.$

---

Just getting started; I don't have a whole argument yet.

We have that $ f(y) \equal{} f(uy)\plus{}f\left(\frac{v}{y}\right)$.
Hence, for $ z : \equal{}\frac{v}{y}$, $ f\left(\frac{v}{z}\right) \equal{} f\left(\frac{uv}{z}\right)\plus{}f\left( z\right)$, so $ f(uz) \equal{}\minus{}f\left(\frac{uv}{z}\right)$.
Putting $ \frac{v}{t}: \equal{} uz$ we get that $ f\left(\frac{v}{t}\right) \equal{}\minus{}f\left( u^{2}t\right)$.
Therefore, $ f(y) \equal{} f(uy)\minus{}f\left( u^{2}y\right)$, which implies $ f\left( uy\right) \equal{} f\left( u^{2}y\right)\minus{}f\left( u^{3}y\right)$, i.e. $ f(y) \equal{}\minus{}f\left( u^{3}y\right)$.

If $ u \equal{} 1$, then $ f(y) \equal{} f(y)\plus{}f\left(\frac{v}{y}\right)$, so $ f\equiv 0$.
Else, $ f(y) \equal{}\minus{}f\left( u^{3}y\right) \equal{} f\left( u^{6}y\right) \equal{}\minus{}f\left( u^{9}y\right) \equal{}\ldots$.
As Kent Merryfield showed, that has limit zero, so that's all.

The only solution is $ f\equiv 0$.

PS: If I made any mistakes, you'll have to excuse me. It's really late here. :blush:


advanced fields
advanced fields unsolved

He,
I  have a problem with L-series and Dirichlet characters:

Let $ \chi$ be  a non-principal charcter modulo $ q$ with $ \chi(\minus{}1)\equal{}1$. Express  $ L(2,\chi)$  as a finite sum.

I have shown that $ L(2, \chi)\equal{}L(2,\chi^{'})\prod_{p|q} ((1\minus{}\chi^{'}(p)p^{\minus{}2})$ where $ \chi^{'}$ is a primitive
character that induces $ \chi$.

Now I have to show that $ L(2,\chi^{'})$ is a finite product/sum... so I can put things together and solve the problem.

Anyone?


calculus
integration
function
algebra open
algebra

Is there any (interesting) relation between two functions [tex]f[/tex] and [tex]g[/tex] such that [tex]f, g \in C^{\infty}[/tex] and [tex]f, g : R \rightarrow R[/tex] and:
[tex]\begin{eqnarray*}
  \int_{g ( a )}^{g ( b )} f ( x ) \tmop{dx} = \int_{f ( a )}^{f ( b )} g ( x
  ) \tmop{dx} &  &
\end{eqnarray*}[/tex]
Can we ''find'' [tex]f[/tex] and [tex]g[/tex] ?

Thanks !

:blush: I just realized I made an awfull mistake (the post has been edited)
By differentiating both terms i got the relation [tex]f(g(b))=g(f(b))[/tex] for any real [tex]b[/tex] (I suppose the [tex]b[/tex] in the statement is any real [tex]b[/tex]). That's all !


We have a triangle $ABC$ such that $AB=13$, $BC=14$, and $CA=15$. Let $M$ be the midpoint of $BC$, and let $N$ be the midpoint of $AB$. What is the value of $\frac{[AMN]}{[ABC]}$? Express your answer as a common fraction (in lowest terms).

[quote="anirudh"]what do you mean by $[]$? is it area? is it angle :maybe:[/quote] It's common notation for area. :D

Is there a common notation for volume?

[hide]$[AMN]=\frac12 [ABM]=\frac14 [ABC]$

$\boxed{\frac{1}{4}}$[/hide]

[quote="anirudh"]i have attached the sol[/quote]
Um, that isn't a sol, that is just a picture...


Let $ a,b,c>0$ . Prove that :

\[ \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{9abc}{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}\ge 7\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Prove that :
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\ge 7\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} v^2$ and $ abc \equal{} w^3.$
Hence, $ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\ge 7\Leftrightarrow$
$ \Leftrightarrow\frac {9uv^2 \minus{} 3w^3}{w^3} \plus{} \frac {9w^3}{9uv^2}\geq7\Leftrightarrow$
$ \Leftrightarrow 9u^2v^4 \minus{} 10uv^2w^3 \plus{} w^6\geq0\Leftrightarrow(uv^2\minus{}w^3)(9uv^2\minus{}w^3)\geq0,$ which is obvious.

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Prove that :
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\ge 7\]
[/quote]
Not really strong ...

Just use AM-GM's inequality :

$ LHS \equal{}\left( \frac {a \plus{} b}{2c} \plus{} \frac {b \plus{} c}{2a} \plus{} \frac {c \plus{} a}{2b} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \right)$ $ \plus{}\left( \frac {a \plus{} b}{2c} \plus{} \frac {b \plus{} c}{2a} \plus{} \frac {c \plus{} a}{2b} \right)$ $ \geq 4\sqrt [4]{\frac {9(a \plus{} b)(b \plus{} c)(c \plus{} a) }{8(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} } \plus{} 3 \geq 7$

[quote="arqady"]The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {81abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\ge 15\]
[/quote]
Rewrite the inequality as
$ \frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{abc}\plus{}\frac {81abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \ge 18$
Which is clearly true according to the AM-GM ineq for 18 numbers.

Nice proof, leedt26!
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers, $ 0\leq k\leq6\plus{}6\sqrt3\cos10^{\circ}.$ Prove that:
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b}\ge2\sqrt{9\plus{}\frac{k(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)}{ab\plus{}ac\plus{}bc}}\]
where $ k\equal{}6\plus{}6\sqrt3\cos10^{\circ}$ is the best value, for which this inequality is true for all positives $ a,$ $ b$ and $ c.$

The following is also true :

\[ \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\ge 7\]

with $ a,b,c> 0$

[quote="Pain rinnegan"]The following is also true :
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\ge 7\]
with $ a,b,c > 0$[/quote]

we have

$ \sum_{cyc}\frac{a\plus{}b}{c}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\minus{}6\equal{}\sum_{cyc}\left(\frac{a}{b}\plus{}\frac{b}{a}\minus{}2\right)\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2}{ab}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge\sum_{cyc}\frac{(a\minus{}b)^2}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$
$ \equal{}\frac{a^2\plus{}b^2\plus{}c^2\plus{}(a^2\plus{}b^2\plus{}c^2\minus{}ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \ge 1$

[quote="Pain rinnegan"]The following is also true :
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\ge 7\]
with $ a,b,c > 0$[/quote]
Here is stronger:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {4\sqrt2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}\ge 6\plus{}4\sqrt2\]

[quote="arqady"][quote="Pain rinnegan"]The following is also true :
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\ge 7\]
with $ a,b,c > 0$[/quote]
Here is stronger:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {4\sqrt2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}\ge 6 \plus{} 4\sqrt2\]
[/quote]

I think $ k\equal{}4\sqrt{2}$ is the best and it's easy by S.O.S  :lol:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that $ S_3$ is not the direct product of any family of its proper subgroups. The same is true of $ \mathbf{Z}_{p^n}$ (p prime, n $ \geq 1$) and $ \mathbb{Z}$. (Hungerford 1.8.1)

First, let me make sure I understand the question. They want me to prove that there is no isomorphism between any direct product of proper subgroups and the original group right?

Second, here is my attempt for $ S_3$. Let me know if I am on the right track. There are 6 elements in $ S_3$ so we need a subgroup of size two and a subgroup of size three. The only subgroups of size two are the transpositions and the only subgroups of size three are the derangements. So, WLOG any candidate for such a direct product would be isomorphic to $ < (2,1,3) > \times < (3,1,2) >$. However, this direct product has only one element of order two namely $ (2,1,3) \times (1,2,3)$ unlike $ S_3$ therefore it cannot be isomorphic.

Yes you understood the question.
Your argument is ok. Note that the group you obtained is $ \mathbb Z_6$ (as this one is cyclic and $ S_3$ is not, this gives another proof).
By the way, for the second part, you may want to show that no cyclic group is a product of proper subgroups (assuming that proper means: not the trivial group; if you only exclude the whole group, you would have $ \mathbb Z \cong (2 \mathbb Z) \times \{0\}$).

[quote="ZetaX"] you may want to show that no cyclic group is a product of proper subgroups.[/quote]

Wait, that's not true: $ Z_2 \times Z_3 \equal{} Z_6$.

for the $ \mathbb{Z}_{p^n}$ part, i would use the fundamental theorem of finite abelian groups, but this may be a bigger hammer than is needed, or has been handed to you yet.

It is easy to prove that if finite cyclic group is a direct product of its subgroups then orders of that subgroups are pairwise relatively prime.

For the third part - every nontrivial subgroup of $ \mathbb{Z}$ is isomorphic to $ \mathbb{Z}$, and for every $ k \equal{} 2,3,\ldots$, $ \ \mathbb{Z}$ is not isomorphic to $ \mathbb{Z}^{k}$, since $ \mathbb{Z}^{k}$ is not cyclic.


LaTeX
induction
graph theory
combinatorics solved
combinatorics

Let's color each of the edges of the complete graph with $n$ vertices either in red or blue.
Determine a close formula for the least number $f(n)$ of monochromatic triangles that appear in the figure.

Pierre.

Don't you use the classical theorem about the number of edges, I can't remember it now :(, but it's the formula for f(k) and f(n(n-1)/2 - k) minimum of them ?

???Sorry but I don't see what you are talking about...

Pierre.

hmm... what about $[\frac {n-3} 3]$? It is a lower bound, but probably not very close.

Far from best possible... :cool:

Pierre.

That's what I thought...the result has been found in 1959 :(
In aonther hand I've less than 50 years late, thus maybe with some extra efforts...

Pierre.

A much better lower bound is $n(n-1)(n-2)/120$. It is obtained just by the fact that in a $K_6$ there is always a monochromatic triangle. now that we may choose $n(n-1)(n-2)(n-3)(n-4)(n-5)/720$ (what is the Latex-command for a binomial coefficient?) 6-element subsets of our graph, we get that much triangles, and every triangle is counted at most $(n-3)(n-4)(n-5)/6$ times, so we get that lower bound. I don't know if it's the best one, but it looks really good I think.

Peter

Yes I can, since I proposed it  :D
I'll give a solution later, if no one solved it :(

Pierre.

Ok Pascual, since I've got a little time, I can give you my solution now. As always it won't be a short post so I will divide it into two or three parts to avoid computer bugs....

Let $A_1,...,A_n$ be the vertices.
For $i,j,k$ pairwise distinct in $ \{1,...,n \}$, let $p_i(j,k) = 2$ if the edges $A_iA_j$ and $A_iA_k$ have the same color, and $p_i(j,k)=-1$ otherwise.

Note that if the triangle $A_iA_jA_k$ is monochromatic then $p_i(j,k)+p_j(k,i)+p_k(i,j)=6$ and $p_i(j,k)+p_j(k,i)+p_k(i,j)=0$ otherwise.

Let $p_i= \sum p_i(j,k)$ (the sum is over $1 \leq j < k \leq n$ and $j,k \neq i$
and let $p= \sum_{i=1}^n p_i$.

Let $b$ (resp. $r$) denotes the number of blue (resp. red) triangles.
Since each pair of adjacent edges belongs to one and only one triangle, we then have :
$p=6(b+r)$. (1)

In another hand, for a given $i$, let $k$ be the number or red edges with endpoint $A_i$. Thus, there are $n-1-k$ blue edges with endpoint $A_i$. We then have :
${p_i = 2C_k^2 + 2C_{n-1-k}^2 - C_k^1C_{n-1-k}^1 = 3(k - \frac {n-1} 2} )^2 + \frac {(n-5)(n-1)} {4}$. (2)

First case : If $n=2a+2$ is even.
Then, from (2), we deduce that $p_i \geq \frac {3} {4} + \frac {(n-5)(n-1)} {4} = a(a-1)$ for each $i$.
Using (1), it follows that $b+r \geq \frac {1} {6} na(a-1) = \frac {a(a+1)(a-1)} {3}$.
Since both $f(2a+2)$ and $ \frac {a(a+1)(a-1)} {3}$ are integers, we have $f(2a+2) \geq \frac {a(a+1)(a-1)} {3}$.

Conversely, let's consider the complete graph whose edges $A_iA_j$ are colored red if $i=j$ mod[2], and blue otherwise.
It is easy to verify that this graph has exactly $2C_{a+1}^3$ monochromatic triangles, which provs that $f(2a+2) \leq \frac {a(a+1)(a-1)} {3}$.
Thus $f(2a+2) = \frac {a(a+1)(a-1)} {3}$ as claimed.

Pierre.

Now, let's turn to the second case, $n =2a+1$ is odd :

Using (2), we deduce that $p_i \geq \frac {(n-5)(n-1)} {4} = a(a-2)$ for each $i$.
Using (1), it follows that $b+r \geq \frac {1} {6} na(a-2) = \frac {a(2a+1)(a-2)} {6}.$
Since $f(2a+1)$ is an integer, we deduce that $f(2a+1) \geq [ \frac {a(2a+1)(a-2)} {6} ]$ where $[x]$ denotes the least integer greater than or equal to $x$ (and not the integer part....).
This is equivalent to
$f(4a+1) \geq \frac {2a(4a+1)(a-1)}{3}$ (which is an integer), and
$f(4a+3) \geq \frac {(2a+1)(4a+3)(2a-1)+3} {6}$ (because $\frac {(2a+1)(4a+3)(2a-1)} {6}$ is the half an odd integer).

Conversely :
- If $n = 4a+1$.
First, for $a=1$, we consider the coloring of the edges of the complete graph $K_5$ seen as all the line segments joining two of the vertices of a regular convex pentagon, where the sides of the pentagon are blues and the diagonals are red. In that case, there is no monochromatic triangle, so that $f(5)=0$.
Note that in this graph, from each vertex there are the same number of red edges than blue ones. Such a graph will be called a balanced graph.

Now, assume that for a given $a \geq 2$, we have a $K_{4(a-1)+1}$ balanced edge bicolored complete graph.
Let's give a complete graph $K_{4a+1}$. And color its edges as follows :
First we construct a balanced $K_{4(a-1)+1}$ for the subgraph whose vertices are $A_5,A_6,...,A_{4a+1}$, from the induction hypothesis.
Then, color the edges of the subgraph whose vertices are $A_1,A_2,...,A_5$ as in the regular pentagon above.
At last, for each $i \geq 6$, give color of $A_5A_i$ to $A_1A_i$ and to $A_2A_i$, and the opposite ne to $A_3A_i$ and $A_4A_i$.
It is easy to verify that such a bicoloring gives us a balanced $K_{4a+1}$.

For such a complete graph, for each $i$ the number of red edges from $A_i$ is equal to $k=2a$, from which we deduce (using (1)) that $p_i = \frac {(n-5)(n-1)} {4}$. It easily follows from above that $b+r= \frac {n(n-5)(n-1)} {4}$ which gives $f(4a+1) \leq \frac {(2a(4a+1)(a-1)} {3}$, and we are done.

- If $n=4a+3$ :
Let's color the edges of $K_{4a+3}$ as follows :
First color the edges of the subgraph $K_{4a+1}$ whose vertices are $A_3,...,A_{4a+3}$ to obtain a balanced graph as above.
Then for each $i \geq 4$ give the color $A_3A_i$ to $A_1A_i$ and the opposite one to $A_2A_i$.
Now give the color blue to $A_1A_2$ and red to $A_1A_3$ and $A_2A_3$.
It is easy to verify that for each $i \neq 3$, we have exactly $k=2a+1$ red edges from $A_i$, so that $p_i = \frac {(n-5)(n-1)} {4}$.
Moreover the number of red edges from $A_3$ is $2a$, so that $p_3 = 3+ \frac {(n-5)(n-1)} {4}$.
It follows that $p = 3 + \frac {n(n-5)(n-1)} {4}$ and $f(4a+3) \geq b+r = \frac
{(2a+1)(4a+3)(2a-1)} {6} + \frac {1} {2}$, and we are done.

Pierre.

I never said thanks.... :?

So it is time: Thanks Pierre for such a wonderful solution!

I think it is a theorem called Goodman.What else,I wonder what Pierre means in French,could you tell me? :blush:


Let be given a [u][b]acute-triangle[/b][/u] $ABC$. Prove that: $\left(\sum\cos A\right)^{2}\le\sum\sin^{2}A$

[hide]
Writing $\sin^{2}A=1-\cos^{2}A$ etc. and using the formula
\[\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1\]
it is equivalent to
\[2\cos A\cos B+2\cos B\cos C+2\cos C\cos A-4\cos A\cos B\cos C\leq 1\]
WLOG let $\angle A\leq \angle B\leq \angle C$, so that $\angle C\geq 60^\circ$ and $\cos C\leq \frac{1}{2}$. The left hand side of the above becomes
\[2\cos A\cos B(1-2\cos C)+2\cos C(\cos A+\cos B)\]
As $2\cos A\cos B=\cos (A+B)+\cos (A-B)=\cos (A-B)-\cos C\leq 1-\cos C$, it remains to prove that
\[(1-\cos C)(1-2\cos C)+2\cos C(\cos A+\cos B)\leq 1\]
But the left hand side equals $1+\cos C(2\cos A+2\cos B+2\cos C-3)$, and $\cos A+\cos B+\cos C\leq 3\cos \left( \frac{A+B+C}{2}\right) =\frac{3}{2}$ by Jensen.
[/hide]

We have: $2\cos B\cos C=\sqrt{\sin 2A\cot A\sin 2B\cot B}\le\frac{1}{2}(\sin 2A\cot B+\sin 2B\cot A)$
Hence:
\begin{eqnarray*}2\sum\cos B\cos C&\leq&\frac{1}{2}\sum\sin 2A\cot B+\sin 2B\cot A\\ &=&\frac{1}{2}\sum\cot A(\sin 2B+\sin 2C)\\ &=&\frac{1}{2}\sum\cot A.2\sin A\cos(B-C)\\ &=&-\frac{1}{2}\sum 2\cos(B+C)\cos(B-C)\\ &=&-\frac{1}{2}\sum\cos 2B+\cos 2C\\ &=&3-2\sum\cos^{2}A \end{eqnarray*}
\[\Longleftrightarrow\left(\sum\cos A\right)^{2}\le 3-\sum\cos^{2}A=\sum\sin^{2}A \]
[b]Remark.[/b] Use: $\sum\cos^{2}A=1-2\prod\cos A$ we have:
\[\boxed{\left(\sum\cos A\right)^{2}\le\sum\sin^{2}A\Longleftrightarrow\sum(\cos B+\cos C)^{2}\le 3\Longleftrightarrow 2\sum\cos B\cos C\le 1+4\prod\cos A}\]

[b] Remark 2.[/b] Use: $\sum\cos A=1+\frac{r}{R}$ we have:
\[\boxed{\left(\sum\cos A\right)^{2}\le\sum\sin^{2}A\Longleftrightarrow 4(R+r)^{2}\le a^{2}+b^{2}+c^{2}}\]

I've just found it at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14567]trig question[/url]. And I'm very interesting with [b]treegoner's solution[/b], therefore I'll rewrite [b]treegoner's solution[/b] in particular $m=n=p=1$ at here.
[b]Lemma 1.[/b] $2\cos A.\sin B.\sin C=\sin^{2}B+\sin^{2}C-\sin^{2}A$
[b]Lemma 2.[/b] $\sum\frac{\cos A}{\sin B\sin C}=2$
Come back with our problem:
We have:
\begin{eqnarray*}\left(\sum\cos A\right)^{2}&=&\left(\sum\sqrt{\cos A\sin B\sin C}\cdot\sqrt{\frac{\cos A}{\sin B\sin C}}\right)^{2}\\ &\leq& \left(\sum \cos A\sin B\sin C\right)\left(\sum\frac{\cos A}{\sin B\sin C}\right)\\ &=&\sum\sin^{2}B+\sin^{2}C-\sin^{2}A\\ &=&\sum\sin^{2}A \end{eqnarray*}


Prove that if $ a,b,c>0$ then

$ \sqrt{abc}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}(a\plus{}b\plus{}c)^{2}\geq 4\sqrt{3abc(a\plus{}b\plus{}c)}$

From Schur's inequality and AM-GM we get
\[ a^2\plus{}b^2\plus{}c^2\plus{}\sqrt{abc}(\sqrt a\plus{}\sqrt b\plus{}\sqrt c)\ge 2(ab\plus{}bc\plus{}ca)\]
Therefore it remains to show that
\[ ab\plus{}bc\plus{}ca\ge\sqrt{3abc(a\plus{}b\plus{}c)}\]
which is just AM-GM.


[quote="Dr. No"][quote="jimli"]x = y
x² = xy
2x²-2xy = x²-xy
2(x²-xy) = x²-xy
2 = 1[/quote]

Step 4, i.e. $2(x^2-xy)=x^2-xy$ implies division by zero.

It's not a puzzle, it's a mistake.[/quote]

I knew it is a mistake.  However, division by zero can be "done"(cough).

1/1=1/2+1/2^2+1/2^3...
1/2=1/3+1/3^2+1/3^3...
1/3=1/4+1/4^2+1/4^3...

1/0=1/1+1/1^2+1/1^3...= :inf:

:lol: You're proof basically proves that Infinity = Infinity.

[quote="shobber"]there is post in the High School--Other Problem Solving forum talking about $1=2$, there are many proofs. You may find it interesting. :D[/quote]

I take exception to the use of the word [i]proof[/i] in this context. The possibility of division by zero is itself founded upon a contradiction. Therefore, any result that follows from division by zero will be logically inconsistent by definition. Such so called proofs are not clever, they're just ignorant.

jimli,

I moved this from Classroom Math to the Basics as we try to leave Classroom Math for ordinary middle school curriculum while the Basics serves as the forum for learning the mechanics of math that eventually leads to advanced high school curriculum.

if x=y, then x 2 -xy is 0. dividing by 0 is not allowed


geometry
incenter
circumcircle
analytic geometry
trigonometry
geometry proposed

Let $ABC$ a triangle and his cirmumcircle. Let  $I$ the incenter of the triangle and let
$D,E,F$ the of intersection of the lines $AI,BI,CI$ with the circumcircle.

Prove that $\frac{AI}{ID}+\frac{BI}{IE}+\frac{CI}{IF}\geq 3$

$l_a$ - the length of the bisector of the angle $\widehat{A}\ ,\ X\in BC\cap AD\Longrightarrow$
$\frac{IA}{b+c}=\frac{IX}{a}=\frac{l_a}{2p}$. It is well-known the relation $l_a=\frac{2}{b+c}\sqrt{bcp(p-a)}$.
Thus, $IA=\frac{b+c}{2p}\cdot l_a\Longrightarrow IA^2=\frac{bc(p-a)}{p}=\frac{abc(p-a)}{ap}=\frac{4Rrp(p-a)}{ap}\Longrightarrow$
$IA^2=2Rr\cdot \frac{b+c-a}{a}\Longrightarrow \sum IA^2=2Rr\cdot \left(-3+\sum \frac{b+c}{a}\right)\ .$
$IA\cdot ID=IB\cdot IE=IC\cdot IF=2Rr\ ;\ \sum \frac{b+c}{a}\ge 6\ .$
$==================================$
$\sum \frac{IA}{ID}=\sum \frac{IA^2}{IA\cdot ID}=\frac{1}{2Rr}\cdot\sum IA^2=$
$-3+\sum \frac{b+c}{a}\ge -3+6=3\ .$Therefore, $\sum \frac{IA}{ID}\ge 3\ .$

[b]Remark.[/b] For a inner point $M$ denote the second intersections $D$, $E$, $F$
of the circumcircle with the lines $AM$, $BM$, $CM$ respectively. For $M: =O$,
is evidently $\sum \frac{OA}{OD}=3$; also, for $M: =G$, $\sum \frac{GA}{GD}=3$.
For any inner point $M$ is true the relation $\sum \frac{MA}{MD}\ge 3\ ?$
[u]Silouan, now you have a research them ![/u]
[hide="Here is a extension."]Denote the intersections $X\in AM\cap BC$, $Y\in BM\cap CA$, $Z\in MC\cap AB$. Let $ABC$ be a triangle and let $M$ be a inner point with
the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$, i.e. $x>0$, $y>0$, $z>0$, $x+y+z=1$ and $\frac{x}{[BMC]}=$ $\frac{y}{[CMA]}=$ $\frac{z}{[AMB]}$. Denote $p=a^2yz+b^2zx+c^2xy$. Then $MA\cdot MD=MB\cdot ME=MC\cdot MF=p$, $(y+z)^2AX^2=(b^2z+c^2y)(y+z)-a^2yz$, $MA=(y+z)AX$ $\Longrightarrow$ $AM^2=c^2y+b^2z-p$. Therefore, $\sum \frac{MA}{MD}=$ $\sum \frac{MA^2}{MA\cdot MD}=$ $\frac 1p\left[\sum (b^2z+c^2y)-3p\right]=$ $-3+\frac 1p\sum a^2(y+z)$ a.s.o.[/hide]

Nice solution mr Virgil .My solution was ungly with trigonometry but this is qute and small. Thanks

I have another solution:
We have $AI=4Rsin\frac{B}{2}sin\frac{c}{2}$.
$DI=BD=2Rsin\frac{A}{2}$.
Thus $\frac{AI}{DI}=\frac{2sin\frac{B}{2}sin\frac{C}{2}}{sin\frac{A}{2}}$ $=$ $2(\frac{1}{1-tg\frac{B}{2}tg\frac{C}{2}} -1)$.
It suffice to prove that $\sum\frac{1}{1-tg\frac{B}{2}tg\frac{C}{2}}-3 \geq  \frac{3}{2}$
$\leftrightarrow$ $\sum\frac{1}{1-tg\frac{B}{2}tg\frac{C}{2}} \geq  \frac{9}{2}$,which is true because $\sum tg\frac{B}{2}tg\frac{C}{2} =1$.

Moreover,we have $IA^2+IB^2+IC^2 \leq  ID^2+IE^2+IF^2$.


ceiling function
logarithms
floor function
number theory
least common multiple
number theory proposed

I'm finding difficult to understand the following (for reasons i will explain), which is part
of the "PRIMES is in P" paper by AKS,
[url]http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf[/url].
Maybe someone can help me pointing out where my misunderstanding is.

Lemma1: if $ n \geq 7$, $ lcm(1,2,\ldots,n)\geq 2^n$
Define $ o_r(n)$ to be the order of $ n$ in $ (\mathbb{Z}/r\mathbb{Z})^*$, for $ (n,r)=1$.
Claim: Suppose n>2. Then there exist an $ r\leq \lceil \log_2^5 n \rceil$ s.t. $ o_r(n)>\log_2^2n$
Proof: Let $ r_1,\ldots,r_t$ be all numbers s.t. either $ o_{r_i}(n)\leq \log_2^2 n$ or $ r_i|n$. Then each $ r_i$
divide the product
$ n\cdot \prod_{i=1}^{\lfloor log_2^2 n\rfloor}(n^i-1)<n^{\log_2^4 n}\leq 2^{\log_2^5 n}$.
By the Lemma1 the lcm of first $ \lceil log_2^5 n\rceil$ numbers is at least $ 2^{\lceil
log_2^5 n\rceil}$ therefore there must exist a number $ s\leq\lceil \log_2^5 n\rceil$ s.t.
$ s \not\in\{r_1, \ldots, r_t\}$. If $ (s, n)=1$ we are done.
[b] If $ (s, n)>1$, then since $ s$ does not divide $ n$ and $ (s,n)\in \{r_1, \ldots, r_t\}$, $ r=\frac{s}{(s,n)}\not\in \{r_1,\ldots,r_t\}$ and so $ o_r(n)>
\log_2^2 n$[/b]

My problem is on [b]bolded[/b] text. Consider $ n=6$ and $ s=4$. We have $ s\leq\lceil \log_2^5 n\rceil$
and $ s\not | n$, and $ o_r(n)$ not defined as $ (s,n)>1$, hence $ s \not\in\{r_1, \ldots, r_t\}$.
But ${ r=\frac{s}{(s,n)}=2 | n}$, so it is not possible that $ o_r(n)>log_2^2 n$, nor $ r\not\in\{r_1, \ldots, r_t\}$. ?!?

I think you are right, the proof seems to be wrong...  :huh:

However, it can be fixed, by choosing $ r$ that does not divide
\[ P \equiv n^2\cdot \prod_{i \equal{} 1}^{\lfloor \log_2^2 n\rfloor}(n^i \minus{} 1).
\]
Then the original argument works. Assume that $ r \equal{} s/(s,n) \in \{r_1,\dots,r_t\}$. Then either (i) $ r$ divides $ n$ or (ii) $ r$ is coprime to $ n$. In the first case, $ s$ divides $ n^2$; in the second case, $ r$ divides $ \prod_{i \equal{} 1}^{\lfloor \log_2^2 n\rfloor}(n^i \minus{} 1)$, and so $ s$ divides $ P$. We get a contradiction.

Update: in fact, what this proves is that $ r \equal{} s/(s,n)$ is not in $ \{r_1,\dots, r_t\}$. But $ r$ is not necessarily coprime to $ n$.  :(

[quote]In the first case,  $ s$ divides $ n^2$ ;[/quote]

Uhm, can you please explain this a little more ? Is the claim:
$ r\in \{r_1, \ldots, r_t\}\wedge (r,n) > 1 \Rightarrow r \equal{} \frac {s}{(s, n)}|n^2 \Rightarrow r|n \Rightarrow s|n^2$ ?
I suspect the $ 2$ in $ n^2$ is not enough... :(

Anyway, thank you for your reply  :)

Yes!!!!! I got it! They've fixed it in this new version, I've just found on agrawal homepage:
[url]http://www.cse.iitk.ac.in/~manindra/algebra/primality_v6.pdf[/url]

Just replace the "big product", $ P$, by
$ n^{\lfloor \log_2 \lceil \log_2^5 n\rceil\rfloor}\cdot\prod_{i \equal{} 1}^{\lfloor \log_2^2 n\rfloor}(n^i \minus{} 1)$

So the journal version has definitely this minor mistake... good to know  :D
Anyway thank you Yuri for having pointed out that we must somehow take a bigger power of $ n$.


Bueno, llego la hora de las apuestas!, esta comunidad es supuestamente iberoamericana, entonces pues me gustaria que en este tema apostemos los resultados de las olimpiadas venideras, Centro, Imo e Ibero, en cuanto aa pais iberoamericano que ganara. Y tambien pues para que vallamos posteando los equipos que se vallan escogiendo:

Mis predicciones son:
Centro: Mexico
IMO: Colombia
Ibero: Colombia

En cuanto al equipo de colombia les cuento que ya empezo la olimpiada colombiana, de la gente que ya ha viajado este año parece repetiran 3 ( Jose Gabriel Acevedo (imo2004), Esteban Gonzales (Imo 2003, ibero 2004), Y yo (imo 2003, 2004, ibero 2003). De resto parece ser gente nueva, entonces no creo que el equipo este tan fuerte como el año pasado que iban 4 ex imos.

En cuanto a otros paises, supongo Manuel Y Leonardo van a ir...(leo Chavez=chang?), pues hay un argentino nuevo en el foro con el nombre Severius o algo asi, y creo que esta trabajando bien, supongo que tambien ira, se tambien que Sherry (como se escriba) parece que va a participar con los EEUU, de resto ni idea.

Otra cosa es que creo que venezuela este año va a tener un equipo mas fuerte por lo visto en el entrenamiento de enero aqui en colombia donde participaron, y tambien parece que muchos paises de centroamerica van a ir a imo, que bien!

Espero pongan sus predicciones y vallamos discutiendo los equipos y todas estas cosas!

Suerte

I also noticed new strong member Severius. :) I am very happy, because my IMO was held in Argentina and I like this country very much :)

I think Colombia will have at least one gold medal ;)

Good luck to all participants!

Bueno lo lamento Pascual pero la ibero este año la va a ganar Argentina!! a pesar de ke sean locales, si si si... con respecto a la IMO weno, este año no se te tiene ke escapar el oro eh!
 Centroamericana no tengo idea kienes participan ni nada, pero supongo ke debe ser algo parecido a la del Cono Sur...

btw talking about Severius, i have to say he surprised everyone here (and hope he still doing that!) because he first participated in the Argentina National Olympiad in 2004, and he won in his level, then got a silver medal at the Rioplatense Math Olympiad  :lol:

Asi ke supongo ke el ira a la IMO, acompañado de Gabriel Carvajal (imo2004,ibero2004), Maxi Camporino(Cono sur 2004), Ignacio Rossi(imo2004), y weno la verdad no estoy muy seguro de los demas, es muy incierto el selectivo de estte año...

 Lo bueno de ke la IMO sea en Mexico este año, es ke van a invitar seguramente a muchos paises de latinoamerica ke usualmente no van a la IMO...
 lo malo es ke yo ya no puedo ir :S

Suerte a todos!

RaMlaF...

Pascual le tienes a ganar a sherry por mas de 5 puntos pa bajarle el guille! (asi se dice alla en colombia jaj?)

Jaja, Ramiro: yo crei que usted iba! ademas la ibero se queda en casa, esperamos llevar este año un equipo IMO a la Ibero, jaja, aunque nada es oficial, vamos a ver si por lo menos me dejan ir a mi!

Manuel: jaja, claro, ademas esa vieja ya no va a representar su pais!, supongo que toco ganarle, no?

Resto de gente: hagan sus pollas!

Myth: I see you know some spanish! Great! :D  :D

[quote="Pascual2005"]Jaja, Ramiro: yo crei que usted iba! cagada que nO!, ademas la ibero se queda en casa, esperamos llevar este año un equipo IMO a la Ibero, jaja, aunque nada es oficial, vamos a ver si por lo menos me dejan ir a mi!
[/quote]

jaja weno.. no es ke sea tan viejo (17), pero termine el colegio en diciembre y no me dejan ir al selectivo  :(  ...  desde ya ke de todas formas los ke vayan van a tener ke traerme un tekila de regalo  :D

si es que no no lo tomamos todos! jaja!

Djimenez: han tratado esos paises de buscar patrocinio con las empresas locales? a nosotros nos ha ayudado!

I think that brazilian time is the best.

[quote="Pascual2005"]Djimenez: han tratado esos paises de buscar patrocinio con las empresas locales? a nosotros nos ha ayudado![/quote]

Lo ignoro. A quienes conozco trabajan mas que todo en la parte de entrenamiento, no tanto en la parte administrativa.

jaja manuel no sea despectivo orale! jaja! solo esperemos que no nos partan como siempre haciendo problemas!

Hey, kiubo!, como andan todos. De los equipos de Venezuela para las olimpiadas venideras solo tengo informacion de la IMO,  al parecer los candidatos mas fuertes somos Roland Hablutzel (Ibero 2004), Andres Guzman (Ibero 2004), y yo (Ibero 2003-IMO 2004). Al parecer puede que Rodrigo (IMO 2003-IMO 2004) no participe este año, pero nada se sabe aun. Ya comenzaron las clases, yo  me incorporare pronto, porque ahora ando en US

Por cierto, quien es chavez????? no es chang o si??

Are you also going to Clay Academy, Leonardo?

No Myth, he is not going, he is living in the USA right now, because his father is there i think, and he is going back to Venezuela soon.

Leo, le mande un PM a Chavez porque me imagino que con ese nick seguramente es Chang (por lo que he oido, no?), sin embargo no ha respondido! suerte

[quote="Dave Vather"]Mis predicciones:
IMO: México
Centro: México
Ibero: México

con suerte, los veo en la IMO[/quote]

Dejame adivinar... Sos de Mexico... Yujuuu, pegue! (Digo, la verdad es que tanta seguridad, pues sorprende). Entre los paises latinoamericanos que van a la IMO, pues yo apostaria hacia Argentina, Brasil o Colombia antes que a Mexico. Con referencia a la Ibero, pues igual, apostaria a Argentina, Brasil o Mexico antes que a Mexico. Con respecto a la Centroamericana, pues talvez Colombia, talvez si, Mexico!!

Si los mexicanos no apostáramos por México, ¿Quién lo haría?
Creo que si hay un año para ser el primer país Iberoamericano en la IMO debería ser este. La competencia por llegar a la IMO está más reñida que en años pasados. La Ibero todavía está lejos pero creo que la podemos ganar, y la centro la llevamos ganando por 3 años seguidos, la pasada por más de 20 puntos y los tres primeros lugares. No la podemos perder.
Si consigo ir a la IMO estoy dispuesto a apostar (al menos, por mí)

[quote="Dave Vather"]Si consigo ir a la IMO estoy dispuesto a apostar (al menos, por mí)[/quote]

Buena Suerte!!

[quote="marko avila"]jajajajajaja pascual q malas predicciones¡¡¡¡ mexiko por todos lados será kien gane. yo konozco a los integrantes del equipo tenemos a varios repetidores. y eso q en los años próximos años el equipo mexikano quedaré mucho mejor xk tendrá a integrantes komo yo. todavía no me konocen a nivel nacional jajaja aunque  he ganado 2 bronces en los 2 nacionales q  he ido peo las kosas este año jejejeje seran mucho mejor tenemos en mi estado a varios integrantes del equipo q  son mui buenos jejejeje nimodo pascual pa la proxima¡¡¡¡[/quote]

Espero no sonar mala gente o grocero (aunque probablemente lo suene) pero si querés hacer cosas bien en Olimpiadas, por qué no empezás con tu ortografía y tu redacción. Tanto aquí como en [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=200957#p200957]este otro post[/url] dejan muchisimo que desear.

Por otro lado, dependiendo de la edad que tengás, dos bronces en una olimpiada nacional no son un tan buen augurio, en especial por que de ir a la IMO o a la Ibero, te encontrarás con gente que gano oros en sus respectivas olimpiadas nacionales durante cuatro o cinco anios consecutivos.

Dicen que una persona debe dejar que sus hechos hablen por él y no él por sus hechos, en especial si estos aun no han ocurrido!

Jaja parece que los de mexico se tienen mucha fe, pero no se acuerdan quien los voltio en atenas :D  :D  :D !! jaja mentiras, aun asi marko, lo que dice david es verdad, lo que uno halla hecho en olimpiadas no le asegura una futura buena marca en imo...por ejemplo...yo nunca gane la olimpaida nacional colombiana pero si fui el mejor en imo, y el que gano la olimpiada no clasifico al equipo ;)  ;)  De todas maneras la confianza es buena y me alegra esten tan motivados para las futuras olimpiadas...Dave, enserio creo que este año les vamos a ganar, usted dice que apuesta por usted...en ese caso yo apuesto por mi...jaja solo por seguirle la corriente pues la verdad la imo no lo es todo...hay objetivos mas grandes...aunque tengo que reconocer en la imo se pasa muy rico! :D  :D

El topic ha sido partido, solamente he movido los posts que estaban mas o menos fuera de lugar en este tema y las he colocado en otro. Sigamos haciendo predicciones sobre las distintas olimpiadas del 2005. :D   :P

He borrado lo que escribi, para no generar otra discucion en este foro :)

Ahhh!!!

Olvide dejar aqui una indicacion. Toda la discusion sobre ortografia fue puesta en un post aparte. Lo que pasa es que parece que mientras yo hice esto y Pascual le dio submit a su mensaje solo unos segundos despues. Estoy tratando de ver como pasar un solo mensaje a otro post.

Bueno, sigamos aqui solo con olimpiadas.

[quote="djimenez"][quote="RaMlaF"] Lo bueno de ke la IMO sea en Mexico este año, es ke van a invitar seguramente a muchos paises de latinoamerica ke usualmente no van a la IMO...
[/quote]

Lo malo es que segun parece, todos los paises centroamericanos que van este anio, no van a ir en los anios posteriores, o al menos eso es lo que he escuchado de varios de los entrenadores. El asunto es que, como en todo lado, para estos eventos escatiman la plata, y pues, ni modo, van este anio por ser en Mexico y quedar bien cerca, pero el otro, no les dan plata ni para los pasajes. Que triste la historia del tercer mundo  :(

Bueno, nos vemos,[/quote]

Hey!!!!! Eso no suena bien, de hecho, en teoría, una vez que hemos sido invitados, podemos seguir participando en la IMO  :D ademas, yo opino que si ha habido dinero (que no sé de dónde lo sacaron) para mandar delegación a Argentina en el 2003, también les alcanzará para mandar a Eslovenia el otro año,  pues creo Argentina esta igual de lejos que Eslovenia de aquí jajajaja  :)

Pues mi amigo, te recomiendo que revises tu mapa, por que estas bastante equivocado, puesto que El Salvador y Argentina estan a algo asi como 5500 kilometros de distancia, mientras que Eslovenia esta a mas de 10000. Y por otro lado, a Argentina se voló en Setiembre, temporada baja, y por el continente, que es muchisimo mas barato. Haciendo una busqueda por algunas agencias de viajes por Internet, mandar a un grupo de 6 personas (4 participantes + 1 tutor + 1 delegado) de San Salvador a Buenos Aires, costaria, en setiembre, al rededor de $950 por cabeza, osea, al rededor de $5700 por todos. Ahora, mandar a 8 personas a Eslovenia, pues hay que hacer un vuelo interatlantico, que por lo general no son baratos, ademas, en temporada alta. Haciendo la misma cuenta, pues el viaje de San Salvador, El Salvador a Ljubljana, Slovenia saldria al rededor de $2850 por cabeza, y dado que el equipo puede ser de 6, mas el tutor y el delegado, pues estamos hablando de ocho tiquetes, pues es casi $23000 por todos.  No mi amigo, no es lo mismo mandar a un equipo a la Ibero en Argentina, que a la IMO en Slovenia.

:(  :(  :(  qué triste...................... bueno, con un poco de suerte quizá podamos participar........... en fin, prefiero dejar que el destino siga su curso....................

Por cierto, mi nombre es Gabriel, y como puedes ver (cómo no, no se mira la banderita :mad:) soy de El Salvador (si te preguntas cómo es que llegué hasta aquí, es una historia muy larga y triste, jajajajajajaja  :) )

Gabriel lo unico que necesitamos los optimistas son ganas, nada de cuentas! asi que espero verlos en emxico y en las olimpiadas venideras

Pues si, lo que hace falta es solamente ganas de hacer mate. Lastimosamente muchas veces los paises no participan en este tipo de actividades por que el boleto es muy caro, aun cuando todo lo demas es gratis, pero pueden mandar a un equipo completo de Futbol al otro lado del mundo en primera clase y pagarles los mejores hoteles... a veces eso me da rabia!! :mad:

Por cierto, bienvenido Gabriel. Si queres hacer una presentacion algo mas extensa, pues podes ir a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30603]este thread[/url] y contarnos alli todo lo que queras de vos.

Por cierto, lo de la banderita lo podemos arreglar... Tenes alguna imagen de la bandera de el Salvador en las dimensiones apropiadas (usualmente son 32x20 pixeles, pero mientras tenga ese radio de 8/5 entre el largo y el ancho, se puede cambiar las dimensiones). Si la tenes, mandamela en un PM y trato de que la cambien!!

Ciao,

Bueno, todos sabemos que la inversion que se tiene que hacer para mandar un equipo a una olimpiada por lo general no es barata. Puede pasarle como a Venezuela, que comenzo participando en la IMO de 1981 (un momento de auge economico aca), pero mandaban en general gente que no estaba bien preparada y por lo general salian mal. Para el 83 (un golpe muy fuerte en la economia venezolana, comienzo de una crisis que se ha sentido hasta ahora  :( ), dejamos de participar, y no fue sino hasta 1998 (creo) que volvimos a ir. Desde entonces, han habido años en donde por la falta de dinero se manda solo 1 participante....cosa que es medio triste, pero a la vez, es de cierto modo una manera de mantenernos llendo. Me imagino que eso lo podrian hacer los paises latinoamericanos que van por primera vez a esta IMO, y que les parece que mandar un equipo completo a Slovenia es demasiado caro. Seria interesante ver mas de nuestra gente en olimpiadas del mundo :)

Por cierto, todavia no tengo noticias de la seleccion del equipo de mi pais, pero tengo noticias del de Argentina. Por si a alguien le interesa

Equipo Argentino - 46° Olimpíada Matemática Internacional  - México
Gabriel Carvajal - Zárate - Buenos Aires
Germán Gieczewski - San Isidro - Buenos Aires
Maximiliano Camporino - Cdad.  de Buenos Aires
Nicolás Romero - Posadas -Misiones
Ignacio Rossi - Garin - Buenos Aires
Tomás Ibarlucia - Tandil - Buenos Aires

Alguien mas ya sabe quienes van de su pais?

Hola a todos, soy Lucas Andisco (IMO 2003, IMO 2004, Ibero 2004), para los que no saben quién es.... Germán Gieczewski=Severius (el apodo se lo puse yo!, jaja, ahora nadie sabe que se llama Germán, todo el mundo le dice Severius, jaja). Gaby y Nacho fueron a la IMO 2004 y Gabriel también fue a la Ibero 2004. Maxi y Nacho fueron a la Cono 2004. Tomás había quedado suplente para el equipo Imo del año anterior y Nico para el de Cono, creo.
Seguro que alguien ya lo dijo antes, el equipo de España se conoce, de los que me acuerdo están Elisa (IMO e Ibero 2004) y Teixido (IMO 2004).
Creo que el de Brasil ya debe estar también, pero no estoy seguro. El que podía participar es Gaby Boujokas, que le fue muuuuy bien en IMO 2004, no sacó oro por el problema 1, y tal vez Fabio Diaz Moreira, que (no estoy seguro) por lo que dijeron andaba con ganas de hacer un año más de olimpíadas.
¿Cómo le va Pascual? Cuidado usted con proposiciones extrañas para las damas de esta IMO.... Ahhh, ¿vieron que la linda suizita también va a la IMO?
Pascual y Leo pónganse las pilas para traer excelentes medallas en esta IMO... pero para la ibero guarda eh, el equipo de Argentina pinta bien. Bueno, un saludo a todos.

Lucas, como sabe que la suizita va??? jaja....

¿Por qué casi todos los que van de Argentina son de Buenos Aires?
¿Está muy centralizada (la olimpiada) allá o es simplemente que los de Buenos Aires son mejores?.
Por cierto México hace estas dos semanas sus últimos 5 exámenes que deciden quien va y quien no. Les informo yo antes que nadie

Pascual, los que estamos en la 'hermandad del salamitrosky' tenemos 'contactos', jajaja
En la prueba no se descuide con albanitas y esas cosas, usted ya tiene experiencia amigo, jeje.
Con respecto a lo de Bs As, hay muchos factores, sin embargo el principal, para la olimpíada y las demás cosas, es que la provincia de Bs As es por muuuuuuucho la provincia con más habitantes del país. Después hay otros motivos, pero ése es en líneas grales el más importante.

Uy como asi, es que ya montaron hermandad y todo?

Fresco que este año ya me volvi muy juicioso con las muejres :)

México tiene ya a 5/6 partes de su selección:
Pablo Soberón de Morelos (6a OMCC)
Joshua Hernández de Coahuila (5a OMCC)
Guevara Guevara (si, su nombre es igual a su apeído) de Zacatecas
Isaac Buenrostro de Jalisco (6a OMCC)
David Torres de Guanajuato (yo  :lol: ) (6a OMCC)
y el sexto es alguno de:
Mario Huicochea (o algo por el estilo) del DF
Daniel García (45 IMO) de Chihuahua
Gonzalo Montealván (45 IMO y 5a OMCC) de Puebla.
Por cierto, los 5 que estamos seguros todos repetimos y respecto a lo de Buenos Aires,
el estado más poblado de México es el estado de México (no confundir con DF) y no
es muy bueno que digamos.

a toda la comunidad olimpica .... un gran saludo desde tierras mexicanas y aun mas especifico desde merida....... ciudad donde ene stos dias se lleva a cabo la 46 IMO .... es para  mi un placer estar con ustedes amantes de la matematica y del razonamiento para resolver problemas ... porque ... ¿ que otra cosa es las matematicas sino resolver problemas?....... .....
reciban desde aqui felicitaciones por participar o ganar medallas para sus respectivos paises........ felicidades para los mexicanos.... que ni moda tendremos que esperar otro año para ganar la primer medalla de oro en estas competencias.......... a prepararse para eslovenia.........
aun asi felicidades por ser los mejores del pais.......... listod para campeche?...... si alguien sabe quien repite aparte del harry potter joshua ( quien estuvo en ixtapa , sabe a que me refiero)........
seguimos en contacto por este fabuloso foro. y a prepararse para las otras olimpiadas

En primer lugar , felicitar a la selección colombiana que ha mostrando un buen nivel en esta 41a IMO - Mexico, y quisiera que los miembros de esta comunidad, comparta y comente los resultados de los equipso de sus paises... en vista de las proximas ediciones...

Pascual... podrias decirme cual fue el equipo que represento a colombia este año?..
segun la codificacion que se maneja en el certamen (COL1, COL2, ETC.)

Bueno .. y hablando de Peru...  ha sido la primera vez que se ha enviado un equipo completo en los ultimos 5 años ......
Que pena que ....se nos escaparon dos platas este año..   :|

Sim embargo... hay muchas esperanzas para las proximas ediciones puesto que este equipo es relativamente joven y varios de los chicos tienes para dos o tres años más en estas competiciones...

Salu2.

Carlos Bravo...   :)
Lima - Peru

Bueno al parecer Pascual va ganando las apuestas, sólo falta ver como queda la ibero, yo también le apuesto a Colombia aunque no se sabe todavía el equipo. El equipo de Colombia para la imo este año fue:

COL1  José Gabriel Acevedo (centro 03, imo 04) (Yo!)
COL2  Daniel Blandón (centro 04)
COL3  Esteban Gonzalez (centro 02, ibero 02, imo 03, ibero 04)
COL4  Miguel Moreno
COL5  David Quinche (centro 04)
COL6  Pascual Restrepo (imo 03, ibero 03, imo 04)

felicidades a todos los colombianos y a los demas latinos!!

por si acaso yo soy manuel de puerto rico imo 2004 pero no fui a esta imo, pero seguramente imo 2006!!!

[quote="matemario"]
aun asi felicidades por ser los mejores del pais.......... listod para campeche?...... si alguien sabe quien repite aparte del harry potter joshua ( quien estuvo en ixtapa , sabe a que me refiero)........
seguimos en contacto por este fabuloso foro. y a prepararse para las otras olimpiadas[/quote]
Pues de hecho es más fácil decir quién no repite. Unicamente Washa (Héctor) de Chihuahua no repite. Los demás podemos repetir.
Por cierto, parece que Colombia y México vamos empatados pues México ganó la centro pero Colombia la IMO (aunque claro cuenta más haber ganado la IMO  :mad: ) Pero espero que en la Ibero nos desquitemos. Por cierto, felicidades a Pascual por su plata, a los peruanos (se veían chiquitos pero todos me ganaron  :( ) y a todos los latinos que fueron a la IMO

he he Gracias y por supuesto estare en cartagena defendiendo mi territorio!  :D


Post Youtube video on this thread, but please, do not put pointless ones, and try to put funny videos, and don't put ones that will probably get this thread locked.

This Italian band Rhapsody of Fire's hilariously terrible music video for the song "Unholy Warcry"
[i]Playing time is 10:32... watch it all or don't watch it. Also, let it time to load first, pauses kill it[/i]

[youtube]ZYpxQEPzpJk[/youtube]

More Weird Al.

[youtube]uKrCAZsJsAI[/youtube]

[youtube]ILHlrx_oRfI[/youtube]

There are more on YouTube...


modular arithmetic
number theory proposed
number theory

hello!
prove that not exist $ (a,b,c)\in{IN^{3}}$ / $ a^{2} + b^{2} + c^{2}\equiv7[8]$.

any square is $ 0$ or $ 1$ or $ 4$ modulo $ 8$, therefore the sume of three squares is never $ 7$ modulo $ 8$.

Pierre.

Squares are equal to $ 0,1,4\pmod{8}$ so looking at the different cases you can conclude the claim
In fact it is easy to see than the numbers of the form $ 4k$, when squared are of the form $ 8l$, the number of the form $ 4k \plus{} 2$ are of the form $ 8l \plus{} 4$ and finally any odd number squared:
$ (2x \plus{} 1)^2 \equal{} 4x^2 \plus{} 4x \plus{} 1 \equal{} 4(x(x \plus{} 1) \plus{} 1$ The first summand is clearly of the form $ 8l$ so the odd numbers are of the form $ 8l \plus{} 1$ when squared

Daniel

EDIT:Pierre was faster  :D


prove that if the equation \[ x^3  \plus{} ax{}^2 \plus{} bx \plus{} c \equal{} 0
\]
has three distinet real roots, then so is
\[ x^3  \plus{} ax{}^2 \plus{} ( \minus{} a^2  \plus{} 4b)x \plus{} a^3  \minus{} 4ab \plus{} 8c \equal{} 0
\]


If 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?
[hide="Answer"]
2 kilograms[/hide]

[quote="ZhangPeijin"][quote="DonkeyKong"]If 74 hens lay 74 dozen eggs in 73 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?
[hide="Answer"]
2 kilograms[/hide][/quote]
I'll assume you meant 74 days because it doesn't work out that way.
1 hen eats 1 kg of wheat in 37 days.
1 hen lays 1 dozen eggs in 74 day.
1 hen eats 2 kg of wheat in 74 days.[/quote]
I am sorry for the typo. I couldn't see the simple logical explaination behind this problem. Thanks.

by your logic:

37 hens => 37kg in 37 days

1 hen => 1kg in 1 day

1 hen => 37 kg in 37 days

37 hens => 37*37kg in 37 days

lol...


Find the area of the isosceles trapezoid in square feet. Express your answer in simplest radical form.
[asy]draw((0,0)--(18,0)--(15,6.3)--(3,6.3)--cycle);
label("7'",(1.5,3.15),WNW);
label("12'",(9,6.3),N);
label("18'",(9,0),S);[/asy]

we know that these are isoceles so we know can drop 12

and the "remaining side is 6 so we divide that by 2 and get 3 on each side

hypotenuse theorem: a^2+b^2=c^2

we get 2sqrt 10 for the height

solving for area =(A1+A2)HEIGHT/2

we get the answer of 30sqrt10

P.S. can someone tell me how to write exponents like 5^2=25

if you can thank you so much!!


Find all primes p for which (2^(p-1)-1)/p is a square

That has been solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=1055 .

  darij

As I know it's for Iran Olympiad . :huh:  ;)


Let x,y,z is positive integer number, Find all x,y,z
\[ xyz \equal{} 4(x \plus{} y \plus{} z)
\]


I'm given 2 matrices, 4*4 and 4*2.  If I'm asked to find whether the span of A equals the span of B, could I just create one 4*6 matrix and solve for linear independence?

Sorry if this is vague.  If needed, I'll put an example into mathtype and post it.

What exactly do you mean by the span of $ A?$ That's the part that is unclear about your question.

One possible meaning: consider $ A$ as the operator $ x\mapsto Ax,$ and ask about its range. This is the same as the column space of $ A$ or span of the columns of $ A.$ It's fairly likely that this is what you're talking about.

Now, what would you do with that one $ 4\times 6$ matrix? Its column space is the sum (as subspaces) of the column spaces of $ A$ and $ B$. If those two are equal, then the column space of the combined matrix is exactly the same.

This will work. We do three row reductions, just for the purpose of counting pivots (that is, determining rank). If $ \text{rank}(A)\equal{}\text{rank}(B)\equal{}\text{rank}[A\mid B],$ then $ \text{col}(A)\equal{}\text{col}(B).$

Thanks for the quick response.

I was given 2 matrices, A and B, and asked if Span[A] = Span[B].  I think I see what you mean, even though a lot of this stuff is still confusing to me.[/code]


Tha points $ A_1$,$ B_1$,$ C_1$ are chosen on the sides $ AB$,$ BC$,$ CA$ of the triangle $ ABC$ s.t the lines $ AA_1$,$ BB_1$ and $ CC_1$ are concurrent at the point $ D$.The lines $ A_1C_1$ and $ BB_1$ are intersecting at the point $ E$.Prove that if $ BD\equal{}2B_1D$ then $ BE\equal{}B_1E$.

the problem is a particolar case of this:

Lemma: Let a triangle ABC, a line $ r$ and a point $ P \in r$, call $ E: BP \cap AC$, $ F: CP \cap AB$ and $ D: AP \cap EF$. So the locus of the point D when P move on the line $ r$ is a line.

Sorry, but how to prove your lemma? :wallbash_red:

Call $ Q: PA \cap BC$, so you can use that D is the harmonic cojugate of Q w.r.t. AP  :wink:

Are there any other solutions,please help! :(


This is what I got: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115798

I was just about to post it now.

Big picture
[img]
http://i96.photobucket.com/albums/l167/adasarathy/edit.jpg[/img]

Zoom of problem:

[img]http://i96.photobucket.com/albums/l167/adasarathy/editbig.jpg[/img]

This is a problem.

I can't edit the above post, so I will post again.

ERROR
[img]http://i96.photobucket.com/albums/l167/adasarathy/edit.jpg[/img]

I can't edit anything now!!! :o

I get the selfsame problem. Wonder if it was somebody hacking?

[quote="ZetaX"]It's possible that you already realised or are on fixing it, but:

Farenhajt today told me that he wasn't allowed to edit one of his posts.
I first checked in Number Theory on some posts and had no problem. By that I thought it's a temporary bug only hitting him.
But I tried in the test forum now, and I've the same problem (I don't want to try through all forums ;) ).

In fact, I think it affects everyone not being a moderator of that special forum.[/quote]
Wow, the same thing is happening to me (anywhere except MathCounts, where I am a moderator). I can't edit any of my posts at all!  :o

But this is likely (and hopefully) just a temporary setting the admins have turned on for some reason (either that or a really weird bug  :wink: ).

Should be fixed now: time for editing 24 hours.

It doesn't work. My latest post (excluding this one) is 4 hours old and I still can't access it.

Same here, even for newly posted posts (but I can delete posts everywhere).

EXACTLY THE SAME HERE!  :(

Won't let me edit, says I can't edit my posts anymore. oh well

Hm, I can't edit that post above either.

Now it's finally fixed :) Small undefined variable, computed the ellapsed time starting 1970 :D Obviously all posts were more than 1 day old :P

[quote="Valentin Vornicu"]No bug guys. Editing is now limited to $x$ hours ([i]the amount is open for debate[/i] - I think 24 hours is enough). After that only moderators can edit the posts in the specific forums (ie you have to be a moderator of the forum where the post lies so that you can edit it).

This is to stop people from going back and deleting their posts using edit method (people have done and are doing that for different reasons).

[POST EDIT :D] Okay there's a bug in the code :) There should be 24 hours available for free editing (after that you can ask the moderator of the forum if it's a really important edit, or post a new post :) ).[/quote]

But why do we need this?

It seems to be working properly now. I can edit a post from 16 hours ago, but not one from 30 hours ago.

I agree with 24 hours as the time limit. After that, a new post is usually a better idea.

[quote="Valentin Vornicu"]I said why we need it right in the middle of the post. Read it carefully ;)[/quote]

Yes, I just did.  :( So, if we were going to keep a scoreboard in a competition, we can't edit it though.  :(

EDIT: Okay. I'll just post a new list.   :)

[quote="anirudh"][quote="Valentin Vornicu"]I said why we need it right in the middle of the post. Read it carefully ;)[/quote]

Yes, I just did.  :( So, if we were going to keep a scoreboard in a competition, we can't edit it though.  :([/quote]
ok, now read the last part of that post


I'm in a high school honors algebra class & have a question about 2 problems I am stuck on! This stuff's really really hard. Any help would be appreciated!

z-7/z-8 = 1/z-8. Solve for z.

Also,

5x+3/x-4 + 3x/4-x.

$ \text{\LaTeX}$ would make this a lot easier to understand.  Is the first problem
$ \frac{z-7}{z-8}=\frac{1}{z-8}$ or $ z-\frac{7}{z-8}=\frac{1}{z-8}$ or something else entirely?

I don't get what you're asking in the second problem.  That's not an equation.

It's the first setup you mentioned. I am so lost on that one. Any help would be appreciated.

hello, multipying for $ z \ne 8$ both sides of the equation by $ z\minus{}8$ we get
$ z\minus{}7\equal{}1$ and adding $ 7$ on both sides we get the solution $ z\equal{}8$ which is impossible, so your equation has no solutions.
Sonnhard.

hello, is your second term $ \frac{3\plus{}5x}{x\minus{}4}\plus{}\frac{3x}{4\minus{}x}$?
Sonnhard.

[quote="gal220"]

z-7/z-8 = 1/z-8. Solve for z.

[/quote]

gal220,

as long as you are typing horizontally, place I) parentheses**
around the numerators that are more than one monomial,
and place II) parentheses around the denominators that are
a) more than a monomial, b) are the product of at least
two constants, c) product of at least two variables, and/or
d) are the product of at least a variable and a constant##
because of the Order of Operations.

You could have had this for a proper equation:

(z - 7)/(z - 8) = 1/(z - 8)

** or other grouping symbols

##
(Specific example) . . . classification(s) from above

Ex. 1/(z - 8) . . . IIa

Ex.  (x - 2)/(5e) . . . I, IIb

Ex.  7/(3*4) . . . IIb

Ex.  10y/(abc) . . . IIc

Ex.  (ab + 1)(x^4 + 4x)/(-3w) . . . I, IId

I think that "+" in the second one is supposed to be an "=" (the one that's surrounded by spaces).

Assuming that's true, $ 5x\plus{}\frac{3}{x}\minus{}4\equal{}\frac{3x}{4}\minus{}x\implies 20x^2\plus{}12\minus{}16x\equal{}3x^2\minus{}4x^2$.

$ 21x^2\minus{}16x\plus{}12\equal{}0$ has no real solutions.

I think he was trying to say (5x+3)/(x-4)=(3x)/(4-x)

Oh yeah, maybe. In that case the only answer is $ \minus{}\frac{3}{8}$.

hello, why don't you use $ \text{\LaTeX}$? it looks much better
Find all real solutions of the equation $ \frac{5x+3}{x-4}=\frac{3x}{4-x}$.
Sonnhard.


limit
trigonometry
calculus
calculus computations

Calculate :

1.$ \lim_{x\rightarrow 0}\frac{1\minus{}\cos^5x\cdot \cos^3 2x\cdot \cos^33x}{x^2}$

2.$ \lim_{n\rightarrow \infty} n(\sqrt[3]{n^3\plus{}3n^2\plus{}2n\plus{}1}\plus{}\sqrt{n^2\minus{}2n\plus{}3}\minus{}2n)$

3.$ \lim_{x\rightarrow 0}\frac{1\minus{}(1\minus{}\sin \ x)(1\minus{}\sin \ 2x)^2...(1\minus{}\sin \ nx)^n}{x}\ , \ n\in \mathbb{N}^*$

2)

$ \lim_{n\rightarrow \infty }n\left( \root{3}\of{n^{3}\plus{}3n^{2}\plus{}2n\plus{}1}\plus{}\sqrt{n^{2}\minus{}2n\plus{}3}\minus{}2n\right)$

$ \equal{}\lim_{n\rightarrow \infty }n^{2}\left( \root{3}\of{1\plus{}\frac{3}{n}\plus{}\frac{2}{n^{2}}\plus{}\frac{1}{n^{3}}}\plus{}\sqrt{1\minus{}\frac{2}{n}\plus{}\frac{3}{n^{2}}}\minus{}2\right)$

$ \equal{}\lim_{x\rightarrow 0\plus{}}\frac{1}{x^{2}}\left( \root{3}\of{1\plus{}3x\plus{}2x^{2}\plus{}x^{3}}\plus{}\sqrt{1\minus{}2x\plus{}3x^{2}}\minus{}2\right)$

$ \equal{}\lim_{x\rightarrow 0\plus{}}\frac{\root{3}\of{1\plus{}3x\plus{}2x^{2}\plus{}x^{3}}\plus{}\sqrt{1\minus{}2x\plus{}3x^{2}}\minus{}2}{x^{2}}$

$ \equal{}\lim_{x\rightarrow 0\plus{}}\frac{\frac{2(\minus{}3\plus{}3x\plus{}5x^{2})}{9\left( 1\plus{}3x\plus{}2x^{2}\plus{}x^{3}\right) ^{5/3}}\allowbreak \plus{}\frac{2}{\left( \left( 1\minus{}2x\plus{}3x^{2}\right) \right) ^{3/2}}}{2}$

$ \equal{}\frac{2}{3}$


algebra
polynomial
quadratics
algebra unsolved

It's a homogeneous bivariate polynomial, so you can use the change $x=zy$. Luckily, the resulting univariate polynomial in $z$ can be factored with the usual trick to find a rational root ($z=-2$) and then with the quadratic formula.


the probability of A speaking the truth is 3/4 and that of B speaking the truth is 7/10. they both assert that a white ball has been drawn out of a bag containing 6 balls all of different colours. what is the probability of the truth of their statement?

what I did:

let W denote the event of drawing a white ball and A the event that both A & B state that a white ball has been drawn. we need to find P(W|A).

$ P(W|A)$ = $ \frac {P(WA)}{P(A)}$

           =$ \frac {P(A|W) P(W)}{P(A|W) P(W) \plus{} P(A|W^C) P(W^C)}$

           = $ \frac {\frac {3}{4}\frac {7}{10} \frac {1}{6}}{\frac {3}{4}\frac {7}{10} \frac {1}{6} \plus{} \frac {1}{4} \frac {3}{10} \frac {5}{6}}$

           = $ \frac {21}{36}$ = $ \frac {7}{12}$.

but the answer in the book is $ \frac {35}{36}$

where did I go wrong? :maybe:

[quote="JBL"][quote="Attila the Hun"]where did I go wrong?[/quote]  In assuming the answer in your book must be correct ;)[/quote]

just what I was thinking.

I agree that you should not always believe what is written in a book, but in this case you should.
In fact I claim that both your answer and the answer in the book are correct, it is just that you have
different interpretations of the problem. I also claim that the answer in the book makes more sense,
since intuitively you should expect that if both A and B come up with the same color,
it is more likely that they speak the truth. So what goes wrong?

The problem is that it is not clearly specified what A and B say when they lie,
and there are several possible interpretations of that.

One possibility is that the lie is to say "white" if the ball is not white and one of the other colors if the ball is white.
This is what your are assuming in your calculation, but somehow this assumption is very artificial since it
implies that A and B have agreed on a certain type of lie before the experiment, and the color white is treated
differently from all the other colors.

A more natural assumption would be that if sombedy lies about the drawn color, he chooses one of the other 5 colors
at random. With this assumption you have $ P(A|W^c) \equal{} \frac 1 4 \frac 1 5 \frac 3 {10} \frac 1 5 \frac 5 6$, which gives
$ P(W|A) \equal{} \frac{35}{36}$.

This kind of problem of interpretation occurs frequently when you are dealing with conditional probabilities;
the most famous example is probably the monty hall problem, where the different interpretations concern
the strategy of the game show host.


Sa se determine toate functiile polinomiale $f: R\rightarrow R$ astfel incat $\int_{x}^{x+1}f(t)\: dt=x^{2},x\in\: R$

Fie $F\in \int f(x)dx$. Atunci conditia se scrie $F(x+1)-F(x)=x^{2}$. Pentru $n\in N$ obtinem $F(n)=F(0)+\frac{(n-1)n(2n-1)}{6}$. Fie $G(x)=F(0)+\frac{(x-1)x(2x-1)}{6}$ un polinom. Cum $F(x)=G(x)$ pentru o infinitate de valori (pe multimea numerelor naturale) rezulta ca $F(x)=G(x),\forall x \in R$. Deci $f(x)=\frac{1}{6}(6x^{2}-6x+1)$


Let $a,$ $b,$ $c,$ $d$ and $e$ are positive numbers. Prove that
$\frac{a}{a+2b+2c}+\frac{b}{b+2c+2d}+\frac{c}{c+2d+2e}+\frac{d}{d+2e+2a}+\frac{e}{e+2a+2b}\geq1.$

[hide="Try it."] Use Cauchy inequality[/hide]

Intersting, I think I've seen this inequality before also:
\[\sum_{cyc}\frac{a}{a+2b+c}\geq 1 \]
for 4 varialbes $a,b,c,d$.

Edit: ---------------------------------- The following is for arqady's problem :)

Using Cauchy we obtain:
\[LHS \geq \frac{(\sum a)^{2}}{\sum a^{2}+2\sum ab}\]
but the latter equals one.

I don't believe it!!!
I'm so sorry, look what I just found (I hadn't check the inequlities forum in quite a while) http://www.mathlinks.ro/Forum/viewtopic.php?t=134422

The ineq. I remembered was from an old Mildorf problems packet.... didn't realize that it was just discussed!!

hikaru123, me@home, see here my first post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134422
I think, it's harder. :wink:


In probability, do you only directly multiply P's together when each P's sample space is not equal to the other P's sample space?

I have noticed that when multiplying probabilities together in which sample spaces are the same set of elements, one must multiply the product of the P's by the number of permutations that the events can occur in.

However, when the sample spaces are not exactly the same, multiplication by permutation of P's is not needed. Is this idea correct?

Does anyone know or have a correction to this?


geometry
circumcircle
geometric transformation
rotation
geometry proposed

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given, with circumcircle $ (O)$ and $ \angle A \equal{} 60^{o}$ and let be the points $ D\equiv AC\cap BO$ and $ E\equiv AB\cap CO,$ where $ O$ is the center of $ (O).$ We denote as $ M,$ the midpoint of the side-segment $ BC$ and let be the point $ N\equiv (O)\cap MO,$ in to the same side of $ BC,$ as vertex $ A.$ Prove that $ PM \equal{} PN,$ where $ P$ is the midpoint of the segment $ DE.$[/color][/size][/b]

Kostas Vittas.

Obviously $ NBC$ is an equilateral triangle. Let's call $ P \equiv CN \cap BO$ and $ Q \equiv BN \cap CO$; $ P$ and $ Q$ are midpoints of $ CN$ and $ BN$. As the angles $ \angle ABN \equal{} \angle ACN$, the isosceles triangles $ CDN$ and $ BEN$ are equal. From these triangles we get $ RD \equal{} QE$; as from the equilateral triangle $ BCN$ we have $ OR \equal{} OQ$.
If $ P \equiv DE \cap RQ$, applying Ceva to the triangle $ DEO$ with the transversal $ RPQ$, we get $ PE \equal{} PD$, i.e. $ P$ belongs to $ QR$, the midline of the triangle $ BCN$, hence $ PN \equal{} PM$.

[i] Remark: In my figure the points $ P$ and $ R$ are interchanged[/i]

Best regards,
sunken rock

In $ \bigtriangleup ABC$ with $ \angle A=60^\circ$, circumcenter $ O$, let $ D\equiv AC\cap BO$ and $ E\equiv AB\cap CO$. Let $ M$ and $ P$ be the midpoints of $ BC$ and $ DE$ respectively. Denote circumcircles of triangles $ AED$ and $ ABC$ by $ C_1$ and $ C_2$ respectively

We observe that $ AEOD$ is cyclic. The center of the spiral similarity ($ T$) that carries $ ED$ into $ BC$ is found to be the intersection of $ C_1$ and $ C_2$ (different from $ A$, and in case the two circles are tangent, $ A$ is obviously the center of central similarity) . Let $ S%Error. "neqA" is a bad command.
$ be the intersections of these two circles.

$ S$ is also the center of spiral similarity($ T'$) that carries $ EB$ into $ DC$. But observe that $ EB=DC$ . Hence, $ T'$ is simply a rotation and we get that $ SE=SD$ and $ SB=SC$

Denote by $ N$ the center of $ C_1$. Since $ SNP$ is carried into $ SOM$ by $ T$, $ \bigtriangleup SNO$ is similar to $ \bigtriangleup SPM$ and since $ SN=NO$, $ SP=PM$

Here is my solution for this problem
[b]Solution[/b]
We have: $\widehat{DOE}$ = $\widehat{BOC}$ = $120^0$ = $180^0$ $-$ $\widehat{BAC}$
So: $A$, $D$, $O$, $E$ lie on a circle
Then: $\widehat{ODE}$ = $\widehat{OAE}$ = $\widehat{OBE}$
Hence: $\triangle$ $BDE$ is $E$ - isosceles
So: $BE$ = $DE$
Similarly: $DE$ = $CD$
Then: $BE$ = $CD$ = $DE$
Hence: ($ADE$) passes through $N$ is midpoint of $\stackrel\frown{BAC}$ of ($O$)
Then: there exist a spiral similar: $S^{k; \alpha}_N$: $E$ $\mapsto$ $B$; $F$ $\mapsto$ $C$ (With $k$ = $\dfrac{NB}{NE}$; $\alpha$ = $\widehat{BNE}$)
But: $P$, $M$ are midpoint of $DE$, $BC$
So: $S^{k; \alpha}_N$: $P$ $\mapsto$ $M$
Hence: $\triangle$ $NBE$ $\sim$ $\triangle$ $NMP$
We also have: $\widehat{EBN}$ + $\widehat{BNE}$ = $\widehat{AEN}$ = $\widehat{AON}$ = 2 . $\widehat{EBN}$
So: $\widehat{EBN}$ = $\widehat{BNE}$
Then: $\triangle$ $BNE$ is $E$ - isosceles
Hence: $\triangle$ $MNP$ is $P$ - isosceles
Therefore: $PM$ = $PN$


I know only one language fluently: Java. I have seen many java games but i don't even know where to start with making one. can someone please help me? like attach some source code or something?

I'm pretty good with Java (currently taking the AP class), and I've written a couple of games in it. For example, I wrote a decent, albeit very simple, version of Tetris from scratch in two days, and created a wheel of fortune program with a partner for last year's final. I still have the code for both of these in case you're interested (although the Tetris has been heavily modified and is now much better).

If you give at least a broad idea of what kind of game you were thinking of, I can try to provide an example to serve as a reference. :)

It's not hard to code simple games in Java (which I mostly use as well), but often you need other languages like C++, Pascal, or Python to run more complex games. I am currently self-studying C/C++; it's much harder to code a functional C program, but it is DEFINITELY faster.

Well there are two main parts to making a game- the thing itself and then representing it.  If you've been programming for a little bit you probably already know how to make the game itself.  Take a simple game like pong.  You would start by creating objects for the paddles and the ball and the playing field.  You would run a loop which changes the ball's position over time, and looks for collisions.  You would look for input from the keyboard and change the position of the paddle.  You might have some AI to control the other paddle, yadda yadda.

This would technically be a game, but it wouldn't be very fun.  For graphics in Java, you'd need to use applets.  There are tons of resources and documentations that will tell you how to create applets and what functions are available to you.  You can start messing around creating shapes and seeing what you can do; this part is pretty simple.

Once you're comfortable with that you can have your program get the positions of the paddles and the ball, throw some rectangles and a circle up on the screen, have it keep updating and you're good to go.

that's a simple example, but it has many elements common to any game.  Accepting user input, manipulating variables inside objects, having an AI algorithm, displaying your information as graphics, etc.

I may be able to show you some old examples if you like.  Good luck!


calculus
integration
calculus computations

This is a horribly ugly indefinite integral; are there any limits you left out?

You know, we do have a calculus forum. A moderator might consider moving this.


All of them. :wink:

They all have neat problems.

[quote="boo"]Which year's AMC/AHSME's should I take as practice?

Since calculators are not allowed, do you think that this year's tests will be more like the previous few years' tests or older AHSME's?[/quote]
If you've noticed, very few problems on AMC10/12 need a calculator, and for very few of them does a calculator actually speed it up significantly enough to help...

I don't recommend taking any test more than couple, namely from recent years (2007-2005).

Instead, I say definitely do the ones that require you to add/subtract because that's where most stupid errors are going to occur. You have a week left so don't expect to suddenly have an epiphany and learn everything. That's just not possible. Relax and just sharpen up your add/subtract skills so you can do that correctly and if you can, that saves time as well.

That was my problem last year, lol. I was like i'll just learn everything the day before >_>


calculus
integration
trigonometry
geometry
geometric transformation
rotation
calculus computations

Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\leq t\leq \pi$,

$ C: \left\{\begin{array}{ll} x \equal{} 2r(t \minus{} \sin t\cos t) & \quad \\
y \equal{} 2r\sin ^ 2 t & \quad \end{array} \right.$

When the point $ P$ moves on the curve $ C$,

(1) Find the magnitude of acceleralation of the point $ P$ at time $ t$.

(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\leq t\leq \pi$.

(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.

Edited.

[quote="kunny"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\leq t\leq \pi$,

$ C: \left\{\begin{array}{ll} x = 2r(t - \sin t\cos t) & \quad \\
y = 2r\sin ^ 2 t & \quad \end{array} \right.$

When the point $ P$ moves on the curve $ C$,

(1) Find the acceleralation of the point $ P$ at time $ t$.

(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\leq t\leq \pi$.

(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]

second problem,note that the  length of the locus
 $ L=\int_{0}^{\pi}\sqrt{x'^2+y'^2}dt=\int_{0}^{\pi}\sqrt{(2r)^2[1-(\cos^2 t-\sin^2 t)]^2+(2r2\sin t\cos t)^2}dt
=2r\int_{0}^{\pi}\sqrt{(1-\cos^2 t+\sin^2 t)^2+4\sin^2 t(1-\sin^2 t)}dt=2r\int_{0}^{\pi}2\sin tdt=8r$

[quote="kunny"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\leq t\leq \pi$,

$ C: \left\{\begin{array}{ll} x = 2r(t - \sin t\cos t) & \quad \\
y = 2r\sin ^ 2 t & \quad \end{array} \right.$

When the point $ P$ moves on the curve $ C$,

(1) Find the acceleralation of the point $ P$ at time $ t$.

(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\leq t\leq \pi$.

(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]
the first problem,you need calculate$ \frac{d^2 y}{d x^2},
\frac{dy}{d x}=\frac{dy/dt}{dx/dt}=\frac{4r\sin t\cos t}{2r[1-(\cos^2t-\sin^2t)]}=\cot t
\frac{d^2 y}{d x^2}=\frac{d\cot t}{dx}=\frac{d\cot t}{dt}\frac{1}{\frac{dx}{dt}}
=(-1-\cot^2 t)\frac{1}{4r\sin^2 t}$

[quote="kunny"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\leq t\leq \pi$,

$ C: \left\{\begin{array}{ll} x \equal{} 2r(t \minus{} \sin t\cos t) & \quad \\
y \equal{} 2r\sin ^ 2 t & \quad \end{array} \right.$

When the point $ P$ moves on the curve $ C$,

(1) Find the acceleralation of the point $ P$ at time $ t$.

(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\leq t\leq \pi$.

(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]
the third problem volume $ V\equal{}\int_{0}^{\pi}\pi y^2 \frac{d x}{dt}dt\equal{}\int_{0}^{\pi}\pi(2r\sin^2 t)^2 4r\sin^2 t\equal{}16\pi r^3\int_{0}^{\pi}\sin^6 t dt\equal{}5\pi^2 r^3$

Yes, those answers are correct, sorry I have just edited context of (1).


In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?

there are $ 7!$ ways to arrange the chairs. now, lets subtract when the two are sitting next to eachother. making them one person, there are $ 6$ ways to arrange them, and then $ 5!$ to arrange the rest. however, they can be arranged in $ 2$ ways themselves. thus, the answer is $ 7!\minus{}6\cdot2\cdot5!\equal{}5040\minus{}12\cdot120\equal{}5040\minus{}1440\equal{}\boxed{3600}$.


inequalities
probability
induction
inequalities unsolved

[quote="mathmanman"]Let $f(x) = (1-x^n)^m+(1-(1-x)^m)^n$, $\forall x \in [0,1]$.
We have : $f(0) = 2$ and $f(1) = 1$.
To show that $f(x) \ge 1$ in $[0,1]$, we just have to show that $f$ decreases on this interval.
$f'(x)= -n \cdot m \cdot x^{n-1} \cdot (1-x^n)^{m-1}+n \cdot m \cdot (1-x)^{m-1} \cdot (1-(1-x)^m)^{n-1} < 0$
$\iff \left(\frac{1-(1-x)^m}x \right)^{n-1} < \left(\frac{1-x^n}{1-x} \right)^{m-1}$,
or, if we develop :
$(1-x+..+m(-x)^{n-1})^{m-1} < (1+x+..+x^{m-1})^{n-1}$.

:)[/quote]

.. and why is the last inequaity true?? :?

That is an easy one inequality solving in not very famous strategy.

Anyway it was not proposed in Czech MO... the source you wrote is incorrect.

Probabilistic interpretation:

Let $x$ be the probability of flipping a coin heads, so $1-x$ is the chance of flipping tails.  Note $1-x^n$ is the probability that at least one tail will appear so $(1-x^n)^m$ is the probability that after $mn$ flips, at least one tail will appear in each of the intervals flips $1$ to $n$, $n+1$ to $2n$, etc $(m-1)n+1$ to $mn$.  Similarly, $(1-(1-x)^m)^n$ can be interpreted as the probability that at least one heads will appear in each of set of flip #s $\{1,n+1,\dots (m-1)n+1\}$, $\{2,n+2,\dots,(m-1)n+2\}$ etc.  At least one of these must occur -- for instance, if the first doesn't occur, there exists a $k$ so that flips $kn+1, \dots , (k+1)n$ are all heads and the second event occurs.  If the second event doesn't occur, one of the sets of flips $\{\ell,n+\ell,\dots (m-1)n+\ell\}$ contains all tails... so we're done!!

[quote="M4RI0"]Let m,n $ \in N, x \in$ [0,1]. Show that

$ (1 \minus{} x^n)^m \plus{} (1 \minus{} (1 \minus{} x)^m)^n \geq 1$[/quote]

if x=1, 0: the inequality is true. Let's assume x is not equal to 0 or 1.
define: $ f(x)\equal{}1\minus{}(1\minus{}x)^m$. The inequality is then equivalent to $ f(x)^{n}\geq f(x^n)$. I will show this by induction:
n=1: trivial
n=2:
$ (1\minus{}(1\minus{}x)^m)^2 \geq 1\minus{}(1\minus{}x^2)^m \equal{}> \frac{(1\minus{}x^2)^m\plus{}((1\minus{}x)^2)^m}{2} \geq (1\minus{}x)^m \equal{}> \frac{(1\plus{}x)^m\plus{}(1\minus{}x)^m}{2} \geq 1$
and this is true by Power Mean (for every $ m \geq 1$).
Let's assume that it's true for some integer n=k>1. for n=k+1:
$ f(x)^{k\plus{}1}\equal{}f(x)f(x)^k \geq f(x)f(x^{k})$
it is sufficient to show that $ f(x)f(x^k)\geq f(x^{k\plus{}1})$, which is equivalent to:
$ (1\minus{}x^{k\plus{}1})^m\plus{}(1\minus{}x\minus{}x^k\plus{}x^{k\plus{}1})^m \geq (1\minus{}x^k)^m\plus{}(1\minus{}x)^m$
which is true by Karamata\Majorization.
So it is true for every $ m \geq 1$, and natural n.

ThAzN1 - how did you come up with this interpretation? :o

xirti: you are wrong, because f(0)=1 and so it doesn't decrease in the interval.

[color=green]Hnm, the general problem was posted by me.  :) [/color]


did any of you know that 0.9999999..... (re-occuring 9) equals 1.  So far I've seen 2 proofs of this. 1 of them:

1/3=.3333333
multiply each side by 3
3/3 or 1 = .999999

  Ive known this for sometime now, have you? :huh:

[quote="A+MATH"]Doesn't work. That requires knowing that 1/3 = .33333333333333333333333.

A better way is this:

[hide]x = .999999...
10x = 9.99999...
Subtract
9x = 9.000000000000000...
x = 1.000000000000.....[/hide][/quote]

Thats the one I learned in my math class. That's pretty cool.

Yea i also knew about the substarction method but i didnt bother postign it.

[quote="A+MATH"]Doesn't work. That requires knowing that 1/3 = .33333333333333333333333.

A better way is this:

[hide]x = .999999...
10x = 9.99999...
Subtract
9x = 9.000000000000000...
x = 1.000000000000.....[/hide][/quote]

its just the numbers repeating thingy for like .432432432=432/999 and like .222222=2/9 so then .99999 must equal 9/9 or just 1

Yeah, I learned that 2 years ago in algebra 1.


Let x_i, y_i >0 and $ \sum^n x_i \equal{}\sum^n y_i \equal{}n$.
Prove that

$ 2n \sum^n \frac{1}{x_i\plus{}y_i} \ge \frac{n^2}{2} \plus{}  \frac{1}{n}  \sum^n \frac{x_i y_i}{x_i\plus{}y_i}$

[quote="4865550150"]Let x_i, y_i >0 and $ \sum^n x_i \equal{} \sum^n y_i \equal{} n$.
Prove that

$ 2n \sum^n \frac {1}{x_i \plus{} y_i} \ge \frac {n^2}{2} \plus{} \frac {1}{n} \sum^n \frac {x_i y_i}{x_i \plus{} y_i}$[/quote]

$ 2n\sum^{n}\frac{1}{x_i\plus{}y_i}\equal{}\sum^{n}(x_i\plus{}y_i)*\sum^{n}\frac{1}{x_i\plus{}y_i}\geq n^{2}$

$ 4ab\leq (a\plus{}b)^{2}$ <--> $ \frac{ab}{a\plus{}b}\leq\frac{a\plus{}b}{4}$

$ \frac{n^2}{2} \plus{} \frac {1}{n} \sum^n \frac {x_i y_i}{x_i \plus{} y_i}\leq\frac {n^2}{2} \plus{} \frac {1}{n} \sum^{n}\frac{x_i\plus{}y_i}{4}\equal{}\frac {n^2}{2} \plus{} \frac {1}{n}*\frac{2n}{4}\equal{}\frac{n^{2}\plus{}1}{2}$

$ 2n \sum^n \frac {1}{x_i \plus{} y_i}\geq n^{2}\geq\frac{n^{2}\plus{}1}{2}\geq\frac{n^2}{2} \plus{} \frac {1}{n} \sum^n \frac {x_i y_i}{x_i \plus{} y_i}$

Is this inequality true?

$ \sum^n \frac {2 \minus{} x_i y_i}{x_i \plus{} y_i} \ge \frac {n}{2}$

Yes of course, as the above solution shows.

$ \sum^{n}\frac{2}{x_{i}\plus{}y_{i}}\ge n$ and

$ \sum^{n}\frac{x_{i}y_{i}}{x_{i}\plus{}y_{i}}\leq\sum^{n}\frac{x_{i}\plus{}y_{i}}{4}\equal{}\frac{n}{2}$

I suppose this is what you meant in the original (i.e. putting $ n$ instead of $ \frac 1n$ in the RHS)  :)

[quote="spanferkel"]Yes of course, as the above solution shows.

$ \sum^{n}\frac {2}{x_{i} \plus{} y_{i}}\ge n$ and

$ \sum^{n}\frac {x_{i}y_{i}}{x_{i} \plus{} y_{i}}\leq\sum^{n}\frac {x_{i} \plus{} y_{i}}{4} \equal{} \frac {n}{2}$

I suppose this is what you meant in the original (i.e. putting $ n$ instead of $ \frac 1n$ in the RHS)  :)[/quote]

yes, thx. :)


Angie walks and jogs as a part of her normal workout. One day she walked for 20 minutes at 2 mph, jogged for 30 minutes at 4 mph, and cooled down by walking for 10 minutes at 2 mph. How many miles did she travel during the hour workout?

Using $ d\equal{}rt$, we have $ \frac{1}{3}\times{2}\plus{}\frac{1}{2}\times{4}\plus{}\frac{1}{6}\times{2}\equal{}\frac{2}{3}\plus{}2\plus{}\frac{1}{3}\equal{}\boxed{3}$.


The finite sums involved do not contain the same number of terms; is that intentional?


very easy!
1) What three numbers between o and 10 can be multiplied together to make a product that equals their sum?
its so easy
really easy!
mean it
2) The average of five numbers is 20. The largest number is 18. What is the sum of the other numbers?
lol this one is easy too
really really easy!

[quote="LuCy4EvA"]1.[hide] eh.. i guessed, since you said it was easy, i guess 1, 2, 3 and guess what? lucky me... hehe[/hide]

2. [hide]5*20-18 = 82... but i have a question, if my answer is correct, then 18 can't be the largest number.... unless i'm totally mis-interrupting the problem... again[/hide][/quote]

I think ur right too. 18 cant be the largest number.

Yo! My sister just make a mistake. The highest number is 28.   :D

1.) 1x2x3=6 and
1+2+3=6
So the answer is 1,2,3

2.) Yeah, the largest number can't be 18. :huh:

o ya sorry about the mistake!
as my brother said ealier the
largest number is 28!
sry bout tat
it was a typin mistake
 :blush:


modular arithmetic
search
number theory unsolved
number theory

(IMO-1990) Determine all integers n>1 such that

$ \frac{2^n\plus{}1}{n^2}$ is an integer.

what notation is this: $ r\equal{}ord_q{2}$

That would be the smallest integer $ r$ such that $ 2^r\equiv 1\mod q$.

For any integer $ k$ with $ 2^k\equiv 1\mod q$, we must have $ r|k$.

I didn't understand why $ 9\not|n$.
Why does r must be even?
why $ 2^n\equiv1 \pmod{q}$ is a contradiction? shouldn't it be a contradiciton if it was like this $ 2^n\equiv1 \pmod{n^2}$

Why $ 9\not|n$.

Lemma. If $ 2^k\plus{}1$ is divisible by $ 3^t$, then $ 3^{t\minus{}1}$ divides $ k$. I'll show this by induction. The base case $ t\equal{}1$ is easy; clearly $ 3^{0}\equal{}1$ divides anything. Now suppose $ t\ge 2$. Then $ k$ is divisible by $ 3$ (just look at $ 2^k\plus{}1\mod 9$). So let $ k\equal{}3k'$.

$ 2^{3k'}\plus{}1\equal{}(2^{k'}\plus{}1)(2^{2k'}\minus{}2^{k'}\plus{}1)$

We can quickly check that $ a^2\minus{}a\plus{}1\equiv 0\mod 9$ has no solutions, so $ 2^{2k'}\minus{}2^{k'}\plus{}1$ is divisible by at most $ 3$. Since $ 3^t|2^{3k'}\plus{}1$, we know that $ 3^{t\minus{}1}|2^{k'}\plus{}1$. Thus $ 3^{t\minus{}2}|k'$ by the inductive hypothesis, so $ 3^{t\minus{}1}|k$, so the lemma is proved.

Hey I just realized this also follows from [url=http://www.mathlinks.ro/viewtopic.php?search_id=345802527&t=205928]lifting the exponent[/url].

So suppose $ 3^{t}$ is the largest power of $ 3$ which divides $ n$, and assume $ t\ge 2$. Then if $ \frac{2^n \plus{} 1}{n^2}$ is an integer, we must have $ 3^{2t}$ dividing $ 2^n \plus{} 1$. By the lemma, $ 3^{2t\minus{}1}$ divides $ n$. But if $ t\ge 2$ then $ 2t\minus{}1>t$, and we assumed that $ 3^t$ was the largest power of $ 3$ dividing $ n$, a contradiction.

Why $ r$ is even.

So he says $ 2^{2n}\equiv 1\mod q$. Now we know $ r|2n$, but if $ r$ is odd, then we would have $ r|n$. Then we would have $ 2^{n}\equiv 1\mod q$.

But we know that $ 2^{n}\equiv \minus{}1\mod n^2$, which means $ 2^{n}\equiv \minus{}1\mod q$, since $ q$ is a divisor of $ n^2$. This contradicts $ 2^{n}\equiv 1$, so we must have $ r$ is even.

So you proved that for any n that is a power of three with $ t>\equal{}2$ there will be no solution. But how do I assure that there will be no solutions for $ n\equal{}5,7,11...$ ? I mean a number that is not a power of three.

[quote="danilorj"]So you proved that for any n that is a power of three with $ t > \equal{} 2$ there will be no solution. But how do I assure that there will be no solutions for $ n \equal{} 5,7,11...$ ? I mean a number that is not a power of three.[/quote]

Well I showed that $ 9$ cannot divide $ n$, and then the remaining cases are handled later in that proof in Fermat's Little Turtle's link...

[quote="Fermat's Little Turtle"] http://www.mathlinks.ro/Forum/viewtopic.php?p=366466#366466 [/quote]
Have any of you understood what [url=http://www.mathlinks.ro/viewtopic.php?p=1239228#1239228]#3[/url] in this link mean?


If $ 1 \minus{} y$ is used as an approximation to the value of $ \frac {1}{1 \plus{} y}$, $ |y| < 1$, the ratio of the error made to the correct value is:

$ \textbf{(A)}\ y \qquad \textbf{(B)}\ y^2 \qquad \textbf{(C)}\ \frac {1}{1 \plus{} y} \qquad \textbf{(D)}\ \frac {y}{1 \plus{} y} \qquad \textbf{(E)}\ \frac {y^2}{1 \plus{} y}\qquad$

[quote=TachyonPulse][hide="Solution"]
$ \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}$

Now we need to find the ratio of the error make to the correct value.

$ \frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2$

The value I got isn't one of the choices so I'll go with $ \boxed{(B) y^2}$.
[/hide]

Can anyone tell me what I did wrong or what I still need to do?[/quote]

$$\frac{\frac{1}{1+y}-{1+y}}{\frac{1}{1+y}}$$ is the ratio of the error made to the correct value. But otherwise, your answer is correct.  :D


Any idea when the Round 1 Solutions will be up?

They are up.  Go to the transcripts of old Math Jams, and one of the very recent ones is "USAMTS Round 1".  Each problem is explained.

I know, but I mean the official solutions on the USAMTS site.


calculus
integration
function
calculus computations

[quote="problem_solver"]Can somebody help me get $\frac{(n+1)(1-t^{2})^{n+1}}{(-2t)!}$  to $\frac{(2^{2}n)(n!)^{2}}{(2n+1)!}$[/quote]
No, we can't help you do that - because the first expression makes no sense. Whatever you think you've done already can't possibly be correct.

I can think of several ways to do this. Here's one:

Use that this is an even function to write the integral as

$\frac12\int_{-1}^{1}(1-t^{2})^{n}\,dt= \frac12\int_{-1}^{1}(1+t)^{n}(1-t)^{n}\,dt$

Now integrate that by parts $n$ times.

Thank you Kent Merryfield, that was very helpfull   :)

I actually wrote this out and it took a lot of paper...and it eventually worked..

I see this sub forum has many interesting integrals for me to solve... :idea:


analytic geometry
modular arithmetic
geometry
rectangle
AMC
counting

I went ahead and added the diagram DPatrick made.

Cool. I edited it a little because I'm crazy and want it to look as much like the diagram on the page.

[quote="Math Geek"][quote="tinytim"]According to someone that I know, the answer can be obtained by computing $ (1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} 4^2 \plus{} 3^2 \plus{} 2^2 \plus{} 1^2) \plus{} (1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} 6^2 \plus{} 7^2)$. I understand how the stuff in the first parentheses is gotten but don't understand the second part...

Another person I know used an inccorect method that should have gotten him answer of 325 but didn't add correctly and got 225 which is the correct answer!!! :o[/quote]

I didn't solve this question correctly, but I'm going to guess that the second part comes from squares that do not have integer lengths.  I'm probably wrong, though.

A person I know brute-forced the answer and got it right. o.O[/quote]

Your guess is correct. :) The solution is misrepresented there, though.

Can somebody post the official solution?

This has to be the fastest solution:

[hide="Beastly Solution"]Clearly each square must have one vertex in each quadrant.  Let $ A, B, C, D$ be the vertices in quadrants 1, 2, 3, 4, respectively.  Let $ B', C', D'$ be the points in quadrant 1 corresponding to $ B, C, D$, respectively.  It should be intuitively clear that $ AB'C'D'$ is also a square (we can prove this using transformations or vectors).  Now if $ A, B', C', D'$ are all the same point, this point corresponds to one square, which gives us 25 squares.  If $ A, B', C', D'$ are not all the same point, then they form a small square.  This small square corresponds to eight squares, since we can choose a vertex of the square to be A and then an order to label the other vertices of the small square, clockwise or counterclockwise.  Therefore, the number of squares is equivalent to $ 25\equiv 1\pmod{8}$.  The only answer choice that satisfies this is E, so we can bubble E and move on.  Or turn in the test.  :) [/hide]

For some reason I get 220.
The method I use is to add up all the squares from one to five, , reasoning that this corresponds to side lengths from 6-10.  Then I multiply by 4 to fall short 5 from the answer. Could someone tell me if this is just serendipitous or how my method should count the next 5 squares?

[quote="matt276eagles"]   This small square corresponds to eight squares [/hide][/quote]

it's 4, not 8

Math1331Math, please draw all possible squares of side length $10$. Now look at the first quadrant, take a red marker, and mark each vertex red. You will end up with a rectangle. In other words, if we can draw in a square in this grid, see how much we can shift it horizontally ($x$) and vertically ($y$), this tells us there are $xy$ congruent squares. We can uniquely define a square by the pair $(x, y)$.

Hey, I made a graph on Desmos that could help! The solution 2 on AoPS (site: https://www.artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 ) may seem confusing, and I didn't understand the part that said all squares in $G_1$ could be mapped to four squares in $G$. Here is a graph for everyone that needs help and doesn't understand that part either:

[hide=Click to reveal hidden link... if you dare!]https://www.desmos.com/calculator/tmyuq5zxru[/hide]

(Make sure that you show the black dots. This is simple to do:

On the graph that has $x_2$ and $y_2$:
click anywhere on the $y_2$ column except for the circle
..........|
..........|
.........\/
$x_2$| O $y_2$
$3$| $6$
$6$| $7$
$7$| $4$
etc.| etc.

That will make the black dots visible.)

[quote=Math1331Math][quote=thkim1011][hide="Solution"]Note that the set of all congruent squares in the same orientation form a rectangle in the first quadrant with their first point. Every different rectangle forms a different set of congruent squares so we have
\[1\cdot1 + 1\cdot2 + \cdots + 1\cdot5 + 2\cdot1 + \cdots + 5\cdot 5 = (1+2+3+4+5)^2 = \boxed{225}.\][/hide][/quote]
This seems cool but I don't entirely get it. Can somebody elaborate what they mean by and how they got the final value[/quote]

Each congruent squares first vertex can be shifted around among certain points which end up forming a rectangle, and each shift forms another square.  Each rectangle corresponds to one set of congruent squares, and all the rectangles have a set of squares so we need to count how many points in each of all the different rectangles.  This is the first long sum, which is equivilant to $(1+2+3+4+5)^2$ due to the expansion of $(v+w+x+y+z)^2$.

Hope that cleared things up

~Mudkipswims42


Find two four digit numbers whose product is $4^{8}+6^{8}+9^{8}$.

Can anyone give me a hint on this problem?  I got

$4^{8}+6^{8}+9^{8}=2^{16}+3^{8}2^{8}+3^{16}$ and $(2^{8}+3^{8})^{2}=2^{16}+2\times2^{8}3^{8}+3^{16}$.   please don't tell me how to solve it, just give an example or a formula or something.

thanks :)
n how do you delete your own post?

[size=59][color=red]b-flat[/color][/size]

[hide="try"]$a^{2}-b^{2}$ with what you have written over there.[/hide]

On these types of problems, try to add zero creatively :D

oh i was so close, thanks for the hint guys.  n thanks for correcting my latex, b-flat.


number theory
relatively prime
number theory unsolved

$n, m$ are positive integers of different parity, and $n > m$.  Find all integers $x$ such that $\frac{x^{2^n} - 1}{x^{2^m} - 1}$ is a perfect square.

It's not hard to see that
$\frac{x^{2^n}-1}{x^{2^m}-1}=(x^{2^m}+1)(x^{2^{m+1}}+1)\cdots (x^{2^{n-1}}+1)=A$.

First case, if $x$ is odd then $v_2(A)=n-m\equiv_2 1$ so $A$ can't
be perfect square. If $x$ is even, then we will prove that
$(x^{2^a}+1,x^{2^b}+1)=1$.

Let $(x^{2^a}+1,x^{2^b}+1)=d$ then
$d|x^{2^b}+1-(x^{2^a}+1)=x^{2^a}(x^{2^a(2^{b-a}-1)}-1)$. How
$(x^{2^a}+1,x^{2^a})=1$ then it's $d|x^{2^a(2^{b-a}-1)}-1$.

Number $2^{b-a}-1$ is odd so
$d|x^{2^a}+1|(x^{2^a})^{2^{b-a}-1}+1=x^{2^a(2^{b-a}-1)}+1$.

So $d|x^{2^a(2^{b-a}-1)}+1 - (x^{2^a(2^{b-a}-1)}-1) =2$ but number
$x^{2^a}+1$ is odd, because $x$ is even, so it has to be $d=1$.

Now we proved that numbers $x^{2^m}+1, x^{2^{m+1}}+1, \cdots x^{2^{n-1}}+1$ are relatively prime in pairs, and their product is
suppose to be perfect square, so they are all squares, and that is
not possible because $(x^{2^{m-1}})^2<x^{2^m}+1<(x^{2^{m-1}}+1)^2$,
so there is no such $x$.